url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3
values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93
values | snapshot_type stringclasses 2
values | language stringclasses 1
value | language_score float64 0.06 1 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
http://www.meritnation.com/ask-answer/question/if-ab-is-a-chord-of-a-circle-with-centre-o-aoc-is-diamet/circles/3620954 | 1,462,176,938,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860128071.22/warc/CC-MAIN-20160428161528-00163-ip-10-239-7-51.ec2.internal.warc.gz | 671,273,079 | 31,063 | 011-40705070 or
Select Board & Class
• Select Board
• Select Class
from Amity International School, asked a question
Subject: Math , asked on 18/1/13
If AB is a chord of a circle with centre O, AOC is diameter and AT is the tangent at the point A.
Prove that ∠BAT=∠ACB
Let ∠ACB = x and ∠BAT = y.
A tangent makes an angle of 90 degrees with the radius of a circle,
so we know that ∠OAB + y = 900 ……..(1)
The angle in a semi-circle is 90, so ∠CBA = 900 .
∠CBA + ∠OAB + ∠ACB = 18 0 0 (Angle sum property of a triangle)
Therefore, 90 + ∠OAB + x = 1800
So, ∠OAB + x = 9 0 0………….(2)
But OAB + y = 900
Therefore, ∠OAB + y = ∠OAB + x ………….[From (1) and (2)]
x = y.
Hence ∠ACB = ∠BAT.
This conversation is already closed by Expert
• 15
View More
## Popular Chapters
Start a Conversation
You don't have any friends yet, please add some friends to start chatting
Unable to connect to the internet. Reconnect
friends:
{{ item_friends["first_name"]}} {{ item_friends["last_name"]}} {{ item_friends["subText"] }} | 383 | 1,072 | {"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-2016-18 | latest | en | 0.800761 |
https://philoid.com/question/109702-what-happens-to-a-function-fx-at-x-a-if | 1,718,394,323,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861568.20/warc/CC-MAIN-20240614173313-20240614203313-00623.warc.gz | 422,776,720 | 7,817 | ##### What happens to a function f(x) at x = a, if
If f(x) is a function defined in its domain such that
f(x) become continuous at x=a
1 | 41 | 139 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-26 | latest | en | 0.929716 |
http://stackoverflow.com/questions/6358031/how-to-split-the-list-on-the-pairwise-elements/6358735 | 1,432,990,563,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207931085.38/warc/CC-MAIN-20150521113211-00185-ip-10-180-206-219.ec2.internal.warc.gz | 241,191,044 | 16,747 | # How to split the list on the pairwise elements? [duplicate]
Possible Duplicate:
How do you split a list into evenly sized chunks in Python?
Let us have a list, there is always an even number of elements. We must break it down by pairing. Example: list['1','2','3','4'] need 1,2 and 3,4
-
## marked as duplicate by Sven Marnach, wheaties, unutbu, Thomas Wouters, DonutJun 15 '11 at 14:12
-
If you want two halfs of a list.
``````l = [1,2,3,4]
print l[:len(l)/2], l[len(l)/2:]
>>> [1, 2] [3, 4]
``````
If you want split a list by pairs then your question is exact duplicate.
-
and if have more elements [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] ? =)) – Denis Jun 15 '11 at 13:13
``````>>> L = [1, 2, 3, 4]
>>> pairs = zip(L[::2], L[1::2])
>>> print pairs
[(1, 2), (3, 4)]
``````
Hope this helps
-
You can use something like this too:
``````lVals = xrange(1,101)
size = len(lVals)
output = ((lVals[i], lVals[i+1] if size > i+1 else None) for i in xrange(0, size, 2))
``````
-
I don't think this one works, try it out! – juanchopanza Jun 15 '11 at 13:19
@juanchopanza i have tried is several times and it works for me - if use 'res = list(output)' you will get a list of tuples for example: res[2] == (3,4) – Artsiom Rudzenka Jun 15 '11 at 13:31
It gives (1,2),(2,3),(3,4) and so on, instead of (1,2),(3,4),(5,6) are required. You need output = ((lVals[i], lVals[i+1] if size > i+1 else None) for i in xrange(0,size,2)) – juanchopanza Jun 15 '11 at 13:34
@juanchopanza - yes, you right, my fault. Thanks for notifying me. – Artsiom Rudzenka Jun 15 '11 at 13:51
``````from itertools import izip_longest
data = range(6)
data_iters = [iter(data)] * 2
pairs = izip_longest(*data_iters)
[pair for pair in pairs]
>>> [(0, 1), (2, 3), (4, 5)]
``````
The clever part is that the two elements of data_iters refer to the same object. Izip_longest alternately consumes from the two iterators passed as arguments, but since they're referring to the same object, it effectively pairs the elements in the iterator.
I take no credit for being clever here, upvote the comment I linked to if you liked my answer. :)
- | 718 | 2,109 | {"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-2015-22 | latest | en | 0.801239 |
https://math.stackexchange.com/questions/3030015/a-simple-equation-with-a-complicated-property | 1,547,706,164,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583658844.27/warc/CC-MAIN-20190117062012-20190117084012-00444.warc.gz | 589,210,994 | 33,675 | # A simple equation with a complicated property
Let, $$\Bbb{P}$$ denote the set of all odd prime numbers and $$\Bbb{N}$$ be the set of all natural numbers. Let, $$2a,2b$$ be two even numbers both greater than $$4$$.
Define, $$A=\{(p,q)\in\Bbb{P}\times\Bbb{P}:p+q=2a\}$$ and $$B=\{(r,s)\in\Bbb{P}\times\Bbb{P}:r+s=2b\}$$. Let, $$X'=\{A,B\}$$
Define, $$n:X'\to \Bbb{N}$$ by, $$n(Y)=\max_{(x,y)\in Y}\{x^2+y^2\},\forall Y\in X'=\{A,B\}$$ My claim is, $$n(A)=n(B)\implies A=B$$
I have no proof as well as no counterexample for my claim. If anyone has any idea about how to prove this or have seen any paper on this or it is very trivial to prove or has a counterexample for this claim please give.
P.S.: I am "in danger of being blocked from asking any more".
• the notation is a bit unnecessarily complicated and obtuse. The entire three paragraph description can be simplified to, "fix even $a,b\in\mathbb Z^+$ each greater than 4. If the sum of the squares of two primes whose sum is each of $a,b$ are equal, are $a$ and $b$ equal?" – YiFan Dec 7 '18 at 16:37
• If you have such kinds of conjectures you should write a program that tests this for a lot of numbers. Testing for all n<1000 should not be a problem. – miracle173 Dec 7 '18 at 17:38
• can you give an example of your claim? – miracle173 Dec 7 '18 at 18:05
• @YiFan Thank you for your kind response. But the statement you made in the comment is a bit wrong. Take $a=31b=7$ and $p=29,q=13$ then $p^2+q^2=a^2+b^2=1010$ but $p\ne a,q\ne b$. That's why I take the maximum of them. – Sujit Bhattacharyya Dec 8 '18 at 2:31
• @miracle173 I am having same trouble with the examples. Assuming $n(A)=n(B)\implies \max\{a^2+b^2\}=max\{p^2+q^2\}\implies a^2+b^2=p^2+q^2$ for some $(a,b)\in A,(p,q)\in B$ where the maximum attained. From here if I am able to show that $a=p,b=q$ then we are done. But still no proof is found. – Sujit Bhattacharyya Dec 8 '18 at 2:41
I'm afraid the question is not well defined; or at least, not proven to be well defined. In your definition for the sets $$A$$ and $$B$$, you essentially say they are the sets of all pairs of odd primes which sum to $$2a$$ and $$2b$$, which are even numbers. The Goldbach Conjecture asserts that every even number ($$>4$$) can be expressed as the sum of two odd primes, and this unfortunately hasn't been proven. Hence, if the Goldbach Conjecture happens to be false, your sets $$A$$, $$B$$ become null sets, and so the question becomes ill-defined and doesn't make sense; indeed, asserting the well-definition of $$A$$, $$B$$ would be asserting Goldbach! (This is of course unless you're willing to define $$\max(x^2+y^2)$$ to mean something else in this case, in which case your conjecture comes down to asserting that there is exactly one even positive integer that can't be expressed as a sum of odd primes, an equally tough problem to prove.)
• If I redefine the functional $n$ as $n(\emptyset)=0$ will it be then well defined? – Sujit Bhattacharyya Dec 8 '18 at 2:38
• @SujitBhattacharyya yes, you're right. If the stated conjecture is true and $A,B\neq\emptyset$ in particular, then goldbach is true. – YiFan Dec 8 '18 at 7:15 | 1,005 | 3,153 | {"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": 20, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2019-04 | longest | en | 0.835285 |
https://docs.rs/rand/0.5.5/rand/ | 1,547,874,126,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583662124.0/warc/CC-MAIN-20190119034320-20190119060320-00538.warc.gz | 510,386,518 | 8,787 | # Crate rand[−][src]
Utilities for random number generation
Rand provides utilities to generate random numbers, to convert them to useful types and distributions, and some randomness-related algorithms.
# Basic usage
To get you started quickly, the easiest and highest-level way to get a random value is to use `random()`.
```let x: u8 = rand::random();
println!("{}", x);
let y = rand::random::<f64>();
println!("{}", y);
if rand::random() { // generates a boolean
}```
This supports generating most common types but is not very flexible, thus you probably want to learn a bit more about the Rand library.
# The two-step process to get a random value
Generating random values is typically a two-step process:
• get some random data (an integer or bit/byte sequence) from a random number generator (RNG);
• use some function to transform that data into the type of value you want (this function is an implementation of some distribution describing the kind of value produced).
Rand represents the first step with the `RngCore` trait and the second step via a combination of the `Rng` extension trait and the `distributions` module. In practice you probably won't use `RngCore` directly unless you are implementing a random number generator (RNG).
There are many kinds of RNGs, with different trade-offs. You can read more about them in the `rngs` module and even more in the `prng` module, however, often you can just use `thread_rng()`. This function automatically initializes an RNG in thread-local memory, then returns a reference to it. It is fast, good quality, and secure (unpredictable).
To turn the output of the RNG into something usable, you usually want to use the methods from the `Rng` trait. Some of the most useful methods are:
• `gen` generates a random value appropriate for the type (just like `random()`). For integers this is normally the full representable range (e.g. from `0u32` to `std::u32::MAX`), for floats this is between 0 and 1, and some other types are supported, including arrays and tuples. See the `Standard` distribution which provides the implementations.
• `gen_range` samples from a specific range of values; this is like `gen` but with specific upper and lower bounds.
• `sample` samples directly from some distribution.
`random()` is defined using just the above: `thread_rng().gen()`.
## Distributions
What are distributions, you ask? Specifying only the type and range of values (known as the sample space) is not enough; samples must also have a probability distribution, describing the relative probability of sampling each value in that space.
In many cases a uniform distribution is used, meaning roughly that each value is equally likely (or for "continuous" types like floats, that each equal-sized sub-range has the same probability of containing a sample). `gen` and `gen_range` both use statistically uniform distributions.
The `distributions` module provides implementations of some other distributions, including Normal, Log-Normal and Exponential.
It is worth noting that the functionality already mentioned is implemented with distributions: `gen` samples values using the `Standard` distribution, while `gen_range` uses `Uniform`.
## Importing (prelude)
The most convenient way to import items from Rand is to use the prelude. This includes the most important parts of Rand, but only those unlikely to cause name conflicts.
Note that Rand 0.5 has significantly changed the module organization and contents relative to previous versions. Where possible old names have been kept (but are hidden in the documentation), however these will be removed in the future. We therefore recommend migrating to use the prelude or the new module organization in your imports.
## Examples
```use rand::prelude::*;
// thread_rng is often the most convenient source of randomness:
if rng.gen() { // random bool
let x: f64 = rng.gen(); // random number in range [0, 1)
println!("x is: {}", x);
let ch = rng.gen::<char>(); // using type annotation
println!("char is: {}", ch);
println!("Number from 0 to 9: {}", rng.gen_range(0, 10));
}```
# More functionality
The `Rng` trait includes a few more methods not mentioned above:
For more slice/sequence related functionality, look in the `seq` module.
There is also `distributions::WeightedChoice`, which can be used to pick elements at random with some probability. But it does not work well at the moment and is going through a redesign.
# Error handling
Error handling in Rand is a compromise between simplicity and necessity. Most RNGs and sampling functions will never produce errors, and making these able to handle errors would add significant overhead (to code complexity and ergonomics of usage at least, and potentially also performance, depending on the approach). However, external RNGs can fail, and being able to handle this is important.
It has therefore been decided that most methods should not return a `Result` type, with as exceptions `Rng::try_fill`, `RngCore::try_fill_bytes`, and `SeedableRng::from_rng`.
Note that it is the RNG that panics when it fails but is not used through a method that can report errors. Currently Rand contains only three RNGs that can return an error (and thus may panic), and documents this property: `OsRng`, `EntropyRng` and `ReadRng`. Other RNGs, like `ThreadRng` and `StdRng`, can be used with all methods without concern.
One further problem is that if Rand is unable to get any external randomness when initializing an RNG with `EntropyRng`, it will panic in `FromEntropy::from_entropy`, and notably in `thread_rng()`. Except by compromising security, this problem is as unsolvable as running out of memory.
# Distinction between Rand and `rand_core`
The `rand_core` crate provides the necessary traits and functionality for implementing RNGs; this includes the `RngCore` and `SeedableRng` traits and the `Error` type. Crates implementing RNGs should depend on `rand_core`.
Applications and libraries consuming random values are encouraged to use the Rand crate, which re-exports the common parts of `rand_core`.
# More examples
For some inspiration, see the examples:
## Modules
distributions Generating random samples from probability distributions. prelude Convenience re-export of common members prng Pseudo-random number generators. rngs Random number generators and adapters for common usage: seq Functions for randomly accessing and sampling sequences.
## Structs
AsciiGenerator [Deprecated] Iterator which will continuously generate random ascii characters. Error Error type of random number generators Generator [Deprecated] Iterator which will generate a stream of random items.
## Enums
ErrorKind Error kind which can be matched over.
## Traits
AsByteSliceMut Trait for casting types to byte slices CryptoRng A marker trait used to indicate that an `RngCore` or `BlockRngCore` implementation is supposed to be cryptographically secure. FromEntropy A convenience extension to `SeedableRng` allowing construction from fresh entropy. This trait is automatically implemented for any PRNG implementing `SeedableRng` and is not intended to be implemented by users. Rand [Deprecated] A type that can be randomly generated using an `Rng`. Rng An automatically-implemented extension trait on `RngCore` providing high-level generic methods for sampling values and other convenience methods. RngCore The core of a random number generator. SeedableRng A random number generator that can be explicitly seeded.
## Functions
random Generates a random value using the thread-local random number generator. sample [Deprecated] DEPRECATED: use `seq::sample_iter` instead. thread_rng Retrieve the lazily-initialized thread-local random number generator, seeded by the system. Intended to be used in method chaining style, e.g. `thread_rng().gen::()`, or cached locally, e.g. `let mut rng = thread_rng();`. weak_rng [Deprecated] DEPRECATED: use `SmallRng` instead. | 1,703 | 7,951 | {"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-2019-04 | latest | en | 0.838895 |
https://www.studyrankers.com/2020/08/mcq-questions-for-class-8-maths-comparing-quantities.html | 1,718,449,571,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861586.40/warc/CC-MAIN-20240615093342-20240615123342-00715.warc.gz | 898,653,382 | 46,604 | MCQ Questions for Class 8 Maths: Ch 8 Comparing Quantities
1. Find the ratio of speed of a cycle 15 km per hour to the speed of scooter 30 km per hour.
(a) It is 1:2
(b) It is 1:3
(c) It is 2:1
(d) It is 3:1
► (a) It is 1:2
2. The ratio of 10m to 10 km is:
(a) 1/10
(b) 1/100
(c) 1/1000
(d) 1000
► (c) 1/1000
3. The price of a house was Rs 34,00,000 last year. It has increased by 20% this year. What is the price now?
(a) Rs 40,40,000
(b) Rs 40,80,000
(c) Rs 30,40,000
(d) Rs 30,80,000
► (b) Rs 40,80,000
4. Calculate compound interest on Rs 10,800 for 3 years at 12.5% per annum compounded annually.
(a) 13,377.34
(b) 4577.34
(c) 14,377.34
(d) None of these
► (b) 4577.34
5. An item marked at Rs 840 is sold for Rs 714. What is the discount %?
(a) 20%
(b) 10%
(c) 15%
(d) none of these
► (c) 15%
6. A sum of money, at compound interest, yields Rs. 200 and Rs. 220 at the end of first and second years respectively. What is the rate percent?
(a) 20%
(b) 15%
(c) 10%
(d) 5%
► (c) 10%
7. A picnic is being planned in a school. Girls are 60% of the total number of students and are 300 in number. Find the ratio of the number of girls to the number of boys in the class.
(a) It is 3:2
(b) It is 3:1
(c) It is 2:3
(d) It is 2:1
► (a) It is 3:2
8. The price of a scooter was Rs 34,000 last year. It has increased by 20% this year. What is the price now?
(a) Rs 30,800
(b) Rs 30,400
(c) Rs 40,800
(d) Rs 40,400
► (c) Rs 40,800
9. A shopkeeper purchased 500 pieces for Rs 20 each. However 50 pieces were spoiled in the way and had to be thrown away. The remaining were sold at Rs 25 each. Find the gain or loss %.
(a) 18%
(b) 15%
(c) 12.5%
(d) none of these
► (c) 12.5%
10. The present population of a town is 25000. It grows at 4%, 5% and 8% during first year, second year and third year respectively. Find the population after 3 years.
(a) 29484
(b) 28696
(c) 24576
(d) 30184
► (a) 29484
11. Find the ratio of speed of a cycle 20 km per hour to the speed of scooter 30 km per hour.
(a) It is 3:1
(b) It is 2:1
(c) It is 2:3
(d) It is 1:3
► (c) It is 2:3
12. The population of a city was 20,000 in the year 1997. It increased at the rate of 5% p.a. Find the population at the end of the year 2000.
(a) 25153
(b) 24153
(c) 23153
(d) None of these
► (c) 23153
13. Sohan bought a washing machine for Rs 40,000, then spent Rs 5,000 on its repairs and sold it for Rs 50,000. Find his loss or gain per cent.
(a) Loss 10%
(b) Loss 20%
(c) Profit 11%
(d) none of these
► (c) Profit 11%
14. ________ means comparing two quantities.
(a) Ratio
(b) Proportion
(c) Percent
(d) None of these
► (a) Ratio
15. Find selling price (SP) if a profit of 5% is made on a cycle of Rs 700 with Rs 50 as overhead charges.
(a) Rs 600
(b) Rs 787.50
(c) Rs 780
(d) None of these
► (b) Rs 787.50
16. Find the ratio of speed of a car 50 km per hour to the speed of scooter 40 km per hour.
(a) It is 4:5
(b) It is 4:1
(c) It is 5:4
(d) It is 1:5
► (c) It is 5:4
17. The sale price of a shirt is Rs.176. If a discount of 20% is allowed on its marked price, what is the marked price of the shirt?
(a) Rs.160
(b) Rs.180
(c) Rs. 200
(d) Rs. 220
► (d) Rs. 220
18. The difference in S.I. and C.I. on a certain sum of money in 2 years at 15% p.a. is Rs.144. Find the sum.
(a) Rs. 6000
(b) Rs. 6200
(c) Rs. 6300
(d) Rs. 6400
► (d) Rs. 6400
19. Rohan bought a second hand refrigerator for Rs 2,500, then spent Rs 500 on its repairs and sold it for Rs 3,300. Find his loss or gain per cent.
(a) Loss 15% 2a
(b) Loss 10%
(c) Profit 10%
(d) None of these
► (c) Profit 10%
20. The value of an article which was purchased 2 years ago, depreciates at 12% per annum. If its present value is Rs.9680, what is the price at which it was purchased?
(a) Rs.10000
(b) Rs.12500
(c) Rs.14575
(d) Rs.16250
► (b) Rs.12500
21. Find the ratio of 5 m to 10 km.
(a) It is 1:3
(b) It is 3000:1
(c) It is 2000:1
(d) It is 1:2000
► (d) It is 1:2000
22. A sum of money, at compound interest, yields Rs. 200 and Rs. 220 at the end of first and second years respectively. What is the rate percent?
(a) 20%
(b) 15%
(c) 10%
(d) 5%
► (c) 10%
23. A shopkeeper purchased 300 bulbs for Rs 10 each. However 10 bulbs were fused and had to be thrown away. The remaining were sold at Rs 12 each. Find the gain or loss %.
(a) 15%
(b) 13%
(c) 16%
(d) none of these
► (c) 16%
24. In what time will Rs.1000 amount to Rs.1331 at 10% p.a. compounded annually?
(a) 4 years
(b) 3 years
(c) 2 years
(d) 1 year
► (b) 3 years | 1,703 | 4,456 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2024-26 | latest | en | 0.906549 |
https://newbedev.com/why-does-chocolate-ice-cream-melt-faster-than-others | 1,696,451,206,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511406.34/warc/CC-MAIN-20231004184208-20231004214208-00421.warc.gz | 453,995,120 | 7,383 | # Chemistry - Why does chocolate ice cream melt faster than others?
## Solution 1:
This appears to be physics more than chemistry. A dark color means it emits less visible light but rather absorbes it with temperature rise. Have your next ice cream in a darker place as to visible and ir emisions to test this.
## Solution 2:
It's a complicated problem and I'm going to add a hand-waving physics answer.
For most materials, at a given wavelength the albedo and and emissivity sum to roughly unity. So if something is close to white with an albedo of 0.9, it's emissivity in the visible range is about 0.1, and if something is dark, those are reversed with albedo ~0.1 and emissivity ~0.9.
However, in the thermal infrared many non-metallic materials have an emissivity of very roughly 0.9 no matter what color they are in the visible. Even glass and water which are transparent in the visible have a high infrared emissivity.
In direct sunshine a surface perpendicular to the light will receive about 1100 W/m^2 in the visible, and so chocolate ice cream will be heated much faster than vanilla. Strawberry is pink because it absorbs shorter wavelengths but there's not a lot of power there from the Sun's black-body spectrum, so strawberry is like vanilla in the Sun.
However on an overcast day, or indoors, the the light is roughly 50 to 100 time dimmer, or about 11 to 22 W/m^2
(You can prove that by shooting a photo at f/16 with a shutter speed of 1/ISO or 1/ASA. I'm dating myself, but if had ASA 64 film you'd shoot f/16 at 1/60th second outdoors. You'd get nothing if you left it that way and shot indoors (Sunny 16 rule)
So now we're only down to say 11 or 22 W/m^2 in the visible, but that's not all the light! The Stefan-Boltzmann law tells us
$$P = \sigma \epsilon T^4.$$
If you plug in 273K and 298K (0° and 25° C) and $$\epsilon=0.9$$, you get about 280 and 400 W/m^2. That means that you, the walls ceiling or clouds are heating your ice cream at 400 W/m^2 while it is only radiating 280 W/m^2 back, it's a difference of 100 W/m^2 or way more than the heating at visible wavelengths.
So I would propose that if the phenomenon happens indoors that it's not related to the color, and could be related to the recipe.
Maybe chocolate ice cream has more salt, depressing the melting point, and so it starts melting sooner because it melts at a lower temperature.
So I've just asked Does chocolate ice cream tend to have more salt than strawberry or vanilla? | 619 | 2,482 | {"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": 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": 2, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2023-40 | latest | en | 0.943439 |
https://www.scienceblogs.com/startswithabang/2010/11/12/why-does-matter-take-up-space | 1,675,540,163,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500151.93/warc/CC-MAIN-20230204173912-20230204203912-00337.warc.gz | 1,012,235,940 | 22,266 | # Why does Matter take up Space?
"A vacuum is a hell of a lot better than some of the stuff that nature replaces it with." -Tennessee Williams
Matter -- everything you know, love, hate, see, taste, and feel -- takes up space.
Even air must take up space. Not just wind, but still, stationary air takes up space. We've known this since Empedocles, in the 5th Century B.C., who had a clepsydra -- a hollow gourd with many holes in the bottom and a single hole at the top -- demonstrated it.
You plunge the gourd into a stream, lake or river, and it fills with water. If you lift the gourd up, the water leaks out of the bottom. But place your thumb or hand over the top, and the water will stop flowing out of the bottom. What prevents it from falling? It's got to be the air beneath the holes, exerting a pressure and taking up space!
You can even build one for yourself using (just like on Monday) a 2-liter bottle.
But why is this? Why -- at a fundamental level -- does air, or matter of any type, have to take up space at all?
Another way of asking this same question is, "why can't more than one object be in the same place at the same time?"
No matter how hard I try, I can't be in the same place at the same time as any other object in the Universe. And it isn't just me; I can't take any particle that makes up matter -- a molecule, an atom, even a proton, neutron, or electron -- and put an arbitrarily large number of them in a finite amount of space.
It didn't have to be that way! You can take photons -- particles of light -- and put an infinitely large number of them in an arbitrarily small space. Same deal with the 2nd, 3rd, and 4th heaviest fundamental particles in the Universe: the Higgs Boson, the Z-boson, and -- if you can overcome their charges -- even the W-bosons.
So why are protons, neutrons, and electrons -- the stuff that makes up all our normal matter in the Universe -- limited in this way?
It's the Pauli Exclusion Principle! Just like in baseball, where you can't have two runners on the same base, the Pauli Principle tells us that you can't have two identical fermions (one of the two basic types of particles, along with bosons) in the same state!
When I was younger, I used to think this was some minuscule technical detail of physics, good for little more than explaining the chemical properties of atoms due to their electron cloud structure.
Big deal, I can't put two identical fermions in the same quantum state.
But it's a much bigger deal than that. If I either cooled the temperature down to absolute zero or compressed matter with an arbitrarily large amount of force, I could squeeze any number of bosons into an arbitrarily small space.
But normal matter is made out of protons, neutrons and electrons, all of which are fermions. And this simple principle means that there's a finite volume that -- once it's occupied by one of these matter particles -- it's off-limits to the others!
And that's why matter takes up space, no matter whether it's charged or neutral, and regardless of temperatures or pressures or any other physical properties!
There are some spectacular astrophysical consequences of this, and two of my favorites are what happens to stars when they die.
A white dwarf star -- somewhere around the mass of the Sun but the physical size of the Earth -- is made of plain old atoms, same as we are. But a white dwarf is about 300,000 times denser than we are! Yet, despite that incredible gravity compressing the white dwarf, the atoms refuse to buckle. Why?
Because the atoms deep inside the star have their electrons bumping up against each other, and the electrons refuse to buckle and let other electrons any closer!
In fact, in the most extreme case, the electrons, rather than let another electron into the same state and violate the Pauli rule, would rather fuse with the protons, producing a neutron (and a neutrino), collapsing all the way down to a neutron star!
But neutrons are fermions, too, and even a star made entirely out of neutrons refuses to collapse! These are the densest known objects in the Universe that are still made of matter, and yet, they take up space!
So you can keep your bosons to yourself; my matter takes up space, and I've got the Pauli Exclusion Principle to thank for it! On the other hand, dark matter could be either bosonic or fermionic; we don't know yet, but my money's on bosonic, and that it truly doesn't take up any space! (How's that for some wonderment headed into the weekend?!)
Tags
### More like this
From the blog:
"It says space is a existence and it is empty"
"Then, by present science knowledge everyone says FORCE is required to move a thing."
Nope, FORCE is required to accelerate a thing. Even then it has to be Net Force.
"An existence speed will be greater than zero"
More ill-presented wordage.
"Empty space has always zero average density."
Ah, Douglas Adams used this years before you in HHGTTG where the guide proves that there is no life in the universe.
Pretty much all your blog is exhaustively terrible. Learn to write. Of course I'm supposing that the problem is your writing rather than the thinking you're trying to express.
@Ethan - Why is your money on Bosonic? Or should I just be patient and wait for a future post to explain your position?
I've heard this explanation before, but I'm not sure I buy it for most of the normal matter we have here on earth. I've heard physicists much smarter than me say it, but I don't understand it. Certainly the Fermi Pressure is what holds up white dwarf stars, but I don't think it's the major thing for "normal density" matter.
Ferinstance, the H_2 molecule is bigger than an H atom, but that can't be because of Pauli exclusion between the two nuclei (since protons are spin-1/2, they can form a spin-antisymmetric state and have a symmetric spatial wavefunction) or between the two electrons (same reason).
The reason why the nuclei don't "overlap" in the hydrogen molecule is just electrostatics and "distinguishable-particle" quantum mechanics. Similarly, I don't think Pauli exclusion can be the only (or even major) thing preventing solid matter made of atoms (charged nuclei) from collapsing or having two things at the same place at the same time.
What am I missing here? Am I screwing up H2, or is there something about all other matter that's different than H2?
By Anonymous Coward (not verified) on 12 Nov 2010 #permalink
What happens in black holes? I thought that was matter that had collapsed on itself.
Anonymous Coward: It's not Fermi pressure per se, but that is the reason why there can only be two electrons in each orbit (one with spin +1/2, one with spin -1/2).
Were it not for the Pauli Exclusion Principle, all the electrons would happily drop down to the lowest-energy orbit, and the various elements would probably all look pretty much the same.
@Ethan - Great Post. But we still don't know why fermions cannot be in the same place. Is there any way to demonstrate the Pauli Exclusion Principle? Or anything more fundamental than that and not just descriptive?
By Lucas Parisi (not verified) on 12 Nov 2010 #permalink
Halfway through the story I started wondering: When did physicists start to distinguish between fermions and bosons?
The exclusion principle looks like a definition of fermions - but when did we start to call them that?
Uh oh, this is entirely wrong. It is *not* the Pauli exclusion principle that keeps matter from collapsing on itself. It is the zero point energy, which applies to all particles both boson and fermions when they are confined in a small region.
Use the example of two protons in an orbit with a symmetric spatial wave function (e.g. s-wave). They have a binding energy, an attraction which holds them together. But the binding energy has the zero point energy to overcome. And to hold them infinitely close together would take an infinitely strong force. Consequently they remain a finite distance apart.
What about a Bose-Einstein condensate? All the particles are in the same quantum state. They can make those out of whole atoms nowadays. How does that relate to the Pauli Exclusion?
I'm no physicist, but my limited Science-Channel understanding of black holes is that all the mass was converted to energy and the energy used to warp space-time. The mass-energy is now stored in warped space-time itself. There is no actual mass or matter inside the black hole, so the Pauli Exclusion principle doesn't hold. Maybe?
Wow, what a long post to avoid answering the question. Saying that 'the Pauli Exclusion Principle' keeps two 'fermions' from occupying the same point in space is only a description of the phenomenon. It doesn't answer the question you asked, to wit: WHY does matter take up space? WHY are the physical laws governing matter the way they are? I'll give you a hint; you can find the answer in Genesis 1:1.
mad the swine: Really, you are going with a god-of-the-gaps answer here? At this level? We are talking about interactions between fundamental particles, and why they behave the way they do...and you posit it's because of some passage in Genesis? To be frank, everything we know, from particle physics to cosmology, is because we discarded the god of the gaps, and instead said, "no, really, why is this the way it is?"
Ugh, can't believe I'm going to say this, Ethan, but I agree with the creationist. Seems to me that saying the PEP is the reason things can't be in the same place is like saying Newton's laws are the reason there is gravity. Both strike me as more of a description of the phenomenon that fermions can't be in the same place. It doesn't seem like it gets to the deeper question, like relativity did for gravity. What gives?
Also @11 you are wrong. The answer can be found at www.venganza.org Seriously, it explains EVERYTHING!
I'm just a layman so excuse the post...
1) I read this as "matter takes up space because fermions take up space and matter is made of fermions"... Which doesn't really answer much. Why do fermions take up space?
2) I thought black holes (not neutron stars) were the densest things in the universe?
3) If bosons don't take up any space, how can they be said to exist?
RE #11 - "God done it" is not an answer, it's an assumption with zero explanatory power.
I'm with Ougaseon @13. The math seems like it's just a description of the physical phenomenon. Or is it the other way around, and at the quantum level the physical phenomenon is just an expression of the math?
I don't understand much (out of my league here)but I can answer part of the layman question in #14:
My undestanding of Black Holes is that the gravity and energy (gravity energy?) contained in a Black Hole is so great that it is another universe at the end of a black hole. All the laws of physics of this universe no longer apply when gravity is great enough to collapse matter into a singularity.
If I am wrong on this - I hope that some person with more knowledge will enlighten me. Please!
I think your description of the clepsydra by Empedocles has a bit of a twist here, the operation described in the Wiki states a gourd thrust into the water with top covered will not fill and that this is the justification for the premises of matter existing, though as you may be hinting, if you fill such container and cover the top, water will still run out the bottom⦠until an equalization occurs between the matter (water/air)left in the gourd and the barometric air pressure on the outside, given the state of mass properties of the gravity metric.
So from this lesson of late (490-430BC) we could state that normal matter seeks equilibrium and there are functions that can be described processes, such as the Pauli exclusion principleâ¦
There is a fundamental difference by which matter is defined in dark matter, and black hole matter, though even within the different states of different generations of matter, they should still be defined as seeking equalization within their generation.
I also strongly suspect the three generations of matter â¦Dark, Normal, Black
(not to be confused with the local generations of the standard model) also seek an equilibrium not only within but between their generations, while the energy of these forces always seek the lowest possible state, whether it be an atom or the universe as a whole, it is the dynamics of this attempted equilibrium that we call space. Time would therefore be lengthened by the expansion of the universe and the slow creation of ever denser forms of matter and would only be recognizable as we know it in the normal generation.
I suspect even black holes in a (hypothetical)pure vacuum would seek equilibrium by warping the distance between them.
And all of this leads one to *wonder* as you have so craftily done, Perhaps there is no such thing as identical fermions in cold dark matter and does the Pauli exclusion principle stand or is it warped? Does Gamma radiation resonance effect the field in which cold Dark matter resides?
It may be that the boson force is hidden by the Fermionâs configuration caused by a slight feeble feed back mechanism between black and dark matter, perhaps the large gamma radiation bubbles above and below the galactic plane serve in a Pregeometry dynamics of cold dark matter.
I would speculate Mass term hierarchy would not remain equivalent for these different generations of matter. D/N/B
Well thatâs how I see it, but then who am I?
My bet is that it still involves both constituents, Bosons and Fermions only the rules of engagement are not quite the same as with our local reality of normal matter.
By Sphere Coupler (not verified) on 13 Nov 2010 #permalink
There can only be one reason why matter occupies and curves space - matter IS curved space.
The idea is simple but realization not so much. But I believe we will get there one day.
What? No music for the weekend diversion,only wonder? I miss your picks.
Surely we can relate the PEP to some artistic medium.
I'll give it a shot...As one Fermion said to the other Fermion;
"Wish you were here".
copy/paste
By Sphere Coupler (not verified) on 14 Nov 2010 #permalink
Henk @7 asks: When did physicists start to distinguish between fermions and bosons?
Shortly after they discovered that particles have spin angular momentum in integer multiples of hbar/2. If the spin s (defined such that the spin angular momentum is hbar * s) is an integer, the particle is a boson; if s is a half-integer, the particle is a fermion. It was soon noticed that if you exchange two otherwise identical particles with arbitrary states, the two-particle wavefunction gets multiplied by (-1)^(2*s). The only way you can do this if the two fermions (for which the multiplier is -1) are in the same quantum state is for the wavefunction to be identically zero, which is the mathematical statement of the Pauli exclusion principle. No such restriction applies to bosons because the particle exchange has no net effect on the wavefunction.
By Eric Lund (not verified) on 14 Nov 2010 #permalink
In a universe in which fermions or their equivalent could be packed ever more densely, nothing would stop gravity compressing matter to a density incompatible with stars and planets as we understand them. That would also be a universe incompatible with life as we know it: a universe in which we would not be present to make observations.
These properties of matter are basically inescapable in a universe in which we could exist.
Thanks Eric Lund @21 for bringing this comment thread back into the realm of science. It was really going off the rails there! I was about to bring up the quantum field theoretic explanation for HOW the exclusion principle works, but I don't think I could've explained it as well.
Ethan has provided an excellent scientific explanation (science quibbling aside) of our familiar physical world.
There is always another question, a deeper mystery; that is a wonder of science.
The self referential aspect of language or mathematics means; that to understand a new insight, we must look with open eyes and mind at the wonder of our physical world (not just look at the words or math of a theory).
Religious myths, e.g. Genesis, generally make pretty lousy science theory; but most do not read religious texts to understand caterpillars or comets.
By AngelGabriel (not verified) on 15 Nov 2010 #permalink
Re: Brian at #5
I agree that Pauli exclusion determines what the occupied ground state orbitals are, and is largely responsible for the chemical diversity of the elements.
But that's not at all the question Ethan is addressing here.
I agree with Ethan that for things like neutron stars and white dwarfs (dwarves?) that Fermi pressure (i.e. Pauli exclusion) is the dominant mechanism preventing collapse and answers the question as to why matter takes up space.
However, I still disagree that for "normal" charged matter on earth it's the dominant mechanism. The dominant mechanisms are electrostatics and what BillK calls "zero-point energy": the energy associated with localizing electron wavefunctions.
Pauli exclusion can't explain why the two nuclei in a hydrogen molecule don't collapse onto each other (see my above comment). My less-subtantiated claim is that, similarly, it's not the primary reason why my hand doesn't go through my desk.
By Anonymous Coward (not verified) on 15 Nov 2010 #permalink
What you've done (and what the Pauli Exclusion Principle does) is explain and describe HOW matter takes up space.
There is no explanation as to WHY it does so.
In an article on matter, you post a picture of a mountain and it's not the Matter-horn?! I'm beside myself with grief. In fact I'm so beside myself that I just accidentally poked myself in the eye. Excuse me for a minute will you?
Re: Ambitwistor at #26
I stand corrected on my "less-substantiated" claim. Thank you very much for the links!
By Anonymous Coward (not verified) on 17 Nov 2010 #permalink
Juice,
Fundamental physics doesn't answer "why" questions. That's the purview of philosophy (and it's questionable how well philosophy can "answer" such questions either). The job of physics is to describe phenomena in terms of laws, not explain why the laws themselves are true. If you ask physics "why does X happen", the answer is always "because X is predicted from the laws of physics, which themselves are supported by data Y".
See Feynman struggle to explain that physics can't answer questions like "why do magnets attract each other".
By Ambitwistor (not verified) on 18 Nov 2010 #permalink
Or, perhaps one might say, physics answers "why" questions, but not in the way that laymen seem to expect.
By Ambitwistor (not verified) on 18 Nov 2010 #permalink
Strictly speaking, matter for which the Pauli Exclusion Principle is the only thing keeping its constituents from coming together any closer is called degenerate matter. Most ordinary, everyday matter is not degenerate: other forces keep matter particles such as atoms from coming together too closely, e.g. electrical repulsion between electrons in atoms accounts for most typical matter taking up the space it does, and the fact that these electrons repel each other is the main reason why everyday solid objects are solid. Degenerate matter can only be formed under extreme pressures. Metallic hydrogen is one example, but it requires pressures greater by far than any achievable in any modern laboratory as of this writing.
The only form of dark matter we know of and can detect as of this writing is fermionic: the neutrinos. They're spin 1/2 just like most normal matter.
By stormwyrm (not verified) on 25 Nov 2010 #permalink
that was fie rigth there .
By akeria howard (not verified) on 13 May 2013 #permalink
I Think That The mass Is Diffrent
OK, it's now massively different. If it had been an improvement, this would have been a good thing...
Starting off, it goes into a valueless point. There being no consensus (its lack or measure has not been presented) does not indicate anything of import. Certainly not that your subsequent issues are indicated as plausible.
You make up a rule of nature next. Try one agreed by consensus. Failing that, one that is coherent and self contained enough to indicate whether your idea has any bearing on it.
Your next step is to insist that matter is 100% solid, when we know it's 99.99999999999999%+ empty space.
This isn't going to go well and it would require far more work on my part to correct you than you deserve on this second attempts' failure to consider whether you need to read up on the basics BEFORE making claims about science. | 4,580 | 20,682 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2023-06 | latest | en | 0.935805 |
https://math.stackexchange.com/questions/433956/proof-of-riemann-lebesgue-lemma-what-does-integration-by-parts-in-each-variabl | 1,623,986,737,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487634616.65/warc/CC-MAIN-20210618013013-20210618043013-00064.warc.gz | 355,244,871 | 37,266 | # Proof of Riemann-Lebesgue lemma: what does “integration by parts in each variable” mean?
I was reading the proof of the Riemann-Lebesgue lemma on Wikipedia, and something confused me. It says the following:
What does the author mean by "integration by parts in each variable"? If we integrate by parts with respect to $x$, then (filling in the limits, which I believe are $-\infty$ and $\infty$) I get $$\hat{f}(z) = \left[\frac{-1}{iz}f(x)e^{-izx}\right]_{-\infty}^{\infty} + \frac{1}{iz}\int_{-\infty}^{\infty}e^{-izx}f'(x)dx.$$
I think I am missing something here...it's not clear to me why the limit at $-\infty$ of the first term should exist. Can anyone clarify this for me?
Thanks very much.
• $f$ is supposed to have compact support, so $f(x) = 0$ for $\lvert x\rvert \geqslant K$. – Daniel Fischer Jul 1 '13 at 20:30
• Ah, I see. Any clue what they meant by "in each variable"? – Eric Auld Jul 1 '13 at 20:54
• I guess it is just a leftover from when it was at least planned to write it down in higher dimensions. – Daniel Fischer Jul 1 '13 at 20:57 | 329 | 1,065 | {"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} | 2.8125 | 3 | CC-MAIN-2021-25 | latest | en | 0.907498 |
https://forums.oscommerce.com/topic/412333-weight-based-table-rate-shipping/ | 1,553,111,785,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202450.86/warc/CC-MAIN-20190320190324-20190320212324-00484.warc.gz | 489,504,026 | 16,191 | #### Archived
This topic is now archived and is closed to further replies.
# Weight Based Table Rate Shipping
## Recommended Posts
Hello,
I'm setting up our office merchant page, when come to the shipping price on weight based, seems like the table rate is not working. I'm listing the shipping rate like this 0.001 - 0.005 = 5.95(follow the rate from our national courier). Is that something worng with shipping value i've added?any ideas to solve this problems?
Regards,
Hizazy.
##### Share on other sites
The syntax for table rate is shown as an example in the module:
The shipping cost is based on the total cost or weight of items. Example: 25:8.50,50:5.50,etc.. Up to 25 charge 8.50, from there to 50 charge 5.50, etc
where units/decimal places are as introduced in products weight and price in default currency
##### Share on other sites
wHAT IS THE MEANING OF Example: 25:8.50,50:5.50,etc.. Up to 25 charge 8.50, from there to 50 charge 5.50, etc
##### Share on other sites
Let me explain
Here is my weight based table 100:2.00,250:2.30,345:3.40,2000:4.40,5000:7.50
It equates to;
0-100 grams £2.00
101-250 grams £2.30
251-345 grams £3.40
346-2000 grams £4.50
2001-5000 grams £7.50
I hope this helps.
Martin
OsC 2.3.4.1 CE Frozen PHP 7.2 MySQL 10.1.36-MariaDB-cll-lve
##### Share on other sites
0.1:2,0.25:2.3,0.345:3.4,2:4.5,5:7.50
product weight only in kilograms
##### Share on other sites
You can have it as grams if you wish
In includes/languages/english/modules/shipping/zones.php change line 16
`define('MODULE_SHIPPING_ZONES_TEXT_UNITS', 'grams');`
On the product listing if it is 186 grams input 186.00
OsC 2.3.4.1 CE Frozen PHP 7.2 MySQL 10.1.36-MariaDB-cll-lve
##### Share on other sites
What is the data
Configuration -> Shipping/Packaging
##### Share on other sites
29 minutes ago, ruden said:
What is the data
Configuration -> Shipping/Packaging
No
It is in Modules-> Shipping and then install the Zones module
Mine looks like this
OsC 2.3.4.1 CE Frozen PHP 7.2 MySQL 10.1.36-MariaDB-cll-lve
##### Share on other sites
Ok
Configuration -> Shipping/Packaging
Enter the Maximum Package Weight you will ship SHIPPING_MAX_WEIGHT
Package Tare weight SHIPPING_BOX_WEIGHT
Larger packages - percentage increase SHIPPING_BOX_PADDING
These parameters are used in calculation
In includes/classes/shipping.php 54-62
##### Share on other sites
6 minutes ago, ruden said:
Ok
Configuration -> Shipping/Packaging
Enter the Maximum Package Weight you will ship SHIPPING_MAX_WEIGHT
Package Tare weight SHIPPING_BOX_WEIGHT
Larger packages - percentage increase SHIPPING_BOX_PADDING
These parameters are used in calculation
In includes/classes/shipping.php 54-62
I am confused. I thought I was helping you not you teaching me. I know how to set up all the parameters you mention. I was trying to clarify Raiwa's post and help cmjain735
OsC 2.3.4.1 CE Frozen PHP 7.2 MySQL 10.1.36-MariaDB-cll-lve | 862 | 2,983 | {"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-2019-13 | latest | en | 0.802312 |
http://serc.carleton.edu/quantskills/teaching_resources/activities.html?q1=sercvocabs__40%253A12 | 1,386,440,536,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163055516/warc/CC-MAIN-20131204131735-00093-ip-10-33-133-15.ec2.internal.warc.gz | 150,961,094 | 6,100 | Quantitative Skills > Teaching Resources > Activities
# Activities
## Materials for Lab and Class
Help
Results 1 - 10 of 10 matches
Estimating Exchange Rates of Water in Embayments using Simple Budget Equations. part of Quantitative Skills:Activity Collection
Keith Sverdrup, University of Wisconsin-Milwaukee
Simple budgets may be used to estimate the exchange of water in embayments that capitalize on the concept of steady state and conservation principals. This is especially true for bays that experience a significant exchange of freshwater. This exchange of freshwater may reduce the average salt concentration in the bay compared to seawater if it involves addition of freshwater from rivers, R, and/or precipitation, P. Alternatively, it may increase the average salt concentration in the bay compared to seawater if there is relatively little river input and high evaporation, E. Since freshwater input changes the salt concentration in the bay, and salt is a conservative material, it is possible to combine two steady state budgets for a bay, one for salt and one for water, to solve for the magnitude of the water flows that enter and exit the bay mouth. Students will make actual calculations for the inflow and outflow of water to Puget Sound, Washington and the Mediterranean Sea and compare them to actual measured values.
Continental Crust Mass Balance Calculation part of Quantitative Skills:Activity Collection
Jennifer Wenner, University of Wisconsin-Oshkosh
A quantitative skills-intensive exercise using data from the Mineral Mountains, Utah, to calculate mass balance and to address the "space problem" involved with emplacing plutons into the crust.
Calculation of the Magnitude of Lunar and Solar Tidal Forces on the Earth part of Quantitative Skills:Activity Collection
Randal Mandock, Clark Atlanta University; Randal Mandock, Clark Atlanta University
Project in which students calculate the magnitude of lunar and solar tidal forces on the earth. They calculate the solar tidal effect relative to the lunar tidal effect and the relative solar tidal effect for spring-tide conditions.
Investigating dimensions of the solar system part of Quantitative Skills:Activity Collection
Francisco San Juan, Elizabeth City State University; Steven Schafersman, University of Texas of the Permian Basin, The; Michael Stewart, University of Illinois at Urbana-Champaign
Planetary data are used to investigate and evaluate the Nebular Hypothesis.
What is the fate of CO2 produced by fossil fuel combustion? part of Quantitative Skills:Activity Collection
Paul Quay
A box model is used to simulate the build up of carbon dioxide in the atmosphere during the industrial era and predict the future increase in atmospheric CO2 levels during the next century.
Density of the Earth - How to Solve It part of Quantitative Skills:Activity Collection
Len Vacher, Dept of Geology, University of South Florida
This module addresses the real problem of determining the density of the Earth and invites the student to figure out how to solve the problem.
Keplers Third Law - The Equation part of Quantitative Skills:Activity Collection
Len Vacher, Dept of Geology, University of South Florida
In this module, students are asked to devise a simple relationship between the sidereal period and orbital radius.
Investigating an ore deposit: from metal reserve to waste disposal part of Quantitative Skills:Activity Collection
Mona Sirbescu, Central Michigan University
How many sand grains on a beach? part of Quantitative Skills:Activity Collection
Alan Whittington, University of Missouri-Columbia
Short exercise designed to give students practice in determining what information is needed to answer a question, estimating an answer, and calculating an answer (including unit conversions and scientific notation). Emphasizes the relevance of large numbers to society (population, debt, etc).
Modeling: (1) Revenue Neutral Carbon Taxes; (2) Accelerated atmospheric C02 concentrations part of SISL:2012 Sustainability in Math Workshop:Activities
Martin Walter
Design a revenue neutral carbon tax and a plan for implementation; together with a model for what happens if we do not institute such a tax-system. | 834 | 4,220 | {"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-2013-48 | latest | en | 0.849416 |
https://forum.code.org/t/cheese-gone-crazy-lesson-27-project/38152 | 1,695,823,995,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510300.41/warc/CC-MAIN-20230927135227-20230927165227-00350.warc.gz | 282,498,803 | 5,146 | # Cheese gone crazy Lesson 27 Project
My student is trying to make the cheese move slowly at random locations across the screen with no luck. I know I’ve seen this problem before but I can’t remember the fix.
Right now, the cheese moves to a random location each time the draw loop passes. That is happening very quickly (at least for a human eye).
What if you created a counter variable at the top of the code and each time through the draw loop had the counter increase by 1. Then, you could have the cheese change positions only when the counter hit multiples of 20…
The code would be something like:
``````if (counter%20==0) {
cheese1.x = randomNumber(100, 400);
cheese1.y = 0;
}
}
``````
This would make the cheese only move once per 20 passes through the draw loop. The % is the modulo operator and the translation of the conditional statement is:
If the remainder of the math problem counter/20 is equal to 20, then do the actions.
Is this what you are after?
Mike
1 Like | 232 | 988 | {"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-2023-40 | latest | en | 0.933304 |
https://www.physicsforums.com/threads/surface-area-of-a-sphere-pi-a-2-b-2-c-2-d-2.1015595/ | 1,713,125,801,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816893.9/warc/CC-MAIN-20240414192536-20240414222536-00698.warc.gz | 883,427,739 | 21,774 | # Surface area of a sphere ##= \pi * (a^2+b^2+c^2+d^2)##
• Trysse
In summary, the OP's formula for area does not hold in the case where the radius of the sphere increases as it decreases.
Trysse
Homework Statement
This is not a homework assignment. It is a problem with which I came up for myself. However, I think it best belongs here.
Imagine the following figure: You have a point ##O## from which four rays point outward. The rays are arranged, so that the angle between the rays is ##arccos - \frac{1}{3}## or ##\approx 109.47°##. You get this figure for example by taking a regular tetrahedron and drawing rays from the center of the tetrahedron through the apices.
Next, imagine you define one arbitrary point on each of the rays. These four arbitrary points always describe a sphere. I can denote this sphere with the 4-tuple ##(a,b,c,d)##, where ##a##, ##b##, ##c##, and ##d## are the distances of the points from the point O along each of the rays.
If I now choose ##a=b=c=d## then ##a##, ##b##, ##c##, and ##d## are the radius of the sphere. On the other side if I define only ##a>0## and ##c=b=d=0## , ##a## is the diameter ##a=d## of the sphere or twice the radius ##a=2r##.
As the surface area of a sphere is ##A=4* \pi * r^2## I wonder, if
$$A=\pi *(a^2+b^2+c^2+d^2)$$
for the above figure. What would a proof look like, that can either prove or disprove the above equation?
Relevant Equations
##A=4* \pi * r^2##
I am not very good at proofs. The only thing I have come up with is the following regularity. However, I am not sure how this can be related to the above problem.
Given a sphere ##S_a## with a center ##C## and a diameter of ##a##. I can now construct a line segment ##b## with the endpoints ##B## and ##B'## so that ##b## is tangent to the sphere and the center of ##b## touches the sphere. Next, I construct a sphere around ##C## with a radius of ##CB=CB'##. According to the theorem of Pythagoras, the surface area of the second sphere is ##\pi * (a^2+b^2)##. In the same way, I can "add" two more line segments to create a sphere where ##a^2 + b^2 + c^2 +d^2##.
Last edited:
PeroK and Delta2
If I take it right r the radius of sphere you say is
$$r=a=b=c=d$$ so
$$4\pi r^2=\pi(a^2+b^2+c^2+d^2)$$
Delta2
anuttarasammyak said:
View attachment 301992If I take it right r the radius of sphere you say is
$$r=a=b=c=d$$ so
$$4\pi r^2=\pi(a^2+b^2+c^2+d^2)$$
Yes, this is one of the two special cases, where the formula for area holde.
Trysse said:
As the surface area of a sphere is ##A=4* \pi * r^2## I wonder, if
$$A=\pi *(a^2+b^2+c^2+d^2)$$
No. As a simple way to visualize this, first set a, b, c and d equal, and note that the three points A, B and C define a unique circle: call this ABC. Now imagine decreasing d a bit: the sphere will shrink and its centre will move away from D. This keeps happening until the centre of the sphere is at the centre of the circle ABC which are now on a great circle of the sphere. Now if we decrease d any more, the sphere can't get any smaller (it can't have a radius smaller than the circle ABC) so it must get larger. If A sometimes gets larger as d gets smaller this contradicts your assertion ## A=\pi *(a^2+b^2+c^2+d^2) ##.
PeroK and Delta2
anuttarasammyak said:
As for the other special case you say
$$a=2r, b=0,c=0,d=0$$
No, at most one of {a, b, c, d} can be zero otherwise there are no longer 4 separate points, however it is easy to prove that in the limit as e.g. {b, c, d} all tend to zero (or alternatively a tends to infinity) the area tends to ## \pi a^2 ##, so these are the two other special cases where the equation is true.
Last edited:
Delta2
Trysse said:
Given a sphere ##S_a## with a center ##C## and a diameter of ##a##. I can now construct a line segment ##b## with the endpoints ##B## and ##B'## so that ##b## is tangent to the sphere and the center of ##b## touches the sphere. Next, I construct a sphere around ##C## with a radius of ##CB=CB'##. According to the theorem of Pythagoras, the surface area of the second sphere is ##\pi * (a^2+b^2)##. In the same way, I can "add" two more line segments to create a sphere where ##a^2 + b^2 + c^2 +d^2##.
You have not thought this through and there are many problems with it. I'll pick just one:
Trysse said:
Next, I construct a sphere around ##C## with a radius of ##CB=CB'##
What makes you think that this sphere will go through the point defined by ## a ##?
Delta2
pbuk said:
No. As a simple way to visualize this, first set a, b, c and d equal, and note that the three points A, B and C define a unique circle: call this ABC. Now imagine decreasing d a bit: the sphere will shrink and its centre will move away from D. This keeps happening until the centre of the sphere is at the centre of the circle ABC which are now on a great circle of the sphere. Now if we decrease d any more, the sphere can't get any smaller (it can't have a radius smaller than the circle ABC) so it must get larger. If A sometimes gets larger as d gets smaller this contradicts your assertion ## A=\pi *(a^2+b^2+c^2+d^2) ##.
If ##a = b = c## then we can take the triangle ABC in the x-y plane, with the centroid at the origin and the apex (D) on the z-axis. Whichever point we choose on the z-axis, defines a sphere with A, B, C and D on the surface (*).
If we choose any values for ##a, b, c, d##, then I can't see whether the points must lie on a sphere, or whether we have a counterexample.
(*) PS I assume that ##d > 0##, which means the point must be beyond the centroid.
PPS If ##d <1##, then the radius of the sphere increases as ##d## decreases. So, the OP's formula cannot hold in that case.
Last edited:
Trysse said:
Next, imagine you define one arbitrary point on each of the rays. These four arbitrary points always describe a sphere.
Can you prove that?
PeroK said:
If we choose any values for ##a, b, c, d##, then I can't see whether the points must lie on a sphere, or whether we have a counterexample.
Provided the four points are not coplanar (ie at least one of { a, b, c, d } is non-zero) then they define a sphere. This is a well-known result.
PeroK
pbuk said:
Provided the four points are not coplanar (ie at least one of { a, b, c, d } is non-zero) then they define a sphere. This is a well-known result.
Ah, of course, any four points lie on a sphere!
pbuk
I meant at most one is zero.
PeroK said:
PPS If ##d <1##, then the radius of the sphere increases as ##d## decreases. So, the OP's formula cannot hold in that ccase.
No, initially the radius decreases until it reaches the critical point where ABC is a great circle as in #4.
PeroK
pbuk said:
As a simple way to visualize this, first set a, b, c and d equal, and note that the three points A, B and C define a unique circle: call this ABC. Now imagine decreasing d a bit: the sphere will shrink and its centre will move away from D. This keeps happening until the centre of the sphere is at the centre of the circle ABC which are now on a great circle of the sphere.
I do not agree with the statement that the sphere's center will be at the center of the triangle ABC. You forgot about the point ##O## in which the four rays originate. If you decrease ##d## to ##0## I.e. you move one of the points inward as far as possible, the points ##O## and ##D## coincide. However, ##O## is not at the center of the triangle ##ABC##.
I have tried to visualize this in GeoGebra. However, this is cumbersome as I cannot construct a sphere from four points in GeoGebra. As a workaround I have constructed four circles instead.
The red ray (top) has the point ##A##, the orange (left front) has the point ##B##, the green ray (right front) has the point ##C## and the blue ray (in the back) has the point ##D##.
In the picture below I have moved the point ##A## all the way down so ##AO=0##. I think you can see that the pink circle going through the points ##BCD## is not a great circle.
You can take a look at the original GeoGebra file here https://www.geogebra.org/classic/fdjwwpy7
I found, that the strange angle between the axes and the use of four axes makes this problem very counterintuitive.
I think I have to reformulate my problem...
pbuk said:
No, initially the radius decreases until it reaches the critical point where ABC is a great circle as in #4.
To do something useful. If we let ##a = b = c = 1##, then I get $$r = \frac{d^2 + \frac 2 3 d +1}{2d + \frac 2 3} = \frac 1 2(d + \frac 1 3) + \frac 4 9(d + \frac 1 3)^{-1}$$And if we set ##d = 1## we get ##r = 1##, as expected.
Setting ##r'(d) = 0## gives ##(d + \frac 1 3) = \frac{\sqrt{8}}{3}##, hence:
$$r_{min} = \frac{\sqrt{8}}{3}$$And if the distance from the centre of the tetrahedron to a vertex is ##1##, then that is indeed the distance from a vertex to the centroid of the associated equilateral triangle. And we have a sphere with ABC on a great circle.
pbuk
Trysse said:
I do not agree with the statement that the sphere's center will be at the center of the triangle ABC. You forgot about the point ##O## in which the four rays originate.
No I didn't.
Trysse said:
If you decrease ##d## to ##0## I.e. you move one of the points inward as far as possible, the points ##O## and ##D## coincide. However, ##O## is not at the center of the triangle ##ABC##.
I did not suggest that you decreased ##d## to ##0##, and I did not suggest that ##O## would be at the center of the triangle ##ABC##: it is obvious that O, A, B and C are fixed and O is not in the same plane as triangle ABC so cannot be its centroid. The centre of the sphere is only at ## O ## when ## { a, b, c, d } ## are equal.
What I said was if you decrease ## d ## until A, B and C lie on a great circle then the centre of the sphere lies at the centroid of triangle ABC.
Trysse said:
In the picture below I have moved the point ##A## all the way down so ##AO=0##. I think you can see that the pink circle going through the points ##BCD## is not a great circle.
So don't move it down that far, only move it as far as you need to so it is a great circle.
Trysse said:
I found, that the strange angle between the axes and the use of four axes makes this problem very counterintuitive.
It might help if you look at it more simply - because the disposition of the 'axes' is symmetrical there must be a position for the fourth point where the other three points (held fixed in an equilateral triangle) lie on a great circle of radius ## r ## with the fourth point ## r ## from the centroid and at this point the sphere can't get any smaller.
PeroK
And in general, for ##a=b = c##, we have$$r = \frac 1 2(d + \frac a 3) + \frac {4a^2}{ 9}(d + \frac a 3)^{-1}$$
PeroK said:
hence:
$$r_{min} = \frac{\sqrt{8}}{3}$$And if the distance from the centre of the tetrahedron to a vertex is ##1##, then that is indeed the distance from a vertex to the centroid of the associated equilateral triangle. And we have a sphere with ABC on a great circle.
It's nice when the numbers confirm what must be true by symmetry (or is it the other way round ).
Let's now figure out why :
SA*( $$S^2$$) = $$4 \pi r^2$$= $$(d/dr)( \frac {4}{3} \pi r^3)$$ ;).
*Surface Area ; $$S^2$$ is the 2-Sphere.
## 1. What is the formula for finding the surface area of a sphere?
The formula for finding the surface area of a sphere is ##\pi * (a^2+b^2+c^2+d^2)##, where ##a, b, c, d## are the lengths of the four sides of the sphere.
## 2. How is the surface area of a sphere different from its volume?
The surface area of a sphere is the total area of the outer surface, while the volume is the amount of space inside the sphere.
## 3. Can the surface area of a sphere be negative?
No, the surface area of a sphere cannot be negative as it is a measure of the total area and cannot have a negative value.
## 4. How do you calculate the surface area of a sphere with a given radius?
To calculate the surface area of a sphere with a given radius, you can use the formula ##4*\pi*r^2##, where ##r## is the radius of the sphere.
## 5. What units are typically used to measure the surface area of a sphere?
The surface area of a sphere is typically measured in square units, such as square meters or square feet.
• Precalculus Mathematics Homework Help
Replies
13
Views
2K
• General Math
Replies
5
Views
1K
• Calculus and Beyond Homework Help
Replies
4
Views
287
• Calculus
Replies
33
Views
3K
• Special and General Relativity
Replies
9
Views
976
• Precalculus Mathematics Homework Help
Replies
11
Views
1K
• Introductory Physics Homework Help
Replies
6
Views
1K
• Introductory Physics Homework Help
Replies
4
Views
876
• Precalculus Mathematics Homework Help
Replies
7
Views
890
• Calculus and Beyond Homework Help
Replies
3
Views
534 | 3,564 | 12,669 | {"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": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2024-18 | latest | en | 0.863474 |
https://math.answers.com/Q/How_many_number_of_straight_sides_does_pyramid_have | 1,653,093,673,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662534693.28/warc/CC-MAIN-20220520223029-20220521013029-00797.warc.gz | 463,910,495 | 32,444 | 0
# How many number of straight sides does pyramid have?
Wiki User
โ 2013-05-20 18:03:30
A pyramid has 4 straIGHT SIDES
Wiki User
โ 2013-05-20 18:03:30
Study guides
20 cards
## A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials
โก๏ธ
See all cards
3.74
โโ
โโ
โโ
โโ
โโ
834 Reviews | 146 | 365 | {"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-2022-21 | latest | en | 0.672856 |
https://mathexamination.com/lab/indeterminate-forms.php | 1,606,667,575,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141201836.36/warc/CC-MAIN-20201129153900-20201129183900-00674.warc.gz | 406,985,500 | 8,613 | ## Take My Indeterminate Forms Lab
As mentioned over, I made use of to create an easy and also straightforward mathematics lab with only Indeterminate Forms However, the easier you make your lab, the easier it ends up being to obtain stuck at the end of it, after that at the beginning. This can be extremely frustrating, and all this can occur to you since you are utilizing Indeterminate Forms and/or Modular Equations incorrectly.
With Modular Formulas, you are currently using the incorrect formula when you obtain stuck at the beginning, if not, after that you are most likely in a stumbling block, and also there is no feasible escape. This will only worsen as the issue comes to be much more complex, but then there is the concern of how to proceed with the issue. There is no way to appropriately tackle resolving this kind of mathematics issue without being able to instantly see what is going on.
It is clear that Indeterminate Forms and Modular Formulas are tough to discover, and also it does take technique to establish your very own sense of instinct. However when you want to address a math trouble, you need to utilize a device, and the devices for finding out are utilized when you are stuck, and they are not made use of when you make the incorrect step. This is where lab Aid Solution is available in.
For instance, what is wrong with the question is incorrect ideas, such as obtaining a partial worth when you do not have sufficient working parts to finish the entire work. There is an excellent factor that this was wrong, and also it refers reasoning, not intuition. Logic enables you to adhere to a step by step procedure that makes sense, and also when you make an incorrect move, you are normally forced to either attempt to go forward as well as correct the mistake, or try to go backward and also do a backwards action.
Another circumstances is when the pupil does not recognize a step of a process. These are both logical failings, and also there is no other way around them. Even when you are embeded an area that does not enable you to make any kind of step, such as a triangular, it is still essential to understand why you are stuck, so that you can make a much better relocation and also go from the action you are stuck at to the following area.
With this in mind, the best way to fix a stuck scenario is to simply take the step forward, as opposed to trying to step. The two procedures are different in their strategy, yet they have some standard resemblances. However, when they are tried with each other, you can quickly inform which one is better at solving the problem, and you can likewise inform which one is more effective.
Let's discuss the initial instance, which associates with the Indeterminate Forms mathematics lab. This is not too complex, so let's first discuss exactly how to begin. Take the adhering to process of affixing a component to a panel to be utilized as a body. This would certainly need three dimensions, and would be something you would certainly need to affix as part of the panel.
Currently, you would have an additional measurement, however that doesn't suggest that you can simply maintain that dimension and also go from there. When you made your first step, you can conveniently ignore the dimension, and afterwards you would need to go back and backtrack your steps.
Nonetheless, instead of keeping in mind the additional measurement, you can use what is called a "psychological faster way" to aid you bear in mind that additional measurement. As you make your initial step, picture yourself taking the measurement and affixing it to the component you want to affix to, and after that see just how that makes you feel when you duplicate the procedure.
Visualisation is a very effective method, and also is something that you ought to not miss over. Envision what it would certainly seem like to actually connect the component and be able to go from there, without the dimension.
Now, allow's take a look at the 2nd instance. Let's take the very same process as before, and now the student has to remember that they are mosting likely to move back one step. If you tell them that they need to return one action, yet then you eliminate the suggestion of needing to move back one action, after that they will not recognize exactly how to proceed with the problem, they won't recognize where to try to find that action, as well as the procedure will certainly be a mess.
Instead, utilize a psychological shortcut like the mental representation to emotionally reveal them that they are mosting likely to return one action. and place them in a position where they can move forward from there. without needing to consider the missing out on a step.
## Hire Someone To Do Your Indeterminate Forms Lab
" Indeterminate Forms - Need Help with a Mathematics lab?" However, several students have had an issue comprehending the concepts of linear Indeterminate Forms. Luckily, there is a new layout for linear Indeterminate Forms that can be used to teach straight Indeterminate Forms to students who have problem with this idea. Students can use the lab Aid Service to help them learn new strategies in direct Indeterminate Forms without encountering a hill of troubles and also without needing to take a test on their concepts.
The lab Assist Solution was created in order to help battling trainees as they move from college and also secondary school to the college and also work market. Numerous students are not able to manage the stress and anxiety of the understanding procedure as well as can have very little success in realizing the ideas of linear Indeterminate Forms.
The lab Help Solution was created by the Educational Screening Solution, that uses a selection of various online examinations that pupils can take and exercise. The Test Aid Solution has actually assisted many trainees improve their scores and also can help you boost your scores as well. As students relocate from university and secondary school to the college and job market, the TTS will assist make your pupils' change much easier.
