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http://www.jiskha.com/display.cgi?id=1173065087 | 1,462,110,181,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860116173.76/warc/CC-MAIN-20160428161516-00078-ip-10-239-7-51.ec2.internal.warc.gz | 601,207,178 | 3,859 | Sunday
May 1, 2016
# Homework Help: proof by induction
Posted by Jennie on Sunday, March 4, 2007 at 10:24pm.
proof by mathmatical induction that the sum of the first n natural numbers is equal n(n+1)/2
It's true for n = 1. Assume that it is true for some n. Then the sum of the first n+1 natural integers can be obtained by dding the last number n+1 to n(n+1)/2. So, the assumption that it is true for n leads to the conclusion that for n+1 the sum must be:
n+1 + n(n+1)/2.
If the formula is correct for n+1 also, then this must be the same as:
(n+1)(n+2)/2.
Expanding out the last bracket gives:
(n+1)(n+2)/2 = (n+1)*n/2 + (n+1)*2/2 =
n+1 + n(n+1)/2.
Thank you Count Iblis | 230 | 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} | 3.9375 | 4 | CC-MAIN-2016-18 | longest | en | 0.918468 |
https://www.jiskha.com/display.cgi?id=1363221939 | 1,501,117,326,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549426693.21/warc/CC-MAIN-20170727002123-20170727022123-00612.warc.gz | 786,660,233 | 4,052 | # heritage hihg school
posted by .
list the 13 prefixes and numerical value (exponent) in order from largest to smallest. (ex;a5 where a is the prefix and 5 is the exponent!)for u use u!
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Post a New Question | 752 | 2,353 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2017-30 | longest | en | 0.839296 |
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Jun 15, 2016
${\lim}_{x \to 1} \frac{{x}^{n} - 1}{x - 1} = n$
#### Explanation:
${\lim}_{x \to 1} \frac{{x}^{n} - 1}{x - 1} = {\lim}_{x \to 1} \frac{\left(x - 1\right) \left({x}^{n - 1} + {x}^{n - 2} + \ldots + x + 1\right)}{x - 1}$
$= {\lim}_{x \to 1} \left({x}^{n - 1} + {x}^{n - 2} + \ldots + x + 1\right)$
$= {\lim}_{x \to 1} {\sum}_{i = 0}^{n - 1} {x}^{i}$
$= {\sum}_{i = 0}^{n - 1} {1}^{i}$
$= {\sum}_{i = 0}^{n - 1} 1$
$= n$ | 279 | 506 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 7, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2024-33 | latest | en | 0.411106 |
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• March 22nd, 2011, 04:10 PM
marijana_ristova
I need help
hey,
I`m new in this forum and this is my first experiance in such forums....
Something about me: my name is Marijana, I`am a 19 years old and i am study...and don`t know what to say another...
Now I need help from you.My code below is for multipling two binary number in IEEE 735 standard and I have this
Code Java:
```package proekt2; public class ArrayUtils{ static String[] lookupTable = { "0+0+0=00", "0+0+1=01", "0+1+0=01", "0+1+1=10", "1+0+0=01", "1+0+1=10", "1+1+0=10", "1+1+1=11", }; public static String lookup(char b1, char b2, char c) { String formula = String.format("%c+%c+%c=", b1, b2, c); for (String s : lookupTable) { if (s.startsWith(formula)) { return s.substring(s.indexOf("=") + 1); } } throw new IllegalArgumentException(); } public static String zeroPad(String s, int length) { while (s.length() < length) { s = "0" + s; } return s; } public static String add(String s1, String s2) { int length = Math.max(s1.length(), s2.length()); s1 = zeroPad(s1, length); s2 = zeroPad(s2, length); String result = ""; char carry = '0'; for (int i = length - 1; i >= 0; i--) { String columnResult = lookup(s1.charAt(i), s2.charAt(i), carry); result = columnResult.charAt(1) + result; carry = columnResult.charAt(0); } if (carry == '1') { result = carry + result; } return result; } public static String multiply(String s1, String s2) { String result = ""; String zeroSuffix = ""; for (int i = s2.length() - 1; i >= 0; i--) { if (s2.charAt(i) == '1') { result = add(result, s1 + zeroSuffix); } zeroSuffix += "0"; } return result;} }```
i call in another class with in this way
Code Java:
```String proiz=null; int pBrojac=0; System.out.print(ArrayUtils.add(binaren, binaren2)); System.out.println(); System.out.print(ArrayUtils.zeroPad(proiz, pBrojac)); System.out.println(); System.out.print(ArrayUtils.multiply(binaren, binaren2));```
and the mistake is this
at proekt2.ArrayUtils.lookup(ArrayUtils.java:24)
at proekt2.Kraj.main(Kraj.java:188)
Java Result: 1
• March 23rd, 2011, 04:45 AM
JavaPF
Re: I need help
Hello Marijana.
Welcome to the Java Programming Forums.
Don't forget you need to have the binaren and binaren2 variables..
Code Java:
``` String proiz = null; int pBrojac = 0; String binaren = "binaren"; String binaren2 = "binaren2"; System.out.print(ArrayUtils.add(binaren, binaren2)); System.out.println(); System.out.print(ArrayUtils.zeroPad(proiz, pBrojac)); System.out.println(); System.out.print(ArrayUtils.multiply(binaren, binaren2));```
Looking at your code, you are throwing the IllegalArgumentException() which would suggest there is a problem here:
Code Java:
``` public static String lookup(char b1, char b2, char c) { String formula = String.format("%c+%c+%c=", b1, b2, c); for (String s : lookupTable) { if (s.startsWith(formula)) { return s.substring(s.indexOf("=") + 1); } } throw new IllegalArgumentException(); }```
• March 26th, 2011, 04:07 AM
marijana_ristova
Re: I need help
yes, i know that there is the problem but u don`t know how to repair it:( | 931 | 3,096 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-34 | longest | en | 0.406605 |
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`AdaOpt` is a probabilistic classifier based on a mix of multivariable optimization and a nearest neighbors algorithm. More details about it are found in this paper. When reading the paper, keep in mind that the algorithm is still very new; only time will allow to fully appreciate all its features. Plus, its performance on this dataset is not an indicator of its future performance, on other datasets.
Currently, the package containing `AdaOpt`, `mlsauce`, can be installed from the command line as:
```pip install git+https://github.com/thierrymoudiki/mlsauce.git
```
In this post, we’ll use `mlsauce`’s `AdaOpt` on a handwritten digits dataset from UCI Machine Learning repository.
The model is firstly trained on a set of digits – to distinguish between a “3”, or a”6”, etc.:
```from time import time
from tqdm import tqdm
import mlsauce as ms
import numpy as np
from sklearn.metrics import classification_report
from sklearn.model_selection import train_test_split
from sklearn.datasets import load_digits
Z = digits.data
t = digits.target
# Split data in training and testing sets
np.random.seed(2395)
X_train, X_test, y_train, y_test = train_test_split(Z, t,
test_size=0.2)
learning_rate=0.3,
reg_lambda=0.1,
reg_alpha=0.5,
eta=0.01,
gamma=0.01,
tolerance=1e-4,
row_sample=1,
k=3)
# Teaching AdaOpt to recognize digits
start = time()
obj.fit(X_train, y_train)
print(time()-start)
```
```0.03549695014953613
```
Then, `AdaOpt` is tasked to recognize new, unseen digits `(X_test, y_test)`, based on what it has seen on the training set `(X_train, y_train)`:
```start = time()
print(obj.score(X_test, y_test))
print(time()-start)
```
```0.9944444444444445
0.19525575637817383
```
The accuracy is high on this dataset. Additional error metrics are presented in the following table:
```preds = obj.predict(X_test)
print(classification_report(preds, y_test))
```
``` precision recall f1-score support
0 1.00 1.00 1.00 31
1 1.00 0.97 0.99 40
2 1.00 1.00 1.00 36
3 1.00 1.00 1.00 45
4 1.00 1.00 1.00 37
5 0.97 1.00 0.98 29
6 1.00 0.98 0.99 42
7 1.00 1.00 1.00 35
8 0.97 1.00 0.99 33
9 1.00 1.00 1.00 32
accuracy 0.99 360
macro avg 0.99 1.00 0.99 360
weighted avg 0.99 0.99 0.99 360
```
Ad here is a confusion matrix:
At test time, `AdaOpt` uses a nearest neighbors algorithm. Which means, a task with quadratic complexity (a large number of operations). But there are a few tricks implemented in `mlsauce`’s `AdaOpt` to alleviate the potential burden on very large datasets, such as: instead of comparing the testing set to the whole training set, comparing it to a stratified subsample of the training set.
`row_sample == 0.1` for example in the next figure, means that 1/10 of the training set is used in the nearest neighbors procedure at test time. The figure represents a distribution of test set accuracy:
We also have the following timings in seconds (current, could be faster in the future), as a function of `row_sample`:
The paper contains a more detailed discussion of how these figures are obtained, and a description of `AdaOpt`.
Note: I am currently looking for a gig. You can hire me on Malt or send me an email: thierry dot moudiki at pm dot me. I can do descriptive statistics, data preparation, feature engineering, model calibration, training and validation, and model outputs’ interpretation. I am fluent in Python, R, SQL, Microsoft Excel, Visual Basic (among others) and French. My résumé? Here!
To leave a comment for the author, please follow the link and comment on their blog: T. Moudiki's Webpage - Python . | 1,102 | 4,056 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-30 | latest | en | 0.818283 |
http://stackoverflow.com/questions/16371317/simultaneous-sorting-removing-data-from-a-list | 1,394,915,603,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394678699570/warc/CC-MAIN-20140313024459-00096-ip-10-183-142-35.ec2.internal.warc.gz | 133,697,290 | 18,653 | # Simultaneous Sorting/Removing data from a List
Currently I have a list of integers. This list contains index values that point to "active" objects in another, much larger list. If the smaller list of "active" values becomes too large, it triggers a loop that iterates through the small list and removes values that have become inactive. Currently, it removes them by simply ignoring the inactive values and adding them to a second list (and when the second list gets full again the same process is repeated, placing them back into the first list and so on).
After this trigger occurs, the list is then sorted using a Quicksort implementation. This is all fine and dandy.
-------Question---------
However, I see a potential gain of speed. I am imagining combining the removal of inactive values while the sorting is taking place. Unfortunately, I cannot find a way to implement quicksort in this way. Simply because the quicksort works with pivots, which means if values are removed from the list, the pivot will eventually try to access a slot in the list that does not exist, etc etc.. (unless I'm just thinking about it wrong).
So, any ideas on how to combine the two operations? I can't seem to find any sorting algorithms as fast as quicksort that could handle this, or perhaps I'm just not seeing how to implement it into a quicksort... any hints are appreciated!
Code for better understanding of whats currently going on: (Current Conditions: values can range from 0 to 2 million, no 2 values are the same, and in general they are mostly sorted, since they are sorted every so often)
``````if (deactive > 50000)//if the number of inactive coordinates is greater than 50k
{
for (int i = 0; i < activeCoords1.Count; i++)
{
if (largeArray[activeCoords[i]].active == true)//if coordinate is active, readd to list
{
}
}
//clears the old list for future use
activeCoords1.Clear();
deactive = 0;
//sorts the new list
Quicksort(activeCoords2, 0, activeCoords2.Count() - 1);
}
static void Quicksort(List<int> elements, int left, int right)
{
int i = left, j = right;
int pivot = elements[(left + right) / 2];
while (i <= j)
{
// p < pivot
while (elements[i].CompareTo(pivot) < 0)
{
i++;
}
while (elements[j].CompareTo(pivot) > 0)
{
j--;
}
if (i <= j)
{
// Swap
int tmp = elements[i];
elements[i] = elements[j];
elements[j] = tmp;
i++;
j--;
}
}
// Recursive calls
if (left < j)
{
Quicksort(elements, elements, left, j);
}
if (i < right)
{
Quicksort(elements, elements, i, right);
}
}
``````
-
if the list is mostly sorted it is not a good idea to use quick sort, because on sorted lists quick sort become O(n^2). – mehdi.loa May 4 '13 at 7:00
Because the number range is between 0 and 2000000 you can use counting sort, and it seems that there is no need too merge the operations. – mehdi.loa May 4 '13 at 7:11
Unfortunately, I believe counting sort would undo certain connections, since values seem to be regenerated instead of moved. I need to keep each value reference intact, as the objects also need to keep their pointers pointing towards the correct value... its a bit obscure, but it needs to stay that way. :/ – Colton May 4 '13 at 7:46
There's something I'm missing - if the list from which you are removing items (`activeCoords1`) is already sorted, the list to which you are copying items (`activeCoords2`) will also be sorted because you are adding items to it in the same order that they were in `activeCoords`. What am I missing? – Matthew Watson May 4 '13 at 8:13
Also: (1) Your Quicksort sample won't compile (what is `oldList?`) and (2) it's using a `List<int>` so the elements to compare are `ints`, but you have the huge overhead of calling `.CompareTo()` to compare them - why not just do a direct comparison? Oh and (3) if you really, really need to make it as fast as possible, you'll have to use plain arrays rather than `List<T>`... – Matthew Watson May 4 '13 at 8:16
It sounds like you might benefit from using a red-black tree (or another balanced binary tree), your search, insert and delete time will be O(log n). The tree will always be sorted so there will be no one off big hits incurred to re-sort.
What is your split in terms of types of access (search, insert, delete) and what are your constraints for reach?
-
Currently, with just a list, there is no need for search, I only add values to the end, and as for delete, with the two list setup I have, there is no need for delete. Instead it just re-adds values to the opposite list that are still active... I'll have to pick up on this later unfortunately. Sorry about that, but I will get back to you as soon as I have a chance to read about your tree! – Colton May 4 '13 at 7:41
I was referring to the needs of the consumer of your list not of the list itself. – Slugart May 4 '13 at 7:42
Yup, that is what I'm explaining for the most part. The "consumer" is just a for loop that moves from 0 to max, grabbing values at the index. So there is no need for searching, etc. Is that what you mean? – Colton May 4 '13 at 7:50
Ok, so you only do traversal of the list and the list must always be sorted when you traverse it. – Slugart May 4 '13 at 7:51
Yes, and it is mostly sorted simply to keep memory calls closer together. So when it traverses through the list, the next value found should be the nearest in the array. (So I only need to sort every few seconds, but when I do, the extra speed would be a big help.) – Colton May 4 '13 at 7:56
I would use a `List<T>` or a `SortedDictionary<TKey, TValue>` as your data structure.
As your reason for sorting ("micro optimization based on feelings") is not a good one, I would refrain from it. A good reason would be "it has a measurable impact on performance".
In that case (or of you just want to do it), I recommend a SortedDictionary. All the sorting stuff is already done for you, no reason to reinvent the wheel.
There is no need to juggle with two Lists if one appropriate data structure suffices. A red-black-tree seems appropriate and is apparently used in the SortedDictionary according to this
-
I think you meant SortList<T>? Nevertheless, I am juggling two lists because otherwise, I would have no fast removal of data. Instead of removing one piece of data, waiting for the list to "move everything down", and repeating, I can simply take what I need, and in essence the "move everything down" process only needs to be completed once. – Colton May 4 '13 at 17:07
Also, jumping around up to 2 million memory spaces will always be faster if they are done in order. On previous tests I've saved more than half the time, so it is most likely worth it, and having them in order also provides a better level of manipulation when the time comes for it (say, if updating from lowest to highest becomes an issue). – Colton May 4 '13 at 17:46
Okay, in that case go with the sorted data structure. SortedList and SortedDictionary are similar in purpose but not in performance. With the former you get faster manipulations (add/remove in O(log n) ) and the latter uses less memory but has O(n) manipulations. (geekswithblogs.net/BlackRabbitCoder/archive/2011/06/16/…) – DasKrümelmonster May 4 '13 at 18:07 | 1,795 | 7,171 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2014-10 | latest | en | 0.901182 |
https://nl.mathworks.com/help/finance/analyze-sets-of-numbers-using-matrix-functions.html | 1,600,939,897,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400214347.17/warc/CC-MAIN-20200924065412-20200924095412-00226.warc.gz | 479,199,671 | 14,869 | ## Analyze Sets of Numbers Using Matrix Functions
### Introduction
Many financial analysis procedures involve sets of numbers; for example, a portfolio of securities at various prices and yields. Matrices, matrix functions, and matrix algebra are the most efficient ways to analyze sets of numbers and their relationships. Spreadsheets focus on individual cells and the relationships between cells. While you can think of a set of spreadsheet cells (a range of rows and columns) as a matrix, a matrix-oriented tool like MATLAB® software manipulates sets of numbers more quickly, easily, and naturally. For more information, see Matrix Algebra Refresher.
### Key Definitions
#### Matrix
A rectangular array of numeric or algebraic quantities subject to mathematical operations; the regular formation of elements into rows and columns. Described as a “m-by-n” matrix, with m the number of rows and n the number of columns. The description is always “row-by-column.” For example, here is a `2`-by-`3` matrix of two bonds (the rows) with different par values, coupon rates, and coupon payment frequencies per year (the columns) entered using MATLAB notation:
```Bonds = [1000 0.06 2 500 0.055 4] ```
#### Vector
A matrix with only one row or column. Described as a “`1`-by-n” or “m-by-`1`” matrix. The description is always “row-by-column.” For example, here is a `1`-by-`4` vector of cash flows in MATLAB notation:
```Cash = [1500 4470 5280 -1299] ```
#### Scalar
A `1`-by-`1` matrix; that is, a single number.
### Referencing Matrix Elements
To reference specific matrix elements, use (row, column) notation. For example:
`Bonds(1,2)`
```ans = 0.06```
`Cash(3)`
```ans = 5280.00 ```
You can enlarge matrices using small matrices or vectors as elements. For example,
```AddBond = [1000 0.065 2]; Bonds = [Bonds; AddBond] ```
adds another row to the matrix and creates
```Bonds = 1000 0.06 2 500 0.055 4 1000 0.065 2 ```
Likewise,
```Prices = [987.50 475.00 995.00] Bonds = [Prices, Bonds] ```
```Bonds = 987.50 1000 0.06 2 475.00 500 0.055 4 995.00 1000 0.065 2 ```
Finally, the colon (`:`) is important in generating and referencing matrix elements. For example, to reference the par value, coupon rate, and coupon frequency of the second bond:
`BondItems = Bonds(2, 2:4)`
```BondItems = 500.00 0.055 4 ```
### Transposing Matrices
Sometimes matrices are in the wrong configuration for an operation. In MATLAB, the apostrophe or prime character (`'`) transposes a matrix: columns become rows, rows become columns. For example,
```Cash = [1500 4470 5280 -1299]' ```
produces
```Cash = 1500 4470 5280 -1299 ``` | 721 | 2,634 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2020-40 | latest | en | 0.808317 |
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Helper IV
## Can someone help me fix this DAX formula?
YTD Total W/O & Disc = Calculate(TKD[Total Adjust])+TKD[Agreed Disc.]),FILTER(Dates,Dates[YTD Flag]="Show YTD")
Getting the Error "The syntax for ',' is incorrect error. I'm not sure what's wrong here. Any help would be greatly appreciated!
Thanks,
Jordan
1 ACCEPTED SOLUTION
Resolver II
Hi @jmays86 ,
Try this:
YTD Total W/O & Disc = Calculate(TKD[Total Adjust]+TKD[Agreed Disc.], Dates[YTD Flag]="Show YTD")
Thanks,
Dheeraj
If this post helps, then please consider Accept it as the solution and give thumbs up to help the other members find it more quickly.
2 REPLIES 2
Super User II
YTD Total W/O & Disc = Calculate(SUM(TKD[Total Adjust])+SUM(TKD[Agreed Disc.]),FILTER(Dates,Dates[YTD Flag]="Show YTD"))
In doing so, you are also helping me. Thank you!
Proud to be a Super User!
Resolver II
Hi @jmays86 ,
Try this:
YTD Total W/O & Disc = Calculate(TKD[Total Adjust]+TKD[Agreed Disc.], Dates[YTD Flag]="Show YTD")
Thanks,
Dheeraj
If this post helps, then please consider Accept it as the solution and give thumbs up to help the other members find it more quickly.
Announcements | 332 | 1,198 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2021-31 | latest | en | 0.795645 |
https://convertoctopus.com/358-cubic-centimeters-to-cubic-inches | 1,632,793,741,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058589.72/warc/CC-MAIN-20210928002254-20210928032254-00529.warc.gz | 233,938,807 | 8,043 | ## Conversion formula
The conversion factor from cubic centimeters to cubic inches is 0.061023743836666, which means that 1 cubic centimeter is equal to 0.061023743836666 cubic inches:
1 cm3 = 0.061023743836666 in3
To convert 358 cubic centimeters into cubic inches we have to multiply 358 by the conversion factor in order to get the volume amount from cubic centimeters to cubic inches. We can also form a simple proportion to calculate the result:
1 cm3 → 0.061023743836666 in3
358 cm3 → V(in3)
Solve the above proportion to obtain the volume V in cubic inches:
V(in3) = 358 cm3 × 0.061023743836666 in3
V(in3) = 21.846500293526 in3
The final result is:
358 cm3 → 21.846500293526 in3
We conclude that 358 cubic centimeters is equivalent to 21.846500293526 cubic inches:
358 cubic centimeters = 21.846500293526 cubic inches
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 cubic inch is equal to 0.045773921981285 × 358 cubic centimeters.
Another way is saying that 358 cubic centimeters is equal to 1 ÷ 0.045773921981285 cubic inches.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that three hundred fifty-eight cubic centimeters is approximately twenty-one point eight four seven cubic inches:
358 cm3 ≅ 21.847 in3
An alternative is also that one cubic inch is approximately zero point zero four six times three hundred fifty-eight cubic centimeters.
## Conversion table
### cubic centimeters to cubic inches chart
For quick reference purposes, below is the conversion table you can use to convert from cubic centimeters to cubic inches
cubic centimeters (cm3) cubic inches (in3)
359 cubic centimeters 21.908 cubic inches
360 cubic centimeters 21.969 cubic inches
361 cubic centimeters 22.03 cubic inches
362 cubic centimeters 22.091 cubic inches
363 cubic centimeters 22.152 cubic inches
364 cubic centimeters 22.213 cubic inches
365 cubic centimeters 22.274 cubic inches
366 cubic centimeters 22.335 cubic inches
367 cubic centimeters 22.396 cubic inches
368 cubic centimeters 22.457 cubic inches | 545 | 2,168 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2021-39 | latest | en | 0.678493 |
https://www.oreilly.com/library/view/table-of-integrals/9780123849335/B9780123849335000060_14.xhtml | 1,566,779,217,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027330913.72/warc/CC-MAIN-20190826000512-20190826022512-00246.warc.gz | 931,101,368 | 14,850 | ## With Safari, you learn the way you learn best. Get unlimited access to videos, live online training, learning paths, books, tutorials, and more.
No credit card required
## 6.9 Mathieu Functions
Notation: k2 = q. For definition of the coefficients Ap(m) and Bp(m) see section 8.6.
### 6.91 Mathieu functions
6.911
1.
$\underset{0}{\overset{2\pi }{\int }}\text{c}{e}_{m}\left(z,q\right)\text{c}{e}_{p}\left(z,q\right)\mathrm{d}z=0$
MA
$[m≠p]$
2.
$\underset{0}{\overset{2\pi }{\int }}c{e}_{2n}\left(z,q{\right)\right]}^{2}\mathrm{d}z=2\pi {\left[{A}_{0}^{\left(2n\right)}\right]}^{2}+\pi \sum _{r=1}^{\infty }{\left[{A}_{2r}^{\left(2n\right)\right]}\right]}^{2}=\pi$
MA
3.
$\begin{array}{l}\underset{0}{\overset{2\pi }{\int }}{\left[\text{c}{\text{e}}_{2n+1}\left(z,q\right)\right]}^{2}\text{d}z=\pi \sum _{r=0}^{\infty }{\left[{A}_{2r+1}^{\left(2n+1\right)}\right]}^{2}=\pi \end{array}$
ET II 411(50)
4.
$\begin{array}{l}\underset{0}{\overset{2\pi }{\int }}\left[{\text{se}}_{m}\left(z,q\right){\text{se}}_{p}\text{(}z,q\right)\text{d}z=0\end{array}$
ET II 411(50) ...
## With Safari, you learn the way you learn best. Get unlimited access to videos, live online training, learning paths, books, interactive tutorials, and more.
No credit card required | 496 | 1,273 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 5, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2019-35 | latest | en | 0.480963 |
http://www.top-law-schools.com/forums/viewtopic.php?f=6&t=216076 | 1,527,260,424,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867094.5/warc/CC-MAIN-20180525141236-20180525161236-00168.warc.gz | 460,855,821 | 8,554 | ## PT67-S4-Q17
Prepare for the LSAT or discuss it with others in this forum.
jitsubruin
Posts: 23
Joined: Sun May 08, 2011 1:56 pm
### PT67-S4-Q17
I am having trouble understanding why B is properly inferred. I have already read Manhattan's forum and none of the responses address my reasoning for throwing out B. B says: "Retail stores that distribute coupons generally compensate..." While the stimulus says: "...retail stores that distribute and accept store coupons as a way of discounting the prices on certain products..."
Isn't it possible that retail stores that just distribute coupons differ from retail stores that both distribute and accept coupons as a way of discounting prices? I guess it doesn't make much sense for a retail store to just distribute coupons and not accept them. Maybe I am being too strict with the language or missing something.
Silvermanlsat
Posts: 12
Joined: Thu Sep 05, 2013 8:06 pm
### Re: PT67-S4-Q17
The answer to this question is entirely based on the fact that the stimulus provides that retail stores that distribute and accept coupons charge more for their products on average than retail stores that do not distribute and accept coupons charge for those same products.
Let's assume we have retail store A, and that retail store does distribute and accept products. Assume they sell only 5 products.
Product 1: \$5;
Product 2: \$5
Product 3: \$6
Product 4: \$6
Product 5: \$7
Now assume retail store B sells the same products but does not accept distribute and accept coupons. Assume its prices are:
Product 1 :\$6
Product 2: \$6
Product 3: \$7
Product 4: \$7
Product 5: \$8
If the above were true, it would be impossible for retail store A, on average, to charge more for its products because computing the average merely entails adding up all the individual prices, and dividing by the number of items. To be specific, the average price of products in retail store A in the example above would be 5+5+6+6+7/5=\$5.80, while the average price of products in retail store B would be 6+6+7+7+8/5=\$6.80.
As per the above, if all the prices in retail store B are greater than all the prices in retail store A, it will not be true that the average price of products in retail store A is higher. So, if the average price of products in retail store A is to be higher (and we are told that the store that distributes coupons has a higher average product price) then it must be true that stores that distribute coupons charge higher prices than stores that do not distribute coupons for (at least) some of its products (choice B).
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### Who is online
Users browsing this forum: Google Adsense [Bot] and 5 guests | 648 | 2,711 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-22 | latest | en | 0.927387 |
https://www.scirra.com/forum/distance-between-2-different-objects_t90434 | 1,521,791,539,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257648198.55/warc/CC-MAIN-20180323063710-20180323083710-00102.warc.gz | 863,068,047 | 9,345 | # distance between 2 different objects
Get help using Construct 2
### » Thu Oct 24, 2013 1:26 pm
I used this value (below) as the value of an instance variable (being set "Every tick")
distance(Players.X, PirateSwordStand.X, Players.Y, PirateSwordStand.Y)
I only want the horizontal distance (x axis) between "players" and "PirateSwordStand".
If the horizontal distance between the 2 objects is less than 20, PirateSwordStand will stop walking (simulated, not by pressing buttons) and go to "Stance" animation.
In what units are the distances being measured? I used 20 as the value but I don't have a clue about how long 20 units of distance is.
The PirateSwordStand DOES NOT go to Stance animation and/or Attack animation.
Also, how do I stop the walking of the platform ?
Edit:
PirateSwordStand is an enemy, so the movements/controls of the GoonBox are simulated.
The codes shown in the pic results in the PirateSwordStand instances walking back and forth.
What I want is for them to stop walking, change animation to Stance, then change to Attack animation, if and only if when the player is near them (per instance)sgn152013-10-24 13:28:54
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### » Thu Oct 24, 2013 2:53 pm
It should be distance(x1, y1, x2, y2) you have distance(x1, x2, y1, y2)
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### » Thu Oct 24, 2013 3:11 pm
Wiki Exspression
Your Distance Calculation is false it should be (x1, y1, x2, y2)
But if you want just check the horizontal distance better do:
+ System compare abs(x1 - x2) >= 20
+ Trigger once while true
you dont need any every tick for thatDarklinki2013-10-24 15:12:38
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Posts: 512
Reputation: 5,555 | 506 | 1,692 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2018-13 | latest | en | 0.837487 |
https://riemannianhunger.wordpress.com/category/abstract-algebra-2/commutative-algebra/ | 1,568,540,778,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514571027.62/warc/CC-MAIN-20190915093509-20190915115509-00112.warc.gz | 642,342,669 | 20,063 | # Study Plan, tentatively, + Algebraic Geometry Exercises
So I think it’s probably best to have a rotating study plan schedule that allows me to do certain topics on certain days. So far, I’m thinking of having a rotation that looks something like:
Differential Geometry -> Algebra -> Clifford Stuff -> Algebraic Topology (optional),
and since yesterday was (unofficially) differential geometry day, I’m going to spend today doing algebra.
First order of business: Eisenbud and Harris. And, since I’ve been meaning to write down some of the solutions to exercises I’ve passed, I guess I’ll do that here.
# Study plans, or Why it’s embarrassingly late into the summer and I still haven’t finalized a good way to learn mathematics
So it’s now creeping into the third (full) week of June. School got out for me during the first (full) week of May. Regardless of how woeful you may consider your abilities in mathematics, I’m sure you can deduce something very clear from these facts:
Summer is about half over.
Generally, that fact in and of itself wouldn’t be too terrible. I mean, big deal: Half the summer’s over, and I’ve been working throughout. How big of a failure can that really be?
In this case, it’s actually a pretty big one.
Despite my having read pretty much nonstop since summer began, I haven’t really made it very far into anything substantial. Compounded onto that is the fact that I’ve had to abandon a handful of reading projects after making what appeared to be pretty not-terrible progress into them because of various hindrances (usually, a lack of requisite background knowledge).
It’s been a pretty frustrating, pretty not successful summer, objectively.
# Yesterday, Today, and Forever
Yesterday was a day filled with reading.
Also, by and large, yesterday was a day consisting entirely of (differential) geometry / topology, so it’s really no surprise that – again – my dreams were all math related and tied to that general realm of theory. More precisely, I spent my entire sleep cycle pondering the Poincaré Conjecture (can we call it the Perelman Theorem yet?) and Ricci Flows. That’s certainly a night well spent.
Unsurprisingly, my day today will be largely similar. I downloaded a bunch of resources concerning the aforementioned topics (Poincaré-Perelman and Ricci Flows), as well as some (more) texts on Riemanninan Geometry (which I started perusing yesterday). Also in the works: A colleague of mine (who I’ll call DW2) and I have decided to work through Atiyah and MacDonald’s Introduction to Commutative Algebra, and I’m pretty sure if I don’t spend a significantly-larger amount of time on my professor’s Clifford paper, I’m going to have zero things about which to ever talk with him…
…then there’s the algebraic geometry stuff I’m working on in Eisenbud and Harris / Dummit and Foote, and the material from the seven or so other books I’m reading through concurrently right now….
Every day I’m huss-uh-lin’….
I have some things I want to write here later – expository things and what not – but for now, it’s just this check-in. Auf Wiedersehen! | 694 | 3,101 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2019-39 | latest | en | 0.952175 |
http://www.mentorforbankexams.com/2017/07/boats-and-streams-practice-problems-set-1.html | 1,566,222,341,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027314752.21/warc/CC-MAIN-20190819134354-20190819160354-00085.warc.gz | 285,230,242 | 40,647 | # Boats and Streams Practice Problems Set – 1
Boats and Streams Practice Problems Set – 1
1. The current of a stream runs at the rate of 4 km/hr. A boat goes 6 km and back to the starting point in 2 hours. The speed of the boat in still water is ?
a) 6 km/hr
b) 7.5 km/hr
c) 8 km/hr
d) 6.8 km/hr
e) 7.2 km/hr
2. A boat moves upstream at the rate of 1 km in 10 minutes and downstream at the rate of 1 km in 6 minutes. The speed of the current is ?
a) 1 km/hr
b) 1.5 km/hr
c) 2 km/hr
d) 2.5 km/hr
e) 3 km/hr
3. A man rows upstream 16 km and downstream 28 km talking 5 hours each time. The velocity of the current is ?
a) 2.4 km/hr
b) 1.2 km/hr
c) 3.6 km/hr
d) 1.8 km/hr
e) 1.6 km/hr
4. A man can row at 5 km/hr in still water and the velocity of current is 1 km/hr. It takes him 1 hour to row to a place and back. How far is the place ?
a) 2.5 km
b) 2.4 km
c) 3 km
d) 3.6 km
e) 3.8 km
5. A man can row 44 km downstream in 4 hours. If the man's rowing rate in still water is 8 km/hr, then find in what time will he cover 25 km upstream ?
a) 5 hours
b) 6 hours
c) 4.5 hours
d) 4 hours
e) 5.5 hours
6. A boat travels upstream from B to A and downstream from A to B in 3 hours. If the speed of the boat in still water is 9km/hr and the speed of the current is 3 km/hr the distance between A and B is ?
a) 4 km
b) 6 km
c) 8 km
d) 12 km
e) 15 km
7. A man can swim 3 km/hr in still water. If the velocity of the stream be 2 km/hr, the time taken by him to swim to a place 10 km upstream and back is ?
a) 8 1/3 hrs
b) 9 1/5 hrs
c) 10 hrs
d) 12 hrs
e) 12 2/5 hrs
8. A man can row three quarters of a kilometre against the stream in 11 1/ 4 minutes and return in 7 1/2 minutes. The speed of the man in still water is ?
a) 2 km/hr
b) 3 km/hr
c) 4 km/hr
d) 5 km/hr
e) 6 km/hr
9. A man can row 9 1/3 in still water and he finds that it takes him thrice as much time to row up than as to row down the same distance in river. The speed of the current is ?
a) 3 1/3 km/hr
b) 3 1/2 km/hr
c) 1 1/4 km/hr
d) 4 2/3 km/hr
e) 1 1/2 km/hr
10. The current of a stream runs at 1 km/hr. A motor boat goes 35 km upstream and back again to the starting point in 12 hours. The speed of motor boat in still water is ?
a) 6 km/hr
b) 7 km/hr
c) 8.5 km.hr
d) 8 km/hr
e) 9 km/hr | 853 | 2,251 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-35 | latest | en | 0.901108 |
https://sourceforge.net/p/libmesh/mailman/message/29208264/ | 1,511,030,631,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805008.39/warc/CC-MAIN-20171118171235-20171118191235-00222.warc.gz | 727,772,260 | 16,407 | ## Re: [Libmesh-users] PETSc/3.2-SNESVI
Re: [Libmesh-users] PETSc/3.2-SNESVI From: Ataollah Mesgarnejad - 2012-05-02 13:21:27 On May 1, 2012, at 10:16 PM, Dmitry Karpeev wrote: > Yeah, it's a bit of a hack right now -- DMCreateMatrix simply increfs and return the matrix from the underlying NonlinearImplicitSystem. The reason is that initially the matrix is preallocated by unassembled, and MatDuplicate > doesn't want to duplicate an unassembled matrix. I can fix that, but I'm a bit puzzled by the fact that KSP messes > with the matrix. Can you run with -snes_view and send me the output? > Dmitry, Here it is: ------------- with the flag ----------------------- Linear solve converged due to CONVERGED_RTOL iterations 44 Calculating surface energy... done Calculating fracture energy... done elastic energy:13.74054,fracture energy:-0.00000,total:13.74054 applying variable bounds ... done NL step 0, |residual|_2 = 2.41046e-01 SNES Jacobian assembly... done Linear solve converged due to CONVERGED_RTOL iterations 3 NL step 1, |residual|_2 = 5.10144e-07 Linear solve did not converge due to DIVERGED_BREAKDOWN iterations 1 SNES Object: 1 MPI processes type: virs maximum iterations=50, maximum function evaluations=10000 tolerances: relative=1e-08, absolute=1e-35, solution=1e-08 total number of linear solver iterations=3 total number of function evaluations=2 KSP Object: 1 MPI processes type: gmres GMRES: restart=30, using Classical (unmodified) Gram-Schmidt Orthogonalization with no iterative refinement GMRES: happy breakdown tolerance 1e-30 maximum iterations=10000, initial guess is zero tolerances: relative=1e-05, absolute=1e-50, divergence=10000 left preconditioning using PRECONDITIONED norm type for convergence test PC Object: 1 MPI processes type: ilu ILU: out-of-place factorization 0 levels of fill tolerance for zero pivot 2.22045e-14 using diagonal shift to prevent zero pivot matrix ordering: natural factor fill ratio given 1, needed 1 Factored matrix follows: Matrix Object: 1 MPI processes type: seqaij rows=153, cols=153 package used to perform factorization: petsc total: nonzeros=2401, allocated nonzeros=2401 total number of mallocs used during MatSetValues calls =0 not using I-node routines linear system matrix = precond matrix: Matrix Object: 1 MPI processes type: seqaij rows=153, cols=153 total: nonzeros=2401, allocated nonzeros=2401 total number of mallocs used during MatSetValues calls =0 not using I-node routines SNESLineSearch Object: 1 MPI processes type: basic maxstep=1e+08, minlambda=1e-12 tolerances: relative=1e-08, absolute=1e-15, lambda=1e-08 maximum iterations=1 Fracture system solved at nonlinear iteration 1 , final nonlinear residual norm: 5.10144e-07 --- without the flag ------ applying variable bounds ... done NL step 0, |residual|_2 = 2.41046e-01 SNES Jacobian assembly... done Linear solve converged due to CONVERGED_RTOL iterations 3 NL step 1, |residual|_2 = 5.10144e-07 SNES Jacobian assembly... done Linear solve converged due to CONVERGED_RTOL iterations 3 NL step 2, |residual|_2 = 1.61153e-12 SNES Object: 1 MPI processes type: virs maximum iterations=50, maximum function evaluations=10000 tolerances: relative=1e-08, absolute=1e-35, solution=1e-08 total number of linear solver iterations=6 total number of function evaluations=3 KSP Object: 1 MPI processes type: gmres GMRES: restart=30, using Classical (unmodified) Gram-Schmidt Orthogonalization with no iterative refinement GMRES: happy breakdown tolerance 1e-30 maximum iterations=10000, initial guess is zero tolerances: relative=1e-05, absolute=1e-50, divergence=10000 left preconditioning using PRECONDITIONED norm type for convergence test PC Object: 1 MPI processes type: ilu ILU: out-of-place factorization 0 levels of fill tolerance for zero pivot 2.22045e-14 using diagonal shift to prevent zero pivot matrix ordering: natural factor fill ratio given 1, needed 1 Factored matrix follows: Matrix Object: 1 MPI processes type: seqaij rows=153, cols=153 package used to perform factorization: petsc total: nonzeros=2401, allocated nonzeros=2401 total number of mallocs used during MatSetValues calls =0 not using I-node routines linear system matrix = precond matrix: Matrix Object: 1 MPI processes type: seqaij rows=153, cols=153 total: nonzeros=2401, allocated nonzeros=2401 total number of mallocs used during MatSetValues calls =0 not using I-node routines SNESLineSearch Object: 1 MPI processes type: basic maxstep=1e+08, minlambda=1e-12 tolerances: relative=1e-08, absolute=1e-15, lambda=1e-08 maximum iterations=1 Fracture system solved at nonlinear iteration 2 , final nonlinear residual norm: 1.61153e-12 > Incidentally, there _was_ a reference counting bug that led to a breakdown very much like what you reported. > I'm attaching that patch and a small improvement to DMVariableBounds. I need to figure out how to go about creating matrices from DM_libMesh. > Let me know if you found anything or if I can be of any help. Thanks, Ata > Dmitry. > > On Tue, May 1, 2012 at 6:21 PM, Ataollah Mesgarnejad wrote: > Dmitry, > > Is there a way so SNES doesn't go calculate the Jacobian for every iteration (my Jac does't change)? I used to use a flag to check if it's assembled but when I do that with petsc-dm, KSP associated with SNES gives DIVERGED_BREAKDOWN for the second iteration! looks like petsc-dm doesn't make a copy of the matrix so when it's used in KSP solve it gets jumbled up. > > Here is how it looks > > -with the flag: > > applying variable bounds ... done > NL step 0, |residual|_2 = 8.90674e-04 > SNES Jacobian assembly... done > Linear solve converged due to CONVERGED_RTOL iterations 6 > NL step 1, |residual|_2 = 2.90987e-05 > Linear solve did not converge due to DIVERGED_BREAKDOWN iterations 1 > > this happens for every SNES solve. > ----------------------- > > -without it: > > applying variable bounds ... done > NL step 0, |residual|_2 = 6.08030e-02 > SNES hessian assembly... done > Linear solve converged due to CONVERGED_RTOL iterations 6 > NL step 1, |residual|_2 = 2.61557e-07 > SNES hessian assembly... done > Linear solve converged due to CONVERGED_RTOL iterations 6 > NL step 2, |residual|_2 = 1.35354e-12 > Fracture system solved at nonlinear iteration 2 , final nonlinear residual norm: 1.35354e-12 > > ps: these are not from the same time step but you get the idea. > > Best, > Ata > > On May 1, 2012, at 8:49 AM, Dmitry Karpeev wrote: > >> As an fyi: I just built and ran miscellaneous_ex7 in parallel with the latest petsc-dev: >> http://petsc.cs.iit.edu/petsc/petsc-dev/rev/12d4a1ea0aca >> If you decide you want to upgrade to this (or later) petsc-dev, you'd need a small patch for >> petsc_dm_nonlinear_solver.C to account for the latest API changes (attached). >> I am working on a bigger patch that adds more functionality to DM_libMesh and streamline >> miscellaneous_ex7. That will take some time to percolate up into the libMesh tree. In the >> meantime you should be able to run with the small fix I've attached. >> >> Let me know how it works. >> Dmitry. >> >> >> On Tue, May 1, 2012 at 8:35 AM, Ataollah Mesgarnejad wrote: >> Dmitry, >> >> I believe that the error is indeed due to petsc-dev version. I'm going to check this during the next few hours and get back to you. >> >> Thanks, >> Ata >> >> >> On Tue, May 1, 2012 at 8:29 AM, Dmitry Karpeev wrote: >> Ata, >> I can't reproduce the problem: miscellaneous_ex7 runs fine for me on 2 procs with a relatively recent version >> of petsc-dev. Can you send me the full PETSc error output? >> >> Thanks. >> Dmitry. >> >> >> On Mon, Apr 30, 2012 at 8:56 PM, Ataollah Mesgarnejad wrote: >> >> On Apr 30, 2012, at 8:39 PM, Dmitry Karpeev wrote: >> >>> Ata, >>> example7 segfaults when you run in parallel? >>> Does it do it on 2 procs? >> Yes and yes. >> >> here is the stack: >> (gdb) where >> #0 0x00000038fbaaada0 in __nanosleep_nocancel () from /lib64/libc.so.6 >> #1 0x00000038fbaaac30 in sleep () from /lib64/libc.so.6 >> #2 0x00002ae457b35c40 in PetscSleep(double) () >> from /share/apps/petsc-dev/Linux-intel12.1-mef90-O/lib/libpetsc.so >> #3 0x00002ae457b7dffa in PetscAttachDebugger() () >> from /share/apps/petsc-dev/Linux-intel12.1-mef90-O/lib/libpetsc.so >> #4 0x00002ae457b7e95f in PetscAttachDebuggerErrorHandler(int, int, char const*, char const*, char const*, int, PetscErrorType, char const*, void*) () >> from /share/apps/petsc-dev/Linux-intel12.1-mef90-O/lib/libpetsc.so >> #5 0x00002ae457b817be in PetscError(int, int, char const*, char const*, char const*, int, PetscErrorType, char const*, ...) () >> from /share/apps/petsc-dev/Linux-intel12.1-mef90-O/lib/libpetsc.so >> #6 0x00002ae457b82ea7 in PetscDefaultSignalHandler(int, void*) () >> from /share/apps/petsc-dev/Linux-intel12.1-mef90-O/lib/libpetsc.so >> #7 0x00002ae457b82883 in PetscSignalHandler_Private () >> from /share/apps/petsc-dev/Linux-intel12.1-mef90-O/lib/libpetsc.so >> #8 >> #9 0x00002ae458189340 in DMCoarsenHookAdd(_p_DM*, int (*)(_p_DM*, _p_DM*, void*), int (*)(_p_DM*, _p_Mat*, _p_Vec*, _p_Mat*, _p_DM*, void*), void*) () >> from /share/apps/petsc-dev/Linux-intel12.1-mef90-O/lib/libpetsc.so >> #10 0x00002ae45835f431 in DMSNESGetContext(_p_DM*, _n_SNESDM**) () >> from /share/apps/petsc-dev/Linux-intel12.1-mef90-O/lib/libpetsc.so >> #11 0x00002ae45835ea34 in DMSNESGetFunction(_p_DM*, int (**)(_p_SNES*, _p_Vec*, ---Type to continue, or q to quit--- >> _p_Vec*, void*), void**) () >> from /share/apps/petsc-dev/Linux-intel12.1-mef90-O/lib/libpetsc.so >> #12 0x00002ae458393d1c in SNESGetFunction(_p_SNES*, _p_Vec**, int (**)(_p_SNES*, _p_Vec*, _p_Vec*, void*), void**) () >> from /share/apps/petsc-dev/Linux-intel12.1-mef90-O/lib/libpetsc.so >> #13 0x00002ae45839b9eb in SNESSetUp(_p_SNES*) () >> from /share/apps/petsc-dev/Linux-intel12.1-mef90-O/lib/libpetsc.so >> #14 0x00002ae45839a6b0 in SNESSolve(_p_SNES*, _p_Vec*, _p_Vec*) () >> from /share/apps/petsc-dev/Linux-intel12.1-mef90-O/lib/libpetsc.so >> >> #15 0x00000000010976ab in libMesh::PetscDMNonlinearSolver::solve (this=0x35506d0, jac_in=..., x_in=..., r_in=...) >> at src/solvers/petsc_dm_nonlinear_solver.C:450 >> #16 0x0000000000cd604f in libMesh::NonlinearImplicitSystem::solve ( >> this=0x353b7e0) at src/systems/nonlinear_implicit_system.C:178 >> #17 0x0000000000426082 in Biharmonic::step (this=0x34f2000, dt=@0x7fff95edca28) >> at biharmonic.C:110 >> #18 0x0000000000426373 in Biharmonic::run (this=0x34f2000) at biharmonic.C:152 >> warning: Range for type (null) has invalid bounds 0..-111 >> #19 0x0000000000436cf5 in main (argc=12, argv=0x7fff95edcc18) >> at miscellaneous_ex7.C:64 >> >> Ata >> >>> >>> Dmitry. >>> >>> On Mon, Apr 30, 2012 at 8:31 PM, Ataollah Mesgarnejad wrote: >>> Dmitry, >>> >>> I had a very good experience on single processor but on multiple processors I get segmentation error at line where we call solve (I tried both my code and example7) . I tried your example without -vi --use-petsc-dm and it works just fine. Right now I'm trying to compile a debug version and look into it; meanwhile I thought it would be a good idea if I let you know. >>> >>> Thanks, >>> Ata >>> >>> On Apr 30, 2012, at 1:30 PM, Dmitry Karpeev wrote: >>> >>>> >>>> >>>> On Mon, Apr 30, 2012 at 1:28 PM, Ataollah Mesgarnejad wrote: >>>> >>>> On Apr 30, 2012, at 1:13 PM, Dmitry Karpeev wrote: >>>> >>>>> >>>>> >>>>> On Mon, Apr 30, 2012 at 1:03 PM, Ataollah Mesgarnejad wrote: >>>>> Dmitry, >>>>> >>>>> I had time to come back and work on the SNESVI/Libmesh program. The way you setup your code is particularly hard for me to follow in my own code since it requires a significant amount of extra work (to define classes and setup the inheritances..). What I did instead was to try add a TransientNonlinearSystem to my EquationSystems object: >>>>> >>>>> TransientNonlinearImplicitSystem & fracture_system = equation_system.add_system("fracture_system"); >>>>> >>>>> and then to add the bounds and Objective and Residual functions: >>>>> >>>>> fracture_system.attach_init_function(init_ic_fracture); >>>>> fracture_system.nonlinear_solver->jacobian= PsiFormJacobian; >>>>> fracture_system.nonlinear_solver->residual= PsiFormResidual; >>>>> fracture_system.nonlinear_solver->bounds = PsiFormBounds; >>>>> >>>>> what I see now is that SNES solve does not apply the bounds. So here is my questions: >>>>> >>>>> 1- Is there inherently wrong with setting up the system specially the bounds the easy way like I did? >>>>> Ata, >>>>> Doing things this "simple" way is just fine: there is no reason to follow the relatively convoluted structure I used in that >>>>> example. When I have time, I'll rewrite it to simplify things. >>>>> 2- Where is the right SNES type is chosen ? I guess this should be hard wired somewhere in NonlinearImplicitSystem! >>>>> 3- When I use -snes_view (with your example or my own code ) I get snes_type ls? shouldn't this be VI ? >>>>> >>>>> What's likely missing is the command-line option --use-petsc-dm. It's necessary since libMesh wants to preserve the usual behavior as the default. In particular, because SNESVI and the DM hookup aren't really possible prior to petsc-3.2. >>>>> In fact, it might not be possible without a recent petsc-dev (let me verify that, once I get to my computer). >>>>> >>>>> In the meantime, could you give --use-petsc-dm a try? >>>>> >>>> >>>> I just saw that in the example help line and used it. It works like a charm. >>>> Great! >>>> Dmitry. >>>> I'll go on check for the convergence and see what I can get. >>>> >>>> Thanks, >>>> Ata >>>>> Thanks. >>>>> >>>>> Dmitry. >>>>> >>>>> Thanks, >>>>> Ata >>>>> >>>>> On Apr 12, 2012, at 10:00 AM, Dmitry Karpeev wrote: >>>>> >>>>>> Convergence is something we might need to work on -- I noticed some chatter in a simple bounded biharmonic heat equation (a stripped down version of Cahn-Hilliard). >>>>>> Thanks. >>>>>> Dmitry. >>>>>> >>>>>> On Thu, Apr 12, 2012 at 9:55 AM, Ataollah Mesgarnejad wrote: >>>>>> Thanks Dmitry, >>>>>> >>>>>> It would be wonderful. I made a wrapper for SNESVI in myself for my program but had difficulty achieving good convergence. Right now I am occupied with other issues but I'll get back to this as soon as I get a chance. >>>>>> >>>>>> Ata >>>>>> >>>>>> On Apr 10, 2012, at 9:55 PM, Dmitry Karpeev wrote: >>>>>> >>>>>>> Ata, >>>>>>> There is now a way to use SNESVI from libMesh. >>>>>>> If interested, pull a recent revision from sourceforge and take a look at examples/miscellaneous/miscellaneous_ex7. >>>>>>> Let me know, if you have any questions. >>>>>>> >>>>>>> Thanks. >>>>>>> Dmitry. >>>>>>> >>>>>>> On Wed, Dec 14, 2011 at 12:45 PM, Dmitry Karpeev wrote: >>>>>>> Okay, thanks for your feedback. >>>>>>> I'll keep you updated on this work. >>>>>>> >>>>>>> Dmitry. >>>>>>> >>>>>>> >>>>>>> On Wed, Dec 14, 2011 at 12:42 PM, Ataollah Mesgarnejad wrote: >>>>>>> >>>>>>> >>>>>>> On Wed, Dec 14, 2011 at 12:35 PM, Dmitry Karpeev wrote: >>>>>>> >>>>>>> >>>>>>> On Wed, Dec 14, 2011 at 12:19 PM, Ataollah Mesgarnejad wrote: >>>>>>> >>>>>>> >>>>>>> On Wed, Dec 14, 2011 at 11:59 AM, Dmitry Karpeev wrote: >>>>>>> Ata, >>>>>>> >>>>>>> A couple of things: >>>>>>> 1. SNESVI solves a VI under what's known as "box constraints": the Vec x that the residual operates on is entrywise between Vecs xlow and xhigh. In order to enable >>>>>>> constraints on a linear combination of the Vec entries (as you suggest), the problem >>>>>>> has to be augmented with extra variables -- these linear combinations. This, of course, >>>>>>> changes the Jacobian, preconditioner, etc. The user provides only the Jacobian for the original system, so libMesh would have to automatically construct the Jacobian for the extended system, and precondition it using the user provided PC for the original subsystem. This is not necessarily the unreasonable thing to do in many circumstances, but will take more thinking than a straighforward nodal constraint. The user API for specifying the constraints, however, becomes rather straighforward. >>>>>>> >>>>>>> 2. In terms of FEM convergence, constraining at the nodes is sufficient (see, for example, >>>>>>> Glowinski's book: http://books.google.com/books/about/Numerical_analysis_of_variational_inequa.html?id=Pf4ed2mtbx4C) >>>>>>> You can show that if you impose the nodal constraints, in the limit of mesh size going to 0 the inequality is satisfied a.e. >>>>>>> >>>>>>> In this case no funny manipulation of the Jacobian or the PC is necessary, but the user API to specify the constraint is a bit amiguous: how do you constrain the node? From a containing element? Then there are multiple, possibly incompatible, constraints. >>>>>>> Or you can simply go through all the nodes in the system once using node iterator. Here is how I think you can do it: >>>>>>> >>>>>>> ------------------------------------- >>>>>>> //make two PetscVector ub,lb from two variables you already added to the system >>>>>>> NumericVector &ub_sys = fracture_system.get_vector("ub"); >>>>>>> NumericVector &lb_sys = fracture_system.get_vector("lb"); >>>>>>> PetscVector&lb = libmesh_cast_ptr*>(&ub_sys); >>>>>>> PetscVector&ub = libmesh_cast_ptr*>(&lb_sys); >>>>>>> >>>>>>> then in the bound routine do: >>>>>>> >>>>>>> // Get a constant reference to the mesh object. >>>>>>> const MeshBase& mesh = es.get_mesh(); >>>>>>> //Get the iterator >>>>>>> MeshBase::const_node_iterator nd = mesh.local_nodes_begin(); >>>>>>> const MeshBase::const_node_iterator end_nd = mesh.local_nodes_end(); >>>>>>> for (; nd != end_nd; nd++){ >>>>>>> const Node* node = *nd; >>>>>>> //get the degree of freedom of node >>>>>>> unsigned int dn = node->dof_number(0,0,0); // system 0 , var 0, component 0 >>>>>>> //set the ub,lb >>>>>>> ub(dn)= A; >>>>>>> lb(dn) = B; >>>>>>> } >>>>>>> //close the vectors >>>>>>> ub.close(); >>>>>>> lb.close(); >>>>>>> >>>>>>> You could do that. That's probably the mode we can support almost immediately. >>>>>>> The issue here is that you lose some of the "context" by processing each node by >>>>>>> itself rather as part of an element. For example what if each element is a different material? >>>>>>> I guess in that case that property should be set on the nodes as well. >>>>>>> >>>>>>> Exactly one can always pass mat prop as a Constant MONOMIAL (e.g. system.add_variable("mat_prop",CONSTANT,MONOMIAL);) variable with values on each node. >>>>>>> >>>>>>> Ata >>>>>>> >>>>>>> Dmitry. >>>>>>> ----------------------------------------------------------------- >>>>>>> Ata >>>>>>> >>>>>>> >>>>>>> Dmitry. >>>>>>> >>>>>>> >>>>>>> On Wed, Dec 14, 2011 at 11:17 AM, Ataollah Mesgarnejad wrote: >>>>>>> I hope I understood the whole thing correctly. >>>>>>> >>>>>>> On Wed, Dec 14, 2011 at 10:56 AM, Dmitry Karpeev wrote: >>>>>>> >>>>>>> >>>>>>> On Wed, Dec 14, 2011 at 10:50 AM, Ataollah Mesgarnejad wrote: >>>>>>> I don't think setting the variable bound on each node separately is a good idea in general. >>>>>>> I'm not sure what you mean by this. Could you clarify? I don't think I'm advocating this. >>>>>>> >>>>>>> Bounds should be set as a sum on each elements degrees of freedom; this way there should not be any nodal inconsistencies. >>>>>>> I'm not sure what you mean here either. A bound is usually not a sum (e.g., a bound on a concentration is 0 <= c <=1). >>>>>>> Perhaps your application expresses the total bound as a sum? I'd be very interested in learning more about that case. >>>>>>> >>>>>>> OK here is my thought process: one optimally wants to bound the variable over the whole Element domain between lb,ub. So we would have something like this: >>>>>>> >>>>>>> lb \leq \sum_{i=1}^{n=NDOF} \phi_{i}(x) \hat{u}_{i} \leq ub, \forall x \in El (Eq-1) >>>>>>> >>>>>>> so it's not exactly that the nodal values should be bounded (\hat{u}_{i}) unless it's a FD application where \phi_i(x) is delta function !! That said I think given a normalized basis, bounding nodal values is stricter requirement that (Eq-1). if that's what you meant by inconsistencies between node bounds then the stronger bound would be my first priority. >>>>>>> >>>>>>> Ata >>>>>>> >>>>>>> >>>>>>> Of course it might run into a situations with semi-smooth VI. >>>>>>> The type of VI solver is independent of how the bounds are specified: once we know the upper and lower bounds for each element of a Vec, we can solve the VI with either a semismooth method or an active set method. >>>>>>> >>>>>>> I really appreciate your feedback. I hope we can get to the point where things are as clear as possible so that you can >>>>>>> profitably use the new capabilities. >>>>>>> >>>>>>> Dmitry. >>>>>>> >>>>>>> Thanks, >>>>>>> Ata >>>>>>> >>>>>>> On Wed, Dec 14, 2011 at 9:47 AM, Dmitry Karpeev wrote: >>>>>>> Are you familiar with SNESVI? The main thing I'm looking at right now is the libMesh API for specifying >>>>>>> the bounds on the variables. It seems to me that the user would ideally want to specify the bounds on the variable >>>>>>> one element at a time, since that's the default traversal mode for libMesh. Then, of course, many nodal values >>>>>>> are traversed many time, allowing for an inconsistency. The question is, what you, as a user, expect in that situation? >>>>>>> Should the latest bound that the user has set be applied or the tightest of the bounds the user has supplied? >>>>>>> >>>>>>> Dmitry. >>>>>>> >>>>>>> >>>>>>> On Wed, Dec 14, 2011 at 9:28 AM, Ataollah Mesgarnejad wrote: >>>>>>> Thanks Dmitry, I look forward to use it. Also if it's available as a dev version I would love to try to work with it. >>>>>>> >>>>>>> Ata >>>>>>> >>>>>>> >>>>>>> On Wed, Dec 14, 2011 at 8:24 AM, Dmitry Karpeev wrote: >>>>>>> We (ANL) are working on an patch to libMesh to enable this functionality. >>>>>>> The motivation is to use it through MOOSE, but it will be exposed as a libMesh API. >>>>>>> Dmitry. >>>>>>> >>>>>>> On Wed, Dec 14, 2011 at 7:53 AM, Ataollah Mesgarnejad wrote: >>>>>>> Dear all, >>>>>>> >>>>>>> I wonder if there is compatible versionIs it possible to >>>>>>> use PETScNonlinearSolver class with SNESVI (PETSc/3.2) to solve a >>>>>>> variational inequality problem? >>>>>>> >>>>>>> Thanks, >>>>>>> Ata Mesgarnejad >>>>>>> ------------------------------------------------------------------------------ >>>>>>> Cloud Computing - Latest Buzzword or a Glimpse of the Future? >>>>>>> This paper surveys cloud computing today: What are the benefits? >>>>>>> Why are businesses embracing it? What are its payoffs and pitfalls? >>>>>>> http://www.accelacomm.com/jaw/sdnl/114/51425149/ >>>>>>> _______________________________________________ >>>>>>> Libmesh-users mailing list >>>>>>> Libmesh-users@... >>>>>>> https://lists.sourceforge.net/lists/listinfo/libmesh-users >>>>>>> >>>>>>> >>>>>>> >>>>>>> >>>>>>> -- >>>>>>> A. Mesgarnejad >>>>>>> PhD Student, Research Assistant >>>>>>> Mechanical Engineering Department >>>>>>> Louisiana State University >>>>>>> 2203 Patrick F. Taylor Hall >>>>>>> Baton Rouge, La 70803 >>>>>>> >>>>>>> >>>>>>> >>>>>>> >>>>>>> -- >>>>>>> A. Mesgarnejad >>>>>>> PhD Student, Research Assistant >>>>>>> Mechanical Engineering Department >>>>>>> Louisiana State University >>>>>>> 2203 Patrick F. Taylor Hall >>>>>>> Baton Rouge, La 70803 >>>>>>> >>>>>>> >>>>>>> >>>>>>> >>>>>>> >>>>>>> >>>>>>> -- >>>>>>> A. Mesgarnejad >>>>>>> PhD Student, Research Assistant >>>>>>> Mechanical Engineering Department >>>>>>> Louisiana State University >>>>>>> 2203 Patrick F. Taylor Hall >>>>>>> Baton Rouge, La 70803 >>>>>>> >>>>>>> >>>>>>> >>>>>>> >>>>>>> -- >>>>>>> A. Mesgarnejad >>>>>>> PhD Student, Research Assistant >>>>>>> Mechanical Engineering Department >>>>>>> Louisiana State University >>>>>>> 2203 Patrick F. Taylor Hall >>>>>>> Baton Rouge, La 70803 >>>>>>> >>>>>>> >>>>>> >>>>>> >>>>> >>>>> >>>> >>>> >>> >>> >> >> >> >> >> >> -- >> A. Mesgarnejad >> PhD Student, Research Assistant >> Mechanical Engineering Department >> Louisiana State University >> 2203 Patrick F. Taylor Hall >> Baton Rouge, La 70803 >> >> > > > <0001-PETSc-DM-incref-bug-fix.patch><0002-PETSc-DM-initializes-bounds-with-Inf-now.patch> | 6,592 | 24,021 | {"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} | 2.75 | 3 | CC-MAIN-2017-47 | latest | en | 0.700081 |
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# Puritan fanatics brought to civil and military affairs a
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22 Nov 2006, 15:19
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Puritan fanatics brought to civil and military affairs a coolness of judgment and mutability of purpose that some writers have though inconsistent with their religious zeal, but which was in fact a natural outgrowth of it.
