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https://www.jiskha.com/display.cgi?id=1193337784 | 1,503,074,762,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886104704.64/warc/CC-MAIN-20170818160227-20170818180227-00637.warc.gz | 915,261,882 | 4,089 | # physics
posted by .
a telephone engineer feeds a sound signal into the mouthpiece of a telephone.As part of the tests he is making he incresesthe frequency of the sound.HOW WILL THIS EFFECT TYE ELECTICAL SIGNAL PASSING ALONG THE TELEPHONE WIRE?
• physics -
It depends. If it is an analoge signal, the frequency. But if it is digital, as on a multiplexed wire, or fiber optics, there will be no difference on the transmitted signal along the "wire".
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More Similar Questions | 736 | 2,927 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2017-34 | latest | en | 0.857689 |
http://www.nickgmorgan.com/math/plotting.html | 1,516,411,027,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084888341.28/warc/CC-MAIN-20180120004001-20180120024001-00639.warc.gz | 538,192,908 | 1,910 | # Plotting Points
Parts of the Coordinate Plane
The coordinate plane is a way to express numbers and equations graphically. It contains two axes, the x and the y. The x axis is the horizontal (left and right) and the y axis is the vertical (up and down). The center of the plane is a point called the origin. Numbers on the x-axis count up in positive numbers to the right of the origin and down in negative numbers to the left of the origin. Numbers on the y-axis count up in positive numbers above the origin and down in negative numbers below the origin.
Ordered Pairs
Every point on the coordinate plane is expressed as an ordered pair. The first digit of the ordered pair is the x-coordinate. That is how far to the left or right the point is on the x-axis. The second number of the ordered pair is the y-coordinate. That is how far up or down on the y-axis the point is. All ordered pairs are written in the format (x,y). For example, if you want to plot the point (3, 2), you would count to the right 3 spaces on the x-axis, then count up 2 spaces on the y-axis.
If the point is (-5,3) you would count to the left 5, then up 3.
Distance Formula
If you have two points that you need to find the distance between, you have a distance formula available on your formula sheet. The distance formula is:
Designate your two points as point 1 and 2. Then plug the x-coordinate of your first point in place of the x1, then the x-coordinate of your second point in place of the x2, then the y-coordinates in place of y1 and y2. Then solve for d. | 381 | 1,564 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-05 | latest | en | 0.852861 |
https://discourse.matplotlib.org/t/cape-and-cin-calculation-in-python/18541 | 1,653,285,811,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662555558.23/warc/CC-MAIN-20220523041156-20220523071156-00290.warc.gz | 264,304,683 | 5,888 | # CAPE and CIN calculation in Python
Hello,
Lately, I am working on plotting sounding profiles on a SkewT/LogP diagram. The SkewT package which is located at https://github.com/tchubb/SkewT has a nice feature to lift a parcel on dry/moist adiabats. This is very useful to demonstrate the regions of CIN and CAPE overlaid with the full sounding.
However, the package misses these diagnostic calculations. This is the only step holding me back to use Python only (migrating from NCL) for my plotting tasks. I am aware that these calculations are usually performed in fortran. Are there any routines wrapped in Python to calculate CAPE and CIN parameters? Any suggestions or comments would be really appreciated.
···
Gökhan
You can easily visualize the CAPE and CIN with matplotlib using fill_between() on the environmental and parcel temperature curves. As for actually calculating it though, I don’t know of a way to do it directly from matplotlib. There are probably several other python packages out there that can, but I am not familiar with them. In any case, why not just write your own function for calculating the CAPE and CIN? It is a bit surprising that this functionality isn’t be included in the SkewT package, but since you can use it to get the parcel temperature curve, you should be able to calculate the CAPE and CIN rather easily by simply discretizing their respective formulas. Here’s a rough example:
import numpy as np
cape = 9.8 * np.sum(dz * (Tp - T) / T)
Where Tp and T are the parcel and environmental temperature arrays respectively, and dz are the height differences between layers. You would of course need to perform the sum from the LFC to EL for CAPE, so the arrays would have to to be subsetted. With numpy the easiest way to do this is with fancy indexing, eg:
levs = (z >= LFC) & (z <= EL)
Tp = Tp[levs]
T = T[levs]
where z is your array of heights (or pressure levels).
Does this help?
Alex
···
On Sat, Mar 29, 2014 at 4:32 PM, Gökhan Sever <gokhansever@…287…> wrote:
Hello,
Lately, I am working on plotting sounding profiles on a SkewT/LogP diagram. The SkewT package which is located at https://github.com/tchubb/SkewT has a nice feature to lift a parcel on dry/moist adiabats. This is very useful to demonstrate the regions of CIN and CAPE overlaid with the full sounding.
However, the package misses these diagnostic calculations. This is the only step holding me back to use Python only (migrating from NCL) for my plotting tasks. I am aware that these calculations are usually performed in fortran. Are there any routines wrapped in Python to calculate CAPE and CIN parameters? Any suggestions or comments would be really appreciated.
Gökhan
Matplotlib-users mailing list
Matplotlib-users@lists.sourceforge.net
https://lists.sourceforge.net/lists/listinfo/matplotlib-users
Alex Goodman
Department of Atmospheric Science
Thanks Alex. This seems to be the easiest way to approach the problem. However, for the sake of reproducibility, I am looking for a way to interface to one of the common fortran routines for the task. Your suggested approach will require some sort of interpolation at the very least to make the estimation number of data point insensitive. Putting this in my TODO list.
···
On Sat, Mar 29, 2014 at 7:56 PM, Alex Goodman <alex.goodman@…4442…> wrote:
You can easily visualize the CAPE and CIN with matplotlib using fill_between() on the environmental and parcel temperature curves. As for actually calculating it though, I don’t know of a way to do it directly from matplotlib. There are probably several other python packages out there that can, but I am not familiar with them. In any case, why not just write your own function for calculating the CAPE and CIN? It is a bit surprising that this functionality isn’t be included in the SkewT package, but since you can use it to get the parcel temperature curve, you should be able to calculate the CAPE and CIN rather easily by simply discretizing their respective formulas. Here’s a rough example:
import numpy as np
cape = 9.8 * np.sum(dz * (Tp - T) / T)
Where Tp and T are the parcel and environmental temperature arrays respectively, and dz are the height differences between layers. You would of course need to perform the sum from the LFC to EL for CAPE, so the arrays would have to to be subsetted. With numpy the easiest way to do this is with fancy indexing, eg:
levs = (z >= LFC) & (z <= EL)
Tp = Tp[levs]
T = T[levs]
where z is your array of heights (or pressure levels).
Does this help?
Alex
Gökhan
On Sat, Mar 29, 2014 at 4:32 PM, Gökhan Sever <gokhansever@…287…> wrote:
Hello,
Lately, I am working on plotting sounding profiles on a SkewT/LogP diagram. The SkewT package which is located at https://github.com/tchubb/SkewT has a nice feature to lift a parcel on dry/moist adiabats. This is very useful to demonstrate the regions of CIN and CAPE overlaid with the full sounding.
However, the package misses these diagnostic calculations. This is the only step holding me back to use Python only (migrating from NCL) for my plotting tasks. I am aware that these calculations are usually performed in fortran. Are there any routines wrapped in Python to calculate CAPE and CIN parameters? Any suggestions or comments would be really appreciated.
Gökhan
Matplotlib-users mailing list
Matplotlib-users@lists.sourceforge.net
https://lists.sourceforge.net/lists/listinfo/matplotlib-users
Alex Goodman
Department of Atmospheric Science
As for the rest of the two suggestions: [https://github.com/pmarshwx/SHARPpy/tree/gui
https://github.com/PyAOS/aoslib]
there doesn’t seem be routines included for CAPE and CIN calculations.
I am working on idealized orographic precipitation simulations in WRF that will hopefully be presented in the Mountain Meteorology conference in August. Perhaps I am extra cautious, but I need a good control especially on CAPE calculation (independent of data-points, being able to specify where to lift the parcel and specify mixed layer depth etc.).
This could be a good subject to discuss on the coming SciPy conference. The abstract deadline is tomorrow. I might join if I can get some funding to attend the conference. Is there any symposium planned for atmospheric science people? Thanks again.
···
On Sat, Mar 29, 2014 at 6:32 PM, Gökhan Sever <gokhansever@…287…> wrote:
Hello,
Lately, I am working on plotting sounding profiles on a SkewT/LogP diagram. The SkewT package which is located at https://github.com/tchubb/SkewT has a nice feature to lift a parcel on dry/moist adiabats. This is very useful to demonstrate the regions of CIN and CAPE overlaid with the full sounding.
However, the package misses these diagnostic calculations. This is the only step holding me back to use Python only (migrating from NCL) for my plotting tasks. I am aware that these calculations are usually performed in fortran. Are there any routines wrapped in Python to calculate CAPE and CIN parameters? Any suggestions or comments would be really appreciated.
Gökhan
Gökhan | 1,673 | 7,066 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2022-21 | latest | en | 0.9259 |
http://stackoverflow.com/questions/2760334/what-short-query-will-return-an-vector-a-one-column-column-table-representing | 1,406,083,022,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1405997872002.22/warc/CC-MAIN-20140722025752-00208-ip-10-33-131-23.ec2.internal.warc.gz | 355,449,454 | 15,272 | # What short query will return an vector (a one-column column table) representing a set of integers from x1 to x2?
Assuming we have no actual table to select the data from, what do we need to SELECT to return a table consisting of one integer column containing all integer numbers starting with x1 and finishing with x2 and sorted in ascending order?
-
you can use this function (it is based on Itzik Ben-Gan's function from an interview I did with him here: Interview With Itzik Ben-Gan Author Of Inside Microsoft SQL Server 2005: T-SQL Querying)
``````CREATE FUNCTION dbo.fn_numbers(@Start AS BIGINT,@End AS BIGINT) RETURNS TABLE
AS
RETURN
WITH
L0 AS(SELECT 1 AS c UNION ALL SELECT 1),
L1 AS(SELECT 1 AS c FROM L0 AS A, L0 AS B),
L2 AS(SELECT 1 AS c FROM L1 AS A, L1 AS B),
L3 AS(SELECT 1 AS c FROM L2 AS A, L2 AS B),
L4 AS(SELECT 1 AS c FROM L3 AS A, L3 AS B),
L5 AS(SELECT 1 AS c FROM L4 AS A, L4 AS B),
Nums AS(SELECT ROW_NUMBER() OVER(ORDER BY c) AS n FROM L5)
SELECT n FROM Nums
WHERE n between @Start and @End;
GO
``````
Here is how you run it
``````-- Test function
SELECT * FROM dbo.fn_numbers(1,100) AS F;
GO
-- Test function
SELECT * FROM dbo.fn_numbers(200,700) AS F;
GO
``````
-
One option, (in T-SQL)
`````` Declare @X1 Integer Set @x1 = 45
Declare @X2 Integer Set @x2 = 87
Declare @I Integer Set @I = @x1
Declare @Outs Table (val integer Primary Key Not Null)
While @I <= @x2 Begin
Insert @Outs(Val) Values (@I)
Set @I = @I + 1
End
Select Val From @Outs
``````
-
You've always a table with data in SQL Server to use a base
``````WITH cDummy AS
(
SELECT
ROW_NUMBER() OVER (ORDER BY c1.id) AS number
FROM
sys.columns c1 CROSS JOIN sys.columns c2 CROSS JOIN sys.columns c3
)
SELECT
number
FROM
cDummy
WHERE
number BETWEEN @x AND @x * 2
``````
- | 558 | 1,784 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2014-23 | latest | en | 0.566393 |
http://www.chegg.com/homework-help/questions-and-answers/power-crossing-surface-perpendicular-equal-theintensity-surface-times-area-surface-since-t-q796079 | 1,464,398,626,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464049277286.54/warc/CC-MAIN-20160524002117-00118-ip-10-185-217-139.ec2.internal.warc.gz | 422,272,640 | 14,071 | The power crossing a surface perpendicular is equal to theintensity at the surface times the area of the surface. Since thesource emits uniformly in all directions, the light intensity isthe same at all points on the imaginary spherical surface.Moreover, the light crosses this surface perpendicularly.
S =cε0E2
relates the average light intensity at the surface to its rmselectric field strength (which is known), and the area of thesurface can be found from a knowledge of its radius.
A light bulb emits light uniformly in all directions. The averageemitted power is 150.0 W. At a distance of 4.70 m from the bulb, determine (a) the averageintensity of the light, (b) the rms value of the electric field,and (c) the peak value of the electric field.
(a) S = 1 2 ---Select--- W/m^2 J/m3 J W (b) Erms = 3 4 ---Select--- N/C J W T N (c) E0 = 5 6 ---Select--- N J N/C W T | 231 | 871 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2016-22 | latest | en | 0.880675 |
http://bootmath.com/lim_x-to-0fractan-x-cdot-sqrt-tan-x-sin-x-cdot-sqrtsin-xx3-sqrtx.html | 1,534,539,566,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221212910.25/warc/CC-MAIN-20180817202237-20180817222237-00116.warc.gz | 55,207,248 | 6,082 | # $\lim_{x \to 0^+}\frac{\tan x \cdot \sqrt {\tan x}-\sin x \cdot \sqrt{\sin x}}{x^3 \sqrt{x}}$
Calculate:
$$\lim_{x \to 0^+}\frac{\tan x \cdot \sqrt {\tan x}-\sin x \cdot \sqrt{\sin x}}{x^3 \sqrt{x}}$$
I don’t know how to use L’Hôpital’s Rule.
I tried to make $\tan x =\frac{\sin x}{\cos x}$ for the term ${\sqrt{\tan x}}$.
#### Solutions Collecting From Web of "$\lim_{x \to 0^+}\frac{\tan x \cdot \sqrt {\tan x}-\sin x \cdot \sqrt{\sin x}}{x^3 \sqrt{x}}$"
You can first remove a few factors
$$\lim_{x \to 0^+}\frac{\tan x \cdot \sqrt {\tan x}-\sin x \cdot \sqrt{\sin x}}{x^3 \sqrt{x}}\\ =\lim_{x \to 0^+}\frac{\tan x \cdot \sqrt{\tan x}}{x \sqrt{x}}\lim_{x \to 0^+}\frac{1-\cos x\sqrt{\cos x}}{x^2}\\ =\lim_{x \to 0^+}\frac{1-\cos x\sqrt{\cos x}}{x^2}.$$
Then multiply by the conjugate
$$=\lim_{x \to 0^+}\frac{1-\cos^3 x}{x^2(1+\cos x\sqrt{\cos x})},$$
evaluate the finite factor at denominator
$$=\frac12\lim_{x \to 0^+}\frac{1-\cos x(1-\sin^2 x)}{x^2},$$
use trigonometric identitites
$$=\frac12\lim_{x \to 0^+}\frac{2\sin^2\frac x2+\cos x\sin^2 x}{x^2},$$
and conclude
$$=\left(\frac12\right)^2+\frac12.$$
We used
$$\frac{\tan x}x=\frac{\sin x}x\frac1{\cos x}\to 1,$$
$$\frac{\sin ax}x=a\frac{\sin ax}{ax}\to a.$$
\begin{align} \lim_{x\to0^{+}}\frac{\tan x \cdot \sqrt {\tan x}-\sin x \cdot \sqrt{\sin x}}{x^3 \sqrt{x}} &=\lim_{x\to0^{+}}\frac{\tan^{3/2}x-\sin^{3/2} x}{x^3 \sqrt{x}}\\ &=\lim_{x\to0^{+}}\frac{\sin^{3/2}x}{x^{3/2}}\frac{\sec^{3/2}x-1}{x^2}\\ &=\lim_{x\to0^{+}}\left(\frac{\sin x}{x}\right)^{3/2}\cdot \lim_{n\to\infty}\frac{\sec^{3/2}x-1}{x^2}\\ &=\lim_{x\to0^{+}}\frac{\sec^{3/2}x-1}{x^2}\\ &=\lim_{x\to0^{+}}\frac{\left(1+\frac12x^2+\frac1{24}x^4+\cdots\right)^{3/2}-1}{x^2}\\ &=\lim_{x\to0^{+}}\frac{\left(1+\frac34x^2+\cdots\right)-1}{x^2}\\ &=\frac34 \end{align}
$$\frac{\tan x\sqrt{\tan x}-\sin x\sqrt{\sin x}}{x^3\sqrt x}=\left(\frac{\sin x} x\right)^{3/2}\cdot\frac{\frac1{\cos^{3/2}x}-1}{x^2}=$$
$$=\left(\frac{\sin x} x\right)^{3/2}\frac{1-\cos^{3/2}x}{x^2\cos^{3/2}x}=\left(\frac{\sin x} x\right)^{3/2}\frac{1-\cos^2x+\cos^{3/2}x(\cos^{1/2}x-1)}{x^2\cos^{3/2}x}=$$
$$=\left(\frac{\sin x} x\right)^{3/2}\left[\frac1{\cos^{3/2}x}\left(\frac{\sin x}x\right)^2+\frac{\cos^{1/2}x-1}{x^2}\right]=$$
$$=\left(\frac{\sin x} x\right)^{3/2}\left[\frac1{\cos^{3/2}x}\left(\frac{\sin x}x\right)^2+\frac{\cos x-1}{x^2(\cos^{1/2}x+1)}\right]=$$
$$=\left(\frac{\sin x} x\right)^{3/2}\left[\frac1{\cos^{3/2}x}\left(\frac{\sin x}x\right)^2-\frac2{\cos^{1/2}x+1}\frac{\sin^2\frac x2}{x^2}\right]=$$
$$=\left(\frac{\sin x} x\right)^{3/2}\left[\frac1{\cos^{3/2}x}\left(\frac{\sin x}x\right)^2-\frac2{\cos^{1/2}x+1}\cdot\frac14\cdot\left(\frac{\sin\frac x2}{\frac x2}\right)^2\right]\xrightarrow[x\to0^+]{}$$
$$\rightarrow1\left[1\cdot1-\frac22\cdot\frac14\cdot1\right]=1-\frac14=\frac34$$ | 1,348 | 2,818 | {"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} | 4.25 | 4 | CC-MAIN-2018-34 | latest | en | 0.297796 |
https://enacademic.com/dic.nsf/enwiki/945910/Out_of_This_World_%28card_trick%29 | 1,579,792,252,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250610919.33/warc/CC-MAIN-20200123131001-20200123160001-00361.warc.gz | 426,978,766 | 14,256 |
# Out of This World (card trick)
Out of This World is a card trick created by magician Paul Curry in 1942. Many performers have devised their own variations of this trick. It is often billed as "the trick that fooled Winston Churchill" due to a story describing how it was performed for him during World War II. The method behind the trick is simple and essentially self-working, and can be enhanced by the presentation of the performer and the use of other principles of magic.
## Effect
1. The performer takes a deck of cards, and places on the table two face-up "marker" cards, one black and one red; the black on the left and the red on the right. The performer tells the subject that he or she is going to deal cards face-down from the deck and the object of the exercise is for the subject to use the power of their mind to identify whether each card in the deck is black or red.
2. The performer takes one card at a time from the deck, face down, and asks the subject to attempt to divine whether it is black or red. The subject states their choice, and the performer then places the card in line with the appropriately coloured marker card, overlapping it at the bottom.
3. About halfway through the deck, the performer stops and announces that it is necessary to switch sides, in order to prevent a possible preference for one side over another from confusing the results. The performer deals two new marker cards onto the existing lines: a red one on the left, and a black one on the right.
4. The performer then continues as before, dealing cards face-down from the deck onto the subject's choice of the black or red line.
5. When the deck is exhausted, the performer instructs the subject to gather up and somehow reveal the left-hand line of cards; the performer does the same for the right-hand line.
6. The exposed lines reveal that every one of the subject's guesses was correct, and the black and red cards have been exactly sorted by colour.
## Method
The principle behind the trick is simple: the deck in use is either stacked or partially stacked prior to performance, or manipulated in the course of performance to form a partial or full stack, in which there is a separation between the red and black cards, with a run of each colour comprising either half the deck or a substantial portion of the stack near the top.
In the first round of dealing face-down cards, all of the cards dealt will be of one colour. The switching of sides occurs at the halfway point through the deck because that is the point at which the first colour cards will be exhausted and the second colour of cards will begin to be dealt.
To understand why this enables the trick to work, think about what will be in the two lines of cards at the end of the trick. The left-hand line will begin with (for example) a black marker card (placed at the start), followed by the face-down black cards from the top of the deck, then the red marker card (placed at the side switch), then the face-down red cards from the bottom of the deck. That's (black marker) (black cards) (red marker) (red cards). In other words, this line is exactly right, and is given to the subject to gather up.
The right-hand line will begin with the red marker card (placed at the start), followed by the face-down black cards from the top of the deck, then the black marker card (placed at the side switch), then the face-down red cards from the bottom of the deck. That's (red marker) (black cards) (black marker) (red cards). Although incorrect, this can be instantly made to appear correct by moving the red marker card from one end of the line to the other, or by reversing the order of all other cards in the line. This must be performed by the magician during the act of gathering up the right-hand line of cards, while the subject is distracted by gathering up their own line. Since the cards are gathered into a stack at that point, this is straightforward.[1]
Chapter 13 of the book Magician's Magic by Paul Curry describes a dinner party during World War II at which a magician named Harry Green performed the trick for Winston Churchill. According to the book, Churchill insisted that the trick be performed for him half a dozen times, and was left thoroughly baffled.[2]
## References
1. ^ Longe, Bob (1992). World's Best Card Tricks. Sterling Publishing Company Inc.. p. 88. ISBN 0806982330.
2. ^
Wikimedia Foundation. 2010.
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• 1986 FIFA World Cup — Mexico 86 1986 FIFA World Cup official logo Tournament details Host country … Wikipedia | 1,497 | 6,727 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2020-05 | latest | en | 0.959682 |
https://studylib.net/doc/8374398/stat-611-solutions-to-selected-problems-from-assignment-%23... | 1,660,587,220,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572198.93/warc/CC-MAIN-20220815175725-20220815205725-00475.warc.gz | 506,850,944 | 13,581 | # STAT 611 Solutions to Selected Problems from Assignment #9 8.5 c
```STAT 611
Solutions to Selected Problems from Assignment #9
8.5 c. Under H0 , T has the same distribution as
"
log
n
Y
#
Yi / min Yi ,
i
i=1
where Y1 , . . . , Yn are iid from density f (y) = y −2 I[1,∞) (y). Letting Y(1) < Y(2) < · · · < Y(n)
£Qn
¤
denote the ordered Yi ’s, T can be expressed as log i=2 (Y(i) /Y(1) ) . The probability
distribution of T can thus be found from the joint distribution of the ordered Yi ’s. The
Qn
joint density of Y(1) , . . . , Y(n) is n! i=1 f (ui ), where 1 < u1 < · · · < un . Define the random
variables Z1 , . . . , Zn by Z1 = Y(1) and Zi = Y(i) /Y(1) , i = 2, . . . , n. Using the standard
change of variables technique, the joint density of Z1 , . . . , Zn is
fZ1 ,...,Zn (z1 , . . . , zn ) =
−(n+1)
n!z1
n
Y
zi−2 ,
for z1 > 1, 1 < z2 < · · · < zn ,
i=2
and 0 otherwise. Integrate out z1 to get the joint density of Z2 , . . . , Zn . Upon doing so,
we see that Z2 , . . . , Zn have the same density as the order statistics in a random sample of
size n − 1 from the density f defined above. Together with previous conclusions, the last
statement implies that T has the same probability distribution as
S=
n−1
X
log W(i) =
i=1
n−1
X
log Wi ,
i=1
where W(1) , . . . , W(n−1) are the order statistics of a random sample W1 , . . . , Wn−1 from
density f . Finally, it is easy to verify (using independence of the Wi ’s) that S has a χ2
distribution with 2(n − 1) degrees of freedom.
8.33 a. It is easy to show that k = 1 − α1/n .
b. Clearly, if θ ≥ 1 − α1/n , then the power is equal to 1. So, let θ < 1 − α1/n , and consider
Power = Pθ (Y1 ≥ k ∪ Yn ≥ 1)
= Pθ (Y1 ≥ k) + Pθ (Yn ≥ 1) − Pθ (Y1 ≥ k ∩ Yn ≥ 1).
Now,
Pθ (Y1 ≥ k) = Pθ (all Xi ’s ≥ k)
= (θ + 1 − 1 + α1/n )n
= (θ + α1/n )n ,
1
and
Pθ (Yn ≥ 1) = Pθ (at least one Xi ≥ 1)
= 1 − Pθ (all Xi ’s < 1)
= 1 − (1 − θ)n .
Using the expression for the joint density of Y1 and Yn on p. 230 of Casella and Berger,
Z
θ+1
Z
θ+1
Pθ (Y1 ≥ k ∩ Yn ≥ 1) =
Z
k
1
1
θ+1
Z
=
k
n(n − 1)(yn − y1 )n−2 dyn dy1
1
Z
n(n − 1)(yn − y1 )n−2 I(y1 ,θ+1) (yn ) dyn dy1
θ+1
Z
θ+1
+
1
n(n − 1)(yn − y1 )n−2 dyn dy1
y1
= (θ + α1/n )n − α.
It follows that
½
Power(θ) =
1 + α − (1 − θ)n , θ < 1 − α1/n ,
1,
θ ≥ 1 − α1/n .
c. When θ ≥ 1, the power of the test in question is 1, and hence must be UMP for θ ≥ 1.
Now let 0 < θ1 < 1 and consider testing the hypotheses
H0 : θ = 0
vs. H1 : θ = θ1 .
Show that the rejection region of the Neyman-Pearson test of these hypotheses has the
form
Y1 ≥ k or Yn ≥ 1.
Since k is chosen to make the test have size α, k has nothing to do with θ1 . It follows from
Theorem 8 of the class notes that the test is UMP for all θ.
d. The power of the test is 1 for every θ ≥ 1 − α1/n . Since 1 − α1/n ≤ 1, it is true by
default that the power is always greater than 0.8 when θ > 1.
2
``` | 1,215 | 2,978 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2022-33 | latest | en | 0.714284 |
https://forum.freecodecamp.org/t/javascript-basic-algorithm/441200 | 1,713,839,544,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818452.78/warc/CC-MAIN-20240423002028-20240423032028-00831.warc.gz | 235,447,100 | 6,672 | # JavaScript basic algorithm
Tell us what’s happening:
Good evening people! i am in serious need of help here,I have been trying this algorithm and i can’t seem to get it right. Can someone please tell me what is wrong with my code.
here is it!!
``````function factorialize(num) {
var counter =0;
for(let i=1; i<=num; i++){
i *= counter;
}
return counter;
}
// function factorialize(num) {
// return num;
// }
factorialize(5);
``````
User Agent is: `Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/87.0.4280.141 Safari/537.36`.
Challenge: Factorialize a Number
I’ve edited your post for readability. When you enter a code block into a forum post, please precede it with a separate line of three backticks and follow it with a separate line of three backticks to make it easier to read.
You can also use the “preformatted text” tool in the editor (`</>`) to add backticks around text.
See this post to find the backtick on your keyboard.
Note: Backticks (`) are not single quotes (’).
Just at a quick glance, you seem a little confused about which variable is holding the product, and which you should be returning.
`````` i *= counter;
``````
Is that where you want to store your running product? You are telling it to multiply the current contents of i by counter and store that into i.
``````return num;
``````
Is that what you want to return?
Arg sorry thats my mistake it shouldn’t be “num”, it should be “counter”.
Right, that solves one problem, the other problem is still here:
`````` for(let i=1; i<=num; i++){
i *= counter;
}
``````
The variable i is your loop control variable - you have to be very careful messing with those in an active loop. It this case, it creates and infinite loop. But that’s OK, because that line needs to be changed anyway.
When I make those two changes, the code works.
Also, please don’t change code/text in previous posts in the thread - it creates confusion for people that try to read the thread later.
Thanks a lot for your feedback i really appreciate it.
This topic was automatically closed 182 days after the last reply. New replies are no longer allowed. | 532 | 2,166 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | latest | en | 0.89605 |
https://math.stackexchange.com/questions/1405001/minimum-of-the-sum-of-two-functions | 1,643,255,885,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305052.56/warc/CC-MAIN-20220127012750-20220127042750-00426.warc.gz | 432,259,605 | 32,611 | # Minimum of the sum of two functions
I want to show that trying to find the minimum of the sum of two or more functions of two different groups is a not convex problem. For example: $\min\limits_{Y,Z} f(X,Y,Z)=...$. Moreover the values $X,Y,Z$ are matrices. My idea is to show that the Hesse-Matrix of the sum of those added functions is not always positive semidefinit by finding a point x, where it is not psd?
$\operatorname{H}_f({Y},{Z})= \left(\frac{\partial^2f}{\partial c_i\partial c_j}({Y},{Z})\right) \begin{pmatrix} \frac{\partial^2 f}{\partial c_1\partial c_1}({X},{Y})&\frac{\partial^2 f}{\partial c_1\partial c_2}({X},{Y})&\cdots&\frac{\partial^2 f}{\partial c_1\partial c_n}({X},{Y})\\[0.5em] \frac{\partial^2 f}{\partial c_2\partial c_1}({X},{Y})&\frac{\partial^2 f}{\partial c_2\partial c_2}({X},{Y})&\cdots&\frac{\partial^2 f}{\partial c_2\partial c_n}({X},{Y})\\ \vdots&\vdots&\ddots&\vdots\\ \frac{\partial^2 f}{\partial c_n\partial c_1}({X},{Y})&\frac{\partial^2 f}{\partial c_n\partial c_2}({X},{Y})&\cdots&\frac{\partial^2 f}{\partial c_n\partial c_n}({X},{Y}) \end{pmatrix}$
with $c_i \in X,Y$, for example $c_i=x_{11}$ Then I pick one $X,Y,Z$ and show the the eigenvalues are $<0$. P.S. log is elementwise
• Why should the sum of non-convex functions be convex in general? I don't understand your question. Aug 21 '15 at 15:08
• I just want to proof that there is no combination for the sum of my non convex function and some of my other functions so that the result is convex. (I think it can happen that the sum of a non convex and another non convex or convex function might be convex) Aug 21 '15 at 15:30
• if you only want to show that your example is not convex, please edit your question accordingly. Aug 21 '15 at 15:33
• No, it is not only for the example, my functions are much more difficults. I deleted it in the question ;) Aug 21 '15 at 15:38
• Well, but for your question, it is only about your example isn't it? Notice that there are some sums of non convex function which are convex, and some which are non convex. So in full generality your question makes little sense to me. Aug 21 '15 at 15:44 | 688 | 2,140 | {"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} | 3.21875 | 3 | CC-MAIN-2022-05 | latest | en | 0.849318 |
https://dresspatternmaking.com/garment-elements/stylelines/empire-line-e/ | 1,718,720,670,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861762.73/warc/CC-MAIN-20240618140737-20240618170737-00306.warc.gz | 184,191,638 | 41,796 | # Empire Line
This page gives instructions for creating the pattern pieces for the Empire line Bodice shown in this image (not the skirt). The instructions given are limited to the style-line, as that is the theory being covered: by this I mean I do not give the instructions for elements such as facing, adding seam allowance, marking the placement of the zip, etc. The theory here is limited to creating an Empire Line Bodice. Step-by-step instructions covering creating a pattern from A-Z (facings, lining, seam allowance, checking seam lengths, etc) can be found in the Patterns Menu: Dress 002.
### Block Used
I will be using the Bodice Blocks for these instructions – Bodice Block Front & Bodice Block Back..
## Outcome
The image below shows the block being used (Bodice Block Front & Back) on the left, and on the right are the final pattern pieces.
Note: The actual pattern would need seam allowance and cutting instructions added, etc. A complete pattern would also need facing, lining or other finishing details. As mentioned above, this is not covered here).
## Instructions (Empire Line)- Figure 1
Firstly we will make Empire Line markings on the block. When we have finished putting all the markings on the block we will trace to create the four pattern pieces (two front, two back).
• Draw the bust mound circle on your Bodice Front with a compass, with the Bust Point as the center. Use your personal measurements.
• Align the Front and Back Blocks at the side seam.
• Mark the points A to J as shown”
• A = CF waist
• B = dart leg on Bodice Front
• C = dart leg on Bodice Front
• D = waist side seam on Bodice Front
• E = waist side seam on Bodice Back
• F = dart leg on Bodice Back
• G = dart leg on Bodice Back
• H = CB waist
• I = point where dart leg B crosses the dart mound circle*
• J = point where dart leg C crosses the dart mound circle*
* B to I should equal C to J.
## Instructions (Empire Line) – Figure 2
• Draw a line from the CF (point K), at right angles to the CF line, to touch point I.
• Measure from A to K: write down the value = X.
## Instructions (Empire Line) – Figure 3
• Measure up from the back waist point H using the sum of the equation (X), where X is: (A to K) minus 1.25-inches. Label the point L.
• Draw a line from L at right angles to the CB line, draw the line some distance past the second dart leg. (The solid orange line in the image).
• Draw a line up and down from the dart point – up to the Across Back line and down to the line L. ***
*** Very important note: Raising the back dart point to the across back will make the design very fitting. Scroll down to Figure 7 see how you lose some width in the bust line by doing this. You can choose to leave the dart point where it is if you prefer a more relaxed fit.
In this example: X = 3.63 inches. Therefore X = 3.69 -1.25 = 2.44-inches).
## Instructions (Empire Line) – Figure 4A: Standard Dart Adjustment
Now we want to increase the width of both the front and back darts; this is because the Empire Line is fitting under the bust. In this image I am using the standard increase amount of 0.5-inches to the back dart (the width increases a total of 0.5-inch, which is 0.25 inch at each side of the dart on the line L), and .75-inch increase to the front dart, which is 0.38-inch on each side of the dart on the bust mound circle. See the image for how the darts are redrawn. Although I am using some standard amounts, you should use your personalized measurement. See Image 5 for instructions. After doing the calculations outlined in Image 4B, you should draw the new darts as in this image, then proceed to the instructions in Image 5.
## Instructions (Empire Line) – Figure 4B: Personalized Dart Adjustment
• Measure (#1) from L to the first dart leg. Measurement = P
• Measure (#2) from the second dart leg (on the L-line) to point J. Measurement = Q
• Measure (#3) from point I to K. Measurement = R
• Add together: P + Q + R = S. Multiply S x 2, since we have measured the half-blocks.
• Now measure your under-bust circumference, letting the back drop down a little, it doesn’t need to be really tight. This measurement = T.
• Take your Personal Body Under-bust measurement T, add 2 inches ease, then divide by 2. This total = U.
• Calculate: U minus S. The final value = V, and is the amount that needs to be added to the two darts. Use 2/5 of the measurement to add to the back dart, and 3/5 add to the front dart.
## Instructions (Empire Line) – Figure 5
You should now have your darts widened to your personalized measurements; as per the red lines in the image below.
• Measure up from E/D for the value of: (A to K) minus 0.75-inch. Mark point M.
## Instructions (Empire Line) – Figure 6
In this image and the next, we’re going to finish off the Empire Line Markings. It is best to do this in a colored marker so they are obvious.
• Draw a line from L to G2.
• Draw a line from F2 to J2, through M in the side seam.
• Draw a line from I2 to K.
….continued in the next image…
## Instructions (Empire Line) – Figure 7
• Draw lines from the new back dart point (on the across back line), down to G2 and F2.
• Draw lines from G2 to G, and from F2 to F.
• Draw lines from the Bust Point on the Bodice Front to J2 and I2. (Note that a dart point should be created and the dart legs drawn; I won’t go through that here).
• Draw lines from J2 to C and from I2 to B.
The Blocks are now ready to use to trace out and create the four pattern pieces for the Empire Line pattern.
## Instructions (Empire Line) – Figure 8
Now we will take the Bodice Front and create the two pattern pieces: the Bodice and the ????.
• Trace around the block from point M on the side seam to point K on the CF. Mark the Bust point and dart width points.
## Instructions (Empire Line) – Figure 9
• Trace the first pattern piece – shown in the image in orange, remembering to mark the Bust Point and two dart width points.
• Move your block and finish the pattern piece.
• You would need to add seam allowance (not shown in image).
## Instructions (Empire Line) – Figure 10
• Put your Front block on the paper and trace the second pattern piece – the midriff section. Pattern piece #2 is shown on the block shaded in green. When tracing, remembering to mark the two dart width points at the top and bottom (J2 & I2 and C and B).
• Move your block and draw the internal lines of the pattern piece (B). Cut the resultant piece out of the paper, then cut out the dart value portion (shaded in green).
• Stick the two pieces together (C).
• You need to stick the resultant midriff pattern piece back onto paper and blend a smooth curve for the top and bottom of the piece.
• You would need to add seam allowance (not shown here).
## Instructions (Empire Line) – Figure 11
• Follow the same steps for the Bodice Back as for the Bodice Front to create pattern pieces #3 and #4.
## Instructions (Empire Line) – Figure 12
Here are the four pattern pieces, without seam allowance or cutting instructions, etc. </a
### 5 Responses
1. Zinat says:
Oh thank God the site is working now. Phewwww I’m going to need to print things to avoid loosing this wealth of knowledge you have provided.
So I made the pattern for this today and omg it will is coming along nicely. It is perfect thanks to you I did it. Will send pictures once I’ve completed the top xx
2. Maria says:
Hello again Zinat
Glad you are finding the information useful. Please do send me a photo of anything you are making, and also upload it here into the comments.
I have actually made about ten dresses since the beginning of the year, here are two of them (not the best photos!).
3. Zinat Owode oyelaja says:
Wow Maria your dresses look awesome on you. Here’s my make
4. Maria says:
Hello Zinat
That looks very stylish, I really like the sleeves. Is it for you or for a client?
5. Veronica Bamfield says:
I have not sewn for years so I’m a little rusty but the instructions seem incredibly detailed and clear. I will try this and let you know how I get on.
### 5 Responses
1. Zinat says:
Oh thank God the site is working now. Phewwww I’m going to need to print things to avoid loosing this wealth of knowledge you have provided.
So I made the pattern for this today and omg it will is coming along nicely. It is perfect thanks to you I did it. Will send pictures once I’ve completed the top xx
2. Maria says:
Hello again Zinat
Glad you are finding the information useful. Please do send me a photo of anything you are making, and also upload it here into the comments.
I have actually made about ten dresses since the beginning of the year, here are two of them (not the best photos!).
3. Zinat Owode oyelaja says:
Wow Maria your dresses look awesome on you. Here’s my make
4. Maria says:
Hello Zinat
That looks very stylish, I really like the sleeves. Is it for you or for a client?
5. Veronica Bamfield says:
I have not sewn for years so I’m a little rusty but the instructions seem incredibly detailed and clear. I will try this and let you know how I get on. | 2,261 | 9,097 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.859861 |
https://github.com/tresata/ganitha | 1,532,265,624,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676593223.90/warc/CC-MAIN-20180722120017-20180722140017-00066.warc.gz | 663,125,967 | 20,237 | tresata/ganitha
scalding powered machine learning
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Ganitha
Tresata is proud to release Ganitha, our first open-source library. Ganitha (derived from the Sanskrit word for mathematics, or science of computation) is a Scalding library with a focus on machine-learning and statistical analysis.
The current pieces to be open-sourced are our integration of Mahout vectors into Scalding, our clustering (K-Means) implementation, and Naive-Bayes classifiers.
Ganitha-Mahout
To make mahout vectors usable in Scala/Scalding we did the following:
• Pimp Mahout vectors: We used the pimp-my-library pattern in Scala to make Mahout vectors more friendly to use (see `RichVector.scala`). Note that we decided to not implement `RichVector` as an `IndexedSeq`, but rather as an `Iterable`, for our first iteration. We also didn't implement `IterableLike` (or `IndexedSeqLike`) and `CanBuildFrom`, which would allow operations like `vector.take(3)` and `vector.map(f)` to return new Mahout vectors. The reason for this was that we were not happy with the interaction between the builders and the pimp-my-library pattern. We might still add these features in the future. As an alternative, we provided the `vectorMap` methods that return Mahout vectors. Thanks to `RichVector`, you can now do things like:
``````scala> import com.tresata.ganitha.mahout._
scala> import com.tresata.ganitha.mahout.Implicits._
scala> // create a sparse vector of size 6 with elements 1, 3 and 5 non-zero
scala> val v = RichVector(6, List((1, 1.0), (3, 2.0), (5, 3.0)))
v: org.apache.mahout.math.Vector = {5:3.0,3:2.0,1:1.0}
``````
(We can see it's using a Mahout `RandomAccessSparseVector` with `v.getClass`). Likewise, we can create a dense vector with 3 elements as follows:
``````scala> val v1 = RichVector(Array(1.0,2.0,3.0))
v1: org.apache.mahout.math.Vector = {0:1.0,1:2.0,2:3.0}
``````
We can also perform basic math and vector operations (map, fold, etc.) on the vectors. Elements inside the vectors can be accessed and set (since Mahout vectors are mutable), however, this is not encouraged.
``````scala> (v + 2) / 2
res1: org.apache.mahout.math.Vector = {5:2.5,4:1.0,3:2.0,2:1.0,1:1.5,0:1.0}
scala> v.map(x => x * 2).sum
res2: Double = 12.0
scala> v.fold(0.0)(_ + _)
res3: Double = 6.0
scala> v(3)
res4: Double = 2.0
scala> v(3) = 3.0
scala> v
res5: org.apache.mahout.math.Vector = {5:3.0,3:3.0,1:1.0}
scala> v * v
res6: org.apache.mahout.math.Vector = {5:9.0,3:4.0,1:1.0}
``````
The `nonZero` method provides access to the non-zero elements as a scala `Iterable`.
``````scala> v.nonZero.toMap
res7: scala.collection.immutable.Map[Int,Double] = Map(5 -> 3.0, 3 -> 3.0, 1 -> 1.0)
``````
Dense vectors can be converted to sparse, and vice versa.
``````scala> v1.toSparse.getClass
res8: java.lang.Class[_ <: org.apache.mahout.math.Vector]
= class org.apache.mahout.math.RandomAccessSparseVector
``````
The `vectorMap` operation provides access to the assignment operation on a Mahout vector, but as a non-mutating operation (it creates a copy first).
``````scala> v.vectorMap(x => x * 2)
res9: org.apache.mahout.math.Vector = {5:6.0,3:4.0,1:2.0}
``````
• Make serialization transparent: Mahout's vectors come with a separate class called `VectorWritable` that implements `Writable` for serialization within Hadoop. The issue with this is that you cannot just register `VectorWritable` as a Hadoop serializer and be done with it. If you did this then you would have to constantly wrap your Mahout vectors in a `VectorWritable` to make them serializable. To make the serialization transparent we added `VectorSerializer`, a Kryo serializer that defers to `VectorWritable` for the actual work. All one has to do is register `VectorSerializer` with Kryo, and serialization works in Scalding. For example, if you are using a `JobConf` you can write:
`VectorSerializer.register(job)`
The same applies to a Scalding `Config` (which is a `Map[AnyRef, AnyRef]`):
`VectorSerializer.register(config)`
Naive-Bayes classifying
A Naive-Bayes classifier is a probabilistic classifier used in machine-learning that involves the application of Bayes' theorem. The underlying model is "naive" because of the assumption that the attributes are conditionally independent of each other. Naive-Bayes learning is suprisingly effective in a wide range of applications, given the simplifying assumption of feature independence. Though not as powerful as decision-tree learning, it is considerably less computationally complex than many other forms of classifiers, and in many cases, the naive assumption has little impact on the quality of predictions.
Ganitha supplies three of the more popular forms of Naive-Bayes classifiers: Gaussian, Multinomial, and Bernoulli. In gaussian Naive-Bayes, a type of classifier used for continuous data, we are making the assumption that the features associated with each class lie along a normal distribution. In a multinomial or Bernoulli event model, we are dealing with discrete features, a common example being the classification of a document given the presence of words (features) in the text. In this case, each word has a score assigned to it for each label, or class. In multinomial Naive-Bayes, each feature vector relates to the term frequency of the words found in the document or class. We make the 'bag-of-words' assumption, in which documents are represented as a multiset of words, disregarding grammar or word order. In Bernoulli Naive-Bayes, features represent binary occurences, and in this classification model, the absence of a word/feature has an effect on the calculated probabilities.
Each classifier consists of a training phase, where an `NBModel` is constructed from the training set of data, and a classifying, or predicting, phase. In the classifying phase, each data point that is to be classified is given a probability (in this case a log probability is used) for each label, and the label with the highest, or maximum a posteriori probability is assigned to the data point.