There are a few various ways that you can make the most of the lab Aid Service. The main manner in which pupils make use of the lab Aid Service is with the Response Managers, which can aid trainees learn methods in linear Indeterminate Forms, which they can use to help them do well in their programs.
There are a variety of troubles that pupils experience when they first use the lab Assist Solution. Trainees are commonly overloaded and don't comprehend how much time they will need to dedicate to the Solution. The Response Managers can assist the pupils examine their principle discovering and also help them to evaluate all of the product that they have actually already discovered in order to be planned for their next program job.
The lab Help Service works similarly that a professor does in regards to assisting trainees realize the ideas of linear Indeterminate Forms. By giving your students with the devices that they need to discover the essential ideas of linear Indeterminate Forms, you can make your students more successful throughout their studies. As a matter of fact, the lab Help Service is so effective that several students have actually changed from standard mathematics class to the lab Help Solution.
The Task Manager is made to assist pupils handle their homework. The Task Manager can be established to set up how much time the student has readily available to complete their designated homework. You can additionally set up a custom-made amount of time, which is a wonderful attribute for trainees who have a hectic schedule or an extremely active senior high school. This feature can help students avoid feeling overwhelmed with mathematics assignments.
An additional beneficial function of the lab Assist Solution is the Trainee Aide. The Trainee Aide helps students handle their work as well as provides an area to publish their research. The Student Aide is helpful for trainees that do not wish to obtain overwhelmed with responding to several inquiries.
As pupils get more comfortable with their jobs, they are urged to get in touch with the Task Supervisor and also the Trainee Aide to get an on the internet support system. The on the internet support group can aid pupils keep their focus as they answer their projects.
All of the tasks for the lab Help Service are included in the bundle. Trainees can login and complete their assigned work while having the pupil aid available in the background to help them. The lab Help Solution can be a fantastic help for your pupils as they start to browse the challenging college admissions and also work searching waters.
Pupils must be prepared to obtain made use of to their assignments as promptly as possible in order to reach their main goal of getting involved in the university. They need to strive enough to see outcomes that will permit them to stroll on at the following level of their researches. Obtaining used to the procedure of finishing their tasks is really vital.
Trainees are able to find various methods to help them learn how to use the lab Assist Solution. Discovering exactly how to use the lab Help Service is vital to pupils' success in college and job application.
## Pay Someone To Take My Indeterminate Forms Lab
Indeterminate Forms is used in a great deal of schools. Some instructors, nevertheless, do not use it really properly or utilize it improperly. This can have an unfavorable influence on the pupil's discovering.
So, when appointing tasks, make use of a great Indeterminate Forms help service to assist you with each lab. These services provide a variety of practical solutions, consisting of:
Projects might need a great deal of assessing as well as browsing on the computer system. This is when making use of a help service can be a terrific benefit. It allows you to get even more work done, enhance your understanding, and prevent a great deal of anxiety.
These kinds of homework solutions are a superb method to start dealing with the best type of help for your requirements. Indeterminate Forms is among the most hard subjects to master for trainees. Dealing with a solution, you can make sure that your demands are met, you are instructed appropriately, and also you comprehend the product correctly.
There are so many ways that you can educate yourself to function well with the course and succeed. Utilize a correct Indeterminate Forms help service to lead you as well as get the work done. Indeterminate Forms is one of the hardest courses to discover but it can be conveniently understood with the right assistance.
Having a research solution likewise helps to improve the trainee's grades. It permits you to add extra credit scores along with raise your Grade Point Average. Getting extra credit score is often a massive benefit in many colleges.
Pupils who don't take full advantage of their Indeterminate Forms course will end up continuing of the remainder of the class. Fortunately is that you can do it with a quick and also very easy solution. So, if you want to continue in your class, make use of a great help solution. One point to bear in mind is that if you truly wish to increase your grade level, your program work requires to obtain done. As high as possible, you require to understand and deal with all your troubles. You can do this with a good aid service.
One benefit of having a research solution is that you can aid on your own. If you don't feel confident in your capability to do so, then a great tutor will certainly have the ability to aid you. They will certainly be able to fix the issues you deal with and aid you recognize them in order to get a far better grade.
When you graduate from senior high school and go into university, you will certainly require to work hard in order to stay ahead of the various other students. That indicates that you will need to strive on your homework. Making use of an Indeterminate Forms service can assist you get it done.
Maintaining your grades up can be challenging since you typically need to examine a great deal as well as take a lot of examinations. You don't have time to deal with your grades alone. Having an excellent tutor can be an excellent aid since they can assist you and your research out.
An assistance solution can make it less complicated for you to manage your Indeterminate Forms class. On top of that, you can learn more concerning on your own and assist you do well. Locate the best tutoring service and you will have the ability to take your study abilities to the following degree. | 2,465 | 12,613 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2020-50 | longest | en | 0.974092 |
http://gaming.stackexchange.com/questions/92266/how-do-i-increase-the-coin-income-per-tap/92278 | 1,469,570,797,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257825124.55/warc/CC-MAIN-20160723071025-00019-ip-10-185-27-174.ec2.internal.warc.gz | 107,257,305 | 16,938 | # How do I increase the coin income per tap?
In his presentation of the Curiosity app, Peter Molyneux said that there are ways (based of rhythm, maybe?) to increase the amount of coins you earn, per tap.
Also, in the "info screen" from the Curiosity app, it is said that
there are skill mechanics that you will need to master
Tapping one cube brings one coin, basically.
What are the ways to increase that amount? Are there combo mechanics, rhythm-based scoring?
-
So far I've discovered:
Combo multipliers: you get a multiplier increasing, as long as you tap without missing, and you tap one new block in the next 2 seconds after tapping one.
So far, I don't manage to find the logic behind the actual number of cubes for the multiplier steps, but here goes for the first ones:
• 2x: 12 cubes
• 3x: 26 cubes
• 4x: 42 cubes
• 5x: 61 cubes
• etc...
"Clear Screen Bonus" - What it says on the tin. Clear your whole screen of cubes and you'll get some coins as a bonus.
The amount is equal to the score you earned on clearing those cubes in the first place (including the bonus from the multiplier), multiplied by 10.
-
After spending far too much time, I came to about the same conclusions, with more numbers for the combo. Didn't know it was proportional to the zoom level for the clean screen, though, I've just checked that, and indeed, I see it now. – Gnoupi Nov 6 '12 at 21:21
I added what I found to your answer, feel free to change things from it. – Gnoupi Nov 6 '12 at 21:34
Ah, found it, it's not proportional to the zoom level, it's simply multiplying the score, including multipliers, by 10. So what seemed proportional was only because of the increased combo chain. – Gnoupi Nov 7 '12 at 6:45
I just played a series and the best combo method I found without any bonuses is methodical and slow tapping to clear screens. Large screens seem not to be worth it, as a full screen combos at 8,000 maybe at 18x multiplier, but 6 cubes value at 600-900 at the same range. small screens that are quick clears are more desirable once you have a large multiplyer I think. Easier to track each tap as well with more white space if they are small groupings. I played at the max zoom for this usually just because I didn't want to risk losing the combo, and it because it did seem to have good returns. Further zooms might be better once you have apick.
watch out for the FB or tool icon to pop up as it will kill your combo timer.
*edit - apparently on the twitter account they suggest the best way to get rid of the FB icon is to log into facebook.... clever.
I got 200,000 coins in about 5-10 minutes by going carefully into a semi-cleared section and screen clearing and not missing a combo. my total tap meter was at about 17,000 and during that stretch I got 200,000 coins. I imagine that if I could buy the chisel at 300,000 that my screen clear time would go much faster and thus my multiplier would add up faster on the clear combo. I'd guess the 25 square tool would super bump the screen clear.
I think the best way to go is to get the pick tools and clear quickly while maximizing combos so that you can afford the tool the next time + coin gain.
-
Yes, indeed, it's better to take the time to tap the whole screen without missing a cube, and under the 2 seconds between each cube, for a much higher score, than some frenzy with tapping in every direction, with all fingers. Although if you manage to find a place without blanks, unzooming then tapping with all fingers can give good results for a start, as the smallness of the cubes makes it that your finger will almost always touch something, until you dig bigger holes. – Gnoupi Nov 6 '12 at 21:37
1. Combo + Clear Screen until 1 Mil.
2. Combo + Clear Screen + Iron chisel
In the 7 Min with the chisel you make around 1.5 - 2.5 Mil. (Net: 05. - 1.5 Mil)
But be careful: coins are getting lost at the moment
-
## protected by StrixVaria♦Nov 7 '12 at 22:55
Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count). | 1,021 | 4,163 | {"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-2016-30 | latest | en | 0.959854 |
https://www.thebalance.com/get-out-of-debt-with-debt-avalanche-4140758 | 1,618,929,382,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039398307.76/warc/CC-MAIN-20210420122023-20210420152023-00304.warc.gz | 1,122,287,093 | 59,094 | # What Is the Debt Avalanche Strategy?
## The Debt Avalanche Strategy Explained
The debt avalanche strategy involves paying off debt starting with the highest interest rate loan. When that is paid off, you apply that payment to the loan with the next highest interest rate.
## Definition and Examples of the Debt Avalanche Strategy
With the debt avalanche strategy, you pay off what you owe by prioritizing loans and credit card balances with the highest interest rates. The goal is to minimize the amount of interest you pay, and this approach might help you pay off debt faster than other strategies, like the debt snowball
## How the Debt Avalanche Works
Here’s how to save on interest with the debt avalanche method:
• Take inventory: Gather a list of everything you owe. List the debts in order of the interest rate on each loan or credit card, starting with the highest rate and working down to the lowest.
• Pay your minimums: Keep making minimum payments on all of your loans or credit card balances. You’ll focus on one balance at a time, but it’s important to stay current on the others to avoid fees and damage to your credit score.
• Pay extra on the highest rate: With any additional money you have available each month, pay extra on the highest interest rate loan. This reduces the amount you owe at that high rate.
• Build momentum: After paying off a loan, cross it off the list and redirect the amount you were paying on that loan to the loan with the next highest interest rate.
Calculate your debt avalanche pay-off schedule using a spreadsheet. If you want to create your own schedule, you can use our loan calculator to quickly determine how much interest you'll pay until a card or account is paid off.
Consider this example. Assume you owe money on the loans detailed below. Based on your monthly budget, you know that you have an extra \$150 available each month for debt elimination. Which loan should be paid off first?
List each of your loans in order of the interest rate, with the highest rate at the top.
With the debt avalanche method, the additional \$150 goes toward the credit card payment because that loan has the highest interest rate. As a result, you pay \$630 to your credit card issuer (the \$480 minimum payment plus the \$150 extra) monthly.
After paying off the credit card, that minimum payment goes away, so you have even more cash flow available on a monthly basis. The \$630 you were paying to your credit card company can now go toward the personal loan. As a result, you pay \$669.60 (\$630 plus your required \$39.60), which quickly eliminates the remaining loan balance.
Next, fold what you were paying on the personal loan into your additional payments, resulting in an additional \$669.60 per month on your student loan. The total amount you then send to the loan servicer is \$853.34 (\$669.60 plus the required \$183.74). Continue the process until you are debt-free.
## Benefits of the Debt Avalanche Strategy
The debt avalanche is an effective strategy because it focuses on interest rates. On most loans, a portion of each monthly payment goes toward interest charges, and the remainder reduces your loan balance. With high rates, you need to pay more to cover interest costs, and your payment might only make a small dent in your loan balance. By minimizing your overall interest rate, you waste less money on interest.
## Do I Need to Use the Debt Avalanche Strategy?
The debt avalanche might be a good fit if you:
• Want to minimize your total cost of borrowing
• Believe in the logic behind the strategy (paying as little interest as possible)
• Have the discipline to keep paying extra on a sizeable debt for an extended period without the satisfaction of seeing it paid off quickly
• Don’t need positive reinforcement frequently (or early in the process)
• Are motivated by facts and figures
## Debt Avalanche Strategy vs. Debt Snowball
A debt avalanche is an excellent strategy for minimizing costs and getting out of debt, but it might not be right for everyone.
Another option is the debt snowball method. With the debt snowball, you pay off your debts in order of size, from smallest to largest. The idea is that having small wins early on helps motivate you to stick with your debt reduction plan, but this method could end up costing you more in total interest.
For some, a debt snowball strategy might be a better option. That’s particularly true if you’re likely to lose motivation during your debt elimination journey. The debt snowball method provides small victories early in the process, which can help you stay disciplined and keep hope alive. With a debt avalanche, you need to trust that it’s best to pay down loans with high rates first, and it might take a long time before you pay off a loan.
Using the earlier example, with a debt avalanche, there would be no rush to pay off the interest-free medical debt because it doesn’t cost any interest. But the debt snowball would instruct you to pay off that loan first because it has the smallest loan balance.
## What to Do If You Need Help
While the debt avalanche method can be a beneficial solution for some, others may need additional guidance. Consider the following options.
### Credit Counseling
If you’re struggling with your payments, and you don’t have extra money to put toward a debt avalanche, consider asking for help. Nonprofit credit counseling agencies can provide guidance and education to help you take control of your debt. They might even set up a debt management plan (CMP), which can offer relief in the form of lower monthly payments and lower rates.
Credit counseling agencies typically charge a modest monthly fee for helping you manage your debt.
### Debt Settlement
Debt settlement is another option, although it’s a more extreme solution. With debt settlement, you attempt to pay creditors less than you owe them with help from a for-profit company. There’s no guarantee that your lender will be willing to negotiate, and debt settlement programs can lead to lower credit scores. As a result, it’s ideal to try other solutions before you try debt settlement.
### Key Takeaways
• With the debt avalanche strategy, you pay off what you owe by prioritizing loans and credit card balances with the highest interest rates.
• The debt avalanche is best for those who want to minimize interest.
• The debt snowball strategy pays off debt starting with the smallest balance, which offers small wins along the way. With this method, you may pay more in interest.
• If you can’t afford your minimum payments, speak with a nonprofit credit counseling agency, and consider debt settlement as a last resort. | 1,349 | 6,684 | {"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-2021-17 | longest | en | 0.941799 |
https://researchpaperswriter.com/1-the-waiting-time-t-between-successive-occurrences-of-an-event-e-in-a-discrete-time-renewal-proces/ | 1,643,440,537,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300573.3/warc/CC-MAIN-20220129062503-20220129092503-00223.warc.gz | 520,934,886 | 12,051 | # 1. The waiting time T between successive occurrences of an event E in a discrete time renewal proces
1. The waiting time T between successive occurrences of an event E in a discrete time renewal process has the probability distribution
P(T =1) = 0.7, P(T = 2) = 0.3.
(iii) If an observer arrives after the renewal process has been running for a long time, what is the approximate probability that an event occurs at the next time point?
2. The waiting time T between successive occurrences of an event E in a discrete-time renewal process has the probability distribution
P(T =1) = 0.7, P(T = 2) = 0.3.
(ii) The waiting time to the sixth occurrence of E is denoted by W6. Find the probability P(W6 = 10).
3. An animal leaves its burrow near a river bank and moves up and down along the river bank foraging for food. Its distance upstream from its burrow after t minutes is denoted by X(t), and may be reasonably modeled as an ordinary Brownian motion {X(t); t ≥ 0} with diffusion coefficient σ2 = 3 (meters)2 per minute.
(i) Find the probability that after 3 minutes the animal is not further than 5 metres away from its burrow.
4.
An animal leaves its burrow near a river bank and moves up and down along the river bank foraging for food. Its distance upstream from its burrow after t minutes is denoted by X(t), and may be reasonably modeled as an ordinary Brownian motion {X(t); t ≥ 0} with diffusion coefficient σ2 = 3 (meters)2 per minute.
(iii) If the animal is observed to be 30 meters upstream from its burrow after an hour, find the probability that it was downstream from its burrow after 20 minutes
5.
An animal leaves its burrow near a river bank and moves up and down along the river bank foraging for food. Its distance upstream from its burrow after t minutes is denoted by X(t), and may be reasonably modeled as an ordinary Brownian motion {X(t); t ≥ 0} with diffusion coefficient σ2 = 3 (meters)2 per minute.
(ii) If the animal is observed to be 6 meters upstream from its burrow after 5 minutes, find the probability that the animal will be less than 2 meters upstream from the burrow after 15 minutes | 523 | 2,131 | {"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} | 3.671875 | 4 | CC-MAIN-2022-05 | latest | en | 0.967031 |
https://homeworkacetutors.com/generate-a-spreadsheet-with-100-trials-replications-that-reflect-the-following-empirical/ | 1,603,393,476,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107880014.26/warc/CC-MAIN-20201022170349-20201022200349-00167.warc.gz | 370,887,348 | 23,405 | # Generate a spreadsheet with 100 trials (replications) that reflect the following empirical
2
Problem #1 7 of 32 points
Genetrounce a spreadsheet with 100 trials (replications) that animadvert the coercionthcoming experimental distribution
(undivided month = 30 days):
Middle # holders transshipped / hour Observed # days/month
100 – 199 6
200 – 299 12
300 – 399 9
400 – 499 3
1. Mathematically, how divers holders would you await to be transshipped each hour, on
average?
2. Delight interpretation the average and measure failure athwart your 100 trials to indicate how divers
containers the harbor can await to transship per hour, on middle, and how proud the variability
of that compute faculty be. Delight contribute a culm file of the values you observed (e.g., “5
to 10”). Interpretation the midpoints of the categories (e.g., 150 coercion the file 100-199) coercion these
calculations.
3. Construct a histogram. When you re-run your trials (by urgent-compulsory F9), what happens to your
mean, measure failure and the model of the distribution as shown in the diagram?
happens to the middle, measure failure and histogram? What does that average really?
Problem #2 12 of 32 points
A coal barge arrives at the generating station’s harbor sum half hour, on middle. It takes abquenched 3 hours
and 18 microscopics to unimpeach a barge, a compute that varies with the bulk of the impeach. There is very little
space in the canal coercion pursuit arriving ships. The harbor currently has 8 lodgings, and the compute of ships in
the harbor is poor to the compute of lodgings.
The harbor antecedent is regarding re-sizing the harbor by changing the compute of lodgings conducive (fewer
or over than currently), to a climax of 10. It has enumerated on you, as a Master of Supply Chain
Management, to succor indicate how divers lodgings to reconfigure to.
1. Using queueing anatomy, estimate whole manageable utilization trounces, amid the file of lodgings
considered. What is the stint compute of lodgings needed?
2. (Delight smooth up your answers to these questions to the direct integer.) How fur period would
barges consume in sum? In sum (indetermination plus lodging)? How divers barges would be indetermination in front
of the harbor, and how divers would be in the harbor, on middle?
3. What would you applaud?
3
Formulas: Wq = Lq / l
W = Wq + 1/µ = Lq/ + 1/. = Wq + Ws
L = Lq + l/µ = Lq + Ls
r = l / Mµ
Do Referable Copy or Post
This muniment is identified coercion interpretation barely by MARKUS BIEHL until October 2010. Copying or posting is an
infringement of copyright. Permissions@hbsp.harvard.edu or 617.783.7860.
Note on the Management of Queues 680-053
13
Exhibit 3 Awaited Compute of People Indetermination in Line (Lq) coercion Values of M and !/µ.
Compute Of Labor Channels, M
!/µ 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
0.10 0.0111
0.15 0.0264 0.0008
0.20 0.0500 0.0020
0.25 0.0833 0.0039
0.30 0.1285 0.0069
0.35 0.1884 0.0110
0.40 0.2666 0.0166
0.45 0.3681 0.0239 0.0019
0.50 0.5000 0.0333 0.0030
0.55 0.6722 0.0149 0.0043
0.60 0.9000 0.0593 0.0061
0.65 1.2071 0.0767 0.0084
0.70 1.6333 0.0976 0.0112
0.75 2.2500 0.1227 0.0147
0.80 3.2000 0.1523 0.0189
0.85 4.8166 0.1873 0.0239 0.0031
0.90 8.1000 0.2285 0.0300 0.0041
0.95 18.0500 0.2767 0.0371 0.0053
1.0 0.3333 0.0454 0.0067
1.2 0.6748 0.9040 0.0158
1.4 1.3449 0.1778 0.0324 0.0059
1.6 2.8444 0.3128 0.0604 0.0121
1.8 7.6731 0.5320 0.1051 0.0227 0.0047
2.0 0.8888 0.1739 0.0398 0.0090
2.2 1.4907 0.2770 0.0659 0.0158
2.4 2.1261 0.4305 0.1047 0.0266 0.0065
2.6 4.9322 0.6581 0.1609 0.0426 0.0110
2.8 12.2724 1.0000 0.2411 0.0659 0.0180
3.0 1.5282 0.3541 0.0991 0.0282 0.0077
3.2 2.3856 0.5128 0.1452 0.0427 0.0122
3.4 3.9060 0.7365 0.2085 0.0631 0.0189
3.6 7.0893 0.0550 0.2947 0.0912 0.0283 0.0084
3.8 16.9366 1.5184 0.4114 0.1292 0.0412 0.0127
4.0 2.2164 0.5694 0.1801 0.0590 0.0189
4.2 3.3269 0.7837 0.2475 0.0827 0.0273 0.0087
4.4 5.2675 1.0777 0.3364 0.1142 0.0389 0.0128
4.6 9.2885 1.4867 0.4532 0.1555 0.0541 0.0184
4.8 21.6384 2.0708 0.6071 0.2092 0.0742 0.0260
5.0 2.9375 0.8102 0.2786 1.1006 0.0361 0.0125
5.2 4.3004 1.0804 0.3680 1.1745 0.0492 0.0175
5.4 6.6609 1.4441 0.5871 1.1779 0.0663 0.0243 0.0085
5.6 11.5178 1.9436 0.6313 0.2330 0.0883 0.0330 0.0119
5.8 26.3726 2.6481 0.8225 0.3032 0.1164 0.0443 0.0164
6.0 3.6828 1.0707 0.3918 0.1513 0.0590 0.0224
6.2 5.2979 1.3967 0.5037 0.1964 0.0775 0.0300 0.0113
6.4 8.0768 1.8040 0.6454 0.2524 0.1008 0.0398 0.0153
6.6 13.7692 2.4198 0.8247 0.3222 0.1302 0.0523 0.0205
6.8 31.1270 3.2441 1.0533 0.4090 0.1666 0.0679 0.0271 0.0105
7.0 4.4471 1.3471 0.5172 0.2119 0.0876 0.0357 0.0141
7.2 6.3135 1.7288 0.6521 0.2677 0.1119 0.0463 0.0187
7.4 9.5102 2.2324 0.8202 0.3364 0.1420 0.0595 0.0245 0.0097
7.6 16.0379 2.9113 1.0310 0.4211 0.1739 0.0761 0.0318 0.0129
7.8 35.8956 3.8558 1.2972 0.5250 0.2243 0.0966 0.0410 0.0168
8.0 5.2264 1.6364 0.6530 0.2796 0.1214 0.0522 0.0220
8.2 7.3441 2.0736 0.8109 0.3469 0.1520 0.0663 0.0283
8.4 10.9592 2.6470 1.0060 0.4288 0.1891 0.0834 0.0361
8.6 18.3223 3.4160 1.2484 0.5236 0.2341 0.1043 0.0459
8.8 40.6824 4.4806 1.5524 0.6501 0.2885 0.1298 0.0577
9.0 6.0183 1.9368 0.7980 0.3543 0.1603 0.0723
9.2 8.3869 2.4298 0.9788 0.4333 0.1974 0.0899
9.4 12.4189 3.0732 1.2010 0.5287 0.2419 0.1111
9.6 20.6160 3.9318 1.4752 0.6437 0.2952 0.1367
9.8 45.4769 5.1156 1.8165 0.7827 0.3588 0.1673
10.0 6.8210 2.2465 0.9506 0.4352 0.2040
4
Problem # 3 13 of 32 points
The board beneath shows pallets of packets arriving at an unloading curtail, concurrently with their look periods.
Some pallets hold smwhole packets (S), others average (M), thus-far others vast packets (L). Using an RFID
reader, it takes 1.8 microscopics to “check in” whole smwhole packets on a pallet, 1.5 microscopics to list whole
average packets on a pallet and 1:00 microscopic to list whole vast packets on a pallet.
Look period
[hh:mm]
Pallet with
packet bulk …
0:00 S
0:02 M
0:03 L
0:05 L
0:06 S
0:09 M
0:10 M
0:11 L
0:16 S
0:18 M
0:19 S
0:20 S
In judgment the answers to the coercionthcoming questions, delight affect loose to interpretation Excel or referable – whatever works
best coercion you.
1. What is the overwhole look trounce, l?
2. What are the look trounces of pallets with smwhole packets, lS, with average packets, lM, and with
vast packets, lL?
3. What are the labor trounces coercion a pallet with smwhole (µS), average (µM) and vast packets (µL)?
What is the overwhole labor trounce, µ?
4. Assuming that the incoming pallets are treated First in-First quenched (FIFO), how divers pallets obtain
enjoy been alunitedly inventoried behind 14 microscopics?
Coercion the coercionthcoming questions, delight pretend that whole of the pallets that are shown in the board above
enjoy already arrived and are conducive coercion processing. I.e., delight overlook the look periods.
5. If you everyot the Shortest Processing Period (SPT) government, how divers pallets obtain enjoy been
alunitedly inventoried behind 14 microscopics?
6. Let’s rehearse that whole pallets with smwhole packets enjoy been staged unitedly in area S, pallets with
average packets in area M and pallets with vast packets in area L. Amid each area, you everyot
the FCFS government. Then picking from the areas is prioritized via the Longest Processing Period government
from the three areas. How divers pallets obtain enjoy been alunitedly processed behind 14 microscopics?
Pages (550 words)
Approximate price: -
Why Work with Us
Top Quality and Well-Researched Papers
We always make sure that writers follow all your instructions precisely. You can choose your academic level: high school, college/university or professional, and we will assign a writer who has a respective degree.
We have a team of professional writers with experience in academic and business writing. Many are native speakers and able to perform any task for which you need help.
Free Unlimited Revisions
If you think we missed something, send your order for a free revision. You have 10 days to submit the order for review after you have received the final document. You can do this yourself after logging into your personal account or by contacting our support.
Prompt Delivery and 100% Money-Back-Guarantee
All papers are always delivered on time. In case we need more time to master your paper, we may contact you regarding the deadline extension. In case you cannot provide us with more time, a 100% refund is guaranteed.
Original & Confidential
We use several writing tools checks to ensure that all documents you receive are free from plagiarism. Our editors carefully review all quotations in the text. We also promise maximum confidentiality in all of our services.
Our support agents are available 24 hours a day 7 days a week and committed to providing you with the best customer experience. Get in touch whenever you need any assistance.
Try it now!
## Calculate the price of your order
Total price:
\$0.00
How it works?
Fill in the order form and provide all details of your assignment.
Proceed with the payment
Choose the payment system that suits you most.
Our Services
No need to work on your paper at night. Sleep tight, we will cover your back. We offer all kinds of writing services.
## Essay Writing Service
No matter what kind of academic paper you need and how urgent you need it, you are welcome to choose your academic level and the type of your paper at an affordable price. We take care of all your paper needs and give a 24/7 customer care support system.
An admission essay is an essay or other written statement by a candidate, often a potential student enrolling in a college, university, or graduate school. You can be rest assurred that through our service we will write the best admission essay for you.
Editing Support
Our academic writers and editors make the necessary changes to your paper so that it is polished. We also format your document by correctly quoting the sources and creating reference lists in the formats APA, Harvard, MLA, Chicago / Turabian.
Revision Support
If you think your paper could be improved, you can request a review. In this case, your paper will be checked by the writer or assigned to an editor. You can use this option as many times as you see fit. This is free because we want you to be completely satisfied with the service offered.
5 to 20% OFF Discount!!
For all your orders at Homeworkacetutors.com get discounted prices!
Top quality & 100% plagiarism-free content. | 3,790 | 10,455 | {"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-2020-45 | latest | en | 0.894777 |
http://www.physicsforums.com/showpost.php?p=3422426&postcount=11 | 1,409,358,352,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1408500833525.81/warc/CC-MAIN-20140820021353-00407-ip-10-180-136-8.ec2.internal.warc.gz | 539,392,161 | 3,489 | View Single Post
P: 460
Quote by Robert1986 Well sure, what we said was a sharper result than Bertrand's Postulate. I still don't see how this is misleading (though I will agree it really doesn't have much to do with the the thread.) The fact that there is ALWAYS a prime between n and 2n for all n is interesting on its own, espicialy when you consider that I can give you an arbirarily long list of consecutive composite integers. I just posted the quote as a joke in the first place.
Yes but that list of r consecutive composite integers must start at some integer n and my gut tells me that n is much larger than r for large r. For instance if r is 100 then n is going to be much larger than 100 and of course 2n-n is the number of integers in this interval. what would be 100/n? my gut tells me it's about 0
so in a simillar fashion r/n goes to zero as r goes to infinity. Remember that r is the number of consecutive composite integers and n is the integer where this sequence starts.
Is my gut mistaken?
It's all good, you don't have to see it my way, my opinions are not 'etched in stone' | 266 | 1,100 | {"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-2014-35 | latest | en | 0.975006 |
https://www.askiitians.com/forums/9-grade-science/define-valency-by-taking-examples-of-silicon-and-o_282204.htm | 1,726,356,994,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651601.85/warc/CC-MAIN-20240914225323-20240915015323-00700.warc.gz | 601,094,986 | 42,726 | # Define valency by taking examples of silicon and oxygen.Define valency by taking examples of silicon and oxygen.
Harshit Singh
3 years ago
Dear
Valency is the combining capacity of an atom.
Atomic number of oxygen = 8
Atomic number of silicon = 14
K L M
Electronic configuration of oxygen = 2 6 –
Electronic configuration of silicon = 2 8 4
In the atoms of oxygen the valence electrons are 6 (i.e., electrons in the outermost shell). To fill the orbit, 2 electrons are required. In the atom of silicon, the valence electrons are 4. To fill this orbit 4 electrons are required.
Hence, the combining capacity of oxygen is 2 and of silicon is 4.
i.e., Valency of oxygen = 2
Valency of silicon = 4
Thanks | 182 | 705 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2024-38 | latest | en | 0.857741 |
https://www.convert-measurement-units.com/convert+Slug+per+second+to+Kilogram+per+minute.php | 1,716,672,246,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058834.56/warc/CC-MAIN-20240525192227-20240525222227-00515.warc.gz | 630,792,816 | 13,441 | Convert slug/s to kg/min (Slug per second to Kilogram per minute)
Slug per second into Kilogram per minute
numbers in scientific notation
https://www.convert-measurement-units.com/convert+Slug+per+second+to+Kilogram+per+minute.php
Convert Slug per second to Kilogram per minute (slug/s to kg/min):
1. Choose the right category from the selection list, in this case 'Mass flow rate'.
2. Next enter the value you want to convert. The basic operations of arithmetic: addition (+), subtraction (-), multiplication (*, x), division (/, :, ÷), exponent (^), square root (√), brackets and π (pi) are all permitted at this point.
3. From the selection list, choose the unit that corresponds to the value you want to convert, in this case 'Slug per second [slug/s]'.
4. Finally choose the unit you want the value to be converted to, in this case 'Kilogram per minute [kg/min]'.
5. Then, when the result appears, there is still the possibility of rounding it to a specific number of decimal places, whenever it makes sense to do so.
With this calculator, it is possible to enter the value to be converted together with the original measurement unit; for example, '256 Slug per second'. In so doing, either the full name of the unit or its abbreviation can be usedas an example, either 'Slug per second' or 'slug/s'. Then, the calculator determines the category of the measurement unit of measure that is to be converted, in this case 'Mass flow rate'. After that, it converts the entered value into all of the appropriate units known to it. In the resulting list, you will be sure also to find the conversion you originally sought. Alternatively, the value to be converted can be entered as follows: '50 slug/s to kg/min' or '32 slug/s into kg/min' or '72 Slug per second -> Kilogram per minute' or '84 slug/s = kg/min' or '87 Slug per second to kg/min' or '9 slug/s to Kilogram per minute' or '94 Slug per second into Kilogram per minute'. For this alternative, the calculator also figures out immediately into which unit the original value is specifically to be converted. Regardless which of these possibilities one uses, it saves one the cumbersome search for the appropriate listing in long selection lists with myriad categories and countless supported units. All of that is taken over for us by the calculator and it gets the job done in a fraction of a second.
Furthermore, the calculator makes it possible to use mathematical expressions. As a result, not only can numbers be reckoned with one another, such as, for example, '(68 * 34) slug/s'. But different units of measurement can also be coupled with one another directly in the conversion. That could, for example, look like this: '256 Slug per second + 768 Kilogram per minute' or '50mm x 62cm x 81dm = ? cm^3'. The units of measure combined in this way naturally have to fit together and make sense in the combination in question.
The mathematical functions sin, cos, tan and sqrt can also be used. Example: sin(π/2), cos(pi/2), tan(90°), sin(90) or sqrt(4).
If a check mark has been placed next to 'Numbers in scientific notation', the answer will appear as an exponential. For example, 6.024 297 476 043 1×1030. For this form of presentation, the number will be segmented into an exponent, here 30, and the actual number, here 6.024 297 476 043 1. For devices on which the possibilities for displaying numbers are limited, such as for example, pocket calculators, one also finds the way of writing numbers as 6.024 297 476 043 1E+30. In particular, this makes very large and very small numbers easier to read. If a check mark has not been placed at this spot, then the result is given in the customary way of writing numbers. For the above example, it would then look like this: 6 024 297 476 043 100 000 000 000 000 000. Independent of the presentation of the results, the maximum precision of this calculator is 14 places. That should be precise enough for most applications. | 924 | 3,946 | {"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} | 3.25 | 3 | CC-MAIN-2024-22 | latest | en | 0.82049 |
https://stacks.math.columbia.edu/tag/02P5 | 1,720,860,957,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514493.69/warc/CC-MAIN-20240713083241-20240713113241-00206.warc.gz | 429,686,689 | 13,259 | ### 42.68.1 Determinants of finite length modules
The material in this section is related to the material in the paper and to the material in the thesis [Joe].
Let $(R, \mathfrak m, \kappa )$ be a local ring. Let $\varphi : M \to M$ be an $R$-linear endomorphism of a finite length $R$-module $M$. In More on Algebra, Section 15.120 we have already defined the determinant $\det _\kappa (\varphi )$ (and the trace and the characteristic polynomial) of $\varphi$ relative to $\kappa$. In this section, we will construct a canonical $1$-dimensional $\kappa$-vector space $\det _\kappa (M)$ such that $\det _\kappa (\varphi : M \to M) : \det _\kappa (M) \to \det _\kappa (M)$ is equal to multiplication by $\det _\kappa (\varphi )$. If $M$ is annihilated by $\mathfrak m$, then $M$ can be viewed as a finite dimension $\kappa$-vector space and then we have $\det _\kappa (M) = \wedge ^ n_\kappa (M)$ where $n = \dim _\kappa (M)$. Our construction will generalize this to all finite length modules over $R$ and if $R$ contains its residue field, then the determinant $\det _\kappa (M)$ will be given by the usual determinant in a suitable sense, see Remark 42.68.9.
Definition 42.68.2. Let $R$ be a local ring with maximal ideal $\mathfrak m$ and residue field $\kappa$. Let $M$ be a finite length $R$-module. Say $l = \text{length}_ R(M)$.
1. Given elements $x_1, \ldots , x_ r \in M$ we denote $\langle x_1, \ldots , x_ r \rangle = Rx_1 + \ldots + Rx_ r$ the $R$-submodule of $M$ generated by $x_1, \ldots , x_ r$.
2. We will say an $l$-tuple of elements $(e_1, \ldots , e_ l)$ of $M$ is admissible if $\mathfrak m e_ i \subset \langle e_1, \ldots , e_{i - 1} \rangle$ for $i = 1, \ldots , l$.
3. A symbol $[e_1, \ldots , e_ l]$ will mean $(e_1, \ldots , e_ l)$ is an admissible $l$-tuple.
4. An admissible relation between symbols is one of the following:
1. if $(e_1, \ldots , e_ l)$ is an admissible sequence and for some $1 \leq a \leq l$ we have $e_ a \in \langle e_1, \ldots , e_{a - 1}\rangle$, then $[e_1, \ldots , e_ l] = 0$,
2. if $(e_1, \ldots , e_ l)$ is an admissible sequence and for some $1 \leq a \leq l$ we have $e_ a = \lambda e'_ a + x$ with $\lambda \in R^*$, and $x \in \langle e_1, \ldots , e_{a - 1}\rangle$, then
$[e_1, \ldots , e_ l] = \overline{\lambda } [e_1, \ldots , e_{a - 1}, e'_ a, e_{a + 1}, \ldots , e_ l]$
where $\overline{\lambda } \in \kappa ^*$ is the image of $\lambda$ in the residue field, and
3. if $(e_1, \ldots , e_ l)$ is an admissible sequence and $\mathfrak m e_ a \subset \langle e_1, \ldots , e_{a - 2}\rangle$ then
$[e_1, \ldots , e_ l] = - [e_1, \ldots , e_{a - 2}, e_ a, e_{a - 1}, e_{a + 1}, \ldots , e_ l].$
5. We define the determinant of the finite length $R$-module $M$ to be
$\det \nolimits _\kappa (M) = \left\{ \frac{\kappa \text{-vector space generated by symbols}}{\kappa \text{-linear combinations of admissible relations}} \right\}$
We stress that always $l = \text{length}_ R(M)$. We also stress that it does not follow that the symbol $[e_1, \ldots , e_ l]$ is additive in the entries (this will typically not be the case). Before we can show that the determinant $\det _\kappa (M)$ actually has dimension $1$ we have to show that it has dimension at most $1$.
Lemma 42.68.3. With notations as above we have $\dim _\kappa (\det _\kappa (M)) \leq 1$.
Proof. Fix an admissible sequence $(f_1, \ldots , f_ l)$ of $M$ such that
$\text{length}_ R(\langle f_1, \ldots , f_ i\rangle ) = i$
for $i = 1, \ldots , l$. Such an admissible sequence exists exactly because $M$ has length $l$. We will show that any element of $\det _\kappa (M)$ is a $\kappa$-multiple of the symbol $[f_1, \ldots , f_ l]$. This will prove the lemma.
Let $(e_1, \ldots , e_ l)$ be an admissible sequence of $M$. It suffices to show that $[e_1, \ldots , e_ l]$ is a multiple of $[f_1, \ldots , f_ l]$. First assume that $\langle e_1, \ldots , e_ l\rangle \not= M$. Then there exists an $i \in [1, \ldots , l]$ such that $e_ i \in \langle e_1, \ldots , e_{i - 1}\rangle$. It immediately follows from the first admissible relation that $[e_1, \ldots , e_ n] = 0$ in $\det _\kappa (M)$. Hence we may assume that $\langle e_1, \ldots , e_ l\rangle = M$. In particular there exists a smallest index $i \in \{ 1, \ldots , l\}$ such that $f_1 \in \langle e_1, \ldots , e_ i\rangle$. This means that $e_ i = \lambda f_1 + x$ with $x \in \langle e_1, \ldots , e_{i - 1}\rangle$ and $\lambda \in R^*$. By the second admissible relation this means that $[e_1, \ldots , e_ l] = \overline{\lambda }[e_1, \ldots , e_{i - 1}, f_1, e_{i + 1}, \ldots , e_ l]$. Note that $\mathfrak m f_1 = 0$. Hence by applying the third admissible relation $i - 1$ times we see that
$[e_1, \ldots , e_ l] = (-1)^{i - 1}\overline{\lambda } [f_1, e_1, \ldots , e_{i - 1}, e_{i + 1}, \ldots , e_ l].$
Note that it is also the case that $\langle f_1, e_1, \ldots , e_{i - 1}, e_{i + 1}, \ldots , e_ l\rangle = M$. By induction suppose we have proven that our original symbol is equal to a scalar times
$[f_1, \ldots , f_ j, e_{j + 1}, \ldots , e_ l]$
for some admissible sequence $(f_1, \ldots , f_ j, e_{j + 1}, \ldots , e_ l)$ whose elements generate $M$, i.e., with $\langle f_1, \ldots , f_ j, e_{j + 1}, \ldots , e_ l\rangle = M$. Then we find the smallest $i$ such that $f_{j + 1} \in \langle f_1, \ldots , f_ j, e_{j + 1}, \ldots , e_ i\rangle$ and we go through the same process as above to see that
$[f_1, \ldots , f_ j, e_{j + 1}, \ldots , e_ l] = (\text{scalar}) [f_1, \ldots , f_ j, f_{j + 1}, e_{j + 1}, \ldots , \hat{e_ i}, \ldots , e_ l]$
Continuing in this vein we obtain the desired result. $\square$
Before we show that $\det _\kappa (M)$ always has dimension $1$, let us show that it agrees with the usual top exterior power in the case the module is a vector space over $\kappa$.
Lemma 42.68.4. Let $R$ be a local ring with maximal ideal $\mathfrak m$ and residue field $\kappa$. Let $M$ be a finite length $R$-module which is annihilated by $\mathfrak m$. Let $l = \dim _\kappa (M)$. Then the map
$\det \nolimits _\kappa (M) \longrightarrow \wedge ^ l_\kappa (M), \quad [e_1, \ldots , e_ l] \longmapsto e_1 \wedge \ldots \wedge e_ l$
is an isomorphism.
Proof. It is clear that the rule described in the lemma gives a $\kappa$-linear map since all of the admissible relations are satisfied by the usual symbols $e_1 \wedge \ldots \wedge e_ l$. It is also clearly a surjective map. Since by Lemma 42.68.3 the left hand side has dimension at most one we see that the map is an isomorphism. $\square$
Lemma 42.68.5. Let $R$ be a local ring with maximal ideal $\mathfrak m$ and residue field $\kappa$. Let $M$ be a finite length $R$-module. The determinant $\det _\kappa (M)$ defined above is a $\kappa$-vector space of dimension $1$. It is generated by the symbol $[f_1, \ldots , f_ l]$ for any admissible sequence such that $\langle f_1, \ldots f_ l \rangle = M$.
Proof. We know $\det _\kappa (M)$ has dimension at most $1$, and in fact that it is generated by $[f_1, \ldots , f_ l]$, by Lemma 42.68.3 and its proof. We will show by induction on $l = \text{length}(M)$ that it is nonzero. For $l = 1$ it follows from Lemma 42.68.4. Choose a nonzero element $f \in M$ with $\mathfrak m f = 0$. Set $\overline{M} = M /\langle f \rangle$, and denote the quotient map $x \mapsto \overline{x}$. We will define a surjective map
$\psi : \det \nolimits _ k(M) \to \det \nolimits _\kappa (\overline{M})$
which will prove the lemma since by induction the determinant of $\overline{M}$ is nonzero.
We define $\psi$ on symbols as follows. Let $(e_1, \ldots , e_ l)$ be an admissible sequence. If $f \not\in \langle e_1, \ldots , e_ l \rangle$ then we simply set $\psi ([e_1, \ldots , e_ l]) = 0$. If $f \in \langle e_1, \ldots , e_ l \rangle$ then we choose an $i$ minimal such that $f \in \langle e_1, \ldots , e_ i \rangle$. We may write $e_ i = \lambda f + x$ for some unit $\lambda \in R$ and $x \in \langle e_1, \ldots , e_{i - 1} \rangle$. In this case we set
$\psi ([e_1, \ldots , e_ l]) = (-1)^ i \overline{\lambda }[\overline{e}_1, \ldots , \overline{e}_{i - 1}, \overline{e}_{i + 1}, \ldots , \overline{e}_ l].$
Note that it is indeed the case that $(\overline{e}_1, \ldots , \overline{e}_{i - 1}, \overline{e}_{i + 1}, \ldots , \overline{e}_ l)$ is an admissible sequence in $\overline{M}$, so this makes sense. Let us show that extending this rule $\kappa$-linearly to linear combinations of symbols does indeed lead to a map on determinants. To do this we have to show that the admissible relations are mapped to zero.
Type (a) relations. Suppose we have $(e_1, \ldots , e_ l)$ an admissible sequence and for some $1 \leq a \leq l$ we have $e_ a \in \langle e_1, \ldots , e_{a - 1}\rangle$. Suppose that $f \in \langle e_1, \ldots , e_ i\rangle$ with $i$ minimal. Then $i \not= a$ and $\overline{e}_ a \in \langle \overline{e}_1, \ldots , \hat{\overline{e}_ i}, \ldots , \overline{e}_{a - 1}\rangle$ if $i < a$ or $\overline{e}_ a \in \langle \overline{e}_1, \ldots , \overline{e}_{a - 1}\rangle$ if $i > a$. Thus the same admissible relation for $\det _\kappa (\overline{M})$ forces the symbol $[\overline{e}_1, \ldots , \overline{e}_{i - 1}, \overline{e}_{i + 1}, \ldots , \overline{e}_ l]$ to be zero as desired.
Type (b) relations. Suppose we have $(e_1, \ldots , e_ l)$ an admissible sequence and for some $1 \leq a \leq l$ we have $e_ a = \lambda e'_ a + x$ with $\lambda \in R^*$, and $x \in \langle e_1, \ldots , e_{a - 1}\rangle$. Suppose that $f \in \langle e_1, \ldots , e_ i\rangle$ with $i$ minimal. Say $e_ i = \mu f + y$ with $y \in \langle e_1, \ldots , e_{i - 1}\rangle$. If $i < a$ then the desired equality is
$(-1)^ i \overline{\lambda } [\overline{e}_1, \ldots , \overline{e}_{i - 1}, \overline{e}_{i + 1}, \ldots , \overline{e}_ l] = (-1)^ i \overline{\lambda } [\overline{e}_1, \ldots , \overline{e}_{i - 1}, \overline{e}_{i + 1}, \ldots , \overline{e}_{a - 1}, \overline{e}'_ a, \overline{e}_{a + 1}, \ldots , \overline{e}_ l]$
which follows from $\overline{e}_ a = \lambda \overline{e}'_ a + \overline{x}$ and the corresponding admissible relation for $\det _\kappa (\overline{M})$. If $i > a$ then the desired equality is
$(-1)^ i \overline{\lambda } [\overline{e}_1, \ldots , \overline{e}_{i - 1}, \overline{e}_{i + 1}, \ldots , \overline{e}_ l] = (-1)^ i \overline{\lambda } [\overline{e}_1, \ldots , \overline{e}_{a - 1}, \overline{e}'_ a, \overline{e}_{a + 1}, \ldots , \overline{e}_{i - 1}, \overline{e}_{i + 1}, \ldots , \overline{e}_ l]$
which follows from $\overline{e}_ a = \lambda \overline{e}'_ a + \overline{x}$ and the corresponding admissible relation for $\det _\kappa (\overline{M})$. The interesting case is when $i = a$. In this case we have $e_ a = \lambda e'_ a + x = \mu f + y$. Hence also $e'_ a = \lambda ^{-1}(\mu f + y - x)$. Thus we see that
$\psi ([e_1, \ldots , e_ l]) = (-1)^ i \overline{\mu } [\overline{e}_1, \ldots , \overline{e}_{i - 1}, \overline{e}_{i + 1}, \ldots , \overline{e}_ l] = \psi ( \overline{\lambda } [e_1, \ldots , e_{a - 1}, e'_ a, e_{a + 1}, \ldots , e_ l] )$
as desired.
Type (c) relations. Suppose that $(e_1, \ldots , e_ l)$ is an admissible sequence and $\mathfrak m e_ a \subset \langle e_1, \ldots , e_{a - 2}\rangle$. Suppose that $f \in \langle e_1, \ldots , e_ i\rangle$ with $i$ minimal. Say $e_ i = \lambda f + x$ with $x \in \langle e_1, \ldots , e_{i - 1}\rangle$. We distinguish $4$ cases:
Case 1: $i < a - 1$. The desired equality is
\begin{align*} & (-1)^ i \overline{\lambda } [\overline{e}_1, \ldots , \overline{e}_{i - 1}, \overline{e}_{i + 1}, \ldots , \overline{e}_ l] \\ & = (-1)^{i + 1} \overline{\lambda } [\overline{e}_1, \ldots , \overline{e}_{i - 1}, \overline{e}_{i + 1}, \ldots , \overline{e}_{a - 2}, \overline{e}_ a, \overline{e}_{a - 1}, \overline{e}_{a + 1}, \ldots , \overline{e}_ l] \end{align*}
which follows from the type (c) admissible relation for $\det _\kappa (\overline{M})$.
Case 2: $i > a$. The desired equality is
\begin{align*} & (-1)^ i \overline{\lambda } [\overline{e}_1, \ldots , \overline{e}_{i - 1}, \overline{e}_{i + 1}, \ldots , \overline{e}_ l] \\ & = (-1)^{i + 1} \overline{\lambda } [\overline{e}_1, \ldots , \overline{e}_{a - 2}, \overline{e}_ a, \overline{e}_{a - 1}, \overline{e}_{a + 1}, \ldots , \overline{e}_{i - 1}, \overline{e}_{i + 1}, \ldots , \overline{e}_ l] \end{align*}
which follows from the type (c) admissible relation for $\det _\kappa (\overline{M})$.
Case 3: $i = a$. We write $e_ a = \lambda f + \mu e_{a - 1} + y$ with $y \in \langle e_1, \ldots , e_{a - 2}\rangle$. Then
$\psi ([e_1, \ldots , e_ l]) = (-1)^ a \overline{\lambda } [\overline{e}_1, \ldots , \overline{e}_{a - 1}, \overline{e}_{a + 1}, \ldots , \overline{e}_ l]$
by definition. If $\overline{\mu }$ is nonzero, then we have $e_{a - 1} = - \mu ^{-1} \lambda f + \mu ^{-1}e_ a - \mu ^{-1} y$ and we obtain
$\psi (-[e_1, \ldots , e_{a - 2}, e_ a, e_{a - 1}, e_{a + 1}, \ldots , e_ l]) = (-1)^ a \overline{\mu ^{-1}\lambda } [\overline{e}_1, \ldots , \overline{e}_{a - 2}, \overline{e}_ a, \overline{e}_{a + 1}, \ldots , \overline{e}_ l]$
by definition. Since in $\overline{M}$ we have $\overline{e}_ a = \mu \overline{e}_{a - 1} + \overline{y}$ we see the two outcomes are equal by relation (a) for $\det _\kappa (\overline{M})$. If on the other hand $\overline{\mu }$ is zero, then we can write $e_ a = \lambda f + y$ with $y \in \langle e_1, \ldots , e_{a - 2}\rangle$ and we have
$\psi (-[e_1, \ldots , e_{a - 2}, e_ a, e_{a - 1}, e_{a + 1}, \ldots , e_ l]) = (-1)^ a \overline{\lambda } [\overline{e}_1, \ldots , \overline{e}_{a - 1}, \overline{e}_{a + 1}, \ldots , \overline{e}_ l]$
which is equal to $\psi ([e_1, \ldots , e_ l])$.
Case 4: $i = a - 1$. Here we have
$\psi ([e_1, \ldots , e_ l]) = (-1)^{a - 1} \overline{\lambda } [\overline{e}_1, \ldots , \overline{e}_{a - 2}, \overline{e}_ a, \ldots , \overline{e}_ l]$
by definition. If $f \not\in \langle e_1, \ldots , e_{a - 2}, e_ a \rangle$ then
$\psi (-[e_1, \ldots , e_{a - 2}, e_ a, e_{a - 1}, e_{a + 1}, \ldots , e_ l]) = (-1)^{a + 1}\overline{\lambda } [\overline{e}_1, \ldots , \overline{e}_{a - 2}, \overline{e}_ a, \ldots , \overline{e}_ l]$
Since $(-1)^{a - 1} = (-1)^{a + 1}$ the two expressions are the same. Finally, assume $f \in \langle e_1, \ldots , e_{a - 2}, e_ a \rangle$. In this case we see that $e_{a - 1} = \lambda f + x$ with $x \in \langle e_1, \ldots , e_{a - 2}\rangle$ and $e_ a = \mu f + y$ with $y \in \langle e_1, \ldots , e_{a - 2}\rangle$ for units $\lambda , \mu \in R$. We conclude that both $e_ a \in \langle e_1, \ldots , e_{a - 1} \rangle$ and $e_{a - 1} \in \langle e_1, \ldots , e_{a - 2}, e_ a\rangle$. In this case a relation of type (a) applies to both $[e_1, \ldots , e_ l]$ and $[e_1, \ldots , e_{a - 2}, e_ a, e_{a - 1}, e_{a + 1}, \ldots , e_ l]$ and the compatibility of $\psi$ with these shown above to see that both
$\psi ([e_1, \ldots , e_ l]) \quad \text{and}\quad \psi ([e_1, \ldots , e_{a - 2}, e_ a, e_{a - 1}, e_{a + 1}, \ldots , e_ l])$
are zero, as desired.
At this point we have shown that $\psi$ is well defined, and all that remains is to show that it is surjective. To see this let $(\overline{f}_2, \ldots , \overline{f}_ l)$ be an admissible sequence in $\overline{M}$. We can choose lifts $f_2, \ldots , f_ l \in M$, and then $(f, f_2, \ldots , f_ l)$ is an admissible sequence in $M$. Since $\psi ([f, f_2, \ldots , f_ l]) = [f_2, \ldots , f_ l]$ we win. $\square$
Let $R$ be a local ring with maximal ideal $\mathfrak m$ and residue field $\kappa$. Note that if $\varphi : M \to N$ is an isomorphism of finite length $R$-modules, then we get an isomorphism
$\det \nolimits _\kappa (\varphi ) : \det \nolimits _\kappa (M) \to \det \nolimits _\kappa (N)$
simply by the rule
$\det \nolimits _\kappa (\varphi )([e_1, \ldots , e_ l]) = [\varphi (e_1), \ldots , \varphi (e_ l)]$
for any symbol $[e_1, \ldots , e_ l]$ for $M$. Hence we see that $\det \nolimits _\kappa$ is a functor
42.68.5.1
$$\label{chow-equation-functor} \left\{ \begin{matrix} \text{finite length }R\text{-modules} \\ \text{with isomorphisms} \end{matrix} \right\} \longrightarrow \left\{ \begin{matrix} 1\text{-dimensional }\kappa \text{-vector spaces} \\ \text{with isomorphisms} \end{matrix} \right\}$$
This is typical for a “determinant functor” (see [Knudsen]), as is the following additivity property.
Lemma 42.68.6. Let $(R, \mathfrak m, \kappa )$ be a local ring. For every short exact sequence
$0 \to K \to L \to M \to 0$
of finite length $R$-modules there exists a canonical isomorphism
$\gamma _{K \to L \to M} : \det \nolimits _\kappa (K) \otimes _\kappa \det \nolimits _\kappa (M) \longrightarrow \det \nolimits _\kappa (L)$
defined by the rule on nonzero symbols
$[e_1, \ldots , e_ k] \otimes [\overline{f}_1, \ldots , \overline{f}_ m] \longrightarrow [e_1, \ldots , e_ k, f_1, \ldots , f_ m]$
with the following properties:
1. For every isomorphism of short exact sequences, i.e., for every commutative diagram
$\xymatrix{ 0 \ar[r] & K \ar[r] \ar[d]^ u & L \ar[r] \ar[d]^ v & M \ar[r] \ar[d]^ w & 0 \\ 0 \ar[r] & K' \ar[r] & L' \ar[r] & M' \ar[r] & 0 }$
with short exact rows and isomorphisms $u, v, w$ we have
$\gamma _{K' \to L' \to M'} \circ (\det \nolimits _\kappa (u) \otimes \det \nolimits _\kappa (w)) = \det \nolimits _\kappa (v) \circ \gamma _{K \to L \to M},$
2. for every commutative square of finite length $R$-modules with exact rows and columns
$\xymatrix{ & 0 \ar[d] & 0 \ar[d] & 0 \ar[d] & \\ 0 \ar[r] & A \ar[r] \ar[d] & B \ar[r] \ar[d] & C \ar[r] \ar[d] & 0 \\ 0 \ar[r] & D \ar[r] \ar[d] & E \ar[r] \ar[d] & F \ar[r] \ar[d] & 0 \\ 0 \ar[r] & G \ar[r] \ar[d] & H \ar[r] \ar[d] & I \ar[r] \ar[d] & 0 \\ & 0 & 0 & 0 & }$
the following diagram is commutative
$\xymatrix{ \det \nolimits _\kappa (A) \otimes \det \nolimits _\kappa (C) \otimes \det \nolimits _\kappa (G) \otimes \det \nolimits _\kappa (I) \ar[dd]_{\epsilon } \ar[rrr]_-{\gamma _{A \to B \to C} \otimes \gamma _{G \to H \to I}} & & & \det \nolimits _\kappa (B) \otimes \det \nolimits _\kappa (H) \ar[d]^{\gamma _{B \to E \to H}} \\ & & & \det \nolimits _\kappa (E) \\ \det \nolimits _\kappa (A) \otimes \det \nolimits _\kappa (G) \otimes \det \nolimits _\kappa (C) \otimes \det \nolimits _\kappa (I) \ar[rrr]^-{\gamma _{A \to D \to G} \otimes \gamma _{C \to F \to I}} & & & \det \nolimits _\kappa (D) \otimes \det \nolimits _\kappa (F) \ar[u]_{\gamma _{D \to E \to F}} }$
where $\epsilon$ is the switch of the factors in the tensor product times $(-1)^{cg}$ with $c = \text{length}_ R(C)$ and $g = \text{length}_ R(G)$, and
3. the map $\gamma _{K \to L \to M}$ agrees with the usual isomorphism if $0 \to K \to L \to M \to 0$ is actually a short exact sequence of $\kappa$-vector spaces.
Proof. The significance of taking nonzero symbols in the explicit description of the map $\gamma _{K \to L \to M}$ is simply that if $(e_1, \ldots , e_ l)$ is an admissible sequence in $K$, and $(\overline{f}_1, \ldots , \overline{f}_ m)$ is an admissible sequence in $M$, then it is not guaranteed that $(e_1, \ldots , e_ l, f_1, \ldots , f_ m)$ is an admissible sequence in $L$ (where of course $f_ i \in L$ signifies a lift of $\overline{f}_ i$). However, if the symbol $[e_1, \ldots , e_ l]$ is nonzero in $\det _\kappa (K)$, then necessarily $K = \langle e_1, \ldots , e_ k\rangle$ (see proof of Lemma 42.68.3), and in this case it is true that $(e_1, \ldots , e_ k, f_1, \ldots , f_ m)$ is an admissible sequence. Moreover, by the admissible relations of type (b) for $\det _\kappa (L)$ we see that the value of $[e_1, \ldots , e_ k, f_1, \ldots , f_ m]$ in $\det _\kappa (L)$ is independent of the choice of the lifts $f_ i$ in this case also. Given this remark, it is clear that an admissible relation for $e_1, \ldots , e_ k$ in $K$ translates into an admissible relation among $e_1, \ldots , e_ k, f_1, \ldots , f_ m$ in $L$, and similarly for an admissible relation among the $\overline{f}_1, \ldots , \overline{f}_ m$. Thus $\gamma$ defines a linear map of vector spaces as claimed in the lemma.
By Lemma 42.68.5 we know $\det _\kappa (L)$ is generated by any single symbol $[x_1, \ldots , x_{k + m}]$ such that $(x_1, \ldots , x_{k + m})$ is an admissible sequence with $L = \langle x_1, \ldots , x_{k + m}\rangle$. Hence it is clear that the map $\gamma _{K \to L \to M}$ is surjective and hence an isomorphism.
Property (1) holds because
\begin{eqnarray*} & & \det \nolimits _\kappa (v)([e_1, \ldots , e_ k, f_1, \ldots , f_ m]) \\ & = & [v(e_1), \ldots , v(e_ k), v(f_1), \ldots , v(f_ m)] \\ & = & \gamma _{K' \to L' \to M'}([u(e_1), \ldots , u(e_ k)] \otimes [w(f_1), \ldots , w(f_ m)]). \end{eqnarray*}
Property (2) means that given a symbol $[\alpha _1, \ldots , \alpha _ a]$ generating $\det _\kappa (A)$, a symbol $[\gamma _1, \ldots , \gamma _ c]$ generating $\det _\kappa (C)$, a symbol $[\zeta _1, \ldots , \zeta _ g]$ generating $\det _\kappa (G)$, and a symbol $[\iota _1, \ldots , \iota _ i]$ generating $\det _\kappa (I)$ we have
\begin{eqnarray*} & & [\alpha _1, \ldots , \alpha _ a, \tilde\gamma _1, \ldots , \tilde\gamma _ c, \tilde\zeta _1, \ldots , \tilde\zeta _ g, \tilde\iota _1, \ldots , \tilde\iota _ i] \\ & = & (-1)^{cg} [\alpha _1, \ldots , \alpha _ a, \tilde\zeta _1, \ldots , \tilde\zeta _ g, \tilde\gamma _1, \ldots , \tilde\gamma _ c, \tilde\iota _1, \ldots , \tilde\iota _ i] \end{eqnarray*}
(for suitable lifts $\tilde{x}$ in $E$) in $\det _\kappa (E)$. This holds because we may use the admissible relations of type (c) $cg$ times in the following order: move the $\tilde\zeta _1$ past the elements $\tilde\gamma _ c, \ldots , \tilde\gamma _1$ (allowed since $\mathfrak m\tilde\zeta _1 \subset A$), then move $\tilde\zeta _2$ past the elements $\tilde\gamma _ c, \ldots , \tilde\gamma _1$ (allowed since $\mathfrak m\tilde\zeta _2 \subset A + R\tilde\zeta _1$), and so on.
Part (3) of the lemma is obvious. This finishes the proof. $\square$
We can use the maps $\gamma$ of the lemma to define more general maps $\gamma$ as follows. Suppose that $(R, \mathfrak m, \kappa )$ is a local ring. Let $M$ be a finite length $R$-module and suppose we are given a finite filtration (see Homology, Definition 12.19.1)
$0 = F^ m \subset F^{m - 1} \subset \ldots \subset F^{n + 1} \subset F^ n = M$
then there is a well defined and canonical isomorphism
$\gamma _{(M, F)} : \det \nolimits _\kappa (F^{m - 1}/F^ m) \otimes _\kappa \ldots \otimes _ k \det \nolimits _\kappa (F^ n/F^{n + 1}) \longrightarrow \det \nolimits _\kappa (M)$
To construct it we use isomorphisms of Lemma 42.68.6 coming from the short exact sequences $0 \to F^{i - 1}/F^ i \to M/F^ i \to M/F^{i - 1} \to 0$. Part (2) of Lemma 42.68.6 with $G = 0$ shows we obtain the same isomorphism if we use the short exact sequences $0 \to F^ i \to F^{i - 1} \to F^{i - 1}/F^ i \to 0$.
Here is another typical result for determinant functors. It is not hard to show. The tricky part is usually to show the existence of a determinant functor.
Lemma 42.68.7. Let $(R, \mathfrak m, \kappa )$ be any local ring. The functor
$\det \nolimits _\kappa : \left\{ \begin{matrix} \text{finite length }R\text{-modules} \\ \text{with isomorphisms} \end{matrix} \right\} \longrightarrow \left\{ \begin{matrix} 1\text{-dimensional }\kappa \text{-vector spaces} \\ \text{with isomorphisms} \end{matrix} \right\}$
endowed with the maps $\gamma _{K \to L \to M}$ is characterized by the following properties
1. its restriction to the subcategory of modules annihilated by $\mathfrak m$ is isomorphic to the usual determinant functor (see Lemma 42.68.4), and
2. (1), (2) and (3) of Lemma 42.68.6 hold.
Proof. Omitted. $\square$
Lemma 42.68.8. Let $(R', \mathfrak m') \to (R, \mathfrak m)$ be a local ring homomorphism which induces an isomorphism on residue fields $\kappa$. Then for every finite length $R$-module the restriction $M_{R'}$ is a finite length $R'$-module and there is a canonical isomorphism
$\det \nolimits _{R, \kappa }(M) \longrightarrow \det \nolimits _{R', \kappa }(M_{R'})$
This isomorphism is functorial in $M$ and compatible with the isomorphisms $\gamma _{K \to L \to M}$ of Lemma 42.68.6 defined for $\det _{R, \kappa }$ and $\det _{R', \kappa }$.
Proof. If the length of $M$ as an $R$-module is $l$, then the length of $M$ as an $R'$-module (i.e., $M_{R'}$) is $l$ as well, see Algebra, Lemma 10.52.12. Note that an admissible sequence $x_1, \ldots , x_ l$ of $M$ over $R$ is an admissible sequence of $M$ over $R'$ as $\mathfrak m'$ maps into $\mathfrak m$. The isomorphism is obtained by mapping the symbol $[x_1, \ldots , x_ l] \in \det \nolimits _{R, \kappa }(M)$ to the corresponding symbol $[x_1, \ldots , x_ l] \in \det \nolimits _{R', \kappa }(M)$. It is immediate to verify that this is functorial for isomorphisms and compatible with the isomorphisms $\gamma$ of Lemma 42.68.6. $\square$
Remark 42.68.9. Let $(R, \mathfrak m, \kappa )$ be a local ring and assume either the characteristic of $\kappa$ is zero or it is $p$ and $p R = 0$. Let $M_1, \ldots , M_ n$ be finite length $R$-modules. We will show below that there exists an ideal $I \subset \mathfrak m$ annihilating $M_ i$ for $i = 1, \ldots , n$ and a section $\sigma : \kappa \to R/I$ of the canonical surjection $R/I \to \kappa$. The restriction $M_{i, \kappa }$ of $M_ i$ via $\sigma$ is a $\kappa$-vector space of dimension $l_ i = \text{length}_ R(M_ i)$ and using Lemma 42.68.8 we see that
$\det \nolimits _\kappa (M_ i) = \wedge _\kappa ^{l_ i}(M_{i, \kappa })$
These isomorphisms are compatible with the isomorphisms $\gamma _{K \to M \to L}$ of Lemma 42.68.6 for short exact sequences of finite length $R$-modules annihilated by $I$. The conclusion is that verifying a property of $\det _\kappa$ often reduces to verifying corresponding properties of the usual determinant on the category finite dimensional vector spaces.
For $I$ we can take the annihilator (Algebra, Definition 10.40.3) of the module $M = \bigoplus M_ i$. In this case we see that $R/I \subset \text{End}_ R(M)$ hence has finite length. Thus $R/I$ is an Artinian local ring with residue field $\kappa$. Since an Artinian local ring is complete we see that $R/I$ has a coefficient ring by the Cohen structure theorem (Algebra, Theorem 10.160.8) which is a field by our assumption on $R$.
Here is a case where we can compute the determinant of a linear map. In fact there is nothing mysterious about this in any case, see Example 42.68.11 for a random example.
Lemma 42.68.10. Let $R$ be a local ring with residue field $\kappa$. Let $u \in R^*$ be a unit. Let $M$ be a module of finite length over $R$. Denote $u_ M : M \to M$ the map multiplication by $u$. Then
$\det \nolimits _\kappa (u_ M) : \det \nolimits _\kappa (M) \longrightarrow \det \nolimits _\kappa (M)$
is multiplication by $\overline{u}^ l$ where $l = \text{length}_ R(M)$ and $\overline{u} \in \kappa ^*$ is the image of $u$.
