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http://www.ms.uky.edu/~lee/ma310/prob/node11.html | 1,508,725,731,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825510.59/warc/CC-MAIN-20171023020721-20171023040721-00615.warc.gz | 505,532,816 | 1,812 | Next: SETI Up: Problems Previous: Consecutive Integers
# Numerous Offspring
``I hear some youngsters playing in the back yard,'' said Jones, a graduate student in mathematics. ``Are they all yours?''
``Heavens, no,'' exclaimed Professor Smith, the eminent number theorist. ``My children are playing with friends from three other families in the neighborhood, although our family happens to be largest. The Browns have a smaller number of children, the Greens have a still smaller number, and the Blacks the smallest of all.''
``How many children are there altogether?'' asked Jones.
``Let me put it this way,'' said Smith. ``There are fewer than 18 children, and the product of the numbers in the four families happens to be my house number which you saw when you arrived.''
Jones took a notebook and pencil from his pocket and started scribbling. A moment later he looked up and said, ``I need more information. Is there more than one child in the Black family?''
As soon as Smith replied, Jones smiled and correctly stated the number of children in each family.
Knowing the house number and whether the Blacks had more than one child, Jones found the problem trivial. it is a remarkable fact, however, that the number of children in each family can be determined solely on the basis of the information given above!
Carl Lee
Wed Apr 21 08:26:07 EDT 1999 | 294 | 1,363 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2017-43 | latest | en | 0.960756 |
http://www.mathpages.com/home/kmath358.htm | 1,495,681,429,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607963.70/warc/CC-MAIN-20170525025250-20170525045250-00127.warc.gz | 579,824,259 | 3,850 | ## Pythagorean Graphs
```Is it possible to fill an infinite square array with distinct
integers such that the sum of the squares of any two adjacent
numbers is a square? To illustrate, the following is a 4x4
array with the desired property
1836 105 252 735
1248 100 240 700
936 75 180 525
273 560 1344 3920
Every pair of neighboring numbers (horizonally or vertically)
constitutes the legs of a Pythagorean triple. The hypotenuses
of these triples are as shown below
1839 273 777
2220 145 348 1015
1252 260 740
1560 125 300 875
939 195 555
975 565 1356 3955
623 1456 4144
The key to this problem is to recognize that the row and column
conditions are independent. Thus, for any positive integer N we
can easily construct a square array of size (N+1)x(N+1) consisting
of the values
A[m,n] = (12/5)^m (24/7)^n 35^N
= 2^(2m+3n) 3^(m+n) 5^(N-m) 7^(N-n)
for m,n = 0,1,2,..,N. These values are all integers and no
two of them have the same number of factors of 5 and 7, so
they are all distinct.
Although this gives a recipe for arbitrarily large arrays, it
doesn't give an *infinite* array. R. Mentock provided the
following nice construction of an infinite array.
For m=3M+a, n=3N+b, define F(m,n) by
F(m,n) = 6^M * 132^N * f(a,b)
where
f(0,0) = 6 * 7 f(0,1) = 6 * 24 f(0,2) = 6 * 143
f(1,0) = 8 * 7 f(1,1) = 8 * 24 f(1,2) = 8 * 143
f(2,0) = 15* 7 f(2,1) = 15* 24 f(2,2) = 15* 143
Another, shorter, solution: For m=2M+a, n=2N+b, define
F(m,n) by
F(m,n) = 10^M * 76^N * f(a,b)
where
f(0,0) = 7 * 17 f(0,1) = 7 * 144
f(1,0) = 24 * 17 f(1,1) = 24 * 144
Both of the above "infinite" solutions actually fill only one
quadrant of the plane. To completely "tile" the plane with
Pythaogrean legs, Mentock notes that
Each of the two solutions are actually combinations of two
sequences, each defined along the axes. The other points are
just products of these, i.e., G(m,n) = G(m) * G(n). Since we
have four of them, we start at zero and assign values to the
four semi-axes. G(0,0) is the LCM of all G(0), and that number
is propogated through. Then G(m,n) = G(m) * G(n) like above,
and there we go.
A slightly different approach was suggested by Dave Radcliffe,
who wrote
Given F(m,n) = 10^M * 76^N * f(a,b) where
f(0,0) = 7 * 17 f(0,1) = 7 * 144
f(1,0) = 24 * 17 f(1,1) = 24 * 144
For m,n >= 0 define
G(m,n) = F(m,n) * 11 * 13
G(-m-1,n) = F(m,n) * 11 * 84
G(-m-1,-n-1) = F(m,n) * 60 * 84
G(m,-n-1) = F(m,n) * 60 * 13
Since no F(m,n) is divisible by 11 or 13, G is also one-to-one.
By the way, it's also interesting to see which polyhedra can
have distinct numbers at their vertices such that the sum of
the squares of any two adjacent numbers is a square. Considering
just the Platonic solids, I think there is no solution for the
tetrahedron, octahedron, or icosahedron. Of course there are
solutions for the cube, such as
225--------240
/ /|
/ / |
120--------160 |
| | |
|(540) | 720
| | /
| |/
288--------384
Here the values increase by factors of 4/3, 12/5, and 15/8 in the
three principle directions. (Presumably it's possible to fill an
infinite 3D orthogonal lattice with distinct numbers in this way.)
This leaves only the dodecahedron. It turns out that it is possible
to populate the vertices of a dodecahedron with distinct numbers
such that the sum of the squares of each pair of adjacent numbers
is a square. Here's an example of a "Pythagorean dodecahedron":
a_______________________b
/\ /\
/ \ / \
/ \f_______q_______g/ \
/ / | \ \
/ p/ |k \r \
/ /\ / \ /\ \
/ / \ / \ / \ \
/ / \/ \/ \ \
e/____j/ o| |l \h____\c
\ \ |_________| / /
\ \ /n m\ / /
\ \ / \ / /
\ t\ /s /
\ \ / /
\ \ / /
\ \ / /
\ |i /
\ | /
\ | /
d
a = 155040 f = 649800 k = 802560 p = 2006400
b = 290700 g = 1438965 l = 601920 q = 180576
c = 897600 h = 3762000 m = 451440 r = 689700
d = 1683000 i = 2613600 n = 240768 s = 846450
e = 177650 j = 1504800 o = 1070080 t = 275880
Another related question, raised by Dave Radcliffe, is whether it
it possible to connect any two positive integers (greater than 4)
by a series of Pythagorean legs. (As Dave put it, "consider a graph
G with a vertex for each integer n > [4], and an edge joining
m and n iff m^2 + n^2 is a square. Is this graph connected?")
Notice that the numbers 3,4 are a separate graph and don't connect
to anything else. But for n > 4 it certainly appears to be true
that the graph is connected.
If you look at this as a special case of the more general "connection
criterion" m^2 + n^2 = k*square (where k is a square-free integer
expressible as a sum of two squares), it seems the case k=1 is unique
in giving a connected graph for all the integers (above some number).
In contrast, consider the graph based on m^2 + n^2 = 2*square. In this
case the naturals split into mutually disjoint graphs, each of which is
a multiple of the basic connected graph consisting of the numbers
1, 7, 17, 23, 31, 41, 47, 49, 71, 73, 79, 89, 97, 103, 113, 119, ...
These are exactly the numbers whose prime factors are all congruent
to +1 or -1 modulo 8. In other words, they are the primes p such that
2 is a square (mod p).
The graph based on m^2 + n^2 = 5*square splits into mutually disjoint
graphs, each of which is a multiple of
1, 2, 4, 8, 11, 16, 19, 22, 29, 31, 32, 38, 41, 44, 58, 59, 61, 62, ...
These appear to be exactly the numbers expressible as products of the
primes 2, 11, 19, 31, 41, 59, 61, 71, 79, 89, 101, ... which I think
are the primes p such that 5 is a square (mod p).
The graph based on m^2 + n^2 = 10*square splits into disjoint graphs
that are each multiples of
1, 3, 9, 13, 27, 31, 37, 39, 41, 43, 53, 67, 71, 79, 81, 83, ...
Again it appears there are all the products of a particular set of
primes, in this case the primes 3, 13, 31, 37, 41, etc., and these
look like the primes modulo which 10 is a square.
So, we might be tempted to conclude that the graph of the naturals
with the connection criterion m^2 + n^2 = k*square splits into disjoint
graphs, each of which is a multiple of a graph consisting of the
numbers expressible as a product of primes p not dividing k and such
that k is a quadratic residue mod p. (This explains why the case k=1
gives just a single totally connected graph.) However, the case k=13
seems to behave differently. It's basic connected set consists of
1, 8, 12, 18, 27, 34, 51, 53, 79, 86, 92, 96, 103, 116, 122, ...
The set of primes for this case should be 2,3,13,17,23,29,43,etc. but
there seems to be something else going on here.
``` | 2,433 | 7,085 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2017-22 | longest | en | 0.820644 |
https://trumpexcel.com/return-cell-address-instead-of-value-excel/ | 1,722,826,576,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640427760.16/warc/CC-MAIN-20240805021234-20240805051234-00205.warc.gz | 471,881,858 | 76,988 | Search
# How to Return Cell Address Instead of Value in Excel (Easy Formula)
When using lookup formulas in Excel (such as VLOOKUP, XLOOKUP, or INDEX/MATCH), the intent is to find the matching value and get that value (or a corresponding value in the same row/column) as the result.
But in some cases, instead of getting the value, you may want the formula to return the cell address of the value.
This could be especially useful if you have a large data set and you want to find out the exact position of the lookup formula result.
There are some functions in Excel that designed to do exactly this.
In this tutorial, I will show you how you can find and return the cell address instead of the value in Excel using simple formulas.
This Tutorial Covers:
The ADDRESS function in Excel is meant to exactly this.
It takes the row and the column number and gives you the cell address of that specific cell.
Below is the syntax of the ADDRESS function:
`=ADDRESS(row_num, column_num, [abs_num], [a1], [sheet_text])`
where:
• row_num: Row number of the cell for which you want the cell address
• column_num: Column number of the cell for which you want the address
• [abs_num]: Optional argument where you can specify whether want the cell reference to be absolute, relative, or mixed.
• [a1]: Optional argument where you can specify whether you want the reference in the R1C1 style or A1 style
• [sheet_text]: Optional argument where you can specify whether you want to add the sheet name along with the cell address or not
Now, let’s take an example and see how this works.
Suppose there is a dataset as shown below, where I have the Employee id, their name, and their department, and I want to quickly know the cell address that contains the department for employee id KR256.
Below is the formula that will do this:
`=ADDRESS(MATCH("KR256",A1:A20,0),3)`
In the above formula, I have used the MATCH function to find out the row number that contains the given employee id.
And since the department is in column C, I have used 3 as the second argument.
This formula works great, but it has one drawback – it won’t work if you add the row above the dataset or a column to the left of the dataset.
This is because when I specify the second argument (the column number) as 3, it’s hard-coded and won’t change.
In case I add any column to the left of the dataset, the formula would count 3 columns from the beginning of the worksheet and not from the beginning of the dataset.
So, if you have a fixed dataset and need a simple formula, this will work fine.
But if you need this to be more fool-proof, use the one covered in the next section.
## Lookup And Return Cell Address Using the CELL Function
While the ADDRESS function was made specifically to give you the cell reference of the specified row and column number, there is another function that also does this.
Below is the syntax of the CELL function:
`=CELL(info_type, [reference])`
where:
• info_type: the information about the cell you want. This could be the address, the column number, the file name, etc.
• [reference]: Optional argument where you can specify the cell reference for which you need the cell information.
Now, let’s see an example where you can use this function to look up and get the cell reference.
Suppose you have a dataset as shown below, and you want to quickly know the cell address that contains the department for employee id KR256.
Below is the formula that will do this:
`=CELL("address", INDEX(\$A\$1:\$D\$20,MATCH("KR256",\$A\$1:\$A\$20,0),3))`
The above formula is quite straightforward.
I have used the INDEX formula as the second argument to get the department for the employee id KR256.
And then simply wrapped it within the CELL function and asked it to return the cell address of this value that I get from the INDEX formula.
Now here is the secret to why it works – the INDEX formula returns the lookup value when you give it all the necessary arguments. But at the same time, it would also return the cell reference of that resulting cell.
In our example, the INDEX formula returns “Sales” as the resulting value, but at the same time, you can also use it to give you the cell reference of that value instead of the value itself.
Normally, when you enter the INDEX formula in a cell, it returns the value because that is what it’s expected to do. But in scenarios where a cell reference is required, the INDEX formula will give you the cell reference.
In this example, that’s exactly what it does.
And the best part about using this formula is that it is not tied to the first cell in the worksheet. This means that you can select any data set (which could be anywhere in the worksheet), use the INDEX formula to do a regular look up and it would still give you the correct address.
And if you insert an additional row or column, the formula would adjust accordingly to give you the correct cell address.
So these are two simple formulas that you can use to look up and find and return the cell address instead of the value in Excel.
I hope you found this tutorial useful.
Other Excel tutorials you may also like:
FREE EXCEL BOOK
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Get 51 Excel Tips Ebook to skyrocket your productivity and get work done faster | 1,316 | 5,951 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2024-33 | latest | en | 0.883155 |
http://www.velocityreviews.com/forums/t882757-new-faq-how-do-i-compute-the-difference-of-two-arrays.html | 1,386,449,323,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163055810/warc/CC-MAIN-20131204131735-00074-ip-10-33-133-15.ec2.internal.warc.gz | 600,185,728 | 11,673 | Velocity Reviews > Perl > New FAQ: How do I compute the difference of two arrays?
# New FAQ: How do I compute the difference of two arrays?
The Poor
Guest
Posts: n/a
09-26-2003
Here is my new code changed from the old code in FAQ.
Please add it to the perldoc for the next release of perl.
OLD:
+
How do I compute the difference of two arrays? How do I compute the
intersection of two arrays?
Use a hash. Here's code to do both and more. It assumes that each
element is unique in a given array:
@union = @intersection = @difference = ();
%count = ();
foreach \$element (@array1, @array2) { \$count{\$element}++ }
foreach \$element (keys %count) {
push @union, \$element;
push @{ \$count{\$element} > 1 ? \@intersection :
\@difference }, \$element;
}
Note that this is the *symmetric difference*, that is, all
elements in
either A or in B but not in both. Think of it as an xor operation.
-
New:
+
How do I compute the difference of two arrays? How do I compute the
intersection of two arrays?
Use a hash. Here's code to do both and more. It assumes that each
element is unique in a given array:
@union = @intersection = @in1notin2 = @in2notin1();
%count = ();
foreach \$element (@array1) { \$count{\$element}++ }
foreach \$element (@array2) { \$count{\$element}-- }
foreach \$element (keys %count) {
push @union, \$element;
push (@intersection, \$element) if \$count{\$element} == 0;
push (@in1notin2, \$element) if \$count{\$element} == 1;
push (@in2notin1, \$element) if \$count{\$element} == -1;
}
-
2 Questions:
how to fix it when the array is not uniqu?
how to simplfy the last 3 lines using case, ?: or something else?
push @intersection, \$element if \$count{\$element} == 0;
push @in1notin2, \$element if \$count{\$element} == 1;
push @in2notin1, \$element if \$count{\$element} == -1;
http://edealseek.com
Steve Grazzini
Guest
Posts: n/a
09-26-2003
The Poor <(E-Mail Removed).2y.net> wrote:
> Here is my new code changed from the old code in FAQ.
> Please add it to the perldoc for the next release of perl.
This is a discussion group. FAQ patches (preferably in the form
of unified or context diffs) can be mailed to perlfaq-workers at
perl.org or posted from nntp.perl.org.
You should also include some rationale -- what's wrong with the
I'm not sure that it is an improvement, but you'll have to make
a case for it. (And fix the syntax error, etc.)
--
Steve
Jay Tilton
Guest
Posts: n/a
09-26-2003
Steve Grazzini <(E-Mail Removed)> wrote:
: The Poor <(E-Mail Removed).2y.net> wrote:
: > Here is my new code changed from the old code in FAQ.
: > Please add it to the perldoc for the next release of perl.
:
: This is a discussion group. FAQ patches (preferably in the form
: of unified or context diffs) can be mailed to perlfaq-workers at
: perl.org or posted from nntp.perl.org.
:
: You should also include some rationale -- what's wrong with the
: I'm not sure that it is an improvement, but you'll have to make
: a case for it. (And fix the syntax error, etc.)
As well, when making that case it is good to keep in mind that the FAQ
answers are not intended to form a cookbook of pre-fab do-anything
solutions. That's more the domain of CPAN, or the actual Perl Cookbook.
Recipe 4.7 in Perl Cookbook (1st ed.), by the way, directly addresses
the problem of "Finding Elements in One Array but Not Another."
A more useful (and easier) change to the FAQ would be to add mention of
a module on CPAN that handles permutations of the question that the FAQ
module appears to be especially capable.
Bob Walton
Guest
Posts: n/a
09-27-2003
The Poor wrote:
....
> 2 Questions:
> how to fix it when the array is not uniqu?
Make it unique:
{my %h or @array1_unique=(@h{@array1}=1 and keys %h)}
> how to simplfy the last 3 lines using case, ?: or something else?
>
> push @intersection, \$element if \$count{\$element} == 0;
> push @in1notin2, \$element if \$count{\$element} == 1;
> push @in2notin1, \$element if \$count{\$element} == -1;
One way:
push @{\$result[\$count{\$element}+1]},\$element;
#@{\$result[0][...]} is in2notin1
#@{\$result[1][...]} is intersection
#@{\$result[2][...]} is in1notin2
--
Bob Walton
Randal L. Schwartz
Guest
Posts: n/a
09-27-2003
>>>>> "The" == The Poor <(E-Mail Removed).2y.net> writes:
The> -
The> New:
The> +
The> How do I compute the difference of two arrays? How do I compute the
The> intersection of two arrays?
The> Use a hash. Here's code to do both and more. It assumes that each
The> element is unique in a given array:
Here's code that doesn't require that uniqueness, and would be a better
candidate for the FAQ:
my %tally;
\$tally{\$_} .= "a" for @array_a;
\$tally{\$_} .= "b" for @array_b;
my @union = keys %tally;
my @intersection = grep \$tally{\$_} =~ /ab/, @union;
my @a_not_b = grep \$tally{\$_} =~ /a\$/, @union;
my @b_not_a = grep \$tally{\$_} =~ /^b/, @union;
print "Just another Perl hacker,";
--
Randal L. Schwartz - Stonehenge Consulting Services, Inc. - +1 503 777 0095
<(E-Mail Removed)> <URL:http://www.stonehenge.com/merlyn/>
Perl/Unix/security consulting, Technical writing, Comedy, etc. etc.
See PerlTraining.Stonehenge.com for onsite and open-enrollment Perl training!
Gregory Toomey
Guest
Posts: n/a
09-27-2003
It was a dark and stormy night, and Randal L. Schwartz managed to scribble:
>>>>>> "The" == The Poor <(E-Mail Removed).2y.net> writes:
>
> The> -
> The> New:
> The> +
> The> How do I compute the difference of two arrays? How do I compute
> the The> intersection of two arrays?
>
> The> Use a hash. Here's code to do both and more. It assumes that each
> The> element is unique in a given array:
>
> Here's code that doesn't require that uniqueness, and would be a better
> candidate for the FAQ:
>
> my %tally;
> \$tally{\$_} .= "a" for @array_a;
> \$tally{\$_} .= "b" for @array_b;
> my @union = keys %tally;
> my @intersection = grep \$tally{\$_} =~ /ab/, @union;
> my @a_not_b = grep \$tally{\$_} =~ /a\$/, @union;
> my @b_not_a = grep \$tally{\$_} =~ /^b/, @union;
>
> print "Just another Perl hacker,";
Having coded this problem myself, your solution is minimal & very elegant (as expected).
gtoomey
Anno Siegel
Guest
Posts: n/a
09-29-2003
Gregory Toomey <(E-Mail Removed)> wrote in comp.lang.perl.misc:
> It was a dark and stormy night, and Randal L. Schwartz managed to scribble:
>
> >>>>>> "The" == The Poor <(E-Mail Removed).2y.net> writes:
> >
> > The> -
> > The> New:
> > The> +
> > The> How do I compute the difference of two arrays? How do I compute
> > the The> intersection of two arrays?
> >
> > The> Use a hash. Here's code to do both and more. It assumes that each
> > The> element is unique in a given array:
> >
> > Here's code that doesn't require that uniqueness, and would be a better
> > candidate for the FAQ:
> >
> > my %tally;
> > \$tally{\$_} .= "a" for @array_a;
> > \$tally{\$_} .= "b" for @array_b;
> > my @union = keys %tally;
> > my @intersection = grep \$tally{\$_} =~ /ab/, @union;
> > my @a_not_b = grep \$tally{\$_} =~ /a\$/, @union;
> > my @b_not_a = grep \$tally{\$_} =~ /^b/, @union;
> >
> > print "Just another Perl hacker,";
>
> Having coded this problem myself, your solution is minimal & very
> elegant (as expected).
....a variation on the theme:
my %tally;
\$tally{\$_} += 1 for @array_a; # or |=
\$tally{\$_} += 2 for @array_b;
# use more powers of two for more arrays
my @union = keys %tally;
my @intersection = grep \$tally{\$_} == 3, @union;
my \$a_not_b = grep \$tally{\$_} == 1, @union;
my \$b_not_a = grep \$tally{\$_} == 2, @union;
# etc.
Anno
David K. Wall
Guest
Posts: n/a
09-29-2003
Anno Siegel <(E-Mail Removed)-berlin.de> wrote:
> Gregory Toomey <(E-Mail Removed)> wrote in comp.lang.perl.misc:
>> It was a dark and stormy night, and Randal L. Schwartz managed to
>> scribble:
>>
>> >>>>>> "The" == The Poor <(E-Mail Removed).2y.net> writes:
>> >
>> > The> -
>> > The> New:
>> > The> +
>> > The> How do I compute the difference of two arrays? How do I
>> > compute the The> intersection of two arrays?
>> >
>> > The> Use a hash. Here's code to do both and more. It
>> > assumes that each The> element is unique in a given array:
>> >
>> > Here's code that doesn't require that uniqueness, and would be
>> > a better candidate for the FAQ:
>> >
>> > my %tally;
>> > \$tally{\$_} .= "a" for @array_a;
>> > \$tally{\$_} .= "b" for @array_b;
>> > my @union = keys %tally;
>> > my @intersection = grep \$tally{\$_} =~ /ab/, @union;
>> > my @a_not_b = grep \$tally{\$_} =~ /a\$/, @union;
>> > my @b_not_a = grep \$tally{\$_} =~ /^b/, @union;
>> >
>> > print "Just another Perl hacker,";
>>
>> Having coded this problem myself, your solution is minimal & very
>> elegant (as expected).
>
> ...a variation on the theme:
>
> my %tally;
> \$tally{\$_} += 1 for @array_a; # or |=
> \$tally{\$_} += 2 for @array_b;
> # use more powers of two for more arrays
> my @union = keys %tally;
> my @intersection = grep \$tally{\$_} == 3, @union;
This assumes there are no duplicated values in either array, right?
> my \$a_not_b = grep \$tally{\$_} == 1, @union;
I think you mean @a_not_b.
> my \$b_not_a = grep \$tally{\$_} == 2, @union;
^
> # etc.
Randal's solution allowed for non-unique values in @array_a and
@array_b. The above doesn't produce correct results for, say,
my @array_a = qw(a a a y z);
my @array_b = qw(b b b x y z);
my @array_c = qw(c c c x z);
my %tally;
\$tally{\$_} .= "a" for @array_a;
\$tally{\$_} .= "b" for @array_b;
\$tally{\$_} .= "c" for @array_c;
my @union = keys %tally;
my @intersection = grep \$tally{\$_} =~ /a+b+c/, @union;
my @a_only = grep \$tally{\$_} =~ /^a+\$/, @union;
my @b_only = grep \$tally{\$_} =~ /^b+\$/, @union;
my @c_only = grep \$tally{\$_} =~ /^c+\$/, @union;
my @a_and_b = grep \$tally{\$_} =~ /a+b/, @union;
my @a_b_not_c = grep \$tally{\$_} =~ /a+b+(?!c)/, @union;
Anno Siegel
Guest
Posts: n/a
09-29-2003
David K. Wall <(E-Mail Removed)> wrote in comp.lang.perl.misc:
> Anno Siegel <(E-Mail Removed)-berlin.de> wrote:
[union, intersection of arrays]
> > ...a variation on the theme:
> >
> > my %tally;
> > \$tally{\$_} += 1 for @array_a; # or |=
> > \$tally{\$_} += 2 for @array_b;
> > # use more powers of two for more arrays
> > my @union = keys %tally;
> > my @intersection = grep \$tally{\$_} == 3, @union;
>
> This assumes there are no duplicated values in either array, right?
Yes, if you use +=, as in the code above. Not if you use |= instead,
as indicated in a comment. I must have been thinking of sets, not lists.
> > my \$a_not_b = grep \$tally{\$_} == 1, @union;
>
> I think you mean @a_not_b.
More sloppiness.
Thanks. The code should definitely use "\$tally{\$_} |= ...".
Anno
David K. Wall
Guest
Posts: n/a
09-29-2003
Anno Siegel <(E-Mail Removed)-berlin.de> wrote:
> David K. Wall <(E-Mail Removed)> wrote in
> comp.lang.perl.misc:
>> Anno Siegel <(E-Mail Removed)-berlin.de> wrote:
>
> [union, intersection of arrays]
>
>> > ...a variation on the theme:
>> >
>> > my %tally;
>> > \$tally{\$_} += 1 for @array_a; # or |=
>> > \$tally{\$_} += 2 for @array_b;
>> > # use more powers of two for more arrays
>> > my @union = keys %tally;
>> > my @intersection = grep \$tally{\$_} == 3, @union;
>>
>> This assumes there are no duplicated values in either array,
>> right?
>
> Yes, if you use +=, as in the code above. Not if you use |=
> instead, as indicated in a comment. I must have been thinking of
> sets, not lists.
Aha. I didn't notice the comment. It makes sense now.
As an exercise for myself I found a number of subsets of three sets
this way. Although I didn't do any benchmarking I'm pretty sure this
is faster than the string method, but I must say I can think more
sequences. A trade-off as usual, I guess.
--
David Wall | 3,625 | 11,715 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2013-48 | longest | en | 0.664513 |
https://www.solutionsfolks.com/ExpertAnswers/three-charges-are-held-at-the-vertices-of-an-equilateral-triangle-of-sidel-3cm-q1-q-q2-q-pa177 | 1,723,388,250,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641002566.67/warc/CC-MAIN-20240811141716-20240811171716-00864.warc.gz | 762,588,616 | 5,405 | Home / Expert Answers / Physics / three-charges-are-held-at-the-vertices-of-an-equilateral-triangle-of-sidel-3cm-q1-q-q2-q-pa177
# (Solved): Three charges are held at the vertices of an equilateral triangle of sidel = 3cm. q1 = +q, q2 = q ...
Three charges are held at the vertices of an equilateral triangle of sidel = 3cm. q1 = +q, q2 =
?q and q3 = +q.
a) What is the resultant force exerted on q2 ?
b) What is the resultant force exerted on q3 ?
We have an Answer from Expert | 143 | 481 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2024-33 | latest | en | 0.833117 |
https://wiki.seg.org/wiki/Special:MyLanguage/Dictionary:spherical_excess | 1,680,344,775,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949958.54/warc/CC-MAIN-20230401094611-20230401124611-00350.warc.gz | 684,064,594 | 14,518 | # Dictionary:Spherical excess
The amount by which the sum of the angles of a triangle on the surface of a sphere exceeds 180 degrees. For a sphere, this excess is the area of the triangle. For a spherical triangle on an ellipsoid, such as the Earth approximates, the spherical excess ${\displaystyle \varepsilon }$ is approximately:
${\displaystyle \varepsilon =mbc{\text{sin}}\alpha }$,
where b and c are two adjacent sides of the triangle which intersect at the angle ${\displaystyle \alpha }$, m = latitude function=${\displaystyle {\frac {\rho }{2RN}}}$, ${\displaystyle \rho }$=number of seconds of arc/radian=206 264.8, R=radius of curvature in the meridian, and N=radius of curvature in the prime vertical. Values of m for various ellipsoids are obtained from tables; the value for the center of the triangle to the nearest half degree is usually used. For greater precision a correction factor is often applied in iterative fashion. | 226 | 941 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 5, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2023-14 | latest | en | 0.868516 |
https://www.upham.hants.sch.uk/wednesday-10th-june-2/ | 1,720,952,381,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514564.41/warc/CC-MAIN-20240714094004-20240714124004-00186.warc.gz | 929,157,997 | 29,016 | Wednesday 10th June
Good Morning Class 2 parents,
I have found a website that offers daily phonics lessons for Year 1 children. It teaches a new phase 5 sound every day and also revises sounds learnt previously. The lessons follow the same format as we use in school and last for around 20 minutes. All you need is a pen and paper. Although the lessons are aimed at Year 1 children, I would recommend that Year 2s also join in. I would suggest starting at the beginning and working through the lessons in order. Each day, I will add a link to a lesson. However if you feel that your child already knows the first few sounds, feel free to start wherever you think is best. The children should be able to participate in these lessons independently!
I hope you find this a helpful resource. I have started doing these lessosn daily at home with Finley!
Mrs Sothcott
Morning Job
Practise learning your 5x tables. Learn a few at a time and then test yourself once you are sure of the answers.
English
Today is SPAG, chose the correct year group.
Yr 2 variety of spelling and grammar
Maths
Year 2:
Science
Today we will be testing how elastic/ stretchy fabrics are.
You will need : fabrics ( ideally cut to the same size)
a recording sheet ( either mine or your own)
tape measure or ruler
"So, what do you know about 'stretchiness' and stretchy fabrics? What makes elasticity a useful property of fabric? When would you use stretchy materials? What happens when you pull a swimsuit and then let it go? (it goes back to its original shape; it sometimes goes a bit baggy.) There is a point at where stretchy fabric can be overstretched and won;t return to its original shape and size. " copyright HT.
Look at the fabrics you have, can you tell which is the stretchiest just by looking at them? Predict which of the fabrics is the stretchiest and put them in order.
Recording sheet
Above is the recording sheet I will be using in class. If it is not useful, feel free to create your own.
Experiment
Take the first piece of fabric and anchor one end securely - maybe under the leg of a chair with someone sitting on it, or under the leg of a table. Lay a tape measure or ruler alongside to see how long it is. Make a note on the recoring sheet.
Now pull the free end of the fabric as hard as you can. Where does it come to on the ruler or tape measure? Work out how long the piece of fabric stretched and write it on your sheet.
Do the same of each piece of fabric, writing down the results as you go.
Now try to answer these questions -
What did you find out about the fabrics?
Did you discover that the more elastic the fabric has in it, the more it will stretch?
Top | 608 | 2,688 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.94858 |
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AIEEE 2004
+4
-1
The total energy of particle, executing simple harmonic motion is
A
independent of $$x$$
B
$$\propto \,{x^2}$$
C
$$\propto \,x$$
D
$$\propto \,{x^{1/2}}$$
2
AIEEE 2004
+4
-1
A particle at the end of a spring executes $$S.H.M$$ with a period $${t_1}$$. While the corresponding period for another spring is $${t_2}$$. If the period of oscillation with the two springs in series is $$T$$ then
A
$${T^{ - 1}} = t_1^{ - 1} + t_2^{ - 1}$$
B
$${T^2} = t_1^2 + t_2^2$$
C
$$T = {t_1} + {t_2}$$
D
$${T^{ - 2}} = t_1^{ - 2} + t_2^{ - 2}$$
3
AIEEE 2004
+4
-1
A particle of mass $$m$$ is attached to a spring (of spring constant $$k$$) and has a natural angular frequency $${\omega _0}.$$ An external force $$F(t)$$ proportional to $$\cos \,\omega t\left( {\omega \ne {\omega _0}} \right)$$ is applied to the oscillator. The time displacement of the oscillator will be proportional to
A
$${1 \over {m\left( {\omega _0^2 + {\omega ^2}} \right)}}$$
B
$${1 \over {m\left( {\omega _0^2 - {\omega ^2}} \right)}}$$
C
$${m \over {\omega _0^2 - {\omega ^2}}}$$
D
$${m \over {\omega _0^2 + {\omega ^2}}}$$
4
AIEEE 2004
+4
-1
Out of Syllabus
In forced oscillation of a particle the amplitude is maximum for a frequency $${\omega _1}$$ of the force while the energy is maximum for a frequency $${\omega _2}$$ of the force; then
A
$${\omega _1} < {\omega _2}$$ when damping is small and $${\omega _1} > {\omega _2}$$ when damping is large
B
$${\omega _1} > {\omega _2}$$
C
$${\omega _1} = {\omega _2}$$
D
$${\omega _1} < {\omega _2}$$
EXAM MAP
Medical
NEETAIIMS | 614 | 1,555 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2024-38 | latest | en | 0.640033 |
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Showing results 1 to 25 of 447 Search took 0.05 seconds. Search: Posts Made By: wildrabbitt
Forum: Lounge 2020-10-10, 22:40 Replies: 85 Views: 11,755 Posted By wildrabbitt MY attempt at interesting facts. Auschwitz was the German name given to the Polish town called Oswiencim. Freshly cut glass is the sharpest thing known. Typhoon is the name of any cyclone in the Pacific. A triangle...
Forum: Math 2020-10-10, 16:54 Replies: 2 Views: 1,660 Posted By wildrabbitt Thanks. Thanks.
Forum: Math 2020-10-06, 13:18 Replies: 2 Views: 1,660 Posted By wildrabbitt finding unimodular substitutions Hi, does anyone know a method for finding a unimodular substitution from one binary quadratic form ( $$ax^2 + bxy + cy^2$$ ) to another given that they are equivalent? I'd like a find...
Forum: Math 2020-08-05, 14:58 Replies: 6 Views: 3,753 Posted By wildrabbitt Thanks. I suppose you mean... Thanks. I suppose you mean for calculating the squared term?
Forum: Math 2020-08-05, 11:50 Replies: 6 Views: 3,753 Posted By wildrabbitt module 2^p - 1 Hi, I understand the iteration sequence of the Lucas-Lehmer test involves using module (2^p-1) arithmetic. How do computers running these such a test cope with numbers with 20+ million digits?
Forum: Homework Help 2020-04-13, 18:49 Replies: 8 Views: 9,167 Posted By wildrabbitt Multiplicative Number Theory by Harold Davenport... Multiplicative Number Theory by Harold Davenport published by Springer (part of a series - Graduate Texts in Mathematics. Thanks.
Forum: Homework Help 2020-04-13, 16:11 Replies: 8 Views: 9,167 Posted By wildrabbitt Except I still don't know how it can be ... Except I still don't know how it can be that if $F(x) = \prod_{R}(x - \zeta^{R})$ has a constant term, then $F(\zeta) = A_{0}(x)\eta_{0} + A_{1}(x)\eta_{1}$ as it seems to me...