K-Means clustering
K-means clustering consists of partitioning data points into k 'clusters' where each point belongs to the cluster with the nearest mean. The process of refining the centers of the clusters is commonly known as Lloyd's algorithm, however there exist heuristic algorithms to seed the initial selection of cluster centers in order to improve the rate of convergence of Lloyd's algorithm. K-Means++ offers an improvement over random initial selection, and more recently, K-Means|| offers an initialization technique that greatly cuts down on the number of iterations needed to determine initial clusters, a very desirable optimization in Hadoop applications, where significant overhead is involved in each iteration.
Ganitha provides an extensible interface for handling vector operations using different representations for data points, including Mahout vectors (which can contain categorical and textual features in addition to numerical). The `VectorHelper` trait can be used to specify how vectors are defined from the input and how distances are calculated between vectors.
K-Means in Ganitha (currently) reads vectors from Cascading Sequence files, and the algorithm writes a list of vectorid-clusterid pairs to a Tap, as well as a list of cluster ids with coordinates.
References
K-Means: Lloyd, S., "Least squares quantization in PCM". IEEE Trans. Information Theory, 28(2):129-137, 1982.
K-Means++: Arthur, D. and Vassilvitskii, S. (2007). "k-means++: the advantages of careful seeding". Proc. ACM-SIAM Symp. Discrete Algorithms. pp. 1027–1035.
K-Means||: Bahmani, B. et al. (2012). "Scalable k-means++". Proceedings of the VLDB Endowment, 5(7), 622-633.
Getting started with Ganitha and K-Means
Ganitha uses sbt for generating builds. To create a runnable jar distribution, run `sbt update` and `sbt assembly`. Unit tests are included and can be run using `sbt test`.
To run K-Means clustering on a test set of data, stored as a comma-separated values file with a header (in this example, with a file on Hadoop named 100kPoints.csv with the header (`id,x,y`), run the following command from within the ganitha directory:
``````hadoop jar ganitha-ml/target/scala-2.10/ganitha-ml-assembly-0.1-SNAPSHOT.jar com.twitter.scalding.Tool com.tresata.ganitha.ml.clustering.KMeansJob --hdfs --vecType StrDblMapVector --distFn euclidean --k 100 --id id --features x y --input 100kPoints.csv --vectors 100kVectors --vectorOutput vectorAssignments --clusterOutput centroids
``````
This will use the `id` columns as the vector id, and will encode the coordinates(`x` and `y`) as `Map[String, Double]` vectors (using the `StrDblMapVector` VectorHelper), under a Euclidean space, and run the algorithm on k=100 clusters. The output is written to a `vectorAssignments` file on Hadoop, with the cluster centroids written to `centroids`. The `vectors` argument specifies a location for the Cascading Sequence file that serves as the input for `KMeans`. | 2,360 | 9,115 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2018-30 | latest | en | 0.845585 |
https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10B_Problems/Problem_12&diff=prev&oldid=107754 | 1,642,892,406,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303917.24/warc/CC-MAIN-20220122224904-20220123014904-00062.warc.gz | 175,084,553 | 11,439 | During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made.
# Difference between revisions of "2003 AMC 10B Problems/Problem 12"
## Problem
Al, Betty, and Clare split $\textdollar 1000$ among them to be invested in different ways. Each begins with a different amount. At the end of one year, they have a total of $\textdollar 1500$ dollars. Betty and Clare have both doubled their money, whereas Al has managed to lose $\textdollar100$ dollars. What was Al's original portion?
$\textbf{(A)}\ \textdollar 250 \qquad \textbf{(B)}\ \textdollar 350 \qquad \textbf{(C)}\ \textdollar 400 \qquad \textbf{(D)}\ \textdollar 450 \qquad \textbf{(E)}\ \textdollar 500$
## Solution
For this problem, we will have to write a three-variable equation, but not necessarily solve it. Let $a, b,$ and $c$ represent the original portions of Al, Betty, and Clare, respectively. At the end of one year, they each have $a-100, 2b,$ and $2c$. From this, we can write two equations.
$$a+b+c=1000$$\\ $$2a+2b+2c=2000$$\\ \\ $$a-100+2b+2c=1500\\ a+2b+2c=1600$$
Since all we need to find is $a,$ subtract the second equation from the first equation to get $a=400.$
Al's original portion was $\boxed{\textbf{(C)}\ \textdollar 400}$. | 381 | 1,230 | {"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": 14, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2022-05 | latest | en | 0.868674 |
http://mathhelpforum.com/advanced-applied-math/208049-cycles-nodes-random-graph-theory-print.html | 1,508,262,649,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822145.14/warc/CC-MAIN-20171017163022-20171017183022-00462.warc.gz | 219,903,107 | 3,099 | # cycles of nodes in random graph theory
• Nov 20th 2012, 04:57 AM
xaosman
cycles of nodes in random graph theory
Could someone help me out with some basic graph theory?
(a cycle consisting of four nodes connected by four links to a cycle. A cycle has no start or end point nor a direction.)
I want to show that in the random graph G(n, p) (where n is the number of nodes nodes and p is the probability of the existence of a link between two nodes), that there are ((n (n-1)(n-2)(n-3))/8 possible cycles of length 4 (consisting of only 4 nodes that is)
I know from examples that there are (1/6)n(n -2)(n -1) possible triangles from n number of nodes. But when I use the same method for the number of connected nodes with length 4 I get:
(from n choose 4) = (1/24)n(n -3)(n -2)(n -1)
Which is not correct and I would need to multiply it with 3 to make it right and I don't get why. could someone help me out?
I'm so sorry if anything is unclear. English is not my frist language.
• Nov 20th 2012, 06:36 AM
Plato
Re: cycles of nodes in random graph theory
Quote:
Originally Posted by xaosman
I want to show that in the random graph G(n, p) (where n is the number of nodes nodes and p is the probability of the existence of a link between two nodes), that there are ((n (n-1)(n-2)(n-3))/8 possible cycles of length 4 (consisting of only 4 nodes that is)
(from n choose 4) = (1/24)n(n -3)(n -2)(n -1)
Which is not correct and I would need to multiply it with 3 to make it right and I don't get why. could someone help me out?
Where did you get this "that there are ((n (n-1)(n-2)(n-3))/8 possible cycles of length 4"
Clearly it should be $\binom{n}{4}=\frac{n(n-1)(n-2)(n-3)}{4!}$.
There must be other restrictions you have not included. | 507 | 1,744 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-43 | longest | en | 0.944037 |
https://www.applewriters.com/daily-energy-balance-2/ | 1,669,524,522,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710192.90/warc/CC-MAIN-20221127041342-20221127071342-00288.warc.gz | 694,145,405 | 15,378 | Select Page
Subject: Physics / General Physics
Question
Name____________________________________________________Section________________Date___________
Week 5: Daily Energy Balance
Invitation to Inquiry
Many kinds of foods are marketed to those who participate in various kinds of sports. The implication is that these foods have additional nutrients or higher quantities of nutrients need by the athlete. Go to a store and read the ingredient label on one of these products. Compare it to an equivalent product that is not marketed in such a manner. For example, you could compare a sports drink to a soft drink or orange juice. You could compare a “nutrient bar” to an equivalent candy bar or snack food. Look specifically at quantities of calories, fats, proteins, sodium, and potassium. How are they different? What other foods could you eat that would provide the same calories and nutrients?
Background
The theoretical biological sciences of biochemistry, anatomy, cell biology, and physiology are brought together in the practical biological field of nutrition. The science of nutrition is the study of the processes involved in taking in, assimilating, and utilizing nutrients. The amount of food and drink consumed by a person from day to day is a person’s diet. There has been an increased interest in diet and personal nutrition as more information concerning these subjects becomes available through the popular press, scientific publications, health clubs, and schools. Not only are people “counting calories” and concerned with the grams of fat they consume, but they are becoming scientifically literate enough to ask significant questions to their physicians, teachers, food manufacturers, and government officials. With a minimal amount of nutrition information, it is possible to get a better handle on your own nutritional status.
In this exercise, you determine your daily basal metabolic rate, voluntary muscular activity, and specific dynamic action per day. These are used to estimate your total energy requirements per day in kilocalories (kcal). You then calculate your total daily kcal intake. By comparing these two figures, you can determine whether or not your present diet should result in your maintaining, losing, or gaining weight.
You will determine your:
1. basal metabolic rate;
2. voluntary muscular activity level;
3. specific dynamic action;
4. kilocalorie intake per day by adding your BMR, activity level, and SDA;
5. total energy requirements and compare to your energy kcal intake per day; and
6. energy balance.
Procedure
Your BMR (basal metabolic rate) is the rate at which kcals are used for maintenance activities and can be measured on a daily basis. This is also the total amount of energy per kilogram per hour expended after a 12-hour fast. Energy is measured in kilocalories, the amount of energy needed to raise the temperature of 1 kg of water 1?C. BMR can be estimated by using a short formula that is based on 1.0 kcal per kilogram of body weight per hour for men, or 0.9 kcal per kilogram of body weight for women. Even though this is a crude method, it does give some idea of the BMR.
Body weight ? BMR factor = Estimated BMR (kcal/kg/hour)
For example: If a male weighs 150 lbs, his mass in kilograms will be 68 kg. Therefore, the estimated BMR is
68 kg ? 1.0 kcal/kg/hr = 68 kcal/kg/hr
24 hours/day ? Estimated BMR/hour = Estimated energy expenditure/day
68 kcal/kg/hr ? 24 hours = 1632 kcal/kg/day
If a female weighs 120 lbs, her mass in kilograms will be 55 kg. Therefore, the estimated BMR is
55 kg ? 0.9 kcal/kg/hr = 49 kcal/kg/hr
24 hours/day ? Estimated BMR/hour = Estimated energy expenditure/day
49 kcal/kg/hr ? 24 hours = 1176 kcal/kg/day
These are estimated basal metabolic rates for these two people. Using this method, calculate your own BMR:
Body weight in kg ? BMR factor in kcal/kg/hr = Estimated energy expenditure in kcal/kg/hr
24 hours/day ? Your estimated energy expenditure/hour = kcal/day (your estimated energy expenditure/day or kcal/day used while at rest)
For a more accurate determination of your BMR, use standard tables from your text and calculate your skin surface area from your height and weight. A table of kilocalories per day per square meter of skin lists the kilocalories expended by a female or male by age group. This kilocalorie figure should be multiplied by your skin surface area to determine your BMR more accurately.
Skin surface area ? Kilocalories per day per square meter of skin = BMR
Your skin surface area ? Kilocalories per day per square meter of your skin = kcal/day
Estimating Your Energy Output per Day
Energy output per day is an estimate of your voluntary muscular activity per day. For a person who engages in only sedentary activities such as desk work, the estimated energy output is approximately 50% of his or her already determined BMR. For example, if the male in the previous example were a typist, his voluntary muscular activity level for the day would be:
0.50 ? 1632 kcal/day = 816 kcal/day
For a person who engages in light activities such as standing, talking, and minor amounts of walking, the estimated energy output is approximately 60% of his or her already determined BMR. For example, if the female in the previous example were a teacher, her voluntary muscular activity level for the day would be:
0.60 ? 1176 kcal/day = 706 kcal/day
For a person who engages in moderate activities that exceed those described as light, the estimated energy output is approximately 70% of his or her already determined BMR. For example, if the male described were a nurse, his voluntary muscular activity level for the day would be:
0.70 ? 1632 kcal/day = 800 kcal/day
Those participating in heavy activities are estimated to use an equivalent of their BMR per day. For a person engaged in heavy lifting and moving or a daily workout of an hour or more:
1.00 ? 1632 kcal/day = 1,632 kcal/day
Estimate your voluntary muscular energy expenditure per day:
Percent of BMR based on activity level ? BMR = kcal/day
Estimating Your Specific Dynamic Action (SDA)
The specific dynamic action (SDA) is the amount of energy needed to metabolize food for the day. This is approximately 10% of a person’s total daily basal expenditure and total daily voluntary muscular activity. For example, the male nurse had a voluntary muscular activity level for the day of 800 kcal, and his daily expenditure was 1,632 kcal. Therefore, his SDA for the day would be anestimated 243 kcal. (Your basal energy expenditure + Your voluntary daily physical activity) ? 0.10 = kcal/day (your SDA) Add together your BMR and your voluntary muscular energy expenditure. BMR + voluntary muscular activity = kcal/day Total Daily Energy Requirements Your total daily energy requirements are the sum of your BMR + voluntary muscular activity + SDA = kcal/day
Before you can draw any conclusions about your daily energy balance, you must determine your total energy intake. This can be estimated by recording your total consumption of nutrients and determining their kcal values. Fill in Table 5.1 beginning with the first meal of your day and ending with the last snack you consume. The estimate of your kcal intake can be determined by using the tables found in many supermarkets, bookstores, and libraries. Such tables are usually referred to as pocket calorie counters.
Compare your total daily energy requirement with your actual kilocalorie intake per day:
Total daily energy requirement – Actual kilocalorie intake = Gain/maintenance/loss
If these two figures are the same, you are meeting your energy requirements and maintaining your weight. If your total daily caloric intake is greater than your caloric requirement, you are exceeding your energy requirements and, therefore, gaining weight. The opposite is true if you are not meeting your caloric requirement. Therefore, you are losing weight.
Table 5.1 Total Energy Intake Food Food Serving Size Energy kcal Breakfast
Lunch
Dinner
Other
Total = ________________
Name ___________________________________________ Lab section____________________
Your instructor may collect these end-of-exercise questions. If so, please fill in your name and lab section.
End-of-Exercise Questions
1. What is basal metabolic rate?
2. Do basal metabolic rates differ between males and females? On what evidence can you base your answer?
3. What factors are involved in accurately determining your BMR?
4. What is specific dynamic action?
5. If your total kilocalorie intake per day is higher than your total kilocalorie requirements, what happens to your weight?
6. What can you do to bring about an energy balance?
7. Other than kilocalories, what information is important in determining whether or not you are consuming a healthy diet?
8. What resources are available to help you develop a balanced diet?
Part 2 – BMI
A: Define BMI-You must define BMI and not just as a general definition indicating its relationship to height and weight. Please also do not forget to define the actual term BMI!
B: Calculate BMI- Do no use an online calculator – show your actual math calculations. Please include all stages of the math calculations associated with determining your BMI.
C: Tie it all together. Consider the diet plan that you created in Table 5.1 for the determining total energy input. Based on your own BMI and how you fall according to the standard tables (underweight, normal, overweight, obese), evaluate your dietary plan. Build on the caloric intake you’ve already listed to include an exercise plan, being as specific as possible. The goal of this section is to develop a plan that incorporated all of the components of a sound fitness program (flexibility, endurance, strength training, body fat, and cardiovascular) as well as discuss specific exercises that you planned on doing.
Order Now | 2,101 | 9,896 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2022-49 | latest | en | 0.938804 |
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Summing inverses II (Posted on 2012-03-16)
The sum of the reciprocal of the square root of all the positive integers up to n is denoted by F(n), that is:
F(n) = 1+1/√2 + 1/√3 +...+ 1/√n
Determine the maximum value of n such that the integer part of the base ten expansion of F(n) DOES NOT exceed 2012.
*** For an extra challenge, solve this puzzle without using a computer program.
No Solution Yet Submitted by K Sengupta No Rating
Comments: ( Back to comment list | You must be logged in to post comments.)
My approximation | Comment 4 of 12 |
integral of x^(-1/2) is 2*sqrt(x)
evaluating from 0 to n we get :
2*sqrt(n)<2012
n<1006^2=1012036
n=1012035
QUICK , BUT DEFINITELY NOT PRECISE
Posted by Ady TZIDON on 2012-03-16 14:37:00
Search: Search body:
Forums (0) | 273 | 884 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2019-04 | latest | en | 0.716721 |
https://proofwiki.org/wiki/Rearrangement_Operation_as_Replacement_Operations | 1,717,028,799,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059412.27/warc/CC-MAIN-20240529230852-20240530020852-00212.warc.gz | 412,093,267 | 11,139 | # Rearrangement Operation as Replacement Operations
Jump to navigation Jump to search
## Example of use of Replacement Operation
Let $\left({a, b, c, d}\right)$ be an ordered quadruple consisting of four variables whose values are to be rearranged into the order $\left({b, c, d, a}\right)$.
This can be implemented using replacement operations.
Let $t$ be a new variable which has been established for this purpose.
Then the sequence of replacement operations:
$t \gets a$
$a \gets b$
$b \gets c$
$c \gets d$
$d \gets t$
performs the task.
## Proof
Observing the values of the variables after each replacement operation:
Operation $a$ $b$ $c$ $d$ $t$
$t \gets a$ $a$ $b$ $c$ $d$ $a$
$a \gets b$ $b$ $b$ $c$ $d$ $a$
$b \gets c$ $b$ $c$ $c$ $d$ $a$
$c \gets d$ $b$ $c$ $d$ $d$ $a$
$d \gets t$ $b$ $c$ $d$ $a$ $a$
Hence the result.
Notice how the sequence:
$a \gets b, b \gets c, c \gets d, d \gets a$
does not do the job.
This is because, when $d \gets a$ is performed, $a$ no longer contains its original value, and the resulting ordered quadruple is $\left({b, c, d, b}\right)$.
$\blacksquare$ | 350 | 1,111 | {"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} | 3.46875 | 3 | CC-MAIN-2024-22 | latest | en | 0.685171 |
https://www.mbaskool.com/business-concepts/human-resources-hr-terms/15437-work-sampling.html | 1,726,856,901,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701419169.94/warc/CC-MAIN-20240920154713-20240920184713-00110.warc.gz | 800,400,637 | 16,042 | # Work Sampling
This article covers meaning & overview of Work Sampling from HRM perspective.
## What is Work Sampling?
Work sampling is the statistical technique for determining the proportion of time spent by workers in various defined categories of activity (e.g. setting up a machine, assembling two parts, idle etc). The process of making a number of observations in a random manner at random intervals of time over a particular period (of time) as well group of employees, processes, machines etc. The calculations are done on a percentage basis wherein we record the happenings at every instant & take into consideration of the percentage of activities taking place in that time frame.
Is a sampling method and makes use of the theory of probability. We use a sample which represents the population is used for the purpose of sampling.
Applications of Work Sampling
1. Used for comparing between the idle and working time, this is the foundation for ratio studies.
2. Widely used as a work measurement technique.
3. Used for preparing performance index.
Main Features of Work Sampling: Dependent on work conditions.
(a) Availability of time for carrying out the particular process/ study.
(b) This method can only be used when we need to study the actions performed by a number of people.
(c) Long cycle time: the job which has to be studied should be spread over a longer time span.
(d) The job should be non repetitive so that people don’t get bored of the job.
Steps involved in sampling study
1. Defining the tasks which need to be analyzed.
2. Defining different elements of the task & breaking down that task by defining various categories like idle time, waiting time and absence.
3. Designing the type of study by designing, recording and estimating the shifts, days etc. needed for the study.
4. Identifying the people who will be involved in sampling and other study and starting with the study.
5. After conducting the study analyzing the results.
Example: A worker working in shifts does work for some part of time and remains idle for rest of the time. Let us suppose that there are 60 observations, out of which there were 57 working and 3 idle observations.
Here, idle time % = (3/60) = 5
Working Time % = (57/60) = 95
If a worker works in a shift of 8 hrs (480 min), idle time is 24 min and working time is 456 min.
1. It’s an economical method of doing the time study of work.
2. It is a highly flexible method wherein data can be collected anytime without affecting the results.
3. Chance of day to day or weekly variation are reduced since the time span of taking observation is wide.
4. These methods are less tedious and fatiguing.
1. It’s not an economical process when a single operator/ machine is involved.
2. Doesn’t provide any elemental time data.
3. The process of work study is difficult to understand compared to time study.
4. The working pattern may be changed by the workers and then it becomes difficult and results may be erroneous.
Hence, this concludes the definition of Work Sampling along with its overview.
This article has been researched & authored by the Business Concepts Team which comprises of MBA students, management professionals, and industry experts. It has been reviewed & published by the MBA Skool Team. The content on MBA Skool has been created for educational & academic purpose only.
Browse the definition and meaning of more similar terms. The Management Dictionary covers over 1800 business concepts from 5 categories. | 725 | 3,519 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2024-38 | latest | en | 0.933624 |
http://www.bowlandmaths.org.uk/materials/projects/online/save_a_baby_kangaroo/start.htm | 1,506,054,114,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818688208.1/warc/CC-MAIN-20170922041015-20170922061015-00440.warc.gz | 403,842,772 | 2,469 | ### Overview
Mathematics can save lives
This case study is an authentic context in which an adult in the Australian Outback attempts to save a young orphaned kangaroo. Through video clips, photographs and data, pupils become familiar with a range of data about the several different species of kangaroo. They use the data to identify which kangaroo has been assigned to them and develop a feeding programme in order to save its life.
### Mathematical content
Key stage 3 National Curriculum areas covered include:
• Key processes – pupils are expected to identify for themselves the mathematics; put together their own chains of reasoning and use mathematical procedures in combination; create convincing arguments from their evidence; consider their assumptions and the accuracy of their results and communicate effectively; reflect on their approaches and results.
• Number and algebra – using rational numbers and their properties and their different representations; rules of arithmetic applied to calculations and manipulation with rational numbers; application of ratio and proportion.
• Geometry and measure – points, lines and shapes in 2D coordinate system; units, compound measures and conversions.
• Statistics – presentation and analysis of grouped data; the handling data cycle.
### Organisation and pedagogy
There are four teacher-led activities and one pupil-led presentation lesson, each of one hour. It is possible that some of the lessons will take longer than an hour. Each lesson is presented in two formats: as a detailed lesson format for those teachers unfamiliar with extended problem solving tasks and also as a lesson outline for more experienced teachers. A mixture of class, individual and group work is involved. The materials are appropriate for all pupils in Key Stage 3. Each lesson has a small homework task. Access to computers is highly desirable, but not absolutely necessary.
### Resources
The resources are contained in two interactive resources:
Pupils' browser - containing all the information, videos etc. needed by pupils
Teacher browser - containing all the teachers resources and lesson plans
Additionally, each pupil will need a paper copy of the following:
Adobe Reader is required to view the pdf files contained within this case study.
Microsoft Excel is needed for some activities. A graph plotting programme is desirable. | 432 | 2,383 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-39 | latest | en | 0.942201 |
https://www.hackmath.net/en/example/3683 | 1,558,934,199,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232261326.78/warc/CC-MAIN-20190527045622-20190527071622-00505.warc.gz | 793,549,438 | 6,979 | # Minute
Two boys started from one place. First went north at velocity 3 m/s and the second to the east with velocity 4 m/s. How far apart they are after minute?
Result
x = 300 m
#### Solution:
Leave us a comment of example and its solution (i.e. if it is still somewhat unclear...):
Be the first to comment!
#### To solve this example are needed these knowledge from mathematics:
Pythagorean theorem is the base for the right triangle calculator.
## Next similar examples:
1. Mail train
The speed of mail train is 1370 meter per minute. Express it in miles per hour correct to three significant figure . Given that 1 meter =39.37 inches.
2. Cars 6
At 9:00 am two cars started from the same town and traveled at a rate of 35 miles per hour and the other car traveled at a rate of 40 miles per hour. After how many hours will the cars be 30 miles apart?
3. Round-trip
A woman works at a law firm in city A which is about 50 miles from city B. She must go to the law library in city B to get a document. Find how long it takes her to drive round-trip if she averages 40 mph.
4. Four ropes
TV transmitter is anchored at a height of 44 meters by four ropes. Each rope is attached at a distance of 55 meters from the heel of the TV transmitter. Calculate how many meters of rope were used in the construction of the transmitter. At each attachment.
8.3 meters long ladder is leaning against the wall of the well, and its lower end is 1.2 meters from this wall. How high from the bottom of a well is the top edge of the ladder?
Ladder 6.4 meters long is positioned in the well such that its lower end is distanced from the wall of the well 1.2 m. The upper part of the ladder is supported on the upper edge of the well. How high is the well?
7. Square
Rectangular square has side lengths 183 and 244 meters. How many meters will measure the path that leads straight diagonally from one corner to the other?
8. Windbreak
A tree at a height of 3 meters broke in the windbreak. Its peak fell 4.5 m from the tree. How tall was the tree?
9. Broken tree
The tree was 35 meters high. The tree broke at a height of 10 m above the ground. Top but does not fall off it refuted on the ground. How far from the base of the tree lay its peak?
10. Broken tree
The tree is broken at 4 meters above the ground and the top of the tree touches the ground at a distance of 5 from the trunk. Calculate the original height of the tree. | 596 | 2,420 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2019-22 | latest | en | 0.951492 |
http://perplexus.info/show.php?pid=3017&cid=48084 | 1,534,276,480,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221209562.5/warc/CC-MAIN-20180814185903-20180814205903-00260.warc.gz | 336,878,613 | 4,363 | All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
perplexus dot info
Go for short! (Posted on 2005-04-19)
Looking at the "Square of an Odd" puzzle that asks to prove that the square of an odd number is always 1 more than a multiple of 8, a professor gave this four parts proof: "All odd numbers are of the form 8K+1, 8K+3, 8K+5 or 8K+7. Squaring these numbers produces 8M+1, 8M+9, 8M+25 or 8M+49, which are all of the form 8N+1. QED"
Another professor came by, and gave a single line proof. Can you manage it?
Note: no one who answered the original problem produced either the four parts solution, or the single line one.
See The Solution Submitted by e.g. Rating: 2.0000 (2 votes)
Comments: ( Back to comment list | You must be logged in to post comments.)
my way | Comment 7 of 8 |
7 years after it was posted:
x^2-1=(x-1)*(x+1) ... one line to look at
if x is odd then we have a product of two succesive even numbers, and one of them must be divisible by 4.
qed
Posted by Ady TZIDON on 2012-08-07 12:55:59
Search: Search body:
Forums (0) | 337 | 1,096 | {"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} | 3.359375 | 3 | CC-MAIN-2018-34 | latest | en | 0.893448 |
https://electricalacademia.com/basic-electrical/series-circuit-applications-troubleshooting/ | 1,721,129,503,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514745.49/warc/CC-MAIN-20240716111515-20240716141515-00110.warc.gz | 199,993,001 | 30,635 | # Series Circuit Applications and Troubleshooting
There are numerous applications for the principles of series circuits. This section demonstrates how to apply those principles and shows you the specific example of an airfield lighting system. In addition, troubleshooting series circuits is discussed.
## Applying Ohm’s Law to a Series Circuit
Ohm’s law can be applied to any individual component of a series circuit. Look at Figure 1. The Ohm’s law formulas are noted at each location. If you know any two values at an individual location, you can apply Ohm’s law to find the third value.
Figure 1. Ohm’s law can be applied at each location in the circuit. Note that when the current value is found, it can be applied anywhere in the circuit.
Example: If R1 has a resistance value of 10 ohms and a voltage drop of five volts, then applying Ohm’s law to the R1 location will give you a current value of 0.5 amperes.
${{I}_{1}}=\frac{{{V}_{1}}}{{{R}_{1}}}=\frac{5V}{10\Omega }=0.5A$
Could you go on from this point and calculate the amount of power used by R1?
Airfield Lighting System
An airfield lighting system has many miles of circuit cable. The system has miles of cable originating from the source of power to the runway, plus the distance around the runway itself.
The only practical way to light a runway and taxiway system is to use the series circuit principles. Figure 2 illustrates how large the system may be. The voltage losses caused by the resistance of the copper lines do not permit a practical application of a parallel circuit.
The lighting consists of a series circuit with a transformer located at each individual light location. The circuit is connected to a voltage regulator that maintains a constant amperage applied to the circuit. Since the circuit has a constant amperage, and each transformer and lighting unit has equal resistance, each lamp will have the same voltage drop value.
Figure 2. An airport lighting system can be over 50 miles in length for all circuits.
At times, the number of lamps in the circuit will change due to lamp failure. When the number of lamps changes, the amount of applied voltage from the regulator will change in order to maintain the constant 6.6 amperes applied to the circuit.
By maintaining a constant 6.6 amperes to the circuit, each lamp will burn at the same brightness, regardless of how many lamps are lit at once, and regardless of the voltage drop along the length of the conductor. See Figure 3.
Figure 3. A typical airfield lighting system consists of a series of lamps connected to a voltage regulator. Each lamp will glow at equal brightness
## Troubleshooting a Series Circuit Using a Voltmeter
A series circuit is really easy to troubleshoot. In Figure 4, three resistors are connected in series to the power supply. A fuse is used for circuit protection and a SPST switch is used to control the current to the resistors.
One of the most common circuit faults in a series circuit is an open. A voltmeter, combined with the knowledge of the laws of voltages, is a quick and easy method of analysis. There are several ways to locate a circuit fault. The following step-by-step presentation is not the only way to locate the fault.
Figure 4. An open switch in a series circuit produces a reading on a voltmeter equal to the source voltage.
Lesson in Safety:
Remember, good safety is a habit that needs to be developed over a period of time. When troubleshooting even low-voltage circuits, practice all safety techniques. Safety must become second nature to a technician when working on electrical circuits.
To begin the troubleshooting process, measure to determine if there is voltage at the supply. Remember, a bad battery can still have a high voltage reading when it is not connected to a load. When checking the battery for proper voltage, make sure the switch (S1) is closed.
If the voltage at the battery is good, move to the fuse. Take a voltage drop reading across the fuse. A good fuse will not produce a voltage drop. A blown fuse will produce a voltage drop that is equal to the source voltage.
Also check the voltage drop across the switch. The closed switch should not produce a voltage drop, Figure 5. A single- pole single-throw (SPST) switch that produces a voltage drop when closed is defective.
Next, check to see if there is a voltage drop across the individual resistors. If a resistor connected in a series circuit is open, it will have a voltage drop equal to the source voltage. The other resistors will have no voltage drop at all.
Figure 5. When the series circuit is complete, there will be voltage drops across each load component in the circuit.
A short circuit is another possible problem. If one of the resistors is shorted, that resistor will show a voltage drop of zero. The other resistors in the circuit will drop the entire source voltage. | 1,042 | 4,878 | {"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} | 4.4375 | 4 | CC-MAIN-2024-30 | latest | en | 0.908167 |
http://www.jiskha.com/display.cgi?id=1368113623 | 1,495,997,861,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463610374.3/warc/CC-MAIN-20170528181608-20170528201608-00416.warc.gz | 679,438,062 | 3,639 | # Math
posted by on .
convert 1./27 convert repeating decimal to a/b
• Math - ,
I will assume you mean 1.272727..
that is 1.27 with a bar over the 27
= 1 + 27/99
= 1 + 3/11
= 14/11
verify by doing | 76 | 202 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-22 | latest | en | 0.832355 |
http://stackoverflow.com/questions/15045920/how-can-i-find-the-smallest-double-divisor-that-wont-give-an-infinite-result | 1,427,399,494,000,000,000 | text/html | crawl-data/CC-MAIN-2015-14/segments/1427131292567.7/warc/CC-MAIN-20150323172132-00074-ip-10-168-14-71.ec2.internal.warc.gz | 214,290,281 | 16,240 | # How can I find the smallest `Double` divisor that won't give an infinite result?
``````Welcome to Scala version 2.9.2 (Java HotSpot(TM) 64-Bit Server VM, Java 1.6.0_26).
scala> 1.0 / Double.MinPositiveValue
res0: Double = Infinity
``````
Oh. Annoying. I was hoping I could do something like:
``````def f(x: Double) = 1.0 / (x + Double.MinPositiveValue)
``````
...and avoid `Infinity` for `f(0.0)`. Let's try to find a slightly bigger number:
``````scala> val xs = Iterator.iterate(Double.MinPositiveValue)(_ + Double.MinPositiveValue)
xs: Iterator[Double] = non-empty iterator
scala> xs.take(10).toList
res1: List[Double] = List(4.9E-324, 1.0E-323, 1.5E-323, 2.0E-323, 2.5E-323, 3.0E-323, 3.5E-323, 4.0E-323, 4.4E-323, 4.9E-323)
``````
OK. Good. It's increasing. How about:
``````scala> xs.map(1.0 / _).take(10).toList
res2: List[Double] = List(Infinity, Infinity, Infinity, Infinity, Infinity, Infinity, Infinity, Infinity, Infinity, Infinity)
``````
Hum... maybe it's going to take a while. Let's try:
``````scala> xs.find(x => !(1.0 / x).isInfinite)
``````
...and I'm still waiting on this one. Doesn't seem like it's going to terminate anytime soon.
How can I find the smallest `Double` divisor that won't give an infinite result?
-
After a little trial and error, I'm almost sure the number is `5.56268464626801E-309`. – Kendall Frey Feb 23 '13 at 21:51
Why are you looking for this magical number? – Randall Schulz Feb 23 '13 at 22:01
@RandallSchulz See my definition of `f`. A bit more context: I'm weighting things by the inverse of their distance to some other thing. If that distance is zero, I want a very big weight, but not infinite either... – Nicolas Payette Feb 23 '13 at 22:08
If you're going to search, at least search with bisection which should take no more than about 1024 iterations since that's 2(#bits in exponent).
But it turns out that you don't need to because you can find it by trial even faster. It should be very close to `1/Double.MaxValue`:
``````scala> Double.MaxValue
res35: Double = 1.7976931348623157E308
scala> 1/res35
res36: Double = 5.562684646268003E-309
scala> 1/res36
res37: Double = Infinity
scala> 1/(res36+math.ulp(res36))
res38: Double = 1.7976931348623143E308
scala> res36+math.ulp(res36)
res39: Double = 5.56268464626801E-309
``````
The thing I find a little bit disturbing is that `1 / ( 1 / Double.MaxValue ) == Double.MaxValue` is `false`. But I guess it's a floating point operation thingymaging :) – Emil H Feb 23 '13 at 22:08 | 806 | 2,508 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2015-14 | latest | en | 0.899247 |
https://foreach.id/EN/fluids/flowratemass/hectogram%7Csecond-to-ton_metric%7Csecond.html | 1,620,691,956,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991553.4/warc/CC-MAIN-20210510235021-20210511025021-00431.warc.gz | 268,076,020 | 8,793 | # Convert hectogram/second to ton (metric)/second (hg/s to t/s)
Batch Convert
• hectogram/second [hg/s]
• ton (metric)/second [t/s]
Copy
_
Copy
• hectogram/second [hg/s]
• ton (metric)/second [t/s]
## Hectogram/second to Ton (metric)/second (hg/s to t/s)
### Hectogram/second (Symbol or Abbreviation: hg/s)
Hectogram/second is one of mass flow rate units. Hectogram/second abbreviated or symbolized by hg/s. The value of 1 hectogram/second is equal to 0.1 kilogram/second. In its relation with ton (metric)/second, 1 hectogram/second is equal to 0.0001 ton (metric)/second.
#### Relation with other units
1 hectogram/second equals to 0.1 kilogram/second
1 hectogram/second equals to 100 gram/second
1 hectogram/second equals to 6,000 gram/minute
1 hectogram/second equals to 360,000 gram/hour
1 hectogram/second equals to 8,640,000 gram/day
1 hectogram/second equals to 6,000,000 milligram/minute
1 hectogram/second equals to 360,000,000 milligram/hour
1 hectogram/second equals to 8,640,000,000 milligram/day
1 hectogram/second equals to 6 kilogram/minute
1 hectogram/second equals to 360 kilogram/hour
1 hectogram/second equals to 8,640 kilogram/day
1 hectogram/second equals to 1e-16 exagram/second
1 hectogram/second equals to 1e-13 petagram/second
1 hectogram/second equals to 1e-10 teragram/second
1 hectogram/second equals to 1e-7 gigagram/second
1 hectogram/second equals to 0.0001 megagram/second
1 hectogram/second equals to 10 dekagram/second
1 hectogram/second equals to 1,000 decigram/second
1 hectogram/second equals to 10,000 centigram/second
1 hectogram/second equals to 100,000 milligram/second
1 hectogram/second equals to 100,000,000 microgram/second
1 hectogram/second equals to 0.0001 ton (metric)/second
1 hectogram/second equals to 0.006 ton (metric)/minute
1 hectogram/second equals to 0.36 ton (metric)/hour
1 hectogram/second equals to 8.64 ton (metric)/day
1 hectogram/second equals to 0.39683 ton (short)/hour
1 hectogram/second equals to 0.22046 pound/second
1 hectogram/second equals to 13.228 pound/minute
1 hectogram/second equals to 793.66 pound/hour
1 hectogram/second equals to 19,048 pound/day
### Ton (metric)/second (Symbol or Abbreviation: t/s)
Ton (metric)/second is one of mass flow rate units. Ton (metric)/second abbreviated or symbolized by t/s. The value of 1 ton (metric)/second is equal to 1000 kilogram/second. In its relation with hectogram/second, 1 ton (metric)/second is equal to 10000 hectogram/second.
#### Relation with other units
1 ton (metric)/second equals to 1,000 kilogram/second
1 ton (metric)/second equals to 1,000,000 gram/second
1 ton (metric)/second equals to 60,000,000 gram/minute
1 ton (metric)/second equals to 3,600,000,000 gram/hour
1 ton (metric)/second equals to 86,400,000,000 gram/day
1 ton (metric)/second equals to 60,000,000,000 milligram/minute
1 ton (metric)/second equals to 3,600,000,000,000 milligram/hour
1 ton (metric)/second equals to 86,400,000,000,000 milligram/day
1 ton (metric)/second equals to 60,000 kilogram/minute
1 ton (metric)/second equals to 3,600,000 kilogram/hour
1 ton (metric)/second equals to 86,400,000 kilogram/day
1 ton (metric)/second equals to 1e-12 exagram/second
1 ton (metric)/second equals to 1e-9 petagram/second
1 ton (metric)/second equals to 0.000001 teragram/second
1 ton (metric)/second equals to 0.001 gigagram/second
1 ton (metric)/second equals to 1 megagram/second
1 ton (metric)/second equals to 10,000 hectogram/second
1 ton (metric)/second equals to 100,000 dekagram/second
1 ton (metric)/second equals to 10,000,000 decigram/second
1 ton (metric)/second equals to 100,000,000 centigram/second
1 ton (metric)/second equals to 1,000,000,000 milligram/second
1 ton (metric)/second equals to 1,000,000,000,000 microgram/second
1 ton (metric)/second equals to 60 ton (metric)/minute
1 ton (metric)/second equals to 3,600 ton (metric)/hour
1 ton (metric)/second equals to 86,400 ton (metric)/day
1 ton (metric)/second equals to 3,968.3 ton (short)/hour
1 ton (metric)/second equals to 2,204.6 pound/second
1 ton (metric)/second equals to 132,280 pound/minute
1 ton (metric)/second equals to 7,936,600 pound/hour
1 ton (metric)/second equals to 190,480,000 pound/day
### How to convert Hectogram/second to Ton (metric)/second (hg/s to t/s):
#### Conversion Table for Hectogram/second to Ton (metric)/second (hg/s to t/s)
hectogram/second (hg/s) ton (metric)/second (t/s)
0.01 hg/s 0.000001 t/s
0.1 hg/s 0.00001 t/s
1 hg/s 0.0001 t/s
2 hg/s 0.0002 t/s
3 hg/s 0.0003 t/s
4 hg/s 0.0004 t/s
5 hg/s 0.0005 t/s
6 hg/s 0.0006 t/s
7 hg/s 0.0007 t/s
8 hg/s 0.0008 t/s
9 hg/s 0.0009 t/s
10 hg/s 0.001 t/s
20 hg/s 0.002 t/s
25 hg/s 0.0025 t/s
50 hg/s 0.005 t/s
75 hg/s 0.0075 t/s
100 hg/s 0.01 t/s
250 hg/s 0.025 t/s
500 hg/s 0.05 t/s
750 hg/s 0.075 t/s
1,000 hg/s 0.1 t/s
100,000 hg/s 10 t/s
1,000,000,000 hg/s 100,000 t/s
1,000,000,000,000 hg/s 100,000,000 t/s
#### Conversion Table for Ton (metric)/second to Hectogram/second (t/s to hg/s)
ton (metric)/second (t/s) hectogram/second (hg/s)
0.01 t/s 100 hg/s
0.1 t/s 1,000 hg/s
1 t/s 10,000 hg/s
2 t/s 20,000 hg/s
3 t/s 30,000 hg/s
4 t/s 40,000 hg/s
5 t/s 50,000 hg/s
6 t/s 60,000 hg/s
7 t/s 70,000 hg/s
8 t/s 80,000 hg/s
9 t/s 90,000 hg/s
10 t/s 100,000 hg/s
20 t/s 200,000 hg/s
25 t/s 250,000 hg/s
50 t/s 500,000 hg/s
75 t/s 750,000 hg/s
100 t/s 1,000,000 hg/s
250 t/s 2,500,000 hg/s
500 t/s 5,000,000 hg/s
750 t/s 7,500,000 hg/s
1,000 t/s 10,000,000 hg/s
100,000 t/s 1,000,000,000 hg/s
1,000,000,000 t/s 10,000,000,000,000 hg/s
1,000,000,000,000 t/s 10,000,000,000,000,000 hg/s
#### Steps to Convert Hectogram/second to Ton (metric)/second (hg/s to t/s)
1. Example: Convert 828 hectogram/second to ton (metric)/second (828 hg/s to t/s).
2. 1 hectogram/second is equivalent to 0.0001 ton (metric)/second (1 hg/s is equivalent to 0.0001 t/s).
3. 828 hectogram/second (hg/s) is equivalent to 828 times 0.0001 ton (metric)/second (t/s).
4. Retrieved 828 hectogram/second is equivalent to 0.0828 ton (metric)/second (828 hg/s is equivalent to 0.0828 t/s).
▸▸ | 2,140 | 6,094 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2021-21 | latest | en | 0.809893 |
http://toughstem.com/problem-answer-solution/449/mass-placed-end-spring-unknown-spring-constant-spring-stretched-beyond-equilibrium-released | 1,521,872,883,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257649931.17/warc/CC-MAIN-20180324054204-20180324074204-00666.warc.gz | 292,197,493 | 18,845 | ToughSTEM
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A mass of 175g is placed on the end of a spring of unknown spring constant. The spring is stretched beyond equilibrium and released. The time to complete one oscillation (1 period) is .725s. What is the spring constant?
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T^2 =4*pi^2 *m/k
K =4*pi^2 *m/T^2 =13.13 N/m =13 N/m (ans)
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Almost done! | 281 | 867 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2018-13 | latest | en | 0.887519 |
https://socratic.org/questions/what-is-x-if-4x-9-x-30 | 1,575,663,804,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540490972.13/warc/CC-MAIN-20191206200121-20191206224121-00161.warc.gz | 565,269,858 | 6,310 | # What is x if -4x+9/x=-30?
Jul 31, 2016
$\frac{15 \pm 3 \sqrt{29}}{4}$
#### Explanation:
Multiply both sides of the equation by x -->
-4x^2 + 9 = - 30x
y = - 4x^2 + 30x + 9 = 0
Solve this equation by the new quadratic formula in graphic form (Socratic Search).
$D = {b}^{2} = {b}^{2} - 4 a c = 900 + 144 = 1044 = 36 \left(29\right)$--> $d = \pm 6 \sqrt{29}$
There are 2 real roots:
$x = - \frac{b}{2 a} \pm \frac{d}{2 a} = - \frac{30}{-} 8 \pm \frac{6 \sqrt{29}}{8} = \frac{15 \pm 3 \sqrt{29}}{4}$
Jul 31, 2016
$x = 7.7889 \mathmr{and} x = - 0.2889$
#### Explanation:
The fact that $x$ is in the denominator already means that we assume it is not equal to 0.
Multiply all the terms by $x$ to get rid of the fraction.
$\textcolor{red}{x \times} - 4 x + \frac{\textcolor{red}{x \times} 9}{x} = \textcolor{red}{x \times} - 30$
$- 4 {x}^{2} + 9 = - 30 x \text{ re-arrange and make} = 0$
$0 = 4 {x}^{2} - 30 x - 9 \text{ does not factorise}$
Use the formula: $a = 4 , b = - 30 , c = - 9$
$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
x = ((-(-30)+-sqrt((-30)^2-4(4)(-9))))/(2(4)
x = (30+-sqrt(900+144))/(8))
$x = \frac{30 \pm \sqrt{1044}}{8}$
$x = 7.7889 \mathmr{and} x = - 0.2889$ | 533 | 1,198 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 16, "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} | 4.65625 | 5 | CC-MAIN-2019-51 | longest | en | 0.644248 |
https://www.physicsforums.com/threads/linear-algebra-proofs.294203/ | 1,511,372,575,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806615.74/warc/CC-MAIN-20171122160645-20171122180645-00357.warc.gz | 818,208,203 | 15,782 | # Linear Algebra Proofs.