Proof. Denote $f_ M \in \kappa ^*$ the element such that $\det \nolimits _\kappa (u_ M) = f_ M \text{id}_{\det \nolimits _\kappa (M)}$. Suppose that $0 \to K \to L \to M \to 0$ is a short exact sequence of finite $R$-modules. Then we see that $u_ k$, $u_ L$, $u_ M$ give an isomorphism of short exact sequences. Hence by Lemma 42.68.6 (1) we conclude that $f_ K f_ M = f_ L$. This means that by induction on length it suffices to prove the lemma in the case of length $1$ where it is trivial. $\square$
Example 42.68.11. Consider the local ring $R = \mathbf{Z}_ p$. Set $M = \mathbf{Z}_ p/(p^2) \oplus \mathbf{Z}_ p/(p^3)$. Let $u : M \to M$ be the map given by the matrix
$u = \left( \begin{matrix} a & b \\ pc & d \end{matrix} \right)$
where $a, b, c, d \in \mathbf{Z}_ p$, and $a, d \in \mathbf{Z}_ p^*$. In this case $\det _\kappa (u)$ equals multiplication by $a^2d^3 \bmod p \in \mathbf{F}_ p^*$. This can easily be seen by consider the effect of $u$ on the symbol $[p^2e, pe, pf, e, f]$ where $e = (0 , 1) \in M$ and $f = (1, 0) \in M$.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar). | 10,629 | 28,214 | {"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": 3, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2024-30 | latest | en | 0.643628 |
https://www.meritnation.com/ask-answer/question/what-is-componendo-and-dividendo-method-plz-explain/trigonometric-functions/2522807 | 1,582,611,962,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146033.50/warc/CC-MAIN-20200225045438-20200225075438-00237.warc.gz | 791,181,994 | 9,861 | # What is componendo and dividendo method? Plz explain.
Componendo and dividendo is a method of simplification in Mathematics.
According to it,
This proof is invalid for a=b and c=d because if such that a = b and c = d
then by application of componendo and dividendo, we get -
[as a=b and c=d]
Since division by zero is not defined. So, this proof is invalid for a = b and c = d
• 542
What are you looking for? | 118 | 418 | {"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-2020-10 | longest | en | 0.893187 |
https://elektroskolarb.ba/palladium-teeth-paxs/5962ec-how-to-convert-aggregate-kg-to-cubic-meter | 1,620,698,315,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991553.4/warc/CC-MAIN-20210510235021-20210511025021-00483.warc.gz | 247,259,062 | 6,290 | Definition of cubic meters of water provided by WikiPedia The cubic meter (in American English) or cubic metre (in British English) is the derived unit of volume. Definition of cubic meters of water provided by WikiPedia The cubic meter (in American English) or cubic metre (in British English) is the derived unit of volume. 1 kg/m3 is equivalent to 0.001g/cm3. 1 yard = 91.44 centimeters = 0.9144 meter (convert yards to meter)1 foot = 30.48 centimeters = 0.3048 meter (convert feet to meter)1 inch = 2.54 centimeters = 0.0254 meter (convert inch to cm)1 meter = 100 centimeters (convert meter to cm)1 millimeter = 0.1 centimeters = 0.001 meter (convert mm to cm)Cubic meter formula for different units Its symbol is m 3 It is the volume of a cube with edges one metre in length. Conversion Table. On: 2. Unit conversion and calculators. Then multiply the amount of Gram/liter you want to convert to Kilogram/cubic Meter, use the chart below to guide you. Concrete, Gravel - 2.40 tonne/cubic meter of cement bags required in 1 cubic meter = 0.2171/0.0347 = 6.25 bags. How to convert Kilograms Per Cubic Meter to Tons [metric] Per Cubic Yard (kg/m3 to t/yd3 [metric])? Also, explore tools to convert ton register or cubic meter to other volume units or learn more about volume conversions. Note: You can use the same formula for calculating cement for other nominal mixes. You can also go to the universal conversion page. Cubic Meter to Cubic Feet Conversion Example Task: Convert 50 cubic meters to cubic feet (show work) Formula: m 3 ÷ 0.0283168466 = ft 3 Calculations: 50 m 3 ÷ 0.0283168466 = 1,765.73333558 ft 3 Result: 50 m 3 is equal to 1,765.73333558 ft 3. First divide the two units variables. About Gravel, loose dry; 1 cubic meter of Gravel, loose dry weighs 1 522 kilograms [kg] 1 cubic foot of Gravel, loose dry weighs 95.01536 pounds [lbs] Gravel, loose dry weighs 1.522 gram per cubic centimeter or 1 522 kilogram per cubic meter, i.e. … Only if you have pure water, the volume of 1 cubic meter weighs of 1 tonne or 1,000 kilogram … »More detailed. wet packed sand has a density of 2.08 kg/L which is 2080 kg/m³ or 2.08 tonne/m³ packed mud has a density of 1.91 kg/L which is 1910 kg/m³ or 1.91 tonne/m³ cubic meters of sand * 2.08 tonne/m³ = tonnes of sand cubic meters of mud * 1.91 tonne/m³ = tonnes of mud You can then convert metric ton (tonne) to U.S. tons on the weight conversion page. "Cubic meter" is a region of space. 12.74 Slugs/Cubic Foot to Kilograms/Cubic Meters 12.74 Kilograms/Cubic Meters to Slugs/Cubic Foot 1.1 Megagrams/Cubic Meter to Kilograms/Cubic Meters 5000000 Tonnes/Cubic Meter to Tons (UK)/Cubic Yard 4 Tons (UK)/Cubic Yard to Tonnes/Cubic Meter 0.0013 Milligrams/Milliliter to Grams/Liter 30000 Milligrams/Liter to Kilograms/Cubic Meters Pound/Gallon : Pound per gallon is a unit of density in the US customary and British (Imperial) systems. Gravel, w/sand, natural - 1.92 tonne/cubic meter. The ton register [ton reg] to cubic meter [m^3] conversion table and conversion steps are also listed. The gravel is assumed clean of dirt and other debris. Liquid water, on the other hand, has a density of just 1000 kg/m^3 at 4 degrees Celsius. Cement required = 01 bag = 50 kg ~ 36 liters or 0.036 cum Instant free online tool for cubic meter to ton register conversion or vice versa. Its official symbol is lb/gal. For one cubic meter of M15 cement, you need 711 kg of fine aggregate and 1460 kg of coarse aggregate. Cubic meters can be abbreviated as m³, and are also sometimes abbreviated as cu m, CBM, cbm, or MTQ.For example, 1 cubic meter can be written as 1 m³, 1 cu m, 1 CBM, 1 cbm, or 1 MTQ. In the US customary system, it is equal to 119.83 kg/m³. Conversion of units between 2040 Kilogram Per Cubic Metre (Si Unit) and Kilogram Per Litre (2040 kg/m3 and kg/L) is the conversion between different units of measurement, in this case it's 2040 Kilogram Per Cubic Metre (Si Unit) and Kilogram Per Litre, for the same quantity, typically through multiplicative conversion factors (kg/m3 and kg/L). Instant free online tool for ton register to cubic meter conversion or vice versa. This is a conversion chart for kilogram per cubic meter (Metric System). December 2020. 1 kg/m3 = 0.35123162882965 t/yd3 [metric]. A cubic meter of typical gravel weighs 1,680 kilograms 1.68 tonnes. 2: Enter the value you want to convert (kilogram per cubic meter). For quick reference purposes, below is a conversion table that you can use to convert from m 3 to ft 3. So let me tell you now How to Convert Sand from CFT (Cubic Feet) to KG (Kilograms). Considering a concrete mix proportion (by volume) of 1:2:4. Instantly Convert Tonnes Per Cubic Meter (t/m 3 ) to Kilograms Per Cubic Meter (kg/m 3 ) and Many More Density Conversions Online. Search flights; Airports; Embassies; Airlines; Home; Convert ton/m3 to kg/m3; 8 Tonnes/Cubic Meter to Kilograms/Cubic Meters; Convert 8 Tonnes/Cubic Meter to Kilograms/Cubic Meters. To convert tonnages into volumes (from tons to cubes) of gravel, sand, or other mediums into cubic yards ("cubes" or yd.³ per ): Find the best description of the medium you are using - in the first column. how much cement in 1 cubic meter brickwork. A square yard of gravel with a depth of 2 in (~5 cm) weighs about 157 pounds (~74 kg). ∴ No. Cubic Meters to Metric Ton Using Gravel Convert and Calculate. 10 Cubic Meters to Cubic Yards = 13.0795: 800 Cubic Meters to Cubic Yards = 1046.3605: 20 Cubic Meters to Cubic Yards = 26.159: 900 Cubic Meters to Cubic Yards = 1177.1556: 30 Cubic Meters to Cubic Yards = 39.2385: 1,000 Cubic Meters to Cubic Yards = 1307.9506: 40 Cubic Meters to Cubic Yards = 52.318: 10,000 Cubic Meters to Cubic Yards = 13079.5062 Then multiply the amount of Kilogram/cubic Meter you want to convert to Gram/milliliter, use the chart below to guide you. Gravel, dry 1/4 to 2 inch - 1.68 tonne/cubic meter. 2,700 kg/m3 to g/cm3 (kilograms/cubic meter to grams/cubic centimeter) g/cm3 to kg/m3 (gram/cubic centimeter to kilogram/cubic meter) 1,025 g/cm3 to kg/m3 (grams/cubic centimeter to kilograms/cubic meter) If there are 10 cubic meters of iron, this represents (10 m^3 * 7850 kg/m^3), which equals 78500 kg. A cubic yard of typical gravel weighs about 2830 pounds or 1.42 tons. Gravel, wet 1/4 to 2 inch - 2.00 tonne/cubic meter. One cubic meter is equal to the volume of a cube with each edge measuring one meter.. The cubic meter, or cubic metre, is the SI derived unit for volume in the metric system. One pound per cubic foot is equivalent to 16.02 kg/m³. How many cubic meter are in One ton of 20 MM coarse aggregates. The kilograms amount 2,406.53 kg - kilo converts into 1 m3, one cubic meter. 1 kg/m3 is equivalent to 0.001g/cm3. To convert between Kilogram/cubic Meter and Gram/milliliter you have to do the following: First divide 1/1 / 0.001/0.000001 = 0.001 . How many kilograms in a cubic meter of 20 MM of aggregate? Its symbol is m 3 It is the volume of a cube with edges one metre in length. Cubic Meters to Metric Ton Using Gravel – OnlineConversion Forums. :- 4530 Kg is weight of 100 cft aggregate. About Stone, crushed; 1 cubic meter of Stone, crushed weighs 1 602 kilograms [kg] 1 cubic foot of Stone, crushed weighs 100.00959 pounds [lbs] Stone, crushed weighs 1.602 gram per cubic centimeter or 1 602 kilogram per cubic meter, i.e. Is there a calculation formula? Convert 1 ton aggregate to cft and m3. It is defined by mass in kilograms divided by volume in cubic metres. How much does a cubic meter of gravel weigh? Re: Cubic Meters to Metric Ton Using Gravel. »More detailed. 1 kg/m3 = 0.001000 ton/m3. If you are an engineer of the Quantity Survey on-site, then it becomes very important to know this. Then click the Convert Me button. Cement = 1440 kg/cum ; Fine Aggregate = 1600 kg/cum ; Coarse Aggregate = 1450 kg/cum ; Water = 1000 kg/cum. Convert 8 Tonnes/Cubic Meter to Kilograms/Cubic Meters with our online conversion. How to convert 2 cubic meters (m3) of concrete into kilograms (kg - kilo)? To switch the unit simply find the one you want on the page and click it. Convert 1 ton aggregate to cubic meter: we know that density of aggregate = 1520 kg/m3,it means 1m3 aggregate weight is 1520kg, 1 ton = 1000kg, 1 ton aggregate to cubic meter = 1000/1520 = 0.6578 m3,so 1 ton aggregate is equal to 0.6578 m3 ( cubic meter). 8 Tonnes/Cubic Meter (ton/m3) = 8,000 Kilograms/Cubic Meters (kg/m3) 1 ton/m3 = 1,000 kg/m3. Hello friends, in today’s post we will learn How to Convert Sand from CFT (Cubic Feet) to KG (Kilograms) on construction sites. 1 x 0.35123162882965 t/yd3 [metric] = 0.35123162882965 Tons [metric] Per Cubic Yard. Pound/Cubic Foot : Pound per cubic foot is a unit of density which is a US customary units. Gravel, loose, dry - 1.52 tonne/cubic meter. If there are 10 cubic meters of water, this represents a mass of (10 m^3 * 1000 kg/m^3), which equals 10000 kg water, a much smaller mass than that of iron in the same amount of volume. I found the following density data for gravel. KILOGRAM/CUBIC METER TO GRAM/MILLILITER (kg/m³ TO g/ml) CHART . Conversion of units between 5515 Kilogram Per Cubic Metre (Si Unit) and Kilogram Per Litre (5515 kg/m3 and kg/L) is the conversion between different units of measurement, in this case it's 5515 Kilogram Per Cubic Metre (Si Unit) and Kilogram Per Litre, for the same quantity, typically through multiplicative conversion factors (kg/m3 and kg/L).
What Is Morality In Ethics, Tree Container Sizes, Medford Ma To Downtown Boston, Neurosurgery Residency Requirements, Prince Lionheart Booster Squish Canada, Deliberate Indifference In Law Enforcement, Raccoon Poop Bleach, Axa Global Care Review, Hebrews 13:5 Msg, | 2,818 | 9,695 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2021-21 | latest | en | 0.738771 |
https://chemistrycalc.com/chemistry/grams-to-moles-calculator/ | 1,720,849,226,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514490.70/warc/CC-MAIN-20240713051758-20240713081758-00353.warc.gz | 146,200,279 | 12,008 | Created By : Rina Nayak
Reviewed By : Phani Ponnapalli
Last Updated : Apr 03, 2023
Make use of free g to mol conversion calculator to convert the mass of a substance in grams to moles easily. All you need to do is enter the substance weight in grams, select the chemical substance name, type and click the calculate button to avail the result within no time.
Chemical
Mass
### How to Calculate Moles from Grams?
Have a look at the simple process of the conversion of grams to moles easily.
• Check the mass of the substance in grams.
• Find the molecular weight of the substance.
• Divide the substance mass in grams by the molecular weight to obtain the molar mass in moles.
### Grams to Moles Formula
The formula to convert the grams to moles of a substance is given here:
Moles = mass of the substance in grams/molar mass
Example:
Question: Convert 56g of water into moles?
Given that
Mass of the substance = 56 g
Molecular mass of water = 18.015 g/mol
Moles = mass in grams/molar mass of water
= 56/18.015
= 3.10 mol
Therefore, the mass of water in moles is 3.10
If you would like to learn more about the other chemical calculator that gives instant results, stay tuned to Chemistrycalc.Com
### FAQ’s on Grams to Moles Calculator
1. How to convert grams to moles?
Get the mass and molecular weight of the substance. Divide the mass in grams by the molecular weight to know how many moles are there in the substance.
2. What is the formula for grams to moles conversion?
The simple formula to convert g to mol is given here.
Moles = mass of the substance in grams/molecular weight.
3. How to use grams to moles calculator?
You have to select the type of chemical substance, substance name and give its mass in grams in the specified input field. Hit the calculate button to check the mass in moles value.
4. What is a mole?
In chemistry, a mole is defined as the amount of substance that contains exactly 6.02214076 x 1023 elementary entities of the given substance. | 476 | 1,999 | {"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-2024-30 | latest | en | 0.888715 |
https://howkgtolbs.com/convert/97.7-kg-to-lbs | 1,656,306,023,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103328647.18/warc/CC-MAIN-20220627043200-20220627073200-00392.warc.gz | 360,121,581 | 12,070 | # 97.7 kg to lbs - 97.7 kilograms to pounds
Do you want to learn how much is 97.7 kg equal to lbs and how to convert 97.7 kg to lbs? Here you go. This whole article is dedicated to kilogram to pound conversion - both theoretical and practical. It is also needed/We also want to point out that all this article is devoted to a specific number of kilograms - this is one kilogram. So if you want to learn more about 97.7 kg to pound conversion - read on.
Before we move on to the practice - it means 97.7 kg how much lbs calculation - we want to tell you some theoretical information about these two units - kilograms and pounds. So let’s start.
How to convert 97.7 kg to lbs? 97.7 kilograms it is equal 215.391629974 pounds, so 97.7 kg is equal 215.391629974 lbs.
## 97.7 kgs in pounds
We are going to begin with the kilogram. The kilogram is a unit of mass. It is a basic unit in a metric system, formally known as International System of Units (in short form SI).
From time to time the kilogram could be written as kilogramme. The symbol of this unit is kg.
The kilogram was defined first time in 1795. The kilogram was described as the mass of one liter of water. First definition was not complicated but difficult to use.
Later, in 1889 the kilogram was described by the International Prototype of the Kilogram (in short form IPK). The IPK was made of 90% platinum and 10 % iridium. The International Prototype of the Kilogram was in use until 2019, when it was substituted by another definition.
Today the definition of the kilogram is build on physical constants, especially Planck constant. The official definition is: “The kilogram, symbol kg, is the SI unit of mass. It is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015×10−34 when expressed in the unit J⋅s, which is equal to kg⋅m2⋅s−1, where the metre and the second are defined in terms of c and ΔνCs.”
One kilogram is exactly 0.001 tonne. It can be also divided into 100 decagrams and 1000 grams.
## 97.7 kilogram to pounds
You know some facts about kilogram, so now we can move on to the pound. The pound is also a unit of mass. It is needed to highlight that there are more than one kind of pound. What does it mean? For instance, there are also pound-force. In this article we want to concentrate only on pound-mass.
The pound is in use in the British and United States customary systems of measurements. Naturally, this unit is used also in other systems. The symbol of the pound is lb or “.
There is no descriptive definition of the international avoirdupois pound. It is equal 0.45359237 kilograms. One avoirdupois pound is divided into 16 avoirdupois ounces and 7000 grains.
The avoirdupois pound was enforced in the Weights and Measures Act 1963. The definition of the pound was written in first section of this act: “The yard or the metre shall be the unit of measurement of length and the pound or the kilogram shall be the unit of measurement of mass by reference to which any measurement involving a measurement of length or mass shall be made in the United Kingdom; and- (a) the yard shall be 0.9144 metre exactly; (b) the pound shall be 0.45359237 kilogram exactly.”
### How many lbs is 97.7 kg?
97.7 kilogram is equal to 215.391629974 pounds. If You want convert kilograms to pounds, multiply the kilogram value by 2.2046226218.
### 97.7 kg in lbs
The most theoretical section is already behind us. In next section we are going to tell you how much is 97.7 kg to lbs. Now you know that 97.7 kg = x lbs. So it is time to know the answer. Have a look:
97.7 kilogram = 215.391629974 pounds.
It is a correct outcome of how much 97.7 kg to pound. It is possible to also round off the result. After rounding off your result will be as following: 97.7 kg = 214.94 lbs.
You learned 97.7 kg is how many lbs, so look how many kg 97.7 lbs: 97.7 pound = 0.45359237 kilograms.
Of course, this time you can also round off the result. After it your result will be as following: 97.7 lb = 0.45 kgs.
We are also going to show you 97.7 kg to how many pounds and 97.7 pound how many kg results in tables. Look:
We will start with a table for how much is 97.7 kg equal to pound.
### 97.7 Kilograms to Pounds conversion table
Kilograms (kg) Pounds (lb) Pounds (lbs) (rounded off to two decimal places)
97.7 215.391629974 214.940
Now look at a chart for how many kilograms 97.7 pounds.
Pounds Kilograms Kilograms (rounded off to two decimal places
97.7 0.45359237 0.45
Now you know how many 97.7 kg to lbs and how many kilograms 97.7 pound, so it is time to go to the 97.7 kg to lbs formula.
### 97.7 kg to pounds
To convert 97.7 kg to us lbs a formula is needed. We will show you two versions of a formula. Let’s begin with the first one:
Number of kilograms * 2.20462262 = the 215.391629974 result in pounds
The first formula give you the most correct outcome. In some cases even the smallest difference could be considerable. So if you need an exact outcome - first formula will be the best solution to calculate how many pounds are equivalent to 97.7 kilogram.
So move on to the second formula, which also enables conversions to learn how much 97.7 kilogram in pounds.
The shorter formula is as following, see:
Amount of kilograms * 2.2 = the outcome in pounds
As you can see, this version is simpler. It can be the best choice if you want to make a conversion of 97.7 kilogram to pounds in easy way, for example, during shopping. Just remember that your outcome will be not so correct.
Now we want to show you these two versions of a formula in practice. But before we are going to make a conversion of 97.7 kg to lbs we are going to show you another way to know 97.7 kg to how many lbs totally effortless.
### 97.7 kg to lbs converter
Another way to know what is 97.7 kilogram equal to in pounds is to use 97.7 kg lbs calculator. What is a kg to lb converter?
Calculator is an application. It is based on first version of a formula which we gave you above. Thanks to 97.7 kg pound calculator you can effortless convert 97.7 kg to lbs. Just enter amount of kilograms which you need to convert and click ‘calculate’ button. The result will be shown in a second.
So try to convert 97.7 kg into lbs with use of 97.7 kg vs pound converter. We entered 97.7 as an amount of kilograms. This is the outcome: 97.7 kilogram = 215.391629974 pounds.
As you see, this 97.7 kg vs lbs calculator is user friendly.
Now let’s move on to our main issue - how to convert 97.7 kilograms to pounds on your own.
#### 97.7 kg to lbs conversion
We are going to begin 97.7 kilogram equals to how many pounds conversion with the first version of a formula to get the most exact outcome. A quick reminder of a formula:
Number of kilograms * 2.20462262 = 215.391629974 the outcome in pounds
So what have you do to learn how many pounds equal to 97.7 kilogram? Just multiply amount of kilograms, in this case 97.7, by 2.20462262. It is equal 215.391629974. So 97.7 kilogram is 215.391629974.
You can also round off this result, for instance, to two decimal places. It is equal 2.20. So 97.7 kilogram = 214.940 pounds.
It is high time for an example from everyday life. Let’s calculate 97.7 kg gold in pounds. So 97.7 kg equal to how many lbs? And again - multiply 97.7 by 2.20462262. It gives 215.391629974. So equivalent of 97.7 kilograms to pounds, if it comes to gold, is 215.391629974.
In this case you can also round off the result. Here is the outcome after rounding off, this time to one decimal place - 97.7 kilogram 214.94 pounds.
Now let’s move on to examples converted with a short version of a formula.
#### How many 97.7 kg to lbs
Before we show you an example - a quick reminder of shorter formula:
Amount of kilograms * 2.2 = 214.94 the result in pounds
So 97.7 kg equal to how much lbs? As in the previous example you have to multiply amount of kilogram, in this case 97.7, by 2.2. Look: 97.7 * 2.2 = 214.94. So 97.7 kilogram is 2.2 pounds.
Let’s make another calculation with use of this version of a formula. Now calculate something from everyday life, for example, 97.7 kg to lbs weight of strawberries.
So let’s calculate - 97.7 kilogram of strawberries * 2.2 = 214.94 pounds of strawberries. So 97.7 kg to pound mass is 214.94.
If you know how much is 97.7 kilogram weight in pounds and can calculate it with use of two different versions of a formula, let’s move on. Now we are going to show you all results in tables.
#### Convert 97.7 kilogram to pounds
We know that outcomes presented in tables are so much clearer for most of you. We understand it, so we gathered all these results in tables for your convenience. Due to this you can quickly make a comparison 97.7 kg equivalent to lbs outcomes.
Begin with a 97.7 kg equals lbs table for the first formula:
Kilograms Pounds Pounds (after rounding off to two decimal places)
97.7 215.391629974 214.940
And now see 97.7 kg equal pound chart for the second formula:
Kilograms Pounds
97.7 214.94
As you see, after rounding off, if it comes to how much 97.7 kilogram equals pounds, the outcomes are the same. The bigger number the more considerable difference. Remember it when you need to make bigger amount than 97.7 kilograms pounds conversion.
#### How many kilograms 97.7 pound
Now you learned how to calculate 97.7 kilograms how much pounds but we want to show you something more. Are you interested what it is? What do you say about 97.7 kilogram to pounds and ounces conversion?
We will show you how you can convert it little by little. Let’s start. How much is 97.7 kg in lbs and oz?
First thing you need to do is multiply amount of kilograms, this time 97.7, by 2.20462262. So 97.7 * 2.20462262 = 215.391629974. One kilogram is 2.20462262 pounds.
The integer part is number of pounds. So in this case there are 2 pounds.
To check how much 97.7 kilogram is equal to pounds and ounces you have to multiply fraction part by 16. So multiply 20462262 by 16. It is exactly 327396192 ounces.
So final outcome is equal 2 pounds and 327396192 ounces. You can also round off ounces, for instance, to two places. Then final result will be exactly 2 pounds and 33 ounces.
As you can see, conversion 97.7 kilogram in pounds and ounces quite simply.
The last conversion which we are going to show you is calculation of 97.7 foot pounds to kilograms meters. Both foot pounds and kilograms meters are units of work.
To convert it it is needed another formula. Before we give you it, let’s see:
• 97.7 kilograms meters = 7.23301385 foot pounds,
• 97.7 foot pounds = 0.13825495 kilograms meters.
Now look at a formula:
Amount.RandomElement()) of foot pounds * 0.13825495 = the outcome in kilograms meters
So to convert 97.7 foot pounds to kilograms meters you need to multiply 97.7 by 0.13825495. It gives 0.13825495. So 97.7 foot pounds is 0.13825495 kilogram meters.
You can also round off this result, for instance, to two decimal places. Then 97.7 foot pounds will be exactly 0.14 kilogram meters.
We hope that this conversion was as easy as 97.7 kilogram into pounds calculations.
We showed you not only how to make a calculation 97.7 kilogram to metric pounds but also two other calculations - to know how many 97.7 kg in pounds and ounces and how many 97.7 foot pounds to kilograms meters.
We showed you also other solution to do 97.7 kilogram how many pounds conversions, this is with use of 97.7 kg en pound calculator. It is the best option for those of you who do not like calculating on your own at all or need to make @baseAmountStr kg how lbs conversions in quicker way.
We hope that now all of you can do 97.7 kilogram equal to how many pounds calculation - on your own or with use of our 97.7 kgs to pounds calculator.
It is time to make your move! Let’s convert 97.7 kilogram mass to pounds in the best way for you.
Do you want to make other than 97.7 kilogram as pounds conversion? For instance, for 15 kilograms? Check our other articles! We guarantee that calculations for other amounts of kilograms are so easy as for 97.7 kilogram equal many pounds.
### How much is 97.7 kg in pounds
We want to sum up this topic, that is how much is 97.7 kg in pounds , we prepared one more section. Here you can find all you need to know about how much is 97.7 kg equal to lbs and how to convert 97.7 kg to lbs . Have a look.
How does the kilogram to pound conversion look? It is a mathematical operation based on multiplying 2 numbers. Let’s see 97.7 kg to pound conversion formula . See it down below:
The number of kilograms * 2.20462262 = the result in pounds
How does the result of the conversion of 97.7 kilogram to pounds? The exact result is 215.391629974 pounds.
It is also possible to calculate how much 97.7 kilogram is equal to pounds with second, shortened version of the equation. Let’s see.
The number of kilograms * 2.2 = the result in pounds
So this time, 97.7 kg equal to how much lbs ? The result is 215.391629974 lb.
How to convert 97.7 kg to lbs quicker and easier? It is possible to use the 97.7 kg to lbs converter , which will make whole mathematical operation for you and give you a correct answer .
#### Kilograms [kg]
The kilogram, or kilogramme, is the base unit of weight in the Metric system. It is the approximate weight of a cube of water 10 centimeters on a side.
#### Pounds [lbs]
A pound is a unit of weight commonly used in the United States and the British commonwealths. A pound is defined as exactly 0.45359237 kilograms. | 3,548 | 13,499 | {"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} | 4.25 | 4 | CC-MAIN-2022-27 | latest | en | 0.946049 |
https://indusladies.com/community/threads/some-logic-thought.15725/ | 1,628,135,542,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046155322.12/warc/CC-MAIN-20210805032134-20210805062134-00035.warc.gz | 329,947,092 | 18,886 | # Some Logic Thought
Discussion in 'Jokes' started by gsaikripa, Nov 12, 2007.
1. ### gsaikripaGold IL'ite
Messages:
4,933
177
Trophy Points:
170
Gender:
Female
Whenever I find the key to success, someone changes the lock.
_____
To Err is human, to forgive is not a COMPANY policy.
_____
The road to success??.. Is always under construction.
_____
Alcohol doesn't solve any problems, but if you think again, neither does Milk.
_____
In order to get a Loan, you first need to prove that you don't need it.
_____
All the desirable things in life are either illegal, expensive or fattening.
_____
Since Light travels faster than Sound, people appear brighter before you hear them speak.
_____
Everyone has a scheme of getting rich?.. Which never works.
_____
If at first you don't succeed?. Destroy all evidence that you ever tried.
_____
You can never determine which side of the bread to butter. If it falls down, it will always land on the buttered side.
_____
Anything dropped on the floor will roll over to the most inaccessible corner.
_____
***** 42.7% of all statistics is made on the spot. *****
_____
As soon as you mention something?? if it is good, it is taken?. If it is bad, it happens.
_____
He who has the gold, makes the rules ---- Murphy's golden rule.
_____
If you come early, the bus is late. If you come late?? the bus is still late.
_____
Once you have bought something, you will find the same item being sold somewhere else at a cheaper rate.
_____
When in a queue, the other line always moves faster and the person in front of you will always have the most complex of transactions.
_____
If you have paper, you don't have a pen??. If you have a pen, you don't have paper?? if you have both, no one calls.
_____
Especially for engg. Students----
If you have bunked the class, the professor has taken attendance.
_____
You will pick up maximum wrong numbers when on roaming.
_____
The door bell or your mobile will always ring when you are in the bathroom.
_____
After a long wait for bus no.20, two 20 number buses will always pull in together and the bus which you get in will be crowded than the other.
_____
If your exam is tomorrow, there will be a power cut tonight.
_____
Irrespective of the direction of the wind, the smoke from the cigarette will always tend to go to the non-smoker | 561 | 2,337 | {"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-2021-31 | longest | en | 0.932186 |
https://converter.ninja/length/meters-to-decimeters/214-m-to-dm/ | 1,719,064,539,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862396.85/warc/CC-MAIN-20240622112740-20240622142740-00129.warc.gz | 151,802,763 | 5,344 | # 214 meters in decimeters
## Conversion
214 meters is equivalent to 2140 decimeters.[1]
## Conversion formula How to convert 214 meters to decimeters?
We know (by definition) that: $1\mathrm{m}=10\mathrm{dm}$
We can set up a proportion to solve for the number of decimeters.
$1 m 214 m = 10 dm x dm$
Now, we cross multiply to solve for our unknown $x$:
$x\mathrm{dm}=\frac{214\mathrm{m}}{1\mathrm{m}}*10\mathrm{dm}\to x\mathrm{dm}=2140\mathrm{dm}$
Conclusion: $214 m = 2140 dm$
## Conversion in the opposite direction
The inverse of the conversion factor is that 1 decimeter is equal to 0.000467289719626168 times 214 meters.
It can also be expressed as: 214 meters is equal to $\frac{1}{\mathrm{0.000467289719626168}}$ decimeters.
## Approximation
An approximate numerical result would be: two hundred and fourteen meters is about two thousand, one hundred and forty decimeters, or alternatively, a decimeter is about zero times two hundred and fourteen meters.
## Footnotes
[1] The precision is 15 significant digits (fourteen digits to the right of the decimal point).
Results may contain small errors due to the use of floating point arithmetic. | 323 | 1,179 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 6, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2024-26 | latest | en | 0.768313 |
https://www.physicsforums.com/threads/sin-x-vs-sin-2-x.160957/ | 1,653,691,002,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663006341.98/warc/CC-MAIN-20220527205437-20220527235437-00099.warc.gz | 1,075,614,403 | 14,414 | # Sin x vs sin^2 x
## Homework Statement
Find any critical numbers of the function
sin^2 x + cos x
## The Attempt at a Solution
I actually have a sort of silly question. Woud sin^2 in the equation be solved using the differeniation rule of d/dx[sin x] = cos x or d/dx [sin u] = (cos u)d/dx u?
d/dx (sin^2 x)= d/dx (sin x * sin x)
= d/dx (sin x) * sin x + sin x * d/dx (sin x)
=2 d/dx (sin x) sin x
=2cos x sin x
So I follow that thinking, but that would be the same as d/dx[sin u] right?
No- I think you're referring to a case like this:
A=d/dx sin(f(x))
substitue u=f(x)
A=du/dx * d/du sin(u)
=[d/dx f(x) ] cos u
=[d/dx f(x) ] cos(f(x))
Remember, sin^2(x) does not equal sin (sin (x)) or sin (x^2)
rather sin^2(x)= sin (x) * sin (x)
Thank you. | 276 | 756 | {"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-2022-21 | latest | en | 0.762812 |
https://community.myfitnesspal.com/en/discussion/comment/45492249 | 1,660,462,374,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571996.63/warc/CC-MAIN-20220814052950-20220814082950-00772.warc.gz | 192,559,055 | 24,271 | # Faith in people and machines
2»
## Replies
• Posts: 9,456 Member
globalc00 wrote: »
How many grams in a "slice" of bread? How many grams in any given bun/roll? USDA database thinks a regular bun is 56 g on average, a slider bun in 28 g, and a "roll" is 48 g. Meanwhile, it thinks a "slice" of white bread is 139 g.
Maybe i'm calculating my bread wrong. I was under the assumption that you add up all your ingridient calories and divide it by your final product. 25g of my bread is 71 calories.
ingridents are
9 g olive oil
33 g unsalted butter
9 g sugar
6 g salt
4 g yeast
20 g milk powder.
makes a 1.5 loaf bread which comesout to be 576g.
Well, for starters, 1.5 lbs is 680 g, not 576. Do you mean it's a recipe for a 1.5 lb, but you only get 576 g?
That aside, 71 calories for a 25 g slice of bread seems reasonable, and your method for calculating the calories in your bread, as I understand it (add up calories in all ingredients, weigh final cooked product, and distribute the total calories across some designated serving size), is the right way to go about it.
But my point is that it's meaningless to compare "a slice" to "a roll" without knowing the weights of both.
• Posts: 103 Member
The point is if 1 slice of 25 gram bread is 71 calories, its shocking that people can make hamburger buns that are 80 calories for top and bottom piece.
• Posts: 25,278 Member
dewit wrote: »
@AnnPT77 , I always love your posts! 💐 Now, by throwing in the law of large numbers, you have my eternal admiration 🌻.
Thanks!
In literal terms, they're probably only semi-large numbers**, but the principle still applies. 😉
** . . . although, when one's been doing this for 5+ years like I have . . . ! 😆
• Posts: 9,456 Member
globalc00 wrote: »
The point is if 1 slice of 25 gram bread is 71 calories, its shocking that people can make hamburger buns that are 80 calories for top and bottom piece.
It wouldn't shock me if the Five Guys buns are only 80 cals. They're so thin and flimsy they dissolve on contact with a little burger grease, much less condiments.
Yeah, I said it. Take your best shot, Five Guys fans.
• Posts: 5,265 Member
We can make ourselves barking crazy over this stuff. I have days (today) where my little fingers will seize up in indecision over entering 1 tsp versus 1.33 of honey crystals on my evening pudding. That’s 5 calories difference!
It’s good you’re paying attention and thinking about this stuff. It was failure to do so that got us in this shared boat in the first place.
Healthy Life makes a terrific line of low cal breads. I particularly like them because the ingredients are “normal” and don’t include cellulose or HFCs. 35 calories a slice, makes great sandwiches, French toast, grilled cheese or regular toast.
In all honesty, when It comes down to it, it’s just regular commercial bread sliced way thinner. Their hamburger buns are slightly smaller than standard and very fluffy.
That’s perfectly fine with me. They do the intended job and taste great.
I will be making a loaf of homemade bread for dinner. Mine calls for 500gr flour, salt, yeast and water. I’ve been fascinated by the variation in weight for the completed loaf. Anywhere from 701gr to 770’ish. Strangely, it was the loaf that didn’t rise properly and was very dense that was the lightest.
I’m wondering if the weight doesn’t drop more as it cools and releases steam. (As if it would ever last that long!)
So many questions about everything. But curiosity is what makes life interesting, even if it is about the piddly stuff.
Today I continue my experiment with instant clear gel. I’m determined to make a sugar free pudding that doesn’t taste bitter, like the boxed stuff has started to.
• Posts: 1,192 Member
globalc00 wrote: »
The point is if 1 slice of 25 gram bread is 71 calories, its shocking that people can make hamburger buns that are 80 calories for top and bottom piece.
How is it shocking? Did you see my post!? Hundreds of brands have hamburger buns for 80-90 calories and slices of bread for 35-50 calories. You use butter, milk powder, oil, sugar which are high calories. None of these breads that are lower contain any of that or they use lower calorie replacements.
Exactly, and a lot of hamburger buns don't weigh very much, as explained.
• Posts: 9,456 Member
kshama2001 wrote: »
globalc00 wrote: »
It just seems like a lot of the "i'm not losing weight post" nobody bring up the fact that if they are eating out, their calorie count probably isn't what they think it is. People usually say, it takes time, fluxuation or the it's not linear.
If one is overestimating exercise and underestimating calories, and most people want variety in the days.... then the question of how much deficit if you are indeed in one can wildly vary.
I would say it also depends on the context. If I rarely eat at a restaurant or get take out, let's say once a month (which is pretty accurate right now), then the fact that the calories in that one restaurant meal might not be 100% correct is not the most relevant factor in why I'm not losing weight. If, however, I ate out 4 times a week, it would be a good question to ask.
I've read many threads on these forums in which the inaccuracy of calories in food prepared outside has actually been mentioned, so it is discussed sometimes.
In every thread I've ever seen where the OP says they aren't losing as expected and eat out a lot, someone, usually many someones, points to the eating out as a likely culprit.
Especially when the OP in question doesn't bother logging meals out because it's "impossible," or logs random user-created entries that they make no attempt to verify, or makes an attempt at deconstructing the meal that looks woefully underestimated on amounts for main ingredients and makes no allowances for added fats.
• Posts: 7,893 Member
globalc00 wrote: »
The point is if 1 slice of 25 gram bread is 71 calories, its shocking that people can make hamburger buns that are 80 calories for top and bottom piece.
How is it shocking? Did you see my post!? Hundreds of brands have hamburger buns for 80-90 calories and slices of bread for 35-50 calories. You use butter, milk powder, oil, sugar which are high calories. None of these breads that are lower contain any of that or they use lower calorie replacements.
globalc00 wrote: »
The point is if 1 slice of 25 gram bread is 71 calories, its shocking that people can make hamburger buns that are 80 calories for top and bottom piece.
How is it shocking? Did you see my post!? Hundreds of brands have hamburger buns for 80-90 calories and slices of bread for 35-50 calories. You use butter, milk powder, oil, sugar which are high calories. None of these breads that are lower contain any of that or they use lower calorie replacements.
Exactly, and a lot of hamburger buns don't weigh very much, as explained.
This and this. Breads vary a lot by density. When I make homemade bread (which I don't really do anymore, but I used to), it was typically denser than any low cal storebought bread. That's what Lynn was getting at with the question about weight too -- the weights of different bread items are going to vary a lot, depending both on the volume and how dense they are.
If I'm making a homemade burger, the bun isn't important to me, so I usually get a low cal one. Never stopped me from losing and I have no reason to think the cals are wrong.
• Posts: 417 Member
zamphir66 wrote: »
globalc00 wrote: »
We all know CICO is key to everyone’s weight loss goal. I often hear people say scale that show body fat are inaccurate. And fitness machine like treadmill or activity tracker over estimate calories burned.
But it seems like people tend to believe the calories in their food that the restaurant puts up. The portions are cut by machines and cooked by different cook every day. The amount of oil used or how big of a scoop of sides you got all varies.
Why would you believe one but not the other?
In every restaurant I've worked at, and I've worked at a few, the portions and quantities of everything are scientifically controlled and managed. Frozen burger patties are the same size. Buns are all the same size. You're trained to put X pickles on something, no more no less. Condiment squirts are precision controlled. It's not just any old bottle of something, it's designed to be precise. Restaurant managers, as much as anything, are making sure no one in the back goes "cowboy" and puts 3.5 scoops of cheese on a pizza when it's only supposed to be 3. All of this is for revenue purposes. And as a consequence, for any reasonably successful franchise restaurant, you can indeed trust their calorie counts to within a reasonably small margin of error.
I came here to say this. Restaurants operate on a very slim profit margin. Food costs are calculated abs watched very closely against what actually gets used/sold. Restaurants are way more accurate then people think.
• Posts: 363 Member
globalc00 wrote: »
I know there are buns out there that claim 80 calories. However I bought a breadmaker to make my own bread, and using as basic ingridents as possible to make bread, 80 calories would be essentially 1 slice of bread.
I make my own bread. A 'slice of bread' is as big as you cut it! | 2,195 | 9,258 | {"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-2022-33 | latest | en | 0.948879 |
https://gmat.magoosh.com/lessons/1030-binomial-situation/?utm_source=gmatblog&utm_medium=blog&utm_campaign=gmatlessons&utm_term=inline&utm_content=3-month-gmat-study-schedule-for-beginners | 1,701,426,275,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100286.10/warc/CC-MAIN-20231201084429-20231201114429-00557.warc.gz | 319,106,899 | 13,069 | This is a free sample lesson. Sign up for Magoosh to get access to over 200 video lessons.
## Binomial Situation
### Transcript
We've already covered the And rule and the Or rule, two very important rules. Now we'll talk about another rule, the Binomial Situation. The first thing I wanna make clear, this is a problem type unlikely to appear on the test except as a very hard problem. So this is very important to understand about the material in this video. If you are the type of student who's aiming for an elite math score, if you really are trying to get almost.
Every math problem right on the test, well, then, yes. You may see this, you need to know this. But, if you're the type of student, you're just trying to man, manage the basics. You know you're not gonna get the hardest math problems right. You just want to make sure that you know the basic material so you get a respectable math score.
Well then definitely you should know the n rule. Definitely you know, you should know the or rule those show up all over place. This you may not even see, so you could probably completely disregard this especially if you start watching this video and you find it to difficult. Don't worry just let it go, it's not something that you're gonna have to worry about.
So I wanna make that very clear before we dive into this material because it is little more difficult, one make very clear, you should be deciding for yourself whether you should be wrestling with this or not. So, what is the binomial situation? Binomial situations concerns multiple trials of the same things.
So we have some simple event, like rolling a die, or flipping a coin, and we're going to do that event several times. Times, and so of course one of the assumptions is that all the trials will be independent, that's always true with dice and coins, and notice that it doesn't matter at all whether I roll one die five times or roll five dice at once, whether I flip one coin seven times or flip seven coins at once.
Both of those are completely identical as far as the probability is concerned. So the probability of success in an individual trial is obvious, is given or obvious from context. I think almost all the problems I've ever seen in the binomial situation on the test, have been either about dice or coins, so it's usually obvious. What's the probability of flipping ten coins, and getting this many heads.
So heads is a success. What's the probability role in this many dice in getting this many fives or if it wouldn't five, is a success. So success is determined in something else. Whatever they were asking the probability of, that is what we're referring to as quote unquote a success.
Two, the number of trials is decided before hand, so, in other words, that's just the way that the problem is framed. The problem frames, alright, ten coins are gonna be flipped so that's not up for debate. There gonna be ten trials. That's already fixed.
Those trials are independent and then the question is within those 10 trials, how often does something happen. So for some number r that could be less than n or equal to n. What is the probability of exactly our successes and end trials. So for example, what is the probability of flipping a coin ten times and getting exactly five heads.
What is the probability of rolling seven dice and getting exactly two sixes, something like that. These would the binomial situations, now how do we calculate, let's look at a couple examples, first of all a very easy example, three fair coins are flipped, so of course these are all going to be independent, very easy, what is the probability of exact, getting exactly two heads?
So, exactly two heads would mean that we get to the coin showing heads, and one of them is not heads, so one is tails, and so the probability of heads is one half, the probability of tails is one half. All these are independent, so we can multiply and that product equals an eighth.
Now, naively, you might think, okay, that's our answer, one eighth. But we have to be very careful here. What are the different ways that the scenario two heads can play out? For example, I could get heads, heads, tails, or I could get heads, tails, heads, or I could get tails, heads, heads. Those would be the three ways it would play out if we were flipping one coin three times in a sequence.
Another way to think about it, suppose we were flipping 3 completely different coins. Suppose we were flipping a nickel, a dime, and a quarter. When we flip those three coins, there's three different ways that we can get two heads with the single tail showing up either on the nickel or the dime or the quarter, and so there are three different ways for it to play out.
These are mutually exclusive, each one has a probability of of one eighth, as we just figured out, so we add them, and so this adds up to a grand total of 3/8, and that is the probability of flipping three fair coins and getting exactly two heads. Let's look at another that's a little more challenging. Practice problem two.
Ten dice, each fair with six sides rolled simultaneously. So we're doing ten different trials. Again, all of these are independent. What is the probability of getting exactly two fives among them. So in other words, what we're gonna get, we're gonna get one die is five, the other die is five, and then we're gonna have eight more.
See if I can make this very neat. Eight more that are something other than fives. So, the probability of getting a five, of course there are six sides on the dice, so the probability of getting any number, the probability of getting a five is one sixth. Of course the probability of getting a five on this one is one sixth.
The probability of getting something other than a five, getting any of the other five faces, well that would be five sixths, so that is the probability for each one of these. Now, course, all of these are independent, so we can multiply. So one way to say this is this is one sixths, times one sixths, times five sixths, times five sixths, times five sixths, times five sixths, times five sixths, times 5/6, let's say times 5/6 times 5/6.
Okay, we could write all that out. Let's obtain that in the key step. Much easy to say of this way. One sixth to the power of two, five sixth to the power of eight Now, no one in the right line is gonna expect you to calculate these fractions. In a fact what prob we going to happen is that the answer choices will just be written in these forms.
Which printed like this. So you don't even have to do this calculation you just have to recognize the right set up. Notice again, this is not the answer. Now we have to deal with the problem of, how many different ways can those two successes, those two fives be distributed among the ten coins.
Well, you may recognize that as a counting question. Now if this is not familiar you may want to go back and look at the combination video and the counting section. But essentially what we have here is ten choose two. Now of course we could figure that out relatively easily. But typically on the test, what I've noticed they don't even expect you to do this.
What they just do is they write the answer with all of this together, and this would be way the answer is printed on the multiple choices, so as long as you can recognize that; then you can answer the question. So they save you the work of doing the calculation on something like this. So now, let's just look at the general formula to summarize.
The general binomial formula, we're going to say lower case p is the probability of success on one trial. So for a coin, that's one half for any face on a die that is one sixth. N is the number of trials, R is the number of successes, so we have r successes, so that's why we have p to the r, we have success happening r times. If we have rs successes, that means we have n-r trials that are not successes, and so we have whatever is not a success, happening in minus r times, and, finally, we have to figure out how many different ways.
Can those R successes be distributed among end trials, that's why we need the combination number. So I would strongly suggest that you don't blindly memorize this formula, but you actually think through the logic
Read full transcript | 1,827 | 8,277 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2023-50 | latest | en | 0.954247 |
https://orangekitchens.net/boil/how-long-does-it-take-to-boil-2-liters-of-water.html | 1,627,442,516,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153521.1/warc/CC-MAIN-20210728025548-20210728055548-00176.warc.gz | 455,485,593 | 19,724 | # How long does it take to boil 2 liters of water?
Contents
## How long does it take to boil off 1 Litre of water?
3. Heat source. The heat source also plays a significant role. If, for instance, you attempt to boil say, 1 liter using a regular stove, it will take about 10-15 minutes for the water to reach boiling point.
## How long will it take to heat 2 Litres of water?
The heat is lost to the atmosphere at constant rate 160J/s, when its lid is open. In how much time will water heated to 77∘C with the lid open ? (specific heat of water = 4.2kJ/∘C-kg) ∴Time required =QP=500s=8min20s. Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.
## How long does it take to boil 1 Litre of water in a kettle?
1 liter of water boils in an electric kettle in about 4 to 5 minutes. 1 liter of water boils in a microwave in about 3 to 4 minutes.
## How long does it take for water to boil?
Only when the water is brought to a boil does the one minute timer begin. Thus, the whole process might take something more like five or ten minutes.
## What happens to the water as it boils?
When water is boiled, the heat energy is transferred to the molecules of water, which begin to move more quickly. Eventually, the molecules have too much energy to stay connected as a liquid. When this occurs, they form gaseous molecules of water vapor, which float to the surface as bubbles and travel into the air.
## How much water should you boil?
Most health organizations, including the Center for Disease Control, recommend that you boil water vigorously for 1 minute up to elevations of 2,000 meters (6,562 feet) and 3 minutes at elevations higher than that.
## How much energy does it take to heat 1000 Litres of water?
24 calorie per second. It takes 1000 calories to raise 1 liter of water one degree C. So apply about 4000 watts for one second to do this.
## How long does it take for a 250 l hot water system to heat up?
This is when your water heater will go through a reheating cycle. A tank can take between 2 to 4 hours to fully reheat.
## How many watts does it take to heat water?
A typical 50-gallon electric tank runs at 4500 watts.
From above, heating a gallon of water by 1 F takes 8.33 BTU. Heating from 68 F to 104 F would by a 36 F rise, or 36 x 8.33 = 300 BTU to heat 1 gallon of water.
IT IS INTERESTING: Do you boil shrimp frozen or thawed?
## How many watt hours does it take to boil water?
So 0 degrees C is 273 degrees Kelvin. What c in the equation is, the specific heat capacity. So in water it takes 1200 joule’s to raise a Kg of water by 1 degree K or in degrees C. If this takes 1 min to boil the power required to boil the water equals 96kilo joules/60 seconds =1600 watts or 1.6 kilo watts.
## Does water boil faster with a lid?
Does covering the pot really make water boil faster? When you heat water in an open pot, some of the energy that could be raising the temperature of the liquid escapes with the vapor. … Covering the pot prevents water vapor from escaping, enabling the temperature to rise more quickly.
## Does a kettle boil quicker with less water?
Originally Answered: Why does less water boil faster than more water? Simple: you’re putting a constant flow of energy into a smaller quantity of water, which means that its temperature rises more quickly than the same amount of energy (per minute) being applied to a larger quantity.
## What will make water boil faster?
Truth: Hot water boils faster.
If you’re in a hurry, turn your tap to the hottest setting, and fill your pot with that hot tap water. It’ll reach boiling a bit faster than cold or lukewarm water. You can also get the water even hotter by using your electric kettle.
## What is the quickest way to boil water?
Fill your pot with hot tap water, rather than cold. It’ll give you a jump start and get you to the boiling stage about 1 1/2 to 2 minutes quicker, depending on the amount of water in your pot. NOTE: this is not recommended in homes with older pipes as the hot water can leach lead and other funky stuff into the water.
IT IS INTERESTING: Best answer: How long does it take to cook beyond meat?
## Do little bubbles count as boiling?
The initial bubbles will occur at much lower temperatures, and long before the water simmers/boils. So, to add to BobMcGee’s answer, 0 is bubbles forming on the bottom – not a simmer, 1 – small bubbles rising towards the surface is a simmer, 2 is a boil, 3 is a rolling boil. | 1,103 | 4,501 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2021-31 | latest | en | 0.950059 |
https://corsi.unige.it/en/off.f/2023/ins/68852 | 1,712,930,743,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816024.45/warc/CC-MAIN-20240412132154-20240412162154-00458.warc.gz | 157,249,901 | 21,370 | CODE 60504 2023/2024 6 cfu anno 2 INGEGNERIA NAUTICA 8721 (L-9) - LA SPEZIA MAT/07 Italian LA SPEZIA 2° Semester Per sostenere l'esame di questo insegnamento è necessario aver sostenuto i seguenti esami: Pleasure Craft Engineering 8721 (coorte 2022/2023) GENERAL PHYSICS 56987 2022 MATHEMATICAL ANALYSIS + GEOMETRY 98339 2022 INGEGNERIA E DESIGN PER LA NAUTICA 10134 (coorte 2022/2023) MATHEMATICS + GEOMETRY 56973 2022 GENERAL PHYSICS 56987 2022 Questo insegnamento è un modulo di: AULAWEB
AIMS AND CONTENT
LEARNING OUTCOMES
The module aims to provide knowledge of mechanics of multi -degree of freedom systems . The case of the rigid body e'trattato in detail .
AIMS AND LEARNING OUTCOMES
After the course completion the student should be familiar with the statics and the dynamics of mechanical systems with finite degrees of freedom (particles systems and systems composed by rigid bodies)
TEACHING METHODS
60 standard teaching hours in attendance
SYLLABUS/CONTENT
Elements of Vector Algebra:
Free and applied vectors. Vector quantities. Geometric representation of vector quantities. Vector structure of the space of free vectors. Scalar product of vectors. Orthonormal bases. Vector, triple scalar and triple vector product of vectors and their component representations. Orthogonal matrices. Change of orthonormal bases. Euler angles. Linear operators. Linear symmetric and skew-symmetric operators. Vector functions. Elements of geometric theory of a curve.
Absolute Kinematics:
Observer. Absolute Space and time. Frame of reference. Velocity, acceleration and their Cartesian and intrinsic representations. Rectilinear, uniform and uniformly accelerated motion. Circular motion. Harmonic motion. Ballistics problems. Central motions and Binet’s formula. Polar, cylindrical and spherical coordinates.
Relative kinematics:
Relative motion of frames of reference. Angular velocity. Poisson formulae. Theorem on composition of angular velocities. Transportation motion. Theorems on composition of velocities and accelerations.
Dynamics:
Newton’s first law. Inertial mass. Momentum of a particle. Momentum conservation for isolated systems. Newton’s second and third laws. Kinetic energy. Work and power of a force. Theorem of energy. Conservative forces. Potential of a conservative force. Theorem on conservation of energy.
Relative Dynamics:
Transportation inertial force. Coriolis inertial force. Earth Mechanics.
Mechanics of a particle:
Motion of a free particle. Friction laws. Motion of a particle along a curve. Motion of a particle on a surface.
Mechanics of systems:
Systems of applied vectors. Resultant and resultant moment of a system of vectors. Scalar invariant. Central axis. Reducible and irreducible systems of vectors. Centre of parallel vectors and centre of gravity. Mechanical quantities of a system. Konig’s theorem. Momentum and angular momentum theorems. Theorem of energy for systems. Conservation laws for systems.
Mechanics of a rigid body:
The body-fixed reference frame of a rigid body. Rigid motion. Velocities and accelerations of the particles of a rigid body. Translational and rotational motions of a rigid body. Composition of rigid motions. Mechanical quantities of a rigid body. Inertia Tensor and its properties. Moment of a rigid body with respect to an Axis. Moments and products of Inertia. Inertia matrices. Huygens and parallel axes theorems. Momentum and angular momentum theorems for a rigid body. Power of a system of forces acting on a rigid body. Energy theorem for a rigid body. Motion of a free rigid body. Ideal constraints applied to a rigid body. Rotational motion of a rigid body about a fixed axis. Rotational motion of a rigid body about a fixed point. Poinsot motions. Elementary theory of a gyroscope and its application to the gyroscopic compass.
Outlines of Lagrangian Mechanics:
Principle of the stationary potential for the equilibrium of a conservative holonomic system (without proof). Lagrange equations for a conservative holonomic system (without proof) .
Enrico Massa, Elementi di Meccanica Razionale, dispense Università di Genova.
T. Levi-Civita and U. Amaldi, Lezioni di Meccanica Razionale, Zanichelli, Bologna (1984).
B. Finzi, Meccanica Razionale, Vol. II, Zanichelli, Bologna, (1965).
G. Grioli, Lezioni di Meccanica Razionale, Edizioni Libreria Cortina, Padova, (1985).
P. Biscari, T. Ruggeri. G. Saccomandi and M. Vianello, Meccanica Razionale per l'Ingegneria, Monduzzi Editore S.p.A., Bologna, (2008)..
TEACHERS AND EXAM BOARD
Exam Board
STEFANO VIGNOLO (President)
VALENTINA BERTELLA (President Substitute)
ROBERTUS VAN DER PUTTEN (President Substitute)
LESSONS
Class schedule
The timetable for this course is available here: Portale EasyAcademy
EXAMS
EXAM DESCRIPTION
A written and an oral test, after passing the written test with a mark greater then or egual to 16.
ASSESSMENT METHODS
The assignment of the exam grade will take into account: knowledge and understanding of the covered topics, ability and clarity of exposition, ability to solve problems related to the covered topics
Exam schedule
Data Ora Luogo Degree type Note
12/01/2024 10:00 LA SPEZIA Orale
09/02/2024 10:00 LA SPEZIA Orale
10/06/2024 14:30 LA SPEZIA Scritto
14/06/2024 10:00 LA SPEZIA Orale
28/06/2024 14:30 LA SPEZIA Scritto
02/09/2024 14:30 LA SPEZIA Scritto | 1,324 | 5,369 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-18 | latest | en | 0.724609 |
https://ximpledu.com/en-us/counterexample/ | 1,603,547,880,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107883636.39/warc/CC-MAIN-20201024135444-20201024165444-00534.warc.gz | 981,942,485 | 4,152 | # Counterexample
How to find the counterexample of the given statement: definition, 2 examples, and their solutions.
## Definition
### Definition
A counterexample is an example
that makes a statement false.
Only one counterexample is needed
to show that the statement is false.
## Example 1
### Solution
The given statement is a conditional statement.
p: x is a prime number.
q: x is an odd number.
Recall that
if p is true and q is false,
then p → q is false.
Conditional Statement: Truth Value
So, to find the counterexample,
find the x that makes
p true (= x is a prime number)
and q false (= x is not an odd number).
Set x = 2.
If x = 2,
p: 2 is a prime number.
This is true.
If x = 2,
q: 2 is an odd number.
This is false.
If x = 2,
p is true
and q is false.
Then p → q is false.
x = 2 makes
the given conditional statement false.
So x = 2 is the counterexample
of the given statement.
So [x = 2] is the answer.
## Example 2
### Solution
Find the case
that makes the given statement false.
Think of the case when
three points are on the same line.
As you can see,
these three points
cannot determine a triangle.
So the given statement is false.
[Three points on the same line]
makes the given statement false.
So [three points on the same line]
is the counterexample
of the given statement.
So [three points on the same line] | 352 | 1,363 | {"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} | 4.53125 | 5 | CC-MAIN-2020-45 | longest | en | 0.863063 |
https://www.physicsforums.com/threads/frictional-force-and-uk-problem.196463/ | 1,511,088,256,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805541.30/warc/CC-MAIN-20171119095916-20171119115916-00150.warc.gz | 870,502,297 | 16,291 | # Frictional Force and µk problem
1. Nov 6, 2007
### Senseiofj
1. The problem statement, all variables and given/known data
A brick is on a 121 cm plank, and the plank is lifted gradually until the brick begins to slide, when the plank is lifted 48cm in the air at an angle of 30º. I need to find the Frictional Force, and µ kinetic.
How do I find these things knowing only these three things?
2. Relevant equations
Ff=(µk)(Fn) let Fn= Normal Force
F=ma
g=9.81 m/s/s
Fn=mg(Cosø)
3. The attempt at a solution
I tried using the first equation I mentioned, but was unable to figure how to get both values.
Help is much appreciated.
2. Nov 6, 2007
### Staff: Mentor
Draw a diagram showing all the forces acting on the brick at the moment just before it begins to slide. What must the net force be in any direction?
3. Nov 6, 2007
### GTrax
You already have the force normal to the surface. You need the other component as well.
4. Nov 6, 2007
### Senseiofj
I think I should do this working purely without numbers, so I can just plug my figures in, I suppose.
I have the height of the end of the plank when the brick started sliding, the length of the plank, Speed of the brick, Mass, Weight, Fn, and the Fp [the force in the plane], I also have µ static. How do I find Ff and µk?
5. Nov 6, 2007
### Staff: Mentor
Identify the forces acting on the brick. Apply Newton's 2nd law to the forces parallel to the incline.
6. Nov 6, 2007
### GTrax
Er - speed of brick? no. The question is "What is the frictional force?"
As the plane is tipped steeper, there is a force component in the direction down the plane. There is another force component in the direction normal to the plane. They both derive from the force (from gravity) acting vertically downwards.
The solution is very easy once you take the major step of understanding the forces. You already have the component normal to the slope. This component, multiplied by the coefficient of friction usually called $$\mu$$ will be the force of friction resisting movement down that slope. It is the bit you do not have!
You do have the way to get at the frictional force. It has to equal the force down that slope, (because the brick stays put!!) Eventually, you discover its maximum at the un-stick angle. Get that, and you have one of the required answers. Getting at the (unknown) coefficient of friction then follows.
The static coefficient is the one that decides where the brick un-sticks. The dynamic coefficient a lesser value, (not so steep slope required). This is when the brick is moving. A special case is when the speed is steady. If accelerating, the available force to do that accelerating is what is left over after subtracting the friction.
Now we ask that you actually get to grips with the forces diagram, and demonstrate that you can take that step. It really is the one necessary and key bit of knowledge this question requires.
Last edited: Nov 6, 2007
7. Nov 6, 2007
### Senseiofj
Thanks a lot, man. I get it now.
8. Nov 7, 2007
### GTrax
OK then - so post the answer of what you think the force is down the slope, or better, the formula expression that you used to figure it. | 800 | 3,175 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2017-47 | longest | en | 0.936847 |
https://www.intmath.com/functions-and-graphs/undefined-slope.php | 1,723,232,361,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640768597.52/warc/CC-MAIN-20240809173246-20240809203246-00486.warc.gz | 639,162,372 | 22,918 | Search IntMath
Close
# What is Undefined Slope in Geometry?
Slope is an important concept in both algebra and geometry. It measures the steepness of a line and is used to help solve problems related to linear equations. When it comes to geometry, the slope of a line is determined by two points on the line, the rise and the run. The rise is the vertical distance between the two points, and the run is the horizontal distance between the two points. In this article, we'll discuss what it means when the slope of a line is undefined.
## What Does "Undefined" Mean?
In geometry, the term "undefined" means that the slope of a line cannot be determined. This often occurs when the line is either vertical or horizontal. A vertical line is a line that runs straight up and down and has no horizontal distance. A horizontal line is a line that runs straight across and has no vertical distance. In either of these cases, the rise and the run of the line will both be zero, and so the slope of the line cannot be calculated.
For example, take a look at the lines below. The first line is horizontal, and the second line is vertical. The run of the first line is zero, and the rise of the second line is zero. Therefore, the slopes of both of these lines are undefined.
## How Is Undefined Slope Used in Algebra?
In algebra, undefined slope is often used to solve linear equations. When an equation has two variables, such as x and y, the slope of the line that is formed by graphing the equation can be calculated using the formula m = (y2 - y1) / (x2 - x1). However, if the line that is formed by graphing the equation is either vertical or horizontal, the slope of the line will be undefined.
For example, take a look at the equation below. This equation can be graphed to form a horizontal line. Since the line is horizontal, the slope of the line is undefined. Therefore, the equation cannot be solved using the slope formula.
y = 7
## Practice Problems
Let's practice calculating the slope of lines and determining if the slope is undefined. In each of the problems below, draw the line on a coordinate plane and then calculate the slope of the line. If the slope of the line is undefined, write "undefined" as the answer.
1. x = 3
2. y = -4
3. y = 4x + 2
4. 3x - 2y = 6
1. undefined
2. undefined
3. m = 4
4. m = -3/2
## Summary
In geometry, undefined slope means that the slope of a line cannot be determined. This often occurs when the line is either vertical or horizontal. In algebra, undefined slope is often used to solve linear equations. When an equation forms a vertical or horizontal line, the slope of the line will be undefined and the equation cannot be solved using the slope formula. Practice calculating the slope of lines to become more familiar with this concept.
## FAQ
### What is a undefined slope in geometry?
In geometry, an undefined slope, sometimes referred to as an "infinite slope," is a term used to describe a line that has no slope. This means that the line is either vertical or horizontal.
### What is an example of an undefined slope?
A vertical line is an example of an undefined slope. A vertical line is always straight up and down, so its slope is undefined since it does not have a rise over a run.
### How do you know if slope is undefined?
You can determine if a slope is undefined by looking at the line. If the line is either vertical or horizontal, then the slope is undefined.
### How do you write undefined slope?
Undefined slope is typically written as either "no slope" or "infinite slope." | 801 | 3,563 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2024-33 | latest | en | 0.952283 |
https://physics.stackexchange.com/questions/658420/baker-campbell-formula/658451 | 1,712,928,904,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816024.45/warc/CC-MAIN-20240412132154-20240412162154-00397.warc.gz | 393,184,864 | 35,174 | # Baker-Campbell formula [closed]
If $$B a B^† = a\cos(\theta)+ib\sin(\theta)$$ then can I write $$B a^n B^† = [{a\cos(\theta)+ib\sin(\theta)}]^{n} ?$$
where $$B = B=e^{i\theta(a^\dagger b+b^\dagger a)}$$
This is my calculation for $$B a B^†$$
$$B a B^†= a+\theta[(a^\dagger)b+(b^\dagger) a),a]+ (\theta)^2/2[(a^\dagger)b+(b^\dagger) a[(a^\dagger)b+(b^\dagger) a),a]]$$ $$=a\cos(\theta)+ib\sin(\theta)$$
• It would help if you provide more information. Where does $\theta$ come from? Aug 10, 2021 at 4:04
• B=eθ((a†)b+(b†)a) Aug 10, 2021 at 4:12
• Note \cos and \sin are valid commands in LaTeX. Aug 10, 2021 at 4:20
• I agree Vishaka. Thanks for the info Aug 10, 2021 at 4:20
• I suggest thinking about what the inverse of $B$ is. Aug 10, 2021 at 6:41
$$B$$ is unitary, hence $$B^{\dagger}B = I = BB^{\dagger}$$.
Let us now compute $$[B a B^{\dagger}]^{n}$$. First consider the case $$n=2$$.
$$[B a B^{\dagger}]^{2} = B a B^{\dagger}BaB^{\dagger} = Ba^{2}B^{\dagger}.$$
No assume that for $$n=k$$
$$[B a B^{\dagger}]^{k} = Ba^{k}B^{\dagger}.$$
Finally multiplying the kth case by $$BaB^{\dagger}$$ we get the following.
$$[B a B^{\dagger}]^{k} BaB^{\dagger}= Ba^{k}B^{\dagger}BaB^{\dagger} = Ba^{k+1}B^{\dagger}.$$
By induction we conclude that $$$$[BaB^{\dagger}]^{n} = Ba^{n}B^{\dagger} (1).$$$$
If $$BaB^{\dagger} = a\cos(\theta)+ib\sin{\theta}$$ then (1) yields the result you are after. i.e. $$[a\cos(\theta)+ib\sin{\theta}]^{n} = Ba^{n}B^{\dagger}.$$ | 617 | 1,471 | {"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": 18, "wp-katex-eq": 0, "align": 0, "equation": 1, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2024-18 | latest | en | 0.681827 |
https://essaycube.com/2017/11/22/itcm-310-plane-surveying-3/ | 1,553,561,638,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912204736.6/warc/CC-MAIN-20190325234449-20190326020449-00132.warc.gz | 487,944,728 | 13,591 | # ITCM 310 – Plane Surveying
ITCM 310 – Plane Surveying
1.   90% Error – A distance was taped six times with the following results: 85.67, 86.33, 85.90, 85.95, 86.16, and 85.92 feet. Compute the 90 percent error and give the maximum and minimum range of accuracy for the survey.