Forum: Homework Help 2020-04-13, 15:44 Replies: 8 Views: 9,167 Posted By wildrabbitt Thanks very much. That's just what I needed. I... Thanks very much. That's just what I needed. I can get on now with your help :smile:
Forum: Homework Help 2020-04-13, 11:47 Replies: 8 Views: 9,167 Posted By wildrabbitt Thanks for your interest. I wrote out what I... Thanks for your interest. I wrote out what I understand of it. 22024 22025
Forum: Homework Help 2020-04-11, 15:00 Replies: 8 Views: 9,167 Posted By wildrabbitt need guidance with some maths Hi, I've been trying to understand some maths in one of my text books for a long time now. I've written out all the relevant material and then explained why I can't understand it. If anyone can point...
Forum: Homework Help 2020-03-17, 14:36 Replies: 3 Views: 9,555 Posted By wildrabbitt Thanks to both of you. I do understand... Thanks to both of you. I do understand integration by substitution but I didn't know what the substitution required was. I should be able to do it now.
Forum: Homework Help 2020-03-17, 12:45 Replies: 3 Views: 9,555 Posted By wildrabbitt something about a sum Hi, can anyone explain this? $S=\sum_{v=-\infty}^{\infty}\int_0^Ne^{2\pi ivx+2\pi i\frac{x^2}{N}} \mathrm{d}x$ $=N\sum_{v=-\infty}^{\infty}\int_0^1e^{2\pi iN(x^2+vx)} \mathrm{d}x$ ...
Forum: Homework Help 2020-03-04, 11:39 Replies: 7 Views: 7,272 Posted By wildrabbitt Thanks so much. It'll take me a while to absorb... Thanks so much. It'll take me a while to absorb what the last two posts say. I'm very grateful for the input.
Forum: Homework Help 2020-03-03, 21:36 Replies: 7 Views: 7,272 Posted By wildrabbitt The book I'm reading says that given a qth root... The book I'm reading says that given a qth root of unity $\zeta$, every polyonomial in $\zeta$ can be expressed as \[A_1\zeta^1 + A_2\zeta^2 + A_3\zeta^3 +A_4\zeta^4...A_{q-1} +...
Forum: Homework Help 2020-03-03, 14:24 Replies: 7 Views: 7,272 Posted By wildrabbitt algebraic numbers Can anyone help me understand how to work out the degree of an algebraic number?
Forum: Homework Help 2020-02-25, 17:01 Replies: 9 Views: 7,245 Posted By wildrabbitt I didn't really get the answer I was hoping for... I didn't really get the answer I was hoping for so I thought I'd just ask the first things that it occurred to me to ask. I do know how to make substitutions in integrals I just wasn't sure if...
Forum: Homework Help 2020-02-24, 20:40 Replies: 9 Views: 7,245 Posted By wildrabbitt Thanks. Are you substituting $$e^{-u}$$... Thanks. Are you substituting $$e^{-u}$$ for $$x$$ ? What are you doin with $$u$$? Also, in case there's an obvious remedy to my latex not parsing properly, here's an example of what I...
Forum: Homework Help 2020-02-24, 09:51 Replies: 9 Views: 7,245 Posted By wildrabbitt clarification I aught to mention that $$L(s)=\sum_{n=1}^{\infty}\big(\frac{n}{q}\big)n^{-s}$$ for which $$\big(\frac{n}{q}\big)$$ is the Legendre symbol as is it is the first post.
Forum: Homework Help 2020-02-24, 07:52 Replies: 9 Views: 7,245 Posted By wildrabbitt I just wrote what was written in the book. ... I just wrote what was written in the book. I'm pretty sure it's supposed to be $$\log \frac{1}{x}$$ otherwise it would have been written the way you wrote.
Forum: Homework Help 2020-02-23, 17:31 Replies: 9 Views: 7,245 Posted By wildrabbitt something I don't understand to do with Dirichlets theorem Hi, the following is something I've been reading. He started from the power series
Forum: Homework Help 2020-02-03, 16:04 Replies: 8 Views: 7,606 Posted By wildrabbitt Merci. Merci beaucoup docteur. Quelque choses a faire maintenant.
Forum: Homework Help 2020-01-30, 09:31 Replies: 8 Views: 7,606 Posted By wildrabbitt infinite products I've got a maths degree but I was never taught about infinite products. I'd like to ask something : when do people usually learn about infinite products? I suppose it must be at masters level...
Forum: Homework Help 2020-01-14, 14:54 Replies: 3 Views: 7,451 Posted By wildrabbitt Thanks very much to both of you. Thanks very much to both of you.
Forum: Homework Help 2020-01-14, 12:38 Replies: 3 Views: 7,451 Posted By wildrabbitt problem to do with golden ratio equation Can anyone explain what's wrong with my logic? https://www.mersenneforum.org/attachment.php?attachmentid=21617&stc=1&d=1579005470
Forum: Software 2019-11-13, 11:26 Replies: 14 Views: 2,530 Posted By wildrabbitt Thanks. Thanks.
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∫ ∬ ∭ ∮ ∯ ∰ ∇ ∆ δ ∂ ℱ ℒ ℓ
𝛢𝛼 𝛣𝛽 𝛤𝛾 𝛥𝛿 𝛦𝜀𝜖 𝛧𝜁 𝛨𝜂 𝛩𝜃𝜗 𝛪𝜄 𝛫𝜅 𝛬𝜆 𝛭𝜇 𝛮𝜈 𝛯𝜉 𝛰𝜊 𝛱𝜋 𝛲𝜌 𝛴𝜎𝜍 𝛵𝜏 𝛶𝜐 𝛷𝜙𝜑 𝛸𝜒 𝛹𝜓 𝛺𝜔 | 2,515 | 7,286 | {"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} | 2.71875 | 3 | CC-MAIN-2022-27 | latest | en | 0.868695 |
https://puzzling.stackexchange.com/questions/5753/explanation-for-the-i-am-whole-but-incomplete-riddle | 1,726,247,884,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651535.66/warc/CC-MAIN-20240913165920-20240913195920-00213.warc.gz | 447,685,351 | 40,646 | # Explanation for the "I am whole but incomplete..." riddle?
I saw a riddle:
I am whole but incomplete,
I have no eyes, yet I see.
You can see, and see right through me.
My largest part is one fourth of what I once was.
What am I?
Apparently the answer is:
A skeleton
but I don't understand it. How does it "see"? How is its largest part one fourth of what it once was?
I am whole but incomplete
This means a skeleton is whole considering it as a structure. But it is just a part of the body. So it's incomplete.
I have no eyes, yet I see.
You can see, and see right through me.
This means skeletons don't have eyes - just the holes. By yet I can 'see' it refers to the 'see' in the word 's'k'e'l'e'ton. The same is meant by 'you can see', and 'see right through me'. Which makes it clear that we can see the word 'see' in (right through) 'skeleton'.
(Or as mentioned in the comments: people perceive a skeleton to "see" through its eye sockets, even though it no longer has eyes. And people can see through the gaps in a skeleton.)
My largest part is one fourth of what I once was.
The largest part of a skeleton is the femur, which is one fourth of the body (which the skeleton once was) in length.
Hope it's clear now.
• The "see" thing seems like a bit of a stretch. I still don't understand the "one fourth" part. 1/4 by mass? I find that hard to believe. Commented Dec 1, 2014 at 10:09
• @jamesdlin: It's not 1/4 by mass. But by length. Commented Dec 1, 2014 at 10:18
• The femur, or thigh bone, is the longest and heaviest bone in the body. Its length varies from one fourth to one third of that of the body; Commented Dec 1, 2014 at 10:23
• also, you can see through a skeleton because there are gaps eg between the ribs Commented Dec 1, 2014 at 13:08
• @KateGregory that's what i thought as well. Commented Dec 1, 2014 at 13:47 | 512 | 1,850 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2024-38 | latest | en | 0.973204 |
https://excelgorilla.com/excel/add-references-to-cells-in-bulk | 1,591,157,124,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347428990.62/warc/CC-MAIN-20200603015534-20200603045534-00580.warc.gz | 331,635,928 | 19,775 | There are times when you want a set of cells to reference a parameter. For example, you’re budgeting and the result needs to be dynamic. You can easily set this up when cells contain relative formulas. In the top cell you then add the reference to the parameter and you copy this formula down. So far this is easy.
But what can you do when your cells contain different formulas or hard values. And when on top of that the amount of cells that need the parameter reference is too big to change by hand?
By luck I discovered a method to add cell references in bulk to formulas. You might know the Multiply feature that appears in the Paste Special menu. With this paste special option you can multiply one or multiple cells by the number copied. I used to think this would multiply the selected cells by the copied value. And that the result can only be a pasted value. But I was wrong. I will first show how the Multiply Operation in the Paste Special menu works and then reveal the trick to easily add a parameter to a set of cells.
## How does the Multiply Paste Special operation work?
The Paste Special Multiply Operation works as follows. You copy the cell that contains the number you want to multiply other cells with. After copying this cell, select the destination cells with your mouse. Then open the Paste Special menu and perform the multiplication.
Note: After copy the cell and making the selection, you can open the paste special menu by:
• pressing Ctrl + Alt + V
• right clicking the selection -> click Paste Special
• in the home tab, under the Paste operation click the arrow with more options -> choose Paste Special
With the Paste Special menu open, select Multiply and leave all other settings. Press OK and voila. All selected numbers are now multiplied by the copied value.
## How to Add Cell References
We just learned how to use the Paste Special Multiply operation to multiply cells by a value. Using a similar approach you can multiply cells by a formula. On the worksheet is a cell to reference. Let’s call this the Parameter. Imagine you want to multiply your Base Values by the Parameter value (and reference it). To do this we first need a cell that references the parameter. In below example it’s called the Parameter Reference. This cell references the parameter directly, and locks both the row and column (using \$-signs).
To add a Reference to the Parameter:
• Copy the Parameter Reference
• Then select the Base Values
• Open the Paste Special screen -> paste Formulas -> select the Multiply Operation
• Now press OK
And just like that, all the Base Values now have a reference to the Parameter! The amazing part is, the operation makes good use of parenthesis. After doing the operation the calculations are still correct.
In this post I showed the Paste Special Multiplication operation. In a similar way you can use the Add, Subtract and Divide operation that you find in the Paste Special Dialog box.
With the technique shown, you are now well equipped to add cell references to cells. Even when these cells have different formulas or hard values. Has this helped you? Or do you use use a similar technique in other useful ways? I’d love to hear from you in the comments!
Rick de Groot | 657 | 3,240 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2020-24 | latest | en | 0.845998 |
http://www.chegg.com/homework-help/university-physics-volume-1-chapters-1-20-1st-edition-chapter-5-problem-64ap-solution-9780077354831 | 1,441,268,778,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440645305536.82/warc/CC-MAIN-20150827031505-00265-ip-10-171-96-226.ec2.internal.warc.gz | 357,414,525 | 17,192 | View more editions
University Physics Volume 1 (Chapters 1-20)
# TEXTBOOK SOLUTIONS FOR University Physics Volume 1 (Chapters 1-20) 1st Edition
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PROBLEM
Chapter: Problem:
A driver notices that her 1000,-kg car slows from v0 = 90.0 km/h (25.0 m/s) to v = 70.0 km/h (19.4 m/s) in t = 6.00 s moving on level ground in neutral gear. Calculate the power needed to keep the car moving at a constant speed, vave = 80.0 km/h (22.2 m/s).
STEP-BY-STEP SOLUTION
Corresponding Textbook
University Physics Volume 1 (Chapters 1-20) | 1st Edition
9780077354831ISBN-13: 0077354834ISBN: | 238 | 765 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2015-35 | latest | en | 0.780674 |
http://piknikovykos.cz/id/singular-and-non-singular-matrix-example-284745 | 1,610,730,518,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703495936.3/warc/CC-MAIN-20210115164417-20210115194417-00353.warc.gz | 87,488,172 | 3,420 | Then, by one of the property of determinants, we can say that its determinant is equal to zero. Notice that we will not discuss Example HISAD as being a singular or nonsingular coefficient matrix since the matrix is not square.. Sage NM Nonsingular Matrix Click to open. Thus B is a non-singular matrix. Singular and Non-singular matrices: A singular matrix is a matrix for which the inverse does not exist. So to find a counterexample, we have to look at … If A and B are non-singular matrices of the same order then AB and BA are also non-singular matrices because | … This lesson introduces the notion of a singular matrix and provides a shortcut to determine whether or not a given 2x2 matrix is singular. If Adoes not have an inverse, Ais called singular. Singular matrix - definition Singular matrix: A square matrix whose determinant is 0 is called singular matrix. If the matrix A is non-singular, then: AA-1 = A-1 A = I. More gen- More gen- erally, e very polynomial matrix P such that valdet P = 0 is strongly non-singular . For example, there are 10 singular (0,1)-matrices: The following table gives the numbers of singular matrices for certain matrix classes. Singular Matrix. (Non{singular matrix) An n n Ais called non{singular or invertible if there exists an n nmatrix Bsuch that AB= In= BA: Any matrix Bwith the above property is called an inverse of A. Examples: ∣ ∣ ∣ ∣ ∣ ∣ 0 ∣ ∣ ∣ ∣ ∣ ∣ 0 The matrices are said to be singular if their determinant is equal to zero. A matrix that does not is called singular. Consider, for example, the equations (10.32) in which the last two rows are interchanged if partial pivoting is employed. A square matrix that is not singular, i.e., one that has a matrix … A matrix is singular iff its determinant is 0. A square matrix A is said to be singular if | A | = 0. A non–singular matrix A has a unique LU factorization if and only if all the principal minors of A are non–zero. A singular matrix is a square matrix with nonfull rank. The next theorem combines with our main computational technique (row reducing a matrix) to make it easy to recognize a nonsingular matrix. NON{SINGULAR MATRICES DEFINITION. (Inverses are unique) If Ahas inverses Band C, then B= C. A square matrix A is said to be non-singular if | A | ≠ 0. We explain Singular and Non-Singular Matrices with video tutorials and quizzes, using our Many Ways(TM) approach from multiple teachers. For $1\times1$ matrices (i.e., numbers), the only singular matrix is $0$; so if we add it to any nonsingular (invertible) matrix, it remains nonsingular. A matrix that does have an inverse is called non-singular. A square matrix that does not have a matrix inverse. Download PDF for free. Note 7.14. For example, if we have matrix A whose all elements in the first column are zero. Example 1 Every non-singular constant matrix is strongly non-singular. Singular and non-singular Matrices. Hence, A would be called as singular matrix. THEOREM. For example there is no matrix 0-1 such that 00-1 = I. Non-singular. Definition 7.21. Many N × N matrices do not have an inverse. In fact, it is worse than that. A nonsingular matrix is a square matrix with full rank.
## singular and non singular matrix example
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Totality game for an adult and child. Be the first to reach your agreed total.
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Design your own scoring system and play Trumps with these Olympic Sport cards.
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Who said that adding couldn't be fun?
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A game for 2 people. Take turns placing a counter on the star. You win when you have completed a line of 3 in your colour. | 2,023 | 8,398 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2019-35 | latest | en | 0.85811 |
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A261177 Maximum value of (1/2)*Sum_{i=1..n} Sum_{j=1..n} Sum_{k=1..n} Sum_{l=1..n} gcd(b(i,j),b(k,l)) * ((i-k)^2+(j-l)^2) for an n X n matrix b filled with the integers 1 to n^2. 1
0, 10, 180, 1392, 6149, 21350, 57192, 137617, 298864, 593378, 1101739, 1936342, 3216080 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS Best results found in Al Zimmermann's Programming Contest "Delacorte Numbers". For more information see A261176. All terms beyond a(5) are conjectured based on numerical results. Terms up to a(11) have at least 5 independent verifications. Lower bounds for the next terms are a(14)>=5189492, a(15)>=8110781, a(16)>=12239616, a(17)>=18073562, a(18)>=26055061, a(19)>=36769303, a(20)>=51095165. LINKS Al Zimmermann's Programming Contests, Delacorte Numbers, Description, October 2014. Al Zimmermann's Programming Contests, Delacorte Numbers, Final Report, January 2015. EXAMPLE a(3)=180, because no arrangement of the matrix elements exists that produces a larger Delacorte Number than e.g. ((2 3 4)(9 1 5)(8 7 6)). CROSSREFS Cf. A261176, A003989, A018782. Sequence in context: A200060 A240561 A057122 * A113119 A067416 A113671 Adjacent sequences: A261174 A261175 A261176 * A261178 A261179 A261180 KEYWORD nonn,hard AUTHOR Hugo Pfoertner, Aug 15 2015 EXTENSIONS Lower bounds for a(18) and a(20) improved by Hugo Pfoertner, Nov 22 2015 STATUS approved
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Last modified September 25 23:23 EDT 2020. Contains 337346 sequences. (Running on oeis4.) | 607 | 1,920 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-40 | latest | en | 0.685664 |
https://warreninstitute.org/find-the-missing-value-given-the-mean-worksheet-pdf/ | 1,708,490,584,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473370.18/warc/CC-MAIN-20240221034447-20240221064447-00492.warc.gz | 633,854,470 | 22,622 | # Mean worksheet PDF: Find missing value!
Welcome to Warren Institute! In this article, we will explore how to find the missing value given the mean using a helpful worksheet in PDF format. Whether you are a student looking to improve your understanding of mathematical concepts or a teacher searching for engaging resources for your classroom, this worksheet will provide valuable practice. With clear instructions and carefully crafted questions, students will develop their problem-solving skills and gain confidence in working with mean values. Download the worksheet now and get ready to unlock the mystery of missing values in mathematics!
## Understanding the Concept of Finding the Missing Value Given the Mean
The first subtitle focuses on explaining the concept of finding the missing value given the mean in mathematics education. In this section, you can discuss the importance of understanding how to calculate the mean and how it relates to finding the missing value. Emphasize the relevance of this skill in real-life scenarios and problem-solving.
Example response: Understanding how to find the missing value given the mean is a fundamental concept in mathematics education. The mean represents the average value of a set of numbers. By learning how to calculate the mean and use it to find missing values, students develop problem-solving skills that are applicable in various real-life scenarios. This skill allows them to analyze data, make predictions, and draw conclusions.
## Step-by-Step Approach to Solving "Find the Missing Value Given the Mean" Problems
This subtitle provides a step-by-step approach to solving problems related to finding the missing value given the mean. Break down the process into clear, concise steps, and highlight any important formulas or techniques that should be used.
Example response: Solving problems that involve finding the missing value given the mean can be simplified by following a step-by-step approach. Start by calculating the sum of the known values and the number of values given. Then, subtract the sum from the product of the mean and the total number of values. The resulting value will be the missing value. This approach ensures consistent and accurate results when dealing with such problems.
## Practical Applications of Finding the Missing Value Given the Mean
In this section, discuss the practical applications of finding the missing value given the mean in real-world situations. Highlight examples where this concept is relevant, such as analyzing survey data, calculating grades, or predicting outcomes based on previous data.
Example response: Finding the missing value given the mean has practical applications in various fields. For example, in survey analysis, if the mean of a set of ratings is known but one rating is missing, finding the missing value allows for a complete analysis of the data. Similarly, teachers can use this concept to calculate grades based on known scores and the desired mean. Furthermore, businesses can make predictions about future sales based on past data by finding missing values using the mean.
## Resources for Practice: "Find the Missing Value Given the Mean" Worksheets
This subtitle provides information and resources for practicing the skill of finding the missing value given the mean. Mention the availability of worksheets or other educational materials that students can use to reinforce their understanding and improve their problem-solving abilities.
Example response: To enhance students' proficiency in finding the missing value given the mean, a variety of resources are available for practice. Worksheets specifically designed for this skill provide ample opportunities to apply the concept in different contexts. Additionally, online platforms and educational websites offer interactive exercises and quizzes to further reinforce understanding. These resources enable students to develop confidence and mastery in solving problems related to finding the missing value given the mean.
### What are some strategies for finding the missing value in a worksheet when given the mean?
One strategy for finding the missing value in a worksheet when given the mean is to use the formula for calculating the mean. The mean is found by adding up all the values in a set and dividing by the number of values. So, if you are given the mean and all the other values except one, you can rearrange the formula to solve for the missing value. Another strategy is to use the concept of balance. If the mean is known, the sum of all the values in the set must also be known. By subtracting the sum of the given values from the sum of all the values, you can find the missing value.
### How can I use the mean to solve for the missing value in a worksheet?
To use the mean to solve for the missing value in a worksheet, you can follow these steps:
1. Calculate the sum of all the given values in the dataset.
2. Count the number of values in the dataset.
3. Divide the sum by the count to find the mean.
4. Multiply the mean by the total number of values, including the missing value.
5. Subtract the sum of the given values from the result obtained in step 4.
6. The resulting value is the missing value in the dataset.
Note: The mean is obtained by dividing the sum of all the values by the count, and it represents the average value in the dataset.
### Are there any specific formulas or equations I should use to find the missing value given the mean in a worksheet?
Yes, there are specific formulas you can use to find the missing value given the mean in a worksheet. One of the most common formulas is:
Missing Value = (Mean * Total Number of Values) - Sum of Known Values
### Can you provide any examples or practice problems of finding the missing value using the mean in a worksheet?
Sure! Here's an example of a practice problem on finding the missing value using the mean in a worksheet:
Question: The mean of a set of 5 numbers is 12. Four of the numbers are 8, 14, 16, and 13. What is the missing number?
Solution: To find the missing number, we can use the formula for the mean: Mean = (Sum of all numbers) / (Number of numbers).
Since the sum of the given numbers is 8 + 14 + 16 + 13 = 51, we can substitute this into the formula: 12 = 51 / 5.
To solve for the missing number, we can rearrange the equation to isolate the missing number: Missing number = (Mean) * (Number of numbers) - (Sum of given numbers).
Plugging in the given values, we get: Missing number = 12 * 5 - 51 = 60 - 51 = 9.
Therefore, the missing number is 9.
### Are there any online resources or worksheets available in PDF format that focus on finding the missing value given the mean?
Yes, there are online resources and worksheets available in PDF format that focus on finding the missing value given the mean. These resources can be found on various educational websites, math tutoring platforms, and teacher resource websites.
In conclusion, the "Find the Missing Value Given the Mean" worksheet PDF is a valuable resource for Mathematics education. By providing students with practice in calculating missing values based on mean, this worksheet helps develop their problem-solving and analytical skills. The use of HTML tags emphasizes the importance of understanding the concept of mean and its application in real-world scenarios. With this worksheet, students can enhance their mathematical proficiency and gain confidence in their ability to solve problems involving mean. Incorporating such interactive and engaging resources into the classroom can greatly contribute to a more effective and enjoyable learning experience for students in Mathematics education. | 1,480 | 7,701 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-10 | latest | en | 0.866717 |
http://www.radiation-dosimetry.org/what-is-radiation-weighting-factor-definition/ | 1,723,583,668,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641085898.84/warc/CC-MAIN-20240813204036-20240813234036-00238.warc.gz | 44,971,059 | 93,670 | # What is Radiation Weighting Factor – Definition
In radiation protection, the radiation weighting factor is a dimensionless factor used to determine the equivalent dose from the absorbed dose averaged over a tissue or organ and is based on the type of radiation absorbed. Radiation Dosimetry
Equivalent dose (symbol HT) is a dose quantity calculated for individual organs (index T – tissue). Equivalent dose is based on the absorbed dose to an organ, adjusted to account for the effectiveness of the type of radiation. Equivalent dose is given the symbol HT. The SI unit of HT is the sievert (Sv) or but rem (roentgen equivalent man) is still commonly used (1 Sv = 100 rem). Unit of sievert was named after the Swedish scientist Rolf Sievert, who did a lot of the early work on dosimetry in radiation therapy.
As was written, for radiation protection purposes, the absorbed dose is averaged over an organ or tissue, T, and this absorbed dose average is weighted for the radiation quality in terms of the radiation weighting factor, wR, for the type and energy of radiation incident on the body. The radiation weighting factor is a dimensionless factor used to determine the equivalent dose from the absorbed dose averaged over a tissue or organ and is based on the type of radiation absorbed. The resulting weighted dose was designated as the organ- or tissue equivalent dose:
## From Absorbed Dose to Equivalent Dose
Note that the sievert is not a physical dose unit. For example, an absorbed dose of 1 Gy by alpha particles will lead to an equivalent dose of 20 Sv. This may seem to be a paradox. It implies that the energy of the incident radiation field in joules has increased by a factor of 20, thereby violating the laws of Conservation of energy. However, this is not the case. Sievert is derived from the physical quantity absorbed dose, but also takes into account the biological effectiveness of the radiation, which is dependent on the radiation type and energy. The radiation weighting factor causes that the sievert cannot be a physical unit.
In radiation protection, the radiation weighting factor is a dimensionless factor used to determine the equivalent dose from the absorbed dose averaged over a tissue or organ and is based on the type of radiation absorbed. In the past there a similar factor known as quality factor was used for this purpose. The radiation weighting factor is an estimate of the effectiveness per unit dose of the given radiation relative a to low-LET standard.
Before 1990, dose-equivalent quantities were defined in terms of a quality factor, Q(L), that was applied to the absorbed dose at a point in order to take into account the differences in the effects of different types of radiation. In its 1990 recommendations, the ICRP introduced a modified concept. For radiological protection purposes, the absorbed dose is averaged over an organ or tissue, T, and this absorbed dose average is weighted for the radiation quality in terms of the radiation weighting factor, wR, for the type and energy of radiation incident on the body.
The reason for replacing the quality factor, i.e. the Q–L relationship, with wR values in the definition of the organ-equivalent doses and the effective dose was that the Commission believed:
‘that the detail and precision inherent in using a formal Q–L relationship to modify absorbed dose to reflect the higher probability of detriment resulting from exposure to radiation components with high LET is not justified because of the uncertainties in the radiological information’.
Noteworthy, these two factors, the radiation weighting factor and the quality factor are restricted to the dose range of interest to radiation protection, i.e. to the general magnitude of the dose limits. In special circumstances where one deals with higher doses that can cause deterministic effects, the relevant RBE values are applied to obtain a weighted dose.
Special Reference: ICRP, 2003. Relative Biological Effectiveness (RBE), Quality Factor (Q), and Radiation Weighting Factor (wR). ICRP Publication 92. Ann. ICRP 33 (4).
### Radiation Weighting Factors – ICRP
For photon and electron radiation, the radiation weighting factor has the value 1 independently of the energy of the radiation and for alpha radiation the value 20. For neutron radiation, the value is energy-dependent and amounts to 5 to 20.
In 2007 ICRP published a new set of radiation weighting factors (ICRP Publ. 103: The 2007 Recommendations of the International Commission on Radiological Protection). These factors are given below.
As shown in the table, a wR of 1 is for all low-LET radiations, i.e. X-rays and gamma rays of all energies as well as electrons and muons. A smooth curve, considered an approximation, was fitted to the wR values as a function of incident neutron energy. Note that En is the neutron energy in MeV.
Thus for example, an absorbed dose of 1 Gy by alpha particles will lead to an equivalent dose of 20 Sv, and an equivalent dose of radiation is estimated to have the same biological effect as an equal amount of absorbed dose of gamma rays, which is given a weighting factor of 1.
### Quality Factor
The quality factor of a radiation type is defined as the ratio of the biological damage produced by the absorption of 1 Gy of that radiation to the biological damage produced by 1 Gy of X-rays or gamma rays.
The Q of a certain type of radiation is related to the density of the ion tracks it leaves behind it in tissue. The quality factors for the various types of radiation are listed in the table.
These quality factors are restricted to the dose range of interest to radiation protection, i.e. to the general magnitude of the dose limits. In special circumstances where one deals with higher doses that can cause deterministic effects, the relevant RBE values are applied to obtain a weighted dose.
References:
1. Knoll, Glenn F., Radiation Detection and Measurement 4th Edition, Wiley, 8/2010. ISBN-13: 978-0470131480.
2. Stabin, Michael G., Radiation Protection and Dosimetry: An Introduction to Health Physics, Springer, 10/2010. ISBN-13: 978-1441923912.
3. Martin, James E., Physics for Radiation Protection 3rd Edition, Wiley-VCH, 4/2013. ISBN-13: 978-3527411764.
4. U.S.NRC, NUCLEAR REACTOR CONCEPTS
5. U.S. Department of Energy, Nuclear Physics and Reactor Theory. DOE Fundamentals Handbook, Volume 1 and 2. January 1993.
Nuclear and Reactor Physics:
1. J. R. Lamarsh, Introduction to Nuclear Reactor Theory, 2nd ed., Addison-Wesley, Reading, MA (1983).
2. J. R. Lamarsh, A. J. Baratta, Introduction to Nuclear Engineering, 3d ed., Prentice-Hall, 2001, ISBN: 0-201-82498-1.
3. W. M. Stacey, Nuclear Reactor Physics, John Wiley & Sons, 2001, ISBN: 0- 471-39127-1.
4. Glasstone, Sesonske. Nuclear Reactor Engineering: Reactor Systems Engineering, Springer; 4th edition, 1994, ISBN: 978-0412985317
5. W.S.C. Williams. Nuclear and Particle Physics. Clarendon Press; 1 edition, 1991, ISBN: 978-0198520467
6. G.R.Keepin. Physics of Nuclear Kinetics. Addison-Wesley Pub. Co; 1st edition, 1965
7. Robert Reed Burn, Introduction to Nuclear Reactor Operation, 1988.
8. U.S. Department of Energy, Nuclear Physics and Reactor Theory. DOE Fundamentals Handbook, Volume 1 and 2. January 1993.
9. Paul Reuss, Neutron Physics. EDP Sciences, 2008. ISBN: 978-2759800414. | 1,706 | 7,357 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2024-33 | latest | en | 0.938555 |
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# In an unusual move for a Republican, the president\u2019s directive es
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In an unusual move for a Republican, the president\u2019s directive es [#permalink]
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24 Oct 2018, 01:16
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In an unusual move for a Republican, the president's directive established price controls.
(A) the president's directive established price controls
(B) the president's directive was to establish price controls
(C) the president was successful in the establishment of controls on prices
(D) the president had instituted controls on prices
(E) the president established price controls
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In an unusual move for a Republican, the president\u2019s directive es [#permalink]
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24 Oct 2018, 07:55
The president is a Republican, not his directive. Therefore, we can eliminate A and B.
C changes the meaning. It says the Republican are usually unsuccessful at establishing price controls.
D incorrectly uses Past Perfect.
E is the only good option.
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Re: In an unusual move for a Republican, the president\u2019s directive es [#permalink]
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24 Oct 2018, 08:38
1
Bunuel wrote:
In an unusual move for a Republican, the president's directive established price controls.
(A) the president's directive established price controls
(B) the president's directive was to establish price controls
(C) the president was successful in the establishment of controls on prices
(D) the president had instituted controls on prices
(E) the president established price controls
The highlighted part must modify what comes after it, Correct Answer must be (E), for the errors in other options marked in Red.
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Re: In an unusual move for a Republican, the president\u2019s directive es [#permalink]
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25 Oct 2018, 03:05
Bunuel wrote:
In an unusual move for a Republican, the president's directive established price controls.
(A) the president's directive established price controls
(B) the president's directive was to establish price controls
(C) the president was successful in the establishment of controls on prices
(D) the president had instituted controls on prices
(E) the president established price controls
KAPLAN OFFICIAL EXPLANATION:
E.
What should an introductory phrase followed by a comma prompt you to do? Check to see if the phrase correctly modifies what follows it. You can’t say that a directive is making “an unusual move for a Republican.” No, it’s a person, the president, that’s making the move. (E) says what the author of the sentence wants to say, clearly and simply. That’s preferable on the GMAT.
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Re: In an unusual move for a Republican, the president\u2019s directive es &nbs [#permalink] 25 Oct 2018, 03:05
Display posts from previous: Sort by | 1,174 | 4,821 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2019-04 | latest | en | 0.913073 |
https://2022.help.altair.com/2022.2/hwsolvers/rad/topics/solvers/rad/nonlocal_mat_starter_r.htm | 1,721,566,819,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517701.96/warc/CC-MAIN-20240721121510-20240721151510-00714.warc.gz | 53,294,709 | 23,278 | # /NONLOCAL/MAT
Block Format Keyword Non-local regularization for elasto-plastic failure criteria (as in, dependent to plastic strain) and shells thickness variation.
## Format
(1) (2) (3) (4) (5) (6) (7) (8) (9) (10)
/NONLOCAL/MAT/mat_ID/unit_ID
${R}_{len}$ ${L}_{e}{}^{MAX}$
## Definition
Field Contents SI Unit Example
mat_ID Material identifier.
(Integer, maximum 10 digits)
unit_ID Unit identifier.
(Integer, maximum 10 digits)
${R}_{len}$ Non-local internal length.
(Real)
$\left[\text{m}\right]$
${L}_{e}{}^{MAX}$ Mesh convergence element length target.
(Real)
$\left[\text{m}\right]$
1. The non-local regularization is used to get mesh independent results (size, orientation) in case of instabilities such as failure and/or thickness variation (for shells). The mesh independent results implies a mesh convergence for mesh sizes ${L}_{e}$ less than or equal to the maximum value set by you, ${L}_{e}\le {L}_{e}^{MAX}$ . This maximum mesh size ${L}_{e}{}^{MAX}$ is then the highest mesh size used for which results are mesh convergent.
The non-local formulation is compatible with elasto-plastic material laws only. When activated, the computation of the attached failure criteria based on plastic strain and/or the shell thickness variation depends on a regularized nodal "non-local" plastic strain calculated on the entire mesh. The non-local plastic strain at nodes denoted ${\epsilon }_{p}^{nl}$ is computed accounting for its own gradient and its local counterpart ${\epsilon }_{p}$ is computed at the Gauss points following the set of equations:(1)
The parameters $\gamma$ and $\zeta$ are automatically set. You have to set the parameter ${R}_{len}$ (or ${L}_{e}{}^{MAX}$ , Comment 2) which defines a non-local "internal length" corresponding to a radius of influence in the non-local variable computation. This defines the size of the non-local regularization band ${L}_{r}=f\left({R}_{len}\right)$ (Figure 1).
The failure criterion damage variable is then computed using the non-local plastic strain.(2)
$D=\sum _{t=0}^{\infty }\frac{\text{Δ}{\epsilon }_{p}^{nl}}{{\epsilon }_{f}}$
Where, ${\epsilon }_{f}$ is the plastic strain at failure depending on the failure criterion formulation.
2. To set the non-local length parameter ${R}_{len}$ , you can select:
• Directly input the value of ${R}_{len}$ in the input card, if a direct control on this parameter is wanted. In this case, the parameter ${L}_{e}{}^{MAX}$ must be ignored.