(A) but which was in fact a natural outgrowth of it
(B) but which were in fact a natural outgrowth of it
(C) but which were in fact natural outgrowths of it
(D) but it was in fact a natural outgrowth of them
(E) which was in fact a natural outgrowth of it
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22 Nov 2006, 21:52
If which refers to " coolness of judgment and mutability of purpose " and it refers to "religious zeal", then I will go for B
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23 Nov 2006, 03:04
C for me!
'a coolness of judgment and mutability of purpose ' --> plural
hence 'was' and 'a natural outgrowth' are not accurate.
'it' should refer to religious zeal.
(A), (B), (D) & (E) were eliminated because of the above.
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23 Nov 2006, 03:11
I think that refers to mutability of purpose only..... therefore should be singular.
Thus i choose A. Very cool question....
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23 Nov 2006, 09:59
mst wrote:
If which refers to " coolness of judgment and mutability of purpose " and it refers to "religious zeal", then I will go for B
You've literally put words in my mouth..that's exactly the reason I chose B as well.
In fact both THAT and WHICH seem to refer to the compound COOLNESS OF JUDGMENT AND MUTABILITY OF PURPOSE.
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23 Nov 2006, 10:02
dwivedys wrote:
mst wrote:
If which refers to " coolness of judgment and mutability of purpose " and it refers to "religious zeal", then I will go for B
You've literally put words in my mouth..that's exactly the reason I chose B as well.
In fact both THAT and WHICH seem to refer to the compound COOLNESS OF JUDGMENT AND MUTABILITY OF PURPOSE.
except of course that B still seems incorrect because of singular "outgrowth" whereas plural seems warranted. So by that logic C would be correct.
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23 Nov 2006, 13:17
C
COOLNESS OF JUDGMENT and MUTABILITY OF PURPOSE are OUTGROWTHS of their RELIGIOUS ZEAL
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23 Nov 2006, 13:58
C gang wins. OA is C...thanks for such a nice discussion!
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23 Nov 2006, 13:58
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# Pritesh Shah
### Symbiosis International University
1,427 2011 이후 총 참여 횟수
MATLAB Certified Associate Programmer
Assistant Professor, Department of Electronics and Telecommunication, Symbiosis Institute of Technology, Lavale, Pune
Professional Interests: Control System
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3년 이하 전 | 1,534 | 5,005 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2020-16 | latest | en | 0.677013 |
https://brilliant.org/problems/a-trigonometric-inequality/ | 1,529,341,964,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267860684.14/warc/CC-MAIN-20180618164208-20180618184208-00466.warc.gz | 589,136,849 | 11,286 | # A trigonometric inequality
Algebra Level 5
Find the smallest integer $$N$$ such that for all positive integers $$n$$ $\lvert \sin \frac{1}{100}+\sin \frac{2}{100}+...+\sin \frac{n}{100} \rvert <N$
Details and assumptions
You may use the fact that $$\pi$$ is an irrational number.
Clarification: The measurements are done in radians, not degrees.
× | 100 | 355 | {"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} | 2.671875 | 3 | CC-MAIN-2018-26 | latest | en | 0.74289 |
jeffbeamish.wordpress.com | 1,500,892,894,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424846.81/warc/CC-MAIN-20170724102308-20170724122308-00441.warc.gz | 672,710,782 | 39,173 | # #WxGeekSpeak: Precipitable Water
Every week, the blog introduces you to a technical term from the American Meteorology Society’s “Glossary of Meteorology”. Welcome to #WxGeekSpeak!
What if we were able to wring the atmosphere dry like a sponge & get every last raindrop? Believe it or not, there’s a meteorological calculation for that. Click the video above to learn about precipitable water.
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The total precipitable water is that contained in a column of unit cross section extending all of the way from the earth’s surface to the “top” of the atmosphere. Mathematically, if x(p) is the mixing ratio at the pressure level, p, then the precipitable water vapor, W, contained in a layer bounded by pressures p1 and p2 is given by
where ρ represents the density of water and g is the acceleration of gravity. In actual rainstorms, particularly thunderstorms, amounts of rain very often exceed the total precipitable water vapor of the overlying atmosphere. This results from the action of convergence that brings into the rainstorm the water vapor from a surrounding area that is often quite large. Nevertheless, there is general correlation between precipitation amounts in given storms and the precipitable water vapor of the air masses involved in those storms. | 328 | 1,612 | {"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} | 2.71875 | 3 | CC-MAIN-2017-30 | longest | en | 0.914355 |
https://community.weatherflow.com/t/wind-arrow-direction-symbol/22790 | 1,701,449,427,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100290.24/warc/CC-MAIN-20231201151933-20231201181933-00052.warc.gz | 227,024,333 | 9,181 | # Wind arrow direction symbol
I think i have most of the tempest symbols figured out save one. on my IPhone screen that shows the weather info by day I see the hour of the day an icon showing cloudy, rain, sunny etc. then temperature then the rain probability % followed by a number with an arrow. What is that last symbol. At first I thought it was wind or gust speed but it has no correlation to the info provided by my weaher station. Can someone help me out? Thanks
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If you are talking about the circled items,
they are showing predicted average wind speed and direction for that hour.
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Also remember that the arrows indicate winds FROM, because real meteorologistsā¦
use āFrom-iesā not āTo-iesā i.e. from the NW at 22 mph . . . or from 320Ā° at 24 kts.
Aha. Thank you all so much for the clarification. So in the example the first one (on the 10th): overcast, 39Ā°, 0% probability of rain, predicted average wind speed of 5mph from the NW. Is that correct ?
All correct. . .eXcept the wind direction. For the WeatherFlow | Tempest wind direction icon. . .it is somewhat misleading to the non-meteorological community. For their direction icon. . .although it Points to the NW. . .it actually means 180 degs opposite of that. In other words. . .the wind forecast in your first example would be FROM the SE at 5 mph. In the 3rd example the wind would be FROM the SSE at 6 mph. A quick way to remember usng the wind direction icons. . .is to use the large āopenā end of the wind icon as the āFROM-ieā. In other words . . . wind is from the SE or SSE at 5-7 mph. (to which way the wind is blowing TO (or TOwards) . .is irrellavent! Here an example winds from one of my one of Tempest devices in Bellevue, NE
Most all of the winds are FROM the NW at 13 mph. . .diminishing in speed to 2 mph from the NW. The 4th wind from the right shows wind becoming FROM the WSW at 2 mph. . .then shifting to become FROM the South at 3-4 mph by the end of the period. Just remember the wind direction is FROM (the large opening). . .not to which way is is pointing to. If you have a weather radio handy. . .you can listen to an hourly weather report.
Hereās how they term it in the hour weather observation roundup:
āAt 2pm in Omaha. . .it was mostly sunny, The temperature was 37 degrees; dewpoint was 15 degrees giving a Relative Humidity of 40%. The wind was from the N at 15 gusting to 24 mph. The pressure was 30.49 inches and Falling.ā
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Again, thank you all. I love keeping track of the info on the weather station. I am now a little smarter.
Joe P.
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ok - so does the arrow point to where the wind is going ātoā (ala smoke would blow thattaway) or does it point to where it is coming āfromā (ala how a weathervane would appear) ?
This wind icon in this example is FROM-ie the South-Southwest (or slightly South of SW) because the ālarge openingā is From-ie the SSW. It is indeed pointing to the NNE. . .but assume for a moment that the pointer has a heavy weight attached to it. . .and the large opening at the other end is of very light weight material. The yellow arrow is being blown along the line of the windflow which is blowing FROM the SSW to the NNE.
Wind direction is a very misunderstood concept. A direction vane will always point into the wind like this. . .
. . .but itās actually pointing to the direction that wind is coming from because the tip has a weight in it and the tail-fin is extremely lighter The Tail fin (being extremely light-weight) is actually what tells the heavily-weighted pointer which way the wind is coming from. In image shown above. . the pointer is pointing into the wind. . .but it means the wind is from that direction. So for the image of the vane. . .the wind is actually blowing from Left to Right. . or blowing from the West. Quite simply. . .a wind vane always shows the direction from which the wind is blowing. In a Light and Variable wind. . .the vane will spin all over the place because the flow is so weak.
Hereās another example:
Q: Is a west wind coming from the west or blowing to the west?
A: The direction the wind is blowing is the direction the wind is coming FROM. A west wind means wind is blowing from the west to the east. Hereās the misnomer: The wind vane will point into the wind because that weighted tip is heavier than the tail fin. Even though it points to the west. . . .it is actually pointing in the direction the wind is coming From-ie the west.
Now one might say. . .ābut your two images donāt make sense. First you say the yellow arrow is pointing into the wind. . .so the wind should be From that direction (NE)ā. No. . .the yellow arrow in the red box is an icon and you use the same exact principle as the black wind vaneā¦ For the yellow arrow in the red box. . .assume the pointer has a heavily weighted tip attached to it. But for the icon. . .you have to think Two Dimensionally as opposed to a Three Dimension physical wind vane. The large opening (the āTail Finā) tells the direction the wind is from.
The Tempest engineers took care of that difficult concept for us. They created the 4 Ultrasonic Transducers underneath the top cap of the device . .so that Tempest determines the direction the wind is from . . .and its speed!
1 Like
You didnāt really answer my question or I got lost in all the wordsā¦
Does the arrow (in the WF app and/or webpage) point to where the wind is going ātoā (ala smoke would blow thattaway) or does it point to where it is coming āfromā (ala how a weathervane would appear) ?
My understanding is that the arrow in the WF is āwhere the wind is goingā However in meteorologic speak, they donāt speak that way. they say where the wind is coming from. . So if the arrow is pointing NW, they say the wind is NE at blah blah MPH. The whole weathervane thing just confused things.
You still arenāt answering the question! Is the arrow in the Tempest app a vector arrow or a weather vane arrow?
In other words, is the predicted wind direction for the 1200-1500 times in my example above generally a North wind or a South wind, meteriologically speaking.
1 Like
Or removing any possibility of jargon-related misunderstanding, is the wind āfromā the NNW ?
Generally a south wind, a vector arrow.
The smoke would blow in that direction, from the wide end of the arrow towards the pointy end.
The arrow does not behave like a weathervane would.
For my personally presented graphs to help with the confusion I include dots to indicate direction and my arrows are the same as Weatherflow displays and several other wind Apps.
cheers Ian
1 Like | 1,692 | 6,675 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | latest | en | 0.927213 |
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Write a polynomial function f of least degree that has rational coefficients, a leading coefficient of 1, and the given zeros. Write the polynomial in standard form. 2i, 1-i
xleb123
We are given a polynomial function f of least degree that has rational coefficients, a leading coefficient of $1$, and the given zeros $2i,1-i.$
As we know that the polynomial function has rational coefficients.
so any complex root always appear in pair.
so as $1-i$ is a root of $f\left(x\right)$ then it's complex conjugate $1+i$ is also an root of $f\left(x\right).$
Hence $f\left(x\right)$ could be represented as:
$f\left(x\right)=\left(x-2i\right)\left[x-\left(1-i\right)\right]\left[x-\left(1+i\right)\right]$
Simplify each term.
Apply the distributive property.
$f\left(x\right)=\left(x-2i\right)\left(x-1\cdot 1+i\right)\left(x-\left(1+i\right)\right)$
Multiply −1 by 1.
$f\left(x\right)=\left(x-2i\right)\left(x-1+i\right)\left(x-\left(1+i\right)\right)$
Multiply −1 by -1.
$f\left(x\right)=\left(x-2i\right)\left(x-1+1i\right)\left(x-\left(1+i\right)\right)$
Multiply $i$ by $1$.
$f\left(x\right)=\left(x-2i\right)\left(x-1+i\right)\left(x-\left(1+i\right)\right)$
Expand $\left(x-2i\right)\left(x-1+i\right)$ by multiplying each term in the first expression by each term in the second expression.
$f\left(x\right)=\left(x\cdot x+x\cdot -1+xi-2ix-2i\cdot -1-2ii\right)\left(x-\left(1+i\right)\right)$
$f\left(x\right)=\left({x}^{2}-x+xi-2ix+2i+2\right)\left(x-\left(1+i\right)\right)$
Subtract $2ix$ from $xi.$
$f\left(x\right)=\left({x}^{2}-x-xi+2i+2\right)\left(x-\left(1+i\right)\right)$
Simplify each term.
$f\left(x\right)=\left({x}^{2}-x-xi+2i+2\right)\left(x-1-i\right)$
Expand $\left({x}^{2}-x-xi+2i+2\right)\left(x-1-i\right)$ by multiplying each term in the first expression by each term in the second expression.
$f\left(x\right)={x}^{2}x+{x}^{2}\cdot -1+x2\left(-i\right)-x\cdot x-x\cdot -1-x\left(-i\right)-xix-xi\cdot -1-xi\left(-i\right)+2ix+2i\cdot -1+2i\left(-i\right)+2x+2\cdot -1+2\left(-i\right)$
Simplify terms.
$f\left(x\right)={x}^{3}-2{x}^{2}+{x}^{2}\left(-i\right)+xi-{x}^{2}i+xi+2ix-2i+2x-2i$
Subtract ${x}^{2}i$ from ${x}^{2}\left(-i\right).$
$f\left(x\right)={x}^{3}-2{x}^{2}-2{x}^{2}i+xi+xi+2ix-2i+2x-2i$
Add $xi$ and $xi.$
$f\left(x\right)={x}^{3}-2{x}^{2}-2{x}^{2}i+2xi+2ix-2i+2x-2i$
Add $2xi$ and $2ix.$
$f\left(x\right)={x}^{3}-2{x}^{2}-2{x}^{2}i+4xi-2i+2x-2i$
Subtract $2i$ from $-2i.$
$f\left(x\right)={x}^{3}-2{x}^{2}-2{x}^{2}i$
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https://www.bartleby.com/solution-answer/chapter-38-problem-10e-single-variable-calculus-concepts-and-contexts-enhanced-edition-4th-edition/9781337687805/392d8e37-cc8e-40d7-b831-0ffd360b6251 | 1,632,138,928,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057036.89/warc/CC-MAIN-20210920101029-20210920131029-00703.warc.gz | 704,875,522 | 62,943 | The maximum height reached by the ball.
Single Variable Calculus: Concepts...
4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
Single Variable Calculus: Concepts...
4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
Solutions
Chapter 3.8, Problem 10E
(a)
To determine
Expert Solution
Answer to Problem 10E
The maximum height reached by the ball is 100ft .
Explanation of Solution
Given information:
The height function of the ball is given by,
s=80t16t2
The maximum height of the ball is achieved when the velocity is zero.
The time required to reach the maximum height of the ball is achieved when the velocity is zero is calculated as,
s=80t16t2dsdt=8032t0=8032tt=2.5s
The maximum height reached by the ball is calculated as,
s=80t16t2=80(2.5)16(2.5)2=100ft
Therefore, the maximum height reached by the ball is 100ft .
(b)
To determine
Expert Solution
Answer to Problem 10E
The velocity of the ball on its way up and down are 16m/s and 16m/s .
Explanation of Solution
Given information:
The height function of the ball is given by,
s=80t16t2
The time required to reach the maximum height of the ball is achieved when the velocity is zero is calculated as,
s=80t16t2dsdt=8032t0=8032tt=2.5s
The time required by the ballto reach 96ft is calculated as,
s=80t16t296=80t16t2t=2and 3s
The velocity of the ball on its way up is calculated as,
s=80t16t2dsdt=8032tv=8032(2)v=16m/s
The velocity of the ball on its way down is calculated as,
s=80t16t2dsdt=8032tv=8032(3)v=16m/s
Therefore, the velocity of the ball on its way up and down are 16m/s and 16m/s .
Have a homework question?
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Phet Energy Forms And Changes Simulation Answer Key. Phet energy forms and changes simulation answer key › free this collection includes resources to support teachers and students as they engage in the topics outlined in the 8th grade nyc science scope & sequence. Answer keys are often not available because the activities are inquiry based and lead to multiple.
Bond energy calculations with phet! You are also able to work with a system where you can manipulate the energy input, observe the process of electrical energy generation and manipulate the output. Adición de vectores publicado el septiembre 30, 2015 septiembre 30, 2015 por noeaguiper esta es una aplicación desarrollada por la universidad de colorado question 3:
You Are Also Able To Work With A System Where You Can Manipulate The Energy Input, Observe The Process Of Electrical Energy Generation And Manipulate The Output.
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Teachers Do Not Post Keys On The Phet Website, You May Try To Email The Author And See If He/She Has One To Share.
How did you use this activity or change it for your class? Change friction and see how it affects the motion of objects. Build your own system, with energy sources, changers, and users.
Energy Forms & Changes Simulation In This Simulation, You Will Be Able To “See” Several Different Forms Of Energy And The Changes (Transfers) That Can Occur Between Them.
Discover the declaration phet energy form and change simulation answers that you are looking for. Click on the “energy systems” tab. Track and visualize how energy flows and changes through your system.
Explore How Heating And Cooling Iron, Brick, Water, And Olive Oil Adds Or Removes Energy.
Bond energy calculations with phet! However below, afterward you visit this web page, it will be so certainly simple to get as well as download guide phet energy form and change simulation answers it will not endure many mature as we tell before. Create an applied force and see how it makes objects move.
This Is An Exploratory Lab Activity That Has Students Build Molecules And Disassemble Them To Form Products Using An Interactive Phet Simulation While Showing Them How To Calculate The Energy Change Using A Bond Energy Table.
You are also able to work with a system where you can manipulate the energy input, observe the process of electrical energy generation and manipulate the output. 2/19/2022 energy forms & changes simulation in this simulation, you will be able to “see” several different forms of energy and the changes (transfers) that can occur between them. Energy forms changes simulation answer key. | 605 | 2,950 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.675113 |
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# GMAT Math / Verbal split important
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Intern
Joined: 03 Feb 2003
Posts: 1
GMAT Math / Verbal split important [#permalink]
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07 Feb 2003, 06:21
Hi,
Just to let you know the composition of maths score and verbal score of the GMAT overall score is important:
INSEAD has rejected my otherwise good quality application (their words) with a 660 score not because of my overall score is not enough but because the math score is too low: The split of my 660 score was very uneven: Maths 36 (48%) and Verbal 46 (97%) :D . Something that was not made clear in the application requirements.
Now, I am resitting the GMAT in less than three weeks and I need to bring up my maths score to 70-75% according to INSEAD to still be able to get an interview. I did study maths at uni, but its all a long time ago and GMAT math is a different issue alltogether. Does anyboday out there have any good strategies on how to improve the GMAT math score ?
Founder
Joined: 04 Dec 2002
Posts: 15153
Location: United States (WA)
GMAT 1: 750 Q49 V42
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07 Feb 2003, 19:32
Hey, sorry it took a while to get to your message....
I'd suggest your case is a pretty good one.
Improving math takes less time than verbal. I assume you have done some math preparation, so you pretty much know the basics and I will omit those.
Since you have 3 weeks, I would suggest concentrating on Arithmetic and Word Problems - these account for about 70% of questions on the GMAT. (my guess). A lot of data sufficience is Arithmetic. Word problems dominate the PS section. Arithemtic, actually, is very often underestimated - one needs to brush up and learn all those little properties of numbers, and zeros, and negatives, etc. THey are the basics. A good source would be Kaplan's Math Workbook. The review sections are very short and are follwed by a large number of exercises. If you don't have time to spend solviing all of them, here are the numbers of the questions you should do for sure. (the intense course)
Kaplan's Math Book 2nd Edition.
Number properties test: 8, 13, 14, 17, 19
Averages test: 4, 6
Ratios: 22
Percents: 4, 7, 8, 22
Powers and Roots: 5, 12
Algebra test 2: 2, 6
Basic Word Problem Exercise (p. 131): 6, 8, 9
Word Problems Test 1: 4, 19, 24
Percents, Ratios, and Rates: 34
General WP Test: 31, 35, 40
Geometry
Triangles Exercise: 7, 11
Triangles Test: 2, 15, 18
Quadraliterals and other polygons: 14, 16, 17
Circles: 2, 5, 10
Multiple Figures: 2, 6, 9, 14
Solids: 5
DS1: 12, 14, 16, 17, 18, 19, 20, 22, 24
DS2: 3, 7, 9, 10, 15, 21, 23, 24
This is for the second edition. I think the third is the second with a few errors corrected, yet I don't guarantee it will be exactly the same. In any case, I do recommend this book a lot. I have used it for my prep.
ONE THING THAT YOU SHOULD DO FOR SURE: Review the rules and properties. Make sure you know the theory before you jump to solve the problems. Don't try to take GMAT sample tests now, concentrate on specific topics.
On this site I have Arithmetic 1 Posted, you can check it out if you have time. Here is the link: http://www.gmatclub.com/members/courses/quant/arithmetic1.shtml
Also, time management on GMAT: http://www.gmatclub.com/members/courses/quant/timing.shtml
I will also email you a few selections I have made while tuturing.
It will be great if you can get in.
Bogdan
Does anybody else have any suggestions for math?
here [#permalink] 07 Feb 2003, 19:32
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## Presentation on theme: "“FENDER” AUTOMATIC MEMORY FENCE INFERENCE Presented by Michael Kuperstein, Technion Joint work with Martin Vechev and Eran Yahav, IBM Research 1."— Presentation transcript:
“FENDER” AUTOMATIC MEMORY FENCE INFERENCE Presented by Michael Kuperstein, Technion Joint work with Martin Vechev and Eran Yahav, IBM Research 1
p0: flag[0] := true while flag[1] = true { if turn ≠ 0 { flag[0] := false while turn ≠ 0 { } flag[0] := true } // critical section turn := 1 flag[0] := false p1: flag[1] := true while flag[0] = true { if turn ≠ 1 { flag[1] := false while turn ≠ 1 { } flag[1] := true } // critical section turn := 0 flag[1] := false Specification: mutual exclusion over critical section Dekker’s Algorithm 2
p0: flag[0] := true while flag[1] = true { if turn ≠ 0 { flag[0] := false while turn ≠ 0 { } flag[0] := true } // critical section turn := 1 flag[0] := false p1: flag[1] := true while flag[0] = true { if turn ≠ 1 { flag[1] := false while turn ≠ 1 { } flag[1] := true } // critical section turn := 0 flag[1] := false Beyond Textbooks: Weak Memory Models Re-ordering of operations Non-atomic stores 3
Memory Fences Enforce order, at a cost! Fences are expensive 10s-100s of cycles Example: removing a single fence yields 3x speedup in a work-stealing queue [Michael, et al. PPoPP ’09] Where should we put fences? Required fences depend on memory model Different kinds of fences 4
Goal “Correct and efficient fencing for the masses” A tool to help the programmer place fences For non-trivial finite-state programs Under a realistic memory model Safe Efficient 5
Easy! p0: flag[0] := true fence while flag[1] = true { if turn ≠ 0 { flag[0] := false while turn ≠ 0 { } flag[0] := true } // critical section turn := 1 flag[0] := false p1: flag[1] := true fence while flag[0] = true { if turn ≠ 1 { flag[1] := false while turn ≠ 1 { } flag[1] := true } // critical section turn := 0 flag[1] := false 6
Chase-Lev Work-Stealing Queue 1 int take() { 2 long b = bottom – 1; 3 item_t * q = wsq; 4 bottom = b 5 long t = top 6 if (b < t) { 7 bottom = t; 8 return EMPTY; 9 } 10 task = q->ap[b % q->size]; 11 if (b > t) 12 return task 13 if (!CAS(&top, t, t+1)) 14 return EMPTY; 15 bottom = t + 1; 16 return task; 17 } 1 void push(int task) { 2 long b = bottom; 3 long t = top; 4 item_t * q = wsq; 5 if (b – t >= q->size – 1) { 6 wsq = expand(); 7 q = wsq; 8 } 9 q->ap[b % q->size] = task; 10 bottom = b + 1; 11} 1 int steal() { 2 long t = top; 3 long b = bottom; 4 item_t * q = wsq; 5 if (t >= b) 6 return EMPTY; 7 task = q->ap[t % q->size]; 8 if (!CAS(&top, t, t+1)) 9 return ABORT; 10 return task; 11} 7
In Practice - Hard This is a real problem Finding the best placement for fences is hard Classical trade-off: correctness vs. efficiency Existing tools are insufficient CheckFence [Alur et al. PLDI ’07] 8
Our Approach: Overview P’ satisfies the specification S under M (Finite-State) Program P (Finite-State) Program P (Safety) Specification S Memory Model M Memory Model M Program P’ with Fences 9
Our Approach: Recipe Compute reachable states for the program Bad news: Reachability problem undecidable even for finite-state programs running under sufficiently weak MM [Atig et al. POPL ’10] So sometimes use an additional bound Compute constraints that guarantee that all “bad states” are avoided The constraints restrict non-determinism allowed by the memory model Implement the constraints with fences 10
Our Approach: Ingredients Operational semantics for weak memory models An algorithm for finding order constraints An algorithm for implementing constraints as fences in the program 11
Classification due to Adve et al. IEEE Computer ‘95 12 Operational Semantics for WMM
Model store buffers Model instruction reordering (execution buffers) Variety of re-ordering rules 13
States and Transitions Processor B: B 1 : R2 = Y B 2 : R1 = X Processor A: A 1 : X = 1 A 2 : Y = 1 Initially X = Y = R1 = R2 = 0 A 2 :Y = 1 A 1 :X = 1 B 2 :R1 = X B 1 :R2 = Y X = 0 Y = 0 R1 = 0 R2 = 0 A 1 :X = 1B 2 :R1 = X B 1 :R2 = Y X = 0 Y = 1 R1 = 0 R2 = 0 A2A2 14
Compute Reachable States (0,0,0,0) (1,0,0,0)(0,1,0,0)(0,0,0,0) (0,1,0,0)(0,1,0,1)(0,1,0,0) (1,1,0,1)(0,1,0,1) (1,1,1,1)(1,1,0,1) A1A1 A2A2 B2B2 B1B1 A1A1 A1A1 A1A1 B1B1 B2B2 B2B2 B2B2 A2A2 A2A2 Error state (x,y,r1,r2) EB1 EB2 legend Specification at final state ¬ (R1 = 0 R2 = 1) initial 15 A1 A2 B1 B2 A2 B1 B2A1 B1 B2 A1 A2B2 A1 A2B1 A1B2A1B2 B1 B2 A1B2 B 1 : R2 = Y B 2 : R1 = X A 1 : X = 1 A 2 : Y = 1
Avoiding states To avoid a state Avoid all incoming transitions To avoid an incoming transition Either avoid the transition itself Or avoid the source state 16
Avoidable Transitions Execution buffer is ordered A transition not executing first instruction in the execution buffer can be avoided By forcing a different transition to execute A 4 :W = 1 A 3 :Z = 1 A 2 :Y = 1 A 1 :X = 1 Processor A A 1 : X = 1 A 2 : Y = 1 A 3 : Z = 1 A 4 : W = 1 A1A1 A2A2 A3A3 A4A4 17
Avoidable Transitions To avoid A 3 in this state Force A 1 to execute before A 3 Or force A 2 to execute before A 3 Language of ordering constraints [A 1 < A 3 ] [A 2 < A 3 ] A 4 :W = 1 A 3 :Z = 1 A 2 :Y = 1 A 1 :X = 1 Processor A A 1 : X = 1 A 2 : Y = 1 A 3 : Z = 1 A 4 : W = 1 A1A1 A2A2 18 A3A3 A4A4
Computing Avoid Formulae Ordering constraint [l 1 < l 2 ] l 2 may not be reordered with l 1 Associate a propositional variable with each constraint “Avoid formulas” are (positive) propositional formulas over ordering constraints Fixed-point computation computes an avoid formula for every state Final constraint formula is the conjunction of avoiding all “bad states” 19
Back to our example (0,0,0,0) false A1A1 (0,1,0,0) A1 < A2 (0,0,0,0) B1 < B2 (1,0,0,0) B1 < B2 (0,1,0,0) A1 < A2 || B1 < B2 (0,1,0,1) A1 < A2 (1,1,0,0) A1 < A2 (1,1,0,0) B1 < B2 (1,1,0,1) [] A1 < A2 && B1 < B2 A1A1 A1A1 A2A2 A2A2 A2A2 B1B1 B1B1 B1B1 B2B2 B2B2 B2B2 20 A1 A2 B1 B2 A1 B1 B2 A1 A2B1 A1B2A1B1A2B1 A1 B1 B 1 : R2 = Y B 2 : R1 = X A 1 : X = 1 A 2 : Y = 1
Fence Placement Processor B B 1 : R2 = Y fence(“load-load”) B 2 : R1 = X Processor A A 1 : X = 1 fence(“store-store”) A 2 : Y = 1 [A 1 < A 2 ] [B 1 < B 2 ] 21
Fence Placement Trivial in the previous example Satisfying assignment to the avoid formula Every satisfied constraint realized as a fence Only had to choose fence type More complicated in practice Which satisfying assignment to chose? 22
Data Structures Treiber’s Stack Michael & Scott’s Non-Blocking Queue Idempotent Work-Stealing Queue Chase & Lev’s Work-Stealing Queue Found a missing fence in an implementation used for an earlier paper. … 23
Sample Results: Michael-Scott Queue Used the results from [Alur et al. PLDI ’07] as a reference Reference contains 7 fences RMO*: 3 found 2 unneeded due to environment issues (memory management) 2 unneeded due to lack of speculation PSO: 1 found, TSO: No fences required 24
Results 25
Summary Fence inference Finite-state programs Safe and optimal Work in progress Scalability Abstraction Over-approximation instead of bounding 26
Download ppt "“FENDER” AUTOMATIC MEMORY FENCE INFERENCE Presented by Michael Kuperstein, Technion Joint work with Martin Vechev and Eran Yahav, IBM Research 1."
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https://answerpail.com/index.php/55196/how-many-miles-does-the-average-person-drive?show=55205 | 1,720,846,822,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514484.89/warc/CC-MAIN-20240713020211-20240713050211-00274.warc.gz | 78,278,736 | 6,844 | # How many miles does the average person drive?
How many miles does the average person drive?
answered Mar 28, 2020 by (102,160 points)
The average miles that people drive per year is around 12,000 miles which would mean they drive around 1,000 miles per month.
Some people drive as much as 100 miles per day to commute to and from work which would mean they put nearly 3000 miles on their vehicle per month or 36,000 miles on their vehicle per year especially if they drive on road trips other than just driving too and from work.
Some people only drive 500 miles per month which would be 6,000 miles per year which is pretty low.
Myself without road trips I drive at most 500 to 600 miles per month.
I work from home so do not need to commute to work.
However the grocery store and other shopping are 60 miles round trip and sometimes I just like to get out and drive.
But the average person puts between 6,000 miles per year to 15,000 miles per year on their vehicle.
Some people may only put 3,000 miles per year on their vehicle depending on how often they drive and how far they drive the vehicle. | 256 | 1,112 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2024-30 | latest | en | 0.984079 |
https://geo.libretexts.org/Courses/Mt._San_Jacinto_College/Environmental_Science_101/03%3A_Species_and_Evolution/3.03%3A_Speciation | 1,726,214,301,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651510.65/warc/CC-MAIN-20240913070112-20240913100112-00611.warc.gz | 256,601,295 | 33,007 | # 3.3: Speciation
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The biological definition of species, which works for sexually reproducing organisms, is a group of actually or potentially interbreeding individuals. According to this definition, one species is distinguished from another by the possibility of matings between individuals from each species to produce fertile offspring. There are exceptions to this rule. Many species are similar enough that hybrid offspring are possible and may often occur in nature, but for the majority of species this rule generally holds. In fact, the presence of hybrids between similar species suggests that they may have descended from a single interbreeding species and that the speciation process may not yet be completed.
Given the extraordinary diversity of life on the planet there must be mechanisms for speciation: the formation of two species from one original species. Darwin envisioned this process as a branching event and diagrammed the process in the only illustration found in On the Origin of Species (Figure below). For speciation to occur, two new populations must be formed from one original population, and they must evolve in such a way that it becomes impossible for individuals from the two new populations to interbreed. Biologists have proposed mechanisms by which this could occur that fall into two broad categories. Allopatric speciation, meaning speciation in “other homelands,” involves a geographic separation of populations from a parent species and subsequent evolution. Sympatric speciation, meaning speciation in the “same homeland,” involves speciation occurring within a parent species while remaining in one location.
Biologists think of speciation events as the splitting of one ancestral species into two descendant species. There is no reason why there might not be more than two species formed at one time except that it is less likely and such multiple events can also be conceptualized as single splits occurring close in time.
A geographically continuous population has a gene pool that is relatively homogeneous. Gene flow, the movement of alleles across the range of the species, is relatively free because individuals can move and then mate with individuals in their new location. Thus, the frequency of an allele at one end of a distribution will be similar to the frequency of the allele at the other end. When populations become geographically discontinuous the free-flow of alleles is prevented. When that separation lasts for a period of time, the two populations are able to evolve along different trajectories. Thus, their allele frequencies at numerous genetic loci gradually become more and more different as new alleles independently arise by mutation in each population. Typically, environmental conditions, such as climate, resources, predators, and competitors, for the two populations will differ causing natural selection to favor divergent adaptations in each group. Different histories of genetic drift, enhanced because the populations are smaller than the parent population, will also lead to divergence.
Given enough time, the genetic and phenotypic divergence between populations will likely affect characters that influence reproduction enough that were individuals of the two populations brought together, mating would be less likely, or if a mating occurred, offspring would be non-viable or infertile. Many types of diverging characters may affect the reproductive isolation (inability to interbreed) of the two populations. These mechanisms of reproductive isolation can be divided into prezygotic mechanisms (those that operate before fertilization) and postzygotic mechanisms (those that operate after fertilization). Prezygotic mechanisms include traits that allow the individuals to find each other, such as the timing of mating, sensitivity to pheromones, or choice of mating sites. If individuals are able to encounter each other, character divergence may prevent courtship rituals from leading to a mating either because female preferences have changed or male behaviors have changed. Physiological changes may interfere with successful fertilization if mating is able to occur. Postzygotic mechanisms include genetic incompatibilities that prevent proper development of the offspring, or if the offspring live, they may be unable to produce viable gametes themselves as in the example of the mule, the infertile offspring of a female horse and a male donkey.
If the two isolated populations are brought back together and the hybrid offspring that formed from matings between individuals of the two populations have lower survivorship or reduced fertility, then selection will favor individuals that are able to discriminate between potential mates of their own population and the other population. This selection will enhance the reproductive isolation.
Isolation of populations leading to allopatric speciation can occur in a variety of ways: from a river forming a new branch, erosion forming a new valley, or a group of organisms traveling to a new location without the ability to return, such as seeds floating over the ocean to an island. The nature of the geographic separation necessary to isolate populations depends entirely on the biology of the organism and its potential for dispersal. If two flying insect populations took up residence in separate nearby valleys, chances are that individuals from each population would fly back and forth, continuing gene flow. However, if two rodent populations became divided by the formation of a new lake, continued gene flow would be unlikely; therefore, speciation would be more likely.
Biologists group allopatric processes into two categories. If a few members of a species move to a new geographical area, this is called dispersal. If a natural situation arises to physically divide organisms, this is called vicariance. Scientists have documented numerous cases of allopatric speciation taking place. For example, along the west coast of the United States, two separate subspecies of spotted owls exist. The northern spotted owl has genetic and phenotypic differences from its close relative, the Mexican spotted owl, which lives in the south (Figure below). The cause of their initial separation is not clear, but it may have been caused by the glaciers of the ice age dividing an initial population into two.
In some cases, a population of one species disperses throughout an area, and each finds a distinct niche or isolated habitat. Over time, the varied demands of their new lifestyles lead to multiple speciation events originating from a single species, which is called adaptive radiation. From one point of origin, many adaptations evolve causing the species to radiate into several new ones. Island archipelagos like the Hawaiian Islands provide an ideal context for adaptive radiation events because water surrounds each island, which leads to geographical isolation for many organisms (Figure below). The Hawaiian honeycreeper illustrates one example of adaptive radiation. From a single species, called the founder species, numerous species have evolved, including the eight shown in Figure below.
The honeycreeper birds illustrate adaptive radiation. From one original species of bird, multiple others evolved, each with its own distinctive characteristics. Notice the differences in the species’ beaks in Figure above. Change in the genetic variation for beaks in response to natural selection based on specific food sources in each new habitat led to evolution of a different beak suited to the specific food source. The fruit and seed-eating birds have thicker, stronger beaks which are suited to break hard nuts. The nectar-eating birds have long beaks to dip into flowers to reach their nectar. The insect-eating birds have beaks like swords, appropriate for stabbing and impaling insects. Darwin’s finches are another well-studied example of adaptive radiation in an archipelago.
## Speciation without Geographic Separation
One form of sympatric speciation can begin with a chromosomal error during meiosis or the formation of a hybrid individual with too many chromosomes. Sympatric speciation may also take place in other ways. For example, imagine a species of fish that lived in a lake. As the population grew, competition for food also grew. Under pressure to find food, suppose that a group of these fish had the genetic flexibility to discover and feed off another resource that was unused by the other fish. What if this new food source was found at a different depth of the lake? Over time, those feeding on the second food source would interact more with each other than the other fish; therefore they would breed together as well. Offspring of these fish would likely behave as their parents and feed and live in the same area, keeping them separate from the original population. If this group of fish continued to remain separate from the first population, eventually sympatric speciation might occur as more genetic differences accumulated between them.
This scenario does play out in nature, as do others that lead to reproductive isolation. One such place is Lake Victoria in Africa, famous for its sympatric speciation of cichlid fish. Researchers have found hundreds of sympatric speciation events in these fish, which have not only happened in great number, but also over a short period of time. Figure below shows this type of speciation among a cichlid fish population in Nicaragua. In this locale, two types of cichlids live in the same geographic location; however, they have come to have different morphologies that allow them to eat various food sources. | 3,563 | 13,996 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2024-38 | latest | en | 0.196444 |
http://mathhelpforum.com/algebra/114156-i-need-help-evaluating-numbers.html | 1,524,722,600,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125948089.47/warc/CC-MAIN-20180426051046-20180426071046-00122.warc.gz | 201,359,626 | 11,111 | # Thread: I need help evaluating this numbers
1. ## I need help evaluating this numbers
x=3/10
evaluate
y=5x^2-3x+4
y=5(3/10)^2 -3(3/10)-1 (How do you evaluate this part, step by step)
y=71/20
2. Originally Posted by Cyberman86
x=3/10
evaluate
y=5x^2-3x+4
y=5(3/10)^2 -3(3/10)-1 (How do you evaluate this part, step by step)
$\displaystyle y=5\left(\frac{3}{10}\right)^2 -3\left(\frac{3}{10}\right)+4$
$\displaystyle y=5\left(\frac{3^2}{10^2}\right) -3\left(\frac{3}{10}\right)+4$
$\displaystyle y=5\left(\frac{9}{100}\right) -3\left(\frac{3}{10}\right)+4$
$\displaystyle y=\frac{5\times 9}{100} -\frac{3\times 3}{10}+4$
$\displaystyle y=\frac{45}{100} -\frac{9}{10}+4$
To finish the problem you must create a common denominator
$\displaystyle y=\frac{45}{100} -\frac{90}{100}-\frac{400}{100}$
$\displaystyle y=\frac{45-90-400}{100}$
Now finish and simplfy.
3. Originally Posted by pickslides
$\displaystyle y=\frac{45-90-400}{100}$
Last term should be + 400/100
4. Originally Posted by Wilmer
Last term should be + 400/100
Thanks Wilmer
I did notice that it could be the case but after reviewing this
Originally Posted by Cyberman86
x=3/10
evaluate
y=5x^2-3x+4
y=5(3/10)^2 -3(3/10)-1 (How do you evaluate this part, step by step)
I was not sure which sign it could be? How can a +4 become a -1?
Posters should be more careful if they are serious about their education.
5. Originally Posted by pickslides
$\displaystyle y=\frac{45}{100} -\frac{9}{10}+4$
To finish the problem you must create a common denominator
$\displaystyle y=\frac{45}{100} -\frac{90}{100}-\frac{400}{100}$
I only looked at these 2 lines...hence noticed the switch from + to -
6. I think the poster should re-supply the question.
7. Originally Posted by pickslides
Thanks Wilmer
I did notice that it could be the case but after reviewing this
I was not sure which sign it could be? How can a +4 become a -1?
Posters should be more careful if they are serious about their education.
I would say + because 45-90 < 0 and so for the answer to be > 0 it would need to be +400 | 695 | 2,067 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2018-17 | latest | en | 0.835692 |
http://articles.bplans.com/writing-a-business-plan/break-even-analysis | 1,394,399,927,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394010354479/warc/CC-MAIN-20140305090554-00011-ip-10-183-142-35.ec2.internal.warc.gz | 11,942,256 | 19,242 | # Break-even Analysis
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The Break-even Analysis lets you determine what you need to sell, monthly or annually, to cover your costs of doing business—your break-even point. Illustration 1 shows the Break-even Analysis table:
Illustration 1: Break-even analysis
The Break-even Analysis table calculates a break-even point based on fixed costs, variable costs per unit of sales, and revenue per unit of sales.
Understanding break-even analysis
The break-even analysis is not our favorite analysis because:
• It is frequently mistaken for the payback period, the time it takes to recover an investment. There are variations on break even that make some people think we have it wrong. The one we do use is the most common, the most universally accepted, but not the only one possible.
• It depends on the concept of fixed costs, a hard idea to swallow. Technically, a break-even analysis defines fixed costs as those costs that would continue even if you went broke. Instead, you may want to use your regular running fixed costs, including payroll and normal expenses. This will give you a better insight on financial realities. We call that “burn rate” these post-Internet days.
• It depends on averaging your per-unit variable cost and per-unit revenue over the whole business.
However, whether we like it or not, this table is a mainstay of financial analysis. You may choose to leave it out, but really, a business plan would not be complete without it. And, although there are some other ways to do a Break-even Analysis, this is the most standard.
The Break-even Analysis depends on three key assumptions:
1. Average per-unit sales price (per-unit revenue):
This is the price that you receive per unit of sales. Take into account sales discounts and special offers. Get this number from your Sales Forecast. For non-unit based businesses, make the per-unit revenue \$1 and enter your costs as a percent of a dollar. The most common questions about this input relate to averaging many different products into a single estimate. The analysis requires a single number, and if you build your Sales Forecast first, then you will have this number. You are not alone in this, the vast majority of businesses sell more than one item, and have to average for their Break-even Analysis.
2. Average per-unit cost:
This is the incremental cost, or variable cost, of each unit of sales. If you buy goods for resale, this is what you paid, on average, for the goods you sell. If you sell a service, this is what it costs you, per dollar of revenue or unit of service delivered, to deliver that service. If you are using a Units-Based Sales Forecast table (for manufacturing and mixed business types), you can project unit costs from the Sales Forecast table. If you are using the basic Sales Forecast table for retail, service and distribution businesses, use a percentage estimate, e.g., a retail store running a 50% margin would have a per-unit cost of .5, and a per-unit revenue of 1.
3. Monthly fixed costs:
Technically, a break-even analysis defines fixed costs as costs that would continue even if you went broke. Instead, we recommend that you use your regular running fixed costs, including payroll and normal expenses (total monthly Operating Expenses). This will give you a better insight on financial realities. If averaging and estimating is difficult, use your Profit and Loss table to calculate a working fixed cost estimate—it will be a rough estimate, but it will provide a useful input for a conservative Break-even Analysis.
Illustration 2 shows a Break-even chart. As sales increase, the profit line passes through the zero or break-even line at the break-even point.
Illustration 2: Break-even chart
The illustration shows that the company needs to sell approximately 1,222 units in order to cross the break-even line. This is a classic business chart that helps you consider your bottom-line financial realities. Can you sell enough to make your break-even volume?
The break-even analysis depends on assumptions made for average per-unit revenue, average per-unit cost, and fixed costs. These are rarely exact. We recommend that you do the break-even table twice: first, with educated guesses for assumptions, as part of the initial assessment, and later on, using your detailed Sales Forecast and Profit and Loss numbers. Both are valid uses.
(10 votes, average: 4.40 out of 5)
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Tim Berry is the founder of Palo Alto Software and Bplans.com. Follow him on Twitter @Timberry. More »
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• Anand
Hi Bplan,pls suggest how to plan a start up wholesale distribution business for a telecom operator.Also would like to know the calculation of break-even analysis.