2.   Relative Accuracy – A distance of 345.75 feet is measured by a survey crew. The true distance is know to be 345.85 feet. What is the relative accuracy of the measurement?
3.   Maximum Error of Closure – What is the maximum error of closure in a measurement of 2500 feet if the relative accuracy is 1:5000?
Trigonometric Review Problems
4.   The hypotenuse of a right triangle is 106.32 ft long and one of the other sides has a length equal to 69.66 ft.  Find the angle opposite to the 69.66 ft. side.
5.   The three sides of a triangle are 60, 80, and 100 ft.  Determine the magnitude of the interior angles.
6.   A surveyor measures an inclined distance and finds it to be 1642.5 ft.  In addition, the angle between the horizontal or level plane and the inclined line is measured at 2°56’30â€. Determine the horizontal distance and the difference in elevation between the two ends of the line.
7.   A section of a road is to be paved. This section has a constant 3% slope or grade (i.e. 3 ft. vertically rise or fall for each 100 ft. horizontally) is to be paved. If the road is 24.000 ft wide and its total horizontal length is 900.0 ft., compute the road area to be paved.
# Our Service Charter
1. ### Excellent Quality / 100% Plagiarism-Free
We employ a number of measures to ensure top quality essays. The papers go through a system of quality control prior to delivery. We run plagiarism checks on each paper to ensure that they will be 100% plagiarism-free. So, only clean copies hit customers’ emails. We also never resell the papers completed by our writers. So, once it is checked using a plagiarism checker, the paper will be unique. Speaking of the academic writing standards, we will stick to the assignment brief given by the customer and assign the perfect writer. By saying “the perfect writer” we mean the one having an academic degree in the customer’s study field and positive feedback from other customers.
2. ### Free Revisions
We keep the quality bar of all papers high. But in case you need some extra brilliance to the paper, here’s what to do. First of all, you can choose a top writer. It means that we will assign an expert with a degree in your subject. And secondly, you can rely on our editing services. Our editors will revise your papers, checking whether or not they comply with high standards of academic writing. In addition, editing entails adjusting content if it’s off the topic, adding more sources, refining the language style, and making sure the referencing style is followed.
3. ### Confidentiality / 100% No Disclosure
We make sure that clients’ personal data remains confidential and is not exploited for any purposes beyond those related to our services. We only ask you to provide us with the information that is required to produce the paper according to your writing needs. Please note that the payment info is protected as well. Feel free to refer to the support team for more information about our payment methods. The fact that you used our service is kept secret due to the advanced security standards. So, you can be sure that no one will find out that you got a paper from our writing service.
4. ### Money Back Guarantee
If the writer doesn’t address all the questions on your assignment brief or the delivered paper appears to be off the topic, you can ask for a refund. Or, if it is applicable, you can opt in for free revision within 14-30 days, depending on your paper’s length. The revision or refund request should be sent within 14 days after delivery. The customer gets 100% money-back in case they haven't downloaded the paper. All approved refunds will be returned to the customer’s credit card or Bonus Balance in a form of store credit. Take a note that we will send an extra compensation if the customers goes with a store credit.
We have a support team working 24/7 ready to give your issue concerning the order their immediate attention. If you have any questions about the ordering process, communication with the writer, payment options, feel free to join live chat. Be sure to get a fast response. They can also give you the exact price quote, taking into account the timing, desired academic level of the paper, and the number of pages.
Excellent Quality
Zero Plagiarism
Expert Writers
or | 1,102 | 4,628 | {"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} | 3.953125 | 4 | CC-MAIN-2019-13 | longest | en | 0.880573 |
http://www.gamedev.net/index.php?app=forums&module=extras§ion=postHistory&pid=4920595 | 1,386,837,663,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164572870/warc/CC-MAIN-20131204134252-00016-ip-10-33-133-15.ec2.internal.warc.gz | 354,075,204 | 13,833 | • Create Account
### #ActualDragonsoulj
Posted 09 March 2012 - 12:52 AM
Your system should have a constant g, should it not? Your entire world would have some form of gravity, and your character that is finished will probably be using the same constant. At mid jump, the apex, your V0y would be 0. At the start and the finish, your V0y would be 0. Mid-jump, your V0y would be your previous Vy. I may be wrong on that very last statement, but I don't think so. If V0y was to consider your initial jump's starting velocity, the character is not moving, so that would fully eliminate V0y * t.
EDIT: As an afterthought, and due to it being late, V0y would be a function included in your equation for y, and each point during the jump would recalculate the equation with a different t: y = (g * t + V0y) * t - ((g * t^2) / 2) where V0y is the very first velocity you want to give the character to jump.
-- To clarify: (g * t + V0y) is the velocity the character is at at the end of the last update.
### #2Dragonsoulj
Posted 09 March 2012 - 12:50 AM
Your system should have a constant g, should it not? Your entire world would have some form of gravity, and your character that is finished will probably be using the same constant. At mid jump, the apex, your V0y would be 0. At the start and the finish, your V0y would be 0. Mid-jump, your V0y would be your previous Vy. I may be wrong on that very last statement, but I don't think so. If V0y was to consider your initial jump's starting velocity, the character is not moving, so that would fully eliminate V0y * t.
EDIT: As an afterthought, and due to it being late, V0y would be a function included in your equation for y, and each point during the jump would recalculate the equation with a different t: y = (g * t + V0y) * t - ((g * t^2) / 2) where V0y is the very first velocity you want to give the character to jump.
### #1Dragonsoulj
Posted 09 March 2012 - 12:42 AM
Your system should have a constant g, should it not? Your entire world would have some form of gravity, and your character that is finished will probably be using the same constant. At mid jump, the apex, your V0y would be 0. At the start and the finish, your V0y would be 0. Mid-jump, your V0y would be your previous Vy. I may be wrong on that very last statement, but I don't think so. If V0y was to consider your initial jump's starting velocity, the character is not moving, so that would fully eliminate V0y * t.
PARTNERS | 666 | 2,459 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2013-48 | latest | en | 0.972875 |
http://toc.123doc.org/document/970070-7-debt-value-adjustment-dva-if-something-sounds-too-good-to-be-true.htm | 1,503,148,199,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105451.99/warc/CC-MAIN-20170819124333-20170819144333-00398.warc.gz | 423,106,357 | 11,109 | Tải bản đầy đủ
7 Debt Value Adjustment (DVA): If Something Sounds Too Good to Be True . . .
# 7 Debt Value Adjustment (DVA): If Something Sounds Too Good to Be True . . .
Tải bản đầy đủ
WEBC12
11/25/2013
15:50:58
Page 275
Correlation and Basel II and III
275
Allows an entity to adjust the value of its portfolio by taking its own
default probability into consideration.
The Basel accord prefers the term CVA liability instead of DVA.
However, we will refer to it as DVA.
In Figure 12.5 we displayed credit exposure and concluded that credit
exposure can only be bigger or equal zero. Credit exposure for entity a with
counterparty c exists if the counterparty c is a net debtor to a. If we allow
recognizing negative credit exposure or debt exposure, Figure 12.5 would
change to Figure 12.13.
This debt exposure of a with respect to c could theoretically be taken into
consideration when evaluating a portfolio. In particular, debt exposure could
be recognized in derivatives transactions. This debt exposure in derivatives
transactions is the netted negative derivatives portfolio value of entity a with
. This is weighted, i.e. reduced by the probability of default of
respect to c, Da;c
entity a. Including a recovery rate of a, we derive in analogy to equation (12.12)
+
´ PDc )(1 − Rc ),
for CVA, which is: CVAa;c = (Da;c
DVAa;c = (Da;c
´ PDa )(1 − Ra )
Credit exposure of entity a
with respect to c
Netted portfolio value
from the viewpoint of
a with respect to c
Debt exposure of entity a
with respect to c
FIGURE 12.13 Debt Exposure when the Netted Portfolio Value of Entity a is
Negative with Respect to Entity c (i.e., a is a net debtor for c)
(12.18)
WEBC12
11/25/2013
15:50:58
Page 276
CORRELATION RISK MODELING AND MANAGEMENT
276
where
DVAa,c: debt value adjustment of entity a with respect to entity c
: netted negative derivatives portfolio value of a with respect to c (i.e.,
Da;c
a is a debtor to c)
PDa: default probability of entity a
Ra: recovery rate of entity a
Importantly, let’s now consider that in the event of default of entity a,
only the recovery rate of a’s debt is paid out. If this is accounted for,
this decreases a’s debt and increases the book value (defined as
assets minus debt) of a. If we apply this concept to a derivatives
portfolio, the derivatives portfolio value increases and equation (12.10)
expands to
Value of
Derivatives
Portfolio
=
Default-Free
Value
CVA
+
DVA
(12.19)
However, there are two critical problems with DVA:
1. An entity such as a would benefit from its own increasing default
probability PDa, since a higher default probability would increase
DVA via equation (12.18), which in turn increases the value of the
derivatives portfolio via equation (12.19).
2. Entity a could realize the DVA benefit only if it actually defaults.
Both properties defy financial logic. Therefore the Basel accord has
principally refrained from allowing DVA to be recognized. In 2008 several
financial firms had actually reported huge increases in their derivatives
portfolios due to DVA. This is no longer possible.
A further recent development relating to CVA and DVA is funding value
WEBC12
11/25/2013
15:50:58
Page 277
Correlation and Basel II and III
277
An adjustment to the price of a transaction due to the cost of funding
for the transaction or the related hedge.
Funding cost had not been a major issue in derivatives pricing in the past.
However, in 2008, when interest rates especially for poor credits increased
sharply, funding cost could no longer be ignored.
There has been quite a spirited debate in 2012 and 2013 about whether
the cost of funding should be taken into consideration when pricing a
derivative. Hull and White as well as Duffie (Risk 2012(a) and 2012(b))
argue that adding funding costs violates the risk-neutral derivatives pricing
principle. It would lead to arbitrage opportunities, since the same derivative
would have different prices. However, derivatives traders argue that their
treasury departments charge them the funding costs. Hence funding costs
exist in reality and cannot just be ignored. The funding cost should be priced
in and passed through to the end user. See “The FVA Debate” in Risk, July
2012, and “Traders v. Theorists” in Risk, September 2012, for further
details.
Let’s look at the issue of cost of funding. The cost of funding of an entity
is mainly a function of the default probability of the entity. Hence we have
FVAa = f (PDa ; . . . ; )
(12.20)
There is a positive relationship between FVA and PD, since the higher the
default probability, the higher is the cost of funding:
∂FVAa
>0
∂PDa
(12.21)
If the cost of FVA is taken into account, the value of a derivatives
portfolio is reduced. Hence equation (12.19) then changes to
Value of
Derivatives
Portfolio
=
Default-Free
Value
CVA
+
DVA
FVA
(12.22)
As we see from equations (12.18) and (12.21), both DVA and FVA
increase if the probability of default increases; hence credit quality decreases.
WEBC12
11/25/2013
15:50:58
278
Page 278
CORRELATION RISK MODELING AND MANAGEMENT
In a 2012 response to the Basel proposal, the International Swaps and
Derivatives Association (ISDA) has suggested that “CVA liability [i.e.
DVA] should be deducted only to the extent that it exceeds the increase in
FVA.”10 In this case there would be no benefit (i.e., no increase in the value of
a derivatives portfolio) if the default probability of an entity increases, as we
can see from equation (12.22), (since DVA is added only up to the amount
that FVA is subtracted).
12.9 SUMMARY
In this chapter we discussed the way correlation risk is addressed in the Basel
II and Basel III frameworks. The Basel committee has recognized the significance of correlation risk and has suggested several approaches to managing
correlation risk.
Correlation risk is a critical factor in managing credit risk. In the Basel II
and III accords, credit risk of a portfolio is quantified with the credit value at
risk (CVaR) concept. CVaR measures the maximum loss of a portfolio due to
credit risk with a certain probability for a certain time frame. Basel II and
Basel III derive CVaR on the basis of the one-factor Gaussian copula (OFGC)
correlation model, which we discussed in Chapter 6.
The required capital to be set aside for credit risk is the CVaR minus the
average probability of default of the debtors in the portfolio. This is because
the Basel committee assumes that banks cover the expected loss (approximated as the average probability of default) with their own provisions such as
the interest rate that they charge.
Interestingly, the Basel committee requires an inverse relationship
between the default correlation of the debtors in a portfolio with respect
to the default probability of the debtors: The lower the default probability of debtors in a portfolio, the higher is the default correlation between
the debtors. This is reasonable, since debtors with a low default probability are more prone to default for systematic reasons; that is, they more
often default together in a recession. Conversely, low rated debtors
with a high default probability are more affected by their own idiosyncratic factors and less by systematic risk. Hence the default risk of low
rated debtors is assumed to be less correlated. This is supported by
empirical data.
10. See ISDA, “ISDA and Industry Response to BCBS Paper on Application of Own
Credit Risk Adjustments to Derivatives,” 2012, www2.isda.org/functional-areas/riskmanagement/page/3.
WEBC12
11/25/2013
15:50:59
Page 279
Correlation and Basel II and III
279
A further aspect, which has become a critical factor in credit risk
management, is credit value adjustment (CVA). CVA is a capital charge
to address credit risk, mainly in derivatives transactions. CVA has a market
risk component (the netted derivatives value) and a credit risk component (the
probability of default of the counterparty). Importantly, these market risk
and credit risk components are typically correlated! This results in the
correlation concept of wrong-way risk (WWR). The Basel committee defines
two types of wrong-way risk:
1. General wrong-way risk arises when the probability of default of
counterparties is positively correlated with general market risk factors.
2. Specific wrong-way risk exists if the exposure to a specific counterparty is
positively correlated with the counterparty’s probability of default due to
the nature of the transaction with the counterparty.
The Basel committee requires financial institutions to address wrong-way
risk: Financial institutions have to increase their credit exposure value
(calculated without wrong-way risk) by 40%. Financial institutions that
use their own internal models can apply a 20% increase. This is conservative,
since banks report a numerical value for wrong-way risk of 1.07 to 1.1.
The Basel committee also realizes the risk reduction that is achieved when
a credit exposure is hedged with a credit default swap (CDS). The Basel
committee allows banks to address the credit risk reduction of a CDS in two
ways: (1) the substitution approach, which allows banks to use the typically
lower default probability of the guarantor (CDS seller) in the credit exposure
calculation, and (2) the double default approach, which derives the joint
probability of the obligor and the guarantor defaulting. This joint default
probability is typically much lower than the individual default probability of
the obligor, lowering the overall credit exposure value.
The concept of CVA has recently been extended by the concepts debt
adjustment (DVA) allows an entity to adjust the value of a position (such as a
loan or a derivative) in a portfolio by taking its own default probability into
consideration. If an entity applies DVA (i.e., takes its own default probability
into consideration), this actually reduces the credit exposure of the entity.
This is highly controversial and has been banned by the Basel committee.
Funding value adjustment (FVA) is an adjustment to the price of a
transaction, typically a derivative, due to the cost of funding the transaction.
FVA has been quite controversially debated in 2012 and 2013. Finance
professors argue that it creates arbitrage opportunities, since different FVA
funding costs are substantial and have to be included in the transaction price.
WEBC12
11/25/2013
280
15:50:59
Page 280
CORRELATION RISK MODELING AND MANAGEMENT
PRACTICE QUESTIONS AND PROBLEMS
1. What information does credit value at risk (CVaR) give us?
2. Why don’t we just apply the market value at risk (VaR) concept to value
credit risk?
3. Which correlation concept underlies the CVaR concept of the Basel II and
III approach?
4. In the Basel committee CVaR approach, what follows for the relationship
between the CVaR value and the average probability of default, if we
assume the correlation between all assets in the portfolio is zero?
5. Suppose Deutsche Bank has given loans to several companies in the
amount of \$500,000,000. The average 1-year default probability of
the companies is 2%. The copula default correlation coefficient between
the companies is 3%. What is the 1-year CVaR on a 99.9% confidence level?
6. In the Basel committee CVaR model, the default correlation is an inverse
function of the average probability of the default of the assets in the
portfolio. Explain the rationale for this relationship.
7. In the Basel committee approach, the required capital to be set aside for credit
risk is the CVaR minus the average probability of default. Explain why.
8. CVA is an important concept of credit risk. What is CVA? Why is it
important?
9. Why can CVA be considered a complex derivative?
10. How can CVA without correlation between market risk and credit risk be
calculated?
11. Including the correlation between market risk and credit, the concept of
wrong-way risk (WWR) arises. What is general wrong-way risk, and
what is specific wrong-way risk?
12. Name two examples of specific wrong-way risk.
13. How does the Basel committee address wrong-way risk?
14. What is DVA? Should DVA be allowed to be applied in financial practice?
15. What is FVA? Should FVA be included in the pricing of derivatives?
BCBS. 2003. “Annex (to Basel II).” www.bis.org/bcbs/cp3annex.pdf.
BCBS. 2005a. “The Application of Basel II to Trading Activities and the Treatment of
Double Default Effects.” www.bis.org/publ.bcbs116.pdf.
BCBS. 2005b. “International Convergence of Capital Measurement and Capital
Standard: A Revised Framework.” November 2005. www.bis.org/publ.bcbs118
.pdf.
WEBC12
11/25/2013
15:50:59
Page 281
Correlation and Basel II and III
281
BCBS. 2011. “Basel III: A Global Regulatory Framework for More Resilient Banks
and Banking Systems.” June 2011, 1, www.bis.org/publ/bcbs189.htm.
Cepedes, J., J. A. Herrero, D. Rosen, and D. Saunders. 2010. “Effective Modeling of
Wrong Way Risk, Counterparty Credit Risk Capital and Alpha in Basel II.”
Journal of Risk Model Validation 4(1): 71–98.
Gordy, M. 2003. “A Risk-Factor Model Foundation for Ratings-Based Bank Capital
Rules,” Journal of Finanical Intermediation 12(3): 199–232.
Hull, J. 2012. Risk Management and Financial Institutions. 3rd ed. Wiley Finance
Series. Hoboken, NJ: John Wiley & Sons, Chapter 12.
Hull, J., and A. White. 2011. “CVA and Wrong Way Risk.” University of Toronto
Working paper.
ISDA. 2012. “ISDA and Industry Response to BCBS Paper on Application of Own
Credit Risk Adjustments to Derivatives.” www2.isda.org/functional-areas/riskmanagement/page/3.
Meissner, G. 2005. Credit Derivatives—Application, Pricing, and Risk Management.
Oxford: Wiley-Blackwell Publishing.
Risk. 2012a. “The FVA Debate.” July, 83–85.
Risk. 2012b. “Traders v. Theorists.” September, 19–22.
Vasicek, O. 1987. “Probability of Loss on a Loan Portfolio.” KMV Working paper.
Results published in Risk magazine with the title “Loan Portfolio Value,”
December 2002.
WEBC12
11/25/2013
15:50:59
Page 282
WEBC13
11/25/2013
16:1:43
Page 283
CHAPTER
13
The Future of Correlation
Modeling
Solving the right problem numerically beats solving the wrong
problem analytically every time.
—Richard Martin
n this chapter we discuss new developments in financial modeling that can
be extended to correlation modeling. We address the application of
graphical processing units (GPUs), which allow fast parallel execution
of numerically intensive code without the need for mathematical solvency.
We also discuss some new artificial intelligence approaches such as neural
networks, genetic algorithms, as well as fuzzy logic, Bayesian mathematics,
and chaos theory.
I
13.1 NUMERICAL FINANCE: SOLVING FINANCIAL
PROBLEMS NUMERICALLY WITH THE HELP OF
GRAPHICAL PROCESSING UNITS (GPUs)
Some problems in finance are quite complex so that a closed form solution is
not available. For example, path-dependent options such as American-style
options principally have to be evaluated on a binomial or multinominal tree,
since we have to check at each node of the tree if early exercise is rational. In
risk management, especially in credit risk management, thousands of correlated default risks have to be evaluated. While there are simple approximate
measures to model counterparty risk in a portfolio such as the Gaussian
copula model (see Chapter 6), it is more rigorous to model counterparty risk
283
WEBC13
11/25/2013
16:1:43
Page 284
284
CORRELATION RISK MODELING AND MANAGEMENT
on a multifactor approach using numerical methods such as Monte Carlo
simulation.
In the recent past, the increase of computer power has made numerical
finance an alternative to analytical solutions. Let’s define it:
NUMERICAL FINANCE
Attempts to solve financial problems with numerical methods (such as
Monte Carlo simulation), without the need of mathematical solvency.
Other terms for numerical finance are statistical finance, computational
finance, and also econophysics. More narrowly defined, econophysics is the
combination of physical concepts and economics. However, the economic
concepts include stochastic processes and their uncertainty, which are also an
essential part of finance.
Why waste good technology on science and medicine?
—Lighthearted phrase of gamers on GPU technology
13.1.1 GPU Technology
Graphical processing units (GPUs) are the basis for a technology that alters
memory in a parallel execution of commands to instantaneously produce
high-resolution three-dimensional images. The GPU technology was derived
in the computer gaming industry, where gamers request high-resolution,
instant response for their three-dimensional activities at low cost. This caught
the attention of the financial industry, which is paying millions of dollars to
receive real-time response for valuing complex financial transactions and risk
management sensitivities.
Hence, over time, financial software providers have started to rewrite
their mathematical code to make it applicable for the GPU environment.
Companies such as Murex, SciComp, Global Valuation Limited,
Hanweck Associates, BNP Paribas, and many others have implemented
GPU-based infrastructures to numerically solve complex derivatives
transactions and calculate risk parameters. The academic environment
has also responded. More than 600 universities worldwide offer courses
in GPU programming. | 4,224 | 17,303 | {"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-2017-34 | latest | en | 0.859815 |
https://file.scirp.org/Html/5-1770175_61606.htm | 1,714,011,734,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296820065.92/warc/CC-MAIN-20240425000826-20240425030826-00149.warc.gz | 220,519,343 | 18,355 | Preferred Economic Dispatch of Thermal Power Units
Journal of Power and Energy Engineering
Vol.03 No.11(2015), Article ID:61606,23 pages
10.4236/jpee.2015.311005
Preferred Economic Dispatch of Thermal Power Units
Selvaraj Durai, Srikrishna Subramanian, Sivarajan Ganesan
Department of Electrical Engineering, Annamalai University, Chidambaram, India
Received 12 October 2015; accepted 27 November 2015; published 30 November 2015
ABSTRACT
Economic Dispatch (ED) problem is one of the main concerns of the power generation operations which are basically solved to generate optimal amount of power from the generating units in the system by minimizing the fuel cost and by satisfying the system constraints. The accuracy of ED solutions is highly influenced by the fuel cost parameters of the generating units. Generally, the parameters are subjected to transform due to aging process and other external issues. Further the parameters associated with the transmission line modelling also change due to aforementioned issues. The loss coefficients which are the functions of transmission line parameters get altered from the original value over a time. Hence, the periodical estimation of these coefficients is highly essential in power system problems for obtaining ideal solutions for ED problem. Estimating the ideal parameters of the ED problem may be the best solution for this issue. This paper presents the Teaching Learning Based Optimization (TLBO) algorithm to estimate the parameters associated with ED problem. The estimation problem is formulated as an error minimization problem. This work provides a frame work for the computation of coefficients for quadratic function, piece- wise quadratic cost function, emission function, transmission line parameters and loss coefficients. The effectiveness of TLBO is tested with 2 standard test systems and an Indian utility system.
Keywords:
Parameter Estimation, Cost Coefficients, Emission Coefficients, Error Estimation, Transmission Line Parameters, Teaching Learning Based Optimization
1. Introduction
1.1. Ideal Economic Dispatch
The economical operation of power system needs more accurate representation of fuel cost coefficients and transmission loss coefficients. Due to changing weather conditions and other external issues, the aforesaid coefficients are not ideal and hence the existing parameters with ED solutions also change. Therefore, the objective of this article is to revise the existing parameters for the Preferred Economic Dispatch (PED) solutions.
1.2. Literature Review
Various optimization techniques have been proposed by many researchers to deal with the parameter estimation problems. To obtain the exact parameters for ED problems different parameter estimation rehearsal has been projected to solve estimation problems in power systems. State estimation based technique such as Least Error Square (LES) and least absolute value methods have been demonstrated. Among the two techniques, LES technique has been the most famous static estimation technique and in use for a long time as the preferred technique for optimum estimation of parameters. In general, some limitations and disadvantages are associated with this approach. El-Hawary and Mansour conducted performance analysis of LES, Bard algorithm, Marquardt algorithm and Powell regression algorithm for estimating coefficients [1] . The methods based on the least absolute value approximations and curve fitting techniques have been reported for fuel cost coefficients estimation [2] . Two polynomial curve fitting methods, Gram-Schmidt orthonormalization and least square are also applied to evaluate the fuel cost coefficients [3] . Further Henry Y. K. Chen and Charles E. Postel applied sequential regression technique to online parameter estimation of input-output curves for thermal units [4] .
Research endeavours indicate that the field of Transmission Line Parameters (TLP) estimation in power system studies is less focused. Traditionally, tower and conductor geometric parameters, conductor type, assumed ambient conditions etc., are utilized for estimating TLP [5] - [7] . The sending and receiving end voltage and current phasors are utilized to derive the TLP and propagation constant [8] . The synchronized phasor measurements at both ends of the transmission line emphasises the online parameter estimation [9] - [12] . Wagenaars et al., have proposed the measurement method, based on a pulse response measurement, to determine the transmission line parameters of the shield-to-phase and phase-to-phase modes [13] . Researchers have reported Newton-Raphson based method for estimating TLP parameters utilizing the measurements of voltage, current and power [14] .
A new category of classical optimization techniques has emerged to cope with some of the traditional algorithms in the field of power system estimation problems. The heuristic techniques such as Genetic Algorithm (GA) [15] , Particle Swarm Optimization (PSO) [16] [17] , Artificial Bee Colony (ABC) [18] , Artificial Neural Network (ANN) [19] [20] , Fuzzy Logic (FL) [21] , Ant Colony Optimization (ACO) [22] and Bacterial Foraging Algorithm (BFA) [23] - [25] have been used for solving various estimation problems such as load forecasting, state estimation and induction motor parameters.
1.3. Optimization Techniques
The heuristic techniques outperform the mathematical methods but their solution quality is sensitive to the algorithmic controlling parameters like population size and number of generations. Besides common control parameters, different algorithms require their own algorithm specific control parameters. Major disadvantage of ABC, PSO and GA are the presence of various parameters that need to be carefully tuned to reach acceptable estimation performance. Recently, TLBO, a nature inspired algorithm that is based on the effect of influence of a teacher on the output of learners in a class [26] [27] is proposed. It is a powerful evolutionary algorithm that maintains a population of students, where each student represents a potential solution to an optimisation problem. Each searching generation includes initializing of class, teacher phase, learner phase and termination criterion. The advantage of TLBO are simple, easy implementation and necessitates few control parameters for tuning that make it suitable to implement for power system parameter estimation problems. TLBO is an algorithm-specific, parameter-less algorithm that does not require any algorithm-specific parameters to be tuned [28] . This algorithm can find the global solution for nonlinear constrained optimization problems with less computation effort and high consistency [29] . The authors have proposed the TLBO algorithm for estimating the accurate parameters of the input output characteristics of thermal units [30] . The adaptable properties of this algorithm encourage the authors to use TLBO as a parameter estimator.
1.4. Research Gap and Motivation
The most important issue in ED problem is to have an accurate estimate of parameters. The exact ED solution in power generation and the estimation of parameters in transmission systems is an important task in power systems. The demonstration of literature review reveals that the estimation of accurate fuel cost coefficients and transmission line parameters are carried out independently. The impact of change in transmission line parameters on the loss coefficients of ED problem is seldom carried out. These points motivate us to contribute research in estimation of cost and emission coefficients, transmission line parameters and transmission loss coefficients for the exact ED solution. The TLBO algorithm is implemented for solving parameter and ED problems on different scale of test systems.
1.5. Highlights
・ Accurate parameters of different generator cost functions and transmission lines are estimated.
・ The preferred economic dispatch is carried out.
・ A 19 unit practical Indian utility system is considered.
・ A nature inspired TLBO is applied for both estimation of parameters and for solving ED problems.
1.6. Paper Organization
The rest of the paper is structured as follows: Problem formulation is explored in Section 2. Section 3 describes the implementation for preferred ED. The detailed discussions about numerical results achieved by various test systems are detailed in Section 4. Section 5 describes the potential verification of TLBO. Finally, Section 6 presents the conclusions of this article.
2. Problem Formulation
2.1. State of the Art Model
The state of the art multi objective ED model is presented as follows.
(1)
is a existing fuel cost function, N is the number of generating units and Pi is the power generated by ith generating unit.
, , are the existing cost coefficients, , are the existing valve point coefficients and, , are the existing emission coefficients .
W1 and W2 are the weights of the function.
ri is the error associated with the ith equation.
Loss function: (2)
indicates existing transmission loss.
, , are the existing transmission loss coefficients.
2.2. Proposed Parameter Estimation Problems
For preferred ED solution, the accurate parameters for fuel cost, emission, and transmission line and transmission loss are needed. Those parameters can be estimated by solving the following problems.
Cost coefficients Estimation, FCEst = Estimate [a, b, c, e, f]
Emission coefficients Estimation, EEst = Estimate [e0, e1, e2]
Transmission line parameters Estimation, TLPEst = Estimate [R, X, B]
Transmission loss coefficients Estimation, KCEst = Estimate [Bij, B0i, B00]
In general to estimate the accurate parameters, the following function are used
(3)
2.3. Accurate Model Using Estimated Parameters
The cost function of generating unit is expressed using estimated coefficients as follows,
(4)
where is an estimated fuel cost.
, , , , , , are the estimated cost coefficients, valve point coefficients and emission coefficients respectively.
Loss function: (5)
, , are the estimated transmission loss coefficients and is the estimated transmission loss.
2.4. Error Minimization
As detailed in [30] , the error is estimated for fuel cost and emission coefficients as follows
The error at each step i can be calculated as,
(6)
(7)
(8)
The % error for each step k of transmission line parameter is expressed as,
(9)
The objective function is minimised subjected to physical limits of each coefficients and TLP.
3. Implementation of TLBO for Preferred Economic Dispatch (PED)
This section details the computational flows for parameter estimation and PED problems.
3.1. Parameter Estimation Procedure
Step 1: Read the required data for estimating the parameters, population size, maximum number of iterations (Itermax), maximum and minimum limits of parameters (PAmin,).
Step 2:Randomly generate parameters using the following equation,
(10)
Step 3: Check for the constraint violation.
Step 4: If the limit is violated, then upgrade the parameters of the units using the following equation
(11)
Step 5: Calculate FCEst by using Equations (4).
Step 6: The error and % error associated with each measurement can be calculated by using Equations (6), (7), (8) and (9).
Step 7: Teacher phase: Compute the difference between existing mean result and best mean by utilizing teaching factor.
Step 8: Learner phase: Evaluate the learners’ parameter values with the help of teacher’s parameter values.
Step 9: Update the parameters subject to constraints.
Step 10: Stopping Criterion: If Iter ≥ Itermax print the optimal results, otherwise go to step 7.
3.2. Computational Flow of PED
Hence the estimated parameters are suitably achieved. The ED solution is carried out and the computational flow is as follows in Figure 1.
4. Numerical Simulation Results and Discussion
This section presents three different test systems such as 10 unit, IEEE 30 bus and a practical Indian Utility system to demonstrate the efficacy of the proposed solution technique for perfect estimate of parameters and preferred ED solutions. The algorithm is implemented in Matlab package and simulations are carried out on a personal computer having the configuration of Intel(R) core i3 CPU with 2 GB RAM. The following algorithmic parameters are chosen for all cases in order to validate the performance of the algorithm: Population size = 10;
Figure 1. Flowchart for preferred economic dispatch using TLBO.
Maximum number of iterations = 120.
4.1. Test System 1: 10 Unit System with Piecewise Quadratic Functions
This test system consists of 10-generating units in which each unit has two or three fuel options like coal, oil and gas. As more than one fuel is used, the generator input-output characteristic is expressed as a piecewise quadratic function. The valve point effects are also considered. The test system data is available in [31] .
The generator characteristic is expressed as a piecewise quadratic function including valve point effects increases the number of variables to be estimated. In this test system, the generator 1 has two fuel options and the remaining units have three fuel options. It is specified that the fuel 2 is uneconomical for generator 9. The simulation is carried out using the proposed TLBO which acquires new values for cost coefficients and are presented in the Table 1. The estimation process has been performed for each generation level and the results of existing
Table 1. Estimation of piecewise quadratic cost function coefficients using TLBO-10 unit system.
fuel cost (FCExi) and estimated fuel cost (FCEst) are presented independently for each fuel type in the Table 1. The error is also computed by comparing FCExi and FCEst using Equation (6). The obtained error using the proposed algorithm is compared with ABC algorithm which clearly illustrates the superiority of TLBO over other algorithm.
The ED is carried out separately considering both existing and estimated fuel cost coefficients. The obtained results for various demands are presented in Table 2 along with % error, which clearly illustrates the impact of estimated coefficients on ED problem.
4.2. Test System 2: IEEE 30 Bus System
The TLBO algorithm has been executed on the test system consists of six generating units interconnected with 41 branches of transmission lines. Generator data, emission characteristics and line data are obtained from [32] . TLBO algorithm is applied to estimate the cost, emission coefficients and transmission line parameters. For each unit, the fuel cost and emission coefficients are computed. Table 3 and Table 4 show the TLBO outcomes of the
Table 2. Economic dispatch with improved parameters for 10 unit system.
Table 3. Estimation of quadratic cost function coefficients by TLBO-IEEE 30 bus system.
Table 4. Estimation of emission function coefficients by TLBO-IEEE 30 bus system.
estimated coefficients. Existing fuel cost and existing emission for three different power generations are compared with its FCEst and EEst and it is found that those values are very close to each other. The new values of coefficients and their related total error are presented in the Table 3 and Table 4. From the results the error obtained by TLBO for all specified output of generating units is less which shows the accurate estimation of the system parameters.
TLP are obtained using Equations (A.2-A.5) for each line and the estimated TLP results are listed in Table 5. The estimated TLP values are compared with base case values and the percentage error is computed using Equation (9). The % error values for transmission lines are almost close to zero except for the lines 24 and 29 for which the TLP values are above 1. The efficiency of transmission lines 11-16, 37 are high as compared to other transmission lines, due to the presence of three winding transformer in the transmission lines. The estimated R, X, B values for each transmission line are presented in Table 5. The estimated transmission loss coefficients using Equations (A.10-A.11) are detailed in the Table 6.
By using existing parameters, economic emission dispatch is carried out and the results are listed in Table 7. The dispatch results are obtained for three different load demands of 275 MW, 284.4 MW and 300 MW. With the new estimated parameters for fuel cost function, emission and loss, the multi-objective ED problem is solved and the results are compared with the existing results. From the results, the percentage deviation of fuel cost, emission and network loss for estimated parameters are slightly elevated as compared to the existing parameters. The % error of all these values reveals the impact of accurate parameter estimation for ED problems.
4.3. Test System 3: 62 Bus Indian Utility System
In this case, a 62 bus Indian utility system is considered for estimating the improved parameters by using the proposed algorithm. The chosen test system consists of nineteen power producers, eighty nine (220KV) lines
Table 5. Estimation of transmission line parameters for IEEE 30 bus system (in p.u).
nl: start bus of a line; nr: end bus of a line; Re, Xe and Be: Errors in TLP.
Table 6. Estimated B-Coefficients for IEEE 30 bus system.
Table 7. Economic emission dispatch with improved parameters for IEEE 30 unit system.
with eleven tap changing transformers. The bus, line and generator data of the test system are taken from [32] .
The proposed method of determining the improved parameters is applied for the chosen test system and the obtained results are listed in Table 8. For each unit the fuel costs using improved parameters for three different real power outputs are also specified. In comparison with base case values the error is calculated for each generating unit. The obtained error is small which reveals that the estimation of the coefficient is accurate.
Further the estimation of TLP is carried out using A.2-A.5. The obtained TLP values by TLBO algorithm are listed in Table 9. The % error values for each transmission line are close to zero, except for few lines, which shows that the obtained TLP values are accurate. With the improved TLP values, the transmission line efficiency is calculated and is listed in Table 9. It is also inferred from Table 9 that the efficiency of each transmission line obtained using TLBO is nearly the same as compared to the base case.
The PED is carried out with improved cost coefficients of generators and improved TLP for three different load demands of 2900 MW, 2967 MW, 3200 MW are obtained and the results are presented in Table 10. The estimated transmission loss coefficients using Equations (A.10-A.11) are presented in the Table 11. The results obtained are compared with the existing fuel cost and loss and the % deviation of fuel cost and transmission loss is computed and presented. The obtained results clearly illustrate the impact of improved parameters on ED problem.
5. Potential Verification of TLBO
5.1. Solution Quality
Using the proposed algorithm several runs have been carried out by varying the population size.In this analysis there is no larger difference in the average fitness value above the population size of 10. Hence the population size of 10 is preferred for all estimation process. The simulations are performed to estimate the quadratic cost function coefficients, piecewise quadratic valve point coefficients, emission coefficients, transmission line parameters and transmission loss coefficients using three different test cases. For these cases the existing coefficients at the time of installation are only available, hence the obtained improved parameters are compared with only these values. From the obtained simulation results, the cost coefficients, emission coefficients and transmission line parameters are very close to the existing values and are presented in Tables 2-6 and Tables 8-10. The comparisons of total error in each case clearly indicate that the proposed method provides the best estimate of accurate coefficients.
Table 8. Estimation of quadratic cost function coefficients using TLBO?62 Bus systems.
5.2. Statistical Analysis
The reliability of the algorithm in finding the best solution can be viewed by performing statistical analysis. The convergence property and the robustness characteristics are focused primly for the statistical analysis. The standard statistical parameters of the objective functions such as best, worst and average values in addition to standard deviation and solution iter are also worked out and analyzed and are given in Table 12. The standard deviation value is small which indicates the accuracy of estimate.
Convergence: The TLBO algorithm is implemented on different scale of test systems and for all cases, the convergence characteristics are plotted. The convergence pattern illustrates the searching ability of algorithm.
Figure 2 shows the convergence characteristics of 10 unit multiple fuels system; the proposed approach converged to a good solution within the 50 iterations for the three different fuels. For the IEEE 30 bus and 62 bus Indian utility systems, the convergence graphs are shown in Figure 3. Figure 2 and Figure 3 show that the TLBO has good convergence property, thus resulting in good evaluation value and low error.
Robustness: Many trials with different initial population have been carried out to test the consistency of the TLBO algorithm. The algorithm is executed for 50 trials and the obtained objective function values are presented in Figure 4. The mean value is close to the optimal value for all the cases. This description clears that the TLBO provides great searching ability, higher solution quality and the best estimate.
5.3. Observations
・ Accurate parameters are estimated for three different test systems involving a practical test system.
・ For the first time in the literature, both the parameter estimation and economic dispatch are carried out
Table 9. Enhanced parameters for Indian Utility 62 Bus system (in p.u).
Figure 2. Cost coefficients convergence curve (a) fuel 1; (b) fuel 2 and (c) fuel 3.
Figure 3. Convergence curve: (a) IEEE 30 bus cost coefficients; (b) IEEE 30 bus emission coefficients; (c) 62 bus cost coefficients.
Table 10. Economic dispatch with improved parameters for Indian utility 62 bus system.
concurrently.
・ The % error estimation of parameters reveals the rate of change in parameters over the years due to physical conditions like aging.
・ The results clearly signify the impact of accurate parameters on ED problems on fetching the accurate dispatch results over existing results.
6. Conclusion
In this paper, the TLBO algorithm is successfully implemented for both the operation of parameter estimations and ED solution. The operational aspects of the generators such as valve-point loading, emission, multiple fuel options and transmission loss are considered. The three different test systems involving a practical test system are selected for this estimation and dispatch problem. The results obtained by satisfying all the constraints for all cases by the proposed algorithm are always comparable or better than the other methods. It is revealed that the
Table 11.Estimated B-Coefficients for 19 unti Indian utility system.
Figure 4. Robustness characteristics: (a) 10 unit system for fuel 1; (b) 10 unit system for fuel 2; (c) 10 unit system for fuel 3; (d) IEEE 30 bus system for cost function; (e) IEEE 30 bus system for emission function; (f) 62 bus system for cost function.
Table 12. Performance indices by TLBO for various test systems.
TLBO possesses better convergence characteristics and robustness. The proposed TLBO approach has shown merits such as better results, easy implementation for the accurate parameter estimation for PED problems. It can be concluded that the estimated parameters obtained show that the proposed method can be used as a very accurate tool for estimating the fuel cost, emission and loss coefficients.
Acknowledgements
The authors gratefully acknowledge the authorities of Annamalai University, Annamalai Nagar, Tamilnadu, India for the facilities provided to carry out this research work.
Cite this paper
SelvarajDurai,SrikrishnaSubramanian,SivarajanGanesan, (2015) Preferred Economic Dispatch of Thermal Power Units. Journal of Power and Energy Engineering,03,47-69. doi: 10.4236/jpee.2015.311005
References
1. 1. El-Hawary, M.E. and Mansour, S.Y. (1982) Performance Evaluation of Parameter Estimation Algorithms for Economic Operation of Power Systems. IEEE Transaction on Power Apparatus Systems, 101, 574-582.
http://dx.doi.org/10.1109/TPAS.1982.317270
2. 2. Soliman, S.A., Emam, S.E.A. and Christensen, G.S. (1991) Optimization of the Optimal Coefficients of Non-Monotonically Increasing Incremental Cost Curves. Electric Power System Research, 21, 99-106.
http://dx.doi.org/10.1016/0378-7796(91)90023-G
3. 3. Liang, Z.X. and Glover, J.D. (1991) Improved Cost Functions for Economic Dispatch Computations. IEEE Transactions of Power System, 6, 821-829.
http://dx.doi.org/10.1109/59.76731
4. 4. Chen, H.Y.K. and Postel, C.E. (1986) On-Line Parameters Identification of Input Output Curves for Thermal Units. IEEE Transaction Power System, 1, 221-224.
http://dx.doi.org/10.1109/TPWRS.1986.4334933
5. 5. Grainger, J. and Stevenson, W. (1994) Power System Analysis. McGraw-Hill., New York.
6. 6. Chan, S.M. (1993) Computing Overhead Line Parameters. Computer Applications and Power, 6, 43-45.
http://dx.doi.org/10.1109/67.180436
7. 7. Dommel, H.W. (1985) Overhead Line Parameters from Handbook Formulas and Computer Programs. IEEE Transactions on Power Apparatus and Systems, 4, 366-372.
http://dx.doi.org/10.1109/TPAS.1985.319051
8. 8. Thorp, J.S., Phadke, A.G., Horowitz, S.H. and Begovic, M.M. (1988) Some Applications of Phasor Measurements To Adaptive Protection. IEEE Transaction Power System, 3, 791-798.
http://dx.doi.org/10.1109/59.192936
9. 9. Chen, C.S., Liu, C.W. and Jiang, J.A. (2002) A New Adaptive PMU Based Protection Scheme for Transposed/ Untransposed Parallel Transmission Lines. IEEE Transactions on Power Delivery, 17, 395-404.
http://dx.doi.org/10.1109/61.997906
10. 10. Kim, I.-D., and Aggarwal, R.K. (2006) A Study on the On-Line Measurement of Transmission Line Impedances for Improved Relaying Protection. Electric Power and Energy System, 28, 359-366.
http://dx.doi.org/10.1016/j.ijepes.2006.01.002
11. 11. Liao, Y. and Kezunovic, M. (2009) Online Optimal Transmission Line Parameter Estimation for Relaying Applications. IEEE Transactions on Power Delivery, 24, 96-102.
http://dx.doi.org/10.1109/TPWRD.2008.2002875
12. 12. Sivanagaraju, G., Chakrabarti, S. and Srivastava, S.C. (2014) Uncertainty in Transmission Line Parameters Estimation and Impact on Line Current Differential Protection. IEEE Transactions on Instrumentation and Measurement, 63, 1496-1504.
http://dx.doi.org/10.1109/TIM.2013.2292276
13. 13. Wagenaars, P., Wouters, P.A.A.F., van der Wielen, P.C.J.M. and Steennis, E.F. (2010) Measurement of Transmission Line Parameters of Three-Core Power Cables with Common Earth Screen. IET Science Measurement and Technology, 4, 146-155.
http://dx.doi.org/10.1049/iet-smt.2009.0062
14. 14. Indulkar, C.S. and Ramalingam, K. (2008) Estimation of Transmission Line Parameters from Measurements. Electric Power and Energy Systems, 30, 337-342.
http://dx.doi.org/10.1016/j.ijepes.2007.08.003
15. 15. Al-Kandari, A.M. and El-Naggar, K.M. (2006) A Genetic-Based Algorithm for Optimal Estimation of Input-Output Curve Parameters of Thermal Power Plants. Electrical Engineering, 89, 585-590.
http://dx.doi.org/10.1007/s00202-006-0047-x
16. 16. El-Naggar, K.M. and Alrashidi, M.R. and Al-Othman, A.K. (2009) Estimating the Input-Output Parameters of Thermal Power Plants Using PSO. Energy Conversion and Management, 50, 1767-1772.
http://dx.doi.org/10.1016/j.enconman.2009.03.019
17. 17. Alrashidi, M.R., El-Naggar, K.M. and Al-Othman, A.K. (2009) Particle Swarm Optimization Based Approach for Estimating the Fuel-Cost Function Parameters of Thermal Power Plants with Valve Loading Effects. Electric Power Components and Systems, 37, 1219-1230. http://dx.doi.org/10.1080/15325000902993589.
18. 18. Sonmez, Y. (2013) Estimation of Fuel Cost Curve Parameters for Thermal Power Plants Using the ABC Algorithm. Turkish Journal of Electrical Engineering and Computer Science, 21, 1827-1841.
http://dx.doi.org/10.3906/elk-1203-10
19. 19. Sinha, A.K. and Mandal, J.K. (1999) Hierarchical Dynamic State Estimator Using ANN-Based Dynamic Load Prediction. IET Generation Transmission and Distribution, 146, 541-549. http://dx.doi.org/10.1049/ip-gtd:19990462
20. 20. Kumar, D.M.V. and Srivastava, S.C. (1999) Power System State Forecasting Using Artificial Neural Networks. Electric Machine and Power System, 27, 653-664. http://dx.doi.org/10.1080/073135699269091
21. 21. Lin, J.M., Huang, S.J. and Shih, K.R. (2003) Application of Sliding Surface Enhanced Fuzzy Control for Dynamic State Estimation of a Power System. IEEE Transaction Power System, 18, 570-577.
http://dx.doi.org/10.1109/TPWRS.2003.810894
22. 22. Bhuvaneswari, R., Subramanian, S. and Madhu, A. (2008) A Novel State Estimation Based on Minimum Errors between Measurements Using Ant Colony Optimization Technique. International Journal of Electric Engineering, 15, 457-568.
23. 23. Sakthivel, V.P., Bhuvaneswari, R. and Subramanian, S. (2010) Design Optimization of Three-Phase Energy efficient Induction Motor Using Adaptive Bacterial Foraging Algorithm. International Journal of Computation and Mathematics in Electrical and Electronics Engineering, 29, 699-726.
24. 24. Sakthivel, V.P., Bhuvaneswari, R. and Subramanian, S. (2010) Non-Intrusive Efficiency Estimation Method for Energy Auditing and Management of In-Service Induction Motor Using Bacterial Foraging Algorithm. IET Electric Power Applications, 4, 579-590.
http://dx.doi.org/10.1049/iet-epa.2009.0313
25. 25. Sakthivel, V.P., Bhuvaneswari, R. and Subramanian, S. (2011) An Accurate and Economical Approach for induction motor Field Efficiency Estimation Using Bacterial Foraging Algorithm. Measurement, 44, 674-684.
http://dx.doi.org/10.1016/j.measurement.2010.12.008
26. 26. Rao, R.V., Savsani, V.J. and Vakharia, D.P. (2011) Teaching-Learning-Based Optimization: A Novel Method for Mechanical Design Optimization Problems. Computer Aided Design, 43, 303-315.
27. 27. Rao, R.V. and Waghmare, G.G. (2014) Complex Constrained Design Optimisation Using an Elitist Teaching-Learning-Based Optimisation. International Journal of Metaheuristics, 3, 81-102.
http://dx.doi.org/10.1504/IJMHEUR.2014.058863
28. 28. Rao, R.V., Savsani, V.J. and Vakharia, D.P. (2012) Teaching-Learning-Based Optimization: An Optimization Method for Continuous Non-Linear Large Scale Problems. Information Science, 183, 1-15.
http://dx.doi.org/10.1016/j.ins.2011.08.006
29. 29. Rao, R.V., Savsani, V.J. and Vakharia, D.P. (2011) Teaching-Learning-Based Optimization: A Novel Method for Mechanical Design Optimization Problems. Computer-Aided Design, 43, 303-315.
30. 30. Durai, S., Subramanian, S. and Ganesan, S. (2015) Improved Parameters for Economic Dispatch Problems by Teaching Learning Optimization. Electric Power and Energy System, 67, 11-24.
http://dx.doi.org/10.1016/j.ijepes.2014.11.010
31. 31. Chiang, C.-L. (2005) Improved Genetic Algorithm for Power Economic Dispatch of Units with Valve-Point Effects and Multiple Fuels. IEEE Transaction Power System, 20, 1690-1699.
http://dx.doi.org/10.1109/TPWRS.2005.857924
32. 32. Tamilnadu Electricity Board Statistics at a Glance (1999-2000), Compiled by Planning Wing of Tamilnadu Electricity Board, Chennai, India.
Appendix
Parameter Estimation Problems
The vector equation describing the relationships between the measured values y, the unknown parameters m, the system matrix D and the residual due to the change in values r are as follows,
(A.1)
The parameter estimation problem for estimating the cost coefficients a, b, c, e, f, emission coefficients e0, e1, e2 and long line transmission parameters R, X, B are formulated as follows:
Let, k-Number of measurements, y(k)-Measured value of the kth measurement, D-System matrix
r-Error vector that relates y(k) to m.
There are k equations available to represent the parameter estimation that forms measurement matrix y(k).
TLBO is used as an estimator to find the unknown values of [m]. These estimates are used to recalculate parameters using consequent equations at each time step. The calculation procedure for the evaluation of parameters such as a, b, c, e, f; e0, e1, e2 and Bij, B0i, B00are detailed in Section 2.2, and the transmission line parameters such as R, X, B are presented in (A.2-A.5).
(A.2)
(A.3)
(A.4)
(A.5)
Next, using the Newton-Raphson (NR) method, the following four non-linear equations of the form F(x) = 0,
where and are solved for the unknown X, R, B,
(A.6)
(A.7)
(A.8)
(A.9)
Transmission Line Loss Parameters
The network loss is a function of transmission line parameters and network configurations. Thus accuracy in the transmission line model and B coefficients are necessary. The transmission line model consists of R, X and B parameters and to determine the existing network loss these parameters must be accurate. The estimation of transmission line parameters is mathematically formulated as an optimization problem (Indulkar and Ramalingam, 2008) and the accurate TLP can be determined by using an optimization technique.
(A.10)
are the estimated transmission line parameters
The Kron’s coefficients (,and) are dependent of network parameters and configuration.
(A.11)
―Estimated loss coefficients
The ideal transmission loss can be determined using by Equation (5). | 7,604 | 33,836 | {"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 | 0.872621 |
https://docsbay.net/doc/219480/guide-to-matlab-programs-for-erlangb-engset-bcq | 1,722,683,523,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640365107.3/warc/CC-MAIN-20240803091113-20240803121113-00519.warc.gz | 168,246,833 | 5,683 | ## Guide to Matlab programs for ErlangB, Engset, BCQ
Xuan Zheng and M. Veeraraghavan, March 30, 2004
Updated by Xiuduan Fang and Eric Humenay Nov 26, 2006
1. erlangb.m
The function [Pb, U]=erlangb(ρ, m)calculates the blocking probability and utilization for an M/M/m/m system with an infinite customer population. The arguments for erlangb.m are as follows:
1)ρ, offered traffic load (defined as the aggregate call arrival rate λdivided by the service rate μ); a positive number. The unit of ρ is Erlang.
2)m, the number of servers in the system; an positive integer.
The function erlangb returns two values:
1)Pb, call blocking probability.
2)U, utilization.
For example, entering the command [Pb,U]=erlangb(1,5), we get
Pb =0.0031 and U=0.1994, which means that the call blocking probability is 0.31% and the utilization is 19.94% whenρ=1erlang and m =5.
1. plot_Pb_util.m
The function plot_Pb_util(starting_rho,ending_rho,step_rho, m, testno) calls the function erlangb toplot call blocking probability and utilizationagainst load for a given value of m. It saves the calculated results into a txt file and the final figure into a Matlab figure. The arguments for plot_Pb_util are as follows:
1)starting_rho: lower bound for the load - can be > 1
2)ending_rho: upper bound for the load - can be > 1
3)step_rho: set step_rho a smaller value if you want finer granularity
4)m: the number of servers
5)testno: the id of the test; a number which is attached to the filename string (i.e., plot_Pb_util for this program) to create the file names for saving figures and data.For example, if the input testno is 2, then the output figure will be “plot_Pb_util 2.fig” and the output txt file will be “plot_Pb_util 2.txt”.
An example command to run this program is plot_Pb_util(1,20,1,20,1).
1. find_m_util.m
The function [m, U]=find_m_util.m(P_b, load) finds the m needed to meet a given Pb for a given load and calculates corresponding utilization too. The outputs are m, the number of servers needed and the utilization, U.
1. plot_m_util.m
The function plot_m_util(starting_rho, ending_rho, step_rho, Pb, testno) plot m and utilization against load for a given value of Pb. It saves the calculated results into a txt file and the final figure into a Matlab figure.
For example, plot_m_util(1, 100, 1, 0.01, 1).
1. bcc.m
The function [Pl, Pb]=bcc(load,server,customer) calculates the blocking probability for a finite number of sources queuing system using the Engset equation.
The arguments for bcc.m are as follows:
1)ρ, offered traffic load (defined as the per-customer call arrival rate λdivided by the service rate μ). The unit of ρ is Erlang.
2)m, the number of servers in the system.
3)n, the number of customers.
The function bcc returns two values:
4)Pl, call congestion probability.
5)Pb, time congestion probability.
For example, calling[Pl, Pb]=bcc(0.5,10, 30), we get call congestion probability Pl=0.2407 and time congestion probability Pb=0.2617 whenρ=0.5erlang,m =10, and n=30.
1. plot_bcc.m
The function plot_bcc(starting_rho, ending_rho, step_rho, server, customer, testno) calls the function [Pl, Pb]=bcc(load,server,customer) to plot call congestion probability, Pl and time congestion probability Pbagainst load for a given number of servers and a given number ofcustomers.
For example, plot_bcc(1/3, 10, 0.5, 3, 4, 1)
1. find_server.m
The function [server, realPl, realPb]=find_server(customer, load, Pl) finds the number of servers needed to meet a given call congestion probability, Pl, provided that we know the number of customer and the offer traffic load for a bcc queueuing system. The arguments for find_server.m are as follows:
1)customer: the number of customers, M
3)Pl: call congestion probability
The function find_server returns three values:
1)server: the number of servers, N
2)realPl: the actual call congestion probability
3)realPb: the actual time congestion probability
1. bcq.m
The function result=bcq(λ, μ, m, n)calculates the mean waiting time for an M/M/m/n system with a finite customer population. The arguments for bcq.m are as follows:
1)λ, the per-customer call arrival rate. The unit of λ is 1/s.
2)μ, the service rate. The unit of λ is 1/s.
3)m, the number of servers in the system.
4)n, the number of customers.
For example, a call bcq(0.5, 0.8, 10, 30) means the mean waiting time when λ=0.5, μ=0.8, m =10, and n=30.
1. plot_bcq.m
The function plot_bcq(lambda, starting_mu, ending_mu, step_mu, server, customer, testno) calls the function result=bcq(lambda,mu,server,customer) to plot call waiting time against call holding time for a given value of lambda, a given number of servers, and a given number of customers.
For example, plot_bcq(3, 1, 10, 1, 3, 10, 1)
1. demo.m
This file calls plot_Pb_util, plot_m_util, plot_bcc, and plot_bcq. | 1,346 | 4,847 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2024-33 | latest | en | 0.748541 |
https://www.rasch.org/rmt/rmt53d.htm | 1,695,399,813,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506420.84/warc/CC-MAIN-20230922134342-20230922164342-00517.warc.gz | 1,054,784,330 | 5,827 | # Log-Odds in Sherwood Forest
Once upon a time Robin Hood held an Archery competition. The targets were an oak, a larch and a pine. Robin Hood, Little John, Will Scarlet and Friar Tuck participated. Maid Marion kept score. First, they all shot 12 arrows at the Oak. Robin had 10 hits, John had 9, Will had 6, and Tuck had 4. Firing so many arrows took a long time, so, to speed things up, they shot at the other trees simultaneously. Robin started firing at the Larch and Tuck at the Pine. But the competition came to a sudden halt when the Sheriff of Nottingham was sighted, and everyone ran for cover. Will Scarlet asked Maid Marion for the final score. "Robin and Tuck both fired 7 more arrows. Robin hit the Larch 5 times, but Tuck only hit the Pine once". "How would I have fared on those other trees?" mused Will. Can we answer his question?
Thinking only about the Oak, we might say that Robin is 4 hits better than Will and 6 better than Tuck. Then, since Robin hit the Larch 5 times, we might expect Will to hit it 5-4 = 1 time. But then Tuck would hit it 5-6 = -1 times. This line of reasoning gives a nonsensical result because the worst Tuck could do is hit it 0 times.
Another approach could be based on proportions of success. Robin hit the oak 10 times to Will's 6 times, i.e. Will hits the oak 6/10 the times of Robin. Then, since Robin hits the Larch 5 times, we might expect Will to hit 5x(6/10) = 3 times. This sounds reasonable. But turn it around. Robin missed the Oak 12-10 = 2 times. Will missed the Oak 12-6 = 6 times. So Will misses the Oak 6/2 = 3 times more than Robin. Robin missed the Larch 7-5 = 2 times, so we might expect Will to miss it 2x3 = 6 times, and so only hit it 7-6 = 1 time. What a paradox! When we think of success, we expect Will to hit the Larch 3 times. When we think of failure, we expect Will to hit the Larch only once.
The accuracy of an archer is a combination of success and failure. At one extreme, there is all hits and no misses. At the other extreme, all misses and no hits. In the middle are half hits and half misses. 3 hits and 1 miss would seem as accurate as 6 hits and 2 misses.
What is twice as good as 6 hits and 2 misses? Reasonable answers are 6 hits and 1 miss, or 12 hits and 2 misses. We need to combine hits and misses in such a way that 3 hits and 1 miss give the same index of accuracy as 6 hits and 2 misses, but 6 hits and 1 miss or 12 hits and 2 misses are twice as good. The only simple solution is that 3/1 = 6/2 = 3. Then a performance twice as good is 6/1 or 12/2 = 6, which is twice 3. It follows that the useful index of accuracy is hits/misses, known as the "odds of success".
Now compare Robin and Will. On the Oak, Robin's odds of success are 10 hits/2 misses = 5, as shown in the Table. On the Larch, Robin's odds of success are 5 hits/2 misses = 2.5. So Robin's odds of success were halved from 5 to 2.5, implying that the Larch is twice as difficult to hit as the Oak.
Will's odds of success on the Oak are 6 hits/6 misses = 1. If Will's odds on the Larch are also halved then his odds become 0.5. So, if Will shot 12 arrows at the Larch, we would expect 4 hits and 8 misses.
What about Will and Tuck? Tuck's odd's of success on the Oak are 4 hits/8 misses = 1/2. His odds of success on the Pine are 1 hit/6 misses = 1/6, one third of his odds of success on the Oak. So the Pine must be 3 times more difficult to hit than the Oak.
Will's odds of success on the Oak were 6 hits/6 misses = 1, twice that of Tuck. So we expect Will's odds of success on the Pine to be 2x1/6 = 1/3. If Will shot 12 arrows at the Pine, we would expect 3 hits and 9 misses.
These odds of success are useful, but they are on a ratio scale. Their arithmetic is multiplicative, not additive. Robin is 5 times as accurate as Will, and Will is 2 times as accurate as Tuck. So Robin's accuracy compared with Tuck's is not 5+2 = 7, but 5x2 = 10 times. It's usually more convenient to think with numbers we can add and subtract, i.e. interval measures. These would be the logarithms of the odds, the log-odds. Then Robin's accuracy compared with Tuck's would be loge(5)+loge(2) = loge(10) in log-odds units (logits). The interval scale makes it clear that Little John is closer to Robin Hood than Friar Tuck is to Will Scarlet.
The additive Rasch model combines this information into one convenient formula:
Will's log-odds of success on the larch =
Will's log-odds of success on the Pine relative to Robin (his ability)
- Larch log-odds accuracy relative to Pine for Robin (its difficulty)
or, more generally,
The log-odds of success by an object on an agent =
the log-odds of success by the object on an agent at the origin of the scale (its ability)
- the log-odds of failure on the agent by an object at the origin of the scale (its difficulty)
Log-Odds in Sherwood Forest, J Linacre … Rasch Measurement Transactions, 1991, 5:3 p. 162-163
Rasch Publications
Rasch Measurement Transactions (free, online) Rasch Measurement research papers (free, online) Probabilistic Models for Some Intelligence and Attainment Tests, Georg Rasch Applying the Rasch Model 3rd. Ed., Bond & Fox Best Test Design, Wright & Stone
Rating Scale Analysis, Wright & Masters Introduction to Rasch Measurement, E. Smith & R. Smith Introduction to Many-Facet Rasch Measurement, Thomas Eckes Invariant Measurement: Using Rasch Models in the Social, Behavioral, and Health Sciences, George Engelhard, Jr. Statistical Analyses for Language Testers, Rita Green
Rasch Models: Foundations, Recent Developments, and Applications, Fischer & Molenaar Journal of Applied Measurement Rasch models for measurement, David Andrich Constructing Measures, Mark Wilson Rasch Analysis in the Human Sciences, Boone, Stave, Yale
in Spanish: Análisis de Rasch para todos, Agustín Tristán Mediciones, Posicionamientos y Diagnósticos Competitivos, Juan Ramón Oreja Rodríguez
Forum Rasch Measurement Forum to discuss any Rasch-related topic
Go to Top of Page
Go to index of all Rasch Measurement Transactions
AERA members: Join the Rasch Measurement SIG and receive the printed version of RMT
Some back issues of RMT are available as bound volumes
Subscribe to Journal of Applied Measurement
Go to Institute for Objective Measurement Home Page. The Rasch Measurement SIG (AERA) thanks the Institute for Objective Measurement for inviting the publication of Rasch Measurement Transactions on the Institute's website, www.rasch.org.
Coming Rasch-related Events
Aug. 11 - Sept. 8, 2023, Fri.-Fri. On-line workshop: Many-Facet Rasch Measurement (E. Smith, Facets), www.statistics.com
Aug. 29 - 30, 2023, Tue.-Wed. Pacific Rim Objective Measurement Society (PROMS), World Sports University, Macau, SAR, China https://thewsu.org/en/proms-2023
Oct. 6 - Nov. 3, 2023, Fri.-Fri. On-line workshop: Rasch Measurement - Core Topics (E. Smith, Facets), www.statistics.com
June 12 - 14, 2024, Wed.-Fri. 1st Scandinavian Applied Measurement Conference, Kristianstad University, Kristianstad, Sweden http://www.hkr.se/samc2024 | 1,886 | 7,040 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2023-40 | latest | en | 0.959453 |
http://www.carrawaydavie.com/how-many-once-in-a-gallon/ | 1,560,848,500,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998708.41/warc/CC-MAIN-20190618083336-20190618105336-00168.warc.gz | 213,834,228 | 8,750 | ## What is a how many once in a gallon
What is a how many once in a gallon as well as the US?
We will consider many other values needed in everyday life, such as how many once in a gallon, how many ounces in a gallon, gallons to an unce and many others.
There are 1000 milliliters in a liter (1000 mL in 1 liter).
Liters to fluid flows (fl oz) volume units conversion factors are listed below. To find out how much the liter is in the x liters, multiply the liter by the conversion factor.
1 Liter = 33.814022 US Fluid Ounces
1 Liter = 35.195079 Imperial Fluid Ounces
So, how many once in a gallon!
## What is a liquid ounce (fl oz) how many once in a gallon
A liquid ounce is an imperial as well as US customary measurement system volume unit. 1 US fluid ounce equals to 29.5735 mL and 1 royal (UK) fluid ounce equals to 28.4131 mL. So, how many once in a gallon?
There are 33.8140226 ounces in a liter because one liter (liter) is defined as the volume of 1 kilogram of water and there are 33.814 fluid ounces in the same volume of water.
There are 35.195079 imperial (UK) fl. ounces in a liter, because 1 fl. an ounce is 1.04084 imperial fl. ounce, and that makes 1.04084 * 33.814 (US fl oz in a liter)= 35.195079 imperial fl. oz in a liter. how many once in a gallon
## We need to know how many once in a gallon
1 gal is equal to 128 oz
Convert fluid formula to this formula:
pints = fluid ounces × 0.0625
1 Gallon (US, Fluid) = 128 Ounces (US, Fluid)
1 Gallon (UK, Fluid) = 160 Ounces (UK, Fluid)
how many once in a gallon
In Summary :.
1 gallon = 4 quarts = 8 pints = 16 mugs = 128 fluid ounces.
how many once in a gallon | 472 | 1,633 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2019-26 | latest | en | 0.913182 |
https://www.beddingandbeyond.club/products/3d-brown-imitation-brick-theme-photo-wallpaper-custom-mural | 1,544,392,090,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823183.3/warc/CC-MAIN-20181209210843-20181209232843-00391.warc.gz | 915,298,549 | 21,712 | ## 3D Brown Imitation Stone Theme Photo Wallpaper Custom Mural
• \$39.95
Product Details
• Features: Waterproof, Smoke-Proof, Moisture-Proof, Fireproof, Mold Proof, Soundproof, Sound-Absorbing, Heat Insulation, Anti-static, Mildew Resistant, Extra Thick, Environment friendly.
• Pattern: Imitation Brown Stone
• Type: Non-woven, Wood Fiber Embossed
• Style: Modern
Materials: Straw texture (default material) / 3D Relief / Imitation leather / Waterproof Oil Canvas / Waterproof Silk Cloth
Important Please Read! \$39.95 is for 1 piece. only. It is very small, about 4'7" x 2'4". To cover a complete wall you will need more pieces. The example pictures show a wall that is fully covered with wallpaper. You need to measure your wall and calculate how many pieces you need to order. When you know, go to checkout and select the number you wish to purchase. Total will be calculated for you.
How to calculate how many pieces you need:
1). Measure your wall. Convert your measurements to centimeters (you can use any online metric converter). 2). If your wall is 360cm width x 240cm height for example, multiply those numbers: 360 x 240 = 8.6. So you would need to buy 9 pieces.