1. Feb 21, 2009
### killpoppop
Hey people. I find myself getting through my course but currently with not as much understanding as I would like. We've got to some proofs and i either vaguely understand them or do not know how to prove them.
1. The problem statement, all variables and given/known data
The first would be to prove the Dimension theorem that.
dimU + dimV = dim(U + V) + dim( U intersection V )
The second is if U,V are finite dimensional vector spaces over a field, and T: V -> W is a bijection
Show dimU = dimW
The last is to do with the Rank-Nullity Theorem and a basic linear transformation T: Rn -> Rm
That if m < n prove that T is not injective
And if m = n prove that T is injective iff T is surjective.
Some general insight into these proofs would be very very helpful. All feedback would be very much appreciated.
Thanks.
2. Feb 22, 2009
### kurnimaha
Let {u1, ... , uk} be a basis for U and {v1, ... , vp} a basis for V. Now consider the intersection U$$\cap$$V. If U$$\cap$$V = {0}, what can you say about the linear depence of the set {u1, ... , uk, v1, ... , vp}? How about the situation U$$\cap$$V $$\neq$$ {0}?
If T: V -> W is a bijective map, then what can you say about its range and kernel? How is this related to the first problem?
Again, consider range and kernel.
3. Feb 22, 2009
### HallsofIvy
Staff Emeritus
I would do this: subtract dim(U intersect V) from both sides to get the equivalent
dim(U+ V)= dimU + dimV- dim( U intersection V )
Choose a basis for U intersect V, then extend it to a basis of U. Extend that same basis for U intersect V to a basis for V. Show that the union of those two bases is a basis for U+ V.
Choose a basis for U, {u1, u2, ..., un} and show that {Tu1, Tu2, ... Tun} are independent. Thus, $dimU\le dimW$. Choose a basis for W, {w1, w2, ..., wm} and show that {T-1w1, T-1w2, ..., T-1wn} are independent.
Again choose a basis for Rn, {v1, v2, ..., vn} and look at {Tv1, Tv2, ..., Tvn}.
Last edited: Feb 22, 2009
4. Feb 22, 2009
### killpoppop
Thanks for the feedback it all helps. But still doesn't clear a lot up for me.
For this questions i've proved U+V is a subspace. What your saying is the next step but it doesn't make much sense to me. Can you explain further?
5. Feb 22, 2009
### killpoppop
Also can anyone suggest a decent book or website where i can read up on these? | 706 | 2,398 | {"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} | 3.96875 | 4 | CC-MAIN-2017-47 | longest | en | 0.922261 |
https://f123movies.live/4-9-to-centimeters/ | 1,675,340,808,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500017.27/warc/CC-MAIN-20230202101933-20230202131933-00165.warc.gz | 262,238,321 | 24,421 | #### 4 9 To Centimeters
4 9 To Centimeters. Web 4.9 inches to centimeters. 9 inches equals to 22.86 centimeters. Web usage of fractions is recommended when more precision is needed. A 4 feet and 9 inches height equals 145 centimeters or 1.45 meters. Web how to convert 4 feet and 9 inches in centimeters. Note that to enter a mixed number like 1 1/2, you show leave a space. 4 foot 9 and a half equals 146.05 cm.
And feet to centimeters conversion you need to multiply. The distance d in centimeters (cm) is equal to the distance d in inches (″) times. Note that to enter a mixed number like 1 1/2, you show leave a space. 5 1 / 5 meters = 520 centimeters: How many centimeters in 4 feet and 9 inches? Multiply the value in inches by the conversion factor '2.54'. 9 inches equals to 22.86 centimeters.
## Multiply the value in centimeters by the conversion factor '0.39370078740204'.
Web 4 feet 6 inches. A 4 feet and 9 inches height equals 145 centimeters or 1.45 meters. How many centimeters in 4 feet and 9 inches? To convert 4.9 cm to in use direct conversion formula below. Web what is 4'9 in centimeters or meters. So, 4.9 centimeters = 4.9 × 0.39370078740204 =. — 4.9 inch = 12.446 cm; Web how to convert 4 feet + 9 inches to cm or m?
### Web How To Convert 4 Feet And 9 Inches In Centimeters.
1, 4, 0.5, 1.9, 1/2, 3 1/2, etc. How to convert 4.9 cm to inch? Web how to convert from 4 1/9 inches to cm? Note that to enter a mixed number like 1 1/2, you show leave a space. So, 4.9 centimeters = 4.9 × 0.39370078740204 =. If we want to calculate how many centimeters are 4.9 feet we have to multiply 4.9 by 762 and divide the.
## 9.4 Inches Equal 23.876 Centimeters (9.4In = 23.876Cm).
First, we need to convert the fraction part 1/9 inches to decimal. How many centimeters in 4.9 inches? Web 4ft9in equals 144.78 cm. Web to convert 4.9 feet into centimeters we have to multiply 4.9 by the conversion factor in order to get the length amount from feet to centimeters. Web one inch equals 2.54 centimeters, in order to convert 4 x 9 inches to cm we have to multiply each amount of inches by 2.54 to obtain the length and width in centimeters. Multiply the value in centimeters by the conversion factor '0.39370078740204'.
## Conclusion of 4 9 To Centimeters.
Web to convert 4/9 inches from a fraction to a decimal, we simply divide the numerator 4 by the denominator 9 to get a decimal inch of 0.444444 inches. Web to convert 4.9 feet into centimeters we have to multiply 4.9 by the conversion factor in order to get the length amount from feet to centimeters. Multiply the value in feet by 30.48 to get the result of the conversion of feet in cm: 4 ft 9 and a quarter equals.. 4 ft 9 and a quarter equals. Multiply the value in centimeters by the conversion factor '0.39370078740204'.
Source | 836 | 2,807 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2023-06 | latest | en | 0.78683 |
https://www.thecrazyprogrammer.com/2012/07/c-program-to-calculate-area-of-circlea.html | 1,721,749,199,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518058.23/warc/CC-MAIN-20240723133408-20240723163408-00393.warc.gz | 882,211,643 | 27,894 | # C++ program to calculate area of a circle,a rectangle or a triangle depending upon user's choice
#include<iostream.h>
#include<conio.h>
#include<math.h>
void main()
{
clrscr(); //to clear the screen
float a,b,c,s,r,area;
int ch;
cin>>ch;
switch(ch)
{
case 1:
{
cin>>r;
area=3.14*r*r;
break;
}
case 2:
{
cin>>a>>b;
area=a*b;
break;
}
case 3:
{
cout<<“nEnter three sides of the triangle:”;
cin>>a>>b>>c;
s=(a+b+c)/2;
area=sqrt(s*(s-a)*(s-b)*(s-c));
break;
}
default: cout<<“nWrong choice…!!!”;
break;
}
cout<<“Area=”<<area;
getch(); //to stop the screen
}
### 11 thoughts on “C++ program to calculate area of a circle,a rectangle or a triangle depending upon user's choice”
1. That was great.
Can you please tell me what I did wrong here??
//Give the Area of a Square or a Triangle.
#include // tells the compiler to include this file
#include // tells the compiler to include this file
#include// tells the compiler to include this file
#include// tells the compiler to include this file
using namespace std; //use the standart namespace
float findSArea(float s); // function that calculates Square area
float findTArea(float b, float h); // function that calculates Triangle area
int main() // start of main function
{
float side, base, height; //input: data for calc
float square, triangle;
char letter;
char S,T;
float num;
cout<<"Please enter S for Square or T for Triangle (in capital letters): "<>letter;
// decide what to do
if ((letter==S) || (letter==T)) {
if (letter==S) {
cout<>side;
num=findSArea(s); // function call to calculate the Square area
cout<>num;
}
if (letter==T) {
cout<>base>>height;
num=findTArea(b,h); // function call to calculate the Triangle area
cout<>num;
}
}
else
{
cout<< "invalid character! ";
}
system ("pause");
return 0; }
// Compute the area of a Square with side s
//Pre: s is defined
//Post: returns area
float findSArea(float s)
{return (pow (s,2)); }
// Compute the area of a triangle with base b & height h
//Pre: b & H are defined
//Post: returns area
float findTArea(float b,h)
{return (0.5 * b * h); }
2. Write a c++ program that will:
• Provide 2 options to the user as below:
 Enter 1 to calculate the area of Rectangle
 Enter 2 to calculate the area of a Trapezoid
• If the user enters 1, then the program should ask the user to enter the width and length of the rectangle and then show the result after calculation. Formula for calculating Area of rectangle is
Area = width x Length.
• If the user enters 2, then the program should ask the user to enter the base1, base2, and height of the trapezoid and then show the result after calculation. Formula for calculating Area of trapezoid is
Area =
• You must implement two different functions for calculating areas of rectangle and trapezoid.
• After taking the choice from the user in form of 1 or 2, the relevant function should be called.
• After showing the output to the user, you need to ask the user if he/she wants to do another calculation. If the user presses y or Y, then the program should ask the user again to enter the choice of shape otherwise the program should be terminated.
Screenshot of the program execution is given below for both shapes:
Sample output:
3. please bro tell me the solution of this problem
i dont wanna code but instructions please
4. a c++ program to calculate area of the circle , square and rectangle using runtime polymorphism anyone plz reply for it thanks in advance
5. #include
using namespace std;
void main()
{int a,b,c,d,e;
for(a=1;a<=4;a++)
{switch(a)
{case 1:
cout<<a<<endl;
case 2:
for(b=2;b<=3;b++)
{cout<<b<<"\t";}
for(e=1;e<=4;e++)
{cout<<"*";}
cout<<endl;break;
case 3:
for(c=4;c<=6;c++)
{cout<<c<<"\t";}
cout<<endl;break;
case 4:
for(d=7;d<=10;d++)
{cout<<d<<"\t";}
cout<<endl;break;
default:
cout<<"lol"<<endl;
}
}} (why case 2 is printing 2 times)
6. Create a program that allows the user to select a shape he/she wants to compute the area. The program should accept either a lowercase or uppercase.
Note: Area should be presented in two decimal places.
Sample Output:
Area of Shapes
[S] – Square
[C] – Circle
[T] – Triangle
[R] – Rectangle | 1,123 | 4,152 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-30 | latest | en | 0.549354 |
https://sigmatricks.com/numbers/ | 1,718,998,120,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862157.88/warc/CC-MAIN-20240621191840-20240621221840-00508.warc.gz | 443,158,826 | 9,192 | ## Rational Numbers: Definition & Examples
What are rational numbers Numbers are core of mathematics. Almost all of mathematics parts consist of numbers. There are some types of numbers. One of them is rational numbers. Rational numbers were invented by Greek mathematician, Phytagoras. Rational numbers consist of natural numbers and integers. Rational numbers symbolized as boldface Q (Q). It is type
## Fibonacci Sequence
About Fibonacci (The Man) His real name was Leonardo Pisano Bogollo. He lived in Italy between 1170 and 1250. “Fibonacci” was his nickname, which roughly means “Son of Bonacci”. Fibonacci was not the first to know about the sequence, it was known in India hundreds of years before! As well as being famous for the Fibonacci
## Even and Odd Numbers
Even Numbers Any integer that can be divided exactly by 2 is an even number. The last digit is 0, 2, 4, 6 or 8 Example: −10, -4, 0, 8, and 32 are all even numbers Odd Numbers Any integer that cannot be divided exactly by 2 is an odd number. The last digit is 1, 3,
## Cardinal and Ordinal Numbers Chart
Definitions A Cardinal Number is a number that says how many of something there are, such as 1, 2, 3, 4. An Ordinal Number is a number that tells the position of something in a list, such as 1st, 2nd, 3rd, 4th. Most ordinal numbers end in “th” except for: one ⇒ first (1st) two
## Absolute Value
Definition of Absolute Value Absolute value or modulus |x| of a real number x is the non-negative value of x without regard to its sign. |x| = x for a positive x, |x| = −x for a negative x (in which case −x is positive) |0| = 0. For example, the absolute value of 5 is
## Multiplication Tables
15 Times Table x 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 2 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 3 3 6 9
## Numbers
A number is a mathematical object used to measure, count, & label. In common usage, number may refer to mathematical object, a word or phrase, or a symbol. The notion of number has been extended over the centuries to include 0, negative numbers, rational numbers such as 2/5 and −1/3, real numbers such as √2
## Reciprocal of a Fraction
Fractions A Fraction (such as 2/3) has two numbers: Numerator: the top number Denominator: the bottom number We call the top number the Numerator, it is the number of parts we have. We call the bottom number the Denominator, it is the number of parts the whole is divided into. Reciprocal of a Fraction To
## Least Common Multiple
Common Multiple A common Multiple number is a multiple of two or more numbers. Example Multiple of 3 and 4: Number Multiples 3 3, 6, 9, 12, 15, 18, 21, 24, … 4 4, 8, 12, 16, 20, 24, … So, the common multiples of 3 and 4 are: 12, 24, … Least Common Multiple
## Greatest Common Factor
Greatest Common Factor (GCF) or Highest Common Factor (HCF) is the largest of the common factors. Greatest Common Factor is useful to simplify a fraction, because it is the largest number we can divide both numerator and denominator of a fraction. How to Find the Greatest Common Factor Method 1 find all factors of both | 873 | 3,109 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2024-26 | latest | en | 0.929707 |
https://www.jiskha.com/display.cgi?id=1299824077 | 1,503,264,202,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886106990.33/warc/CC-MAIN-20170820204359-20170820224359-00350.warc.gz | 924,419,685 | 4,277 | Calculus
posted by .
derivative of ((Sqrt(x+3y)+sqrt(3xy))=(sqrt(21)+sqrt(90)) at point (6,5)google
• Calculus-implicit diff. -
Check that (6,5) lies on the curve by substituting x=6,y=5 and see if the equality holds. (yes, it holds).
Differentiate:
((Sqrt(x+3y)+sqrt(3xy))=(sqrt(21)+sqrt(90))
(1/2)/sqrt(x+3y)*(1+3y')+(1/2)/sqrt(3xy)*(3y+3xy')=0
transpose and cross multiply:
-sqrt(x+3y)*(1+3y')=sqrt(3xy)*(3y+3xy')
Solve for y'
y'(x,y)=-(sqrt(3*y+x)+3^(3/2)*y*sqrt(x*y))/(3*sqrt(3*y+x)+3^(3/2)*x*sqrt(x*y))
so
y'(6,5)=(-5*3^(3/2)*sqrt(30)-sqrt(21))/(2*3^(5/2)*sqrt(30)+3*sqrt(21))
=-0.796....
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When I solve the inquality 2x^2 - 6 < 0, I get x < + or - sqrt(3) So how do I write the solution?
4. math,algebra,help
Directions are simplify by combining like terms. x radiacal 18 -3 radical 8x^2 can someone show me how to do these types of problems. thanks I cant determine the second term. For the first, I think you meant x sqrt(18) which reduces …
5. Math
How do you find a square root of a number that's not a perfect square?
6. Mathematics
sqrt 6 * sqrt 8 also sqrt 7 * sqrt 5 6.92820323 and 5.916079783 So you can see the steps — sqrt 6 * sqrt 8 = sqrt 48 sqrt 7 * sqrt 5 = sqrt 35 I hope this helps a little more. Thanks for asking.
Could someone show me how to solve these problems step by step.... I am confused on how to fully break this down to simpliest terms sqrt 3 * sqrt 15= sqrt 6 * sqrt 8 = sqrt 20 * sqrt 5 = since both terms are sqrt , you can combine …
8. Calculus
Please look at my work below: Solve the initial-value problem. y'' + 4y' + 6y = 0 , y(0) = 2 , y'(0) = 4 r^2+4r+6=0, r=(16 +/- Sqrt(4^2-4(1)(6)))/2(1) r=(16 +/- Sqrt(-8)) r=8 +/- Sqrt(2)*i, alpha=8, Beta=Sqrt(2) y(0)=2, e^(8*0)*(c1*cos(0)+c2*sin(0))=c2=2 …
9. Math/Calculus
Solve the initial-value problem. Am I using the wrong value for beta here, 2sqrt(2) or am I making a mistake somewhere else?
10. Algebra
Evaluate sqrt7x (sqrt x-7 sqrt7) Show your work. sqrt(7)*sqrt(x)-sqrt(7)*7*sqrt(7) sqrt(7*x)-7*sqrt(7*7) sqrt(7x)-7*sqrt(7^2) x*sqrt 7x-49*x ^^^ would this be my final answer?
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# Standard normal distribution table
Do not type or enter data in this area ! ! upper tail lower tail 2-tail central probability probability probability probability 0.05% 0.05% 0.1% 99.9% 0.5% 0.5% 1.0% 99.0% 1.0% 1.0% 2.0% 98.0% 2.5% 2.5% 5.0% 95.0% 5.0% 5.0% 10.0% 90.0% 7.5% 7.5% 15.0% 85.0% 10.0% 10.0% 20.0% 80.0% 15.0% 15.0% 30.0% 70.0% 20.0% 20.0% 40.0% 60.0% 25.0% 25.0% 50.0% 50.0%
value of z 3.29 2.58 2.33 1.96 1.64 1.44 1.28 1.04 0.84 0.67
IF YOU KNOW THE PROBABILITY, ENTER IT HERE Enter probabilities here: upper tail probability lower tail probability 2-tail probability central probability 1.0% 1.0% 5.0% 95.0% +
## Prepared by William Lepowsky for the Laney College Math Department
INSTRUCTIONS: Enter values in the red outlined boxes that look like this: Do NOT enter values in the purple boxes that look like this:
IF YOU KNOW THE VALUE OF Z, ENTER IT HERE Enter the value of z here: 3 upper tail probability (to the right of z) 0.1% 2-tail probability 0.3% central probability (from z to z) 99.7%
Here are more detailed values of the probabilities: 0.134989803% 0.269979606% 99.73002039%
## Student's t-distribution table
Enter the number of degrees of freedom here: Do not type or enter data in this area ! ! upper tail lower tail 2-tail central probability probability probability probability 0.05% 0.05% 0.1% 99.9% 0.5% 0.5% 1.0% 99.0% 1.0% 1.0% 2.0% 98.0% 2.5% 2.5% 5.0% 95.0% 5.0% 5.0% 10.0% 90.0% 7.5% 7.5% 15.0% 85.0% 10.0% 10.0% 20.0% 80.0% 15.0% 15.0% 30.0% 70.0% 20.0% 20.0% 40.0% 60.0% 25.0% 25.0% 50.0% 50.0% 15 = d.f.
value of the t-statistic 4.07 2.95 2.60 2.13 1.75 1.52 1.34 1.07 0.87 0.69
IF YOU KNOW THE PROBABILITY, ENTER IT HERE Enter probabilities here: upper tail probability lower tail probability 2-tail probability central probability 6.0% 2.0% 1.3% 99.0% + value of the t-statistic 1.65 2.25 2.82 2.95
## Prepared by William Lepowsky for the Laney College Math Department
Thank you for entering the number of degrees of freedom INSTRUCTIONS: Enter values in the red outlined boxes that look like this: Do NOT enter values in the purple boxes that look like this:
IF YOU KNOW THE VALUE OF THE T-STATISTIC, ENTER IT HERE Enter the value of the t-statistic here: -3.1 lower tail probability (to the left of t) 0.4% central probability (from t to t) 99.3%
## 2-tail probability 0.7%
Here are more detailed values of the probabilities: 0.365861529% 0.731723057% 99.26827694%
## Chi-square distribution table
Enter the number of degrees of freedom here: 6 = d.f.
Do not type or enter data in this area ! ! value of the upper tail chi-square probability statistic 22.46 0.1% 18.55 0.5% 16.81 1.0% 14.45 2.5% 12.59 5.0% 10.64 10.0% 9.45 15.0% 8.56 20.0% 7.84 25.0%
INSTRUCTIONS:
## Do NOT enter values in the purp
IF YOU KNOW THE PROBABILITY, ENTER IT HERE Enter the upper tail probability here: 10.0% value of the chi-square statistic 10.64
IF YOU KNOW THE VALUE OF T Enter the value of the chisquare statistic here: 16.75
## Thank you for entering the number of degrees of freedom
alues in the red outlined boxes that look like this: enter values in the purple boxes that look like this:
## KNOW THE VALUE OF THE CHI-SQUARE STATISTIC, ENTER IT HERE
upper tail probability 1.0% Here is a more detailed value of the probability: 1.024730537% | 1,251 | 3,329 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-22 | latest | en | 0.732444 |
https://medtox.org/label-the-axes-phases-phase-changes-and-important-points-on-the-phase-diagram-below/ | 1,643,139,304,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304872.21/warc/CC-MAIN-20220125190255-20220125220255-00161.warc.gz | 449,563,217 | 4,979 | Phase diagram is a graphical depiction of the physical says of a substance under different conditions of temperature and also pressure. A usual phase diagram has actually pressure on the y-axis and temperature ~ above the x-axis. As we cross the present or curves on the phase diagram, a phase readjust occurs. In addition, two claims of the problem coexist in equilibrium ~ above the currently or curves.
You are watching: Label the axes, phases, phase changes and important points on the phase diagram below.
## Introduction
A phase shift is the change from one state of matter to another. There space three says of matter: liquid, solid, and also gas.
Liquid: A state of matter that is composed of loose, cost-free moving particles which type the shape collection by the limits of the container in i m sorry the fluid is in. This happens because the activity of the individual particles in ~ a fluid is much less limited than in a solid. One may notice that some liquids circulation readily whereas part liquids circulation slowly. A liquid"s family member resistance to flow is viscosity. Solid: A state of matter with tightly pack particles which execute not readjust the shape or volume that the container the it is in. However, this go not median that the volume the a solid is a constant. Solids deserve to expand and also contract when temperatures change. This is why as soon as you look up the thickness of a solid, the will suggest the temperature at which the worth for thickness is listed. Solids have strong intermolecular pressures that keep particles in nearby proximity come one another. An additional interesting thing to think about is the all true solids have crystalline structures. This means that your particles room arranged in a three-dimensional, orderly pattern. Solids will certainly undergo phase alters when lock come throughout energy changes. Gas: A state of matter where particles room spread out v no definite form or volume. The particles of a gas will certainly take the shape and fill the volume of the container that it is put in. In a gas, there are no intermolecular pressures holding the particles of a gas together due to the fact that each bit travels at its own speed in its very own direction.
See more: The Probability Assigned To Each Experimental Outcome Must Be
The corpuscle of a gas are frequently separated by good distances.
Phase diagrams highlight the variations between the states of issue of facets or compounds together they relate to pressure and also temperatures. The adhering to is an instance of a step diagram because that a share single-component system:
api/deki/files/8721/Phase_Diagram_CO2.jpg?revision=1" />
Figure (PageIndex2a): phase diagrams for CO2
Figure (PageIndex2b):Phase diagrams for H2O | 553 | 2,776 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2022-05 | latest | en | 0.941638 |
https://math.stackexchange.com/questions/2996613/tiling-circles-intersection-with-squares | 1,568,884,200,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573465.18/warc/CC-MAIN-20190919081032-20190919103032-00027.warc.gz | 605,032,251 | 29,790 | # Tiling circles intersection with squares
Is there an algorithm to calculate the amount of $$1\times 1$$ tiles that the intersection of $$2$$ circles lies into, given 2 circles in a $$30\times 30$$ grid with their radiuses and centers?
What I tried was approximating by using the line between the intersection points but this doesn't work when the intersection is wider than $$1$$ line of tiles.
Another version of this question is how can I check if a tile with given coordinates lies in the intersection of $$2$$ circles.
• Do the 1x1 tiles have to be contained completely in the intersection or do they just have to touch the intersection? – Neo Nov 13 '18 at 11:21
• @Neo a part of the tile has to lie in the intersection. So not completely. – Kristijan Nov 13 '18 at 11:24 | 190 | 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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 5, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2019-39 | latest | en | 0.948029 |
http://www.chegg.com/homework-help/questions-and-answers/4-your-market-research-group-estimated-the-following-demand-curve-for-gadgets-the-product--q3527337 | 1,368,951,767,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368696400149/warc/CC-MAIN-20130516092640-00080-ip-10-60-113-184.ec2.internal.warc.gz | 385,305,410 | 8,326 | 4. Your market research group estimated the following demand curve for gadgets, the product your company produces and sells. Qd = 4,000 – 40P If this relationship between quantity demanded and prices continues to hold true in the future, a. How many gadgets will be demanded at $10,$20, and $30? b. What is the arc price elasticity between$10 and $20; between$20 and $30? c. What is the point price elasticity at each of the three prices? d. If your company sold 3,000 gadgets last year, what is the price it charged? Answers (3) • http://dc361.4shared.com/doc/Oevg3E1F/preview.html • http://www.chegg.com/homework-help/questions-and-answers/your-market-research-group-estimated-the-following-demand-curve-for-gadgets-the-product-yo-q2408103 • a) The quantity demanded when: i) p = 10, Qd = 4000 - 40*10 = 3600 ii) p = 20, Qd = 4000 - 40*20 = 3200 iii) p = 30, Qd = 4000 - 40*30 = 2800 b) The formula for arc elasticity is: i) Ed = {(10 + 20)/(3600 + 3200)}*-400/10 = -0.1765 ii) E= {(20 + 30)/(3200 + 2800)}*-400/10 = -0.3333 c) The formula for point elasticity is Now δQd/δP = -40 i) when P = 10, E= -40*10/3600 = -0.1111 ii) when P = 20, E= -40*20/3200 = -0.25 iii) when P = 10, E= -40*30/2800 = -0.4286 d) We know that Qd = 3000, thus the price is : 3000 = 4000 - 40*P P=$25
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More than 200 experts are waiting to help you now... | 463 | 1,349 | {"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} | 3.890625 | 4 | CC-MAIN-2013-20 | latest | en | 0.819619 |
https://www.physicsforums.com/threads/help-with-the-3d-gaussian-function.541664/ | 1,511,302,491,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806426.72/warc/CC-MAIN-20171121204652-20171121224652-00244.warc.gz | 832,368,466 | 14,395 | # Help with the 3D gaussian function
1. Oct 18, 2011
### PythagoreLove
Hi,
I need help with the gaussian function in 3D. I'm using the form:
z=A*exp( (x-xo)^2/(2σx^2)+(y-yo)^2/(2σy^2))
I know that A is the amplitude and xo,yo are the center coordinate.
I found that formula on http://en.wikipedia.org/wiki/Gaussian_function and they say that σx and σy are the spread of the blob. But if I put σx=σy=1, the function is from -3 to 3... Why is that ?
2. Oct 18, 2011
### phyzguy
They only plotted it from -3 to 3. It extends to +/-infinity in both x and y, but once |x-x0|/sigmax is much larger than 3.0, the value of the function is very close to zero.
3. Oct 18, 2011
### PythagoreLove
But is there a pretty good way to find σx and σy, graphically speaking. I have a gaussian curve and want to find what is the function.
4. Oct 18, 2011
### phyzguy
A good way is to use FWHM(Full Width at Half Maximum). In other words, you measure the width of the curve at half of the peak value. This is then easily converted into sigma using the formula on this page:
http://en.wikipedia.org/wiki/Fwhm
Since you have a 2D Gaussian, you need to measure the FWHM in the x-direction with y=y0 to find sigmax, then do the same in the y-direction with x=x0 to find sigmay.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook | 405 | 1,369 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2017-47 | longest | en | 0.905173 |
https://plainmath.net/38439/write-the-solution-set-of-the-given-homogeneous-system-in | 1,638,314,792,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964359082.76/warc/CC-MAIN-20211130232232-20211201022232-00160.warc.gz | 517,717,792 | 8,508 | # Write the solution set of the given homogeneous system in
Write the solution set of the given homogeneous system in parametric vector form.
$$\displaystyle{x}_{{1}}+{3}{x}_{{2}}-{5}{x}_{{3}}={0}$$
$$\displaystyle{x}_{{1}}+{4}{x}_{{2}}-{8}{x}_{{3}}={0}$$
$$\displaystyle-{3}{x}_{{1}}-{7}{x}_{{2}}+{9}{x}_{{3}}={0}$$
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The given system is
$$\displaystyle{x}_{{1}}+{3}{x}_{{2}}-{5}{x}_{{3}}={0}$$
$$\displaystyle{x}_{{1}}+{4}{x}_{{2}}-{8}{x}_{{3}}={0}$$
$$\displaystyle-{3}{x}_{{1}}-{7}{x}_{{2}}+{9}{x}_{{3}}={0}$$
The augmented matrix is
$\begin{bmatrix}1&3&-5&0\\1&4&-8&0\\-3&-7&9&0\end{bmatrix}$
Reduce matrix to row echelon form
$\begin{bmatrix}-3&-7&9&0\\0&\frac{5}{3}&-5&0\\0&\frac{2}{3}&-2&0\end{bmatrix}\begin{matrix}R_3-2/5R_2\to R_3\\3/5R_2\to R_2 \end{matrix}\sim\begin{bmatrix}-3&-7&9&0\\0&1&-3&0\\0&0&0&0 \end{bmatrix}$
$\begin{bmatrix}-3&-7&9&0\\0&1&-3&0\\0&0&0&0 \end{bmatrix}\begin{matrix}R_1+7R_2\to R_1\\-1/3R_1\to R_1\end{matrix}\begin{bmatrix}1&0&4&0\\0&1&-3&0\\0&0&0&0\end{bmatrix}$
Thus we get
$$\displaystyle{x}_{{1}}+{4}{x}_{{3}}={0}\Rightarrow{x}_{{1}}=-{4}{x}_{{3}}$$
$$\displaystyle{x}_{{2}}-{3}{x}_{{3}}={0}\Rightarrow{x}_{{2}}={3}{x}_{{3}}$$
Hence, the given homogeneous system in parametric vector form is
$x=\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}=x_3\begin{bmatrix}-4\\3\\1\end{bmatrix}$ | 691 | 1,532 | {"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} | 4.25 | 4 | CC-MAIN-2021-49 | latest | en | 0.390469 |
http://www.mathcs.emory.edu/~cheung/Courses/255/Syllabus/5-repr/values.html | 1,513,422,812,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948587577.92/warc/CC-MAIN-20171216104016-20171216130016-00737.warc.gz | 409,872,645 | 3,702 | CS255 Sylabus
Numeric values
• Numeric value are something intrinsic
One would ideally represent "numeric values" in a universal manner, such as:
• This is obviously very clumsy (try larger values :-)
• So humanoids have invented many different representations for numerical values
(This practice is obviously very important for their survival...)
• The most popular representation NOW (that was not always the case !) for numerical value is the decimal number system
This system is based on the following ten familiar looking symbols:
``` 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 ```
I am sure you are thoroughly familiar with this decimal number system, in fact, so familiar that you do not even think about what decimal numbers actually mean...
• There are other representations for numerical values invented by humanoids.
A famous example is the number system invented by a class of humanoids that we call Romans
Their number system goes like:
``` I, II, III, IV, V, VI, VII, VIII, IX, X, XI, XII, XIII, XIV, XV, .... ```
Chinese numbers:
BTW, notice there is no symbol for ZERO. Chinese character for ZERO is:
A not-so-famous system is the Egyptian number system:
= 1 = 10 = 100 = 1000 = 10000 = 100000 = 1000000
Note:
Repesentations for numbers is an example of a code !!!
There are many other humanoids who have invented their own representation systems for numerical values, among others: Greeks (they use the Greek alphabet), Chinese (I'll show you in class...), etc.
Here is a copy of a page from a book of my 6 yr old first grader (in 2003) that show a number of number systems used in the other cultures: click here
• It is important to know that:
• A value does not depends on the representation system used:
• If you see that there are 4 students in the classroom, no matter how you represent this number, there will be 4 students, no more and no less
• As I mentioned above: a value is an intrinsic property....
• Roman Arithmetic....
• How can I use Roman numerals to do arithmetic problems?
Let's start with an addition problem: 23 + 58. In Roman numerals, that's XXIII + LVIII. We'll begin by writing the two numbers next to each other: XXIII LVIII. Next, we rearrange the letters so that the numerals are in descending order: LXXVIIIIII. Now we have six I's, so we'll rewrite them as VI: LXXVVI. The two Vs are the same as an X, so we simplify again and get LXXXI, or 81, as our final answer. (We can check this answer using Arabic numerals.)
• Complex arithmetic in Roman era:
• When Romans wanted to do complicated arithmetic problems, they used a special counting board or an abacus:
An abacus represents values using a positional representation:
The right most column has weight = 1 The second right most column has weight = 10 And so on.
Notice that this is an encoding method !!!
It is an agreement on how to represent a value
• Positional representation system
• Positional (value) representation:
• A position representation system uses the same symbol to represent different values
• The value that is represented by a certain symbol depends on:
The symbol itself, and The position in which that symbox is found !!!
• Example:
The symbol 1 in the number 111 represents the value * (= 1 dot). The symbol 1 in the number 111 represents the value ********** (= 10 dot).
In contrast:
The symbol V will represent the value ***** (5) not matter where you find it in a Roman number !!!
• When humans started to use positional system (based on 10), we can teach children to add any two numbers by:
• The base 10 addition table:
``` | 1 2 3 4 5 6 7 8 9 ----+------------------------------------- 1 | 2 3 4 5 6 7 8 9 10 2 | 3 4 5 6 7 8 9 10 11 3 | 4 5 6 7 8 9 10 11 12 4 | 5 6 7 8 9 10 11 12 13 5 | 6 7 8 9 10 11 12 13 14 6 | 7 8 9 10 11 12 13 14 15 7 | 8 9 10 11 12 13 14 15 16 8 | 9 10 11 12 13 14 15 16 17 9 | 10 11 12 13 14 15 16 17 18 ``` | 1,060 | 3,904 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2017-51 | longest | en | 0.908297 |
https://www.expertsmind.com/library/probability-through-discrete-probability-distribution-519531.aspx | 1,708,553,104,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473558.16/warc/CC-MAIN-20240221202132-20240221232132-00177.warc.gz | 789,962,551 | 11,494 | Probability through discrete probability distribution
Assignment Help Basic Statistics
Reference no: EM1319531
Consider the probability experiment of flipping a fair coin 6 times. Let X be the number of heads. The probability distribution is given in the following table:
X P(X) Frequency (f) 0 0.016 1 1 0.094 6 2 0.234 15 3 0.313 20 4 0.234 15 5 0.094 6 6 0.016 1
a. Draw the histogram.
b. Calculate the population mean and population standard deviation.
c. What is the probability of flipping a coin 6 times and getting exactly 3 heads?
d. What is the probability of flipping a coin 6 times and getting 4 or more heads?
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The following data from a sample of 100 families show the record of college attendance by fathers and their oldest sons: in 22 families, both father and son attended college; in 31 families, neither father nor son attended college; in 12 families,..
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Findout the value of X using normal distribution which gives standard deviation also mean. A math teacher gives two different tests to measure students' aptitude for math.
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The booster club has gotten several businesses and car dealers to donate gasoline and cars for the college students to use to make a maximum of 575 personal contacts daily during the fund-raising drive. The college student will donate a total of 2..
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Let p represent the proportion of registered voters in the state that would vote for the Republican candidate. The standard error for the proportion of those who phoned in who answered "yes" is
Point estimate for the population mean
What is a point estimate for the population mean µ? | 619 | 2,933 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2024-10 | latest | en | 0.885563 |
https://www.convert-measurement-units.com/convert+kcal+g+to+kJ+Kg.php | 1,723,226,305,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640768597.52/warc/CC-MAIN-20240809173246-20240809203246-00875.warc.gz | 555,553,932 | 13,520 | Convert kcal/g to kJ/Kg (Specific energy)
## kcal/g into kJ/Kg
numbers in scientific notation
https://www.convert-measurement-units.com/convert+kcal+g+to+kJ+Kg.php
# Convert kcal/g to kJ/Kg (Specific energy):
1. Choose the right category from the selection list, in this case 'Specific energy'.
2. Next enter the value you want to convert. The basic operations of arithmetic: addition (+), subtraction (-), multiplication (*, x), division (/, :, ÷), exponent (^), square root (√), brackets and π (pi) are all permitted at this point.
3. From the selection list, choose the unit that corresponds to the value you want to convert, in this case 'kcal/g'.
4. Finally choose the unit you want the value to be converted to, in this case 'kJ/Kg'.
5. Then, when the result appears, there is still the possibility of rounding it to a specific number of decimal places, whenever it makes sense to do so.
With this calculator, it is possible to enter the value to be converted together with the original measurement unit; for example, '397 kcal/g'. In so doing, either the full name of the unit or its abbreviation can be used. Then, the calculator determines the category of the measurement unit of measure that is to be converted, in this case 'Specific energy'. After that, it converts the entered value into all of the appropriate units known to it. In the resulting list, you will be sure also to find the conversion you originally sought. Alternatively, the value to be converted can be entered as follows: '64 kcal/g to kJ/Kg' or '73 kcal/g into kJ/Kg' or '46 kcal/g -> kJ/Kg' or '28 kcal/g = kJ/Kg'. For this alternative, the calculator also figures out immediately into which unit the original value is specifically to be converted. Regardless which of these possibilities one uses, it saves one the cumbersome search for the appropriate listing in long selection lists with myriad categories and countless supported units. All of that is taken over for us by the calculator and it gets the job done in a fraction of a second.
Furthermore, the calculator makes it possible to use mathematical expressions. As a result, not only can numbers be reckoned with one another, such as, for example, '(64 * 46) kcal/g'. But different units of measurement can also be coupled with one another directly in the conversion. That could, for example, look like this: '1 kcal/g + 82 kJ/Kg' or '28mm x 10cm x 91dm = ? cm^3'. The units of measure combined in this way naturally have to fit together and make sense in the combination in question.
The mathematical functions sin, cos, tan and sqrt can also be used. Example: sin(π/2), cos(pi/2), tan(90°), sin(90) or sqrt(4).
If a check mark has been placed next to 'Numbers in scientific notation', the answer will appear as an exponential. For example, 2.734 345 654 129 8×1020. For this form of presentation, the number will be segmented into an exponent, here 20, and the actual number, here 2.734 345 654 129 8. For devices on which the possibilities for displaying numbers are limited, such as for example, pocket calculators, one also finds the way of writing numbers as 2.734 345 654 129 8E+20. In particular, this makes very large and very small numbers easier to read. If a check mark has not been placed at this spot, then the result is given in the customary way of writing numbers. For the above example, it would then look like this: 273 434 565 412 980 000 000. Independent of the presentation of the results, the maximum precision of this calculator is 14 places. That should be precise enough for most applications. | 848 | 3,572 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2024-33 | latest | en | 0.867923 |
https://fred.stlouisfed.org/graph/?chart_type=scatter&s[1][id]=AWHI | 1,480,724,897,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698540798.71/warc/CC-MAIN-20161202170900-00497-ip-10-31-129-80.ec2.internal.warc.gz | 848,544,952 | 21,174 | # Index of Aggregate Weekly Hours: Production and Nonsupervisory Employees: Total Private Industries (AWHI) Excel (data) CSV (data) Image (graph) PowerPoint (graph) PDF (graph)
Observation:
Nov 2016: 113.2
Updated: 8:36 AM CST
Units:
Index 2002=100,
Frequency:
Monthly
1Y | 5Y | 10Y | Max
EDIT PLOT 1
(a) Index of Aggregate Weekly Hours: Production and Nonsupervisory Employees: Total Private Industries, Index 2002=100, Seasonally Adjusted (AWHI)
Indexes of aggregate weekly hours are calculated by dividing the current month's aggregate hours by the average of the 12 monthly figures, for the base year. For basic industries, the hours aggregates are the product of average weekly hours and employment of workers to which the hours apply (all employees or production and nonsupervisory employees). At all higher levels of industry aggregation, hours aggregates are the sum of the component aggregates.
The series comes from the 'Current Employment Statistics (Establishment Survey).'
The source code is: CES0500000034
Index of Aggregate Weekly Hours: Production and Nonsupervisory Employees: Total Private Industries
Select a date that will equal 100 for your custom index:
to
#### Customize data:
Write a custom formula to transform one or more series or combine two or more series.
You can begin by adding a series to combine with your existing series.
Now create a custom formula to combine or transform the series.
Need help? []
Finally, you can change the units of your new series.
Select a date that will equal 100 for your custom index:
#### Add data series to graph:
FORMAT GRAPH
Log scale:
NOTES
Release: Employment Situation
#### Notes:
Indexes of aggregate weekly hours are calculated by dividing the current month's aggregate hours by the average of the 12 monthly figures, for the base year. For basic industries, the hours aggregates are the product of average weekly hours and employment of workers to which the hours apply (all employees or production and nonsupervisory employees). At all higher levels of industry aggregation, hours aggregates are the sum of the component aggregates.
The series comes from the 'Current Employment Statistics (Establishment Survey).'
The source code is: CES0500000034
#### Suggested Citation:
US. Bureau of Labor Statistics, Index of Aggregate Weekly Hours: Production and Nonsupervisory Employees: Total Private Industries [AWHI], retrieved from FRED, Federal Reserve Bank of St. Louis; https://fred.stlouisfed.org/series/AWHI, December 2, 2016.
RELATED CONTENT
RELEASE TABLES
Retrieving data.
Updating graph. | 580 | 2,597 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-50 | latest | en | 0.874732 |
http://www.slideserve.com/galya/interactive-probability-instruction | 1,493,134,805,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917120461.7/warc/CC-MAIN-20170423031200-00259-ip-10-145-167-34.ec2.internal.warc.gz | 691,354,664 | 17,898 | # Interactive Probability Instruction - PowerPoint PPT Presentation
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Interactive Probability Instruction. Ivo Dinov, Michigan Dennis Pearl, OSU Kyle Seigrist , UAH www.distributome.org/meetings/JMM_2014 www.distributome.org/V3. Login. Username = TBD Password = TBD. A Paradigm. statistics. assumptions. approximations.
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Interactive Probability Instruction
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## Interactive Probability Instruction
Ivo Dinov, Michigan
Dennis Pearl, OSU
Kyle Seigrist, UAH
www.distributome.org/meetings/JMM_2014
www.distributome.org/V3
statistics
assumptions
approximations
### Questions to ask (to understand how well probability statements reflect reality)
• What assumptions are being made? Are they realistic?
• Can model deviations from reality be quantified?
• What approximations are being made?
• Are they accurate? Are they necessary?
• Do data agree with model predictions?
• Does the model need to be refined or simplified?
### Jigsaw Review
• Teams work together on solving a set of problems.
• New teams are formed taking one member from each original team.
• Members of the solving team must explain the solution to their problems to their new team.
### What’s the distribution? A
• X = how much ice will be used on the next full Southwest Airline flight from Denver, CO to Raleigh/Durham, NC.
• X = how long until the next 2001 Toyota Prius hybrid goes through a toll booth on the Golden Gate Bridge.
• X = how many randomly chosen people will it take to find 10 who are left-handed.
• X= how many randomly chosen people will it take find one who is left-handed.
### What’s the distribution? B
• X = the number of red-flowering plants in 100 crosses between pink flowering plants
• X = the position (degrees clockwise from you) of a fish swimming in a cylindrical tank.
• X = median position of the same fish wrt the first person visiting the aquarium each day this week.
• X = the average amount of time that a sample of 500 regular subscribers to the New York Times spend on the Sunday crossword puzzle.
### What’s the distribution? C
• X = how many 2001 Toyota Prius hybrids go across the Golden Gate bridge tomorrow.
• X = the average age of the quarters in the pockets of 100 men at a NY Yankees baseball game.
• Four participants in this workshop will be picked to form the first team for a “jigsaw review”. X = how many graduate students are on that team.
• X = In a sample of just one person – how many favor imposing a uniform sales tax on internet purchases.
### What’s the distribution? D
• X = how many hours until the next diet coke is purchased from a vending machine.
• X = how many hours until five separate people purchase a diet coke from a vending machine.
• X = how many of the next ten parties that a waiter serves will leave a tip of more than 20% of the bill.
• X = the birth month of the next person to board a plane from Honolulu to Los Angeles.
### Follow-ups
• A: Worldwide 11% of people are left-handed. Are the workshop participants unusual in this regard?
• B: An icthyologist suggests that fish will tend to shy away from the visitor. What median angles would give strong evidence for this theory?
• C: Is the number of grad students in your team unusual? What about the highest number in any team?
• D: Do you think it’s likely to have at least two months with no birthdays for the people in the room?