• Stanley Clark
I need a sample copy ..on how to compute a break-even…how do you calculate it
• krishna
i want a sample copy of a break even point
and how to calculate it..
• mare
Fixed costs don not change with a level of activity for a particular activity. e.g. rates insurance cost of machinery, overheads.
you have a coffee bar
-the price for each cupof coffee is 3-50 =selling price (SP)
-milk, coffe beans and sugare cost 50cents = varieable costs (VC)
- Fixed costs are 10000 (FC)
-How many cups of coffee do you have to sell to breakeven in the first month.
Breakeven = FC_
CM (contribution margin)
= \$10000 (fixed cost) over
3.50 minus 50cents
= 3.00
= 3.00 divided into 10000 (FC)
= 3,333 cups of coffee
• Gary Alexander
How do you work out breakevens when your entire business is services rather than tangible goods sold?
• jookem
well gary, you would take into account the salaries(expenses) paid employees doing the services, and what you charge the clients (profit) for them.
• doris
rham,
The basic formula for the break even analysis is as follow:
P(x)=A+B(X)
where P is the Per unit sales price (per unit revenue)
A is the Fixed cost and B is the Variable cost per unit.
If you basic algebra…you shoul be all set solving the equation…Good luck!
• Hamida
please i want to know why you don’t construct your total cost line when you are drawing the break even chart.
• http://www.ArticlesForKnowledgeSharing.com Pankaj Trivedi
Doris, you have simplified it nicely.
-Pankaj
• Mo
How do you calculate the break even point for a service company?
• http://www.businessingeneral.com Chelle Parmele
Mo,
Have you tried using our Break-Even Calculator? http://www.bplans.com/business_calculators/break_even_analysis.cfm
That might be super useful in determining your break even for your company.
• Natalia
Hi, Please help, how do I count a gross profit for each sale if it’s a day care center. THe wages and rent and utilities are included inthe fixed costs, does it mean I would have to show a 100% gross profit for each child? If not then I would have different number of kids and as a result I would need different number of employees, do I have to make a brake even for each number of kids or count the muximum I can have?
Thank you
• Luke
If I receive a loan to purchase inventory, how do I factor the loan figure into my variable per unit cost? If I express my per unit cost as a variable cost and my principle loan amount as a fixed cost, am I not purchasing my inventory twice? Once with the variable expense and one with the fixed expense? So confused
————————————–
The main point to keep in mind here is that the details of your loan should be kept separate from the inventory that you’ll be purchasing with that money.
If you’ve chosen the option to manage inventory in your plan, Business Plan Pro will use the Cost of Sales information from your Sales Forecast to automatically estimate Inventory Purchases based on a few other assumptions that the program will ask you about. Those Inventory Purchases are automatically added to the Accounts Payable in your plan and flow through into the Cash Flow as part of the Bill Payments as they should be.
The loan amount should be entered into the appropriate tables (Start-up Funding or Cash Flow depending on timing) and the Principal Payments for that loan should be entered into the Cash Flow table. Business Plan Pro will automatically calculate the Interest Expense on the loan and place that value at the bottom of the Profit & Loss statement for you.
For additional information, please review:
Cost of Sales
Inventory Detail
FAQ: Handling of Loans, Interest and Repayment
Sean Serrels
Product Trainer/Evangelist
Palo Alto Software
• http://www.B-School.com Bill
I run a internet startup and was thrilled to hear the analogy of burn rate. I’m not trying to convert everything to a post internet term and create a new age bplan
Your explanations are very helpful
Previous post: | 1,890 | 8,686 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2014-10 | latest | en | 0.927961 |
https://forum.freecodecamp.org/t/learn-how-to-work-with-numbers-and-strings-by-implementing-the-luhn-algorithm-step-35/685174 | 1,716,827,123,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059044.17/warc/CC-MAIN-20240527144335-20240527174335-00301.warc.gz | 219,679,062 | 5,901 | # Learn How to Work with Numbers and Strings by Implementing the Luhn Algorithm - Step 35
### Tell us what’s happening:
I am not able to this can someone help me plz I have been trying but notable to do
``````
# User Editable Region
def verify_something_valid(something_valid):
sum_of_odd_digits = 0
something_valid_reversed = something_valid[::-1]
odd_digits = something_valid_reversed[::2]
for digit in odd_digits:
sum_of_odd_digits += int(digit)
sum_of_even_digits = 0
even_digits = something_valid_reversed[1::2]
for digit in even_digits:
number = int(digit) * 2
if number >= 10:
number = (number // 10) + (number % 10)
sum_of_even_digits += number
total = sum_of_odd_digits + sum_of_even_digits
def main():
card_number = '4111-1111-4555-1141'
card_translation = str.maketrans({'-': '', ' ': ''})
translated_something_valid = something_valid.translate(card_translation)
if verify_something_valid(translated_something_valid):
print('VALID!')
else:
print('INVALID!')
main()
# User Editable Region
``````
User Agent is: `Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/123.0.0.0 Safari/537.36`
### Challenge Information:
Learn How to Work with Numbers and Strings by Implementing the Luhn Algorithm - Step 35
As a final step, remove the `print` call from the `verify_card_number` function, and change the `card_number` back to something valid.
The instructions may not have been clear.
The function and variable names are to stay the same.
Change the value in `card_number`
You correctly removed the `print` call.
Happy coding
I am not able to do this can you give me solutions plz
Nope, that’s not what the forum is for. If you have a question that you can ask about anything that’s confusing, feel free to ask and we will help. | 463 | 1,791 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.735015 |
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Solved
Posted on 2003-12-01
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I'm trying to learn more about graph implementation using Adjacency tables/matrix, adjacency list, and linked lists. I have been searching but can't really find a SIMPLE example. I'm basically wanting to see how something is inserted into graph (vertices/edges) and printing it out. Not sure how the code is going to look for this type of thing. Thanks for any help/links/etc. I plan to eventually write up something similar but the text book is not the best.
0
Question by:killer455
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LVL 11
Accepted Solution
ID: 9854956
Hey, killer455:
(1) What text book? (Just curious)
(2) I can give you the typical descriptions of what an adjacency matrix and an adjacency list are. I tend to represent graphs using two classes, node and edge, where each edge has pointers at two nodes and each node has two lists of edges (in arc and out arcs); your choice among the various representations depends on what you want to do.
Adjacency matrix: Square matrix with one row (and one column) for each node in the graph. The entry at [i][j] in the matrix is 0 if there is no edge between node[i] and node[j], 1 if there is such an edge.
Adjacency list: Since the size of an adjaceny matrix is quadratic in the number of nodes (O(n^2) where n is the number of edges), if the graph is not densely connected, keeping an array of nodes, each with a list of the nodes they are adjacent to, can be more space efficient. Each edge is represented in two lists (so if there is an edge from node[i] to node[j], the edge is represented by a "j" appearing in list[i] and a "i" appearing in list[j]).
(3) Printing out:
Adjacency matrix is easy to print, fairly compact, and you can do it with a pair of for loops that move the indices over the whole square.
Adjaceny lists can be harder to print (one count-controlled loop, one loop to traverse a list) because you typically need separators between successive entries in any one list (so you can tell the difference between node 1 and node 2 appearing in the list and node 12 appearing in the list).
(4) Code structure
Regardless of structure you should think about the interface of a graph object. You need constructor/destructor. you need to be able to add a node (with whatever information a node requires) and you need to be able to add an edge (this requires being able to specify two nodes already in the graph). Then printing the graph rounds out the basic set of functions you need in the interface.
To handle insertion of edges you probably need to be able to find a node by "name" (some sort of node identifier) but that is a private function.
The actual data members of the graph class depend on the graph representation you select.
Hope this helps, -bcl
0
Author Comment
ID: 9862463
Ok well i chose to start implementing the graph as a linked list.
Here are some general questions I have:
1) is this correct for the linked list implementation? any tips?
2) what would be the difference with an adjacency list?
Basically in the end im wanting three implementations:
Current Code: http://www.geocities.com/bjohnson045/program/
0
LVL 11
Expert Comment
ID: 9862638
(1) How you store the vertex list with an adjacency list is really immaterial so using a linked list of vertices is as good as a vector of vertices or even an array of vertices.
(2) What are you going to DO with your graphs? Saying you will have three implementations of graphs is all fine and well but it is similar to saying I will have three implementations of cars. Without knowing what the cars will be used for it is hard to decide between a gas-guzzling monster truck or a pressboard Traubant.
I will try to take a look at your code tomorrow morning (EST; in about 13 hours). I don't really have any words of wisdom other than: Make a good, easy to use graph interface. Make sure it hides the implementation so that changing the implementation is easy.
Good luck, -bcl
0
Author Comment
ID: 9862965
To tell you the truth this is an assignment im working on.
The three implementations are a requirement. To tell you the truth I dont know why we need to do all three.
As listed above they are:
I'm guessing that if I can get one implementation the others will be easier correct? As you said make a easy to use graph interface and the implementation changes should be easy.
0
LVL 11
Expert Comment
ID: 9863506
(1) Interface is the key. Make sure that your Graph class does not expose any "private" information. You don' t want the internal representation of vertices and edges to be visible outside the class; use functions like addNode() and addEdge() that take "outside" information as parameters.
What I mean is, think about a graph being used to represent airplane routes. The nodes are named cities and edges connect cities with direct connecting flights and are labeled with their weight, the cost of a ticket between the two cities. I want to be able to use it as in the following:
Graph G;
cout << G << endl;
Note that each node is known by its name OUTSIDE the graph class. Each edge is represented by two node names and the weight.
The underlying graph could be a digraph (in which case this grpah is not connected) or an undirected graph (in which case this graph IS connected).
Think carefully about how you will name nodes and edges across the interface.
-bcl
0
Author Comment
ID: 9864073
I have it setup as you mentioned.
0
Author Comment
ID: 9868381
Did you get a chance to check out my code thats up there?
Just wanting to see what you think.
0
LVL 11
Expert Comment
ID: 9868513
I downloaded it when I got to work and then I started into doing stuff and forgot to look at it. It looks pretty good. I am not sure why Edge and Vertex are classes outside of Graph (you have a find function that returns a pointer at a Vertex inside the graph. That is very, very dangerous because I can do bad things to the internal pointers of that vertex, things that bork your graph; once you return copies of the data or some such there is no need to have Edge or Vertex appear outside of Graph so you can move them into the private section of the structure).
I tend to put two end pointers on an edge, both the head and the tail vertices. I also tend to have Graph keep track of all edges in it as well as all vertices (permits certain types of checks on the number of edges versus the number of edges there could be in a complete graph and stuff like that).
It looks pretty good if I am just making simple design suggestions.
-bcl
0
Author Comment
ID: 9868683
I guess im pretty confused at this point. So I should only have one class? I'm wanting to get this as simple as i can, so obviously i can understand it a little better. Right now im lost. The three classes...
class Vertex, lass Edge, class Graph were examples in the text.
0
LVL 11
Expert Comment
ID: 9869187
Only difference in my suggestion is that class Vertex and class Edge ARE INSIDE class Graph:
class Graph {
private:
class Vertex {
// ... as you have it now
};
class Edge {
// ... as you have it now
};
Vertex * allTheVerticesList;
public:
Graph();
// ... as you have it now
};
Change in .cpp file is that the names of Vertex and Edge are now Graph::Vertex and Graph::Edge
Hope this makes it clearer.
-bcl
0
Author Comment
ID: 9870584
Ive been working a bit on the adjacency table implementation but it looks pretty bad.
Am I headed in any direction at all?
#ifndef _TABLEGRAPH_H
#define _TABLEGRAPH_H
template <int max_size>
class Digraph {
public:
void write(max_size);
private:
int count;
};
#endif
Graph::Graph()
{
count = 0;
}
{
"Entr connections:";
int i, j;
while(cin >>i >> j) // read a connection and set the edges
{
}
}
int main()
{
int verts;
cout <<"Enter the number of vertices:";
cin >> verts;
Digraph <verts> mygraph;
mygraph.write();
return 0;
}
0
LVL 11
Expert Comment
ID: 9870698
Stop!
The signature of your graph classes should be the same. I mean exactly the same. You should be able to use one main.cpp and only change the header included and, possibly, the other file linked to use any one of the three graph implementations.
You might argue about that since you want to use an integer template parameter for compile time size determination but if you use dynamic memory allocation AND hack the constructor in every graph to take the maximum number of nodes (only enforced in the table version) you can use the same code across the board.
<Note: I wrote my dissertation on graph-based algorithms and graph-based data storage and I came to it from a hypertext rather than a graph-theoretic direction. That means I call them nodes and not vertices. Also means I sometimes call edges links. Read your favorite name everywhere below as I may well switch back and forth. My apologies in advance for any confusion this may cause.>
Having said that, how would you write addVertex for this graph implementation? Well, since you need to associate a name with a row/column in the table you need some mechanism to match the two, say an array of node data and a getNodeIndesFromName and getNodeNameFromIndex functions (implementation functions so they are NOT in the interface for Graph). Thus addNode updates the relationship between names and indices.
What does addEdge need to do in this representation? Well it looks up a couple of indices and if the named nodes exist it adds an edge...wonder what that actually entails.
Finally, how would you print this implementation out? Probably as the matrix.
Hope this helps, -bcl
0
Author Comment
ID: 9870705
Well you dont have to say anything about the stupid mistakes. I see those already but mostly im wondering about the following.
1) why cant you use a variable for the template?? (verts)
2) I'm wanting a read function that takes in from the terminal the number of vertices. Should this be taken into count? I'm confused on what determines how big the overall matrix is? Isnt its size the number of vertices? I dunno im confused here.
0
Author Comment
ID: 9870708
Oh i didnt see your most recent post, let me look it over.
0
LVL 11
Expert Comment
ID: 9870723
You can't use a template if you expect the user to provide the number of vertices in the graph at RUNTIME. You basically have to use dynamic memory yourself or use an STL container that does dynamic memory allocation for you under the hood.
-bcl
0
Author Comment
ID: 9870727
Well ok, I should have mentioned this. I'm wanting to make all of this a bit more simple. So I am basically starting over with the adjacency table and am going to re-do my linked list version. All I really want is a read function that takes in the number of vertices (nodes) from the terminal, and then takes in the connections (edges/links) from the terminal. I also want a write function that writes graph info to the terminal.
0
Author Comment
ID: 9870742
Well specifying the number of vertices is confusing to me, considering I'm using part of the code from the text. It gives us...
template <int max_size>
class Digraph {
int count;
};
But the problem statement is to have number of vertices determined from user input from the terminal.
0
Author Comment
ID: 9870749
The only thing i could think of is that it wants use to determine the max size of the matrix, but then only part of it is used?
0
LVL 11
Expert Comment
ID: 9870789
Okay. That is fine. So the interface has changed to read and write:
class Graph {
public:
void write();
private:
// The devil is in the details
};
Okay, what does read do? I would guess that it is something like this (based on your code with a little bit more abstraction):
int i, j, n;
cout << "Number of nodes: ";
cin >> n;
if (0 < n) {
setNumberOfNodes(n);
while (cin >> i >> j) {
}
}
}
And write? How about ALWAYS printing it out like it is an adjacency table (so you can compare your implementations for correctness):
void Graph::write() {
for (int i = 0; i < numberOfNodes; ++i) {
for (int j = 0; j < numberOfNodes; ++j)
if (isEdge(i, j))
cout << "1 ";
else
cout << "0 ";
cout << endl;
}
}
Then you just write addEdge(int, int) (easy in this case, just the two lines from inside your loop) and isEdge(int, int) (how would you implement this?).
Then the only hard part is the constructor and the setNubmerOfNodes routines. Constructor is easy (does nothing; don't know size until setNumberOfNoces) and setNumberOfNodes allocates a dynamic two dimensional array.
Hope this helps some. If I am pushing you where you don't want to go, let me know and I will help with what you need help on. I think a single, unified design will help make the rest of the coding task easier.
-bcl
0
Author Comment
ID: 9870871
I appreciate any direction you help to move me in. The help is greatly appreciated.
Well here is kinda what I came up with. I beleive my private section of the class is not right.
class Digraph {
public:
void write( );
private:
int numberOfNodes;
Graph( );
setNumberOfNodes( );
};
int i, j, n;
cout << "Number of nodes: ";
cin >> n;
if (0 < n) {
setNumberOfNodes(n);
while (cin >> i >> j) {
}
}
}
void Graph::write( ) {
for (int i = 0; i < numberOfNodes; ++i) {
for (int j = 0; j < numberOfNodes; ++j)
if (isEdge(i, j))
cout << "1 ";
else
cout << "0 ";
cout << endl;
}
}
Graph::Graph( )
{}
{
}
bool isEdge(int i, int j)
{
{ return 1; }
else
{ return 0; }
}
Graph::setNumberOfNodes(int n)
{
int numberOfNodes = n;
}
0
LVL 11
Expert Comment
ID: 9870945
(1) Graph's constructor is public.
(2) Let's look at what is happening inside of setNumberOfNodes:
Graph::setNumberOfNodes(int n)
{
int numberOfNodes = n;
}
Well, from the point of view of the Graph object, NOTHING is happening inside here (it also won't compile but that is not the point right now). Note that two local variables are declared inside of setNumberOfNodes and both local variables go away at the end of the function. The object-level variable numberOfNodes is NOT changed by this function (inside this function numberOfNodes is a local variable that hides this->numberOfNodes).
So, step 1 is to remove the declaration of a new integer.
Step 2 is to remove the local declaration of adj (it, too, should be an object-level variable). The problem is that the declaration you use doesn't compile. There are two solutions:
(a) Pick a maximum number of nodes in the graph and only use part of the adjacency matrix.
(b) Dynamically allocate the space for the array.
(a) is easier so we'll use the maxNodes variable for this purpose:
private:
const static int maxNodes = 100;
(code not tested with compiler; I am pretty sure a const static can be used in this fashion).
Almost the same as the template parameter from the text's version.
0
Author Comment
ID: 9871053
So all the setNumberOfNodes function is doing is setting numberOfNodes = n ??
0
LVL 11
Expert Comment
ID: 9871081
Sure.
Told you (a) was easier than (b)
0
Author Comment
ID: 9871147
Sooner or later im going to do a depth and breadth first search function, doesnt it need to be a directed graph to do this?
0
LVL 11
Expert Comment
ID: 9871208
(1) No, depth, breadth, and order first (like finding the shortest path) are all well defined on digraphs AND undirected graphs. Assuming the code you have posted works you have an undirected graph that is one line away from being a digraph (hint: it is removing one line).
0
Author Comment
ID: 9871252
Oh ok, gotcha. Well ok ill keep you posted on how im coming along. (im sure more problems will arise). I feel I have a long way to go to where I want to get, but hopefully it wont be too hard with this code as a starter. I plan on implemented all three implementations (adj table, linked vertex list w/ linked list, linked list) along with depth/breadth for each, and Prim's, Kruskals, and Dijkstra's algorithms for each.
0
Author Comment
ID: 9871267
Meant to say contigous vertex list of linked adjacency lists for the second implementation.
0
Author Comment
ID: 9871276
Another quick question, if the graph is undirected, won't the vertices need some type of weight or name to determine how the breadth and depth work?
0
LVL 11
Expert Comment
ID: 9871409
Depth-first and breadth-first searches don't depend on edge weights; the other algorithms you mention do (Dijkstra's is an in-order search; it is possible to write one routine that does depth, breadth, and Dijkstra's simply by changing one data structre from a stack to a queue to a priority queue). Notice that the weights are traditionally on the edges and not on the nodes themselves.
-bcl
0
Author Comment
ID: 9872145
So basically im going to have to make edges into something like a struct, giving it a weight?
0
LVL 11
Expert Comment
ID: 9873869
Yep but when you have a class representing an edge (as you had in early code) then you can just add weight. What to do in the adjacecny matrix...
Wait a minute, if all you need is a weight and a weight is a number and you have a matrix of numbers indexed by the nodes in the graph...insiight like the preceding leads to the use of a weight matrix in place of an adjacency matrix. Need to work on isEdge and decide what value to use for element [i][i] and non-edges but it is fairly easy to accomodate weights.
Take a look at the text and I'll bet they show how to do weights using an adjacency list.
-bcl
0
Author Comment
ID: 9875484
Ah so basically instead of using a bool (1 or 0) matrix the number will represent the weight of the edge?
0
LVL 11
Expert Comment
ID: 9875722
Roger that.
0
Author Comment
ID: 9876128
Ok sounds good. Well im working on getting this to compile. However im stuck. Doesnt isEdge and addEdge need to be a part of graph since they are using the matrix? It also is giving me an error concerning the constant for adj[ ][ ]. Along with numberOfNodes, being out of scope (another reason i thought isEdge and addEdge needed to be a part of graph. Well here is the code if you want to take a look.
#ifndef _TABLEGRAPH_H
#define _TABLEGRAPH_H
#include "tablegraph.cpp"
class Graph {
public:
Graph();
~Graph();
void write();
private:
int numberOfNodes;
const static int maxVertices;
bool isEdge(i, j);
};
#endif
#include <iostream>
Graph::Graph()
{}
{
int i, j, n;
cout << "Number of vertices: ";
cin >> n;
if(0 < n){
setNumberOfNodes(n);
while(cin >> i >> j){
}
}
}
void Graph::write()
{
for(int i=0; i < numberOfNodes; i++){
for(int j=0; j < numberOfNodes; j++){
if(isEdge(i, j))
cout << "1 ";
else
cout << "0 ";
}
cout << endl;
}
}
void Graph::setNumberOfNodes(int n)
{
numberOfNodes = n;
}
{
}
bool Graph::isEdge(int i, int j)
{
return 1;
else
return 0;
}
#include <iostream>
#include "tablegraph.h"
using namespace std;
int main()
{
Graph mygraph;
mygraph.write();
return 0;
}
0
LVL 11
Expert Comment
ID: 9876192
How about including tablegraph.h in tablegraph.cpp?
I think that is missing
0
LVL 11
Expert Comment
ID: 9876255
Okay, I misread what you were doing:
Don't include the .cpp in the .h UNLESS you are implementing a template. If you are implementing a template then include the .cpp at the very END of the .h
Give maxVertices a value INSIDE the class definition:
const static int maxVertices = 100;
This makes no sense to the compiler:
THIS does:
Still should include tablegraph.h in tablegraph.cpp
Need using namespace std in tablegraph.cpp (or qalify cout and cin and all of that).
Hope this helps, -bcl
0
Author Comment
ID: 9878384
Ok well that works. Thanks for the tips. On to dfs and bfs. I started on bfs. The stl queue can be used correct? Here is what I have so far, but im a little confused near the end. Any critiquing?
void Graph::bfs(??)
{
Queue q;
bool visited[maxVertices];
for(int i=0; i<maxVertices; i++)
visited[i]=false;
for(int i=0; i<maxVertives; i++){
if(!visited[i]){
q.append(i);
while(!q.empty()){
q.retrieve(w);
if(!visited[w]){
visited[w]=true;
for(all vertices adj to w) // ??
q.append(x);
}
q.serve();
}
}
}
}
0
LVL 11
Expert Comment
ID: 9879106
(1) A breadth first search starts from somewhere. Thus the parameter to the function is a vertex. The driving loop is then whether or not there are nodes to be visited (whether the queue is empty or not), NOT a count controlled loop on the number of vertices. Why? Because you don't know, a priiori, whether or not the graph is connected.
The basic algorithm is:
bfs(v)
q.enqueue(v);
while !q.empty
x = q.front
visit x
for each child of x
if !visited child
q.enqueue(child)
So that strips off your outer loop (you are doing that so that you end up visiting all nodes even if the graph is not connected?).
How would you find all the children (nodes adjacent to) a given node w (or x)? You would have to go across a row of the adjacency/weight matrix OR use isEdge...using isEdge means that you can write the bfs in an implementation neutral manner (no need to rewrite it for other implementations).
for (int child = 0; child < numberOfNodes; ++child) {
if (!isEdge(w, child)) continue;
// do the real work here
Question: Why not use a vector<char> for your vistited array? (You don't want to use vector<bool> for a lot of technical reasons and one not so technical reason: it is BROKEN in the standard.) The vector means you don't have to set aside space with a constant (you can use numberOfNodes). Makes it easier to be implementation neutral.
-bcl
0
Author Comment
ID: 9879377
So you're saying that the nodes that are not connected should not be visited? The BFS i wrote a post back I wrote basing it on the psuedo code in the book, so they had it going to every node?
0
LVL 11
Expert Comment
ID: 9879475
The Dictionary of Algorithms and Data Strucutres defines it in terms of nodes connected to a single starting node (http://www.nist.gov/dads/HTML/breadthfirst.html) and some other on-line notes work from a single node (http://www.ics.uci.edu/~eppstein/161/960215.html) and I have always seen it done that way. This is particularly important if you plan to compare depth-first, breadth-first, and best-first algorithms since Dijkstra's algorithm will find no path to nodes outside the connected component of the starting node.
Thus with appeal to authority I shall carry the argument.
Actually, with some consideration, I guess you could expand depth and bread first definitions to deal with non-connected graphs (in some way other than just visiting nodes in one connected component). I think you should ask your instructor what form they want it to take. If you write the single connected component version you can have it return the number of nodes it visited (or, better yet, a vector of the nodes in the order visited). Then, if you need to do the "all" case you can put a wrapper function around the single connected component one that finds an unvisited node and calls the connected component version.
The more I think about it the less I AM convinced that visiting all of the connected components makes any sense. What do I know, though. What text book are you using? Sedgewick? He is sometimes less than clear for algorithms you've never seen before.
In any case, I would ask but I think you start at a given root.
-bcl
0
Author Comment
ID: 9879628
Im using the Druse and Ryba text, Data Structures and Program Design in C++. Our professor is a little unclear and what she wants, and even when asked is sometimes unsure. It is her first year teaching. Her study is based in graphic design, and it just seems she lacks experience in teaching this subject. The reason i included each node to be visited is simply because the text had the basic algorithm coded that way. I agree that it should start at a single vertex and if nodes are not connected they should not be visited.
Couple questions:
1) Why in the code below do you have !isEdge, if there is not a edge nothing happens. Shouldnt it simple be isEdge, and then if true, push child onto the queue?
for (int child = 0; child < numberOfNodes; ++child) {
if (!isEdge(w, child)) continue;
// do the real work here
2) Im confused on why visited[] should not be bool? Isnt it simply going to be true or false, always? not sure what you mean by "it is BROKEN in the standard."
0
LVL 11
Expert Comment
ID: 9879697
(1) Sure, the positive if statement is clearer and correct. Don't know what I was thinking.
(2) If you use vector<T> (the STL vector class), T should not be bool. Templates support explicit specialization which means that you can write a special version of the template for a particular template parameter (so you could have a template that did one thing if instantiated with char and another if instantiated with string, for example). The exact syntax is unimportant to the story. vector<bool> has a specialization (it was hoped during the standardization process that vector<bool> could be implemented as a collection of bits with set and clear operations); that specialization was not well described during the standards process and it should never have been included (many on the standards committee now agree). So, I was warning you that if you switch to vector (which I think you should) you would have to change the element type of the container.
-bcl
0
Author Comment
ID: 9880051
Ok but you would still use 1 and 0 for visited and not visited correct? Even if the type is string, char, int it will still work the same. Well ok here is what i get from all of that :)
template <class T>
void Graph::bfs(v)
{
Queue q;
vector<T> visited;
for(int i=0; i<numberOfNodes; i++){
visited[i]=0;
}
q.push(v);
while(!q.empty()){
w = q.front();
if(visited[w]==0){
visited[w]=1;
for(int child = 0; child < numberOfNodes; ++child){
if(isEdge(w, child)){
q.push(child);
}
}
}
}
}
0
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Expert Comment
ID: 9881778
No need for bfs to be templated. visited is a vector of char/bytes and yes, I would use 0/1 for true/false (as C++ will treat them that way).
void Graph::bfs(NodeNdx v) // where NodeNdx is probably really int but it is the index of the starting node
{
Queue q; // Did you mean this to be the STL queue? If so, queue is a templated type
vector<unsigned char> visited; // signed/unsigned is really unimportant; vector of char IS important
// Initialize contents of visited; remember that it is currently empty:
for (int i = 0; i < numberOfNodes; ++i)
visited.push_back(false); // Note that C++ translates this to a char (value of '\0') for you
The rest looks right: push ALL of the children and worry about visitedness (not really a word) when they come off the queue.
-bcl
0
Author Comment
ID: 9884077
Oh yeah that was another question, where in here do they come off the queue, I would say when they are visited correct?
0
LVL 11
Expert Comment
ID: 9884355
Your code for setting visted[w] looks like it is in the right place. If you want to do more (as in visit the node), do it right before you set visited[w] to 1.
-bcl
0
Author Comment
ID: 9884758
Well doesnt the element have to come off of the queue somewhere? or else q.front( ) is always going to be the same thing?
void Graph::bfs(int v)
{
queue<int> q;
vector<unsigned char> visited;
for(int i=0; i<numberOfNodes; i++){
visited.push_back(false);
}
q.push(v);
while(!q.empty()){
int w = q.front();
if(visited[w]==0){
visited[w]=1;
q.pop();
for(int child = 0; child < numberOfNodes; ++child){
if(isEdge(w, child)){
q.push(child);
}
}
}
}
}
0
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Expert Comment
ID: 9884790
Sorry; I don't think in STL terms. Basically right after q.front() you want q.pop(); I read front as taking off and returning the front value.
-bcl
0
Author Comment
ID: 9885283
Does this work as a recursive algorithm for dfs? Or would it be smart to stick with using a stack? This actually isnt working, for some reason its not breaking, but just wanting to see if this is one way I can go about dfs.
void Graph::dfs(int v)
{
vector<unsigned char> visited;
for(int i=0; i<numberOfNodes; i++){
visited.push_back(false);
}
visited[v]=1;
cout << v << " ";
for(int child = 0; child < numberOfNodes; ++child){
if(isEdge(v, child)){
if(!visited[child]){
dfs(child);
}
else
break;
}
}
}
0
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Expert Comment
ID: 9885334
(1) Use a stack; it is just like bfs with a stack for a queue. Really.
(2) Note that visited is initialized each time the recursive function is called. If you want to do this recursively all the calls need to share a single instance of the visited vector.
-bcl
0
Author Comment
ID: 9885430
Ok cool. I had a couple questions concerning the other implementations I plan on doing.
I'm a bit confused on what this is actually trying to say.
1) A linked vertex list with linked adjacency lists. Is that going to be something like this? Or is it going to be entirely linked lists?
___ ______________
|___|---|__|__|__|__|__|
|___|
|
|
_V_ ______________
|___|---|__|__|__|__|__|
|___|
2) A contiguous vertex list of linked adjacency lists. Something like this?
___ ______________
|___|-->|__|__|__|__|__|
|___|-->|__|__|__|__|__|
|___|-->|__|__|__|__|__|
|___|-->|__|__|__|__|__|
0
LVL 11
Expert Comment
ID: 9886098
I honestly don't recognize the difference between linked and contiguous in this instance (I know what the word means and I think your drawing makes sense) but haven't used that distinction before.
Since the nubmer of adjacent vertices is unknown, that is typically some dynamically allocated structure (say a linked list or a vector<>).
Note also that if you are using weights in the vertex list you need to have pairs in the lists associated with any given vertex.
0
Author Comment
ID: 9886314
Ok so for the contiguous basically use a vertex or array or even a list for the vertexes and then have linked nodes for the adjaceny list.
Something like:
struct Node{
int v;
Node* next;
};
class Graph{
public:
void write();
private:
const static int maxVertices=50;
Graph();
};
0
LVL 11
Expert Comment
ID: 9887906
Sure BUT what you are calling a Node in this case is REALLY an edge. So if you want to extend this model with edge weights THAT is where you need to look.
-bcl
0
Author Comment
ID: 9889788
Am I doing the addEdge right here? I'm a bit confused on how I would check to see if there is a edge??
#ifndef _TABLEGRAPH_H
#define _TABLEGRAPH_H
class Graph{
public:
Graph();
void write();
void bfs(int v);
void dfs(int v);
private:
int numberOfNodes;
const static int maxVertices=50;
bool isEdge(int i, int j);
void setNumberOfNodes(int n);
};
#endif
#include <iostream>
using namespace std;
Graph::Graph()
{
for(int i=0; i<maxVertices; i++){
}
}
{
int i, j, n;
cout << "Number of vertices: ";
cin >> n;
if(0 < n){
setNumberOfNodes(n);
while(cin >> i >> j){
cout << "Adding Edge between " << i << " and " << j << "." << endl;
}
}
cout << endl;
}
void Graph::write()
{}
void Graph::setNumberOfNodes(int n)
{
numberOfNodes=n;
}
{
}
bool Graph::isEdge(int i, int j)
{
if(edge between i and j)
return 1;
else
return 0;
}
0
Author Comment
ID: 9889948
Oops, put the wrong header file up there.
struct Node{
int num;
Node* next;
Node():num(0),next(0)
{}
Node(int x, Node* t){
num=x;
next=t;
}
};
class Graph{
public:
Graph();
void write();
private:
int numberOfNodes;
const static int maxVertices=50;
isEdge(int i, int j);
void setNumberOfNodes(int n);
};
#endif
0
Author Comment
ID: 9890450
I figured maybe something like this for isEdge??
void Graph::isEdge(int i, int j)
for(Node* t = adj[i]; t!=0; t=t->next){
if(t == j)
return 1;
else
break;
}
return 0;
}
0
Author Comment
ID: 9890527
Ok well i got it working somewhat. It finds basically half of the adjacent nodes. Do I have it coded somehow that this is a directed graph or something? I'm using the code above with one revision:
for isEdge, if(t->num == j)
Not sure whats causing this, if u need me to pose the entire code again let me know, but if you get a chance do u see anything im missing??
0
LVL 11
Expert Comment
ID: 9891920
The bug is in your isEdge routine:
1: void Graph::isEdge(int i, int j) {
2: for(Node* t = adj[i]; t!=0; t=t->next){
3: if(t->num == j)
4: return 1;
5: else
6: break;
7: }
8: return 0;
9: }
What happens when we have the following graph:
Graph G;
and now we call G.isEdge(0,2)? The answer SHOULD be 1, right? It IS 0. Why?
The linked list off of adj[0] is {1, 2} (they are nodes with next pointers but that is the list, in order).
At line 2 t is initialize to point at the node containing 1. Line 3 1 is compared to 2 (the number sought) and doesn't match. The else clause (line 6) is executed, the loop terminated, line 8 returns 0.
The question becomes what is the logic of having a break in loop?
Hope this helps, -bcl
0
Author Comment
ID: 9898088
Ah ok i see where that was messing up. Well cool seems I got those two implementations working. However the linked list version is giving me trouble. I should only have to change the addEdge, isEdge, and constructor right?
0
Author Comment
ID: 9898089
Ah ok i see where that was messing up. Well cool seems I got those two implementations working. However the linked list version is giving me trouble. I should only have to change the addEdge, isEdge, and constructor right?
0
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Expert Comment
ID: 9898174
That is, basically, the point of having a good interface.
-bcl
0
Author Comment
ID: 9900101
Well where should the vertices/nodes be created? Am I going to need a addVertex method also?
0
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Expert Comment
ID: 9900128
Yes; way back at the top of this thread we worked on an interface that consisted of addNode, addEdge, isEdge, constructor and destructor. You will want those interface functions in all of your versions so the graphs are interchangeable. It may be that addNode just increments a counter (in the first implementation) or pushes something back on a vecotor of pointers (second implementation) or allocates a new node from the heap (third implementation); the client program and functions like bfs don't need to know anything about that.
-bcl
0
Author Comment
ID: 9900177
Ok, thats what i figured, i didnt end up putting addVertex in the others, but now i see its needed.
Ok i kinda have an idea of what im trying to do here, am i going about the addVertex, addEdge methods correctly? The isEdge i really am not sure of yet, so no code yet.
0
Author Comment
ID: 9900836
For the adjacency matrix the nodes were created by:
const static int maxVertices=50;
setNumberOfNodes simply created a boundry on this determined by the user
For the adjacency list the nodes were created here:
const static int maxVertices=50;
set nodes does the same thing here
I dont understand where u are saying a counter is incremented or a vector?
0
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Expert Comment
ID: 9900993
I tend to write my adjacency matrix versions to increment the number of nodes each time addNode is called. This is because I usually don't refer to nodes by nubmer but rather by some sort of name. Thus the addNode funciton I was thinking of looks like this:
nameVector.push_back(nodeName);
++numberOfNodes;
return true; // should check for too many nodes and return true or false
}
That way I can write the addEdge function to take node names:
bool addEdge(const &string source, const &string target) {
int sourceNdx = getNdx(source);
int targetNdx = getNdx(target);
if ((sourceNdx < 0) || (targetNdx < 0))
return false;
adj[sourceNdx][targetNdx] = 1; // only one; this is a digraph
}
In the other instance I mentally used a vector to hold the list of nodes, a vector of pair<string, Edge *> (where string is the name of the node as above and Edge * points at the list of adjacent nodes).
So your addNode routines can be NOP (do nothing) routines in those cases or you can avoid setNumberOfNodes and just add them one at a time. With your implementations it would mean to add one to numberOfNodes each time a new vertex is added.
I won't have a chance to look at your code until tomorrow morning. Hope this helped clear up what I was talking about.
-bcl
0
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Expert Comment
ID: 9903645
(1) Make a constructor for Edge and a constructor for Vertex so that you make sure your pointer fields are all initialized properly.
(2) Why does first_vertex get pointed to a new Vertex in Graph::Graph? Are you using a dummy head node on your linked list? That is, is an "empty" list one with a single, not really used, element in it? If not, why set first_vertex. If so I still say you should have a constructor for Vertex that does all of the work you're doing here.
(3) In addEdge, the following declaration doesn't do what you think it does:
Vertex *first_position, second_position;
You have declared a pointer at a Vertex (first_position) and a Vertex (second_position); the "*" only modifies the very next variable. Three ways to fix it (in increasing desireability):
(a) Vertex *first_position, *second_position; // fixed but hard to maintain
(b) Vertex *first_position;
Vertex *second_position; // one delcaration per line reminds you to add the star
(c) // earlier in the program:
typedef Vertex * PVertex; // create a new type pointing at a Vertex
VertexP first_position, second_position; // one type applies across whole line
(4) I don't understand addEdgeat all (sorry, but I am pretty sure you don't, either). Without paying too much attention to detail, what does the function need to do in this implementation?
first_position = find_vertex(i) // find_vertex returns a pointer at the given vertex or NULL if there isn't one
second_position = find_vertex(j);
if ((first_position != NULL) && (second_position != NULL)) {
// Create a new edge
// new edge's end point is second_position
// insert new edge in the list of edges pointed to by first_position->first_edge
}
Yes, you have to write a private member funciton find_vertex(int k); you have the guts of it in the beginning of addEdges but by making it a function you will make what you are doing in the function clearer and you will not get the two loops confused (which I believe you have).
Also: NEVER have a parameter and a local variable with the same name! I think the compiler would catch this one but don't use i as a parameter and as a local count control variable.
(5) addVertex should create a new vertex and insert it into the list. It is just a tail insert into a linked list. You seem to be confused by having multiple types of pointers. Make sure you work with one set of pointers (next_vertex or next_edge) in any given loop.
Hope this helps
0
Author Comment
ID: 9909044
Yeah you are right im having a hard time with this. Well does this look any better??
void Graph::find_vertex(int n)
{
Vertex *temp;
temp = first_vertex;
while(temp->next_vertex != NULL){
if(temp->value == n){
return temp;
}
else{
temp=temp->next_vertex;
}
}
return NULL;
}
{
Vertex *first_position, *second_position;
first_position = find_vertex(v1);
second_position = find_vertex(v2);
if((first_position != NULL) && (second_position != NULL)){
Edge* temp;
temp = first_position->first_edge;
temp->end_point = second_position;
// not sure about insert new edge
}
0
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Expert Comment
ID: 9909368
find_vertex does not check the value in the last entry in the linked list. Look at the logic inside the loop.
if((first_position != NULL) && (second_position != NULL)){
Edge* temp;
temp = first_position->first_edge;
temp->end_point = second_position;
// not sure about insert new edge
}
Okay, temp is the first edge in the out edges of *first_position. I will call *first_position "firstVertex" since it is a vertex and is pointed to by first_position; similarly with secondVertex.
So, again, temp points at the first edge OUT of firstVertex. You then point the edge at secondVertex. There are two obvious problems here: You are changing an edge that already points somewhere to point at secondVertex (means you lose whatever information was in the edge before) and you never check whether or not temp actually points at anything. Thus temp->end_point could, easily, be dereferencing a NULL pointer and dereferencing a NULL pointer is very, very bad (program crashes and dies bad).
So, have you written code to insert something into a linked list? If so pack it up into a function or, better yet, a linked-list class and just insert a new edge into the linked list. In anycase inserting a new edge means MAKING a new edge (means dynamically allocating it) AND inserting it into the list of outgoing edges in firstVertex. You must ADD the new edge to the list, not replace other edges in the list. You COULD check for a duplicate edge (standard graph theory does not permit duplicate edges in a digraph; the extended structure with multiple edges and self edges is called a multigraph though that term has several incompatible definitions).
Hope this helps, -bcl
0
Author Comment
ID: 9909883
I dont understand how this isnt what you meant?
// new edge's end point is second_position
temp->end_point = second_position;
0
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Expert Comment
ID: 9911922
The question is WHAT IS temp POINTING AT?
You do the right thing to the wrong pointer.
-bcl
0
Author Comment
ID: 9914819
Its point at the first_positions's first edge right?
0
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Expert Comment
ID: 9914901
Yes, it is. But is that WHERE it should be pointing. Consider again the following sequence of calls:
Graph G;
First, the program as written will crash because
temp = first_position->first_edge;
sets temp to NULL and then
temp->end_point = second_position;
is equivalent to
NULL->end_point = ...
Okay. Now let us say you insert a new node into first_position->first_edge and then run your code for the second addEdge. What happens?
Well, after the assignment of temp we have:
temp->end_point->data == 1 (the first edge inserted pointed from 0 to 1).
temp->next_edge == NULL
now we run
temp->end_point = second_position;
but second_position points at a DIFFERENT node than it did the last time this was run so NOW
temp->end_point->data == 2
temp->next_edge == NULL
There is only one edge out of node 0 and it points at 2.
Hope this helps,
-bcl
0
Author Comment
ID: 9915251
Ok im really confused now... i dont know what you meant by the second half of your last post. I can't seem to follow what you are saying.
Where did data come from?
0
Author Comment
ID: 9915302
Basically I have never written a program using linked list, which is obviously a problem because now that im faced with sometime like this i certainly don't have a slight clue about a lot of what im doing.
0
LVL 11
Expert Comment
ID: 9915376
.data is the number of the node. Since we have a linked list of nodes there is no obvious "index" of the node so I have added a "data" field to keep track of what number the node has (so they can be found by the value in data).
If you want to ignore that, try the following (from the line "Well, after the assignment of temp we have:" above):
========================================================================
temp->end_point == find_vertex(1) // the vertex numbered 1
temp->next_edge == NULL
now we run
temp->end_point = second_position;
but second_position points at a DIFFERENT node than it did the last time this was run so NOW
temp->end_point == find_vertex(2) // the vertex numbered 2
temp->next_edge == NULL
There is only one edge out of node 0 and it points at 2.
========================================================================
Hopefully that clears that up.
Question: Do you have to write your own linked list implementation for this? If the answer is yes I urge you to write a simple templated linked list that you can use to keep track of edges and nodes (where each type is appropriate). I can help you with that.
If the answer is "what other choice do I have?" then think about std::list. This is a STL container with performance characteristics that mean that every implementation I know of is a doubly-linked list under the hood.
Just a thought.
-bcl
0
Author Comment
ID: 9915542
Like ive said before im not sure sure what we are required to do, it is never made clear. But i will ask.
Ok the code you list above in between the === signs. Is this giving me an example of how my previous code will not work?
0
LVL 11
Expert Comment
ID: 9915588
Yes.
The problem is that you don't add a new edge to the list of outgoing edges from the node. Instead you point a pointer at the first element in such a list and change the values stored in it.
Again, consider the code I gave a couple of posts back. How many edges should there be out of each of the nodes 0, 1, and 2? The answer is 2 edges out of 0 (one going to 1, the other to 2) and none out of 1 or 2 (assumption is that this is a digraph). That is, after the insertion the adjacency matrix would look like this:
0 1 1
0 0 0
0 0 0
Your code would leave at most one edge out of 0 and that would be the last one added:
0 0 1
0 0 0
0 0 0
This is why the code has problems.
-bcl
0
Author Comment
ID: 9916429
I guess we are able to use std::list, havent used it before, but im assuming this will make it a lot easier.
0
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Expert Comment
ID: 9916462
Lots easier. Use lists of pointers (most likely).
Good luck and ask any questions you have.
-bcl
0
Author Comment
ID: 9916707
So this is the same thing as the STL list?? If so how would you go about having the double lists?
0
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Expert Comment
ID: 9916773
The STL std::list is implemented as a doubly linked list. You don't need to do anything to have a doubly linked list. You just declare an approriate variable just as you would a vector:
#include <list>
using namespace std;
...
list<Node *> allNodes;
allNodes.push_back(new Node(nodeNumber));
}
bool addEdge(int source, int target) {
Node * s = findNode(source);
Node * t = findNode(target);
s->edges.push_back(new Edge(target));
}
You can use almost everything you know about vectors with std::list. The big differences are that list has no operator[] (you can't treat it like an array) and list has a push_front() operator.
-bcl
0
Author Comment
ID: 9924482
Does list have a find feature? I'm looking at some documentation on it but not sure what it is saying, use an iterator? or does find( ) work? on addEdge, do you have to declare edges somewhere? I'm looking at how to do is edge, is there a easy way to see if there is a edge between two vertices?
0
LVL 11
Expert Comment
ID: 9925170
Given one vertex you can iterate down the list of outgoing edges to see if one "points" to the other node. Something like this:
bool isEdge(int i, int j) {
Node * iN = findNode(i);
for (list<Edge *>::iterator it = iN->edges.begin(); it != iN->edges.end(); ++it) {
if ((*it)->end_point == j)
return true;
}
return false;
}
This assumes that edges is of type list<Edge *> and that an Edge has a field called end_point that is the index of a Node. It also assumes that findNode returns a pointer to a Node.
Note that there is, in fact, a find algorithm in the <algorithm> header file but it is not really suited to finding an element in a container by a field in the class.
-bcl
0
Author Comment
ID: 9925217
So you still have to define the node and edge?
0
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Expert Comment
ID: 9925245
Oh yes. std::list is JUST LIKE std::vector in that it is a templated container. You have to decide what gets stored inside it and Node and Edge are the best candidates in your system.
At a minimum, Node contains a list of edges. Edge may only have one or two fields in it (say end_point and, perhaps, weight). Neither has any pointers to other Nodes or Edges; std::list hides the whole linked list thing from you. That is a good thing.
-bcl
0
Author Comment
ID: 9925735
So is findNode going to work the same as we last worked on it, or can you use the iterator for the list, and traverse the node list using that until its found, and then return it?
0
LVL 11
Expert Comment
ID: 9927736
Iterate the list is probably the easiest. You can use push_back to add elements to the two lists in addEdge and addNode.
-bcl
0
Author Comment
ID: 9927968
Don't the nodes/vertices need some type of value to distinguish themselves from the others??, i know when i had them there is a vertex number but where does this go??
void Graph::findVertex(int n)
{
for(list<Vertex *>::iterator it = allVerts.begin(); it != allVerts.end(); ++it){
if(what iterator points to's value == n)
return true
}
return false;
}
0
LVL 11
Expert Comment
ID: 9928092
Vertex should have a node number data member. Then you can look it up easily with iterators.
That make addVertex something like this:
Node * n = new Node(v); // now n->vertex_number == n
allVerts.push_back(n);
}
-bcl
0
Author Comment
ID: 9928168
ok one question, when u say Node(v), is this getting set through the constructor?
Also does list require it to be Node or can it be named vertex if i choose?
0
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Expert Comment
ID: 9928258
Sorry. My Node/Vertex problem emerged again. You can have a list of anything that has a copy constructor and an assignment operator (int, pointer, char, class, struct, ...) (I am pretty sure the only requirements are those two things; don't feel like looking it up in the standard right now but I am pretty sure that is correct).
My assumption was that the Vertex constructor would set the vertex_number field in the newly created Vertex object (there, I used Vertex everywhere I could in that sentence).
-bcl
0
Author Comment
ID: 9929650
Well ok i got some code together but getting several errors referring to the allVerts and edges not being declared right?
Code is here if you want to take a look:
http://www.geocities.com/bjohnson045/program/
0
LVL 11
Expert Comment
ID: 9929758
(1) You need to have list defined inside your graph.h file. Just inside the #ifndef stuff put
#include <list>
and then refer to list as std::list inside the header file. It is considered bad form to have using lines inside of header files. Since you only use it a couple of times you can just put in the qualification.
(2) Vertex has no constructor (in the class declaration inside of graph.h). It needs one.
(3) When it has a constructor, the name of the constructor OUTSIDE the class is
Graph::Vertex::Vertex(/* parameters go here */)
You are missing one of the ::Vertex parts
(4) findVertex needs to return a value.
That is as far as I got in working on compiler errors; there are a lot more.
Hope this helps
-bcl
0
Author Comment
ID: 9934835
What does "void value not ignored as it ought to be" usually mean?
Doesnt findVertex need to return a pointer or null? So would it return a Vertex*.
Note: it will not let me increment points past 500, ill make a seperate points post afterwards.
0
LVL 11
Expert Comment
ID: 9935013
(1) Whatever with points.
(2) Yes, findVertex should return a pointer or NULL if there is no such vertex. Can you show me the routine where you're getting the error message? I don't recognize the error.
-bcl
0
Author Comment
ID: 9937746
Basically these are the errors im getting. Ill post the up to date code here
www.geocities.com/bjohnson045/program
bash-2.05\$ g++ -o test graph.cpp main.cpp
graph.cpp: In method `void Graph::addEdge (int, int)':
graph.cpp:87: void value not ignored as it ought to be
graph.cpp:88: void value not ignored as it ought to be
graph.cpp: In method `bool Graph::isEdge (int, int)':
graph.cpp:95: void value not ignored as it ought to be
graph.cpp: At top level:
graph.cpp:104: syntax error before `*'
graph.cpp:106: syntax error before `!='
0
LVL 11
Expert Comment
ID: 9938005
graph.h:32 you declare findVertex to return void:
void findVertex(int n);
Bet it should return Vertex *.
Also: Outside of the Graph class the name of the class is Graph::Vertex (in graph.cpp, for example).
Also, findVertex is flawed. It will only find the matching vertex if it is the very first one in the list. The problem is with the logic of you if statement.
-bcl
0
Author Comment
ID: 9938198
>> "Outside of the Graph class the name of the class is Graph::Vertex (in graph.cpp, for example)."
I don't see where your saying i dont have this??
Why do i get syntax errors here:
Vertex* Graph::findVertex(int n) // get syntax here before *
{
for(list<Vertex *>::iterator it = allVerts.begin(); it != allVerts.end(); ++it){ // get syntax before !=
if((*it)->number == n)
return *it;
}
return NULL;
}
0
LVL 11
Expert Comment
ID: 9938243
The return type of that function is NOT Vertex *. If Vertex is defined INSIDE of the class Graph then the name of the return type is Graph::Vertex *.
There is a new extended lookup in the C++ standard (new as in 3 years old) and I think it handles the local variables inside of the functions (it permits context sensitive lookup). If you are not having any problems with compiling then don't worry about it. I had a problem with the return type name.
-bcl
0
Author Comment
ID: 9938281
Ok when I make that change then it tells me that allVerts, Vertex* are all undeclared. I am guessing I dont have the header file setup correctly but don't see what i need to change.
0
LVL 11
Expert Comment
ID: 9938336
I have the code you posted compiling. Only changed the header to have findVertex return Vertex * and the return type in graph.cpp is changed to Graph::Vertex *. Other than that I have made no changes (okay, I changed the file names but that is immaterial). As poste, graph.h:32 and graph.cpp:104, respectively.
-bcl
0
Author Comment
ID: 9938341
Im getting errors in findVertex with everything not being declared.... i posted my new code.
bash-2.05\$ g++ -o test graph.cpp main.cpp
graph.cpp: In function `Graph::Vertex *findVertex (int)':
graph.cpp:106: `Vertex' undeclared (first use this function)
graph.cpp:106: (Each undeclared identifier is reported only once for
each function it appears in.)
graph.cpp:106: parse error before `>'
graph.cpp:106: `it' undeclared (first use this function)
graph.cpp:106: `allVerts' undeclared (first use this function)
0
LVL 11
Expert Comment
ID: 9938491
It appears the g++ is not using context sensitive lookup OR that it doesn't apply inside of template parameters (I thought it did but my copy of the standard isn't on this computer so I can't look it up). The fix is to use the fully qualified name for Vertex inside of the name of the type for the loop control variable:
list<Graph::Vertex *>::iterator it /* and so on */
-bcl
0
Author Comment
ID: 9938551
Its still not recognizing allVerts?
0
LVL 11
Expert Comment
ID: 9939511
I am sorry but I don't have the error and don't recognize the error. Try this->allVerts (should not be necessary but ...).
Hope this helps, -bcl
0
LVL 11
Expert Comment
ID: 9942444
I just looked at your most recently posted code. The name of the function is Graph::findVertex. I don't know if I added that or if it was somehow removed during your modifications. That would explain why it could not recognize allVerts.