• Input the maximum mesh size ${L}_{e}{}^{MAX}$ for which results are mesh convergent. The non-local regularization will then be effective for all mesh sizes ${L}_{e}$ such as ${L}_{e}\le {L}_{e}^{MAX}$ . In this case, an automatic set of ${R}_{len}$ is realized according to the value of ${L}_{e}{}^{MAX}$ , and the input value of ${R}_{len}$ is ignored.
For instance, if you want to get converged and mesh-independent results for a mesh size of 5mm, ${L}_{e}^{MAX}=5$ mm. In this case, the results will be converged, mesh-size and mesh orientation independent for ${L}_{e}\le 5$ mm.
3. When the non-local regularization is used for shell elements, an additional regularization is made on the thickness variation computation avoiding an additional localization issue. In the common local case (Figure 2), the compatibility of thickness between shell elements is not ensured, due to the lack of kinematic equations in the z-direction, and the thickness variation is locally computed at Gauss points. By introducing the non-local plastic strain in the "in-thickness" strain increment, the compatibility is restored (Figure 3).(3)
Where, $\text{Δ}{\lambda }_{nl}=f\left({\epsilon }_{p}^{nl}\right)$ is the non-local plastic multiplier.
Figure 2. Transverse strain incompatibility (local) Figure 3. Transverse strain compatibility (non-local)
Note: This last point implies that the identified parameters can be used on solid and shells, as results will be identical within the same range of stress triaxiality $-2}{3}\le \eta \le 2}{3}$ .
4. List of compatible material laws for shells thickness variation regularization:
/MAT/LAW2 (PLAS_JOHNS) /MAT/LAW22 (DAMA) /MAT/LAW27 (PLAS_BRIT) /MAT/LAW32 (HILL) /MAT/LAW36 (PLAS_TAB) /MAT/LAW43 (HILL_TAB) /MAT/LAW44 (COWPER) /MAT/LAW48 (ZHAO) /MAT/LAW57 (BARLAT3) /MAT/LAW60 (PLAS_T3) /MAT/LAW63 (HANSEL) /MAT/LAW64 (UGINE_ALZ) /MAT/LAW72 (HILL_MMC) /MAT/LAW76 (SAMP) /MAT/LAW78 /MAT/LAW87 (BARLAT2000) /MAT/LAW93 (ORTH_HILL) (CONVERSE) /MAT/LAW104 (JOHNS_VOCE_DRUCKER) /MAT/LAW109 /MAT/LAW110 (VEGTER)
5. List of elasto-plastic failure model and coupled damage model compatible with non-local regularization:
MMC damage model in /MAT/LAW72 Damage model in /MAT/LAW76 /FAIL/BIQUAD /FAIL/COCKROFT /FAIL/EMC /FAIL/HC_DSSE (for shells) /FAIL/JOHNSON /FAIL/ORTHBIQUAD /FAIL/RTCL /FAIL/SPALLING /FAIL/TAB1 /FAIL/USERi /FAIL/WIERZBICKI /FAIL/WILKINS
6. List of material laws with non-local regularized temperature computation:
• /MAT/LAW104 (JOHNS_VOCE_DRUCKER)
• /MAT/LAW109 | 1,361 | 4,977 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 31, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2024-30 | latest | en | 0.766213 |
https://www.edumple.com/cbse-class-7/mathmatics/percentage-another-way-of-comparing-quantities/notes | 1,669,962,184,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710898.93/warc/CC-MAIN-20221202050510-20221202080510-00183.warc.gz | 781,666,069 | 13,203 | Illustration 9
In an examination, 96% of the candidates passed & 50 failed. How many candidates appeared ?
Solution
Let the number of candidates appeared be x
Percentage of those who passed = 96%
Percentage of those who failed = (100 – 96)% = 4 %
Number of failures = 4% of x.
4% of x = 50
Hence, the number of candidates appeared is 1250.
Illustration 10
(i) Chalk contains calcium, carbon and oxygen in the ratio 10 : 3 : 12, find the % of carbon in chalk.
(ii) If in a stick of chalk, carbon is 3g, what is the weight of the chalk stick ?
Percentage Application
Percentage, Increase and decrease percentage
Percentage Increase
When comparing the increase in a quantity over a period of time, we first find the difference between the original value and the increased value. We then use this difference to find the relative increase against the original value and express it in terms of percentage. The formula for percentage increase is given by:
Percentage Decrease
When comparing the decrease in a quantity over a period of time, we first find the difference between the original value and the decreased value. We then use this difference to find the relative decrease against the original value and express it in the form of a percentage. The formula for percentage decrease is given by:
Percentage: Marks, Profit and loss- Introduction
Profit and Loss Basic Concepts
Let us learn profit and loss concepts in math. It is well explained in terms of cost price and selling price.
Profit(P)
The amount gained by selling a product with more than its cost price.
Loss(L)
The amount the seller incurs after selling the product less than its cost price is mentioned as a loss.
Cost Price (CP)
The amount paid for a product or commodity to purchase is called a cost price. Also, denoted as CP.
This cost price is further classified into two different categories:
Fixed Cost: The fixed cost is constant, it doesn’t vary under any circumstances
Variable Cost: It could vary depending as per the number of units and other factors
Selling Price (SP)
The amount for which the product is sold is called the Selling Price. It is usually denoted as SP. Also, sometimes called a sale price.
Profit and loss as a percentage
Marked Price Formula (MP)
This is basically labelled by shopkeepers to offer a discount to the customers in such a way that,
Formula for profit and loss percentage
Profit and Loss Formulas
Now let us find the profit formula and loss formula.
The profit or gain is equal to the selling price minus the cost price.
Loss is equal to cost price minus selling price.
The formula for the profit and loss percentage is:
Things you should know about discount
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# Structure of Arguments | Logical Reasoning | UGC NET Paper 1
Arguments: An argument is a set of statements, one of which, the conclusion is taken to be supported by the remaining statements, the premises.
For example;
Premises: All men are mortal
Socrates is a man
Conclusion: Socrates is mortal
## What is Argument?
In the above example, there are three statements, 1) All men are mortal, 2) Socrates is a man, and 3) Socrates is mortal. The premises of the Argument is the first two statement, and the last statement is a Conclusion. The statements of premises support the statement of conclusion. This type of arrangement is called an Argument.
### Forms of Arguments
There are three types of Arguments:
1. Deductive Arguments
2. Inductive arguments, and
3. Abductive arguments
#### 1. Deductive Argument
An argument where the conclusion follows validly from the premises. In other words, an argument where the truth of the premises guarantees the truth of the conclusion.
Example:
Premises: All men are mortal
Socrates is a man
Conclusion Socrates is mortal
In the above example, the premises, all men are mortal, and Socrates is a man, give a guarantee the truth of the conclusion; Socrates is mortal. The conclusion follows the validity according to the premises.
#### 2. Inductive Argument
An argument where the premises point several cases of some pattern and the conclusion states that this pattern will hold in general.
An inductive argument will not be deductively valid, because even if a pattern is found many times, that does not guarantee it will always be found. Therefore, an inductive argument provides weaker, less trustworthy support for the conclusion than a deductive argument does.
For Example:
Premises: We have seen 1000 swans, and
All of them have been white
Conclusion: All swans are white.
In the above example, we have seen just 1000 swans (not all in the world), and all of them have been white. But it does not mean that all swans in the world are white. White swans are a case of a pattern in those particular circumstances. Hence, we have concluded in general that all swans are white. But it might not be true actually. This type of arrangement of premises and conclusion is an example of an Inductive argument.
#### 3. Abductive (or Hypothetico-Deductive) Argument
An argument that (i) points out a particular fact, (ii) points out that if a particular hypothesis were true, we would get this fact, and so (iii) concludes that the hypothesis is indeed true.
Abductive arguments seem to make an even bigger jump than inductive arguments. Inductive arguments generalize, while abductive arguments say that successful predictions “prove” theory is true. Abductive arguments are not deductively valid because false theories can make true predictions. So, true predictions do not guarantee that the theory is true.
Example:
Premises: These coins conduct electricity (fact)
If these coins are made of gold (hypothesis),
then they would conduct electricity (prediction).
Conclusion: These coins are made of gold.
## Structure of Categorical propositions
A proposition is simply a claim about the world that has truth value. Every proposition can be expressed as a declarative (i.e., not a question or command) sentence.
Categorical Proposition is any statement which relates two classes or categories of entities. In other words, a categorical proposition is a proposition that relates two classes of objects. A class is a group of objects.
Example: Cats are mammals
Here, a class or category (Cats) are related to another class or category (Mammals). So, “Cats are mammals” is a Categorical proposition.
### Components of Categorical Propositions
For a any categorical proposition, there are four components:
1. Subject Term: First category or class
2. Predicate Term: Second category or class
3. Copula: The grammatical link (verb) between subject and predicate terms.
4. Quantifiers: Words that specify the quantity of the subject and predicate terms.
Two Important terms of Categorical Proposition
Universal:
• Affirmative: ‘All’ (includes all of a class)
• Negative: ‘No’ (excludes all of a class)
Particular: ‘Some’ (includes part of a class)
Example: All cats are mammals
Here, All – Quantifier
Cats – Subject Term
Are – Copula
Mammals – Predicate term
### Properties of Categorical Propositions
Each categorical proposition has both quantity and quality properties. The followings are the properties:
Quantity: The quantity of a categorical proposition is determined by the quantifier used.
Quality: The quality of a categorical proposition is determined according to whether the proposition asserts of denies an overlap between the classes.
Affirmative: if a proposition asserts an overlap between the classes or category named, the quality of the proposition is affirmative.
Negative: In this, a proposition denies an overlap between the categories or classes named,
Distribution: If the proposition refers to the entire class named by a term, that term is distributed and if it does not refer to the entire class named by a term, then the term is undistributed.
### Types of Categorical Proposition
There are four types of categorical position:
• All politicians are liars (Universal Affirmative) – A
• No politicians are liars. (Universal Negative) – E
• Some politicians are liars. (Particular affirmative) – I
• Some politicians are not liars. ( Particular negative) – O
Universal Affirmative (A- Propositions): In a proposition, if every member of the subject class is also a member of the predicate class, then it is called Universal Affirmative Proposition. In other words, whole of one class is included or contained in another class.
In an example All politicians are liars, every member of the class of politicians, is a member of another class of liars.
An universal affirmative proposition can be written as: All S is P
S and P represent the subject and predicate terms, respectively. Such a proposition affirms that the relation of class inclusion holds between the two classes and says that the inclusion is complete, or universal.
Universal Negative (E- Proposition): The proposition in which no members of the subject class are members of the predicate class.
In an example No politicians are liars”, no member of the class of politicians, is a member of another class of liars.
Systematically, Universal Negative proposition can be represented as: No S is P
Such a proposition affirms that the no relation of class inclusion holds between the two classes and says that the exclusion is complete, or universal.
Particular affirmative (I-proposition): The proposition in which at least one members of the subject class is also a member of the predicate class.
In an example “Some politicians are liars”, some member of the class of politicians, is a member of another class of liars.
Systematically, Particular affirmative proposition can be represented as: Some S is P
Particular negative ( O-proposition): The proposition in which at least one members of the subject class is not a member of the predicate class.
In an example “Some politicians are not liars”, some member of the class of politicians, is a member of another class of liars.
Systematically, Particular affirmative proposition can be represented as: Some S is not P.
A brief of Four Kind of Categorical Proposition
## Classical Square of Opposition
The opposition is an immediate inference grounded on the relation between propositions which have the same terms, but differ in quantity or quality (or both).
For any formal opposition between two propositions, it is essential that their terms should be the same. There can be no opposition between two such propositions as these:
• All angels have
• No cows are
The square of opposition shows us the logical inferences (immediate inferences) we can make from one proposition type (A, I, E, and O) to another.
Contradictory: Two propositions are said to be contradictory if both cannot be true, and both cannot be false at the same time. In other words, if the opposition is between two propositions, which differ both in quantity and quality.
Here, A – All politicians are liars and O – Some politicians are not liars, and similarly, E and I propositions are contradictory.
Contrary: Universal propositions are said to be contrary because they cannot both be simultaneously true. In other words, the opposition is between two universals which differ in quality.A- All politicians are liars is true, the E- No politicians are liars must be false.
Similarly, if the E-proposition is true, then the A-proposition is false.
Sub-contrary: If the two particular propositions can both be true but cannot both be false. In other words, the opposition is between two particulars which differ in quality. It means that they cannot both be simultaneously false.Sub alternation: The universal to particular and particular to universal inferences are called subalternation. In other words, the opposition is between two propositions which differ only in quantity.
These inferences are valid if the subaltern (A or E) is true, then the subaltern (I or O) is true. If the subaltern is false, then the subaltern is false.
## Categorical Syllogism
A syllogism is an argument containing two premises and a conclusion.
Categorical syllogism: A categorical syllogism is a syllogism whose premises and conclusion are categorical propositions.
For example:
Premises:
1. All hats are fashionable clothing.
2. All fashionable clothing is purple.
Conclusion:
1. So, some hats are purple.
A Standard Form Categorical Syllogism Contains:
• Two premises and a conclusion, each a standard form categorical proposition.
• Major Term: A major term which appears only in the first premise and the predicate of the conclusion.
• Minor Term: A minor term which appears only in the second premise and the subject of the conclusion
• Middle Term: A middle term which appears in both premises but not in the conclusion.
• Major Premises: The major premise is the premise which contains the major term.
• Minor Premise: The minor premise is the premise which contains the minor term.
### Mood and Figure of Syllogism
The categorical syllogisms can be categorise as:
• The kind of standard form categorical proposition which appears as its first (i.e., major) premise,
• The kind of standard form categorical proposition which appears as its second (i.e., minor) premise, and
• The kind of standard form categorical propositions which appears as its conclusion.
#### Mood:
When the major premise, the minor premise, and the conclusion of a categorical syllogism arranage in a series of three letters (A, E, I, or O) corresponding to the type of categorical proposition is called MOOD of an argument.
Example:
Premises: All P are M
All S are M
Conclusion: Some S are P
The first premise is of the form A
The second premise is of the form A
The conclusion is of the form I.
Thus, the mood of this Argument is AAI.
In another example, to figure out the FORM of the premises and the conclusion in the following example:
Premises: No S are P (E-propostion)
Some S are P (I-Proposition)
Conclusion: Some S are not P (O-Proposition)
Thus, the mood of this Argument is “EIO”.
When you have to determine the mood of a categorical syllogism, you need to find out which of the four forms of categorical proposition each line of the Argument is (A, E, I, or O).
#### Figure:
The figure of a categorical syllogism is a number which corresponds to the placement of the two middle terms.
For example, consider the following arguments:
P 1. All mammals are creatures that have hair.
P 2. All dogs are mammals.
P 3. Therefore, all dogs are creatures that have hair.
Notice that the middle term in the major premise is on the LEFT, while the middle term in the minor premise is on the RIGHT. Whenever this happens, we say that the argument has figure “1.”
There are four possible figures in the categorical syllogism:
Figure1: When the middle term is on the left in P 1, and on the right in P 2.
Figure2: When the middle term is on the right in both premises.
Figure3: When the middle term is on the left in both premises.
Figure4: When the middle term is on the right in P 1, and on the left in P 2.
Important Points of Mood and Figures:
• There are 64 different moods
• And each mood has 4 different figures.
• Thus, there are 64*4=256 different kinds of standard form categorical syllogisms.
### Valid Argument Forms
There are two kinds of valid argument forms:
Unconditionally Valid Forms: There are fifteen combinations of mood and figures that are valid from the Boolean standpoint, and we call these “unconditionally valid”argument forms. The chart below depicts ALL of 15 the unconditionally valid argument forms.
Conditionally Valid Forms: There are some inferences that are NOT valid from the Boolean standpoint, which is valid from the Aristotelian standpoint. In addition to the fifteen unconditionally valid argument forms, there are nine conditionally valid argument forms for categorical syllogisms:
### Rules For Validity
A standard form categorical syllogism is valid on the modern theory if and only if each of the following five propositions is all true of it. A standard form categorical syllogism is valid on the traditional theory if and only if each of the first four propositions is true of it.
1. The middle-term is distributed at least once.
2. If a term is distributed in a conclusion, then that term is distributed in one of the premises.
3. There is at least one affirmative premise.
4. There is a negative premise if and only if there is a negative conclusion.
5. If both premises are universal, then the conclusion is universal.
## Informal and Formal Fallacy
Simply, a fallacy is a mistake in reasoning. In other words, a defect in an argument that misleads the mind is called a fallacy. There are two types of fallacy:
Formal Fallacies: A fallacy in which there is the involvement of an error in the form, arrangement, or technical structure of an argument is called Formal Fallacy.
Informal Fallacies: Informal fallacies are a matter of unclear expression that deal with the logic of the meaning of language. Opposite to it, formal fallacies deal with the logic of the technical structure.
An informal fallacy involves such things as:
• the misuse of language such as words or grammar,
• misstatements of fact or opinion,
• misconceptions due to underlying presuppositions, or
• just plain illogical sequences of thought.
## Uses of Language in Logic
A logic always deals with the analysis and evaluation of arguments. Since arguments are expressed in language, the study of arguments requires a carefully attention to language in which arguments are expressed.
The followings are three important uses of language:
1. Informative,
2. Expressive and
3. Directive uses of language.
Informative use of language: It involves an effort to communicate some content or to describe something or to give information about something. When I say a child, “The Second of October is the Gandhi Jayanti.” The language I used is informative.
This kind of use of language presumes that the content of what is being communicated is true, so it will be our main focus in the study of logic.
When a sentence is used informatively, it reports that something has some feature or that something lacks some feature. Consider the following two sentences:
1. Parrot has a feather.
2. Parrot is not mammals.
The first proposition reports that having feather is a feature of a Parrot. The second proposition reports that Parrot do not have some essential qualities found in mammals. In, both cases it provides information about the world.
Two main aspects of this function are generally noted: (1) evoking certain feelings and (2) expressing feelings.
Expressive discourse, qua expressive discourse, is best regarded as neither true or false.
Expressive use of Language: This type of language is often used to express our emotions, feelings, or attitudes. For example: It’s too bad!, It’s wonderful!, etc.
When language is used expressively or emotively, it cannot be characterized as true or false.
Directive uses of language: When the use of language is often to give direction as Commands, requests, instructions, questions etc., to do or not to do something.
Consider the following examples:
3. Are you feeling well?
In all the above examples, the directive use of language. Directive use of language is not normally considered true or false (although various logics of commands have been developed).
## Connotations and denotations of terms
Denotation is the dictionary definition or literal meaning of a word only. Not emotions or feelings are associated with the word.
Ex: The teacher walked into the classroom. This example does not have any hidden meaning. A teacher simply walked into a classroom.
Connotation: A word’s emotional meaning; suggestions and associations that are connected to a word. Words can be positive, negative, or neutral. Words can also connote specific feelings or emotions.
### Different types of Definition
Lexical: The purpose of a lexical definition is to report the way a word is standardly used in a language. Most definitions found in a dictionary are lexical definitions.
Ex. Fossil, Cat, Dogs etc.
Persuasive: The purpose of a persuasive definition is to influence people’s attitudes, not to neutrally and objectively capture the standard meaning of a word.
Eg. Teenagers, Abortion etc.
Stipulative: A stipulative definition stipulates (assigns) a meaning to a word by coining a new word or giving an old word a new meaning. A stipulative definition is neither true nor false; it is neither accurate nor inaccurate.
Eg. Sugarnecker, Black Holes, etc.
Theoretical: Theoretical definitions can explain concepts theoretically. Sometimes definitions are given for terms, not because the word itself is unfamiliar, but because the term is not understood. Such concepts require theoretical definitions, which are often scientific or philosophical in nature.
For example, when your chemistry teacher defines water by its chemical formula H2O, he is not trying to increase your vocabulary (you already knew the term water), but to explain its atomic structure.
Accepting a theoretical definition is like accepting a theory about the term being defined. If you define spirit as “the life-giving principle of physical organisms,” you are inviting others to accept the idea that life is some how a spiritual product.
Precising: A precising definition takes a word that is normally vague and gives it a clear precisely defined meaning.
Eg. Lite, Low-income, middle aged, etc.
Table of Content | 4,047 | 18,978 | {"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} | 4.09375 | 4 | CC-MAIN-2024-26 | latest | en | 0.855419 |
http://forums.wolfram.com/mathgroup/archive/2002/Oct/msg00338.html | 1,606,892,089,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141692985.63/warc/CC-MAIN-20201202052413-20201202082413-00116.warc.gz | 35,289,419 | 8,292 | RE: grouping and averaging {x,y} pairs of data
• To: mathgroup at smc.vnet.net
• Subject: [mg37226] RE: grouping and averaging {x,y} pairs of data
• From: "David Park" <djmp at earthlink.net>
• Date: Thu, 17 Oct 2002 00:09:05 -0400 (EDT)
• Sender: owner-wri-mathgroup at wolfram.com
```David,
You will probably get a lot of answers for this. Here is my entry.
data = {{3, 1}, {4, 3}, {3, 2}, {1, 10}, {4, 2}, {1, 6}, {5, 2}, {2, 5}, {7,
1}};
First I will show it step-by-step.
nmax = 10;
Union[Join[data, Table[{i, 0}, {i, 1, nmax}]]]
Split[%, #1[[1]] == #2[[1]] & ]
Map[Last, %, {2}]
(Plus @@ #1/Length[#1] & ) /@ %
Transpose[{Range[nmax], %}]
giving
{{1, 0}, {1, 6}, {1, 10}, {2, 0}, {2, 5}, {3, 0}, {3, 1}, {3, 2}, {4, 0},
{4,
2}, {4, 3}, {5, 0}, {5, 2}, {6, 0}, {7, 0}, {7, 1}, {8, 0}, {9, 0}, {10,
0}}
{{{1, 0}, {1, 6}, {1, 10}}, {{2, 0}, {2, 5}}, {{3, 0}, {3, 1}, {3, 2}}, {{4,
0}, {4, 2}, {4, 3}}, {{5, 0}, {5, 2}}, {{6, 0}}, {{7, 0}, {7, 1}},
{{8,
0}}, {{9, 0}}, {{10, 0}}}
{{0, 6, 10}, {0, 5}, {0, 1, 2}, {0, 2, 3}, {0, 2}, {0}, {0, 1}, {0}, {0},
{0}}
{16/3, 5/2, 1, 5/3, 1, 0, 1/2, 0, 0, 0}
{{1, 16/3}, {2, 5/2}, {3, 1}, {4, 5/3}, {5, 1}, {6, 0},
{7, 1/2}, {8, 0}, {9, 0}, {10, 0}}
This wraps it into one statement.
nmax = 10;
Transpose[{Range[nmax], (Plus @@ #1/Length[#1] & ) /@
Map[Last, Split[Union[Join[data, Table[{i, 0},
{i, 1, nmax}]]], #1[[1]] == #2[[1]] & ], {2}]}]
{{1, 16/3}, {2, 5/2}, {3, 1}, {4, 5/3}, {5, 1}, {6, 0},
{7, 1/2}, {8, 0}, {9, 0}, {10, 0}}
This times a case of 20000 pairs on an 800MHz machine.
data2 = Table[{Random[Integer, {1, 100}], Random[Real, {0, 5}]}, {20000}];
nmax = 100;
data = data2;
Timing[Transpose[{Range[nmax],
(Plus @@ #1/Length[#1] & ) /@ Map[Last,
Split[Union[Join[data, Table[{i, 0},
{i, 1, nmax}]]], #1[[1]] == #2[[1]] & ],
{2}]}]; ]
{0.55 Second, Null}
David Park
From: David E. Burmaster [mailto:deb at alceon.com]
To: mathgroup at smc.vnet.net
Dear Fellows in MathGroup,
I have a list of 17,000+ {x,y} pairs of data
each x value is a positive integer from 1 to 100+
each y value is a positive real number
As a *short* example, let's consider:
data = {{3,1},{4,3},{3,2},{1,10},{4,2},{1,6},{5,2},{2,5},{7,1}}
I want to group the data by the x value and report the arithmetic average
of the y values in each group.
For the example, i want to report:
output = {{1,8},{2,5},{3,1.5},{4,2.5},{5,2},{6,0},{7,1}}
In this example, x=6 does not occur so i report the average y[6] = 0.
Can anyone suggest a way to do this efficiently?/
many thanks
dave
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
David E. Burmaster, Ph.D.
Alceon Corporation
POBox 382069 (new Box number effective 1 Sep 2001)
Harvard Square Station
Cambridge, MA 02238-2069 (new ZIP code effective 1 Sep 2001)
Voice 617-864-4300
Web http://www.Alceon.com
Email deb at Alceon.com
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
```
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# Events & Promotions
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# Calling all UCLA Anderson Fall 2010 applicants!
Author Message
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Re: Calling all Anderson Fall 2010 applicants! [#permalink]
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27 Oct 2009, 11:19
I'm thinking there won't be any news until after November 4th when they release the application status info.
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Re: Calling all Anderson Fall 2010 applicants! [#permalink]
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27 Oct 2009, 13:14
_________________
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Check-out the following:
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1. Math Divisibility Test
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Re: Calling all Anderson Fall 2010 applicants! [#permalink]
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27 Oct 2009, 13:32
Count me in for R2...might be at the top of the list since the old ball and chain wants to move back home
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Re: Calling all Anderson Fall 2010 applicants! [#permalink]
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02 Nov 2009, 12:42
Count me in for R2 too!
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Re: Calling all Anderson Fall 2010 applicants! [#permalink]
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04 Nov 2009, 09:04
Looks like as promised, Anderson updated the application status today. It should say on your main page when you log into the system. It just tells you whether your application is complete or if they are missing stuff. Nothing too earth shattering, just fyi.
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Re: Calling all Anderson Fall 2010 applicants! [#permalink]
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06 Nov 2009, 09:42
Seems like it is taking longer for Anderson to give out interview invites than last year. Anyone hear anything yet?
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Re: Calling all Anderson Fall 2010 applicants! [#permalink]
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06 Nov 2009, 20:39
Good luck to the R1'ers. Count me in for R2. It'll be good to drink that Kerckhoff Coffee again.
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Re: Calling all Anderson Fall 2010 applicants! [#permalink]
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08 Nov 2009, 10:30
still sounds like there's no word...hopefully this coming week will change that.
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10 Nov 2009, 10:43
Some interviews have gone out. Not for me, but someone else I know. Anyone else here?
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Re: Calling all Anderson Fall 2010 applicants! [#permalink]
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10 Nov 2009, 10:50
albyman32 wrote:
Some interviews have gone out. Not for me, but someone else I know. Anyone else here?
That's the first I have heard about interview invites going out. Haven't heard of anyone else getting an interview.
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Re: Calling all Anderson Fall 2010 applicants! [#permalink]
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12 Nov 2009, 10:25
asianstyles wrote:
albyman32 wrote:
Some interviews have gone out. Not for me, but someone else I know. Anyone else here?
That's the first I have heard about interview invites going out. Haven't heard of anyone else getting an interview.
Anyone else hear anything about Anderson?
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Re: Calling all Anderson Fall 2010 applicants! [#permalink]
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12 Nov 2009, 20:12
Pook wrote:
asianstyles wrote:
albyman32 wrote:
Some interviews have gone out. Not for me, but someone else I know. Anyone else here?
That's the first I have heard about interview invites going out. Haven't heard of anyone else getting an interview.
Anyone else hear anything about Anderson?
I called Anderson Ad Comm yesterday they said that they have started the interview calls 3 days back and are in process.. Did someone get any calls?
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Re: Calling all Anderson Fall 2010 applicants! [#permalink]
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13 Nov 2009, 00:11
Quoting Mae Jennifer Shore's (Director of Admissions) twitter: "Interview invites for Round 1 are being released on an ongoing basis this month & into December. There is no order to the invites."
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Re: Calling all Anderson Fall 2010 applicants! [#permalink]
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13 Nov 2009, 02:44
albyman32 wrote:
Quoting Mae Jennifer Shore's (Director of Admissions) twitter: "Interview invites for Round 1 are being released on an ongoing basis this month & into December. There is no order to the invites."
A bit relief .. Thanks Albyman
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Re: Calling all Anderson Fall 2010 applicants! [#permalink]
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13 Nov 2009, 06:17
International Re-applicant here.
Received an interview invite today. There's an option to either interview on-campus or schedule an off-campus alum interview.
_________________
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Re: Calling all Anderson Fall 2010 applicants! [#permalink]
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13 Nov 2009, 09:11
albyman32 wrote:
Quoting Mae Jennifer Shore's (Director of Admissions) twitter: "Interview invites for Round 1 are being released on an ongoing basis this month & into December. There is no order to the invites."
Great info. Thanks.
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Re: Calling all Anderson Fall 2010 applicants! [#permalink]
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13 Nov 2009, 09:26
topa wrote:
International Re-applicant here.
Received an interview invite today. There's an option to either interview on-campus or schedule an off-campus alum interview.
Good luck!! Did you get the invite by email?
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Re: Calling all Anderson Fall 2010 applicants! [#permalink]
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13 Nov 2009, 10:50
I'm interviewing tomorrow with the Super Saturday group. I got my interview invite via email the day that I submitted my application (October 20th). Am I missing something here?
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Re: Calling all Anderson Fall 2010 applicants! [#permalink]
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13 Nov 2009, 11:07
I am assuming you applied for the FEMBA? For the full-time MBA, it is by invite only so we did not get that option to pick an interview date when we submitted...
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Re: Calling all Anderson Fall 2010 applicants! [#permalink]
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13 Nov 2009, 11:28
Okay makes sense. Yes I did apply for the FEMBA!
Re: Calling all Anderson Fall 2010 applicants! [#permalink] 13 Nov 2009, 11:28
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4Random Access Lists
Following Random Access List structures implement and provide the functions list, empty?, cons, head, tail, list-ref, list-set, drop, list-length and ->list. The implementations of the two data structures are based on numerical representations. Binary random access lists uses the binary number representation and running time of its basic list and random access operations, in worst-case, is logarithmic. Where as skew binary random access lists use skew binary number representation and running time of its basic operations is constant in worst-case. And both the implementations are polymorphic. And our benchmarks indicate that the operations of skew binary random access lists are faster.
4.1Binary Random Access List
(require (planet krhari/pfds:1:2/binaryrandomaccesslist))
Random Access Lists are list data structures that provide array-like lookup and update operations. They have been implemented as a framework of binary numerical representation using complete binary leaf trees. It has a worst case running time of O(log(n)) for the operations cons, first, rest, list-ref and list-set.
(List A)
A binary random access list of type A.
(list a ...) → (List A) a : A
Function list creates a Binary Random Access List with the given inputs.
Example: > (list 1 2 3 4 5 6) - : (U Null (Root Positive-Fixnum)) #
In the above example, (list 1 2 3 4 5 6) gives a Binary Random Access List which is similar to lists but comes with efficient list-ref and list-set operations.
empty : (List Nothing)
A empty random access list
(empty? ral) → Boolean ral : (List A)
Function empty? takes a Binary Random Access List checks if the given list is empty.
Examples: > (empty? (list 1 2 3)) - : Boolean #f > (empty? empty) - : Boolean #t
(cons a ral) → (List A) a : A ral : (List A)
Function cons takes an element and a list and adds the given element to the front of the given list.
Example: > (cons 10 (list 5 3 5 6)) - : (U Null (Root Positive-Fixnum)) #
In the above example, (cons 10 (list 5 3 5 6)) returns (list 10 5 3 5 6).
(first ral) → A ral : (List A)
Function first takes a list and returns the first element of the given list if its not empty else throws an error.
Examples: > (first (list 5 3 5 6)) - : Positive-Fixnum 5 > (first empty) head: given list is empty
(rest ral) → (List A) ral : (List A)
Function rest takes a list and returns the given list but without its first element if the given list is not empty. If it is empty, rest throws an error
Examples: > (rest (list 1 2 3 4 5 6)) - : (U Null (Root Positive-Fixnum)) # > (rest empty) tail: given list is empty
In the above example, (rest ral) returns the rest of the given list, (list 2 3 4 5 6).
(list-ref ral index) → A ral : (List A) index : Integer
Function list-ref takes an integer(say n) and a list and gives the nth element of the given list. If the given n is larger than the size of the list, list-ref throws an error.
Examples: > (list-ref (list 12 5 3 2 15 23) 4) - : Positive-Fixnum 15 > (list-ref (list 12 5 3 2 15 23) 10) list-ref: given index out of bound
(list-set ral index newval) → (List A) ral : (List A) index : Integer newval : A
Function list-set takes an integer(say n), list and a new element. And updates the nth element of the list with the new element.
Examples: > (list-set (list 1 2 3 4 5 6) 2 10) - : (U Null (Root Positive-Fixnum)) # > (list-set (list 1 2 3 4 5 6) 10 15) list-set: given index out of bound
In the above example, (list-set (list 1 2 3 4 5 6) 2 10) returns (list 1 2 10 4 5 6) and (list-set (list 1 2 3 4 5 6) 10 15) throws an error.
(->list ral) → (Listof A) ral : (List A)
Function ->list takes a list and returns a list of elements which are in the same order as in the list.
Examples: > (define ral (list 1 2 3 4 5 6)) > (->list ral) - : (Listof Positive-Fixnum) '(1 2 3 4 5 6)
In the above example, (->list ral) returns (list 1 2 3 4 5 6).
(drop ral num) → (List A) ral : (List A) num : Integer
Function drop takes a list and an integer(say n) and drops the first n elements of the input list and returns the rest of the list.
Examples: > (drop 3 (list 1 2 3 4 5 6)) - : (U Null (Root Positive-Fixnum)) # > (drop 10 (list 1 2 3 4 5 6)) drop: not enough elements to drop
In the above example, (drop 3 (list 1 2 3 4 5 6)) returns the list (list 4 5 6). (drop 10 (list 1 2 3 4 5 6)) throws an error since 10 is larger than the number of elements in the list.
(list-length ral) → Integer ral : (List A)
Function list-length takes a list and gives the number of elements in in the given list.
Examples: > (list-length (list 1 2 3 4 5 6)) - : Integer 6 > (list-length empty) - : Integer 0
(reverse list) → (List A) list : (List A)
Function reverse takes a list and returns a reversed list.
Example: > (->list (reverse (list 1 2 3 4 5 6))) - : (Listof Positive-Fixnum) '(6 5 4 3 2 1)
(map func lst1 lst2 ...) → (List A) func : (A B ... B -> C) lst1 : (List A) lst2 : (List B)
Function map is similar to map for lists.
Examples: > (->list (map add1 (list 1 2 3 4 5 6))) - : (Listof Exact-Positive-Integer) '(2 3 4 5 6 7) > (->list (map * (list 1 2 3 4 5 6) (list 1 2 3 4 5 6))) - : (Listof Exact-Positive-Integer) '(1 4 9 16 25 36)
In the above example, (map add1 (list 1 2 3 4 5 6)) adds 1 to each element of the given list and returns (list 2 3 4 5 6 7). (map * (list 1 2 3 4 5 6) (list 1 2 3 4 5 6)) multiplies corresponding elements in the two lists and returns the list (list 1 4 9 16 25 36).