• ALL GROUPS: suggest your own follow-up that can be investigated by students | 922 | 4,020 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2017-17 | longest | en | 0.898627 |
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With surface temperatures estimated at minus 230 degrees Fahrenheit, J
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Joined: 09 Jan 2016
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With surface temperatures estimated at minus 230 degrees Fahrenheit, J [#permalink]
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05 Apr 2017, 08:19
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With surface temperatures estimated at minus 230 degrees Fahrenheit, Jupiter's moon Europa has long been considered far too cold to support life, and with 60 square miles of water thought to be frozen from top to bottom.
A. Europa has long been considered far too cold to support life, and with
B. Europa has long been considered far too cold to support life, its
C. Europa has long been considered as far too cold to support life and has
D. Europa, long considered as far too cold to support life, and its
E. Europa, long considered to be far too cold to support life, and to have
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Re: With surface temperatures estimated at minus 230 degrees Fahrenheit, J [#permalink]
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05 Apr 2017, 09:50
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This particular question has already been covered in detail here: with-surface-temperatures-estimated-at-minus-230-degrees-142362.html
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Re: With surface temperatures estimated at minus 230 degrees Fahrenheit, J &nbs [#permalink] 05 Apr 2017, 09:50
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https://post.lurk.org/@mathr/102761988945272640 | 1,582,093,960,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875144058.43/warc/CC-MAIN-20200219061325-20200219091325-00241.warc.gz | 503,434,592 | 9,023 | trying to do some maths in Maxima, it is telling me "if a < 0 then b else if a = 0 then b else if a > 0 then b". b is my original input that I'm trying to get it to expand/rearrange. this is not useful. it shouldn't matter if it's even real or not, is just rearranging algebra (bivariate series)
got something to expand by limiting summation ranges to 3 instead of \infty. would have been nicer to get summations with binomial coefficients or what have you. this is an expansion of something done to $\sum{i,j} a_{i,j} c^i z^j$, and I want to collect up all the powers and find expressions for the new $a_{i,j}$ in terms of the previous ones. maybe I should try some other software, maxima is a pain
That expression is for A_n → A_{n+1}, which I already have code for (thanks to knighty) but what I'm really after today is something for A_n → A_{2n}.
Suppose
$$A_n(c,z) = \sum a_{n,i,j} c^i z^j$$ and
$$A_{2n} = A_n(c, A_n(c, z))$$
find expressions for
$$a_{2n,i,j}$$
in terms of
$$a_{n,i,j}$$
nevermind. that approach violates the precondition that z (as argument to the outer A_n) is small when n is not a multiple of the periodic return to 0.
I think I want something more like $A_n(c, B_m(c, z))$ but I can't figure out if the order of A and B is conducive to what I'm trying to do... I don't think it commutes.
Welcome to post.lurk.org, an instance for discussions around cultural freedom, experimental, new media art, net and computational culture, and things like that. | 420 | 1,482 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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 | 3 | CC-MAIN-2020-10 | latest | en | 0.950601 |
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# Elliott Engines Inc. produces three products—pistons, valves, and cams—for the heavy equipment industry. Elliott Engines has...
Elliott Engines Inc. produces three products—pistons, valves, and cams—for the heavy equipment industry. Elliott Engines has a very simple production process and product line and uses a single plantwide factory overhead rate to allocate overhead to the three products. The factory overhead rate is based on direct labor hours. Information about the three products for 20Y2 is as follows:
Budgeted Volume (Units) Direct Labor Hours Per Unit Price Per Unit Direct Materials Per Unit Pistons 7,000 0.30 \$35 \$17 Valves 22,000 0.15 9 3 Cams 4,000 0.20 47 20
The estimated direct labor rate is \$20 per direct labor hour. Beginning and ending inventories are negligible and are, thus, assumed to be zero. The budgeted factory overhead for Elliott Engines is \$223,200.
If required, round all per unit answers to the nearest cent.
a. Determine the plantwide factory overhead rate.
\$ per dlh
b. Determine the factory overhead and direct labor cost per unit for each product.
Direct Labor Hours Per Unit Factory Overhead Cost Per Unit Direct Labor Cost Per Unit Pistons dlh \$ \$ Valves dlh \$ \$ Cams dlh \$ \$
c. Use the information above to construct a budgeted gross profit report by product line for the year ended December 31, 20Y2. Include the gross profit as a percent of sales in the last line of your report, rounded to one decimal place. Enter all amounts as positive numbers, except for a negative gross profit/gross profit percentage of sales.
Elliot Engines Inc. Product Line Budgeted Gross Profit Reports For the Year Ended December 31, 20Y2 Pistons Valves Cams \$ \$ \$ Product Costs \$ \$ \$ Total Product Costs \$ \$ \$ Gross profit \$ \$ \$ Gross profit percentage of sales % % %
d. What does the report in (c) indicate to you?
Valves have the gross profit as a percent of sales. Valves may require a price or cost to manufacture in order to achieve the same profitability as the other two products.
a. Budgeted direct labor hours = 7,000 x 0.30 + 22,000 x 0.15 + 4,000 x 0.20 = 6,200
Plantwide factory overhead rate = Budgeted Factory Overhead / Budgeted Direct Labor Hours = \$ 223,200 / 6,200 = \$ 36 per direct labor hour.
b.
Direct Labor Hours per Unit Factory Overhead Cost per Unit Direct Labor Cost per Unit Pistons 0.30 dlh \$ 10.80 \$ 6.00 Valves 0.15 dlh 5.40 3.00 Cams 0.20 dlh 7.20 4.00
c.
Elliot Engines Inc. Product Line Budgeted Gross Profit Reports For the year ended December 31, 20Y2 Pistons Valves Cams \$ \$ \$ Budgeted Sales Revenue \$ 245,000 \$ 198,000 \$ 188,000 Product Costs Direct Materials 119,000 66,000 80,000 Direct Labor 42,000 66,000 16,000 Factory Overhead 75,600 118,800 28,800 Total Product Costs 236,600 250,800 124,800 Gross Profit 8,400 (52,800) 63,200 Gross Profit Percentage to Sales 3.43 % ( 26.67 % ) 33.62 %
d. Valves have a negative gross profit as a percencent of sales. Valves may require a higher price or lower cost to manufacture in order to achieve the same profitability as the other two produts. | 811 | 3,131 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-33 | latest | en | 0.85163 |
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# In an a.c. circuit, the r.m.s. value of current, irms is related to the peak current, i0 by the relationa)b)c)d)Correct answer is option 'D'. Can you explain this answer? Related Test: Test: Alternating Current - From Past 28 Years Questions
## Class 12 Question
By Chetan · Jul 08, 2020 ·Class 12
Bindaas Boy answered Jul 08, 2020
This discussion on In an a.c. circuit, the r.m.s. value of current, irmsis related to the peak current, i0 by the relationa)b)c)d)Correct answer is option 'D'. Can you explain this answer? is done on EduRev Study Group by Class 12 Students. The Questions and Answers of In an a.c. circuit, the r.m.s. value of current, irmsis related to the peak current, i0 by the relationa)b)c)d)Correct answer is option 'D'. Can you explain this answer? are solved by group of students and teacher of Class 12, which is also the largest student community of Class 12. If the answer is not available please wait for a while and a community member will probably answer this soon. You can study other questions, MCQs, videos and tests for Class 12 on EduRev and even discuss your questions like In an a.c. circuit, the r.m.s. value of current, irmsis related to the peak current, i0 by the relationa)b)c)d)Correct answer is option 'D'. Can you explain this answer? over here on EduRev! Apart from being the largest Class 12 community, EduRev has the largest solved Question bank for Class 12. | 371 | 1,421 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-34 | latest | en | 0.910163 |
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# Is a-3b an even number? 1). b=3a+3 2). b-a is an odd number
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Is a-3b an even number? 1). b=3a+3 2). b-a is an odd number [#permalink]
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26 Sep 2006, 06:06
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Is a-3b an even number?
1). b=3a+3
2). b-a is an odd number
When the question description is like this without informing anything about whether a and b are integers or real numbers, how do we proceed. What does GMAT/OG say when nothing is given on what type of numbers we should use by default.
Solve the above problem and highlight below for the Question.
When a and b are integers you wil get the answer as D. However when you consider a and b as real numbers you will get a different answer. For ex: Try b=5 and a = 2, b = 3.5 and a = 1.5, your choice D fails.
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26 Sep 2006, 06:13
For a number to be even it has to be a integer, either pos or negative.
any integers chosen here result in a odd integer.
(D)
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Re: DS- Even or Odd - Very important [#permalink]
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26 Sep 2006, 06:49
withme wrote:
Is a-3b an even number?
1). b=3a+3
2). b-a is an odd number
(1) a-3b=a-3(3a+3)= -8a-9 which may be even, odd, integer, non-integer, rational, irrational... NOTE THAT WE ARE NOT TOLD ABOUT THE PROPERTIES OF a. NOT SUFF
(2) b=(2k+1)+a where k is an integer
a-3b= a-3((2k+1)+a)= -2a -6k-3 NOT SUFF for the same reason
Together we see that -8a-9=-2a-6k-3 for some integer k
-6a=-6k+6=-6(k+1) => a=k+1
Thus a is an integer, either odd or even
(2) tells us that b is also an integer and that exactly one of {a,b} is even
If a is even and b is odd, a-3b is odd
If b is even and a is odd a-3b is odd
(1) and (2) combined tell us that a-3b is an odd number, not an even one
SUFF
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### Show Tags
26 Sep 2006, 06:59
GMATT73 is right
only integers can be considered positive or negative. therefore the answer must be D
statement A:
b=3a+3
so 3a-b=-3, which is odd
that means a or b must be odd and the other even, so a-3b= odd
SUFF
statement B
b-a=odd
same as above, one of them must be odd and the other even, therefore a-3b=odd
SUFF
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26 Sep 2006, 07:41
The question is asking to find whether a-3b is even but did not say anything on what should a and b can assume.
For example: take a = 0.5 and b = 3.5
which will give you a-3b = 0.5 - 10.5 = -10, an even number
26 Sep 2006, 07:41
Display posts from previous: Sort by | 1,231 | 3,793 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2017-04 | latest | en | 0.858706 |
https://www.indiabix.com/non-verbal-reasoning/classification/discussion-262 | 1,723,214,796,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640767846.53/warc/CC-MAIN-20240809142005-20240809172005-00573.warc.gz | 625,461,222 | 8,290 | # Non Verbal Reasoning - Classification - Discussion
Discussion Forum : Classification - Section 1 (Q.No. 3)
Directions to Solve
In each problem, out of the five figures marked (1), (2), (3), (4) and (5), four are similar in a certain manner. However, one figure is not like the other four. Choose the figure which is different from the rest.
3.
Choose the figure which is different from the rest.
(1) (2) (3) (4) (5)
1
2
3
4
5
Explanation:
In all other figures, the arrow and the V sign lie towards the black end of the main figure.
Discussion:
33 comments Page 1 of 4.
Bumsi said: 1 week ago
I think it is 3 because the orientation of the addition sign and the arrow are different than all of the others.
Saqib Tanveer said: 2 years ago
@All.
I agree with your answer but if we see that arrow in all figures is towards the empty glass shapes except 4 because it is opposite and in the direction of the glass's lower side.
(1)
Jiya said: 2 years ago
Now I get it, thanks everyone for explaining.
(1)
Papa said: 3 years ago
@Muzamil,
As per your view then every figure is odd because every figure has its own side like figure - 1 bold or dark side is facing down except any other, the correct explanation is, except 4 all arrows are opposite to black side.
(3)
Muzamil said: 3 years ago
4 is the only one which is horizontal.
(1)
Aman Kumar said: 3 years ago
The 4th one where the arrow is unfortunately towards the side shaded dark.
i.e , Here it matters is the *arrow pointing direction* and *the shaded side.
Vaish said: 4 years ago
All the figures has + and arrow is towards the side of black shading. But 4th option is opposite.
(1)
Shiv said: 4 years ago
I don't understand, please explain to me in detail.
Adarsh Pathak said: 5 years ago
SIMPLE THING the arrow is always away from the side painted black except the 4th one where the arrow is unfortunately towards the side shaded dark i.e., the smaller side though the smaller or larger side doesn't matter what matters is the *arrow pointing direction* and *the shaded side*
(1)
Shiv kumar raghav said: 5 years ago
Plus and arrow is white side but in all figure plus and arrow is the black side. | 579 | 2,207 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.933103 |
https://iwritehomework.com/solved-homework-suppose-fxr-rightarrow-r-be-a-function-given-by-find-maximum-and-minimum-of-functional-values-using-an/ | 1,686,175,560,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224654016.91/warc/CC-MAIN-20230607211505-20230608001505-00022.warc.gz | 357,698,587 | 11,110 | # (Solved Homework): Suppose f(x):R rightarrow R be a function given by Find maximum and minimum of functional values using an
Suppose f(x):R rightarrow R be a function given by Find maximum and minimum of functional values using an algorithm with programming code. Find complexity of the algorithm.
Algorithm :
Step1 : start
Step2 : function declaration of max
Step3 : function declaration of min
Step4 : array declaration
Step5 : for(x=0;x<8;x++)
Step6 : print the elements of the F
Step7 : calling function to max
Step8 : calling function to min
Step9 : print the maximum element of the F
Step10 : print the minimum element of the F
Step11 : called function of max
Step12 : maxInt=F[0];
Step13 : for(x=0; x<8; x++)
Step14 : temp=F[x];
Step15 : if(temp>maxInt)
Step16 : maxInt=temp;
Step17 : return maximum value
Step18 : called function of min
Step19 : minInt=F[0];
Step20 : for(x=0; x<8; x++)
Step21 : temp=F[x];
Step22 : if(temp<minInt)
Step23 : minInt=temp;
Step24 : return minimum value
Step25 : stop
Program :
#include <iostream>
using namespace std;
int findMax(int F[], int index); // function declaration
int findMin(int F[], int index); // function declaration
int main()
{
int F[8]={33,25,71,83,22,47,51,36}; // F declaration
int x,max,min;
for(x=0;x<8;x++)
{
cout<<“F[“<<x<<“] = “<<F[x]<<endl; // print the elements of the F
}
max=findMax(F,8); // calling function
min=findMin(F,8); // calling function
cout<<“The maximum element is “<<max<<endl; // print the maximum element of the F
cout<<“The minimum element is “<<min<<endl; // print the minimum element of the F
return 0;
}
int findMax(int F[], int index) // called function
{
int maxInt,temp,x;
maxInt=F[0];
for(x=0; x<8; x++) // loop to find maximum element of an F
{
temp=F[x]; // assign F element in temp
if(temp>maxInt) // if the F element is less than the maximum value
{
maxInt=temp; // change maximum value
}
}
return maxInt; // return maximum value
}
int findMin(int F[], int index) // called function
{
int minInt,temp,x;
minInt=F[0];
for(x=0; x<8; x++) // loop to find minimum element of an F
{
temp=F[x]; // assign F element in temp
if(temp<minInt) // if the F element is less than the minimum value
{
minInt=temp; // change minimum value
}
}
return minInt; // return minimum value
}
Output :
Complexity of algorithm :
• Here for loop for n times.
• The time complexity of for loop is O(n).
• Since the for loop consumes more time which is O(n) and remaining code consumes less than O(n).
So Time complexity of algorithm is O(n).
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## Essay Writing Service
No matter what kind of academic paper you need and how urgent you need it, you are welcome to choose your academic level and the type of your paper at an affordable price. We take care of all your paper needs and give a 24/7 customer care support system. | 1,185 | 4,663 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-23 | latest | en | 0.271284 |
https://www.crazy-numbers.com/en/31377 | 1,656,207,464,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103036363.5/warc/CC-MAIN-20220626010644-20220626040644-00644.warc.gz | 793,916,365 | 6,296 | Discover a lot of information on the number 31377: properties, mathematical operations, how to write it, symbolism, numerology, representations and many other interesting things!
## Mathematical properties of 31377
Is 31377 a prime number? No
Is 31377 a perfect number? No
Number of divisors 4
List of dividers 1, 3, 10459, 31377
Sum of divisors 41840
Prime factorization 3 x 10459
Prime factors 3, 10459
## How to write / spell 31377 in letters?
In letters, the number 31377 is written as: Thirty-one thousand three hundred and seventy-seven. And in other languages? how does it spell?
31377 in other languages
Write 31377 in english Thirty-one thousand three hundred and seventy-seven
Write 31377 in french Trente et un mille trois cent soixante-dix-sept
Write 31377 in spanish Treinta y uno mil trescientos setenta y siete
Write 31377 in portuguese Trinta e um mil trezentos setenta e sete
## Decomposition of the number 31377
The number 31377 is composed of:
2 iterations of the number 3 : The number 3 (three) is the symbol of the trinity. He also represents the union.... Find out more about the number 3
1 iteration of the number 1 : The number 1 (one) represents the uniqueness, the unique, a starting point, a beginning.... Find out more about the number 1
2 iterations of the number 7 : The number 7 (seven) represents faith, teaching. It symbolizes reflection, the spiritual life.... Find out more about the number 7
Other ways to write 31377
In letter Thirty-one thousand three hundred and seventy-seven
In roman numeral
In binary 111101010010001
In octal 75221
In US dollars USD 31,377.00 (\$)
In euros 31 377,00 EUR (€)
Some related numbers
Previous number 31376
Next number 31378
Next prime number 31379
## Mathematical operations
Operations and solutions
31377*2 = 62754 The double of 31377 is 62754
31377*3 = 94131 The triple of 31377 is 94131
31377/2 = 15688.5 The half of 31377 is 15688.500000
31377/3 = 10459 The third of 31377 is 10459.000000
313772 = 984516129 The square of 31377 is 984516129.000000
313773 = 30891162579633 The cube of 31377 is 30891162579633.000000
√31377 = 177.13554132359 The square root of 31377 is 177.135541
log(31377) = 10.353830419422 The natural (Neperian) logarithm of 31377 is 10.353830
log10(31377) = 4.4966114177172 The decimal logarithm (base 10) of 31377 is 4.496611
sin(31377) = -0.94162469789598 The sine of 31377 is -0.941625
cos(31377) = 0.33666441497774 The cosine of 31377 is 0.336664
tan(31377) = -2.7969237495987 The tangent of 31377 is -2.796924 | 776 | 2,522 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2022-27 | latest | en | 0.701764 |
http://www.chegg.com/homework-help/questions-and-answers/question-1-value-x-following-code-executed-5-points-int-x-int-xptr-x-xptr-17-3-answer-5-qu-q3986663 | 1,475,328,081,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738662856.94/warc/CC-MAIN-20160924173742-00136-ip-10-143-35-109.ec2.internal.warc.gz | 382,408,149 | 16,587 | Question 1
What is the value of x after the following code is executed? (5 points)
int x;
int *xPtr = &x;
*xPtr = 17 / 3;
answer=5
Question 2
What is the value of x[2,2] after the following code is executed? (5 points
int *x = new int[3 * 3];
for (int iRow=0; iRow<3; iRow++)
{
for (int iCol=0; iCol<3; iCol++)
*(x+iRow * 3 + iCol) = iRow * iCol
}
Question 3
Write a line of code that would go after the closing brace in problem 2 that shows how to display the value of x[2,2]. Remember that you can’t use that notation; you need to dereference. (5 points
Question 4
What is the value of x after the following code executes (5 points)?
int x;
int *xPtr = &x;
int **xPtrPtr = &xPtr;
**xPtrPtr = 17 % 5;
answer=2
Question 5
Write a function that allocates a one-dimensional array of a given size and initializes the first element to the given size, the second element to the size -1, and so on, to the last element being initialized to 1. Return the array as the value of the function. Thus you must write the function header and parameters, and the body. A call would look like this: int *b = q5(12); (Assuming that q5 is the name of the function) (12 points.
Question 6
What is wrong with the following code? (5 points)
char str1[] = “I understand â€;
char str2[] = “pointersâ€;
strcat(str2, str1);
Question 7
Write a function that sums the digits in a string containing digits. Assume the string contains only digits. That is, calling the function on “123†would return 1+2+3=6. The function takes an integer argument and returns an integer value. Hint: your function should be around 12 lines of code, including braces on a separate line. Remember to initialize your local variables. A sample call would look like this: (12 points)
string s = â€123456â€;
cout << q7(s);
Question 8
Write a function that takes a string argument and returns a string with the characters of the input string in reverse order. A sample call follows. (12 points) Hint: the substr member function of the string class takes as arguments the starting position of a character and the number of characters and returns a string value. The append function appends one string to another.
string str1 = “abcâ€;
string str2 = reverse(str1); // this will print cba
Question 9
Write a single line of code that converts an int value to a C_String. Explain why using this function could crash your program if you’re not careful.
Question 10
Define an enumerated type for the five kinds of vertebrates (fish, amphibians, reptiles, birds, and mammals). Use correct case. (5 points)
Question 11
Define a struct for a pet (called pet) that includes the name of the pet, the number of feet (e. g. birds have 2, etc.), and the kind, using the enum you defined above. Make sure you get the syntax correct. (5 points).
Question 12
Given the following struct and the array of student structs that follows, write a function that returns a pointer to the array element (not the array index) of the student with the highest average grade. Assume that the array has been filled with student information and that everyone has all six grades. Also assume that no two people have the same average. The call shows that there are 41 students, and that number is also passed to your function. (14 points) Hint: this can be done with nested loops in about 16 lines of code.
struct student
{
string name;
int credits;
double grades[6];
};
student s[41];
A call to the function would look like this:
student *h = highestGrade(s[], 41);
Question 13
After the function call in question 12 returns, write a single line of code that prints the name of the student with the highest average. (5 points)
The call looked like this:
student *h = highestGrade(s[], 41);
Question 14
Given that the numeric values of the upper-case letters start with 65 decimal, what is printed by the following code? (5 points)
unionu
{
int iPart;
char cPart[4];
};
u test;
test.cPart[0] = 'E';
cout << "int value: " << test.iPart; | 1,038 | 4,029 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2016-40 | latest | en | 0.8328 |
https://careercup.com/question?id=14452675 | 1,603,674,854,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107890108.60/warc/CC-MAIN-20201026002022-20201026032022-00667.warc.gz | 252,584,238 | 12,128 | ## Qualcomm Interview Question
• 0
Country: United States
Comment hidden because of low score. Click to expand.
1
of 1 vote
Ans:280 comparisons in worst case.This case will be of Radix Sort.
7*4*10=280(7=no. of numbers,4=no. of digits,10=size of decimal(0-9)).
Comment hidden because of low score. Click to expand.
0
of 0 vote
Didn't get the exact question here.
Are the 4 digits of each number not taken as number on a whole?
Is digit by digit comparison required?
Comment hidden because of low score. Click to expand.
0
of 0 vote
First MSB will be sorted using counting or bucket sort (both are stable sorts) total number of buckets will be 10 (decimal base 10) so 10 comparisons. similarly for other digit positions.
Total comparison 4*10 =40
Comment hidden because of low score. Click to expand.
0
THEY ARE ASKING ABOUT COMPARISONS AND COUNTING SORT IS NOT A COMPARISON SORT ALGORITHM.
Comment hidden because of low score. Click to expand.
0
I don't think they are asking answer for comparison sort. Option doesn't make sense in that case. Even worst of comparison sorts in worst of cases will give less number of comparison than the given option
Comment hidden because of low score. Click to expand.
0
of 0 vote
merge sort....
only 21 comparisons
Comment hidden because of low score. Click to expand.
0
Wrong! Merge sort needs even less number of comparisons.
It needs n ⌈lg n⌉ - 2^⌈lg n⌉ + 1 in worst case.
Which is 14 here.
Comment hidden because of low score. Click to expand.
0
of 0 vote
we can also use count sort in place of bucket sort, count sort is stable and requires only 7 operations, and applying this we will get 28.
please correct me if i am wrong.
Comment hidden because of low score. Click to expand.
0
Number of buckets will be 10.
Comment hidden because of low score. Click to expand.
0
of 2 vote
If Radix Sort is used then Maximum no of comparisons is 280. For each number of form ad ad-1 ... ak...a1,a0 we have to compare each digit ak with bucket digit(here 0 to 9) so for each numner 10*4=40 comaprisons are required. For 7 such numbers 7*40=280 comparisons are required
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0
You don't need to compare with each bucket. 1 will go in bucket 1 , 2 will go in bucket 2 .....
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0
Ok. Then have a look at "MCQsnin Computer Science: Fourth Edition
By Timothy J. Williams". This question is there with solution. If there is fault in my logic then please correct.
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0
radix sort is not a comparison sort mind you!!!!
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0
Well I don't have the book.
Think in this way
My buckets will be link lists in an array say,
buck[10]
when I need to push a digit say i, in its corresponding bucket
the bucket will be buck[i]. What comparison I did?? ith bucket is mapped to ith location.
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0
Radix sort is not a comperasion based sort, it uses counting sort which sort numbers in O(max(n,d)) where d is difference between min and maximum number, here we sort each digit column which contain number from {0,1,2,3,4,5,6,7,8,9} only, we apply same for each column of digits msd to lsd. it will give O(max(n,d)*r), if r is length of each number , here n = 7, d will be 10, r = 4. but remember we are not doing any comparisons here, we just count number of numbers in each buckets only, it require 7 int space, but no comparison.
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0
of 0 vote
QUESTION IS CLOSED
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0
of 0 vote
plz give an explanation to this!!
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0
of 0 vote
I think, comparisions will depend upon sorting method, and also if you are comparising each digit or whole number.
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0
of 0 vote
If the focus is only on comparisons then we can also prefer Radix sort in which comparisons are close to zero but the number of copies needed for seven 4 digit numbers would actually be 2*N*K (K being the number of digits) so here copies would be 56
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0
of 0 vote
I don't understand this question. Can anyone please elaborate this question?
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0
of 0 vote
I guess for Radix sort complexity:
O(#digits*(#elements+base of each digit)) = O(d(n+b))
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0
of 2 vote
We need to sort the numbers into the buckets based on every digit from LSB to MSB. So, we get (7*10*4=280) -> 7 numbers, 4 digits/no, 10 buckets (0-9)
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0
of 0 vote
nlogn where n=7.
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0
for quick sort it will take 20 comaprisons(6*5*4*3*2) and for each comparison -> as there is 4 digit number 14 bits is required to compare so for each number comparison 14 bit comparison and hence 20*14=280
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CareerCup's interview videos give you a real-life look at technical interviews. In these unscripted videos, watch how other candidates handle tough questions and how the interviewer thinks about their performance. | 1,420 | 5,580 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2020-45 | latest | en | 0.86078 |
http://physicistfarmer.blogspot.com/2010/10/calculus-and-yogurt-with-just-smidgen.html | 1,532,019,585,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591150.71/warc/CC-MAIN-20180719164439-20180719184439-00126.warc.gz | 300,917,254 | 25,553 | ## Monday, October 18, 2010
### Calculus and yogurt, with just a smidgen of algebra.
Mmm, Yogurtland. I went there again today, and once again it is delicious (‘ono in Hawaiian). As I write this, I'm savoring a mix of Pomegranate & Raspberry Tart, Arctic Vanilla, Blueberry Tart, and Double Cookie Crumble flavors, topped with kiwi slices, cookie dough minis, chocolate chips, strawberry chunks, sliced macadamia nuts, honey, white chocolate sauce, and Ghirardelli caramel, which pretty much filled up the container I was using. The taste...er, tastes, are myriad, varied, and impossible to describe. The part of my brain that handles taste is probably forging new neural connections right now in order to handle the tsunami of new combinations of sensations flooding it. I hadn't noticed the honey the first time I was there, and it really adds a certain je ne sais quoi, an undefinable essence that permeates every bite and reminds me of the pleasant days back when I had a hive or two of my own, and how relaxing it was to be out there, working with the bees, watching them go about their little lives oblivious to me, inhaling the rich, intoxicating odor that comes only from the inside of a busy and active hive, the smell of little insects so full of life…ahhh…good times, good times. Up until the Varroa mites and the Colony Collapse Disorder killed ‘em.
But to keep this post from going somewhere very different from where it started, I've decided to do a little integral calculus, just for fun. A friend recently wondered aloud how much a certain large glass bottle filled with water weighed. Having had very few chances to use the calculus that I love so much all this semester (I've mostly only used algebra, which I despise), I decided to do a little volume integration and see if I couldn't figure it out. (hmmm…getting a little sleepy now after ingesting ~10 ounces of yogurt and toppings!)
I've never seen the bottle in question myself, so I only have some measurements and photos to go on. I've got that the outside diameter of the bottle is about 8 inches and its height is about 24 inches. This is corroborated by performing measurements on a (poor) picture of the bottle, which also shows its neck width to be about 2 inches. The glass is supposed to be about 3/8 of an inch thick, and it has a 5-inch deep V-shaped dent in the bottom for resisting internal pressure.
I played around with equations in Deadline (a nice little graphing program good for calculus) for a while until I found one that seems to fit pretty well. The structure of the bottle immediately reminded of a hyperbolic tangent function, and while it's probably impossible to get a perfect fit due to the perspective in the photos I used, the equation $$(3/2)\tanh(x/4)+5/2$$ seems to work quite well. You can see it graphed in the picture below (the red line is the graph Deadline put out, the green parts are my attempt to show what the bottle is approximately like):
Now, without delving too deeply into the theory of calculus, let me explain briefly what I will be doing. Imagine trying to find the volume of a cylinder. You know its volume is merely the area of its base ($$\pi r^2$$) times its height. Now instead of doing it in one step, suppose you divide the cylinder into a number of slices. It seems obvious that the volume of the cylinder is equal to the volume of all of those slices put together, and the volume of each slice can be found by using the volume equation for a cylinder using the new height of the slice. This is all well and good when considering a cylinder, but what if you have a cone? Any slices you make will not be cylinders, and thus cannot have the cylinder volume formula applied to them. After pondering this problem for a while, you may suddenly find yourself thinking “Aha! What if I make the slices infinitely thin? Then they will actually be little cylinders again, and I can find their volumes and sum them all up.”
On the face of it, this seems utterly preposterous, at least to me. Taking an infinite number of infinitely thin slices and adding them up to get a volume? Ridiculous! And yet the amazing thing is, it works. Isn't that incredible? Doesn't it just send shivers up and down your spine? To think that you, a mere mortal, can harness the concept of infinity for your own purposes! It's hard to convey the awe this fills me with whenever I ponder it. To me, doing these sorts of problems always gives me a feeling of flying, soaring high and away above the boring, mundane world of algebra and cutting right to the heart of a problem with eagle-like fleetness. Of course, to actually get an answer will usually require some algebra, which always feels like crash landing to me.
Anyway, the integral of integral calculus is simply a way to sum up an infinite number of slices of an object. Technically, what it adds up is the height of the function you are integrating at any given point. If we take the height of the function as the radius of a cylinder, we can square it and multiply by π to get the area of that particular slice. Adding all these areas up will give us the volume.
So! To begin...here is the equation we need to integrate:
$V=\pi\int_{-10}^{18}\left(\frac{3}{2}\tanh\Big(\frac{x}{4}\Big)+\frac{5}{2}\right)^2dx$ (note how pretty $$\LaTeX$$ makes everything!) Notice we have $$\pi$$ multiplied by the sum of all the little slices from -10 to 18, which makes a total of 28 inches (the limits here are slightly strange, even for calculus, but it's for convenience with Deadline). The little dx at the end is an important part of calculus, but we don't need to bother with it now. Now, since this equation, while not difficult, would take quite a lot of algebra to solve, I'm just going to cheat and have Deadline do the integration for us. This gives us the result $$\pi\times81.96\approx257.5$$ cubic inches. The bottle, however, is hollow. It has a thickness estimated at 3/8 of an inch. To a good approximation, the volume inside the bottle is simply the same function with 3/8 subtracted from it, like this:
\begin{align}V&=\pi\int_{-10}^{18}\left(\frac{3}{2}\tanh\Big(\frac{x}{4}\Big)+\frac{5}{2}-\frac{3}{8}\right)^2dx\\
&=\pi\int_{-10}^{18}\left(\frac{3}{2}\tanh\Big(\frac{x}{4}\Big)+\frac{17}{8}\right)^2dx\end{align} Although no more difficult in principle than the first integration, this is still a lot of writing to evaluate by hand, so again we call upon Deadline to get an answer of $$\pi\times71.46\approx224.5$$ cubic inches. This is a reassuring answer, as subtracting it from the previous answer gives 33 cubic inches, which means that a little over 92% of the bottle's volume is actually available volume, not glass (i.e, the glass the bottle is made of occupies 33 cubic inches). However, we have not taken into account the V-shaped indentation in the bottom. Assuming that it spreads to the outside diameter minus $$2\times(3/8)=3/4$$ inches, a little contemplation gives the equation: $V=\pi\int_0^5(0.62x)^2dx$(the 0.62 comes from taking the inverse tangent of (29/8)/5 and converting to radians) At this point, things start getting somewhat complicated, so we're going to simply assume that there is a cone with this volume taken out of the bottom of the bottle. Obviously, this is not perfectly correct, and may lead to the final weight being slightly on the low side, but we weren't perfectly careful with the ends of the bottle either which ought to bring it back up slightly. The volume given by Deadline is $$\pi\times7.75\approx24.3$$ cubic inches. Subtracting this from the volume found previously gives ≈ 200 cubic inches available for holding water.
So, let's recap: on some ever-so-slightly shaky assumptions we've found that the bottle has 33 cubic inches of glass and 200 cubic inches of holding space for water, assuming it's full right to the top. This translates to a carrying capacity of about 3.3 liters. Converting both values into cubic centimeters, we get ≈ 540 cc's of glass and ≈ 3274 cc's of water. The density of water at 25 °C (room temperature, roughly) is 0.997 g/cm$$^3$$. The density of glass, unfortunately, varies widely depending on the type, from less dense than aluminum to more dense than iron. According to Encyclopedia Britannica, 1971 edition, “common” glass has densities ranging from $$2.4-2.8$$ g/cm$$^3$$. Since this is tinted glass we're dealing with, and I have no idea how common it is, and usually less “common” glasses tend to be denser, I'm going to go with the heavier side and use 2.8. Multiplying the volumes times the densities gives the masses, which turn out to be ≈ 3264 g of water and 1512 g of glass (interestingly, this implies the glass makes up just a bit less than 1/3 of the total mass of the full bottle). Multiplying the masses (in kg) of the water and glass times the acceleration due to gravity (9.8 m/s$$^2$$) at the Earth's surface gives the weight, in Newtons: ≈ 32 N for the water, and ≈ 15 N for the glass. Together this gives a weight of 47 Newtons, which is about 10 and a half pounds. This works out to about 68% of the weight being water and 32% being glass, in agreement with our earlier estimate.
Error analysis: possible error: quite large. I've never held or even seen one of these bottle in person, so I'm relying on estimated measurements by other people. Function fitting was rudimentary and qualitative: if I wanted to be really through, I could have done a least-squares-fit regression analysis to determine the ‘best’ fit to the bottle (I don't actually know how to do that, but it never seemed that difficult in theory). Of course, a more thorough analysis of the complicated bottom of the bottle could be done; this reminds me of the “just assume everything is a frictionless, homogeneous sphere in vacuum” joke in physics. Personally, I think 3/8 of an inch is a bit thick for the glass; I hazard a guess that 1/4 would be closer to the mark, which would raise the weight a bit by having more volume for water, but again I've never handled the glass to see about this, and it probably varies throughout the bottle, a factor I didn't take into account. The glass may be denser than the value I assumed; it looks like old glass, which could be of (significantly) higher density than the value I used.
I suppose, in the end, the slightly depressing thing is that for all my mathematical tricks the quickest and easiest way remains simply to weigh the thing. If such a thing is ever done, I would appreciate being informed of just how off I was. Oh, and if you spot any errors in my math, please let me know! | 2,524 | 10,584 | {"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} | 2.515625 | 3 | CC-MAIN-2018-30 | latest | en | 0.950045 |
https://oeis.org/A261138 | 1,508,563,760,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824570.79/warc/CC-MAIN-20171021043111-20171021063111-00093.warc.gz | 762,832,284 | 3,743 | This site is supported by donations to The OEIS Foundation.
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A261138 The concatenation of 123456...n and reverse of this number. 0
11, 1221, 123321, 12344321, 1234554321, 123456654321, 12345677654321, 1234567887654321, 123456789987654321, 1234567891001987654321, 12345678910111101987654321, 123456789101112211101987654321, 1234567891011121331211101987654321 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS All terms are palindromic numbers by construction. Starts as A259937, but A259937 generates non-palindromic terms at n>9. LINKS FORMULA a(n) = concatenate( A007908(n), A138793(n) leading zeros kept). MATHEMATICA Table[d = Flatten[IntegerDigits /@ Range@ n]; FromDigits@ Flatten[{d, Reverse@ d}], {n, 13}] (* Michael De Vlieger, Aug 20 2015 *) CROSSREFS Cf. A259937, A002477. Sequence in context: A233012 A019524 A274766 * A259937 A078269 A015009 Adjacent sequences: A261135 A261136 A261137 * A261139 A261140 A261141 KEYWORD nonn,base AUTHOR Umut Uludag, Aug 10 2015 STATUS approved
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Last modified October 21 00:56 EDT 2017. Contains 293679 sequences. | 440 | 1,420 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2017-43 | latest | en | 0.560841 |
https://anthrogenica.com/showthread.php?327-E-V13&s=178ee57ac2b54488b5e12b2a671f5ab7 | 1,660,426,690,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571987.60/warc/CC-MAIN-20220813202507-20220813232507-00752.warc.gz | 120,318,752 | 14,787 | 1. ## E-V13
My son-in-law (Josh), 211910, is E-V13. At FTDNA he has no matches (as in zero) at 67 or 37 markers. He has 2 matches at 25 with a GD of 2. We have been unable to confirm the European country from where Josh's Floyd line emigrated from but two candidates are Ireland or Wales. Any insight into our exploration is warmly welcomed.
Best, Roy
2. Hi Roy,
Josh actually matches a cluster in the E-M35 project that we call E-V13*-D. You can see the cluster at the following link:
http://www.haplozone.net/e3b/project/cluster/43
The reason he has so few matches in his myFTDNA account is because he has a RecLOH at DYS464, which caused him to have 7 values at that marker instead of the normal 4. The comparison for DYS464 is not as straight-forward as other markers -- rather than a straight marker-to-marker comparison (DYS464a to DYS464a, DYS464b to DYS464b, etc), FTDNA compares all values and then counts a GD of 1 for any values that don't match.
For example, let's compare Josh's DYS464 with Lewis (11508):
Josh has: 14 14 15 15 16 16 17
Lewis has: 14 15 16 16 17
You'll see that Lewis has 5 values at DYS464 -- so he also has more than the standard 4. When you look at the actual values -- they both have one 14, one 15, two 16s and one 17. Then Josh has two values left over that don't match up with Lewis: the second 14 & the second 15. All other markers in the first 25 match, so Josh and Lewis end up with a GD of 2 in the first 25 markers, and that's within the threshold that FTDNA displays for matches.
Now Josh and Lewis *should* be a close match at 37 markers too, but there's another multi-copy marker in the 26-37 panel which was affected by the RecLOH -- CDY. Josh has values 30-30 and Lewis has values 30-36. The comparison for CDY isn't treated the same way as DYS464 -- instead of counting the difference at CDYb as a GD of 1, it's counted as a GD of 6. They also differ by 1 on two additional markers in the 26-37 panel.
So altogether, Josh and Lewis have GD=10 at 37 markers, which is beyond FTDNA's matching threshold for the 37-marker level (GD=4).
The rest of the guys in the cluster all have the standard 4 values at DYS464 -- so at minimum, they're GD=3 from Josh at 25 markers, which is beyond the matching threshold, and therefore none of them show up on Josh's Matches page.
If Josh didn't have these extra and doubled values caused by the RecLOH, then many of the people in his cluster would have shown up on his Matches page. The many different surnames in this cluster was probably caused by this lineage not using inherited surnames until relatively recently (past few hundred years), and when the different branches of the family did finally take surnames, they ended up with different ones. This type of situation is very common among communities and populations that used patronymics until relatively recently (past few hundred years). It's especially common for Ashkenazi Jews and Scandinavians, and I've come across several of these tight clusters of different-surname people with Welsh and Irish ancestry.
Elise
3. Elise ~
Thanks a bunch for your reply. You are correct, Lewis is Josh's closest match, albeit at 25 markers a GD of 2. I did some exploration about RecLOH at 464 after reading your post. I see this is an odd marker and prone to RecLOH. And, although what you say about excluding the spurious 464 and CDY values makes sense, when I compare Josh's 67 marker results at the E3b and E-M35 Project with his closest matches, the GD still isn't close. E3b project, just above and below Floyd: Evans a GD 8 and Johnson a GD of 9. At the E M35 project is Plant GD 17 and Manolis GD 20. These represent a GD without the 464 and CDY values.
I did ask for Josh to join the Null 425 project.
Thanks again. Roy
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• | 1,016 | 3,925 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2022-33 | latest | en | 0.948895 |
https://forum.openoffice.org/en/forum/viewtopic.php?p=386673& | 1,558,398,652,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256184.17/warc/CC-MAIN-20190521002106-20190521024106-00021.warc.gz | 500,584,828 | 10,030 | ## [Solved] SpellNumber in Calc Basic
Keyboard macros or custom scripts
### [Solved] SpellNumber in Calc Basic
I got a VBA from Excel which works well in Excel. I want to put it in Calc macro. The function works except the case where dollar is zero and or cents is zero. Macro does not recognize the case when dollar or cent is "". Below is the macro. Hope that someone can help
Code: Select all Expand viewCollapse view
`REM ***** BASIC *****'Main FunctionFunction SpellNumber(ByVal MyNumber) Dim Dollars, Cents, Temp Dim DecimalPlace, Count ReDim Place(9) As String Place(2) = "Thousand " Place(3) = "Million " Place(4) = "Billion " Place(5) = "Trillion " ' String representation of amount. MyNumber = Trim(Str(MyNumber)) ' Position of decimal place 0 if none. DecimalPlace = InStr(MyNumber, ".") ' Convert cents and set MyNumber to dollar amount. If DecimalPlace > 0 Then Cents = GetTens(Left(Mid(MyNumber, DecimalPlace + 1) & _ "00", 2)) MyNumber = Trim(Left(MyNumber, DecimalPlace - 1)) End If Count = 1 Do While MyNumber <> "" Temp = GetHundreds(Right(MyNumber, 3)) If Temp <> "" Then Dollars = Temp & Place(Count) & Dollars If Len(MyNumber) > 3 Then MyNumber = Left(MyNumber, Len(MyNumber) - 3) Else MyNumber = "" End If Count = Count + 1 Loop Select Case Dollars Case "" Dollars = Str(Dollars) Case "One " Dollars = "One Dollar " Case Else Dollars = Dollars & " Dollars " End Select If Dollars = "" Then Select Case Cents Case "" Cents = "" Case "One " Cents = "One Cent Only" Case Else Cents = Cents & "Cents Only" End Select Else Select Case Cents Case "" Cents = "Only" Case "One " Cents = "and One Cent Only" Case Else Cents = "and " & Cents & "Cents Only" End Select End If SpellNumber = Dollars & CentsEnd Function ' Converts a number from 100-999 into textFunction GetHundreds(ByVal MyNumber) Dim Result As String If Val(MyNumber) = 0 Then Exit Function MyNumber = Right("000" & MyNumber, 3) ' Convert the hundreds place. If Mid(MyNumber, 1, 1) <> "0" Then Result = GetDigit(Mid(MyNumber, 1, 1)) & "Hundred " End If ' Convert the tens and ones place. If Mid(MyNumber, 2, 1) <> "0" Then Result = Result & GetTens(Mid(MyNumber, 2)) Else Result = Result & GetDigit(Mid(MyNumber, 3)) End If GetHundreds = ResultEnd Function ' Converts a number from 10 to 99 into text.Function GetTens(TensText) Dim Result As String Result = "" ' Null out the temporary function value. If Val(Left(TensText, 1)) = 1 Then ' If value between 10-19... Select Case Val(TensText) Case 10: Result = "Ten " Case 11: Result = "Eleven " Case 12: Result = "Twelve " Case 13: Result = "Thirteen " Case 14: Result = "Fourteen " Case 15: Result = "Fifteen " Case 16: Result = "Sixteen " Case 17: Result = "Seventeen " Case 18: Result = "Eighteen " Case 19: Result = "Nineteen " Case Else End Select Else ' If value between 20-99... Select Case Val(Left(TensText, 1)) Case 2: Result = "Twenty " Case 3: Result = "Thirty " Case 4: Result = "Forty " Case 5: Result = "Fifty " Case 6: Result = "Sixty " Case 7: Result = "Seventy " Case 8: Result = "Eighty " Case 9: Result = "Ninety " Case Else End Select Result = Result & GetDigit _ (Right(TensText, 1)) ' Retrieve ones place. End If GetTens = ResultEnd Function ' Converts a number from 1 to 9 into text.Function GetDigit(Digit) Select Case Val(Digit) Case 1: GetDigit = "One " Case 2: GetDigit = "Two " Case 3: GetDigit = "Three " Case 4: GetDigit = "Four " Case 5: GetDigit = "Five " Case 6: GetDigit = "Six " Case 7: GetDigit = "Seven " Case 8: GetDigit = "Eight " Case 9: GetDigit = "Nine " Case Else: GetDigit = "" End SelectEnd Function`
Last edited by LaMancha on Fri May 13, 2016 10:37 am, edited 2 times in total.