Here are some fixed sizes:
Quantity 1 : 1 square meter = 140cm(W) x 70cm(H) (4'7" x 2'4")
Quantity 2 : 2 square meter = 200cm(W) x 100cm(H) (6'7" x 3'3")
Quantity 3 : 3 square meter = 220cm(W) x 140cm(H) (7'3" x 4'7")
Quantity 4 : 4 square meter = 250cm(W) x 160cm(H) (8'2" x 5'3")
Quantity 5 : 5 square meter = 280cm(W) x 180cm(H) (9'2" x 5'11")
Quantity 6 : 6 square meter = 300cm(W) x 200cm(H) (9'10" x 6'7")
Quantity 7 : 7 square meter = 330cm(W) x 210cm(H) (10'10" x 6'11")
Quantity 8 : 8 square meter = 360cm(W) x 230cm(H) (11'10" x 7'6")
Quantity 9 : 9 square meter = 380cm(W) x 240cm(H) (12'5" x 7'10")
Quantity 10 : 10 square meter = 400cm(W) x 250cm(H) (13'1" x 8'2")
Quantity 11 : 11 square meter = 420cm(W) x 260cm(H) (13'9'' x 8'6'')
Quantity 12 : 12 square meter = 440cm(W) x 270cm(H) (14'5" x 8'10")
Quantity 13 : 13 square meter = 460cm(W) x 280cm(H) (15'1'' x 9'2'')
Quantity 14 : 14 square meter = 480cm(W) x 290cm(H) (15'9'' x 9'6'')
Quantity 15 : 15 square meter= 500cm(W) x 300cm(H) (16'5'' x 9'10'')
Quantity 16 : 16 square meter=500cm(W) x 320cm(H) (16'5" x 10'6")
We Also Recommend | 858 | 2,330 | {"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-2018-51 | latest | en | 0.496787 |
https://www.biketinker.com/tag/graph/ | 1,720,894,872,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514512.50/warc/CC-MAIN-20240713181918-20240713211918-00419.warc.gz | 612,635,422 | 13,046 | ## Optimal Tire Pressure for bicycles
EDIT 2022:
Just use Jan’s calculator:
Run it for your bikes and write the pressure on the tires or the rims.
Ignore all this old stuff:
Here’s a short overview of tire pressure on the Problem Solver’s blog. It’s a little more concise than this page, and has more pictures.
## The Calculator
The pressure calculator was created by Dave Adams, who sent it to me because of a tire pressure post I’d put up with an extended chart. You can open the spreadsheet in Open Office, Excel, or Google Docs and use it for free. Just remember “Dave Adams” when you use it.
## How to Use It
Fill in the yellow fields in the spreadsheet (tire width; bike and rider weight; percent of weight on each tire). Go pump your tires. You can make (copy) a tab for each of your bikes, and make the tires and rider fatter or thinner over time (or adjust for touring loads).
## The Science
According to Frank Berto and Jan Heine (of Bicycle Quarterly magazine), two top bicycle science guys, the most efficient bicycle tire pressure is one that gives you a 15% drop in tire height when you get on the bike. “This tire has too little air!” It’s squishy and hard to turn. “This tire has too MUCH air!” It loses energy bouncing off small surface irregularities. “This tire is JUST RIGHT!” Fifteen percent is the Mama Bear of tire drop.
Given that you want each tire to ‘drop’ 15%, and bikes don’t weight the front as much as the rear, you don’ t want the same amount of air in each tire. It seems obvious when you think about it, but it was revelation to me.
Most bikes put 60% of the weight on the rear, for a 40-60 fore-aft weight distribution. The “Quickbeam” tab in the spreadsheet is set up this way, and I actually weighed the bike with me on it to get the split. I could have saved some work and trusted to Bicycle Quarterly, but I like to check things for myself.
Low-trail French Randonneuring bikes are different, with only 55% on the rear. The “Ross” tab in the spreadsheet is for my low-trail Ross Super Grand Tour fixed gear tourer. Also weighed out accurately to confirm the BQ numbers.
Dave says his equation looks like this:
PSI = 153.6 * Weight / (TireSize^1.5785) – 7.1685
## History
A few years ago, in the Spring of 2007, Bicycle Quarterly had a “Tire Drop” article based on Frank Berto’s research about proper inflation for best efficiency, in which they published a very useful graph, and instructions on how to set up your tires.
It didn’t have a line for 35mm tires, which I used at the time, so I added another line, and extended it to allow for heavy loads. Jan Heine gave me permission to republish the Bicycle Quarterly graph with my additions, and it turned out to be pretty useful for some people. I actually had an internet friend send it to me, not knowing I’d posted it in the first place.
Dave Adams saw the post in his research on the same subject, and sent me a copy of his spreadsheet, which I’ve been using ever since. With his permission, I posted it to the RBW (Rivendell Bicycle Works) Google Group, where it’s also archived. | 737 | 3,094 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2024-30 | latest | en | 0.937771 |
http://www.mathworks.com/help/physmod/sdl/ref/discbrake.html?requestedDomain=true&nocookie=true | 1,519,502,837,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891815934.81/warc/CC-MAIN-20180224191934-20180224211934-00546.warc.gz | 498,472,588 | 14,467 | # Disc Brake
Frictional brake with pressure-applying cylinder and pads
• Library:
• Simscape / Driveline / Brakes & Detents / Rotational
## Description
The Disc Brake block represents a brake arranged as a cylinder applying pressure to one or more pads that can contact the shaft rotor. Pressure from the cylinder causes the pads to exert friction torque on the shaft. The friction torque resists shaft rotation.
### Disc Brake Model
This figure shows the side and front views of a disc brake.
A disc brake converts brake cylinder pressure from the brake cylinder into force. The disc brake applies the force at the brake pad mean radius.
The equation that the block uses to calculate brake torque, depends on the wheel speed, Ω, such that when $\Omega \ne 0$,
`$T=\frac{{\mu }_{k}P\pi {D}_{b}{}^{2}{R}_{m}N}{4}.$`
However when $\Omega =0,$ the torque applied by the brake is equal to the torque that is applied externally for wheel rotation. The maximum value of the torque that the brake can apply when $\Omega =0,$ is
`$T=\frac{{\mu }_{s}P\pi {D}_{b}{}^{2}{R}_{m}N}{4}.$`
In any case, $Rm=\frac{Ro+Ri}{2}$.
In the equations:
• T is the brake torque.
• P is the applied brake pressure.
• Ω is the wheel speed.
• N is the number of brake pads in disc brake assembly.
• μs is the disc pad-rotor coefficient of static friction.
• μk is the disc pad-rotor coefficient of kinetic friction.
• Db is the brake actuator bore diameter.
• Rm is the mean radius of brake pad force application on brake rotor.
### Thermal Model
You can model the effects of heat flow and temperature change by selecting a thermal block variant. Selecting a thermal variant:
• Exposes port H, a conserving port in the thermal domain.
• Enables the Thermal mass parameter, which allows you to specify the ability of the component to resist changes in temperature.
• Enables the Temperature variable, which allows you to set the priority and initial target values for the block variables before simulating. For more information, see Set Priority and Initial Target for Block Variables (Simscape).
To select a thermal variant, right-click the block in your model and, from the context menu, select Simscape > Block choices. Select a variant that includes a thermal port.
## Ports
### Input
expand all
Physical signal port associated with cylinder pressure.
### Conserving
expand all
Rotational mechanical conserving port associated with the shaft.
The thermal conserving port is optional and is hidden by default. To expose the port, select a variant that includes a thermal port.
#### Dependencies
Selecting a thermal variant enables thermal parameters.
## Parameters
expand all
#### Geometry
Diameter of the piston.
#### Friction
Independent vector for determining static and kinetic frictions. The values in the vector, in K, must increase from left to right.
#### Dependencies
Selecting a thermal block variant for the Block choice parameter enables this parameter.
Coefficient of static friction. The value that you specify for this parameter must be greater than the value that you specify for the Coulomb friction coefficient parameter.
#### Dependencies
If you select a thermal block variant for the Block choice parameter, specify this parameter using a vector. The vector must contain the same number of elements as the temperature vector. Each value must be greater than the value of the corresponding element in the kinetic friction coefficient vector.
Coefficient of kinetic friction. The value that you specify for this parameter must be greater than `0`.
#### Dependencies
If you select a thermal block variant for the Block choice parameter, specify this parameter as a vector. The vector must contain the same number of elements as the temperature vector.
Angular speed at which friction switches from static to kinetic.
Coefficient of viscous friction.
#### Thermal Port
Thermal energy required to change the component temperature by a single degree. The greater the thermal mass, the more resistant the component is to temperature change.
#### Dependencies
Selecting a thermal block variant for the Block choice parameter enables this parameter. | 883 | 4,188 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 6, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2018-09 | latest | en | 0.800479 |
http://www.jiskha.com/display.cgi?id=1374040520 | 1,498,620,562,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128322320.8/warc/CC-MAIN-20170628032529-20170628052529-00030.warc.gz | 574,989,845 | 4,698 | # Chemistry
posted by .
A 10.0 mL solution of 0.300 M NH3 is titrated with a 0.100 M HCl solution. Calculate the pH after the addition of 20.0 mL of the HCl solution.
(potentially useful info: Ka of NH4+ = 5.6 x 10−10)
• Chemistry -
millimoles NH3 = 10 x 0.3 = 3
mmols HCl = 20 x 0.1 = 2
NH3 + HCl => NH4Cl
So you have 2 mmols NH4^+
and 1 mmol NH3.
pH = pKa + log (base)/(acid)
• Chemistry -
Thanks, I got the answer 8.95. So why do I not need to divide the mmols by the total volume?
• Chemistry -
The Henderson-Hasselbalch equation is
pH = pKa + log(base)/(acid)
With regard to the (base) and (acid), that means CONCENTRATION of base and CONCENTRATION of acid. Technically, then, one must divide millimoles/milliliter to get concn. But if you will put in the numbers (let's use made up numbers of 2 mmols for base and 3 mmols for acid and a volume of 10 mL. So this log ratio will come up with
(2/10)/(3/10) but notice that the 10 cancels and you have simply 2/3. Since it is the SAME volume in the solution for both acid and base you will note that the volume will ALWAYS cancel so you can simply use mmols base/mmols acid. When I taught this years ago I always counted off a few points because I told the students it was (concn/concn) not (mmoles/mmoles) and never mind that the answer was the same. So if you want to be technically correct you will use (mmols/mL)/(mmols/mL). If you want to do it quick and dirt you can simply use mmols because the answer will be the same. In fact, when I taught this I NEVER actually divided mmols/mL and plugged in that number; I always wrote mmols and mL, canceled the mL and went from there. Fast forward 25 years: I note some of the profs now use mmols directly and don't insist on the concn stage. I think it's ok to use that procedure as long as we understand what we are doing. If I were a student in a class these days I would ask the prof how s/he wanted it done on exams.
• Chemistry -
Thank you for the explanation! | 561 | 1,977 | {"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-2017-26 | latest | en | 0.929247 |
http://sandbaronfalseriver.com/epub/algebra-i-chapters-1-3 | 1,582,339,890,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145648.56/warc/CC-MAIN-20200222023815-20200222053815-00266.warc.gz | 127,110,970 | 8,411 | # Algebra I: Chapters 1-3 by N. Bourbaki
By N. Bourbaki
This softcover reprint of the 1974 English translation of the 1st 3 chapters of Bourbaki’s Algebre offers an intensive exposition of the basics of common, linear, and multilinear algebra. the 1st bankruptcy introduces the elemental gadgets, similar to teams and earrings. the second one bankruptcy reports the homes of modules and linear maps, and the 3rd bankruptcy discusses algebras, specially tensor algebras.
Best linear books
LAPACK95 users' guide
LAPACK95 is a Fortran ninety five interface to the Fortran seventy seven LAPACK library. it's suitable for a person who writes within the Fortran ninety five language and wishes trustworthy software program for easy numerical linear algebra. It improves upon the unique user-interface to the LAPACK package deal, profiting from the massive simplifications that Fortran ninety five permits.
Semi-Simple Lie Algebras and Their Representations (Dover Books on Mathematics)
Designed to acquaint scholars of particle physics already accustomed to SU(2) and SU(3) with innovations appropriate to all uncomplicated Lie algebras, this article is principally fitted to the research of grand unification theories. topics comprise basic roots and the Cartan matrix, the classical and unheard of Lie algebras, the Weyl workforce, and extra.
Lectures on Tensor Categories and Modular Functors
This booklet provides an exposition of the family one of the following 3 issues: monoidal tensor different types (such as a class of representations of a quantum group), three-dimensional topological quantum box idea, and 2-dimensional modular functors (which evidently come up in 2-dimensional conformal box theory).
Additional info for Algebra I: Chapters 1-3
Example text
Similarly for 3. INVERTIBLE ELEMENTS DEFINITION 6. Let E be a unital magma, T its law of composition, e its identity element and x and x' two elements of E. x' is called a left inverse (resp. right inverse, resp. inverse) ofx ifx' T x = e (resp. x T x' = e, resp. x' T x = x T x' =e). An element x ofE is called left invertible (resp. right invertible, resp. invertible) if it has a left inverse (resp. right inverse, resp. inverse). A monoid all of whose elements are invertible is called a group. 15 ALGEBRAIC STRUCTURES I Symmetric and symmetrizable are sometimes used instead of inverse and invertible.
Right cancellable) elements of an associative magma is a submagma. Ifyx and y 11 are injective so is YxTU &xnr = Yx o y 11 (Proposition 1). Similarly for 3. INVERTIBLE ELEMENTS DEFINITION 6. Let E be a unital magma, T its law of composition, e its identity element and x and x' two elements of E. x' is called a left inverse (resp. right inverse, resp. inverse) ofx ifx' T x = e (resp. x T x' = e, resp. x' T x = x T x' =e). An element x ofE is called left invertible (resp. right invertible, resp. invertible) if it has a left inverse (resp.
Ii) If x is invertible, there exists a unique homomorphism g ofZ into E such that g( 1) = x and g coincides with f on N . Writingf(n) = T x for all n eN, the formulae Tx = e and (-T x) T (;. x) = mTn x (no. 1) express the fact thatfis a homomorphism ofN into E and obviously f(l) = x. Ifj' is a homomorphism of N into E such thatf'(l) = x, then f = j', by§ 1, no. 4, Proposition 1, (iv). Suppose now that x is invertible. By no. 3, Corollary 2 to Proposition 4, . : 0. By construction, Z is the group of differences ofN and hence (no. | 913 | 3,463 | {"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-2020-10 | latest | en | 0.858876 |
https://www.mrexcel.com/board/threads/lookup-date-data-and-return-value-between-2-date-ranges.229479/ | 1,627,410,570,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153474.19/warc/CC-MAIN-20210727170836-20210727200836-00669.warc.gz | 943,783,071 | 17,867 | # Lookup date data and return value between 2 date ranges
#### dahlia94040
##### New Member
This is my first time posting a question in this forum. I'm desparate for help! I tried searching to see if anyone has asked the same question but I haven't been successful yet.
Sheet 1 contains a transaction table. I need to do a lookup pertaining to that customer & product for that transaction date and have it return a rebate value from sheet two which may not contain that actual date since it only has a start & end date.
Sheet 1 example:
Customer Product Trans.Date Rebate Earned
AAAAAA 111111 01/01/06 ?
BBBBBB 222222 04/15/06 ?
EEEEEE 333333 02/03/06 ?
AAAAAA 333333 01/01/06 ?
EEEEEE 333333 03/01/06 ?
Sheet 2 contains the following:
Customer Product Start Date End Date Rebate
AAAAAA 111111 12/23/05 01/15/06 0.10
AAAAAA 111111 01/16/05 10/26/06 0.05
AAAAAA 222222 01/02/06 12/31/20 0.02
AAAAAA 333333 01/01/05 01/01/06 0.01
AAAAAA 333333 01/02/06 12/31/20 0.03
BBBBBB 222222 04/05/06 04/11/06 0.15
BBBBBB 222222 04/12/06 04/30/06 0.13
EEEEEE 333333 03/04/05 02/15/06 0.55
As you can see from above the start & end dates are not constant. I need the formula to return these rebate values in the last column:
Customer Product Trans.Date Rebate Earned
AAAAAA 111111 01/01/06 0.10
BBBBBB 222222 04/15/06 0.13
EEEEEE 333333 02/03/06 0.55
AAAAAA 333333 01/01/06 0.01
EEEEEE 333333 03/01/06 #N/A
### Excel Facts
VLOOKUP to Left?
Use =VLOOKUP(A2,CHOOSE({1,2},\$Z\$1:\$Z\$99,\$Y\$1:\$Y\$99),2,False) to lookup Y values to left of Z values.
#### Krishnakumar
##### Well-known Member
Hi,
=INDEX(Sheet2!\$E\$2:\$E\$9,MATCH(1,IF(Sheet2!\$A\$2:\$A\$9=A2,IF(Sheet2!\$B\$2:\$B\$9=B2,IF(C2>=Sheet2!\$C\$2:\$C\$9,IF(C2<=Sheet2!\$D\$2:\$D\$9,1)))),0))
Confirmed with Ctrl+Shift+Enter.
HTH
#### dahlia94040
##### New Member
Thanks for your quick reply Kris. I tried it and I got #N/A.
#### Krishnakumar
##### Well-known Member
This is an array formula.
Hold down CTRL and SHIFT keys while you pressing ENTER key.
See help files on array formula.
HTH
#### dahlia94040
##### New Member
I thought I did that but I clicked on the formula and held down CTRL and SHIFT while pressing ENTER again and it worked! Thank you!
If I need to copy that formula down (32 thousand rows) do I have to do that each time.
#### dahlia94040
##### New Member
Never mind Kris. Next time I should try it before I ask dumb questions. Thank you soooo much again!
#### Krishnakumar
##### Well-known Member
IIf I need to copy that formula down (32 thousand rows) do I have to do that each time.
Using array formulas in 32k rows will affect the performance of the system. Consider VBA.
Replies
5
Views
35
Replies
5
Views
180
Replies
2
Views
94
Replies
11
Views
139
Replies
4
Views
65
1,136,512
Messages
5,676,289
Members
419,618
Latest member
Grisego
### We've detected that you are using an adblocker.
We have a great community of people providing Excel help here, but the hosting costs are enormous. You can help keep this site running by allowing ads on MrExcel.com.
### Which adblocker are you using?
1)Click on the icon in the browser’s toolbar.
2)Click on the icon in the browser’s toolbar.
2)Click on the "Pause on this site" option.
Go back
1)Click on the icon in the browser’s toolbar.
2)Click on the toggle to disable it for "mrexcel.com".
Go back
### Disable uBlock Origin
Follow these easy steps to disable uBlock Origin
1)Click on the icon in the browser’s toolbar.
2)Click on the "Power" button.
3)Click on the "Refresh" button.
Go back
### Disable uBlock
Follow these easy steps to disable uBlock
1)Click on the icon in the browser’s toolbar.
2)Click on the "Power" button.
3)Click on the "Refresh" button.
Go back | 1,183 | 3,728 | {"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} | 3.40625 | 3 | CC-MAIN-2021-31 | latest | en | 0.643616 |
https://www.unitsconverters.com/en/Meterpersquaredecisecond-To-Accelerationoffreefallonvenus/Unittounit-7542-3438 | 1,713,690,469,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817729.87/warc/CC-MAIN-20240421071342-20240421101342-00246.warc.gz | 940,047,647 | 44,428 | Formula Used
1 Meter per Square Second = 0.01 Meter per Square Decisecond
1 Meter per Square Second = 0.1128004660374 Acceleration of free fall on Venus
1 Meter per Square Decisecond = 11.28004660374 Acceleration of free fall on Venus
## Meter per Square Deciseconds to Acceleration of free fall on Venuss Conversion
m/ds² stands for meter per square deciseconds and g stands for acceleration of free fall on venuss. The formula used in meter per square deciseconds to acceleration of free fall on venuss conversion is 1 Meter per Square Decisecond = 11.28004660374 Acceleration of free fall on Venus. In other words, 1 meter per square decisecond is 12 times bigger than a acceleration of free fall on venus. To convert all types of measurement units, you can used this tool which is able to provide you conversions on a scale.
## Convert Meter per Square Decisecond to Acceleration of free fall on Venus
How to convert meter per square decisecond to acceleration of free fall on venus? In the acceleration measurement, first choose meter per square decisecond from the left dropdown and acceleration of free fall on venus from the right dropdown, enter the value you want to convert and click on 'convert'. Want a reverse calculation from acceleration of free fall on venus to meter per square decisecond? You can check our acceleration of free fall on venus to meter per square decisecond converter.
How to convert Meter per Square Decisecond to Acceleration of free fall on Venus?
The formula to convert Meter per Square Decisecond to Acceleration of free fall on Venus is 1 Meter per Square Decisecond = 11.28004660374 Acceleration of free fall on Venus. Meter per Square Decisecond is 11.28 times Bigger than Acceleration of free fall on Venus. Enter the value of Meter per Square Decisecond and hit Convert to get value in Acceleration of free fall on Venus. Check our Meter per Square Decisecond to Acceleration of free fall on Venus converter. Need a reverse calculation from Acceleration of free fall on Venus to Meter per Square Decisecond? You can check our Acceleration of free fall on Venus to Meter per Square Decisecond Converter.
How many Meter per Square Second is 1 Meter per Square Decisecond?
1 Meter per Square Decisecond is equal to 11.28 Meter per Square Second. 1 Meter per Square Decisecond is 11.28 times Bigger than 1 Meter per Square Second.
How many Kilometer per Square Second is 1 Meter per Square Decisecond?
1 Meter per Square Decisecond is equal to 11.28 Kilometer per Square Second. 1 Meter per Square Decisecond is 11.28 times Bigger than 1 Kilometer per Square Second.
How many Micrometer per Square Second is 1 Meter per Square Decisecond?
1 Meter per Square Decisecond is equal to 11.28 Micrometer per Square Second. 1 Meter per Square Decisecond is 11.28 times Bigger than 1 Micrometer per Square Second.
How many Mile per Square Second is 1 Meter per Square Decisecond?
1 Meter per Square Decisecond is equal to 11.28 Mile per Square Second. 1 Meter per Square Decisecond is 11.28 times Bigger than 1 Mile per Square Second.
## Meter per Square Deciseconds to Acceleration of free fall on Venuss Converter
Units of measurement use the International System of Units, better known as SI units, which provide a standard for measuring the physical properties of matter. Measurement like acceleration finds its use in a number of places right from education to industrial usage. Be it buying grocery or cooking, units play a vital role in our daily life; and hence their conversions. unitsconverters.com helps in the conversion of different units of measurement like m/ds² to g through multiplicative conversion factors. When you are converting acceleration, you need a Meter per Square Deciseconds to Acceleration of free fall on Venuss converter that is elaborate and still easy to use. Converting Meter per Square Decisecond to Acceleration of free fall on Venus is easy, for you only have to select the units first and the value you want to convert. If you encounter any issues to convert, this tool is the answer that gives you the exact conversion of units. You can also get the formula used in Meter per Square Decisecond to Acceleration of free fall on Venus conversion along with a table representing the entire conversion.
Let Others Know | 933 | 4,293 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-18 | latest | en | 0.714585 |
http://stackoverflow.com/questions/9350852/can-someone-explain-this-snippet-of-code-c/9350912 | 1,449,020,586,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398525032.0/warc/CC-MAIN-20151124205525-00320-ip-10-71-132-137.ec2.internal.warc.gz | 222,120,435 | 18,026 | # Can someone explain this snippet of code? - C++ [duplicate]
Possible Duplicate:
Using Recursion to raise a base to its exponent - C++
``````int raisingTo(int base, unsigned int exponent)
{
if (exponent == 0)
return 1;
else
return base * raisingTo(base, exponent - 1);
}
``````
I wrote this code for raising an exponent to its base value using values passed by value from the `main()`. This function uses recursion to do this. Can someone explain how it returns a value each time it calls itself? I need a detailed explanation of this code.
-
## marked as duplicate by Oliver Charlesworth, talonmies, Ben Voigt, Michael Krelin - hacker, BartFeb 19 '12 at 16:30
Did you try just doing it on paper? – Christian Jonassen Feb 19 '12 at 16:20
@Bart writing code that works but you don't understand is not uncommon for new programmers – Seth Carnegie Feb 19 '12 at 16:20
Does copying the code from a school assignment count as "wrote it"? – Alan Feb 19 '12 at 16:24
And the explanation of how it works is explicitly given there as well... – Bart Feb 19 '12 at 16:25
@MichaelKrelin-hacker No, but how would I have known he copied it? I don't automatically assume people are lying when they don't understand something they say they wrote (I've done it plenty of times). – Seth Carnegie Feb 19 '12 at 16:32
It is best illustrated by doing iterations manually (as suggested in the comments). Suppose we have `base = 2` and `exponent = 2`.
• During the first iteration the function returns `2 * (whatever function yields when called with the arguments 2 and (2 - 1), which is 1)`.
• The second iteration with the arguments 2 and 1 gets the result `2 * (whatever the next iteration with arguments 2 and 0 returns)`.
• The thrid iteration will also be the last one since the function is set to return 1 when exponent is 0.
Now we have the full chain 2 * 2 * 1, therefore the result of the calculation is 4.
-
Thank you so much Malcolm! – Ram Sidharth Feb 19 '12 at 16:43
@Malcolm.I understand it now. Thanks Malcolm. – Ram Sidharth Feb 19 '12 at 16:47
We use the equation: `x^n = x * x^(n-1)` which is true for all real numbers.
So that we use it to create recursive function. The bottom of recursion is when the exponent == 0.
For example `2^4 = 2 * 2^3`; `2^3 = 2 * 2^2`; `2^2 = 2 * 2^1`; `2^1 = 2 * 2^0` and `2^0 = 1`.
-
+1, you wrote the code the OP "wrote", so you get to the second level ;-) – Michael Krelin - hacker Feb 19 '12 at 16:28
@MichaelKrelin-hacker.Thank you for your most enlightening comment. If you are willing to answer my question, please do. Thank you. – Ram Sidharth Feb 19 '12 at 16:42 | 746 | 2,612 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2015-48 | latest | en | 0.896606 |
https://www.justanswer.com/math-homework/5g1gy-assume-population-heights-female-college-students.html | 1,606,727,576,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141211510.56/warc/CC-MAIN-20201130065516-20201130095516-00267.warc.gz | 717,223,883 | 50,145 | Math Homework
Related Math Homework Questions
1. Consider a population with u=74.5 and Q=3.04 . (A) Calculate
1. Consider a population with u=74.5 and Q=3.04 . (A) Calculate the z-score for x=75.7 from a sample of size 18. (B) Could this z-score be used in calculating probabilities using Table 3 in Appendix B… read more
Osbert12
Master's Degree
2,468 satisfied customers
"for David" Assume that the population of heights of female
"for David" Assume that the population of heights of female college students is approximately normally distributed with mean m of 66.24 inches and standard deviation s of 5.07 inches. A random sample … read more
David
Post-Doctoral Degree
Post-Doctoral Degree
2,310 satisfied customers
For Steve Assume that the population of heights of female
For Steve Assume that the population of heights of female college students is approximately normally distributed with mean of 64.64 inches and standard deviation of 6.02 inches. A random sample of… read more
Stevewh
Teacher
Bachelor's Degree
6,427 satisfied customers
For Steve Assume that the population of heights of female
For Steve Assume that the population of heights of female college students is approximately normally distributed with mean µ of 67.26 inches and standard deviation ơ of 5.96 inches. A random sample of… read more
Stevewh
Teacher
Bachelor's Degree
6,427 satisfied customers
Assume that the population of heights of female college students
Assume that the population of heights of female college students is approximately normally distributed with mean µ of 66.24 inches and standard deviation s of 5.07 inches. A random sample of 72 height… read more
SusanAthena
Master's Degree
Master's Degree
41 satisfied customers
Assume that the population of heights of female college students
Assume that the population of heights of female college students is approximately normally distributed with mean m of 64.58 inches and standard deviation s of 3.82 inches. A random sample of 89 height… read more
Chirag
Master's Degree
10,329 satisfied customers
1. Assume that the population of heights of female college
1. Assume that the population of heights of female college students is approxiamtely normally distributed with mean µ of 68.66 inches and standard deviation ? of 5.13 inches. A random sample of 67 hei… read more
Sandhya
Master's Degree
5,928 satisfied customers
3. A sample of 93 golfers showed that their average score on
3. A sample of 93 golfers showed that their average score on a particular golf course was 89.33 with a standard deviation of 6.43. Answer each of the following (show all work and state the final answe… read more
Osbert12
Master's Degree
2,468 satisfied customers
Statistics Homework I desperately need help with these two
Statistics Homework I desperately need help with these two problems. I am completely lost. 4. Assume that the population of heights of female college students is approximately normally distributed wit… read more
Sandhya
Master's Degree
5,928 satisfied customers
1. Consider a population with 77.6 and 5.63. (A) Calculate
1. Consider a population with µ= 77.6 and s = 5.63. (A) Calculate the z-score for x ¯ = 76.1 from a sample of size 52. (B) Could this z-score be used in calculating probabilities using Table 3 in Appe… read more
Osbert12
Master's Degree
2,468 satisfied customers
Assume that the population of heights of male college students
Assume that the population of heights of male college students is approximately normally distributed with mean m of 68 inches and standard deviation s of 3.75 inches. A random sample of 16 heights is … read more
Ray Atkinson
Bachelor's Degree
1,687 satisfied customers
Assume that the population of heights of female college students
Assume that the population of heights of female college students is approximately normally distributed with mean m of 44 inches and standard deviation s of 4.00 inches. A random sample of 20 heights i… read more
mhasan420
Bachelor's Degree
1,768 satisfied customers
Assume that the population of heights of female college students
Assume that the population of heights of female college students is approximately normally distributed with mean ? of 69 inches and standard deviation ? of 4.75 inches. A random sample of 15 heights i… read more
Sandhya
Master's Degree
5,928 satisfied customers
Thanks Scott, One last question Assume that the population
Thanks Scott, One last question Assume that the population of heights of female college students is approximately normally distributed with a mean of 72 inches and standard deviation of 3.75 inches. A… read more
Scott
Master's Degree
17,585 satisfied customers
. Consider a binomial distribution with 15 identical trials,
. Consider a binomial distribution with 15 identical trials, and a probability of success of 0.4 i. Find the probability that x = 2 using the binomial tables ii. Use the normal approximation to find t… read more
abozer
Bachelor's Degree
6,158 satisfied customers
Assume that the population of heights of female college students
Assume that the population of heights of female college students is approximately normally distribution with mean u-,population mean- of 69 inches and standard deviation-sigma sign,population standard… read more
Chirag
Master's Degree
10,329 satisfied customers
3. The diameters of grapefruits in a certain orchard are normally
3. The diameters of grapefruits in a certain orchard are normally distributed with a mean of 6.85 inches and a standard deviation of 0.80 inches. Show all work. (A) What percentage of the oranges in t… read more
Stevewh
Teacher
Bachelor's Degree
6,427 satisfied customers
2.Determine whether each of the distributions given below represents
2.Determine whether each of the distributions given below represents a probability distribution. Justify your answer. (6 pts) a) x 1 2 3 4 P(x) 1/4 5/12 1/3 1/6 b) x 3 6 8 P(x) 0.1 3/5 0.3 c) x 20 30 … read more
Stevewh
Teacher
Bachelor's Degree
6,427 satisfied customers
Disclaimer: Information in questions, answers, and other posts on this site ("Posts") comes from individual users, not JustAnswer; JustAnswer is not responsible for Posts. Posts are for general information, are not intended to substitute for informed professional advice (medical, legal, veterinary, financial, etc.), or to establish a professional-client relationship. The site and services are provided "as is" with no warranty or representations by JustAnswer regarding the qualifications of Experts. To see what credentials have been verified by a third-party service, please click on the "Verified" symbol in some Experts' profiles. JustAnswer is not intended or designed for EMERGENCY questions which should be directed immediately by telephone or in-person to qualified professionals.
Ask-a-doc Web sites: If you've got a quick question, you can try to get an answer from sites that say they have various specialists on hand to give quick answers... Justanswer.com.
...leave nothing to chance.
Traffic on JustAnswer rose 14 percent...and had nearly 400,000 page views in 30 days...inquiries related to stress, high blood pressure, drinking and heart pain jumped 33 percent.
Tory Johnson, GMA Workplace Contributor, discusses work-from-home jobs, such as JustAnswer in which verified Experts answer people’s questions.
I will tell you that...the things you have to go through to be an Expert are quite rigorous.
## What Customers are Saying:
Wonderful service, prompt, efficient, and accurate. Couldn't have asked for more. I cannot thank you enough for your help.
Mary C.Freshfield, Liverpool, UK
This expert is wonderful. They truly know what they are talking about, and they actually care about you. They really helped put my nerves at ease. Thank you so much!!!!
AlexLos Angeles, CA
Thank you for all your help. It is nice to know that this service is here for people like myself, who need answers fast and are not sure who to consult.
GPHesperia, CA
I couldn't be more satisfied! This is the site I will always come to when I need a second opinion.
JustinKernersville, NC
Just let me say that this encounter has been entirely professional and most helpful. I liked that I could ask additional questions and get answered in a very short turn around.
EstherWoodstock, NY
Thank you so much for taking your time and knowledge to support my concerns. Not only did you answer my questions, you even took it a step further with replying with more pertinent information I needed to know.
RobinElkton, Maryland
He answered my question promptly and gave me accurate, detailed information. If all of your experts are half as good, you have a great thing going here.
DianeDallas, TX
< Previous | Next >
## Meet the Experts:
rajeevanpillai
Master's Degree
1,048 satisfied customers
M.Sc Statistics MPhil Statistics
Manal Elkhoshkhany
Bachelor's Degree
239 satisfied customers
I have completed my BA degree in 1988, and graduated with a GPA of 4.0.
Jabi
Tutor
155 satisfied customers
I've been an Expert on JustAnswer since July 2010
Mr. Gregory White
Master's Degree
113 satisfied customers
M.A., M.S. Education / Educational Administration
Dr Arthur Rubin
Doctoral Degree
76 satisfied customers
Ph.D. in Mathematics, California Institute of Technology
JACUSTOMER-yrynbdjl-
Master's Degree
49 satisfied customers
I have taught college Mathematics for the past 10 years.
SusanAthena
Master's Degree
41 satisfied customers
Tutor for Algebra, Geometry, Statistics. Explaining math in plain English.
< Previous | Next >
Disclaimer: Information in questions, answers, and other posts on this site ("Posts") comes from individual users, not JustAnswer; JustAnswer is not responsible for Posts. Posts are for general information, are not intended to substitute for informed professional advice (medical, legal, veterinary, financial, etc.), or to establish a professional-client relationship. The site and services are provided "as is" with no warranty or representations by JustAnswer regarding the qualifications of Experts. To see what credentials have been verified by a third-party service, please click on the "Verified" symbol in some Experts' profiles. JustAnswer is not intended or designed for EMERGENCY questions which should be directed immediately by telephone or in-person to qualified professionals.
Show MoreShow Less | 2,366 | 10,352 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2020-50 | latest | en | 0.89908 |
http://thawom.com/intro.html | 1,675,410,021,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500044.16/warc/CC-MAIN-20230203055519-20230203085519-00560.warc.gz | 45,032,395 | 4,494 | # Introduction
T
he problem with most mathematics textbooks is that they don’t teach much mathematics. They teach you how to apply some procedures and formulas, but that’s not what mathematicians do; that’s what computers do. In a world where you can look up any formula in seconds, and computers can be programmed to apply them, we don’t need people who have memorised how to solve a bunch of problems; we need people who can tackle problems that no-one has solved before, problems that require creativity and cunning. That’s what mathematics is about. That’s what this book is about.
Mathematical problems make for delightful puzzles which you might solve just for the fun of it, but there is also a practical side to mathematics. Why not practise mathematical problem solving by deriving all the important mathematical facts that you’d normally learn by rote, by answering questions asked throughout history by mathematicians trying to understand the heavens, help people through better technology, manage their family’s finances, or get to the fundamental truth and beauty of the universe?
This book leads you to discover all the mathematical rules and formulae normally taught in schools. Not only does this make you a better problem solver, but also satisfies your curiosity about why mathematical formulae work, and how people figured them out.
I’m not claiming that discovery is the most efficient way to learn how to apply a formula though. If I ask you to figure out how to find the area of a triangle, it isn’t because that will make you better at calculating areas of triangles; it’s to give you practice in mathematical problem solving. If you just want to know the formula you can skip over the discovery questions entirely. Any important conclusions will be written in a box so you can easily find them.
I’ve tried to give you as much choice as possible in what you learn and in what order. You can let your curiosity and individual needs guide you. You don’t have to read this book in the order it’s written. I’ve been frustrated by other textbooks that don’t let you do this – they don’t tell you which parts you need to read before other parts, and which parts you can skip. If you skip ahead in this book and come across a symbol you don’t recognise, you can look it up in Appendix B: Symbols.
You can also choose how hard the problems should be. I’ve made them very challenging, but you can use the hints to make them easier. Sometimes you might need to read the start of the answer to get some ideas, or just treat it as a worked example and read through the whole answer.
If you’re new to mathematical problem solving then the best advice I can give you is to give yourself permission to be wrong. In solving a problem, you’ll come up with ideas that don’t get you anywhere, and that’s perfectly normal. You just need to keep playing around. If you put pressure on yourself to get to the perfect solution as quickly as possible then it stifles creativity. Instead, treat it as a game and let your imagination run wild.
### Practice Questions
There are many facts and procedures that need to be memorised in mathematics – everything from times tables to how to differentiate trigonometric functions. Fortunately there has been a lot of research on the most effective and efficient ways to memorise information, and we now know that one of the best ways is to try to recall the information, check the answer, and then repeat that process some time later depending on how hard the fact was to remember.
Many students use ‘spaced repetition software’ to schedule revisions. I myself use it to learn foreign language vocabulary. You’re supposed to open the program every day and it will ask you the questions that you need to revise that day.
There is a strict distinction in this book between the regular questions and practice questions. The regular questions are usually quite complicated and require a lot of creative problem solving, so they have hints and full solutions. Practice questions on the other hand should be as simple as possible – ideally they should only require recalling a single fact.
You would normally look at a regular question until you understand it, then you don’t need to go over it again. But you need to return to practice questions over and over again, for many weeks, until you know them by heart. This process is made so much easier by spaced repetition software because it can figure out the ideal day to revise a fact so that you don’t waste your time by revising too early, or forget the fact completely by revising too late.
It’s a bit difficult to use spaced repetition software to practise mathematics, because you often need a slightly different variation of the question each time. For example, if you want to remember that $(a+b)^2 = a^2 + 2ab + b^2$, and you only practise with the question ‘Expand $(a+b)^2$’ then you probably won’t be able to expand $(x+y)^2$ very quickly. You really need to practise the procedure on different variables. Or to take a more basic example, if you only practise remembering $8\times 3$, you might be stumped when you come across $3\times 8$. It’s the same fact (that the numbers $8$ and $3$ in any order have a product of $24$), but you need to practise each variation of it.
At the moment, I don’t know of any spaced repetition software that allows you to make variations of a question without a custom plugin. Making different ‘cards’ for each variation isn’t a good idea because then you’d be practising the fact too often. Rather than getting people to download the software and the plugin and work out how to install it, I decided to make a web-based spaced repetition program.
If you log in to http://thawom.com/practice, it decides what questions to ask you that day, and cycles through the different variations. You add questions by clicking on the ‘Add these questions to my cards’ link whenever you come across a new fact on the website.
Of course you can do your revision any way you like; you don’t have to use spaced repetition software. But I recommend doing so, and practising for a few minutes every day. It’s a good idea to decide in advance when you’ll be practising, for example ‘while I’m waiting for the bus in the morning’, or ‘after dinner’. Or you can use some sort of alarm to remind you. | 1,315 | 6,325 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2023-06 | latest | en | 0.955137 |
http://qcsalon.net/en/yahtzee? | 1,547,903,269,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583667907.49/warc/CC-MAIN-20190119115530-20190119141530-00025.warc.gz | 187,555,527 | 4,444 | # Yahtzee
Yahtzee, sometimes Yatzee without the h, is certainly one of the most famous dice game. It is played with 5 dice and the goal is to make as many points as possible by judiciously filling your score sheet.
## Game rules
One play in turn each after the other. At your turn, after having rolled the 5 dice for the first time, you can decide to roll again one, more or even all of your dice to improve your hand. When you are satisfied, or when you have exhausted the three rolls you are allowed to do at maximum, you must choose a row on your score sheet to put your points in.
The points you get depends on the combination you managed to obtain with your dice, and the row you chose. In general, you get 0 points if you didn't success in making the required combination for your selected row. Sometimes, you are obliged to select a row knowing that you will mark 0 because you have no other choice, or simply because some of the combinations are very difficult to have and you wish to protect some other rows. It's especially with the Yahtzee category. Each row is only used once, you aren't allowed to come back to an already marked row. A game is therefore composed of exactly 13 turns, after which the 13 rows of the sheet are full.
### Score sheet and combinations
Your score sheet is composed of two main parts :
• The upper part contains the rows « Group 1 » to « Group 6 ». For these six rows, only the dice showing the corresponding group number are counting. For example, the hand 2 4 4 4 6 gives 12 point if it is put on group 4 because tehre are 3 dice showing the value 4, giving 4 points each. If it is put on group 2, it gives only 2 points because tehre is only a single die showing the value 2, and on group 1 it would be worth 0 points because tehre isn't any die showing number 1.
• The lower part contains different combinations mostly imspired from poker, that you must realise in order to effectively score points :
• Three of a kind: you obtain the sum of your 5 dice if at least three are identical. For example the hand 1 3 3 3 4 would count for 14 points.
• Four of a kind: with the same principle as above, you obtain the sum of your 5 dice if at least 4 are identical.
• Full, shorten for full house: at least 25 points are given for a combination containing three identical dice plus two identical dice of another value. For example 1 1 1 5 5. You can get more than 25 points by doing something like 6 6 6 5 5, but you will always at least get 25 points, even for 1 1 1 2 2, as long as you succeeded in making the combination correctly.
• Yahtzee: a forfait of 50 points is given if you succeed in obtaining 5 identical dice. Note that it is usually hard to get it, and, subsequently, this row is often sacrified.
• Small straight: a straight of 4 consecutive dice gives 30 points. There are only 3 possible small straight: 1 2 3 4, 2 3 4 5 and 3 4 5 6.
• Large straight: a straight of 5 consecutive dice gives 40 points. Only two large straights exist: 1 2 3 4 5 and 2 3 4 5 6. Be careful, it isn't so easy to do as it looks like.
• Chance: the sum of your 5 dice are counted without any restriction. This row is especially useful to score an unsatisfying hand which wouldn't go in another place.
A bonus of 35 points is given if the total score obtained in the upper part of the sheet (groups 1 to 6) is greater or equal than 63 points. These 63 points corresponds to 3+6+9+12+15+18, what is equivalent to say that to get the bonus, you must have got at least three dice of the correct value for each of the groups. An interesting strategy is to try to sacrify or compense weaknesses in smaller groups (1 or 2), which anyway never give many points, by better successes in greater groups (4, 5 and 6).
### Variants
#### Additionnal rows in score sheets
This variant adds three rows in the score sheets, making a game last for a total of 16 turns unstead of 13 :
• Pair: the sum of the 5 dice are counted as soon as there are two identical.
• Double pair: the sum of the 5 dice is given if there are two groups of two identical dice. For example 1 3 3 5 5 would give 17 points.
• Misere: miser gives 36 points minus the sum of the 5 dice without any condition. You can see it as being the opposite of the chance, because the more the dice are big in value, the less points it can give. Note that you can count a maximum of 31 points for misere, but only 30 for chance.
## Keyboard shortcut summary
• Enter: roll dice or marking score
• 1 to 6: select dice to roll again
• Shift + 1 to 6: unselect dice to roll again
• Space: repeat dice currently selected
• D: repeat latest roll obtained
• S: announce scores
• Shift+S: see full detailled score sheets
• T: announce who is playing | 1,207 | 4,731 | {"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-2019-04 | latest | en | 0.946697 |
https://origin.geeksforgeeks.org/category/geeksforgeeks-initiatives/truegeek/page/4/ | 1,638,242,024,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358903.73/warc/CC-MAIN-20211130015517-20211130045517-00531.warc.gz | 529,294,885 | 22,872 | Category Archives: TrueGeek
Basically, you did the last commit in your local repository and have not pushed back to the public repository or simply you want to change,… Read More
Azure is Microsoft’s cloud platform, just like Google has its Google Cloud and Amazon has its Amazon Web Service or AWS.000. Generally, it is a… Read More
Tombstone Diagrams or generally referred to as T- Diagrams are basically fragments used to denote compilers and language processing like some puzzle pieces. They demonstrate… Read More
In this article, we will discuss Java’s equivalent implementation of the upper_bound() method of C++. This method is provided with a key-value which is searched… Read More
The percentage of market share in the global desktop browser of Internet Explorer(IE) is less than 3%. If we are developing a site that includes… Read More
In this article, we will see the use of IN Operator with a SubQuery in SQL. IN operator is used to compare the column values… Read More
The term “combination” is thrown around loosely, and usually in the wrong way. Things like, “Hey, what’s the suitcase lock combination?” are said But what… Read More
Elements are chemically occurring natural substances that may be found in a free state or combined state. They are extracted from ores. ores are found… Read More
Probability is a mathematical branch that deals with calculating the likelihood of occurrence of a random event. Its value ranges between 0 (event will never… Read More
Probability is the likelihood of something happening. When we say the probability of something, it means how likely that something is. Some events have a… Read More
In this article, we will see, how to delete certain rows based on comparisons with other tables. We can perform the function by using a… Read More
Mensuration is a Greek word that means ‘measurement’. Mensuration is a branch of mathematics that involves the calculation of geometric figures like squares, rectangles, cones,… Read More
In context to the scaling of the MongoDB database, it has some features know as Replication and Sharding. Replication can be simply understood as the… Read More
Statistics is the study of the collection, analysis, interpretation, presentation, and organization of data. It is a method of collecting and summarising the data. This… Read More
Future technology basically deals with the blurring lines between humans and their advancement in technology. In the coming years, technology will dominate the workplace with… Read More | 514 | 2,508 | {"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} | 2.53125 | 3 | CC-MAIN-2021-49 | latest | en | 0.924224 |
https://electronics.stackexchange.com/questions/279497/slew-rate-limits | 1,723,205,778,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640763425.51/warc/CC-MAIN-20240809110814-20240809140814-00302.warc.gz | 179,837,297 | 42,360 | # Slew Rate Limits
I'm trying to calculate slew rate of OPAMP.
1. Applied square wave with 3V peak value to VIN+ of OPAMP.
2. Connected VIN- to VOUT as a feedback.
3. Graphic of VOUT voltage can be seen below.
1. Calculate Peak value of VOUT.
2. Find the time value of %90 (Time%90) and %10 (Time%10) of VOUT.
3. SR = (Vpeak*9/10-Vpeak*1/10)/(Time%90-Time%10)
I want to ask two basic question:
1. Is my slew rate calculation TRUE?
2. My OPAMP has the values of VDD +5V and VSS -5V. So should I apply square wave to OPAMP with 5V peak value? Or is it enough to apply square wave with 3V peak value.
• I don't think that's right... you'll be conflating GBW and SR. Just hit your opamp with a big (fast) step and measure how fast the output voltage rises... Volts/ uS. (They also list the value in the spec sheets... typically.) Commented Jan 10, 2017 at 16:16
• So how can I calculate Slew Rate? Commented Jan 10, 2017 at 16:18
• Do you have a 'scope? You can read the slew rate directly, as the rate of rise when you hit your opamp buffer with a large amplitude square wave. Commented Jan 10, 2017 at 16:21
• I'm using Pspice to simulate Commented Jan 10, 2017 at 16:22
• Well then plot the voltage vs. time... (just like a 'scope). I don't know much about opamp modeling in spice, but isn't the slew rate one of the input parameters Commented Jan 10, 2017 at 16:27 | 416 | 1,369 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-33 | latest | en | 0.900235 |
https://www.aqua-calc.com/calculate/mole-to-volume-and-weight/substance/methanol | 1,606,274,254,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141180636.17/warc/CC-MAIN-20201125012933-20201125042933-00217.warc.gz | 593,825,622 | 7,889 | # Moles of Methanol
## methanol: convert moles to volume and weight
### Volume of 1 mole of methanol
centimeter³ 40.49 milliliter 40.49 foot³ 0 oil barrel 0 Imperial gallon 0.01 US cup 0.17 inch³ 2.47 US fluid ounce 1.37 liter 0.04 US gallon 0.01 meter³ 4.05 × 10-5 US pint 0.09 metric cup 0.16 US quart 0.04 metric tablespoon 2.7 US tablespoon 2.74 metric teaspoon 8.1 US teaspoon 8.21
### Weight of 1 mole of methanol
carat 160.21 ounce 1.13 gram 32.04 pound 0.07 kilogram 0.03 tonne 3.2 × 10-5 milligram 32 042
### The entered amount of methanol in various units of amount of substance
centimole 100 micromole 1 000 000 decimole 10 millimole 1 000 gigamole 1 × 10-9 mole 1 kilogram-mole 0 nanomole 1 000 000 000 kilomole 0 picomole 1 000 000 000 000 megamole 1 × 10-6 pound-mole 0
#### Foods, Nutrients and Calories
COFFEE BROWNIE BLISS GREEK YOGURT WITH BISCOTTI COOKIES, MILK CHOCOLATE, and MOCHA BROWNIES, UPC: 818290014689 contain(s) 120 calories per 100 grams or ≈3.527 ounces [ price ]
#### Gravels, Substances and Oils
CaribSea, Freshwater, Instant Aquarium, Torpedo Beach weighs 1 505.74 kg/m³ (94.00028 lb/ft³) with specific gravity of 1.50574 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Dioxosilane [SiO2] weighs 2 320 kg/m³ (144.83287 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | mass and molar concentration | density ]
Volume to weightweight to volume and cost conversions for Refrigerant R-13, liquid (R13) with temperature in the range of -95.56°C (-140.008°F) to 4.45°C (40.01°F)
#### Weights and Measurements
The ounce per cubic decimeter density measurement unit is used to measure volume in cubic decimeters in order to estimate weight or mass in ounces
The linear density (μ) of a one-dimensional object, also known as linear mass density, is defined as the mass of the object per unit of length.
lb/US qt to sl/US c conversion table, lb/US qt to sl/US c unit converter or convert between all units of density measurement.
#### Calculators
Weight to Volume conversions for sands, gravels and substrates | 694 | 2,272 | {"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-2020-50 | longest | en | 0.630005 |
http://boomeria.org/grades/demos/greatdemos.html | 1,550,511,980,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247487595.4/warc/CC-MAIN-20190218155520-20190218181520-00124.warc.gz | 43,916,426 | 12,264 | PHYSICS DEMOS FROM
THE WOODROW WILSON PHYSICS INSTITUTE
compiled by Pat Cannan
submitted by Preston "The Boom" Boomer
Physics Institute
Woodrow Wilson National Fellowship Foundation
Box 642
Princeton, NJ 08542
Banana Drop:
When introducing acceleration of gravity, discuss it in terms of a falling banana (or rutabaga, or whatever). Demonstrate the fall and then compare to a heavy banana (filled with lead shot and rubber latex or aquarium sealant). Drop both bananas at once by quickly pulling a book out from under them.
Conclusion: All bananas accelerate at the same rate. This can then be quoted for the rest of the year to remind students of the demonstration.
And/or another variation:
Galileo's home country-- Italy. National fruit of Italy-- Grapes.
So all grapes fall at the same rate whether dropped individually or in a bunch. Show it. Bunching them makes no difference! Each atom accelerates at g regardless of its companions.
Ellipses:
Kepler's dad was a plumber. So take two plumber's helpers (preferably with short handles) and stick them onto the chalkboard for the foci of the ellipse. Then using a looped string and chalk, draw the ellipse around the two helpers.
How can an astronaut distinguish between a lead banana and one that is just a hollowed out (and reinforced) peel? By sensing their resistance to motion by shaking them. Pass bananas around so students can FEEL the mass.
Gravity discovered-- the real story.
One day, young Isaac Newton, then in his mid-twenties, was sitting under the banana tree in his back yard...
Centripetal hang-ups:
Bend a coat hanger and its hook so that a penny will balance on the upturned hook. Hold the hanger by your index finger and swing it in a circle. The penny will (with practice) remain in place.
Swingin' big scare:
Suspend a small (25cm diam) board from three strings so it can be vertically swung around. Place objects on the board and scare everyone! Practice this before trying beakers of water etc.
Car Parts:
Cars come equipped with a positive accelerator, a negative accelerator, and a change sign lever. When the lever is in the + position, the positive and negative accelerators work as designed.
When the lever is in the - position, the positive accelerator produces a negative acceleration, the negative accelerator a positive acceleration. (for those doubting students, what would have happened in the second case if the - accelerator had indeed accelerated the car negatively?)
TP Rip-off:
Single-ply toilet paper takes a force of about 10 newtons to separate. A rapid linear acceleration of the paper takes advantage of the rotational inertia of the roll to help stretch and tear the paper. The build-up to the breaking point must occur quickly so that angular velocity of the roll is kept small and paper is not dumped onto the floor. As the roll is used up, the moment of inertia decreases making it increasingly difficult to get paper off with one hand.
Place a new roll of TP and an almost empty roll on a bar held by two students. Give the new roll a yank, and the paper should tear nicely. Give the small roll a yank, and it should unravel onto the floor.
Discuss the moment of inertia. The new roll approximates a disk, the old roll a hoop.
Sound Thinking:
With a small transistor radio blaring away, enclose it in a cage of wire mesh. The Faraday cage will shield the radio from any electric fields and hence will shield it from radio waves. (The electric waves of light enter and leave the cage because their wavelengths are much smaller than the mesh size.)
When Chocolate Chip Speaks, Students Listen:
Take about 50 turns of fine, insulated wire and tape to the back of an ice cream carton (or whatever), leaving the two leads of the wire to attached to the output of an amplifier. Bring a large magnet up to the back of the voice coil when the amplifier is signaling appropriate music.
You may construct the speaker in class, discussing it in abstract terms so students are taken by surprise. If you do not have a carton, the daily bulletin or and administrative pronouncement will do.
Mixed Nuts:
Tie large steel hex nuts (of varying mass) on a string at spaces of S=1/2gt2 where t=.1, .2, .3, etc. Release the string onto a noisy flat plate and listen for a constant rat-a-tat-tat as th nut hit. First drop a string with evenly spaced nuts.
Instant Parabola:
Put tape on a meter stick at intervals similar to the Mixed Nuts. Mark off a parabola on the board by moving equal horizontal distances as you mark off vertical positions from the tape.
Challenge students to toss an object that matches the parabola, or drag a student in a wagon while she tosses an object straight up.
The Beat Goes On:
Drafting supply stores have stick-on tape with parallel bars. By Xeroxing this and shrinking it various amounts, waves of different length are obtained. Make transparencies and place them over each other to project beats or group velocity onto the screen.
By choosing l = fl (l= lambda) where f is a simple fraction, beat frequencies are harmonic with the generating frequencies and give a pleasing sound. (Thank you, Pythagoras).
Running Interference:
Concentric ring patterns may be purchases from drafting suppliers of about \$3 per sheet. Make different wave lengths by enlarging or reducing the pattern. With these made into overheads, you may demonstrate (1) 2-slit interference, (2) the effect of changing slit spacing or wavelength (3) n-slit interference (4) diffraction grating (5) effect of telescope aperture and incident wavelength on resolving power.
Golden Rule:
Measuring the wavelength of light with a diffraction grating demands and act of faith-- are there really all those lines on the grating? You can diminish those concerns by using a metal ruler with scored divisions of less than a millimeter or so (\$2.50 at a hardware store). Allow laser light to reflect off it at a grazing angle and project the pattern onto a wall.
Kitchen Scale Equilibrium:
Take a two-meter or so 2X4, mark it at 30 cm intervals and show your class its weight on the scale. Now support one end on the scale, one on a block, and ask the class to predict weight on the balance before you actually release the board. Then reverse block and scale and ask again. Try various locations of the block and scale and even add extra weights to the beam. This demonstrates moments under a variety of conditions.
Back to Normalcy:
Clamp a weight to the scale and tilt it to show normal force variance with angle. In general, you will find a kitchen scale to be a frequently used piece of apparatus for all sorts of phenomena. They are available with metric readouts.
Reflections on the Wave Nature of Light:
Reflect a laser beam off a flat mirror, make incident and reflected beams visible with clouds of chalk dust. Then reflect light off a good quality diffraction grating. Ask students what is going on.
Polarizing Influence:
Kids do not outgrow the desire to take something home with them. Diffraction gratings and polarizing materials are so cheap that they should be given to every student. Tape 1cm X 1cm pieces of each behind punched holes on a card. Challenge students to write down observations when they look at clouds, reflected light, neon or sodium lamps, stars, etc.
Trapped in the corner:
A corner reflector may be made by cementing three mirrors together at right angles to each other. Use aquarium sealant for adhesive and drafting triangles to insure accurate right angles.
Reflect laser beam off the interior of the cube. Students will be able to see how three reflections are required for it to work, and they will actually be able to follow the light path. Rock the cube around so it is clear that the return path is not dependent upon orientation.
Bubble Dome:
Make a soap solution as follows: 70ml of Joy, 200ml glycerin, 230 ml water. Roll a cone from a piece of paper and blow a large bubble onto a glass plate on an overhead projector. Ignore projection on screen and look at beautiful, iridescent interference on the bubble itself. With the right mixture of bubble soap the bubble should get thin enough to become totally transparent to reflected light, just before it breaks.
Canned music:
Reflect a strong light off a soap film across the end of a can. With a lens of appropriate focal length, focus an image of the bubble on a screen. A series of spectra characteristic of a thin film is visible. Now bring a speaker to the back of the can, and interesting distortions of the image will occur. If you hook up a signal generator, very the frequencies to get resonance on the soap film.
The Swing Era:
Hang several pendula of different lengths from a semi-rigid support. Challenge students to get a particular pendulum swinging to the exclusion of the others by pulling on a rubber band attached to the support.
Spring String:
The classic demonstration of a mass suspended between two strings, protecting the upper string from breaking by its inertia does not communicate the importance of a stretchable string. If the string were absolutely unyielding, the upper string would break every time.
By replacing the upper string with a spring, a slow motion of the mass downward stretches the spring and visibly puts tension on it. A rapid jerk on the string breaks it without significant stretch of the spring.
The Big Attraction:
With a charged lucite rod, rubber rod, or golf tube, attract an empty pop can. Balance objects on the dome of a watch glass and observe the effects-- everything up to a 2-meter 2X4 will work.Small charged objects may be discharged with an anti-static gun available at a record store. The gun has a piezoelectric crystal connected to a sharp pin. The potential developed on the point creates ions that stream off the point and discharge whatever.
Funneling Momentum:
Suspend a large funnel from a support so that it can spin freely. Fill it with sand and release it, giving it a small initial angular velocity.
Soda Straw Symphony:
Clip the flattened end of a drinking straw to a point, forming a double reed. Pinch the reed end slightly with your lips as you blow HARD to get something like a high-pitched duck call. (No self-respecting beaver would respond to such a noise). While blowing, clip the other end off to change resonant frequencies. Remind the class that shorter tube lengths produce higher resonant frequencies.in the straw just as it did on the soap bubble. Connect multiple straws and flex straws for good bass notes, insert smaller straw to make a slide trombone, cut holes for advanced work. Cut appropriate lengths so class can play school fight song, Christmas carols, etc.
No Strings Attached:
Poke a hole 2/3 of the way into a Nerf ball and imbed a 30g sinker attached to a string in the middle of the foam. Swing the ball in a circle over your head and ask students at what moment in its path you should release it to hit a target. The demonstration is more forceful if some of the student predictions result in the ball flying into the class.
Shifting to Doppler:
Get a code oscillator circuit (e.g. Radio Shack #20-115), a 5cm speaker, a small switch and a 9-volt battery clip. With a sharp knife slice into a Nerf ball and imbed all parts inside. Turn on the switch and throw the ball to students in the class. Pitch will change noticeably depending on whether the ball is approaching or receding.
Electric Washtub:
Mount a length of piano wire under tension (from a spring or weights). Place a large magnet over it and hook the ends of the wire to the input of an amplifier. Plucking the wire will induce currents that will amplify as musical sounds.
Splitting Hairs:
A human hair held in the laser beam will produce a single-slit interference pattern. (The hair forms a single thin barrier.) The width of the hair can be determined by measuring the spacing of the secondary maxima and using the single-slit equation.
Learning the Ropes:
A convincing session in vectors: Have two burley guys pull a rope between them as tight as they can. Then have your smallest kid pull sideways in the center of the rope. He will have no trouble pulling the burleys toward each other.
Bubble Battle:
If two soap bubbles (or balloons) are connected at opposite ends of a pipe, the smaller bubble (or balloon) will force air into the larger one. The pressure inside a bubble varies inversely as the radius of the bubble. It's neat to have a valve in the pipe and set up the bubbles or balloons and ask the kids to hypothesize on what will happen when the valve is opened and why.
Soap Film Trampoline:
Use a coat hanger or a large wire frame. Dip it into a bubble solution. With practice you can cause the large soap film to undergo many interesting modes of vibration.
Football Spin:
Try spinning the following objects on a bare floor or on a smooth table top: Small toy football, hollow egg-shaped plastic container, hard-boiled egge, full-size football. Although the football begins its rotations about its short axis, it reorients itself to a lower energy state by standing up and rotating about its longer axis.
The Levitating Screwdriver:
When various objects are individually placed in a narrow stream of fast moving air, they seem to float. Objects which have been used include: golf balls, small footballs, styrofoam balls, rubber balls, steel balls, hollow egg-shaped plastic containers, and smooth handled screwdrivers.
Pat the Pipe:
Pat the end of the pipe with the flat palm of your hand. Use two distinctly different motions: 1. Leave the hand against the end of the pipe after striking it. 2. Quickly remove the hand away from the end of the pipe immediately after striking it. If you listen carefully, you should hear two different octaves because an open pipe has antinodes at both ends while a closed pipe has a node at one end.
Singing Rod:
Hold on to the midpoint of a solid aluminum rod (18mm diameter is good) with the thumb and forefinger of one hand. Stroke the rod with the thumb and forefinger of the other hand. A LOUD tone emerges from the longitudinal oscillations set up in the rod. By holding the rod at other locations, higher harmonics are heard. It is helpful (essential to get resin, stick-um, or some kind of frictional material on the rod. Consult your local sports store.
Baffle the Speaker: Purchase an ear phone attachment for a cassette player. Cut off the ear piece and in its place solder a small (5cm) speaker (Radio Shack). Plug this speaker into the cassette player and listen to the musical sounds before and after the speaker is placed near the opening of each of the following objects: plastic pipe, bottomless styrofoam cup, a sheet of 60cm square cardboard with a 5cm hole cut in the center.
Light My Balloon:
Fill one balloon with water. Fill another balloon with air. Place a lighted match beneath each of the balloons. Explain.
Use a bicycle wheel or a circular disk (such as a disk stroboscope). Attach a styrofoam ball and arrows radially inward and tangential to the circle. Use shadow projection and watch the length of the arrows shadows simulate the acceleration and velocity vectors of a body in simple harmonic motion. A 1cm hole in an aluminum slide in the projector helps narrow the beam of light.
An Uplifting Experience:
Use duct tape to attach a plastic garbage bag to the outlet hose of an air blower (vacuum cleaner). Have someone sit on the bag as you turn on the air. Watch the person being lifted as the bag fills with air. An alternative is to attach several hoses and have classmates blow up the bag with lung pressure. This method is sometimes used to lift automobiles in accident situations. Big total force from a small pressure over a large area.
Smoke Cannon:
How to construct a smoke cannon: Remove one end of a cardboard box. Attach long rubber bands to each of the remaining 4 corners. Use duct tape to attach a sheet of flexible plastic across the open end of the box. Attach each of the four free ends of the rubber bands to a knob near the middle of the plastic sheet. Cut a circular hole in the side of the box opposite the plastic sheet. Saturate the interior of the box with smoke and blow smoke rings across the room.
Vortex Generator:
A simpler version of the Smoke Cannon. Use a plastic milk bottle. Whap it on the side. Nice vortex rings shoot out. Fill the bottle with methane from your gas jets. Shoot the vortex rings at a burning burner. Watch the rings ignite with nice sound action!
Better still super great Vortex Blammer:
Take a plastic bucket (10 liter) and cut a 25cm hole in the bottom. Cover the top with rubber (wet suit) material (surfers can usually supply you with old wet suits). Tie or clamp the rubber on (with hose clamp stock). It should be tight as a drum. Bam the rubber and WOW! Great vortex rings. Shoot the kids across the room. Fill it with methane, shoot the Bunsen Burner and oooohhhhhh.... NEAT! Big roaring rings of fire!
Diet Lite:
Drop two full cans of pop into a tank of water. One a can of diet pop, the other regular. The diet drink will float. The sugar content of the regular drink will increase its density sufficiently to sink it.
Instant Recycling:
Take an empty pop can, put in a few ml of water. Heat over a burner until steam emerges vigorously from the opening. Quickly invert the can into a pan of cold water. Condensing of the steam is so rapid that the low pressure created allows the atmosphere to crush the can instantly. (Unless you are a real man, you'll want to hold the can in tongs). The pressure of the atmosphere is 1kg/cm2 (9.8n)/cm2.
Sweet Success:
The sugars in Karo syrup rotate the plane of polarized light and rotate different colors different degrees. Place a bottle of it on a polarizing filter on the overhead. Students looking at the bottle through their own polarizers see spectacular colors.
The trick may also be done by stretching scotch tape and layering it randomly on a glass plate. The stretching is essential to straighten out long coiled organic molecules.
The Silver Lining (or a cloud in a 4-liter jug):
Take a 4-liter jug (you may first have to consume the contents the previous weekend), swirl a few ml of water around in it and pour out the excess to raise the humidity. Introduce some smoke from a freshly blown out match for hydroscopic particles. With your mouth over the mouth of the jug (mouth-to mouth) blow hard into the jug and then release the pressure. Ah, fog in a 4-liter jug!
Or connect a rubber tube to the jug with a one-hole stopper and glass tube and blow or suck on the tube. Nice adiabatic warming and cooling.
Discuss capacity of air, absolute humidity, relative humidity, and the variance of the capacity with temperature. And do not forget the dew point.
Sticky Situation:
Mix corn starch (from the grocer) with water to make a fluid paste. Pour it into a beaker and challenge students to QUICKLY punch their finger into it and withdraw. Under rapidly applied force the mixture becomes an elastic solid. Scoop out some of the fluid and squeeze it into a ball. As soon as you quit squeezing it will become a liquid in your hands. This is completely washable, so you can really get into it.
Vanishing Charge:
Take a commercial disectible Layden jar and charge it by holding the base in your hand and reaching toward a Van de Graaf generator with the central plate. Be sure your body is grounded or it won t charge. With a wire between the inside and outside plate, show it is charged. Great Spark! Set it on a table and with faked care lift the center plate out with an insulated rod and hand it to a reluctant student. Remove the glass and with the insulated rod hand the outer cup to another student. Ask them to touch hands. Nothing happens.
Now touch the cups to show there is no charge. Nothing happens. Reconstruct the jar and VOILA there is a spark. You can make your own jar with the bottom half of a soda can in a peanut butter jar with a steel can on the outside. The charge resides on the glass, not the metal.