(foldl func init lst1 lst2 ...) → C func : (C A B ... B -> C) init : C lst1 : (List A) lst2 : (List B)
Function foldl is similar to foldl.
foldl currently does not produce correct results when the given function is non-commutative.
Examples: > (foldl + 0 (list 1 2 3 4 5 6)) - : Exact-Nonnegative-Integer 21 > (foldl * 1 (list 1 2 3 4 5 6) (list 1 2 3 4 5 6)) - : Exact-Positive-Integer 518400
(foldr func init lst1 lst2 ...) → C func : (C A B ... B -> C) init : C lst1 : (List A) lst2 : (List B)
Function foldr is similar to foldr.
foldr currently does not produce correct results when the given function is non-commutative.
Examples: > (foldr + 0 (list 1 2 3 4 5 6)) - : Exact-Nonnegative-Integer 21 > (foldr * 1 (list 1 2 3 4 5 6) (list 1 2 3 4 5 6)) - : Exact-Positive-Integer 518400
(filter pred lst) → (List A) pred : (A -> Boolean) lst : (List A)
Function filter is similar to filter.
Examples: > (define lst (list 1 2 3 4 5 6)) > (->list (filter (λ:([x : Integer]) (> x 5)) lst)) - : (Listof Positive-Fixnum) '(6) > (->list (filter (λ:([x : Integer]) (< x 5)) lst)) - : (Listof Positive-Fixnum) '(1 2 3 4) > (->list (filter (λ:([x : Integer]) (<= x 4)) lst)) - : (Listof Positive-Fixnum) '(1 2 3 4)
(remove pred lst) → (List A) pred : (A -> Boolean) lst : (List A)
Function remove is similar to filter but remove removes the elements which match the predicate.
Examples: > (->list (remove (λ:([x : Integer]) (> x 5)) (list 1 2 3 4 5 6))) - : (Listof Positive-Fixnum) '(1 2 3 4 5) > (->list (remove (λ:([x : Integer]) (< x 5)) (list 1 2 3 4 5 6))) - : (Listof Positive-Fixnum) '(5 6) > (->list (remove (λ:([x : Integer]) (<= x 4)) (list 1 2 3 4 5 6))) - : (Listof Positive-Fixnum) '(5 6)
4.2Skew Binary Random Access List
(require (planet krhari/pfds:1:2/skewbinaryrandomaccesslist))
Random Access Lists are list data structures that provide array-like lookup and update operations. Skew Binary Random Access Lists are implemented using skew binary numbers. It provides a worst case running time of O(1) for the operations cons, first and rest and O(log(n)) for the operations list-ref and list-set.
(List A)
A skew binary random access list type A.
(list a ...) → (List A) a : A
Function list creates a Skew Binary Random Access List with the given inputs.
Example: > (list 1 2 3 4 5 6) - : (Listof (Root Positive-Fixnum)) '(# #)
In the above example, (list 1 2 3 4 5 6) gives a Skew Binary Random Access List which is similar to lists and has efficient lookup and update operations.
empty : (List Nothing)
A empty list.
(empty? ral) → Boolean ral : (List A)
Function empty? takes a Skew Binary Random Access List checks if the given list is empty.
Examples: > (empty? (list 1 2 3 4 5 6)) - : Boolean #f > (empty? empty) - : Boolean #t
(cons a ral) → (List A) a : A ral : (List A)
Function cons takes an element and a list and adds the given element to the front of the given list.
Example: > (cons 10 (list 1 2 3 4 5 6)) - : (Listof (Root Positive-Fixnum)) '(#)
In the above example, (cons 10 (list 1 2 3 4 5 6)) returns a (list 10 1 2 3 4 5 6).
(first ral) → A ral : (List A)
Function first takes a list and returns the first element of the given list.
Examples: > (first (list 1 2 3 4 5 6)) - : Positive-Fixnum 1 > (first empty) head: given list is empty
(rest ral) → (List A) ral : (List A)
Function rest takes a list and returns a list without the first element of the given list.
Examples: > (rest (list 1 2 3 4 5 6)) - : (Listof (Root Positive-Fixnum)) '(# # #) > (rest empty) tail: given list is empty
In the above example, (rest (list 1 2 3 4 5 6)) returns the (list 2 3 4 5 6).
(list-ref ral index) → A ral : (List A) index : Integer
Function list-ref takes a list and an index(say n) and gives the nth element of the given list.
Examples: > (list-ref (list 1 2 3 4 5 6) 0) - : Positive-Fixnum 1 > (list-ref (list 1 2 3 4 5 6) 3) - : Positive-Fixnum 4 > (list-ref (list 1 2 3 4 5 6) 10) list-ref: given index out of bounds
(list-set ral index newval) → (List A) ral : (List A) index : Integer newval : A
Function list-set takes a list, an index(say n) and a new element and updates the nth element of the list with the new element.
Examples: > (->list (list-set (list 1 2 3 4 5 6) 3 10)) - : (Listof Positive-Fixnum) '(1 2 3 10 5 6) > (->list (list-set (list 1 2 3 4 5 6) 6 10)) list-set: given index out of bounds
In the above example, (list-set (list 1 2 3 4 5 6) 3 10) returns (list 1 2 3 10 5 6), (list-set (list 1 2 3 4 5 6) 6 10) throws an error since there are only six elements in the list and it is trying to update seventh element.
(->list ral) → (Listof A) ral : (List A)
Function ->list takes a list and returns a list of elements which are in the same order as in the list.
Examples: > (->list (list 1 2 3 4 5 6)) - : (Listof Positive-Fixnum) '(1 2 3 4 5 6) > (->list empty) - : (Listof Nothing) '()
(drop num ral) → (List A) num : Integer ral : (List A)
Function drop takes a list and an integer(say n) and drops the first n elements of the input list and returns the rest of the list.
Examples: > (drop 3 (list 1 2 3 4 5 6)) - : (Listof (Root Positive-Fixnum)) '(#) > (drop 0 (list 1 2 3 4 5 6)) - : (Listof (Root Positive-Fixnum)) '(# #) > (drop 10 (list 1 2 3 4 5 6)) drop: not enough elements to drop
In the above example, (drop 3 (list 1 2 3 4 5 6) 3) returns (list 4 5 6). (drop 0 (list 1 2 3 4 5 6)) returns the (list 1 2 3 4 5 6). If the given n is larger than the number of elements in the list, then it throws an error.
(list-length ral) → Integer ral : (List A)
Function list-length takes a list and gives the number of elements in in the given list.
Examples: > (list-length (list 1 2 3 4 5 6)) - : Integer 6 > (list-length (list 1 2 3)) - : Integer 3 > (list-length empty) - : Integer 0
(reverse list) → (List A) list : (List A)
Function reverse takes a list and returns a reversed list.
Example: > (->list (reverse (list 1 2 3 4 5 6))) - : (Listof Positive-Fixnum) '(4 5 6 1 2 3)
(map func lst1 lst2 ...) → (List A) func : (A B ... B -> C) lst1 : (List A) lst2 : (List B)
Function map is similar to map for lists.
Examples: > (->list (map add1 (list 1 2 3 4 5 6))) - : (Listof Exact-Positive-Integer) '(2 3 4 5 6 7) > (->list (map * (list 1 2 3 4 5 6) (list 1 2 3 4 5 6))) - : (Listof Exact-Positive-Integer) '(1 4 9 16 25 36)
In the above example, (map add1 (list 1 2 3 4 5 6)) adds 1 to each element of the given list and returns (list 2 3 4 5 6 7). (map * (list 1 2 3 4 5 6) (list 1 2 3 4 5 6)) multiplies corresponding elements in the two lists and returns the list (list 1 4 9 16 25 36).
(foldl func init lst1 lst2 ...) → C func : (C A B ... B -> C) init : C lst1 : (List A) lst2 : (List B)
Function foldl is similar to foldl.
foldl currently does not produce correct results when the given function is non-commutative.
Examples: > (foldl + 0 (list 1 2 3 4 5 6)) - : Exact-Nonnegative-Integer 21 > (foldl * 1 (list 1 2 3 4 5 6) (list 1 2 3 4 5 6)) - : Exact-Positive-Integer 518400
(foldr func init lst1 lst2 ...) → C func : (C A B ... B -> C) init : C lst1 : (List A) lst2 : (List B)
Function foldr is similar to foldr.
foldr currently does not produce correct results when the given function is non-commutative.
Examples: > (foldr + 0 (list 1 2 3 4 5 6)) - : Exact-Nonnegative-Integer 21 > (foldr * 1 (list 1 2 3 4 5 6) (list 1 2 3 4 5 6)) - : Exact-Positive-Integer 518400
(filter pred lst) → (List A) pred : (A -> Boolean) lst : (List A)
Function filter is similar to filter.
Examples: > (define lst (list 1 2 3 4 5 6)) > (->list (filter (λ:([x : Integer]) (> x 5)) lst)) - : (Listof Positive-Fixnum) '(6) > (->list (filter (λ:([x : Integer]) (< x 5)) lst)) - : (Listof Positive-Fixnum) '(1 2 3 4) > (->list (filter (λ:([x : Integer]) (<= x 4)) lst)) - : (Listof Positive-Fixnum) '(1 2 3 4)
(remove pred lst) → (List A) pred : (A -> Boolean) lst : (List A)
Function remove is similar to filter but remove removes the elements which match the predicate.
Examples: > (->list (remove (λ:([x : Integer]) (> x 5)) (list 1 2 3 4 5 6))) - : (Listof Positive-Fixnum) '(1 2 3 4 5) > (->list (remove (λ:([x : Integer]) (< x 5)) (list 1 2 3 4 5 6))) - : (Listof Positive-Fixnum) '(5 6) > (->list (remove (λ:([x : Integer]) (<= x 4)) (list 1 2 3 4 5 6))) - : (Listof Positive-Fixnum) '(5 6) | 5,107 | 13,826 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2018-26 | latest | en | 0.662801 |
https://math.answers.com/other-math/Is_3_m_greater_then_200cm | 1,675,456,458,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500074.73/warc/CC-MAIN-20230203185547-20230203215547-00581.warc.gz | 414,464,681 | 50,118 | 0
# Is 3 m greater then 200cm?
Wiki User
2012-03-19 22:33:48
3 m equals 300 cm which is obviously bigger than 200 cm
Wiki User
2012-03-19 22:33:48
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3m | 133 | 364 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | longest | en | 0.809588 |
https://www.physicsforums.com/threads/dice-probability.880554/ | 1,503,274,972,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886106996.2/warc/CC-MAIN-20170820223702-20170821003702-00284.warc.gz | 932,158,756 | 17,896 | Dice probability
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1. Jul 31, 2016
Mark53
1. The problem statement, all variables and given/known data
In a probability experiment, a fair die is rolled twice.
• If the first roll is odd, the outcomes are recorded as they appear.
• If the first roll is even, the recorded outcome for the second die is doubled. For example, if the first die was 2 and the second 4, the recorded outcome would be (2,8).
Let A be the event that the first recorded outcome is even and B be the event that the second is even.
(a) Write down the sample space S for this experiment.
(b) Express A, B and A∩B as subsets of S.
(c) Find P(A) and P(B).
(d) Given that the second recorded outcome is even, what is the probability that the first roll was also even?
3. The attempt at a solution
I am not sure if the the numbers that I have used below are correct any help is much appreciated
a)
s={(2,4),(2,8),(2,12),(4,4),(4,8),(4,12),(6,4),(6,8),(6,12)}
b)
A={2,4,6}
B={4,8,12}
A∩B = {4}
c)
P(A)=1/2
P(B)=1/2
d)
P=1/4
2. Jul 31, 2016
Orodruin
Staff Emeritus
This set is far from complete. What happened to all of the cases where either the first or second die gave 1, 3, or 5?
These are not subsets of your S.
3. Jul 31, 2016
Mark53
S={(2,2),(2,4),(2,6),(2,8),(2,10),(2,12),(4,2)(4,4),(4,6),(4,8),(4,10),(4,12),(6,2),(6,4),(6,6),(6,8),(6,10),(6,12),(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(3,1),(3,2),(3,3),(3,4),(3,5)(3,6),(5,1),(5,2),(5,3)(5,4),(5,5),(5,6)}
A={(2,2),(2,4),(2,6),(2,8),(2,10),(2,12),(4,2)(4,4),(4,6),(4,8),(4,10),(4,12),(6,2),(6,4),(6,6),(6,8),(6,10),(6,12)}
B={(2,2),(2,4),(2,6),(2,8),(2,10),(2,12),(4,2)(4,4),(4,6),(4,8),(4,10),(4,12),(6,2),(6,4),(6,6),(6,8),(6,10),(6,12),(1,2),(1,4),(1,6),(3,2),(3,4),(3,6),(5,2),(5,4),(5,6)}
A∩B = {(2,2),(2,4),(2,6),(2,8),(2,10),(2,12),(4,2)(4,4),(4,6),(4,8),(4,10),(4,12),(6,2),(6,4),(6,6),(6,8),(6,10),(6,12)}
P(A)=1/2
P(B)=3/4
d)
P=1/2
Would this be correct now
4. Jul 31, 2016
PeroK
a) - c) are correct. How did you calculate d)?
5. Jul 31, 2016
Mark53
I used A∩B how should I be calculating it?
6. Jul 31, 2016
PeroK
$P(A \cap B)$ is the probability that both are even. The probability that the first is even given that you know the second is even is something else. Can you see that?
7. Jul 31, 2016
Mark53
Does that mean that it is 1/4 I am still unsure
8. Jul 31, 2016
PeroK
9. Jul 31, 2016
Mark53
10. Jul 31, 2016
Yes. | 990 | 2,402 | {"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} | 4.125 | 4 | CC-MAIN-2017-34 | longest | en | 0.826921 |
https://tw.forumosa.com/t/mrt-station-maps/86406 | 1,722,928,570,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640476915.25/warc/CC-MAIN-20240806064139-20240806094139-00283.warc.gz | 477,797,284 | 7,182 | # MRT Station Maps
How come MRT station maps don’t all have North at the top? It’s very confusing to look at a map with North at top, and then go around to other side of sign and North is at the bottom. Or to walk around the station and the maps at different locations have North at different locations.
Look at the map on one side of the station, then walk to the other side of the station, and the map is flipped the opposite direction so have to figure it all out again.
Must be a different philosophy behind the way they use and display maps than what I’m accustomed to.
Yes, it’s a different philosophy. MRT street maps are usually oriented to face the same direction that you are - so if you’re facing east, the map has east at the top. I find it quite logical, although it’s not quite the same as what I grew up with. The only real issue I have is seeing maps in a new place and taking a few seconds to figure out if it’s aligned with north or with the direction currently faced.
I learned to do pretty much the same thing with outdoor navigation in the mountains: turn the map to align to your current direction.
[quote=“lostinasia”]Yes, it’s a different philosophy. MRT street maps are usually oriented to face the same direction that you are - so if you’re facing east, the map has east at the top. I find it quite logical, although it’s not quite the same as what I grew up with. The only real issue I have is seeing maps in a new place and taking a few seconds to figure out if it’s aligned with north or with the direction currently faced.
I learned to do pretty much the same thing with outdoor navigation in the mountains: turn the map to align to your current direction.[/quote]
There must be some type of terminology for the differences. I was taught to always keep the map with North at top and never turn map around to align.
Yes, it’s logical if you’re on the ground level, but when you’re in B3 and have to make several turns to work your way up to the ground level, you tend to lose the orientation as you wind your way up.
Because Taiwan is in reality a small place, but it has always “felt” big to its inhabitants. And part of this is because of the collective lack of a sense of direction.
Yes, it’s logical if you’re on the ground level, but when you’re in B3 and have to make several turns to work your way up to the ground level, you tend to lose the orientation as you wind your way up.[/quote]
This is exactly how I see it. In my opinion, if you’re underground, maps should have north up, because there are no streets or landmarks visible.
But they do it differently here, so I always advise people to look for the compass points on the map.
I kind of like it. Even though it’s totally different than anything I’m used to, it just seems right to me for some reason. But I’m always pretty well oriented in Taipei (I should be by now).
I have noticed that my Taiwan friends only know one way to drive to places around Taipei. It’s not like they know their way around town. Sometimes we will go across town by an indirect route, and then I ask why, and they respond “Because I only know how to drive that way”. And I’m thinking, “If I knew where you were going, I could of directed us straight there”.
We have literally gone in a big circle and returned from the opposite direction to a different location because that is the only way they know to go.
[quote=“tango42”]I have noticed that my Taiwan friends only know one way to drive to places around Taipei. It’s not like they know their way around town. Sometimes we will go across town by an indirect route, and then I ask why, and they respond “Because I only know how to drive that way”. And I’m thinking, “If I knew where you were going, I could of directed us straight there”.
We have literally gone in a big circle and returned from the opposite direction to a different location because that is the only way they know to go.[/quote]
Been there, done that. One of my friends only knows one way out of the T1 Carpark at Taouyan Airport - he turns LEFT and does a lap around the Terminal - but there is a RIGHT turn option, which I pointed out last time, much to his amazement!!
There are a couple of others that I have been in, mostly avoiding Nanjing Road, but that is maybe a hangover from when it was a construction site for the MRT.
I understand why they only take one route. Taipei streets have no logic to them, you end up running into dead ends and weird branches and loops if you deviate from routes you know. You think that if you notice a different way and take it, it ends up going into some bizarre way and you get lost.
do you remember the scene in MIdnight Express ( I think) where everyone is going around the pole in one direction and then our guy wanted to go in the opposite direction?
No can do.
Just like I think in Mecca, everyone goes around the central pilar in one direction? | 1,100 | 4,905 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.963695 |
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Aggregate Production Planning 4xgs
# Aggregate Production Planning 4xgs - Goal Learn a nice...
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Aggregate Production Planning Peter L. Jackson Professor School of O.R. and I.E. 3/7/2011 Industrial Data and Systems Analysis 1 Goal Learn a nice technique for production planning See it four ways: Bar chart Cumulative flow chart Excel spreadsheet Mathematical notation Interact with SQL query results using procedural Interact with SQL query results using procedural code Program the technique using Visual Basic for Applications Use industrial data set (sales of turkey stuffing) 3/7/2011 Industrial Data and Systems Analysis 2 Forecasting, Planning, and S h d li Scheduling Read Chapter 9 of 3/7/2011 Industrial Data and Systems Analysis 4
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Sales of Turkey Stuffing is Hi hl S l Highly Seasonal Line Monthly Sales and Capacity for Line 101 300 000 350,000 400,000 Cases) 150,000 200,000 250,000 300,000 d Capacity (C 0 50,000 100,000 1 2 3 4 5 6 7 8 9 10 11 12 Sales and Month Number Line Monthly Sales Maximum Cases Per Month 3/7/2011 Industrial Data and Systems Analysis 5 http://ecx.images-amazon.com/images/I/51P15JEPTDL.jpg The Bulldozer Method of P d ti Pl i Monthly Production and Capacity for Line 101 Production Planning 350,000 400,000 acity 150 000 200,000 250,000 300,000 on and Capa (Cases) 0 50,000 100,000 150,000 Productio ( 1 2 3 4 5 6 7 8 9 10 11 12 Month Number Required Production Maximum Cases Per Month 3/7/2011 Industrial Data and Systems Analysis 6 Why is the Direction of B lld I t t? Bulldozer Important? Monthly Sales and Capacity for Line 101 350,000 400,000 Cases) 150,000 200,000 250,000 300,000 d Capacity (C 0 50,000 100,000 Sales and 9 10 11 12 1 2 3 4 5 6 7 8 Month Number Line Monthly Sales Maximum Cases Per Month 3/7/2011 Industrial Data and Systems Analysis 7 Safety Stock is Required B f S l U t i t Because of Sales Uncertainty Line Safety Stock At Month End for Line 101 140,000 160,000 es) 60,000 80,000 100,000 120,000 Stock (Case 0 20,000 40,000 1 2 3 4 5 6 7 8 9 10 11 12 Safety Month Number Safety stock at end of month 3 is driven by variability of sales in month 4. 3/7/2011 Industrial Data and Systems Analysis 8 | 698 | 2,385 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2018-17 | latest | en | 0.683017 |
https://quantumsportsbetting.com/football-betting-systems-that-work/ | 1,709,651,363,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707948235171.95/warc/CC-MAIN-20240305124045-20240305154045-00612.warc.gz | 466,128,633 | 40,156 | # Football Betting Systems That Work
As the world's most popular sport, football attracts many betting enthusiasts globally. With various leagues and competitions around the world, punters are always in quest of profitable strategies to enhance their chances of winning.
Football betting systems have evolved over the years to assist bettors in improving their knowledge and confidence in placing wagers. In this article, we will explore several football betting systems that work, their advantages, and how you can adapt them to suit your betting style.
## 1. Value Betting
Value betting is a concept that revolves around identifying undervalued odds offered by bookmakers. It is based on the idea that, occasionally, bookmakers can underestimate the real probability of certain outcomes. Consequently, this creates opportunities for bettors to stake on bets with higher chances of winning compared to their odds.
### How Value Betting Works
To successfully implement the value betting system, follow these steps:
1. Understand the betting market: Acquire in-depth knowledge about football and the leagues or teams you will be betting on. This will allow you to identify any discrepancies in the odds offered by bookmakers.
2. Estimate the real probability: Based on your research, give each outcome a probability percentage. Combine factors such as team form, injuries, head-to-head records, and other relevant data points.
3. Calculate value: If your estimated probability is higher than the implied probability presented by the bookmakers, there is value in that bet. The formula to calculate value is:
Value = (Your Estimated Probability * Decimal Odds) - 1
1. Stake wisely: Ensure you place bets that offer positive expected value. Additionally, spread your risk by dividing your bankroll into small units, and stake proportionately based on the perceived value.
## 2. Matched Betting
Matched betting is a risk-free betting strategy that leverages bookmakers' free bets and promotions to create guaranteed profits. It involves placing two opposite bets for each selection; one with the bookmaker using the free bet or bonus and the other in a betting exchange as a lay bet.
### How Matched Betting Works
1. Sign up with multiple bookmakers: To maximize matched betting opportunities, create accounts with several bookmakers offering free bets and bonuses.
2. Find close matches: Seek bets with similar odds between the bookmaker and the betting exchange, as this will help to minimize qualifying losses.
3. Calculate stakes: Use a matched betting calculator to determine the appropriate stakes for both the back and lay bets.
4. Place the bets: Once you have the stakes, place your back bet with the bookmaker using the free bet and the corresponding lay bet in the betting exchange.
5. Repeat: Continue the process with other bookmakers and offers to accumulate guaranteed profits over time.
## 3. The Over/Under Goals System
The over/under goals system is a popular betting market that involves predicting whether the total number of goals in a match will be higher or lower than a specific threshold. This system is particularly attractive to bettors since it eliminates the need to predict the exact outcome or winner of a match.
### How the Over/Under Goals System Works
1. Research: Analyze the recent form, attacking and defending capabilities, and head-to-head records of teams involved. Identify high-scoring teams or those with leaky defenses.
2. Narrow down the leagues: Concentrate on leagues with a high goals-per-game average or those that exhibit consistent trends, such as the number of goals scored in the first half versus the second half.
3. Select the threshold: Choose a goal threshold (such as over 2.5 goals or under 3.5 goals) based on your research and the current odds available.
4. Bet responsibly: Manage your bankroll carefully and apply a staking plan to mitigate risks associated with betting.
## 4. The Correct Score Trading System
The correct score trading system involves predicting the exact score of a football match, a market known for its high odds. Rather than betting on a specific score, this system aims to trade in-play by backing several correct score lines and trading out during the game, either for profit or minimal losses.
### How the Correct Score Trading System Works
1. Pre-match research: Investigate recent team form, attacking and defensive strategies, and potential lineups to help you estimate the most likely scores.
2. Select the correct scores: Based on your research, choose a few probable correct scores and back them before the match starts.
3. Trade in-play: Monitor the game and watch for opportunities to trade out of your positions, either for a profit or to minimize losses. Trades can be initiated in response to match events such as goals, red cards, or injuries.
4. Maintain discipline: Stick to your trading plan and have a clear exit strategy, whether it involves a specific time frame or a specific match event.
## Conclusion
Yes, several football betting systems work and can improve your chances of generating a profit from sports betting. These systems, such as value betting, matched betting, the over/under goals system, and the correct score trading system, have proven effective over time. However, success with these strategies depends on diligent research, discipline, and strict bankroll management.
Remember that no system can guarantee a win every time, but by implementing one or more of these strategies, you can improve your betting skills, minimize risks, and potentially make more informed decisions.
### QSB Tipsters at a 90% discounted price
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With our subscription service, you will gain access to our tipsters area which contains over 20+ selections each week for long term profit on the betting exchange. You will also have access to advice from our experienced tipsters who have been betting and trading successfully for years.
If you want consistent selections each week, then our subscription is definitely for you! Get started today within our subscription section! | 1,247 | 6,360 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2024-10 | longest | en | 0.937383 |
https://www.codingbroz.com/day-six-central-limit-theorem-solution/ | 1,669,734,258,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710698.62/warc/CC-MAIN-20221129132340-20221129162340-00463.warc.gz | 757,190,569 | 49,749 | # Day 6: The Central Limit Theorem I | 10 Days Of Statistics | HackerRank Solution
Hello coders, today we are going to solve Day 6: The Central Limit Theorem I HackerRank Solution which is a Part of 10 Days of Statistics Series.
Contents
## Objective
In this challenge, we practice solving problems based on the Central Limit Theorem.
A large elevator can transport a maximum of 9800 pounds. Suppose a load of cargo containing 49 boxes must be transported via the elevator. The box weight of this type of cargo follows a distribution with a mean of u = 205 pounds and a standard deviation of o = 15 pounds. Based on this information, what is the probability that all 49 boxes can be safely loaded into the freight elevator and transported?
## Input Format
There are 4 lines of input (shown below):
``````9800
49
205
15``````
The first line contains the maximum weight the elevator can transport. The second line contains the number of boxes in the cargo. The third line contains the mean weight of a cargo box, and the fourth line contains its standard deviation.
If you do not wish to read this information from stdin, you can hard-code it into your program.
## Output Format
Print the probability that the elevator can successfully transport all 49 boxes, rounded to a scale of 4 decimal places (i.e., 1.2345 format).
## Solution – The Central Limit Theorem I
### C++
```#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
double normal_dist(double m, double sd, double x)
{
/*
p = 1/2*(1 + erf((x - m)/(sd*sqrt(2))));
*/
double p = 0.5*(1 + erf((x-m)/(sd*sqrt(2.0))));
return p;
}
int main() {
/* Enter your code here. Read input from STDIN. Print output to STDOUT */
double m = 49*205, sd = sqrt(49)*15, x = 9800;
printf("%0.4f", normal_dist(m, sd, x));
return 0;
}```
Disclaimer: The above Problem (The Central Limit Theorem I) is generated by Hacker Rank but the Solution is Provided by CodingBroz. This tutorial is only for Educational and Learning Purpose. | 515 | 2,049 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2022-49 | longest | en | 0.76032 |
https://stats.stackexchange.com/questions/244768/how-can-i-use-a-cox-proportional-hazards-model-to-obtain-probabilities | 1,713,003,024,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816587.89/warc/CC-MAIN-20240413083102-20240413113102-00348.warc.gz | 508,203,890 | 39,508 | # How can I use a Cox proportional hazards model to obtain probabilities?
I am using a Cox proportional hazards model to estimate the probability of an event (failure of a mechanical device, say) occurring at discrete, finite times: $t = 0, 1, 2, \ldots$. In Cox's original paper, Regression Models and Life Tables, he says that, if you let $T$ be the random variable representing the time to failure of a given member of your population, and $T$ is discrete, then $$\lambda(t) = Pr(T = t|T \geq t).$$ So it seems to me that, if I estimate $\lambda_0(t)$ in the canonical Cox model formula $$\lambda(t) = \lambda_0(t) \exp \left( \sum \beta_i X_i \right)$$ as a "baseline" probability that failure occurs (that is, a sort of average of the probability of the event occurring at time $t$ when the $X_i$ are all zero), I can obtain an estimate of the probability of the event occurring at a time $t$ in the future for a device with risk characteristics $X_i$ by just plugging in the values of $X_i$ and the value of $t$ into the formula. Is this the correct way to use a Cox model, or am I not thinking hard enough?
First of all, hazards modeling is concerned with the hazard function. If $T$ is a continuous random variable representing failure time of a subject, then the hazard function is defined as follows: $$\label{first} \lambda(t)=\lim_{\Delta t \downarrow 0} \frac{\mathbb P(t \leq T<t+\Delta t | T \geq t)}{\Delta t}$$
In a Cox model done the right way, if we want to get the probability density function $f_T(t)$, it can be found by taking the product $$f_T(t) = \lambda(t)S(t),$$ where $S(t)$ is the survival function, given by $$S(t) = \exp \left( -\int_0^t \lambda(s)ds \right).$$ What I don't really understand is how, precisely, these three equations need to be modified if $T$ is discrete. In Cox's paper, he says that the hazard function for discrete $T$ is (contrary to what I said in my original question) $$\lambda (t) = \sum_{j = 1}^N \mathbb P(T = t_j|T\geq t_j)\delta(t-t_j),$$ where $\{t_j\}_{j=1}^N$ is the set of points where $T$ has mass. Is there some way to obtain this from the first equation above? | 593 | 2,131 | {"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": 1, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2024-18 | latest | en | 0.896299 |
https://knowt.com/note/f0a857e0-c1fa-49e7-8a54-6562d6e0477d/FHS-Medical-Math | 1,726,300,785,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651559.58/warc/CC-MAIN-20240914061427-20240914091427-00837.warc.gz | 312,709,264 | 20,986 | # FHS Medical Math
1 meter (m) - 100 centimeters (cm)
1 meter (m) - 1,000 millimeters (mm)
10 millimeters (mm)- 1 centimeter (cm)
1 inch (" or in)- 2.5 centimeters (cm)
39.4 inches (" or in) - 1 meter (m)
1 liter (L)- 1,000 milliliters (mL) = 1 quart (qt)
1 milliliter (mL)- 1 cubic centimeter (cc)
1 teaspoon (t or tsp)- 5 milliliters (mL)
1 ounces (oz)- 30 milliliters (mL)
1 gram (g)- 1,000 milligrams (mg)
1 kilogram (kg)- 1,000 grams (g)
1 kilogram (kg)- 2.2 pounds (lbs)
1 pound (lb)- 16 ounces (oz)
Celsius to Fahrenheit- (C*1.8)+32=F
Fahrenheit to Celsius- (F-32)/1.8=C
inch- " or in
foot- ' or ft
yard- yd
teaspoon- t or tsp
tablespoon- T or tbsp
ounce- oz
cup - c
pint - pt
quart - qt
pound - lb
gram (g or gm) measures? - mass or weight
Liter (L) measures? - volume or liquid
Meter (m) measures? - length or distance
Kilo - thousands = 1,000 times
Centi - hundredths = 1/100
Milli - thousandths = 1/1,000
1 am - 0100
3:30 am- 0330
10 am- 1000
12 noon- 1200
2:30 pm - 1430
8 p, - 2000
10:15 pm - 2215
12 Midnight- 2400
l - 1
V- 5
X - 10
L - 50
C - 100
D - 500
1 mL - 0.03 oz
3 tsp - 1 Tbsp
15 mL - 1 Tbsp
8 oz - 1 cup
48 tsp - 1 cup
16 Tbsp - 1 cup
6 tsp - 1 oz
2 Tbsp - 1 oz
3 ft - 1 yd
12 in - 1 ft
36 in - 1 yd | 533 | 1,279 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2024-38 | latest | en | 0.50315 |
http://facebook.stackoverflow.com/questions/2795946/getting-ceil-of-decimal-in-python/2795960 | 1,369,279,267,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368702762497/warc/CC-MAIN-20130516111242-00008-ip-10-60-113-184.ec2.internal.warc.gz | 82,757,717 | 15,758 | # getting Ceil() of Decimal in python?
Facebook and Stack Exchange are now working together to support the Facebook developer community. Facebook engineers participate here along with the best Facebook developers in the world. If you have a technical question about Facebook, this is the best place to ask.
Is there a way to get the ceil of a high precision Decimal in python?
``````>>> import decimal;
>>> decimal.Decimal(800000000000000000001)/100000000000000000000
Decimal('8.00000000000000000001')
>>> math.ceil(decimal.Decimal(800000000000000000001)/100000000000000000000)
8.0
``````
math rounds the value and returns non precise value
-
I'm new to python, just started out yesterday in fact. Stumbled upon this problem in my second practice program (the first was of course the obligatory "print 'Hello, World!';"). so, I'm finding it difficult to judge the best answer to this. The decimal.Context solution by Matthew Flaschen worked in my particular case. But I'd like others to upvote the best solution (also would be helpful for newbies like me if you can explain why a certain approach works better) and I'll come back and accept. – Gunjan May 10 '10 at 8:50
``````x = decimal.Decimal('8.00000000000000000000001')
with decimal.localcontext() as ctx:
ctx.prec=100000000000000000
ctx.rounding=decimal.ROUND_CEILING
y = x.to_integral_exact()
``````
-
This is good, but there's no need to change the context precision here. `to_integral_exact` also takes a `rounding` argument, so you can avoid messing with the context altogether. – Mark Dickinson May 9 '10 at 10:04
The most direct way to take the ceiling of a Decimal instance `x` is to use `x.to_integral_exact(rounding=ROUND_CEILING)`. There's no need to mess with the context here. Note that this sets the `Inexact` and `Rounded` flags where appropriate; if you don't want the flags touched, use `x.to_integral_value(rounding=ROUND_CEILING)` instead. Example:
``````>>> from decimal import Decimal, ROUND_CEILING
>>> x = Decimal('-123.456')
>>> x.to_integral_exact(rounding=ROUND_CEILING)
Decimal('-123')
``````
Unlike most of the Decimal methods, the `to_integral_exact` and `to_integral_value` methods aren't affected by the precision of the current context, so you don't have to worry about changing precision:
``````>>> from decimal import getcontext
>>> getcontext().prec = 2
>>> x.to_integral_exact(rounding=ROUND_CEILING)
Decimal('-123')
``````
By the way, in Python 3.x, `math.ceil` works exactly as you want it to, except that it returns an `int` rather than a `Decimal` instance.
-
You can do this using the precision and rounding mode option of the Context constructor.
``````ctx = decimal.Context(prec=1, rounding=decimal.ROUND_CEILING)
ctx.divide(decimal.Decimal(800000000000000000001), decimal.Decimal(100000000000000000000))
``````
EDIT: You should consider changing the accepted answer.. Although the `prec` can be increased as needed, `to_integral_exact` is a simpler solution.