-bcl
0
LVL 11
Expert Comment
ID: 10013362
No big deal but you might want to close this question (I am assuming from the lack of chatter that you have finished your work).
Hope all went well for you.
-bcl
0
LVL 5
Expert Comment
ID: 10330205
No comment has been added lately, so it's time to clean up this TA.
I will leave a recommendation in the Cleanup topic area that this question is:
migoEX
EE Cleanup Volunteer
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Statistics 371-001, Sp 09 Lecture #182 31 Mar 09 Chapter 10. Analysis of Categorical Data 1. Chi-Square Goodness of Fit Test Hypothesis testing Categorical data Multinomial instead of binomial Parameters – mean & SD, most common ways to describe a population. Nonparametric test, Chi-square, χ 2 Other questions; estimate proportions or relative frequencies of subjects in response categories of a distribution vs probability of a success. Eg., Number of women dentist compared to number of men dentist. Which cola is most preferred by Americans, Coke or Pepsi? Proportion of college students majoring in business today vs 10 yrs ago. Use sample data to test hypotheses about the shape or proportions of a population distribution. Chi-square Goodness of Fit Test Observed frequencies – number of items from sample classified in a particular category. Each subject or observational unit is counted in one and only one category. Data; relatively simple. Expected frequencies - number of items in each category predicted from the null hypothesis and the sample size, n . How well do observed frequencies fit frequencies specified by H 0 ?? Expected frequencies can specify H 0 of No Preference. Eg., Among leading brands of soft drinks, are there preferences? Coke Pepsi Mt. Dew H 0 : Expected frequencies can specify H 0 of No Difference from a Known Population. Eg., Same distribution eye color exits in newborns in Wisconsin in 2009 as in 1959? Lecture 18 page 1
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QuickTime and a TIFF (Uncompressed) decompressor are needed to see this picture. Brown Blue Green Other H 0 : 20 10 4 4 Hypothetical Problem: Volunteers at a teen hotline are assigned based on the assumption that the primary issue raised by 40% of calls is drugs, for 25% of calls the primary issue is related to sex, for 25% it’s stress, and for 10% it’s educational issues. Is the assumption regarding the distribution of primary topic issue appropriate? 120 calls were randomly selected and categorized according to the primary issue addressed in the call.
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# An office supply room is used to store paper. What is the maximum numb
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An office supply room is used to store paper. What is the maximum numb [#permalink]
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Updated on: 15 Feb 2017, 06:37
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An office supply room is used to store paper. What is the maximum number of stacks of paper that the room can hold?
1. Currently there are 20 stacks of paper.
2. If 5 stacks of paper are removed when the room is half full, then the number of stacks of paper will decrease by 1/3.
Originally posted by sauravleo123 on 15 Feb 2017, 05:39.
Last edited by Bunuel on 15 Feb 2017, 06:37, edited 1 time in total.
Renamed the topic and edited the question.
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An office supply room is used to store paper. What is the maximum numb [#permalink]
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Updated on: 15 Feb 2017, 22:49
sauravleo123 wrote:
An office supply room is used to store paper. What is the maximum number of stacks of paper that the room can hold?
1. Currently there are 20 stacks of paper.
2. If 5 stacks of paper are removed when the room is half full, then the number of stacks of paper will decrease by 1/3.
(1) clearly insuff
(2) let R be max. number of stacks of paper that the room can hold
then R/2-5 = R/2(1-1/3)
R/2-5 =R/3
thus R=30
Ans B
Originally posted by rohit8865 on 15 Feb 2017, 06:59.
Last edited by rohit8865 on 15 Feb 2017, 22:49, edited 1 time in total.
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Re: An office supply room is used to store paper. What is the maximum numb [#permalink]
### Show Tags
15 Feb 2017, 21:40
sauravleo123 wrote:
An office supply room is used to store paper. What is the maximum number of stacks of paper that the room can hold?
1. Currently there are 20 stacks of paper.
2. If 5 stacks of paper are removed when the room is half full, then the number of stacks of paper will decrease by 1/3.
St I
Currently there are 20 stacks of paper. No other information is provided ----------Insufficient
St II
If 5 stacks of paper are removed when the room is half full, then the number of stacks of paper will decrease by 1/3
Let x be the max number of stacks of paper.
Therefore (x/2)-5 = x*(1/3)
(x/2)-(x/3)=5
x/6=5
x=30 ----------Sufficient
Hence Option B is correct.
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An office supply room is used to store paper. What is the maximum numb [#permalink]
### Show Tags
15 Feb 2017, 21:41
1
rohit8865 wrote:
sauravleo123 wrote:
An office supply room is used to store paper. What is the maximum number of stacks of paper that the room can hold?
1. Currently there are 20 stacks of paper.
2. If 5 stacks of paper are removed when the room is half full, then the number of stacks of paper will decrease by 1/3.
(1) clearly insuff
(2) let R be max. number of stacks of paper that the room can hold
then R/2-5 = R/2(1-1/3)
R/2-5 =R/3
thus R=15
Ans B
hi @rohit8864, just a minor correction R = 30 and not 15.
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Re: An office supply room is used to store paper. What is the maximum numb [#permalink]
### Show Tags
15 Feb 2017, 22:49
0akshay0 wrote:
rohit8865 wrote:
sauravleo123 wrote:
An office supply room is used to store paper. What is the maximum number of stacks of paper that the room can hold?
1. Currently there are 20 stacks of paper.
2. If 5 stacks of paper are removed when the room is half full, then the number of stacks of paper will decrease by 1/3.
(1) clearly insuff
(2) let R be max. number of stacks of paper that the room can hold
then R/2-5 = R/2(1-1/3)
R/2-5 =R/3
thus R=15
Ans B
hi @rohit8864, just a minor correction R = 30 and not 15.
Edited!!!!! +1 to u
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Re: An office supply room is used to store paper. What is the maximum numb [#permalink]
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16 Feb 2017, 11:15
sauravleo123 wrote:
An office supply room is used to store paper. What is the maximum number of stacks of paper that the room can hold?
1. Currently there are 20 stacks of paper.
2. If 5 stacks of paper are removed when the room is half full, then the number of stacks of paper will decrease by 1/3.
We need to determine the maximum number of stacks of paper that a room can hold.
Statement One Alone:
Currently there are 20 stacks of paper.
Statement one alone does not provide sufficient information to determine the maximum stacks of paper that the room can hold.
Statement Two Alone:
If 5 stacks of paper are removed when the room is half full, then the number of stacks of paper will decrease by 1/3.
We can let x = the maximum stacks of paper that the room can hold and create the following equation:
(1/2)x - 5 = (2/3)(1/2)x
(1/2)x - 5 = (1/3)x
Multiplying the entire equation by 6 gives us:
3x - 30 = 2x
x = 30
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Re: An office supply room is used to store paper. What is the maximum numb &nbs [#permalink] 16 Feb 2017, 11:15
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# Problem 232. Project Euler: Problem 2, Sum of even Fibonacci
Solution 2812249
Submitted on 9 Aug 2020 by Tillman Sherwood
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x =4000000 y_correct = 4613732; assert(isequal(euler002(x),y_correct))
x = 4000000
2 Pass
x =97455000 y_correct = 82790070; assert(isequal(euler002(x),y_correct))
x = 97455000
3 Pass
x =597455000 y_correct = 350704366; assert(isequal(euler002(x),y_correct))
x = 597455000
4 Pass
x =666576 y_correct = 257114; assert(isequal(euler002(x),y_correct))
x = 666576
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# Measurement Error And Standard Error
## Contents
National Center for Health Statistics (24). You want to be confident that your score is reliable,i.e. Of course, T / n {\displaystyle T/n} is the sample mean x ¯ {\displaystyle {\bar {x}}} . In other words, it is the standard deviation of the sampling distribution of the sample statistic. http://facetimeforandroidd.com/standard-error/measurement-error-standard-deviation.php
Viewed another way, the student can determine that if he took a differentedition of the exam in the future, assuming his knowledge remains constant, hecan be 95% (±2 SD) confident that Scenario 2. This formula may be derived from what we know about the variance of a sum of independent random variables.[5] If X 1 , X 2 , … , X n {\displaystyle In the diagram at the right the test would have a reliability of .88. https://en.wikipedia.org/wiki/Standard_error
## Standard Error Of Measurement Formula
Standard error of the mean Further information: Variance §Sum of uncorrelated variables (Bienaymé formula) The standard error of the mean (SEM) is the standard deviation of the sample-mean's estimate of a As a result, we need to use a distribution that takes into account that spread of possible σ's. American Statistician.
Consider a sample of n=16 runners selected at random from the 9,732. For any random sample from a population, the sample mean will usually be less than or greater than the population mean. With n = 2 the underestimate is about 25%, but for n = 6 the underestimate is only 5%. Standard Error Of Measurement Interpretation The sample standard deviation s = 10.23 is greater than the true population standard deviation σ = 9.27 years.
The mean age was 23.44 years. Standard Error Of Measurement Example However, if our hypothetical student had only scored a 199 (with a standard error of 3) on the second test administration, our conclusions would be much less certain. It is rare that the true population standard deviation is known. https://en.wikipedia.org/wiki/Standard_error In general, the precision of observed MAP scores can be boosted (i.e., SEMs decreased) in two ways: increasing the number of items within a test event, and by including only items
This estimate may be compared with the formula for the true standard deviation of the sample mean: SD x ¯ = σ n {\displaystyle {\text{SD}}_{\bar {x}}\ ={\frac {\sigma }{\sqrt {n}}}} Standard Error Of Measurement Spss In this example, the change from fall to spring (17 points) is relatively large compared to the standard error of the change score(4.24), so we can be very comfortable in concluding Hyattsville, MD: U.S. ISBN 0-8493-2479-3 p. 626 ^ a b Dietz, David; Barr, Christopher; Çetinkaya-Rundel, Mine (2012), OpenIntro Statistics (Second ed.), openintro.org ^ T.P.
## Standard Error Of Measurement Example
Bence (1995) Analysis of short time series: Correcting for autocorrelation. http://segmeasurement.com/content/what-standard-error-assessment Between +/- two SEM the true score would be found 96% of the time. Standard Error Of Measurement Formula The following expressions can be used to calculate the upper and lower 95% confidence limits, where x ¯ {\displaystyle {\bar {x}}} is equal to the sample mean, S E {\displaystyle SE} Standard Error Of Measurement And Confidence Interval A medical research team tests a new drug to lower cholesterol.
For example, the sample mean is the usual estimator of a population mean. check my blog Notice that the population standard deviation of 4.72 years for age at first marriage is about half the standard deviation of 9.27 years for the runners. As will be shown, the standard error is the standard deviation of the sampling distribution. National Center for Health Statistics typically does not report an estimated mean if its relative standard error exceeds 30%. (NCHS also typically requires at least 30 observations – if not more Standard Error Of Measurement Calculator
The researchers report that candidate A is expected to receive 52% of the final vote, with a margin of error of 2%. Please try the request again. Michael Dahlin 9Dr. this content Standard error of mean versus standard deviation In scientific and technical literature, experimental data are often summarized either using the mean and standard deviation or the mean with the standard error.
The graph below shows the distribution of the sample means for 20,000 samples, where each sample is of size n=16. Standard Error Of Measurement Excel Larger sample sizes give smaller standard errors As would be expected, larger sample sizes give smaller standard errors. As the sample size increases, the sampling distribution become more narrow, and the standard error decreases.
## The graph below shows the distribution of the sample means for 20,000 samples, where each sample is of size n=16.
If σ is not known, the standard error is estimated using the formula s x ¯ = s n {\displaystyle {\text{s}}_{\bar {x}}\ ={\frac {s}{\sqrt {n}}}} where s is the sample The standard deviation of all possible sample means of size 16 is the standard error. The next graph shows the sampling distribution of the mean (the distribution of the 20,000 sample means) superimposed on the distribution of ages for the 9,732 women. Standard Error Of Measurement Vs Standard Deviation And for the most part, when you look at the group, they tend to balance each other out.
A quantitative measure of uncertainty is reported: a margin of error of 2%, or a confidence interval of 18 to 22. In other words, it is the standard deviation of the sampling distribution of the sample statistic. Ecology 76(2): 628 – 639. ^ Klein, RJ. "Healthy People 2010 criteria for data suppression" (PDF). http://facetimeforandroidd.com/standard-error/mean-standard-deviation-and-standard-error-calculator.php Because these 16 runners are a sample from the population of 9,732 runners, 37.25 is the sample mean, and 10.23 is the sample standard deviation, s.
If σ is not known, the standard error is estimated using the formula s x ¯ = s n {\displaystyle {\text{s}}_{\bar {x}}\ ={\frac {s}{\sqrt {n}}}} where s is the sample Generated Tue, 18 Oct 2016 23:26:46 GMT by s_ac5 (squid/3.5.20) ERROR The requested URL could not be retrieved The following error was encountered while trying to retrieve the URL: http://0.0.0.9/ Connection If σ is known, the standard error is calculated using the formula σ x ¯ = σ n {\displaystyle \sigma _{\bar {x}}\ ={\frac {\sigma }{\sqrt {n}}}} where σ is the The standard deviation of all possible sample means of size 16 is the standard error.
About the Author Michael Dahlin is a Research Scientist at NWEA, where he specializes in research and reporting on college readiness, and school accountability policy. The standard error estimated using the sample standard deviation is 2.56. in Biology from Pomona College. Standard error of the mean Further information: Variance §Sum of uncorrelated variables (Bienaymé formula) The standard error of the mean (SEM) is the standard deviation of the sample-mean's estimate of a
Standard error From Wikipedia, the free encyclopedia Jump to: navigation, search For the computer programming concept, see standard error stream. Standard errors provide simple measures of uncertainty in a value and are often used because: If the standard error of several individual quantities is known then the standard error of some The standard deviation of all possible sample means is the standard error, and is represented by the symbol σ x ¯ {\displaystyle \sigma _{\bar {x}}} . In each of these scenarios, a sample of observations is drawn from a large population.
The SEM is an estimate of how much error there is in a test. The proportion or the mean is calculated using the sample. Mike holds a Ph.D. Because the 9,732 runners are the entire population, 33.88 years is the population mean, μ {\displaystyle \mu } , and 9.27 years is the population standard deviation, σ.
Or decreasing standard error by a factor of ten requires a hundred times as many observations. | 1,829 | 8,118 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2019-04 | latest | en | 0.887686 |
https://www.enotes.com/homework-help/f-x-4x-x-2-1-find-critical-numbers-function-521364 | 1,686,287,142,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224655247.75/warc/CC-MAIN-20230609032325-20230609062325-00710.warc.gz | 796,021,994 | 17,237 | # `f(x) = (4x)/(x^2 + 1)` Find the critical numbers of the function.
Given `f(x)=(4x)/(x^2+1)`
Find the derivative using the Quotient Rule. The set the derivative equal to zero and solve for the critical x value(s).
`f'(x)=[(x^2+1)(4)-(4x)(2x)]/(x^2+1)=0`
`f'(x)=4x^2+4-8x^2=0`
`f'(x)=-4x^2+4=0`
`f('x)=-4(x^2-1)=0`
`x=1,x=-1` | 155 | 333 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2023-23 | latest | en | 0.458171 |
https://mlpack.org/doc/mlpack-git/doxygen/eigenvalue__ratio__constraint_8hpp_source.html | 1,660,839,776,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573242.55/warc/CC-MAIN-20220818154820-20220818184820-00640.warc.gz | 365,594,401 | 5,151 | eigenvalue_ratio_constraint.hpp
Go to the documentation of this file.
1
12 #ifndef MLPACK_METHODS_GMM_EIGENVALUE_RATIO_CONSTRAINT_HPP
13 #define MLPACK_METHODS_GMM_EIGENVALUE_RATIO_CONSTRAINT_HPP
14
15 #include <mlpack/prereqs.hpp>
16
17 namespace mlpack {
18 namespace gmm {
19
28 {
29 public:
36 EigenvalueRatioConstraint(const arma::vec& ratios) :
37 // Make an alias of the ratios vector. It will never be modified here.
38 ratios(const_cast<double*>(ratios.memptr()), ratios.n_elem, false)
39 {
40 // Check validity of ratios.
41 if (std::abs(ratios[0] - 1.0) > 1e-20)
42 Log::Fatal << "EigenvalueRatioConstraint::EigenvalueRatioConstraint(): "
43 << "first element of ratio vector is not 1.0!" << std::endl;
44
45 for (size_t i = 1; i < ratios.n_elem; ++i)
46 {
47 if (ratios[i] > 1.0)
48 Log::Fatal << "EigenvalueRatioConstraint::EigenvalueRatioConstraint(): "
49 << "element " << i << " of ratio vector is greater than 1.0!"
50 << std::endl;
51 if (ratios[i] < 0.0)
52 Log::Warn << "EigenvalueRatioConstraint::EigenvalueRatioConstraint(): "
53 << "element " << i << " of ratio vectors is negative and will "
54 << "probably cause the covariance to be non-invertible..."
55 << std::endl;
56 }
57 }
58
62 void ApplyConstraint(arma::mat& covariance) const
63 {
64 // Eigendecompose the matrix.
65 arma::vec eigenvalues;
66 arma::mat eigenvectors;
67 covariance = arma::symmatu(covariance);
68 if (!arma::eig_sym(eigenvalues, eigenvectors, covariance))
69 {
70 Log::Fatal << "applying to constraint could not be accomplished."
71 << std::endl;
72 }
73
74 // Change the eigenvalues to what we are forcing them to be. There
75 // shouldn't be any negative eigenvalues anyway, so it doesn't matter if we
76 // are suddenly forcing them to be positive. If the first eigenvalue is
77 // negative, well, there are going to be some problems later...
78 eigenvalues = (eigenvalues[0] * ratios);
79
80 // Reassemble the matrix.
81 covariance = eigenvectors * arma::diagmat(eigenvalues) * eigenvectors.t();
82 }
83
88 void ApplyConstraint(arma::vec& diagCovariance) const
89 {
90 // The matrix is already eigendecomposed but we need to sort the elements.
91 arma::uvec eigvalOrder = arma::sort_index(diagCovariance);
92 arma::vec eigvals = diagCovariance(eigvalOrder);
93
94 // Change the eigenvalues to what we are forcing them to be. There
95 // shouldn't be any negative eigenvalues anyway, so it doesn't matter if we
96 // are suddenly forcing them to be positive. If the first eigenvalue is
97 // negative, well, there are going to be some problems later...
98 eigvals = eigvals[0] * ratios;
99
100 // Reassemble the matrix.
101 for (size_t i = 0; i < eigvalOrder.n_elem; ++i)
102 diagCovariance[eigvalOrder[i]] = eigvals[i];
103 }
104
106 template<typename Archive>
107 void serialize(Archive& ar, const uint32_t /* version */)
108 {
109 // Strip the const for the sake of loading/saving. This is the only time it
110 // is modified (other than the constructor).
111 ar(CEREAL_NVP(const_cast<arma::vec&>(ratios)));
112 }
113
114 private:
116 const arma::vec ratios;
117 };
118
119 } // namespace gmm
120 } // namespace mlpack
121
122 #endif
EigenvalueRatioConstraint(const arma::vec &ratios)
Create the EigenvalueRatioConstraint object with the given vector of eigenvalue ratios.
static util::PrefixedOutStream Warn
Definition: log.hpp:99
Linear algebra utility functions, generally performed on matrices or vectors.
The core includes that mlpack expects; standard C++ includes, Armadillo, cereal, and a few basic mlpa...
void ApplyConstraint(arma::vec &diagCovariance) const
Apply the eigenvalue ratio constraint to the given diagonal covariance matrix (represented as a vecto...
static util::PrefixedOutStream Fatal
Definition: log.hpp:105
Given a vector of eigenvalue ratios, ensure that the covariance matrix always has those eigenvalue ra...
void serialize(Archive &ar, const uint32_t)
Serialize the constraint.
void ApplyConstraint(arma::mat &covariance) const
Apply the eigenvalue ratio constraint to the given covariance matrix. | 1,144 | 4,100 | {"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} | 2.640625 | 3 | CC-MAIN-2022-33 | latest | en | 0.322101 |
https://essaywriter.nyc/education-questions/studying-abroad/homework-help-please-algebra/ | 1,709,172,593,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474775.80/warc/CC-MAIN-20240229003536-20240229033536-00149.warc.gz | 240,890,748 | 23,603 | A few days ago
frnchfries2000
# Homework help please: Algebra?
(4-m) /m >=5
When I solve I get:
4-m >=5m
4 >=6m
2/3 >= m
If m = 1/2, it works, but if m= -1 it doesn’t.
Can someone explain why?
A few days ago
Aquaboy
Your answer is somewhat correct – it’s actually incomplete…
Let’s consider scenarios where m>0 and m<0. Note that m cannot equal zero because then the denominator of the left side of the inequality would be zero: If m is positive, (m > 0), then:
(4 – m) / m >= 5
4 – m >= 5m
4 >= 6m
4 / 6 >= m
m <= 2/3 If m is negative, (m < 0), then you'll want to know, "for which values of m is the following statement true?" (4 - (-m)) / (-m) >= 5
(4 + m) / (-m) >= 5
But notice how, (when m is negative), the left hand side of the inequality will always have a positive numerator and a negative denominator, which means that the left hand side of the inequality is always negative when m is negative. In other words, the left hand side of the inequality cannot be greater than or equal to 5 if m is negative; therefore, m cannot be negative.
In other words, for this inequality: (4 – m) / m >= 5,
0 < m <= 2/3.
0
A few days ago
I LIKE CHOCOLATE MILK!!!
your answer is correct! I’m gonna do this in decimal form, if that’s OK w/ u??
4-.66 /.66>=5
3.34/.66>=5
5.06>=5
CORRECT!!!
0 | 411 | 1,310 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2024-10 | latest | en | 0.871857 |
https://codegolf.stackexchange.com/questions/37167/extraction-of-bits/37170 | 1,571,032,004,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986649232.14/warc/CC-MAIN-20191014052140-20191014075140-00352.warc.gz | 519,586,528 | 39,436 | # Extraction of Bits
Your task is to devise a programme that extracts the bits, in order, represented by the first argument based on the set bits represented by the second argument (the mask).
Formal Definition:
• Let the first argument x represent the binary string S=(xn)(xn-1)...(x1) with n>=0 and x1 being the least significant digit and xn = 1.
• Let the second argument y represent the binary string T=(ym)(ym-1)...(y1) with m>=0 and y1 being the least significant digit and ym = 1.
• Let the result represent 0 if x represents 0 or y represents 0.
• Let the number of 1 digits in T be p. Let {di}, i in [1, p] be the set of binary strings of the form (dim)(di(m-1))...(d1) defined by (for all i, (for all k, |{dik such that dik = 1}| = 1)) and (if 1 <= i < j <= p then 0 < di < dj) and (dp | dp-1 | ... | d1 = T). In other words, {di} is the ordered set of strings with a single 1 digit that makes up the string T when | is applied.
• Let the result of the programme represent the following binary string:
((dp & S) << (p - 1))| ... | ((d2 & S) << 1) | (d1 & S)
Examples:
• 1111, 110 -> 11
• 1010, 110 -> 01
• 1010, 10110 -> 01
• 1101010111, 11111000 -> 01010
Input and Output:
• The two arguments may be taken in any way
• The two arguments need not be binary strings. They can also be integers which represent binary strings.
• Answers which do not take input representing binary strings of arbitrary length are permitted.
Winning: Shortest code wins.
Clarifications:
• If your output gives a bit string, order of bits is not important. Simply state the order in your answer. As stated before, the output of your programme need only represent the bits required.
• What's your objective winning criterion, i.e. how do you determine who wins? Shortest code, or something else? – Doorknob Sep 1 '14 at 12:19
• I assumed that shortest code was the default criterion. I will add this to the post. – justinpc Sep 1 '14 at 12:50
• Is there a maximum number of bits in each binary string? For example, can we assume that both inputs will be less than 100 characters long? – Doorknob Sep 1 '14 at 12:52
• Your clarification allows reversed output. I assume it also allows reversed input? – John Dvorak Sep 1 '14 at 13:51
• Yes, if the input involves bit strings. – justinpc Sep 1 '14 at 14:05
# CJam, 9 bytes
eaz{)~*}%
This reads the bit strings (LSB first) as command line arguments and prints the result (also LSB first).
To try this code in the CJam interpreter, change ea to qS% and separate the strings with a space.
$cjam <(echo 'eaz{)~*}%') 1110101111 00011111; echo 01011 ### How it works ea " Read both command line arguments and collects them in an array. "; z " Interleave the characters of both strings. "; { }% " For each pair of characters: "; )~ " Pop the last character and interpret it as an integer. "; * " Repeat the remaining strings that many times. "; • If the mask character is 0, the string consisting of the corresponding character is repeated zero times. • If the mask character or the corresponding character do not exist, popping a character will result in an empty string to be repeated. In either case, the result will be an empty string. • I'm accepting this as it's the shortest, and doesn't use any loopholes. Sorry it took half a decade. – justinpc Feb 8 at 17:29 # Edit: APL, 1 character /, the compression operator (Documentation). Example of use: 0 0 1 1 1 1 1 0 0 0/1 1 0 1 0 1 0 1 1 1 0 1 0 1 0 Try it yourself at http://ngn.github.io/apl/web/index.html The mask is left, and the input bitstring is right. Disclaimer: I'm not an expert in APL and have never used it; I only submitted this to substitute the disqualification of my machine-code solution below. I admit that a limitation of the code above is that the left and right bitstrings must be of equal length in APL, and lack the APL skills to fix this. # Rejected as loophole: # x86-64 Machine Code (BMI2, Haswell+), 5 bytes c4 e2 f2 f5 d0 is the hex representation of the raw bytes of the program as they would appear in an executable x86-64 binary program. This represents the single instruction pext %rax, %rcx, %rdx, which accepts S in rax, T in rcx and stores the result in rdx. Max 64 bits. Runs in 3 clock cycles on Haswell, has throughput of 1 per clock cycle. xxx@xxx:~/Documents/SO> cat > bmi2.s .text pext %rax, %rcx, %rdx xxx@xxx:~/Documents/SO> gcc bmi2.s -c -o bmi2.o xxx@xxx:~/Documents/SO> objdump -d bmi2.o bmi2.o: file format elf64-x86-64 Disassembly of section .text: 0000000000000000 <.text>: 0: c4 e2 f2 f5 d0 pext %rax,%rcx,%rdx Of course, requires a pretty modern assembler. The documentation for PEXT is much clearer than OP's description, so here it is: • @immibis The CPU has to execute the raw bytes in order for them to do what they must do. When I want to say "take these values and consider them as raw bytes" I generally use "binary", in the sense of "raw binary data". Is there a better way to convey that meaning? I can't paste the actual bytes c4 e2 f2 f5 d0, so I've given their hex expression, but they must be actually "binary" when executed. – Iwillnotexist Idonotexist Sep 1 '14 at 23:11 • This is a [standard loophole](). I've updated it to not be as Mathematica-specific, but it equally applies here. – Isiah Meadows Sep 2 '14 at 0:11 • @impinball How can I invite you to chat? I'd like to discuss your characterization of my solution as a mere loophole. – Iwillnotexist Idonotexist Sep 2 '14 at 0:51 • @impinball downvoted the loophole and it's now just below the threshold. If a one-op solution exists, it's now officially a fault of the question, not of the answer. – John Dvorak Sep 2 '14 at 3:44 • If we're allowed APL's / on its own, then surely we're allowed J's # on its own? :) – justinpc Sep 2 '14 at 5:58 # Haskell, 68 characters f[c:d,'0':b]=f[d,b];f[c:d,_:b]=c:f[d,b];f _="";main=interact$f.words
Prints nothing if the input isn't exactly two words.
The first argument can contain any characters. So can the filter; only 0 is treated as "not select".
If arguments are of different length, they are taken left-aligned (LSB first; explicitly allowed since revision #4).
f(c:d)('0':b)=f d b;f(c:d)(_:b)=c:f d b;f _ _=""
Subtract two more characters if "function of two integer arrays" is an acceptable signature
# Golfscript, 54 51 characters
'0':x;{.,64\-x*\+}:b~:m;'X':x;b{m(\:m;48={;}*}%'X'-
Only works for binary strings of length 64 or less.
Explanation:
'0':x; store '0' in the x variable (which will be used in the following)
{.,64\-x*\+} block that pads strings to a length of 64 with the character stored in x
:b~ store the block in b and execute, second argument (y) is now padded with '0's
:m; store second argument as mask (m)
'X':x;b pad the first argument with 'X's (will be removed later)
{...}% for each character in the first argument (x), ...
m(\:m; grab and delete the first character of m (mask)
48={...}* if this character is equal to 48 (0 in ASCII)...
; remove the digit. Hence, this if statement keeps a digit only if the mask is 1.
'X'- finally, remove the 'X's added in step 5
• 49={}{;}if -> 48={;}* – John Dvorak Sep 1 '14 at 13:31
• @JanDvorak Thanks; that's pretty interesting. Execute the block n times, where n is a boolean (0 or 1), meaning that it will only execute the block if the boolean is true. Great trick! – Doorknob Sep 1 '14 at 13:33
• Taking the input LSB-first is now allowed. Can it shorten your code? – John Dvorak Sep 1 '14 at 13:50
# J, 16 Characters:
(#=@i.@#)@],@:#[
Usage:
1 0 1 0 1 1 ((#=@i.@#)@],@:#[) 0 1 1 1 0 0 -> 0 1 0
Explanation:
First argument is left, second argument is right (mask).
(#=@i.@#)@],@:#[ -> (# =@i.@#)@] ,@:# [
• Take right argument, produce identity matrix of size length of right argument.
• Use set bits of right argument to choose rows of produced identity matrix.
• Use chosen rows of identity matrix to choose digits of left argument.
• Concatenate the digits chosen from left argument.
# J, 13 Characters
According to the rules, the form of output is not important. Therefore, if we omit the concatenation of the output digits, we can have the following shorter answer.
(#=@i.@#)@]#[
# J, 1 character
#, the tally operator (Documentation).
Example of use:
0 0 1 1 1 1 1 0 0 0#1 1 0 1 0 1 0 1 1 1
0 1 0 1 0
This solution does not cater for the case when the length of the mask is less than that of the string. Given that the above might be seen as a loophole, we have the following solution.
# J, 11 characters
[#~],0$~-&# This solution is like the original #, but also caters for the case when the length of the mask is less than that of the string. Note here that the mask is the right argument, and not the left argument as in the first solution. x ([ #~ ] , 0$~ -&#) y
x #~ (y , (0 $~ (x -&# y))) x #~ (y , (0$~ ((# x) - (# y))))
x #~ (y , (((# x) - (# y)) $0)) x #~ (y , 0 0 ... 0 0) x #~ y 0 0 ... 0 0 y 0 0 ... 0 # x • [ and ] choose x and y respectively. • x (u&v) y -> (v x) u (v y). • (x -&# y) -> (# x) - (# y). • # x -> length of x. • x v~ y -> y v x. • n$ 0 -> list of n 0s.
• (0 $~ (x -&# y)) -> m 0s, where m is the difference between the length of the string and the length of the mask. • (y , 0 0 ... 0 0) -> the list y 0 0 ... 0 0 such that its length is equal to that of the string. • a # b -> the elements of b chosen by the binary mask a. Even though this solution is shorter than my other solution of 16 characters, I think it still makes unfair use of #. In any case, I think it's still interesting. • +1. The question somewhat begs for the use of compression/reduction/fold/inject operators, but I find it interesting how you pad the mask with 0s in the APL-derived languages. – Iwillnotexist Idonotexist Sep 2 '14 at 12:00 • @IwillnotexistIdonotexist The fun thing is that the # operator can actually do many more things, like representing a number in an arbitrary base. – FUZxxl Sep 2 '14 at 13:16 # Haskell, 33 characters ((map snd.filter((>0).fst)).).zip # Usage (((map snd.filter((>0).fst)).).zip) [1, 0, 1, 0] [1, 0, 0, 1] [1, 0] (((map snd.filter((>0).fst)).).zip) [1, 1] [1, 1, 0, 1] [1, 1] # J, 10 characters, no use of tally # This solution does not cater for input where mask is shorter than string. ;@:(<@$"0)
Usage
1 0 0 1 (;@:(<@$"0)) 1 0 1 0 1 0 Explanation • This makes no use of #, which on its own can be used to solve this golfing problem. • This solution works by interleaving both sides, one element by one. This is expressed by "0, which makes <@$ applied to each pair of elements.
• x $y means x ys as a list. 3$ 4 -> 4 4 4.
• < x puts x in a "box". A box can contain an object of any dimension (list, matrix, nothing). Boxes can be concatenated into lists.
• 1 $x -> one-dimensional list consisting of x. • 0$ x -> nothing
• 1 (<@$"0) x -> one-dimensional list consisting of x inside a box. • 0 (<@$"0) x -> nothing in a box.
• ; takes a list of boxes and concatenates its contents, throwing away empty boxes.
• u@:v composes u and v in such a way that u is applied only after v has been applied to the whole of the arguments. In this case ; is applied only after the right side of the @: has been applied to each interleaved pair x and y made from the arguments. | 3,394 | 11,262 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2019-43 | latest | en | 0.847721 |
https://www.gradesaver.com/textbooks/math/algebra/algebra-1/chapter-10-radical-expressions-and-equations-10-3-operations-with-radial-expressions-practice-and-problem-solving-exercises-page-616/12 | 1,534,779,679,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221216475.75/warc/CC-MAIN-20180820140847-20180820160847-00563.warc.gz | 916,856,359 | 14,001 | ## Algebra 1
$-3\sqrt2$
$4\sqrt2-7\sqrt2\longrightarrow$ use distributive property =$\sqrt2(4-7)\longrightarrow$ simplify =$-3\sqrt2$ | 49 | 134 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2018-34 | longest | en | 0.314016 |
https://www.lnunesart.com/paintings/question-how-many-paintings.html | 1,653,416,533,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662573189.78/warc/CC-MAIN-20220524173011-20220524203011-00413.warc.gz | 1,003,215,314 | 11,043 | # Question: How Many Paintings?
## How many paintings exist in the world?
All it means is that painting needs a marketing push since if there are 15 billion paintings in the world, that is really only 2-3 for every person.
## How many paintings should be in a series?
The goal in this step is to get enough photos, sketches, and compositional ideas to create at least ten complete paintings. To do that, you might have to do some things which will put you outside of your comfort zone.
## How many paintings did Picasso?
However, during Picasso’s long life — he died in 1973 at age 91 — he is estimated to have completed 13,500 paintings and around 100,000 prints and engravings. A comprehensive retrospective of his work and the numerous artistic traditions it spanned, is a massive undertaking.
## How many paintings does Van Gogh have?
Despite only working for 10 years – from the age of 27 up until his early demise at 37 – van Gogh was incredibly prolific. He produced more than 900 paintings and many more drawings and sketches, which works out at nearly a new artworks every 36 hours. Phew!
You might be interested: Who Started Word Paintings Or Text Portraits?
## How many paintings in museums are fake?
Some statistics have said that up to 20 percent of the paintings in major museums are fake, but Charney says this number is false.
## What is a series of paintings called?
A collection of artworks and art pieces make up an art series. A set of drawings, paintings, sculptures, and even photographs create what can be called an art series.
## How do you label a series of artwork?
1. Artist’s name.
2. Nationality, birth year (Optional.
3. Title of the artwork (in bold or italic), year created.
4. Medium used to create (ex: crayon on paper)
5. Brief description (This is where you can include any information about the artist, why they created the piece, how they created the piece, etc.)
## What makes a series of paintings?
A series is essentially a collection of paintings that when viewed leaves no doubt the same artist created them all. The theme running through the work is stated and restated in different yet interconnected ways, and the viewer can look at the collection and understand more easily what the artist is trying to convey.
## Why are Picasso paintings so expensive?
Picasso’s masterpieces are now in short supply and therefore getting increasingly expensive. This is especially true for paintings from his “Blue” and “Rose” periods, early Cubist works, and pieces that are intimately linked to the artist’s private life.
## What is the most expensive Picasso painting?
Most Expensive Painting All it took was a little over 8 minutes to set the record price for a piece of art sold at auction. On May 4, Christie’s sold Pablo Picasso’s Nude, Green Leaves and Bust, a painting created in the span of a single day in 1932, for \$106.5 million dollars.
You might be interested: Question: What Was The Original Purpose Of Paintings?
## What killed Picasso?
It is impossible to place a value on such a famous and treasured work of art, though other works by Van Gogh have sold for more than 80 million dollars at auction. As arguably Van Gogh’s most famous work of art, it is safe to estimate the value of Starry Night at well over 100 million dollars.
## Where are the real Van Gogh paintings?
In addition to the Van Gogh Museum in Amsterdam, and his works can be found in the Yale University Art Gallery, New Haven, Connecticut; Albright-Knox Art Gallery, in Buffalo, New York; the Pushkin State Museum of Fine Arts, Moscow; and the Museum of Modern Art, New York, which is home to his resplendent Starry Night,
## How old is the earliest known human artwork?
In September 2018, scientists reported the discovery of the earliest known drawing by Homo sapiens, which is estimated to be 73,000 years old, much earlier than the 43,000 years old artifacts understood to be the earliest known modern human drawings found previously. | 872 | 3,988 | {"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} | 2.640625 | 3 | CC-MAIN-2022-21 | latest | en | 0.974211 |
https://myelectrical.com/notes/tag/mining | 1,719,026,907,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862249.29/warc/CC-MAIN-20240622014659-20240622044659-00483.warc.gz | 358,882,771 | 11,567 | ## Posts Tagged 'mining'
We are pleased to present below all posts tagged with 'mining'. If you still can't find what you are looking for, try using the search box.
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## Have some knowledge to share
If you have some expert knowledge or experience, why not consider sharing this with our community.
By writing an electrical note, you will be educating our users and at the same time promoting your expertise within the engineering community.
To get started and understand our policy, you can read our How to Write an Electrical Note | 497 | 2,452 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2024-26 | latest | en | 0.894397 |
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13 years ago | 1 download | | 807 | 3,172 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2022-05 | latest | en | 0.765083 |
https://discuss.interviewbit.com/t/why-is-this-a-greedy-algortihm-problem/45169 | 1,610,784,567,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703505861.1/warc/CC-MAIN-20210116074510-20210116104510-00652.warc.gz | 317,475,399 | 3,532 | # Why is this a Greedy algortihm problem?
#1
why is this a Greedy algorithm problem? it’s more of a brain teaser.
the fact that i am switching on the first off bulb is greedy nature? Is there a non-greedy way of solving it?
My sol:
class Solution:
# @param A : list of integers
# @return an integer
def bulbs(self, A):
count = 0
flag = True
# if any bulb switched twice, it will come back to original state
for bulb in A:
if bulb == 0 and flag: #this bulb in original state
count += 1
flag = False
elif bulb ==1 and not flag: #this bulb not in orig state, i.e now in off mode
count += 1
flag = True
`` return count``
#2
This is a greedy problem because you cannot make a decision later about an OFF bulb if you move ahead without turning it ON.
The problem demands every bulb to be ON, so when you encounter an OFF bulb, you MUST turn it ON, simple.
Hope that clarifies. | 237 | 879 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2021-04 | latest | en | 0.844492 |
https://m.hanspub.org/journal/paper/41571 | 1,624,258,729,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488268274.66/warc/CC-MAIN-20210621055537-20210621085537-00108.warc.gz | 347,291,721 | 17,231 | 认知无线电协同传输在车联网中的应用及性能分析
# 认知无线电协同传输在车联网中的应用及性能分析Application and Performance analysis of Cognitive Radio Cooperative Transmission in Internet of Vehicles
Abstract: With the rapid development of communication and smart antenna technology, the Internet of Vehicles (IoV) has received extensive attention and research. The application of IoV is hampered by lack of spectrum and limited communication coverage. In this paper, cognitive radio cooperative transmission is introduced into the IoV system to expand the transmission range without increasing the transmission power. The outage probability of end-to-end transmission is calculated by analyzing and deducing the signal to interference plus noise ratio. Numerical simulation results show that compared with direct communication, the use of cognitive radio cooperative transmission can effectively reduce the outage probability and improve the performance of the IoV.
1. 引言
[8] 将最佳中继协同传输引入到认知无线电中，推导了对主网络的干扰小于或等于阈值的情况下，精准的中断概率表达式。但 [8] 中的协同中继转发策略依赖大量信道状态信息，实现起来并没有那么容易。在 [9] 中提出来一种新的中继协同传输方法，通过选择部分具有固定增益的中继，减少对于信道信息的需求。 [8] [9] 都为单中继协同传输策略， [10] 提出了多中继的协同传输策略，它们以牺牲一定的带宽来获得更高的信干噪比。
2. 系统模型及信干噪比分析
2.1. 系统模型
Figure 1. System model
2.2. 信干噪比分析
${y}_{1}=\sqrt{{P}_{1}}{h}_{1}{x}_{1}+{n}_{0}+\sqrt{{P}_{3}}{h}_{3}{x}_{2}$ (1)
${y}_{2}=\sqrt{{P}_{2}}{h}_{2}{x}_{1}+{n}_{0}$ (2)
${\zeta }_{1}=\frac{{P}_{1}{|{h}_{1}|}^{2}}{{N}_{0}+{P}_{3}{|{h}_{3}|}^{2}}=\frac{\frac{{P}_{1}{|{h}_{1}|}^{2}}{{N}_{0}}}{1+\frac{{P}_{3}{|{h}_{3}|}^{2}}{{N}_{0}}}=\frac{{\gamma }_{1}}{1+{\varphi }_{1}}$ (3)
${\zeta }_{2}=\frac{{P}_{2}{|{h}_{2}|}^{2}}{{N}_{0}}={\gamma }_{2}$ (4)
$x\ge 1$ 时， ${F}_{{\varphi }_{1}+1}\left(x\right)=\mathrm{Pr}\left\{{\varphi }_{1}+1\le x\right\}=\mathrm{Pr}\left\{{\varphi }_{1}\le x-1\right\}={F}_{{\varphi }_{1}}\left(x-1\right)$${\varphi }_{1}+1$ 分布函数为：
${F}_{{\varphi }_{1}+1}\left(x\right)=1-{\text{e}}^{-{\mu }_{3}\left(x-1\right)}$ (5)
${f}_{{\varphi }_{1}+1}\left(x\right)={{F}^{\prime }}_{{\varphi }_{1}+1}\left(x\right)={\mu }_{3}{\text{e}}^{-{\mu }_{3}\left(x-1\right)}$ (6)
$x<1$ 时， ${F}_{{\varphi }_{1}+1}\left(x\right)=0$${f}_{{\varphi }_{1}+1}\left(x\right)=0$
$\begin{array}{l}\because {F}_{{\zeta }_{1}}\left(x\right)=\mathrm{Pr}\left\{\frac{{\gamma }_{1}}{1+{\varphi }_{1}}\le x\right\}={\int }_{0}^{\infty }\mathrm{Pr}\left\{{\gamma }_{1}\le xt\right\}\cdot {f}_{{\varphi }_{1}+1}\left(t\right)\text{d}t\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\int }_{0}^{\infty }\left(1-{\text{e}}^{-{\mu }_{1}\left(xt\right)}\right)\cdot \left({\mu }_{3}{\text{e}}^{-{\mu }_{3}\left(t-1\right)}\right)\text{d}t={\text{e}}^{{\mu }_{3}}-\frac{{\mu }_{3}{\text{e}}^{{\mu }_{3}}}{{\mu }_{1}x+{\mu }_{3}}\end{array}$ (7)
(7)式为车辆S到车辆R链路的信干噪比分布函数，又因为 ${\zeta }_{2}=\frac{{P}_{2}{|{h}_{1}|}^{2}}{{N}_{0}}={\gamma }_{2}~\mathrm{exp}\left({\mu }_{2}\right)$
${\zeta }_{2}$ 的分布函数为：
${F}_{{\zeta }_{2}}\left(x\right)=1-{\text{e}}^{-{\mu }_{2}x}$ (8)
${\zeta }_{e2e}=\mathrm{min}\left\{{\zeta }_{1},{\zeta }_{2}\right\}$
$\begin{array}{c}{F}_{{\zeta }_{e2e}}\left(x\right)=\mathrm{Pr}\left\{\mathrm{min}\left\{{\zeta }_{1},{\zeta }_{2}\right\}\le x\right\}=1-\mathrm{Pr}\left\{\mathrm{min}\left\{{\zeta }_{1},{\zeta }_{2}\right\}>x\right\}\\ =1-\mathrm{Pr}\left\{{\zeta }_{1}>x\right\}\cdot \mathrm{Pr}\left\{{\zeta }_{2}>x\right\}=1-\left(1-{F}_{{\zeta }_{1}}\left(x\right)\right)\cdot \left(1-{F}_{{\zeta }_{2}}\left(x\right)\right)\end{array}$ (9)
${F}_{{\zeta }_{e2e}}\left(x\right)=1-\left(1-{\text{e}}^{{\mu }_{3}}+\frac{{\mu }_{3}{\text{e}}^{{\mu }_{3}}}{{\mu }_{1}x+{\mu }_{3}}\right)\cdot \left({\text{e}}^{-{\mu }_{2}x}\right)$ (10)
3. 性能分析及数值模拟
3.1. 中断概率
${P}_{out}=\mathrm{Pr}\left(\frac{1}{2}{\mathrm{log}}_{2}\left(1+{\zeta }_{e2e}\right)\le r\right)={F}_{{\zeta }_{e2e}}\left({2}^{2r}-1\right)$
${P}_{out}=1-\left(1-{\text{e}}^{{\mu }_{3}}+\frac{{\mu }_{3}{\text{e}}^{{\mu }_{3}}}{{\mu }_{1}\left({2}^{2r-1}\right)+{\mu }_{3}}\right)\cdot \left({\text{e}}^{-{\mu }_{2}\left({2}^{2r-1}\right)}\right)$ (11)
3.2. 数值模拟
Figure 2. The influence curve of ${\gamma }_{1}$ on ${P}_{out}$
Figure 3. The influence curve of ${\varphi }_{1}$ on ${P}_{out}$
4. 结论
[1] Haykin, S. (2005) Cognitive Radio: Brain-Empowered Wireless Communications. IEEE Journal on Selected Areas in Communications, 23, 201-220.
https://doi.org/10.1109/JSAC.2004.839380
[2] 曹华孝, 鲜永菊, 徐昌彪. 认知无线电技术的国内外发展与研究现状[J]. 数字通信, 2009, 36(2): 62-68.
[3] 刘元, 彭端, 陈楚. 认知无线电的关键技术和应用研究[J]. 通信技术, 2007, 40(7): 87-90.
[4] Kusaladharma, S. and Tellambura, C. (2012) Aggregate Interference Analysis for Underlay Cognitive Radio Networks. IEEE Wireless Communication Letters, 1, 641-644.
https://doi.org/10.1109/WCL.2012.091312.120600
[5] Suoping, L., Wei, L., Jaafar, G., Kejun, J. and Fan, W. (2020) Design and Analysis of Adaptive Full-Duplex Cognitive Relay Cooperative Strategy Based on Primary System Behavior. Wireless Networks, 26, 6237-6252.
https://doi.org/10.1007/s11276-020-02433-w
[6] Zou, Y., Zhu, J., Zheng, B. and Yao, Y. (2010) An Adaptive Cooperation Diversity Scheme with Best-Relay Selection in Cognitive Radio Networks. IEEE Transactions on Signal Processing, 58, 5438-5445.
https://doi.org/10.1109/TSP.2010.2053708
[7] Simeone, O., Bar-Ness, Y. and Spagnolini, U. (2007) Stable Throughput of Cognitive Radios with and without Relaying Capability. IEEE Transactions on Communications, 55, 2351-2360.
https://doi.org/10.1109/TCOMM.2007.910699
[8] Ashrafinia, S., Pareek, U. and Naeem, M. (2011) Biogeography-Based Optimization for Joint Relay Assignment and Power Allocation in Cognitive Radio System. 2011 IEEE Symposium on Swarm Intelligence, Paris, 11-15 April 2011, 1-8.
https://doi.org/10.1109/SIS.2011.5952589
[9] Elalem, M., Zhao, L. and Liao, Z. (2010) Interference Mitigation Using Power Control in Cognitive Radio Networks. Vehicular Technology Conference, Taipei, 16-19 May 2010, 1-5.
https://doi.org/10.1109/VETECS.2010.5494050
[10] Anghel, P.A. and Kaveh, M. (2004) Exact Symbol Error Probability of a Cooperative Network in a Rayleigh-Fading Environment. IEEE Transactions on Wireless Communications, 3, 1416-1421.
https://doi.org/10.1109/TWC.2004.833431
[11] Yu, S., Lee, J., Park, K., Das, A.K. and Park, Y. (2020) IoV-SMAP: Secure and Efficient Message Authentication Protocol for IoV in Smart City Environment. IEEE Access, 8, 167875-167886.
https://doi.org/10.1109/ACCESS.2020.3022778
[12] 程刚, 郭达. 车联网现状与发展研究[J]. 移动通信, 2011, 35(17): 23-26.
[13] Li, S.P., Wang, F., Gaber, J. and Chang, X.K. (2020) Throughput and Energy Efficiency of Cooperative ARQ Strategies for VANETs Based on Hybrid Vehicle Communication Mode. IEEE Access, 8, 114287-114304.
https://doi.org/10.1109/ACCESS.2020.3003813
[14] Islam, S.N., Sadeghi, P. and Durrani, S. (2013) Error Performance Analysis of Decode-and-Forward and Amplify-and-Forward Multi-Way Relay Networks with Binary Phase Shift Keying Modulation. IET Communications, 7, 1605-1616.
https://doi.org/10.1049/iet-com.2013.0284
[15] Ikki, S. and Ahmed, M.H. (2007) Performance Analysis of Cooperative Diversity Wireless Networks over Nakagami-m Fading Channel. IEEE Communications Letters, 11, 334-336.
https://doi.org/10.1109/LCOM.2007.348292
Top | 3,045 | 7,404 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 55, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2021-25 | latest | en | 0.538635 |
https://www.teacherspayteachers.com/Product/Multiplying-Decimals-Printable-Packet-Assessment-and-Center-Activity-3204273 | 1,534,456,890,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221211185.57/warc/CC-MAIN-20180816211126-20180816231126-00594.warc.gz | 1,010,789,164 | 19,722 | # Multiplying Decimals - Printable Packet, Assessment and Center Activity
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This printable packet is a very useful resource for any classroom that is learning about multiplying decimals by whole numbers and by other decimals. The printables in this packet are great options for classroom grades, homework, review or a as a unit assessment covering this topic!
Included in this printable packet you will find:
1. Multiplying Decimals - Whole Numbers by Tenths and Hundredths: This introductory printable shows examples and helpful guidelines on how to relate multiplication of whole numbers by decimals to the tenths and hundredths place. The students will use basic multiplication facts along a number sequence to determine decimal multiplication products.
2. Multiplying Multi-Digit Decimals by Whole Numbers: This is a 16 question printable that is preceded by explanations and tips along with two visual examples that will help your students realize that decimal multiplication works exactly like multiplication with whole numbers in a foundational way, and will explain to the students how to place the decimal in the answer when computation is finished.
3. Multiplication of Decimals with Estimation: Estimation is a very helpful skill when multiplying decimals and this printable page with 12 questions will help students learn how useful this skill really is all while allowing them to practice their multiplication of multi-digit numbers.
4. Multiplying Decimals to Tenths and Hundredths - Review: This 13 question printble homework or classwork page will serve as a great review of concepts that students must know to effectively and accurately multiply decimals by whole numbers and by other decimals (to tenths and hundredths).
5. Multiplying Decimals Assessment: This 17 question assessment will test your students skills covering decimal multiplication to tenths and hundredths. The 13 questions on this page are strictly computation skills and will ask students to multiply decimals by other decimals as well as by whole numbers.
6. Assessment (page 2): This assessment continues onto page two with four more word problems that seek to aim at real world application of this skill as well as mimic questions that students may receive on a state test.
7. Center Activity / Game: Spin It To Win It: Students will work in pairs or individually to take two spinners (a paperclip anchored by a pencil works great) and find two numbers that they will multiply to see how close they can get their product to the number 50. There are ten rounds and whoever gets closest to 50 the most times is the winner. This will be a fun game for them to do in your class that will allow them to practice multiplying decimals as well as comparing decimals.
*******If you have purchased my Decimals Skills Mega Bundle do not buy this product! You already have it!*******
All of these resources come with answer keys so these printables will fit into your classroom with ease. If you need some extra grades for your students in assessing their knowledge of decimal multiplication, or if you simply would like to add some more excellent resources to your classroom in this area, this product will greatly benefit your classroom! Thank you so much!
********************************************************************************************************************************************************
You may also be interested in these excellent resources in practicing decimal skills:
Decimal Place Value Charts
Decimals and Place Vale
Comparing and Ordering Decimals
Rounding Decimals
Adding Decimals
Subtracting Decimals
Decimal Multiplication with Powers of Ten
Decimal Division with Powers of Ten
Dividing Decimals
Decimal Task Cards
Decimals Center Activities
********************************************************************************************************************************************************
Total Pages
14 pages
Answer Key
N/A
Teaching Duration
N/A
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Topic closed. 10 replies. Last post 10 years ago by RJOh.