Change in the Wind:
Place a dime about 3 cm in from the edge of a table. Set a can at an angle near the coin. Challenge students to get the coin into the can without touching either. Solution is to blow SHARPLY over the top of the coin. The Bernoulli effect will lift it and pop it into the can. A real expert can lift a quarter.
Milk, Motion, and Molecules:
Dilute homogenized milk with about 10 parts water to 1 part milk. Place a drop on a microscope slide, cover and view under high power. The tiny fat particles will be seen to dance continually from the collisions of the water molecules. The effect is called Brownian Motion and was crucial evidence for the existence of molecules. Einstein did the math on it.
Absorbing Interest:
Dissolve about 10g of erbium chloride or any rare earth chloride in a large test tube. Hold the tube in front of a showcase bulb while students look through diffraction gratings. They will see a beautiful absorption spectrum.
The Incredible Inverting Pin:
Take a plastic film container and poke a small pinhole in the bottom center. Then push the pin through the side near the front so the head is centered in the opening. Hold it up to you eye and look at the pin silhouetted in the little circle of light. It will appear to be inverted.
Bloogle:
Take a short piece of flexible plastic hose (60 cm long) and swing it briskly in a circle. Three or four resonant frequencies can be heard. With a little practice you can play Taps on it.
Acoustical hang-ups:
Tie about 1.5m of thread or very fine wire to a coat hanger. Loop it over your head as you lean forward, and with your index fingers stick the tread into you ears. Have a friend strike the hanging hanger with a pencil, and revel in the sound produced. Ah, GreatTom of Westminster!
Physics Transferred:
Take two PSSC air core solenoids. Attach lamp cord directly to the terminals of one. to the other attach the terminals from a low-watt household light bulb. Through the first solenoid place a bundle of straightened coat hanger wires. Plug the assembly into the 120 volt line. The coil will jump a little onto the iron core.
Pressure Situation:
To find the pressure on a balloon, simply press the inflated balloon down onto a balance until it flattens to a circle of known radius (cm). Read the balance for force and calculate pressure. P = fA.
Eggstra Eggsitement:
Two students hold a sheet by its four corners. Someone throws a raw egg into the sheet as hard as possible. It does not break. Be careful not to miss the sheet.
Tubular Resonance:
The is a good follow-up to the closed-tube resonance one gets from reflecting tuning-fork sound into a graduated cylinder with water in it. Just take two cardboard tubes that fit one inside the other. Adjust the total length by sliding them in and out until resonance with the tuning fork is found. Your open tube will resonate at twice the length of the closed tube.
Look Look!
Print DICK JANE on a card. DICK in blue, JANE in red. When viewed through a water-filled test tube, JANE appears inverted, while DICK appears normal. Let the kids hypothesise this one. The fact is that they are both inverted, but DICK looks the same either way.
Lake Level:
Famous Archimedian problem. Ask class... A row boat has an anchor in it as it floats on a lake. When the anchor is thrown overboard, will the lake level rise, fall, or stay the same?
Collect student answers and justifications. Then try this experiment. Float a 400 or 500 ml beaker with mass in it (say 200g) in a larger beaker of water (say 2000 ml beaker). Mark the water level. Then remove the mass and place it in the bottom of the larger beaker (the lake). The lake lowers because in the boat, the anchor displaces its weight of water but in the lake it displaces its volume of water.
Pin-Point Discharge:
Get a Pom-Pon from a cheerleader. Place it on the Van de Graaf machine and charge it up. Approach the charged pom pon with a sharp pointed object. It will discharge fast. Ben Franklin found that charge accumulates on sharp points... The Lightning Rod.
Galloping Gourmet:
Cook a hot dog by inserting nail electrodes at each end and connecting to the 120 v line. Watch out for shock hazard! it takes only a few moments to cook.
Electric Jump Rope:
Connect a long wire to the terminals on a galvanometer. Swing it like a jump rope. In the earth s magnetic field it will generate current. Swing only one loop of the wire so that the opposite loop won't cancel the current. | 5,363 | 24,590 | {"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} | 3.328125 | 3 | CC-MAIN-2019-09 | longest | en | 0.918449 |
https://electronics.stackexchange.com/questions/533890/r-cd-parallel-damping-during-power-on | 1,716,426,912,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058575.96/warc/CC-MAIN-20240522224707-20240523014707-00464.warc.gz | 195,917,024 | 41,477 | # R-Cd parallel damping during power on
I am researching the PI filter. There're a number of great articles on how to calculate values for inductor and capacitors, the response curves and frequencies. Examples are: Input Filter Design for Switching Power Supplies, Ferrite Beads Demystified, Optimal Single Resistor Damping of Input Filters etc etc.
The circuit below works great, suppressing the ringing during fast power supply voltage rise. I will not include simulation pics as it is fairly easy to draw and simulate it.
However I can not find any information on the damping resistor power rating. The steeper voltage rise, the higher immediate current flows through the resistor while its series capacitor is not charged. Taking damping resistor as 2 Ohm, it will anyway be much larger than tantalum cap or MLCC ESR, thus during power on most power will be seen on the resistor. I simulated it to be up to 45 Watt...
Here's my test circuit:
L2 is a ferrite bead BLM21PG221, consider it working in its inductive region. (however during fast transients, including power on, may it be more capacitive/resistive rather than inductive?)
The only statement about damping resistor current I found in third source I cited:
Capacitor Cd blocks dc current, to avoid significant power dissipation in R.
Yes, in static it is. But I am sure people also considered dynamic performance - in particular during power on.
So what do I miss here and what would be the power rating on the damping resistor?
There's a second question, very closely coupled with the above one: what is practically achievable power rail rise time? The answer to this question will define the answer to former one. Yes, ATX power supplies have strict standard on the power level rise time, what about other type of power supplies (switching and AC-DC)? What about turning switch on closing the circuit, theoretically happening in a nanosecond causing very sharp rise time? Does the first capacitor on the power wire path makes a difference "absorbing" the initial voltage strike?
• The power rating will be what it needs to be to suit the supply voltage and the number of instances per second it switches on and off. There is no generic answer... it depends... on the application. Nov 24, 2020 at 13:47
The supply is supposed to be steady, or having small variations, so the only major events should be turn on and turn off. Then, the snubber acts as a differentiator for the voltage across it, so the main factor contributing to the current will be the variation of the voltage, or (in the case of on/off) the rising and falling times. Which means that the current through the branch will be the same for 2, or 20 Ω (there is a higher limit and it's given by the time constant). This, in turn, means that a higher resistor will dissipate more power, so the 2 Ω will dissipate 10x less than the 20 Ω. It will also damp less.
But these events do not happen too often, and the dissipated power is an integral, thus the instantaneous power will be temporally smeared. So for this case, the dissipated power will not be the resistor's nominal value, e.g. 0.5 W, it will be the graph you see as the maximum permissible peak power. For example, here is the graph for some metal film resistors from Vishay:
This graph has meaning for the instantaneous power, and only now you can plot it to see whether the resistor is suitable (note: the pulse is 48 V):
The series resistance is stepped through the values of 2, 20, 200, and 2000 Ω, and the bottom plot shows the instantaneous power vs the output voltage (upper plot). Unfortunately, the range involves three orders of magnitude, and the logarithmic scale didn't help very much, so it doesn't look very well, but you can still see that for 2 Ω, there is a small peak (black trace); there is another one for the falling edge, but it's buried. At any rate, the measurements show that for all the values except 2000, the powers are relatively proportional to the falling and the rising edges of the voltage, but the 2 kΩ one (green trace) forms a time constant of 20 ms which, compared to the falling edge of V(x) of ~11 ms is too great and the snubber now has adverse effects. In fact, even the 200 Ω (red trace) is a bit too much, though rather borderline.
The measurements for the peak powers are:
Measurement: pmax1
step MAX(v(y)*i(r1)) FROM TO
1 0.0790655 0 0.08
2 0.720628 0 0.08
3 2.41864 0 0.08
4 0.806291 0 0.08
Measurement: pmax2
step MAX(v(y)*i(r1)) FROM TO
1 0.00417929 0.08 0.2
2 0.0417876 0.08 0.2
3 0.397681 0.08 0.2
4 0.587792
Looking over the graph which shows that for a 10 ms duration the maximum allowed power for a 50% time of the total pulse duration is ~800 mW. The 2nd trace (20 Ω) has the 1st peak at ~721 mW, so it should be safe. 200 Ω is too much, though it should be safe for the 2nd pulse. This implies the power supply to be turned on for 10 ms, and then off, every 20 ms, and it will be at the limit, but should be safe. Of course, since the graph is titled "maximum permissible peak pulse power", I wouldn't test the limits, but I would go until half, two thirds, or so.
If you consider the averages of the pulses, then you must consider the rated dissipated power, in this case, for SFR16S, 0.5 W.
If you want to test for yourself, these are the .meas lines:
.meas Pmax1 max V(y)*I(R1) from 0 to 80m
.meas Pmax2 max V(y)*I(R1) from 80m to 0.2
.meas tr1 find time when V(y)*I(R1)=Pmax1/100 rise=1
.meas tf1 find time when V(y)*I(R1)=Pmax1/100 fall=1
.meas tr2 find time when V(y)*I(R1)=Pmax2/100 rise=2
.meas tf2 find time when V(y)*I(R1)=Pmax2/100 fall=2
.meas Pd1 avg V(y)*I(R1) from tr1 to tf1
.meas DeltaT1 param tf1-tr1
.meas Pd2 avg V(y)*I(R1) from tr2 to tf2
.meas DeltaT2 param tf2-tr2
• Excellent, thank you very much! When rise or fall times are 2 ms there's no need in damping resistor at all as there's no ringing; problems start when these times are < 100 us, and critical when < 1 us (e.g. hot plug event without precharge circuit). If we consider damping resistor, then its value seem to be in range of 0.5 to 1 Ohms. For 0.7 Ohm resistor, 12 V supply and 1 us rise time immediate power simulated is about 180 Watts. If I am reading graphs correctly, this is maximum for single pulse for SFR16S at 10^(-6) time.Derating as you advised means to use SFR25H. Nov 24, 2020 at 17:30
• I can not find appropriate SMD resistor to use, Yageo 1206 has max 15 W for single pulse. Vishay RCL1225 seems to be a candidate, but starts with 1 Ohm. Probably WFMB2512 (current sensing resistor) of 0.5 Ohm is also a candidate... Nov 24, 2020 at 17:43 | 1,783 | 6,657 | {"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} | 2.65625 | 3 | CC-MAIN-2024-22 | latest | en | 0.939632 |
https://www.coursehero.com/file/226967/week3notes/ | 1,518,918,327,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891811243.29/warc/CC-MAIN-20180218003946-20180218023946-00533.warc.gz | 873,686,108 | 302,928 | {[ promptMessage ]}
Bookmark it
{[ promptMessage ]}
week3notes
# week3notes - Random Variables Example We roll a fair die 6...
This preview shows pages 1–5. Sign up to view the full content.
week 3 1 Random Variables Example: We roll a fair die 6 times. Suppose we are interested in the number of 5’s in the 6 rolls. Let X = number of 5’s. Then X could be 0, 1, 2, 3, 4, 5, 6. X = 0 corresponds to the 56 elements of our 66 elements of . X = 1 corresponds to the elements etc. X is an example of a random variable. Probability models often stated terms of random variables. E.g. - model for the # of H’s in 10 flips of a coin. - model for the height of a randomly chosen person. - model for size of a queue.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
week 3 2 Discrete Probability Spaces ( , F , P )
week 3 3 Discrete Random Variable Definition: A random variable X is said to be discrete if it can take only a finite or countably infinite number of distinct values. A discrete random variable X maps the sample space onto a countable set. Define a probability mass function (pmf) or frequency function on X such that Where the sum is taken over all possible values of X . Note that there is a theorem that states that there exists a probability triple and random variable whenever we have a function p such that Definition: The probability distribution of a discrete random variable X is represented by a formula, a table or a graph which provides the list of all possible values that X can take and the pmf for each value
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
week 3 4 Examples of Discrete Random Variables Discrete Uniform Distribution We roll a fair die.
This is the end of the preview. Sign up to access the rest of the document.
{[ snackBarMessage ]}
### Page1 / 15
week3notes - Random Variables Example We roll a fair die 6...
This preview shows document pages 1 - 5. Sign up to view the full document.
View Full Document
Ask a homework question - tutors are online | 503 | 2,110 | {"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} | 3.53125 | 4 | CC-MAIN-2018-09 | latest | en | 0.832994 |
https://discuss.leetcode.com/topic/70046/two-solutions-with-bucket-sort-o-n-and-priority-queue-o-nlogk | 1,511,306,486,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806438.11/warc/CC-MAIN-20171121223707-20171122003707-00261.warc.gz | 609,228,883 | 12,821 | # Two solutions with Bucket Sort O(N) and Priority Queue O(NlogK)
• I believe this problem is waiting a Priority Queue solution, as it says
Your algorithm's time complexity must be better than O(n log n), where n is the array's size.
So I implement a minimum heap first, with time complexity O(NlogK).
`````` # O(NlogK) using minimum priority queue
def top_k_frequent(nums, k)
counter = nums.reduce( Hash.new(0) ) do |ha, num|
ha[num] += 1
ha
end
pg = PriorityQueue.new
pair = Struct.new(:key, :value)
counter.each do |num, count|
node = pair.new(count, num)
if pg.size < k
pg.insert node
else
next if node.key <= pg.get_min.key
pg.delete_min
pg.insert node
end
end
# puts "pg: #{pg.array} size: #{pg.size}"
# generate results
results = []
while pg.size > 0
results.unshift pg.delete_min
end
# puts "results: #{results}"
results.map(&:value)
end
# A minimum priority queue
class PriorityQueue
attr_accessor :array, :size
def initialize
@array = [nil]
@size = 0
end
def insert(node)
self.size += 1
array[size] = node
swim(size)
end
def delete_min
res = get_min
swap(1, size)
self.size -= 1
sink(1)
res
end
def get_min
array[1]
end
private
def swim(k)
while k/2 >= 1
j = k/2
break if !less(k, j)
swap(j, k)
k = j
end
end
def sink(k)
while 2*k <= size
j = 2*k
j += 1 if j+1 <= size && less(j+1, j)
break if !less(j, k)
swap(j, k)
k = j
end
end
def less(j, k)
array[j].key < array[k].key
end
def swap(j, k)
t = array[j]; array[j] = array[k]; array[k] = t
end
end
``````
Then I noticed that Bucket Sort can also solve the problem, within linear time. Good!
`````` # O(N) using bucket sort
def top_k_frequent(nums, k)
counter = nums.reduce( Hash.new(0) ) do |ha, num|
ha[num] += 1
ha
end
# puts "counter: #{counter}"
max_count = 0
frequency = counter.reduce( Hash.new{|hh,kk| hh[kk] = Array.new } ) do |ha, (num, count)|
max_count = count if max_count < count
ha[count] << num
ha
end
# puts "frequency: #{frequency}"
results = []
max_count.downto(1) do |count|
break if results.count == k
while !frequency[count].empty? && results.count < k
results << frequency[count].shift
end
end
results
end
``````
P.S. Quick Select could also work, estimated time complexityO(KN)
Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect. | 712 | 2,299 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2017-47 | longest | en | 0.550802 |
https://forums.halite.io/t/what-does-1-rating-point-represent/825/2 | 1,542,490,908,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743854.48/warc/CC-MAIN-20181117205946-20181117231946-00394.warc.gz | 650,082,366 | 3,525 | # What does 1 rating point represent?
#1
I was looking at TheDuck314s profile and couldn’t help but admire the increase in his rating, I was wondering what percentage of games a ‘1’ change in rating represents, is it a 60% win rate, 80%?
Current (26) 91.28 1 111 Still Playing
25 90.25 1 381 07/11/18 15:02
24 88.47 1 203 04/11/18 17:57
23 87.61 1 242 03/11/18 17:27
22 86.23 1 270 02/11/18 13:33
21 85.52 1 355 01/11/18 12:48
20 84.34 2 226 30/10/18 16:14
#2
If you win a game you rating increase (or decrease if you lose) by some value which is f(game_played_of_your_current_version, opponent_rating) (the more games you played the lesser rank is changed). So it’s just an estimination of the skill of your bot and i think it has no maximum value.
Also see this How Ratings Get Calculated for Halite III
#3
I understand this, but say in chess a 500elo difference in rating gives you about a 95% chance of winning. the scales are different here, so I was wondering approximately what a one point improvement is worth. | 330 | 1,026 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2018-47 | latest | en | 0.932766 |
https://realpythonproject.hashnode.dev/day12-anonymous-functions-in-python | 1,624,474,243,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488539764.83/warc/CC-MAIN-20210623165014-20210623195014-00137.warc.gz | 430,763,793 | 29,017 | #Day12 - Anonymous Functions in Python
Subscribe to my newsletter and never miss my upcoming articles
Today, we will be talking about anonymous functions in Python. Anonymous functions are also known as lambda functions. They are one-liners and look elegant. Below is the syntax
``````lambda parameters: expression
``````
Since they have no names, they are called anonymous functions. However, to actually use them, we need to assign them to a variable
``````variable = lambda parameters: expression
``````
To call a lambda function, we just call the variable and pass the parameters
``````variable(parameters)
``````
Let's look at an example and compare a function written as a normal function and as a lambda function
Normal
``````def square(x):
result = x*x
return result
ans = square(5)
print(and)
``````
Lambda
``````square = lambda x: x*x
ans = square(5)
print(and)
``````
Both print the same result. However, the lambda expression looks cleaner.
Lambda functions with multiple parameters
``````add = lambda x,y: x+y
``````
Using lambda functions with the sorted function
``````pairs = [(1,2) , (3,14) , (5,26) , (10,20) ]
sorted_pairs = sorted(pairs, key = lambda pair: max(pair[0],pair[1]))
print(sorted_pairs)
``````
We have a list of tuples and want to sort it in ascending order based on the maximum number in the pair.
Parameterless lambda function
``````func = lambda : return "String"
print(func())
``````
Lambda Function without variable name
``````print( (lambda x,y: x+y)(4,5))
``````
```
#programming-languages#python#codenewbies#100daysofcode | 407 | 1,587 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2021-25 | latest | en | 0.813861 |
http://www.jiskha.com/display.cgi?id=1193348964 | 1,495,791,810,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608652.65/warc/CC-MAIN-20170526090406-20170526110406-00338.warc.gz | 685,117,149 | 3,860 | # Algebra
posted by on .
A statue is 8ft. tall. The display case for a model of the statue is 18 inches tall. Which scale allows for the tallest model of the statue that will fit in the display case?
A. 1 inch: 2 inches
B. 1 inch: 7 inches
C. 1 inch: 5 inches
D. 1 inch: 10 inches
• Algebra - ,
The statue is 8 ft. tall. The model can't be any more than 1.5 feet tall. The scale must be less than 19%.
Which of your choices shows a ratio of 19% or less?
• Algebra - ,
Thanks!
• Algebra - ,
Idk
### Related Questions
More Related Questions
Post a New Question | 166 | 571 | {"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} | 3.328125 | 3 | CC-MAIN-2017-22 | latest | en | 0.883811 |
https://rdrr.io/cran/Sim.PLFN/man/cuts.to.PLFN.html | 1,628,028,118,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154471.78/warc/CC-MAIN-20210803191307-20210803221307-00396.warc.gz | 493,535,648 | 7,837 | # cuts.to.PLFN: Convert cuts to Piecewise Linear Fuzzy Number In Sim.PLFN: Simulation of Piecewise Linear Fuzzy Numbers
## Description
This function can easily covert a cuts matrix into a Piecewise Linear Fuzzy Number.
## Usage
`1` ```cuts.to.PLFN(cuts) ```
## Arguments
`cuts` A matrices, with knot.n raw and 2 column, which contains all information about a Piecewise Linear Fuzzy Number.
## Value
This function returned a Piecewise Linear Fuzzy Number.
## References
Gagolewski, M., Caha, J. (2015) FuzzyNumbers Package: Tools to deal with fuzzy numbers in R. R package version 0.4-1, https://cran.r-project.org/web/packages=FuzzyNumbers
Gagolewski, M., Caha, J. (2015) A guide to the FuzzyNumbers package for R (FuzzyNumbers version 0.4-1) http://FuzzyNumbers.rexamine.com
DISTRIB FuzzyNumbers FuzzyNumbers.Ext.2 Calculator.LR.FNs
## Examples
``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17``` ```if(!require(FuzzyNumbers)){install.packages("FuzzyNumbers")} library(FuzzyNumbers) knot.n = 2 T <- PLFN( knot.n=knot.n, X.dist="norm", X.dist.par=c(0,1), slX.dist="exp", slX.dist.par=3, srX.dist="beta", srX.dist.par=c(1,3) ) T plot(T, type="b") CUTS <- PLFN.to.cuts(T, knot.n) CUTS T2 = cuts.to.PLFN(CUTS) plot(T2, type="b", col=2, add=TRUE, lwd=3, lty=3) ```
Sim.PLFN documentation built on May 2, 2019, 5:51 a.m. | 442 | 1,330 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2021-31 | latest | en | 0.553277 |
https://r4r.co.in/mcqs/learn.php?test_id=1985&subid=656&subcat=NCERT%20Class%2012&test=MCQ%20Questions%20for%20Class%2012%20Physics%20Chapter%2013%20Nuclei | 1,656,630,197,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103915196.47/warc/CC-MAIN-20220630213820-20220701003820-00258.warc.gz | 530,595,800 | 5,972 | # NCERT Class 12/MCQ Questions for Class 12 Physics Chapter 13 Nuclei Sample Test,Sample questions
1.mesons
2.photons
3.electrons
4.positrons
## Question: `Alpha particle emitted from a radioactive material are:`
1. Helium nuclei
2.Hydrogen nuclei
3.Lithium nuclei
4.None of the above
## Question: `An alpha particle is emitted from 88Ra226, then the product nuclei has:`
1. Z = 84, A = 224
2. Z = 86, A = 224
3.Z = 86, A = 222
4.Z = 82, A = 222
1. 0.53 MeV
2.1.06 MeV
3.2.12 MeV
4.zero
## Question: `Chadwick was awarded the Nobel prize in Physics in 1935 for his discovery of:`
1.electron
2.proton
3.neutron
4.None of the above
1. -13.6 MeV
2.-13.6 eV
3. -13.6 J
4.13.6 J
## Question: `Fusion takes place at high temperature because:`
1.Atom are ionised at high temperature
2.Molecules break up at high temperature
3.Nuclei break up at high temp.
4.Kinetic energy is high enough to overcome repulsion between nuclei
1.92
2.146
3.238
4.0
## Question: `If radio active nuclei emits β-particle, then mass-number:`
1.increased by 1 unit
2.decreases by 1 unit
3. increases by 2 unit
4.decreases by 2 unit
## Question: `Nuclear force is:`
1.strong, short range and charge independent force
2. charge independent, attractive and long range force
3. strong, charge dependent and short range attractive force
4. long range, change dependent and attractive force
## Question: `Rutherford is the unit of:`
2.energy
3.photoelectric current
4.magnetic field
## Question: `The Bohr model of atom:`
1.assumes that the angular momentum of electron is quantized
2. uses Einstein’s photoelectric equation
3. predicts continuous emission spectra for atoms
4. predicts the same emission spectra for all types of atoms
1.1015 kg m-3
2.1018 kg m-3
3.1017 kg m-3
4.1016 kg m-3
1.0 eV
2.-13.6 eV
3.2 eV
4.-27.2 eV
1.0.6930
2. 0.0693
3.10.6930
4.0.3070
1. two protons
2. two electrons
3.two neutrons
4.six nucleons
1. M – Z
2. M
3.Z
4.M + Z
1.234
2.235
3.236
4. 238
## Question: `The phenomena of radioactivity is:`
1. Exothermic change with increase or decrease with temperature
2.increases on applied pressure
3.nuclear process does not depend on external factors
4.None of the above
1. 3.125%
2.6.25%
3..33%
4.31%
## Question: `When the mass of a sample of a radioactive substance decreases, the mean life of the sample:`
1.increases
2.decreases
3.remain unchanged
4.first decreases then increases
1.Becqueral
2. Marie curie
3.Roengton
4.Vanlaw
## More MCQS
##### R4R Team
R4Rin Top Tutorials are Core Java,Hibernate ,Spring,Sturts.The content on R4R.in website is done by expert team not only with the help of books but along with the strong professional knowledge in all context like coding,designing, marketing,etc! | 878 | 2,797 | {"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-2022-27 | latest | en | 0.751975 |
https://smartasset.com/financial-advisor/accounting-rate-of-return | 1,721,032,051,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514680.75/warc/CC-MAIN-20240715071424-20240715101424-00045.warc.gz | 476,044,066 | 64,659 | # Understanding Accounting Rate of Return (ARR)
Share
Accounting rate of return is a tool used to decide whether it makes financial sense to proceed with a costly equipment purchase, acquisition of another company or another sizable business investment. It is the average annual net income the investment will produce, divided by its average capital cost. If the result is more than the minimum rate of return the business requires, that is an indication the investment may be worthwhile. If the accounting rate of return is below the benchmark, the investment won’t be considered.
## Calculating Accounting Rate of Return
To calculate the accounting rate of return for an investment, divide its average annual profit by its average annual investment cost. The result is expressed as a percentage. For example, if a new machine being considered for purchase will have an average investment cost of \$100,000 and generate an average annual profit increase of \$20,000, the accounting rate of return will be 20%.
Average annual profit increase \$20,000 / Average investment cost \$100,000 = 0.20
The ARR on this investment is 0.20 x 100 or 20%.
To calculate accounting rate of return requires three steps, figuring the average annual profit increase, then the average investment cost and then apply the ARR formula.
To arrive at a figure for the average annual profit increase, analysts project the estimated increase in annual revenues the investment will provide over its useful life. Then they subtract the increase in annual costs, including non-cash charges for depreciation.
To get average investment cost, analysts take the initial book value of the investment plus the book value at the end of its life and divide that sum by two.
For example, say a company is considering the purchase of a new machine that will cost \$100,000. It will generate a total of \$150,000 in additional net profits over a period of 10 years. After that time, it will be at the end of its useful life and have \$10,000 in salvage (or residual) value.
First, calculate average annual profit:
Minus depreciation (purchase cost minus salvage value): \$90,000
Total profits after depreciation: \$60,000
Average annual profit over 10 years: \$6,000
Second, calculate average investment costs:
Average investment (\$100,000 first year book value plus \$10,000 last year book value) / 2 = \$55,000
Now apply the accounting rate of return formula:
\$6,000 / \$55,000 = 0.109
The ARR for this investment would be 0.109 x 100 or 10.9%.
## ARR Pros and Cons
Managers can decide whether to go ahead with an investment by comparing the accounting rate of return with the minimum rate of return the business requires to justify investments. For example, a business may require investments to return at least 15%. In the above case, the purchase of the new machine would not be justified because the 10.9% accounting rate of return is less than the 15% minimum required return.
Accounting rate of return is also sometimes called the simple rate of return or the average rate of return. Accounting rate of return can be used to screen individual projects, but it is not well-suited to comparing investment opportunities. One reason is that it does not consider the time value of money. Different investments may involve different time periods, which can change the overall value proposition.
Unlike other widely used return measures, such as net present value and internal rate of return, accounting rate of return does not consider the cash flow an investment will generate. Instead, it focuses on the net operating income the investment will provide. This can be helpful because net income is what many investors and lenders consider when selecting an investment or considering a loan. However, cash flow is arguably a more important concern for the people actually running the business. So accounting rate of return is not necessarily the only or best way to evaluate a proposed investment.
## Bottom Line
Accounting rate of return is a simple and quick way to examine a proposed investment to see if it meets a business’s standard for minimum required return. Rather than looking at cash flows, as other investment evaluation tools like net present value and internal rate of return do, accounting rate of return examines net income. However, among its limits are the way it fails to account for the time value of money.
## Tips for Evaluating Capital Investments
• Consider working with an experienced financial advisor if you are evaluating a proposed investment. SmartAsset’s free tool matches you with up to three vetted financial advisors who serve your area, and you can interview your advisor matches at no cost to decide which one is right for you. If you’re ready to find an advisor who can help you achieve your financial goals, get started now.
• If the ARR calculation of a proposed capital investment or acquisition looks weak, it might make more sense to outsource. Outsourcing is a complicated issue so it’s good to have grasped the basic arguments for and against outsourcing before you take that alternative step. | 1,032 | 5,129 | {"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-2024-30 | latest | en | 0.927703 |
http://movies.pop-cult.com/toq3as/act-prep-math-worksheets-pdf-463737 | 1,610,721,913,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703495901.0/warc/CC-MAIN-20210115134101-20210115164101-00281.warc.gz | 82,765,468 | 7,366 | # act prep math worksheets pdf
A die has six faces, so the odds of rolling any particular number is \$1/6\$. How likely is it that you’ll flip either heads or tails for each toss? You're free! 0000045627 00000 n We are told that the 10th graders have a ratio of 86:255 to the school population and the 11th graders have a ratio of 18:51 to the total student population. 0000045192 00000 n This means that we can eliminate answer choices H, J, and K. As you can see, no matter which method you use, you can find the right solution. 0000029929 00000 n So there are 86 10th graders, 90 11th graders, and the remaining students are 12th graders. 0000043237 00000 n The odds of either two or more events occurring will be greater than the odds of one of the events alone. We can see that the 11th graders have a reduced ratio, so we must multiply each side of the ratio by the same amount in order to equal the total number of students as the 10th graders’ ratio (255). An “either/or” question asks whether or not one of the multiple events occurs (no matter which event is was). A good way to remember this is to remember that a combined probability question will ultimately have a lower probability than the that of just one (or either) event occurring. The current probability of selecting a red marble is: Now, we are adding a certain number of red marbles and only red marbles. "What are the odds of two or more events both/all happening?" In this one throw, there is one possible chance of getting heads. Their sum will become the probability of either event happening. hbspt.cta._relativeUrls=true;hbspt.cta.load(360031, '999536b9-3e8d-43b1-bb4b-469b84affecc', {}); Courtney scored in the 99th percentile on the SAT in high school and went on to graduate from Stanford University with a degree in Cultural and Social Anthropology. See what makes a "good" score and how you can get the most out of your studying time to reach your target goal. ACT Math Prep Book. We can also see that the larger the number we add to both the numerator and the denominator, the larger our probability will be (you can test this by plugging in answer choice J or K—for K, if you add 40 to both 12 and 32, your final probability fraction will be \$52/72\$ => \$13/18\$, which is even larger than \$2/3\$.). And always feel free to fall back on your PIA or PIN, as needed. We've got guides on all your individual math needs, from trigonometry to slopes and more. 0000040254 00000 n 0000007288 00000 n On the other hand, an impossible event will have a probability of \$0/x\$ or 0. What are the odds that Jenny will roll a pair of dice and get six on both? Look to our guide on how to maximize your time and your score in the hour allotted. This kind of probability question is called a combined probability and there is a good chance you’ll see a question of this type in the later half of the ACT math section. ACT Writing: 15 Tips to Raise Your Essay Score, How to Get Into Harvard and the Ivy League, Is the ACT easier than the SAT? 0000009598 00000 n 2. 0000052291 00000 n You are combining forces to increase your odds of getting a desirable outcome. Once you get used to working with probabilities, you’ll find that probability questions are often just fancy ways of working with fractions and percentages. Because each coin toss is independent of another coin toss. 0000012888 00000 n So we must increase our red marbles (and, consequently, the total number of marbles) by 12 in order to get a probability of \$⅗\$ of selecting a red marble. Now that you've stacked the odds in your favor on your probability questions, it's time to make sure you're caught up with the rest of your ACT math topics. Almost always, the ACT will use the word “probability,” but make sure to note that these words are all interchangeable. Complete this test, take it seriously, and you’re sure to be much more at ease on exam day. Our free PDF can … How likely is it that your first AND second coin tosses will BOTH be heads? If you liked this Math lesson, you'll love our program. Understand that probabilities are simply fractional relationships of desired outcomes over all potential outcomes, and you’ll be able to tackle these kinds of ACT math questions in no time. So let’s look again at our earlier example with Mara and her beads. 0000040377 00000 n Formulas, definitions, and concepts . 0000015529 00000 n Now, we must multiply the 11th grade ratio by 5 on each side to even out the playing field. The alternative method is to use plugging in answers. We must first set these ratios to an equal number of total students in order to determine the number of students in each class. 0000029753 00000 n Check out how to get a perfect score on the ACT math, written by a 36-scorer. If you are in your sophomore year and you are taking the Pre-ACT, you might prefer our Pre ACT book instead. 0000047919 00000 n SAT® is a registered trademark of the College Entrance Examination BoardTM. \${\probability \of \either \event = [{\outcome A}/{\total \number \of \outcomes}] + [{\outcome B}/{\total \number \of \outcomes}]\$, (Special note: this is called a “non-overlapping” probability. startxref Along with more detailed lessons, you'll get thousands of practice problems organized by individual skills so you learn most effectively. This is another example of an altering probability question and, again, we have two choices when it comes to solving it. A Comprehensive Guide. Check out our best-in-class online ACT prep program. 0000005700 00000 n Always take a moment to think about probability questions logically so that you don’t multiply when you should add, or vice versa. This answer is a little bit too large. 0000053803 00000 n 0000056745 00000 n 0000030205 00000 n 0000005413 00000 n We are assuming for now that there are 255 students total (there may be \$255 *2\$ or \$255 * 3\$, and so forth, but this will not affect our final outcome; all that matters is that we choose a total number of students that is equal for all grades/ratios.). On this web page "ACT math practice problems worksheet pdf" we have given plenty of worksheets on ACT math. It can be easy to make a mistake with probability ratios, or to mix up an either/or probability question with a both/and question. %PDF-1.4 %���� A probability ratio is the exact same thing as a question that simply asks you for a ratio. On the other hand, an either/or probability question will have higher odds than the probability of just one of its events happening. Let’s go through both the algebra/proportion method and PIA. The probability of drawing this hand is less than 0.0000004%, so I'm gonna go ahead and go all in. To find the probability of a combined probability question, we must multiply our probabilities. If you liked this Math lesson, you'll love our program. 0000056704 00000 n Supplemental ACT Prep. For a refresher on ratios, check out our guide to ACT fractions and ratios. And there are still \$5 + 10 + 15 + 20 = 50\$ beads total for our denominator. 0000038824 00000 n 0000013464 00000 n (If it helps to picture, you can rephrase the question as: “What are the odds that BOTH his first coin tosses were heads? Get the latest articles and test prep tips! 0000055233 00000 n We are asked to find an additional number of red marbles that we must add to the total number of marbles in order to find a new probability. 0000004586 00000 n 0000005133 00000 n This means that this is a simple matter of determining our desired outcome over the number of total outcomes. To find the probability of an “either/or” question, we must add our probabilities. 0000046448 00000 n Simply use the understandings we learned above and you’ll be able to solve these kinds of questions without issue. The odds are 7 in 10 (\$7/10\$) that Mara will draw any color bead except green. Again, if you need a refresher on ratios, check out our guide to ACT fractions and ratios. Flash Cards. Pin On Grade Math Worksheets & Sample Printables 0000003478 00000 n First of all, you will know if you are being asked for a probability question on the ACT because, somewhere in the problem, it will ask you for the "probability of," the "chances of," or the "odds of" one or more events happening.
Yamaha A1r Acoustic-electric Guitar, Ibanez String Height, Luxa Polish Alter Ego, 20 Action Words With Sentences, Trader Joe's Curry Powder Ingredients, Marmite Chicken Without Frying, Is Quantum Physics The Hardest Subject, Fanta Pina Colada, 925 Sterling Silver, | 2,077 | 8,491 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2021-04 | latest | en | 0.942587 |
http://www.ohiorc.org/standards/ohio/grade/search/mathematics/grade10/387.aspx?page=1 | 1,386,640,731,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164004837/warc/CC-MAIN-20131204133324-00033-ip-10-33-133-15.ec2.internal.warc.gz | 463,442,459 | 6,194 | Ohio Resource Center
# Ohio's Academic Content Standards in Mathematics
Results 1 - 9 of 9: Return to Grade 10 Indicators Ohio Standard: Patterns, Functions and Algebra StandardGrade Level Indicator: (Grade: 10) 1. Define function formally and with f(x) notation.
1
Successive Discounts
ORC# 7866
Resource Information
Resource Type: Lessons
Discipline: Mathematics
Professional Commentary: Students?model successive discounts as a composition of functions and show that, in general, function composition is not commutative. They study the graphs of the discount functions to find the values for which?different orders of successive discounts yield equal results.?An activity sheet, discussion questions, lesson extensions, suggestions for assessment, and prompts for teacher reflection are included....
2
Domain Representations
ORC# 7707
Resource Information
Resource Type: Lessons
Discipline: Mathematics
Professional Commentary: Students use graphs, tables, number lines, verbal descriptions, and symbolic representations to analyze the domains of various functions. An activity sheet, discussion questions, lesson extensions, and suggestions for assessment are included....
3
NAEP Assessment Item, Grade 12: Given a Graph, Find the Value
ORC# 2065
Resource Information
Resource Type: Assessments
Discipline: Mathematics
Professional Commentary: Students must find the value of a composition of functions by looking at the graphs of the functions. This multiple-choice question is a sample test item used in grade 12 in the 1990 National Assessment of Educational Progress (see About NAEP)....
4
NAEP Assessment Item, Grade 12: Given a Graph, Find the Value
ORC# 2064
Resource Information
Resource Type: Assessments
Discipline: Mathematics
Professional Commentary: Students must determine the value of a function by looking at its graph. This multiple-choice question is a sample test item used in grade 12 in the 1990 National Assessment of Educational Progress (see About NAEP)....
5
ORC# 2009
Resource Information
Resource Type: Assessments
Discipline: Mathematics
Professional Commentary: Students are asked to evaluate a quadratic function for a noninteger value of x. They have the option of using a calculator....
6
NAEP Assessment Item, Grade 12: Solve a Function Using Substitution
ORC# 2000
Resource Information
Resource Type: Assessments
Discipline: Mathematics
Professional Commentary: This item asks students to evaluate the composition of a quadratic function with a linear function for x=2. This multiple-choice question is a sample test item used in grade 12 in the 1992 National Assessment of Educational Progress (see About NAEP)....
7
NAEP Assessment Item, Grade 12: Determine composition of two functions
ORC# 11292
Resource Information
Resource Type: Assessments
Discipline: Mathematics
Professional Commentary: Students are asked to find the composition of two functions. This constructed-response question is a "hard"?test item used in grade?12 of the 2005 National Assessment of Educational Progress (see About NAEP)....
8
Pre-Calculus
ORC# 1122
Resource Information
Resource Type: Content Supports -- Reference materials
Discipline: Mathematics | 672 | 3,174 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2013-48 | latest | en | 0.820315 |
https://physics.gurumuda.net/dynamics-object-connected-by-cord-over-pulley-atwood-machine-problems-and-solutions.htm | 1,656,695,116,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103943339.53/warc/CC-MAIN-20220701155803-20220701185803-00401.warc.gz | 506,031,468 | 15,093 | Basic Physics
# Dynamics, object connected by cord over pulley, atwood machine – problems and solutions
1. Block A with a mass of 5 kg, placed on a smooth horizontal plane. Block B with a mass of 3 kg hanging at one end of the cord connected with block A over a pulley. Acceleration due to gravity is 10 m/s2. What is the acceleration of both blocks?
Known :
Mass of block A (mA) = 5 kg
Mass of block B (mB) = 3 kg
Acceleration due to gravity (g) = 10 m/s2
Weight of block B (wB) = mB g = (3)(10) = 30 Newton
Wanted: Acceleration of both blocks (a)
Solution :
The horizontal plane is smooth so there is no friction force. The force that accelerates both blocks is the weight of the block B.
ΣF = m a
wB = (mA + mB) a
30 = (5 + 3) a
30 = 8 a
a = 30 / 8
a = 3.75 m/s2
Read : Carnot engine (application of the second law of thermodynamics) - problems and solutions
2.
Based on the figure above,
(1) acceleration of object = 0
(2) the object moves at a constant velocity
(3) object at rest
(4) the object moves if the weight of the object is smaller than the force that pulls the object.
Solution :
(1) The acceleration of the object = 0.
Net force :
F = m a –> acceleration (a) = 0
F = 0
F1 + F2 – F3 = 12 + 24 – 36 = 36 – 36 = 0 N
(2) The object moves at a constant velocity
No acceleration means object at rest or moves at a constant velocity.
(3) object at rest
No net force means object at rest.
(4) the object moves if the weight of the object is smaller than the force that pulls the object.
Weight acts on the vertical direction, while the pull force acts on the horizontal direction.
The object moves in a horizontal direction so only the horizontal forces that act on the object
Read : Electric charge stored in capacitor – problems and solutions
3. If the coefficient of kinetic friction between the block A and the table surface is 0.1. Acceleration due to gravity is 10 m/s2, then what is the force acts on the block A so that the system moves to leftward in 2 m/s2.
Known :
Mass of block A (mA) = 30 kg
weight of block A (wA) = (30 kg)(10 m/s2) = 300 kg m/s2 or 300 Newton
Mass of block B (mB) = 20 kg
weight of block B (wB) = (20 kg)(10 m/s2) = 200 kg m/s2 or 200 Newton
Acceleration due to gravity (g) = 10 m/s2
Coefficient of kinetic friction (μk) = 0.1
Acceleration of system (a) = 2 m/s2 (to leftward)
Force of kinetic friction (fk) = μk N = μk wA = (0.1)(300) = 30 Newton
Wanted : The magnitude of force F
Solution :
Newton’s second law :
ΣF = m a
The object A moves to leftward :
F – fk – wB = (mA + mB) a
F – 30 – 200 = (30 + 20)(2)
F – 230 = (50)(2)
F – 230 = 100
F = 230 + 100
F = 330 Newton
4. Two objects, A = 2 kg and B = 6 kg, attached at one end of cord over a pulley, as shown in figure below. If acceleration due to gravity is 10 ms-2 then what is the acceleration of the object B.
Known :
Mass of object A (mA) = 2 kg, mB = 6 kg, g = 10 m/s2
weight of object A (wA) = (mA)(g) = (2)(10) = 20 N
weight of object B (wB) = (mB)(g) = (6)(10) = 60 N
Wanted : Acceleration of object b (system’s acceleration).
Solution :
wB > wA so that object B moves downward, object A moves upward
ΣF = m a
wB – wA = (mA + mB) a
60 – 20 = (2 + 6) a
40 = (8) a
a = 5 m/s2
Read : Average velocity – problems and solutions
5. Two objects connected by a cord over a smooth pulley, as shown in figure below. If m1 = 1 kg, m2 = 2 kg, and acceleration due to gravity is 10 ms-2, then what is the tension force T.
Known :
Mass of object 1 (m1) = 1 kg
Mass of object 2 (m2) = 2 kg
Acceleration due to gravity (g) = 10 m/s2
weight of object 1 (w1) = m1 g = (1 kg)(10 m/s2) = 10 kg m/s2 or 10 Newton
weight of object 2 (w2) = m2 g = (2 kg)(10 m/s2) = 20 kg m/s2 or 20 Newton
Wanted : The tension force (T) ?
Solution :
w2 > w1 so m2 moves downward, m1 moves upward.
ΣF = m a
w2 – w1 = (m1 + m2) a
20 – 10 = (1 + 2 ) a
10 = (3) a
a = 3.3 m/s2
System’s acceleration = 3.3 m/s2.
m2 moves downward :
w2 – T2 = m2 a
20 – T2 = (2)(3.33)
20 – T2 = 6.66
T2 = 20 – 6.66
T2 = 13.3 Newton
m1 moves upward :
T1 – w1 = m1 a
T1 – 10 = (1)(3.3)
T1 – 10 = 3.33
T1 = 10 + 3.33
T1 = 13.3 Newton
The tension force (T) = 13.3 Newton.
Read : Inclined plane – problems and solutions
6. Mass of m1 = 6 kg and mass of m2 = 4 kg. The horizontal surface is smooth. Acceleration due to gravity is 10 m/s2. What is the system’s acceleration.
Known :
Mass of m1 = 6 kg
mass of m2 = 4 kg
acceleration due to gravity (g) = 10 m/s2
weight of w1 = m1 g = (6 kg)(10 m/s2) = 60 kg m/s2 or 60 Newton
weight of w2 = m2 g = (4 kg)(10 m/s2) = 40 kg m/s2 or 40 Newton
Wanted : System’s acceleration (a)
Solution :
m1 on a smooth horizontal plane without friction so that the system accelerated by the weight of the block 2.
Apply Newton’s second law :
F = m a
w2 = (m1 + m2) a
40 N = (6 kg + 4 kg) a
40 N = (10 kg) a
a = 40 N / 10 kg
a = 4 m/s2
7. Two blocks, each block has the mass of 2 kg, connected by a cord over a pulley, as shown in the figure below. The horizontal plane and pulley are smooth. If the block B is pulled by a horizontal force of 40 Newton, then what is the acceleration of block. Acceleration due to gravity is 10 m/s2.
Known :
mass of block A (mA) = mass of block B (mB) = 2 kg
Acceleration due to gravity (g) = 10 m/s2
Force of F = 40 N
weight of object A (wA) = m g = (2)(10) = 20 N
Wanted : System’s acceleration (a) ?
Solution :
Apply Newton’s second law :
F = m a
F – wA = (mA + mB) a
40 – 20 = (2 + 2) a
20 = (4) a
a = 20 / 4
a = 5 m/s2
Read : Magnetic field produced by two parallel current-carrying wires - problems and solutions
8. Mass of block A = 2 kg and mas of block B = 1 kg. Block B initially at rest, then accelerated downward until it hits ground. Acceleration due to gravity is 10 m/s2. What is the magnitude of the tension force.
Known :
Mass of block A (mA) = 2 kg
Mass of block B (mB) = 1 kg
Acceleration due to gravity (g) = 10 m/s2
Weight of block B (wB) = mB g = (1)(10) = 10 Newton
Wanted : The magnitude of the tension force (T)
Solution :
Ignore the friction force.
System’s acceleration (a)
F = m a
wB = (mA + mB) a
10 = (2 + 1) a
10 = 3 a
a = 10/3
The tension force (T)
The tension force on the block A :
F = m a
T = mA a = (2)(10/3) = 20/3 = 6.7 Newton
The tension force on the block B :
F = m a
wB – T = mB a
10 – T = (1)(10/3)
10 – T = 3.3
T = 10 – 3.3 = 6.7 Newton
The tension force (T) = 6.7 Newton
9. Mass of block A = 2 kg and mass of block B = 1 kg. The friction force between object A with the horizontal plane = 2.5 Newton. Ignore friction on pulley and cord. What is the acceleration of both blocks.
Known :
Mass of block A (mA) = 2 kg
Mass of block B (mB) = 1 kg
Friction force between lock a and the horizontal plane (fkA) = 2.5 Newton
Acceleration due to gravity (g) = 10 m/s2
Weight of block (wB) = mB g = (1)(10) = 10 Newton
Wanted : Acceleration of both blocks (a)
Solution :
Apply Newton’s second law :
F = m a
wB – fk = (mA + mB) a
10 – 2.5 = (2 + 1) a
7.5 = 3 a
a = 7.5 / 3
a = 2.5 m/s2
Read : Electromagnetic induction, induced EMF – problems and solutions
10. Mass of block a = 30 kg, rest on a horizontal plane connected with block B with mass of 10 kg over a pulley. What is the system’s acceleration. Acceleration due to gravity is 10 ms-2.
Known :
Mass of block A (mA) = 30 kg
Mass of block B (mB) = 10 kg
Acceleration due to gravity (g) = 10 m/s2
weight of block B (wB) = mB g = (10)(10) = 100 Newton
Wanted : System’s acceleration (a)
Solution :
F = m a
wB = (mA + mB) a
100 = (30 + 10) a
100 = 40 a
a = 100 / 40
a = 2.5 m/s2
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 2,625 | 7,743 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2022-27 | latest | en | 0.875713 |
https://www.teacherspayteachers.com/Product/Multiplication-Facts-Activities-and-Games-for-the-Factor-10-2516792 | 1,495,583,390,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607704.68/warc/CC-MAIN-20170523221821-20170524001821-00604.warc.gz | 955,997,837 | 22,853 | # Main Categories
Total:
\$0.00
Whoops! Something went wrong.
# Multiplication Facts Activities and Games for the Factor 10
Product Rating
4.0
3 ratings
File Type
PDF (Acrobat) Document File
Be sure that you have an application to open this file type before downloading and/or purchasing.
15 MB|10 pages
Product Description
Multiplication Fact Practice for Factor 10! Multiplication Activities and Math Games!
Do your students need more practice multiplying one digit numbers by 10?
This packet contains fun practice activities just focusing around the factor ten! Your students can have multiplication FACT Power with this packet of fun practice!!
It is Super Hero Spectacular!
This easy to use packet is highly effective to help students master multiplication factor 10 facts. Use these multiplication pages as morning work, homework or math centers.
This Multiplication Factor 10 is a sample of a larger resource. The larger resource has 31 pages. This freebie contains the following 4 Activities:
★ A Journal Cover to put together as a booklet for students
★ Circles the Multiplies of 10 Practice Page
★ Fact Family Number Puzzle
★ True and False Superhero Fun (cut/sort/glue)
★ Bump Games (1 color 1 b&w)
★ Answer Keys included
Skills Covered in this packets are:
★multiples of 10
★fact families of 10
★fact fluency
---------------------------------------------------------------------------------------------------------
Teachers Who Have Used This Resource Have Said:
♦"Great multiplication practice! Kids will love this fun practice! I bet the full resource is just wonderful!"
(Thank you Believe to Achieve by Anne Rozell)
---------------------------------------------------------------------------------------------------------
Fair Warning (or Not-At-All-Fine-Print)
Use these teaching tools often enough and you might just find yourself with a math lover on your hands! Which is just a fun way of sayin’ - you might want check out my other multiplication activity packets. Especially if you want to make yourself some math lovers!
★Four Reasons to Follow Me★
Followers are the First to hear about new products!
Followers often get Fifty percent of on the First Day! Not kidding.
Followers get Freebies
Followers inspire me to do more!
Click the star to become a follower! Thanks! You’re a SuperStar!★
Thank you for stopping by my store!
Tricia :)
Products you may also like...
Multiplication Facts Packet: Multiplying by 4
Multiplication Facts Packet: Multiplying by 11
Multiplication Facts Packet: Multiplying by 12
100th Day of School Freebie
Multiplication Bundle
Connect with Me!
Follow Tricia's Terrific Teaching Trinkets Blog
Follow Tricia's Terrific Teaching Trinkets on Facebook
Follow Tricia's Terrific Teaching Trinkets Pinterest
Follow Tricia's Terrific Teaching Trinkets on Instagram
Follow Tricia's Terrific Teaching Trinkets on Bloglovin'
Total Pages
10 pages
Included
Teaching Duration
Lifelong tool
• Comments & Ratings
• Product Q & A
FREE
FREE | 648 | 2,990 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2017-22 | longest | en | 0.837627 |
https://bytes.com/topic/python/answers/898835-how-do-you-do-binary-addition-python | 1,575,714,567,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540497022.38/warc/CC-MAIN-20191207082632-20191207110632-00446.warc.gz | 307,719,224 | 12,831 | 440,095 Members | 1,586 Online
Need help? Post your question and get tips & solutions from a community of 440,095 IT Pros & Developers. It's quick & easy.
# How do you do binary addition on python?
P: 32 I'm having trouble trying to add these together correctly. I have kind of ran into a brick wall. I have gotten it converted and that's it as of now. My Code: Expand|Select|Wrap|Line Numbers class BinaryNumber: '''A Binary Number Class''' #pass def __init__(self, bitString = ''): '''Constructor''' #pass self._bits = bitString def __str__(self): '''string representation''' if self._bits: return self._bits else: return '0' def __int__(self): '''conversion''' #pass total = 0 for pos in range(len(self._bits)): if self._bits[pos] == '1': total += 2 ** (len(self._bits) - 1 - pos) return total def __add__(self, other): '''addition: implementation of the + operator''' pass I am not allowed to import math either. Any help would be much appreciated! :] Oct 15 '10 #1
14 Replies
Expert Mod 2.5K+ P: 2,851 All you need is a conversion of a decimal integer to binary. Then your magic method __add__ would be: Expand|Select|Wrap|Line Numbers def __add__(self, other): '''addition: implementation of the + operator''' return self.dec2bin((int(self)+int(other))) Have you tried to write code for it? Oct 15 '10 #2
P: 32 I have tried a similar code that used "import math" and tried changing it to get it to work, but it didn't. This is all of my code as of right now for my project: Expand|Select|Wrap|Line Numbers # Program: ex6_18.py # Authors: Robert Mineweaser # # This example is discussed in Chapter 6 of the book # Object-Oriented Programming in Python # class BinaryNumber: '''A Binary Number Class''' #pass def __init__(self, bitString = ''): '''Constructor''' #pass self._bits = bitString def __str__(self): '''string representation''' if self._bits: return self._bits else: return '0' def __int__(self): '''conversion''' #pass total = 0 for pos in range(len(self._bits)): if self._bits[pos] == '1': total += 2 ** (len(self._bits) - 1 - pos) return total def __add__(self, other): '''addition: implementation of the + operator''' #pass '''carry = 0 result = '' # Work from right to left. for i in range(0, len(self._bits))[::-1]: tmp = int(((a[i]) + int(b[i]) + carry)/2) res = str(int(a[i]) + int(b[i]) + carry - 2 * tmp) result += res carry = tmp result = (result + str(carry))[::-1] try: return result[result.index('1'):] except ValueError, ex: return '0' ''' return self.dec2bin((int(self)+int(other))) def __subtract__(self, other): '''subtraction: implementation of the - operator''' # Optional pass def __lt__(self, other): '''less: implementation of the < operator''' #pass if int(a) <= int(b): return True else: return False def __gt__(self, other): '''great: implementation of the > operator''' #pass if int(a) >= int(b): return True else: return False def __eq__(self, other): '''equal: implementation of the = operator''' #pass if int(a) == int(b): return True else: return False if __name__ == '__main__': a = BinaryNumber('01111001') # binary numbers must start with 0 b = BinaryNumber('00110101') # a = 121, b = 53 print 'a = ', a, int(a) print 'b = ', b, int(b) print 'a + b', a + b, int(a + b) #print 'a - b', a - b, int(a - b) # optional print 'a < b', a < b print 'a > b', a > b print 'a == b', a == b print 'Press the Enter key to finish' raw_input() If you couldn't tell, subtraction is optional and I commented the code i tried to change to get to work. Any advice on what I could do to fix this. Oct 15 '10 #3
Expert Mod 2.5K+ P: 2,851 Function dec2bin() would be a class method. Check out this thread. Oct 16 '10 #4
P: 32 Alright. After it is added together, I need my output to show as " a + b (binary result) (decimal result). How would I make it add them together and have it show that as the result of the two? Oct 16 '10 #5
Expert Mod 2.5K+ P: 2,851 Modify the string returned by __add__ and use string formatting. Untested: Expand|Select|Wrap|Line Numbers return "%s + %s (%s) (%s)" % (self, other, self.dec2bin((int(self)+int(other))), (int(self)+int(other))) Oct 16 '10 #6
P: 32 I am still a little confused. All I need to do is add 2 binary numbers together and have the output be " a + b (binary result) (decimal result) ". The conversion was to just give me the decimal result. So how would I go about adding this all together to get the given output, " a + b (binary result) (decimal result) " ? If you don't understand what I am trying to ask please feel free to tell me. This was my original code for " __add__" without any additions from me: Expand|Select|Wrap|Line Numbers class BinaryNumber: '''A Binary Number Class''' #pass def __init__(self, bitString = ''): '''Constructor''' #pass self._bits = bitString def __str__(self): '''string representation''' if self._bits: return self._bits else: return '0' def __int__(self): '''conversion''' #pass total = 0 for pos in range(len(self._bits)): if self._bits[pos] == '1': total += 2 ** (len(self._bits) - 1 - pos) return total def __add__(self, other): '''addition: implementation of the + operator''' pass def __subtract__(self, other): '''subtraction: implementation of the - operator''' # Optional pass def __lt__(self, other): '''less: implementation of the < operator''' #pass if int(a) <= int(b): return True else: return False def __gt__(self, other): '''great: implementation of the > operator''' #pass if int(a) >= int(b): return True else: return False def __eq__(self, other): '''equal: implementation of the = operator''' #pass if int(a) == int(b): return True else: return False if __name__ == '__main__': a = BinaryNumber('01111001') # binary numbers must start with 0 b = BinaryNumber('00110101') # a = 121, b = 53 print 'a = ', a, int(a) print 'b = ', b, int(b) print 'a + b', a + b, int(a + b) #print 'a - b', a - b, int(a - b) # optional print 'a < b', a < b print 'a > b', a > b print 'a == b', a == b print 'Press the Enter key to finish' raw_input() Oct 16 '10 #7
Expert Mod 2.5K+ P: 2,851 See my first post in this thread. Oct 16 '10 #8
P: 32 Yes, but even when I put that in, I get an error, "AttributeError: BinaryNumber instance has no attribute 'dec2bin'". How exactly would I have to make this code to look to get it to add the binary numbers together correctly and have it give me the output, "a + b (binary result) (decimal result)" ? Oct 16 '10 #9
Expert Mod 2.5K+ P: 2,851 You must add a BinaryNumber method dec2bin(). The method must return the decimal number converted to binary. Oct 16 '10 #10
P: 32 Is this the only way to do it? Or is there another way possible? If not, how would I go about doing this? Oct 16 '10 #11
Expert Mod 2.5K+ P: 2,851 That is the way I would do it, because that's the easiest for me. There are other ways though. You can write the code to do binary addition and subtraction in methods __add__() and __sub__() respectively. Check out this website. Oct 16 '10 #12
P: 32 I understand the concepts of binary addition, but I just have trouble putting it into code. How would I write the code for binary addition in the method __add__? Oct 16 '10 #13
Expert Mod 2.5K+ P: 2,851 I am sorry, but I cannot write the code for you. Please show some effort on your own, then we can help. Since you understand the concepts of binary addition, you may know more than me about it. Even if you do the binary addition, I think the code should go into a separate function. Oct 18 '10 #14
Expert 100+ P: 391 Hi It seems to me that, since you're implementing your own version of this, you might as well do the addition in binary. Your bitString is just a string of 1s and 0s, if I'm not mistaken? So, I would first implement something that returns a single digit in a given position. Something like: Expand|Select|Wrap|Line Numbers def binDigit[self,position]: if position>len(self._bits): return 0 else: return self._bits[-position] if self._bits="101", then position[1] = 1, position[2]=0, position[3]=1 and position[4]=0 etc. Now create something that does the addition bit by bit (literally!): Expand|Select|Wrap|Line Numbers def __add__[self,other]: v=max([len(self._bits),len(other._bits)])+1 ans='' carry=0 for i in range(1,v+1): a=self.position(i) b=self.position(i) t=carry+int(a)+int(b) if t<=1: carry=0 else: carry=1 if t%2==0: ans='0'+ans else: ans='1'+ans return BinaryNumber(ans) I haven't checked any of this, but it should give you some idea. Oct 25 '10 #15 | 3,297 | 9,884 | {"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-2019-51 | latest | en | 0.470099 |
https://www.chemeurope.com/en/news/155747/exploring-the-physics-of-a-chocolate-fountain.html | 1,642,742,868,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320302723.60/warc/CC-MAIN-20220121040956-20220121070956-00315.warc.gz | 692,385,856 | 20,053 | 26-Nov-2015 - Institute of Physics
# Exploring the physics of a chocolate fountain
A mathematics student has worked out the secrets of how chocolate behaves in a chocolate fountain, answering the age-old question of why the falling 'curtain' of chocolate surprisingly pulls inwards rather than going straight downwards.
"Chocolate fountains are just cool, aren't they!" says Adam Townsend, an author on the paper, based on his MSci project. "But it's also nice that they're models of some very important aspects of fluid dynamics."
The conundrum of the converging curtain was solved by looking at some classic work on 'water bells'.
"You can build a water bell really easily in your kitchen" says Dr Helen Wilson, the other author of the paper, and supervisor during Townsend's MSci project. "Just fix a pen vertically under a tap with a 10p coin flat on top and you'll see a beautiful bell-shaped fountain of water."
The physics of the water bell is exactly the same as the falling curtain of chocolate; and the reason the chocolate falls inwards turns out to be primarily surface tension.
They also looked at the flow up the pipe to the top of the fountain, and the flow over the plastic tiers that form the distinctive chocolate fountain shape.
"Both the chocolate fountain and water bell experiments are surprisingly simple to perform." Dr Wilson continues. "However they allow us to demonstrate several aspects of fluid dynamics, both Newtonian and non-Newtonian."
The researchers were also pleased to see that their work allowed them to engage with the public.
"It's serious maths applied to a fun problem." continues Adam Townsend. "I've been talking about it at mathematics enrichment events around London for the last few years. If I can convince just one person that maths is more than Pythagoras' Theorem, I'll have succeeded. Of course, the same mathematics has a wide use in many other important industries - but none of them are quite as tasty as chocolate."
Townsend and Wilson don't consider the chocolate fountain licked; there is lots more to learn from looking at the way the curtain changes over time.
"This was only an undergraduate project - modelling the chocolate fountain completely would need a lot more detail. Thankfully, individual strands - like the screw-pump flow up the pipe - have applications well beyond chocolate, and international teams are working on them now." concludes Wilson.
Townsend is now finishing a PhD investigating suspensions of solid particles in fluids, supervised by Wilson in the Department of Mathematics at University College London.
### Institute of Physics
Recommend news PDF version / Print Add news to watchlist
Share on
Facts, background information, dossiers
More about Institute of Physics
• News
Graphene microphone outperforms traditional nickel and offers ultrasonic reach
The researchers, based at the University of Belgrade, Serbia, created a vibrating membrane - the part of a condenser microphone which converts the sound to a current - from graphene, and were able to show up to 15 dB higher sensitivity compared to a commercial microphone, at frequencies up ... more
Using magnetic permeability to store information
Scientists have made promising steps in developing a new magnetic memory technology, which is far less susceptible to corruption by magnetic fields or thermal exposure than conventional memory. The findings, which report the use of magnetic permeability - how easily a magnetic field will ma ... more
'RoboClam' hits new depths as robotic digger
A digging robot inspired by the unique mechanisms employed by the Atlantic razor clam has been created by a group of researchers in the US. The robot, dubbed RoboClam, is able to dig with extreme efficiency by transforming the surrounding soil from a solid into a liquid, and could have a va ... more | 761 | 3,858 | {"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-2022-05 | latest | en | 0.969419 |
https://fpnotebook.com/Lung/Lab/PkExprtryFlwRt.htm | 1,708,535,580,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473524.88/warc/CC-MAIN-20240221170215-20240221200215-00075.warc.gz | 273,774,187 | 7,661 | #### II. Calculation: Estimated PEFR in Children
1. PEFR = (Ht - 100) x5 + 100
2. Where
1. Ht is standing height in centimeters
2. PEFR is Peak Expiratory Flow in liters per minute
3. References
1. Balasubramanian (2002) Indian Pediatr 39:104-6 [PubMed]
#### III. Calculation: Estimated PEFR in Adult Females
1. PEFR = (((Height x3.72) +2.24) - (Age x 0.03)) x 60
2. Where Height in meters or (inches x 0.0254)
#### IV. Calculation: Estimated PEFR in Adult Males
1. PEFR = (((Height x5.48) +1.58) - (Age x 0.041)) x 60
2. Where Height in meters or (inches x 0.0254)9
#### V. Calculation: Estimated Peak Flow in children based on height
1. Height 43 inches: 147 L/m
2. Height 44 inches: 160 L/m
3. Height 45 inches: 173 L/m
4. Height 46 inches: 187 L/m
5. Height 47 inches: 200 L/m
6. Height 48 inches: 214 L/m
7. Height 49 inches: 227 L/m
8. Height 50 inches: 240 L/m
9. Height 51 inches: 254 L/m
10. Height 52 inches: 267 L/m
11. Height 53 inches: 280 L/m
12. Height 54 inches: 293 L/m
13. Height 55 inches: 307 L/m
14. Height 56 inches: 320 L/m
15. Height 57 inches: 334 L/m
16. Height 58 inches: 347 L/m
17. Height 59 inches: 360 L/m
18. Height 60 inches: 373 L/m
19. Height 61 inches: 387 L/m
20. Height 62 inches: 400 L/m
21. Height 63 inches: 413 L/m
22. Height 64 inches: 427 L/m
23. Height 65 inches: 440 L/m
24. Height 66 inches: 454 L/m
25. Height 67 inches: 467 L/m
26. References
1. Polgar (1971) PFTs in Children, Saunders
#### VI. Calculation: Estimated Peak Flow in Men based on Age and height
1. Height 65 inches (165 cm)
1. Age 25 years: 520 L/m
2. Age 30 years: 510 L/m
3. Age 40 years: 489 L/m
4. Age 50 years: 468 L/m
5. Age 60 years: 447 L/m
2. Height 70 inches (178 cm)
1. Age 25 years: 592 L/m
2. Age 30 years: 581 L/m
3. Age 40 years: 561 L/m
4. Age 50 years: 543 L/m
5. Age 60 years: 519 L/m
3. Height 75 inches (190 cm)
1. Age 25 years: 664 L/m
2. Age 30 years: 653 L/m
3. Age 40 years: 632 L/m
4. Age 50 years: 611 L/m
5. Age 60 years: 590 L/m
#### VII. Calculation: Estimated Peak Flow in Women based on Age and height
1. Height 60 inches (152 cm)
1. Age 25 years: 365 L/m
2. Age 30 years: 357 L/m
3. Age 40 years: 342 L/m
4. Age 50 years: 327 L/m
5. Age 60 years: 312 L/m
2. Height 65 inches (165 cm)
1. Age 25 years: 401 L/m
2. Age 30 years: 394 L/m
3. Age 40 years: 379 L/m
4. Age 50 years: 364 L/m
5. Age 60 years: 349 L/m
3. Height 70 inches (178 cm)
1. Age 25 years: 439 L/m
2. Age 30 years: 431 L/m
3. Age 40 years: 416 L/m
4. Age 50 years: 401 L/m
5. Age 60 years: 386 L/m | 995 | 2,519 | {"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-2024-10 | latest | en | 0.562617 |
https://lingpipe-blog.com/2009/08/26/diagnostic-precision-from-sensitivity-specificity-and-prevalence/?replytocom=5342 | 1,638,946,921,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363445.41/warc/CC-MAIN-20211208053135-20211208083135-00374.warc.gz | 438,205,398 | 23,357 | ## Diagnostic Precision from Sensitivity, Specificity and Prevalence
Suppose you have a diagnostic for your favorite binary condition (YFBC). For the sake of concreteness, let’s consider the mammogram test for breast cancer. The American College of Preventive Medicine provides an estimate of between 0.75 and 0.90 sensitivity and 0.9 to 0.95 specificity. Let’s give the mammogram the benefit of the doubt and assume the high end, with sensitivity of 0.90 and specificity of 0.95.
### Sensitivity
Recall that sensitivity is the accuracy of the test on women who have breast cancer. In other words, if a woman who has breast cancer gets a mammorgram, the test will come back positive 90% of the time because the test has a 0.9 sensitivity. Cases where the condition is present and the test is positive are called true positives (TP). For 10% of the women with breast cancer, the mammogram test result will be a false negative (FN), where the test misdiagnoses a woman who has the disease.