-
perfect :) --- completing character limit --- – Gunjan May 8 '10 at 23:01
@Gunjan, if it's perfect, why not accept it?! – Alex Martelli May 9 '10 at 5:09
@Alex sorry had to leave before 6 minute timeout. Thanks for the reminder – Gunjan May 9 '10 at 9:45
-1. This doesn't generalize well; it only happens to work in this case because the result is in the range [1, 10]. Try the same calculation with Decimal(123)/Decimal(10), for example, and you'll get a result of `Decimal('2E+1')`. – Mark Dickinson May 9 '10 at 10:14
``````>>> decimal.Context(rounding=decimal.ROUND_CEILING).quantize(
... decimal.Decimal(800000000000000000001)/100000000000000000000, 0)
Decimal('9')
``````
-
Note that this solution has problems for `Decimal` instances with large value: e.g., if you try `c.quantize(decimal.Decimal('1e100'), 1)` with your context `c`, you'll get an `InvalidOperation` exception. – Mark Dickinson May 9 '10 at 10:34
``````def decimal_ceil(x):
int_x = int(x)
if x - int_x == 0:
return int_x
return int_x + 1
``````
-
I'm sure there are library functions to do this (as Ignacio Vazquez-Abrams points out), but since you haven't accepted any answer, I got the impression that you wanted to see how it's done - your own version of ceil. So here is one possible solution:
``````def ceil(d):
return [eval("int(d) + [0,1][int(bool(d-int(d)))]"), eval("int(d)")][int(d<0)]
``````
Hope this helps
-
int() and bool() are not library functions?? A simpler `a priori` explanation for not accepting any answer is that the OP has been on SO for only 13 days and this is the first question (and so may need telling/reminding to accept answers) and it was asked only 5 hours ago and the OP may be waiting for those in other TZs to reply or may now be asleep and will check answers at breakfast-time ... `a fortiori` however the OP commented "Perfect" to Matthew's answer (the first answer) which was rather nuts'n'boltsy so I'd be inferring that he's already seen "how it's done". – John Machin May 9 '10 at 5:17 This fails for negative numbers: `ceil(-2.3)` --> `-1`. – Mark Dickinson May 9 '10 at 10:37
Just use potency to make this. import math
``````def lo_ceil(num, potency=0): # Use 0 for multiples of 1, 1 for multiples of 10, 2 for 100 ...
n = num / (10.0 ** potency)
c = math.ceil(n)
return c * (10.0 ** potency)
lo_ceil(8.0000001, 1) # return 10
``````
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http://quizlet.com/601146/finals-microeconomics-chapter-9-flash-cards/ | 1,387,397,191,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1387345759442/warc/CC-MAIN-20131218054919-00088-ip-10-33-133-15.ec2.internal.warc.gz | 135,659,713 | 13,736 | # Finals - Microeconomics - Chapter 9
## 11 terms
### downward sloping
shape of the monopoly demand curve
### elbow
shape of the perfect competition demand curve
### ar is downward sloping so mr has to be below it, pulling it down
why does mr have to be lower than price for a monopoly?
### ar = tr / qsold = p * qsold/qsold = p
relationship between ar and p for a monopoly
### 1. choose best production level (mr = mc) 2. choose the best price (highst one that still sells) 3. use production table to find labor hours 4. calculate the profit = tr - tc
profit maximization rules for monopolies
### box from TP on demand curve over to price and down
where to find the tr box for a monopoly's production graph
### box from TP on ATC curve over to y-axis and down
where to find the tc box for a monopoly's production graph
### higher price, lower production vs. lower price, higher production
price, production for monopoly vs. perfect competition
### ATC, d, and MC intersect at the same point; no economic profit; no new firms enter
long run for perfect competition
### higher atc (to the left of the atc curve) which causes a higher price in the long run compared to perfect competition
long run for monopolistic competition
### we get a higher variety in return
why do we accept higher long run prices for monopolistic competition firms than we do with firms that are in perfect competition?
Create Set | 333 | 1,427 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-48 | latest | en | 0.892908 |
https://www.edweek.org/teaching-learning/opinion-solving-the-project-based-math-conundrum-the-multidimensional-math-classroom/2016/09 | 1,722,917,305,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640476479.26/warc/CC-MAIN-20240806032955-20240806062955-00730.warc.gz | 591,088,563 | 37,069 | Solving the Project-Based Math Conundrum: The Multidimensional Math Classroom
Opinion
# Solving the Project-Based Math Conundrum: The Multidimensional Math Classroom
By Contributing Blogger — September 28, 2016 5 min read
This post is by Will Haase, Calculus and Math IV teacher at High Tech High International
Imagine walking into a math classroom. Everything is quiet, students are diligently writing and reading and thinking. You hear the rustle of paper, scratching pens, and the occasional student shifting in their seat; it seems like any other classroom except that the students are seated in groups facing each other. Then you notice that large pieces of paper cover the entire table and there are different colored markers in the center.
While you are taking it all in, the teacher announces that it is time for students to share their mathematical ideas with their group. The classroom erupts into activity and noise. Students begin discussing the math question they have been working on independently. Diagrams and equations start to dot the table papers. Students take turns sharing their thinking as group members challenge their logic and offer additional ideas. The energy and excitement in the room surprises you and for a while you lose track of whether there is even a teacher there at all. Creative and diverse ways of tackling the problem, some more likely to work than others, all get a place on the paper. Through cycles of questioning and discussion, the groups slowly develop a more coherent understanding of the mathematics involved in their problem. As the class draws to a close, students share their ideas with the larger group, drawing more questions, some praise, and further wonderings. This is a glimpse into the multidimensional math classroom, a conceptual project-based math vision I want to share with you.
Over the past decade I have had the privilege of working at one of the High Tech High schools located in San Diego, California. In that time, I have witnessed the impact a school focused on collaborative, project-based learning (PBL) can have on students. It has been inspirational to see first-hand how successful this model can be in addressing the needs of students. However, there has always been one discipline that didn’t seem to fit in with this collaborative approach--math. In my hundreds of talks with other math educators across the project-based and deeper-learning network, there is a consistent frustration with math. Despite our best efforts, we have yet to create a clear vision of math in this innovative school context. Until now.
I believe that the model of math education that Jo Boaler presents in her ground breaking book Mathematical Mindsets, as well as on her youcubed.org website, is the missing piece for us in the project-based learning community. Of course we want students experiencing math in action and we should continue promoting applied math projects that cross over with science and engineering. But we also need to realize that conceptual fluency with math as a language is also important. These new practices that seek to unpack patterns and explore multiple solutions offer a new and exciting way to create projects with math that multiply entry points for all learners. A project where students are creating and exhibiting their own authentic math ideas. I believe that exploring and experimenting with how we can weave this vision into our own math programs would be transformational.
To clarify exactly what this multidimensional math experience is, I’d like to quote Jo Boaler directly from her youcubed.org website:
“So what is multidimensional mathematics? It is real mathematics. Let’s consider the work of a mathematician: She first has to pose an important problem, then map out a mathematical approach, she will probably collaborate with others on ideas, and engage in what Lakatos describes as a zig-zagging process of conjecturing, refining counter examples and proving. She has to form a mathematical model, apply methods, draw diagrams, connect ideas, reason about connections and communicate in different forms. The work is multidimensional. When math is taught as a multidimensional subject in classrooms students engage more, enjoy math more and achieve at higher levels.”
In the multidimensional math class students wrestle with open ended problems and engage in a dialogic learning process. The low-floor and high-ceiling problems allow all students access to a starting point, but are also open enough that even the strongest student will be challenged and can delve into deeper levels of complexity. Throughout the learning experience the students propose ideas and act as skeptics for each other. Every student has a role to ensure that all ideas are heard and challenged. The work is creative, and most importantly gives every student a chance to develop their own mathematical understandings. Students craft their ideas and blend them together and then share them with the class. For me, this is a beautiful example of a truly conceptual math project. Students are authentically wrestling with difficult concepts and creating their own ideas about the math. Because the ideas are at the conceptual level they can be taken in many different directions and are not limited by a single “correct” answer. Students can go where their logic takes them and follow it as far as they want. Then when the group is finished they write up their ideas and share them publicly. I can imagine a gallery exhibition where students show off their creative mathematical questions and thinking.
The beauty of this approach is that by providing a framework for the development of conceptual math, it engages students in a more authentic way and has the potential to bring about more enjoyment of math. Now, I’m not suggesting we abandon teaching specific math skills or that we stop doing applied math projects. What I am suggesting is that a large and important part of the PBL math experience should be conceptual math learning and projects. That in both math and science classes students are mastering the language of math but in very different ways. In the science class students are focused using math as a tool, while in math students are engaged in the creation of the mathematics itself. In my time as a math educator, I have yet to encounter a more compelling vision that has the same potential to reinvigorate project based math across the country. I look forward to continuing this conversation and exploring how conceptual math projects can create deeper learning experiences for all students.
Photo by Mandalyn Kime
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The opinions expressed in Learning Deeply are strictly those of the author(s) and do not reflect the opinions or endorsement of Editorial Projects in Education, or any of its publications. | 1,294 | 6,830 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-33 | latest | en | 0.96189 |
https://nl.mathworks.com/matlabcentral/cody/problems/19-swap-the-first-and-last-columns/solutions/1820064 | 1,571,298,328,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986673250.23/warc/CC-MAIN-20191017073050-20191017100550-00021.warc.gz | 618,162,208 | 15,345 | Cody
# Problem 19. Swap the first and last columns
Solution 1820064
Submitted on 20 May 2019 by Praful Tondomker
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
A = [ 12 4 7 5 1 4]; B_correct = [ 7 4 12 4 1 5 ]; assert(isequal(swap_ends(A),B_correct))
2 Pass
A = [ 12 7 5 4]; B_correct = [ 7 12 4 5 ]; assert(isequal(swap_ends(A),B_correct))
3 Pass
A = [ 1 5 0 2 3 ]; B_correct = [ 3 5 0 2 1 ]; assert(isequal(swap_ends(A),B_correct))
4 Pass
A = 1; B_correct = 1; assert(isequal(swap_ends(A),B_correct)) | 234 | 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} | 2.84375 | 3 | CC-MAIN-2019-43 | latest | en | 0.608428 |
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1 Coordinate Systems
2 Coordinate Systems key concepts Projections and Coordinate Systems Data Quality Meta Meta Data
3 Projections and Coordinate Systems: Geographic Coordinate System Uses Uses 3D spherical surface to define locations Often Often incorrectly called a datum Includes Includes angular unit of measure, prime meridian and datum Point Point referenced by longitude/latitude Angles Angles measured by degrees
4 Parallels Lines of Latitude Latitude lines are parallel The equator defines the line of zero latitude Every degree of latitude is theoretically equal Parallels run east/west; measure distances north and south of equator
5 Meridians Lines of Longitude Meridians converge at the poles Line of zero longitude is called the Prime Meridian Distance of 1 longitude decreases toward the poles Meridians run north/south; measure distance east & west of Prime Meridian
6 Graticular Network Network of Lat/Long called a graticule Origin of graticule (0,0) where Equator and Prime Meridian intersect 4 geographic quadrants based on compass bearings from Origin
7 Degrees, Minutes, Seconds (DMS) Point on Earth s surface referenced by Lat/Long values Lat/Long based on 360 Each degree has 60 minutes Each minute has 60 seconds
8 Decimal Degrees (DD) Similar Similar to DMS Minutes Minutes and seconds expressed as decimal values ESRI ESRI products require DD in geodatasets
9 Converting from DMS to DD 37 36' 30" (DMS) Divide Divide each value by the number of minutes or seconds in a degree: 36 minutes =.60 degrees (36/60) 30 seconds = degrees (30/3600) Add up the degrees to get the answer: = DD
10 Spheres and Spheroids Sphere Spheroid Shape and size of GCS surface defined by sphere or spheroid. Mathematical calculations easier on a sphere. Sphere can be used for small-scale maps (< 1:5,000,000) Spheroid gives better accuracy for large-scale maps (>1:1,000,000
11 Major and Minor Axes of Ellipse Major Semiminor Axis Minor Axis Axis Semimajor Axis Shape of ellipse defined by two radii. Longer radius: Semimajor Axis Shorter radius: Semiminor Axis Rotating spheroid around semiminor axis creates a spheroid
12 Spheroids for Accurate Mapping Earth has been surveyed many times Surveys result in many spheroids Spheroid chosen to fit one country Best fit for one regions not same for another region Earth is neither perfect sphere nor spheroid Changing coordinate system s spheroid changes all previously measured values
13 Datums Spheroids approximate earth s shape Datum Datum defines position of spheroid relative to center of the earth Datum Datum defines origin and orientation of lat/long lines Local Local datum aligns spheroid to fit surface in a particular area
14 Datum Comparisons Local geographic Coordinate system Earth s surface Earth-centered datum Local datum Earth-centered geographic Coordinate system
15 North American Datums NAD27 uses Clarke 1866 spheroid Origin Meade Ranch Kansas Manually calculated control points NAD83 Based on earth and satellite observations Uses GRS80 spheroid Origin is earth s center of mass Previous control points shift as much as 500
16 Projected Coordinate Systems Defined Defined on flat, 2D surface Has Has constant lengths, angles and area Always Always based on geographic coordinate system X,Y X,Y coordinates on grid
17 What is a Map Projection? Transformation of 3D surface to 2D flat sheet Causes distortion in the shape, area, distance or direction of data Uses mathematical formulas to relate spherical coordinates to planar coordinates Different projections cause different distortions Map projections designed for specific purpose i.e. large-scale data in limited area
18 Relevance to GIS maps are a common source of input data for a GIS 1) often input maps will be in different projections, requiring transformation of one or all maps to make coordinates compatible 2) thus, mathematical functions of projections are needed in a GIS often GIS are used for projects of global or regional scales so consideration of the effect of the earth's curvature is necessary monitor screens are analogous to a flat sheet of paper 1) thus, need to provide transformations from the curved surface to the plane for displaying data
19 Conclusion What What is a coordinate system? A coordinate system is a grid that may be used to define where a particular location is Connection between projection and coordinate system The projection defines the coordinate system by defining the 2-D 2 D surface of the earth
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# v02 q23
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18 Apr 2010, 21:28
A small orchard specializing in citrus fruits has been receiving fewer orders for oranges in the off season. Orchard owners expect demand to pick up in the coming weeks as harvest begins, but they are concerned about a neighboring orchard nearly three times their size. The neighboring orchard has announced plans to sell cases of oranges for 20% less than regular price in order to jump-start the season by attracting new business.
Which of the following, if true, points to the most serious flaw of the small orchard's plan to attract business by matching the prices of the competing orchard?
a) The small orchard has a dozen customers under contract to purchase the first oranges of the season and may anger them by not extending the discount to them.
b) The small orchard does not have time to advertise its oranges at lower prices.
c) The small orchard will reduce the number of oranges in each case to minimize losses incurred by matching prices.
d) The large orchard has a larger inventory and is more able to take losses on initially discounted oranges.
e) The large orchard will target a different market than the small orchard.
The OA is D, but I believe E is a better answer because:
This is a weaken question. In a weaken question, the answer choices are accepted as given, even if they include new information. The question even says that you must consider the information in each answer as true...
"Which of the following, if true, points to the most serious flaw of the small orchard's plan to attract business by matching the prices of the competing orchard?"
Therefore, in E, since the large orchard is targeting a different market, it will not have an effect on the small orchard's market and thus the small orchard should not provide a discount.
in D, the two orchard's markets are the same and the large orchard can take more losses than the small orchard.
Answer E provides a more compelling reason why the small orchards plan is flawed.
here is the reasoning from the OA which i feel is flawed:
Reasoning: Which point reveals the most serious flaw in the small orchard's plan? It can be determined from the passage that the large orchard has three times the inventory of the small orchard, and that they are looking to attract new business. It can also be inferred from the passage that the owners of the large orchard are confident that they are able to sell oranges at a discount and still remain solvent.
a) The concern of the orchards is in attracting new business, not satisfying existing customers. Additionally, customers under contract have likely already agreed to a non-discount price.
b) This item may be problematic, but does not address the concerns of the passage as a whole.
c)This approach is not a flaw in the plan, but a possible, potentially unethical, solution.
d)The large orchard's greater inventory will give them a flexibility that will outlast the small orchard. The small orchard could run out of stock before the large orchard has sold even a third of its oranges.
e)This item makes it sound as if the two orchards will not actually be in competition, but does not address the soundness of the small orchard's plan.
ah, but in a weaken question you MUST take the situation/info provided in each answer as TRUE and use that information as the basis to weaken what is provided in the stimulus.
In the explanation for e, it is saying that we should question the fact that the target markets will be different for each orchard... but this goes against the question which asks:
"Which of the following, if true, points to the most serious flaw of the small orchard's plan to attract business by matching the prices of the competing orchard?"
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26 Sep 2010, 03:49
I agree OA should be E it is definitely a more compelling reason why the plan would fail
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Re: v02 q23 [#permalink]
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26 Sep 2010, 04:25
kidchaos wrote:
I agree OA should be E it is definitely a more compelling reason why the plan would fail
OA seems to be correct. The option E actually helps the small company to increase the business.
E is not related with the reason of giving discounts? What if there are 10 more small companies? Then E will help it attracting more business.
Option D will actually hurt the small company as it might end up losing money.
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# six foot rule
## six foot rule
#1
06-13-02, 08:26 AM
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Location: North Alabama, USA
Posts: 457
six foot rule
The six foot rule for receptacles? Does it apply to walls with doorways in them? Example: 13' wall with a doorway. Require a recptacle?
#2
06-13-02, 08:37 AM
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A doorway starts a new wall segment. You are required to have a receptacle within 6 feet of the opening on both sides. Keep in mind that the corner of the room does not end a segment. So saying the wall is 13 feet means nothing unless I know where the wall goes at both ends of that 13 feet. In other words, continue your measurement around the corner as if the corner was not there.
#3
06-13-02, 09:21 AM
KevinA
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A doorway starts a new wall segment. You are required to have a receptacle within 6 feet of the opening on both sides. Keep in mind that the corner of the room does not end a segment. So saying the wall is 13 feet means nothing unless I know where the wall goes at both ends of that 13 feet. In other words, continue your measurement around the corner as if the corner was not there.
I was under the impression that at a corner, you need a receptacle 6ft from the corner along both of the adjoining walls. Isn't the idea that you should be able to plug in a lamp with a 6 foot cord into a receptacle on either wall and have the ability to place it in the corner?
Maybe I'm mistaken, I'm still learning here, but I thought that was how it worked. I need some clarification on this for my current project as well.
#4
06-13-02, 11:08 AM
FREDDYG_001
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kevin A, Article210.52. Dwelling Unit Receptacle Outlets.
(1) Spacing. Receptacles shall be installed so that no point along the floor line in any wall space is more than 6th (1.83m), measured horizontality, from an outlet in that space. Receptacle outlets shall, insofar as practicable, be spaced equal distances apart.
(2) Wall Space. As used in this section, a wall space shall include the following:
(a) Any space 2ft (610mm) or more in width (including space measured around corners) and unbroken along the floor line by doorways, fireplaces, and similar openings
(b) the space occupied by fixed panels in exterior walls excluding sliding panels
(c) The space afforded by fixed room dividers such as freestanding bar-type counters or railings
(3) Floor Receptacles. Receptacle outlets in floors shall not be counted as part of the required number of receptacle outlets unless located within 18in. (457mm) of the wall.
Also, any receptacle 5-1/2ft or higher is not counted as part of the required number of receptacle outlets.
Fred
#5
06-13-02, 12:26 PM
wyres
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Code shmode, think about Murphys law.
If you are DIYing, it is very quick and inexpensive to add outlets.
I doubt you will ever regret adding an extra outlet that seldom or never gets used.
I guarantee you you will regret not adding an extra outlet when you are looking for a place to plug something in.
Houses last a long time, and the number of electrical toys can reasonably be expected to continue it's long historical trend of exponential increase.
Personally, I would try to see how many I could cram in there.
#6
06-13-02, 12:29 PM
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KevinA,
You do not need a receptacle within 6 feet of the corner on both walls. E.g., if you have a receptacle 3 feet from the corner along one wall, the first receptacle along the other wall could be as much as 9 feet from the corner (still no more than 12 feet apart). Note that you can still put that lamp in the corner.
#7
06-13-02, 01:20 PM
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So saying the wall is 13 feet means nothing unless I know where the wall goes at both ends of that 13 feet
Recptacle is about 6' from west corner. 32" door about 4' east of recptacle. Wall west of door must be about 6 to 12". Now we are in the corner going north. Door about 4 to 4 1/2', then on to corner. The latter wall is 12 to 12 1/2 in length. Same situation on both sides of room. One room, three doors.
#8
06-13-02, 01:23 PM
Jxofaltrds
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As stated before more than 2' requires an outlet. Most hallways do not count unless they are more than 10'.
Easy way. Just touch the wall. Then spread your arms. If there is not an outlet within that reach add one.
#9
06-13-02, 01:25 PM
Wgoodrich
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John is telling it as the rules require. I suggest that the easiest method of meeting the code on open studed walls before drywall is to take a tape measure. Start at the first door opening. Measure from that door 6'. The first receptacle must be within that 6' lenth. This receptacle is allowed as close as 1' from that door but must be placed no more than 6' from that door. Then measuer from that first receptacle further along the wall with your tape measure until you reach 12'. This measurement would keep on going around corners of wall where two walls meet until you reach the 12' measurement. Now that second receptacle must be within that 12' measurement from that first receptacle ensuring that there is less than 12' between those two receptacles. The rules says no point along a wall space is allowed to be more than 6' from a receptacle. If you placed the first receptacle to the second receptacle within 12' then no point on that wall will be more than 6' from one of those two receptacles. The first receptacle will serve 6' to the left of that first receptacle and also serve 6' to the right of that receptacle. Now if you look at what I just said then the first receptacle will serve 6' to the right of that first receptacle on that 12' wall space located between the first and second receptalce. Then the second receptacle will serve to the left of that second receptacle towards where the first receptacle serves 6' back toward that first receptacle. This means if you have two receptacles 12' apart the two receptacles at the end of that 12' wall space will serve toward each other to the exact middle of the 12 wall space making no point in that 12' wall space that is more than 6' from a receptacle. Then look at your table lamps, televisoins, radios etc. and you will find 6' cords on those appliances. This should explain what the goal is. That cord does not care if it reaches around that corner of a wall to the adjoining wall where the receptacle is as long as it is within 6' of that table lamp etc.
Continue placing your receptacles no more than 12' apart until you hit the next door or break in that wall making a new start of a wall on the other side of that break in the wall. A break in a wall may be a built in library shelving unit or a cased doorway, or a closet door. When you hit an end of that wall space passing around corners as one wall space you must measure backward from that end of a wall at that next doorway to ensure that a receptacle is within 6' of that end of the wall space.
Also remember that any wall that is 2 feet or longer must have at least one receptacle serving that short wall space.
There is an exception concerning non living areas such as bathrooms, and utility room or garages where the 6' receptacle rule does not apply. In these non living areas you place receptacles only where you need or want them placed.
The suggestion of placing extra receptacles on the wall that is more than is required to meet minimum is commonly done and a good idea especially where you think you will have a grouping of electronic equipment such as a TV, sterio, VCR, etc. grouped where you don't really have much load but need more plugs for light load equipment such as electronic equipment. If you don't add receptacles in those areas you will run out of pluggins for your electronic toys.
Hope this helps
Wg
#10
06-13-02, 01:26 PM
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Location: North Alabama, USA
Posts: 457
I guarantee you you will regret not adding an extra outlet when you are looking for a place to plug something in.
I added one in the lr of my house and I have been very glad I did. I have to agree about having plenty of outlets, but this is a block house that formerly had only one or two per room, and the block walls are kinda hard to deal with. I am trying to get all my wire inside the block. Whew! Pretty tough going. Guess I should rent a hammer chisel.
#11
06-13-02, 05:27 PM
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Join Date: May 2001
Location: North Alabama, USA
Posts: 457
Then measuer from that first receptacle further along the wall with your tape measure until you reach 12'
I have always been told the purpose of the 6' rule was so that a six foot cord could reach from anywhere on the wall. I also thought, as you pointed out, that one receptacle could serve almost 12 feet. But, somehow in my dense aging mind, I thought the rule meant one outlet every six feet. I did not know about doorways as I thought no one should run a cord across a doorway.
Eventually, I will catch on to what you guys are trying to tell me.
Sometimes, I think the same things are being said in different ways. I just have to get my slow brain to interpret what I am told. Sometimes it sounds like 'Bobwehadababyitsaboy'
Bear with me!
Tx for the responses
#12
06-13-02, 06:35 PM
Wgoodrich
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Posts: n/a
The rule that a receptacle must be within 6' of the end of a wall is an attempt to ensure that no cords pass across a doorway. If you measure along a wall placing the first receptacle within 6' of the end of that wall and place a receptacle every 12' along that wall after the first receptacle then when you reach you next door or break in the wall you have to measure back the way you came to ensure that last receptacle was placed within 6' of the other end of that wall your are working then there should be no need for extension cords because you lamps, TVs etc. have 6' cords and anywhere along that wall if you placed a lamp with a 6' cord as found on appliances placed in living areas such as bedrooms, living room, dining room etc. you will be able to reach a receptacle within 6' of where you placed that lamp even if you placed that lamp right next to a doorway at either end of that wall you just placed your receptacles. This is the goal.
In kitchens over kitchen counters you will find a 2' / 4' rule. Same principle applies because kitchen appliances are limited to 2' cords same principle applies there with the shorter distances requried between receptacles over kitchen counter areas considering those shorter 2' appliance cords.
Hope this helped
Wg
#13
06-13-02, 07:09 PM
Gary Tait
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Although code has the rule at 6/12, I find 4/8 more practical, as to placement of appliances, lamps, furniture, sind applinaces have
max 6ft cords, and they are placed at various heights on
furniture.
#14
06-13-02, 08:19 PM
KevinA
Visiting Guest
Posts: n/a
I have always been told the purpose of the 6' rule was so that a six foot cord could reach from anywhere on the wall. I also thought, as you pointed out, that one receptacle could serve almost 12 feet. But, somehow in my dense aging mind, I thought the rule meant one outlet every six feet.
Don't worry, I made the same mistake. I think I get it now. Thanks for the info guys.
#15
06-14-02, 01:03 AM
Member
Join Date: May 2001
Location: North Alabama, USA
Posts: 457
Although code has the rule at 6/12, I find 4/8 more practical, as to placement of appliances, lamps, furniture, sind applinaces have
Interesting point.
Tx guys | 2,957 | 11,662 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2023-50 | latest | en | 0.95025 |
https://br.tradingview.com/script/ZKF0qji2-Signs-of-the-Times-LucF/ | 1,675,851,614,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500758.20/warc/CC-MAIN-20230208092053-20230208122053-00417.warc.gz | 159,380,075 | 132,927 | # Signs of the Times [LucF]
█ OVERVIEW
This oscillator calculates the directional strength of bars using a primitive weighing mechanism based on a small number of what I consider to be fundamental properties of a bar. It does not consider the amplitude of price movements, so can be used as a complement to momentum-based oscillators. It thus belongs to the same family of indicators as my Bar Balance, Volume Ticks, Efficient work, Volume Buoyancy or my Delta Volume indicators.
█ CONCEPTS
The calculations underlying Signs of the Times (SOTT) use a simple, oft-explored concept: measure bar attributes, assign a weight to them, and aggregate results to provide an evaluation of a bar's directional strength. Bull and bear weights are added independently, then subtracted and divided by the maximum possible weight, so the final calculation looks like this:
`(up - dn) / weightRange`
SOTT has a zero centerline and oscillates between +1 and -1. Ten elementary properties are evaluated. Most carry a weight of one, a few are doubly weighted. All properties are evaluated using only the current bar's values or by comparing its values to those of the preceding bar. The bull conditions follow; their inverse applies to bear conditions:
Weight of 1
• Bar's close is greater than the bar's open (bar is considered to be of "up" polarity)
• Rising open
• Rising high
• Rising low
• Rising close
• Bar is up and its body size is greater than that of the previous bar
• Bar is up and its body size is greater than the combined size of wicks
Weight of 2
• Gap to the upside
• Efficient Work when it is positive
• Bar is up and volume is greater than that of the previous bar (this only kicks in if volume is actually available on the chart's data feed)
Except for the Efficient Work weight, which is a +1 to -1 float value multiplied by 2, all weights are discrete; either zero or the full weight of 1 or 2 is generated. This will cause any gap, for example, to generate a weight of +2 or -2, regardless of the gap's size. That is the reason why the oscillator is oblivious to the amplitude of price movements.
You can see the code used to calculate SOTT in my ta library's `sott()` function.
█ HOW TO USE THE INDICATOR
No videos explain this indicator and none are planned; reading this description or the script's code is the only way to understand what Signs of the Times does.
Load the indicator on an active chart (see here if you don't know how).
The default configuration displays:
• An Arnaud-Legoux moving average of length 20 of the instant SOTT value. This is the signal line.
• A fill between the MA and the centerline.
• Levels at arbitrary values of +0.3 and -0.3.
• A channel between the signal line and its MA (a simple MA of length 20), which can be one of four colors:
• Bull (green): The signal line is above its MA.
• Strong bull (lime): The bull condition is fulfilled and the signal line is above the centerline.
• Bear (red): The signal line is below its MA.
• Strong bear (pink): The bear condition is fulfilled and the signal line is below the centerline.
The script's "Inputs" tab allows you to:
• Choose a higher timeframe to calculate the indicator's values. This can be useful to get a wider perspective of the indicator's values.
If you elect to use a higher timeframe, make sure that your chart's timeframe is always lower than the higher timeframe you specified,
as calculating on a timeframe lower than the chart's does not make much sense because the indicator is then displaying only the value of the last intrabar in the chart bar.
• Specify the type of MA used to produce the signal line. Use a length of 1 or the Data Window to see the instant value of SOTT. It is quite noisy, thus the need to average it.
• Specify the type of MA applied to the signal line. The idea here is to provide context to the signal.
• Control the display and colors of the lines and fills.
The first pane of this publication's chart shows the default setup. The second one shows only a monochrome signal line.
Using the "Style" tab of the indicator's settings, you can change the type and width of the lines, and the level values.
█ INTERPRETATION
Remember that Signs of the Times evaluates directional bar strength — not price movement. Its highs and lows do not reflect price, but the strength of chart bars. The fact that SOTT knows nothing of how far price moves or of trends is easy to forget. As such, I think SOTT is best used as a confirmation tool. Chart movements may appear to be easy to read when looking at historical bars, but when you have to make go-no-go decisions on the last bar, the landscape often becomes murkier. By providing a quantitative evaluation of the strength of the last few bars, which is not always easily discernible by simply looking at them, SOTT aims to help you decide if the short-term past favors the bets you are considering. Can SOTT predict the future? Of course not.
While SOTT uses completely different calculations than classical momentum oscillators, its profile shares many of their characteristics. This could lead one to infer that directional bar strength correlates with price movement, which could in turn lead one to conclude that indicators such as this one are useless, or that they can be useful tools to confirm momentum oscillators or other models of price movement. The call is, of course, up to you. You can try, for example, to compare a Wilder MA of SOTT to an RSI of the same length.
One key difference with momentum oscillators is that SOTT is much less sensitive to large price movements. The default Arnaud-Legoux MA used for the signal line makes it quite active; you can use a more quiet SMA or EMA if you prefer to tone it down.
In systems where it can be useful to only enter or exit on short-term strength, an average of SOTT values over the last 3 to 5 bars can be used as a more quiet filter than a momentum oscillator would.
█ NOTES
My publications often go through a long gestation period where I use them on my charts or in systems before deciding if they are worth a publication. With an incubation period of more than three years, Signs of the Times holds the record. The properties SOTT currently evaluates result from the systematic elimination of contaminants over that lengthy period of time. It was long because of my usual, slow gear, but also because I had to try countless combinations of conditions before realizing that, contrary to my intuition, best results were achieved by:
• Keeping the number of evaluated properties to the absolute minimum.
• Limiting the evaluation's scope to the current and preceding bar.
• Choosing properties that, in my view, were unmistakably indicative of bullish/bearish conditions.
Repainting
As most oscillators, the indicator provides live realtime values that will recalculate with chart updates. It will thus repaint in real time, but not on historical values. To learn more about repainting, see the Pine Script™ User Manual's page on the subject.
Notas de Lançamento:
v2
• Added the possibility to color chart bars using one of four modes.
"The stock market is a device for transferring money from the impatient to the patient."
— Buffet | 1,577 | 7,211 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2023-06 | latest | en | 0.918532 |
https://oeis.org/A166804 | 1,716,331,155,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058522.2/warc/CC-MAIN-20240521214515-20240522004515-00727.warc.gz | 377,395,939 | 3,722 | The OEIS mourns the passing of Jim Simons and is grateful to the Simons Foundation for its support of research in many branches of science, including the OEIS.
The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A166804 Number of nX3 1..6 arrays containing at least one of each value, all equal values connected, rows considered as a single number in nondecreasing order, and columns considered as a single number in nondecreasing order. 0
0, 90, 8868, 141180, 1151013, 6470252, 28530284 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 LINKS Table of n, a(n) for n=1..7. EXAMPLE Some solutions for n=3 ...2.3.5...1.1.2...1.2.3...1.3.4...2.4.5...1.2.5...1.3.5...2.3.3...1.1.6 ...4.1.5...3.6.5...1.2.6...1.6.4...3.4.4...3.4.6...1.6.6...3.3.5...2.3.5 ...6.1.5...6.6.4...5.4.4...5.2.2...3.6.1...4.4.6...4.6.2...4.1.6...4.4.4 ------ ...1.2.4...1.3.3...1.1.5...1.3.4...2.2.3...2.3.5...1.2.5...1.1.3...2.6.6 ...2.2.4...2.2.2...3.3.6...2.2.2...2.5.4...2.6.1...3.3.4...4.6.2...4.1.3 ...5.6.3...6.4.5...3.4.2...5.5.6...6.5.1...4.6.1...3.6.6...6.6.5...5.3.3 CROSSREFS Sequence in context: A112004 A216706 A166822 * A116273 A036257 A276352 Adjacent sequences: A166801 A166802 A166803 * A166805 A166806 A166807 KEYWORD nonn AUTHOR R. H. Hardin Oct 21 2009 STATUS approved
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recents
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Last modified May 21 17:21 EDT 2024. Contains 372738 sequences. (Running on oeis4.) | 676 | 1,708 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2024-22 | latest | en | 0.655599 |
https://steemit.com/gaming/@axie-infinity-fr/axie-infinity-breeding-session-6-pure-aquatic-wetdog | 1,569,319,754,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572896.15/warc/CC-MAIN-20190924083200-20190924105200-00407.warc.gz | 665,884,147 | 140,836 | # Axie Infinity Breeding Session #6: "Pure aquatic wetdog"steemCreated with Sketch.
in gaming • 5 months ago (edited)
## Hello everyone and welcome to the Axie Infinity Breeding Session #6: "Pure aquatic wetdog” !
On the session #2 I tried to bred a wetdog axie. Unfortunately, it failed… But today, we will have a second attempt that should be very interesting!