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Member #1888
July 20, 2003
184 Posts
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Posted: October 21, 2006, 8:57 pm - IP Logged
3,20,21,24,25,29
Member #1888
July 20, 2003
184 Posts
Offline
Posted: October 21, 2006, 10:07 pm - IP Logged
so there you have it folks winning numbers 2,4,25,29,41,43 i had the 25,29 once agaias promised 2 numbers
Sunny California
United States
Member #40295
May 31, 2006
7712 Posts
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Posted: October 21, 2006, 10:19 pm - IP Logged
so there you have it folks winning numbers 2,4,25,29,41,43 i had the 25,29 once agaias promised 2 numbers
So how do you get the other 4 again?
Member #1888
July 20, 2003
184 Posts
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Posted: October 21, 2006, 10:25 pm - IP Logged
thats what i dont know the system guarantees every draw 2 to 3 winning numbers every draw and like i said before all of a sudden my odds have dropped to only needing 3 to 4 numbers
Member #1888
July 20, 2003
184 Posts
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Posted: October 21, 2006, 10:47 pm - IP Logged
i'am tired and sleepy and my judgement isnt good so here's the link www.winthelotterysoftware.com
mid-Ohio
United States
Member #9
March 24, 2001
19816 Posts
Online
Posted: October 21, 2006, 11:05 pm - IP Logged
With 6 numbers you have 15 possible combinations of two's to use as key numbers. Your odds of having the other 4 number correct in a 6/49 game are 1:211,876. multiply that by the 15 possible sets of key numbers and your odds of having all six numbers correct are 1:3,178,140 providing two of those six numbers are winners. I think you could do just a well using the most active six numbers for your sets of key numbers.
* you don't need to buy more tickets, just buy a winning ticket *
Member #1888
July 20, 2003
184 Posts
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Posted: October 21, 2006, 11:10 pm - IP Logged
i have been pondering this same thing using these 6 numbers in a whelel with other numbers but i would continue to play the sets as provided because his jacpots were produced that way
rainbow lake
Member #25177
November 2, 2005
10762 Posts
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Posted: October 22, 2006, 12:23 am - IP Logged
Getting 2 numbers is generally no problem even playing Random. As I am sure you all know.
The Key is getting the first 2 correct Then your on your way. If the first number is wrong you have already lost once the second number is drawn and its wrong then your in the lower prizes b,but theres nothing wrong with a four win,
But i can get 2 numbers 60 % of the time with 30 combinations and can almost guarantee a 3 hit with 81 combinations, But you still have 3 left!!!
That is where you need the luck!!!
I will post a easy method later bed timeZZZZZZZZZZZZZ
Member #1888
July 20, 2003
184 Posts
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Posted: October 22, 2006, 1:11 am - IP Logged
rudy shilli system guarantees 2to 3 100% of the time in 1 to 4 combos
Member #1888
July 20, 2003
184 Posts
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Posted: October 22, 2006, 12:36 pm - IP Logged
i know i will hit the lotto using his system why not join me for the price of a large pizza you could too!!!
mid-Ohio
United States
Member #9
March 24, 2001
19816 Posts
Online
Posted: October 22, 2006, 1:17 pm - IP Logged
i know i will hit the lotto using his system why not join me for the price of a large pizza you could too!!!
I don't know what a large pizza cost where you live, but I can buy 9-15 lottery tickets where I live for the price of a large pizza and that depends on the number of toppings. If Rudy can prove he can hit the lotto for the price of a large pizza then I'm interested.
I'm not talking about buying \$10-\$15 worth of West Virgina Cash25 or Ohio Rolling Cash5 and winning back a dollar, I can do that now with my system or QPs. There are lots of games like these that have overall odds of winning a prize of 1:10 that have payouts as small as \$1.
* you don't need to buy more tickets, just buy a winning ticket *
Page 1 of 1 | 1,245 | 4,014 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-50 | latest | en | 0.853427 |
http://www.proapod.com/calculator/free/o_ber.php | 1,503,475,448,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886117911.49/warc/CC-MAIN-20170823074634-20170823094634-00528.warc.gz | 661,712,724 | 4,170 | Explain
Close
Shows the ratio (as a percentage) between a property's combined operating expenses plus debt service (outflow) to it's gross operating income (inflow).
"Gross operating income is all rental and other income produced less an allowance for vacancy. Debt service is the annual principal and interest loan payment."
Example
Close
You project a first-year rental income of \$61,400, 5.5% vacancy, and 42% operating expenses. You want to know whether the break-even ratio is viable if you finance \$368,400 at 4.75% for 30 years.
"Operating expenses are computed as a percent of gross operating income for entries 1 - 100."
"For best results, use annual amounts for all entries."
Break-Even Ratio
Solves the break-even ratio.
*Required | 171 | 752 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-34 | longest | en | 0.934939 |
https://goprep.co/q20-if-a-lll-1-0-2-0-2-1-2-0-3-then-show-that-a-is-a-root-of-i-1nla33 | 1,606,691,967,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141203418.47/warc/CC-MAIN-20201129214615-20201130004615-00477.warc.gz | 321,497,458 | 33,404 | # If then show that A is a root of the polynomial f(x) = x3 – 6x2 + 7x + 2.
Given: and f(x) = x3 – 6x2 + 7x + 2
To find the value of f(A)
We will substitute x = A in the given equation we get
f(A) = A3 – 6A2 + 7A + 2I……………..(i)
Here I is identity matrix
Now, we will find the matrix for A2, we get
[as cij = ai1b1j + ai2b2j + … + ainbnj]
Now, we will find the matrix for A3, we get
So, Substitute corresponding values from eqn(i) and (ii) in equation f(A) = A3 – 6A2 + 7A + 2I, we get
[as rij = aij + bij + cij],
Hence the A is the root of the given polynomial.
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I was enjoying a Diet Mountain Dew today and happened to notice the ingredients list, and the first two ingredients in the list (and I assume, by volume?), are carbonated water and orange juice concentrate. How is it physically and chemically possible that the highest sugared fruit juice concentrate, that I can think of, is the second ingredient in an advertised, zero calorie beverage?
How is that even possible?
Same way they were once able to advertise an olive as a low fat food. There’s less than 1 gram of fat in one single olive. :eek::dubious: (They can’t do that anymore, note)
There’s not a lot of OJ in those 8oz, likely just enough to not add a calorie. Or perhaps there might be a calorie or two, they are allowed some rounding.
Or even more mysteriously, how it has 0 g of sugar?
I’m going ta guess it must have 4.9 mg of sugar? Rounding down from 1 gram?
While I agree with the general sentiment, I believe 499 mg would make more sense. (1 g = 1000 mg)
Why of course. Gotta get all logical on me, and all. I’m simply rounding up.
I figure it would be what, 4.9 decagrams? Me, I’m a pounds and ounces guy, I’ll guess your weight but it will be in archaic weights and measures. The Metric system is slow to culturize.
No, guess it would be more than .049 kilograms. Maybe 49 kilograms to .049 of a metric tonne.
Slow to acculturate.
Or maybe it’s 499 Kg of OJ Concentrate to a metric tonne.
Note that those figures are per serving. And they define a serving as something less than a can, so the figures are small, especially when rounded.
Pepsi didn’t arbitrarily choose to report the information on less than a can; you can request info on larger sizes. The pulldown menus to the left side of that page let you check nutritional information on larger servings if you wish-- a 12 oz. can (which is also considered a full serving) contains 0 calories, too.
The 20 oz bottle, on the other hand, contains 10 calories, which lends credence to the theory that there’s just some rounding going on.
::checks one of the many Diet Mountain Dew cans lying around::
Yep, Serving size: 1 can.
Ironically, it may be the case that they are legally required to round to the nearest 5 calories, in order to disincentivize subtly toying around with the serving size or ingredient list to have “just” slightly lower calories than the competitor. But it leads to abfurd situations like a 3 calorie beverage having zero calories.
(OK, I admit, I’m halfway making that answer up, and I too have wondered how diet mt dew has zero calories while having some OJ in it. I guess we have the answer now: it does indeed have more than 1 calorie in it.)
or Maybe, it’s 14 Kilocalories of Grapefruit in a Fizzy Pepsi Drink?
No, you’re right. Calories are rounded to the nearest 5. So 2.4 calories is still 0.
I also note that the label (according to the OP’s link) says “concentrated orange juice” which is different from “orange juice concentrate.” Unsweetened orange juice doesn’t have anywhere near the calories or sugar of sweetened juices or concentrate.
The FDA regulation states that a food may be labled “calorie free” if it contains less than 5 calories per serving, with a few stipulations: http://frwebgate1.access.gpo.gov/cgi-bin/TEXTgate.cgi?WAISdocID=138108203894+2+1+0&WAISaction=retrieve. This is similar to the regulation that allows a food to be labeled fat free if it contains less than 0.5g fat per serving, or “sugar free” if it has less than 0.5g sugar per serving: http://frwebgate4.access.gpo.gov/cgi-bin/TEXTgate.cgi?WAISdocID=137823306717+2+1+0&WAISaction=retrieve. Even artificial sweeteners have SOME caloric content, it’s just that you use so little of it, it’s negligible.
Keep in mind that just because something is listed as the second ingredient in a product doesn’t mean there’s much of it – it just means there’s more of it than of the first ingredient, and more of it than everything else listed after. In this case, it’s essentially just a flavoring.
LOL!!!
For the first time in years, I can sleep tonight.
I’m not sure whether to laugh or open fire. Some dead things ought to remain dead.
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http://www.chegg.com/homework-help/questions-and-answers/solid-ball-radius-015-m-starts-rest-rolls-downa-5-m-long-board-inclined-30-degrees-horizon-q406674 | 1,472,603,761,000,000,000 | text/html | crawl-data/CC-MAIN-2016-36/segments/1471983077957.84/warc/CC-MAIN-20160823201117-00286-ip-10-153-172-175.ec2.internal.warc.gz | 357,475,133 | 13,616 | A solid ball of radius 0.15 m starts from rest and rolls downa 5 m long board inclined at 30 degrees above the horizontal. Theball rolls smoothly onto a flat table top and then off the edge ofthe table. The ball lands 1.3 m from the table’s edge.
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1. ## sum across various conditions
Hi All,
I'm trying to get a formula for cell D4 (and, by extension, all of D4:F403) of the Family-Recital sheet. You can see my failed attempts in D6 and D7 of that sheet (there were many others that got nowhere that have since been deleted).
What I'm looking for in D4 is the number of times that any student that's part of the family identified on the row (4 in this case) appears in any class that is part of the recital identified in the column (D in this case). That way, the parents, grandparents, etc. know if they need/want to go to that recital if one or more of their children are in that recital regardless of which class(es) those kids are in.
Students are matched to families in the Student-Family sheet while Students belong or not to a class roster (with the associated Class #) according to the Class Rosters sheet. Finally the Class Names sheet says whether the Class # is part of Recital 1, 2, and/or 3 in columns G-I (there are other ways to get this last piece of info but those are harder to search).
As an intermediate step in developing what I needed, I created 3 columns in the Student-Family sheet (see columns S-U) to sum up how many times each student was in Recital 1, 2, 3. I haven't developed that since I was looking for the "all-in-one" formula per above. Failing that (which I have so far), I figured I could do a SUMIF searching the Family # in column A of the Family-Recital sheet in column C of the Student-Family sheet and SUMIF on column S for recital 1 in the Family-Recital sheet, on column T for recital 2 in the Family-Recital sheet, etc.
The info in the above paragraph (how many times each student appears in each recital) may or may not be useful to the users of this spreadsheet, so I didn't want go down that path unless necessary. I've taken my formula from D7 of the Family-Recital sheet and, with some modifications to focus just on the student, added that to S2 of the Student-Family sheet; still no good.
Attached is the spreadsheet as it stands now (with a few unnecessary sheets deleted to get the file down to acceptable size). The formulas in D6 and D7 of the Family-Recital sheet don't seem to be working. They are pretty much the same with 1 small inconsequential difference and both return 0 (which is incorrect). When I try to partially evaluate the formulas with F9, I sometimes get a "Formula Too Long" message - yet the formula as a whole does seem to give a value.
TIA
Fred
2. Hi Fred
Well, it was a simple request I guess.
Yeah right!
And in Excel2003 too!
I have attached a file with my solution as a starting point for you.
Essentially, the formula I placed in [D4] was
=countAppearances(D\$3,\$A4)
..where D\$3 = "Recital #1"
..and \$A4 = Family #
Hope this gives you a start.
zeddy
3. Hi Fred
Further to my previous post:
Yes, I used some Custom Functions in my proposed solution.
In my attached file, I actually placed a longer formula, by checking first whether there are any members of a Family #, i.e. like this:
in cell [D4]
Code:
`=IF(ISNA(MATCH(\$A4,'Student-Family'!\$C\$2:\$C\$401,FALSE)),"",countAppearances(D\$3,\$A4))`
The Custom Function uses the Recital Number and Family number as inputs.
=countAppearances(recital, family#)
What this does is:
1. From the family number, get the list of students in the family
(I used another Custom Function to get this - see cells in column [I] on sheet [Familty-Recital]
2. get the list of Classes that are involved in the Recital.
This is given in column [D] of sheet [Recital #1-Routine Info]
To simplify things, I assigned a range name for this recital1_Classes
(You will need to create recital2_Classes and recital3_Classes for the other Recitals etc etc)
3. Check if each family member is in any class involved in the Recital.
If they are in more than one class in the Recital, count each class as an 'appearance'.
Well, its a starting point for you.
To simplify my testing of the reported Appearances, on sheet [Class Rosters] I highlighted the class columns that related to Recital#1.
(blue cells in row 1); I hid some of the class columns (you can unhide them again).
This allowed me to easily add some 'additional' students (blue cells) to test the counts.
Please note, I am travelling to New York this week.
It will be a no-computer trip.
So, unless you report back quickly, it may be some while before I can repond again.
zeddy
4. Hi Zeddy,
Thks for the reply. Will not have a chance to take a look at your suggestions at least until tonight (8am here on the East Coast of the US) so you may be gone.
But if you do get this before you leave AND if you have time while in NYC, take a chance on "A Gentleman's Guide to Love and Murder" on Broadway. I just saw it 2 days ago, knowing almost nothing about. it. It was funny, clever - a lot of fun (more than I had trying to get that formula).
Will take a look ASAP.
BTW: Who said this was easy? If I could do it in a DB program, I would. Person for whom this is does not have a DB program (yet).
Fred
5. Hi Fred
Leaving for NYC on Thursday, so will be here in Newcastle UK tomorrow.
Sounds like fun. I have a busy schedule but will check it out.
zeddy
6. Hi Fred
NYC was fantastic! Had no time to see any shows but had the best steaks ever.
Now, did you get the chance to look at my posted file?
It may look complicated, but if there is anything you want me to explain I'd be very happy to go through it step-by-step.
..Or maybe some others on this forum could look at other possible solutions.
zeddy
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• | 1,438 | 5,741 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-04 | longest | en | 0.955364 |
http://betterlesson.com/lesson/resource/2807715/scaffolding-gaps-in-place-value-knowledge-wmv | 1,477,120,244,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988718840.18/warc/CC-MAIN-20161020183838-00559-ip-10-171-6-4.ec2.internal.warc.gz | 27,939,763 | 23,738 | ## Scaffolding gaps in place value knowledge.wmv - Section 2: Warm up
Scaffolding gaps in place value knowledge.wmv
# The Cup Half Full (day 1 of 3)
Unit 8: Exploring Rational Numbers
Lesson 12 of 20
## Big Idea: When the decimal "moves" to the left the digits are actually moving to the left because their value is 10 times less.
Print Lesson
4 teachers like this lesson
Standards:
Subject(s):
Math, Number Sense and Operations, prior knowledge, percent of a number, Mental Math, fraction sense, decimal sense, Flexibility, rational numbers
54 minutes
### Erica Burnison
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Environment: Suburban | 315 | 1,344 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2016-44 | longest | en | 0.815958 |
https://answers.yahoo.com/question/index?qid=20080116192130AAb6oYh | 1,585,464,540,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370493818.32/warc/CC-MAIN-20200329045008-20200329075008-00408.warc.gz | 354,115,002 | 20,279 | Anonymous
# in the equation PV=NkT, what does k represent?????
Relevance
k is a proportionality constant that relates pressure volume and temperature. Since N is counting number ("atoms") this means k has dimensions of Temperature/(Pressure*Volume) = Temperature/force*area*volume = Km/N = K/J in MKS. See the answer above.
Boltzmann's constant = 1.38 x 10^23 JK^-1
Boltzmann's constant
Source(s): I cheated | 109 | 413 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2020-16 | latest | en | 0.903964 |
https://worksheets.tutorvista.com/decimal-worksheets.html | 1,563,728,005,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195527089.77/warc/CC-MAIN-20190721164644-20190721190644-00172.warc.gz | 591,624,615 | 12,894 | Decimal Worksheets | Problems & Solutions
# Decimal Worksheets
Decimal Worksheets
• Page 1
1.
The table shows the cost of items.
Item Cost Cheeseburger $4.15 Soda$0.75 Fruit Juice $1.63 Hamburger$2.15
Anna has \$3.50. Which 2 items can she buy?
a. Cheeseburger and Soda b. Hamburger and Soda c. Hamburger and Cheeseburger d. Hamburger and Fruit Juice
2.
Which of the models represents the number 1.30?
a. Model 1 b. Model 2 c. Model 3 d. Model 4
#### Solution:
In the number 1.30, the integer value is 1. So, 1 is represented by a complete grid.
The digit in the tenths position is represented by 10 hundredths, that is 10 colored pieces.
In the number, the digit in the tenths place is 3 and so, its value is represented by 3 tenths, which is 3 × 10 = 30 hundredths.
So, the model in the Model 2 represents the number 1.30.
3.
Represent 0.92 in the form of a grid model.
a. Model 1 b. Model 2 c. Model 3 d. Model 4
#### Solution:
In each grid there are a total of 100 pieces. So each piece represents one hundredth.
In the number, the integer part is zero. So, there should not be a completely shaded grid. In all the four choices, there is no completely shaded grid.
The digit in the tenths position is 9. So, it is represented by 9 tenths, that is 9 × 10 = 90 shaded pieces.
The digit in the hundredths place, which is 2 is represented by 2 hundredths, that is 2 shaded pieces.
So, 0.92 is represented by 90 + 2 = 92 hundredths. Thus, Model 1 is the correct answer.
4.
Which of the following represents the model?
a. sixty two hundredths b. sixty four hundredths c. thirty six hundredths d. forty hundredths
#### Solution:
The above figure represents a hundredths model formed by drawing ten equally spaced vertical lines, each divided by ten equally spaced horizontal lines.
Thirty six out of 100 pieces of the model are shaded.
The model represents thirty six hundredths.
5.
Determine whether the decimals represented by the two models are equal or not.
a. No b. Yes
#### Solution:
The figure A represents a hundredths model. Thirty out of 100 square pieces are shaded. The model A represents = 0.30
The figure B represents a tenths model. 3 out of 10 strips are shaded. The model B represents = 0.3
Since zero after the decimal has no significance, 0.30 = 0.3
6.
What is the number represented by the grid model?
a. 1.55 b. 55 c. 0.55 d. 0.5
#### Solution:
In the above given grid model, 55 out of 100 squares are filled.
Each square represents one hundredth, 55 squares represent 55 hundredths.
It can be read as "Fifty five hundredths" and represented by 0.55.
7.
What is the number represented by the grid model?
a. 4.8 b. 0.48 c. 0.048 d. 48
#### Solution:
In the above figure, 48 out of 100 squares are shaded.
Each shaded square represent one hundredth.
48 shaded squares represent 48 hundredths and is written as 0.48.
8.
How many hundredths are equivalent to 7 tenths?
a. 7 hundredths b. 70 hundredths c. 7 thousandths d. None of the above
#### Solution:
Seven tenths = 0.7
Draw the model that represents 7 tenths.
Divide the above model into hundredths and shade 70 squares. This represents 0.70
70 hundredths = 7 tenths
[Since zero after the decimal has no value, 0.70 = 0.7]
9.
A square cake is cut horizontally and vertically into 100 equal pieces. Chris got 11 pieces of the cake. Write the decimal equivalent.
a. 0.101 b. 0.11 c. 0.01 d. 0.011
#### Solution:
Each piece represents a hundredths place.
Eleven out of 100 pieces represents eleven hundredths = 11 / 100 = 0.11
So, Chris got 0.11 part of the cake.
10.
Which of the models represents 0.4?
a. Model 1 b. Model 2 c. Model 3 d. Model 4
#### Solution:
One colored strip in the tenths model represents one tenth of the tenths model = 0.1.
0.4 in the tenths model should have 4 colored strips out of ten equally spaced strips.
From the models, model 3 have the four colored strips. | 1,089 | 3,929 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2019-30 | latest | en | 0.900415 |
https://icsesolutions.com/ml-aggarwal-class-6-solutions-for-icse-maths-chapter-3-ex-3-3/ | 1,722,678,240,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640365107.3/warc/CC-MAIN-20240803091113-20240803121113-00451.warc.gz | 248,729,066 | 12,503 | ## ML Aggarwal Class 6 Solutions for ICSE Maths Chapter 3 Integers Ex 3.3
Question 1.
Evaluate the following, using the number line:
(i) 4 – (-2)
(ii) -4 – (-2)
(iii) 3 – 6
(iv) -3 – (-5)
Solution:
(i) Start from 4 on the number line.
Move 2 units to the digits we reach at 6
∴ 4 – (-2) = 4 + 2 = 6
(ii) Start from -4 on the number line.
Move 2 units to the right, we reach at -2
∴ -4 – (-2) = —4 + 2 = -2
(iii) Start from 3 on the number line.
Move 6 units to the left, we reach at -3
3 – 6 = -3
(iv) Start from -3 on the number line.
Move 5 units to the right, we reach at 2
-3 – (-5) = -3 + 5 = 2
Question 2.
Subtract :
(i) -6 from 9
(ii) 6 from -9
(iii) -6 from -9
(iv) -725 from -63
(v) -376 from 10
(vi) 92 from -620
Solution:
(i) 9 – (-6) = 9 + 6 = 15
(ii) -9 – 6 = -15
(iii) -9 – (-6) = -9 + 6 = -3
(iv) -63 – (-725) = -63 + 725 = +662
(v) 10 – (-376) = 10 + 376 = 386
(vi) -620 – 92 = -712
Question 3.
Evaluate the following:
(i) -237 – (+ 1884)
(ii) -346 – (- 1275)
(iii) -190 – (-3512)
(iv) -2718 – (+ 6827)
Solution:
(i) -237 – (+ 1884)
= -237 – 1884
= -(237 + 1884) = -2121
(ii) -346 -(- 1275)
= -346 + 1275
= 1275 – 346 = 929
(iii) -190 – (-3512)
= -190 + 3512
= 3512- 190 = 3322
(iv) 2718 – (+ 6827)
= -2718 – 6827
= -(2718 + 6827) = -9545
Question 4.
The sum of two integers is 17. If one of them is -35, find the other.
Solution:
One number = -35
Sum of two integers =17
Second number = Sum of integers – (The given number)
= 17 – (-35)
= 17 + 35 = 52
Question 5.
What must be added to -23 to get -9?
Solution:
Let the number to be added = x
∴ -23 + x = -9
∴ The required number = -9 – (-23)
= -9 + 23 = 14
Question 6.
Find the predecessor of 0.
Solution:
Predecessor of 0 = 0 – 1 = -1
Question 7.
Find the successor and the predecessor of the following integers:
(i) -31
(ii) -735
(iii) -240
Solution:
(i) Successor of -31 = -31 + 1 = -30
Predecessor of -31 = -31 – 1 = -32
(ii) Successor of -735 = -735 + 1 = -734
Predecessor of-735 = -735 – 1 = -736
(iii) Successor of -240 = -240 + 1 = -239
Predecessor of -240 = -240 – 1 = -241 | 852 | 2,062 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.84375 | 5 | CC-MAIN-2024-33 | latest | en | 0.704138 |
https://www.dw-math.com/ac/static/B52.html | 1,716,364,011,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058534.8/warc/CC-MAIN-20240522070747-20240522100747-00859.warc.gz | 647,839,252 | 4,841 | Custom math worksheets at your fingertips
For all problems there are free worksheets and solution sheets available for instant download. Just click on the link for a detail view.Is there a type of problem missing that you need? Register here, and add it as an item to our wish list!
# All problems for the topic of Whole numbers
If you remember the quickname of a problem, you can add it to your worksheet very quickly: On the bottom right of the worksheet editor, there is a quick entry form field.
Name
Description
Quickname with link for detail view
Add and sub sibling numbers in the range from 1 to 100
Add and subtract spliting one digit number
Addition and subtraction by splitting one digit number
Add and subtract terms with incrementing summand
In a series of addition or subtraction terms, one number changes stepwise.
Add and subtract times table product
Addition and subtraction of times table products
Add and subtract times table product with same factor
Addition and subtraction of times table products for a single number one to ten
Add multiples of powers of ten
Multiples of powers of ten added to natural numbers.
Numbers with or without decimal places have to be added.
Add two numbers with specific number of places
Addition of two numbers with controlled number of digits for each number.
Addition of powers of ten, fill blanks
What multiple of a power of ten has been added?
Addition or subtraction, insert correct symbol
In a series of addition or subtraction problems, the plus or minus symbol is missing.
Addition up to ten in graphical form
Plus problems up to ten are to be described using marbles.
Blanks in columnar addition to be filled in correctly.
Columnar subtraction
Divide a fraction by a whole number
A fraction has to be divided by a whole number.
Division by multiples of power of ten
Divison by a power of ten or a multiple thereof.
Division series 1x1 with factor 10
Division problem series with divisor and dividend alternating with factor ten.
Division times table 1-10
Division problems from the multiplication tables
Double a given number
Double a number with or without decimal places.
Factor or multiple?
Determine whether a number is a factor or multiple of another.
Factors of a number
For a given number, all factors have to be listed.
Frames & Arrows w/ elementary arithmetic - add & sub
Simple frames and arrows chain task with one elementary arithmetic operation, results are to be filled in blank spaces.
Frames & Arrows w/ elementary arithmetic - varying ops
In a frames and arrows calculation chain task, the results are to be filled in the blank spaces.
Halve numbers
Divide numbers by two, whole numbers or with decimal places.
Insert relational operator - simple
Insert the correct relational symbol between two numbers.
Insert relational operator for subtraction terms
The correct relational operator has to be inserted for subtraction terms.
Insert relational operator for sum terms
The correct relational operator has to be inserted for addition terms.
Insert the arithmetic operator
In an equation with two operands, insert the correct arithmetic symbol.
Insert two arithmetic operators
In an equation with three operands, insert two operators.
Long multiplication to be performed for two factors
Two factors have to be multiplied using the long multiplication written form.
Magic triangle with sums to be filled in correctly
In a magic triangle, six numbers have to be filled in correctly.
Matrix of addition and subtraction problems
Table of addition and subtraction problems
Mean and median to be determined
For a series of numbers, the mean or median is to be found.
Median and mean - decide which number is which
For two given numbers for a number series, decide which is the median and which is the mean.
Median or mean value - Which one is it?
Decide whether a number is the median or mean of a given series of whole numbers.
Multiples of a number list first n
For a given number, list the first n multiples.
Multiples of a number up to a limit
For a given number, the multiples up to a given limit have to be listed.
Multiplication by power of ten
Multiplication of a whole numbers with powers of ten or multiples of powers of ten
Multiplication of two natural numbers
Multiplication problems with defined number range
Multiplication tables with mistakes
In a number of tasks from the multiplication tables, mistakes have to be found.
Multiply fractions by whole numbers
A fraction has to multiplied by a whole number.
Negative whole numbers add and subtract
Negative whole numbers are to be added and subtracted
Negative whole numbers add and subtract choose result
For an addition or subtraction problem with negative numbers, the correct result must be chosen.
Number wall sums of whole numbers
In the number wall missing sums have to be added.
Number wall times table products
In a number wall, products from the times tables 1-10 are to be filled in.
Place value board table and number words
In a table, place values and number words have to be added
Relational operator insert digit for true statement
Add a digit in a relational statement to make it true.
Relational symbol with times table terms and number
Compare Times table multiplication terms to numbers, decide what is the correct relational symbol.
Subtract mupltiples of powers of ten
Multiples of powers of ten subtracted from natural numbers.
Subtract numbers in range
Multiple numbers with or without decimal places have to be subtracted.
Subtraction by filling up place values
Calculate difference by stepping up each place value.
Subtraction power of ten fill blanks
What multiple of a power of ten has been subtracted?
Subtraction term graphical representation
A series of marbles represents a subtraction term - empty spaces to be filled in term.
Subtraction two numbers with specific no of places
Subtraction of two numbers with controlled number of digits for each number.
Table of multiplication problems
Matrix of multiplication problems with whole numbers
Test for divisibility
A row of numbers is presented. Delete those that are not divisible by a given divisor.
Times table fill in blanks
Simple times table problems where the product and a factor are given.
Times table with factor ten
Multiplication problem series introducing factor of ten
Times table:Multiplication problems in range one to ten
Multiplication problems from the 1-10 times tables.
These informational pages with samples describe math problems that can be combined on custom math worksheets with solutions for home and K-12 school use. | 1,319 | 6,562 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2024-22 | latest | en | 0.884625 |
https://www.arithmix.com/content/how-to-calculate-levered-cash-flow-in-odoo | 1,722,962,056,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640497907.29/warc/CC-MAIN-20240806161854-20240806191854-00851.warc.gz | 528,358,773 | 17,121 | # How To Calculate Levered Cash Flow in Odoo | Arithmix
Learn how to calculate levered cash flow in Odoo with our comprehensive guide. Understand the key concepts and steps involved in determining levered cash flow for your business. Boost your financial analysis skills and make informed decisions with Odoo.
Levered cash flow is an important financial metric that helps businesses understand their financial health and make informed decisions about investments and financing. In simple terms, it is the amount of cash flow that is available to investors and lenders after all the expenses and debts have been paid off. Calculating levered cash flow can be a bit tricky, but with the right tools and techniques, it can be done easily and accurately.
## What Is Levered Cash Flow?
Levered cash flow is a measure of a company's financial performance that takes into account the impact of debt and other financial obligations. It is calculated by subtracting the company's interest expenses and other debt payments from its operating cash flow. The resulting figure represents the amount of cash flow that is available to investors and lenders after all the expenses and debts have been paid off.
Levered cash flow is an important metric for investors and lenders because it provides a more accurate picture of a company's financial health than other measures such as net income or earnings per share. By taking into account the impact of debt and other financial obligations, levered cash flow provides a more realistic view of a company's ability to generate cash and pay off its debts.
## When Is It Valuable To Calculate Levered Cash Flow?
Calculating levered cash flow is valuable in a variety of situations. For example, it can be useful when evaluating the financial health of a company that is considering taking on additional debt or issuing new shares of stock. By calculating levered cash flow, investors and lenders can get a better sense of the company's ability to generate cash and pay off its debts.
Levered cash flow can also be useful when evaluating the financial health of a company that is considering making a major investment or acquisition. By calculating levered cash flow, investors and lenders can get a better sense of the potential return on investment and the risks associated with the investment or acquisition.
In addition, calculating levered cash flow can be useful when evaluating the financial health of a company that is experiencing financial difficulties. By calculating levered cash flow, investors and lenders can get a better sense of the company's ability to generate cash and pay off its debts, and can make informed decisions about whether to provide additional financing or support.
In conclusion, levered cash flow is an important financial metric that can provide valuable insights into a company's financial health and performance. By understanding how to calculate levered cash flow, investors and lenders can make informed decisions about investments and financing, and can better evaluate the potential risks and rewards of different opportunities.
## How Do You Calculate Levered Cash Flow in Odoo
Odoo itself isn’t naturally geared towards letting you calculate complex metrics like Levered Cash Flow. As an alternative, teams typically use products like Arithmix to import data from Odoo and build out dashboards.
## What is Arithmix?
Arithmix is the next generation spreadsheet - a collaborative, web-based platform for working with numbers that’s powerful yet easy to use. With Arithmix you can import data from systems like Odoo, combine it with data from other systems, and create calculations like Levered Cash Flow.
In Arithmix, data is organized into Tables and referenced by name, not by cell location like a spreadsheet, simplifying calculation creation. Data and calculations can be shared with others and re-used like building blocks, vastly streamlining analysis, model building, and reporting in a highly scalable and easy to maintain platform. Data can be edited, categorized (by dimensions) and freely pivoted. Calculations are automatically copied across a dimension - eliminating copy and paste of formulas. | 790 | 4,192 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-33 | latest | en | 0.956732 |
https://tbc-python.fossee.in/convert-notebook/Engineering_Mechanics_by_R.K_Bansal/chapter_1.ipynb | 1,708,842,182,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474581.68/warc/CC-MAIN-20240225035809-20240225065809-00556.warc.gz | 561,685,558 | 39,434 | # Chapter 1:Fundamental Of Engineering Mechanics¶
## Example 1.1,Page No.8¶
In [2]:
import math
from math import sin, cos, tan, radians, pi
#Declaration of Variables
P=10 #N #Force1
Q=8 #N #Force2
alpha=60 #Degrees #Angle Between Two Forces
#Calculations
#Magnitude of Resultant Force
R=(P**2+Q**2+2*P*Q*cos(alpha*pi*180**-1))**0.5 #N
#Result
print"Magnitude of Resultant Force",round(R,2),"N"
Magnitude of Resultant Force 15.62 N
## Example 1.2,Page No.8¶
In [3]:
import math
#Declaration of Variables
alpha=60 #Degrees #Angle between Forces
R=20*(3)**0.5
#Let P & Q be the Two forces
#As Two Forces are equal i.e P=Q
#Magnitude of Resultant Force
#R=(P**2+Q**2+2*P*Q*cos(alpha*pi*180**-1))**0.5 #N
#After Sub values and Furhter simplifying above equations we get
#R=2*P*cos(alpha*2**-1*pi*180**-1)
#Further on Simplifying we get
P=R*((3)**0.5)**-1 #N
#Result
print"Magnitude of Force is",round(P,2),"N"
Magnitude of Force is 20.0 N
## Example 1.3,Page No.9¶
In [4]:
import math
import numpy as np
#Declaration of Variables
#Case-1
R1=14 #N #Resultant1
alpha1=60 #Degrees #Angle between two forces
#Case-2
R2=(136)**0.5
alpha2=90 #Degrees #Angle between two Forces
#Let P And Q be the two forces
#R=(P**2+Q**2+2*P*Q**cos(alpha))
#Now For case-1,we get Resultant as
#P**2+Q**2+P*Q=196 ............................(1)
#For case-2,we get Resultant as
#P**2+Q**2=136 ...................................(2)
#Subtracting Equation 2 from equation 1 we get
#P*Q=60 .........................................(3)
#Multiplying abovw equation by 2 we get
#2*P*Q=120 .......................................(4)
#Adding equation 4 to equation 2 we get
#P**2+Q**2+2*P*Q=256
#After Further simplifying we get
#P=16-Q ..........................(5)
#Sub value of P in equation 3 we get
#Q**2-16*Q+60=0
a=1
b=-16
c=60
X=b**2-4*a*c
Q1=(-b+X**0.5)*(2*a)**-1
Q2=(-b-X**0.5)*(2*a)**-1
#Now sub value of Q in equation 5 we get
P1=16-Q1
P2=16-Q2
#Result
print"Magnitude of two Forces is:P2",round(P2,2),"N"
print" :Q1",round(Q2,2),"N"
Magnitude of two Forces is:P2 10.0 N
:Q1 6.0 N
## Example 1.4,Page No.10¶
In [2]:
import math
from math import sin, cos, tan, radians, pi
import numpy as np
P=50 #N #Force acting at pt O
Q=100 #N #force acting at pt O
alpha=30 #DEgree #Angle Between Two Forces
#Calculations
#MAgnitude of Resultant
R=(P**2+Q**2+2*P*Q*cos(alpha*pi*180**-1))**0.5 #N
#Angle Made by resultant with the direction of P
X=(Q*sin(alpha*180**-1*pi)*(P+(Q*cos(alpha*pi*180**-1)))**-1)
theta=np.arctan(X)*(180*pi**-1) #Degrees
#Angle made by resultant with x-axis is
Y=theta+alpha*2**-1 #Degrees
#Result
print"Resultant in the Magnitude is",round(R,2),"N"
print"Resultant in the Direction is",round(Y,2),"Degrees"
Resultant in the Magnitude is 145.47 N
Resultant in the Direction is 35.1 Degrees
## Example 1.5,Page No.10¶
In [6]:
import math
#Declaration of Variables
R=1500 #N #REsultant of two Forces
alpha=90 #Degrees #Angle between two Forces
theta=36 #Degrees #Angle Made by Resultant with one Force
#Calculations
#Now From Equation of Direction of Resultant,we get
#tan(theta)=(Q*sin(alpha))*(P+Q*sin(alpha))**-1
#After Further sub values and simplifying we get
#Q=0.726*P .......................................(1)
#Now From Equation of Resultant
#R=(P**2+Q**2+2*P*Q*cos(alpha))
#After sub values and further simplifying we get
#R=1.527*P**2
#Therefore,we get value of P After simplifying above equation
P=(R**2*(1.527)**-1)**0.5
#Sub value of P in equation 1 we get
Q=0.726*P
#Result
print"Magnitude Of Forces:P",round(P,2),"N"
print" :Q",round(Q,2),"N"
Magnitude Of Forces:P 1213.87 N
:Q 881.27 N
## Example 1.6,Page No.11¶
In [7]:
import math
from math import sin, cos, tan, radians, pi
import numpy as np
#Declaration of Variables
#P+Q=120
R=180 #N #Resultant of two Forces
theta=90 #Degrees #Angle between force and Resultant
#Calculations
#Now From Equation of Direction of Resultant,we get
#Tan(theta)=Q*sin(alpha)*(P+Q*sin(alpha))**-1
#After Further ssub values a in above equation and further simplifying we get
#P=-Q*cos(alpha) ..........(1)
#Now From equation of Resultant we get
#R=(P**2++Q**2+2*P*Q*cos(alpha))**0.5
#After sub values and further simplifying
#Q-P=120 ......................................(1)
#P+Q=270 ......................................(2)
#After Adding above equations i.e equations 1 and 2 we get
Q=390*2**-1 #N
P=270-Q #N
#Value of angle alpha
alpha=np.arccos(-P*Q**-1)*(180*pi**-1) #Degrees
#Result
print"Magnitude of Each Force:P",round(P,2),"N"
print" :Q",round(Q,2),"N"
print"Angle between Two Forces",round(alpha,2),"Degrees"
Magnitude of Each Force:P 75.0 N
:Q 195.0 N
Angle between Two Forces 112.62 Degrees
## Example 1.7,Page No.13¶
In [8]:
import math
from math import sin, cos, tan, radians, pi
#Declaration of Variables
W=1000 #N #Weight
#Angles
CAB=30 #Degrees
CBA=CBD=60 #Degrees
ACB=90 #Degrees
#Calculations
#Angle
#In Right-Angle Triangle,Angle BDC
#Angle
BCD=90-CBD #Degrees
ACE=180-ACB-90-60 #DEgrees
BCE=180-ACE-ACB #DEGrees
#Applying LAmi's Theorem at Point C
#T1*(sin150)**-1=T2*(sin(120)**-1=1000*sin(90)**-1
#After Further simp;ifying we get
T1=W*sin(150*pi*180**-1) #N
T2=W*sin(120*pi*180**-1) #N
#Result
print"Tension in Chain is:T1",round(T1,2),"N"
print" :T2",round(T2,2),"N"
Tension in Chain is:T1 500.0 N
:T2 866.03 N
## Example 1.8,Page No.18¶
In [9]:
import math
from math import sin, cos, tan, radians, pi
import numpy as np
#Declaration of Variables
W=900 #N #Weight at C
#Length
AC=4 #m
BC=3 #m
AB=5 #m
#Calculations
#In Triangle ABC
X=AC**2+BC**2
Y=AB**2
#Therefore,
#X=Y
#Therefore,
#Triangle ABC is Right Angle Triangle,In Which Angle ACB=90 Degrees
alpha=np.arcsin(BC*AB**-1)*(180*pi**-1)
Beta=90-alpha
theta1=90-alpha
#In Right Angle Triangle BDC,
theta2=90-Beta
#Now,Angles
ACE=180-theta1
BCE=180-theta2
#Now applying ami's Theorem
#T1*(sin(BCE))**-1=T2*(sin(ACE))**-1=W*(sin(90))**-1
#Tensions in chains
T1=W*sin(BCE*180**-1*pi) #N
T2=W*sin(ACE*180**-1*pi) #N
#Result
print"Tension in Chains are:T1",round(T1,2),"N"
print" :T2",round(T2,2),"N"
#Answer in hte book For T2 is incorrect
Tension in Chains are:T1 540.0 N
:T2 720.0 N
## Example 1.9,Page No.18¶
In [10]:
import math
from math import sin, cos, tan, radians, pi
#Declaration of Variables
W=15 #N #Weight at Pt C
OAC=FAC=60 #Degrees
CBD=BCF=45 #Degrees
FCA=90-FAC #Degrees
#Calculations
#Using Lami's theorem,
#W*(sin(BCA))**-1=T1*(sin(ACE))**-1=T2*(sin(ACE))**-1
#Angles
BCA=BCF+FCA #Degrees
ACE=180-FCA #Degrees
BCE=180-BCF #Degrees
#Force's in the string AC
T1=W*sin(ACE*180**-1*pi)*(sin(BCA*180**-1*pi))**-1 #N
T2=W*sin(BCE*180**-1*pi)*(sin(BCA*180**-1*pi))**-1 #N
#Result
print"Force's in the string:T1",round(T1,2),"N"
print" :T2",round(T2,2),"N"
Force's in the string:T1 7.76 N
:T2 10.98 N
## Example 1.10,Page No.16¶
In [11]:
import math
from math import sin, cos, tan, radians, pi
import numpy as np
#Declaration of Variables
#Forces
P=50 #N
Q=100 #N
alpha=30 #Angle Between Two Forces
theta=15 #Degrees #Angle Made By Force P with x-axis
theta2=alpha+theta #Degrees
#Calculations
#Sum Of COmponents of forces along X-Axis is
H=P*cos(theta*pi*180**-1)+Q*cos(theta2*pi*180**-1) #N
#Sum Of COmponents of forces along Y-Axis is
V=P*sin(theta*pi*180**-1)+Q*sin(theta2*pi*180**-1) #N
#MAgnitude Of Resultant Force is
R=(H**2+V**2)**0.5 #N
#Let Direction Of Resultant Force be beta
#Direction Of Resultant Force is
beta=np.arctan(V*H**-1)*(180*pi**-1) #Degrees
#Result
print"Magnitude of Resultant Force is",round(R,2),"N"
print"Direction of Resultant Force is",round(beta,2),"Degrees"
Magnitude of Resultant Force is 145.47 N
Direction of Resultant Force is 35.1 Degrees
## Example 1.11,Page No.17¶
In [3]:
import math
from math import sin, cos, tan, radians, pi
import numpy as np
#Declaration of Variables
#Let 3Forces Be
R1=40 #KN
R2=15 #KN
R3=20 #KN
#Angles Made by respective forces with X-Axis
theta1=60 #Degrees
theta2=120 #Degrees
theta3=240 #Degrees
#Calculations
#Now sum of components of all forces along X-Axis
H=R1*cos(theta1*pi*180**-1)+R2*cos(theta2*pi*180**-1)+R3*cos(theta3*pi*180**-1)
#Now sum of components of all forces along Y-Axis
V=R1*sin(theta1*pi*180**-1)+R2*sin(theta2*pi*180**-1)+R3*sin(theta3*pi*180**-1)
#MAgnitude of Resultant Force is
R=(H**2+V**2)**0.5 #N
#Direction of Resultant Force is
theta=np.arctan(V*H**-1)*(pi**-1*180)
#Result
print"Magnitude of Resultant Force is",round(R,2),"KN"
print"Direction of Resultant Force is",round(theta,2),"Degrees"
#Declaration of Variables
#Let 3Forces Be
R1=40 #KN
R2=15 #KN
R3=20 #KN
#Angles Made by respective forces with X-Axis
theta1=60 #Degrees
theta2=120 #Degrees
theta3=240 #Degrees
#Calculations
#Now sum of components of all forces along X-Axis
H=R1*cos(theta1*pi*180**-1)+R2*cos(theta2*pi*180**-1)+R3*cos(theta3*pi*180**-1)
#Now sum of components of all forces along Y-Axis
V=R1*sin(theta1*pi*180**-1)+R2*sin(theta2*pi*180**-1)+R3*sin(theta3*pi*180**-1)
#MAgnitude of Resultant Force is
R=(H**2+V**2)**0.5 #N
#Direction of Resultant Force is
theta=np.arctan(V*H**-1)*(pi**-1*180)
#Result
print"Magnitude of Resultant Force is",round(R,2),"KN"
print"Direction of Resultant Force is",round(theta,2),"Degrees"
Magnitude of Resultant Force is 30.41 KN
Direction of Resultant Force is 85.28 Degrees
Magnitude of Resultant Force is 30.41 KN
Direction of Resultant Force is 85.28 Degrees
## Example 1.12,Page No.18¶
In [13]:
import math
from math import sin, cos, tan, radians, pi
import numpy as np
#Declaration of Variables
#Let 4 Forces Be
R1=10 #KN
R2=15 #KN
R3=20 #KN
R4=40 #KN
#Angles Made by respective forces with X-Axis
theta1=30 #Degrees
theta2=60 #Degrees
theta3=90 #Degree
theta4=120 #Degrees
#Calculations
#Now sum of components of all forces along X-Axis
H=R1*cos(theta1*pi*180**-1)+R2*cos(theta2*pi*180**-1)+R3*cos(theta3*pi*180**-1)+R4*cos(theta4*pi*180**-1)
#Now sum of components of all forces along Y-Axis
V=R1*sin(theta1*pi*180**-1)+R2*sin(theta2*pi*180**-1)+R3*sin(theta3*pi*180**-1)+R4*sin(theta4*pi*180**-1)
#MAgnitude of Resultant Force is
R=(H**2+V**2)**0.5 #N
#Direction of Resultant Force is
theta4=np.arctan(V*H**-1)*(pi**-1*180)
theta=180+theta4 #Degrees
#Result
print"Magnitude of Resultant Force is",round(R,2),"KN"
print"Direction of Resultant Force is",round(theta,2),"Degrees"
Magnitude of Resultant Force is 72.73 KN
Direction of Resultant Force is 93.03 Degrees
## Example 1.13,Page No.19¶
In [14]:
import math
#Declaration of Variables
L=10 #m #Length of beam
#Distances
L_AC=4 #m
L_CB=6 #m
#Calculations
#Let R_A & R_B be the forces acting at A & B
#Taking Moment at A
R_B=(W*L_CB)*L**-1 #N
R_A=W-R_B #N
#Result
print"Beam Reactions are:R_A",round(R_A,2),"N"
print" :R_B",round(R_B,2),"N"
Beam Reactions are:R_A 80.0 N
:R_B 120.0 N
## Example 1.14,Page No.20¶
In [15]:
import math
from math import sin, cos, tan, radians, pi
import numpy as np
#Declaration of Variables
#Let 4 Forces be
F1=10 #N
F2=20 #N
F3=30 #N
F4=40 #N
#Calculations
#Net Forces in Horizontal direction is
H=F1-F3 #N
#Net Forces in Vertical direction is
V=F2-F4 #N
#Resultant Force is given by
R=(H**2+V**2)**0.5 #N
#Direction of resultant Forces
theta=np.arctan(V*H**-1)*(pi**-1*180) #Degrees
#Since H & V are negative theta lies between 180 & 270
theta2=180+theta #Degrees
#Result
print"Magnitude of Force is",round(R,2),"N"
print"Direction of Force is",round(theta2,2),"Degrees"
Magnitude of Force is 28.28 N
Direction of Force is 225.0 Degrees | 4,225 | 11,728 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2024-10 | latest | en | 0.578982 |
https://www.dataunitconverter.com/mebibit-per-day-to-petabyte-per-hour | 1,709,567,659,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476452.25/warc/CC-MAIN-20240304133241-20240304163241-00330.warc.gz | 713,256,575 | 17,252 | # Mibit/Day to PB/Hr → CONVERT Mebibits per Day to Petabytes per Hour
expand_more
info 1 Mibit/Day is equal to 0.0000000000054613333333333333333333333333 PB/Hr
Input Mebibits per Day (Mibit/Day) - and press Enter.
Mibit/Day
Sec
Min
Hr
Day
Sec
Min
Hr
Day
S = Second, M = Minute, H = Hour, D = Day
## Mebibits per Day (Mibit/Day) Versus Petabytes per Hour (PB/Hr) - Comparison
Mebibits per Day and Petabytes per Hour are units of digital information used to measure storage capacity and data transfer rate.
Mebibits per Day is a "binary" unit where as Petabytes per Hour is a "decimal" unit. One Mebibit is equal to 1024^2 bits. One Petabyte is equal to 1000^5 bytes. There are 7,629,394,531.25 Mebibit in one Petabyte. Find more details on below table.
Mebibits per Day (Mibit/Day) Petabytes per Hour (PB/Hr)
Mebibits per Day (Mibit/Day) is a unit of measurement for data transfer bandwidth. It measures the number of Mebibits that can be transferred in one Day. Petabytes per Hour (PB/Hr) is a unit of measurement for data transfer bandwidth. It measures the number of Petabytes that can be transferred in one Hour.
## Mebibits per Day (Mibit/Day) to Petabytes per Hour (PB/Hr) Conversion - Formula & Steps
The Mibit/Day to PB/Hr Calculator Tool provides a convenient solution for effortlessly converting data rates from Mebibits per Day (Mibit/Day) to Petabytes per Hour (PB/Hr). Let's delve into a thorough analysis of the formula and steps involved.
Outlined below is a comprehensive overview of the key attributes associated with both the source (Mebibit) and target (Petabyte) data units.
Source Data Unit Target Data Unit
Equal to 1024^2 bits
(Binary Unit)
Equal to 1000^5 bytes
(Decimal Unit)
The conversion from Data per Day to Hour can be calculated as below.
x 60
x 60
x 24
Data
per
Second
Data
per
Minute
Data
per
Hour
Data
per
Day
÷ 60
÷ 60
÷ 24
The formula for converting the Mebibits per Day (Mibit/Day) to Petabytes per Hour (PB/Hr) can be expressed as follows:
diamond CONVERSION FORMULA PB/Hr = Mibit/Day x 10242 ÷ (8x10005) / 24
Now, let's apply the aforementioned formula and explore the manual conversion process from Mebibits per Day (Mibit/Day) to Petabytes per Hour (PB/Hr). To streamline the calculation further, we can simplify the formula for added convenience.
FORMULA
Petabytes per Hour = Mebibits per Day x 10242 ÷ (8x10005) / 24
STEP 1
Petabytes per Hour = Mebibits per Day x (1024x1024) ÷ (8x1000x1000x1000x1000x1000) / 24
STEP 2
Petabytes per Hour = Mebibits per Day x 1048576 ÷ 8000000000000000 / 24
STEP 3
Petabytes per Hour = Mebibits per Day x 0.000000000131072 / 24
STEP 4
Petabytes per Hour = Mebibits per Day x 0.0000000000054613333333333333333333333333
Example : By applying the previously mentioned formula and steps, the conversion from 1 Mebibits per Day (Mibit/Day) to Petabytes per Hour (PB/Hr) can be processed as outlined below.
1. = 1 x 10242 ÷ (8x10005) / 24
2. = 1 x (1024x1024) ÷ (8x1000x1000x1000x1000x1000) / 24
3. = 1 x 1048576 ÷ 8000000000000000 / 24
4. = 1 x 0.000000000131072 / 24
5. = 1 x 0.0000000000054613333333333333333333333333
6. = 0.0000000000054613333333333333333333333333
7. i.e. 1 Mibit/Day is equal to 0.0000000000054613333333333333333333333333 PB/Hr.
Note : Result rounded off to 40 decimal positions.
You can employ the formula and steps mentioned above to convert Mebibits per Day to Petabytes per Hour using any of the programming language such as Java, Python, or Powershell.
### Unit Definitions
#### What is Mebibit ?
A Mebibit (Mib or Mibit) is a binary unit of digital information that is equal to 1,048,576 bits and is defined by the International Electro technical Commission(IEC). The prefix 'mebi' is derived from the binary number system and it is used to distinguish it from the decimal-based 'megabit' (Mb). It is widely used in the field of computing as it more accurately represents the amount of data storage and data transfer in computer systems.
arrow_downward
#### What is Petabyte ?
A Petabyte (PB) is a decimal unit of digital information that is equal to 1,000,000,000,000,000 bytes (or 8,000,000,000,000,000 bits) and commonly used to measure the storage capacity of enterprise storage arrays and data centers. It is also used to express data transfer speeds and in the context of data storage and memory, the binary-based unit of Pebibyte (PiB) is used instead.
## Excel Formula to convert from Mebibits per Day (Mibit/Day) to Petabytes per Hour (PB/Hr)
Apply the formula as shown below to convert from 1 Mebibits per Day (Mibit/Day) to Petabytes per Hour (PB/Hr).
A B C
1 Mebibits per Day (Mibit/Day) Petabytes per Hour (PB/Hr)
2 1 =A2 * 0.000000000131072 / 24
3
If you want to perform bulk conversion locally in your system, then download and make use of above Excel template.