### Specificity
Specificity is the accuracy of the mammogram for women who do not have breast cancer. That is, if a woman does not have breast cancer, the test will be negative 95% of the time. If a woman does not have breast cancer and the test results are negative, the outcome is called a true negative (TN). The other 5% of the time we get a false positive, which means the test is positive for a woman who does not have breast cancer.
### Prevalence
Before we can calculate diagnostic precision, we need to know the prevalence of the condition, which is the percentage of the population that has the condition. We rarely know population prevalences, but they can be estimated. For breast cancer, the following paper provides estimates of prevalence of breast cancer, which increase with age:
Consulting the chart above, the prevalence of breast cancer is roughly 0.015 for women aged 50.
### Precision (aka Positive Predictive Value)
The precision of a test, also known as its positive predictive value, is the chance that a subject testing positive actually has the condition being tested for. Precision is thus defined as TP/(TP+FP).
### My Mammogram is Positive, Do I Have Breast Cancer?
Now let’s suppose you’re a 50-year old woman whose mammogram was positive. What’s the chance you have breast cancer? Let’s calculate all the probabilities.
• p(TP) = prevalence * sensitivity = 0.015 * 0.9 = 0.014
• p(FN) = prevalence * (1 – sensitivity) = 0.0015
• p(TN) = (1 – prevalence) * specificity = 0.97 * 0.95 = 0.94
• p(FP) = (1 – prevalence) * (1 – specificty) = 0.049
Thus a generous estimate (remember we took the upper bounds for the accuracy estimates) of the chance of a 50-year old having cancer given that she tests positive on a mammogram is only:
p(TP)/(p(TP)+p(FP))=0.21
Now consider a 40-year old woman getting a mammogram. The prevalence is only .005 (less than half a percent), leading to a test precision of 0.083, meaning only 1 in 12 40-year old women with a positive mammogram will actually have breast cancer.
If the sensitivity and specificity are on the low end of the estimate, namely sensitivity is 0.7 and specificity 0.90, precision for 50-year old subjects is only .096 (less than 1 in 10), and that for 40-year old subjects 0.033, or about 1/30.
That’s why you only start getting diagnostic tests when you’re older — they’re just too inaccurate for the young and likely healthy. Or if you’re known to be more likely to get breast cancer, for instance because of incidence in the family or genetic tests; the prevalence for your situation will determine the test’s predictive accuracy.
The U.S. National Cancer Institute recommends “Women in their 40s and older should get a mammogram every 1 to 2 years.” That’s a whole lot of chances for false positives.
Low positive predictive accuracy is also why there are follow-up imaging tests (sonograms) even before biopsies. But keep in mind that the same caveats apply to those imperfect tests.
For more information on other tests (sonogram, MRI, physical exam and various biopsy procedures), do a PubMed Search. This article from the National Cancer Institute looks interesting:
In a survey of 50+ facilities between 1996 and 2002, they found varying values for sensitivity and specificity among facilities and by procedure (specialist in breast cancer, reading, double versus single reading, etc.), and an overall positive predictive value of mammograms of only 4% (meaning 1/25 women who test positive actually have breast cancer).
### 4 Responses to “Diagnostic Precision from Sensitivity, Specificity and Prevalence”
1. Ken Says:
This is fine, provided the threshold for a positive is chosen (if possible) to give sensitivity and specificity that result in sensible outcomes. Missing breast cancer has serious consequences but a false positive simply results in a biopsy which can be easily done, so it is not a problem. This is something that people are researching in other areas, the consequences of each decision, but it requires decent data on what happens in each of the four possible outcomes.
One problem is that many diagnostic tests are designed to be used after some other symptom is found, not as screening tools. A headache without other explanation and with certain symptoms deserves a scan, and anything found can be a possible cause. No symptoms and anything found is probably unimportant, but now there is a dilemma, should they have a look or just wait and see what happens. Not a lot of data on this but neurosurgery can have severe consequences so nobody should do anything unless there is an obvious defect.
2. lingpipe Says:
Great points. Indeed, you often only get tests when there is some reason to believe you fall in a high-prevalence group. Much of the epidemiology literature’s concerned with multiple tests, which requires estimating correlation among the tests to derive sensible final estimates.
Mammograms are screening tests, at least in the U.S. if you’re following NCI’s advice. Even simple needle biopsies are not completely without infection risk and local anesthetic risk. Of course, surgical biopsies are more dangerous, but they also have higher sensitivity. The major risk seems to be false positives resulting in needless major surgery; I have no idea how big a risk that is for breast biopsies.
3. lingpipe Says:
Why am I not surprised by a new set of recommendations from the United States Preventive Services Task Force. Other groups are sticking with a recommendation of 40. There’s more info in:
In the calculations above, we were being fairly generous with diagnostic accuracy estimates.
These numbers need to be constantly updated due to changes in sensitivity and specificity of the test(s).
In a fuller model, the exact age at which it makes sense to be tested will also vary based on predictive features of the patients, such as genetic profiles. These would change the prevalence in the above calculations, and thus the received diagnostic precision for patients with a given set of predictors (e.g. age, genetic profile, health history).
4. Owen M. Says:
Starting a site much like this one got me to get into some research and I found your post to be extremely helpful. My site is centered around the idea curing cancer by stopping the angiogenic process. I wish you good luck with your research in the future and you can be sure I’ll be following it. | 1,638 | 7,411 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2021-49 | longest | en | 0.926338 |
http://html.rhhz.net/jckxjsgw/html/50623.htm | 1,709,521,740,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476409.38/warc/CC-MAIN-20240304002142-20240304032142-00325.warc.gz | 21,310,824 | 14,661 | 水下滑翔机附加质量数值计算
舰船科学技术 2016, Vol. 38 Issue (12): 116-120,134 PDF
Numerical method for added mass of an underwater glider
YANG Lei, CAO Jun-jun, YAO Bao-heng, ZENG Zheng, LIAN Lian
Naval Architecture and Ocean Engineering National Laboratory, Shanghai Jiaotong University, Shanghai 200240, China
Abstract: Added mass of an underwater glider is quite important for the motions of glider. In this paper, the added mass of an arbitrary three-dimensional body is obtained through Hess-Smith method. Then an underwater glider which was designed by our laboratory is meshed by Gambit software in order to obtain its added mass. Besides, the Planar Motion Mechanism (PMM) tests of the glider are simulated by using CFD software, dynamic mesh technique and UDF. By comparing with the Hess-Smith results, the characters and advantages of Hess-Smith method and PMM are analyzed.
Key words: added Mass underwater glider Hess-Smith PMM
0 引言
1 基于面元法的附加质量计算 1.1 基本方程和边界条件
$\Phi =x{{V}_{\infty x}}+y{{V}_{\infty y}}+z{{V}_{\infty z}}+\varphi ,$ (1)
${{\nabla }^{2}}\varphi =0,$ (2)
$\frac{\partial \varphi }{\partial n}=-V\cdot n,$ (3)
$\varphi =0,$ (4)
1.2 速度势离散
$\varphi (p)=\iint\limits_{s}{\frac{\sigma (q)}{{{r}_{pq}}}}\text{d}{{s}_{q}},$ (5)
$2\pi \sigma (p)+\iint\limits_{s-\varepsilon }{\sigma }\left( q \right)\frac{\partial }{\partial {{n}_{p}}}\left( \frac{1}{{{r}_{pq}}} \right)\text{d}{{s}_{q}}=-V\cdot n,$ (6)
$\iint\limits_{s}{\sigma }\left( q \right)\frac{\partial }{\partial {{n}_{p}}}\left( \frac{1}{{{r}_{pq}}} \right)\text{d}{{s}_{q}}\approx \sum\limits_{j=1}^{N}{{{\sigma }_{j}}\iint\limits_{\Delta {{Q}_{j}}}{\frac{\partial }{\partial {{n}_{p}}}}\left( \frac{1}{{{r}_{pq}}} \right)}\text{d}{{s}_{q}},$ (7)
$2\pi \sigma (p)+\sum\limits_{j=1}^{N}{{{\sigma }_{j}}\iint\limits_{\Delta {{Q}_{j}}}{\frac{\partial }{\partial {{n}_{p}}}}\left( \frac{1}{{{r}_{pq}}} \right)}\text{d}{{s}_{q}}=-V\cdot n$ (8)
图 1 水下滑翔机三角形网格划分 Fig. 1 The generation of triangular mesh for an underwater glider
1.3 附加质量求解
${{\lambda }_{ij}}={{\lambda }_{ji}}$ (9)
$\lambda = \left[{\begin{array}{*{20}{c}} {{\lambda _{11}}}&0&0&0&0&0\\[8pt] 0&{{\lambda _{22}}}&0&{{\lambda _{24}}}&0&{{\lambda _{26}}}\\[8pt] 0&0&{{\lambda _{33}}}&0&{{\lambda _{35}}}&0\\[8pt] 0&{{\lambda _{42}}}&0&{{\lambda _{44}}}&0&{{\lambda _{46}}}\\[8pt] 0&0&{{\lambda _{53}}}&0&{{\lambda _{55}}}&0\\[8pt] 0&{{\lambda _{62}}}&0&{{\lambda _{64}}}&0&{{\lambda _{66}}} \end{array}} \right] = \left[{\begin{array}{*{20}{c}} {\frac{1}{2}\rho {L^3}{{X'}_{\dot u}}}&0&0&0&0&0\\[8pt] 0&{\frac{1}{2}\rho {L^3}{{Y'}_{\dot v}}}&0&{\frac{1}{2}\rho {L^4}{{K'}_{\dot v}}}&0&{\frac{1}{2}\rho {L^4}{{N'}_{\dot v}}}\\[8pt] 0&0&{\frac{1}{2}\rho {L^3}{{Z'}_{\dot w}}}&0&{\frac{1}{2}\rho {L^4}{{M'}_{\dot w}}}&0\\[8pt] 0&{\frac{1}{2}\rho {L^4}{{Y'}_{\dot p}}}&0&{\frac{1}{2}\rho {L^5}{{K'}_{\dot p}}}&0&{\frac{1}{2}\rho {L^5}{{N'}_{\dot p}}}\\[8pt] 0&0&{\frac{1}{2}\rho {L^4}{{Z'}_{\dot q}}}&0&{\frac{1}{2}\rho {L^5}{{M'}_{\dot q}}}&0\\[8pt] 0&{\frac{1}{2}\rho {L^4}{{Y'}_{\dot r}}}&0&{\frac{1}{2}\rho {L^5}{{K'}_{\dot r}}}&0&{\frac{1}{2}\rho {L^5}{{N'}_{\dot r}}} \end{array}} \right]{\text {。}}\!\!$ (10)
${{\lambda }_{ij}}=-\rho \text{ }\iint\limits_{s}{{{\varphi }_{i}}}\frac{\partial {{\varphi }_{j}}}{\partial n}\text{d}s=-\rho \iint\limits_{s}{{{\varphi }_{i}}}{{n}_{j}}\text{d}s$ (11)
2 基于 CFD 软件的附加质量计算
Fluent 软件可提供动网格技术，动网格模型可以用来模拟流程形状由于边界运动而随时间改变的问题，利用 UDF 可以定义边界的运动方式。本文则是对水下滑翔机进行 PMM 试验模拟，从而得到附加质量。
2.1 计算方程
$\left\{ \begin{array}{*{35}{l}} \zeta =a\sin (\omega t), \\ \theta =\dot{\theta }=0, \\ w=\dot{\zeta }=aw\cos (\omega t), \\ \dot{w}=\ddot{\zeta }=-a{{w}^{2}}\sin (\omega t). \\ \end{array} \right.$ (12)
图 2 水下滑翔机纯升沉运动示意图 Fig. 2 The heaving motion of an underwater glider
$\left\{ \begin{array}{*{35}{l}} Z={{Z}_{{\dot{w}}}}\dot{w}+{{Z}_{w}}w+{{Z}_{0}}= \\ -a{{\omega }^{2}}{{Z}_{{\dot{w}}}}\sin (\omega t)+a\omega {{Z}_{w}}\cos (\omega t)+{{Z}_{0}}, \\ M={{M}_{{\dot{w}}}}\dot{w}+{{M}_{w}}w+{{M}_{0}}= \\ -a{{\omega }^{2}}{{M}_{{\dot{w}}}}\sin (\omega t)+a\omega {{M}_{w}}\cos (\omega t)+{{M}_{0}}. \\ \end{array} \right.$ (13)
$\left\{ \begin{array}{*{35}{l}} \zeta =a\sin (\omega t), \\ w=\dot{\zeta }=aw\cos (\omega t), \\ \dot{w}=\ddot{\zeta }=-a{{w}^{2}}\sin (\omega t), \\ \theta ={{\theta }_{0}}\cos (\omega t), \\ q=\dot{\theta }=-\omega {{\theta }_{0}}\sin (\omega t), \\ \dot{q}=\ddot{\theta }=-{{\omega }^{2}}{{\theta }_{0}}\cos (\omega t). \\ \end{array} \right.$ (14)
$\left\{ \begin{array}{*{35}{l}} \theta \approx \tan \theta =\frac{w}{U}=\frac{a\omega }{U}\cos (\omega t)={{\theta }_{0}}\cos (\omega t), \\ {{\theta }_{0}}=\frac{a\omega }{U}. \\ \end{array} \right.$ (15)
图 3 水下滑翔机纯俯仰运动示意图 Fig. 3 The pitching motion of an underwater glider
$\left\{ \begin{array}{*{35}{l}} Z={{Z}_{{\dot{q}}}}\dot{q}+{{Z}_{q}}q+{{Z}_{0}}= \\ -{{\theta }_{0}}{{\omega }^{2}}{{Z}_{{\dot{q}}}}\sin (\omega t)+{{\theta }_{0}}\omega {{Z}_{q}}\cos (\omega t)+{{Z}_{0}}, \\ M={{M}_{{\dot{q}}}}\dot{q}+{{M}_{q}}q+{{M}_{0}}= \\ -{{\theta }_{0}}{{\omega }^{2}}{{M}_{{\dot{q}}}}\sin (\omega t)+{{\theta }_{0}}\omega {{M}_{q}}\cos (\omega t)+{{M}_{0}}. \\ \end{array} \right.$ (16)
2.2 CFD 计算准备
图 4 计算域内域和外域网格划分 Fig. 4 The mesh of internal and external flow domain
2.3 结果处理与分析
图 5 纯升沉运动 f = 0.4 时垂向力 Z 的变化 Fig. 5 The changes of force in Z-direction when the frequency of heaving motion is 0.4
图 6 纯升沉运动 f = 0.4 时绕 Z 轴力矩的变化 Fig. 6 The changes of moment around Z-direction when the frequency of heaving motion is 0.4
Origin 软件具有自定义拟合函数的功能，利用式（13）编写出相应的拟合函数，从而得到高精度的拟合结果。
f = 0.4 时，得到拟合力和力矩的曲线函数为：
$\left\{ \begin{array}{*{35}{l}} Z={{Z}_{{\dot{w}}}}\dot{w}+{{Z}_{w}}w+{{Z}_{0}}= \\ 26.037\sin (\omega t)-16.627\cos (\omega t)-0.116, \\ M={{M}_{{\dot{w}}}}\dot{w}+{{M}_{w}}w+{{M}_{0}}= \\ 0.313\sin (\omega t)-2.065\cos (\omega t)-0.002. \\ \end{array} \right.$
$\left\{ \begin{array}{*{35}{l}} {{Z}_{a}}=-\frac{a{{\omega }^{2}}L}{{{V}^{2}}}{{{{Z}'}}_{w}}, \\ {{M}_{a}}=-\frac{a{{\omega }^{2}}L}{{{V}^{2}}}{{{{M}'}}_{w}}. \\ \end{array} \right.$ (17)
图 7 利用最小二乘法计算加速度系数 Fig. 7 The acceleration coefficients applying the least square method
3 结 语
1）对于水下滑翔机，Hess-Smith 面元法程序和 Fluent 仿真得到的附加质量结果均比较精确，满足工程需求。
2）相比于 Fluent 软件仿真需要花费较多时间多次建模与计算，Hess-Smith 方法仅需要重新划分物体的面网格，节省大量时间，但编写程序较为复杂。
3）本文提供了计算水下滑翔机附加质量的 2 种方法，介绍了两者原理与优劣，同时得到精确的附加质量，验证了 2 种方法的可行性和准确性，对水下滑翔机的水动力设计和运动仿真具有重要指导意义和参考价值。
[1] ZHANG S W, YU J C. Spiraling motion of underwater gliders:modeling, analysis, and experimental results[J]. Ocean Engineering , 2013, 60 :1–13. DOI:10.1016/j.oceaneng.2012.12.023 [2] 孟凡豪. 50 kg级水下自航行器整体水动力学性能优化设计[D]. 杭州:中国计量学院, 2014:56-57. MENG Fan-hao. Shape optimization design of 50 kilograms autonomous underwater vehicle based on hydraulic characteristics[D]. Hangzhou:China Jiliang University, 2014:56-57. [3] 林超友, 朱军. 潜艇近海底航行附加质量数值计算[J]. 船舶工程 , 2003, 25 (1) :26–29. LIN Chao-you, ZHU Jun. Numerical computation of added mass of submarine maneuvering with small clearance to sea-bottom[J]. Ship Engineering , 2003, 25 (1) :26–29. [4] 杨勇, 邹早建, 张晨曦. 深浅水中KVLCC船体横荡运动水动力数值计算[J]. 水动力学研究与进展 , 2011, 26 (1) :85–93. YANG Yong, ZOU Zao-jian, ZHANG Chen-xi. Calculation of hydrodynamic forces on a KVLCC hull in sway motion in deep and shallow water[J]. Chinese Journal of Hydrodynamics , 2011, 26 (1) :85–93. [5] 孙铭泽, 王永生, 张志宏, 等. 基于网格变形技术的全附体潜艇操纵性计算[J]. 武汉理工大学学报(交通科学与工程版) , 2013, 37 (2) :420–424. SUN Ming-ze, WANG Yong-sheng, ZHANG Zhi-hong, et al. Numerical simulation of submarine maneuverability based on mesh deformation technology[J]. Journal of Wuhan University of Technology (Transportation Science&Engineering) , 2013, 37 (2) :420–424. [6] 黄劲, 余志毅, 刘朝勋, 等. 基于CFD的两栖车辆水动力导数的计算方法[J]. 车辆与动力技术 , 2014 (2) :39–43. HUANG Jin, YU Zhi-yi, LIU Chao-xun, et al. Numerical method for hydrodynamic coefficients of an amphibious vehicle based on CFD[J]. Vehicle&Power Technology , 2014 (2) :39–43. [7] 马骋, 连琏. 水下运载器操纵控制及模拟仿真技术[M]. 北京: 国防工业出版社, 2009 . MA Cheng, LIAN Lian. Maneuvering control and simulation technology of underwater vehicles[M]. Beijing: National Defense Industry Press, 2009 . [8] 张晓频. 多功能潜水器操纵性能与运动仿真研究[D]. 哈尔滨:哈尔滨工程大学, 2008:25-26. ZHANG Xiao-pin. Research on maneuverability and motion simulation of multifunction vehicle[D]. Harbin:Harbin Engineering University, 2008:25-26. | 3,693 | 8,285 | {"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} | 2.859375 | 3 | CC-MAIN-2024-10 | latest | en | 0.588141 |
https://forum.thethirdmanifesto.com/forum/topic/which-type/?part=9 | 1,686,320,603,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224656737.96/warc/CC-MAIN-20230609132648-20230609162648-00710.warc.gz | 293,134,232 | 14,572 | ## The Forum for Discussion about The Third Manifesto and Related Matters
Forum breadcrumbs - You are here:
Please or Register to create posts and topics.
# Which type?
PreviousPage 9 of 9
So we ended up arguing about whether "values have type" in the other topic I started about if "function/operator(s) can return only one result".
I thought of an example that might (but I rather suspect not) help on this topic, and I thought it should go in the original topic.
It is not known ifย `๐๐` is irrational.
We know that (regardless of how we represent it) `๐๐` is a real number. We know, in short, that it is "a thing".
Therefore I cannotย - unless I have some newly found mathematical proof (and hopefully some nice big mathematical prize)) to hand - correctly say
`๐๐:irrational`
Now, I'm sure someone will say that `๐๐` is ofย (or at if you prefer) (most specific) typeย `real` and hence does not contradict the "all (scalar) values carry with them (at least one) type" claim.
If we discover that `๐๐` is (or is not) `irrational`, does that change what the value `๐๐` is? Is it's type (conceptually or otherwise) part of its value?
BTW, I'm not asking just to be a pain, I (think) I genuinely don't fully understand the "carry with them" concept of the statement "values shall always carry with them, at least conceptually, some identification of the type to which they belong".
I understand the "value at type" idea - a value being the current value of a variable (or inout/output from an operator, or indeed, the value of an attribute in a relational tuple) that is constrained to be a given (named) set of values. The given name being the value in the phrase "value at type".ย I just don't get what it (or "carry with") means when a value appears as-is/standalone. Where is the information held of what the type of a value is?ย A "catalog" relation? But then, that is just a set, and if a type is more than just a set, what information makesย it more than just a set? If that information is "well that set is used in constraints and is the set of input/output values for some operators", fine. But that is quite a wooly definition for something (that should be, if it is in your foundation) claimed to be axiom concept.
If I create a new value, which I uniquely represent by `*Y!` but I fail to give it a type, then someone would say, well, it is of type `notBool` at the very least - hence it does not contradict the "all values carry with them (at least one) type" claim.
If I then say, oh, I've just discovered that `*Y!` is a real, and I add it to the set of (representable) reals, so it's (most specific) type is now `real`, would you say that is not the same value that I started with?
If I say, oh, I've just discover that George Boole was wrong, and there are actually three true values, and `*Y!` is the third truth value...ย well, permission to shoot me frankly. :-)
PreviousPage 9 of 9 | 773 | 2,962 | {"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-2023-23 | longest | en | 0.919587 |
http://color.inria.fr/doc/CoLoR.Util.Relation.RelUtil.html | 1,579,969,723,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251678287.60/warc/CC-MAIN-20200125161753-20200125190753-00420.warc.gz | 39,993,169 | 15,641 | # Library CoLoR.Util.Relation.RelUtil
CoLoR, a Coq library on rewriting and termination. See the COPYRIGHTS and LICENSE files.
• Frederic Blanqui, 2005-02-17
• Adam Koprowski and Hans Zantema, 2007-03
• Joerg Endrullis and Dimitri Hendriks, 2008-07
General definitions and results about relations.
Set Implicit Arguments.
Notations for some relations and operations on relations.
Notation rel := relation.
Notation incl := inclusion.
Infix "<<" := incl (at level 50) : relation_scope.
Notation same := same_relation.
Infix "==" := same (at level 70).
Infix "U" := union (at level 45) : relation_scope.
Notation "x #" := (clos_refl_trans x) (at level 35) : relation_scope.
Notation "x !" := (clos_trans x) (at level 35) : relation_scope.
Open Scope relation_scope.
Properties of incl.
Instance incl_refl A : Reflexive (@incl A).
Ltac incl_refl := apply incl_refl.
Instance incl_trans A : Transitive (@incl A).
Ltac incl_trans S := apply incl_trans with S; try incl_refl.
Instance incl_same A : Proper (same ==> same ==> impl) (@incl A).
Lemma incl_elim A (R S : rel A) : R << S -> forall x y, R x y -> S x y.
Properties of same.
Lemma rel_eq A (R S : rel A) : R == S <-> forall x y, R x y <-> S x y.
Instance same_equiv A : Equivalence (@same A).
Basic meta-theorems.
Lemma eq_incl_refl_rel A (R : rel A) : Reflexive R -> eq << R.
Lemma sym A (R : rel A) x y : Symmetric R -> (R x y <-> R y x).
Morphisms wrt incl and same.
Instance refl_same A : Proper (same ==> impl) (@Reflexive A).
Instance sym_same A : Proper (same ==> impl) (@Symmetric A).
Instance trans_same A : Proper (same ==> impl) (@Transitive A).
Instance equiv_same A : Proper (same ==> impl) (@Equivalence A).
Properties of Proper.
Lemma prop2_incl A B C (f : A -> B -> C) :
Proper (incl --> incl --> incl ==> impl)
(fun R S T => Proper (R ==> S ==> T) f).
Lemma prop3_incl A B C D (f : A -> B -> C -> D) :
Proper (incl --> incl --> incl --> incl ==> impl)
(fun R S T V => Proper (R ==> S ==> T ==> V) f).
Lemma prop4_incl A B C D E (f : A -> B -> C -> D -> E) :
Proper (incl --> incl --> incl --> incl --> incl ==> impl)
(fun R S T V W => Proper (R ==> S ==> T ==> V ==> W) f).
Instance prop_rel_same A (E : rel A) :
Proper (same ==> impl) (Proper (E ==> E ==> impl)).
Lemma prop_subrel A (R S : rel A) :
Equivalence S -> subrelation R S -> Proper (R ==> R ==> impl) S.
Lemma prop_rel_sym A B (f : A -> B -> Prop) R S :
Symmetric R -> Symmetric S
-> Proper (R ==> S ==> impl) f -> Proper (R ==> S ==> iff) f.
Instance equiv_prop A (R : rel A) :
Transitive R -> Symmetric R -> Proper (R ==> R ==> impl) R.
Empty relation.
Definition empty_rel {A} : rel A := fun x y => False.
Relation associated to a boolean function.
Definition rel_of_bool A (f : A->A->bool) : rel A :=
fun x y => f x y = true.
Boolean function associated to a decidable relation.
Section bool_of_rel.
Variables (A : Type) (R : rel A) (R_dec : rel_dec R).
Definition bool_of_rel t u := if R_dec t u then true else false.
Lemma bool_of_rel_true x y : bool_of_rel x y = true <-> R x y.
Lemma bool_of_rel_false x y : bool_of_rel x y = false <-> ~R x y.
Lemma bool_of_rel_refl x : Reflexive R -> bool_of_rel x x = true.
Global Instance bool_of_rel_prop :
Symmetric R -> Transitive R -> Proper (R ==> R ==> Logic.eq) bool_of_rel.
End bool_of_rel.
Definition of some properties on relations.
Definition classic_left_total A B (R : A -> B -> Prop) :=
forall x, exists y, R x y.
Definition left_total A B (R : A -> B -> Prop) :=
forall x, {y | R x y}.
Definition functional A B (R : A -> B -> Prop) :=
forall x y z, R x y -> R x z -> y = z.
Definition finitely_branching A B (R : A -> B -> Prop) :=
forall x, {l | forall y, R x y <-> In y l}.
Definition irreflexive A (R : rel A) := forall x, ~R x x.
Definition asymmetric A (R : rel A) := forall x y, R x y -> ~R y x.
Predicate saying that f is an infinite sequence of R-steps.
Definition IS A (R : rel A) f := forall i, R (f i) (f (S i)).
Definition EIS A (R : rel A) := exists f, IS R f.
Definition NT A (R : rel A) x := exists f, f 0 = x /\ IS R f.
Predicate saying that f and g describe an infinite sequence of R-steps modulo E: for all i, f(i) E g(i) R f(i+1).
Definition ISMod A (E R : rel A) (f g : nat -> A) :=
forall i, E (f i) (g i) /\ R (g i) (f (S i)).
Definition quasi_ordering A (R : rel A) := reflexive R /\ transitive R.
Definition ordering A (R : rel A) :=
reflexive R /\ transitive R /\ antisymmetric R.
Definition strict_ordering A (R : rel A) := irreflexive R /\ transitive R.
Strict part.
Definition strict_part A (R : rel A) : rel A := fun x y => R x y /\ ~R y x.
Intersection.
Definition intersection A (R S : rel A) : rel A := fun x y => R x y /\ S x y.
Lemma intersection_dec A (R S : rel A) (Rdec : rel_dec R) (Sdec : rel_dec S) :
rel_dec (intersection R S).
Finitely branching relations.
Section finitely_branching.
Variables (A : Type) (R : rel A) (FB : finitely_branching R).
Definition sons x := proj1_sig (FB x).
Lemma in_sons_R : forall x y, In y (sons x) -> R x y.
Lemma R_in_sons : forall x y, R x y -> In y (sons x).
End finitely_branching.
Equivalence relation associated to a PreOrder.
Definition inter_transp A (R : rel A) : rel A := fun x y => R x y /\ R y x.
Lemma inter_transp_refl A (R : rel A) :
Reflexive R -> Reflexive (inter_transp R).
Lemma inter_transp_sym A (R : rel A) : Symmetric (inter_transp R).
Lemma inter_transp_trans A (R : rel A) :
Transitive R -> Transitive (inter_transp R).
Lemma inter_transp_incl A (R : rel A) : inter_transp R << R.
Properties of IS.
Instance IS_incl A : Proper (incl ==> Logic.eq ==> impl) (@IS A).
Instance IS_same A : Proper (same ==> Logic.eq ==> impl) (@IS A).
Instance EIS_same A : Proper (same ==> impl) (@EIS A).
Lemma NT_IS_elt A (R : rel A) f k : IS R f -> NT R (f k).
Lemma red_NT A (R : rel A) t u : R t u -> NT R u -> NT R t.
Properties of irreflexive.
Monotony.
Definition monotone A B (R : rel A) (S : rel B) f :=
forall x y, R x y -> S (f x) (f y).
Lemma monotone_transp A B (R : rel A) (S : rel B) f :
monotone R S f -> monotone (transp R) (transp S) f.
Composition.
Definition compose A (R S : rel A) : rel A :=
fun x y => exists z, R x z /\ S z y.
Infix "@" := compose (at level 40) : relation_scope.
Instance compose_incl A : Proper (incl ==> incl ==> incl) (@compose A).
Ltac comp := apply compose_incl; try incl_refl.
Instance compose_same A : Proper (same ==> same ==> same) (@compose A).
Definition absorbs_right A (R S : rel A) := R @ S << R.
Definition absorbs_left A (R S : rel A) := S @ R << R.
Lemma comp_assoc A (R S T : rel A) : (R @ S) @ T << R @ (S @ T).
Lemma comp_assoc' A (R S T : rel A) : R @ (S @ T) << (R @ S) @ T.
Ltac assoc :=
match goal with
| |- (?s @ ?t) @ ?u << _ => incl_trans (s @ (t @ u)); try apply comp_assoc
| |- ?s @ (?t @ ?u) << _ => incl_trans ((s @ t) @ u); try apply comp_assoc'
| |- _ << (?s @ ?t) @ ?u => incl_trans (s @ (t @ u)); try apply comp_assoc'
| |- _ << ?s @ (?t @ ?u) => incl_trans ((s @ t) @ u); try apply comp_assoc
end.
Lemma absorbs_left_rtc A (R S : rel A) : R @ S << S -> R# @ S << S.
Lemma absorbs_right_rtc A (R S : rel A) : S @ R << S -> S @ R# << S.
Properties of union.
Instance union_incl A : Proper (incl ==> incl ==> incl) (@union A).
Ltac union := apply union_incl; try incl_refl.
Instance union_same A : Proper (same ==> same ==> same) (@union A).
Lemma union_commut A (R S : rel A) : R U S == S U R.
Lemma union_assoc A (R S T : rel A) : (R U S) U T == R U (S U T).
Lemma comp_union_l A (R S T : rel A) : (R U S) @ T == (R @ T) U (S @ T).
Lemma comp_union_r A (R S T : rel A) : T @ (R U S) == (T @ R) U (T @ S).
Lemma union_empty_r A (R : rel A) : R U empty_rel == R.
Lemma union_empty_l A (R : rel A) : empty_rel U R == R.
Lemma incl_union_l A (R S T : rel A) : R << S -> R << S U T.
Lemma incl_union_r A (R S T : rel A) : R << T -> R << S U T.
Lemma union_incl_eq A (R R' S : rel A) : R U R' << S <-> R << S /\ R' << S.
Reflexive closure.
Definition clos_refl A (R : rel A) : rel A := eq U R.
Notation "x %" := (clos_refl x) (at level 35) : relation_scope.
Instance rc_incl A : Proper (incl ==> incl) (@clos_refl A).
Instance rc_same A : Proper (same ==> same) (@clos_refl A).
Instance rc_refl A (R : rel A) : Reflexive (R%).
Instance rc_trans A (R : rel A) : Transitive R -> Transitive (R%).
Lemma incl_rc A (R : rel A) : R << R%.
Compatibility closure.
Section comp_clos.
Variables (A : Type) (E : rel A) (E_eq : Equivalence E).
Inductive comp_clos (R : rel A) : rel A :=
| comp_clos_intro u u' v v' : E u u' -> E v v' -> R u' v' -> comp_clos R u v.
Lemma comp_clos_intro_refl (R : rel A) t u : R t u -> comp_clos R t u.
Lemma comp_clos_eq R : same (comp_clos R) (E @ (R @ E)).
Lemma incl_comp_clos R : R << comp_clos R.
Global Instance comp_clos_incl : Proper (incl ==> incl) comp_clos.
Global Instance comp_clos_same : Proper (same ==> same) comp_clos.
Lemma comp_clos_union R S :
same (comp_clos (R U S)) (comp_clos R U comp_clos S).
Global Instance comp_clos_prop R : Proper (E ==> E ==> impl) (comp_clos R).
End comp_clos.
Properties of clos_trans.
Instance tc_trans A (R : rel A) : Transitive (R!).
Instance tc_incl A : Proper (incl ==> incl) (@clos_trans A).
Instance tc_same A : Proper (same ==> same) (@clos_trans A).
Lemma incl_tc A (R S : rel A) : R << S -> R << S!.
Lemma trans_tc_incl A (R : rel A) : Transitive R -> R! << R.
Lemma trans_comp_incl A (R : rel A) : Transitive R -> R @ R << R.
Lemma absorbs_left_tc A (R S : rel A) : R @ S << S -> R! @ S << S.
Lemma tc_absorbs_left A (R S : rel A) : R @ S << S -> R @ S! << S!.
Lemma trans_intro A (R : rel A) : R @ R << R <-> Transitive R.
Lemma comp_tc_idem A (R : rel A) : R! @ R! << R!.
Lemma tc_min A (R S : rel A) : R << S -> Transitive S -> R! << S.
Lemma trans_tc A (R : rel A) : Transitive R -> R! == R.
Lemma tc_idem A (R : rel A) : R!! == R!.
Lemma tc_eq A (R S : rel A) : S << R -> R << S! -> R! == S!.
Lemma tc_incl_trans A (R S : rel A) : Transitive S -> R << S -> R! << S.
Instance tc_prop A (E R : rel A) : Reflexive E ->
Proper (E ==> E ==> impl) R -> Proper (E ==> E ==> impl) (R!).
Lemma tc_prop_iff A (R E : rel A) : Equivalence E
-> Proper (E ==> E ==> iff) R -> Proper (E ==> E ==> iff) (clos_trans R).
Lemma tc_step_l A (R : rel A) x y :
R! x y -> R x y \/ (exists2 z, R x z & R! z y).
Lemma tc_step_r A (R : rel A) x y :
R! x y -> R x y \/ (exists2 z, R! x z & R z y).
Lemma tc_transp A (R : rel A) : transp (R!) == (transp R)!.
Symmetric closure of a relation.
Section clos_sym.
Variable (A : Type) (R : rel A).
Inductive clos_sym : rel A :=
| s_step : forall x y, R x y -> clos_sym x y
| s_sym : forall x y, clos_sym y x -> clos_sym x y.
Global Instance sc_sym : Symmetric clos_sym.
End clos_sym.
Reflexive, transitive and symmetric closure of a relation.
Section clos_equiv.
Variable (A : Type) (R : rel A).
Inductive clos_equiv : rel A :=
| e_step : forall x y, R x y -> clos_equiv x y
| e_refl : forall x, clos_equiv x x
| e_trans : forall x y z, clos_equiv x y -> clos_equiv y z -> clos_equiv x z
| e_sym : forall x y, clos_equiv x y -> clos_equiv y x.
Global Instance ec_equiv : Equivalence clos_equiv.
End clos_equiv.
Instance ec_incl A : Proper (incl ==> incl) (@clos_equiv A).
Lemma ec_min A (R S : rel A) : Equivalence S -> R << S -> clos_equiv R << S.
Properties of clos_refl_trans.
Instance rtc_refl A (R : rel A) : Reflexive (R#).
Instance rtc_trans A (R : rel A) : Transitive (R#).
Instance rtc_incl A : Proper (incl ==> incl) (@clos_refl_trans A).
Instance rtc_same A : Proper (same ==> same) (@clos_refl_trans A).
Lemma incl_rtc A (R : rel A) : R << R#.
Lemma tc_incl_rtc A (R : rel A) : R! << R#.
Lemma tc_split A (R : rel A) : R! << R @ R#.
Lemma rc_incl_rtc A (R : rel A) : R% << R#.
Lemma rtc_split A (R : rel A) : R# << eq U R!.
Lemma rtc_split_eq A (R : rel A) : R# == eq U R!.
Lemma rtc_split2 A (R : rel A) : R# << eq U R @ R#.
Lemma tc_split_inv A (R : rel A) : R# @ R << R!.
Lemma tc_merge A (R : rel A) : R @ R# << R!.
Lemma rtc_transp A (R : rel A) : transp (R#) << (transp R)#.
Lemma incl_rtc_rtc A (R S : rel A) : R << S# -> R# << S#.
Lemma comp_rtc_idem A (R : rel A) : R# @ R# << R#.
Lemma trans_rtc_incl A (R : rel A) : Reflexive R -> Transitive R -> R# << R.
Lemma rtc_invol A (R : rel A) : R # # == R #.
Lemma rtc_intro_seq A (R : rel A) f i : forall j, i <= j ->
(forall k, i <= k < j -> R (f k) (f (1+k))) -> R# (f i) (f j).
Lemma rtc_min A (R S : rel A) : PreOrder S -> R << S -> R# << S.
Instance rtc_sym A (R : rel A) : Symmetric R -> Symmetric (R#).
Quasi-ordering closure.
Section qo_clos.
Variables (A : Type) (E : rel A) (E_eq : Equivalence E).
Inductive qo_clos (R : rel A) : rel A :=
| qo_step : R << qo_clos R
| qo_refl : E << qo_clos R
| qo_trans : Transitive (qo_clos R).
Variable R : rel A.
Global Instance qo_clos_preorder : PreOrder (qo_clos R).
Variable R_prop : Proper (E ==> E ==> impl) R.
Global Instance qo_clos_prop : Proper (E ==> E ==> impl) (qo_clos R).
Global Instance qo_clos_prop_iff : Proper (E ==> E ==> iff) (qo_clos R).
Lemma qo_clos_inv : forall t u,
qo_clos R t u -> E t u \/ exists v, R t v /\ qo_clos R v u.
End qo_clos.
Properties of transp.
Instance transp_incl A : Proper (incl ==> incl) (@transp A).
Instance transp_same A : Proper (same ==> same) (@transp A).
Lemma transp_refl A (R : rel A) : Reflexive R -> Reflexive (transp R).
Lemma transp_trans A (R : rel A) : Transitive R -> Transitive (transp R).
Lemma transp_sym A (R : rel A) : Symmetric R -> Symmetric (transp R).
Lemma transp_invol A (R : rel A) : transp (transp R) == R.
Lemma transp_union A (R S : rel A) : transp (R U S) == transp R U transp S.
Relations between closures, union and composition.
Lemma rtc_comp_permut A (R S : rel A) : R# @ (R# @ S)# << (R# @ S)# @ R#.
Lemma rtc_union A (R S : rel A) : (R U S)# << (R# @ S)# @ R#.
Lemma rtc_comp A (R S : rel A) : R# @ S << S U R! @ S.
Lemma union_fact A (R S : rel A) : R U R @ S << R @ S%.
Lemma union_fact2 A (R S : rel A) : R @ S U R << R @ S%.
Lemma incl_rc_rtc A (R S : rel A) : R << S! -> R% << S#.
Lemma incl_tc_rtc A (R S : rel A) : R << S# -> R! << S#.
Lemma rtc_comp_modulo A (R S : rel A) : R# @ (R# @ S)! << (R# @ S)!.
Lemma tc_union A (R S : rel A) : (R U S)! << R! U (R# @ S)! @ R#.
Commutation properties.
Section commut.
Variables (A : Type) (R S : rel A) (commut : R @ S << S @ R).
Lemma commut_rtc : R# @ S << S @ R#.
Lemma commut_rtc_inv : R @ S# << S# @ R.
Lemma commut_tc : R! @ S << S @ R!.
Lemma commut_tc_inv : R @ S! << S! @ R.
Lemma commut_comp T : R @ (S @ T) << (S @ R) @ T.
Lemma comp_incl_assoc T V : R @ S << T -> R @ (S @ V) << T @ V.
End commut.
Inverse image.
Section inverse_image.
Variables (A B : Type) (R : rel B) (f : A->B).
Definition Rof : rel A := fun a a' => R (f a) (f a').
Lemma Rof_refl : Reflexive R -> Reflexive Rof.
Lemma Rof_trans : Transitive R -> Transitive Rof.
Lemma Rof_sym : Symmetric R -> Symmetric Rof.
Lemma Rof_preorder : PreOrder R -> PreOrder Rof.
Lemma Rof_equiv : Equivalence R -> Equivalence Rof.
Variable F : A -> B -> Prop.
Definition RoF : rel A :=
fun a a' => exists b', F a' b' /\ forall b, F a b -> R b b'.
End inverse_image.
Instance Rof_incl A B : Proper (incl ==> Logic.eq ==> incl) (@Rof A B).
Alternative definition of the transitive closure, more convenient for some inductive proofs.
Inductive clos_trans1 A (R : rel A) : rel A :=
| t1_step : forall x y, R x y -> clos_trans1 R x y
| t1_trans : forall x y z, R x y -> clos_trans1 R y z -> clos_trans1 R x z.
Notation "x !1" := (clos_trans1 x) (at level 35) : relation_scope.
Instance tc1_trans A (R : rel A) : Transitive (R!1).
Lemma tc1_eq A (R : rel A) x y : R!1 x y <-> R! x y.
Alternative definition of the reflexive and transitive closure, more convenient for some inductive proofs.
Inductive clos_refl_trans1 A (R : rel A) : rel A :=
| rt1_refl : forall x, clos_refl_trans1 R x x
| rt1_trans : forall x y z,
R x y -> clos_refl_trans1 R y z -> clos_refl_trans1 R x z.
Notation "x #1" := (clos_refl_trans1 x) (at level 9) : relation_scope.
Instance rtc1_preorder A (R : rel A) : PreOrder (R#1).
Lemma rtc1_eq A (R : rel A) x y : R#1 x y <-> R# x y.
Lemma tc1_incl_rtc1 A (R : rel A) : R!1 << R#1.
Lemma rtc1_union A (R S : rel A) : (R U S)#1 << (S#1 @ R)#1 @ S#1.
Lemma union_rel_rt1_left A (R S : rel A) : R#1 << (R U S)#1.
Lemma union_rel_rt1_right A (R S : rel A) : S#1 << (R U S)#1.
Lemma incl_rt1_union_union A (R S : rel A) : (R!1 U S!1)#1 << (R U S)#1.
Lemma incl_union_rt1_union A (R S : rel A) : (R U S)#1 << (R!1 U S!1)#1.
Lemma comm_s_rt1 A (R S : rel A) :
S@(R!1) << (R!1)@(S!1) -> (S!1)@(R!1) << (R!1)@(S!1).
Lemma comm_s_r A (R S : rel A) :
S@R << (R!1)@(S#1) -> (R U S)#1 @ R @ (R U S)#1 << (R!1) @ (R U S)#1.
Extension to option A of a relation on A so that it is reflexive on None.
Section opt_r.
Variables (A : Type) (R : rel A).
Inductive opt_r : rel (option A) :=
| opt_r_None : opt_r None None
| opt_r_Some : forall a b, R a b -> opt_r (Some a) (Some b).
Global Instance opt_r_refl : Reflexive R -> Reflexive opt_r.
Global Instance opt_r_sym : Symmetric R -> Symmetric opt_r.
Global Instance opt_r_trans : Transitive R -> Transitive opt_r.
Global Instance Some_prop : Proper (R ==> opt_r) (@Some A).
End opt_r.
Extension of a relation on A to option A so that it is irreflexive on None.
Section opt.
Variables (A : Type) (R : rel A).
Inductive opt : rel (option A) :=
| opt_intro : forall x y, R x y -> opt (Some x) (Some y).
Global Instance opt_trans : Transitive R -> Transitive opt.
Global Instance opt_sym : Symmetric R -> Symmetric opt.
Global Instance opt_opt_r E : Proper (E ==> E ==> impl) R ->
Proper (opt_r E ==> opt_r E ==> impl) opt.
Lemma opt_absorbs_right_opt_r E : R @ E << R -> opt @ opt_r E << opt.
Lemma opt_absorbs_left_opt_r E : E @ R << R -> opt_r E @ opt << opt.
End opt.
Instance opt_incl A : Proper (incl ==> incl) (@opt A).
Instance opt_prop A (E1 E2 R : rel A) :
Proper (E1 ==> E2 ==> impl) R -> Proper (opt E1 ==> opt E2 ==> impl) (opt R).
Lemma opt_absorbs_right A (R E : rel A) :
R @ E << R -> opt R @ opt E << opt R.
Lemma opt_absorbs_left A (R E : rel A) :
E @ R << R -> opt E @ opt R << opt R.
Restriction of a relation to some set.
Section restrict.
Variables (A : Type) (P : A -> Prop) (R : rel A).
Definition restrict : rel A := fun x y => P x /\ R x y.
Global Instance restrict_prop E :
Proper (E ==> impl) P -> Proper (E ==> E ==> impl) R ->
Proper (E ==> E ==> impl) restrict.
End restrict.
Lemma restrict_union A (P : A -> Prop) R S :
restrict P (R U S) == restrict P R U restrict P S. | 6,236 | 18,695 | {"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-2020-05 | latest | en | 0.551595 |
https://www.wyzant.com/resources/answers/651/how_many_drops_do_you_take_per_dose | 1,430,977,208,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1430460381476.88/warc/CC-MAIN-20150501060621-00012-ip-10-235-10-82.ec2.internal.warc.gz | 963,181,299 | 12,428 | Search 75,476 tutors
0 0
# how many drops do you take per dose??
your grandmother sends you a remedy for the heebie geebies with the following instructions.
take 1 drop per 10 lbs. of body weight per day divied into 4 doses until the heebie geebies are gone
if u weight 170
how many drops do you take per dose??
you have this information right now
170 lbs
1 drop per day = 10 lbs
1 day = 4 doses
170 lbs x 1 drop = 17 drops x 1 day = 17 drops = 17/4 drops/dose or 4.25 drops/dose
10 lbs 1 day 4 doses 4 doses | 177 | 574 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2015-18 | longest | en | 0.435587 |
https://number.academy/1002670 | 1,679,647,233,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945279.63/warc/CC-MAIN-20230324082226-20230324112226-00342.warc.gz | 498,127,160 | 12,052 | # Number 1002670
Number 1,002,670 spell 🔊, write in words: one million, two thousand, six hundred and seventy , 1.0 million. Ordinal number 1002670th is said 🔊 and write: one million, two thousand, six hundred and seventieth. The meaning of the number 1002670 in Maths: Is it Prime? Factorization and prime factors tree. The square root and cube root of 1002670. What is 1002670 in computer science, numerology, codes and images, writing and naming in other languages
## What is 1,002,670 in other units
The decimal (Arabic) number 1002670 converted to a Roman number is (M)MMDCLXX. Roman and decimal number conversions.
#### Weight conversion
1002670 kilograms (kg) = 2210486.3 pounds (lbs)
1002670 pounds (lbs) = 454808.1 kilograms (kg)
#### Length conversion
1002670 kilometers (km) equals to 623031 miles (mi).
1002670 miles (mi) equals to 1613642 kilometers (km).
1002670 meters (m) equals to 3289560 feet (ft).
1002670 feet (ft) equals 305618 meters (m).
1002670 centimeters (cm) equals to 394752.0 inches (in).
1002670 inches (in) equals to 2546781.8 centimeters (cm).
#### Temperature conversion
1002670° Fahrenheit (°F) equals to 557021.1° Celsius (°C)
1002670° Celsius (°C) equals to 1804838° Fahrenheit (°F)
#### Time conversion
(hours, minutes, seconds, days, weeks)
1002670 seconds equals to 1 week, 4 days, 14 hours, 31 minutes, 10 seconds
1002670 minutes equals to 2 years, 3 weeks, 3 days, 7 hours, 10 minutes
### Codes and images of the number 1002670
Number 1002670 morse code: .---- ----- ----- ..--- -.... --... -----
Sign language for number 1002670:
Number 1002670 in braille:
QR code Bar code, type 39
Images of the number Image (1) of the number Image (2) of the number More images, other sizes, codes and colors ...
## Mathematics of no. 1002670
### Multiplications
#### Multiplication table of 1002670
1002670 multiplied by two equals 2005340 (1002670 x 2 = 2005340).
1002670 multiplied by three equals 3008010 (1002670 x 3 = 3008010).
1002670 multiplied by four equals 4010680 (1002670 x 4 = 4010680).
1002670 multiplied by five equals 5013350 (1002670 x 5 = 5013350).
1002670 multiplied by six equals 6016020 (1002670 x 6 = 6016020).
1002670 multiplied by seven equals 7018690 (1002670 x 7 = 7018690).
1002670 multiplied by eight equals 8021360 (1002670 x 8 = 8021360).
1002670 multiplied by nine equals 9024030 (1002670 x 9 = 9024030).
show multiplications by 6, 7, 8, 9 ...
### Fractions: decimal fraction and common fraction
#### Fraction table of 1002670
Half of 1002670 is 501335 (1002670 / 2 = 501335).
One third of 1002670 is 334223,3333 (1002670 / 3 = 334223,3333 = 334223 1/3).
One quarter of 1002670 is 250667,5 (1002670 / 4 = 250667,5 = 250667 1/2).
One fifth of 1002670 is 200534 (1002670 / 5 = 200534).
One sixth of 1002670 is 167111,6667 (1002670 / 6 = 167111,6667 = 167111 2/3).
One seventh of 1002670 is 143238,5714 (1002670 / 7 = 143238,5714 = 143238 4/7).
One eighth of 1002670 is 125333,75 (1002670 / 8 = 125333,75 = 125333 3/4).
One ninth of 1002670 is 111407,7778 (1002670 / 9 = 111407,7778 = 111407 7/9).
show fractions by 6, 7, 8, 9 ...
### Calculator
1002670
#### Is Prime?
The number 1002670 is not a prime number. The closest prime numbers are 1002653, 1002679.
#### Factorization and factors (dividers)
The prime factors of 1002670 are 2 * 5 * 100267
The factors of 1002670 are 1 , 2 , 5 , 10 , 100267 , 200534 , 501335 , 1002670
Total factors 8.
Sum of factors 1804824 (802154).
#### Powers
The second power of 10026702 is 1.005.347.128.900.
The third power of 10026703 is 1.008.031.405.734.162.944.
#### Roots
The square root √1002670 is 1001,33411.
The cube root of 31002670 is 100,088921.
#### Logarithms
The natural logarithm of No. ln 1002670 = loge 1002670 = 13,818177.
The logarithm to base 10 of No. log10 1002670 = 6,001158.
The Napierian logarithm of No. log1/e 1002670 = -13,818177.
### Trigonometric functions
The cosine of 1002670 is 0,757501.
The sine of 1002670 is -0,652834.
The tangent of 1002670 is -0,861826.
### Properties of the number 1002670
Is a Fibonacci number: No
Is a Bell number: No
Is a palindromic number: No
Is a pentagonal number: No
Is a perfect number: No
## Number 1002670 in Computer Science
Code typeCode value
1002670 Number of bytes979.2KB
Unix timeUnix time 1002670 is equal to Monday Jan. 12, 1970, 2:31:10 p.m. GMT
IPv4, IPv6Number 1002670 internet address in dotted format v4 0.15.76.174, v6 ::f:4cae
1002670 Decimal = 11110100110010101110 Binary
1002670 Decimal = 1212221101221 Ternary
1002670 Decimal = 3646256 Octal
1002670 Decimal = F4CAE Hexadecimal (0xf4cae hex)
1002670 BASE64MTAwMjY3MA==
1002670 MD59da50858c6781b4f0907ee106ee028df
1002670 SHA1fd24d26c9a5ccb556ce8666dc770e2580c372866
1002670 SHA224f30278cdd0f0607bbc0df32c51e08baaa0afe6b1a309060863f36f45
1002670 SHA256c6e8ccf8ce99f5b72c095c6f78564ef9e8ece15dd1346f70203acf944fbfa819
1002670 SHA384555c2a103f701fcd9caeb36592a6b26994cff274579b4b67ec249ab6a52e6e2cae348e682b29627c3d39ecee6770c16e
More SHA codes related to the number 1002670 ...
If you know something interesting about the 1002670 number that you did not find on this page, do not hesitate to write us here.
## Numerology 1002670
### Character frequency in the number 1002670
Character (importance) frequency for numerology.
Character: Frequency: 1 1 0 3 2 1 6 1 7 1
### Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 1002670, the numbers 1+0+0+2+6+7+0 = 1+6 = 7 are added and the meaning of the number 7 is sought.
## № 1,002,670 in other languages
How to say or write the number one million, two thousand, six hundred and seventy in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 1.002.670) un millón dos mil seiscientos setenta German: 🔊 (Nummer 1.002.670) eine Million zweitausendsechshundertsiebzig French: 🔊 (nombre 1 002 670) un million deux mille six cent soixante-dix Portuguese: 🔊 (número 1 002 670) um milhão e dois mil, seiscentos e setenta Hindi: 🔊 (संख्या 1 002 670) दस लाख, दो हज़ार, छः सौ, सत्तर Chinese: 🔊 (数 1 002 670) 一百万二千六百七十 Arabian: 🔊 (عدد 1,002,670) واحد مليون و ألفان و ستمائةسبعون Czech: 🔊 (číslo 1 002 670) milion dva tisíce šestset sedmdesát Korean: 🔊 (번호 1,002,670) 백만 이천육백칠십 Dutch: 🔊 (nummer 1 002 670) een miljoen tweeduizendzeshonderdzeventig Japanese: 🔊 (数 1,002,670) 百万二千六百七十 Indonesian: 🔊 (jumlah 1.002.670) satu juta dua ribu enam ratus tujuh puluh Italian: 🔊 (numero 1 002 670) un milione e duemilaseicentosettanta Norwegian: 🔊 (nummer 1 002 670) en million, to tusen, seks hundre og sytti Polish: 🔊 (liczba 1 002 670) milion dwa tysiące sześćset siedemdziesiąt Russian: 🔊 (номер 1 002 670) один миллион две тысячи шестьсот семьдесят Turkish: 🔊 (numara 1,002,670) birmilyonikibinaltıyüzyetmiş Thai: 🔊 (จำนวน 1 002 670) หนึ่งล้านสองพันหกร้อยเจ็ดสิบ Ukrainian: 🔊 (номер 1 002 670) один мiльйон двi тисячi шiстсот сiмдесят Vietnamese: 🔊 (con số 1.002.670) một triệu hai nghìn sáu trăm bảy mươi Other languages ...
## News to email
I have read the privacy policy
## Comment
If you know something interesting about the number 1002670 or any other natural number (positive integer), please write to us here or on Facebook. | 2,604 | 7,296 | {"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-2023-14 | longest | en | 0.630965 |
https://hub.jmonkeyengine.org/t/optimizing-forest-framerate/39284 | 1,669,735,830,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710698.62/warc/CC-MAIN-20221129132340-20221129162340-00283.warc.gz | 355,907,760 | 12,563 | # Optimizing forest framerate
Hello,
I’ve been working on generating a forest and at this stage I’m trying to optimize the framerate. The forest contains about 10K trees, each of them having around 1k triangles. I’m projecting shadows. The whole thing looks like this
• First thing I did was generating LODs on the meshes (so far only tried with 2 lods), I did see a significant impact on the framerate
• Then I tried using batch geometries. Here, reading:
My experiences with batching objects and the resulting frame rate
I made sure I split the whole forest into smaller nodes (100 nodes of about 100 trees, all trees use the same material) and this is what I see, at medium range, I feel the bulk geometry is actually a bit worse:
in comparison to the non-bulked
I don’t understand how come I still render 64 out of 100 nodes. With the layout of my forest, I would expect a much smaller fraction to show up from this camera angle (in the non-bulked, the fraction of objects showed is much smaller in comparison to the total)
At close range, I can see an improvement (like x2) from the bulking. Still I see that I’m rendering 44 nodes out of 100
vs the non-bulked
At long range, I see the bulk geometry doesn’t use LODs (even though I’m calling it through this
GeometryBatchFactory.optimize(subNode, true);
I don’t really care that much since I won’t be displaying the whole forest very often, but let me know if you see I am missing something)
Do you think there is a way to improve either the bulking, the z-culling or frustrum culling? Is there any other rendering optimization technique that would benefit the framerate?
Thanks!
Yes, there are billboards/impostors. That is a 2d quad with a texture of a tree. You can generate eg. 4-8 different textures from different angles and swap between these. You normally apply these for the furthest trees. They are very fast since 1 quad = 4 vertexes. To render them efficiently you can use instancing. The rotation and selection of texture can be done in a shader also. The generation of the textures can be done by code too.
Also there is the instancing way to do it.
Instancing might work nice with this kind of use case.
Group them into a larger grid. Then as they get further away you could lower the lod until a certain point, then replace them with billboards, instanced billboards, or batched billboards, though you might have to have some shader knowledge first. The point is, when those trees are only a few pixels on screen, it is much better to just have a quad (two triangles) rendered per tree instead of a mesh. The trick is if you want to combine those quads into groups so the renderer can cull out large sections of the scene without having to check each tree individually. Most AAA games do this is some way or another, from skyrim to battlefield etc. You could combine all your tree billboard textures into one big texture atlas and render them all from that with a shader.
1 Like
I tried creating an instancednode for the whole forest but then framerate dropped considerably since all trees have to be rendered all the time
Then I tried to instance each of the 100 sub nodes but then something very strange happened onscreen (like the forest turned into a big black shape with pieces of tree everywhere and framerate dropped to around 1 frame per 10 s) maybe because the 100 nodes share the same instanced material ?
Regarding the billboard, why not and I don’t mind digging the shader, but you were saying there is a way to generate the different billboard textures from a single mesh ? I will take a look already
I don’t know why that happens, but make sure you are putting the trees under a node based on location, not just random location, so that the renderer can check the nodes and cull areas that are offscreen.
You can do offscreen rendering at runtime or prebuilt that will render your tree into a 2d texture, just like taking a screenshot of it. Then you can arrange whatever you want into a texture atlas.
1 Like
Yes, that’s the way the forest is generated (by different nodes, and then I move the nodes around) so each node groups trees belonging to the same area
One thing regarding the culling, does the actual node position matter for this? I have all my 100 nodes centered at 0,0,0 and I just place the trees at their real world coordinates
I’m testing moving the location of the actual subnodes instead of just the trees, it doesn’t seem to help the culling
No, I believe the bounding box for a node is only based on the geometry max extents. You might want to try davidb’s viewer for debugging:
So you should:
1. Have close range trees be Instanced in location grid groups, or close range trees be in a Batch node/geometry batch factory, or close range trees be regular spatials if it is necessary.
2. Transition lods as the grids get further away
3. Eventually switch to a 2d billboard of the tree. The billboards should still be Instanced / batched per grid group, so they can be culled out optimally.
The hard part (that many games still haven’t solved perfectly) is transitioning between these without too much popping or noticeable change to the player. There are many techniques such as fading the object out/ fading in billboard that you can try out. There’s also dithering and rendering different rotations of the trees to fade between based on what direction the player is looking at the tree.
Just putting these here in case you want to dig around in other solutions.
SimArboreal will generate trees that have LOD, etc. built in, including simpler trees as well as full atlases. It supports wind, etc. even for instanced trees. It also supports an in-between LOD where each branch is rendered as a single camera-aligned quad (hard to describe in words).
http://simsilica.github.io/SimArboreal-Editor/
On its own, it doesn’t have paging, etc. built in but you can look at how that was done in the IsoSurface Demos if you are curious:
Video of SimArboreal regular mesh-based LOD (simplified tree geometry):
Video of the aligned-billboard branches thing I was mentioning:
Trees in the wind: (wind effect can be tuned up/down as needed… even at runtime)
Putting it all together in the IsoSurface Demo:
1 Like
it looks pretty neat! I downloaded the zip file but when importing it says I’m missing a bunch of com.simsilica.lemur.* which i can’t find on the simsilica home page, by any chance do you know where to get these?
Also, I have integrated the library to SS Editor
https://bitbucket.org/JavaSabr/ss-editor-tree-generator
1 Like
Also note: the IsoSurfaceDemo has some prebuilt binaries if you just want to try it out:
I don’t remember what state they were in at that time.
SimArboreal Editor also has some prebuilt binaries:
Really sorry about that but I’m still seeing missing references such as below, even from simarboreal itself, for example the package (and I double checked the content from github to make sure it wasn’t just on my side) doesn’t contain the following references:
The class
Simarboreal\src\main\java\com\simsilica\arboreal\AtlasGeneratorState.java
import com.simsilica.arboreal.mesh.BillboardedLeavesMeshGenerator;
import com.simsilica.arboreal.mesh.SkinnedTreeMeshGenerator;
import com.simsilica.arboreal.mesh.Vertex;
import com.simsilica.builder.Builder;
import com.simsilica.builder.BuilderReference;
import com.simsilica.builder.BuilderState;
Then in fileActionState I got an unknown class on TreeParameters at line 251 and 259
then ForestGridState references elements from the lemur node
com.simsilica.lemur.props
which doesn’t exist
and there are a few more like this
Continuing on the idea of the billboarding, I’ve played a bit with the concept and created a control to which you provide a node containing the regular mesh and a simple quad. There is a control attached to each node (each node represents a tree) and that control will swap the two geometries (removing one and adding the other) depending on the distance and rotate the quad accordingly to the camera (based on the demo from the jme3 library)
I am very surprised to see that my map with billboarding is slower than the one that only has meshes. You can see below that the object count is about the same, and as expected the triangle count of the billboard scene is 4M triangles vs 11M for all meshes
I will investigate more why the fps is dropping, but happy to hear if you can think of anything
Well, if you are checking the distance for every tree once per frame, that’ll do it. I was thinking more along the lines of only checking the distance per small chunk of trees, and maybe do some kind of time slicing so you’re only checking one chunk of trees a frame or so. Maybe check every chunk of trees once a second, though not all at once. In addition, computing the billboard rotation for every billboard every frame may be having an impact. Might be better to do it in a vertex shader.
I downloaded the IsoSurfaceDemo for Windows and tried to run it, but I got an error dialog: “The registry refers to a nonexistent Java Runtime Environment installation of the runtime is corrupted. The system cannot find the file specified.”
1 Like
I’m not sure about billboard trees, but flatpoly trees computing only on GPU | 2,072 | 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} | 2.875 | 3 | CC-MAIN-2022-49 | latest | en | 0.927233 |
https://katespadenew.com/math-solver-743 | 1,675,514,015,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500126.0/warc/CC-MAIN-20230204110651-20230204140651-00343.warc.gz | 357,256,756 | 6,090 | Using Permutations to Calculate Probabilities
from math import factorial def permutations(l): permutations=[] length=len(l) for x in xrange(factorial(length)): available=list(l) newPermutation=[] for radix in xrange(length, 0, -1):
• Homework Support Online
If you're struggling with your homework, our Homework Help Solutions can help you get back on track.
• Solve math problem
Algebra can be difficult to wrap your head around, but once you understand the basics, it can be a breeze.
• Figure out math question
The average satisfaction rating for this product is 4.7 out of 5. This product is sure to please!
• Figure out mathematic equations
There's no need to be scared of math - it's a useful tool that can help you in everyday life!
• More than just an app
To solve a math equation, you need to find the value of the variable that makes the equation true.
• Deal with math problem
I can solve the math problem for you.
Easy Permutations and Combinations
1. Some graphing calculators offer a button to help you solve permutations without repetition quickly. It usually looks like nPr. If your calculator has one, hit your n{\displaystyle n} value firs See more
• Solve algebra
• Solve word questions too
• Determine mathematic equations
• Keep time
• Timely Delivery
Combinations and Permutations
For each permutation, calculate the median. Determine the proportion of permutation medians that are more extreme than our observed median. That proportion is our
250 Teachers
15 Years of experience
60657 Delivered assignments
Customers said
There are very good step by step instructions. If you use this app, make sure you understand how it got the answer and not just use it to cheat, otherwise it defeats the purpose, as a student, it can be hard to learn sometimes when you can't check your answers to see if you are right.
Luis Coleman
The best app fore helping you with any math problem it's basically a digital tutor, I love this app, very accurate answers, it works wonders especially when you have just started to learn algebra highly recommended for any if not all kinds of learners of math.
Michael Donohoe
How to Calculate Permutations: 8 Steps (with Pictures)
Avg. satisfaction rating 4.7/5
Deal with mathematic question
Solve mathematic
Passing Rate
Save time
Figure out mathematic problems
How to Calculate a Permutation
Permutations Formula: P ( n, r) = n! ( n − r)! For n ≥ r ≥ 0. Calculate the permutations for P (n,r) = n! / (n - r)!. The number of ways of obtaining an ordered subset of r elements from a set of n elements. [1] Permutation Problem 1
Math knowledge that gets you
Work on the task that is attractive to you
Get the Most useful Homework explanation
Build brilliant future aspects | 611 | 2,748 | {"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-2023-06 | latest | en | 0.903104 |
https://forums.odforce.net/topic/46544-vortex-cloud-simulation/?tab=comments | 1,623,773,645,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487621450.29/warc/CC-MAIN-20210615145601-20210615175601-00442.warc.gz | 268,537,041 | 20,609 | Jump to content
# Vortex Cloud Simulation
## Recommended Posts
Why my cloud push away in the few final frame?~ my velocity field is not change direction~~~~~
Thank you~
cortex_cloud.mp4
• 1
#### Share this post
##### Share on other sites
I solved the problem with divergence~~~but any better solution to do vortec cloud?? thank you
#### Share this post
##### Share on other sites
This is likely a result of linear/tangential velocity increasing while the centripetal force remains the same.
The equation F=mv2/R, where F is centripetal force, m is mass, v is linear/tangential velocity, and R is the radius, shows that if F and m remain constant and v increases then the radius must also increase.
Linear/tangential velocity increases because the force(s) acting on the smoke cause an acceleration, which is why it starts moving in the first place.
I haven't tested this, but if you decrease the force(s) causing it to spin it will take longer to get up to the speed you want, but then I think it will settle at a smaller radius.
• 1
• 1
#### Share this post
##### Share on other sites
Whether to reduce the radius to extend the life？
#### Share this post
##### Share on other sites
On 2020/7/21 at 12:41 AM, eimk said:
This is likely a result of linear/tangential velocity increasing while the centripetal force remains the same.
The equation F=mv2/R, where F is centripetal force, m is mass, v is linear/tangential velocity, and R is the radius, shows that if F and m remain constant and v increases then the radius must also increase.
Linear/tangential velocity increases because the force(s) acting on the smoke cause an acceleration, which is why it starts moving in the first place.
I haven't tested this, but if you decrease the force(s) causing it to spin it will take longer to get up to the speed you want, but then I think it will settle at a smaller radius.