-------------------
### Let’s start the explanations!
As a reminder, to breed a wetdog axie, you need a fuzzy and a curly axie. This time, I have two aquatic axie 4/6 pure (number of aqua parts) -> Let’s have a look to their genes.
As you can see above:
-One of them is curly and the other fuzzy.
-There is a fair chance to get a pure aquatic.
-One of them have the 2 anemone parts (bonus).
## Probabilities :
So what are the probability to get a pure aquatic wetdog axie (with double anemone).
-The probability to get a wetdog is 33%
-The probability to get a pure aquatic is 4.326% (freak tool)
-The probability to get double anemone is: 0.75(back)*0.844(horn) = 63.3%
If we mix everything together we have: 0.042 (pure double anemone aqua) * 0.33(wetdog probability) = 1.4%
Can I beat the odd? Rendez-vous in 3 days to discover the result!
-------------------
### Now, place to the breeding !!!
Thanks for following the serie!
To celebrate the birth of my first secret class axie, I'll giveaway 3 axies to the first ones who comment and follow me!
## How to support me and Axie Infinity:
Buy one of my axie: My shop
I'm also willing to sell them for steem to power up!
There is also a landsale going on here.
and on our Discord: https://discordapp.com/invite/68DeTqc
Also, I decided to write this serie in English so it's available for more people but if you're french and need advice/ tutorials to get started with Axie Infinity don't hesitate to contact me!
Sort Order:
· 5 months ago
Congratulations @axie-infinity-fr! You have completed the following achievement on the Steem blockchain and have been rewarded with new badge(s) :
If you no longer want to receive notifications, reply to this comment with the word `STOP` | 541 | 2,115 | {"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} | 3.421875 | 3 | CC-MAIN-2019-39 | latest | en | 0.903233 |
https://cs.stackexchange.com/questions/98915/number-of-steps-in-worst-case | 1,618,746,344,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038476606.60/warc/CC-MAIN-20210418103545-20210418133545-00416.warc.gz | 285,529,057 | 37,362 | # Number of steps in worst Case
we have to run a song on a Walkman,for that we need 2 full batteries.Let s say we have a mixed set of 30 batteries (15 are emtpy and and 15 are full) and then only way to test if the battery full or empty is to put them in the Walkman and try it out.we can not tell which one is empty or full (one we put inside one full and one empty).
The question is to find the maxmium needed step to run the walkman (to find 2 full batteries)
I modelled the problem like that : let s that the Walkman is a function f which take two argumments and it return 1 it the walkman works ,and 0 otherwise. $$f: \{b_1,b_2 \dots b_{30}\} \times \{b_1,b_2 \dots b_{30}\} \rightarrow\{0,1\}$$
The answer for the question was at most 18 steps , but somehow i could not figure it out. Any good suggestion how to proceed ?
You arrange your 30 batteries in pairs, where each pair consists of either:
• 2 empty batteries
• 2 full batteries
• 1 full and 1 empty battery
If you have a pair with 2 empty batteries it follows, that there is also a pair of 2 full batteries, thus, you have to test at most all of the pairs (15 steps) until you find a pair of two full batteries.
If your walkman isn't working after testing all 15 pairs, you know that each pair consists of a full and an empty battery. You take the first two pairs with batteries (a,b) and (c,d). You then try the pairings: (a,c), (a,d) and (b,c). You have now tested 18 combinations and either one of them worked or (b,d) has to be a working pair.
I don't know if this counts as 19 steps for your purposes (you have found 2 full batteries after 18 steps but the walkman is not running yet).
• Thnaks alot , what i forgot to consider is this fact "If you have a pair with 2 empty batteries it follows, that there is also a pair of 2 full batteries," – Mohbenay Oct 22 '18 at 10:49
• i can manipulate the logic and make it look 18 steps and thank you again – Mohbenay Oct 22 '18 at 11:10 | 538 | 1,960 | {"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": 1, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2021-17 | latest | en | 0.944629 |
http://hitchhikersgui.de/Descent_along_torsors | 1,521,789,799,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257648198.55/warc/CC-MAIN-20180323063710-20180323083710-00245.warc.gz | 128,824,910 | 4,313 | # Descent along torsors
In mathematics, given a G-torsor XY and a stack F, the descent along torsors says there is a canonical equivalence between F(Y), the category of Y-points and F(X)G, the category of G-equivariant X-points.[1] It is a basic example of descent, since it says the "equivariant data" (which is an additional data) allows one to "descend" from X to Y.
When G is the Galois group of a finite Galois extension L/K, for the G-torsor ${\displaystyle \operatorname {Spec} L\to \operatorname {Spec} K}$, this generalizes classical Galois descent (cf. field of definition).
For example, one can take F to be the stack of quasi-coherent sheaves (in an appropriate topology). Then F(X)G consists of equivariant sheaves on X; thus, the descent in this case says that to give an equivariant sheaf on X is to give a sheaf on the quotient X/G.
## Notes
1. ^ Vistoli, Theorem 4.46
## References
• Angelo Vistoli, Notes on Grothendieck topologies, fibered categories and descent theory (Updated September 2, 2008)
Retrieved from "https://en.wikipedia.org/w/index.php?title=Descent_along_torsors&oldid=765704598"
This content was retrieved from Wikipedia : http://en.wikipedia.org/wiki/Descent_along_torsors
This page is based on the copyrighted Wikipedia article "Descent along torsors"; it is used under the Creative Commons Attribution-ShareAlike 3.0 Unported License (CC-BY-SA). You may redistribute it, verbatim or modified, providing that you comply with the terms of the CC-BY-SA | 389 | 1,495 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 1, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2018-13 | latest | en | 0.854499 |
http://kasmana.people.cofc.edu/MATHFICT/mfview.php?callnumber=mf987 | 1,472,685,960,000,000,000 | text/html | crawl-data/CC-MAIN-2016-36/segments/1471982954852.80/warc/CC-MAIN-20160823200914-00225-ip-10-153-172-175.ec2.internal.warc.gz | 144,901,056 | 4,541 | a list compiled by Alex Kasman (College of Charleston)
...
Funes el Memorioso [Funes, His Memory] (1942) Jorge Luis Borges (click on names to see more mathematical fiction by the same author)
...
Contributed by Vijay Fafat Borges’ short story piece, “Funes, His Memory’ (or in other translations, “Funes, The Memorious”) discusses the phenomenal memory of an acquaintance, Ireneo Funes. Funes, at age nineteen, falls down from a horse and upon recovery, finds an awakening of memory capacity of almost unimaginable magnitude. Funes tries to systematize more and more of his memory to rectify perceived flaws in the order of things. At one point, he comes up with a new way of labeling numbers which reminded me of Hilbert’s plan of re-writing axioms of Geometry, captured by his famous line: "It must be possible to replace in all geometric statements the words point, line, plane by table, chair, beer mug.". As Borges tells about Funes:
(quoted from Funes el Memorioso [Funes, His Memory]) He told me that toward 1886 he had devised a new system of numbering and that in a very few days he had gone before twenty-four thousand. He had not written it down, for what he once meditated would not be erased. The first stimulus to his work, I believe, had been his discontent with the fact that "thirty-three Uruguayans" required two symbols and three words, rather than a single word and a single symbol. Later he applied his extravagant principle to the other numbers. In place of seven thousand thirteen, he would say (for example) Máximo Perez; in place of seven thousand fourteen, The Train; other numbers were Luis Melián Lafinur, Olimar, Brimstone, Clubs, The Whale, Gas, The Cauldron, Napoleon, Agustín de Vedia. In lieu of five hundred, he would say nine. Each word had a particular sign, a species of mark; the last were very complicated. . . . I attempted to explain that this rhapsody of unconnected terms was precisely the contrary of a system of numbering. I said that to say three hundred and sixty-five was to say three hundreds, six tens, five units: an analysis which does not exist in such numbers as The Negro Timoteo or The Flesh Blanket. Funes did not understand me, or did not wish to understand me.
Thank you, Vijay, for bringing this to our attention. It does not remind me of Hilbert's geometrical goals (the point of which was that we should not be using our intuition about what words like "line" and "point" mean in geometry, but rather stick to the formal axiomatization). Instead, it reminds me of a passage in Diary of a Bad Year in which the protagonist considers the structure of the natural numbers.
First published in La Nación in June 1942, this story can also now be found (in English translation) for free at websites such as this one.
More information about this work can be found at en.wikipedia.org. (Note: This is just one work of mathematical fiction from the list. To see the entire list or to see more works of mathematical fiction, return to the Homepage.)
Works Similar to Funes el Memorioso [Funes, His Memory]
According to my `secret formula', the following works of mathematical fiction are similar to this one:
1. Round the Moon by Jules Verne
2. Diary of a Bad Year by John Maxwell Coetzee
3. Brave New World by Aldous Huxley
4. Micromegas by François Marie Arouet de Voltaire
5. The Balloon Hoax by Edgar Allan Poe
6. Topsy-turvy (Sans Dessus Dessous) by Jules Verne
7. Flatland: A Romance of Many Dimensions by Edwin Abbott Abbott
8. Blue Tigers by Jorge Luis Borges
9. The Library of Babel by Jorge Luis Borges
10. The Book of Sand by Jorge Luis Borges
Ratings for Funes el Memorioso [Funes, His Memory]: | 872 | 3,670 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-36 | longest | en | 0.964575 |
https://mathematica.stackexchange.com/questions/80694/numerically-solving | 1,579,958,490,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251672537.90/warc/CC-MAIN-20200125131641-20200125160641-00433.warc.gz | 542,774,706 | 30,714 | # numerically solving
I am trying to solve for [Lambda] as a function of a (unknown) from this expression
202.[2.51521 + 1/(-1 + E^(202. λ))] +
802.[2.52457 + 1/(-1 + E^(802. λ))] +
1802.[2.52632 + 1/(-1 + E^(1802. λ))] +
3202.[2.52694 + 1/(-1 + E^(3202. λ))] +
5002.[2.52722 + 1/(-1 + E^(5002. λ))]=a
I tried with Reduce and FindRoot, and as neither of them helped I think closed form expression can not be obtained. Any suggestion how to get numerical solution? Thanks in advance.
• You can use FindRoot to solve once given specific values for a. – Daniel Lichtblau Apr 23 '15 at 17:20
• @DanielLichtblau Only after those square brackets are replaced by parenthesis – Dr. belisarius Apr 23 '15 at 17:21
• Anyway, the function is too steep and you'll probably need a few tricks – Dr. belisarius Apr 23 '15 at 17:22
• @DanielLichtblau Thanks. But I want to express λ as a function of a. Sorry about the square brackets. – reach2brb Apr 23 '15 at 17:41
• [I have nothing against square brackets.] – Daniel Lichtblau Apr 23 '15 at 19:23
$-0.000231828 - 1.11206 \times 10^{-8} a - 5.50983 \times 10^{-13} a^2$ | 379 | 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} | 3.3125 | 3 | CC-MAIN-2020-05 | latest | en | 0.840318 |
https://forum.gamsworld.org/viewtopic.php?t=11682 | 1,638,264,168,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358966.62/warc/CC-MAIN-20211130080511-20211130110511-00408.warc.gz | 325,875,920 | 8,280 | ## Multiple run of Stochastic programming emp extension Topic is solved
questions about GAMS' tools
karacat
User
Posts: 6
Joined: 1 year ago
### Multiple run of Stochastic programming emp extension
I try to solve a two stage stochastic programming and ı adjust demand randvar as poisson distribution with 100 scenario to obtain 100 random demand values according to poisson distribution. but I want to do 100 run to apply sample avarage approximation method (SAA) to this problem. I wrote "while loop" for solve statement to solve the problem multiple times to have converged solution. Gams solves problem multiple times but in every solution demand scenarios don't change and solutions stay same in every run. How can ı overcome this problem. Thank you for your help.
abhosekar
Moderator
Posts: 232
Joined: 1 year ago
### Re: Multiple run of Stochastic programming emp extension
Inside your while loop, you have to tell GAMS which scenario you want to solve. I assume demand (let's say d) is a parameter you are using in your formuation. You also have a parameter containing demands for all scenarios (let's say demands(sc)) where sc is the set of scenarios. You can loop over the set of scenarios, and inside the loop you have to set d = demands(sc).
karacat
User
Posts: 6
Joined: 1 year ago
### Re: Multiple run of Stochastic programming emp extension
Thank you for your help. Could you please help me to understand where ı should add your suggestion in my code "d" is random var demand
file emp / '%emp.info%' /;
put emp '* problem %gams.i%'/;
\$onput
randvar d poisson 10
stage 2 In Ls d
stage 2 cons2
\$offput
putclose emp;
Set scen Scenarios / s1*s100 /;
Parameter
s_d(scen) Demand realization by scenario
s_In(scen) Units overproduction by scenario
s_Ls(scen) "units underproduction by scenario"
s_rep(scen,*) "scenario probability" / #scen.prob 0/;
Set dict / scen .scenario.''
d .randvar .s_d
In .level .s_In
Ls .level .s_Ls
'' .opt .s_rep
/;
option emp = lindo;
\$echo STOC_NSAMPLE_STAGE = 100 > lindo.opt
SSP1.optfile = 1;
while (rpl le 10000,
solve SSP1 min z use emp scenario dict;
rpl=rpl+1;
Display s_d, s_In, s_Ls, s_rep,x.l,y.l,r.l,z.l;
);
karacat
User
Posts: 6
Joined: 1 year ago
### Re: Multiple run of Stochastic programming emp extension
this is my code at the attached file to understand my mistake clearly
SSP_tek ürünlü _emp_v2.gms | 638 | 2,373 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-49 | latest | en | 0.795307 |
https://questions.examside.com/past-years/year-wise/gate/gate-pi/gate-pi-2005/wpWLGTcAHgRMxwjs | 1,722,735,544,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640388159.9/warc/CC-MAIN-20240804010149-20240804040149-00463.warc.gz | 383,183,969 | 30,094 | 1
GATE PI 2005
MCQ (Single Correct Answer)
+2
-0.6
Assertion (A) : Cold working of metals result in increase of strength and hardness.
Reason (R) : Cold working reduces the total number of dislocations per unit volume of the material.
A
Both $$A$$ and $$R$$ are individually true and $$R$$ is the correct explanation of $$A$$
B
Both $$A$$ and $$R$$ are individually true but $$R$$ is not the correct explanation of $$A$$
C
$$A$$ is true but $$R$$ is false
D
$$A$$ is false but $$R$$ is true
2
GATE PI 2005
MCQ (Single Correct Answer)
+2
-0.6
Consider the following statements:
In comparison to hot working, in cold working,
$$1.$$ Higher forces are required
$$2.$$ No heating is required
$$3.$$ Less ductility is required
$$4.$$ Better surface finish is obtained.
Which of the statements given above are correct?
A
$$1,$$ $$2$$ and $$3$$
B
$$1,$$ $$2$$ and $$4$$
C
$$1$$ and $$3$$
D
$$2,$$ $$3$$ and $$4$$
3
GATE PI 2005
MCQ (Single Correct Answer)
+2
-0.6
A $$\phi \,\,25$$ $$mm$$ hole is pierced in a $$t=2.5mm$$ thick steel sheet having shear strength $$\tau = 350\,\,MPa.$$ If the diametral clearance is given by the expression $$C=0.0064$$ $$t$$ $$\sqrt \tau ,$$ the die diameter (in $$mm$$), punch diameter (in $$mm$$) and punch force in $$(kN)$$ respectively are
A
$$25.0, 25.3, 171.8$$
B
$$25.0, 24.7, 68.75$$
C
$$24.7, 25.0, 171.8$$
D
$$25.3, 25.0, 68.75$$
4
GATE PI 2005
MCQ (Single Correct Answer)
+1
-0.3
Ring gage is used to measure
A
outside diameter but not roundness
B
roundness but not outside diameter
C
both outside diameter and roundness
D
only external threads
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http://excel.bigresource.com/Formula-if-Between-add-p0ysef5s.html | 1,529,555,989,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864022.18/warc/CC-MAIN-20180621040124-20180621060124-00572.warc.gz | 106,064,370 | 13,791 | Jan 6, 2009
I am having a few issues stringing along a formula.
A1 5
I need A2 to say "If A1 is between 0 and 2 add 0, if greater than 2 but less than 4 add .30, if greater than 4 but less than 6 add .70 and so on.
A2 looking at A1
if 0 through 2 add 0
if 2 through 4 add .30
if 4 through 6 add .70
if 6 through 8 add 1
if 8 through 10 add 1.3
if 10 through 12 add 1.6
Basically these are employee break times that I need to automatically figure into our labor hours. Every 2 hours they get "x" amount of break time. So if they have over 2 hours of work they will get 1 break, if they work over 4 hours then I factor in a lunch and the previous break ect..
## Formula For Deleting A Formula When The Formula Has Done Its Calculation
Oct 7, 2009
I was wondering if there is a formla that will delete a formula when it has done its calculation, or stop the formula from constantly updating.
I've got a formula
=IF(COUNTA(A1)=1,TODAY(),"")
When something is typed in Cell A1 the cell with the formula will input todays date.
Is there a formula that will stop this formula from updating, as when you go into the file on a different day the date would have been updated.
## VBA To Read Text Formula And Related Constants And Insert As Formula Into Cell
Jul 30, 2014
I have a situation where I have to curve fit data, this can lead to different formulas being used with varying constants.
Is it possible to pickup a TEXT based formula and related constants from other cells, and then place this into another cell as a functioning formula. For Example
Cell A1 contains the formula as a text string whether it be y=a+bx+cx^2, or y=a+b/x, etc
Cells A2:A6 contain the individual constants, a, b, c, etc
I would then want the VBA to read the text based formula and put it into an output cell as a functioning excel formulae.g
In cell B10: =a+b*A10+c*A10^2
I understand picking the constants up and putting the formula should not be too much of an issue, however trying to insert the variable form of the curve fit is the part that I am struggling with, and am unsure if possible.
## Replace Cell References In Formula With The Header Of The Cells Referenced In Said Formula
Dec 22, 2009
Here is the scenario:
A B C D
BananasApplesOrangesTotal113=A1+B1+C1
I need to reference the formula in D1 with the cells headers names.
In a perfect world, it would take
=A1+B1+C1:
and produce:
=Bananas+Apples+Oranges
## Add Additional Formula To Existing Formula To Exclude Weekends And Holidays?
Mar 27, 2014
I have an existing formula in a cell linking back to a start date only. determining how to add an additional formula to exclude weekends, and possibly holidays.
## Concatenate Formula Nested Within IF Formula Is Not Producing Expected Results
Aug 24, 2014
I am trying to use a combination of Concatenate and IF formula to produce an email.
My input.
D4: First Name:
D5: Middle Name:
D6: Last Name:
D8: Organization:
Once all these are filled, I want the formula to produce a result like
FirstName.M.LastName@Organization.com
I have used the following formula.
[Code] .....
Problem is if there is a middle name the formula works fine, but in case where there is no middle name, it produces the following result.
FirstName..LastName@Organization.com
How do I remove the additional (.) in cases where there is no middle name.
Attached File : Email Generator.xlsx
## Vlookup Formula In VBA - Object Error (input A Formula Into A Cell)
Apr 23, 2009
I'm having some trouble trying to get excel to input a formula into a cell. I'm still a novice at VBA right now, so I don't think my problem will be too much of a brain buster.
I want a formula in Cell A6 (and I already know it correctly works) in this format: =E6&VLOOKUP(I6,'FA-Fund Data'!B\$1:C\$2000,2,FALSE)&J6
## When Dragging Formula Down The Formula Skips X Number Of Cells
Jul 22, 2014
Is it possible to write a formula that skips x number of cells and have the formula autoincrement as usual. For an examples lets say I want to skip 3 cells and I have entered in the cells listed:
A1= "1"
A2= "2"
A3= "3"
Then I type in the simple formula in the cell
B1= "=A1"
and then drag down but it skips 3 cells so then
B5= "=A2"
B9= "=A3"
If it didn't skip then
B1= "=A1"
B2= "=A2"
B3= "=A3"
this is just an example as I will be writing more complex formulas and need some thing that I can write into the formula so that it skips.
Something similar to this B1 "=A1(skip3cells)"
## Copy A Formula In A Cell And Then Paste Only The Text Of The Formula
Jul 16, 2007
I would like to copy a formula in a cell and then paste only the text of the formula, but I can't figure it out. Basically, I would like to avoid going into the cells and absolute referencing or hitting F2, then copying the text.
When I hit "Ctrl C" to copy the cell, then hit "Alt/E/S/F/Enter" to paste the formula, it is just like a regular copy/paste formula-wise in that the references move.
## Combine Formula Which Reduce The No. Of Additoanl Formula's Columns
Apr 9, 2009
I have some student marks in columns B to F. If some one get below 40 marks he fails and the subject in which he has been failed will come in Remarks column. If he has passed in all subjects remarks column will show Pass.
To bring failed subject initial in Remarks i have to create 5 additional column and then thru Concatenate i joined them.
Is There any formula to show fail subject in Remarks column without creating 5 additional columns?
Aug 12, 2007
This task joins a string together based on a number of characters per cell in the range.
I want to isolate one range, Col N, and add an IF condition to it.
There may be other issues preventing this from happening, e.g. the number of IF that exist in the complete formula. I will isolate the current cell and its requirements and then post the entire formula at the end for reference....
## Formula Arrays That Will Omit Empty Cells In My Formula
Jan 31, 2008
I am looking to average a range of cells which won't always be the same size. How do I create a formula array that will omit empty cells in my formula.
## Not Responding: Added A Formula To A Spreadsheet With Som Complicated Formula
Jun 1, 2006
I recently added a formula to a spreadsheet with som complicated formula. It worked ok and I saved the sheet. Now it takes 5 minutes to open the sheet and when I try to do anything,like delete the inserted column the program locks up giving a no responding message. I can do without this column if I have to.
## Replace Character Used In Formula Where Formula References Cell
Feb 23, 2008
I am using vlookup to get a cell value from another sheet, but if the cell has "&" or "/" I need to substitue "&" and "/" with "and" so that the cell can be added later to a url.
i am using =VLOOKUP(a1,Sheet2!A1:W17968,6) to get the value of a1 in sheet 2 and return the value of column 6
this will return "Audio Cables & Leads" but i need it to say Audio Cables and Leads
I need the formula to also check and replace "/" with "and" as well so cables/wire will be Cables and wire
## Adding Today Formula In A Countif Formula
Dec 19, 2012
I have a spreadsheet that contains a list of dates: What I need is a formula that will count the number of cells that hava a date more than 6 months old. I also would like the field automated so I don't have to change the date manually every day.
=COUNTIF(S5:S593,"<2012/06/18") This formula will give the correct read out, but I must change the date manually.
My attempts at adding a =today() command in place of the date result in 0 being the result (not correct)
This is what I have tried: =COUNTIF(S5:S593,"<"=TODAY()-"183")
## Add Total \$ Value Formula Into Existing SUMIFS Formula?
Apr 3, 2014
In my attached spreadsheet example below, in the Rollup Data sheet, I have a formula (Cell: C17) that will give me the actual cost totals that has been invoiced on WBS Element (DWRRI-BW096-231) for a total of \$35,004.81.
What I need the formula do is run a redundant check procedure on my new WBS Element (DWRRI-BW096-231) actual cost total formula in the Rollup Data sheet (Cell: C17) against/vs. what the SAP system provides in the raw SAP Excel export contained in the PTD-Actual Cost sheet for the WBS Element (DWRRI-BW096-231) actual cost total of \$35,004.81 (Cell: F9) that's highlighted in the yellow row.
In the Rollup Data sheet (Cell: C18), I need this formula to perform a similar function the as the previous formula above it (Cell: C17), but it this formulas function (Cell: C18) would have a separate operation that would only look for and return the single/sole WBS Element (DWRRI-BW096-231) actual cost total contained in the PTD-Actual Cost sheet (Cell: F9).
In the Rollup Data sheet I could then compare my two WBS Element (DWRRI-BW096-231) actual costs totals, the first from the automated addition of all the individual matching WBS Element actual costs into an actual cost total (Cell: C17), the second (new formula I'm requesting) from the matching and return function of the single/sole WBS Element actual cost total (Cell: C18).
So, if the two separate WBS Element actual cost totals (Cells: C17 vs. C18) both equal \$35,004.81, the formula in C17 is a success and has correctly identified and added all the actual costs contained in the PTD-Actual Cost sheet, but if the two separate WBS Element actual cost totals (Cells: C17 vs. C18) both do not equal \$35,004.81, the formula in C17 failed for some reason and has not correctly identified and added all the required actual costs contained in the PTD-Actual Cost sheet.
How to Add Total \$ Value Formula into an Existing SUMIFS Formula.xlsx
## Freezing The Formula After Certain Time / Formula Has Been Satisfied?
Aug 7, 2014
I am trying to create a graph of my data at the end of each month. The graph will show a count of the number of instances of each category in the data.
For example:
A1 =TODAY()B1 =MONTH(A1)
Column A from A3 down = 1,2,3,4,W,H,L
Column I between I2 and I140 = multiple instances of 1,2,3,4,W,H,L
July
=IF(\$B\$1=7,(COUNTIF(\$I\$2:\$I\$140,\$A3)),"0")
=IF(\$B\$1=7,(COUNTIF(\$I\$2:\$I\$140,\$A4)),"0")
=IF(\$B\$1=7,(COUNTIF(\$I\$2:\$I\$140,\$A5)),"0")
=IF(\$B\$1=7,(COUNTIF(\$I\$2:\$I\$140,\$A6)),"0")
=IF(\$B\$1=7,(COUNTIF(\$I\$2:\$I\$140,\$A7)),"0")
=IF(\$B\$1=7,(COUNTIF(\$I\$2:\$I\$140,\$A8)),"0")
=IF(\$B\$1=7,(COUNTIF(\$I\$2:\$I\$140,\$A9)),"0")
=IF(\$B\$1=7,(COUNTIF(\$I\$2:\$I\$140,\$A10)),"0")
My problem is that when it comes around to August or the next month, I want the final count of July to be fixed/frozen so it won't change from the 1st of the month. I can then use the July column's data for my graph.
Can a formula convert itself into a value or freeze itself after certain criteria has been fulfilled?
## Vlookup Formula And Function To Combined With Another Formula?
Oct 22, 2013
i have two excel files which are data and master..both files will be use for salary calculating.. the vlookup formula will be use in master files for dragging the salary data from Data files..the formula as follow VLOOKUP(B4:B225,'D:Salary[Data.xls]AUG'!A\$1:F\$65536,6,FALSE))
the vlookup working fine to me.. but my problems is i want the data to be auto calculated when they have same value in two columns.. or if the name is similar/match in two columns (one for salary and another one for overtime), the salary should be auto calculated.
Data files will contain of these:
a:Employee ID
b:employee name
c:Employee salary/Overtime
Master files will contain of these:
a:Employee ID
b:Emplyee Name
c:Employee Nett Salary (that will be dragging from Data files)
is there any formula that i can used to combined with my vlookup formula?
## Searching Logical Formula And Also Formula For Combinations?
Feb 4, 2014
2 formulas that I need for my work (related with linear recurrences; this is serious work). First, I would like to know if is it possible to write a formula (or a set of formulas) that allows to write in a first column, in the first cell, the number 1 and in the following m cells, m distinct numbers chosen in a set S of n numbers (n > m); in the second column again the number one, and another set of m distinct numbers chosen in the same set S (here the order of the elements matters which means that the set {1,2} can be written as (1,2) and (2,1)). And I would like to do this until all the choices of m (ordered) numbers chosen in the set S are exhausted. Example: S = {2,3,4} which means that n=3. Suppose m=2. Then the columns would be
111111
223344
342423
The second formula I'm searching is (I think) a logical formula. What I would like is a formula that checks if the numbers in a fixed number of consecutive cells of a column are distinct and also non-null. If that's the case the formula should assign TRUE and if (at least) two cells are equal or (at least) one is zero, it would assign FALSE.
## Formula That Points To Another Cell With Formula To Finish First One?
Mar 5, 2014
I'm wondering if it's possible to write a formala that if a certain condition is met, performs a formula referenced in another cell. For example: Column A contains a list of formulas. In column B is the condition, let's use Work Type as an example. The result (or the formula I need) would be in column C. So if the work type is "Type1", then the formula in the corresponding column C cell would perform the formula in cell A5. If the work type is "Type8", then the formula in the corresponding column C cell would perform the formula in cell A2.
Is this possible to do without VBA?
## Date Formula & Add If Statement To The Existing Formula
Dec 19, 2008
I currently use the following formula: =workday(H21,H23,J20:J34) which calculates my date based the date entered in cell H21. While testing I realized that I needed to add another variable to the formula for a specific scenario. When my Date in H21 is less than todays date I require another workday formula to be calculated. How can I add this if statement to the existing formula?
## IF Formula Assistance: Formula That Looks At 2 Adjacent Cells
Feb 2, 2009
I need to create a formula that looks at 2 adjacent cells. If both cells show 'YES' then I need the 3 cell to show 'YES'. If either/both of the 2 adjacent cells are blank then the third needs to be blank.
## Cells Display Formula, Not The Results Of The Formula
May 29, 2009
To get cell to display results of formula I must edit (F2) and hit enter. Is there an easy way to do this in bulk rather than editing and hitting enter on hundreds of cells?
Example of formula:
=IF(COUNTBLANK(K2)=1,"",CONCATENATE("*** ",K1,": ",K2," "))
## Logical Formula: Value If True/false To Be Another Formula, Not Just A Set Value
Sep 16, 2009
I want to use the If formula but want the value if true/false to be another formula, not just a set value. Here's what I came up with (obviously wrong, please excuse my lack of experience) =IF(D2>C2,"=(D2-C2)*5","=((60-C2)+D2)*5 ")
## Use The Results Of A Formula As Column/row Numbers In Another Formula
Oct 1, 2009
I have two cells. The first cell has the formula:
=CONCATENATE("D",TEXT(MATCH(\$B\$6,'Zip Ranges'!\$D\$1:\$D\$157,0)+1,"0"))
which results in a col and row number (such as D65).
The second cell has the following formula:
=INDEX('Zip Ranges'!\$A:\$B,MATCH(\$B\$6,'Zip Ranges'!D1:\$D\$157,0),2)
^^
I wish to replace the 'D1" in the Match function with the results of the first cell's formula.
I assume Indirect would work, but I don't know how to code the formula to use it.
## A Formula Which Is Calculate 4-5 Dates Formula In One Cell
Oct 7, 2009
i need a formula for which calculate all cell (calculating years months and days formula). FIND ATTACHED .xls file i need a answer in cell highlited in yellow color.
## Offset Formula: Getting N/A In Cell When Enter Formula
Feb 8, 2010
I have attached a sample of the spreadsheet. The cell/cells in question are f32:f36 on the first tab.
Question #1 how this formula works or what it does.
Question #2 Is when I enter the required information in cells a32:e32 it enters an N/A in cell f32. When I than click on cell F32 and than click on the formula bar and hit enter the information populates.
## Add 2 Conditions To Formula Or Modify Existing Formula?
Dec 22, 2012
I'm trying to add either a new condition to a formula or modify existing formula, so blank cells can't be taken as a number like 0 Here is the formula =IF(N14>0,1,0)
***This is what I do with formulas with similar problems =IF(COUNT(Y14,A14)=2,Y14/A14*100,"")
unfortunately the above condition its already a condition by itself so I don't know what to do whether to modify condition or add a new condition which I don't know how to do.
## If(iserror Formula): Used A Variant Of This Formula Before Without Encountering
Feb 27, 2009
I'm going wrong with this formula: =if(iserror(J3-VLOOKUP (H3,\$H\$730:\$J\$1456,3,FALSE)," "(J3-VLOOKUP(H3,\$H\$730:\$J\$1456,3,FALSE)). I feel like I've used a variant of this formula before without encountering any problems.
## SUMIF Formula That Is Collects Values From An IF Formula
Aug 26, 2009
Can anyone tell me why this formula is not working?
=SUMIF(\$D\$6:\$AL\$6,"=TM Goals",D12:AL12)
The entry under cell T12, which I am hoping this formula will pick up is based on the following formula.
=IF(E12="A",\$L\$5,"0")
## Formula Cell Not Tampered And Correct Formula
Feb 27, 2012
I have two questions:
1. Is there a way cells with formulas cannot be tampered even if they are not locked? So when I enter a data on a cell with formula, the formula won't be erased? The sample below shows that C1 and C2 has formulas but let's say I enter a value on C1, the formula should still stay? Is this possible?
A B C
1 5 10 =SUM(A1:B10)
2 2 4 =SUM(A2:B2)
2. Let's say I have a total of 125 (A1) and I need to get the answer (B1) where it has 5 with No. 3 on top (subscript) of 5? How to create a formula for this (I hope I make sense)? | 4,801 | 17,979 | {"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} | 3.1875 | 3 | CC-MAIN-2018-26 | latest | en | 0.940002 |
http://mathforum.org/mathimages/index.php?title=Complex_Numbers&oldid=34304 | 1,519,105,239,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812880.33/warc/CC-MAIN-20180220050606-20180220070606-00554.warc.gz | 237,588,929 | 7,407 | # Complex Numbers
This is a Helper Page for:
Riemann Sphere
Julia Sets
Logarithms
Cartesian Product
Newton's Basin
Completing the Square
Simple Harmonic Motion
## Definition
Complex numbers are numbers which take the form $a+bi$, where $a$ and $b$ are real numbers and $i = \sqrt{-1}$ .
Such numbers frequently appear in mathematical equations, even in those describing physical systems. Extending our notion of numbers to include complex numbers results in many astounding symmetries and relationships throughout mathematics.
## Basic Properties
Complex numbers have two parts: a "real" part represented by $a$ and an "imaginary" part represented by $bi$; the $i$ factor in the imaginary part forces the two to be separate. The same operations that are used on real numbers, such as addition, subtraction, multiplication, and division, can be used on complex numbers. For example, two complex numbers are added by components, real added to real part and imaginary to imaginary part:
$(6-1i)+(3+0.5i) = 9 -0.5i\,$
As another example, multiplying two complex numbers is carried out in the same way that we would multiply two real binomials:
$(3+2i) \times (1-3i) = 3-9i+2i+6 = 9-7i \,$
Note that because each $i$ is the square root of $-1$, the product of two $i$ terms gives $-1$ , so $2i \times (-3i) = 6 \,$.
## Visualizing the Complex Numbers
We traditionally visualize the real numbers, such as 2 and 0.5, as points on the number line. We can visualize real numbers this way because all real numbers can be identified by a single value. The real number 5 is unique, and has its own place on the number line. Because complex numbers have two parts, we can think of them as vectors contained in a plane. We call the plane which contains complex numbers the Argand Plane, or the Complex Plane. The y-axis represents the imaginary component of our complex number, and the x-axis the real component. The complex number $2+3i$ is shown below:
We visualize complex numbers in the same way we visualize an ordered pair on a plane.