## Python Code for Mebibits per Day (Mibit/Day) to Petabytes per Hour (PB/Hr) Conversion
You can use below code to convert any value in Mebibits per Day (Mibit/Day) to Mebibits per Day (Mibit/Day) in Python.
mebibitsperDay = int(input("Enter Mebibits per Day: "))
petabytesperHour = mebibitsperDay * (1024*1024) / (8*1000*1000*1000*1000*1000) / 24
print("{} Mebibits per Day = {} Petabytes per Hour".format(mebibitsperDay,petabytesperHour))
The first line of code will prompt the user to enter the Mebibits per Day (Mibit/Day) as an input. The value of Petabytes per Hour (PB/Hr) is calculated on the next line, and the code in third line will display the result.
## Frequently Asked Questions - FAQs
#### How many Petabytes(PB) are there in a Mebibit(Mibit)?expand_more
There are 0.000000000131072 Petabytes in a Mebibit.
#### What is the formula to convert Mebibit(Mibit) to Petabyte(PB)?expand_more
Use the formula PB = Mibit x 10242 / (8x10005) to convert Mebibit to Petabyte.
#### How many Mebibits(Mibit) are there in a Petabyte(PB)?expand_more
There are 7629394531.25 Mebibits in a Petabyte.
#### What is the formula to convert Petabyte(PB) to Mebibit(Mibit)?expand_more
Use the formula Mibit = PB x (8x10005) / 10242 to convert Petabyte to Mebibit.
#### Which is bigger, Petabyte(PB) or Mebibit(Mibit)?expand_more
Petabyte is bigger than Mebibit. One Petabyte contains 7629394531.25 Mebibits.
## Similar Conversions & Calculators
All below conversions basically referring to the same calculation. | 1,918 | 6,334 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2024-10 | latest | en | 0.710603 |
https://lists.defectivebydesign.org/archive/html/help-octave/2016-08/msg00222.html | 1,695,528,831,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506559.11/warc/CC-MAIN-20230924023050-20230924053050-00347.warc.gz | 423,099,304 | 3,374 | help-octave
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## Re: Help in parameter estimation
From: Nicholas Jankowski Subject: Re: Help in parameter estimation Date: Tue, 30 Aug 2016 11:56:55 -0400
error: leasqr: subscript indices must be either positive integers less than 2^31 or logicals
>
>
>> error: called from
>> leasqr at line 329 column 9
>>
>
Since it's giving that error pointing out line 329 within leasqr I don't think he's shadowing the function.
Bharath:
ok, I removed the definitions for x, p and pin from the function f, as there's no point in passing it values of f if the function is setting them anyway. So:
function F = f(x,p)
typeinfo(x)
m(1,1)=20;
for i=1:length(x)
m(i+1,1)=((p(1)-m(i,1))/(p(2)*p(3))+((p(4)*x(i,2)*(x(i,3)-m(i,1)))+(p(5)*x(i,4))+(p(6)*x(i,5)))/(p(3))+p(7))+m(i,1);
endfor
F=m(2:end);
endfunction
>>x = [19.54533 0.00000 21.33717 0.00000 0.60320;
19.30000 0.00000 21.28817 0.00000 0.66134;
19.30000 0.00000 20.55900 0.00000 0.71948;
19.13750 0.00000 21.21883 0.00000 0.77762;
19.00000 0.00000 20.68650 0.00000 0.83576];
>> p = [20, 1, 100, 1, 0.001, 1, 0.01];
>> pin=[p(1);p(2);p(3);p(4);p(5);p(6);p(7)];
>> f(x,p)
ans = matrix
ans =
20.016
20.032
20.049
20.067
20.084
>> [L,p,cvg,iter]=leasqr(x,y,pin,f)
error: 'y' undefined near line 1 column 24
error: evaluating argument list element number 2
oops, forgot to ask for testvalues for y.
for now, just using y = f(x,p)
>> [L,p,cvg,iter]=leasqr(x,f(x,p),pin,f)
>> leasqr(x,f(x,p),pin,f)
ans = matrix
error: 'x' undefined near line 2 column 12
error: called from
f at line 2 column 3
error: evaluating argument list element number 1
error: called from
f at line 2 column 3
error: evaluating argument list element number 4
NOW, adding the @f as I said before (it also works with 'f' ):
>> [L,p,cvg,iter]=leasqr(x,f(x,p),pin,@f)
ans = matrix
ans = matrix
ans = matrix
ans = matrix
ans = matrix
ans = matrix
ans = matrix
ans = matrix
ans = matrix
ans = matrix
ans = matrix
ans = matrix
ans = matrix
ans = matrix
ans = matrix
ans = matrix
L =
20.016
20.032
20.049
20.067
20.084
p =
2.0000e+001
1.0000e+000
1.0000e+002
1.0000e+000
1.0000e-003
1.0000e+000
1.0000e-002
cvg = 1
iter = 1
so, it works, and once the input arguments were straightened out you get a result. Since I don't know what your data and function really are, I'm not in a position to evaluate whether the output is 'right'. | 932 | 2,454 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2023-40 | latest | en | 0.504728 |
https://devforum.roblox.com/t/function-has-extreme-delay-or-just-sometimes-doesnt-work-as-intended/1928230 | 1,679,458,663,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943749.68/warc/CC-MAIN-20230322020215-20230322050215-00058.warc.gz | 269,687,803 | 5,223 | # Function has extreme delay or just sometimes doesnt work as intended
Idk what happened i swear it was working normally before but now it just sometimes doesnt work / has a long delay (3-4 seconds) on it
``````local ExplosivePart = script.Parent
local Touched = true
local Debug = true
local Vx = 0
local Vy = 0
local Vz = 0
local StartedFall = false
local EndedFall = false
local StartTime = tick()
local EndTime = tick()
local StartPos = Vector3.zero
local TouchedPos = Vector3.zero
local Force
ExplosivePart.Touched:Connect(function(Part)
Touched = true
TouchedPos = ExplosivePart.Position
end)
local V = ExplosivePart.AssemblyLinearVelocity
local X1 = if math.abs(V.X) > 0 then math.abs(V.X) else 0
local Y1 = if math.abs(V.Y) > 0 then math.abs(V.Y) else 0
local Z1 = if math.abs(V.Z) > 0 then math.abs(V.Z) else 0
if X1 > Vx then Vx = X1 end
if Y1 > Vy then Vy = Y1 end
if Z1 > Vz then Vz = Z1 end
if Vy > 10 or Vx > 10 or Vz > 10 then
if not StartedFall then StartTime = tick(); StartPos = ExplosivePart.Position; StartedFall = true end
end
if StartedFall then
print("d")
if Touched then
local hit = false
local V2 = ExplosivePart.Velocity
if Vx-V2.X > Vx*0.80 then hit = true end
if Vy-V2.Y > Vy*0.80 then hit = true end
if Vz-V2.Z > Vz*0.80 then hit = true end
if hit then
EndTime = tick()
local Vctr = (Vx+Vy+Vz)
local Mass = ExplosivePart.Mass
local Distance = (TouchedPos - StartPos).Magnitude
if Distance < 10*(Mass/4) then Distance = 10*(Mass/4) end
Force = Mass * Vctr / (2 * Distance)
if Force >= 5 then
ExplosivePart:FindFirstChildWhichIsA("Humanoid"):TakeDamage(Force)
end
Vx = 0
Vy = 0
Vz = 0
Distance = nil
StartedFall = false
Touched = false
hit = false
if Debug then
print(Force)
print("Part fell for " .. EndTime - StartTime .. " Seconds")
end
else
Touched = false
end
end
end
end
`````` | 583 | 1,831 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.436472 |
http://jefclaes.be/2016/03/functional-one-liner-for-running-totals.html | 1,701,542,718,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100448.65/warc/CC-MAIN-20231202172159-20231202202159-00606.warc.gz | 25,853,307 | 3,553 | Jef Claes
On software and life
Visualizing some data earlier this week I had to compute the running total of a sequence of numbers.
For example, if the input sequence was [ 100; 50; 25 ] the result of the computation would be a new sequence of [ 100; 150; 175 ].
Muscle memory made me take a procedural approach, which works, but made me wonder if I could get away with less lines of code and without mutable state.
``````static IEnumerable<int> ToRunningTotalProcedural(IEnumerable<int> sequence)
{
var runningTotal = 0;
foreach (var item in sequence)
{
runningTotal += item;
yield return runningTotal;
}
}
``````
Although C# doesn’t try very hard to push a functional approach, the BCL does give you some useful tools.
The first thing that comes to mind is using IEnumerable’s Aggregate function, which will apply a function over each item in the sequence and will pass the aggregated partial result the next time the function is applied. Each time the function is applied, we can take the last item (if it exists) of the aggregated partial result and add the current item’s value to it, and append that sum to the aggregated partial result.
``````static IEnumerable<int> ToRunningTotalFunctionalAggregate(IList<int> sequence)
{
return sequence.Aggregate(
ImmutableList<int>.Empty,
(acc, item) => rt.Add(acc.Any() ? acc.Last() + item : item));
}
``````
Another more compact - but less efficient approach - I could think of, is using the index of each element in the sequence, to take subsets and to sum their values.
``````static IEnumerable<int> ToRunningTotalLinqTake(IList<int> sequence)
{
return sequence.Select((item, i) => sequence.Take(i + 1).Sum());
}
``````
Running out of ideas, I ported F#‘s Scan function which allows more compact code, without giving up efficiency. This function, similar to the Aggregate function, applies a function over each item in the sequence. However, instead of passing the aggregated partial result each time the function is applied, the value of the last computation is passed in, to finally return the list of all computations.
With a bit of good will, C# allows you to be more functional too. | 487 | 2,150 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | latest | en | 0.846122 |
https://jp.maplesoft.com/support/help/errors/view.aspx?path=geometry%2FMinorAxis | 1,721,722,647,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518029.81/warc/CC-MAIN-20240723072353-20240723102353-00468.warc.gz | 287,432,491 | 21,456 | MinorAxis - Maple Help
geometry
MinorAxis
find the length of the minor axis of a given ellipse
Calling Sequence center(e)
Parameters
e - an ellipse
Description
• The routine returns the length of the minor axis of the given ellipse e.
• The line segment through the center and perpendicular to the major axis is called the minor axis.
• The command with(geometry,MinorAxis) allows the use of the abbreviated form of this command.
Examples
> $\mathrm{with}\left(\mathrm{geometry}\right):$
> $\mathrm{ellipse}\left(\mathrm{e1},2{x}^{2}+{y}^{2}-4x+4y=0,\left[x,y\right]\right):$
> $\mathrm{MinorAxis}\left(\mathrm{e1}\right)$
${2}{}\sqrt{{3}}$ (1) | 182 | 664 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 4, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.542554 |
uwdatasci.wordpress.com | 1,531,951,261,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590329.62/warc/CC-MAIN-20180718213135-20180718233135-00564.warc.gz | 771,279,890 | 22,485 | # test
This is a test
Is this going to work?
# Spam Filter, Algo Trading, Nonparametric methods etc.
Today’s meeting covered aspects of spam filter, algorithmic trading and clustering.
Gobi started by talking about writing a spam filter using logistic regression. The idea is to use character 4-grams as features, and hash it into p buckets, where p is some large enough prime. Then, use the number of 4-grams landing in each bucket as a predictor in a logistic regression model. This technique apparently did better than Google’s spam filter technology. Another interesting thing is that the researches who came up with this idea used google search results for “spam” subjects and “non-spam” subjects as training data.
David discussed his work for the Rotman International Trading Competition and Market Microstructure. One problem is that when you’re ordering a large number of shares, the market isn’t liquid enough to handle all your orders without drastically increasing price of the shares. It is a challenge to figure out how to buy/sell a large number of stocks in the optimal way possible. Another problem is related to market making: posting both a bid and ask at different prices to pocket the spread. The difficulty here is figuring out what the bid and ask prices should be, and how to handle the risk of the market moving.
John discussed his work at Pitney Bowes mining business related data using economical, demographic, and geographical aspects of a business. It spurred discussions in cluster detection, visualization of high-dimensional data, and cluster validation.
Samson talked briefly about non-parametrical methods and a review of bootstrapping: using empirical data to estimate the probability density function non-parametrically to estimate non-parametric quantities such as mean and variance, and using sampling with replacement to construct an estimate of the variance.
As always meetings are on Mondays at 4:30 in M3-3109.
# WEEK 3: GRAPHING MULTIVARIATE DATA
Visualizing and making sense of multivariate data geometrically in the Euclidean space is very challenging to say the least when more than three variables are in question.
This week, the introduction of Chernoff Faces as a tool for graphing multivariate data made dealing with so many variables bearable, and making sense of the data as a whole more intuitive. Different data dimensions map onto different facial features. The example (taken from “FlowingData”, http://flowingdata.com/2010/08/31/how-to-visualize-data-with-cartoonish-faces/) below will show how Chernoff Faces are constructed and how easier it makes data interpretation.
Parallel Coordinates was another graphical representation of multivariate data that was introduced. In this method, each variable is assigned its own vertical axis and each axis is parallel to the other. A horizontal line between any axes implies positive correlation while an intersection implies negative correlation. It seems to be a good tool for measuring the association between variables.
Star plotting was briefly mentioned at the meeting, and the Iris data set was used to calculate the chances of a reading ( that isn’t isolated nor distinctively associated with any near by clusters of readings) being apart of any cluster surrounding it
# Week 2: d3.js and line detection
This week’s meeting focused on talks given by Lisa and Jeeyoung.
Lisa talked first, discussing her latest project using Data-Driven Documents (d3). The dataset of choice was the Iris dataset, that which any R enthusiast will be familiar with. It contains five variables: Sepal.Length, Sepal.Width, Petal.Length, Petal.Width, and Species.
For those unfamiliar with d3.js, it is a visualization library that provides functionality to bind data to objects. The advantages of using d3.js is that it is extremely fast and flexible.
Here is a simple example of selecting a circle and updating it with new coordinates and size :
var circle = svg.selectAll(“circle”)
.attr(“cy”, 90)
.attr(“cx”, 30)
.attr(“r”, 40);
But what if we want to make it more interesting. Consider a data set of size two (i.e [20, 35]). D3.js makes it easy to add a new circle and visualize the data in such a way that the size and x-coordinate of each circle reflects each data point.
var circle = svg.selectAll(“circle”)
.data([20, 35])
.enter().append(“circle”)
.attr(“cy”, 90)
.attr(“cx”, function(d) {return d;})
.attr(“r”, function(d) {return d;});
Lisa’s project explored multiple visualization strategies including an easy-to-use interface to change the axis of a scatter plot.
Jeeyoung discussed edge detection and line detection. In edge detection, a user applies an algorithm (that uses a threshold on the directional derivatives) to produce an image with the original’s edges. One of the more common edge detection algorithms is Canny edge detection (see below: source Wikipedia).
From there, you can apply line detection. It is easier to do this by mapping each point along the edges to a line in the parameter space (slope-intercept coordinate plane) and looking for intersections. While the exact details will not be mentioned, the transformation to be used is the Hough Transformation.
Note you can implement both using Matlab commands : edge(.) and hough(.)
As usual come join us in the Stats Club Office every Monday at 4:30pm!
Today, the Data Science team kicked off its first meeting of 2012! Welcome to Arthur, David, Paul, and Will for joining the cause! The meeting covered exciting new topics ranging from internet ad bidding to Kaggle competition.
Stay Alert! Kaggle Competition: David gave a walk-through of the Stay Alert! Ford challenge on Kaggle. He described two main findings which propelled his model to finish 6th in the overall rankings. The first was a data visualization method which, in his case, zeroed out the randomly added datapoints. The second was the discovery of interaction effects between variables.
The next meeting will be held next Monday at 4:30pm in the Stats Club Office M3 3109.
# Gaussian Mixture Model, K-means algorithm, and how to solve them.
Introduction
Mixture model is a weighted combinations of probability distributions. Mixture model is a powerful and well-understood tool for various problems in artificial intelligence, computer vision, and statistics. In this post, we will examine Gaussian mixture models, and algorithms to solve them.
Let’s first introduce Gaussian Mixture Model. Let $(\mu_j, \sigma_j)$ be a collection of Gaussian distributions, with some associated weights $\pi_j \in [0,1]$. Let $(x_i)_{i=1}^n$ be set of observations that are generated by the above distributions.
$P(x_i|\sigma_j) \sim \pi_i G(\mu_j, \sigma_j)$
Figure 1. Mixture of two Gaussian distributions, with $\pi = (0.25, 0.75)$, $\mu = (0, 5)$, $\sigma=(1,3)$.
We call this model Mixture of Gaussians. The parameter $\pi$ is called the mixing portion of the Gaussians.
By solving Gaussian Mixture Model, we are given a set of observed points $X=(x_i)_{i=1}^n$ and we want to find the model parameters $(\pi, \mu, \sigma)$ that maximizes the posterior probability $P(X | \pi,\mu,\sigma)$.
Figure 2. Histogram of 10,000 points generated by the above Gaussian mixture model. We want to find the model parameters which maximizes the posterior probability of generating those points.
K-means clustering
K-means clustering solves a different problem, but K-means problem gives an insight over how we can solve the mixture of Gaussian problem. In K-means clustering, a set of observed points $(x_i)_{i=1}^{n}$, a positive integer $k$ are given. We want to cluster the points into $k$ clusters $(S_j)_{j=1}^{k}$ so that it minimizes the within cluster sum of squares (WCSS).
$\underset{S_x}{argmin}\sum_{j=1}^{k} \sum_{x \in S_j} || x - \mu_j ||^2$
$\mu_j = \frac{\sum_{x \in S_j} x}{\#S_j}$ is the mean of the cluster $S_j$.
Figure 3. Example of a K-means clustering in 2 dimensions, K = 2. The observed points are clustered into the two clusters. The X mark denotes the centre of the cluster.
Figure 4. Another example of a K-means clustering in 2 dimensions, K=4.
Expectation Maximization (EM)
Expectation Maximization is an iterative algorithm for solving maximum likelihood problems. Given the set of observed data X, generated by the probability model with parameter 𝜃, we want to find the parameter 𝜃 that maximizes the posterior probability $P(X|\theta)$. In K-means clustering,
• $X$ – observed data $(x_i)_{i=1}^{n}$
• $\theta$ – the clustering of the observed data points $(S_j)_{j=1}^{k}$, and their associated centres $\mu_j$.
EM Algorithm starts with an initial hypothesis for the parameter 𝜃 and iteratively calculate the posterior probability $P(X|\theta)$, and re-calculates the hypothesis 𝜃 each step. The details are going to be different between each EM algorithms, but the following are the approximate steps.
2. (Expectation step) Calculate the posterior probability $P(X | \theta)$.
3. (Maximization step) Categorize the observed data according to the probability, and update the hypothesis accordingly.
4. Repeat this process until the hypothesis converges.
Algorithm – Hard K-means
We assume that the $P(x_i|S_j)$, the likelihood of point $x_i$ belonging to the cluster centered at the point $\mu_j$ is 1 iff $\mu_j$ is the closest cluster.
$P(x_i|S_j) = 1\ if\ j = \underset{k}{argmin} || x_i - \mu_k ||, 0\ otherwise$
During the E-step, the distances $|| x_i - \mu_j ||$ are calculated to determine $P(x_i|S_j)$. During the M-step$S_j$ are determined by clustering each $x_i$ to the cluster closest to it. Finally, each $\mu_j$ are re-calculated by taking the mean of the points in $S_j$.
$\mu_j = \displaystyle\sum\limits_{x \in S_j}\frac{x}{\#S_j}$
Figure 5. Example of a K-means algorithm, for K=2. Initial cluster centres are randomly chosen from the observed points. Each iteration, the observed points are categorized to the nearest cluster centres, and the cluster centres are re-calculated. This process is repeated until the cluster centres converge.
Algorithm – Soft K-means
In Hard-K means. the observed points can only belong in a single cluster. However, it may be useful to consider the probability $P(x_i|S_j)$ during the computation of $\mu$. This is especially true for the points near a boundary between two clusters.
We want to calculate $\mu_j$ via the weighted average of $x_i$ with $P(x_i|S_j)$.
We assume that the likelihood $P(x_i|S_j)$ has exponential distribution with the stiffness factor $\beta > 0$.
$P(x_i|S_j) \sim e^{-\beta || x_i - \mu_j ||}$
This way, $P(x_i|S_j)$ is still monotonically decreasing with respect to the distance away from the centres, but the probability is panellized for the higher distance.
Similar to the Hard K-means algorithm, $P(x_i|S_j)$ is obtained by normalizing this value.
$P(x_i|S_j) = \frac{e^{-\beta||x_i-\mu_j||}}{\sum_l e^{-\beta || x_l - \mu_j ||}}$
$\mu_j$ is obtained by the weighted averages of all $x_i$.
$\mu_j = \frac{\sum_i P(x_i|S_j) x_i}{\sum_i P(x_i|S_j)}$
One thing to note is that Hard K-mean algorithm is equivalent to a Soft K-mean algorithm as $\beta \rightarrow \infty$.
Algorithm – Gaussian K-means
K-means algorithms are great, but the algorithm only reveals information about the cluster membership. However, we can modify the EM algorithm to calculate
We come back to the original assumption, that the likelihood $P(x_i|S_j)$ follows a Gaussian distribution.
$P(x_i|S_j) \sim \pi_j G(||x_i - \mu_j||, \sigma_j)$
Where $G(\bullet, \bullet)$ is the Gaussian probability density distribution. The actual probably is calculated as following.
$P(x_i|S_j) = \frac{\pi_j G(||x_i-\mu_j||, \sigma_j)}{\sum_k \pi_k G(||x_i-\mu_k||, \sigma_k)}$
We have to re-calculate three parameters, $\mu, \sigma, \pi$. Similar to the Soft K-mean algorithm, μ is the weighted averages of all $x_i$. $\pi$ is the normalized ratio fo the sums $\sum_i P(x_i|S_j)$. 𝜎 is the weighted standard deviation.
$\mu_j = \frac{\sum_i P(x_i|S_j) x_i}{R_k}$.
$\sigma_j^2 = \frac{\sum_i P(x_i|S_j)|| x_i - \mu_j || }{I R_k}$
$\pi_j = \frac{R_j}{\sum_k R_k}$
$R_j = \sum_i P(x_i|S_j)$
Where I is the dimensionality of x.
Figure 6. Points sampled from a mixture of two Gaussians. Top left plot is the data points generated by the model parameters $\pi=(\frac{1}{2},\frac{1}{2})$, $\sigma_x = \sigma_y = (0.3, 1)$ $\mu_x = (-2, 0), \mu_y = (0, 0)$. The other plots show the results of Hard K-means, Soft K-means, and Gaussian K-means.
Figure 7. Another example of Gaussian K-means algorithm, with 4 clusters.
References
# Graphing Russia’s Election Fraud
Following Russia’s parliamentary elections on December 4, a link was posted to Reddit reporting an impossibly high turnout (99.51%) and near unanimous support (99.48%) for Putin’s ruling party, United Russia, in the last location one would expect it: the republic of Chechnya. Even if relations with the secessionist region have improved since the Second Chechen War, both the turnout and United Russia’s vote share are a complete joke. This absurdity prompted a more thorough examination of all regions, many of which were also plagued by irregularities. In this post, I will give some detailed visualizations of both region- and precinct- level election data, and point out some highly likely instances of fraud.
# Visualizing 4+ Dimensions
As a student of pure math, I am often asked the question of how to visualize high dimensional objects. This question isn’t as important to pure mathematicians as it is to statisticians, so I write about it here.
…I’ve never had to visualize anything high-dimensional in my pure math classes. Working things out algebraically is much nicer, and using a lower-dimensional object as an example or source of intuition usually works out — at least at the undergrad level.
But that’s not a really satisfying answer, for two reasons. One is that it is possible to visualize high-dimensional objects, and people have developed many ways of doing so. Dimension Math has on its website a neat series of videos for visualizing high-dimensional geometric objects using stereographic projection. The other reason is that while pure mathematicians do not have a need for visualizing high-dimensions, statisticians do. Methods of visualizing high dimensional data can give useful insights when analyzing data….
# Eigenfaces, Data Visualizations and Clustering Methods
Today’s meeting covered everything from classifying whether a certain photo is of Justin Bieber using linear algebra, to understanding how call steering portals really work.
Eigenfaces: Samson explained through principal component analysis, how eigenfaces can be used in face recognition.
Data Visualization: Samson also talked about information density, the data-ink ratio and how to condense visualizations without sacrificing information. He described several methods including sparklines, one of Tufte’s inventions, as well as why the “ideal” length of a x-y graph is dependent on the rate of change of the trendline.
Semantic Clustering: Konrad gave an overview of some of the work he did at Nuance involving semantic and k-means clustering algorithms. He described how the randomness of customer calls, among other variables, makes it difficult to know exactly the appropriate semantic tags one should use in large data sets.
Next meeting will be next Tuesday Nov 29th at 3pm in the Stats Club Office M3 3109.
# Stanford Online Classes in Winter
Once again Stanford is offering online courses next term. Here are some that might be relevant:
Other classes offered are: | 3,781 | 15,633 | {"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": 68, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2018-30 | latest | en | 0.916921 |
http://www.puzzles.com/puzzleplayground/CountingTriangles/CountingTrianglesSol.htm | 1,501,166,592,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549428300.23/warc/CC-MAIN-20170727142514-20170727162514-00249.warc.gz | 540,457,755 | 3,839 | Counting the Triangles Solution
Home / Puzzle Playground / Puzzles / Visual /
There are seven groups of triangles shown in the diagrams above. Each group consists of exactly five triangles with every triangle rotated 72 degrees around the center of the pentagon; one triangle from every group is highlighted in the respective diagram. So the total number of the triangles in the pentagon is 7x5=35.
Last Updated: January 21, 2008 | Posted: June 23, 2001 | 106 | 460 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2017-30 | latest | en | 0.907498 |
https://www.spactronics.com/electronic-component-inductor-l/?v=6c8403f93333 | 1,652,679,353,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662509990.19/warc/CC-MAIN-20220516041337-20220516071337-00730.warc.gz | 1,196,860,723 | 13,897 | ## ELECTRONIC COMPONENT —- INDUCTOR (L)
In this article the information consists of basic knowledge related to the electronic component Inductor. Inductor is the most basic and important electronic component that is invented by Joseph Henry. The Inductor is represented by L in the electronic terminology. The inductance of an Inductor is measured in SI unit as Henry.
INDUCTOR REPRESENTATION IN THE CIRCUIT
WORKING OF AN INDUCTOR
Inductor works on the principle of Electromagnetic Induction.
The inductor is a conductor that is turned into the coiled structure.
Inductor stores the energy in the form of magnetic energy.
According to the law of Electromagnetic Induction, current carrying conductor induces the magnetic field around it.
When the inductor is in the form of coiled structure the magnetic field lines approach in the closest distance and the density of the field lines is high inside the coil.
Since magnetic field lines cannot intersect or coincide over one another, therefore they ensure to move in the closest distance where the density of magnetic field is high.
In this way the Inductor stores the energy in the form of magnetic field.
PROPERTIES AND IMPORTANCE OF INDUCTOR
In case of inductors the current lags behind the voltage by 90 degrees.
The current lags in Inductors because it takes some time to store the current and supply it to the circuit.
Inductor store and supply the energy in very short period of time.
Inductors are the passive element that stores the electrical energy in the form of magnetic energy.
Inductors are generally designed to block AC and allow the DC Current to pass.
Inductors in combination with capacitors acts as the circuit that be used for Tuning and Resonance principle
The equivalent inductance for the inductors that are connected in series in the circuit is the sum of the inductances.
The equivalent inductance for the inductors that are connected in parallel in the circuit is the reciprocal sum of the inductances.
Inductance of the coil depends on the number of turns of the coil, the degree of overlap, area of cross section and the material of the core inside the coil used.
Inductance can be increased by increasing the
1. Number of turns of the coil.
2. Cross section area of the coil.
3. The degree of overlap that is by using tight wound coil.
Thus in this way the principle of inductor helps to store the energy in the form of magnetic field and enhances its importance as an electronic component and helps in the invention and development of different circuits and helps to solve the complex issues related to the electronics domain and even promotes its use in the electronic field related to the robotics. | 566 | 2,716 | {"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} | 2.765625 | 3 | CC-MAIN-2022-21 | latest | en | 0.923187 |
http://bobbyroel.com/horario-forex/black-scholes-put-call-parit-t.php | 1,516,685,659,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891750.87/warc/CC-MAIN-20180123052242-20180123072242-00479.warc.gz | 47,542,836 | 7,092 | # Black scholes put call parit t. To help understand the Black-Scholes formula for call and put options we 2. BLACK-SCHOLES and the payoff at maturity to a digital put option is: pb(T) = 1 if S(T) ≤ K. 0 if S(T) > K. We now show how to value the digital call option. The end price ST There is a simple condition for put call parity for digital options.
## Black scholes put call parit t. M. Spiegel and R. Stanton, 3. Put-Call parity and early exercise s Put-call parity: C = S + P – K / (1+r)T s Put-call parity gives us an important result about exercising American call options. s In words, the value of a European (and hence American) call is strictly larger than the payoff of exercising it today. (). ().KS. r1KS.
Stochastic Processes and Advanced Mathematical Finance. These pages are prepared with MathJax. MathJax is an open source JavaScript display engine for mathematics that works in all browsers. With the additional terminal condition V S , T given, a solution exists and is unique. We observe that the Black-Scholes is a linear equation, so the linear combination of any two solutions is again a solution. By uniqueness, the solutions must be the same, and so.
This relationship is known as the put-call parity principle between the price C of a European call option and the price P of a European put option, each with strike price K and underlying security value S. One can then calculate that the price of a call option with these assumptions is 1 1. At expiration this portfolio always has a value which is the strike price.
This example portfolio has total value 1 0 0. This is an illustration of the use of options for hedging an investment, in this case the extremely conservative purpose of hedging to preserve value.
Consider buying a put and selling a call, each with the same strike price K. But this payout is exactly what we would get from a futures contract to sell the stock at price K. This replicates the futures contract, so the future must have the same price as the initial outlay. Therefore we obtain the put-call parity principle:. Another way to view this formula is that it instructs us how to create synthetic portfolio. This same principle of linearity and the composition of more exotic options in terms of puts and calls allows us to create synthetic portfolios for exotic options such as straddles, strangles, and so on.
As noted above, we can easily write their values in closed form solutions. Knowing any two of S , C or P allows us to calculate the third. Of course, the immediate use of this formula will be to combine the security price and the value of the call option from the solution of the Black-Scholes equation to obtain the value of the put option:.
For the sake of mathematical completeness we can write the value of a European put option explicitly as:. Using the symmetry properties of the c. Value of the put option at maturity.
Now we use the Black-Scholes formula to compute the value of the option before expiration. Value of the call option at various times. Notice two trends in the value from this graph:. We can also plot the value of the put option as a function of security price and the time to expiration as a value surface.
Value surface from the Black-Scholes formula. This value surface shows both trends. This section is adapted from: Financial Derivatives by Robert W. Michael Steele, Springer, New York, , page The result can be plotted as functions of the security price as done in the text. In particular, the calculation of d 1 and d 2 uses broadcasting, also called binary singleton expansion, recycling, single-instruction multiple data, threading or replication. R script for Black-Scholes pricing formula for a put option.
Octave script for Black-Scholes pricing formula for a put option. Institute of Finance, Stochastic Calculus and Financial Applications. I check all the information on each page for correctness and typographical errors. Nevertheless, some errors may occur and I would be grateful if you would alert me to such errors.
Your use of the information from this website is strictly voluntary and at your risk. I have checked the links to external sites for usefulness. Links to external websites are provided as a convenience.
I do not endorse, control, monitor, or guarantee the information contained in any external website. Use the links here with the same caution as you would all information on the Internet. Information on this website is subject to change without notice. Security Call Put Portfolio 80 0 20 90 0 10 0 0 0 0 Security, call and put option values at expiration.
For a particular scripting language of your choice, modify the scripts to create a function within that language that will evaluate the Black-Scholes formula for a put option at a time and security value for given parameters.
For a particular scripting language of your choice, modify the scripts to create a script within that language that will plot the Black-Scholes solution for V P S , t as a surface over the two variables S and t.
More...
1433 1434 1435 1436 1437 | 1,088 | 5,071 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-05 | latest | en | 0.894269 |
https://www.examyear.com/data-structures-mcq-questions-and-answers/ | 1,695,434,028,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506429.78/warc/CC-MAIN-20230922234442-20230923024442-00325.warc.gz | 839,999,846 | 20,906 | # Data Structures MCQ Questions and Answers
The Free download links of Data Structures MCQ Questions and Answers Papers enclosed below. Candidates who are going to start their preparation for the Data Structures MCQ papers can use these links. Download the Data Structures MCQ Papers PDF along with the Answers. Data Structures MCQ Papers are updated here. A vast number of applicants are browsing on the Internet for the Data Structures MCQ Question Papers & Syllabus. For those candidates, here we are providing the links for Data Structures MCQ Papers. Improve your knowledge by referring the Data Structures MCQ Question papers.
## MCQ Questions and Answers on Data Structures
1. The number of vertices of odd degree in a graph is
(a) always even
(b) either even or odd
(c) always odd
(d) always zero
2. A graph in which all nodes are of equal degree is known as
(a) Multigraph
(b) regular graph
(c) non regular graph
(d) complete graph
3. A vertex of degree one is called as
(a) Pendant
(b) null vertex
(c) isolated vertex
(d) coloured vertex
4. The maximum degree of any node in a simple graph with n vertices is
(a) n – 1
(b) n/2
(c) n
(d) n – 2
5. Two isomorphic graphs must have
(a) the same number of vertices
(b) an equal number of vertices
(c) the same number of edges
(d) all of the above
6. The terminal vertices of a path are of degree
(a) One
(b) Zero
(c) Two
(d) more than four
7. If there exists at least one path between every pair of vertices in a graph, the graph is known
(a) complete graph
(b) disconnected graph
(c) connected graph
(d) Eular graph
8. A simple graph with n vertices and k components can have at most
(a) n edges
(b) n-k edges
(c) (n-k) (n-k-1)/2 edges.
(d) (n-k) (n-k + 1)/2 edges
9. A given connected graph G is a Euler graph if and only if all vertices of G are of
(a) same degree
(b) odd degree
(c) even degree
(d) different degree
10. A circuit in a connected graph which includes every vertex of the graph is known as
(a) Eular
(b) Hamiltonian
(c) Unicursal
(d) Clique
11. The length of a Hamiltonian path (if exists) in a connected graph of n vertices is
(a) n-1
(b) n + 1
(c) n
(d) n/2
12. A simple graph in which there exists an edge between every pair vertices is called a (an)
(a) incomplete graph
(b) Eular graph
(c) complete graph
(d) planner graph
13. The total number of edges in a complete graph of n vertices is
(a) n
(b) n(n+1)/2
(c) n²/2
(d) n(n-1)/2
14. A tree with n nodes has
(a) n/2 edges
(b) n edges
(c) n – 1 edges
(d) n + 1 edges
15. The number of paths between any pair of nodes in a tree on n nodes is
(a) 0
(b) 1
(c) (n-1)
(d) n
16. A complete graph with five vertices is
(a) nonplanar
(b) a non-regular graph
(c) planar
(d) a tree
17. Kuratowski’s first graph is the nonplanar complete graph with the smallest number of vertices. The number of vertices is
(a) 4
(b) 6
(c) 5
(d) 7
18. f a graph G does not contain either Kuratowski’s two graphs or any graph isomorphic to them, the graph G is then
(a) planar graph
(b) Eular graph
(c) non planar graph
(d) regular graph
19. The number of regions is a connected planar graph with n vertices and e edges is
(a) n + e
(b) e-n-2
(c) e-n
(d) e/2
20. The number of circuits in a tree with n nodes is
(a) zero.
(b) n-1
(c) 1
(d) n/2
21. A graph is a tree if and only if it
(a) is completely connected
(b) contains a circuit
(c) is minimally connected
(d) is planar
22. The minimum number of spanning trees in a connected graph with n nodes is
(a) 1
(b) n-1
(c) 2
(d) n/2
23. The rank of a graph with n vertices, e edges and k components is
(a) n
(b) n-1
(c) e-n+k
(d) n + k
24. The nullity of a graph with n vertices, e edges and k components is
(a) n
(b) n-k
(c) e-n + k
(d) n + k
25. The number of elements (edges) in a cutset of a tree with n vertices is
(a) 1
(b) n/2
(c) n-1
(d) n
26. Let a graph G has edge connectivity and node connectivity. Then
(a) α < ß
(b) α = ß
(c) α ≥ ß
(d) α ≤ ß
27. What is the edge connectivity of a complete graph with n vertices?
(a) 1
(b) n + 1
(c) n-1
(d) n(n+1)/2
28. If A(G) is an incident matrix of a connected graph G with n vertices, the rank of A(G) is
(a) 1
(b) n-1
(c) 2
(d) n
29. If G is a disconnected graph with n vertices and k components, the rank of the incident matrix A(G) of the graph is
(a) n-k-1
(b) n + k-1
(c) n-k
(d) n + k
30. A graph with n vertices and n-1 edges that is not a tree, is
(a) Connected
(b) Eular
(c) Disconnected
(d) a circuit
31. If B is a circuit matrix of a graph with k components, e edges and n vertices, the rank of B is
(a) n-k
(b) e-n+k
(c) e-n-k
(d) e+n-k
32. Consider a graph G having cutset matrix C(G) and incidence matrix A(G). The rank of C(G) is
(a) same as the rank of A(G)
(b) more than the rank of A(G)
(c) independent of the rank of A(G)
(d) less than the rank of A(G)
33. Let X be the adjacency matrix of a graph G with no self loops. The entries along the principle diagonal of X are
(a) all zeros
(b) both zeros and ones
(c) all ones
(d) different
34. A graph consisting of only isolated n vertices is
(a) k/2
(b) k
(c) k-1
(d) 1
35. If a graph requires k different colours for its proper colouring, then the chromatic number of the graph is
(a) 1-chromatic
(b) 3-chromatic
(c) 2-chromatic
(d) n-chromatic
36. A graph with one or more edges (wothout self loop) is a least
(a) 2-chromatic
(b) n-chromatic
(c) 1-chromatic
(d) 3-chromatic
37. A complete graph with n vertices is
(a) 2-chromatic
(b) (n-1) chromatic
(c) n/2 chromatic
(d) n-chromatic
38. 1f d_{max} is the maximum degree of the vertices in a graph G, the chromatic number of G is
(a) equal to d_{max}
(b) less than or equal to d_{max}
(c) greater than d_{max}
(d) less than or equal to d_{max} + 1
39. If the vertex set V of a graph can be decomposed into two subsets V₁ and V₂ such that every edge in G joins a vertex in V₁ with a vertex in V₂, the graph is called
(a) Bipartite
(b) 1-chromatic
(c) tri-partite
(d) V-partite
40. A covering of an n-vertex graph has at least
(a) n/2 edges
(b) [(n-1)/2] edges
(c) [n/2] edges
(d) n edges
41. The number of colors required to properly color the vertices of every planar graph is
(a) 2
(b) 3
(c) 5
(d) 4
42. 1f A(G) is the incidence matrix of a connected digraph of n vertices the rank of A(G) is
(a) n
(b) 2
(c) n/2
(d) n-1
43. The number of different rooted labeled trees with n vertices is
(a) 2^{n-1}
(b) 2^{n}
(c) n^{n-1}
(d) n^{n}
44. Let A = {1, 2, 3, 4, 5} Which of the following sets equal A?
(a) {4, 1, 2, 3, 5}
(b) {3, 2, 4, 1}
(c) {5, 1, 3, 2, 1}
(d) {3, 4, 5}
45. Which of the following sets are empty?
(a) {x|x is a real number and x² – 1 = 0}
(b) {x|x is a real number and x = 2x + 1}
(c) {x|x is a real number and x² + 1 = 0}
(d) {x|x is a real number and x² = -9}
46. A computer company must hire 25 programmers to handle systems programming tasks and 40 programmers for applications programming. Out of these, 10 will be expected to perform tasks of each type. How many programmers must be hired?
(a) 55
(b) 45
(c) 65
(d) 35
47. In how many ways can six people be seated in a circle?
(a) 720
(b) 120
(c) 360
(d) 12
48. In how many ways can six men and six women be seated in a row if any person may sit next to any other person?
(a) 12!
(b) 6!
(c) 12!/2
(d) 6! /4
49. Which of the following tool is used to specify lexical analyzers for a variety of languages
(a) Yaac
(b) TeX
(c) Lex
(d) EQN
50. Any deterministic finite automata for the regular expression (al b)* a(a| b)(a| b)… (a b), where there are n-1 (a| b)’s at the end, must have at least
(a) n states
(b) n^{2} states
(c) 2^{n} states
(d) 2^{2n} states | 2,482 | 7,599 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2023-40 | latest | en | 0.868855 |
https://www.daniweb.com/programming/software-development/threads/313972/nested-c | 1,506,113,728,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818689192.26/warc/CC-MAIN-20170922202048-20170922222048-00586.warc.gz | 787,433,261 | 15,157 | INPUT's:
1) Name
2) Current date (month, day, year)
3) Birthday (month, day, year)
The output will be the person's age in years, months, and days. See example below:
Input name: shane
Input current date:
Month (1-12): 9
Day (1-31): 25
Year (yyyy): 2010
Input birthday:
Month (1-12): 9
Day (1-31): 4
Year (yyyy): 1978
shane, you are now 32 year(s), 0 month(s), and 21 day(s) old.
5
Contributors
7
Replies
10
Views
6 Years
Discussion Span
Last Post by usep
So?
pls try
``````#include "stdio.h"
#include "conio.h"
int ay,yil,gun,a,y,g,guns,ays,yils;
main()
{
a=9; // todays month //
g=27; // todays day
y=2010; // todays year
printf("enter the day ");scanf("%d",&gun);
printf("enter the month:");scanf("%d",&ay);
printf("enter the year:");scanf("%d",&yil);
if(gun>g) {
g=g+30;
a=a-1;
guns=g-gun;
}
else guns=g-gun;
if(ay>a) {
a=a+12;
y=y-1;a
ays=a-ay;
}
else ays=a-ay;
yils=y-yil;
printf("day difference= %d month difference=%d year difference=%d",guns,ays,yils);
getche();
return 0;
}``````
^ Ouch!
pls try
``````#include "stdio.h"
#include "conio.h"
int ay,yil,gun,a,y,g,guns,ays,yils;
main()
{
a=9; // todays month //
g=27; // todays day
y=2010; // todays year
printf("enter the day ");scanf("%d",&gun);
printf("enter the month:");scanf("%d",&ay);
printf("enter the year:");scanf("%d",&yil);
if(gun>g) {
g=g+30;
a=a-1;
guns=g-gun;
}
else guns=g-gun;
if(ay>a) {
a=a+12;
y=y-1;a
ays=a-ay;
}
else ays=a-ay;
yils=y-yil;
printf("day difference= %d month difference=%d year difference=%d",guns,ays,yils);
getche();
return 0;
}``````
Wow you are such a clever person!
no I found this script in turkish website. it is not mine
I want to help usep
I am learning.
In the interests of getting the OP in trouble with his teacher for plagiarism:
``````#include <cctype>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <iostream>
#include <string>
namespace {
const int MONTHS = 12;
const int MONTH_NAME_MAX = 9;
static const char *whitespace = " \t";
static const char *separators = ".-/";
static const char *months_full[] = {
"january" , "february", "march" , "april",
"may" , "june" , "july" , "august",
"september", "october" , "november", "december"
};
static const char *months_abbr[] = {
"jan", "feb", "mar", "apr",
"may", "jun", "jul", "aug",
"sep", "oct", "nov", "dec"
};
int compare_insensitive(const char *a, const char *b)
{
int x, y;
do {
x = std::tolower((unsigned char)*a++);
y = std::tolower((unsigned char)*b++);
} while (x != '\0' && x == y);
return (unsigned char)x - (unsigned char)y;
}
const char *skip_leading(const char *s, const char *match)
{
while (std::strchr(match, *s) != NULL)
++s;
return s;
}
const char *extract_name(const char *s, char *name, size_t limit)
{
/* Assume limit includes space for a null character */
while (--limit > 0 && std::isalpha((unsigned char)*s))
*name++ = (char)std::tolower((unsigned char)*s++);
*name = '\0';
return s;
}
bool extract_numeric(const char*& s, long& result)
{
char *end;
if (!std::isdigit((unsigned char)*s))
return false;
errno = 0;
result = std::strtol(s, &end, 10);
if (errno)
return false;
s = end;
return true;
}
bool is_abbr_name(const char *name, bool validated)
{
bool result = validated;
/* May is the only month the same abbreviated as not */
if (compare_insensitive(name, "may") == 0)
return false;
if (validated)
return true;
for (std::size_t i = 0; i < MONTHS; i++) {
if (compare_insensitive(name, months_abbr[i]) == 0) {
result = true;
break;
}
}
return result;
}
bool is_valid_month(const char *name, long& month)
{
const char **months = months_full;
bool result = false;
if (std::strlen(name) == 3)
months = months_abbr;
for (std::size_t i = 0; i < MONTHS; i++) {
if (compare_insensitive(name, months[i]) == 0) {
if (month != NULL)
month = i + 1;
result = true;
break;
}
}
return result;
}
bool is_valid_date(long year, long month, long day, std::tm*& result)
{
std::time_t date = std::time(NULL);
std::tm *corrected = std::localtime(&date);
if (year < 100) {
/* Allow years without a century (assume the current one) */
int century = ((int)((1900 + corrected->tm_year) / 100) * 100);
year += century;
if (year > 1900 + corrected->tm_year) {
/* The assumption was probably wrong (future date); use the previous century */
year -= 100;
}
}
corrected->tm_year = year - 1900;
corrected->tm_mon = month - 1;
corrected->tm_mday = day;
date = std::mktime(corrected);
corrected = std::localtime(&date);
/* Verify that the corrected date is the same as our desired date */
if (date == (std::time_t)-1 ||
corrected->tm_year != year - 1900 ||
corrected->tm_mon != month - 1 ||
corrected->tm_mday != day)
{
return false;
}
result = corrected;
return true;
}
bool extract_month(const char*& s, long& month, char *name, std::size_t limit)
{
bool result = false;
/* Check for a month name vs. month number */
if (std::isdigit((unsigned char)*s))
result = extract_numeric(s, month) && (0 < month && month <= MONTHS);
else {
s = extract_name(s, name, limit);
result = !std::isalpha((unsigned char)*s) && is_valid_month(name, month);
}
return result;
}
/*
Parse a date string according to <month> <day> <year> ordering
*/
bool parse_mdy(const char *s, std::tm*& date)
{
char name[MONTH_NAME_MAX + 1] = {0};
long month, day, year;
char separator;
if (!extract_month(s, month, name, sizeof name))
return false;
/*
If the month name is used, restrict separators to whitespace,
otherwise restrict separation to *one* of the same character
*/
if (name[0] != '\0') {
/* Allow abbreviated names to end with a dot */
if (*s == '.' && is_abbr_name(name, true))
++s;
}
else if (std::strchr(separators, *s)) {
separator = *s++;
}
if (!extract_numeric(s, day))
return false;
if (name[0] != '\0') {
/* Allow a trailing comma on the day */
if (*s == ',')
++s;
}
else if (*s == separator) {
++s;
}
return extract_numeric(s, year) && is_valid_date(year, month, day, date);
}
}
int main()
{
std::string name;
std::cout<<"Name: ";
if (!getline(std::cin, name))
return EXIT_FAILURE;
std::string date;
std::tm *tm_date;
std::cout<<"Birthday: ";
if (!getline(std::cin, date) || !parse_mdy(date.c_str(), tm_date))
return EXIT_FAILURE;
else {
std::tm bday = *tm_date;
std::time_t now = std::time(0);
tm_date = std::localtime(&now);
if (tm_date->tm_year == bday.tm_year &&
tm_date->tm_mon == bday.tm_mon &&
tm_date->tm_mday == bday.tm_mday)
{
std::cout<< name<<", you are 0 year(s), 0 month(s), and 0 day(s) old.\n";
}
else {
tm_date->tm_mon -= bday.tm_mon;
tm_date->tm_mday -= bday.tm_mday;
now = std::mktime(tm_date);
tm_date = std::localtime(&now);
std::cout<< name <<", you are "
<< tm_date->tm_year - bday.tm_year <<" year(s), "
<< tm_date->tm_mon <<" month(s), and "
<< tm_date->tm_mday <<" day(s) old.\n";
}
}
return EXIT_SUCCESS;
}``````
See? This assignment was a piece of cake! I can't imagine how any beginner would have trouble with it. ;)
Structure declaration for Student info:
- name
- student id #
- program (course)
- year level
- # of courses (subjects)
- credits (units) for each course
- weighted percentage average (wpa)
Sample run:
(If first time to run the program, create a text file named "1stSem_SY1011.txt")
No. of students: 0
Enter the following student information:
Name: Maui
ID #: 123
Program: BSCS
Year: 1
No. of courses: 3
Course 1: CS111
Credits: 3
Course 2: Math 106
Credits: 6
Course 3: PE 1
Credits: 2
WPA: 89.54
No. of students: 1
Enter another (y/n)?
/*------------- WPA calculation ---------------- */
3 x 85 = 255
6 x 90 = 540
2 x 95 = 190
11 985
WPA: 985 / 11 = 89.54
/*----------------------------------------------------------*/
File format for 1stSem_SY1011.txt
Maui 123 BSCS 1
CS111 3 85
Math106 6 90
PE1 2 95
89.54
---------------------------------------------------- | 2,439 | 7,787 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2017-39 | longest | en | 0.557178 |
https://www.my.freelancer.com/projects/php/nutrition-macros-calculator-for-website/ | 1,555,986,138,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578584186.40/warc/CC-MAIN-20190423015050-20190423041050-00528.warc.gz | 755,490,876 | 37,449 | # Nutrition Macros Calculator for website
I'm a personal fitness trainer and I have a Daily Caloric Intake javascript calculator on my website that allows a user to input his: weight, age, gender, height, and activity level ( from a drop-down list of choices) and it calculates his base daily calorie requirements, and depending upon which activity level he selects it recommends a specified calorie level. This part already works well.
I would like to take that specified calorie recommendation and provide a few more recommendations for my web users.
1.) I would like to provide the users with the number of calories per day necessary to lose weight each week in increments of: -.5 lbs, -1 lb, -1.5 lbs, or -2 lbs per week. It's a basic math equation of expending 3500 calories each day over a 1 week period in order to lose 1 lb. So -3500/7 = -500 therefore ; -500 calories per day are required in order to lose 1 lb per week.
So, If my original calculator had provided a specified calorie recommendation of 2000 calories per day based on the information the user provided, the new "1 lb of weight-loss per week" option would inform the user that their recommended calories per day would be reduced to 1500 in order to reach that goal. There are many websites that do this, so examples are prevalent online.
2.) I'd then like to take the number of calories recommended for weight-loss, for instance 1500 in the example above, and calculate the percentages of those calories that should be allocated to the 3 macro nutrient categories of: Carbohydrates, Proteins and Fats based on the desired percentages the user provides. For example; after the calculator states that 1500 calories per day is recommended to lose one pound per week for a user who's normal daily calorie requirement is 2000 calories per day, the user could input the following percentages into a calculator that calculates the number of calories that should be allocated to each group based on the percentage the user provides. Given the example above, if a user inputs 50% for Carbs - this would output a result of 750 calories based on a 1500 total calorie per day allotment. 25% Protein would = 375 calories, and 25% Fat would = 375 calories as well.
3.) Lastly, in addition to the calorie percentage breakdowns for the macros listed above, I would also like the calculator to list the # of grams associated with each percentage. That is also basic math as the number of grams of protein simply equals the number of calories divided by 4. In the above example 750 calories / 4 = 187.5 g protein. The calculation for Carbs is also divided by 4, so 375 calories / 4 = 93.75 g Carbs. And lastly the calculation for fat grams = the calories divided by 9. Therefore 375/9 = 41.67 g Fat
4.) [login to view URL] - Here is the current calculator on my webpage.
5.) Here is a similar version of a macro calculator - [login to view URL]
Please submit a bid for costs and time to complete
Kemahiran: CSS, HTML, Javascript, jQuery / Prototaip, PHP
Tentang Majikan:
( 4 ulasan ) Mckinney, United States
ID Projek: #19004109
vishwanath89
Hey Can deliver in a day's time. Assuming the inputs will change to get the carbs/fats/protein values % from the user?Also i could see the existing site is hosted on [login to view URL] would i be integrating this in wix or i Lagi
\$100 USD dalam sehari
(73 Ulasan)
6.8
## 17 pekerja bebas membida secara purata \$171 untuk pekerjaan ini
ITLove007
How are you? I read your job detail carefully and understand 100%. I am a senior web expert with 7+ years of experiences. I can complete your project on 2 day and 100\$. First day, I have to understand your source Lagi
\$155 USD dalam 3 hari
(66 Ulasan)
7.5
polarjin2017
Hello? How are you? I have seen the project - "Nutrition Macros Calculator for website." I have been working in these fields((CSS, HTML, Javascript, jQuery / Prototype, PHP)) for 7 yrs as a freelancer. I can work Lagi
\$155 USD dalam 3 hari
(93 Ulasan)
6.8
CrazyWebGuru
Hi. I am a full time developer and have high skills in web develop by CSS, HTML, JQuery, PHP & MySQL. So I can complete this project. My price and time is negotiable. Let's discuss details via chat. I'll provide best s Lagi
\$155 USD dalam 3 hari
(78 Ulasan)
6.6
techplusintl
Hi there, Greetings from TechPlus! I've reviewed requirements & ready to develop calculator for Nutrition Macros. 5+ Years Expertise in : CMS (Wordpress, Magento, Shopify, OpenCart, Joomla, Drupal, Prestashop Lagi
\$155 USD dalam 3 hari
(90 Ulasan)
6.0
schoudhary1553
Hello, I have gone through your job posting and become very much interested to work with you. I am an expert in this field. I have already completed several projects like this. For evidence you can see my profile. Lagi
\$250 USD dalam 3 hari
(47 Ulasan)
5.9
stanciuloredana9
Hello! My name is Loredana and i have 5 years of experience in Web Design and Development! I also know SEO and Online Marketing! Contact me for more informations! Thanks!