OpenOffice 4.1 on Windows XP
LaMancha
Posts: 3
Joined: Thu May 12, 2016 3:25 am
### Re: SpellNumber in Calc Basic
Use a typed variable when you only need a specific type of data.
Here, an empty variant is not equal to an empty string.
Replace the first declaration by:
Code: Select all Expand viewCollapse view
`Dim Dollars As String, Cents As String, Temp As String`
Instead of this naive and inefficient code, use function MONEYTEXT, provided by extension Numbertext.
See this thread : How to format a number to its written form
Bernard
OpenOffice.org 1.1.5 / Apache OpenOffice 4.1.1 / LibreOffice 5.0.5
MS-Windows 7 Home SP1
B Marcelly
Volunteer
Posts: 1160
Joined: Mon Oct 08, 2007 1:26 am
Location: France, Paris area
### Re: SpellNumber in Calc Basic
Thank for your reply. Moneytext did work but it appears the word U.S. dollar rather than dollar only. The problem is just as stated for the variant is not empty string.
OpenOffice 4.1 on Windows XP
LaMancha
Posts: 3
Joined: Thu May 12, 2016 3:25 am
### Re: SpellNumber in Calc Basic
=SUBSTITUTE(MONEYTEXT(A1;"USD";"en-US");"U.S. ";"")
=NUMBERTEXT(A2;"en-US")&" dollars"
Please, edit this topic's initial post and add "[Solved]" to the subject line if your problem has been solved.
Ubuntu 18.04, OpenOffice 4.x & LibreOffice 6.x
Villeroy
Volunteer
Posts: 26626
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Location: Germany
### [Solve] Re: SpellNumber in Calc Basic
The MoneyText function is solved.
OpenOffice 4.1 on Windows XP
LaMancha
Posts: 3
Joined: Thu May 12, 2016 3:25 am
### Re: [Solved] SpellNumber in Calc Basic
I want enter spellcurr macro in calc basic
please help me to addin macro
and also share code for macro number to text
OpenOffice 3.1 on Windows Vista
akhter28
Posts: 5
Joined: Thu Jan 18, 2018 7:03 pm
### Re: [Solved] SpellNumber in Calc Basic
As far as I know spellcurr is an Excel function which is not available in OpenOffice
This extension may help
Numbertext
If the code needs tweaking, you should be able to extract and debug that with little difficulty.
Apache OpenOffice 4.1.6 on Xubuntu 18.04.2 (mostly 64 bit version) and very infrequently on Win2K/XP
RoryOF
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### Re: [Solved] SpellNumber in Calc Basic
How i can apply numbertext function please complete process required
please email me at akhterali28@yahoo.com
OpenOffice 3.1 on Windows Vista
akhter28
Posts: 5
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### Re: [Solved] SpellNumber in Calc Basic
Install extension
Restart office
Use it like an ordinary spreadsheet function
=NUMBERTEXT(A1)
Please, edit this topic's initial post and add "[Solved]" to the subject line if your problem has been solved.
Ubuntu 18.04, OpenOffice 4.x & LibreOffice 6.x
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### Re: [Solved] SpellNumber in Calc Basic
Is Extention a seprate soft ware which i have to installed?
Please share how can i install it
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akhter28
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### Re: [Solved] SpellNumber in Calc Basic
Download the extension (see link provided above).
Open OpenOffice and go to Tools>Extension manager
Click the Add button
Select the extension you've downloaded.
AOO 4.1.6 on Xubuntu 19.04 and 4.1.5 on Windows 7 (with winPenPack port).
Hagar Delest
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### Re: [Solved] SpellNumber in Calc Basic
Link not provided
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akhter28
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### Re: [Solved] SpellNumber in Calc Basic
Link provided in the seventh post in this topic; the text for the link is
Numbertext
on a separate line. please scroll up to find it.
Cheers
David
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# Basic question on synthetics
Discussion in 'Options' started by cdowis, Oct 6, 2006.
1. ### cdowis
Basic question here:
2 x XYX stock at 73, short 2 x 70 calls, and long 5 x 75 calls.
I am trying to understand how this works:
As I understand, the 73 calls "cancel" out the short 70 calls (almost), so you are left with two short 73 puts and 5 x 75 calls.
You are net plus deltas approx 3 x 75 calls (approx)?
Or am I totally confused.
2. ### momoneythansens
Long stock = long CALL + short PUT
It doesn't matter which strike. If you buy a CALL and sell a PUT at any strike, the position should behave like long stock i.e. stock at 73 is irrelevant.
Turn your 2 x stock into:
2 x long CALLs + 2 x short PUTs
(you can decide the strike later but they must be the same for both PUTs and CALLs)
Looking at your existing inventory you can choose to offset the 2 x long CALLs against your existing 2 x short 70 CALLs (this is just one option, but als the most logical) i.e. we have now settled on the strike for the synthetic stock to be 70.
So you are left with:
2 x short 70 PUTs (from synthetic stock at strike 70)
5 x long 75 CALLs
Yes, this is a very bullish position, net long DELTAs.
One possible way to dissect the position to help you understand it's characteristics are:
2 x split-strike combos (risk reversal) 70/75
+3 extra CALLs.
Good luck.
MoMoney.
3. ### spindr0
I think that your question is confusing because you have what I think is a typo. You have "short 2 x 70 calls, and long 5 x 75 calls" and they turn into 73 calls and puts?
Ignoring your conclusion and sticking with the original 70 and 75 calls and keeping it simple, if these options had a very low IV, say about .15, the 5 long 75 calls would "cancel" out the delta of the 2 short 70 calls at the current price of 73 (at higher IV, you'd have a net positive delta from these options).
But what about the 200 delta from the long 200 shares?
This is a bullish position with 200 long delta plus the delta of the option position.
4. ### riskarb
5-wide risk-reversals are roughly 75d at \$73 due to gamma. So long 150d on the 2 lot reversal and another 35d per call on the additional 2 75s. Perhaps +220 deltas at the moment. Don't hold me to that delta fig, as I have no interest in booting a modeling app.
The payoff is nonlinear, so it's nowhere near synthetic stock.
5. ### cdowis
Thanks very much.
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I am a masters student looking for some direction on using neural network on market depth data to help predict market liquidity and bid-ask spreads. Can some of the more experienced people give me some research guidance and guide me to some papers that I should be reading? I have tried Google scholar but can't find something meaningful to kick off my research. Thanks
I agree with all Robert says above, but if you already have the data, and you want to quickly create a neural network model and run the analysis, I would suggest the following:
1. The Heaton Site has a Wiki, links to papers, links to books, a forum, etc. that will help you get started, but you might try the PluralSight course Introduction to Machine Learning with ENCOG 3 if you want to get up to speed very quickly as this course shows you how to use Heaton's free software to create a neural network model.
2. Download ENCOG 3 from the Heaton site. It is a machine learning Framework that runs on a variety of languages (C#, Java, JavaScript, etc.)
3. In a few hours, you can get setup with code that looks something like this and have your first neural network analysis:
• Thanks for the advice. I have the data. That's less of an issue for me thankfully (although the answer would have been different 6 weeks ago). What I am trying to figure out is: 1) what are the virgin territories where market micro-structure data and ML/datamining techniques intersect? 2) Related to that what are the real life questions in which professional folk are interested (think fund managers, bankers, traders and the like). I would really like to answer questions which can help people rather than producing research for the sake of it. Please let me know if you have more thoughts. Thanks Sep 23, 2015 at 12:35
You can try using different approaches. Starting from something not that "heavy" like the NN.
0) Pre study
- you need to prepare your data
(how you will treat a negative spread (i.e. ASK - BID <0),
what will you do if you will have 0 spread and then you will divide some value by it?),
1) Time series models
- check if the bid-ask spread tends to be stationary (you can check it using some extracted moving period from your time series), apply statistical tests
- apply ARMA, ARIMA models for periods which bid-ask spread is stationary
- try to figure out what should you do when the time series stop to be stationary
- try to apply mean reverting models
- after doing this you will have some solid understanding of the characteristics of bid-ask spread models and it will be easier to build NN models.
- this book might be useful http://www.amazon.com/Series-Analysis-James-Douglas-Hamilton/dp/0691042896
2) Neural networks
- try to use different input parameters (i.e. you can smooth the data using moving average)
- try to use also "bid ask volume", "time" variables
- try to use different type of neural networks (different layers) and output neurons and discover which one is the best "out of sample" and "in sample"
BTW what is your goal of the research? Where is the science? What do you want to discover or prove? What is your hypothesis? Answering those questions might help to plan the research in advance.
• Thanks for the reply. I am at a Data Science post grad program. My own past background has been in energy trading. I am based in a place where the local markets (think more emerging markets) still look at fundamental type analysis as far as capital markets are concerned. I am looking to change that and bring some of these more advanced techniques into the area through my research. Sep 3, 2015 at 15:39
• Some of the questions we are trying to answer are: •Where is the largest volume traded on a given day? At a given time? Most importantly why? •Which sort of an investor is behind the trade? An institutional (hedge funds, Asset Management Companies), a broker, or an international trader. •Where is the stock price most likely to head given the latest market stack (buy/sell side) changes? •Is the current standing order book going to change the market tomorrow? And by how much? •Can we detect changes in market liquidity using techniques such as NN or ANN or kNN? Sep 3, 2015 at 15:40
• Can you recommend anything else thinking about the above 2 comments? Regards. Sep 3, 2015 at 15:43
• Where is the largest volume traded on a given day? At a given time? - Try tu use time series models on ask and bid volume. Questions regarding why somthing happened are hard to answer - this is looking for a casuality, so this is more an econometric problem than machine learning. Liquitdy - try to read about market microstruture. Many models how to model liquidity were developed during last 10 years. Sep 3, 2015 at 21:13 | 1,087 | 4,778 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | latest | en | 0.956103 |
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# Expressions
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###### Definition
We’ve already seen plants, snowflakes and coastlines, and here are some more examples. Isolate EF by dividing by 6. In this problem we know the percent and the whole amount, so we can set up our equation as follows. Optional module availability Student demand for optional modules may affect availability. The most common applications of the latter algorithm seems to be producing abstract book covers for books about the universe. Example: the ratio given when a team wins 4 out of 6 games is 4/6, 4:6, four out of six, or 67%. »number bonds to 40Extend your work with number bonds with facts up to 40. Pȋ n Cyrillic spelling пи̑. Mathematics, statistics and computer science, and their career opportunities include. GMAT Online Prep Blog. However, to succeed in the Math Lab program students must have motivation and self discipline. Before you provide additional information to third party providers, we encourage you to review their privacy policies and information collection practices. Note that when E and F are mutually exclusive that is, when EF=Ø, then Eq. Everything is about understanding things for the sake of understanding them. A practical problem is how to fold a map so that it may be manipulated with minimal effort or movements. Equivalent ratios are similar to equivalent fractions. But since it is impossible to move faster than the speed of light, nothing can escape. Example: For a data set containing 2, 19, 25, 32, 36, 38, 31, 42, 57, 45, and 84. The content that the SAT Math section tests can be broken down into four sections: Numbers and Operations; Algebra and Functions; Geometry and Measurement; and Data Analysis, Statistics, and Probability. The median is the middle score of a distribution. For example, the sum of the aliquot divisors of 48, excluding the 1, is 75 while the sum of the aliquot divisors of 75, excluding the 1, is 48. Mathematical discourse in shared storybook reading. Investment decisions can be complex and difficult to predict. , en be the standard basis ∑ n for Rn , so that v = i=1 αi ei and w = ni=1 βi ei. These, along with the general curriculum in Mathematics, provide a strong foundation for the students pursuing a graduate degree in mathematics. Solution: The relation R is antisymmetric as a = b when a, b and b, a both belong to R. So we use this as our common denominator. This can make it hard to write and solve equations. Using a calculator, we can then find. | 4,004 | 18,635 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2022-49 | longest | en | 0.939805 |
https://electronics.stackexchange.com/questions/100312/constant-current-source-circuit-improvisation/100315 | 1,718,418,311,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861578.89/warc/CC-MAIN-20240614235857-20240615025857-00258.warc.gz | 191,805,330 | 42,569 | # Constant Current Source circuit improvisation
simulate this circuit – Schematic created using CircuitLab
I was wondering whether I could use this circuit as a constant current source provided my load circuit impedance was less than Rb/hfe? All points of view are welcome....:)
• hfe varies too much to make this a useful current source. Commented Feb 19, 2014 at 20:03
• what are D1 and D2 doing in this circuit? Commented Feb 19, 2014 at 20:03
• I mistakenly used 2 didodes D1 and D2, only one should have been enough. I used the diode to make sure that the circuit does not retract a current in the opposite direction. But as long as my operating point is in the active region, shouldn't my hfe be constant?? Commented Feb 19, 2014 at 20:06
• hfe varies per device and over temperature. Commented Feb 19, 2014 at 20:11
It's not going to be an accurate constant current source because the voltage-current slope of the BJT is not that great but if you can live with this then that's OK: -
Ideally, once outside the saturation region, the lines would be horizontal to the base line of Vce - horizontal means that no-matter how you change the resistance of the load (which obviously changes the voltage across it and Vce, the current would remain the same. As you can see, the lines are not horizontal but have a gentle upward slope and get progressively less-horizontal for higher base currents. Added to this your circuit won't control the base current as neatly as you think - current through the emitter will slightly reduce the base current. If you want a better constant current source try one of these: -
• How is the BJT circuit u showed in figure 1 or figure 2 going to solve the issue with the non-horizontal character of the Vce vs Ic curve? Could you explain the 2 circuits to me? Commented Feb 19, 2014 at 20:19
• @ubuntu_noob putting an emitter resistor in circuit calms things down a lot - for instance it masks the internal emitter resistance I mentioned in my answer - it would probably be 100 times bigger so that the effect of collector current on the "set-point" is nowhere near as bad. With an emitter resistor you are setting a voltage across it from the base (0.7V difference) and that stabler emitter voltage and that stable emitter resistance makes for a stabler constant current with less slope on the graphs. Commented Feb 19, 2014 at 21:28
• okay. Now it's starting to make some sense, it is probably bcoz of this we were advised to ground the emitter if we wanted to saturate our transistors. Commented Feb 19, 2014 at 22:02
• OK coolio. If you want a really stable current circuit go for the ones with an op-amp controlling the voltage across the emitter resistor. Commented Feb 19, 2014 at 22:06
A more common set up for a current source is as follows:
simulate this circuit – Schematic created using CircuitLab
• The voltage across R1 will be reasonably constant at $2×U_D-U_{BE} = 2× 0.7 - 0.7 = 0.7\text{V}$.
• Program the current by calculating $R_1 = \dfrac{U_D}{I}$
• Pick a current through R2 that is approximately 10 times bigger than through R1. $R_2 = \dfrac{U_{BAT1} - U_{D1} - U_{D2}}{10 × I}$. The rule of thumb for 10 times is that you can safely disregard the base current for calculations when the transistor has a hFE > 100.
This only works for currents in the mA range, otherwise these transistor and diodes will be overloaded.
• and why is this circuit better than the circuit I put up? I just want to know the inspiration behind the modifications....that's all, since I am currently learning Commented Feb 19, 2014 at 20:24
• @ubuntu_noob This circuit is better because in practice h(FE) has a wide spread, in the order of 100 .. 300 and hFE tends to vary with collector current. Als the two diodes in your circuit don't have a function. They drop voltage that the transistor can drop equally well for you. Commented Feb 19, 2014 at 20:31
• but as I change my load, my Vce value would obviously change since my potential drop across my load would be changing and the drop across R1 is constant, hence when Vce changes wouldn't my operating point shift, causing a shift hfe and hence Ic? That would bring me back to square1 again. I know I am missing something, I just cant figure out which part.... Commented Feb 19, 2014 at 20:49
• @ubuntu_noob Why not breadboard your circuit and see what happens. Apart from fiddling with the load, also exchange the transistor for a couple alternatives of the same type. The "something" you are missing is the difference between a simple transistor model, fit for doing quick calculations and the real part with non-ideal behavior. Unfortunately ideal transistors are not sold in shops. Commented Feb 19, 2014 at 21:03
• @ubuntu_noob I was serious about the breadboard. Experimenting with these things give you valuable insight that no text book can compete with. Commented Feb 20, 2014 at 5:50
No, you need to keep the base voltage constant. Try Figure 5 at http://en.wikipedia.org/wiki/Current_source, which uses an LED for this purpose.
You could switch right at the load, if you need to | 1,258 | 5,091 | {"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} | 3.171875 | 3 | CC-MAIN-2024-26 | latest | en | 0.952576 |
http://chatswing.info/problem-solving-dilations-lesson-12-7-11/ | 1,576,025,702,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540529516.84/warc/CC-MAIN-20191210233444-20191211021444-00131.warc.gz | 29,782,299 | 8,700 | ### PROBLEM SOLVING DILATIONS LESSON 12-7
Images coordinates 2 3, 13 to get the coordinates of the image 6, 3 problem 2. Shed the societal and cultural narratives holding you back and let free step-by-step Geometry Common Core Practice and Problem Solving Workbook textbook solutions reorient your old paradigms. Identifying Dilations Tell whether each transformation appears to be a dilation. Find the coordinates of the image only do not graph. At the end of the lesson students reflect on their experiences with the activity and make connections to the types of problem solving they will be doing for the rest of the course. Tell whether the window has line symmetry. Submit Feedback Report Problems x.
No; the triangles are not the same size. Auth with social network: Lines drawn through the pairs of corresponding vertices meet at a point called the center of dilation. At the end of the lesson students reflect on their experiences with the activity and make connections to the types of problem solving they will be doing for the rest of the course. Practice and Problem Solving: My presentations Profile Feedback Log out. The image and the preimage of a figure under a dilation are similar.
Properties of Dilations Practice and Problem Solving: Home Lesson problem solving dilations answers Lesson problem solving dilations answers ThursdayAll photos “Lesson problem solving dilations answers” photos: Dilation is a non-Rigid transformation. Match each rule with the correct translation.
Tell whether the quilt block design. Drawing Dilations Copy the figure and the center of dilation P. Find the perimeter of the rectangle in the projection.
ESSAY TUNGKOL SA KALAMIDAD PAGHANDAAN GUTOM AT MALNUTRISYON AY AGAPAN
# Lesson Problem Solving Ratios In Similar Polygons
Let the actual diameter of the flower be d in. Extend Sequences Multi-Part Lesson 3: Auth with social network: If you dont see any interesting for solvimg, use our search form on bottom. Lesson 1 Skills Practice Congruence and Transformations and dilations. Multiply the coordinates of point D0, -1 by 4. Determine the coordinates of the triangle shown below after a dilation with a scale factor of 4.
The first one is done for you. Find the coordinates of the image only do not graph.
## Problem solving dilations lesson 12-7
Lesson problem solving dilations answers. For example, in the first problem above, there soolving many different places to put the center of dilation in order to meet the constraints specified in the problem. Home Problem solving dilations lesson Problem solving dilations lesson MondayRandom photos “Problem solving dilations lesson ” photos: If you wish to download it, please recommend it to your friends in any social system.
In the second problem, there are many different sequences of. As you sketch the dilation, you do not need to include a center of dilation because you are dilatiions no information about it.
A dilation with a scale factor greater than 1 is an enlargement, or dilatiions. Yes, the figures are similar and the image is not turned or flipped.
EDEXCEL SNAB BIOLOGY A2 COURSEWORK MARK SCHEME
We think you have liked this presentation. Translate the triangle with vertices A 2, —1B 4, 3and C —5, 4 along the vector. An isometry is a transformation that does not change the shape or size of a figure.
Geometry-Chapter 10 Lesson Step 2 On each line, mark twice the distance from Q to the vertex. This lesson is a fun introduction to the open-ended, collaborative, and creative problem solving students will be using over the rest of this unit and course.
# LESSON Problem Solving Dilations
A reflection across the y-axis is followed by a dilation with. If so, draw all the lines of symmetry.
Perform a Dilation of 3 on point A 2, 1 which you can see in the graph below. Step 2 On each line, mark twice the distance from P to the vertex. Reflections, translations, and rotations are all isometries. Recall that a dilation changes the size of a figure by a scale factor, but does not change Sample answer: Take Apart Three-Dimensional Shapes 1. | 887 | 4,082 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-51 | latest | en | 0.844408 |
https://oeis.org/A122567 | 1,686,188,352,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224654031.92/warc/CC-MAIN-20230608003500-20230608033500-00285.warc.gz | 481,384,448 | 3,641 | The OEIS is supported by the many generous donors to the OEIS Foundation.
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A122567 Twin primes modulo 4. 0
3, 1, 3, 3, 1, 1, 3, 1, 3, 1, 3, 3, 1, 3, 1, 1, 3, 3, 1, 1, 3, 1, 3, 3, 1, 3, 1, 1, 3, 3, 1, 3, 1, 1, 3, 1, 3, 3, 1, 3, 1, 3, 1, 3, 1, 1, 3, 1, 3, 1, 3, 3, 1, 1, 3, 1, 3, 3, 1, 1, 3, 1, 3, 3, 1, 1, 3, 1, 3, 3, 1, 3, 1, 1, 3, 1, 3, 3, 1, 3, 1, 1, 3, 1, 3, 1, 3, 1, 3, 3, 1, 3, 1, 3, 1, 1, 3, 3, 1, 3, 1, 3, 1, 3, 1 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 LINKS Table of n, a(n) for n=1..105. EXAMPLE The twin primes are 3,5,7,11,13,17,19,... and mod 4 is 3,1,3,3,1,1,3... MAPLE a[1]:=3:i:=2:for k from 5 to 3000 do if isprime(k) and isprime(k+2) then a[i]:=k mod 4:a[i+1]:=k+2 mod 4: i:=i+2 fi od: seq(a[n], n=1..i-1); CROSSREFS Cf. A001097. Sequence in context: A298262 A359525 A124814 * A122431 A279340 A089607 Adjacent sequences: A122564 A122565 A122566 * A122568 A122569 A122570 KEYWORD nonn AUTHOR Miklos Kristof, Sep 21 2006 STATUS approved
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Last modified June 7 21:29 EDT 2023. Contains 363157 sequences. (Running on oeis4.) | 671 | 1,387 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-23 | latest | en | 0.552851 |
https://blogpedia.web.id/which-of-the-following-could-not-be-a-correlation-coefficient/ | 1,675,476,348,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500080.82/warc/CC-MAIN-20230204012622-20230204042622-00200.warc.gz | 162,885,436 | 14,087 | # Which of the following could not be a correlation coefficient
1, 3 and 6
Step-by-step explanation:
Correlation coefficient denoted as r tells us about the Linear Relationship between two quantitative variables. The value of r always lies between -1 and 1. It is a unit-less quantity.
If the absolute value of correlation coefficient is closer to 1 it indicates a strong Linear relationship between two variables. If the absolute value of r is closer to 0 it indicates a weak Linear Relationship between two variables.
Therefore,
Statement 1 is true.
Statement 2 is False as correlation coefficient has no units.
Statement 3 is true.
Statement 4 is false. As r = 0 indicates no linear relationship at all.
Statement 5 is false. The absolute value of r is close to 1 so it indicates a strong linear relationship
Statement 6 is true. Correlation does not imply causation. This means the two variable may or may not be causing each other but still be co-related. | 208 | 971 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2023-06 | latest | en | 0.93654 |
https://nordstrommath.com/IntroGameTheory/S_EquilibPoints.html | 1,701,259,934,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100081.47/warc/CC-MAIN-20231129105306-20231129135306-00497.warc.gz | 496,772,984 | 7,198 | ## Section2.5Equilibrium Points
In this section, we will try to gain a greater understanding of equilibrium strategies in a game. In general, we call the pair of equilibrium strategies an equilibrium pair, while we call the specific payoff vector associated with an equilibrium pair an equilibrium point.
Determine the equilibrium point(s) for the following games.
1. $\left[\begin{matrix}(2, -2)\amp (-1, 1)\\ (2, -2)\amp (-1, 1) \end{matrix} \right]$
2. $\left[\begin{matrix}(0, 0)\amp (-1, 1)\amp (0, 0)\\ (-1, 1)\amp (0, 0)\amp (-1, 1)\\ (0, 0)\amp (1, -1)\amp (0, 0) \end{matrix} \right]$
What do you notice about the values of the equilibrium points of the games in Exercise 2.5.1?
The big question we want to answer is “Can two equilibrium points for a two-player zero-sum game have different values?” By experimenting with some examples, try to create an example of a game with two equilibrium points where those points have different values for one of the players. If you can successfully create such an example, you will have answered the question. But just because you can't find an example, that doesn't mean one doesn't exist!
After trying several examples, you might be beginning to believe that the answer to the above question is “no.” Now you are ready to try to prove the following theorem:
##### Solution Theorem for Zero-Sum Games.
Every equilibrium point of a two-person zero-sum game has the same value.
Let's start with the $2 \times 2$ case. We will use a proof by contradiction. We will assume the theorem is false and show that we get a logical contradiction. Once we reach a logical contradiction (a statement that is both true and false), we can conclude we were wrong to assume the theorem was false; hence, the theorem must be true. Make sure you are comfortable with the logic of this before moving on.
Since we want to assume the theorem is false, we assume we have a two-player zero-sum game with two different equilibrium values. Since we don't have a specific example of such a game, we want to represent the game in a general form. In particular, we can represent the general game
\begin{equation*} \left[\begin{matrix}(a, -a)\amp (c, -c)\\ (d, -d)\amp (b, -b) \end{matrix} \right]. \end{equation*}
Note that if $a$ is negative, then $-a$ is positive; thus, every possible set of values is represented by this matrix. We want to look at the possible cases for equilibria.
Explain what goes wrong if $(a, -a)$ and $(d, -d)$ are equilibria with $a \neq d\text{.}$
Hint
Think about the different cases, such as $a\lt d\text{,}$ $a>d\text{.}$
Generalize you answer to Exercise 2.5.3 to explain what goes wrong if the two equilibria are in the same column. Similarly, explain what happens if the two equilibria are in the same row.
Does the same explanation hold if the two equilibria are diagonal from each other? (Explain your answer!)
From your last answer, you should see that we need to do more work to figure out what happens if the equilibria are diagonal. So let's assume that the two equilibria are $(a, -a)$ and $(b, -b)$ with $a \neq b\text{.}$ It might be helpful to draw the payoff matrix and circle the equilibria.
Construct a system of inequalities using the fact that a player prefers an equilibrium point to another choice. For example, Player 1 prefers $a$ to $d\text{.}$ Thus, $a > d\text{.}$ List all four inequalities you can get using this fact. You should get two for each player– remember that Player 1 can only compare values in the same column since he has no ability to switch columns. If necessary, convert all inequalities to ones without negatives. (Algebra review: $-5 \lt -2$ means $5 > 2\text{!}$)
Now string your inequalities together. For example, if $a \lt b$ and $b \lt c$ then we can write $a \lt b \lt c\text{.}$ (Be careful, the inequalities must face the same way; we cannot write $a> b \lt c\text{!}$)
Explain why you now have a contradiction (a statement that must be false). We can now conclude that our assumption that $a \neq b$ was wrong.
Explain why you can conclude that all equilibria in a $2 \times 2$ two-player zero-sum game have the same value.
We just worked through the proof, step-by-step, but now you need to put all the ideas together for yourself.
Write up the complete proof for the $2 \times 2$ case in your own words.
Can you see how you might generalize to a larger game matrix? You do not need to write up a proof of the general case, just explain how the key ideas from the $2 \times 2$ case would apply to a bigger game matrix.
Hint
Think about equilibria in (a) the same row, (b) in the same column, or (c) in a different row and column.
We've seen that any two equlibrium points must have the same value. However, it is important to note that just because an outcome has the same value as an equilibrium point, that does not mean it is also an equilibrium point.
Give a specific example of a game matrix with two outcomes that are $(0, 0)\text{,}$ where one is an equilibrium point and the other is not.
Working through the steps of a mathematical proof can be challenging. As you think about what we did in this section, first make sure you understand the argument for each step. Then work on understanding how the steps fit together to create the larger argument.
The next section summarizes all our work with finding equilibrium points for zero-sum games. | 1,361 | 5,390 | {"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} | 4.5 | 4 | CC-MAIN-2023-50 | latest | en | 0.907233 |
http://classblogmeister.com/blog.php?blogger_id=115017&assignmentid=1811 | 1,462,217,138,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860117405.91/warc/CC-MAIN-20160428161517-00037-ip-10-239-7-51.ec2.internal.warc.gz | 54,925,786 | 10,454 | -- Blogmeister
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Article posted February 28, 2008 at 04:09 AM GMT • comment (2) • Reads 80 Key: /=divide #=number -= subtract Hey! My teacher is making us do a division problem!! It also has to be big!! I have to talk you throught the steps too! OK...My problem is:34,424/4=? Ok, so first you have to know how your going to split up the #. Like for 34,000 you could split it up in to 32,000. When a # is not divisible by the # your dividing then you have to find the nearest # that is divisible by the #. Then after that you have to subtract 32,000 from 34,424 which = 2,424! After that you have to the next # divisible by 4..that is 2,400. Then you have to find the next # divisible by 4 which is 24!! Now you have to do this: You have to divide 32/4=8. Then 320/4=80. Then 3,200/44=800. Then 32,000/4=8,000. What I am doing is I am taking a big # and I am making it smaller to divide. Then it gets bigger each time until I get to the # I origianlly had to divide!! Now we have do do the same thing except with different #'s. So now we have to do: 24/4=6. Then 240/4=60. Then 2,400/4=600!! So far we have 8,000 and 600. Now we have to divide 24/4=6!! So now we have to add up all our answers. Which are 8,000 and 600 and 6. The answer is 8,606!!! So thats how you solve a big division problem!! Sincerely, Goldilocks :) Article posted February 28, 2008 at 04:09 AM GMT • comment (2) • Reads 80
Article posted February 27, 2008 at 11:21 PM GMT • comment (1) • Reads 84 This is a wierd way to do division so you can learn how to do long division. 20/4=5 24/4=6 4/4=1 so..... 20,000 2,400 + 4 _________ 22,404 So the problem is 22,404. 20/4=5 200/4=50 2000/4=500 20000/4=5000 24/4=6 240/4=60 2400/4=600 4/4=1 Now all you have to do is add up the answers. 5000 600 + 1 ______ 5601 Article posted February 27, 2008 at 11:21 PM GMT • comment (1) • Reads 84
Article posted February 27, 2008 at 11:07 PM GMT • comment • Reads 120 we have bine lerning about multipalse of 6,4and3? I am not sure about that one if we had lernd about it. So now we have to make a large divison problem and we have to show you how to solve it so on the next blog I am about to make is the divison problem oh yeah bye from contolaf Article posted February 27, 2008 at 11:07 PM GMT • comment • Reads 120
Article posted February 26, 2008 at 07:34 PM GMT • comment • Reads 153 HI HI WORLD, Im going to tell you a divition problem oh yay (NOT) so lets get on with it. 50,424/4=126,10 So we have to find out how much we have left, 50,424-48,000=2,424 2,424-2400=24 48/4=12 480/4=120 4800/4=1200 48000/4=12000 24/4=6 240/4=60 2400/4=600 40/4=10 12000+600+10=126,10 Well thats all bye! Article posted February 26, 2008 at 07:34 PM GMT • comment • Reads 153
Article posted February 26, 2008 at 05:51 PM GMT • comment • Reads 89 This is a wierd way to do division so you can learn how to do long division. 20/4=5 24/4=6 4/4=1 so..... 20,000 2,400 + 4 _________ 22,404 So the problem is 22,404. 20/4=5 200/4=50 2000/4=500 20000/4=5000 24/4=6 240/4=60 2400/4=600 4/4=1 Now all you have to do is add up the answers. 5000 600 + 1 ______ 5601 Article posted February 26, 2008 at 05:51 PM GMT • comment • Reads 89
Article posted February 26, 2008 at 05:51 PM GMT • comment (1) • Reads 88 Professor McGonagall is really getting annoying with the division we are doing:) Today she started us off with 38028/4(She said that the / means divided by so) To make up a really good BIG division problem you have to think of the multiplues of the number that you are dividing by. In this case it's 4 so now you think of the basic facts of 4 The only pairs that I split the number into are 36000/4 2000/4 28/4=7 I got those because 38 can be split into 36 with 2 left over. I got the 20 from the left over 2 from the 38 and the 0 in 38028. Lastly I got the 28 from the last two numbers in 38028. To figure out 36000 I started with 36 and built up. 36/4=9 360/4=90 3600/4=900 36000/4=9000 So to start out with a 9000. Now I do the same with the 2000 and 28. I start with 20 and... 20/4=5 200/4=50 2000/4=500 ...you have a 500 and a 9000=9500 Then you do the same with the 28... 28/4=7 and you have a 9000, 500 and a 7. Add them all up and you get 9507. Do you get it? That is a really long way to division even though long division is doing all that in only a little work. Article posted February 26, 2008 at 05:51 PM GMT • comment (1) • Reads 88
Article posted February 26, 2008 at 05:50 PM GMT • comment (3) • Reads 568 I'm going to show you my division problem. The problem is 38,028 divided by 4 the first step is you find the basic fact in the problem. The second step is when you need to break it up into numbers that can be divided by 4. I subtracted 38,000 from 36,000 and I got 2028. Then I subtracted 2028 from 2000 I got 28. 36,20,28 are all divisible by 4. The third step is when you have to figure out what 36,000 divided by 4 is. What I did is that I turned 36,000 into 36,36 divided by 4 is 9 so you just have to do 36 divided by 4 all the way up to 36,000. Then do that with the other numbers 20 divided by 4 is 5 and 28 divided by 4 is 7 add those answers together and you get 9507 that's what I got. From,Frodo Article posted February 26, 2008 at 05:50 PM GMT • comment (3) • Reads 568 | 2,154 | 6,346 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2016-18 | latest | en | 0.865202 |
http://forums.bistudio.com/showthread.php?126712-Flight-dynamics-(important-issues)/page2 | 1,386,610,402,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163992799/warc/CC-MAIN-20131204133312-00036-ip-10-33-133-15.ec2.internal.warc.gz | 72,069,031 | 22,622 | # Thread: Flight dynamics (important issues)
1. Originally Posted by MD500Enthusiast
agreed to OP, but you forgot the stupidly annoying overly exaggerated ground effect. the ground effect ballooning of the helicopter is still there but it's not as bad as before, i don't think they understand when we tell them it's a cushion of air when we're IGE and that it requires LESS collective/blade pitch to hold a steady hover. i do agree that there is a ballooning effect in helicopter flight during hovering but it's not caused by ground effect. this is the only reason that i have encountered the ballooning effect. say if we do a left pedal turn and do not add more collective which means more power is transferred to the tail rotor to rotate the helicopter and there's less power available for the main rotor so we slowly descend, and when we see that descend we add more power but add too much and now we're gaining altitude and we just keep chasing the collective up and down. same thing for right pedal turns and not decreasing collective because we're using less power on the tail rotor so the main rotor has more power so we raise up a little bit and forget to lower collective, then we lower the collective when we realize we're ascending and we lower too much and we drop down again etc. however none of that is a big springing effect that bounces the helicopter back up 20ft when we're coming to IGE while descending down to the ground. how it should look like is we're descending and maybe 25ft above ground and we slowly increase collective to arrest the descend, as we're coming near the ground but still descending we stop adding collective at around 10ft or so and let the helicopter settling itself into IGE and that extra cushion of air will arrest our descend and we come into a hover. what it does NOT do is bounce us back up into the air unless we've added too much collective in the descending process which would result in us flying straight and level instead. it's not so bad when we're landing into solid ground but when you're trying to land on top of a building it gets ridiculous and totally throws you off your approach. i've landing on top of a building before in real life and there's no bouncing up when you get close to the top of the roof.
Actually, I DID mention the ground effect, I just did not go into as much detail as you did. The reason being, you had already covered it in another forum thread . Read my second paragraph again.
Regarding the torque effects on the tail rotor and the tail rotor on available power, I agree with you that this is not modeled properly, however, I this is a very difficult thing to model (not even Flight Simulator can model it properly, although somehow the Dodo team managed it with some external plug ins) and I don't expect to see that working fully in this game (it is after all, a game, not a simulator), though it WOULD be nice to see. I am trying to focus mainly on the blatantly OBVIOUS things that are annoyingly wrong. Things that contradict their claim of "realistic" flight dynamics.
Yes, it's a game, we know that. But if you claim "realistic" flight dynamics, the program needs to live up to that. I don't expect perfection here, but some of this stuff is very very backwards.
There are many great flying videos on Youtube, including some that I have done. Here's my channel: http://www.youtube.com/user/nightsta1ker1?feature=mhee Watch control movements and how the helicopter responds in some of these and compare it to the game. You can see the obvious differences.
2. Originally Posted by nightsta1ker
Dyssemetry of lift: Modeled on the wrong side of the helicopter. The helicopter seems to have a tendency to roll right, and when I deliberately got over VNE to test retreating blade stall, the helicopter did what it was supposed to, except in the wrong direction.
How to fix it: Get rid of the rolling tendency in forward flight. Real helicopter rotor systems have the ability to flap which nulls the rolling moment. Basically, the dyssimetry is there, but the pilot can't tell because the rotor system is doing the fixing on it's own. Retreating blade stall happens when the dyssimetry of lift becomes too great for the flapping hinge to overcome, THEN you get the rolling/pitch up tendency, but it will roll to the LEFT on counter-clockwise turning rotor systems. So get rid of the rolling tendency below VNE, and then, when VNE is exceeded, have the helicopter roll to the LEFT.
One thing to note here. Real helicopter pilots aren't likely to get in the air and try to test this on a real helicopter. It is a very dangerous thing to test. But when you lose lift because of blade stall on the left side of the rotor disk, the disk doesn't fall on the left. It will fall on the back (90 degree precession thing). Then as the rotor blades come around on the right they are starting to flop back up and have greater attack on the air on the right side. This causes an excess lift on the right but the effect is a rise on the rotor disk in the front. Thus making the situation even worse. The roll to the left comes in when then body of the helicopter is resisting the pitch up effect of the rotor disk. This downward strain on the blades at front cause the disk to rotate down on the left and up stress on the back causes the disk to rotate up on the right. This is left roll you feel. So basically, you get a drastic up and left roll effect and usually serious damage to your rotors blades.
This would be a very UGLY situation to be in.
3. Originally Posted by Bothersome
One thing to note here. Real helicopter pilots aren't likely to get in the air and try to test this on a real helicopter. It is a very dangerous thing to test. But when you lose lift because of blade stall on the left side of the rotor disk, the disk doesn't fall on the left. It will fall on the back (90 degree precession thing). Then as the rotor blades come around on the right they are starting to flop back up and have greater attack on the air on the right side. This causes an excess lift on the right but the effect is a rise on the rotor disk in the front. Thus making the situation even worse. The roll to the left comes in when then body of the helicopter is resisting the pitch up effect of the rotor disk. This downward strain on the blades at front cause the disk to rotate down on the left and up stress on the back causes the disk to rotate up on the right. This is left roll you feel. So basically, you get a drastic up and left roll effect and usually serious damage to your rotors blades.
This would be a very UGLY situation to be in.
It's not that bad... and on some helicopters it's really difficult to get into. Where it becomes dangerous in situations where you are low to the ground, high density altitude, etc where it's hard to recover. The thing about retreating blade stall is, it's self correcting, when the nose pitches up, the helicopter slows down and it recovers itself. There have not been many incedences where this has caused a crash that I am aware of. One such incident happened to a Finnish military MD 500 Defender that was doing high speed NOE training and had the blades stall in a left bank. They rolled right into the ground. Often times, the helicopter does not even have enough power to get going that fast in the first place. You would see it at high density altitudes, where the rotors are less efficient and the blade stall happens at lower airspeeds, like mountain flying, and in descents with high forward airspeed. Also, there is usually a warning first, the helicopter sill start to oscillate on the edge. You really have to be accellerating THROUGH that barrier at a rapid rate to get into full blown RBS.
4. More issues: With a new graphics card that runs everything well so poor hardware performance is no longer an excuse.
First off, everything mentioned before this still applies. I just wanted to tack on some other things I find really irritating and unrealistic.
Pedal turns... What the hell is up with the pedals? They don't want to work! Even with null zone gone and full sensitivity I kick the pedal all the way over and the helicopter slowly, lacadazically turns in whatever direction... Pedals on helicopers are sensitive. You kick that pedal (or fail to kick it in some cases) and you are going to be watching the world spin around you in a blur. The pedals need to be more sensitive. Alot more.
Man they REALLY want to roll over. That stability thing I talked about needs to get fixed pronto.
Control lag. Yes, there is still a control lag. Perhaps those of you not accustomed to flying helos, or even simulators for that matter, don't notice this, but there is a half second lag between my control inputs and the response. This would be a little more realistic in the heavy helicopter, but not the light or even much in the medium.
There is also a weird relationship between the pedal and the cyclic. I cannot quite put my finger on what is going on because it is unpredictable, but when I move the pedals, the helicopter wants to roll and pitch around a bit. Very strange and not lifelike at all. Also, there seems to be a need for left pedal at higher airspeeds... this is backwards from reality. More pedal is needed in a hover, less is needed as the aircraft accelerates.
Also, thought I should note, the jump between trainee and expert is pointless. The helicopter is exponentially harder to fly, but not because it is "more realistic" It's just harder. Nothing that changes from trainee to expert is something that you would see or feel in a real helicopter. I am sure with some time, I could get used to the quirks and learn to fly it, but it's not comparable to a real helo. Too many behaviors are not present, backwards, or otherwise modeled incorrectly. I can see some of this behavior and can recognize what BIS was trying to do, but they misunderstood something critical and got it wrong.
I know this is going to sound rude, and for that I apologize, but in my opinion it's the truth. BIS needs to fix these behavioral issues or they need to take this out of their advertisement of the game: "Authentic Flight Model - Incredibly realistic piloting experience." You failed to deliver.
Now with that said, if there is anything I or any of us real pilots can do to HELP you deliver, please please feel free to pick our brains. We WANT this game to be successful.
5. nightsta1ker really seem to know about this stuff, I just hope BIS will listen and fix the issues...
6. Yes, he does. However I know at least on other real pilot that is not seeing what he is so we need to figure out what is different.
7. I find the medium to be the worst flyier of the 3. The damn right roll is crazy when you add collective.
8. Originally Posted by Daimaju
nightsta1ker really seem to know about this stuff, I just hope BIS will listen and fix the issues...
Agreed, there is a lot of greatly detailed feedback here which would improve the game above any competition.
I'd recommend that these issues should be logged here TOH Community Issue Tracker
That way these are trackable.
9. Originally Posted by Panther353
Yes, he does. However I know at least on other real pilot that is not seeing what he is so we need to figure out what is different.
Chris, it could be something as simple as where we downloaded the game from? Are you using Steam or Sprocket?
Also, what settings do you have? Perhaps I missed something in the setup. I have all my controls configured, with no null zones and medium sensitivity (full sensitivity makes it uncontrollable due to the control lag I am experiencing) with my difficulty on Expert. Are there any other settings that affect the realism?
I'm sorry but I am just absolutely not getting a realistic sensation of helicopter flight from this game. Some of what I am experiencing is just plain WRONG. Backwards, or incorrect, or just plain weird. I would really like to get to the bottom of it. It seems some people are really happy with the game overall, but all of the pilots I have heard from so far are very displeased with the flight characteristics... except you and your friend of course.
Another thing it could be is my controls... they are extremely sensitive. My cyclic alone has 400 increments across its x and y axis (twice to three times what you see on a normal joystick). This makes it great for simming, but maybe not so great for TOH? I will try flying around with my Logitech next time and see if it changes anything for me, but I doubt it will. Controls are controls.... they work just fine in X-plane and FSX. Why wouldn't they work in TOH? I will try it anyway just to see.