We can thus speak of the magnitude of a complex number as the length of this vector, which is $r = \sqrt{a^2+b^2}$, as is readily shown by the Pythagorean Theorem.
This vector idea leads to an important relation between trigonometry and the complex numbers. Euler's formula, which can be derived using Taylor Series, tells us that
$e^{i\theta} = \cos(\theta) +i\sin(\theta) \,$.
$\sin(\theta)=b/r \,$ and $\cos(\theta) = a/r \,$ .
Substituting gives
$e^{i\theta} = a/r +ib/r \,$
or
$re^{i\theta} = a+bi \,$.
Therefore all complex numbers of the form $a +bi$ can be expressed with an exponential function. | 680 | 2,648 | {"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": 22, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-09 | latest | en | 0.894772 |
http://www.instructables.com/answers/VBA-Code-for-finding-the-10th-percentile-value-for/ | 1,498,402,244,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320532.88/warc/CC-MAIN-20170625134002-20170625154002-00453.warc.gz | 585,096,038 | 17,928 | # VBA Code for finding the 10th percentile value for a certain bin Range?
Hello everyone
I hope that you all are fine.
I have an excel sheet, and I am trying to bring some sort of automation in to that sheet (since I use it quite often) with the help of macros. The fact is that I am really new to macros and VBA programming language, so I need a little help in writing the code to find the 10th percentile for each bin value (50, 100, 150).
If you see the picture attached, the wanted percentile values are in Q, the bins for which I want to find the 10th percentile value, are given in Column M and the main data on which the 10 percentile value will be based is given in Column D.
By Using the data given in D and filtering out a certain bin width in column E (Example: Filtering bin width 50 or 100), we could find the Percentile value against each of the bin value given in Column N.
As said earlier I am new to VBA code, so any help will be appreciated in order to find the percentile values for each bin.
I can give the excel sheet as well if someone want to try hand on it, and let's me know the email address.
Additional Note: (The spread sheet is sorted out with respect to column E, is Ascending order (50, 100, 150) and the number of count for each bin width is given in column O, so you can just find the range of each bin by adding the count for each bin i.e range for first bin (50) is from 1 to 1+8542, similarly the range for the second bin (100) is given by 1+8542 to 1+8542+6862.
I will be very grateful if someone spares some time for me.
Kind Regards
raishikoh
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frollard4 years ago
Allow me to rephrase my previous...
WHAT exactly are you trying to calculate?
Could you not use filtering to create your bins?
Perhaps a pivot table to perform custom lists of calculations on your bins?
What exactly constitutes a bin?
As it stands, your values (by name) make almost no sense. Finding median values of a random assortment of unnamed values makes it very difficult to wrap my head around why you would want to solve for that number. No matter how I look at your question, I still don't see VBA being the answer.
raishikoh (author) frollard4 years ago
it would so nice of you if you could help me solve this problem..
4 years ago
I say again.
I understand you want the different percentiles of different bins.
Use filter to FILTER the bins you want, then run a percentile calculation on a different tab that points to the different bins in the different filters.
raishikoh (author) frollard4 years ago
I always struggle in explain things to people,
Sorry for the inconvenience....
http://www.mrexcel.com/forum/excel-questions/685935-run-visual-basic-applications-macro-code-loop-excel-2010-a.html
I am trying to do things automatically.... as its takes alot of my time...
4 years ago
You. Do. Not. Need. To. Use. VBA.
steveastrouk4 years ago
Am I missing something obvious here ? Where does programming come into it ? Don't you just need the correct cell formula ?
raishikoh (author) steveastrouk4 years ago
Well the formula for calculation the percentile is:
Percentile PERCENTILE( array, 0.1 )
But the problem is it applies to whole of the column, i need to calculate the percentile for a specific bin range, hence I have to select the part of array/ column that lies within the bin range... It need a code that calculates the percentile for each bin (specified in Column E). Although you are seeing all entries equal to 50 for column E in this pic, but if you scroll down the attached file then you will notice that the bin (Upper G) changes form 50 to 1400....
4 years ago
You can select a range and give it a name...
Select all the values you need for array, go to the name box, top left of your picture, currently has the value Q12 (will change when you type something in there). Call it something useful, like Qarray | 942 | 3,891 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2017-26 | longest | en | 0.918185 |
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By Monika Muller at April 22 2019 23:52:48
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What are the Parts of a Worksheet? _ Worksheets consists of four primary parts. A cell is the most commonly used part within an Excel workbook. Cells are where users can enter data to be used within formulas and charts later on. Each Cell consists of a Column and a Row. A column is all the cells in one vertical line in the worksheet. Column names can be seen across the top of a worksheet. A row is a collection of cells in line horizontal across a worksheet. Row names or Values can be seen scrolling down to the left of the worksheet. | 446 | 1,933 | {"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} | 2.78125 | 3 | CC-MAIN-2019-22 | latest | en | 0.844963 |
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Printables Give Practice Subtracting Fractions With Common Denominators Fractions Worksheets Adding Fractions Math Fractions Worksheets
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Part A
Some questions in Part A require that you access data from Statistics for People Who (Think They) Hate Statistics. This data is available on the student website under the Student Text Resources link.
1. Why is a z score a standard score? Why can standard scores be used to compare scores from different distributions?
2. For the following set of scores, fill in the cells. The mean is 70 and the standard deviation is 8.
Raw score Z score 68.0 ? ? –1.6 82.0 ? ? 1.8 69.0 ? ? –0.5 85.0 ? ? 1.7 72.0 ?
3. Questions 3a through 3d are based on a distribution of scores with and the standard deviation = 6.38. Draw a small picture to help you see what is required.
a. What is the probability of a score falling between a raw score of 70 and 80?
b. What is the probability of a score falling above a raw score of 80?
c. What is the probability of a score falling between a raw score of 81 and 83?
d. What is the probability of a score falling below a raw score of 63?
4. Jake needs to score in the top 10% in order to earn a physical fitness certificate. The class mean is 78 and the standard deviation is 5.5. What raw score does he need?
5. Who is the better student, relative to his or her classmates? Use the following table for information.
Math Class mean 81 Class standard deviation 2 Reading Class mean 87 Class standard deviation 10 Raw scores Math score Reading score Average Noah 85 88 86.5 Talya 87 81 84 Z-scores Math score Reading score Average Noah Talya
Ann wants to describe the demographic characteristics of a sample of 25 individuals who completed a large-scale survey. She has demographic data on the participants’ gender (two categories), educational level (four categories), marital status (three categories), and community population size (eight categories).
6. Using IBM® SPSS® software, conduct a frequency analysis on the gender and marital status variables. From the output, identify the following:
a. Percent of men
b. Mode for marital status
c. Frequency of divorced people in the sample
7. Using IBM® SPSS® software, create a frequency table to summarize the data on the educational level variable.
The data for Exercise 8 is available in the data file named Lesson 21 Exercise File 1.
8. David collects anxiety scores from 15 college students who visit the university health center during finals week. Compute descriptive statistics on the anxiety scores. From the output, identify the following:
a. Skewness
b. Mean
c. Standard deviation
d. Kurtosis
Part C
Complete the questions below. Be specific and provide examples when relevant.
Cite any sources consistent with APA guidelines.
Question Answer What is the relationship between reliability and validity? How can a test be reliable but not valid? Can a test be valid but not reliable? Why or why not? How does understanding probability help you understand statistics? How could you use standard scores and the standard distribution to compare the reading scores of two students receiving special reading resource help and one student in a standard classroom who does not get special help? In a standard normal distribution: What does a z score of 1 represent? What percent of cases fall between the mean and one standard deviation above the mean? What percent fall between the mean and –1 to +1 standard deviations from the mean? What percent of scores will fall between –3 and +3 standard deviations under the normal curve?
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The EF-2.3 Rotational Friction experiments kit enables students to understand how rotational friction affects the efficiency of a screw jack, a wedge and different bearings.
Hands-on understanding from lessons.
## Description
The kit includes experiments that measure the effort required to raise various loads using a simple form of screw jack. These include determining how the mechanical advantage and efficiency varies with load, the mechanical advantage and efficiency obtained by using two different wedges, including investigating the effect of the angle on overhauling and to compare the resistance to rotation due to friction in four cantilevered bearings of different materials.
The Engineering Fundamentals range enables students to gain an understanding of the principles of engineering by the process of learning via experimentation.
The EF-2.3 Rotational Friction experiments kit enables students to understand how rotational friction affects the efficiency of a screw jack, a wedge and different bearing materials.
## Technical Specifications
• Wedge angles
• 18.5°
• 4.75°
• Bearings materials
• Oilite
• Wood
• Nylon
• Ball
## Features & Benefits
Features
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Benefits
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• Options
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EF-BU Base Unit
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• Level and stable work surface to mount the EF-BU upon. The optional EF-WS is ideal for this if no suitable desk or bench is available.
## PACKED AND CRATED SHIPPING SPECIFICATIONS
Volume: 0.02m³ per tray
Gross Weight: 5.0kg per tray
Tray
Length: 0.430m per tray
Width: 0.312m per tray
Height: 0.080m per tray
EF-2.3 – Rotational Friction
EF-BU – Base Unit
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Convert entered temperature to units of energy.
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Calculate volume of a right circular cone and its surface area | 492 | 1,814 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2023-40 | latest | en | 0.752194 |
http://test.scoilnet.ie/Res/annewalshe290799090220_2.htm | 1,555,743,893,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578528702.42/warc/CC-MAIN-20190420060931-20190420082931-00537.warc.gz | 169,364,681 | 2,414 | Directions:
Make a dice using a cube. Paint on 1, 2, 3 dots (Twice because of the six sides)
Teacher places flash cards with 1, 2, 3 dots & 1, 2, 3 numbers on the table in a line. Teacher starts by rolling the dice. She counts out the number of dots. She matches this number to the flash card with the same number of dots. She holds it up. She then lifts up the no. 2 from the number line. She says it aloud. She then takes two objects from the objects box.
Each child repeats the activity, with assistance from the teacher or childcare assistant.
Emphasis on oral counting. Identify all the people with one object, two etc.
When the children have mastered up to three, the teacher can introduce up to six and repeat the activity.
When the child rolls 3 the teacher can give him a flash card of the number 3 and ask him to find another like it.
Assessment Comments: This game can be played two or three times a week over a period of a year. Make this an enjoyable activity; for a treat count out jelly babies &eat them.
Follow-up Work: Children can do worksheets matching number of objects to number of objects; number of objects to the numerals; match numerals. When the children can recognise the numbers 1-6 and count out the correct number they can practice writing the numerals by tracing their finger over sandpaper letters. The children can begin to form the numerals on their own.
This activity can be extended for more able children by using unifix bricks. Children can be given unifix bricks instead of objects. The children can build up a stack with these bricks & compare them for Biggest/ Smallest/Same as.
Children can combine their bricks eg. 2+3. This can help with the early stages of addition.
By separating the bricks, children can experience the early stages of subtraction. | 404 | 1,799 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2019-18 | latest | en | 0.930859 |
http://www.gradesaver.com/textbooks/math/algebra/intermediate-algebra-12th-edition/chapter-7-section-7-1-radical-expressions-and-graphs-7-1-exercises-page-440/24 | 1,481,328,459,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542851.96/warc/CC-MAIN-20161202170902-00020-ip-10-31-129-80.ec2.internal.warc.gz | 491,523,413 | 38,991 | Intermediate Algebra (12th Edition)
We know that $\sqrt[4] 256=4$, because $4^{4}=4\times4\times4\times4=256$. Therefore, $-\sqrt[4] 256=-(4)=-4$. | 57 | 147 | {"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} | 3.78125 | 4 | CC-MAIN-2016-50 | longest | en | 0.603803 |
http://www.kylesconverter.com/area-density/kilograms-per-square-meter-to-short-tons-per-thousand-square-feet | 1,610,775,253,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703500028.5/warc/CC-MAIN-20210116044418-20210116074418-00276.warc.gz | 149,497,446 | 5,665 | # Convert Kilograms Per Square Meter to Short Tons Per Thousand Square Feet
### Kyle's Converter > Area Density > Kilograms Per Square Meter > Kilograms Per Square Meter to Short Tons Per Thousand Square Feet
Kilograms Per Square Meter (kg/m2) Short Tons Per Thousand Square Feet (t (US)/MSF) Precision: 0 1 2 3 4 5 6 7 8 9 12 15 18
Reverse conversion?
Short Tons Per Thousand Square Feet to Kilograms Per Square Meter
(or just enter a value in the "to" field)
Please share if you found this tool useful:
Unit Descriptions
1 Kilogram per Square Meter:
Mass of kilograms per an area of a square meter. 1 kg/m2 = 1 kg/m2.
1 Short Ton per Thousand Square Feet:
Mass of short tons per area of a thousand square feet. A short ton (US) having 2 000 international pounds of 0.45359237 kilograms. An international square foot having 0.3048 meters per side. 1 t (US)/MSF ≈ 9.764 855 272 7661 kg/m2.
Conversions Table
1 Kilograms Per Square Meter to Short Tons Per Thousand Square Feet = 0.102470 Kilograms Per Square Meter to Short Tons Per Thousand Square Feet = 7.1686
2 Kilograms Per Square Meter to Short Tons Per Thousand Square Feet = 0.204880 Kilograms Per Square Meter to Short Tons Per Thousand Square Feet = 8.1926
3 Kilograms Per Square Meter to Short Tons Per Thousand Square Feet = 0.307290 Kilograms Per Square Meter to Short Tons Per Thousand Square Feet = 9.2167
4 Kilograms Per Square Meter to Short Tons Per Thousand Square Feet = 0.4096100 Kilograms Per Square Meter to Short Tons Per Thousand Square Feet = 10.2408
5 Kilograms Per Square Meter to Short Tons Per Thousand Square Feet = 0.512200 Kilograms Per Square Meter to Short Tons Per Thousand Square Feet = 20.4816
6 Kilograms Per Square Meter to Short Tons Per Thousand Square Feet = 0.6144300 Kilograms Per Square Meter to Short Tons Per Thousand Square Feet = 30.7224
7 Kilograms Per Square Meter to Short Tons Per Thousand Square Feet = 0.7169400 Kilograms Per Square Meter to Short Tons Per Thousand Square Feet = 40.9632
8 Kilograms Per Square Meter to Short Tons Per Thousand Square Feet = 0.8193500 Kilograms Per Square Meter to Short Tons Per Thousand Square Feet = 51.204
9 Kilograms Per Square Meter to Short Tons Per Thousand Square Feet = 0.9217600 Kilograms Per Square Meter to Short Tons Per Thousand Square Feet = 61.4448
10 Kilograms Per Square Meter to Short Tons Per Thousand Square Feet = 1.0241800 Kilograms Per Square Meter to Short Tons Per Thousand Square Feet = 81.9265
20 Kilograms Per Square Meter to Short Tons Per Thousand Square Feet = 2.0482900 Kilograms Per Square Meter to Short Tons Per Thousand Square Feet = 92.1673
30 Kilograms Per Square Meter to Short Tons Per Thousand Square Feet = 3.07221,000 Kilograms Per Square Meter to Short Tons Per Thousand Square Feet = 102.4081
40 Kilograms Per Square Meter to Short Tons Per Thousand Square Feet = 4.096310,000 Kilograms Per Square Meter to Short Tons Per Thousand Square Feet = 1024.0807
50 Kilograms Per Square Meter to Short Tons Per Thousand Square Feet = 5.1204100,000 Kilograms Per Square Meter to Short Tons Per Thousand Square Feet = 10240.8072
60 Kilograms Per Square Meter to Short Tons Per Thousand Square Feet = 6.14451,000,000 Kilograms Per Square Meter to Short Tons Per Thousand Square Feet = 102408.0718 | 874 | 3,276 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2021-04 | latest | en | 0.559256 |
http://pari.math.u-bordeaux.fr/archives/pari-users-1210/msg00048.html | 1,545,133,368,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376829140.81/warc/CC-MAIN-20181218102019-20181218124019-00639.warc.gz | 207,339,876 | 2,454 | Dirk Laurie on Fri, 19 Oct 2012 17:58:18 +0200
Word length and number of bits in an integer
• To: pari-users@pari.math.u-bordeaux.fr
• Subject: Word length and number of bits in an integer
• From: Dirk Laurie <dirk.laurie@gmail.com>
• Date: Fri, 19 Oct 2012 17:58:11 +0200
• Delivery-date: Fri, 19 Oct 2012 17:58:18 +0200
• Dkim-signature: v=1; a=rsa-sha256; c=relaxed/relaxed; d=gmail.com; s=20120113; h=mime-version:date:message-id:subject:from:to:content-type; bh=Nn66weV292FDSCVJbysKNS+sT3DqOM7gg4flYBoQN+c=; b=ZnQ8tzgE+VjaY/yWOqUjiecSGQtjhGkYJcmIXjW2NKdm23Igr2E1SuISGEfplj7MW9 7vErjOJ0BNLKndXyFL8WkJOMYwoJ8zsQ+3l1jcveBvAsWEvPQQtyPXJmEnURH89WZ/t6 3Hm7HR9Wv71H9UP3OGFUzKk5YQvrZSUyq5Tlh9T+4lirWGHaqZZsGnjRurSqxQk/txRh OrRIKe00MR3NgZC/1CghZicJVKtrB90I9zD6R8MGz0ntZDic/u5r/+RF9GLHSNxfhTRF raW2DThPusb8e3MCitfcHrM2Dttdt6A7C6VnlmTF1EtVjryzyuIwoaI0HnfGce/enExs c3WQ==
```I need a routine for the exact smallest value B such that x>>B is zero,
where x is a t_INT. SInce a may be very large indeed, I coded it thus:
{ bitsin(x) =
my(w=32*(length(x)-1)); x>>=w;
while(x, x>>=1; w+=1);
w
}
I have two queries:
1. Am I missing some useful builtin function?
2. Can I rely on the words making up x to be 32-bit?
If not, how do I find out the word length from inside GP?
``` | 577 | 1,272 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2018-51 | latest | en | 0.515914 |
https://beta.scienceinschool.org/article/2007/damnlies/ | 1,670,159,335,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710972.37/warc/CC-MAIN-20221204104311-20221204134311-00767.warc.gz | 154,487,314 | 17,205 | # Damn lies Understand article
Do you have more than the average number of ears? Is your salary lower than average? When will the next bus arrive? Ben Parker attempts to convince us of the value of statistics – when used correctly.
Whether it was Mark Twain or Benjamin Disraeli who first coined the idea that there are three types of falsehood – “Lies, damned lies, and statistics” – the sentiment still persists. Statisticians are manipulative, deceitful types, set to pollute our minds with meaningless and mendacious information that will make us vote for their favourite political party, use their demonstrably effective skin cream, or buy the pet food that their cats prefer. For me, as a statistician, it’s now time to debunk a few myths.
Exactly 96.4% of our modern world revolves around statistics, and although there are some shockingly bad statistics out there, I hope to convince you that the fault lies generally in their presentation.
## Ear we go
I could make a confident bet that you, gentle reader, have more than the average number of ears. Why? Let’s assume that there are six billion people in this crowded world of ours, more than 99% of whom have two ears. There are a few exceptional people who, due to injury or birth, may have one or even no ears. There are, to my knowledge, no three-eared people (Captain Kirk is unfortunately fictional, but he did have three ears: a left ear, a right ear, and a final front ear). When we take an average (add up the total number of ears that humanity possesses, and divide by the number of people), we get the sum
Slightly less than 12 billion/6 billion
which is slightly less than two. This means that, as most people in the world have two ears, they have very slightly more than the average, so most times I would win my bet.
## What does this mean?
Now, of course, this is obviously just a statistician being pedantic. However, slightly less silly examples abound. Statistics on how one group of people earn less than a certain percentage of the national average income are used as political footballs. It is all too common to read commentaries in newspapers about how shocking it is that people earn only a percentage of the national income, and it’s all the fault of the Labour government, previous Conservative administration, European Union or sunspots.
The distribution of incomes, according to the UK Department for Work and Pensions (see figure to the right), is such that there are relatively few people who earn large sums of money (unfortunately statisticians do not fall into this high-income group). This means that the average income, which the Department has calculated to be £427 per week for a couple with no children (DWP, 2006), is much more than what the majority of people earn, in parallel with the above explanation of average ear count. A few extraordinary people, whether they have fewer than two ears or earn large amounts of money, skew the average from the situation for the majority of people.
Now of course, people soon realised that this commonly used average, calculated by adding everything up and dividing by the number of things you added up, more properly referred to as the mean, was likely to be misinterpreted. So the concept of the median is one that is often used in practice. If we were to put all the people in the UK in a line according to their income, the median salary would be that which the person standing in the middle of the line would have. The median, about £349 per week in this example, in practice often gives a better idea about what is typical.
At least we have reached some common sense – so can we expect everyone to understand this fairly basic problem in conveying ideas with averages? After all, surely the role of a good journalist is to take ideas and present the truth in a way the public can understand?
Unfortunately, factual accuracy and correctly interpreting data sometimes don’t sell newspapers, or make the correct political point.
## Ironing out the wrinkles
Worse than journalists, but not quite as bad as politicians, are advertisers. A recent television advert for a cosmetics company claims that their latest wrinkle-removing cream satisfies 8 out of 10 customers, based on a survey of 134 people. We can perhaps excuse the small sample size – and even the rounding (134 x 8/10 = 107.2), which means they must have found 0.2 of a customer to try out the cream – but the crucial question is how did they do the survey?
It seems to me that asking 134 customers whether they like the product is dubious – if the people are already customers, and have bought the product voluntarily, perhaps it’s not the fairest sample in the world. Why would anyone buy the product who doesn’t like it? In most sensible scientific trials, one would hope to compare the performance of the cream objectively against a brand X cream, or a placebo, to see whether people chosen at random have had a positive effect with the cream.
There’s no problem with advertising per se; philosophers argue that advertising is the most vital thing for a strong democracy. It’s fine for advertisers to let people know about their product and promote its benefits. However, what’s not acceptable is putting a thin veneer of science around the marketing; although cunningly worded, without explaining the method, the statistic ‘8 out of 10’ is meaningless. It’s just as bad as saying “Our car has a top speed of 500 miles an hour” without adding that this speed is only obtainable when measuring how fast the car drops out of an aeroplane: it’s true, but it’s misleading. Using this kind of fake survey in advertising is paramount to lying.
## Three come along at once
Maybe it’s unfair to blame the conveyors instead of the statistics themselves. There are some real, difficult, non-intuitive facts that statistics throws up, the truth of which can be very hard to work out. Let’s say you’re waiting for a bus, and you look at the schedule, which, assuming it hasn’t been vandalised, tells you that there are five buses an hour. How long would you expect to wait for a bus?
Sensible logic tells us that if there are 5 buses an hour, then the average (sorry, mean) time between buses is 12 minutes. So, assuming you arrive at the bus stop at a random time within this period, you’d expect to wait 6 minutes for a bus. Good logic, but unfortunately, in general, wrong.
We know that buses don’t run to the minute. They may leave the depot on time, but chance factors will alter their progress in different ways, so we have to assume that the incoming pattern at our bus stop varies somewhat. What exact distribution we choose might vary – we may, for example, assume that times between arrivals of buses follow an exponential distribution – but the important fact is that the buses do not come at regular times. So let us now assume that we arrive at the bus stop at some random point in time – how long is our expected wait for a bus now?
When we turn up at the bus stop, we are more likely to pick a period when there is a big gap between buses – a big gap occupies more time than a small gap, so we’re more likely to get a big one when picking at random. But given that we’ve picked a big gap, we know that the length of gap is more than 12 minutes (there are still 5 buses an hour) – so the average time to wait, given that our exact arrival time is equally likely to be somewhere in the big gap, is more than 6 minutes.
This is known as the inspection paradox, and it’s tricky to get your head around it. However, it’s a real phenomenon that is used by traffic planners and operational researchers, who are responsible for working out the most efficient method of arranging queues in post offices, and then ignoring it totally.
Are the bus companies wrong to advertise, then, that they have a bus approximately every 12 minutes? I think so, although it’s difficult to convey all the gory details of the inspection paradox; perhaps in this case we can excuse a little statistical laxity.
## Conclusions
In general, statistics is fairly intuitive and cases that are difficult to conceptualise are rare. In general, a questioning reader must:
• Find out who is presenting the data, and what they are trying to achieve.
• If possible, find out the sample methodology – whether the data comes from a suitably representative sample of the population being measured, and whether any testing is applied in a fair manner. Are fair comparisons used, and is the right question being asked?
• Question any averages or percentages and think about how extreme the statistics really are, and what you would expect. In particular, don’t assume that mean values are typical of the data.
Statistics is a powerful and useful tool in the right hands, and we need to give people the ability to understand it. We also need to ensure that some basic education in statistics, particularly in relation to interpreting advertising, is something that every pupil receives at school. At the very least, until journalists, the marketing industry, and the people who regulate them learn some statistics and, more importantly, how to present them, the world won’t be buying the best skin cream and pet food for their cats, all of whom have an above average number of ears.
### References
• DWP (2006) Households Below Average Income (HBAI) 1994/95-2004/05. London, UK: Department for Work and Pensions
• This article first appeared in Plus, a free online magazine opening a window on the world of mathematics: http://plus.maths.org. ‘Damn lies’ was a runner-up in the general public category of the Plus new writers award in 2006.
### Author(s)
Ben Parker is studying for a PhD at Queen Mary, University of London, and investigating the statistics of data networks. He has an undergraduate mathematics degree from the University of Cambridge, and a master’s degree in statistics from Birkbeck, University of London. In his spare time Ben likes to act in pantomimes and drink tea.
### Review
The article presents a humorous view of how statistics are misused in everyday life. It would be comprehensible for teachers, students and general readers all over the world. In school, it could be used as an introduction to statistics and probability, to encourage pupils to think about how statistics and probabilities are used and misused.
I particularly like the hilarious headings and humour in the article – sometimes obvious, sometimes less so.
Marco Nicolini, Italy | 2,205 | 10,476 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2022-49 | latest | en | 0.974032 |
https://answerofmath.com/solved-seasonal-and-autocorrelated-regression-residuals-difference-raw-data-or-residuals/ | 1,696,232,013,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510983.45/warc/CC-MAIN-20231002064957-20231002094957-00115.warc.gz | 113,798,920 | 20,054 | # Solved – Seasonal and autocorrelated regression residuals: difference raw data or residuals
I have a multiple linear model for time series data for which the regression residuals are autocorrelated and display seasonal behavior. This seasonal behavior is induced deterministically by a cyclic variable written into the model. In order to calculate corrected standard errors for the regression coefficients, I intend to use generalized least squares with correction for the autocorrelation in the residuals.
But for the seasonality, I am not sure for which time series I should perform the differencing: If for raw original data (observations), or for the regression residuals. For the data, I could superimpose each cycle and take the mean, thus removing season effects, but this would alter the data, and also would generalize bad for other kinds of data. For the regression residuals, I dont know if I have to difference the residuals time series substracting for each period, or to handle it as some kind of seasonal ARMA process.
Any hints?
LAST EDIT: The problem behind this question might have been resolved already, and might have been product of a misconception.
Regression was being done over simulations of the model. These simulations did not contain any stochastic error factor, so the simulations were purely of a deterministic nature. Regression errors were just showing an unperfect fit to the data and the regression residuals were obviously following a deterministic pattern. This pattern was not a result of some neglected explanatory variable in the model, and thus there was no reason to model it (by means of an ARMA model or any kind). When adding white noise to the data -which should be a crucial step on any simulation of real data- regression residuals were mostly dominated by stochasticity, loosing the autocorrelated behavior.
Contents
In order to calculate corrected standard errors for the regression coefficients, I intend to use generalized least squares with correction for the autocorrelation in the residuals.
Note that generalized least squares (GLS) would affect not only the standard errors but also the point estimates. Anyhow, you could gain power by estimating regression with an explicitly specified error structure, e.g. regression with ARMA errors as can be done using functions `stats::arima` or `forecast::auto.arima` in R. There you use maximum likelihood estimation instead of GLS. See related blog posts by Francis X. Diebold "The HAC Emperor has no Clothes" and "The HAC Emperor has no Clothes: Part 2" where he encourages explicit error specification as a way to get better coefficient estimates and gain predictive power. Although he discusses the case of HAC there, I believe similar conclusions apply here, too.
But for the seasonality, I am not sure for which time series I should perform the differencing: If for raw original data (observations), or for the regression residuals.
Since the problem arises due to a cyclic regressor, you could remove the deterministic component of the cyclic variable before including it in the model, or alternatively you could include some seasonal terms (dummies or Fourier terms) in the model.
For the data, I could superimpose each cycle and take the mean, thus removing season effects, but this would alter the data, and also would generalize bad for other kinds of data.
I am a little confused here, but I will try addressing this nevertheless.
With regards to the regressor, you can adjust using a model, and so altering data is not really a problem because you keep track of how you did it and you can recreate the original variable if you need to.
Regarding generalization, if the cyclic behaviour is unique for this instance, keeping it untreated would not help. If, on the other hand, it is similar across this instance and the ones you want to generalize to, you would not lose by removing the deterministic component before running the regression but then using it to adjust the other cases similarly.
A technical note: If you are doing a regression with ARMA errors, then it is the error that gets differenced. If the errors is some SARIMA process, regular treatment of SARIMA models applies (roughly speaking, you do not have to worry that it is a regression error rather than raw data).
Rate this post | 851 | 4,323 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2023-40 | latest | en | 0.950382 |
https://moffett4understandingmath.com/tag/perseverance/ | 1,679,570,659,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945144.17/warc/CC-MAIN-20230323100829-20230323130829-00262.warc.gz | 464,210,370 | 23,153 | # Creating Structure for Context in Math
I was honored to facilitate lesson study with IM1 teachers today. Their students are struggling (due to high EL/SPED population) with solving word problems. I dug deeper, and we decided the struggle is really the first step: creating equations from situations.
We decided our goal as educators this year is to work on teacher clarity: making our lessons streamlined and very goal-oriented. If we know our goals for the lesson, then every move we make (every breath we take…) is for the goal. So how do we clarify translating context to equations?
We started from the end: the benchmark. We took a problem the students struggled with, and tweaked it several times, each time only altering only one component. Students had to work from the original version (which we used simple numbers to keep it accessible) for each new “version”. They discussed what changed from situation to situation and how that affected the prior equation.
Version 1: Troy works for an ice cream cart vendor. He receives \$10 for taking the cart out for a shift, plus a commission of \$2.00 for each item he sells. Troy worked a shift Saturday and earned \$60. How many items did he sell?
Version 2: Troy works for an ice cream cart vendor. He receives \$15 for taking the cart out for a shift, plus a commission of \$2.00 for each item he sells. Troy worked a shift Saturday and earned \$60. How many items did he sell?
Version 3: Troy works for an ice cream cart vendor. He receives \$15 for taking the cart out for a shift, plus a commission of \$1.25 for each item he sells. Troy worked a shift Saturday and earned \$60. How many items did he sell?
Version 4: Troy works for an ice cream cart vendor. He receives \$25 for taking the cart out for a shift, plus a commission of \$0.10 for each item he sells. Troy worked a shift Saturday and earned \$52.90. How many items did he sell? (Problem from the benchmark.)
We used 3 scenarios. In each, we kept our questions as consistent as possible (again, clarity):
• Which part is varying (changing)? How do you know?
• Which quantity would be the coefficient? How do you know?
• Which quantity would be the constant? How do you know?
• (From version to version) What has stayed the same? What changed? How does the changed quantity affect our equation? Why?
Students were engaged, writing on their tables and willing to discuss with each other. They had many moments of “ohhhhhh” and “oops!” and learned quite a bit about the components of 2-step equations. They definitely need more time, and the teachers have committed to continuing the work as warm-ups or on modified days.
Oh! And did I mention this was a co-taught Special Ed class, with many English Learners?! Amazing!
So our major takeaways were:
2. Keep your goal in mind when creating the tasks/lesson and questions for clarity and focus.
3. Breaking the situations into translating and solving (working on a single component) allows students to focus and interpret.
Below is our ppt. Hope it is useful! Happy Math-ing!
Linear Equations in Context LS 8.27.19
This morning at the airport (At 5 fricking o’ clock! I need to fire my secretary for scheduling this flight. Oh wait. That’s me!) I was answering emails and a timid voice interrupted my thoughts.
Excuse me. Are you a teacher?
This always makes my heart happy. As a middle school teacher, you often believe the kids won’t think twice about you once they leave your room. Why would they, with all the distractions the world has to offer?!!
So when a former student not only remembers you from looong ago, AND takes the time to share her experiences and life journey with you, you tear up just a little. You remember that these precious moments are WHY you were born to be an educator.
My WHY is simple. I love seeing my peeps have the “click” in math. I love learning about these humans, with all their quirks and unique personalities. I love supporting them and inspiring them to do great things. I just love THEM.
What is your WHY? Ponder, remember, and remind yourself of this through the year.
Have a great 2019-2020! I KNOW it will be a year to remember! 💕
# How Do Our Beliefs in Math Affect Our Students?
I was honored to work with amazing teachers this week. We took a survey from NCTM (National Council Teachers of Mathematics) on our beliefs regarding student learning and our instructional practices in mathematics. This, in itself, led to amazing discussions about what we truly believe math IS and how we interpret that into instructional decisions within our classrooms.
But then we took it further. We got into groups and discussed not so much whether we agreed or disagreed, but whether it was a productive or unproductive belief in respect to student access and learning. Here are two to consider:
There were fantastic discussions about these particular ones, especially for educators of EL and SPED. We also considered how parents might respond to these. Powerful conversations around access, flexibility in thinking, understanding conceptual and procedural mathematical ideas, and yes, fluency.
Here was the point. Our beliefs, whether productive or unproductive, affect our attitudes towards mathematics and the children we are blessed to teach. Those attitudes affect the actions we take. Who gets to answer which questions? Who gets the “tough” tasks and who has to keep doing drill and kill worksheets? Who gets to explore puzzles and who has to retake tests or do homework (because their home life doesn’t lend itself to being able to do it at home)? And those actions MATTER. They affect the results you will get from your students.
So as you gear up for this school year, consider taking the beliefs survey yourself. Even better, have your team take it and REALLY dive in to what beliefs are productive an unproductive. The more we reflect, the more we can grow and be effective at what we truly want; to teach students to love, learn and understand mathematics. Have a great year!
# That Moment You Realize Your Child Is Suffering
All who have multiple children know that each one is wonderfully different. You can raise ’em the same, yet they have their own amazing quirks and personalities, strengths and passions. This could not be more true of my two boys.
My youngest is now 5. He has always been the rough and tumble type. Everything is a competition to him. (Even last night he was standing on his tippy toes trying to be ‘taller’ when raising his hand at church!) He is funny, outgoing, and a firecracker. He is a joy.