\$100 USD dalam sehari
(51 Ulasan)
4.9
bluebear1888
✅I have done this kind of project. I have got 5+ years of rich experiences in the field which you want and able to work full-time for you. - Web scrapping in python - Python - Web(Codeignitor, Laravel, JSP, Node.js Lagi
\$250 USD dalam 3 hari
(18 Ulasan)
4.4
finaldragon123
Hello,sir I 'd really like to work with you on this one if possible! I have read your description with great interest. I've gone through the post description carefully and ready to start work on this project. I do Lagi
\$200 USD dalam 3 hari
(9 Ulasan)
3.9
BoyVit85
Dear sir. I have experience build like your project. I think this project is for me. I am confident I can exceed your expectations. I can achieve the results that you are asking for. I would love to work on your p Lagi
\$200 USD dalam 3 hari
(14 Ulasan)
4.1
NewFuture338
Hello, dear. I’ve gone through your project description and would like to submit my candidacy for the same. I can finish your task for 1 days. I think budget is \$200. I’ve over six years of experience working with Lagi
\$277 USD dalam sehari
(1 Ulasan)
2.6
abuakar21
Hi, I'm Abubakar and I'm a JavaScript programmer. I can edit your existing macro and add the necessary improvements you need. It will be an honour to work on your project.
\$40 USD dalam 3 hari
(1 Ulasan)
1.5
evonsystems7
"Hi, Hope you are doing well! Thanks for sharing your project requirement with us. It will be our great pleasure to work on your project. I have checked your requirement, yes we can do it, because we already work on si Lagi
\$208 USD dalam 7 hari
(2 Ulasan)
0.0
ruwanarru
hi let me have the job i will make your project more accurate and proper working website have good experienced in this kind of projects
\$222 USD dalam 5 hari
(0 Ulasan)
0.0
princeariv
hello sir i had done projects like this before. i can do it for you at a reasonable price. i am professional developer as well as designer. i have a strong knowledge of JavaScript,php,c#,asp :)...
\$77 USD dalam 4 hari
(0 Ulasan)
0.0
fska
Im fullstack developer from México city I have a lot programing experience making proyects for companies I can make the service in 2 days but need some specification
\$133 USD dalam 2 hari
(0 Ulasan)
0.0
JorgeMag96
I know exactly what you want, if you hire me I guarantee a 100% satisfaction on the work, and maybe we could implement an api of nutritional information of certain foods to print a diet. Relevant Skills and Experience Lagi
\$222 USD dalam 2 hari
(0 Ulasan)
0.0 | 1,931 | 7,549 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2019-18 | latest | en | 0.905295 |
https://ebs.sakarya.edu.tr/DersDetay/DersinDetayliBilgileri/678/36772?Disaridan= | 1,576,498,908,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540565544.86/warc/CC-MAIN-20191216121204-20191216145204-00456.warc.gz | 346,537,959 | 22,920 | Ders Bilgileri
#### Ders Tanımı
Ders Kodu Yarıyıl T+U Saat Kredi AKTS
ADSORPTION AND SURFACE AREA KIM 563 0 3 + 0 3 6
Dersin Dili Türkçe Dersin Seviyesi Yüksek Lisans Dersin Türü SECMELI Dersin Koordinatörü Prof.Dr. MURAT TEKER Dersi Verenler Prof.Dr. MURAT TEKER Dersin Yardımcıları Dersin Kategorisi Dersin Amacı Adsorption events, theory, kinetic approaches, conditions and pace of change to find out how these conditions and thermodynamic properties of Adsorpsiyonu change data to examine factors. Dersin İçeriği United surface solids, pore distribution and classification, the adsorption isotherms, isotherms and adsorption energy, adsorption of gas in the non-porous solid, BET equation, to calculate the surface area, porous and solid micro porous adsorption of the gas, the Gibbs adsorption isotherm, Kemisorpsiyon, adsorption from aqueous solution, Calculation of adsorption heat
Dersin Öğrenme Çıktıları Öğretim Yöntemleri Ölçme Yöntemleri 1 - To investigate the kinetics Adsorpsiyonu 1 - 2 - 3 - 4 - A - C - 2 - Adsorpsiyonu calculation of thermodynamic data 1 - 2 - 3 - 4 - A - C -
Öğretim Yöntemleri: 1:Lecture 2:Question-Answer 3:Discussion 4:Drilland Practice Ölçme Yöntemleri: A:Testing C:Homework
#### Ders Akışı
Hafta Konular ÖnHazırlık
1 Adsorption introduction, history, solids with large surface area
2 Pore distribution, classification, Adsorption energy
4 Adsorption of gases in non porous solid
5 BET isotherm, the surface area calculation
6 Adsorption of gases in porous solids
7 Adsorption of gases on solid micro-porous
8 Gibbs adsorption isotherm and the Gibbs equation
9 Mid. Exam.
10 Isotherms of Langmuir and Freundlich
11 Chemisorption
14 Calculation of heat of adsorption
Ders Notu
Ders Kaynakları
#### Dersin Program Çıktılarına Katkısı
No Program Öğrenme Çıktıları KatkıDüzeyi
1 2 3 4 5
1 In addition to expanding and making deeper the knowledge established on undergraduate education in a same or different area, reaches the knowledge by doing research and also improves his or her knowledge to the expertise level by doing evaluation and practice. X
2 Utilizing incomplete or limited data in his or her study area expands the knowledge by experimental methods and use those obtained knowledge by scientific, social and ethical responsibility.
3 - Constructs a problem with his/her adviser and develops a solution method and by evaluating the results apply them if it is necessary in his/ her area. X
4 -Transfer the current developments and his/her studies to the other related or nonrelated groups as written, orally and visually.
5 - Develops new strategic approach and produces solutions in unexpected and complicated situations in his or her area of practice.
6 - Develops strategic and practice plans and evaluates the obtained results within the framework of quality process in his or her study area.
7 - Has written communication ability in one of well known foreign languages (“European Language Portfolio Global Scale”, Level A1). X
8 -Has knowledge for software and hardware about computers and uses data processing. X
9 Protecting scientific, social, and ethical values, teaches and controls these values in the steps of collection, evaluation and announcement of data in the international arena. X
10 In addition to expanding and making deeper the knowledge established on undergraduate education in a same or different area, reaches the knowledge by doing research and also improves his or her knowledge to the expertise level by doing evaluation and practice. X
11 Utilizing incomplete or limited data in his or her study area expands the knowledge by experimental methods and use those obtained knowledge by scientific, social and ethical responsibility.
12 - Constructs a problem with his/her adviser and develops a solution method and by evaluating the results apply them if it is necessary in his/ her area. X
13 -Transfer the current developments and his/her studies to the other related or nonrelated groups as written, orally and visually.
14 - Develops new strategic approach and produces solutions in unexpected and complicated situations in his or her area of practice.
15 - Develops strategic and practice plans and evaluates the obtained results within the framework of quality process in his or her study area.
16 - Has written communication ability in one of well known foreign languages (“European Language Portfolio Global Scale”, Level A1). X
17 -Has knowledge for software and hardware about computers and uses data processing. X
18 Protecting scientific, social, and ethical values, teaches and controls these values in the steps of collection, evaluation and announcement of data in the international arena. X
19 - Applies his or her digested knowledge and problem solving ability in advanced studies. X
#### Değerlendirme Sistemi
YARIYIL İÇİ ÇALIŞMALARI SIRA KATKI YÜZDESİ
AraSinav 1 50
Odev 1 40
PerformansGoreviSeminer 1 10
Toplam 100
Yıliçinin Başarıya Oranı 60
Finalin Başarıya Oranı 40
Toplam 100
#### AKTS - İş Yükü
Etkinlik Sayısı Süresi(Saat) Toplam İş yükü(Saat)
Course Duration (Including the exam week: 16x Total course hours) 16 3 48
Hours for off-the-classroom study (Pre-study, practice) 16 3 48
Mid-terms 1 10 10
Assignment 2 10 20
Performance Task (Seminar) 1 10 10
Final examination 1 20 20
Toplam İş Yükü 156
Toplam İş Yükü /25(s) 6.24
Dersin AKTS Kredisi 6.24
; ; | 1,316 | 5,368 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-51 | latest | en | 0.612526 |
https://brainmass.com/business/weighted-average-cost-of-capital/npv-irr-wacc-18224 | 1,713,415,771,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817187.10/warc/CC-MAIN-20240418030928-20240418060928-00617.warc.gz | 129,481,354 | 7,700 | Purchase Solution
# NPV, IRR, WACC
Not what you're looking for?
6.20 Here are the cash flows for two mutually exclusive projects:
Project C0 C1 C2 C3
A (\$20,000) \$8,000 \$8,000 \$8,000
B (\$20,000) 0 0 \$25,000
a. At what interest rates would you prefer project A to B?
b. What is the IRR of each project?
7.21 Revenues generated by a new fad product in each of the next 5 years are forecasted as follows:
Year Revenues
1 \$40,000
2 30,000
3 20,000
4 10,000
Thereafter 0
Expenses are expected to be 40 percent of revenues, and working capital required in each year is expected
to be 20 percent of revenues in the following year. The product requires an immediate investment of
\$50,000 in plant and equipment.
a. What is the initial investment in the product? Remember working capital.
b. If the plant and equipment are depreciated over 4 years to a salvage value of zero using straight-line
depreciation, and the firm's tax rate is 40 percent, what are the project cash flows in each year?
c. If the opportunity cost of capital is 10 percent, what is the project NPV?
d. What is the project IRR?
11.10 Find the WACC of William Tell Computers. The total book value of the firm's equity is \$10 million; book value
per share is \$20. The stock sells for a price of \$30 per share, and the cost of equity is 15 percent. The firm's
bonds have a par value of \$5 million and sell at a price of 110 percent of par. The yield to maturity on the bonds
is 9 percent, and the firm's tax rate is 40 percent.
Please see the attachments for the questions. Answers are to be completed in Excel
##### Solution Summary
1) Calculates IRR;
2) Calculates project cash flows, NPV, and IRR;
3) Calculates the Weighted Average Cost of Capital
##### Free BrainMass Quizzes
This quiz is intended to help business students better understand business processes, including those related to manufacturing and marketing. The questions focus on terms used to describe business processes and marketing activities.
##### Organizational Behavior (OB)
The organizational behavior (OB) quiz will help you better understand organizational behavior through the lens of managers including workforce diversity.
##### Managing the Older Worker
This quiz will let you know some of the basics of dealing with older workers. This is increasingly important for managers and human resource workers as many countries are facing an increase in older people in the workforce
##### IPOs
This Quiz is compiled of questions that pertain to IPOs (Initial Public Offerings) | 607 | 2,548 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2024-18 | latest | en | 0.90994 |
https://ctftime.org/writeup/33291 | 1,701,988,584,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100705.19/warc/CC-MAIN-20231207221604-20231208011604-00042.warc.gz | 224,076,289 | 7,607 | Tags: z3 constraint reversing
Rating: 5.0
# ezbolt
## Bypassing Decryption
I really did not feel like reversing the decryption that was laid out in the ezbolt extension.
Therefore instead I added in GDB into the Docker container, built the container, then when running the container, I used GDB to break on the point at which the decrypted text was being passed to execute by PHP. This was how I dumped out the following string:
Which realistically is a lot of bs to look through. However, if you take it apart, it is a bunch of comparisons that occur and are &&'d together. This is their "one liner" way of deciding if the input is correct against their **constraints**.
Oo constraints. Fancy. Lets use Z3.
## The Script
The following script was used to solve the challenge. Let's break it down.
### compare_to_constraint
I am using this function to convert $f[9]*$f[17]-$f[35]+$f[33]==4646 into flag[9]*flag[17]-flag[35]+flag[33]==4646 which we can then use to add constraints into Z3. For those who do not know, z3 is a nice constraint solver which could let you solve multivariate problems given a well constrained problem. You may find something similar in your linear algebra class where you had to solve a system of equations using constraints set forth by the other equations. This is a similar problem; each comparison will give a constraint and at the end, the flag must adhere to all of them. So if I can convert all of these into a comparison that the Z3 solver will understand, I can compaile one big comparison chain then just ask z3 to solve it for me.
### BitVec?
This is a bitvector. All that is going on here is we are creating an array of bit vectors, think of these like arrays of singular bits. You may think of a string as one array of characters. Now think of a string as an array of characters which are each an array of 8 bits. Now you are coming to what is this array of bitvectors. This object is what z3 uses to allow us to add in constraints and it to represent its solve in. We then convert this later into an actual character by doing chr(s.model()[c].as_long()).
#!/usr/bin/env python3
import z3
import re
def compare_to_constraint(comp, flag, solver):
res = int(comp.split(b"==")[1])
args = [m.start() for m in re.finditer(b"\\$f", comp)]
ops = []
indexes = []
for idx in args:
stop = comp[idx:].find(b']') + 1
ops.append(
comp[idx + stop: idx + stop + 1].decode().strip()
)
indexes.append(
int(comp[idx + 3: idx + stop - 1])
)
ops = ops[:-1]
chall = ''
ops.reverse()
indexes.reverse()
try:
while True:
chall += f'flag[{indexes.pop()}]'
chall += f'{ops.pop()}'
except:
pass
chall += f'=={res}'
resp = eval(chall)
s = z3.Solver()
flag = []
for c in range(39):
flag.append(z3.BitVec(f"flag_{c}", 8))
for comp in php:
con = compare_to_constraint(comp, flag, s)
if s.check() == z3.sat:
print(
''.join(chr(s.model()[c].as_long()) for c in flag)
)
else:
print("Failed..")
## End
All and all. Quite simple, but a fun and good intro example for making use of z3. | 784 | 3,018 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2023-50 | latest | en | 0.952755 |
https://teamtreehouse.com/community/please-can-you-help-me-with-this-python-challenge | 1,713,543,619,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817438.43/warc/CC-MAIN-20240419141145-20240419171145-00888.warc.gz | 509,998,831 | 34,863 | ## Welcome to the Treehouse Community
Want to collaborate on code errors? Have bugs you need feedback on? Looking for an extra set of eyes on your latest project? Get support with fellow developers, designers, and programmers of all backgrounds and skill levels here with the Treehouse Community! While you're at it, check out some resources Treehouse students have shared here.
### Looking to learn something new?
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# Please can you help me with this python challenge
This challenge says where is sillycase() as an error message. please can someone tell me what's wrong.
silly.py
```# The first half of the string, rounded with round(), should be lowercased.
# The second half should be uppercased.
# E.g. "Treehouse" should come back as "treeHOUSE"
def sillycase (Treehouse):
leng = (len(Treehouse) / 2)
first = silly[:leng].round()
second = silly[leng:]()
return first + second
```
Hi Andy,
That's a really good attempt! You understand the general solution. We just need to fix some syntax and relocate the `round()` function.
First, let's tackle `round()`. Without it, when calling `len()` on a string with an odd number of letters, it will return a floating point value.
```len("apple") / 2
# equals 2.5
# In order to round this value, we can put this expression
# inside the round function.
round(len("apple") / 2)
# equals 2
# Your variable is called leng, but really it's holding half the length.
# Perhaps consider renaming it to half.
```
Next up is slicing the string. Your code is trying to slice a string inside a undefined variable `silly`. Your method is accepting a variable called `Treehouse`, so you can either change the variable name being sliced to `Treehouse` or change the parameter of the method to `silly` - either way will work.
Finally, change the first half of the string to lowercase and the last half to uppercase.
```"apple".upper() # equals "APPLE"
"TOMATO".lower() # equals "tomato"
# we can chain methods after slicing like this
Treehouse[:half].lower()
# the string inside the variable name Treehouse will be sliced from index 0
# up to but excluding index half and then lowercased.
```
Hope this helps,
Cheers
Hi thankyou for the good feedback and help. It passed I did it a bit different I just needed the round function in the leng variable. | 574 | 2,480 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-18 | latest | en | 0.885313 |
https://couryes.com/real-analysis-dai-xie-uniform-convergence-and-differentiation/ | 1,713,618,035,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817650.14/warc/CC-MAIN-20240420122043-20240420152043-00411.warc.gz | 165,315,963 | 33,410 | # 数学代写|实分析作业代写Real analysis代考|Uniform Convergence and Differentiation
#### Doug I. Jones
Lorem ipsum dolor sit amet, cons the all tetur adiscing elit
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## 数学代写|实分析作业代写Real analysis代考|Uniform Convergence and Differentiation
In this section, we consider the question of interchange of limits and differentiation. Example 8.1.2(e) shows that even if the sequence $\left{f_n\right}$ converges uniformly to $f$, this is not sufficient for convergence of the sequence $\left{f_n^{\prime}\right}$ of derivatives. Example 8.5.3 will further demonstrate very dramatically the failure of the interchange of limits and differentiation. There we will give an example of a series, each of whose terms has derivatives of all orders, that converges uniformly to a continuous function $f$, but for which $f^{\prime}$ fails to exist at every point of $\mathbb{R}$. Clearly, uniform convergence of the sequence $\left{f_n\right}$ is not sufficient. What is required is uniform convergence of the sequence $\left{f_n^{\prime}\right}$.
THEOREM 8.5.1 Suppose $\left{f_n\right}$ is a sequence of differentiable functions on $[a, b]$. If
(a) $\left{f_n^{\prime}\right}$ converges uniformly on $[a, b]$, and
(b) $\left{f_n\left(x_o\right)\right}$ converges for some $x_o \in[a, b]$,
then $\left{f_n\right}$ converges uniformly to a function $f$ on $[a, b]$, with
$$f^{\prime}(x)=\lim _{n \rightarrow \infty} f_n^{\prime}(x)$$
Remarks. (a) Convergence of $\left{f_n\left(x_o\right)\right}$ at some $x_o \in[a, b]$ is required. For example, if we let $g_n(x)=f_n(x)+n$, then $g_n^{\prime}(x)=f_n^{\prime}(x)$, but $\left{g_n(x)\right}$ need not converge for any $x \in[a, b]$. In Exercise 1 you will be asked to show that uniform convergence of $\left{f_n^{\prime}\right}$ is also required; pointwise convergence is not sufficient.
(b) If in addition to the hypotheses we assume that $f_n^{\prime}$ is continuous on $[a, b]$, then a much shorter and easier proof can be provided using the fundamental theorem of calculus. Since $f_n^{\prime}$ is continuous, by Theorem 6.3.2
$$f_n(x)=f_n\left(x_o\right)+\int_{x_o}^x f_n^{\prime}(t) d t$$
for all $x \in[a, b]$. The result can now be proved using Corollary 8.3.2 and Theorem 8.4.1. The details are left to the exercises (Exercise 2).
## 数学代写|实分析作业代写Real analysis代考|The Weierstrass Approximation Theorem
In this section, we will prove the following well known theorem of Weierstrass.
THEOREM 8.6.1 (Weierstrass) If $f$ is a continuous real-valued function on $[a, b]$, then given $\epsilon>0$, there exists a polynomial $P$ such that
$$|f(x)-P(x)|<\epsilon$$
for all $x \in[a, b]$
An equivalent version, and what we will actually prove, is the following:
If $f$ is a continuous real-valued function on $[a, b]$, then there exists a sequence $\left{P_n\right}$ of polynomials such that
$$f(x)=\lim _{n \rightarrow \infty} P_n(x) \quad \text { uniformly on }[a, b] .$$
Before we prove Theorem 8.6.1, we state and prove a more fundamental result that will also have applications later. Prior to doing so, we need the following definitions.
DEFINITION 8.6.2 A real-valued function $f$ on $\mathbb{R}$ is periodic with period $p$ if
$$f(x+p)=f(x) \quad \text { for all } x \in \mathbb{R} .$$
The canonical examples of periodic functions are the functions $\sin x$ and $\cos x$, both of which are periodic of period $2 \pi$. The graph of a periodic function of period $p$ is illustrated in Figure 8.3. The graphs of a periodic function of period $p$ on any two successive intervals of length $p$ are identical. It is clear that if $f$ is periodic of period $p$, then
$$f(x+k p)=f(x) \quad \text { for all } k \in \mathbb{Z} \text {. }$$
# 实分析代写
## 数学代写|实分析作业代写Real analysis代考|Uniform Convergence and Differentiation
8.5.3 将进一步非常戏剧性地证明极限和微分互换的失 败。在那里我们将给出一个序列的例子，它的每一项都 有所有阶数的导数，它一致地收敛到一个连续函数 $f$ ，但 为此 $f^{\prime}$ 不存在于每一点 $\mathbb{R}$. 显然，序列的一致收敛
$$f^{\prime}(x)=\lim {n \rightarrow \infty} f_n^{\prime}(x)$$ $x_o \in[a, b]$ 是必须的。例如，如果我们让 $g_n(x)=f_n(x)+n$ ，然后 $g_n^{\prime}(x)=f_n^{\prime}(x)$ ，但 左 $\left{\mathrm{g} _\mathrm{n}(\mathrm{x}) \backslash\right.$ 右 $}$ 不需要收敛于任何 $x \in[a, b]$. 在练习 1 是必需的；逐点收敛是不够的。 (b) 如果除了假设之外我们假设 $f_n^{\prime}$ 是连续的 $[a, b]$ ，然后 可以使用微积分基本定理提供更短和更容易的证明。自 从 $f_n^{\prime}$ 是连续的，由定理 6.3.2 $$f_n(x)=f_n\left(x_o\right)+\int{x_o}^x f_n^{\prime}(t) d t$$
## 数学代写|实分析作业代写Real analysis代考|The Weierstrass Approximation Theorem
$$|f(x)-P(x)|<\epsilon$$
$$f(x)=\lim _{n \rightarrow \infty} P_n(x) \quad \text { uniformly on }[a, b] \text {. }$$
$$f(x+p)=f(x) \quad \text { for all } x \in \mathbb{R} .$$
$$f(x+k p)=f(x) \quad \text { for all } k \in \mathbb{Z} \text {. }$$
## 有限元方法代写
tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。
## MATLAB代写
MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。
Days
Hours
Minutes
Seconds
# 15% OFF
## On All Tickets
Don’t hesitate and buy tickets today – All tickets are at a special price until 15.08.2021. Hope to see you there :) | 2,435 | 6,122 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2024-18 | latest | en | 0.565966 |
https://docs.interpretable.ai/stable/examples/jury/ | 1,726,853,478,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701419169.94/warc/CC-MAIN-20240920154713-20240920184713-00618.warc.gz | 186,302,753 | 217,628 | # Racial Bias in Jury Selection
This example presents the fully-worked analysis in our published work that uses interpretable methods to investigate the presence of human biases in decision making. In particular, we consider the role of race in jury selection. In 1986, the U.S. Supreme Court ruled that using race as a reason to remove potential jurors is unconsititutional. Despite this ruling, a large disparity in juror strike rates across races appears to remain.
This disparity was the focus of the 2019 U.S. Supreme Court case "Flowers v. Mississippi", where it was ruled that the District Attorney Doug Evans from the Fifth Circuit Court District in Mississippi had discriminated based on race during jury selection in the six trials of Curtis Flowers.
To support the case, APM Reports collected and published court records of jury strikes in the Fifth Circuit Court District and conducted analysis to assess if there was a systematic racial bias in jury selection in this district. The data included information on each trial, juror, and the voir dire answers by the jurors between 1992 and 2017. As part of their analyses, they used a logistic regression model and concluded that there was significant racial disparity in jury strike rates by the State, even after accounting for other factors in the dataset.
We will use our methods to investigate:
1. Whether we reach the same conclusion that there is significant racial disparity in strike rates
2. Whether the racial disparity is the same across the board, or there are specific subgroups where the disparity is most prominent.
## Data Preparation
We follow the same data preparation as the methodology in the report to ensure consistency. First, we prepare the data so that each row corresponds to a juror at a particular trial:
using CSV, DataFrames
data = leftjoin(jurors, trials, on=(:trial__id => :id))
on=[(:id => :juror_id), (:trial__id => :juror_id__trial__id)])
3545×112 DataFrame
Row │ id trial trial__id race gender r ⋯
│ Int64 String Int64 String String S ⋯
──────┼─────────────────────────────────────────────────────────────────────────
1 │ 107 2004-0257--Sparky Watson 3 White Male J ⋯
2 │ 108 2004-0257--Sparky Watson 3 Black Female J
3 │ 109 2004-0257--Sparky Watson 3 Black Female J
4 │ 110 2004-0257--Sparky Watson 3 Black Female J
5 │ 111 2004-0257--Sparky Watson 3 White Male J ⋯
6 │ 112 2004-0257--Sparky Watson 3 Black Female J
7 │ 113 2004-0257--Sparky Watson 3 Black Male J
8 │ 114 2004-0257--Sparky Watson 3 White Male J
⋮ │ ⋮ ⋮ ⋮ ⋮ ⋮ ⋱
3539 │ 262 1994-9942--Suzanne Ilene Tavares 6 White Female J ⋯
3540 │ 1094 2002-0067--Deondray Johnson 22 White Female J
3541 │ 3478 2010-0012--Jerome Patterson 70 White Female J
3542 │ 3485 2010-0012--Jerome Patterson 70 White Female J
3543 │ 3487 2010-0012--Jerome Patterson 70 Black Female J ⋯
3544 │ 2980 1995-2258--Robert Bingham 60 Black Female J
3545 │ 2386 2001-0003--Lawrence Branch 47 White Male J
107 columns and 3530 rows omitted
We are interested in understanding what leads to a juror being struck by the State. For this purpose, we subset to only jurors eligible to be struck by the State.
data = data[(data.strike_eligibility .== "State") .+
(data.strike_eligibility .== "Both State and Defense") .> 0, :]
Next, we assemble the features, which include the juror's gender, race, and the defendant's race. In addition, we have the voir dire answers to 65 questions.
data.is_black = data.race .== "Black"
data.same_race = data.race .== data.defendant_race
categorical_vars = [["is_black", "gender", "defendant_race", "same_race"];
using CategoricalArrays
X = select(data, categorical_vars .=> categorical, renamecols=false)
The target is whether the juror was struck by the State:
y = [v == "Struck by the state" ? "Strike" : "No strike"
for v in data.struck_by]
Finally, we can split into training and testing:
seed = 1
(X_train, y_train), (X_test, y_test) = IAI.split_data(:classification, X, y,
seed=seed)
## Optimal Feature Selection
The first model we apply is Optimal Feature Selection. This is similar to the backward stepwise logistic regression model used in the original study, except that instead of iteratively selecting and removing variables that are insignificant, the Optimal Feature Selection will pick the optimal set of variables in a single step.
We run the Optimal Feature Selection, considering all possible combinations of up to 15 features, and selecting the best combination based on the AUC on a hold-out validation set:
ofs_grid = IAI.GridSearch(
IAI.OptimalFeatureSelectionClassifier(
random_seed=seed,
),
sparsity=1:15,
)
IAI.fit_cv!(ofs_grid, X_train, y_train, validation_criterion=:auc)
IAI.get_learner(ofs_grid)
Fitted OptimalFeatureSelectionClassifier:
Constant: -1.99072
Weights:
accused=true: 2.54637
death_hesitation=true: 1.86823
defendant_race=Unknown: 0.589933
fam_accused=true: 1.47964
fam_crime_victim=true: 0.534126
fam_law_enforcement=true: -0.365057
is_black=true: 1.41219
know_def=true: 1.273
law_enforcement=true: -0.686454
leans_defense=true: 1.76438
married=married: -0.605019
medical=true: 2.2713
no_death=true: 4.29094
same_race=true: 0.374411
(Higher score indicates stronger prediction for class Strike)
We see that is_black is among the 14 features selected in the best model, in addition to other variables such as know_def (juror has prior familiarity with defendant through either personal or professional channels) and fam_accused (the juror has friends or family accused of being involved in criminal activity). This reaffirms the finding of the previous analysis that is_black is a useful feature in the logistic regression model for predicting the probability of strike.
IAI.score(ofs_grid, X_test, y_test, criterion=:auc)
0.8253396158517383
The model also has strong predictive performance, with an out-of-sample AUC of 0.826.
To augment these findings, we can visualize the variable importance across all sparsity levels. The importance is normalized so the most important variable has a value of 1 at each sparsity. The variables are incrementally included from the bottom as they become selected under higher sparsity.
using Plots
plot(ofs_grid, type=:importance, size=(600, 600))
We can see that is_black is evaluated as the most important variable for every level of sparsity, which demonstrates that it has roughly the same predictive power regardless of what other features are added. This gives evidence that it is capturing signal not present in the other variables.
To further confirm that the race of the juror being black is important in explaining the strike decision, we can build a model without the race variables and compare the performance:
ofs_grid_no_race = IAI.GridSearch(
IAI.OptimalFeatureSelectionClassifier(
random_seed=seed,
),
sparsity=1:15,
)
IAI.fit_cv!(ofs_grid_no_race, select(X_train, Not([:is_black, :same_race])),
y_train, validation_criterion=:auc)
IAI.score(ofs_grid_no_race, X_test, y_test, criterion=:auc)
0.6900949125095581
We see that AUC falls by around 12% when we remove race from the model, a very strong indication that the race being black is highly explanatory to being struck, and that we cannot proxy this signal using other features in the dataset.
As a small side note, if the two models had similar performance this would not be sufficient evidence to conclude that race has no impact on the decision, as in that case other variables in the dataset may still be proxying for the race. However, in our case the large decrease in performance that we see upon removing race is strong evidence that race has unique predictive power in explaining the outcome that cannot be proxied with other variables.
## Identifying Subgroups with Disparity
We have strong evidence that the race of the juror plays a strong role in predicting the probability of being struck by the State. Next, we would like to investigate if there are specific subpopulations where this effect is more or less pronounced. To do this, we will move away from linear models and use Optimal Classification Trees as a tool to identify subpopulations with statistically significant differences in strike rate based on race.
To do this, we first train an Optimal Classification Tree without the race variable:
grid = IAI.GridSearch(
IAI.OptimalTreeClassifier(
criterion=:gini,
random_seed=seed,
missingdatamode=:separate_class,
split_features=Not([:is_black, :same_race]),
),
max_depth=2:5,
)
IAI.fit!(grid, X_train, y_train, validation_criterion=:auc)
lnr = IAI.get_learner(grid)
IAI.set_display_label!(lnr, "Strike")
Optimal Trees Visualization
The resulting tree has identified seven subgroups of jurors that have similar probabilities of being struck by the State. Importantly, these subgroups are defined without considering the race of the juror. For example, node 2 contains jurors that have been accused of a crime in the past, who understandably have a very high strike rate of 93%.
Next, we test if there is a significant difference in strike rate between black and non-black jurors in each subgroup using Fisher's Exact Test.
group = [x == true ? "Black" : "Non-black" for x in X.is_black]
outputs = IAI.compare_group_outcomes(lnr, X, y, group, positive_label="Strike")
pvalues = [o.p_value["vs-rest"]["Black"] for o in outputs]
Because we are simultaneously conducting several hypothesis tests, we use the Holm-Bonferroni method to adjust the p-values to avoid false positives.
using MultipleTesting
pvalues = adjust(pvalues, Holm())
We display the strike rate for each group in the node, and the p-value from the test. If a node shows a statistically significant difference between the two groups, it is colored red or green depending on the sign of the difference.
extras = map(1:length(outputs)) do i
summary = outputs[i].summary
p_value = pvalues[i]
node_color = if p_value > 0.05
"#FFFFFF"
elseif summary.prob[1] > summary.prob[2]
else
"#92D86F"
end
node_summary = "Strike rate for Black: $(round(Int, summary.prob[1] * 100))%; " * "For non-Black:$(round(Int, summary.prob[2] * 100))% " *
"(p=\$(round(p_value, digits=3)))"
node_details = IAI.make_html_table(summary)
Dict(:node_summary_include_default => false,
:node_details_include_default => false,
:node_summary_extra => node_summary,
:node_details_extra => node_details,
:node_color => node_color)
end
IAI.TreePlot(lnr, extra_content=extras)
Optimal Trees Visualization
There are two cohorts where there is no statistically significant difference in strike rate between black and non-black jurors:
• Node 2: potential jurors that have been accused of a crime
• Node 12: potential jurors with reservations about the death penalty
However, there is statistically significant distinction in strike rate between black and non-black potential jurors in nodes 5, 6, 9, 10 and 13. For example, Node 5 contains those who have never been accused of a crime, but know the defendant and have family in law enforcement. If a person in this group is black, the strike rate is 73% compared to 0% if this person were white. It appears that this reason for striking jurors is disproportionally applied to black jurors, as none of the 18 non-black jurors meeting these criteria were struck.
Overall, the Optimal Classification Tree finds four major lines of reasoning that are used to exclude jurors from the pool. Of the six resulting groups likely to be struck, the strikes in two of the groups (nodes 2 and 12) are applied regardless of race, but in the others (nodes 5, 6, 9 and 10) the strikes are applied to black jurors significantly more than non-black jurors, which is strongly suggestive of racial biases factoring into the strike decisions of the State.
## Conclusions
The original study used a backward stepwise logistic regression to conclude that there is significant racial disparity in juror strike outcome. This method is only an approximation to variable selection and has the potential to lead to a misleading characterization of variables as important or not.
With Optimal Feature Selection, we still find that race is consistently selected as the most predictive variable of juror strike outcome at all levels of sparsity. This strengthens the conclusion in the original study, as we knowing that the variable selection is exact. Furthermore, we observe a significant decrease in model performance when the race variable is removed, indicating this feature is providing unique signal in the data.
In addition, we used Optimal Classification Trees to systematically identify subgroups of the population with similar chances of being struck, and found that in five of these groups there was a significant disparity in strike rate between black and non-black jurors. These subgroups suggest systemic patterns of racial bias in the strike process, and provide direct characterizations of the situations in which black jurors are likely to have experienced discrimination. | 3,207 | 13,512 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-38 | latest | en | 0.942278 |
https://www.24houranswers.com/technical-tutoring-tips/Median-Mean-and-Mode-in-Math-and-Statistics | 1,708,588,138,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473735.7/warc/CC-MAIN-20240222061937-20240222091937-00439.warc.gz | 648,855,121 | 17,606 | Median, Mean and Mode in Math and Statistics
Sep 17, 2022
Mathematics - Other
What Is the Median?
Median, in math, or to be more exact, in statistics, is the middle number of a given list of numbers that are arranged (whether ascending or descending.)
The technical definition is that a median is a point above and below which 50% of the list of numbers (or a set of observed data) falls.
The median, in statistics, is a type of descriptive statistics, which is a summary statistics that describe features from a collection of data (in this case, a set of numbers.) With this definition, the median represents, or describes, the exact midpoint of the number list.
The median is often compared to other descriptive statistics, mainly the mean (average), mode, and standard deviation.
Median
Median, as we've discussed, is the middle number of a list of numbers. We can find a median by first arranging the numbers (it wouldn't matter if the data is ordered in descending or ascending) and then picking out the number exactly in the middle.
For example, if the data set is "1,2,3,4,5,6,7,8,9", then 5 is the median.
If, however, the data set is "1,2,3,4,5,6,7,8,10", then there are two middle numbers, 5 and 6. In such cases, the median is the mean of these two numbers, which is 5.5.
Mean
Mean is the "average" of the numbers in the list (the data set) and can be discovered by adding all numbers in the set and dividing by the count which is the number of the values being added.
For example, if there are three numbers in the list: 1, 5, and 7, then the mean is (1+5+7) divided by 3 since there are 3 numbers in the set, and we get 4.333 as the mean.
Mean is often denoted with the symbol μ.
Mode
Mode is the most frequent number in the list.
For example, if the list consists of five numbers: 5,3,5,1,4,5, then the mode is 5 because it occurs three times, more than any other number in the list.
Median, mean, and mode in a Normal Distribution (Bell Curve)
The normal distribution, also known as the Gaussian distribution, or the bell curve, is a probability distribution that is perfectly symmetric around the mean: half the values below the mean, and another half above the mean.
The key characteristic of a normal distribution is that the mean, mode, and median are exactly the same. When visualized in a graphical form, it will show a bell-shaped curve, hence the name.
In a typical skewed data set, however, the mean, median, and mode will be different.
Standard deviation
A standard deviation (often denoted with the symbol σ) is the measure of the amount of dispersion (or variation) of a set of numbers or values.
The standard deviation is measured by how dispersed the data is in relation to the mean/average. A low standard deviation means more numbers in the list are clustered around the average number, while a high standard deviation means the numbers are more spread out.
For example, let's say there are six numbers in the set: 2,5,5,5,6,7
To calculate standard deviation, first, we have to determine the mean of this set:
μ=(2+5+5+5+6+7)/6=5
Then, we can calculate the deviations of each number (data point) from the mean, and the square results of each:
(2-5)=(-3)=9
(5-5)=(0)=0
(5-5)=(0)=0
(5-5)=(0)=0
(6-5)=(1)=1
(7-5)=(2)=4
The variance is the mean of these values, and we can get:
σ=(9+0+0+0+1+4)/6=2.33
And the standard deviation is equal to the square root of the variance=
σ=
The importance of median
The simple mean, or average, is the more common descriptive statistic that we use to analyze the characteristics of a data set.
However, using the mean has a major downside in the fact that it often does not depict the typical outcome. For example, in a very large data set, the mean or average may be 5, but the biggest number in the data set is 50, which is very far from the rest of the data. This outcome is called an outlier, and obviously, this outlier outcome will strongly affect the mean.
In this case, the mean number may not represent the whole data point, depending on your purpose in analyzing the data set, and this is where measuring the median can be more useful.
The median is better for describing the typical value in a data point, which is especially useful in observing data points with large standard deviations like house prices and income levels.
For example, assume you are a home-buyer looking to find out how much you should expect to spend on buying a house in Toronto, Canada.
One method you can use is to randomly select ten house prices in the area, and let's say we get these numbers, sorted from cheapest to the most expensive:
1. \$2.8 million
2. \$2.9 million
3. \$3.2 million
4. \$3.5 million
5. \$3.7 million
6. \$4.2 million
7. \$4.3 million
8. \$4.7 million
9. \$4.8 million
10. \$30.8 million
The mean of these house prices is \$6.02 million.
However, although this average mean is accurate, we can see that it doesn't really reflect the price of these ten houses since the house valued at \$30.8 million skewed the average.
This is where measuring the median can be more useful in determining the price of available housing in Toronto. Since there is an even number of outcomes in the list, we take the mean of the middle two: (\$3.7 million+\$4.2 million)/2=\$3.95 million.
The median house price is \$3.95 million, which better reflects what house shoppers should expect to spend.
Many economists also favor the median for reporting a nation's income since it is more representative of the actual distribution than the mean.
Wrapping Up
By following the tutorial above, now you already have a solid knowledge about the median in mathematics, as well as the median's relation to the mean, mode, and standard deviation.
If you need to learn more, 24HourAnswers.com has experienced online math tutors and more tutorials available where you can learn and practice your knowledge more thoroughly.
View Available Mathematics Tutors
24houranswers.com Parker Paradigms, Inc 5 Penn PLaza, 23rd Floor New York, NY 10001
SUBJECTS COVERED | 1,481 | 6,051 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.75 | 5 | CC-MAIN-2024-10 | latest | en | 0.944599 |
http://www.mathemafrica.org/?p=11589 | 1,582,418,718,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145742.20/warc/CC-MAIN-20200223001555-20200223031555-00353.warc.gz | 203,826,024 | 34,999 | I’ve been asked a few times for more practice questions on complex numbers. This is where Wolfram Alpha can be your friend (like it’s not already!).
I’ll just give a few examples of questions from the tut on complex numbers which you could have solved using Wolfram Alpha, and from this you will be able to set up your own questions.
For instance, question 48 c) Find the roots of $z^5=1+\sqrt(3) I$ can be solved in Wolfram Alpha with the command:
Solve[z^5==1+sqrt[3]I,z]
Moreover it will solve this for you, give you the five roots and plot them in the complex plane. So now you can come up with any root question you can possibly think of. There’s an infinite number of questions to start you off. You can thank me later!
If you want to convert between the trigonometric form and the exponential form, you can use the two commands:
TrigToExp[Sin[x]+2 I Cos[x]]
ExpToTrig[Exp[I z+3]]
Though remember the definition of the hypergeometric trig functions from a previous tut.
If you want to get an expression in the normal cartesian form, then you have to be a bit more careful. For instance, $(1+\sqrt(3) I)(1+\sqrt(3)I)$: Here you can’t just plug it in because it doesn’t know the form you’re looking for it in. Here you have to put:
Re((1+sqrt[3] I)(1+sqrt[3]I))+I Im((1+sqrt[3] I)(1+sqrt[3]I))
ie. ask for the real part and the imaginary parts separately.
If you want to find the solutions to: $e^{3z}=1+I$, you have to do this in two steps. First you have to put in:
Solve[Exp[3z]==1+I]
This will return for you:
$z=\frac{1}{3}(2\pi n i+\log(1+I)$
but we know that we don’t want the log of a complex number. So you have to take $\log(1+I)$ and calculate that in terms of the cartesian form:
Re(log(1+I))+I Im(log(1+I))
Now you have all the information you need to solve this question.
To find the argument of a complex number you use the command Arg, and to find the magnitude, you use the command Abs. If you want to write $(\sqrt(3)+I)^{17}$ in modulus argument form, you can calculate:
Abs[(sqrt(3)+I)^17] and Arg[(sqrt(3)+I)^17]
separately and then write it yourself together in whatever form you want.
Note also that the conjugate of a complex number is Conjugate[z].
Hopefully this should be enough information for you to be able to go through the tut questions and write some of your own. Let me know if there are questions which you can’t answer using Wolfram Alpha or your own complex numbers intuition…
How clear is this post? | 661 | 2,468 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 6, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2020-10 | latest | en | 0.876644 |
https://www.gradesaver.com/textbooks/math/algebra/elementary-linear-algebra-7th-edition/chapter-2-matrices-2-3-the-inverse-of-a-matrix-2-3-exercises-page-72/48 | 1,575,987,203,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540527620.19/warc/CC-MAIN-20191210123634-20191210151634-00269.warc.gz | 724,124,969 | 12,028 | Elementary Linear Algebra 7th Edition
Published by Cengage Learning
Chapter 2 - Matrices - 2.3 The Inverse of a Matrix - 2.3 Exercises - Page 72: 48
Answer
(a) $x=1$, $y=1$, $z=1$. (b) $x=1$, $y=0$, $z=1$.
Work Step by Step
(a) The coefficient matrix $A$ is given by $$A=\left[ \begin {array}{ccc} 1&1&-2\\ 1&-2&1 \\ 1&-1&-1\end {array} \right] .$$ Using Gauss-Jordan elimination, one can calculate $A^{-1}$ as follows $$A^{-1}=\left[ \begin {array}{ccc} 1&1&-1\\ \frac{ 2}{3}&\frac{1}{3}&-1 \\ \frac{ 1}{3}&\frac{2}{3}&-1 \end {array} \right] .$$ The solution is given by $$\left[ \begin {array}{cc} {x}\\ {y}\\{z} \end {array} \right]=\left[ \begin {array}{ccc} 1&1&-1\\ \frac{ 2}{3}&\frac{1}{3}&-1 \\ \frac{ 1}{3}&\frac{2}{3}&-1 \end {array} \right]\left[ \begin {array}{cc} {0}\\ {0}\\{-1} \end {array} \right]=\left[ \begin {array}{c} 1\\ 1\\1\end {array} \right].$$ That is, $x=1$, $y=1$, $z=1$. (b) The coefficient matrix $A$ is given by $$A=\left[ \begin {array}{ccc} 1&1&-2\\ 1&-2&1 \\ 1&-1&-1\end {array} \right] .$$ Using Gauss-Jordan elimination, one can calculate $A^{-1}$ as follows $$A^{-1}=\left[ \begin {array}{ccc} 1&1&-1\\ \frac{ 2}{3}&\frac{1}{3}&-1 \\ \frac{ 1}{3}&\frac{2}{3}&-1 \end {array} \right] .$$ The solution is given by $$\left[ \begin {array}{cc} {x}\\ {y}\\{z} \end {array} \right]=\left[ \begin {array}{ccc} 1&1&-1\\ \frac{ 2}{3}&\frac{1}{3}&-1 \\ \frac{ 1}{3}&\frac{2}{3}&-1 \end {array} \right]\left[ \begin {array}{cc} {-1}\\ {2}\\{0} \end {array} \right]=\left[ \begin {array}{c} 1\\ 0\\1\end {array} \right].$$ That is, $x=1$, $y=0$, $z=1$.
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | 735 | 1,746 | {"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} | 4.6875 | 5 | CC-MAIN-2019-51 | latest | en | 0.505985 |
https://www.physicsforums.com/threads/can-i-exploit-johnson-noise-to-violate-the-second-law.176002/ | 1,540,193,865,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583514879.30/warc/CC-MAIN-20181022071304-20181022092804-00213.warc.gz | 1,030,809,703 | 12,717 | # Can i exploit johnson noise to violate the second law?
1. Jul 5, 2007
### parsec
Thermal noise known as “Johnson noise” or “Nyquist noise” is well known and characterized. This noise is a result of thermal agitation of charge carriers inside a conductor. Its power spectral density is given by:
v^2=4kTR
where kB is Boltzmann's constant in joules per kelvin, T is the resistor's absolute temperature in kelvins, and R is the resistor value in ohms.
Imagine a resistor in an adiabatic enclosure (one incapable of heat transfer) connected to a lossless diode. The wires leave the enclosure through sealed ports to power a circuit. Shouldn’t the rectification of this random fluctuation in the voltage across the resistor create a net positive voltage and hence allow us to violate the second law of thermodynamics (creating free energy)?
I once asked a physics professor this question and he excitedly gave me a rushed explanation of why this wouldn’t work. It involved Brownian motion and he quickly digressed into a tangent about proving the continuity between classical (Rayleigh Jeans) and quantum blackbody theory. It involved some esoteric thought experiment where he defiled my adiabatic enclosure with a big parabolic dish. I wasn’t impressed.
edit: I just realised the parallels between this hypothetical and Maxwell's demon, which seems to be still up for debate.
Last edited: Jul 5, 2007
2. Jul 5, 2007
### Cthugha
It is not that easy. What you present here is a version of Smoluchowski's Trapdoor, I think.
The problem is your diode. A diode is not a perfect binary switch. It is subject to thermal noise as well. Therefore you will only notice some net electron movement, if there is a temperature gradient between the diode and your conductor. This thing will work similar to a thermocouple.
Have a look at the Seebeck-effect. It is very similar. | 428 | 1,872 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2018-43 | latest | en | 0.926089 |
https://www.slideserve.com/kiara/chapter-11-liquids-solids | 1,624,391,373,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488519735.70/warc/CC-MAIN-20210622190124-20210622220124-00311.warc.gz | 835,415,547 | 22,682 | Chapter 11: Liquids & Solids
Chapter 11: Liquids & Solids
Chapter 11: Liquids & Solids
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Presentation Transcript
1. Chapter 11: Liquids & Solids Renee Y. Becker Valencia Community College
2. Kinetic Molecular Theory Introduction • Gases • Gas particles act independent of one another • Attractive forces are very weak • Particles are free to move randomly • Occupy whatever space available Liquids and solids are different from gases in that they have strong attractive forces between particles
3. Polar Covalent Bonds • Polar Covalent Bonds • Form between a non-metal/non-metal of different electronegativities EN = ENA – ENB When EN 1.7 ionic bond When EN < 1.7 polar covalent bond When EN < .5 non-polar covalent bond
4. Polar Molecules • Polar Molecules • Just as bonds can be polar, molecules as a whole can be polar • Net sum of individual bond polarities and lone-pair contributions
5. Dipole Moment • Dipole moment, , (ionic and covalent) • Measure of net molecular polarity • The magnitude of the charge Q at either end of the molecular dipole times the distance, r, between the charges = Q x r • Expressed in debyes, D, where 1 D = 3.336 x 10-30 coulomb meters • Q = 1.6 x 10-19 C (electron charge)
6. Dipole Moment % Ionic Character = experimental (100%) calculated a high % IC means that the bond is similar to or is an ionic bond a low % IC means that it is more like a covalent bond
7. Example 1 Chloromethane a) Calculate the dipole moment b) Calculate the % ionic character of the bond Experimentally measured dipole moment = 1.87 D C-Cl bond distance = 178 pm = 178 x 10-12 m If we assume that the contributions of the nonpolar C-H bonds are small, then most of the chloromethane dipole moment is due to the C-Cl bond
8. Example 2 Hydrochloric acid Calculate the % ionic charcater 1. Distance between atoms is 127 pm 2. Experimentally measured dipole moment = 1.03 D
9. Example 3 Tell which of the following compounds are likely to have a dipole moment and show the direction of each. a) SF6 b) CHCl3 c) CH2Cl2 d) CH2CH2
10. Intermolecular Forces • Van der Waals forces – intermolecular forces as a whole, all are electrical in origin and result from the mutual attraction of unlike charge or mutual repulsion of like charges. 4 main types • Dipole-dipole • Ion-dipole • Dispersion forces • Hydrogen bonding
11. Dipole-dipole • Neutral but polar molecules experience dipole-dipole forces as a result of electrical interactions among dipoles on neighboring molecules. • Forces can be attractive or repulsive, depending on the orientation of the molecules. • c) These forces are weak 3-4 kJ/mol and only significant if molecules are close
12. Ion-dipole Result of electrical interactions between an ion and the partial charges on a polar molecule b) Particularly important in aqueous solutions of ionic substances such as NaCl, in which polar water molecules surround the ions
13. London Dispersion Forces a) Result from the motion of electrons b) At any given time more electrons may be in a particular area of the molecule c) This gives the molecule an instantaneous dipole d) This short lived dipole can affect the electron distribution in neighboring molecules and induce temporary dipoles in them e) More electrons a molecule has the stronger the dispersion forces
14. Hydrogen Bonding a) Attractive interaction between a hydrogen atom bonded to a very electronegative atom (O, N, F) and an unshared electron pair on another electronegative atom • Hydrogen bonds arise because O-H, N-H, and F-H bonds are highly polar with partial positive charge on the hydrogen and partial negative on the electronegative atom. • Hydrogen has no core electrons to shield its nucleus and it is small so it can be approached closely by other molecules d) The dipole-dipole attraction between the hydrogen and an unshared electron pair on a nearby atom is usually strong
15. Hydrogen Bonding e) Water is able to form a vast 3D network of hydrogen bonds because each H2O molecule has two hydrogens and two electron pairs
16. Intermolecular Forces
17. Example 4 Identify the likely kinds of intermolecular forces in the following A) HCl B) CH3CH3 C) CH3NH2 D) Kr
18. Example 5 Of the substances Ar, Cl2, CCl4 and HNO3 which has: a) The largest dipole-dipole forces? b) The largest hydrogen-bond forces? c) The smallest dispersion forces?
19. Properties of Liquids Viscosity 1. The measure of a liquids resistance to flow 2. Related to the ease with which individual molecules move around in the liquid and thus to the intermolecular forces present 3. Substances with small non-polar molecules have weak intermolecular forces and low viscosities (free flowing) 4. More polar substances have stronger intermolecular forces and have higher viscosities
20. Properties of Liquids Surface Tension 1. The resistance of a liquid to spread out and increase its surface area • Caused by differences in intermolecular forces experienced by molecules at the surface and the interior • Surface molecules feel attractive forces on only one side and are drawn in toward the liquid 4. Interior molecules are drawn equally in all directions 5. Higher in liquids that have stronger intermolecular forces
21. Phase Changes • Physical form changes but chemical identity does not change Fusion (melting) solid liquid Freezing liquid solid Vaporization liquid gas Condensation gas liquid Sublimation solid gas Deposition gas solid
22. Thermochemistry Free energy change, G 1. All naturally occurring processes, every phase change has a free-energy change 2. G = H - T S 3. H enthalpy, heat flow, positive (from surrounding to system, bond breaking takes energy), negative (from system to surroundings, bond making) 4. S entropy, disorder, positive (ordered to disorderd), negative (disordered to ordered)
23. Thermochemistry
24. Thermochemistry Calculating the temperature at a phase change • G > 0 non-spontaneous G < 0 spontaneous G = 0 equilibrium 2. Set G = 0 0 = H - TS and solve for T T = H/S
25. Heating Curve for H2O
26. Heating Curve for H2O Heating curve for H2O E = molar heat capacity (T) Energy to heat ice from -25C to 0C Molar heat capacity of ice= 36.57 J/molC E = Energy to heat H2O from 0C to 100C Molar heat capacity of liquid H2O = 75.4 J/ molC E =
27. Thermochemistry Heat of fusion, Hfusion The amount of energy required to overcome enough intermolecular forces to convert a solid into a liquid Heat of Vaporization, Hvap The amount of energy necessary to convert a liquid into a gas
28. Example 6 Chloroform has Hvap = 29.2 kJ/mol and Svap = 87.5 J/K mol. What is the boiling point of chloroform?
29. Evaporation, Vapor Pressure, and Boiling Point Vapor Pressure 1. The partial pressure of a gas in equilibrium and at constant temperature with liquid 2. Thepressure exerted by gaseous molecules above a liquid
30. Evaporation, Vapor Pressure, and Boiling Point
31. Evaporation, Vapor Pressure, and Boiling Point The higher the temperature and the lower the boiling point of the substance the greater the fraction of molecules in the sample that have sufficient kinetic energy to break free from the surface of the liquid and escape into the vapor.