10. actually i disagree about medium helicopter to be the worse, i think it's the easiest one to fly. it actually descends some what near what a real helicopter would descend like if the throttle was full down. you can see over torquing on the helicopter and it starts to shake and rattle constatly until the point where it stops flying all together if you're not easy on the controls and just throw the helo around i.e using lots of rudder to assist in high speed turns adds a lot of stress on your tail rotor and transmission or raising the collective from idle to full power really fast.
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• | 2,916 | 13,417 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-48 | latest | en | 0.962867 |
https://rpg.stackexchange.com/questions/153293/how-many-hit-points-does-the-ua-battle-smith-artificers-iron-defender-have?noredirect=1 | 1,631,895,288,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780055684.76/warc/CC-MAIN-20210917151054-20210917181054-00295.warc.gz | 541,828,529 | 43,812 | How many hit points does the UA Battle Smith Artificer's Iron Defender have?
The Iron Defender's stat block for the Battle Smith Artificer (from Unearthed Arcana: The Artificer Returns), under "hit points" says the following:
equal to five times your level in this class + your Intelligence modifier + the Iron Defender’s Constitution modifier
I could read this two ways:
Way 1:
equal to five times (your level in this class) + your Intelligence modifier + the Iron Defender’s Constitution modifier
Or way 2:
equal to five times (your level in this class + your Intelligence modifier + the Iron Defender’s Constitution modifier)
Which of these readings is correct?
• – V2Blast
Aug 9 '19 at 22:06
(5 × Level) + your Int + defenders Con
Interpreting equations like this isn't defined by the game rules and so we use the english meaning. Unfortunately, english (and most other languages AFAIK) isn't very good at it, which is a big reason for using mathematical symbolics. As a good rule; take the simplest reading as you scan through. This gives:
$$5 \times \text{Level} + \text{Your Int} + \text{Defender's Con}$$
which for anyone not fully comfortable with order of operations equals
$$(5 \times \text{Level}) + \text{Your Int} + \text{Defender's Con}$$
If it were to mean
$$5 \times ( \text{Level} + \text{Your Int} + \text{Defender's Con} )$$
it should (in my opinion) been worded as (italics to emphasise change):
equal to five times the sum of your level in this class + your Intelligence modifier + the Iron Defender’s Constitution modifier
As additional support, the same problem of having such equations in D&D rules have appeared for Arcane Ward where lead designer Jeremy Crawford has chimed in on twitter (not official ruling, but gives what the intended meaning is) with the statement that
The ward has hit points equal to twice your wizard level + your Intelligence modifier.
should be read as:
its hit point maximum equals your Intelligence modifier plus twice your wizard level.
Meaning to be read as
$$(2 \times \text{Level}) + \text{Int}$$
similar to that above.
As an addendum: the final version of the subclass clarifies the calculation as:
equal the steel defender’s Constitution modifier + your Intelligence modifier + five times your level in this class
and this is how the feature is implemented in D&D Beyond. (Thanks to @Garret Rooney and @V2Blast)
• Please Email My Dad A Shark... Aug 10 '19 at 0:05
It is likely interpretation 1, due to the initial HP difference between the two interpretations.
Assuming you go to level 20, here is the range of HP the defender could have under the 2 interpretations, with your Intelligence score assumed to be 18 (+4 modifier).
Interpretation 1: (5× Artificer level) + Intelligence modifier (4) + Iron Defender Constitution modifier (2) = 21 (5×3 + 4 +2) HP at level 3, to 106 (5×20 + 4 + 2) HP by level 20.
Interpretation 2: (5× Artificer level + Intelligence modifier [4] + Iron Defender Constitution modifier [2]) = 45 (5×[3+4+2]) HP at level 3, to 130 (5×[20+4+2]) HP by level 20.
As seen by interpretation 2, its HP would likely be higher than the Artificer themselves at first, and would be equal to an average of 2d10 + 4 HP per level, higher than any class.
The way it is printed, it's HP are calculated by (lvl×5) + 2(CON bonus) + (players INT bonus). As to whether the Steel Defender should have more HP then the Artificer, it's a guardian - it darn well should have more HP than what it's protecting. The fact that it might not seems off.
The formula looks off because it gets its CON bonus to HP only once, not per level like every other creature in the game. The only explanation I can come up with is that its hit dice are not equal to the Artificers level, like it seems it should be. It's a zero hit dice creature, but that just doesn't feel right.
• Why are you saying it gets twice its con modifier? I'm failing to see where that comes from. In general, it is a really good idea to quote (and cite) the rules text you are basing yourself of (and explain your basis). That way readers can see and understand why your answer is correct, or more accurately correct any missteps. Feb 2 '20 at 15:22
• Also, Welcome to RPG.SE! Take the tour if you haven't already and see the help center or ask us here in the comments (use @ to ping someone) if you need more guidance. Good Luck and Happy Gaming! Feb 2 '20 at 15:22
When the Artificer was officially released, they reworded the formula for the Iron Defender's Hit Points as follows:
2 + your Intelligence modifier + 5 times your artificer level (the defender has a number of Hit Dice [d8s] equal to your artificer level)
The Iron Defender has a Constitution stat of 2, so this is equivalent to "way 1" as stated in the question.
The wording makes me say it is equal to your level times x 5 + your Intelligence modifier + the Iron Defender's Constitution modifier.
If it were the second option, it would have been worded more like this:
equal to your level times and your Intelligence modifier x 5 + the Iron Defender's Constitution modifier.
• I'm not saying you're wrong, but you should add why you (think you) are right? Rules quotes, citations, logic, etc. Aug 9 '19 at 22:10
• -1. One could assume this, but I'd appreciate it if you backed your answer up with a source. Aug 9 '19 at 22:10
• @Someone_Evil the way it is worded is what makes me say this. If it was the second choice, Unearthed Arcana would have said something more along the lines of 'equal to five times your level in this class, and your Intelligence modifier + the Iron Defender’s Constitution modifier'. Aug 9 '19 at 22:14
• Then explain why the wording makes you come to that conclusion (and say it in the answer). Aug 9 '19 at 22:15
• Gonna second the other recommendations in the comments. Personally speaking, I think this interpretation is correct, but if it were as straight-forward as writing it as a math expression and PEMDAS-ing it, I don't think there would be as much confusion as necessary to generate the question in the first place. So explaining why you think this is the correct explanation (and make sure you put that explanation in the answer itself) would vastly improve this answer. Aug 9 '19 at 22:22 | 1,539 | 6,263 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 4, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2021-39 | longest | en | 0.895273 |
https://metric2011.wordpress.com/2017/05/16/notes-of-yves-cornuliers-cambridge-lecture-16-05-2017/ | 1,500,914,949,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424889.43/warc/CC-MAIN-20170724162257-20170724182257-00614.warc.gz | 681,621,462 | 36,896 | ## Notes of Yves Cornulier’s Cambridge lecture 16-05-2017
Commensurating actions of groups of birational transformations
Joint work with Serge Cantat.
1. Birational geometry
This is more than a century old. Let ${X,Y}$ be irreducible varieties over ${{\mathbb C}}$. A rational map ${X\rightarrow Y}$ is a regular (given by the ratio of two polynomials) function between Zariski dense open subsets, modulo coincidence on a dense open subset. A birational map ${X\rightarrow Y}$ is a regular function between Zariski dense open subsets which is an isomorphism of varieties, again modulo coincidence on a dense open subset.
Example. The map ${{\mathbb C}^2\rightarrow P^1\times P^1}$, ${(x,y)\mapsto ((x:1);(y:1))}$ is birational. Therefore ${P^1\times P^1}$ and ${P^2}$ are birationally equivalent. Therefore a rational map comes in several different models. One often needs to consider all of them simultaneously.
A set of rational functions on ${{\mathbb C}^d}$ forms a birational map to ${{\mathbb C}^d}$ iff it is algebraically free over ${{\mathbb C}}$ and generates the field ${{\mathbb C}(t_1,\ldots,t_d)}$.
Notation. ${Bir(X)}$ is the group of birational equivalences of ${X}$, i.e. the automorphism group of the field ${{\mathbb C}(X)}$ of rational functions on ${X}$.
Example. It turns out that ${Bir({\mathbb C})=Bir(P^1)=Aut(P^1)=PGl(2,{\mathbb C})}$. On the other hand, ${Bir(P^2)}$ contains ${Aut(P^2)=PGl(3,{\mathbb C})}$ but also ${Aut(P^1\times P^1)=PGl(2,{\mathbb C})\times PGl(2,{\mathbb C})}$. However, ${Bir(P^2)}$ contains an abelian group of infinite rank, the set of maps ${(x,y)\mapsto (x+P(y),y)}$, where ${P\in{\mathbb C}[t]}$.
The difficulty of ${Bir}$ is that it comes with quasi-actions which are not quite actions. Some results have been obtained in dimension 2, nearly nothing is known in higher dimensions. For instance, one does not know which finite groups occur in ${Bir(P^3)}$, and it is still possible (but unlikely) that every finite group occurs in ${Bir(P^4)}$.
2. Commensurating actions
2.1. Commensurated sets
Say a group ${G}$ acting on a set ${E}$. Say a subset ${A\subset E}$ is commensurated if ${\forall g\in G}$, ${\ell_A(g):=|A\Delta gA|<\infty}$ (Stallings, Dunwoody,…). Say that ${A}$ is transfixed if there exists a ${G}$-invariant subset ${A'}$ such that ${|A\Delta A'|<\infty}$. It turns out that it is equivalent to ${\ell_A}$ being bounded (proved by several authors in the 1960’s, with good bounds finally obtained by W. Neumann).
Example. Let ${G}$ act on a Schreier graph ${G/H}$. Then a subset is commensurated iff its boundary is finite. A subset is transfixed iff it is finite or its complement is finite. In particular, there exists a commensurated, non-transfixed subset iff the Schreier graph has at least 2 ends.
Example. Let ${G}$ act on set ${E}$, and ${A\subset E}$. Let ${\ell_A^2(E)=1_A+\ell^2(E)\subset{\mathbb R}^E}$ is an affine Hilbert space. It is ${G}$ invariant iff ${A}$ is commensurated. This provides an affine isometric action of ${G}$.
Example. Cubulation. Define a graph whose vertices are subset ${B\subset E}$ such that ${|B\Delta A|<\infty}$. Put an edge between ${B}$ and ${B\cup\{b\}}$. This a median graph, therefore an action on a ${CAT(0)}$ cube complex arises from it. Conversely, every action of ${G}$ on a ${CAT(0)}$ cube complex induces a commensurating action on the set ${E}$ of half-spaces, with ${A}$ being the subset of half-spaces contaning a fixed vertex. This gives huge cube complexes or actions, far from being optimal.
2.2. Property FW
Definition 1 Say ${G}$ has property FW if for every commensurating action of ${G}$ on ${(E,A)}$, ${A}$ is transfixed.
Say ${G}$ has property FW relative to subgroup ${H}$ if for every commensurating action of ${G}$ on ${(E,A)}$, ${A}$ is transfixed by ${H}$.
Example. For every cyclic distorted subgroup ${H, ${G}$ has property FW relative to ${H}$ (Haglund).
This can be used to prove that ${Sl(d,{\mathbb Z})}$, ${d\geq 3}$, or ${Sl(d,{\mathbb Z}[\sqrt{2}])}$, ${d\geq 2}$, has property FW (using bounded generation by exponentially distorted unipotents).
Property (T) implies property FW (use ${\ell^2}$ action).
Proposition 2 or any group ${G}$, the following are equivalent:
1. ${G}$ has property FW.
2. Every isometric action of ${G}$ on a ${CAT(0)}$ cube complex has a fixed point.
3. Every isometric action of ${G}$ on a median graph has a finite orbit.
4. If ${G}$ is finitely generated, every Schreier graph of ${G}$ has at most 1 end.
5. For all actions of ${G}$ on sets ${E}$, ${H^1(G,{\mathbb Z}^{(E)})=0}$ (functions with finite support).
3. Birational groups
Fix a projective variety ${X}$. Let ${Hyp(X)}$ be the set of irreducible hypersurfaces of ${X}$. Every subgroup ${G acts on a set ${E}$ which commensurates ${A=Hyp(X)}$ as follows. The crucial point is the following. If ${X}$ is smooth (in fact, normal is sufficient) and ${f:Y\rightarrow X}$ is a birational morphism, inverse images of hypersurfaces are well defined: among the several subvarieties whose union is ${f^{-1}(H)}$, there is exactly one which maps onto ${H}$. Therefore one can define ${E=\widetilde{Hyp}(X)}$ as the inverse limit of ${Hyp(Y)}$ over all birational morphisms ${Y\rightarrow X}$. Birational self-maps of ${X}$ act on ${E}$ and the subset ${A=Hyp(X)}$ is commensurated.
If ${G}$ acts by automorphism of a Zariski open set of ${X}$, then ${G}$ transfixes ${A}$. This is also true for pseudo-automorphisms, i.e. isomorphisms between proper Zariski open subsets of ${X}$.
Theorem 3 Let ${G\subset Bir(X)}$. Then ${G}$ transfixes ${Hyp(X)}$ iff the ${G}$ action on ${X}$ is conjugate to an action by pseudo-automorphisms.
Example. The monomial action of ${Sl(n,{\mathbb Z})}$ on ${{\mathbb C}^{n^2}}$ raises each coordinate to a power given by a matrix coefficient. These are pseudo-automorphisms.
Theorem 4 Let ${X}$ have dimension 2. Let ${G\subset Bir(X)}$ have property FW. Then the action of ${G}$ on ${X}$ is conjugate to an action by automorphisms on some variety ${Y}$, with a short list of exceptions.
This fails in higher dimensions (see the monomial action). There is a corresponding statement for groups with relative property FW. | 1,834 | 6,212 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 115, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-30 | longest | en | 0.857964 |
http://www.amigosdelmoyate.es/plate/ce9920ac9562078.html | 1,623,917,048,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487629632.54/warc/CC-MAIN-20210617072023-20210617102023-00542.warc.gz | 52,337,630 | 11,730 | # ss40steel plate for bridge thickness
• ## ss40steel plate for bridge thickness
### results for this questionFeedbackBOLTED FIELD SPLICES FOR STEEL BRIDGE FLEXURAL
thickness of each web splice plate should be at least one-half the thickness of the thinner web at the splice plus 1/16 of an inch,and must be equal to or greater than the minimum permitted thickness of 5/16-inches for structural steel (AASHTO LRFD Article 6.7.3). results for this questionHow much does a steel plate weigh?How much does a steel plate weigh?The weight of a 1 inch thick 36 (3 ft) x 96 (8 ft) steel plate with weight 40.8 lb/ft2 (from table above) can be calculated as W = (40.8 lb/ft2) (3 ft) (8 ft) = 979 lb Calculating Steel Plate WeightSteel Plates - Size Weight - Engineering ToolBox
### results for this questionWhat is a Trench plate Weld?What is a Trench plate Weld?Steel trench plate deformation may occur during loading,but if a steel plate is deformed without loading to at least 0.5 inch per 10 feet in length the plate shall be removedand replaced.626.2.2.4 Welding.To hold multiple plates tightly together,plates may require a 6 inch long tack weld for stability.Section 626 Temporary Steel Plate Trench Bridging results for this questionWhat is the weight of hot rolled steel?What is the weight of hot rolled steel?Steel sheets and plates are typical stocked as size 2000 mm x 3000 mm.Weight per sq.meter of hot rolled mild steel plates can be calculated as W = 7.85 t (2)Steel Plates - Size Weight - Engineering ToolBox stainless steel sheet 12 gauge
304 Stainless Steel Sheet,#4 Brushed Finish,Annealed,ASTM A240/ASME SA240,0.03 Thickness,24 Width,24 Length,22 Gauge 4.5 out of 5 stars 76 \$25.00 \$ 25 .00
### 10 Bridge Bearings Design Guide for Bridges for Service
However,in plates of the thickness cur- rently used,this behavior does not occur until the load has reached 5 to 10 times its design value,so plate fracture seldom controls design.Plates could actually be thinner,but they are typically sized in practice to keep them12345NextBOLTED FIELD SPLICES FOR STEEL BRIDGE FLEXURALthickness of each web splice plate should be at least one-half the thickness of the thinner web at the splice plus 1/16 of an inch,and must be equal to or greater than the minimum permitted thickness of 5/16-inches for structural steel (AASHTO LRFD Article 6.7.3).Allowable load on steel road plate - Structural Feb 09,2007·(The 1 thickness is based on a square plate- 8'x8',by the way).That 3 sounds too high.If this is only supported at the ends,not at the sides,you may have sizable deflection as well.RE Allowable load on steel road plate apsix (Structural) 8 Feb 07 21:37.
### Author Bryan GalloupPlate Specification Guide 2015-2016
Plate Types Carbon HSLA Alloy Commercial Grades,Military Armor,Aircraft Quality,Tool Steel Duracorr&(12% Chromium Stainless) Cut-To-Length Plate Floor Plate (Sure-Foot®) Range Most Complete Size Range Thickness Carbon/HSLA 3/16 to 30 Alloy 1/8 to 15 Widths up to 195 wide Lengths up to 1525 and longerBridge Plates on Martin GuitarsBridge Plates on Martin Guitars The bridge plate is a small,thin block of wood,usually made of mahogany,maple,or rosewood,roughly the same size as the bridge,and glued inside the top,opposing the bridge,to stiffen the top and provide a hard surfaceCover Plates - bgstructuralengineeringJul 30,2011·Figure 8.6.2.1 shows a typical bridge beam with cover plates.Figure 8.6.2.2 shows a typical drawing of a W section with cover plates. t is the thickness of the cover plates.For unsymmetrical plates (i.e.the cover plates are of different sizes or a cover plate is applied to only one flange),the Z for the whole section must be recomputed
### Features Application - Steel Plate - Products - POSCO
Steel Plate.Steel plates are hot rolled products that have a relatively high thickness.Products having a thickness of over 4.5 mm are classified as steel plates,and those having a thickness of over 100 mm are called thick steel plates.Slabs of steel plates made through continuous casting are tailored to the dimensions specified by the File Size 790KBPage Count 80Section 626 Temporary Steel Plate Trench BridgingJan 18,2013·the direction of travel,steel plates shall have a minimum thickness of 1 inch.626.2.2.2 Structural Design.For trenches and excavations with spans greater than .four (4.0) feet as measured in the direction of travel,a structural design,to include plate dimensions,thickness,ASTM A 36 or ASTM A 572 steel grade,and minimum shoring orFlat Plates Stress,Deflection Equations and Calculators Engineering Calculators Menu Engineering Analysis Menu.Flat Plates Stress,Deflection Equations and Calculators The follow web pages contain engineering design calculators that will determine the amount of deflection and stress a flat plate of known thickness will deflect under the specified load and distribution..Many of the stress and deflection equations and calculators referenced from
### G 9.1 - Steel Bridge Bearing Design and Detailing
The sole plate should extend transversely beyond the edge of the bottom flange of the girder a minimum of 1 (25 mm) on each side.The minimum thickness of the sole plate should be 1½ (37mm) after beveling if the field weld is directly over the elastomer.Galvanized Pipe SS40 - Fence-itChain Link Catalog Visit our complete line of Quality Chain Link Fence Products below.For the hard to find chain link parts or bundle discounts call (800)878-7829Guidance Note Through thickness properties No.3SCI P185 Guidance notes on best practice in steel bridge construction 3.02/1 Revision 2 Scope This Guidance Note gives advice on the need for steel with improved through thickness properties and on the selection of an appro-priate quality class where such steel is needed.Steel is an anisotropic material Steel plate and sections are
### High Performance Steels - JISF
Longitudinally-proled (LP) steel plate Maximum difference in thickness 25~30 mm,maximum taper gradient 4 mm/m,total length 6~25 m 32 33 High-strength steel wire for bridge cables ST1770 (tensile strength 1,770 N/mm 2 and over),ST1960 (tensile strength 1,960 N/mm 2 and over) 34 36Is Your Flattop's Bridge Plate up to Snuff? Premier GuitarJan 11,2021·Bridge-plate sizes,for example,shifted from small,vintage-style footprints and thinner profiles to a footprint that was twice the size and double the thickness.By the 1970s,some makers moved to plywood bridge plates in an attempt to remedy some instruments structural issues.JIS G3101 SS 400 steel plate standard,specification and SS400steel plate sheet is a kind of steel for general structure,and it is used in ship,house,bridge making and so on.This kind of steel plate sheet SS400 has some features,like its heat condition,excellent freezing,even structure with pressure resistance and its good mechanical properties.
### Lamalar Tearing of steel plates loaded transverse - Bridge
Oct 13,2008·Steel plate lamination processes accumulated deffects in the middle of the steel plate and created points where internal tears could create if plates were loaded transversally to the plate thickness.These days,steel quality is much better and the issue is not so prevalent.Line Rail Clamps Chain Link Fence PartsImages of Ss40 Steel Plate For Bridge Thickness imagesPeople also askWhat is the thickness of a steel plate?What is the thickness of a steel plate?Plates shall be placed perpendicular or parallel to the direction of travel.In all situations,the longitudinal edges of the steel plates shall not be in the wheel path.626.2.2.1 Thickness.For trench widths less than or equal to 4.0 feet as determined in the direction of travel,steel plates shall have a minimum thickness of 1 inch.Section 626 Temporary Steel Plate Trench BridgingPlate Specification Guide 2015-20162015-2016 Plate Steel Specification Guide Page 4 ASTM Specifications continued SPECIFICATION A204 Grade A A204 Grade B A204 Grade C A242s Type 1** Type of Steel Alloy Alloy Alloy Carbon Requirements for Delivery A20 A20 A20 A6 Tensile Strength (ksi) 65/85 70/90 75/95 70 Min.to ¾ incl.; 67 Min.over ¾-1½ incl.;
### Preferred Practices for Steel Bridge Design,Fabrication
Preferred Practices for Steel Bridge Chapter 1Introduction Design,Fabrication,and Erection .1-1 June 2019 .1.Introduction.1.1.Overview .This document provides guidance to help steel bridge designers working on TexasPrevious123456NextGalvanized Pipe SS40 - Fence-itChain Link Catalog Visit our complete line of Quality Chain Link Fence Products below.For the hard to find chain link parts or bundle discounts call (800)878-7829STEEL PLATE TEMPORARY DECK BRIDGING PLANplate thickness table connection and steel see temporary deck plate temporary deck bridging,thickness table see steel plate the registered civil engineer for the project is responsible for the selection and proper application of the component design and any modifications shown.similar for regular holes.plate departure.countersink details
### Section 626 Temporary Steel Plate Trench Bridging
Jan 18,2013·the direction of travel,steel plates shall have a minimum thickness of 1 inch.626.2.2.2 Structural Design.For trenches and excavations with spans greater than .four (4.0) feet as measured in the direction of travel,a structural design,to include plate dimensions,thickness,ASTM A 36 or ASTM A 572 steel grade,and minimum shoring orStainless Steel Gauge Chart Cumberland Diversified MetalsCumberland stocks sheet,plate and coil,both prime and secondary.Contact Us.We Buy Surplus and Excess Metals.If you have material to sell,talk to us first.Contact Us.About Cumberland.Cumberland Diversified Metals (CDM) was founded in 1985.The company has grown exponentially in product offerings,market share,and most importantly our Steel Plate Expansion Joint Systems D.S.BrownSteel Plate Expansion Joint Systems,such as finger joint assemblies and sliding plate and armor joint systems,are still specified on many bridge projects due to proven long-term structural performance.These joint systems are also convenient to install on bridge rehabilitation projects requiring a shallow joint depth and/or staged construction.
### Steel Plate Expansion Joint Systems D.S.Brown
Steel Plate Expansion Joint Systems,such as finger joint assemblies and sliding plate and armor joint systems,are still specified on many bridge projects due to proven long-term structural performance.These joint systems are also convenient to install on bridge rehabilitation projects requiring a shallow joint depth and/or staged construction.Steel Plates Requirements Used in Connection with Steel plates mu st be able to withstand H-20 traffic loading without any movement. Steel plates shall be fabricated to meet ASTM A36 steel requirements. When two or more of plates are used,the plates shall be tack welded together at each corner to reduce or eliminate vertical movement.Steel Plates for Bridge Use and Their ApplicationK 20 (Plate thickness) PCM 0.29 PCM 0.27 PCM 0.20 Conventional 570 N/mm2 class steel Steel for easy fabrication Plate thickness (mm) Preheat temperature,T p (° C) Fig.1 Relation between plate thickness and preheat tem-perature for 570 N/mm2 class steel Table 1 Examples of chemical composition and mechanical properties of steel plate produced
### Steel-girder bridge with variable flange thickness
Mar 27,2019·To obtain the variable flange thickness desired,users should assign non-prismatic girder sections to the bridge deck section.Create new model from template.Select File > New Model > KN,m,C units > Quick Bridge template.Set span lengths to '30; 30',and select Steel Girder as the bridge deck-section type.Define non-prismatic steel girder Tele bridge plate thickness? Page 2 The Gear Page·I think the added sustain is the bridge plate thickness.Groberts.Gold Supporting Member.Messages 3,670.Aug 9,2017 #25 Here is a pic of my Callaham Vintage Bridge assembly with the enhanced saddles.I love how it looks,but moreover,the build quality and tone is incredible!The rosewood bridge plate years.avoid them? - The Apr 15,2007·IMO,the bridge plate material would not slo me down if i was interested.happy huntin.nod aka don Edit:I seem to recall a 1986 D-28 that was remakable,a bluegrass circle,large room,barbeque restraunt type commotion,lots of kids and the like,and that was the D-28 of the note a memorable one.Life is not a journey to the grave with the intention of
### Top 4 Types of Steel Bridges (With Examples)
The types are 1.Rolled Steel Beam Bridges 2.Plated Beam Bridges 3.Plate Girder Bridges 4.Trussed Girder Bridges.Type # 1.Rolled Steel Beam Bridges This is the simplest type steel bridge having RSJ as the girder and steel trough plate filled with concrete or reinforced concrete slab as the bridge deck as shown in Fig.14.1.Trench Plate Tab Data - National Trench Safetyplate size (ft x ft) / weight (lb) hs-20-44 loading nts steel crossing plates tabulated data 15955 w.hardy road,suite 100 houston,tx.77060 toll free (800) 234-9244 fax (832) 200-0989 ntsafety trench plate ac must have stable trench walls section a-a ac span + note 4 note 4 span note 3 note 4 section a-a must have stable trench Updates to the AASHTO Design SpecificationIntroduction of the Unified Effective Width Approach The nominal compressive resistance,Pn,is obtained by multiplying Fcr based on the gross cross-sectional area by an effective area,Aeff.Aeffis generally computed as the summation of effective areas of the cross-section based on reduced effective widths,be,for each slender element in the cross-section (Article 6.9.4.2.2a).
### Wudtone bridge plate Thicker than OEM trem plate
Mar 14,2014·Can anyone tell me if the Wudtone bridge plate is thicker than a typical vintage style OEM trem plate? I'm thinking of getting one and thicker would be a good thing.Stratoblaster,Mar 13,2014 #1.buddhuu Strat-Talker.Messages 489.Joinedsole plate design - Bridge engineering - Eng-TipsNov 30,2004·I am attempting to determine the required thickness for a sole plate to be used at the pier bearing for a 2 span continuous steel multi-girder bridge (177-177).We have designed a circular laminated elastomeric bearing pad that will work at the location.Typically we do not do sole plate | 3,292 | 14,246 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2021-25 | longest | en | 0.878825 |
http://www.gurufocus.com/term/Dividends+Per+Share/BBY/Dividends%2BPer%2BShare/Best%2BBuy%2BCo.%252C%2BInc | 1,484,595,129,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560279248.16/warc/CC-MAIN-20170116095119-00306-ip-10-171-10-70.ec2.internal.warc.gz | 483,505,733 | 28,349 | Switch to:
GuruFocus has detected 3 Warning Signs with Best Buy Co Inc \$BBY.
More than 500,000 people have already joined GuruFocus to track the stocks they follow and exchange investment ideas.
Dividends Per Share
\$1.07 (TTM As of Oct. 2016)
Best Buy Co Inc's dividends per share for the three months ended in Oct. 2016 was \$0.28. Its dividends per share for the trailing twelve months (TTM) ended in Oct. 2016 was \$1.07. Its Dividend Payout Ratio for the three months ended in Oct. 2016 was 0.47. As of today, Best Buy Co Inc's Dividend Yield is 2.54%.
During the past 12 months, Best Buy Co Inc's average Dividends Per Share Growth Rate was 21.60% per year. During the past 3 years, the average Dividends Per Share Growth Rate was 11.70% per year. During the past 5 years, the average Dividends Per Share Growth Rate was 8.30% per year. During the past 10 years, the average Dividends Per Share Growth Rate was 9.50% per year. Please click Growth Rate Calculation Example (GuruFocus) to see how GuruFocus calculates Wal-Mart Stores Inc (WMT)'s revenue growth rate. You can apply the same method to get the average dividends per share growth rate.
During the past 13 years, the highest 3-Year average Dividends Per Share Growth Rate of Best Buy Co Inc was 20.70% per year. The lowest was 0.00% per year. And the median was 5.10% per year.
Definition
Dividends paid to per common share.
Explanation
1. Dividend payout ratio measures the percentage of the companys earnings paid out as dividends.
Best Buy Co Inc's Dividend Payout Ratio for the quarter that ended in Oct. 2016 is calculated as
Dividend Payout Ratio = Dividends Per Share (Q: Oct. 2016 ) / EPS without NRI (Q: Oct. 2016 ) = 0.28 / 0.6 = 0.47
During the past 13 years, the highest Dividend Payout Ratio of Best Buy Co Inc was 0.40. The lowest was 0.13. And the median was 0.19.
2. Dividend Yield measures how much a company pays out in dividends each year relative to its share price.
Best Buy Co Inc Recent Full-Year Dividend History
Amount Ex-date Record Date Pay Date Type Frequency
0.2802016-12-062016-12-082016-12-29Cash Dividendquarterly
0.2802016-09-092016-09-132016-10-04Cash Dividendquarterly
0.2802016-06-102016-06-142016-07-05Cash Dividendquarterly
0.2802016-03-152016-03-172016-04-07Cash Dividendquarterly
Best Buy Co Inc's Dividend Yield (%) for Today is calculated as
Dividend Yield = Most Recent Full Year Dividend / Current Share Price = 1.12 / 43.83 = 2.56 %
Current Share Price is \$43.83.
Best Buy Co Inc's Dividends Per Share for the trailing twelve months (TTM) ended in Today is \$1.12.
During the past 13 years, the highest Dividend Yield of Best Buy Co Inc was 5.76%. The lowest was 0.53%. And the median was 1.87%.
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Related Terms
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Best Buy Co Inc Annual Data
Feb07 Feb08 Feb09 Feb10 Feb11 Feb12 Jan13 Jan14 Jan15 Jan16 Dividends Per Share 0.36 0.46 0.54 0.56 0.58 0.62 0.66 0.68 0.72 0.92
Best Buy Co Inc Quarterly Data
Jul14 Oct14 Jan15 Apr15 Jul15 Oct15 Jan16 Apr16 Jul16 Oct16 Dividends Per Share 0.17 0.19 0.19 0.23 0.23 0.23 0.23 0.28 0.28 0.28
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GuruFocus Affiliate Program: Earn up to \$400 per referral. ( Learn More) | 1,023 | 3,537 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-04 | longest | en | 0.955733 |
http://www.chegg.com/homework-help/questions-and-answers/p-excuse-poor-drawing-serves-s-purpose-160-i-need-analyze-circuit-finding-voltage-current--q2270033 | 1,369,097,883,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368699632815/warc/CC-MAIN-20130516102032-00078-ip-10-60-113-184.ec2.internal.warc.gz | 388,052,970 | 8,542 | ## VERY simple combination circuit problem
<p>First of all, please excuse my poor drawing, but it serves it's purpose!  I need to analyze this circuit, finding voltage and current for each of the three resistors, and so far, this is how I went about it:  First I made an equivalent Parallel circuit out of the 10 and 33 ohm resistors, so I had a 43 and 100 ohm resistor set in parallel.  Then since voltage is the same across parallel, the <span style="background-color: #ffffff;">voltage across the 100 ohm resistor is 5. </span> I used Ohms law on the 43 ohm resistor to get current.  I=V/R=5/43=.116 Amps.  Since current is same in series, then<span style="background-color: #ffffff;"> the 10 ohm and 33 ohm both get .116 amps current.  I used Ohms law again to get voltage for the 10 ohm and 33 ohm....V=IR=10(.116)=1.16 V and 33(.116)=3.828 V.    So, in summary:</span></p>
<p><span style="background-color: #ffffff;">The 10 ohm resistor has 1.16 V and .116 A, The 33 ohm resistor has 3.828 V and .116 A, and the 100 ohm resistor has 5 V and .05 A?   I have a feeling I am doing something wrong here.  Could someone please point me in the right direction?  Thanks.</span></p> | 374 | 1,256 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2013-20 | longest | en | 0.848959 |
https://discourse.julialang.org/t/how-to-convert-the-python-code-below-to-julia/92151 | 1,675,158,535,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499857.57/warc/CC-MAIN-20230131091122-20230131121122-00384.warc.gz | 230,324,028 | 6,764 | # How to convert the python code below to julia?
Hello, good afternoon,
How to convert the below code from python to julia?
``````from numpy import array
# Transformation 1
i_hat1 = array([0, 1])
j_hat1 = array([-1, 0])
transform1 = array([i_hat1, j_hat1]).transpose()
# Transformation 2
i_hat2 = array([1, 0])
j_hat2 = array([1, 1])
transform2 = array([i_hat2, j_hat2]).transpose()
# Combine Transformations
combined = transform2 @ transform1
# Test
print("COMBINED MATRIX:\n {}".format(combined))v = array([1, 2])
print(combined.dot(v))
# [-1, 1]
``````
Could you tell us what this code is meant to accomplish, mathematically? Probably more people here are conversant in math than python.
Does the code form two matrices and then multiply them?
3 Likes
Without knowing what its good for and “best” Julia style, this would be a 1:1 conversion:
``````i_hat1 = [0 1]
j_hat1 = [-1 0]
transform1 = [i_hat1;j_hat1]'
i_hat2 = [1 0]
j_hat2 = [1 1]
transform2 = [i_hat2;j_hat2]'
combined = transform2 * transform1
v = [1, 2]
combined * v
``````
2 Likes
Yes. Is related to algebra linear and matrix multiplication, linear transformation. I know Python, and I’m trying to get the same studies with julia.
I will try it. Did transform1 = [i_hat1, j_hat1]’ but got issues as the result is a Adjoint.
That worked. The result is correct. The issue was in concatenating the two vectors. [x, y] with comas. Using “;” solves the issue. Thank you all
The `;` is doing vertical concatenation of the two row vectors `i_hat1` and `j_hat1`, followed by transposition using`'` as a suffix.
Be careful with the last line. You have used numpy’s dot function which does quite a lot of things depending on the input type. In this case it should give the sum product over the last axes. I have just written a standard matrix-vector product.
1 Like
I had read an thread about the dot with python and was aware of the issues you told, but in some calc with vectors, “,” did the trick. For matrix, needed vertical concatenation as you show. We must pay attention in this detail.
From vector to matrix doing the transpose operation, as the version in python did and the book says to do; ihs and jhs are vector, then better implement they as vectors. Python and numpy are hidding lots of things.
``````begin
ih1 = [0, 1];
jh1 = [-1, 0];
ih2 = [1, 0];
jh2 = [1, 1];
transform1 = [ih1'; jh1']';
transform2 = [ih2'; jh2']';
result = transform2 * transform1;
v1 = [1, 2];
result * v1
end
``````
Try
``````ih1 = [0, 1]
jh1 = [-1, 0]
ih2 = [1, 0]
jh2 = [1, 1]
transform1 = [ih1 jh1]
transform2 = [ih2 jh2]
``````
Should be more efficient and natural for column major arrays.
4 Likes
thats nice solution.
I still think the question and the replies would benefit from an explanation of what the OP is trying to accomplish, mathematically. The way `transform1` and `transform2` are built is unnecessarily complicated. You can get the same in four lines with
``````A = [0 -1; 1 0]
B = [1 1; 0 1]
C = B*A
C*[1;2]
``````
or in one line with
``````[1 1; 0 1] * [0 -1; 1 0] * [1; 2]
``````
The book did this way because we need to observe the basis and the vector that is transformed. Just to be more didact. | 976 | 3,214 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2023-06 | latest | en | 0.89151 |
http://www.ques10.com/p/16477/theoretical-computer-science-question-paper-may--2/ | 1,558,323,936,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232255536.6/warc/CC-MAIN-20190520021654-20190520043654-00510.warc.gz | 323,830,584 | 7,647 | Question Paper: Theoretical Computer Science : Question Paper May 2016 - Computer Engineering (Semester 4) | Mumbai University (MU)
0
## Theoretical Computer Science - May 2016
### Computer Engineering (Semester 4)
TOTAL MARKS: 80
TOTAL TIME: 3 HOURS
(1) Question 1 is compulsory.
(2) Attempt any three from the remaining questions.
(3) Assume data if required.
(4) Figures to the right indicate full marks.
1(a) Explain post correspondence problem.(5 marks) 1(b) Differentiate between NFA and DFA.(5 marks) 1(c) Show that language L = {0i | i is prime number} is not regular(5 marks) 1(d) Compare recursive and recursively enumerable languages.(5 marks) 2(a) Design the DFA to accept all the binary strings over ∑={0, 1} that are beginning with 1 and having its decimal value multiple of 5.(10 marks) 2(b) Design DPDA to accept language L = {x ∈ {1, b}* | N1 (x) > Nb (x)}. Na (x) > Nb (x) means number of a's are greater than number of b's in string x.(10 marks) 3(a) Explain variations and equivalence of Turing machine.(10 marks) 3(b) State and prove pumping lemma for context free languages.(10 marks) 4(a) Design mealy machine to find out 2's complement of a binary number.(10 marks) 4(b) Convert the following NFA to an equivalent DFA
State a b ∈ → q0 {q0, q1} {q1} {} q1 {q2} {q1, q2} {} *q2 {q0} {q2} {q1}
(10 marks) 5(a) Consider the following grammar G = (V, T, P, S), V = {S, X}, T = {a, b} and productions P are
S → aSb | aX
X → Xa | Sa | a
Convert this grammar in Greibach Normal Form (GNF).
(10 marks)
5(b) State and prove Rice's theorem.(10 marks) 6(a) Design a Tuning machine as an acceptor foe the language
{anbm | n , m ≥ 0 and m ≥ n}
(10 marks)
6(b) Design PDA to check even parentheses over ∑={0, 1}.(10 marks) | 537 | 1,736 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2019-22 | latest | en | 0.807529 |
https://blog.analytics-toolkit.com/t/statistical-significance-2/ | 1,725,722,088,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650883.10/warc/CC-MAIN-20240907131200-20240907161200-00573.warc.gz | 122,366,275 | 16,641 | # Articles onstatistical significance
## Stop AbUsing the Mann-Whitney U Test (MWU)
The Mann Whitney U Test (MWU), also known as the Wilcoxon Rank Sum Test and the Mann-Whitney-Wilcoxon Test, continues to be advertised as the go-to test for analyzing non-normally distributed data. In online experimentation it is often touted as the most suitable for analyses of non-binomial metrics with typically non-normal (skewed) distributions such as average […] Read more…
Posted in A/B testing, Statistics |
## Fully Sequential vs Group Sequential Tests
What is the best design for a statistical test with sequential evaluation of the data at multiple points in time? This is a question anyone who has realized that unaccounted for peeking with intent to stop is the bane of A/B testing eventually comes to ask. So how does one go about answering that? This […] Read more…
Posted in AGILE A/B testing, Statistics |
## A/B Testing Statistics – A Concise Guide for Non-Statisticians
Navigating the maze of A/B testing statistics can be challenging. This is especially true for those new to statistics and probability. One reason is the obscure terminology popping up in every other sentence. Another is that the writings can be vague, conflicting, incomplete, or simply wrong, depending on the source. Articles sprinkled with advanced math, […] Read more…
Posted in A/B testing, Statistics |
## P-values and Confidence Intervals Explained
Hundreds if not thousands of books have been written about both p-values and confidence intervals (CIs) – two of the most widely used statistics in online controlled experiments. Yet, these concepts remain elusive to many otherwise well-trained researchers, including A/B testing practitioners. Misconceptions and misinterpretations abound despite great efforts from statistics educators and experimentation evangelists. […] Read more…
|
## Top Misconceptions About Scientific Rigor in A/B Testing
Have you ever thought that statistically rigorous A/B tests are impractical? Or do you have trouble selling the need for rigor in testing to your clients, coworkers, or boss? This article debunks the top five myths about the necessity and difficulties of applying scientific method in online A/B testing. Read more…
## The Effect of Using Cardinality Estimates Like HyperLogLog in Statistical Analyses
This article will examine the effects of using the HyperLogLog++ (HLL++) cardinality estimation algorithm in applications where its output serves as input for statistical calculations. A prominent example of such a scenario can be found in online controlled experiments (online A/B tests) where key performance measures are often based on the number of unique users, […] Read more…
Posted in A/B testing, Google Analytics, Statistics |
## Error Spending in Sequential Testing Explained
Sequential analysis of experimental data from A/B tests has been quite prominent in recent years due to the myriad of Bayesian solutions offered by big industry players. However, this type of sequential analysis is not sequential testing proper as these solutions have generally abandoned the idea of testing and therefore error control, substituting it for […] Read more…
Posted in A/B testing, AGILE A/B testing, Statistics |
## Statistical Methods in Online A/B Testing – the book
The long wait is finally over! “Statistical Methods in Online A/B Testing” can now be found as a paperback and an e-book on your preferred Amazon store. The book is a comprehensive guide to statistics in online controlled experiments, a.k.a. A/B tests, and tackles the difficult matter of statistical inference in a way accessible to […] Read more…
Posted in A/B testing, Conversion optimization, Statistics |
## The A/B Testing Guide to Surviving on a Deserted Island
The secluded and isolated deserted island setting has been used as the stage for many hypothetical explanations in economics and philosophy with the scarcity of things that can be developed as resources being a central feature. Scarcity and the need to keep risk low while aiming to improve one’s situation is what make it a […] Read more…
## Designing successful A/B tests in Email Marketing
The process of A/B testing (a.k.a. online controlled experiments) is well-established in conversion rate optimization for all kinds of online properties and is widely used by e-commerce websites. On this blog I have already written in depth about the statistics involved as well as the ROI calculations in terms of balancing risk and reward for […] Read more…
## Analysis of 115 A/B Tests: Average Lift is 4%, Most Lack Statistical Power
Oct 2022 update: A newer, much larger and likely less biased meta analysis of 1,001 tests is now available! What can you learn from 115 publicly available A/B tests? Usually, not much, since in most cases you would be looking at case studies with very basic data about what was tested and the outcome of […] Read more…
Posted in A/B testing, Conversion optimization |
## Confidence Intervals & P-values for Percent Change / Relative Difference
In many controlled experiments, including online controlled experiments (a.k.a. A/B tests) the result of interest and hence the inference made is about the relative difference between the control and treatment group. In A/B testing as part of conversion rate optimization and in marketing experiments in general we use the term “percent lift” (“percentage lift”) while in […] Read more…
#### Browse by year
The book on user testing
Take your A/B testing program to the next level | 1,106 | 5,558 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-38 | latest | en | 0.916905 |
https://forums.phpfreaks.com/topic/315838-ratings-in-database-ala-amazon-how-many-of-each/#comment-1605131 | 1,721,436,383,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514972.64/warc/CC-MAIN-20240719231533-20240720021533-00799.warc.gz | 229,707,406 | 31,902 | # Ratings in database ala Amazon. How many of each?
Go to solution Solved by requinix,
## Recommended Posts
I can't wrap my head around this one. If you look at the reviews of an item on Amazon, they show a little graph with something like:
5 stars - 107
4 stars - 23
3 stars -3
2 stars - 0
1 star - 3
I have a simple table along those lines.
key rating
1 3
2 5
3 5
4 5
5 2
6 2
7 5
I want to count how many of each, so I do:
`SELECT rating,sum(*) FROM reviews GROUP by rating`
and I'll get something like this:
1 - 2
2 - 1
3 - 1
5 - 4
But I need to get that zero (0) from the 4 star rating (that nobody selected). I want:
1 - 2
2 - 1
3 - 1
4 - 0
5 - 4
Thanks
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• Solution
Does it have to be in your results? It would be really easy to just write a tiny bit of code that uses 0 if there aren't any ratings of a particular star count.