Wow!thank you,I think I should increase the centripetal force as much as the rotational force increases
#### Share this post
##### Share on other sites
I don't know what your setup is, but if you keep the forces constant eventually they will balance out with air resistance and the velocity will become constant too. If the forces keep increasing the sim could get chaotic and difficult to manage.
#### Share this post
##### Share on other sites
7 hours ago, eimk said:
I don't know what your setup is, but if you keep the forces constant eventually they will balance out with air resistance and the velocity will become constant too. If the forces keep increasing the sim could get chaotic and difficult to manage.
here is my file and cache thank you.
odforce.rar
#### Share this post
##### Share on other sites
My computer doesn't have enough RAM to run your sim without Houdini crashing, but my best guess for what could help would be decreasing the strength of your velocity field and/or adding a drag force. Alternatively you could look into the vortex force node, which might be easier to work with than your current setup.
#### Share this post
##### Share on other sites
You are running a pyro sim, which by default aims to make the velocity field divergent free. The velocity field that you have set up is clearly not non-divergent. You are trying to push all the smoke toward the center (i.e. the field's divergence is strongly negative in the middle). So when the pyro sim tries to make the velocity field non-divergent, it has to modify the velocity field near the middle to push the smoke out and away from the center. What you did by specifying the target divergence field is correct. You are telling the pyro sim to make the velocity field's divergence negative in the middle, rather than zero everywhere, which will give you the effect of pushing the smoke toward the middle.
#### Share this post
##### Share on other sites
On 01/08/2020 at 11:44 AM, ziconic said:
You are running a pyro sim, which by default aims to make the velocity field divergent free. The velocity field that you have set up is clearly not non-divergent. You are trying to push all the smoke toward the center (i.e. the field's divergence is strongly negative in the middle). So when the pyro sim tries to make the velocity field non-divergent, it has to modify the velocity field near the middle to push the smoke out and away from the center. What you did by specifying the target divergence field is correct. You are telling the pyro sim to make the velocity field's divergence negative in the middle, rather than zero everywhere, which will give you the effect of pushing the smoke toward the middle.
Awesome internet took me here while I tried to solve my problem with pyro by using AXIOM solver. I have cross product of N to introduce swirl around a mesh, then do want to increase the speed of v but not make the smoke expand in same acceleration speed magnitude value. I did not know about centripetal force until Hristo Velev from FB Grp Hodini Artist point me to some of this knowledge. Thank you.
My solve for this is create divergence field and push it inward (negative value) but solve by TIME STEP but not solve by every frame, in the same time Velocity field will add in with method PULL to act as a drag slowly accelerate to the max value but not add in right away. That was my case.
Here is cross product result:
And here is result from add more speed but can not reduce expand
## Create an account or sign in to comment
You need to be a member in order to leave a comment
## Create an account
Sign up for a new account in our community. It's easy!
Register a new account
## Sign in
Already have an account? Sign in here.
Sign In Now
×
• Donations
• Leaderboard | 1,276 | 5,693 | {"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} | 3.390625 | 3 | CC-MAIN-2021-25 | latest | en | 0.905992 |
https://courseworkheroines.com/healthcare-homework-question-book-used-louis-c-gapenski-healthcare-finance-an-introduction-to-accounting-and-financial-management/ | 1,670,504,008,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711336.41/warc/CC-MAIN-20221208114402-20221208144402-00675.warc.gz | 219,007,017 | 12,097 | # Healthcare Homework Question, book used Louis C. Gapenski Healthcare Finance An Introduction to Accounting and Financial Management
14.6 The director of capital budgeting for Big Sky Health Systems Inc. has estimated the following cash flows in thousands of dollars for a proposed new service:
Year Expected Net Cash Flow
0 (\$100)
1 \$70
2 \$50
3 \$20
The project’s cost of capital is 10%.
a. What is the project’s payback period?
b. What is the project’s NPV?
c. What is the project’s IRR?
d. What is the project’s MIRR?
14.5. Assume that you are the chief financial officer at Porter Memorial Hospital. The CEO has asked you to analyze two proposed capital investments – Project X and Project Y. Each project requires a net investment outlay of \$10,000 and the cost of capital for each project is 12%. The project’s expected net cash flows are as follows:
Year Project X Project Y
0 (\$10,000) (\$10,000)
1 \$6,500 \$3,000
2 \$3,000 \$3,000
3 \$3,000 \$3,000
4 \$1,000 \$3,000
a. Calculate each project’s payback period?
b. Calculate net present value (NPV)?
c. Calculate internal rate of return (IRR)?
d. Which project (or projects is financially acceptable?
### Save your time - order a paper!
Get your paper written from scratch within the tight deadline. Our service is a reliable solution to all your troubles. Place an order on any task and we will take care of it. You won’t have to worry about the quality and deadlines
Order Paper Now
Can these calculation be put on an excel spreadsheet? I need this by 5:00pm Anchorage, Alaska time. Is this possible??? | 391 | 1,575 | {"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} | 2.625 | 3 | CC-MAIN-2022-49 | longest | en | 0.897396 |
http://hoggresearch.blogspot.com/2013/07/flexible-rigid-models.html | 1,516,418,119,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084888878.44/warc/CC-MAIN-20180120023744-20180120043744-00417.warc.gz | 170,462,735 | 23,916 | 2013-07-09
flexible rigid models
I worked on Gaussian Processes with Foreman-Mackey today. In our formulation of the "search Kepler data for transiting planet candidates" problem, what is usually called "the model" of the transit sets (in our formulation) the mean of a Gaussian Process, and the photon noise, stellar variability, and spacecraft-induced variability is all subsumed into a covariance function that sets the variance tensor of that Process. The nice thing about this formulation is that the estimation or inference is just what you would do with linear chi-squared inference (optimization or sampling or whatever you like), but just with a much more complicated covariance matrix. In particular, the covariance matrix is now nothing like diagonal.
We are off the usual path of Gaussian Processes, but only because we have a non-trivial function for the mean of the Process. That isn't much of a change, but it is rarely mentioned in the usual texts and descriptions of GPs. I think the fundamental reason for this is that GPs are usually used as incredibly flexible models for barely understood data. In typical astronomical (and physical) problems, there are parts of the problem that are barely understood and require flexible models, but other parts that are very well understood and require pretty rigid, physically motivated models. In our use of GPs, we get both at once: The mean function is set by our physical model of transiting exoplanet; the variance function is set to capture as well as possible all the variability that we don't (officially) care about.
More soon.
2 comments:
1. Cool. The other thing I've noticed about Gaussian process literature is they often use the GP distribution as a prior for some unknown function, but it sounds like in your case the GP is the sampling distribution with a lot of effects integrated out.
2. There can be numerical problems when integrating out a mean function to build one do-everything covariance. The simplest example is a GP with non-zero but unknown constant mean offset:
c ~ N(0, σ²) # c is a scalar
f ~ N(c1, Σ) # "1" is a vector of ones
which is marginally equivalent to a zero mean process:
f ~ N(0, Σ+σ²) # σ² added to every element
The (Σ+σ²) covariance is poorly conditioned for large σ². I believe there's some discussion in Rasmussen and Williams on how to write well-behaved code with this sort of covariance. | 520 | 2,407 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2018-05 | longest | en | 0.951138 |
http://www.education.com/study-help/article/polygons-triangles/?page=3 | 1,411,016,319,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657125488.38/warc/CC-MAIN-20140914011205-00014-ip-10-196-40-205.us-west-1.compute.internal.warc.gz | 482,573,044 | 27,200 | Education.com
Try
Brainzy
Try
Plus
Polygons and Triangles Study Guide (page 3)
based on 2 ratings
By
Updated on Oct 5, 2011
Right Triangles
Right triangles have a rule of their own. Using the Pythagorean theorem, we can calculate the missing side of a RIGHT triangle. a2+ b2 = c2(c refers to the hypotenuse)
Example: What is the perimeter of the triangle shown at the right?
1. Since the perimeter is the sum of the lengths of the sides, we must first find the missing side. Use the Pythagorean theorem:
2. Substitute the given sides for two of the letters. Remember: Side c is always the hypotenuse:
3. To solve this equation, subtract 9 from both sides:
4. Then, take the square root of both sides. (Note: Refer to Lesson 20 to learn about square roots.) Thus, the missing side has a length of 4 units.
5. Adding the three sides yields a perimeter of 12:
Tip
Sometime today you'll be bored, and doodling is a good way to pass the time. Doodle with purpose: Draw a triangle! After you draw it, examine it closely to determine if it's an equilateral, isosceles, scalene, or right become! Practice with these shapes until you know them all by heart.
Find practice problems and solutions for these concepts at Polygons and Triangles Practice Questions.
150 Characters allowed
Related Questions
Q:
See More Questions
Top Worksheet Slideshows | 331 | 1,354 | {"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} | 4.375 | 4 | CC-MAIN-2014-41 | latest | en | 0.894448 |
https://quant.stackexchange.com/questions/64108/what-does-it-mean-for-a-coupon-bond-to-have-par-value | 1,718,344,159,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861521.12/warc/CC-MAIN-20240614043851-20240614073851-00080.warc.gz | 439,697,121 | 41,262 | # What does it mean for a coupon bond to have "par value"?
I am doing the Interest Rate Models course on Coursera. In the third lecture of the second week, the lecturer provides this lemma:
Lemma 1
A coupon bond has par value at $$T_0$$ if and only if its coupon rates equal the corresponding swap rate:
$$1 = \sum_{i=1}^n P(T_0, T_i)\delta R_{\text{swap}}(T_0)+P(T_0, T_n)\text{.}$$
Proof
Exercise.
My question is what does "par value" mean in this Lemma? I did a google search of par value, and I got this definition from Investopedia:
Par value, also known as nominal value, is the face value of a bond or the stock value stated in the corporate charter.
The definition above makes it sound like all coupon bonds would have a par value, although that par value might be $$0$$. I don't know what that par value would have to do with the coupon rate. It seems to me that it would only make sense to say something like
A coupon bond has par value $$X$$ at $$T_0\ldots$$
so I am not sure what the claim is in the provided Lemma. What does "par value" mean here?
• It simply means that the bond's current clean price is par (i.e., $P = 100$ assuming a notional amount of 100). Commented May 19, 2021 at 16:06
• Bad choice of words by the instructor. Instead of "has par value" he/she should have said "is at par" or "has value equal to par" etc. Commented May 19, 2021 at 17:27
• @Helin if you submit that as an answer, I will mark "accept" it. Commented May 19, 2021 at 17:59
• The definition you found seems also confusing to me by badly choosing words. I agree with the explanation of @noob2 Commented May 20, 2021 at 6:53
## 2 Answers
Sloppy English + no editor.
The lemma really says that if you calculate the fair value of an instrument (FRN, or a floating leg of an interest rate swap..) that pays LIBOR (no spread added to it) by projecting the floating coupon cash flows using swap curve and discounting all the cash flows (coupons and principal) using the same curve, then this fair value is equal to the undiscounted face value of the remaining principal repayments.
The present value of the projected coupons is exactly the difference between the face value of the remaining principal repayments minus the present value of the principal discounted using the swap curve.
When swap rates go up (down), then the present value of your principal repayments will go down (up), but the coupon amount changes exactly to offset the change in the present value of the principal.
However if the coupon is being set in advance, as usually done with LIBOR, then all this is not quite true in the middle of a coupon period - only at the beginning of coupon period. Once current coupon effectively becomes fixed, the instrument's price can deviate a little from par as the short-term rates moves. Related question: Why does the valuation of the floating leg of a swap only use the next payment?
Note that the instrument can be amortizing. There's no need to assume that all the principal is repaid only at maturity.
Par value is the principal payment made at maturity, versus present value which is essentially the price.
• This definition seems to be the the same as the one I provided. I'm not seeing how your definition makes the Lemma make sense. I would think that all bonds "have par value," and that this value might be $0$. I'm not sure what the par value would have to do with the coupon rate. Taking the definition "principle amount payed at maturity" doesn't line up with the the Lemma "A coupon bond has par value at $T_0$..." because $T_0$ is not the time of maturity. Commented May 19, 2021 at 14:27
• I misquoted: "principle payment made at maturity" not "principle amount payed at maturity." Commented May 19, 2021 at 14:34
• i don't see a difference in the language above, those statements mean the same thing to me. An amortizing bond can have a par value of say 100 and a principal payment or near 0 at maturity. In the case of a bullet bond the final principal payment can also include a coupon payment. Commented May 19, 2021 at 14:40 | 1,004 | 4,071 | {"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": 5, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2024-26 | latest | en | 0.911356 |
https://www.geeksforgeeks.org/implementation-of-non-preemptive-shortest-job-first-using-priority-queue/ | 1,695,793,349,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510259.52/warc/CC-MAIN-20230927035329-20230927065329-00228.warc.gz | 854,415,915 | 45,869 | Open In App
Implementation of Non-Preemptive Shortest Job First using Priority Queue
Read here for Shortest Job First Scheduling algorithm for same arrival times.
Shortest job first (SJF) or shortest job next, is a scheduling policy that selects the waiting process with the smallest execution time to execute next.
In this article, we will implement the Shortest Job First Scheduling algorithm (SJF) using a priority queue, so that we can handle processes at different arrival time.
Examples:
```Input: The processes are
Process id Arrival time Burst time
p1 4 3
p2 0 8
p3 5 4
p4 9 2
Output: Process scheduling according to SJF is
Process id Arrival time Burst time
p2 0 8
p1 4 3
p4 9 2
p3 5 4
Input: The processes are
Process id Arrival time Burst time
p1 0 3
p2 0 8
p3 5 4
p4 9 2
Output: Process scheduling according to SJF is
Process id Arrival time Burst time
p1 0 3
p2 0 8
p4 9 2
p3 5 4 ```
In this program, the task is to schedule the processes according to SJF scheduling algorithm, which states that the process with minimum burst time will be given priority, which can simply be implemented by sorting the burst time of the processes in ascending order. The problem arises when we have to handle the processes at different arrival time, then we simply can’t sort the processes according to burst time as we need to consider the arrival time of the process so that the processor doesn’t stay idle.
Example:
If a process with more burst time arrives before a process with less burst time, then we have to allow the processor time to the process that arrived first so that the processor doesn’t stay idle.
Approach:
To handle processes with a different arrival time in case of SJF scheduling:
• First, sort the processes according to the arrival time.
• Maintain a wait queue, which keeps the process with minimum burst time at the top.
• Maintain the current run time, that is the sum of burst times of the executed processes.
• A process enters the wait queue according to it’s arrival time if a new process has arrival time less than
equal to the current running time, it is pushed in the wait queue.
• A process is popped from the wait queue when it is to be executed. It’s burst time is added to the current run time, it’s arrival time is updated to -1 so that it doesn’t enter the wait queue again.
Below is the implementation of the above approach:
CPP
`// C++ implementation of SJF``#include ``using` `namespace` `std;` `// number of process``#define SIZE 4` `// Structure to store the``// process information``typedef` `struct` `proinfo {`` ``string pname; ``// process name`` ``int` `atime; ``// arrival time`` ``int` `btime; ``// burst time` `} proinfo;` `// This structure maintains the``// wait queue, using burst``// time to compare.``typedef` `struct` `cmpBtime {`` ``int` `operator()(``const` `proinfo& a,`` ``const` `proinfo& b)`` ``{`` ``return` `a.btime > b.btime;`` ``}``} cmpBtime;` `// This function schedules the``// process according to the SJF``// scheduling algorithm.``void` `sjfNonpremetive(proinfo* arr)``{`` ``// Used to sort the processes`` ``// according to arrival time`` ``int` `index = 0;`` ``for` `(``int` `i = 0; i < SIZE - 1; i++) {`` ``index = i;`` ``for` `(``int` `j = i + 1; j < SIZE; j++) {`` ``if` `(arr[j].atime`` ``< arr[index].atime) {`` ``index = j;`` ``}`` ``}`` ``swap(arr[i], arr[index]);`` ``}` ` ``// ctime stores the current run time`` ``int` `ctime` `= arr[0].atime;` ` ``// priority queue, wait, is used`` ``// to store all the processes that`` ``// arrive <= ctime (current run time)`` ``// this is a minimum priority queue`` ``// that arranges values according to`` ``// the burst time of the processes.`` ``priority_queue,`` ``cmpBtime>`` ``wait;` ` ``int` `temp = arr[0].atime;` ` ``// The first process is`` ``// pushed in the wait queue.`` ``wait.push(arr[0]);`` ``arr[0].atime = -1;` ` ``cout << ``"Process id"`` ``<< ``"\t"``;`` ``cout << ``"Arrival time"`` ``<< ``"\t"``;`` ``cout << ``"Burst time"`` ``<< ``"\t"``;` ` ``cout << endl;` ` ``while` `(!wait.empty()) {` ` ``cout << ``"\t"``;`` ``cout << wait.top().pname << ``"\t\t"``;`` ``cout << wait.top().atime << ``"\t\t"``;`` ``cout << wait.top().btime << ``"\t\t"``;`` ``cout << endl;` ` ``// ctime is increased with`` ``// the burst time of the`` ``// currently executed process.`` ``ctime` `+= wait.top().btime;` ` ``// The executed process is`` ``// removed from the wait queue.`` ``wait.pop();` ` ``for` `(``int` `i = 0; i < SIZE; i++) {`` ``if` `(arr[i].atime <= ``ctime`` ``&& arr[i].atime != -1) {`` ``wait.push(arr[i]);` ` ``// When the process once`` ``// enters the wait queue`` ``// its arrival time is`` ``// assigned to -1 so that`` ``// it doesn't enter again`` ``// int the wait queue.`` ``arr[i].atime = -1;`` ``}`` ``}`` ``}``}` `// Driver Code``int` `main()``{`` ``// an array of process info structures.`` ``proinfo arr[SIZE];` ` ``arr[0] = { ``"p1"``, 4, 3 };`` ``arr[1] = { ``"p2"``, 0, 8 };`` ``arr[2] = { ``"p3"``, 5, 4 };`` ``arr[3] = { ``"p4"``, 9, 2 };` ` ``cout << ``"Process scheduling "``;`` ``cout << ``"according to SJF is: \n"`` ``<< endl;` ` ``sjfNonpremetive(arr);``}`
Java
`// Java implementation of SJF``import` `java.util.*;` `class` `GFG {`` ``// number of process`` ``static` `int` `SIZE = ``4``;` ` ``// A class to represent the`` ``// process information`` ``static` `class` `proinfo {`` ``String pname; ``// process name`` ``int` `atime; ``// arrival time`` ``int` `btime; ``// burst time`` ``proinfo(String pname, ``int` `atime, ``int` `btime)`` ``{`` ``this``.pname = pname;`` ``this``.atime = atime;`` ``this``.btime = btime;`` ``}`` ``}` ` ``static` `void` `swap(``int` `i, ``int` `idx, proinfo[] arr)`` ``{`` ``proinfo tmp = arr[i];`` ``arr[i] = arr[idx];`` ``arr[idx] = tmp;`` ``}` ` ``// This function schedules the`` ``// process according to the SJF`` ``// scheduling algorithm.`` ``static` `void` `sjfNonpremetive(proinfo[] arr)`` ``{`` ``// Used to sort the processes`` ``// according to arrival time`` ``int` `index = ``0``;`` ``for` `(``int` `i = ``0``; i < SIZE - ``1``; i++) {`` ``index = i;`` ``for` `(``int` `j = i + ``1``; j < SIZE; j++) {`` ``if` `(arr[j].atime < arr[index].atime) {`` ``index = j;`` ``}`` ``}`` ``swap(i, index, arr);`` ``}` ` ``// ctime stores the current run time`` ``int` `ctime = arr[``0``].atime;` ` ``// priority queue, wait, is used`` ``// to store all the processes that`` ``// arrive <= ctime (current run time)`` ``// this is a minimum priority queue`` ``// that arranges values according to`` ``// the burst time of the processes.`` ``PriorityQueue wait`` ``= ``new` `PriorityQueue(`` ``(a, b) -> { ``return` `a.btime - b.btime; });` ` ``// The first process is`` ``// pushed in the wait queue.`` ``wait.add(``new` `proinfo(arr[``0``].pname, arr[``0``].atime,`` ``arr[``0``].btime));`` ``arr[``0``].atime = -``1``;` ` ``System.out.print(``"Process id"`` ``+ ``"\t"``);`` ``System.out.print(``"Arrival time"`` ``+ ``"\t"``);`` ``System.out.println(``"Burst time\t"``);` ` ``while` `(!wait.isEmpty()) {` ` ``System.out.print(``"\t"``);`` ``System.out.print(wait.peek().pname + ``"\t\t"``);`` ``System.out.print(wait.peek().atime + ``"\t\t"``);`` ``System.out.println(wait.peek().btime + ``"\t\t"``);` ` ``// ctime is increased with`` ``// the burst time of the`` ``// currently executed process.`` ``ctime += wait.peek().btime;` ` ``// The executed process is`` ``// removed from the wait queue.`` ``wait.remove();` ` ``for` `(``int` `i = ``0``; i < SIZE; i++) {`` ``if` `(arr[i].atime <= ctime`` ``&& arr[i].atime != -``1``) {`` ``wait.add(``new` `proinfo(arr[i].pname,`` ``arr[i].atime,`` ``arr[i].btime));` ` ``// When the process once`` ``// enters the wait queue`` ``// its arrival time is`` ``// assigned to -1 so that`` ``// it doesn't enter again`` ``// int the wait queue.`` ``arr[i].atime = -``1``;`` ``}`` ``}`` ``}`` ``}` ` ``// Driver Code`` ``public` `static` `void` `main(String[] args)`` ``{`` ``// an array of process info structures.`` ``proinfo[] arr = ``new` `proinfo[SIZE];`` ``arr[``0``] = ``new` `proinfo(``"p1"``, ``4``, ``3``);`` ``arr[``1``] = ``new` `proinfo(``"p2"``, ``0``, ``8``);`` ``arr[``2``] = ``new` `proinfo(``"p3"``, ``5``, ``4``);`` ``arr[``3``] = ``new` `proinfo(``"p4"``, ``9``, ``2``);` ` ``System.out.print(``"Process scheduling "``);`` ``System.out.println(``"according to SJF is: \n"``);`` ``sjfNonpremetive(arr);`` ``}``}``// This code is contributed by Karandeep Singh`
Python3
`# Python 3 implementation of SJF``import` `heapq as hq``# number of process``SIZE``=``4` `# This function schedules the``# process according to the SJF``# scheduling algorithm.``def` `sjfNonpremetive(arr):`` ``# Used to sort the processes`` ``# according to arrival time`` ``index ``=` `0`` ``for` `i ``in` `range``(SIZE ``-` `1``):`` ``index ``=` `i`` ``for` `j ``in` `range``(i ``+` `1``, SIZE) :`` ``if` `(arr[j][``1``] < arr[index][``1``]) :`` ``index ``=` `j`` ` ` ` ` ``arr[i], arr[index]``=``arr[index],arr[i]`` ` ` ``# ctime stores the current run time`` ``ctime ``=` `arr[``0``][``1``]` ` ``# priority queue, wait, is used`` ``# to store all the processes that`` ``# arrive <= ctime (current run time)`` ``# this is a minimum priority queue`` ``# that arranges values according to`` ``# the burst time of the processes.`` ``wait``=``[]` ` ``temp ``=` `arr[``0``][``1``]` ` ``# The first process is`` ``# pushed in the wait queue.`` ``hq.heappush(wait,arr[``0``].copy())`` ``arr[``0``][``1``] ``=` `-``1` ` ``print``(``"Process id"``,end``=``"\t"``)`` ``print``(``"Arrival time"``,end``=``"\t"``)`` ``print``(``"Burst time"``,end``=``"\t"``)` ` ``print``()` ` ``while` `(wait) :` ` ``print``(end``=``"\t"``)`` ``print``(wait[``0``][``2``],end``=` `"\t\t"``)`` ``print``(wait[``0``][``1``],end``=``"\t\t"``)`` ``print``(wait[``0``][``0``],end``=``"\t\t"``)`` ``print``()` ` ``# ctime is increased with`` ``# the burst time of the`` ``# currently executed process.`` ``ctime ``+``=` `wait[``0``][``0``]` ` ``# The executed process is`` ``# removed from the wait queue.`` ``hq.heappop(wait)` ` ``for` `i ``in` `range``(SIZE):`` ``if` `(arr[i][``1``] <``=` `ctime`` ``and` `arr[i][``1``] !``=` `-``1``) :`` ``hq.heappush(wait,arr[i].copy())` ` ``# When the process once`` ``# enters the wait queue`` ``# its arrival time is`` ``# assigned to -1 so that`` ``# it doesn't enter again`` ``# int the wait queue.`` ``arr[i][``1``] ``=` `-``1`` ` ` ` ` ` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:`` ``# an array of process info structures.`` ``arr``=``[``None``]``*``SIZE` ` ``arr[``0``] ``=``[``3``, ``4``, ``"p1"``]`` ``arr[``1``] ``=` `[``8``, ``0``, ``"p2"``]`` ``arr[``2``] ``=` `[``4``, ``5``, ``"p3"``]`` ``arr[``3``] ``=` `[``2``, ``9``, ``"p4"``]` ` ``print``(``"Process scheduling according to SJF is: \n"``)` ` ``sjfNonpremetive(arr)`
Javascript
`// JavaScript implementation of SJF` `// number of process``const SIZE = 4;` `// A class to represent the``// process information``class ProInfo {`` ``constructor(pname, atime, btime) {`` ``this``.pname = pname;`` ``this``.atime = atime;`` ``this``.btime = btime;`` ``}``}` `const swap = (i, idx, arr) => {`` ``const tmp = arr[i];`` ``arr[i] = arr[idx];`` ``arr[idx] = tmp;``};` `// This function schedules the``// process according to the SJF``// scheduling algorithm.``const sjfNonPremetive = arr => {`` ``// Used to sort the processes`` ``// according to arrival time`` ``let index = 0;`` ``for` `(let i = 0; i < SIZE - 1; i++) {`` ``index = i;`` ``for` `(let j = i + 1; j < SIZE; j++) {`` ``if` `(arr[j].atime < arr[index].atime) {`` ``index = j;`` ``}`` ``}`` ``swap(i, index, arr);`` ``}` ` ``// ctime stores the current run time`` ``let ctime = arr[0].atime;` ` ``// wait is used to store all the processes that`` ``// arrive <= ctime (current run time)`` ``// this is an array that arranges values according to`` ``// the burst time of the processes.`` ``let wait = [];` ` ``// The first process is`` ``// pushed in the wait queue.`` ``wait.push(``new` `ProInfo(arr[0].pname, arr[0].atime, arr[0].btime));`` ``arr[0].atime = -1;` ` ``console.log(``"Process\tArrival\tBurst\t"``);` ` ``while` `(wait.length > 0) {`` ``wait.sort((a, b) => a.btime - b.btime);`` ``const process = wait.shift();`` ` ` ``console.log(`\t\${process.pname}\t\t\${process.atime}\t\t\${process.btime}\t\t`);` ` ``// ctime is increased with`` ``// the burst time of the`` ``// currently executed process.`` ``ctime += process.btime;` ` ``for` `(let i = 0; i < SIZE; i++) {`` ``if` `(arr[i].atime <= ctime && arr[i].atime !== -1) {`` ``wait.push(``new` `ProInfo(arr[i].pname, arr[i].atime, arr[i].btime));` ` ``// When the process once`` ``// enters the wait queue`` ``// its arrival time is`` ``// assigned to -1 so that`` ``// it doesn't enter again`` ``// int the wait queue.`` ``arr[i].atime = -1;`` ``}`` ``}`` ``}``};` `// Driver Code``console.log(``"Process scheduling according to SJF is: \n"``);``const arr = [`` ``new` `ProInfo(``"p1"``, 4, 3),`` ``new` `ProInfo(``"p2"``, 0, 8),`` ``new` `ProInfo(``"p3"``, 5, 4),`` ``new` `ProInfo(``"p4"``, 9, 2)];``console.log(``"Process scheduling "``);``console.log(``"According to SJF is: \n"``);``sjfNonPremetive(arr);` `// This code is contributed by anskalyan3`
C#
`// C# implementation of SJF``using` `System;``using` `System.Collections.Generic;` `namespace` `SJF``{`` ``// Structure to store the process information`` ``public` `struct` `proinfo`` ``{`` ``public` `string` `pname; ``// process name`` ``public` `int` `atime; ``// arrival time`` ``public` `int` `btime; ``// burst time`` ``}` ` ``// This structure maintains the wait queue, using burst`` ``// time to compare.`` ``public` `class` `cmpBtime : IComparer`` ``{`` ``public` `int` `Compare(proinfo a, proinfo b)`` ``{`` ``return` `a.btime.CompareTo(b.btime);`` ``}`` ``}` ` ``class` `Program`` ``{`` ``// number of processes`` ``const` `int` `SIZE = 4;` ` ``// This function schedules the process according to the SJF scheduling algorithm.`` ``static` `void` `sjfNonpremetive(proinfo[] arr)`` ``{`` ``// Used to sort the processes according to arrival time`` ``int` `index;`` ``for` `(``int` `i = 0; i < SIZE - 1; i++)`` ``{`` ``index = i;`` ``for` `(``int` `j = i + 1; j < SIZE; j++)`` ``{`` ``if` `(arr[j].atime < arr[index].atime)`` ``{`` ``index = j;`` ``}`` ``}`` ``Swap(``ref` `arr[i], ``ref` `arr[index]);`` ``}` ` ``// ctime stores the current run time`` ``int` `ctime = arr[0].atime;` ` ``// priority queue, wait, is used to store all the processes that arrive <= ctime (current run time)`` ``// this is a minimum priority queue that arranges values according to the burst time of the processes.`` ``var` `wait = ``new` `SortedSet(``new` `cmpBtime());` ` ``int` `temp = arr[0].atime;` ` ``// The first process is pushed in the wait queue.`` ``wait.Add(arr[0]);`` ``arr[0].atime = -1;` ` ``Console.WriteLine(``"Process id\tArrival time\tBurst time\n"``);` ` ``while` `(wait.Count > 0)`` ``{`` ``Console.Write(``"\t"``);`` ``Console.Write(wait.Min.pname + ``"\t\t"``);`` ``Console.Write(wait.Min.atime + ``"\t\t"``);`` ``Console.Write(wait.Min.btime + ``"\t\t"``);`` ``Console.WriteLine();` ` ``// ctime is increased with the burst time of the currently executed process.`` ``ctime += wait.Min.btime;` ` ``// The executed process is removed from the wait queue.`` ``wait.Remove(wait.Min);` ` ``for` `(``int` `i = 0; i < SIZE; i++)`` ``{`` ``if` `(arr[i].atime <= ctime && arr[i].atime != -1)`` ``{`` ``wait.Add(arr[i]);` ` ``// When the process once enters the wait queue its arrival time is`` ``// assigned to -1 so that it doesn't enter again in the wait queue.`` ``arr[i].atime = -1;`` ``}`` ``}`` ``}`` ``}` ` ``// Helper method to swap two Proinfo elements in an array`` ``private` `static` `void` `Swap(``ref` `proinfo a, ``ref` `proinfo b)`` ``{`` ``proinfo temp = a;`` ``a = b;`` ``b = temp;`` ``}`` ``static` `void` `Main(``string``[] args)`` ``{`` ``// an array of process info structures.`` ``proinfo[] arr = ``new` `proinfo[SIZE];` ` ``arr[0] = ``new` `proinfo { pname = ``"p1"``, atime = 4, btime = 3 };`` ``arr[1] = ``new` `proinfo { pname = ``"p2"``, atime = 0, btime = 8 };`` ``arr[2] = ``new` `proinfo { pname = ``"p3"``, atime = 5, btime = 4 };`` ``arr[3] = ``new` `proinfo { pname = ``"p4"``, atime = 9, btime = 2 };`` ` ` ``Console.Write(``"Process scheduling "``);`` ``Console.Write(``"according to SJF is: \n"``);`` ` ` ``sjfNonpremetive(arr);`` ``}`` ``}``}`
Output:
```Process scheduling according to SJF is:
Process id Arrival time Burst time
p2 0 8
p1 4 3
p4 9 2
p3 5 4```
Time Complexity: O(N^2).
Auxiliary Space: O(N). | 6,582 | 19,716 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2023-40 | latest | en | 0.884104 |
https://www.scala-lang.org/files/archive/api/3.1.3/scala/Function$.html | 1,679,800,224,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945381.91/warc/CC-MAIN-20230326013652-20230326043652-00257.warc.gz | 1,069,321,516 | 22,078 | # Function
object Function
A module defining utility methods for higher-order functional programming.
Source:
Function.scala
class Object
trait Matchable
class Any
## Value members
### Concrete methods
def chain[T](fs: Seq[T => T]): T => T
Given a sequence of functions `f1`, ..., `fn`, return the function `f1 andThen ... andThen fn`.
Given a sequence of functions `f1`, ..., `fn`, return the function `f1 andThen ... andThen fn`.
Value parameters:
fs
The given sequence of functions
Source:
Function.scala
def const[T, U](x: T)(y: U): T
The constant function
The constant function
Source:
Function.scala
def tupled[T1, T2, R](f: (T1, T2) => R): ((T1, T2)) => R
Tupling for functions of arity 2.
Tupling for functions of arity 2. This transforms a function of arity 2 into a unary function that takes a pair of arguments.
Note:
These functions are slotted for deprecation, but it is on hold pending superior type inference for tupling anonymous functions.
Source:
Function.scala
def tupled[T1, T2, T3, R](f: (T1, T2, T3) => R): ((T1, T2, T3)) => R
Tupling for functions of arity 3.
Tupling for functions of arity 3. This transforms a function of arity 3 into a unary function that takes a triple of arguments.
Source:
Function.scala
def tupled[T1, T2, T3, T4, R](f: (T1, T2, T3, T4) => R): ((T1, T2, T3, T4)) => R
Tupling for functions of arity 4.
Tupling for functions of arity 4. This transforms a function of arity 4 into a unary function that takes a 4-tuple of arguments.
Source:
Function.scala
def tupled[T1, T2, T3, T4, T5, R](f: (T1, T2, T3, T4, T5) => R): ((T1, T2, T3, T4, T5)) => R
Tupling for functions of arity 5.
Tupling for functions of arity 5. This transforms a function of arity 5 into a unary function that takes a 5-tuple of arguments.
Source:
Function.scala
def uncurried[T1, T2, R](f: T1 => T2 => R): (T1, T2) => R
Uncurrying for functions of arity 2.
Uncurrying for functions of arity 2. This transforms a unary function returning another unary function into a function of arity 2.
Source:
Function.scala
def uncurried[T1, T2, T3, R](f: T1 => T2 => T3 => R): (T1, T2, T3) => R
Uncurrying for functions of arity 3.
Uncurrying for functions of arity 3.
Source:
Function.scala
def uncurried[T1, T2, T3, T4, R](f: T1 => T2 => T3 => T4 => R): (T1, T2, T3, T4) => R
Uncurrying for functions of arity 4.
Uncurrying for functions of arity 4.
Source:
Function.scala
def uncurried[T1, T2, T3, T4, T5, R](f: T1 => T2 => T3 => T4 => T5 => R): (T1, T2, T3, T4, T5) => R
Uncurrying for functions of arity 5.
Uncurrying for functions of arity 5.
Source:
Function.scala
def unlift[T, R](f: T => Option[R]): PartialFunction[T, R]
Turns a function `A => Option[B]` into a `PartialFunction[A, B]`.
Turns a function `A => Option[B]` into a `PartialFunction[A, B]`.
Important note: this transformation implies the original function may be called 2 or more times on each logical invocation, because the only way to supply an implementation of `isDefinedAt` is to call the function and examine the return value. See also scala.PartialFunction, method `applyOrElse`.
Value parameters:
f
a function `T => Option[R]`
Returns:
a partial function defined for those inputs where f returns `Some(_)` and undefined where `f` returns `None`.
scala.PartialFunction, method `lift`.
Source:
Function.scala
def untupled[T1, T2, R](f: ((T1, T2)) => R): (T1, T2) => R
Un-tupling for functions of arity 2.
Un-tupling for functions of arity 2. This transforms a function taking a pair of arguments into a binary function which takes each argument separately.
Source:
Function.scala
def untupled[T1, T2, T3, R](f: ((T1, T2, T3)) => R): (T1, T2, T3) => R
Un-tupling for functions of arity 3.
Un-tupling for functions of arity 3. This transforms a function taking a triple of arguments into a ternary function which takes each argument separately.
Source:
Function.scala
def untupled[T1, T2, T3, T4, R](f: ((T1, T2, T3, T4)) => R): (T1, T2, T3, T4) => R
Un-tupling for functions of arity 4.
Un-tupling for functions of arity 4. This transforms a function taking a 4-tuple of arguments into a function of arity 4 which takes each argument separately.
Source:
Function.scala
def untupled[T1, T2, T3, T4, T5, R](f: ((T1, T2, T3, T4, T5)) => R): (T1, T2, T3, T4, T5) => R
Un-tupling for functions of arity 5.
Un-tupling for functions of arity 5. This transforms a function taking a 5-tuple of arguments into a function of arity 5 which takes each argument separately.
Source:
Function.scala | 1,407 | 4,528 | {"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} | 2.65625 | 3 | CC-MAIN-2023-14 | latest | en | 0.714579 |
http://mathhelpforum.com/algebra/64354-expansion-print.html | 1,506,231,298,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818689874.50/warc/CC-MAIN-20170924044206-20170924064206-00015.warc.gz | 218,368,531 | 2,822 | # expansion
• Dec 10th 2008, 08:44 AM
hmmmm
expansion
how would you expand (a^x+b)^y?
sorry if this is in wrong place but i wasnt sure where to put it.
thanks for any help.
• Dec 10th 2008, 08:49 AM
Jhevon
Quote:
Originally Posted by hmmmm
how would you expand (a^x+b)^y?
sorry if this is in wrong place but i wasnt sure where to put it.
thanks for any help.
look up binomial expansion
$(x + y)^n = \sum_{k = 0}^{n} {n \choose k}x^{n - k}y^k$
• Dec 10th 2008, 08:53 AM
hmmmm
thanks
thanks very much but what happens if the x has a power that isnt 1?
• Dec 10th 2008, 12:36 PM
Jhevon
Quote:
Originally Posted by hmmmm
thanks very much but what happens if the x has a power that isnt 1?
say we write x as $a^x$ in the above formula:
$(a^x + y)^n = \sum_{k = 0}^{n} {n \choose k}(a^x)^{n - k}y^k$ | 297 | 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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 3, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2017-39 | longest | en | 0.949386 |
https://fortunetelleroracle.com/health-diet-products/how-can-bmi-calculator-help-in-tracking-your-body-mass-738402 | 1,679,465,303,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943750.71/warc/CC-MAIN-20230322051607-20230322081607-00726.warc.gz | 308,039,654 | 10,854 | How Can BMI Calculator Help in tracking Your Body Mass
# How Can BMI Calculator Help in tracking Your Body Mass
Here's your chance to show off your creative side. Our BMI Calculator can help you find out your ideal weight for both health and beauty reasons. Whether you're looking to increase your muscle mass or lose weight, our tool can help make a difference. Just input in all of the pertinent data and we'll take care of the rest. So what are you waiting for? Start using our BMI Calculator today!
## How Can BMI Calculator Help in tracking Your Body Mass
Having it in your insight that the amount you gauge and the amount you want to keep up with assumes a basic part in your general prosperity. It is basic to know all about the general proportion of your Height and weight, and it requests a smart numerical calculation.
Fortunately these days, there are such creative projects accessible on online platforms every one can access with one click, which feature online calculators. These are online calculators that consequently sort out the BMI and everything you really want to do is input weight in kg and height in feet. The most awesome aspect of using a BMI calculator is that it will provide you with an estimate regardless of whether your body mass is according to your height.
# Fast steps for computing body mass-BMI
1. To calculate your BMI, you should know your weight in (pounds) or (kg) and height in (feet).
2. Multiply the body mass by 4.88 and divide it by your squared height.
3. It's not so easy to calculate again and again go online and enter your height in your feet and weight in kg or pounds you know your BMI and some precious suggestion for improvements according to your BMI that getting through the BMI calculator.
Besides using BMI calculators, you can utilize a BMI Chart. The outline includes a general weight Chart, in which, you can analyze the fittingness of your body mass. For the most part, in a BMI Chart, body mass is given on the even pivot though Height is given on the upward hub. You should simply draw a parallelism between your Height and weight.
A BMI calculator will cause you to distinguish whether you are underweight, or overweight, depending on your age, weight, and gender. Although BMI Chart for women provides values according to a lady’s body shape or weight and height the to Hence, utilizing a BMI Chart can get awareness and adopt exercise and diet, it is constantly prescribed by specialists to involve a BMI calculator for precise outcomes. It has been seen that now and again individuals who are strong can't properly assess their body fat. The great explanation is that strong weight is heavier than body fat, which leads to giving deceiving results.
All in all, as per specialists, BMI tests play a crucial part in deciding body fat for a deskbound person's way of life. Other than this, this test is prescribed to people who don't exercise consistently, take their exercises sincerely, or more all, remove eating unhealthy food. BMI for the most part invests an outline without getting into the wreck of those modern tests and costly bits of equipment that every last one of us needs to stay away from.
Alsa Pakistan Share Knowledgeable data regarding the BMI calculator in this article. The BMI calculator is utilized as the best gear that assists with dealing with your eating routine arrangement and works on your technique for weig | 686 | 3,418 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2023-14 | latest | en | 0.918841 |
https://stromcv.com/what-is-the-missing-step-in-the-proof-37 | 1,674,916,314,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499634.11/warc/CC-MAIN-20230128121809-20230128151809-00119.warc.gz | 540,632,181 | 4,002 | # What is the missing step in the proof
what is the missing step in this proof? (A)Statement: The slope of The slope of. Reason: Definition of the slope. (B)Statement: The slope of. A: Given that:. One of its implications — a
## Solved Given: AABC with ABS What is the missing step in this
The Number E Is Irrational Fill In All The Missing Steps. The slope is zero since it is a horizontal line and it crosses the y axis at y =1 so the y intercept is (0,1) part 2. We have to
310 Math Teachers
9.1/10 Quality score
24700 Student Reviews | 144 | 541 | {"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-2023-06 | latest | en | 0.896176 |
https://exceljet.net/videos/how-to-look-things-up-with-index | 1,670,105,526,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710941.43/warc/CC-MAIN-20221203212026-20221204002026-00039.warc.gz | 278,834,911 | 10,683 | Practice worksheet included with online video training.
## Transcript
What does the INDEX function do?
Unlike the MATCH function, which gets the position of an item in a list or a table, INDEX assumes you already know the position, and it gets the value of the item at that position.
Let's take a look.
INDEX is a powerful and flexible function that can be used for advanced features like dynamic ranges and two-way lookups.
However, in this example, we're going to use INDEX for the simplest use case possible—retrieving items from a known position.
We're going to use INDEX to look up a city at a given position in this list; let's say the 3rd position.
In its simplest form, INDEX takes just two arguments: an array and a row number.
So in this case, we give INDEX an array, which is just a reference to the city name, and then point back to the position for the row number.
INDEX correctly retrieves Chicago from the list of cities. If I put in "13" we'll get Austin, the 13th city in the list.
So far, we're just using INDEX to retrieve an item from a single column based on its position. Now let's expand on this to use INDEX with multiple columns.
First, to make the formulas easier to read, I'll name the ranges for both the city data and the position. Now, I'll add the first INDEX formula to retrieve the city name.
Because we're now using INDEX on a multi-column table, I need to add one more argument: the column number from which to get the city name. In this case, that's the column number "1".
Now I can copy the formula down and come back and adjust the column numbers for each INDEX formula.
As you can see, we can easily retrieve all the information for a given city with INDEX by supplying the right row and column numbers.
Now, at this point, you're probably feeling a little underwhelmed. So what if you can retrieve something when you know the position, you might be thinking; you hardly ever know the position of anything in a worksheet.
Yes, that's true, and that's where the MATCH function comes in. MATCH can find the position of an item. And, when you use INDEX together with MATCH, you can do everything VLOOKUP can do, and a lot more.
Author
### Dave Bruns
Hi - I'm Dave Bruns, and I run Exceljet with my wife, Lisa. Our goal is to help you work faster in Excel. We create short videos, and clear examples of formulas, functions, pivot tables, conditional formatting, and charts. | 550 | 2,428 | {"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-2022-49 | longest | en | 0.909015 |
https://socratic.org/questions/a-compound-has-an-empirical-formula-of-hco-2-and-a-molecular-mass-of-90-grams-pe | 1,718,273,929,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861372.90/warc/CC-MAIN-20240613091959-20240613121959-00862.warc.gz | 503,232,407 | 5,839 | A compound has an empirical formula of HCO_2 and a molecular mass of 90 grams per mole. What is the molecular formula of this compound?
Feb 25, 2016
The molecular formula is always a multiple of the empirical formula: ("empirical formula")_n ="molecular formula".
Explanation:
So $\left(1.00794 + 12.011 + 2 \times 15.999\right) \cdot g \cdot m o {l}^{-} 1 \times n$ $=$ $90 \cdot g \cdot m o {l}^{-} 1$.
We solve for $n$, and clearly $n$ $=$ $2$.
So empirical formula is ${C}_{2} {H}_{2} {O}_{4}$. | 166 | 504 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 9, "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-2024-26 | latest | en | 0.744716 |
http://reference.wolfram.com/legacy/v7/ref/ImageData.html | 1,505,997,068,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818687766.41/warc/CC-MAIN-20170921115822-20170921135822-00252.warc.gz | 290,419,069 | 10,027 | This is documentation for Mathematica 7, which was
based on an earlier version of the Wolfram Language.
# ImageData
ImageData[image] gives the array of pixel values in image. ImageData[image, "type"]gives the array of pixel values converted to the specified type.
• ImageData[image] by default gives a 2D array whose elements are lists of values for each channel.
• The array generated by ImageData[image] is arranged in the same way that the image is displayed, with the first row corresponding to the top row of pixels in the image.
• For binary images, ImageData[image] returns integer values 0 or 1. For all other images, ImageData[image] returns real values, normally between 0 and 1.
• Possible types specified by ImageData[image, "type"] are:
"Bit" integer 0 or 1 "Byte" integer 0 through 255 "Bit16" integer 0 through 65535 "Real" machine real
• For images of type "Byte" or "Bit16", ImageData[image] always normalizes values to lie between 0 and 1. For images of type "Real", ImageData[image] returns whatever real values are used in the image.
• With the default setting , ImageData returns a 2D array of lists of channel values.
• With , ImageData returns a list of 2D arrays of values for each channel.
• returns in the native form used to store the image.
Channel data for the first five pixels of the first row:
Construct a byte-valued image:
Extract normalized raster data:
Extract raw bytes:
Channel data for the first five pixels of the first row:
Out[1]=
Construct a byte-valued image:
Out[1]=
Extract normalized raster data:
Out[2]=
Extract raw bytes:
Out[3]=
Options (1)
Data of multi-channel images is by default returned in interleaved form:
This gives the pixel data in planar form:
New in 7 | 419 | 1,726 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2017-39 | longest | en | 0.673926 |
https://www.experts-exchange.com/questions/23839236/Calculating-values-using-data-from-grouped-rows.html | 1,480,825,261,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541187.54/warc/CC-MAIN-20161202170901-00078-ip-10-31-129-80.ec2.internal.warc.gz | 951,282,668 | 32,992 | Solved
# Calculating values using data from grouped rows
Posted on 2008-10-22
265 Views
I have a query that returns one or two rows (typically) and would like to be able to calculate the difference and percent change between one of the fields to show it in a report. Specifically, using the TotalSales from row A, which is this year, and TotalSales from row B, which is last year, and having the group values show the difference of A-B and percent change from the same.
0
Question by:Rich
• 7
• 6
LVL 16
Expert Comment
Try In a new column type the following
PercentChange: 100*((sum([rowB])-sum([rowA]))/sum([rowA])
0
Author Comment
The question is how to access just rowA and rowB in the query?
0
LVL 16
Expert Comment
I am not sure that I understand what you are trying to do. I believe you wanted to have a report that shows the percentage change.
In that case just do all you calculation in the query. Then create a report from the query.
You can also create the recordset programatically from within the report but that's a little bit more complicated
0
Author Comment
That is correct: percentage change from valueA in rowA to valueA in rowB
So if table sales has Location, Date and SalesAmount, and records such as:
loc1, 10/10/2007, \$40
loc2, 10/10/2007, \$40
loc3, 10/10/2007, \$60
loc1, 10/10/2008, \$50
loc2, 10/10/2008, \$30
loc3, 10/10/2008, \$70
I want a report that shows:
--- start of report ---
loc1
10/10/2008 \$50
10/10/2007 \$40
\$change = \$10 +25%
loc2
10/10/2008 \$30
10/10/2007 \$40
\$change = -\$10 -33%
loc2
10/10/2008 \$70
10/10/2007 \$60
\$ change - \$10 +17%
----- end of report -----
I can get everything except the \$ change and the percent change, since I do not know how to reference data in the two rows needed as the same time.
0
LVL 16
Expert Comment
OK in that case just perform the calculation from the report text box.
Let's say the \$50 is in textbox1 and the \$40 is in textbox2
then you create the \$ change textbox as an unbounded textbox and set the record source to
= [textbox2]-[textbox1]
You also need an unbounded textbox for the percentage. Its record source will be
=[textbox\$change]/[textbox2] set its format to percentage
0
Author Comment
Are you saying that the returned row values of each line in the report are in a 'addressable' textbox, and that I have access to them? If so, what is the syntax? So if I was to use an OnLoad event for the report, I could refer back to each value in the report and calculate/populate the \$ and % change?
0
LVL 16
Expert Comment
Try the following code for the onload event
Me.textbox\$change = Me.textbox2-Me.textbox1
Me.TextboxPercent = Me.textbox\$change/Me.textbox2
0
Author Comment
The problem is that I don't see how the text box with the value is named textbox1 and textbox2, it is named [Total Sales], without number; there is only one text box on the report, in the repeating detail section, called [Total Sales]. Can multiple (rows) of data in [Total Sales] simply be referenced by adding a 1 or 2 after the name. I don't think so, and I tried your code, but the Me.AmountChange and Me.PercentChange were not populated.
0
LVL 16
Expert Comment
send a screen shoot of your report so that I can have a beter understanding of how its designed
The name textbox1 and textbox2 are just genreic names that I am using as example. Your code must contain the actual name of the relevant fields. You have to find out the name of the field that holds the total sale for this year and last year. To find this open your report in design view. Right click on the textbox that holds these value and click property. Go to the "other" tab. The name of the field will be on the first row.
"I don't think so, and I tried your code, but the Me.AmountChange and Me.PercentChange were not populated. "
Attach a snippet of your code so that I can have a look and try to figure out where you've gone wrong.
0
Author Comment
Thanks for helping. I have added screen shots of the report witht the two fields that I am trying to calculate, [Dollar Change] and [Peercent Change], which should be based on the [Total Sales] values from the two rows above. I have also added a shot of the report design, showing that the values that I am trying to get are in [Total Sales], but from multiple rows, which is what I am having questions about getting.
There is no code to calculate the [Dollar Change] and [Percent Change] values at this time since I do not know how to reference the [Total Sales] variables.
report.gif
report-design.gif
0
LVL 16
Accepted Solution
Sheils earned 500 total points
I see why this approach is not working. The [Total Sale] is the same for both current and previous year. The report from report view was unreadable but the design view has given ne a fairly good idea of what the problem is.
You will need to create a query recordset at the on current event. Then use the field from the recordset to populate the unbounded fields
0
LVL 16
Expert Comment
You are probably not familiar with sql and record sets. There a few websites that might help you out. And do feel free to send a sample/copy of your database and I fix it for you. This might be the most dtraight forward way of solving this. But you will have to send it in the 2003 format
http://www.codecoffee.com/articles/sql1.html
http://www.w3schools.com/sql/sql_syntax.asp
http://articles.techrepublic.com.com/5100-10878_11-1050307.html
Plus some stuffs on recordset attached
Pages-from-Access-2007-VBA-Progr.pdf
0
Author Closing Comment
Thanks for looking into this for me.
Actually, I figured out at least one solution without having to manually create the recordset, which is to add some VBA code to the Format event of the various sections of the report. The VBA collects and summarizes the data just printed on the report in Public variables, and then calculates and fills in the results in unbound variables that I added to the footer sections.
Not sure if this is the best or most effecient way to solve this, and it only works with Print Preview.
0
## Featured Post
In the previous article, Using a Critera Form to Filter Records (http://www.experts-exchange.com/A_6069.html), the form was basically a data container storing user input, which queries and other database objects could read. The form had to remain op…
In a multiple monitor setup, if you don't want to use AutoCenter to position your popup forms, you have a problem: where will they appear? Sometimes you may have an additional problem: where the devil did they go? If you last had a popup form open…
Familiarize people with the process of retrieving data from SQL Server using an Access pass-thru query. Microsoft Access is a very powerful client/server development tool. One of the ways that you can retrieve data from a SQL Server is by using a pa…
In Microsoft Access, when working with VBA, learn some techniques for writing readable and easily maintained code. | 1,766 | 7,007 | {"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} | 2.953125 | 3 | CC-MAIN-2016-50 | longest | en | 0.882829 |
https://simple.m.wikipedia.org/wiki/Transitivity_(mathematics) | 1,720,889,604,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514510.7/warc/CC-MAIN-20240713150314-20240713180314-00314.warc.gz | 433,593,764 | 17,895 | # Transitivity (mathematics)
binary relation R with the property that xRy and yRz implies xRz
In logic and mathematics, transitivity is a property of a binary relation. It is a prerequisite of an equivalence relation and of a partial order.
## Definition and examples
In general, given a set with a relation, the relation is transitive if whenever a is related to b and b is related to c, then a is related to c. For example:
• Size is transitive: if A>B and B>C, then A>C. [1]
• Subsets are transitive: if A is a subset of B and B is a subset of C, then A is a subset of C.
• Height is transitive: if Sidney is taller than Casey, and Casey is taller than Jordan, then Sidney is taller than Jordan.
• Rock, paper, scissors is not transitive: rock beats scissors, and scissors beats paper, but rock doesn't beat paper. This is called an intransitive relation.
Given a relation ${\displaystyle R}$ , the smallest transitive relation containing ${\displaystyle R}$ is called the transitive closure of ${\displaystyle R}$ , and is written as ${\displaystyle R^{+}}$ .[2]
## References
1. "Transitivity". nrich.maths.org. Retrieved 2020-10-12.
2. "Comprehensive List of Algebra Symbols". Math Vault. 2020-03-25. Retrieved 2020-10-12. | 323 | 1,237 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 4, "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-2024-30 | latest | en | 0.891938 |
http://stackoverflow.com/questions/4424374/determining-if-a-number-is-prime?answertab=active | 1,419,632,592,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1419447549808.107/warc/CC-MAIN-20141224185909-00084-ip-10-231-17-201.ec2.internal.warc.gz | 95,853,597 | 25,873 | # Determining if a number is prime
I have perused a lot of code on this topic, but most of them produce the numbers that are prime all the way up to the input number. However, I need code which only checks whether the given input number is prime.
Here is what I was able to write, but it does not work:
``````void primenumber(int number)
{
if(number%2!=0)
cout<<"Number is prime:"<<endl;
else
cout<<"number is NOt prime"<<endl;
}
``````
I would appreciate if someone could give me advice on how to make this work properly.
### Update
I modified it to check on all the numbers in a for loop.
``````void primenumber(int number)
{
for(int i=1; i<number; i++)
{
if(number%i!=0)
cout<<"Number is prime:"<<endl;
else
cout<<"number is NOt prime"<<endl;
}
}
``````
-
All your code does is report if the number is divisible by 2. What general approach would you use to detect primes? Let's start with that, and craft it into executable code. – Michael Petrotta Dec 12 '10 at 22:13
Have you thought about what it means for a number to be prime? Write it out in pseudocode then turn it into real code. – Cameron Skinner Dec 12 '10 at 22:14
possible duplicate of deciding if a number is perfect or prime – Henk Holterman Dec 12 '10 at 22:21
## 12 Answers
You need to do some more checking. Right now, you are only checking if the number is divisible by 2. Do the same for 2, 3, 4, 5, 6, ... up to `number`. Hint: use a loop.
After you resolve this, try looking for optimizations.
-
up to the number's root... – hillel Dec 12 '10 at 22:16
There are then many optimisations. You only need to test up to SQRT(n), since if n can be factored into 2 numbers one of them must be <= SQRT(n). You can skip all even numbers (except 2) since if an even number divides n, then so will 2.... just to get you started. – James Gaunt Dec 12 '10 at 22:17
when it gets to optimization - think of the highest number to test. does it have to be 'number' or can it be a bit less? ;) edit: arrrgh, people are fast today ;) – mad Dec 12 '10 at 22:18
Or even better than skipping even numbers, you can skip all numbers that aren't one less or one greater than a multiple of 6. (multiple of 6 + 2 or 4 is divisible by 2 and multiple of 6 + 3 is divisible by 3, leaving only multiples of 6 + 1 or 5) – user470379 Dec 12 '10 at 22:20
You can also take it a step further and check numbers based on their value modulo 30, which is still feasible to do as an unrolled loop. Though of course, you get diminishing returns as you go higher...(not checking multiples of 2 cuts the number of checks in half, based off 6 only takes another 1/3rd off that, going to 30 only eliminates another 1/5th...) – Anon. Dec 12 '10 at 22:24
``````// PrimeDef.cpp : Defines the entry point for the console application.
//
#include "stdafx.h"
#include<iostream>
#include<stdlib.h>
using namespace std;
const char PRIME = 176;
const char NO_PRIME = 178;
const int LAST_VALUE = 2000;
bool isPrimeNumber( int value )
{
if( !( value / 2 ) ) return false;
for( int i = 1; ++i <= value / 2; )
if( 0 == value % i ) return false;
return true;
}
int main( int argc, char *argv[ ] )
{
char mark;
for( int i = -1, count = 1; ++i < LAST_VALUE; count++ )
{
mark = NO_PRIME;
if( isPrimeNumber( i ) ) mark = PRIME;
cout << mark;
if(i > 0 && !( count % 50 ) ) cout << endl;
}
return 0;
}
``````
-
``````bool isPrime(int number){
if(number < 2) return false;
if(number == 2) return true;
if(number % 2 == 0) return false;
for(int i=3; (i*i)<=number; i+=2){
if(number % i == 0 ) return false;
}
return true;
}
``````
-
``````#define TRUE 1
#define FALSE -1
int main()
{
/* Local variables declaration */
int num = 0;
int result = 0;
/* Getting number from user for which max prime quadruplet value is
to be found */
printf("\nEnter the number :");
scanf("%d", &num);
result = Is_Prime( num );
/* Printing the result to standard output */
if (TRUE == result)
printf("\n%d is a prime number\n", num);
else
printf("\n%d is not a prime number\n", num);
return 0;
}
int Is_Prime( int num )
{
int i = 0;
/* Checking whether number is negative. If num is negative, making
it positive */
if( 0 > num )
num = -num;
/* Checking whether number is less than 2 */
if( 2 > num )
return FALSE;
/* Checking if number is 2 */
if( 2 == num )
return TRUE;
/* Checking whether number is even. Even numbers
are not prime numbers */
if( 0 == ( num % 2 ))
return FALSE;
/* Checking whether the number is divisible by a smaller number
1 += 2, is done to skip checking divisibility by even numbers.
Iteration reduced to half */
for( i = 3; i < num; i += 2 )
if( 0 == ( num % i ))
/* Number is divisible by some smaller number,
hence not a prime number */
return FALSE;
return TRUE;
}
``````
-
I follow same algorithm but different implementation that loop to sqrt(n) with step 2 only odd numbers because I check that if it is divisible by 2 or 2*k it is false. Here is my code
``````public class PrimeTest {
public static boolean isPrime(int i) {
if (i < 2) {
return false;
} else if (i % 2 == 0 && i != 2) {
return false;
} else {
for (int j = 3; j <= Math.sqrt(i); j = j + 2) {
if (i % j == 0) {
return false;
}
}
return true;
}
}
/**
* @param args
*/
public static void main(String[] args) {
for (int i = 1; i < 100; i++) {
if (isPrime(i)) {
System.out.println(i);
}
}
}
}
``````
-
My own IsPrime() function, written and based on the deterministic variant of the famous Rabin-Miller algorithm, combined with optimized step brute forcing, giving you one of the fastest prime testing functions out there.
``````__int64 power(int a, int n, int mod)
{
__int64 power=a,result=1;
while(n)
{
if(n&1)
result=(result*power)%mod;
power=(power*power)%mod;
n>>=1;
}
return result;
}
bool witness(int a, int n)
{
int t,u,i;
__int64 prev,curr;
u=n/2;
t=1;
while(!(u&1))
{
u/=2;
++t;
}
prev=power(a,u,n);
for(i=1;i<=t;++i)
{
curr=(prev*prev)%n;
if((curr==1)&&(prev!=1)&&(prev!=n-1))
return true;
prev=curr;
}
if(curr!=1)
return true;
return false;
}
inline bool IsPrime( int number )
{
if ( ( (!(number & 1)) && number != 2 ) || (number < 2) || (number % 3 == 0 && number != 3) )
return (false);
if(number<1373653)
{
for( int k = 1; 36*k*k-12*k < number;++k)
if ( (number % (6*k+1) == 0) || (number % (6*k-1) == 0) )
return (false);
return true;
}
if(number < 9080191)
{
if(witness(31,number)) return false;
if(witness(73,number)) return false;
return true;
}
if(witness(2,number)) return false;
if(witness(7,number)) return false;
if(witness(61,number)) return false;
return true;
/*WARNING: Algorithm deterministic only for numbers < 4,759,123,141 (unsigned int's max is 4294967296)
if n < 1,373,653, it is enough to test a = 2 and 3.
if n < 9,080,191, it is enough to test a = 31 and 73.
if n < 4,759,123,141, it is enough to test a = 2, 7, and 61.
if n < 2,152,302,898,747, it is enough to test a = 2, 3, 5, 7, and 11.
if n < 3,474,749,660,383, it is enough to test a = 2, 3, 5, 7, 11, and 13.
if n < 341,550,071,728,321, it is enough to test a = 2, 3, 5, 7, 11, 13, and 17.*/
}
``````
To use, copy and paste the code into the top of your program. Call it, and it returns a BOOL value, either true or false.
``````if(IsPrime(number))
{
cout << "It's prime";
}
else
{
cout<<"It's composite";
}
``````
If you get a problem compiling with "__int64", replace that with "long". It compiles fine under VS2008 and VS2010.
How it works: There are three parts to the function. Part checks to see if it is one of the rare exceptions (negative numbers, 1), and intercepts the running of the program.
Part two starts if the number is smaller than 1373653, which is the theoretically number where the Rabin Miller algorithm will beat my optimized brute force function. Then comes two levels of Rabin Miller, designed to minimize the number of witnesses needed. As most numbers that you'll be testing are under 4 billion, the probabilistic Rabin-Miller algorithm can be made deterministic by checking witnesses 2, 7, and 61. If you need to go over the 4 billion cap, you will need a large number library, and apply a modulus or bit shift modification to the power() function.
If you insist on a brute force method, here is just my optimized brute force IsPrime() function:
``````inline bool IsPrime( int number )
{
if ( ( (!(number & 1)) && number != 2 ) || (number < 2) || (number % 3 == 0 && number != 3) )
return (false);
for( int k = 1; 36*k*k-12*k < number;++k)
if ( (number % (6*k+1) == 0) || (number % (6*k-1) == 0) )
return (false);
return true;
}
}
``````
How this brute force piece works: All prime numbers (except 2 and 3) can be expressed in the form 6k+1 or 6k-1, where k is a positive whole number. This code uses this fact, and tests all numbers in the form of 6k+1 or 6k-1 less than the square root of the number in question. This piece is integrated into my larger IsPrime() function (the function shown first).
If you need to find all the prime numbers below a number, find all the prime numbers below 1000, look into the Sieve of Eratosthenes. Another favorite of mine.
As an additional note, I would love to see anyone implement the Eliptical Curve Method algorithm, been wanting to see that implemented in C++ for a while now, I lost my implementation of it. Theoretically, it's even faster than the deterministic Rabin Miller algorithm I implemented, although I'm not sure if that's true for numbers under 4 billion.
-
Your function only accepts arguments of type int. If it accepted bigger numbers, it wouldn't work for numbers > 2^32. – gnasher729 Mar 22 at 1:14
If you are lazy, and have a lot of RAM, create a sieve of Eratosthenes which is practically a giant array from which you kicked all numbers that are not prime. From then on every prime "probability" test will be super quick. The upper limit for this solution for fast results is the amount of you RAM. The upper limit for this solution for superslow results is your hard disk's capacity.
-
working in segments, you only need to store the primes below `sqrt` of your upper limit. The segment array can be further shrinked by using bits, working with odds only, or even 2-3-5-coprimes. – Will Ness Jan 20 '13 at 9:45
Putting everything to bits from bytes you gain a 2^3 multiply of your space, which is insignificant if you are hunting for large primes. And if you store 2,3,5,7 and such primes as integers, you have to store those numbers at least as a 32 bit integer, which is a waste. – karatedog Jan 25 '13 at 22:57
working segment by narrow segment ... storing `sqrt(N)` `int` primes is worth it. – Will Ness Jan 25 '13 at 23:23
There are several different approches to this problem.
The "Naive" Method: Try all (odd) numbers up to (the root of) the number.
Improved "Naive" Method: Only try every 6n ± 1.
Probabilistic tests: Miller-Rabin, Solovay-Strasse, etc.
Which approach suits you depends and what you are doing with the prime.
You should atleast read up on Primality Testing.
-
One of my favorite pastimes, reading on Primality Testing. Always has been my favorite subject in Number Theory. Oh, and you left out Deterministic tests. ECM and AKS seems to be leading that category. – LostInTheCode Dec 12 '10 at 22:59
Yes I didn't list all types of tests. I left that as an excercise for the reader. ;) – Sani Huttunen Dec 13 '10 at 15:50
If you know the range of the inputs (which you do since your function takes an `int`), you can precompute a table of primes less than or equal to the square root of the max input (2^31-1 in this case), and then test for divisibility by each prime in the table less than or equal to the square root of the number given.
-
That could done via the Sieve of Eratosthenes, but if the range is too large, there won't be enough memory for the system to use. And anyways, half of the memory would be wasted, all the even numbers are obviously composite. Guess a map (or a deque? Forgot which is the C++ equivalent of a PHP associative array) would work best, to store all the prime numbers below a number, after you determine the prime numbers using the sieve. If it's not in the array, then it's composite. – LostInTheCode Dec 12 '10 at 23:08
@LostInTheCode There are only approximately 4300 or so numbers you would need in the table. Reread what I wrote. I didn't say to create a table saying whether each number from 1 to 2^31 is prime -- I said to store only the prime numbers from 1 to sqrt(2^31) and test the number for divisibility by each of those numbers. – user470379 Dec 13 '10 at 19:36
This code only checks if the number is divisible by two. For a number to be prime, it must not be evenly divisible by all integers less than itself. This can be naively implemented by checking if it is divisible by all integers less than `floor(sqrt(n))` in a loop. If you are interested, there are a number of much faster algorithms in existence.
-
I would guess taking sqrt and running foreach frpm 2 to sqrt+1 if(input% number!=0) return false; once you reach sqrt+1 you can be sure its prime.
- | 3,693 | 12,987 | {"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} | 3.328125 | 3 | CC-MAIN-2014-52 | latest | en | 0.901921 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.