32. Evaporation, Vapor Pressure, and Boiling Point Numerical value of Vapor Pressure depends on: a) Magnitude of intermolecular forces The smaller the forces the higher the vapor pressure, loosely held molecules escape easily b) Temperature The higher the temperature, the higher the vapor pressure, larger fraction of molecules have sufficient kinetic energy to escape
33. Evaporation, Vapor Pressure, and Boiling Point The Clausius-Clapeyron Equation ln Pvap = - Hvap 1 + C R T Y = m x + b m is the slope and b is the y-intercept Where R is the gas constant 8.314 J/K mol C is a constant characteristic of each specific substance T temperature in Kelvin
34. Evaporation, Vapor Pressure, and Boiling Point The Clausius-Clapeyron Equation ln P2 = Hvap 1 - 1 P1 R T1 T2
35. Evaporation, Vapor Pressure, and Boiling Point • This equation makes it possible to calculate the heat of vaporization of a liquid by measuring its vapor pressure at several temperatures and then plotting the results to obtain the slope • Once the heat of vaporization and the vapor pressure at one temperature are known, the vapor pressure of the liquid at any other temperature can be calculated.
36. Evaporation, Vapor Pressure, and Boiling Point Normal boiling point 1. The temperature at which boiling occurs when the pressure is exactly 1 atm. 2. Boiling point – when the vapor pressure of a liquid rises to the point where it becomes equal to the external pressure
37. Example 7 The vapor pressure of ethanol at 34.7C is 100.0 mm Hg, and the heat of vaporization of ethanol is 38.6 kJ/mol. What is the vapor pressure of ethanol in mm Hg at 65.0C?
38. Example 8 The normal boiling point of benzene is 80.1 C and the heat of vaporization is 30.8 kJ/mol. What is the boiling point of benzene (in C) on top of Mt. Everest where P = 260 mm Hg?
39. Phase Diagrams • Shows which phase is stable at different combinations of pressure and temperature.
40. Phase Diagrams Triple Point: The only condition under which all three phases can be in equilibrium with one another. Critical Temperature (Tc): The temperature above which the gas phase cannot be made to liquefy at any pressure. Critical Pressure (Pc) : The minimum pressure required to liquefy a gas at its critical temp. Supercritical Fluid: Neither true liquid nor true gas Normal boiling and melting point always at 1 atm
41. Example 9 Can you label the following? a) solid region b) Liquid region c) Gas region d) Normal boiling point e) Normal melting point f) Triple point g) Supercritical fluid region h) Critical point, what is the critical pressure and temperature
42. Types of solids 1. Molecular solid -held together by intermolecular forces -H2O(s), CO2(s) 2. Metallic solid -positively charged atomic cores surrounded by delocalized electrons -Fe, Cu, Ag
43. Types of solids 3. Ionic solid -cations and anions held together by electrical attraction of opposite charges -NaCl 4. Covalent network solid -atoms held together in large networks by covalent bonds -diamonds | 2,670 | 10,412 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2021-25 | latest | en | 0.757056 |
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Post subject: Need help pleasePosted: Sun Mar 10, 2013 9:14 pm
Joined: Sun Jan 20, 2013 10:22 am
Posts: 73
Question
Ian is now twice his sister's age. In 4 years' time Ian will be 16. How old will his sister be then?
Ans : 10
It's not clear from the question,which one we have to find, current age of Ian's sister or in 4 years time? Based on the answer, i can see sister age in 4 years will be 10.
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Post subject: Re: Need help pleasePosted: Sun Mar 10, 2013 9:26 pm
Joined: Sat Mar 06, 2010 11:39 pm
Posts: 2080
then = in 4 years' time
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Post subject: Re: Need help pleasePosted: Sun Mar 10, 2013 9:28 pm
Joined: Wed Jan 18, 2012 11:41 am
Posts: 4587
Location: Essex
Seriously, you do have all the info you need to work this one out....
Hint - in 4 years' time, he will be 16, so at the moment he is 16 minus those 4 years. The answer to that is twice the current age of his sister.
_________________
Outside of a dog, a book is a man's best friend. Inside of a dog it's too dark to read.Groucho Marx
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Post subject: Re: Need help pleasePosted: Sun Mar 10, 2013 9:58 pm
Joined: Sun Jan 20, 2013 10:22 am
Posts: 73
Thank you very much for your help.
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Jump to: Select a forum ------------------ FORUM RULES Forum Rules and FAQs 11 PLUS SUBJECTS VERBAL REASONING MATHS ENGLISH NON-VERBAL REASONING CEM 11 Plus GENERAL GENERAL 11 PLUS TOPICS 11 PLUS APPEALS 11 PLUS TUTORS INDEPENDENT SCHOOLS 11 PLUS CDs/SOFTWARE 11 PLUS TIPS PRIMARY SEN and the 11 PLUS EVERYTHING ELSE .... 11 PLUS REGIONS Berkshire Bexley and Bromley Birmingham, Walsall, Wolverhampton and Wrekin Buckinghamshire Devon Dorset Essex Essex - Redbridge Gloucestershire Hertfordshire (South West) Hertfordshire (Other and North London) Kent Lancashire & Cumbria Lincolnshire Medway Northern Ireland Surrey (Sutton, Kingston and Wandsworth) Trafford Warwickshire Wiltshire Wirral Yorkshire BEYOND 11 PLUS Beyond 11 Plus - General GCSEs 6th Form University | 821 | 2,698 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-50 | latest | en | 0.874273 |
https://ivyhawnschool.org/classroom-connect/fourth-grade/mariner-a/e-learning-april-20th-april-24th/ | 1,591,219,155,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347436466.95/warc/CC-MAIN-20200603210112-20200604000112-00207.warc.gz | 397,747,304 | 9,668 | # E-Learning April 20th- April 24th
Teacher Angela Mariner Math and Science 4
Standard(s) Taught
Learning Targets and Learning Criteria
Classroom Activities
Monday- April 20th
Math
Look for YouTube video posted on Bloomz (by 11am)
IXL (80% or higher)- Z.3 Measure angles on a circle
Science
IXL (80% or higher)- O.1 What affects traits? Use observations to support a hypothesis.
Study vocab
Tuesday- April 21st
Math
Look for YouTube video posted on Bloomz (by 11am)
IXL (80% or higher)- Z.4 Measure angles with a protractor
Science
IXL (80% or higher)- O.2 Match offspring to parents using inherited traits
Study vocab
Wednesday-April 22nd
Math
No Video today
IXL- (80% or higher)- (Literally between Z.4 and Z.5) New! Draw angles with a protractor
Science
IXL (80% or higher)- O.3 Identify inherited and acquired traits
Study Vocab
Thursday- April 23rd
Math
Look for YouTube video posted on Bloomz (by 11am)
IXL- (80% or higher)- Z.5 Estimate Angle Measurements
Science
IXL (80% or higher)- O.4 Identify inherited and acquired traits: Use evidence to support a statement
Study Vocab
Friday- April 24th
Math
IXL- (80% or higher)- Z.6 Adjacent angles
i-Ready 45 minutes total for the week, and 2 quizzes
Science
Graded assignment- Animal life cycle vocab
Assignments Due
Study Animal Life Cycle Vocab
Life cycle- the series of changes in the life of an organism.
Instinct- what some animals use in order to eat, drink, or rest in order to meet their needs. These animals do not need to be taught how to do certain things, they just automatically know.
Reproduction- the act, or process, of creating a copy of something, including offspring.
Worker- a female bee
Drone- a male bee
Tadpole- the tailed aquatic larva of an amphibian (frog, toad, newt, or salamander), breathing through gills and lacking legs until the later stages of its development.
Molting- when a caterpillar, or larva, sheds is skin.
Chrysalis- another word for a butterfly pupa where the metamorphic stage takes place.
Nectar- a sugary fluid secreted by plants, especially within flowers to encourage pollination by insects and other animals. It is collected by bees to make into honey.
Honey- a sweet, sticky yellowish-brown fluid made by bees and other insects from nectar collected from flowers.
K-selected- only producing a few offspring.
R-selected- producing many offspring in hopes that some survive.
Offspring- a person or an animal’s child. | 631 | 2,468 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2020-24 | latest | en | 0.901808 |
http://mathhelpforum.com/discrete-math/141691-triangular-count-print.html | 1,511,023,519,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934804976.22/warc/CC-MAIN-20171118151819-20171118171819-00027.warc.gz | 190,081,690 | 2,539 | # triangular count
Printable View
• Apr 27th 2010, 07:22 AM
peaz9482
triangular count
How many equilateral triangles are there in an eight-fold triangular grid. I was able to determine the number of 1x1, 2x2, 3x3, 4x4, 5x5, 6x6, 7x7, and 8x8 size triangles to be 64, 43, 27, 16, 10, 3, 1 respectively and the total number of triangles to be 170. I am having a hard time though coming up with a general formula to show for example how many 4x4 triangles are in a nine-fold triangle. I thought it had something to do with triangular numbers, but the pattern doesn't hold. Thanks for any help you can provide!! | 188 | 609 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2017-47 | longest | en | 0.941273 |
http://nrich.maths.org/public/leg.php?code=32&cl=2&cldcmpid=1114 | 1,503,019,125,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886104204.40/warc/CC-MAIN-20170818005345-20170818025345-00451.warc.gz | 296,620,865 | 9,877 | # Search by Topic
#### Resources tagged with Multiplication & division similar to The Magic Number and the Hepta-tree:
Filter by: Content type:
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Challenge level:
### There are 165 results
Broad Topics > Calculations and Numerical Methods > Multiplication & division
### What's My Weight?
##### Stage: 2 Short Challenge Level:
There are four equal weights on one side of the scale and an apple on the other side. What can you say that is true about the apple and the weights from the picture?
### Which Is Quicker?
##### Stage: 2 Challenge Level:
Which is quicker, counting up to 30 in ones or counting up to 300 in tens? Why?
### A Square of Numbers
##### Stage: 2 Challenge Level:
Can you put the numbers 1 to 8 into the circles so that the four calculations are correct?
### Highest and Lowest
##### Stage: 2 Challenge Level:
Put operations signs between the numbers 3 4 5 6 to make the highest possible number and lowest possible number.
### Factor-multiple Chains
##### Stage: 2 Challenge Level:
Can you see how these factor-multiple chains work? Find the chain which contains the smallest possible numbers. How about the largest possible numbers?
### Divide it Out
##### Stage: 2 Challenge Level:
What is the lowest number which always leaves a remainder of 1 when divided by each of the numbers from 2 to 10?
### All the Digits
##### Stage: 2 Challenge Level:
This multiplication uses each of the digits 0 - 9 once and once only. Using the information given, can you replace the stars in the calculation with figures?
### Multiplication Square Jigsaw
##### Stage: 2 Challenge Level:
Can you complete this jigsaw of the multiplication square?
### Trebling
##### Stage: 2 Challenge Level:
Can you replace the letters with numbers? Is there only one solution in each case?
### Today's Date - 01/06/2009
##### Stage: 1 and 2 Challenge Level:
What do you notice about the date 03.06.09? Or 08.01.09? This challenge invites you to investigate some interesting dates yourself.
### What's in the Box?
##### Stage: 2 Challenge Level:
This big box multiplies anything that goes inside it by the same number. If you know the numbers that come out, what multiplication might be going on in the box?
### Code Breaker
##### Stage: 2 Challenge Level:
This problem is based on a code using two different prime numbers less than 10. You'll need to multiply them together and shift the alphabet forwards by the result. Can you decipher the code?
### Four Go
##### Stage: 2 Challenge Level:
This challenge is a game for two players. Choose two numbers from the grid and multiply or divide, then mark your answer on the number line. Can you get four in a row before your partner?
### Clever Santa
##### Stage: 2 Challenge Level:
All the girls would like a puzzle each for Christmas and all the boys would like a book each. Solve the riddle to find out how many puzzles and books Santa left.
### What Two ...?
##### Stage: 2 Short Challenge Level:
56 406 is the product of two consecutive numbers. What are these two numbers?
### Countdown
##### Stage: 2 and 3 Challenge Level:
Here is a chance to play a version of the classic Countdown Game.
### Tom's Number
##### Stage: 2 Challenge Level:
Work out Tom's number from the answers he gives his friend. He will only answer 'yes' or 'no'.
### The Clockmaker's Birthday Cake
##### Stage: 2 Challenge Level:
The clockmaker's wife cut up his birthday cake to look like a clock face. Can you work out who received each piece?
### ABC
##### Stage: 2 Challenge Level:
In the multiplication calculation, some of the digits have been replaced by letters and others by asterisks. Can you reconstruct the original multiplication?
### Four Go for Two
##### Stage: 2 Challenge Level:
Four Go game for an adult and child. Will you be the first to have four numbers in a row on the number line?
### Being Thoughtful - Primary Number
##### Stage: 1 and 2 Challenge Level:
Number problems at primary level that require careful consideration.
### 1, 2, 3, 4, 5
##### Stage: 2 Challenge Level:
Using the numbers 1, 2, 3, 4 and 5 once and only once, and the operations x and ÷ once and only once, what is the smallest whole number you can make?
### Asteroid Blast
##### Stage: 2 Challenge Level:
A game for 2 people. Use your skills of addition, subtraction, multiplication and division to blast the asteroids.
### One Million to Seven
##### Stage: 2 Challenge Level:
Start by putting one million (1 000 000) into the display of your calculator. Can you reduce this to 7 using just the 7 key and add, subtract, multiply, divide and equals as many times as you like?
##### Stage: 2 Challenge Level:
Use the information to work out how many gifts there are in each pile.
### Zios and Zepts
##### Stage: 2 Challenge Level:
On the planet Vuv there are two sorts of creatures. The Zios have 3 legs and the Zepts have 7 legs. The great planetary explorer Nico counted 52 legs. How many Zios and how many Zepts were there?
### Which Symbol?
##### Stage: 2 Challenge Level:
Choose a symbol to put into the number sentence.
### Function Machines
##### Stage: 2 Challenge Level:
If the numbers 5, 7 and 4 go into this function machine, what numbers will come out?
### Pies
##### Stage: 2 Challenge Level:
Grandma found her pie balanced on the scale with two weights and a quarter of a pie. So how heavy was each pie?
##### Stage: 3 Challenge Level:
If you take a three by three square on a 1-10 addition square and multiply the diagonally opposite numbers together, what is the difference between these products. Why?
### Practice Run
##### Stage: 2 Challenge Level:
Chandrika was practising a long distance run. Can you work out how long the race was from the information?
### Zargon Glasses
##### Stage: 2 Challenge Level:
Zumf makes spectacles for the residents of the planet Zargon, who have either 3 eyes or 4 eyes. How many lenses will Zumf need to make all the different orders for 9 families?
### The Deca Tree
##### Stage: 2 Challenge Level:
Find out what a Deca Tree is and then work out how many leaves there will be after the woodcutter has cut off a trunk, a branch, a twig and a leaf.
### Four Goodness Sake
##### Stage: 2 Challenge Level:
Use 4 four times with simple operations so that you get the answer 12. Can you make 15, 16 and 17 too?
### Rabbits in the Pen
##### Stage: 2 Challenge Level:
Using the statements, can you work out how many of each type of rabbit there are in these pens?
### Route Product
##### Stage: 2 Challenge Level:
Find the product of the numbers on the routes from A to B. Which route has the smallest product? Which the largest?
### Sort Them Out (2)
##### Stage: 2 Challenge Level:
Can you each work out the number on your card? What do you notice? How could you sort the cards?
### Sept03 Sept03 Sept03
##### Stage: 2 Challenge Level:
This number has 903 digits. What is the sum of all 903 digits?
### It Was 2010!
##### Stage: 1 and 2 Challenge Level:
If the answer's 2010, what could the question be?
### Carrying Cards
##### Stage: 2 Challenge Level:
These sixteen children are standing in four lines of four, one behind the other. They are each holding a card with a number on it. Can you work out the missing numbers?
### Clock Face
##### Stage: 2 Challenge Level:
Where can you draw a line on a clock face so that the numbers on both sides have the same total?
### X Is 5 Squares
##### Stage: 2 Challenge Level:
Can you arrange 5 different digits (from 0 - 9) in the cross in the way described?
### How Much Did it Cost?
##### Stage: 2 Challenge Level:
Use your logical-thinking skills to deduce how much Dan's crisps and ice-cream cost altogether.
### Amy's Dominoes
##### Stage: 2 Challenge Level:
Amy has a box containing domino pieces but she does not think it is a complete set. She has 24 dominoes in her box and there are 125 spots on them altogether. Which of her domino pieces are missing?
### Difficulties with Division
##### Stage: 1 and 2
This article for teachers looks at how teachers can use problems from the NRICH site to help them teach division.
### Napier's Bones
##### Stage: 2 Challenge Level:
The Scot, John Napier, invented these strips about 400 years ago to help calculate multiplication and division. Can you work out how to use Napier's bones to find the answer to these multiplications?
### Penta Post
##### Stage: 2 Challenge Level:
Here are the prices for 1st and 2nd class mail within the UK. You have an unlimited number of each of these stamps. Which stamps would you need to post a parcel weighing 825g?
### Multiples Sudoku
##### Stage: 3 Challenge Level:
Each clue in this Sudoku is the product of the two numbers in adjacent cells.
### A First Product Sudoku
##### Stage: 3 Challenge Level:
Given the products of adjacent cells, can you complete this Sudoku?
### Multiplication Squares
##### Stage: 2 Challenge Level:
Can you work out the arrangement of the digits in the square so that the given products are correct? The numbers 1 - 9 may be used once and once only. | 2,139 | 9,122 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2017-34 | latest | en | 0.846572 |
https://mathoverflow.net/questions/194986/integrability-conditions-of-lax-pairs/195956 | 1,618,586,656,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038066981.0/warc/CC-MAIN-20210416130611-20210416160611-00334.warc.gz | 495,864,518 | 31,063 | # Integrability - conditions of lax pairs
I'm trying to understand what the conditions are for the Lax pairs for the zero-curvature representation:
$$\partial_t U - \partial_x V + [U,V]=0$$
where $U=U(x,t,\lambda)$ and $V=V(x,t,\lambda)$ are matrix-valued functions and $\lambda$ is a parameter.
The motivation behind this question is that the Lax pairs for the KdV equation:
$$u_t + 6uu_x - u_{xxx} = 0$$
is given by:
$$U = \begin{pmatrix} 0 & 1 \\ \lambda + u & 0 \end{pmatrix} \text{ and } V = \begin{pmatrix} u_x & 4 \lambda - 2u \\ 4 \lambda^2 + 2 \lambda u + u_{xx} - 2u^2 & - u_x \end{pmatrix}$$
Now, it is not too difficult to verify that this indeed satisfies the zero-curvature representation, but I'm trying to figure out why we cannot use the Lax pairs:
$$U = \begin{pmatrix} 0 & 0 \\ \lambda + u & 0 \end{pmatrix} \text{ and } V = \begin{pmatrix} 0 & 0 \\ \lambda + 3 u^2 - u_{xx} & 0 \end{pmatrix}$$
These matrices clearly satisfy the zero-curvature representation, but for some reason none of the notes I've been reading use them. What is the reason that they are not a valid Lax pair for the KdV equation?
I've also asked this question here (I hope that's ok): http://www.physicsoverflow.org/26475/integrability-conditions-of-lax-pairs
One way to see this, is that you want the zero-curvature representation to be useful and tell you something you didn't know before. Your representation has the problem of being singular, in the sense that the Lax matrices have zero determinant. It would be more eveident if we were really speaking of the Lax representation: $$L_t=[M,L]$$ where all the usefulness comes from having the possibility to say that the eigenvalues (or the coefficients of the characteristic polynomial) of $L$ do not evolve (and hence are integrals of motion). You see that, if you take an $L$ for which the coefficients of the characteristic polynomials are identically zero, you don't achieve much. This very same argument applies to the zero curvature representation if you remember how to pass from it to the Lax representation via the monodromy matrix. In any case, even if you ignore for the time being this problem and try to go through inverse scattering, you very soon hit the same wall. | 605 | 2,238 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2021-17 | latest | en | 0.890539 |
https://proofwiki.org/wiki/Increasing_Union_of_Sequence_of_Ideals_is_Ideal | 1,627,155,990,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046150308.48/warc/CC-MAIN-20210724191957-20210724221957-00677.warc.gz | 476,548,962 | 10,309 | # Increasing Union of Ideals is Ideal/Sequence
## Theorem
Let $R$ be a ring.
Let $S_0 \subseteq S_1 \subseteq S_2 \subseteq \dotsb \subseteq S_i \subseteq \dotsb$ be ideals of $R$.
Then the increasing union $S$:
$\displaystyle S = \bigcup_{i \mathop \in \N} S_i$
is an ideal of $R$.
## Proof
Let $\displaystyle S = \bigcup_{i \mathop \in \N} S_i$.
From Increasing Union of Subrings is Subring, we have that $S$ is a subring of $R$.
Now we need to show that it is an ideal of $R$.
Let $a \in S$.
Then $\exists i \in \N: a \in S_i$.
Let $b \in R$.
Then $a b \in S_i$ and $b a \in S_i$, as $S_i$ is an ideal of $R$.
Thus $a b \in S$ and $b a \in S$.
So $S$ is an ideal of $R$.
$\blacksquare$ | 262 | 705 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2021-31 | latest | en | 0.799863 |
http://mathhelpforum.com/calculus/160174-values-x-sin-1-x-when-x-does-not-equal-0-0-0-point-print.html | 1,526,888,205,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863967.46/warc/CC-MAIN-20180521063331-20180521083331-00619.warc.gz | 187,331,381 | 2,636 | # Values of a for x^a sin(1/x) when x does not equal 0 and (0,0) is a point.
• Oct 18th 2010, 06:34 PM
WorkDayNNight
Values of a for x^a sin(1/x) when x does not equal 0 and (0,0) is a point.
i) Determine "a" values for which f is differentiable at 0 and determine the f'(o) values.
ii) Determine "a" values for which f' is continuous at 0.
iii) Determine "a" values for which f"(0) exists and determine its value(s).
• Oct 19th 2010, 03:57 AM
HallsofIvy
Have you tried anything? Like writing down the formula for the derivative at 0:
$\displaystyle \lim_{h\to 0}\frac{h^a sin(1/h)- 0}{h}= \lim_{h\to 0}h^{a-1}sin(1/h)$
Does that give you any ideas.
(I assume that your "and (0, 0) is a point" means that (0, 0) is a point in the graph of y= f(x)- in other words, that f(0)= 0.) | 283 | 782 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2018-22 | latest | en | 0.870751 |
https://www.doubtnut.com/question-answer/if-the-line-2x-y-10-touches-the-circle-at-the-point-25-and-the-centre-of-the-circle-lies-in-the-line-21406 | 1,623,539,572,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487586465.3/warc/CC-MAIN-20210612222407-20210613012407-00037.warc.gz | 646,455,984 | 78,203 | Class 11
MATHS
The Circle
# If the line 2x-y+1=0 touches the circle at the point (2,5) and the centre of the circle lies in the line x+y-9=0. Find the equation of the circle.
Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.
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Text Solution
Solution :
2x-y+1=0<br> Equation of Normal<br> Y-Y_1=m(x-x_1)<br> Y-S=-1/2(x-2)<br> Y-S=-1/2x+1<br> 2y-10+x-1=0<br> 2y+x-11=0<br> Equation of number.<br> r=sqrt((7-2)^2+(9-5)^2)=sqrt34<br> Equation of line<br> (x-7)^2+(y-2)^2=34.
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5:49 | 514 | 1,165 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2021-25 | latest | en | 0.481971 |
http://www.planetmath.org/exampleofinfinitesimplegroup | 1,529,500,235,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863519.49/warc/CC-MAIN-20180620124346-20180620144346-00137.warc.gz | 484,428,460 | 3,797 | # example of infinite simple group
This fact that finite alternating groups are simple can be extended to a result about an infinite group. Let $G$ be the subgroup of the group of permutations on a countably infinite set $M$ (which we may take to be the set of natural numbers for concreteness) which is generated by cycles of length $3$. Note that any since every element of this group is a product of a finite number of cycles, the permutations of $G$ are such that only a finite number of elements of our set are not mapped to themselves by a given permutation.
We will now show that $G$ is simple. Suppose that $\pi$ is an element of $G$ other than the identity. Let $m$ be the set of all $x$ such that $\pi(x)\neq x$. By our previous comment, $m$ is finite. Consider the restriction $\pi_{m}$ of $\pi$ to $m$. By the theorem of the parent entry (http://planetmath.org/SimplicityOfA_n), the subgroup of $A_{m}$ generated by the conjugates of $\pi_{m}$ is the whole of $A_{m}$. In particular, this means that there exists a cycle of order $3$ in $A_{m}$ which can be expressed as a product of $\pi_{m}$ and its conjugates. Hence the subgroup of $G$ generated by conjugates of $\pi$ contains a cycle of length three as well. However, every cycle of order $3$ is conjugate to every other cycle of order $3$ so, in fact, the subgroup of $G$ generated by the conjugates of $\pi$ is the whole of $G$. Hence, the only normal subgroups of $G$ are the group consisting of solely the identity element and the whole of $G$, so $G$ is a simple group.
Title example of infinite simple group ExampleOfInfiniteSimpleGroup 2013-03-22 16:53:31 2013-03-22 16:53:31 rspuzio (6075) rspuzio (6075) 8 rspuzio (6075) Example msc 20E32 msc 20D06 | 469 | 1,728 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 30, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2018-26 | latest | en | 0.907064 |
https://www.physicsforums.com/threads/uniform-circular-motion-problem-help.889060/ | 1,532,356,350,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676596463.91/warc/CC-MAIN-20180723125929-20180723145929-00593.warc.gz | 953,720,180 | 14,258 | # Homework Help: Uniform Circular Motion Problem. Help!
1. Oct 13, 2016
### aki_23
1. The problem statement, all variables and given/known data
A stone tied to a string with length 1.10 m is whirled in a circle both horizontally and vertically with the same constant speed. In the vertical case the maximum tension in the string is 15.0% larger than the tension that exists when the string was horizontal. Determine the speed of the stone.
2. Relevant equations
a = v2/r
3. The attempt at a solution
for horizontal, my equation was Fc = mv2/r
for vertical, my equation was Fc = Ft - mg, because at the bottom of the vertical circle fg is present
mv2/r = 0.15Ft - mg
I dont know how to take this further
2. Oct 13, 2016
### haruspex
Consider the horizontal case in more detail. There are vertical forces too. You should be able to find a relationship between the speed, the string length, and the tension. | 235 | 914 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-30 | latest | en | 0.900059 |
https://www.mockbank.com/bulletin/bodmas-rule-with-sign-replacement/ | 1,618,439,838,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038078900.34/warc/CC-MAIN-20210414215842-20210415005842-00395.warc.gz | 1,022,252,363 | 22,749 | ## BODMAS Rule with Sign Replacement: Check Here For Simplification Tricks
The number series in the aptitude section is a very important topic for competitive exams. This topic is divided into several small topics and one such topic is a simplification. In this topic the questions asked is related to the BODMAS rule. This rule is very common in mathematics. It stands for Bracket Of Division Multiplication Addition and Subtraction. Thus, today we are going to cover simplification on BODMAS rule. Though this area may have less of direct questions appearing in most of the competitive exams, but its indirect application makes this topic very relevant from the Quant section point of view. Simplification or simplify fractions means to simplify a complicated mathematical expression to get a single or direct answer.
Now our basic understanding of mathematical operations is strong. We know how to add, subtract, multiply and divide even large numbers. But how do we solve a question that involves all of these operations? Come let us find out using the BODMAS rule.
We have shown here 2 tricks to solve the Simplification Question:
#### Trick 1: Sign Replacement
• Multiply replace with subtraction: * => –
• Addition replace with division: + => /
• Division replace with addition: / => +
• Subtraction replace with addition: – =>+
Trick 1: BODMAS Rule
• Bracket (Brackets are solved in order of (), {} and [] respectively.
• Orders (i.e. power and square root)
• Division and Multiplication (Left to right)
• Addition and Subtraction (Left to right)
### Steps to solve BODMAS Question
Step 1: Always remember to solve the expressions that are in the bracket first. Also, while solving these brackets you need to apply the BODMAS rule first.
Step 2: After solving the brackets, you need to solve the mathematical operators ‘order’ and ‘of’ next. Here, ‘of’ means a part of and it is solved by replacing it with a multiplication sign. While ‘order’ is similar to an exponent. After the brackets are solved, powers are solved. This also includes the roots.
Step 3: In the next step, the equations that have multiplication and division sign in it is solved left to right. You need to calculate this after completing the above 2 steps.
Step 4: At the end, you need to solve the equations left to right that contain ‘subtraction’ and ‘addition’. Here you can notice that these steps also followed the BODMAS rule.
#### Example 1: 15+2/800+80×100 =?
• Replace all sign using the trick 1 (i.e. 15/2+800/80-100).
• After that apply BODMAS rule left to right (i.e. 7.5 + 800/80-100)
• Then divide 800 to 80 (i.e. 7.5 + 10 -100)
• Solve left to right (i.e. 17.5-100)
### Example 2: 36+18/12-44×30 =?
• Replace all sign using the trick 1 (i.e. 36/18*12+44-30).
• After that apply BODMAS rule left to right (i.e. 2*12+44-30 )
• Then multiply 2 to 12 (i.e. 24+44-30 )
• Solve left to right (i.e. 68-30 )
### Simplifying the Brackets
The expansion of brackets is known as the simplification of the brackets. To remove the brackets from an expression, you need to expand them by multiplying them. Also, for this step, you can use the distributive law. It is written as
a(b + c) = ab + bc
For brackets also there is a rule that you need to follow while simplifying them. You need to follow the order of (), {}, [ ] for solving the brackets. This means that you solve the equations in the () bracket first and then the other two.
Related Article: | 857 | 3,455 | {"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} | 4.8125 | 5 | CC-MAIN-2021-17 | longest | en | 0.913019 |
http://openstudy.com/updates/50945eafe4b0c875f79e33eb | 1,448,871,181,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398461132.22/warc/CC-MAIN-20151124205421-00096-ip-10-71-132-137.ec2.internal.warc.gz | 174,057,317 | 12,551 | ## janessia 3 years ago If the point (-1, 3) is translated according to the rule (x, y) → (x + 4, y – 2), and then reflected across the x-axis, what are the coordinates of the final resulting image? Choose one answer. a. (-1,3) b. (-3,-1) c. (3,-1) d. (3,1)
1. jim_thompson5910
First use the rule (x, y) → (x + 4, y – 2) how do we do this?
2. janessia
to be honest i really do not know i never learned this
3. jim_thompson5910
When they say (x, y) → (x + 4, y – 2), they mean "add 4 to the x coordinate and subtract 2 from the y coordinate" Ex: (2,3) turns into (6, 1) because we add 4 to the x coordinate to get 2+4 = 6 and subtract 2 from the y coordinate to get 3-2 = 1.
4. thinker1
c?
5. janessia
why do u think that?
6. thinker1
|dw:1352241183944:dw|
7. janessia
ohk thnks
8. thinker1
ur mot welcome | 298 | 820 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2015-48 | longest | en | 0.787937 |
https://www.savespendsplurge.com/how-i-got-little-bun-to-read-and-do-math-by-the-age-of-3/?replytocom=43500 | 1,582,936,787,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875148163.71/warc/CC-MAIN-20200228231614-20200229021614-00137.warc.gz | 876,452,324 | 36,425 | So a little bit of background, Little Bun is 5 at the time of this writing, and in brief:
English – He reads my emails out loud, he understands instructions in his activity books on his own and he can spell pretty well, but longer words like Christmas, he needs help with. He is at a Grade 2/3 level right now.
He can understand about 75% what he is reading too, not just read the words, but get the message. He practices reading comprehension and word problems on emails my mother writes to him and from activity books.
Math – He can do addition and subtraction with 2-digits in his head, he can do simple (up to 12X) multiplication and division in his head, and for long multiplication or division, he does it on paper.
He knows all of his squares up to 12, he knows simple fractions, can do BEDMAS with simple operations, decimals we are working on, knows basic geometry, and we will be doing easy algebra next after he masters long division and decimals.
To get him there, this is what I did.
PLEASE. NOTE. This is just what I did, it doesn’t mean it will work on every child, and that it is the best strategy. I am only talking about what I did and if it works for you too, great!
So before we begin….
# Do not force them to do anything
I never made him study or do anything on a schedule at any age, at any time.
I just suggest things, and sometimes he suggests new games like – Mommy, let’s do the long division game!
If he wants to stop, we stop. I don’t force him, I don’t yell at him, I repeat, I show him the same concept over and over again, and I test him to see if he understood it or not, but I NEVER force him to sit down and do it.
# Understand that each child is different
My child is introverted, quiet, he doesn’t like a lot of stimulation or action, and is overwhelmed by it. Just like we are. Maybe he inherited it.
He likes to read, he asks a lot of questions, he is very curious and playful, and your child is too, but to some level. Do not compare children to each other but play to their strengths as individuals – the same way it is done with us as adults.
My friend tried playing with number cards with her daughter (same age) and she couldn’t care less about it. *shrug* Don’t force them.
# Make it a game
It is fun. I make math fun, I am genuinely excited about it, I love discovering new things in books, and many books make it so easy to learn that I wish I had them as a child.
But make it fun. Funny gestures, funny noises, etc.
He even came up with a Stuffed Animal Math Journey game on his own, where the animal had to go through checkpoints and to make it through he had to do math, score “bonus” points by getting the answer quickly, and hit “high scores”.
ALL OF THIS WORKS. He loved doing equations to get his stuffed animal to the finish line.
# Children are sponges
Their brains until about the age of 5-6 is so open to new concepts and learning that you can teach them ANYTHING and they will pick it up.
I heard this from two math teachers, and I immediately started when he was 18 months because I wanted to see where it could go – couldn’t hurt right?
I was blown away by how much he picked up, and now he has advanced beyond what I ever thought was possible.
They can truly absorb all sorts of concepts, and math is the best time to be introduced at this age so that they are unfettered by what they know of numbers.
I thought he would just know his alphabet and count to 100. This child is doing simple fractions now and once we master that, we will do some basic algebra or geometry (which he already has learned a little)…
# ENGLISH
## Flash Cards
I made my own flash cards in French and English, for each alphabet letter in uppercase and lowercase letters in fun colours with pictures.
A – Abeille (Bee) and A – Apple for instance.
I did not take their cards specifically, because I wanted other words for certain letters in either language, and so I made my own, and printed them at Staples on thick white paper so they would be durable.
I’d ask him to find the matches for each of the letters, A with A, and then using flash cards to spell out titles on books.
So he’d look at GOODNIGHT MOON, and say:
G – Giraffe
O – Octopus
O – Octopus
D – Dog
M – Moon
… etc.
He used those flash card words to identify them with each of the book title letters. This was interminably long, because each time we’d read a book, he’d painstakingly “spell” the titles.
But it’s okay, it worked for him.
## Sound out the letters
This is apparently an outdated form of teaching, but I found it helped him see what the word would sound like if he sounded it out.
You know, A for “AHHH”, B for “BUHHH”…
They don’t do this in schools any more, and kids are supposed to look at the word as a whole and know/memorize it, but it is helpful for larger words for him to know what it might sound like.
It has now become natural for him to sound it out, but not with “buh buh kuh kuh” sounds, just with seeing the syllables in the word as a whole, if that makes sense.
You also read much faster if you read the whole word at once rather than spelling it out.
## Free-Spelling Cards
Later, as he became more advanced, I found a game in a thrift store for \$3, and wrote my own letters in the back in fun colours with a sharpie so that he could free spell his own words, sentences, etc.
He spelled this at .. age 3 ? And we didn’t have enough for DADDY, so he told me to go to the store to “buy more Daddy”. LOL.
## Slowly going through books
From there, we started reading books that were easy, mostly Eric Carle-style books, and I would tap each word slowly, read it out loud, and encourage him to recognize words.
I’d say – The cow jumped over the ..?? … and I’d pause. If he didn’t respond within 10 seconds, I’d say – MOON! .. M-O-O-N spells MOON!
By about age 2, he was spelling at the back of the car, telling me D-O-G spelled DOG. I almost crashed the car when I heard his baby voice tell me different spellings of words!
By the age of 3, he was fully reading on his own with baby books. We did shared reading – I’d do one page, he’d do another.
He still preferred me to read the book, and was having a little trouble with comprehension with new books/words, but practice makes perfect and I’d ask him – did you understand what the book was about? My goodness look at that silly cat! .. and I’d explain the book to him, and then we’d read it again and he would absorb more of the words and their meanings.
By about age 4, he was reading slightly harder books on his own, but then he started understanding what he was reading – not just reading it or pronouncing it, but getting what the message meant.
Each time we came across a word I wasn’t sure he’d know, I’d explain it in another way. He once read: Out of Exile on my playlist in the car, and I launched into a monologue about Exile and what that word meant – not being allowed to come back (that was the gist of my explanation).
Later, I used it on him and told him – Little Bun! If you don’t behave, you’re going to be exiled!!! … and he seemed to “get” what that word meant.
I do this with all words I think are tricky. I explain everything. I talk until I am hoarse, and I am always using new words, explaining, etc.
He asks me sometimes what words mean and I try my best to explain it to him, using real-life examples.
# MATH
For math, we started with him playing on the iPad with my partner. He is the one who kicked off his love for numbers with the Calculator app of all things.
Little Bun LOVED seeing numbers pop up on the calculator, and soon my partner had him saying numbers by the age of 2 or so.
I found little matching number cards at a thrift store for \$1 that looked like fun. I used them to teach him numbers, put them in order,
Same thing – I made a game out of it. I made him put numbers in order, and by the age of 2.5 he was recognizing numbers into the hundreds of thousands.
I’d stack down 3627 and he would say: three thousand, six hundred, twenty-seven.
I repeated over and over again this game (it was fun, not forced), and even up to the hundreds of thousands but he had trouble recognizing the invisible comma between the groups of 3 numbers.
By age 3.5 he was doing simple addition with my partner on the calculator, which I then mirrored and reinforced.
We did addition first, then subtraction.
He was using his fingers a lot between the ages of 3-4. He still sort of reverts to it, but now at age 5, he is doing subtraction in his head with the tens and ones.
You can give him any number like 35 minus 15 and he can do it in his head now.
## Use Visual Methods and add funny gestures
I had a very visual method to teach him math, using paper, and crossing out numbers, etc. I tried to show him with circles, with little pictures, anything.
For addition, I’d teach him to add the ones first, then carry the tens over.
For subtraction, I’d teach him to try and subtract the ones first, then borrow from the tens.
Very basic, simple, fun stuff. I’d add in funny gestures like: UH OH! We don’t have enough ones to subtract! What should we do?
If he made mistakes, I’d correct them and reinforce it with more examples so he learned it properly.
We moved from one digit addition and subtraction to two-digit, then three-digit, then 4-digit until I was satisfied he got it.
After he did that, we moved on to simple multiplication and division.
Again, one-digit, then two-digit, etc.
Multiplication, he straight up memorized them. There is no other way.
He wanted to move on to division because he thought multiplication was too easy and it got to the point where I told him he had to memorize his multiplication COLD before division, because the two are fact families.
So, he studied when I went to work. I know this because he practically slept with his Usborne Times Table Book.
## Try cute concepts like Fact Families
He likes the concept of ‘families’ and ‘fact families’ and now, will occasionally give me an answer in the form of a riddle.
If I ask him: What is 20 plus 5?
(Yes he also memorized all the squares up to 12.)
I tell him that division is the opposite of multiplication, and he loves the idea of how all the numbers fit together in one big puzzle.
## Break down the numbers
With long division, I showed him how to divide it out, and multiplication how to multiply it out and look at the numbers by breaking them down.
25 is two tens and one 5 for instance, and I’d work with him to puzzle it out.
I’d say: What is two times 25? What is two times 5? Add those two together! That’s your answer.
# Now we are currently on fractions and decimals…
He just did this page by himself when I was at work.
Please note his super cute addition of the 3/3 = 2/2 line, to show that they are BOTH one whole fractions. <3
This is quite easy for him now, and I am going to start doing larger fractions and making him break it down with division or to build it up to a common denominator and do some addition or subtraction with it.
I also taught him BEDMAS – Brackets, Exponents, Decimals, Multiplication, Addition, Subtraction, and that DM = equivalent operations, and AS = equivalent operations.
I gave him simple BEDMAS equations to puzzle out and he loves them because they challenge him.
# My book & app recommendations
This whole right side? ALL LITTLE BUN.
I used a mix of books and iPad apps.
Say what you will about screen time, those apps are amazing. Kids love them, they’re fun, interactive, and he learned SO MUCH from it.
When you make learning fun, kids want to do it. If you make learning boring, a chore, and something you have to do, that stresses everyone out and achieves nothing except make the child hate learning… and school.
For activity books, I find Sylvan Learning to be the most fun, with colours and interesting problems, ideas, etc.
For smaller books that are easier to carry or tote around on vacation, you can look at the FlashKids set which is fun, but repetitive to the point of boredom (for me anyway) and without colour.
Lastly, for general fun learning:
## ENGLISH
If you want to see his entire library of books, the full book list of what is on our bookshelves is here.
These apps I found helpful for English:
• Endless Monsters – they have alphabet, Spanish, numbers, the whole range – they are SO MUCH FUN and cute.
• Mobile Montessori – Again also has numbers and shapes, etc.
• Mindsnacks – Lots of languages, English and French are well done, fun games to play with but the pictures can be a bit strange – they’re photos rather than created images for the app (and odd but not creepy ones at that sometimes), so.. be forewarned it may not be as buttoned up as you expect…
## MATH
Hands down, this Usborne trio set on math has been transformational. I know Little Bun is a bit unusual, but he really really really loves these 3 books on math. He also goes wild for this Computers and Coding one (I learned a lot too.)
# Enjoy!
#### Sherry of Save. Spend. Splurge.
I got out of \$60,000 of debt in 18 months using TheBudgetingTool.com. Since then, I have worked 50% of my career (taking 1-2 year breaks), and quadrupled my income within 2 years of graduating, going from \$65K to \$260K (savings rate = 85%). I could retire today if I wanted, but love my work-life balance as a freelancing consultant in STEM (Science, Technology, Engineering, Math). I also post daily on Instagram @saverspender.
#### Ask Sherry: What are some reasonably priced pots & pans to buy?
Posted on November 24, 2017
#### The best and worst things about being a parent
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#### Why I am teaching my kids about money by not giving them any to begin with
Posted on August 27, 2014
1. S
###### Sarah
Tried to comment yesterday, but it looks like the internet are it, boooo.
I thought I got the idea for “Teach Your Child to Read in 100 Easy Lessons” (which is how we taught our son to read at 4). Though daycare had already taught him letter sounds and counting.
We have been slower at math- just takes longer to click. I had a teacher friend complain about her 1st graders knowing multiplication only by rote, so I’m trying to focus on the foundations. Never thought to try fractions!
Who knows how much any of this helps. Like you, I’m trying to make learning fun. Are you still going to try to put him in kindergarten, or jump to 1st grade?
1. ###### Sherry of Save. Spend. Splurge.
Ooo I didn’t know that existed! I ought to check it out.
I honestly just made math fun. And I started with foundations like you said — these are “groups of 2”, and these are “groups of 4″… a lot of the iPad apps I used have really helped in this regard to show me how they make it fun for kids and I expanded on that. 🙂
I am putting him in kindergarten, he needs socialization and it is actually better that he is starting later because he knows how to write a little, read, do math, and will feel confident / smart amongst the classmates and not so stressed/worried he doesn’t get it. I read that somewhere with boys that they develop slower in that regard and that intimidation factor at that age can be crucial
Who knows if it is true but I am hoping it will give him self-esteem and confidence to start in a new school with kids…
I am trying not to get him to memorize everything by rote… but multiplication is the foundation of division so he has to know it, and has been good at it, although he defaults to his Times Tables book 🙂
2. G
###### Gail
You are a natural teacher! What you are doing could help any child to some degree. Schools where I have taught and where my kids and grandkids have gone DO use the letter sounds with a few sight words to supplement. It seems, here in the parts of the U.S.where we have lived, to have come full circle. With your instincts and his mind Little Bun will stay on top of school always.
1. ###### Sherry of Save. Spend. Splurge.
Thank you, this means a lot coming from you! I have my mother to thank as well for helping me be a ‘natural’ teacher, and I just see what works for him and that’s what I go with. If it doesn’t seem to sink in, I try another tactic.
3. S | 3,830 | 16,186 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-10 | longest | en | 0.980731 |
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# Flush penalty cards and trend betting
30 August 2015
QUESTION: In a recent column, you talked about a flush penalty card. What is that?
ANSWER: In video poker, a flush penalty card is one in your initial hand that is the same suit as other cards you are considering holding and would decrease your chances of drawing a flush if you discarded it.
For example, if you're dealt Jack-10-8-4 of hearts and 2 of diamonds, and you're trying to decide whether to hold Jack-10-8 to go for a straight flush, one of the considerations is that even if you don't draw the straight flush, it's possible to draw a flush. Discarding the 4 means there's one fewer card available that could be part of a flush. A reduced chance of drawing a flush is your penalty for discarding the 4. Thus, the 4 is a flush penalty card.
The specific hand referred to in a July column on strategies for 8/5 Bonus Poker with a progressive jackpot for royal flushes. As the jackpot increases, we change strategies to take more chances at a royal.
One example given was holding the potential two-card royal flush, Jack-10, vs. holding the Jack with a King in a different suit, given that one of the discards was going to be a 3 in the same suit as the Jack-10.
In that case, the 3 is a penalty card that makes holding suited Jack-10 less valuable by limiting the flush possibilities. Because of that penalty, it’s a better play to hold the unsuited King-Jack to give you more potential high pair pays provided the jackpot is less than 4,745 coins, but a better play to hold Jack-10 with the jackpot is higher.
Either way, you’re going to discard the 3, but it’s the influence on the potential of the suited Jack-10 that makes it a penalty card.
QUESTION: Why is it that there are boards that show the last 18 or so winning numbers at roulette everywhere I play now? When I started playing 20-some years ago, nobody was giving away that information. Now everybody does. Why the change?
ANSWER: First of all, a tote board doesn’t cost the casino anything other than the price of the equipment and the power to run it. Showing you the most recent numbers doesn’t change the odds of the game one bit. On a double-zero wheel, the chances of any given number showing up are 1 in 38, every time, no matter what has gone before.
Some systems with electronic betting will show you a lot more than that. At the touch of a screen, you can find number history, streaks, hit percentages for each number over the last 1,000 spins – just about anything you could want to know about what’s gone on at that wheel. The casino can freely give away that information, because as long as the wheel is balanced properly, the odds are unchanging and the information doesn’t give players an edge.
But those are reasons the casinos needn’t fear the information, not reasons they should invest in sharing it. That reason is simple: It’s popular. Players like it, and have proven it with their pocketbooks, When the first tote boards went up at roulette tables, casinos found that play increased by 30%.
Nobody in table games management wanted to be left behind as players gravitated toward wheels that shared recent winning numbers. So more and more operators added the tote boards until we reached today’s situation where everybody has them and they’re part of the standard equipment for roulette.
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John Grochowski
John Grochowski is the best-selling author of The Craps Answer Book, The Slot Machine Answer Book and The Video Poker Answer Book. His weekly column is syndicated to newspapers and Web sites, and he contributes to many of the major magazines and newspapers in the gaming field, including Midwest Gaming and Travel, Slot Manager, Casino Journal, Strictly Slots and Casino Player.
Listen to John Grochowski's "Casino Answer Man" tips Tuesday through Friday at 5:18 p.m. on WLS-AM (890) in Chicago. Look for John Grochowski on Facebook and Twitter @GrochowskiJ.
#### Books by John Grochowski:
John Grochowski
John Grochowski is the best-selling author of The Craps Answer Book, The Slot Machine Answer Book and The Video Poker Answer Book. His weekly column is syndicated to newspapers and Web sites, and he contributes to many of the major magazines and newspapers in the gaming field, including Midwest Gaming and Travel, Slot Manager, Casino Journal, Strictly Slots and Casino Player.
Listen to John Grochowski's "Casino Answer Man" tips Tuesday through Friday at 5:18 p.m. on WLS-AM (890) in Chicago. Look for John Grochowski on Facebook and Twitter @GrochowskiJ. | 1,039 | 4,639 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2023-50 | latest | en | 0.950993 |
http://printfu.org/rectangles | 1,590,861,349,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347410284.51/warc/CC-MAIN-20200530165307-20200530195307-00322.warc.gz | 104,278,256 | 6,598 | # Rectangles
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math.springbranchisd.com
### Instructions: Find the area and perimeter of each rectangle.
MATH-DRILLS.COM MATH-DRILLS.COM MATH-DRILLS.COM MATH-DRILLS.COM Area and Perimeter of Rectangles (A) Instructions: Find the area and perimeter of each rectangle.
www.math-drills.com
### Simulating Rectangles
Fathom Summer Institute Simulating Rectangles ' 2000, Professional Development Center, Key Curriculum Press ¥ 1 Simulating Rectangles Exploring Mathematics with Fathom Summer Institute In one way this activity is about the properties and relationships you can find in a collection of rectangles ...
www.keypress.com
### TILING RECTANGLES WITH SQUARES (FOR DUMMIES)
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math.arizona.edu
### Simple Proofs ofa Rectangle Tiling Theorem
Abstract If a finite number of rectangles, everyone of which has at least one integer side, perfectly tileabig rectangle, then the big rectangle also has at least one integer side.
www.inference.phy.cam.ac.uk
### CH 14 − THE PYTHAGOREAN THEOREM AND RECTANGLES
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algebrawithsteve.com
### Traceable Rectangle Page - Help Kids Learn Shapes
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www.kidslearningstation.com
### Exploring Properties of Rectangles and Parallelograms Using ...
Illuminations Internet Mathematics Excursion October 2000 Rectangles and Parallelograms Page 1 of 7 Exploring Properties of Rectangles and Parallelograms Using Dynamic Software Overview: This Internet Mathematics Excursion is based on an E-example from the NCTM Principles and Standards for ...
illuminations.nctm.org
### Factors and Primes in Dimensions of Rectangles
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www.tip.duke.edu
### Prime & Composite Numbers (Rectangle Area Investigation)
Using 2 or more tiles, students are to determine how many unique rectangles (different orientations do not count as different rectangles) they can create with a given area a.
www.erusd.k12.ca.us
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http://web2.0calc.com/questions/the-equation-x-cubed-8x-squared-20-has-two-negative-solutions-between-0-and-8-use-trial-and-improvement-to-find-these-solutions-give | 1,516,529,985,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890514.66/warc/CC-MAIN-20180121100252-20180121120252-00386.warc.gz | 394,024,588 | 5,761 | +0
# The equation x cubed + 8x squared = 20 has two negative solutions between 0 and -8. Use trial and improvement to find these solutions. Give
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The equation x cubed + 8x squared = 20 has two negative solutions between 0 and -8. Use trial and improvement to find these solutions. Give your answer to two decimal places.
Guest Feb 26, 2015
#1
+26405
+5
1. Rewrite as x3 + 8x2 -20 = f(x) where we want the x that makes f(x) = 0.
Plot a graph:
1. Guess x = -1, f(-1) = -13
2. Guess x = -2 f(-2) = 4
3. Use f(-2)/(-2-x) = -f(-1)/(x- -1) to get the next value of x, namely x = -1.77 to 2 decimal places
When x = -1.77 f(-1.77) = -0.482 Not yet close enough to zero. Because it is negative repeat step 3, but replace -1 and f(-1) with -1.77 and f(-1.77) respectively
4. Use f(-2)/(-2-x) = -f(-1.77)/(x- -1.77) to get x = -1.80 to 2 decimal places
When x = -1.80 f(-1.80) = -0.09
Repeat with
5. f(-2)/(-2-x) = -f(-1.80)/(x- -1.80) to get x = 1.80 to 2 decimal places
So one solution is x = 1.80 to 2 decimal places.
Try using the graph above to get two reasonable approximations to the other solution and repeat the above process.
.
Alan Feb 26, 2015
Sort:
#1
+26405
+5
1. Rewrite as x3 + 8x2 -20 = f(x) where we want the x that makes f(x) = 0.
Plot a graph:
1. Guess x = -1, f(-1) = -13
2. Guess x = -2 f(-2) = 4
3. Use f(-2)/(-2-x) = -f(-1)/(x- -1) to get the next value of x, namely x = -1.77 to 2 decimal places
When x = -1.77 f(-1.77) = -0.482 Not yet close enough to zero. Because it is negative repeat step 3, but replace -1 and f(-1) with -1.77 and f(-1.77) respectively
4. Use f(-2)/(-2-x) = -f(-1.77)/(x- -1.77) to get x = -1.80 to 2 decimal places
When x = -1.80 f(-1.80) = -0.09
Repeat with
5. f(-2)/(-2-x) = -f(-1.80)/(x- -1.80) to get x = 1.80 to 2 decimal places
So one solution is x = 1.80 to 2 decimal places.
Try using the graph above to get two reasonable approximations to the other solution and repeat the above process.
.
Alan Feb 26, 2015
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