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Don't confuse sum and count. It works here, but the two functions are very different. Highly recommend you not get used to substituting sum for count as if they are interchangeable. I also always alias computed columns so I can predict what the column name will be when fetched.
`SELECT rating, count(*) AS count_of FROM reviews GROUP BY rating`
Now I heartily agree with @requinix that a simple class or function is best here. There is no way for the database to know the available universe of values with a scheme like this, which might facilitate an outer join solution to your problem.
With that said, here is a SQL solution that makes a number of assumptions that might not be directly applicable to your database structure, but at least illustrates the idea. I don't think this is a good solution for a number of reasons, including the fact, that you have to hardwire the ratings into your SQL statement AND you are doing a distinct query to figure out which values you need to exclude.
There might be a better way to accomplish this (maybe @Barand has something better?
This does work, and I provided everything you would need to verify in a test database.
```create table reviews (id int unsigned PRIMARY KEY AUTO_INCREMENT, rating int);
insert into reviews(rating) values (5), (5), (2), (3), (1), (3), (2), (5), (1);
SELECT * FROM
(SELECT * FROM
(SELECT 1 as rating, 0 as count_of
union
SELECT 2 as rating, 0 as count_of
union
SELECT 3 as rating, 0 as count_of
union
SELECT 4 as rating, 0 as count_of
union
SELECT 5 as rating, 0 as count_of) as t1
WHERE t1.rating NOT IN (SELECT distinct(rating) as rating FROM reviews)
union
select rating, count(*) as count_of FROM reviews
GROUP BY rating
) as u1
ORDER BY rating ASC;```
With this test data you get your desired result:
```rating count_of
1 2
2 2
3 2
4 0
5 3```
Just to reiterate, a php function using range to pre-populate an array you then use to merge with your result, would be a great way to handle this in your PHP code, and that's what I would do here rather than utilize the convoluted solution I demonstrate here.
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Ok, I get it. How would the DB know there are 5 options? That makes total sense. To simplify things, I used Amazon's rating thing. In reality, this is a quiz composed of 15 questions each of varying lengths. Like the first question has 5 options, the 2nd has 4, etc.
All I have to do is pass the number of expected results. Sometimes things are SO simple, they blind you!
Thanks very much both of you!
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12 hours ago, gizmola said:
maybe @Barand has something better?
Create a "valid_ratings" table in your DB containg a row for each valid rating (1-5). Useful for creating dropdowns or radio button lists for the user to choose from.
If you don't want to this, you can create it as a temporary table when you want thse totals
``` #
#
\$pdo->exec("create temporary table valid_rating (rating int unsigned not null primary key)");
\$pdo->exec("insert into valid_rating values (1), (2), (3), (4), (5)");
#
# get ratings totals
#
\$res = \$pdo->query("SELECT v.rating
, count(r.id) as total
FROM valid_rating v
LEFT JOIN
reviews r USING (rating)
GROUP BY rating;
");
printf("<pre>\n| Rating | Total |\n\n");
foreach (\$res as \$r) {
printf("| %6d | %5d |\n", \$r['rating'], \$r['total']);
}
| Rating | Total |
| 1 | 2 |
| 2 | 2 |
| 3 | 2 |
| 4 | 0 |
| 5 | 3 |```
An alternative solution is the one @gizmola alluded to
``` \$results = array_fill_keys(range(1,5), 0); // create array with 0 total for all possible ratings
\$res = \$pdo->query("SELECT rating
, count(*) as total
FROM reviews
GROUP BY rating
");
foreach (\$res as \$r) {
\$results[\$r['rating']] = \$r['total']; // put totals from query into the array
}
printf("<pre>\n| Rating | Total |\n\n");
foreach (\$results as \$rating => \$total) {
printf("| %6d | %5d |\n", \$rating, \$total);
}```
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You can use a subquery to get the list of all possible ratings and then left join it with your reviews table to get the count of each rating. Here's an example query that should work:
```SELECT ratings.rating, COALESCE(reviews.count, 0) as count FROM (
SELECT DISTINCT rating FROM reviews
) as ratings
LEFT JOIN (
SELECT rating, COUNT(*) as count FROM reviews GROUP BY rating
) as reviews
ON ratings.rating = reviews.rating```
This query first gets a list of all distinct ratings from the reviews table, and then left joins it with a subquery that gets the count of each rating from the reviews table. The COALESCE function is used to replace any NULL values in the count column with 0, in case a rating doesn't have any reviews.
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To save time, let's just assume that sometimes you are.
Selecting "DISTINCT rating" from the input file created by gizmola will yield 1, 2, 3, 5 as there are no ratings of "4". This was the initial problem and your solution does nothing to solve it.
```mysql> SELECT ratings.rating, COALESCE(reviews.count, 0) as count FROM (
-> SELECT DISTINCT rating FROM reviews
-> ) as ratings
-> LEFT JOIN (
-> SELECT rating, COUNT(*) as count FROM reviews GROUP BY rating
-> ) as reviews
-> ON ratings.rating = reviews.rating;
+--------+-------+
| rating | count |
+--------+-------+
| 5 | 3 |
| 2 | 2 |
| 3 | 2 |
| 1 | 2 |
+--------+-------+```
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4 hours ago, Barand said:
To save time, let's just assume that sometimes you are.
Sometimes yes, but in this case (and my other posts last night) no since it wasn't my response. Allow me to explain...
I have been experimenting with the ChatGPT AI (Artificial Intelligence). "My" post was 100% generated by the ChatGPT AI based on the full text of the OP. Obviously AI still has a ways to go and can be absolutely wrong or even provide "dangerous" code solutions. Still, it is pretty cool what it can do at this stage. You can check it out here https://openai.com/blog/chatgpt/
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1 hour ago, benanamen said:
no since it wasn't my response
Of course it was your response. You posted it. It has your name on it.
So given how you obtained the "solution", and knowing is was incorrect, do think that posting an ai-produced, misleading solution was the responsible thing to do?
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For starters, the OP basically lied. The OP has ABSOLUTELY NOTHING to do with "Ratings in database ala Amazon". The OP admits it later on..
"In reality, this is a quiz composed of 15 questions each of varying lengths."
And no, I didn't know it was incorrect for what the op was really trying to do. Had it really been a rating system, the query does indeed work to show ratings. Nevertheless, I won't be using AI content in future posts.
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1 minute ago, benanamen said:
And no, I didn't know it was incorrect
That blows your signature's assumption out of the water.
The OP didn't claim to processing Amazon ratings but wanted to show ratings a la Amazon
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1 hour ago, Barand said:
That blows your signature's assumption out of the water.
It's just a signature Barand. We are just arguing at this point. Grab a pint and enjoy your day. See you on another thread. 🙂
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Only 75 emoji are allowed. | 2,216 | 8,591 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.918871 |
http://docs.stack-assessment.org/en/Topics/Unsorted_multi_input/ | 1,722,922,432,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640476479.26/warc/CC-MAIN-20240806032955-20240806062955-00751.warc.gz | 8,382,793 | 10,918 | Quite often we have a number of separate inputs, but we don't know which order the student will choose to enter.
Let's assume we ask a student to break down integration using linearity for example we have where the ? are inputs ans1 and ans2. We might expect ans1=x^2 but of course we could also have ans1=sin(x).
### Use the feedback variables
One option is to define sans:ans1+ans2 in the feedback variables. Then the PRT can check sans is equivalent to the integrand up to commutativity and associativity (or something else if you perefer). The problem with this is the difficulty in awarding partial credit.
### Partial credit for some correct answers
We want to provide some partial credit when student have some, but not all, the inputs correct.
Let's assume we are looking for $n$ different inputs. We define $m$ to be the number of missing, $w$ to be the number of "not wanted" inputs, then we choose the score to be E.g., here if a student gets all $n$ wrong so that $m=w=n$ then the score is zero. If the student types in all the required expressions, $n=0$ and $n$ wrong ones in addition, then $s=\frac{1}{2}$. This function is, of course, a choice of the teacher.
In the above example, put the following in the question variables
ta1:x^2;
ta2:sin(x);
tas:{ta1, ta2};
p:ta1+ta2;
In the feedback variables put
sans:{ans1, ans2};
missing:setdifference(tas, sans);
notwanted:setdifference(sans, tas);
score:max(1-(length(missing)+length(notwanted))/(2*length(tas)),0);
Continuing the above example, in the PRT use
1. The answer test sets perhaps with the quiet option.
2. sans is {ans1,ans2}.
3. tans is tas.
4. Assign the score in both prt branches to be score.
### Dealing with duplicate entries
How do we decide partial credit when there may be duplicates, e.g. eigenvalues with repetition? If the teacher's answer is [1,1,2] then we can't use the above example based on sets.
STACK provides a maxima function list_cancel(l1,l2) which removes any common elements from [l1,l2], with duplication. E.g. use the following in the question variables.
sans:{ans1, ans2};
[missing, notwanted]:list_cancel([sans, tas]);
score:max(1-(length(missing)+length(notwanted))/(2*length(tas)),0);
Note that list_cancel will not establish algebraic equivalence and within this function two expressions are considered the same using maxima's is(ex1=ex2)). Hence, some pre-processing of the lists might be needed, depending on the situation and what you consider is the "same". For example if we have
l1:[x^2,x^3,x^2-1,x+x];
l2:[x^2,x^4,(x-1)*(x+1),2*x];
list_cancel([l1,l2]);
will return Notice the last elements are remove because default simplification takes place but $x^2-1$ and $(x-1)(x+1)$ are not considered the same by is. In this case ratsimp can be applied to the lists first. In other situations functions like trigsimp or trigrat might be needed. | 748 | 2,868 | {"found_math": true, "script_math_tex": 10, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2024-33 | latest | en | 0.852238 |
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Topic: Crank Nikolson scheme for semi linear parabolic equations
Replies: 1 Last Post: Feb 16, 2013 11:51 PM
Search Thread: Advanced Search
AMX Posts: 35 Registered: 8/22/09
Re: Crank Nikolson scheme for semi linear parabolic equations
Posted: Feb 16, 2013 11:51 PM
Plain Text Reply
On Sat, 16 Feb 2013 07:27:19 -0800 (PST), Sandeep Kumar
<searchsandy1712@gmail.com> wrote:
> Can anybody help me with this?
> I am trying to implement a fourth order semi linear parabolic
> equation called as Cahn-Hilliard equation, in MatLab.
> Its given by
> del u/ del t = - epsilonsquare* laplacian^2(u)+ laplacian(u^3)-lapacian(u)
It cannot be in that form. For scalar field u, ?u is a vector
while ?(?u) is a scalar. You cannot add a scalar to a vector.
I've checked it at wiki and there is a bit different form.
> Can anybody please tell me how to deal with these non-linear terms?
In the case of linear PDEs discrete form leads to linear set of
algebraic equations. In the case of nonlinear PDEs it leads to
nonlinear algebaric equations. You have to solve nonlinear set of
equations at each step.
I've never worked with CH equation but first what I think of is
that u is in range (-1, 1), then the term u^3 is small in
comparision to u. Defining new field p=u^3 would lead to iterative
process of searching n-th approximation of u field with
F(u_(n),p_(n-1))=0 with p_0=0. It just the first idea; I do not
know if this works.
AMX
--
adres w rot13
Nyrxfnaqre Znghfmnx r-znk@b2.cy
© The Math Forum at NCTM 1994-2016. All Rights Reserved. | 496 | 1,659 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-40 | longest | en | 0.90244 |
http://www.stumblingrobot.com/2015/07/13/draw-the-graphs-and-find-the-points-of-intersection-of-x-and-x3/ | 1,606,809,522,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141672314.55/warc/CC-MAIN-20201201074047-20201201104047-00317.warc.gz | 161,868,673 | 14,182 | Home » Blog » Draw the graphs and find the points of intersection of x and x^3
# Draw the graphs and find the points of intersection of x and x^3
Let and . Draw the graphs of and and show the points at which they intersect.
First, we can determine algebraically the points of intersection. They are the points such that , i.e.,
The function values at these points are,
Thus, the points of intersection are . The graph is below. | 99 | 432 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-50 | latest | en | 0.941614 |
https://wiki.freecadweb.org/index.php?title=Draft_Ellipse&diff=418475&oldid=66232 | 1,603,842,335,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107894890.32/warc/CC-MAIN-20201027225224-20201028015224-00704.warc.gz | 588,728,571 | 16,606 | # Difference between revisions of "Draft Ellipse"
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Previous: Arc
Next: Polygon
Draft Ellipse
Draft → Ellipse
Workbenches
Draft, Arch
Default shortcut
E L
Introduced in version
0.7
Draft Circle, Draft Arc
## Description
The Ellipse tool creates an ellipse in the current work plane by entering two points, defining the corners of a rectangular box in which the ellipse will fit. It uses the Draft Linestyle set on the Draft Tray.
This tool can also be used to create elliptical arcs by specifying the start and end angles. To create circles and circular arcs use the Draft Circle and Draft Arc tools. You can also approximate an elliptical or circular arc using the Draft BSpline and Draft BezCurve tools.
Ellipse defined by the rectangle corners
## How to use
1. Press the button, or press E then L keys.
2. Click a first point on the 3D view, or type a coordinate and press the button.
3. Click a second point on the 3D view, or type a coordinate and press the button.
The ellipse can be turned into an elliptical arc after creation, by setting its first angle and last angle properties to different values.
## Options
• To enter coordinates manually, simply enter the numbers, then press Enter between each X, Y and Z component. You can press the button when you have the desired values to insert the point.
• Press R or click the checkbox to toggle relative mode. If relative mode is on, the coordinates of the second point are relative to the first one; if not, they are absolute, taken from the origin (0,0,0).
• Press T or click the checkbox to toggle continue mode. If continue mode is on, the Ellipse tool will restart after you finish the shape, allowing you to draw another one without pressing the tool button again.
• Press L or click the checkbox to toggle filled mode. If filled mode is on, the ellipse will create a filled face (DataMake Face `true`); if not, the ellipse will not make a face (DataMake Face `false`).
• Hold Ctrl while drawing to force snapping your point to the nearest snap location, independently of the distance.
• Hold Shift while drawing to constrain your second point horizontally or vertically in relation to the first one.
• Press Esc or the Close button to abort the current command.
## Properties
An Ellipse object shares many properties with a Draft Circle, but some properties only make sense for the ellipse.
### Data
• DataFirst Angle: specifies the angle of the first point of the ellipse; normally 0°.
• DataLast Angle: specifies the angle of the last point of the ellipse; normally 0°.
If the both radius have the same value, the ellipse looks the same as a Draft Circle.
• DataMake Face: specifies if the Ellipse makes a face or not. If it is `true` a face is created, otherwise only the perimeter is considered part of the object. This property only works if the shape is a full ellipse.
For it to be a full ellipse DataFirst Angle and DataLast Angle should have the same value; otherwise, an elliptical arc is displayed. The values 0° and 360° are considered the same.
### View
• ViewPattern: specifies a Draft Pattern with which to fill the face of the shape. This property only works if DataMake Face is `true`, and if ViewDisplay Mode is "Flat Lines".
• ViewPattern Size: specifies the size of the Draft Pattern.
## Scripting
The Ellipse tool can by used in macros and from the Python console by using the following function:
```Ellipse = makeEllipse(majradius, minradius, placement=None, face=True, support=None)
```
• Creates an `Ellipse` object with given major (`majradius`) and minor (`minradius`) radius in millimeters.
• The bigger value will be used for the major radius (X axis) if no other placement is given.
• If a `placement` is given, it is used; otherwise the shape is created at the origin.
• If `face` is `True`, the ellipse will make a face, that is, it will appear filled.
Example:
```import FreeCAD, Draft
Ellipse1 = Draft.makeEllipse(3000, 200)
Ellipse2 = Draft.makeEllipse(700, 1000) | 979 | 4,133 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2020-45 | latest | en | 0.711181 |
https://indinews.com/2022/06/what-is-roulette-wheel-and-how-does-it-work/ | 1,656,904,968,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104293758.72/warc/CC-MAIN-20220704015700-20220704045700-00787.warc.gz | 372,556,227 | 61,120 | Home Roulette What is roulette wheel and how does it work
# What is roulette wheel and how does it work
Have you ever thought about playing and winning at an online roulette game? If you do, you must know how to play this game. Even if you don’t, we are here to help. We will provide you with every detail related to the roulette game. In this article we are going to discuss roulette wheels and their types.
Read more: How to play roulette
## Roulette wheel
A roulette wheel has a spinning disk and a ball. The divisions around the wheel are numbered from 1 to 36 in a random pattern and it is coloured red and black alternatively. Apart from this there is a green division which is zero. In American roulette, there are two green divisions and the second one is known as 00.
The ball is rolled on this wheel. Before this the player bets on which number the ball will stop. This bet is placed in different ways. During this, chips are decorated on a mat. These chips indicate that a bet is being placed. The whole game revolves around this wheel.
## European Roulette Wheels
European Roulette Wheels has a total of 37 numbers instead of 36 taking into account the green zero at the top. So, instead of your odds of winning being 35/1 as one would usually assume, the additional zero reduces those odds to 36/1, giving the casino what is known as a ‘house edge.’
Read more: How to play baccarat
## Zero on roulette wheel
Every roulette wheel has a pocket zero which is identified by its green colour. In American roulette there are 2 zero’s. Any player can place a bet on zero. Although the probability of the ball landing on an individual number selected by a player is very low, the odds of betting on zero are very high and it lures the players to place a bet. The zero pocket on the roulette wheel is special because if the ball lands on it, everyone else loses their money. That means if other players placed their bet on anything, be it red/black, odd/even or row/column, all the best will be forfeit. The odds of green zero on roulette wheels are generally 35/1 i.e. if you put \$5 and you win then your payout will be \$175. This zero gives the house a slight advantage as they take all the money placed on the bets.
## Triple zero roulette wheel
Triple zero roulette wheels are now coming in trend from the past one and half years. It is a special wheel which has three green zero pockets. Apart from this, all the rules of a triple zero roulette are similar to regular roulette. Because of an extra pocket, it gives more opportunities to players to bet. However, It gives casinos a more edge than any other roulette. If the ball lands on any of the green zero pockets of the triple zero roulette wheel, your payout will be 35/1. As every coin has two sides, the triple zero wheel has its own disadvantages as well. Just like wins are big, the losses are also big and there is a chance that you might run out of your bankroll faster. It’s becoming a trend now and many casinos are opting to provide three zero wheels experience to players. Although it has some downsides, it is definitely worth giving it a shot.
As we already discussed, every roulette wheel has a pocket zero which is identified by its green colour. Apart from american roulette, most of the roulette wheels have a single zero. It is the most common and widely used roulette wheel. Any player can place a bet on zero. Although the probability of the ball landing on an individual number selected by a player is very low, the odds of betting on zero are very high and it lures the players to place a bet. The zero pocket on the roulette wheel is special because if the ball lands on it, everyone else loses their money. That means if other players placed their bet on anything, be it red/black, odd/even or row/column, all the best will be forfeit. The odds of green zero on roulette wheels are generally 35/1 i.e. if you put \$5 and you win then your payout will be \$175. This zero gives the house a slight advantage as they take all the money placed on the bets. Single zero roulette is convenient and players have more advantage in single zero wheel than double zero wheel and triple zero wheel as their probability of winning is higher than others.
Read more: How to play Teen Patti online
## 3 wheel roulette
In this, Three roulette wheels are used and 3 balls exist in the play. Players place the same bets in all three wheels and after the game is over, the win and the lose is decided by the three wheels combined. This game is much more tricky but contains a hefty dose of entertainment.
## Multi-Wheel Roulette
A number of casinos, now-a-days, use Multi wheel roulette. It is more fun and a lot more exciting than other roulettes. Many casino wheels are used in this method. You get eight wheels at a time to bet. You place the same bet on all these wheels.
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The online casino industry is booming and from a handful of bookmakers marking their presence a little over a decade. India is now home... | 1,434 | 6,666 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2022-27 | latest | en | 0.960549 |
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# SS Project Truth
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### SS Project Truth - Page Text Content
S: The Search for Truth: Pythagoras
FC: How do we know what is true? World History Project on Greece | Pythagoras
1: - a verified or indisputable fact, proposition, or principle
2: Information about Pythagoras | Pythagoras was an Ionian Greek philosopher, mathematician, and founder of the religious movement called Pythagoreanism. He is often referred to as a great mathematician and scientist, but he is best known for the Pythagorean theorem.
3: Pythagoras' Accomplishments | -discovered even and odd numbers -discovered perfect numbers -developed Pythagorean Theorem -discovered triangular numbers -discovered irrational numbers
4: The Search For Truth | Pythagoras and his followers were interested in exploring the principles of mathematics, the concept of mathematical numbers, and the idea of a proof. Their motivation behind trying to find the truth was to understand higher, a divine reality in numerous fields including mathematics and astronomy. By searching for truth through mathematics, the Pythagoreans made important discoveries in different areas of math and science that are both mathematically correct and continue to be used today.
5: Astronomy Pythagoras made many significant discoveries and inferences in the field of astronomy while he was on his search for truth. He hypothesized that the earth was round because of its reflection on the moon. He also hypothesized that all the planets orbited around a central fire. Finally, he hypothesized that the sun and other planets were all spheres along with Earth. | Truth Over Time All of Pythagoras' inferences have stayed the same and have been proved to be correct. They are the basis of our interaction with space and allow us to understand the ways of the universe.
6: Music | Pythagoras made several interesting conclusions in the field of music. He believed that mathematics and music were very closely related and that each music note had an exact ratio in math. He observed that vibrating instruments produce harmonious sounds when the ratios of the lengths of the instruments were whole numbers. He came to this conclusion when he heard the harmonious sounds of a local blacksmith shop and discovered that the hammers were all of perfect ratio sizes. | Truth Over Time This conclusion has been proved false over time because of the fact that these perfect ratios are only relevant to string length and not to hammer weight or length. Therefore it does not apply to all music, only certain stringed instruments.
7: Math | Pythagoras made the most conclusions in the field of math. One of those conclusions was on number theory. The Pythagoreans applied varying qualities to different numbers, each one having a character and property of its own. They believed that numbers were physical entities that were the physical components of everything in existence. In addition, The Pythagoreans are credited with the construction of the five regular solids. The Pythagoreans also discovered the idea of dimensions in terms of geometry throughout the world. They proved that the sum of the interior angles of a triangle is equal to 180 degrees. Another discovery was the existence of irrational numbers in mathematics. However, the Pythagoreans’ most famous discovery was the Pythagorean Theorem. The Pythagorean Theorem states that “In any right triangle, the sum of the squares of the lengths of the two shorter sides equals the square of the length of the long side.” The Pythagoreans discovered this theorem through geometrical means, by arranging triangles in various ways to prove the conclusion. The formula for the Pythagorean Theorem is A^2 + B^2 = C^2, with A and B being the shorter sides and C being the longest side, or hypotenuse, of the triangle. Pythagoras was not the first person to discover the theorem, but he was the first one who was able to prove it. | Truth Over Time All of Pythagoras' theorems and conclusions except for one have been proved to be correct and are used in mathematics all over the world. His one conclusion that was not proved, that numbers have physical properties, was deemed incorrect and decided that numbers only relate to math, not the entire world. But Pythagoras' other conclusions have all been proved and continue to be used to solve problems in both practical life and in math. The Pythagorean Theorem is now the most important formula when talking about triangles and is used in countless proofs throughout math.
8: Final Thoughts | In | In conclusion, the majority of the truth that Pythagoras found has stayed true throughout time. Many of his theorems and ideas continue to be used today and are applied in many fields including astronomy and mathematics. His theories are the basis for much of geometry and his legacy will live forever because of his discovery of the truth.
9: Consequences | d | Positive: -made contributions to Geometry -discovered the Pythagorean Theorem -helped society gain an understanding of Outer Space -was basis for future searches for truth in the fields of astronomy and mathematics Negative: -people became jealous and often attacked him or his religious buildings -was considered mentally ill by some because of his extreme views about astronomy
10: Citations | 1.http://www.storyofmathematics.com/images2/pythagoras_music.gif 2.http://www.crystalinks.com/pythagsun.gif 3.http://utnsliteracy.pbworks.com/f/1256058634/pythag_thm.gif 4.http://mrfazio.edublogs.org/files/2011/10/pythagorean-theorem-16l2i95.jpg 5.http://home.wlu.edu/~mahonj/Ancient_Philosophers/Pythagoras4.jpg 6.http://s3.amazonaws.com/files.posterous.com/soulache/gz498F9y3xIb7TBBBDmTeVS9f1wejNTazjXKvridxtWWMrcJEWqkkaVGKkg1/truth.gif?AWSAccessKeyId=AKIAJFZAE65UYRT34AOQ&Expires=1328476850&Signature=W2rinvIq1eAO5B%2FhfoGIzagvcdA%3D 7.http://edu.glogster.com/media/5/27/29/31/27293175.jpg 8.http://2.bp.blogspot.com/_SKXGBUeUyQs/SW646vpLZhI/AAAAAAAAAQk/D4hAvxuFiDE/s400/PythagoreanTheorem.gif 9.http://natureofmathematics.files.wordpress.com/2010/05/pythagoras_statue.jpg 10.http://practicalphysics.com/imageLibrary/jpeg463/1932.jpg
11: The End!
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Quiz 2 Question 1 Jack and Jill exercise in a 25.0-m-long swimming pool. Jack swims nine lengths of the pool in 140.1 seconds, while Jill, the faster swimmer, covers ten lengths in the same time interval. Find the average speed of each swimmer. Answers: Selected Answer: d. Jack 1.61 m/s and Jill 1.78 m/s a. Jack 5.60 m/s and Jill 6.22 m/s b. Jack 6.22 m/s and Jill 5.60 m/s c. Jack 1.78 m/s and Jill 1.61 m/s d. Jack 1.61 m/s and Jill 1.78 m/s
Response Feedback: Correc t! Question 2 Kinematics can be defined as the study of motion without considering its causes. Answers:
Response Feedback: Correc t! Question 3 Find the following for path B in Figure 2-59 : (a) The distance traveled. (b) The magnitude of the displacement from start to finish. (c) The displacement from start to finish. Figure 2-59 | 296 | 979 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2021-49 | latest | en | 0.821024 |
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# Earth System Responses to Natural and Human Induced Changes
This unit focuses on local plant species; students learn to identify common species and will examine their life cycle characteristics as evidence of climate change. Through the use of the national citizen science project titled Project BudBurst,... (View More)
# Solar Week Monday: Do the Activity - Measuring Solar Activity
This is an online lesson associated with activities during Solar Week, a twice-yearly event in March and October during which classrooms are able to interact with scientists studying the Sun. Outside of Solar Week, information, activities, and... (View More)
# Explorando mapas de colores: Usando datos de ozono estratosférico
Using the 5E instructional model, students discover the value of using color maps to visualize data. The activity requires students to create a color map of the ozone hole from Dobson data values derived from the Aura satellite. Students then... (View More)
# Exploring Color Maps: Using Stratospheric Ozone Data
Through the use of the 5E instructional model, students discover the value of using color maps to visualize data. The activity requires students to create a color map of the ozone hole from Dobson data values derived from the Aura satellite.... (View More)
# Black Ice: A Slippery Arctic Road
In this lesson students investigate the effects of black carbon on arctic warming and are introduced to a mechanism of arctic warming that is not directly dependent on greenhouse gases in the atmosphere: black carbon deposition on Arctic snow and... (View More)
Audience: High school
Materials Cost: \$10 - \$20 per group of students
# Building for Hurricanes: Engineering Design Challenge
This activity is a short engineering design challenge to be completed by individual students or small teams. A real-world problem is presented, designing buildings for hurricane-prone areas, but in a simulated way that works in a classroom, after... (View More)
# GPM Core Observatory Paper Model
This activity allows participants to build a paper model of the GPM Core Observatory and learn about the technology the satellite uses to measure precipitation from space. Directions explain how to cut, fold and glue the individual pieces together... (View More)
# Mars Math
This book contains 24 illustrated math problem sets based on a weekly series of space science problems. Each set of problems is contained on one page. The problems were created to be authentic glimpses of modern science and engineering issues, often... (View More)
Keywords: Mars
Audience: Middle school, High school
Materials Cost: Free
# Dusty Dilemma
This is a lesson about statistics in science as it applies to the measurement of dust in space. Learners will be introduced to the concepts of error analysis, including standard deviation. They will apply the knowledge of averages (means), standard... (View More)
# Designing a Spectroscopy Mission
This is a math-science integrated unit about spectrographs. Learners will find and calculate the angle that light is transmitted through a holographic diffraction grating using trigonometry. After finding this angle, the students will build their... (View More)
Audience: High school
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https://techstalking.com/is-operator-result-what-is-happening-duplicate/ | 1,657,032,835,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104585887.84/warc/CC-MAIN-20220705144321-20220705174321-00418.warc.gz | 615,417,068 | 16,758 | “is” operator result: What is happening? [duplicate]
Each Answer to this Q is separated by one/two green lines.
I was quite surprised when
``````[] is not []
``````
evaluated to `True`.
What is happening in this code? What really `not` and `is` statements are doing?
`a is not b` is a special operator which is equivalent to `not a is b`.
The operator `a is b` returns True if a and b are bound to the same object, otherwise False. When you create two empty lists you get two different objects, so `is` returns False (and therefore `is not` returns True).
`is` is the identity comparison.
`==` is the equality comparison.
Your statement is making two different lists and checking if they are the same instance, which they are not. If you use `==` it will return true and because they are both empty lists.
The best way to describe WHY that happens is this:
``````>>> x = []
>>> y = []
>>> print(x is y)
... False
``````
`x` and `y` are actually two different lists, so if you add something to `x`, it does not appear in `y`
``````>>> x.append(1)
>>> print(x)
... [1]
>>> print(y)
... []
``````
So how do we make (`x is y`) evaluate true?
``````>>> x = []
>>> y = x
>>> print(x is y)
... True
>>> x.append(10)
>>> print(x)
... [10]
>>> print(y)
... [10]
>>> print(x is y)
... True
``````
if you want to see if two lists have the same contents…
``````>>> x = []
>>> y = []
>>> print(x == y)
... True
>>> x.append(21)
>>> print(x)
... [21]
>>> print(y)
... []
>>> print(x == y)
... False
>>> y = [21]
>>> print(x == y)
... True
``````
`is` means is same instance. It evaluates to true if the variables on either side of the operator point to the same object and false otherwise.
Reference, near the bottom.
is checks for identity. `[]` and `[]` are two different (but equivalent) lists. If you want to check if both the lists are empty you can use their truth value (false for empty strings, collections, and zeros).
``````if not ([] and []):
print 'Spanish Inquisition'
``````
the only time that `is` is guaranteed to return True is for singletons such as None. Like the Highlander, there can be only one instance of None in your program – every time you return None it’s the very same “thing” as the none referred to if you type `print None`.
[], OTOH, is not guaranteed to be anything except an empty list and evaluate to False in a boolean context.
I know I am posting to a pretty old post. however, this might help someone stumble upon this like me.
“is” checks, whether a memory address is same or not, while “==” check if the value is same or not. Would be much clear from the following example
let’s first talk about immutable objects, as it’s easy to understand
``````# could be any immutable object
immutable_a = 10
immutable_b = 10
# prints address of a and b variable
print "address of a is %s" % id(immutable_a)
print "address of a is %s" % id(immutable_b)
# as both addresses is same, following shall be true
print immutable_a is immutable_b
# as the values are also same, following shall be true as well
print immutable_a == immutable_b
``````
now let’s talk about mutable objects
``````# could be any mutable object
mutable_a = [10]
mutable_b = [10]
# prints address of a and b variable
print "address of mutable_a is %s" % id(mutable_a)
print "address of mutable_b is %s" % id(mutable_b)
# as addresses are not same, following shall be false
print mutable_a is mutable_b
# as the values are same, following shall be true
print mutable_a == mutable_b
``````
@Jiaaro is right. Using `is` with immutable data types is dangerous because it is not predictable because of Pythons interpreter optimization.
See this example:
``````10 * "a" is 10 * "a" # True
100 * "a" is 100 * "a" # False
``````
In the second line it is faster to create a new object with a new id for the interpreter. So use the `is` operator only with mutable types.
The answers/resolutions are collected from stackoverflow, are licensed under cc by-sa 2.5 , cc by-sa 3.0 and cc by-sa 4.0 . | 1,066 | 4,019 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.91382 |
https://blog.his.school/lecture/ordering-fractions | 1,722,825,352,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640427760.16/warc/CC-MAIN-20240805021234-20240805051234-00560.warc.gz | 105,231,570 | 6,436 | # Ordering Fractions
Published on
Fraction sorting is a way of arranging fractions in order from smallest to largest or largest to smallest.
When you're sorting fractions, you compare the numbers one at a time, just like you would when comparing whole numbers. The difference is that fractions are made up of two parts: the numerator (top number) and the denominator (bottom number).
For example, let's say you have the following fractions: 1/4, 1/2, 3/4, 1/8
To sort these fractions from smallest to largest, you would start with the first two fractions and compare them. If the first fraction is smaller than the second fraction, you move on to the next two fractions. If the first fraction is larger than the second fraction, you swap them.
Here's how the sorting process would look for our example:
1/8 < 1/4, keep 1/8 and 1/4 in place 1/4 < 1/2, keep 1/4 and 1/2 in place 1/2 < 3/4, keep 1/2 and 3/4 in place
The sorted fractions are now: 1/8, 1/4, 1/2, 3/4
It's important to pay attention to the numerator and denominator of each fraction when sorting fractions. With a little practice, you'll be sorting fractions like a pro!
Copyright
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HiSchool
All rights reserved
We use cookies to improve your experience | 322 | 1,230 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2024-33 | latest | en | 0.922361 |
http://gamblingmy.com/blackjack-strategy-smarts/ | 1,568,619,120,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572491.38/warc/CC-MAIN-20190916060046-20190916082046-00404.warc.gz | 84,822,021 | 7,585 | Why do you need blackjack strategy and smarts?
Blackjack is a game of skill. Players need to make meaningful decisions with their hands in order to have an advantage. The right decisions will of course help you to win or save a few chips rather than a wrong decision which would decrease your chips and in turn, cost you money.
The correct rules of playing blackjack is known as basic strategy. These rules aren’t just guesswork or a bunch of nonsense, they ave all worked out if you look at the mathematics behind it.
The house edge against a blackjack player will vary significantly which means that it can’t a fixed percentage. The only real advantage the house has is when the dealer and the player bust at the same time, then the dealer will win.
There are certain methods a player can employ in order to counter the house’s advantage:
• The player gets paid 3 to 2 for blackjack but if the dealer gets a blackjack, the player would only lose the original bet.
• The player can make his or own playing decisions while the dealer sticks to the house rules.
• The player can double down if the situation is favorable.
• The player can split when there is an advantage.
• The player can take insurance if there is an advantage.
The foundation of Blackjack strategy.
Basic strategy is a set of rules that tells you the right way to play your hand during a match. It work especially well when you have no knowledge of the remaining cards. These rules will reveal when to hit, stand, double split or surrender by taking these factors into account:
• the dealers upcard.
In the long run, basic strategy will help you to win more and lose less. It will maximize your expectation and bring you very close to even with the house. It will e the bet you can do without counting cards. In this case “basic” does not mean beginner or simplified but it means essential because these rules are needed for your success.
Proper basic strategy will always give you the best house edge in order for you to stand a chance. Eventually you should learn the complete blackjack strategy but the basic strategy will do for now. The house edge will still be very low.
Table of basic strategy rules.
Hard Hands.
9 or lower. Hit.
10 or 11. Double down if your total is more than the dealer’s upcard. Hit otherwise.
12 through 16 Hit when the dealer’s upcard is 7 or higher. Stand otherwise.
17 or higher. Stand.
Soft Hands.
13 through 18 Double down when the dealer’s upcard is 5 or 6.
17 or lower. Hit.
18. Hit when the dealer’s upcard is 6 or less. Stand when the dealer’s upcard is 7 or more.
19 or higher. Stand.
Pairs
Always split a pair of aces or 8s.
Never split 4s, 5s or 10s.
Split all other pairs if the dealer’s upcard is 6 or less. | 607 | 2,736 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-39 | latest | en | 0.966754 |
https://robertovitillo.com/monoids/ | 1,610,816,509,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703506832.21/warc/CC-MAIN-20210116165621-20210116195621-00716.warc.gz | 551,842,057 | 71,186 | # Monoids for analytics
January 16, 2016
This is a short post on the elegance of using abstract algebra for analytics in Scala.
A monoid is a set $T$ that is closed under an associative binary operation $append$ with an identity element $zero$ such that $append(a, zero) = a$. In other words, the following 3 properties apply:
• Closure - the result of combining two elements of the set is also an elment of the set:
$\forall a, b \in T: append(a, b) \in T$
• Associativity - when combining more than two elements of a set the order of the pairwise combinations doesn’t matter, which makes a monoid well suited for parallel computations:
$\forall a, b, c \in T: append(append(a, b), c) = append(a, append(b, c))$
• Identity - there is a special element of the set that when combined with any other element of the set yields the same element:
$\exists zero \in T: \forall a \in T: append(zero, a) = append(a, zero) = a$
One way to define a trait for a monoid in Scala is the following:
trait Monoid[T] {
def zero: T
def append(a: T, b: T): T
}
Monoids are everywhere; think of the set of natural numbers and addition or the set of strings and concatenation. Also note that the same set can have multiple “monoidal forms”; for example the set of natural numbers can have both an additive and a multiplicative monoid.
Monoids compose well; for example a tuple of monoids is itself a monoid, as such it’s simple to define a monoid for a complex type once monoids for its constituents types exists. Scalaz and Algebird are two Scala libraries that provide monoids for data types such as List, Set, Option, Map and others. Algebird in particular, which is targeted at building aggregation systems, comes with a set of monoids useful for counting such as DecayedValue for exponential decay, AveragedValue for averaging and HyperLogLog for approximate cardinality counting.
As a concrete example on how monoids provided by Scalaz are used in practice, suppose we have a server that receives sparse histograms from its clients:
import scalaz._
import Scalaz._
case class Histogram(var values: Map[Long, Long], sum: Int, count: Int)
We would like to aggregate histograms over a time window. It’s simple enough to write some code to do that, yet monoids offer an elegant solution. Since Scalaz provides by default an additive monoid for Map and Long, we can easily define one for Histogram as well:
implicit def histogramMonoid: Monoid[Histogram] = new Monoid[Histogram] {
def zero = Histogram(Map(), 0, 0)
def append(a: Histogram, b: => Histogram) = Histogram(a.values |+| b.values,
a.sum |+| b.sum,
a.count |+| b.count)
}
Where |+| invokes the binary operator of the monoid defined over the type of its operators. We can aggregate histograms now that we have a monoid, e.g.:
val h1 = Histogram(Map(10 -> 12, 100 -> 41), 53, 3)
val h2 = Histogram(Map(10 -> 21, 80 -> 14), 35, 17)
h1 |+| h2
yields:
Histogram(Map(10 -> 33, 80 -> 14, 100 -> 41), 88, 20)
Let’s say that we have different histograms for different measurements which are stored in a mapping from strings to histograms. Since there is a monoid defined over Histogram then we can easily aggregate multiple Map[String, Histogram] as well:
val m1 = Map("metric1" -> h1, "metric2" -> h2)
val m2 = Map("metric1" -> h2, "metric3" -> h1)
m1 |+| m2
yields:
Map(metric1 -> Histogram(Map(10 -> 33, 80 -> 14, 100 -> 41), 88, 20),
metric2 -> Histogram(Map(10 -> 21, 80 -> 14), 35, 17),
metric3 -> Histogram(Map(10 -> 12, 100 -> 41), 53, 3))
This is just the tip of the iceberg of the elegance provided by abstract algebra. If this short article caught your interest grab a copy of Functional Programming in Scala, which is without any doubt the best functional programming book I have ever read. The book introduces a variety of abstract algebra concepts with their relative implementations.
Written by Roberto Vitillo
## I am writing a book about distributed systems
Do you write applications that make network calls? If so, congratulations - you are a distributed systems engineer! My book teaches the core principles of distributed systems that will help you design, build, and maintain scalable cloud applications that will stand the test of time.
Want a sneak peek? Subscribe to receive a sample and be notified when a new chapter is released.
I respect your privacy. Unsubscribe at any time. | 1,094 | 4,370 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 7, "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} | 3.296875 | 3 | CC-MAIN-2021-04 | longest | en | 0.898705 |
https://nrich.maths.org/5040 | 1,484,659,071,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560279915.8/warc/CC-MAIN-20170116095119-00383-ip-10-171-10-70.ec2.internal.warc.gz | 827,182,191 | 5,168 | ### Polydron
This activity investigates how you might make squares and pentominoes from Polydron.
### Eight Dominoes
Using the 8 dominoes make a square where each of the columns and rows adds up to 8
### Prime Magic
Place the numbers 1, 2, 3,..., 9 one on each square of a 3 by 3 grid so that all the rows and columns add up to a prime number. How many different solutions can you find?
# Tubular Path
##### Stage: 2 Challenge Level:
Move the green spot along the tube by moving the yellow spot.
Explain how you did it.
Full Screen Version
This text is usually replaced by the Flash movie.
The yellow spot does not follow this grey tube.
I wonder if there is a tube which the yellow spot and the green spot would both follow?
Could you draw one? | 186 | 755 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2017-04 | longest | en | 0.896336 |
https://courses.lumenlearning.com/mathforliberalartscorequisite/chapter/finding-the-slope-of-a-line-from-its-graph/ | 1,642,569,902,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301263.50/warc/CC-MAIN-20220119033421-20220119063421-00265.warc.gz | 217,480,172 | 12,989 | ## Finding the Slope of a Line From Its Graph
### Learning Outcomes
• Given the graph of a line, determine the slope of the line
• Identify the slope of a horizontal line given it’s equation
• Identify the slope of a vertical line given it’s equation
Now we’ll look at some graphs on a coordinate grid to find their slopes. The method will be very similar to what we just modeled on our geoboards.
To find the slope, we must count out the rise and the run. But where do we start?
We locate any two points on the line. We try to choose points with coordinates that are integers to make our calculations easier. We then start with the point on the left and sketch a right triangle, so we can count the rise and run.
### example
Find the slope of the line shown:
Solution
Locate two points on the graph, choosing points whose coordinates are integers. We will use $\left(0,-3\right)$ and $\left(5,1\right)$.
Starting with the point on the left, $\left(0,-3\right)$, sketch a right triangle, going from the first point to the second point, $\left(5,1\right)$.
Count the rise on the vertical leg of the triangle. The rise is $4$ units. Count the run on the horizontal leg. The run is $5$ units. Use the slope formula. $m={\Large\frac{\text{rise}}{\text{run}}}$ Substitute the values of the rise and run. $m={\Large\frac{4}{5}}$ The slope of the line is ${\Large\frac{4}{5}}$ .
Notice that the slope is positive since the line slants upward from left to right.
### Find the slope from a graph
1. Locate two points on the line whose coordinates are integers.
2. Starting with the point on the left, sketch a right triangle, going from the first point to the second point.
3. Count the rise and the run on the legs of the triangle.
4. Take the ratio of rise to run to find the slope. $m={\Large\frac{\text{rise}}{\text{run}}}$
### example
Find the slope of the line shown:
Notice that the slope is negative since the line slants downward from left to right.
What if we had chosen different points? Let’s find the slope of the line again, this time using different points. We will use the points $\left(-3,7\right)$ and $\left(6,1\right)$.
Starting at $\left(-3,7\right)$, sketch a right triangle to $\left(6,1\right)$.
Count the rise. The rise is $−6$. Count the run. The run is $9$. Use the slope formula. $m=\Large\frac{\text{rise}}{\text{run}}$ Substitute the values of the rise and run. $m={\Large\frac{-6}{9}}$ Simplify the fraction. $m=-{\Large\frac{2}{3}}$ The slope of the line is $-{\Large\frac{2}{3}}$.
It does not matter which points you use—the slope of the line is always the same. The slope of a line is constant!
### try it
The lines in the previous examples had $y$ -intercepts with integer values, so it was convenient to use the y-intercept as one of the points we used to find the slope. In the next example, the $y$ -intercept is a fraction. The calculations are easier if we use two points with integer coordinates.
### example
Find the slope of the line shown:
### try it
In the following video we show another example of how to find the slope of a line given a graph. This graph has a positive slope.
In the following video we show another example of how to find the slope of a line given a graph. This graph has a negative slope.
## Finding the Slope of Horizontal and Vertical Lines
Do you remember what was special about horizontal and vertical lines? Their equations had just one variable.
• horizontal line $y=b$; all the $y$ -coordinates are the same.
• vertical line $x=a$; all the $x$ -coordinates are the same.
So how do we find the slope of the horizontal line $y=4?$ One approach would be to graph the horizontal line, find two points on it, and count the rise and the run. Let’s see what happens. We’ll use the two points $\left(0,4\right)$ and $\left(3,4\right)$ to count the rise and run.
What is the rise? The rise is $0$. What is the run? The run is $3$. What is the slope? $m=\Large\frac{\text{rise}}{\text{run}}$ $m={\Large\frac{0}{3}}$ $m=0$
The slope of the horizontal line $y=4$ is $0$.
All horizontal lines have slope $0$ . When the $y$ -coordinates are the same, the rise is $0$ .
### Slope of a Horizontal Line
The slope of a horizontal line, $y=b$, is $0$.
Now we’ll consider a vertical line, such as the line $x=3$, shown below. We’ll use the two points $\left(3,0\right)$ and $\left(3,2\right)$ to count the rise and run.
What is the rise? The rise is $2$. What is the run? The run is $0$. What is the slope? $m=\Large\frac{\text{rise}}{\text{run}}$ $m={\Large\frac{2}{0}}$
But we can’t divide by $0$. Division by $0$ is undefined. So we say that the slope of the vertical line $x=3$ is undefined. The slope of all vertical lines is undefined, because the run is $0$.
### Slope of a Vertical Line
The slope of a vertical line, $x=a$, is undefined.
### example
Find the slope of each line:
1. $x=8$
2. $y=-5$
Solution
1. $x=8$
This is a vertical line, so its slope is undefined.
2. $y=-5$
This is a horizontal line, so its slope is $0$.
### Quick Guide to the Slopes of Lines
The following example shows you how to determine the slope of horizontal and vertical lines that are plotted on the coordinate axes. | 1,397 | 5,201 | {"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} | 4.90625 | 5 | CC-MAIN-2022-05 | latest | en | 0.858034 |
http://stackoverflow.com/questions/13788154/integral-with-variable-limits | 1,432,867,206,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207929832.32/warc/CC-MAIN-20150521113209-00044-ip-10-180-206-219.ec2.internal.warc.gz | 219,689,643 | 16,719 | # Integral with variable limits
Hi im and struggling to solve an integral with a variable as the limit using matlab, the 2 biggest problems I have is that matlab can't find the integral explicitly and a lot of the numerical methods wont except variables
I need to solve
``````0=H/2R - integral (z(x) between b and 1)
z(x)= (((x/((a*x*x)+1-a))^2)-1)^-0.5
b= (sin(t)+sqrt(t^2 + 4a(a-1)))/2a
``````
I know H,R and t and the idea is to solve the integral then solve the nonlinear equation for a, I know to suse fzero/fsolve for the nonlinear equation but I am stuggling to solve the integral
-
You could try a shooting method - guess a value for a and numerically solve from there until you find an a value which solves the last equation. Heres something that should work (though I guessed randomly at the numeric values and didn't get it to converge)
``````function test
a_guess = .1
fzero(@(a) solveWithA(a), a_guess)
function res = solveWithA(a)
t = .9;
H = 1.5;
R = 1.1;
z = @(x) (((x/((a*x*x)+1-a))^2)-1)^-0.5;
b = (sin(t)+sqrt(t^2 + 4*a*(a-1)))/(2*a);
lower_limit = b;
integrand = z;
[T, Y] = ode45(@(t, x) integrand(x),[lower_limit 1],0);
res = norm((H/2/R - Y(end)))
end
end
``````
But an analytic expression for a... I think its pen and paper :) Try doing the indefinite integral by hand, then applying the limits? Though, removing a from the integrand still leaves a nasty result. There's probably a 'trick' for this better posed on math overflow.
-
thanks, it has to be done in matlab, i wasnt sure if shooting could be used as its not really an IVP/BVP, why do you use norm at the end? – user1889524 Dec 9 '12 at 17:30
I used norm so that the imaginary part of the number is included in the error for fsolve :) – ccook Dec 9 '12 at 19:12
BTW - how did this problem come up? – ccook Dec 9 '12 at 19:14
Its to work out surface tension on a compressed cylinder, H is the height, R the radius and i need to solve to the equation for a to work out the surface tension, thats why im not sure whether their should be an imaginary part – user1889524 Dec 9 '12 at 19:34
I wouldn't think it would be imaginary... - I was just getting them with my random input values. Of course make sure the values make physical sense :) – ccook Dec 9 '12 at 19:45 | 664 | 2,273 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-22 | latest | en | 0.907788 |
http://www.kalmstrom.com/Tips/Excel-2016-Course/FindFloorCosts-Exercise.htm | 1,506,326,016,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818690376.61/warc/CC-MAIN-20170925074036-20170925094036-00362.warc.gz | 483,534,983 | 5,491 | SharePoint/Outlook Products Buy FAQ Services Apps Tips Books Contact About us Blog
Home >Tips >Excel 2016 from Scratch >Vertical lookup B Site map
# Excel 2016 from Scratch Exercise Vertical Lookup B
## An Excel tutorial by Peter Kalmstrom
In the previous exercise in the Excel 2016 from Scratch series, Peter Kalmstrom introduced formulas with the vertical lookup (VLOOKUP) function. Here is an exercise, that you can try yourself before you look at the solution in the demo.
As an example Peter uses a calculation of flooring with different materials in rooms of different sizes. The formula looks up the cost of the selected flooring to calculate the cost for each room.
To have a more detailed explanation of the vertical lookup function used in this exercise, please refer to the next article. It uses the same example exercise, but it does not calculate the floor area. It just explains how to use the function guide to create a formula with a vertical lookup.
### Exercise
If you want to try the exercise yourself, please download the Excel file Peter uses in the demo!
### Content
This is what Peter shows in the demo below:
• How to write a formula that calculates a rectangular area by multiplying its width and lenght.
• Brief explanation on how to use the function guide to create a formula with a vertical lookup. See next article for details.
Peter of course also shows how to check a formula, how to fill down a formula and other things that he has shown in several earlier demos too. These things need to be done almost every time!
Peter uses Excel 2016 for his demo, but the Excel calculation functions are the same for earlier versions of Excel.
---------------------------------------------------------------------------------------
Always the latest news in the kalmstrom.com blog | 357 | 1,816 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2017-39 | latest | en | 0.868141 |
https://educationcloset.com/2015/07/23/mathematical-thinking-and-creativity-align/ | 1,590,806,772,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347407001.36/warc/CC-MAIN-20200530005804-20200530035804-00246.warc.gz | 332,916,403 | 17,334 | Gina Cherkowski | July 2015
# Mathematical Thinking and Creativity Align!
Often, people do not mix math and creativity together, but they should. According to the site Creativity at Work, “Creativity is characterised by the ability to perceive the world in new ways, to find hidden patterns, to make connections between seemingly unrelated phenomena, and to generate solutions.” They add that, “Creativity involves two processes: thinking, then producing”.
## Typically, school math focuses on learning procedures to solve routine problems.
The key to being successful in the math classroom is to learn to think inside-the-box. However, real-world Mathematicians think out-side-the-box so they can solve real-world problems as they arise. Kids today need to be creative and innovative out-side-the-box thinkers, like Mathematicians, so they can solve real-world problems and become the innovators of tomorrow.
It has been argued that in the 21st Century, kids need the ability to learn what they do not already know, the ability to solve new and novel problems, and the ability think innovatively. Being a creative thinker is a critical factor for each of these important skills. And yes, we can hone and develop them inside the math classroom. How? First, we need to inspire kids to want to engage in math. Dan Meyer gives math teachers excellent examples of how to get kids engaged in problem solving inside the math classroom. Engagement is key; once students are engaged in a problem, they are more likely to persist through and find a solution. By doing this, students are able to develop the mindset and skills needed to learn what they do not already know.
In addition to being engaged and being able to persist through a problem, math teachers want their students to see and recognize patterns and make important connections. This is an integral part of mathematics and is critical for connecting math to the world beyond the four walls of the classroom. Finally, we want students to be able to look at word problems and mathematical tasks and find more than one solution or method. This requires students to think flexibly and to see problems in more than one way.
## Thinking Mathematically….
In sum, today’s math classroom demands that students develop the ability to think mathematically. This means they have the tools and the mindset to persist through struggles as they solve real-world problems as opposed to merely following routine procedures and solving rote problems. We want students to think and produce but this does not mean produce what they have been carefully guided to produce.
Instead, producing means engaging students in their own learning by capturing their interest, teaching them to see problems in many ways, putting forth conjectures, ideating solutions, juxtaposing different ideas, and evaluating their solutions/ideas. If we look closely, we see that these skills are inherently linked to the creative process. In my graduate course this fall, Creativity Across the Disciplines, we will be exploring these connections more closely. It is an exciting time to be a Mathematics/STEM Educator! | 616 | 3,137 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2020-24 | latest | en | 0.967108 |
https://stat.ethz.ch/pipermail/r-help/2011-November/296382.html | 1,556,084,900,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578633464.75/warc/CC-MAIN-20190424054536-20190424075600-00037.warc.gz | 534,275,668 | 2,308 | # [R] Binned line plot
Jim Lemon jim at bitwrit.com.au
Tue Nov 22 08:51:05 CET 2011
```On 11/22/2011 04:29 PM, Jeffrey Joh wrote:
>
> I have a scatter plot with 10000 points. I would like to add a line that bins every 50 points and connects the average of each bin. I'm looking for something similar to line type "m" in Stata.
>
> With this dataset of 10000 points, I would also like to bin the data and make boxplots at certain intervals, so that I have a set of boxplots to represent each bin. I would also like the width of each box to be proportional to the number of points in each bin.
>
> How can I make these plots? Is there a simple package to use?
>
Hi Jeffrey,
There are three possibilities that come to mind:
1) You want to bin the points based on their order in the data frame.
2) You want to bin the points based on the x or y values of the coordinates.
3) You want to bin the points based on the x _and_ y values of the
coordinates.
Number 1 is trivial and has already been answered (assume a two column
data frame of coordinates named "xypoints").
#first point - set up a loop to get a vector of averages
meanx<-rep(0,200)
meany<-rep(0,200)
for(index in 1:200) {
start<-1+50*(index-1)
meanx[index]<-mean(xypoints[start:(start+49),"x"])
meany[index]<-mean(xypoints[start:(start+49),"y"])
}
plot(meanx,meany,type="l")
Number 2 requires that you sort the pairs based on the value of the one
you want, then apply the same process as 1 to the sorted pairs. Number 3
is somewhat more difficult.
I don't do this much, and some of the people who do map analysis will
probably come up with a much better method.
Find the most extreme point.
Find the 49 points closest to that point to constitute group 1.
Remove those points from the data frame.
Go back to the first step if there are any points left.
You will end up with 200 groups of points that are spatially grouped.
Get the centroids and plot as above.
Another wild guess from
Jim
``` | 524 | 1,965 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2019-18 | latest | en | 0.92186 |
https://electronics.stackexchange.com/questions/665349/laplace-inverse-transform-of-an-exponential-multiplying-another-function | 1,708,477,242,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473360.9/warc/CC-MAIN-20240221002544-20240221032544-00262.warc.gz | 236,608,878 | 39,931 | # Laplace inverse transform of an exponential multiplying another function
I was trying to understand the inverse transform of $$\Y(s)=F(s)e^{-as}\$$, but on different sources I found three different answers (H being the step): $$\f(t-a)\$$, $$\H(t)f(t-a)\$$ and $$\H(t-a)f(t-a)\$$.
The first and the second I guess could be the same, given usually the analysis starts at t = 0, but which inverse transform is the right one?
• While Laplace Transforms are used in EE, given that this reads as pretty much a pure math question, I'm wondering if it would be more at home on math.se. May 3, 2023 at 18:06
• It could be; I posted it here because I was studying this while studying control theory, but you're right it's more abstract than that. Should I move it? May 3, 2023 at 21:00
The Laplace Transform on the interval $$\0 is common used in electronics. Certainly causal signals $$\u(t)=0; t.
We all get lazy when writing equations because often the context is understood.
For example: if the Laplace transform of $$\f(t)=t\$$, then all the values for t<0 get ignored. What we take the transform of is $$\f(t)H(t)\$$ where $$\H(t)\$$ is the unit step function (also called the switching function).
So then the inverse Laplace transform $$\L^{-1}[G(s)] = g(t)H(t)\$$. The multiplication by H(t) is often not written but understood to be there.
When a causal function is shifted to the right by $$\t_0\$$, the values of the function from 0 to $$\t_0\$$ is zero. So in this case the "switching" happens at $$\t_0\$$ instead of 0, so $$\H(t-t_0)\$$ is more descriptive but $$\H(t)\$$ will give the same result.$$L^-{1}[F(s)e^{-t_0 s}]=f(t-t_0)H(t-t_0)\tag{equ 1}$$is the correct and most descriptive form. However the other forms are correct if it is "understood that they mean equ 1
Solutions for the Laplace transform are valid for $$\t \ge 0\$$.
• I edited the title, thanks. Why would $H(t)$ give the same result? I understand the answer you're giving, but I thought that between $H(t)$ and $H(t - t_0)$ there is a difference, hence the question: namely, I thought the latter is 0 also between t = 0 and $t_0$, while the former isn't. Do you mean they are the same if you understand the "switching" really happens at $t_0$? May 3, 2023 at 21:08
• Do the LT of $f(t-t_0)H(t)$ and then of $f(t-t_0)H(t-t_0)$ for any "causal" function $f(t)$ and see for your self. If the function is not causal then you are correct there will be a difference. May 3, 2023 at 21:49
• What about if I have more than one exponential? I was trying to plot the inverse transform of $Y(s)=4(e^{-s}/s^2-e^{-3s}/s^2)$ (that should be $y(t)=4(H(t-1)(t-1)-H(t-3)(t-3))$), but since the right form is that with $H(t-a)$ and I have two different exponentials I'm not sure how I should plot it: the first starts climbing at t = 1, the second at t = 3, so there is an initial ramp then it levels at y = 8? May 4, 2023 at 19:14
• @Mauro yes! that is correct May 4, 2023 at 21:29 | 885 | 2,953 | {"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": 17, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2024-10 | latest | en | 0.917052 |
https://www.physicsforums.com/threads/uniform-circular-motion-centripital-acceleration-vs-acceleration.174131/ | 1,539,739,941,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583510932.57/warc/CC-MAIN-20181017002502-20181017024002-00107.warc.gz | 1,025,115,785 | 13,277 | # Homework Help: Uniform Circular Motion: Centripital Acceleration vs. Acceleration
1. Jun 16, 2007
### Chele
I am very new to physics and am taking a my first University Physics class. This is not a call for a problem to be solved, but a clarification on terminology.
In solving problems for uniform circular motion, some problems call for the acceleration of the object (a=v^2/r) and others the centripital or instantanious acceleration (a=4pi^2r/T^2).
Can you please attempt to explain, in layman's terms, the difference between the two references to acceleration?
2. Jun 16, 2007
### Staff: Mentor
The two formulas are equivalent. (Express the speed in terms of circumference over period and you'll see for yourself.)
For uniform circular motion, the acceleration is centripetal. (Centripetal just means "towards the center".)
3. Jun 16, 2007
### Chele
Wow- I'll need to look at that in further detail.... :yuck:
4. Jun 16, 2007
### nrqed
Indeed, the two are equivalent for UCM. This can be seen easily if you recall that for constant speed, you may use v= distance/time. If you wait for the particle to go through a full circle, it will have covered a distance 2 Pi r, and the time elapsed will be the period T.
So
$v_{ucm} = \frac{2 \pi r}{T}$
Using this formula it is simple to prove that the two equations for acceleration you gave are equal.
5. Jun 17, 2007
### Chele
Okay, thanks. I worked it out and it is exactly the same. Not sure why I didn't see it before. Thanks guys! | 389 | 1,504 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2018-43 | latest | en | 0.912789 |
https://towerrumble.com/qa/what-4-numbers-add-up-to-42.html | 1,611,609,808,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703644033.96/warc/CC-MAIN-20210125185643-20210125215643-00727.warc.gz | 579,291,451 | 8,350 | # What 4 Numbers Add Up To 42?
## What 3 numbers add up to 42?
Therefore, three consecutive integers that add up to 42 are 13, 14, and 15..
## What 4 numbers add up to 40?
So there are 4 sets of 3 integers that are multiples of 4 which sum to 40. They are: 4 8 28 , 4 12 24 , 4 16 20, & 8 12 20.
## What times what gives you 36?
36 times 13636 times 63,3636 times 74,1236 times 84,4836 times 95,2418 more rows
## How do you find three consecutive integers?
Three consecutive even integers can be represented by x, x+2, x+4. The sum is 3x+6, which is equal to 108. Thus, 3x+6=108.
## What are consecutive numbers?
more … Numbers which follow each other in order, without gaps, from smallest to largest. 12, 13, 14 and 15 are consecutive numbers. 22, 24, 26, 28 and 30 are consecutive even numbers.
## What is the sum of all the factors of 42?
Why is 96 the sum of all factors of 42, including 42? The factors of the number 42 are: 1, 2, 3, 6, 7, 14, 21, 42. If you add them up, you will get the total of 96.
## What times what gives me 42?
Answer and Explanation: 6 times 7 is 42.
## What times what gives you 44?
44 = 1 x 44, 2 x 22, or 4 x 11. Factors of 44: 1, 2, 4, 11, 22, 44.
## What is a factor of 165?
Answer : 1,3,5,11,15,33,55,165, Related Links : Is 165 a rational number?
## What is the prime factor of 42?
The only way to write 42 as the product of primes (except to change the order of the factors) is 2 × 3 × 7. We call 2 × 3 × 7 the prime factorization of 42. It turns out that every counting number (natural number) has a unique prime factorization, different from any other counting number.
## What is the greatest common factor for 42?
How to find the greatest common factor (GCF)What are the factors of 12 ?1 , 2 , 3, 4 , 6 , 12What are the factors of 42 ?1 , 2 , 3 , 6 , 7 , 14 , 21 , 42What factors are common (What are factors of both 12 and 42 ?)1 , 2 , 3 , 6The Greatest Common Factor is ….6
## What two numbers add to 42?
For your convenience, we have made a list of all the combinations of two numbers multiplied by each other that will make 42: x 42 = 42. x 21 = 42. x 14 = 42.
## What gives you 40?
40 is a composite number. 40 = 1 x 40, 2 x 20, 4 x 10, or 5 x 8. Factors of 40: 1, 2, 4, 5, 8, 10, 20, 40. Prime factorization: 40 = 2 x 2 x 2 x 5, which can also be written 2³ x 5.
## What number can go into 45?
So, the factors of 45 are 1, 3, 5, 9, 15 and 45.
## What are the prime numbers of 30 and 40?
The prime numbers between 30 and 40 are: 31 and 37. To find this answer, you start with listing the factors of all numbers between 30 and 40.
## What is the sum of 40?
The number series 1, 2, 3, 4, . . . . , 39, 40. Therefore, 820 is the sum of positive integers upto 40.
## What are the prime factors of 45?
Students learn that the prime factorization of a number is the given number written as the product of its prime factors. For example, to find the prime factorization of 45, use a factor tree to find that 45 is 5 x 9, and 9 is 3 x 3. So the prime factorization of 45 is 5 x 3 x 3, or 5 x 3^2.
## What three consecutive numbers equal 51?
So the three numbers whose sum is 51 are 16, 17 and 18. | 1,068 | 3,169 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-04 | latest | en | 0.911633 |
https://www.projectrhea.org/rhea/index.php?title=Ece438f19bhelfrechtbonus&diff=next&oldid=77136 | 1,600,427,777,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400187390.18/warc/CC-MAIN-20200918092913-20200918122913-00128.warc.gz | 1,378,361,010 | 14,236 | Sampling rate conversion: an intuitive view
Background:
In signal processing, sampling is the act of converting a continuous-time (CT) signal into a discrete-time (DT) one. Although it may be easier to mathematically process a CT signal directly, this is not possible in reality since storing a real-world signal would require an infinite amount of memory. Consequently, all signals are sampled before being processed by a computer system. Sampling is performed at a certain rate limited by hardware known as the sampling rate/frequency(samples per second, or Hz), or with a certain time between samples known as the sampling period(seconds). It should be noted that these terms are reciprocals of each other. That is,
$f_s = 1 / T_s$
The connection between these two terms means that when we reduce the sampling rate, we increase the sampling period and vice versa. Therefore, reducing the sampling rate increases the length of time between consecutive measurements.
With this information, we can express the sampling of a CT signal $x(t)$in terms of its sampling period T:
$x(t) β$ sampling with period T $β x(nT) = x[n]$
The result of this expression, $x[n]$, is a discrete time signal where each value is taken from $x(t)$at an integer multiple of the sampling period T. Therefore, the CT signal $x(nT)$ is equivalent to the DT signal $x[n]$, where $n$ is an integer and indexes the DT signal.
If a DT signal is created by sampling a CT signal faster than a certain rate known as the Nyquist rate, the original CT signal can be perfectly reconstructed from the DT sampling. The Nyquist rate, $f_0$, is defined as:
$f_0 = 2f_{max}$, where $f_{max}$ is the greatest frequency in the signal of interest, $x(t)$.
If the sampling rate of a signal is not greater than $f_0$, the sampled signal will exhibit aliasing, which results when two distinct frequency bands overlap in the Fourier domain. In other words, aliasing occurs when the original signalβs frequency profile is altered after sampling. In this case, the CT signal cannot be reconstructed from the DT sampling without error.
For the purpose of this report, we will assume that we have measured a band-limited, real-world signal $x(t)$ with maximum frequency $f_{max}$ (in Hz). During measurement, the CT signal is sampled with sampling rate $f_1 = 1 / T_1$ to produce a DT signal $x(nT_1) = x_1[n]$. Note that the original time-domain signal is not accessible because it cannot be stored by a computer. The question is: can we convert the sampled signal $x_1[n]$ to a different signal $x_2[n]$ with sampling rate $T_2$ that is equivalent to a signal obtained by directly sampling $x(t)$with sampling period $T_2$? The figure below provides a graphical view of this question:
[ILLUSTRATION]
As we will see, it is indeed possible to obtain $x_2[n]$ from $x_1[n]$ by various means--downsampling, decimation, upsampling, and interpolation--collectively known as sampling rate conversion.
Downsampling:
We will begin by providing a motivating example for downsampling. Letβs say that a scientist aboard the International Space Station (ISS) would like to communicate with the engineers at Mission Control. The scientistβs voice has a maximum frequency $f_{max} = 1,000$ Hz, and the high-end analog-to-digital converters in his communication radio sample his speech well above the Nyquist rate at $f_s = 10,000$ Hz. However, a signal sampled this quickly contains a large number of data points, which will greatly increase the transmit power required to send the signal back to Earth. Due to limited power availability aboard the ISS, the scientist would like to keep all communications as low-power as possible. What can the scientist do to reduce his power consumption?
To reduce the required transmit power, the sampling rate of the speech signal must be reduced, or downsampled. Downsampling allows us to extract a second signal $x_2[n]$ with sampling period $T_2$ from another signal $x_1[n]$ with lower sampling period $T_1$. That is, downsampling requires $T_2 > T_1$, where $T_2$ is proportional to $T_1$ by a factor $D$, which is an integer greater than 1. From this, we establish the following relationship:
$T_2 = D * T_1$
or equivalently
$T_1 = T_2 / D$
The second equation may be useful in the future to help distinguish downsampling from upsampling, in that $T_1$ and $T_2$ appear in order left-to-right, and the factor $D$ appears below (down from) $T_2$. Overall, downsampling creates a new signal $x_2[n]$ by taking every $D^{th}$ sample of $x_1[n]$.
In the time domain, downsampling effectively removes all data between each $D^{th}$ sample. In the frequency domain, this process increases the signalβs maximum frequency (and therefore its bandwidth) and decreases its amplitude, both by a factor $D$. Why does this happen? Letβs take a look at an example.
Let $x_{10000}[n]$ be the scientistβs speech signal sampled at 10,000 Hz. Note that 10,000 Hz is the sampling frequency, so the sampling period, $T_s = 1 / f_s$, is 1/10000 seconds (0.1 ms). To save transmission power, he would like to downsample the signal by a factor of $D = 4$ to reduce the sampling rate $f_s$ to 2,500 Hz ($T_s = 1/2500$ seconds, 0.4 ms). From the connection between sampling rate and sampling period, it may seem like this increase in time between samples will decrease the maximum frequency of the original signal since samples are being taken less often (less frequently). However, it is important to note that this is not the case. Instead, the maximum frequency of the downsampled signal actually increases because samples of that frequency are moved closer together when data between them is removed. See the illustration below:
[ILLUSTRATION]
This explains the bandwidth increase seen after downsampling. The amplitude decrease is related to the Fourier transform formula.
Because downsampling increases the bandwidth of the original signal, one must be careful to prevent aliasing. To ensure that no aliasing occurs when downsampling, $x_2[n]$ must satisfy the Nyquist criterion. It can be shown that if $x_2[n]$ has a lower sampling rate than $x_1[n]$, if $x_2[n]$ satisfies Nyquist, then $x_1[n]$ will also. To meet this condition mathematically, we require:
$f_{max} < 1 / 2T_2$
or equivalently
$f_{max} < 1 / 2DT_1$, since $T_2 = D * T_1$
If we look at these requirements in terms of radians/sec which is commonly used when expressing the discrete-time Fourier transform, we get:
$2πT_1 * f_{max} < π / D$
or equivalently
$2πT_2 * f_{max} < π$, since $T_2 = D * T_1$
This conversion comes from the fact that $w = 2πf$ in continuous time. However, a sampled signal introduces an extra factor $T$, which is the sampling period of the signal. From this, we define the discrete time frequency $w = 2πfT$.
Finally, one should note that downsampling a signal with original sampling period $T_0$ by a factor $D$ is equivalent to resampling that signal with sampling period $T_1 = D*T_0$. In terms of the initial sampled signal $x_1[n]$ and its downsampled version $x_2[n]$, we have:
$x_2[n] = x_1[nD]$
Decimation:
Sometimes it is not possible to prevent aliasing in a sampled signal. We will provide an example of this by changing the scenario given above. Letβs say that it is the birthday of one of the engineers at Mission Control and the scientists aboard the ISS want to send them the βHappy Birthdayβ song which has $f_{max} = 3,000$ Hz. As before, the scientists would like to downsample the song by a factor of $D = 4$ before transmission to conserve power, but this will cause aliasing since $f_{max}$ will increase to 12,000 Hz after downsampling. How can the song be sent without introducing aliasing?
In cases where the sampling rate is not sufficient to prevent aliasing after downsampling, decimation can be used instead. This technique will remove high frequency content from the original signal by first passing it through a low pass filter to ensure the final downsampled signal satisfies the Nyquist criterion. See the illustration below:
$x_1[n]$ β Low pass filter with cutoff $π / D$ and gain 1 β Downsample by factor $D β x_2[n]$
This technique is used to band-limit $x1[n]$ to $π / D$ before downsampling by a factor of $D$ which increases the frequency range back to [-π, π]. This process guarantees that no aliasing occurs because the DTFT of a signal with bandwidth 2π (and period 2π, by the definition of the DTFT) will never overlap itself. However, some high frequency content from the original signal is inevitably lost.
For our example the original sampled signal $x1[n]$ sampled at $T_1 = 10,000$ Hz must first be low pass filtered with a cutoff frequency of $π / D = π / 4$ to remove the frequency content that will cause aliasing when its bandwidth is increased by a factor of 4. Downsampling by a factor of 4 will then cause the bandwidth to fill [-π, π]. The result will be $x_2[n]$, sampled at 2,500 Hz. Although some high frequency content was lost, the resulting signal does not show aliasing in the frequency domain, which would have occurred if the low pass filter was not applied.
It is important to note that downsampling and decimation produce equivalent results if the original signal $x_1[n]$ satisfies the Nyquist condition for the sampling period $T_2$ of the downsampled signal $x_2[n]$. In other words, if no aliasing will occur after downsampling (equations XXX and XXX are satisfied), the low pass filter used in decimation will have no effect, and both processes will produce the same result. In the example where the scientist only sought to transmit his voice, both downsampling and decimation would have produced the same result since $f_{max}[itex] after downsampling would be 4 * 1,000 Hz = 4,000 Hz. Since [itex]f_s = 10,000 Hz < 2f_{max, 2} = 8,000$ Hz, the Nyquist criterion is satisfied for $T_2 = 1 / 4,000$.
Below is a summary of downsampling and decimation in pictorial form:
[SUMMARY ILLUSTRATION]
Note the low pass filters used when sampling $x(t)$ with period $T_1$ and $T_2$. We use these filters because we have specified that $f_{max} > 1/2T_1$, so aliasing is guaranteed to occur when sampling with $T_1$. If aliasing will occur with smaller sampling period $T_1$, it is also guaranteed to occur with larger sampling period $T_2$, hence the cutoff frequency of $1/2T_2$ for $x_2[n]$. These filters ensure that no aliasing occurs during the initial sampling of $x(t)$ to create $x_1[n]$ and $x_2[n]$.
Upsampling:
There are some instances in which the sampling rate of a given signal is lower than we would like it to be for processing. To demonstrate this, we will continue our previous scenario by imagining that the engineerβs radio at Mission Control has a fixed playback rate of 10,000 Hz. If he tries to play the signals sent by the scientist in the ISS back at this rate, they will be played 4x too fast. To solve this problem and to play back the audio at the proper speed, the engineer must upsample the signal by a factor of 4 to bring its sampling rate back to 10,000 Hz.
In contrast to downsampling and decimation, upsampling allows us to extract a second signal $x_2[n]$ with sampling period $T_2$ from another signal $x_1[n]$ with higher sampling period $T_1[itex]. In other words, upsampling requires [itex]T_1 > T_2$, where $T_1$ is proportional to $T_2$ by a factor $D$, which is an integer greater than 1. In the time domain, $x(nT_1)$ is sampled less often than $x(nT_2)$. With this, we can establish the following relationship:
$T_1 = D * T_2$
If we compare the above equation to equation XXX, we can see a convenient pattern in upsampling versus downsampling. In the upsampling formula above, $T_1$ and $T_2$ appear in order left-to-right, and the factor $D$ appears in the numerator (up) with $T_2$. Overall, upsampling creates a new signal $x_2[n]$ by placing $D - 1$ zeros between every sample of a given signal $x_1[n]$. This process is also known as zero-padding. Mathematically, we can represent the relationship between $x_1[n]$ and $x_2[n]$ as:
x2[n] = { x1[n / D], if n / D is integer { 0, otherwise
In the time domain, zero-padding causes the samples of $x[n]$ to occur less often, thereby decreasing the frequency content of the signal. In the frequency domain, this compresses the frequency axis by a factor $D$ so that multiple copies of the original signal may appear in the range [-π, π]. The new signal is now periodic with period $2π / D$. This rescaling also increases the amplitude of the upsampled signalβs frequency profile by a factor $D$ by the definition of the Fourier transform.
Similar to the fact that downsampling is not necessarily the best way to decrease the sampling rate of a signal, upsampling is not necessarily the best way to increase the sampling rate of a signal. When the signalβs frequency content is scaled down by a factor $D$, more copies of the signal may appear in the interval [-π, π]. These extra copies contribute to aliasing when the signal is transformed back to the time domain. To fix this, the upsampling equivalent of decimation known as interpolation can be used.
Interpolation:
Aliasing introduced by upsampling can be prevented by low pass filtering the upsampled signal with a cutoff frequency of $π / D$ and gain $D$. This low pass filter eliminates the frequency copies not centered directly around $w = 0$ in the frequency domain. See the diagram below:
$x_1[n]$ β upsample by factor $D$ β low pass filter with cutoff $π / D$ and gain $D β x_2[n]$
The low-pass filter ensures that only the center frequency copy remains in the interval [-π, π] and eliminates aliasing when the inverse Fourier transform is taken.
Summary:
In summary, downsampling, decimation, upsampling, and interpolation can all be used to convert one sampling rate to another. In some cases, care must be taken to prevent aliasing. These methods allow one to convert a given sampled signal into a second signal sampled at a different rate, which can be useful in many situations. | 3,541 | 14,133 | {"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} | 3.859375 | 4 | CC-MAIN-2020-40 | longest | en | 0.931963 |
https://www.idcafe.net/how-do-i-calculate-my-payroll-taxes/ | 1,721,420,670,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514928.31/warc/CC-MAIN-20240719200730-20240719230730-00684.warc.gz | 709,698,966 | 11,363 | # How do I calculate my payroll taxes?
## How do I calculate my payroll taxes?
To calculate Social Security withholding, multiply your employee’s gross pay for the current pay period by the current Social Security tax rate (6.2%). To calculate Medicare withholding, multiply your employee’s gross pay by the current Medicare tax rate (1.45%).
### What percentage is payroll tax in PA?
3.07 percent
About Employer Withholding Taxes: Employers are required to withhold PA personal income tax at a flat rate of 3.07 percent of compensation from resident and nonresident employees earning income in Pennsylvania. This rate remains in effect unless you receive notice of a change from the Department of Revenue.
#### How do I calculate the percentage of taxes taken out of my paycheck?
How do I calculate taxes from paycheck? Calculate the sum of all assessed taxes, including Social Security, Medicare and federal and state withholding information found on a W-4. Divide this number by the gross pay to determine the percentage of taxes taken out of a paycheck.
What is the payroll tax rate for 2020?
4.85 per cent
Rates and thresholds The payroll tax threshold increased to \$1.2 million from 1 July 2020. The current payroll tax rate is 4.85 per cent. View previous rates and thresholds.
How Much Does employer pay in payroll taxes?
In Alberta, businesses must remit 10% in provincial tax on annual taxable income from \$0 to \$131,220.00 — or \$100 of \$1000 in wages.
## How much tax is deducted from a paycheck in Pennsylvania?
Pennsylvania levies a flat state income tax rate of 3.07%.
### How much taxes do you pay on 1000?
The tax rate will be determined by your income. So, for instance, if you make \$42,000 annually and file as single, your federal tax rate is 22%. If you win \$1,000, your total income is \$43,000, and your tax rate is still 22%.
#### What is payroll tax vs income tax?
What’s the Difference Between Payroll and Income Taxes? The key difference is that payroll taxes are paid by employer and employee; income taxes are only paid by employers. However, both payroll and income taxes are required to be withheld by employers when they make payroll.
What percentage is the payroll tax?
The current tax rate for social security is 6.2% for the employer and 6.2% for the employee, or 12.4% total. The current rate for Medicare is 1.45% for the employer and 1.45% for the employee, or 2.9% total.
How do you figure payroll taxes?
How to calculate FICA payroll tax. Social Security withholding. To calculate Social Security withholding, multiply your employee’s gross pay for the current pay period by the current Social Security tax rate (6.2%). This is the amount you will deduct from your employee’s paycheck and remit along with your payroll taxes.
## What is the tax withholding rate in Pennsylvania?
Federal law and some states require that employers use tax tables to figure income tax withholding. Pennsylvania, however, mandates a flat withholding rate of 3.07 percent for state income tax.
### How do you calculate taxes taken from your paycheck?
Add the taxes assessed to determine the total amount of tax to withhold from an employee’s check. Divide the result by gross pay to establish the percentage of the paycheck dedicated to taxes. To establish the total percentage of taxes withheld for all employees, add the taxes taken out of each individual employee’s check and total the result.
#### How do you calculate salary after taxes?
To figure the employee’s after tax income, one would simply subtract the deductible amount from the gross income and multiply the result by the tax rate. In addition to deductions, some tax systems also offer extra complexity to the process of figuring salary after tax. | 800 | 3,764 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-30 | latest | en | 0.940258 |
https://amathew.wordpress.com/tag/number-fields/ | 1,679,730,046,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945317.85/warc/CC-MAIN-20230325064253-20230325094253-00749.warc.gz | 125,855,484 | 23,733 | Class field theory is about the abelian extensions of a number field ${K}$. Actually, this is strictly speaking global class field theory (there is an analog for abelian extensions of local fields), and there is a similar theory for function fields of transcendence degree 1 over finite fields, but we shall not deal with it.
Let us, however, consider the situation for local fields—which we will later investigate more—as follows. Suppose ${k}$ is a local field and ${L}$ an unramified extension. Then the Galois group ${L/k}$ is isomorphic to the Galois group of the residue field extension, i.e. is cyclic of order ${f}$ and generated by the Frobenius. But I claim that the group ${k^*/NL^*}$ is the same. Indeed, ${NU_L = U_K}$ by a basic theorem about local fields that we will prove using abstract nonsense later (but can also be easily proved using successive approximation and facts about finite fields). So ${k^*/NL^*}$ is cyclic, generated by a uniformizer of ${k}$, which has order ${f}$ in this group. Thus we get an isomorphism
$\displaystyle k^*/NL^* \simeq G(L/k)$
sending a uniformizer to the Frobenius. (more…)
As usual, let ${K}$ be a global field. Now we do the same thing that we did last time, but for the ideles.
1. Ideles
First of all, we have to define the ideles. These are only a group, and are defined as the restricted direct product
$\displaystyle J_K = \prod'_v K_v^*$
relative to the unit subgroups ${U_v}$ of ${v}$-units (which are defined to be ${K_v^*}$ if ${v}$ is archimedean). In other words, an idele ${(x_v)_v}$ is required to satisfy ${|x_v|=1}$ for almost all ${v}$.
If ${S}$ is a finite set of places containing the archimedean ones, we can define the subset ${J^S_K = \prod_{v \in S} K_v \times \prod_{v \notin S} U_v}$; this has the product topology and is an open subgroup of ${J_K}$. These are called the ${S}$-ideles. As we will see, they form an extremely useful filtration on the whole idele group.
Dangerous bend: Note incidentally that while the ideles are a subset of the adeles, the induced topology on ${J_K}$ is not the ${J_K}$-topology. For instance, take ${K=\mathbb{Q}}$. Consider the sequence ${x^{(n)}}$ of ideles where ${x^{(n)}}$ is ${p_n}$ at ${v_{p_n}}$ (where ${p_n}$ is the ${n}$-th prime) and 1 everywhere else. Then ${x^{(n)} \rightarrow 0 \in \mathbf{A}_{\mathbb{Q}}}$ but not in ${J_{\mathbb{Q}}}$.
However, we still do have a canonical “diagonal” embedding ${K^* \rightarrow J_K}$, since any nonzero element of ${K}$ is a unit almost everywhere. This is analogous to the embedding ${K \rightarrow \mathbf{A}_K}$. (more…)
Adele – c’est un nom si belle. (Oops, that’s bad French, isn’t it?)
I actually will not be able to finish the proof of the unit theorem here, because I don’t get to the ideles in this post. That will come next time (there are some of the same themes as here).
Let ${K}$ be a global field, i.e. a finite extension of either ${\mathbb{Q}}$ or ${\mathbb{F}_p(t)}$. Then we can consider the set absolute values on ${K}$. In the number field case, these are extensions (up to a power) of the archimedean absolute value on ${\mathbb{Q}}$ or the ${p}$-adic absolute values by a theorem of Ostrowski classifying absolute values on ${\mathbb{Q}}$. In the function field case, we need another result.
Here’s how we define the adele ring. It is the restricted direct product
$\displaystyle \mathbf{A}_K := \prod'_v K_v$
where restricted means that any vector ${(x_v)_{v \in V} \in \mathbf{A}_K}$ is required to satisfy ${|x_v|_v \leq 1}$ for almost all ${v}$. This becomes a topological ring if we take a basis of the form
$\displaystyle \prod_{v \in S} T_v \times \prod_{v \notin S} \mathcal{O}_v$
where ${T_v \subset k_v}$ are open and ${\mathcal{O}_v}$ is the ring of integers, and ${S}$ is a finite set containing the archimedean places. It is clear that addition and multiplication are continuous, and that ${A_K}$ is locally compact. For ${S}$ finite and containing the archimedean absolute values ${S_\infty}$, there is a subring ${\mathbf{A}_K^S = \prod_{v \in S} K_v \times \prod_{v \notin S} \mathcal{O}_v}$, and ${\mathbf{A}_K}$ is the union of these subrings.
Since any ${x \in K}$ is contained in ${\mathcal{O}_v}$ for almost all ${v}$ (this is analogous to a rational function on a curve having only finitely many poles), there is an injective homomorphism ${K \rightarrow \mathbf{A}_K}$.
Next, we may define a Haar measure on ${\mathbf{A}_K^S}$ by taking the product of the Haar measures ${\mu_v}$ on ${K_v}$, normalized such that ${\mu_v(O_v )=1}$ for ${v \notin S_\infty}$. Thus one gets a (i.e., the) Haar measure on ${\mathbf{A}_K}$ itself. (more…)
There is another major result in algebraic number theory that we need to get to! I have this no longer secret goal of getting to class field theory, and if it happens, this will be a key result. The hard part of the actual proof (namely, the determination of the rank of a certain lattice) will be deferred until next time; it’s possible to do it with the tools we already have, but it is cleaner (I think) to do it once ideles have been introduced.
Following the philosophy of examples first, let us motivate things with an example. Consider the ring ${\mathbb{Z}[i]}$ of Gaussian; as is well-known, this is the ring of integers in the quadratic field ${\mathbb{Q}(i)}$. To see this, suppose ${a+ib, a, b \in \mathbb{Q}}$ is integral; then so is ${a-ib}$, and thus ${2a,2b}$ are integers. Also the fact ${(a+ib)(a-ib) = a^2+b^2 }$ must be an integer now means that neither ${a,b}$ can be of the form ${k/2}$ for ${k}$ odd.
What are the units in ${\mathbb{Z}[i]}$? If ${x}$ is a unit, so is ${\bar{x}}$, so the norm ${N(x)}$ must be a unit in ${\mathbb{Z}[i]}$ (and hence in ${\mathbb{Z}}$). So if ${x=a+ib}$, then ${a^2+b^2=1}$ and ${x = \pm 1}$ or ${\pm i}$. So, the units are just the roots of unity.
In general, however, the situation is more complicated. Consider ${\mathbb{Z}[\sqrt{2}]}$, which is again integrally closed. Then ${x=a+b\sqrt{2}}$ is a unit if and only if its norm to ${\mathbb{Q}}$, i.e. ${a^2 - 2b^2}$ is equal to ${\pm 1}$. Indeed, the norm ${N(x)}$ of a unit ${x}$ is still a unit, and since ${\mathbb{Z}}$ is integrally closed, we find that ${N(x)}$ is a ${\mathbb{Z}}$-unit. In particular, the units correspond to the solutions to the Pell equation. There are infinitely many of them.
But the situation is not hopeless. We will show that in any number field, the unit group is a direct sum of copies of ${\mathbb{Z}}$ and the roots of unity. We will also determine the rank. (more…)
This post won’t be as cool as the title sounds. But I will prove something neat, and it will lead to neater things as time goes on (assuming I keep posting on this topic).
We will now return to algebraic number theory, following Lang’s textbook, and study the distribution of points in parallelotopes.
The setup is as follows. ${K}$ will be a number field, and ${v}$ an absolute value (which, by abuse of terminology, we will use interchangeable with “valuation” and “place”) extending one of the absolute values on ${\mathbb{Q}}$ (which are always normalized in the standard way); we will write ${|x|_v}$ for the output at ${x \in K}$.
Suppose ${v_0}$ is a valuation of ${\mathbb{Q}}$; we write ${v | v_0}$ if ${v}$ extends ${v_0}$. Recall the following important formula from the theory of absolute values an extension fields:
$\displaystyle |N^K_{\mathbb{Q}}(x)|_{v_0} = \prod_{v | v_0} |x|_v^{ [K_v: \mathbb{Q}_{v_0} ] }.$
Write ${N_v := [K_v: \mathbb{Q}_{v_0} ] }$; these are the local degrees. From elementary algebraic number theory, we have ${\sum_{v | v_0} N_v = N := [K:\mathbb{Q}].}$ This is essentially a version of the ${\sum ef = N}$ formula.
The above formalism allows us to deduce an important global relation between the absolute values ${|x|_v}$ for ${x \in K}$.
Theorem 1 (Product formula) If ${x \neq 0}$,$\displaystyle \prod_v |x|_v^{N_v} = 1.$
The proof of this theorem starts with the case ${K=\mathbb{Q}}$, in which case it is an immediate consequence of unique factorization. For instance, one can argue as follows. (more…) | 2,388 | 8,144 | {"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": 117, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2023-14 | latest | en | 0.912962 |
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