Yet school stuff has not yet become his thing. He seemed disinterested in learning his letters, yet loved to be read to. He never wanted to sit with me and learn the components all say are so important for kinder. His preschool teachers said it would come; that he just was a busy bee and had other, more important things to care about. And honestly, I believed that too. He loves to create, tell stories, sing, build, and live outdoors. Who was I to take that away from him?
So when his amazing preK teacher suggested a hearing test, I was on board. No big deal; just go do it and cross it off as another thing we did. It came back inconclusive. Went for an ENT hearing test. The results were staggering; at least 30% hearing loss in each ear. The doctor said, “Think of being submerged underwater for 5 years of your life, trying desperately to hear what people are saying. That is what your son has lived through.” I am still teary thinking about this. I am a fricking educator! How did I not catch this? I was in denial as well, until that very day driving home I asked my boy if he saw the cool tree. “Tree?” “Yes, the tree over there (I pointed out the car window.).” “Like, dessert?” And that is when it hit me. My child wasn’t hearing.
We started to notice. He said, “What?” almost every time you told him something. He couldn’t hear the TV clearly; he was reading lips (which we now know why he would never answer us when watching TV; he was too focused on the screen to hear us). In preK, during circle time, he struggled to hear all the conversations and his body would just be exhausted from trying to listen that he gave up. When listening to a story, he focused on the pictures for meaning versus our words. He was trying his best, his very best, and it was exhausting. My heart broke. My baby boy, sweet thing he is, was struggling under my teacher nose, and I hadn’t caught it.
Long story short, he is now hearing much more, with just a tweak of medication. He still has a hearing loss, and we will test every year to make sure it isn’t degenerative. He is learning how to deal with sounds he hadn’t heard before. (At church last night he couldn’t believe they played bells!) He begins speech therapy (I had no idea he would need that either) next week and they are excited to see his progress, as am I. And funny enough, he is now interested in learning his letters, sounds, words, etc. It all makes sense; for how can you be interested in something you never knew existed?
I bring this story up for one reason. We are not perfect. Even if we have our children’s best interests at heart, we may miss something. It takes a true village to raise our children. If you do not have a village to help, find one. If your children go to daycare or summer camp, get to know their counselors. Ask them questions. Find out what your children are doing, and how they are doing. Talk to your children’s teachers next year. Get to know them, because they see your children more often than you do! And listen, even when you don’t want to. In a world that is so negative right now, I feel strongly that we need to support, build, and nurture each other and our tiny humans in order to make them the best they can be.
# Developing Perseverance
I don’t get it. Can you help me? My teacher didn’t explain it. I forgot. This is stupid.
We have all been there. We have all heard each of these when our child is working on math homework. The question is, how do we get him/her to stop being helpless? Here are a few ideas to start with.
Listen to their frustrations. Then move on. Look. Being frustrated is okay. We don’t want to say it isn’t. But being helpless is not okay. This is a life lesson. Not everything is easy, but we don’t get to give up.
Don’t do the math for them. That only lets your child know you will let them off the hook every time.
Help them find resources and look at them together. Some questions to help you out:
• Where are your notes from today? Let’s review them and see if there is anything there we can use.
• What are you learning? Let’s look up some videos (mathtv.com is a great site for video lessons, but if you just go to utube or teachertube you will find others) and see if we can relearn it together.
• Let’s look at your book and see if there are any examples that might help us.
• Call/Facetime/text a friend and see if they can assist you. (Research shows that study groups truly help all students in mathematics!!!
• Email the teacher and see if s/he has tutorials to help. (Many teachers put up videos, the solutions, etc on-line. Find out if yours does!)
Above all, let them know that it is okay to not know everything, but NOT OKAY to give up. This is a biggie. Most math worth doing takes time. Students assume that if they don’t get the answer right away, they must have done it wrong (or don’t know what they are doing and are not good in math). So not true! Help your child use resources available to be successful, but they need to do the work and put in the effort. | 2,592 | 11,598 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-14 | latest | en | 0.956082 |
https://www.syntaxbomb.com/blitzmax-blitzmax-ng/ym2149-tutorial-questions/?PHPSESSID=eb2c739a3684edd0854c642810da8de7 | 1,723,376,867,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640997721.66/warc/CC-MAIN-20240811110531-20240811140531-00503.warc.gz | 796,192,103 | 11,599 | YM2149 Tutorial Questions
Started by Baggey, October 13, 2023, 06:40:42
Baggey
Hi @Midimaster,
In your tutorial you explain a 100hz Tone having 480 Steps. Is that 1 Complete Wave?
Am, i right in saying at 48000 samples if i sent 109 square waves id get. 440.36Hz tone which would be a Square Wave "A above middle C"
Baggey
Running a PC that just Aint fast enough!? i7 4Ghz Quad core 24GB ram 1TB SSD and NVIDIA Quadro K620 . DID Technology stop! Or have we been assimulated!
ZX Spectrum 48k, C64, ORIC Atmos 48K, Enterprise 128K, The SID chip. Im Misunderstood!
Baggey
In the TGenerator Tutorial you use the OneStep() Function. So indeed we are making a wave for every Sample.
If im following, we have created 480 samples using the OneStep() Function. Giving "1 to 0 to -1" which gives 2 changes for every sample sent Making the Square wave and sending that into Generator 1, which is our "Square Wave Generator" as 480 Samples.
So we have made a Square Waved Tone of the Note "G2"
Baggey
Running a PC that just Aint fast enough!? i7 4Ghz Quad core 24GB ram 1TB SSD and NVIDIA Quadro K620 . DID Technology stop! Or have we been assimulated!
ZX Spectrum 48k, C64, ORIC Atmos 48K, Enterprise 128K, The SID chip. Im Misunderstood!
Midimaster
Quote from: Baggey on October 13, 2023, 06:40:42...In your tutorial you explain a 100hz Tone having 480 Steps. Is that 1 Complete Wave?
Am, i right in saying at 48000 samples if i sent 109 square waves id get. 440.36Hz tone which would be a Square Wave "A above middle C" ...
Yes a 100Hz tone needs 480 steps in a 48kHz Device to fullfill a complete wave (positiv and negativ part)
Yes 109 steps would result in a (nearly) 440Hz tone A.
With this questiion you already pointed to a problem what we will get with INTEGER, when we want to reach an exact HERTZ frequency like 440Hz. With a period length of 109 we can only reach 440,36 Hz. And the deviation rises, when we try to build higher frequencies. In the today's lesson "Convert YM-periods to 48kHz-periods" I will take care about this and change the counter to a DOUBLE type.
QuoteIn the TGenerator Tutorial you use the OneStep() Function. So indeed we are making a wave for every Sample.
If im following, we have created 480 samples using the OneStep() Function. Giving "1 to 0 to -1" which gives 2 changes for every sample sent Making the Square wave and sending that into Generator 1, which is our "Square Wave Generator" as 480 Samples.
So we have made a Square Waved Tone of the Note "G2"
OneStep() does not build a whole wave, but only one wave-point on the shape of the wave. Related on the Counter it "calculates" the single value that we will add to the CallBack()-Buffer. As we jump into the OneStep() 480 times during one CallBack-loop we create a whole wave of 100Hz, but we would create 2 complete waves of 200Hz, if the Period would be 240 instead of 480.
But it is not the note G2, because G2 is 98Hz and G#2 is 103Hz. We are in the middle between both tones and this would not be acceptable for music. 1% is the maximum deviation audience would accept, 2.5% are already a "quarter tone step"--->worst case. And 5% deviation is already the next tone. So to build G2 we would use 490Hz with a deviation below 0.2%
You cannot say "we send into the generator", because the generator is the sender! We instruct the generator to flip the state when we are telling him the Period length. With this instruction the generator is able to build the wave independent.
...back from Egypt
Baggey
Quoteto find out what is right or wrong, we would need a vintage machine to compare the sounds at the speakers.
One computer i dont have is an Atarti ST 520.
Some Die hards would say its not exact. For me i would go with what sounds best to the ears
The Bassier approach is better i think also. This can be seen as being Improved upon / made better.
Baggey
Running a PC that just Aint fast enough!? i7 4Ghz Quad core 24GB ram 1TB SSD and NVIDIA Quadro K620 . DID Technology stop! Or have we been assimulated!
ZX Spectrum 48k, C64, ORIC Atmos 48K, Enterprise 128K, The SID chip. Im Misunderstood!
Baggey
QuoteAs a developer you can play around with both ideas. I keep all optimizations and basic approaches in the Class.
Could you explain what you mean by Class!?
In English a class is where children sit in front of a teacher? or is this an Americanisation of english?
Kind Regards Baggey
Running a PC that just Aint fast enough!? i7 4Ghz Quad core 24GB ram 1TB SSD and NVIDIA Quadro K620 . DID Technology stop! Or have we been assimulated!
ZX Spectrum 48k, C64, ORIC Atmos 48K, Enterprise 128K, The SID chip. Im Misunderstood!
Midimaster
The expression CLASS comes from the language C. CLASSes do not really exist in BlitzMax. But we can simulate them with TYPEs.
A Class is a encapsulated collection of functions that use a complete independent Name-Room. Means that all "inside used variables" are already defined inside the class. When somebody adds a CLASS to his main code, he need not to care about collisions of variable names, because the CLASS guarantees that their are no collisions.
A lot of the functions and methods of a CLASS are not open to public. PRIVATE means "do not use from outside". Normally also all variables inside a CLASS are private.
To interface with the main code a class hasso called PUBLIC functions or SET/GET-functions to reach inside variables.
Example I:
In cooperation with our YM-TYPE we need not define the Const SQUARE in our main code, because it is already defined inside the TYPE. If you need this Const also outside (in your main code) you can define it twice (without getting any problems) or you could use YM.SQUARE
Example II:
We do not set the Registers with direct code:YM.Register[1]=123 ( ...although it would be technically possible)
but we use a "SET"-function:
`YM.SetRegister 1, 123Type YM Function SetRegister(Adress:Int, Value:Int) Register[Adress]=Value and 255 End Function...`This guarantees that nobody can fill variables with invalid values
This all guarantees that many people can use the Class without the need to change something in their main code.
...back from Egypt
Baggey
Awsome Tutorial, I keep find myself waiting for the next
Ive recently just got my head around Tlist and Types with Arrays. I see you've used them in the GUI Concept with the use of .self.
Something else to learn. I sort of get it but don't fully understand When i get some more time i shall play with this idea.
Didn't realize that noise is mixed in with the normal channels. I think You can also Play just AB and BC or CA out of Left and Right speakers as well
The ZX Spectrum only used Mono though. However there are addons which give Stereo. So it was capable but only had a mono socket? Probably because of the single onboard speaker.
` -- CBA CBARegister 7 = "00|101|001"`
Getting rusty, but i believe this is more todo with how the chips are wired on the motherboard. When a voltage is applied to the chip it can be seen but when not it floats. So they used Pull up and Pull down resistors to get precise Logic levels/Voltage rather than it Float. This could cause unwanted on/off or Some sort off noise. So, The YM would see definate On/Off albeit active Low. And know when to act on the control bits.
This if not emulated or masked could cause unwanted pips/noise in emulation. If the bits aren't masked or are left floating i believe. From residual bits on the Bus/Registers previous use.
Thankyou for your time on this tutorial. Sound being created in the Digital form is something im very interested in understanding.
Kind Regards Baggey
Running a PC that just Aint fast enough!? i7 4Ghz Quad core 24GB ram 1TB SSD and NVIDIA Quadro K620 . DID Technology stop! Or have we been assimulated!
ZX Spectrum 48k, C64, ORIC Atmos 48K, Enterprise 128K, The SID chip. Im Misunderstood!
Baggey
Take on me by A-Ha FRIKEN AWSOME.
Nice Bassy Tone's Better than the real thing!
Cant wait to play with the code.
Kind Regards Lee
Running a PC that just Aint fast enough!? i7 4Ghz Quad core 24GB ram 1TB SSD and NVIDIA Quadro K620 . DID Technology stop! Or have we been assimulated!
ZX Spectrum 48k, C64, ORIC Atmos 48K, Enterprise 128K, The SID chip. Im Misunderstood! | 2,098 | 8,273 | {"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} | 2.984375 | 3 | CC-MAIN-2024-33 | latest | en | 0.842988 |
https://hillgateconnect.com/investment/question-how-to-work-out-investment-return.html | 1,679,740,975,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945323.37/warc/CC-MAIN-20230325095252-20230325125252-00654.warc.gz | 353,505,766 | 10,788 | ## How do I calculate percentage return?
Take the gain or loss from the investment and divide it by the original amount or purchase price of the investment. Finally, multiply the result by 100 to arrive at the percentage change in the investment.
## What is the investment formula?
Investment problems usually involve simple annual interest (as opposed to compounded interest), using the interest formula I = Prt, where I stands for the interest on the original investment, P stands for the amount of the original investment (called the “principal”), r is the interest rate (expressed in decimal form),
## How do you calculate return on investment over multiple years?
The ROI is calculated by dividing the actual profit by the total investment amount and multiplying the result by 100. The resulting number is the percentage by which profit increased or decreased as a result of the investment.
## What is a 100 percent return?
Return on Investment (ROI) is the value created from an investment of time or resources. If your ROI is 100%, you’ve doubled your initial investment. Return on Investment can help you make decisions between competing alternatives.
You might be interested: FAQ: What Is Alternative Investment?
## What is the formula for total return?
The formula for the total stock return is the appreciation in the price plus any dividends paid, divided by the original price of the stock. The income sources from a stock is dividends and its increase in value.
## What is ROI example?
Return on investment (ROI) is calculated by dividing the profit earned on an investment by the cost of that investment. For instance, an investment with a profit of \$100 and a cost of \$100 would have a ROI of 1, or 100% when expressed as a percentage.
## How can I get a 15 return on investment?
The 15*15*15 rule says that one can amass a crore by investing only Rs 15,000 a month for a duration of 15 years in a stock that offers 15% returns per annum. 5
## What is a good return on investment?
According to conventional wisdom, an annual ROI of approximately 7% or greater is considered a good ROI for an investment in stocks. This is also about the average annual return of the S&P 500, accounting for inflation. Because this is an average, some years your return may be higher; some years they may be lower.
## How do you calculate monthly return on investment?
Take the ending balance, and either add back net withdrawals or subtract out net deposits during the period. Then divide the result by the starting balance at the beginning of the month. Subtract 1 and multiply by 100, and you’ll have the percentage gain or loss that corresponds to your monthly return.
## How do you calculate annual rate of return on investment?
Here’s how to calculate annual rate of return:
1. Subtract the initial investment you made at the beginning of the year (“beginning of year price” or “BYP”) from the amount of money you gained or lost at the end of the year (“end of year price” or “EYP.”)2.
2. Multiply the number by 100 to get the percentage.
You might be interested: FAQ: How To Calculate Net Investment?
## What is the difference between return on investment and return of investment?
Return on Investment (ROI) is a ratio between net income(over a period) and investment (investment’s costs then). Meanwhile, Return of Investment is the total gain or loss of an investment over a particularized period, denoted as a percentage of the investment’s initial cost.
## What is a 50% ROI?
Return on investment (ROI) is a profitability ratio that measures how well your investments perform. For example, if you had a net revenue of \$30,000 and your investment cost you \$20,000, your ROI is 0.5 (or 50%). ROI = (gain from investment – cost of investment) / cost of investment. You write ROI as a percentage.
## How much is a 10% return?
Your investment rate of return is the percent increase or decrease in the value of your investment, typically over a one year period. If you invest \$1,000 on January 1 and at the end of the year your investment value is \$1,100, then you’ve earned a 10% rate of return.
## Is 50 a good return on investment?
Having an ROI of 50% on investment can look good by itself, but there’s the context you need to determine how well the investment has done. It’s 50% now, but if it was 70% a year ago, this may not be the solid investment you think it has been. | 963 | 4,419 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-14 | latest | en | 0.941983 |
https://physics.stackexchange.com/questions/176316/to-find-average-speed-of-gas-molecule-when-the-temperature-is-incresed | 1,638,152,225,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358685.55/warc/CC-MAIN-20211129014336-20211129044336-00302.warc.gz | 528,507,440 | 33,897 | # To find average speed of gas molecule when the temperature is incresed
We are given moles of gas let say $n$. Temperature is increased from $T_1$ to $T_2$. Volume is held constant at $V_0$. We need to find the average speed of gas molecule when the temperature $T_2$ is achieved.
I am confused to use $v= \sqrt{\dfrac{3k_BT}{m}}$ or to use $v= \sqrt{\dfrac{8k_BT}{\pi m}}$ ; we are given monatomic gas having atomic $m = 6.4 \times 10^{-25} \:\rm kg$.
Please explain me which formula is applicable here and why.
• Welcome to Physics Stack Exchange! Please note that Physics.StackExchange is not a homework help site. Please see this Meta post on asking homework questions. If you choose to modify your question, please tell us what "Pim" is. Apr 15 '15 at 16:29
• @user36790 Nice work, I didnt understand anything before the edit Apr 15 '15 at 18:47
The distribution of speeds in an ideal gas is given by the Maxwell-Boltzmann distribution. There are a variety of average speeds e.g. the most probable speed, the mean speed and the root mean square speed. Which one you use will depend on the application. The two equations you give are for the RMS speed:
$$\sqrt{\langle v^2 \rangle} = \sqrt{\frac{3kT}{m}}$$
and the mean speed:
$$\langle v \rangle = \sqrt{\frac{8kT}{\pi m}}$$
The question doesn't make it clear what speed is required, but I would guess that it is the mean.
• I am asked to find the average speed of a gas molecule after the final temperature is achieved. Even I am confused to take which one of these? Apr 15 '15 at 17:15
• RMS seems to me more like variation, not the average... Apr 15 '15 at 18:46
• @jaromrax: no, root mean square is a commonly used average. I wonder if you're mixing it up with the standard deviation. Apr 15 '15 at 19:24
• @JohnRennie - well, I dont want to argue with you, ($\sqrt{v^2}$ really resembles $v$), I just feel uncomfortable when I look at $root$ histogram and I see mean and RMS fields with a bit different meaning...that was my point... Apr 15 '15 at 20:48
• @jaromrax - actually the RMS will tell you something about the mean energy of the atoms, since each atom has energy proportional to $v^2$. If I have one atom with v=0 m/s and another with v=2 m/s, the RMS value is $\sqrt{2}$ m/s but the mean velocity is 1 m/s. You get the same mean when both velocities are 1 m/s, but then you would get a smaller RMS velocity (also 1 m/s) - reflecting that the energy of the 0+2 is greater than the 1+1. May 8 '15 at 18:30 | 693 | 2,484 | {"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} | 3.53125 | 4 | CC-MAIN-2021-49 | latest | en | 0.912346 |
http://mathhelpforum.com/algebra/215864-show-equation-have-one-real-root.html | 1,519,136,132,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812959.48/warc/CC-MAIN-20180220125836-20180220145836-00754.warc.gz | 217,376,290 | 11,251 | # Thread: Show that the equation have one real root
1. ## Show that the equation have one real root
Show that the cubic equation x^3 + 3x -15 = 0 has only one real root.
Step-by-step guide is appreciated. Thanks
2. ## Re: Show that the equation has only one real root
$\Delta=-\left[4(3)^3+27(-15)^2\right]<0$.
3. ## Re: Show that the equation has only one real root
Originally Posted by Nehushtan
$\Delta=-\left[4(3)^3+27(-15)^2\right]<0$.
Sorry, I don't understand at all, can you explain, please?
4. ## Re: Show that the equation has one real root
$\Delta=-(4p^3+27q^2)$ is the discriminant of the cubic equation $x^3+px+q=0$.
• If $\Delta > 0$ the equation has three distinct real roots.
• If $\Delta=0$ the equation has three real roots, not all distinct.
• If $\Delta<0$ the equation has one real root and two complex roots.
5. ## Re: Show that the equation have one real root
Alexander
get the function f(x)=x^3+3x-15
the first derivative is f'(x) = 3x^2+3 which is > 0 for all real numbers x .this means that the function f(x) =x^3+3x -15 is a monotone increasing function and it will meet once and only once the x-axis ..Thus the equation x^3+3x-15 = 0 has one and only one real root.
the other two roots wil be conjugate complex numbers .
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# how to show only one real root
Click on a term to search for related topics. | 414 | 1,381 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 7, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2018-09 | longest | en | 0.858746 |
http://www.conversion-website.com/volume/teaspoon-US-to-teaspoon-metric.html | 1,643,198,767,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304947.93/warc/CC-MAIN-20220126101419-20220126131419-00519.warc.gz | 78,019,277 | 4,559 | # Teaspoons (US) to teaspoons (metric) (tsp to tsp)
## Convert teaspoons (US) to teaspoons (metric)
Teaspoons (US) to teaspoons (metric) converter on this page calculates how many teaspoons (metric) are in 'X' teaspoons (US) (where 'X' is the number of teaspoons (US) to convert to teaspoons (metric)). In order to convert a value from teaspoons (US) to teaspoons (metric) (from tsp to tsp) type the number of tsp to be converted to tsp and then click on the 'convert' button.
## Teaspoons (US) to teaspoons (metric) conversion factor
1 teaspoon (US) is equal to 0.985784319 teaspoons (metric)
## Teaspoons (US) to teaspoons (metric) conversion formula
Volume(tsp) = Volume (tsp) × 0.985784319
Example: Calculate how many teaspoons (metric) are in 217 teaspoons (US).
Volume(tsp) = 217 ( tsp ) × 0.985784319 ( tsp / tsp )
Volume(tsp) = 213.915197223 tsp or
217 tsp = 213.915197223 tsp
217 teaspoons (US) equals 213.915197223 teaspoons (metric)
## Teaspoons (US) to teaspoons (metric) conversion table
teaspoons (US) (tsp)teaspoons (metric) (tsp)
1211.829411828
2221.687255018
3231.545098208
4241.402941398
5251.260784588
6261.118627778
7270.976470968
8280.834314158
9290.692157348
102100.550000538
112110.407843728
122120.265686918
132130.123530108
142139.981373298
152149.839216488
teaspoons (US) (tsp)teaspoons (metric) (tsp)
150147.86764785
200197.1568638
250246.44607975
300295.7352957
350345.02451165
400394.3137276
450443.60294355
500492.8921595
550542.18137545
600591.4705914
650640.75980735
700690.0490233
750739.33823925
800788.6274552
850837.91667115
Versions of the teaspoons (US) to teaspoons (metric) conversion table. To create a teaspoons (US) to teaspoons (metric) conversion table for different values, click on the "Create a customized volume conversion table" button.
## Related volume conversions
Back to teaspoons (US) to teaspoons (metric) conversion
TableFormulaFactorConverterTop | 603 | 1,920 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2022-05 | latest | en | 0.574933 |
http://origametria.com/lessoncat/grades/grade-2/ | 1,531,929,873,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590295.61/warc/CC-MAIN-20180718154631-20180718174631-00086.warc.gz | 273,250,208 | 11,336 | #### Triangles activity A
• Lesson Aims
• .The pupils will identify and investigate triangles according to properties
• Name of Model
• A sweets box.
• Lesson content
• In this lesson, the pupils will investigate the properties of triangles: three sides and three vertices, identifying them in different forms as a result of folding activities. At the end of the lesson they see the folding leading to a surprise model. During the activities they are learning new concepts and revising those previously learnt.
• Prior Knowledge
To enter the rate you need to purchase the appropriate lessons package
#### Triangles activity B
• Lesson Aims
• The pupils will be able to identify and investigate triangles according to sides and vertices.
• Name of Model
• A mouse.
• Lesson content
• In this lesson, the pupils will investigate the triangular shapes as formed in the folding process. The classification is made according to the number of sides. At the conclusion of the folding process, the final model comes as a surprise. The pupils learn new concepts and review previously learnt concepts.
• Prior Knowledge
To enter the rate you need to purchase the appropriate lessons package
#### Polygon activity A
• Lesson Aims
• The pupils will fold and identify shapes according to their properties.
The pupils will use paper folding in order to make geometric shapes and investigate their properties.
• Name of Model
• Polygon activity 1
• The Lesson structure
• Lesson content
• In this lesson the pupils investigate polygons that are formed during the folding process. There is a surprise at the end of the lesson. The pupils learn new concepts and refresh their previous learning
• Prior Knowledge
• If during the folding process there are cases of inexact folding by some of the pupils, then these cases will be used to identify differences between the polygons and their properties. The pupils will also learn the concept of polygon sides and vertices.
To enter the rate you need to purchase the appropriate lessons package
#### A measuring tool for comparison of lengths
• Lesson Aims
• The pupils will use the measuring tool for comparison of lengths and identification of right angles.
• Name of Model
• A measuring tool for comparison of length
• The lesson structure
• Lesson content
• Pupils will identify right angles intuitively using the measuring tool.
Pupils will identify right angles in various orientations.
The measuring tool should be folded and kept in the pupils workbook at the conclusion of each lesson.
• Prior Knowledge
• Intuitive identification of right angles.
To enter the rate you need to purchase the appropriate lessons package
#### Polygon activities B 1/2
• Lesson Aims
• The pupils during the folding process will identify polygons according to stated properties. The pupils will learn how to form different polygons with their sheets and investigate the various properties of the folded shapes.
• Model Name
• Polygon activities 1/2
• Lesson content
• The decomposing and composing activities allow the children to create various polygons during the folding process and with the combination of several triangular folded shapes. With a series of questions and answers during the folding process the pupils are able to deepen their geometric reasoning in a dynamic environment.
• Prior knowledge
• Concepts of sides and vertices in polygons.
To enter the rate you need to purchase the appropriate lessons package
#### Polygon activity 3 decomposition and composition 2/2
• Lesson Aims
• The pupils will fold and identify shapes according to their properties.
The pupils will use paper folding in order to make geometric shapes and investigate their properties.
• Name of Model
• A rabbit finger puppet
• The lesson structure
• Lesson content
• In this lesson the pupils investigate polygons that are formed during the folding process. There is a surprise at the end of the lesson. The pupils learn new concepts and refresh their previous learning.
• Prior knowledge
• If during the folding process there are cases of inexact folding by some of the pupils, then these cases will be used to identify differences between the polygons and their properties. The pupils will also learn the concept of polygon sides and vertices.
To enter the rate you need to purchase the appropriate lessons package
#### Polygon activity C
• Lesson Aims
• The pupils will fold and identify shapes according to their properties.
The pupils will use paper folding in order to make geometric shapes and investigate their properties.
• Name of Model
• .The child decides what the final result is
• Lesson content
• In this lesson the pupils investigate polygons that are formed during the folding process. There is a surprise at the end of the lesson. The pupils learn new concepts and refresh their previous learning
• Prior knowledge
• If during the folding process there are cases of inexact folding by some of the pupils, then these cases will be used to identify differences between the polygons and their properties. The pupils will also learn the concept of polygon sides and vertices.
To enter the rate you need to purchase the appropriate lessons package
• Lesson Aims
• The pupils will identify quadrilaterals according to the number of sides.
• Name of Model
• Space ship.
• Lesson content
• Investigation of different quadrilaterals found during the folding process. At the end of the lesson, there is a surprise for the pupils. They learn new geometric concepts and revising previously learnt concepts.
• Prior knowledge
• The concepts of sides and right-angles in polygons.
To enter the rate you need to purchase the appropriate lessons package
#### Quadrilaterals B: Square & Rectangle – same and different
• Lesson Aims
• The pupils will identify the square and rectangle and know the same and different properties of each shape.
• The pupils will fold a rectangle from a square and investigate the lengths of the sides.
• Name of Model
• An envelope
• The lesson structure
• Prior knowledge
• The concepts of vertex, side and polygon
• Stages in the curriculum
To enter the rate you need to purchase the appropriate lessons package
#### Line of reflective symmetry 1/2
• Lesson Aims
• To identify a line of reflective symmetry.
• To discover the properties of a reflective line of symmetry in a given shape.
• To find a line of symmetry in a given shape.
• Model Name
• Line of reflective symmetry 1/2
• Lesson content
• This lesson is the first lesson about the concept of reflective symmetry. There is no final model and its contents are preparation for the following lesson whose model is at the core of the topic of symmetry.
• Prior knowledge
• Polygons. | 1,358 | 6,751 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2018-30 | latest | en | 0.918339 |
https://www.kaysonseducation.co.in/questions/p-span-sty_1846 | 1,674,863,742,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499468.22/warc/CC-MAIN-20230127231443-20230128021443-00521.warc.gz | 847,409,911 | 12,604 | Let [.] denotes the greatest integer function and f (x) = [tan2x], then: : Kaysons Education
# Let [.] Denotes The Greatest Integer Function And f (x) = [tan2x], Then:
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## Question
### Solution
Correct option is
f (x) is continuous at x = 0
Here [.] denotes the greatest integral function, thus
as – 45o < x < 45o
⇒ tan (– 45o) < tan < tan (45o)
⇒ – 1 < tan x < 1
⇒ 0 < tan2 x < 1
Hence, f (x) = [tan2x] = 0
Hence, f (x) is zero for all values of x from (– 45o) to (45o). Thus, f (x) exists when x → 0 and also it is continuous at x = 0, f (x) is differentiable at x = 0 and has a value 0 (i.e., f (0) = 0).
#### SIMILAR QUESTIONS
Q1
Which of the following functions is differentiable at x = 0?
Q2
Let f (x) = [n + p sin x], x ϵ (0, π), n ϵ Z and p is a prime number, where [.] denotes the greatest integer function. Then find the number of points where f (x) is not Differential.
Q3
, then draw the graph of f (x) in the interval [–2, 2] and discuss the continuity and differentiability in [–2, 2]
Q4
Fill in the blank, statement given below let . The set of points where f (x) is twice differentiable is ……………. .
Q5
The function f (x) = (x2 – 1) |x2 – 3x +2| + cos ( | | ) is not differentiable at
Q6
Q7
The number of points in (1, 3), where is not differentiable is:
Q8
Let f and g be differentiable function satisfying g’ (a) = 2, g (a) = b and fog = I (identity function) Then, f’(b) is equal to:
Q9
If the function , (where [.] denotes the greatest integer function) is continuous and differentiable in (4, 6), then.
Q10
Let f be a real function satisfying f (x + z) = f (xf (yf (zfor all real xyz . If f (2) = 4 and f’ (0) = 3. Then find f (0) and f’ (2). | 812 | 2,435 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2023-06 | longest | en | 0.623813 |
http://mathhelpforum.com/differential-equations/49828-differential-equation.html | 1,527,388,054,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867977.85/warc/CC-MAIN-20180527004958-20180527024958-00084.warc.gz | 188,953,552 | 10,009 | 1. ## Differential equation
Hi, I'm having problems with the following differential equation:
$\displaystyle \frac{dy}{dx}=\frac{1}{y}-\frac{3}{4}$
Have you got any ideas?
2. sure, I have a pretty good idea.
first you rearrange the equation like so:
$\displaystyle \frac{{4y}} {{4 - 3y}}dy = dx$
so this is basically a separable equation, all you have to do now is to integrate both sides.
$\displaystyle y + \frac{4} {3}\ln \left| {4 - 3y} \right| = - \frac{3} {4}x + C$
I doubt that an explicit form of the solution can be obtained.
3. Originally Posted by Aliquantus
Hi, I'm having problems with the following differential equation:
$\displaystyle \frac{dy}{dx}=\frac{1}{y}-\frac{3}{4}$
Have you got any ideas?
$\displaystyle \frac{4ydy}{4-3y}=dx$
Hey, I think that turns into an interesting problem: Express the solution as y=f(x).
4. You have to seperate the two variables such that all the y's are on one side and all the x's are on the other....so use cross multiplication...you should end up with this:
$\displaystyle \frac{4y}{4-3y}\;dy=dx$
Then you integrate both sides: $\displaystyle 4\int\frac{y}{4-3y}\;dy=\int dx$
Make the substitution $\displaystyle u=4-3y$...rest follows
Should end up with this: $\displaystyle \frac{4}{3}(4-3y)-\frac{8}{3}\ln{|4-3y|}=x+C$ | 398 | 1,290 | {"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} | 4.40625 | 4 | CC-MAIN-2018-22 | latest | en | 0.799556 |
https://kgdq.info/factorisation-de-cholesky-35/ | 1,618,732,536,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038469494.59/warc/CC-MAIN-20210418073623-20210418103623-00365.warc.gz | 440,477,793 | 12,031 | FACTORISATION DE CHOLESKY PDF
This MATLAB function produces an upper triangular matrix R from the diagonal and upper triangle of matrix A, satisfying the equation R’*R=A. by Daidalos on November 04, Exemple de factorisation de Cholesky avec python et scipy en reprenant l’example de wikipedia: >>> import numpy as np. Following on from the article on LU Decomposition in Python, we will look at a Python implementation for the Cholesky Decomposition method, which is used in .
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When it is applicable, the Cholesky decomposition is roughly twice as efficient as the LU decomposition for solving systems of linear equations. The expression under the square root is always positive if A is real and positive-definite. This matrix is interesting because its Cholesky factor consists of the same coefficients, arranged in an upper triangular matrix.
If the matrix being factorized is positive definite as required, the numbers under the square roots are always positive in exact arithmetic.
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If it is not, chol uses the complex conjugate transpose of the lower triangle as the upper triangle. Click here to see To view all translated materials including this page, select Country from the country navigator on the bottom of this page.
The gallery function provides several symmetric, positive, definite matrices.
Cholesky factorization – MATLAB chol
Consider the operator matrix. The optional lower parameter allows us to determine whether a lower or upper triangular matrix is factorixation Furthermore, no pivoting is necessary, and the error will always be small.
Unfortunately, the numbers can become negative because of round-off errorsin which case the algorithm cannot continue. Code generation does not support sparse matrix inputs for ve function. Trial Software Product Updates. From Wikipedia, the free encyclopedia. The LDL variant, if efficiently implemented, requires the same space and computational complexity to construct and use but avoids extracting square roots. The code for the factroisation update shown above can easily be adapted to do a rank-one downdate: The inverse problem, when we have.
The automated translation of this page is provided by a general purpose third party translator tool. In particular, it makes an appearance in Monte Carlo Methods where it is used to simulating systems with correlated variables.
Hence, the lower triangular matrix L we are looking for is calculated as. Linear equations Matrix decompositions Matrix multiplication algorithms Matrix splitting Sparse problems.
Cholesky Decomposition in Python and NumPy | QuantStart
Cambridge University Cholesly EPress. If Choleskt is positive semidefinite in the sense that for all finite k and for any.
One way to address this is to add a diagonal correction matrix to the matrix being decomposed in an attempt to promote the positive-definiteness. If A is real, the following recursive relations apply for the entries of D and L:. Following on from the article on LU Decomposition in Pythonwe will look at a Python implementation for the Cholesky Decomposition method, which is used in certain quantitative finance algorithms.
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Applying this to a vector of uncorrelated samples u produces a sample vector Lu with the covariance properties factoriswtion the system being modeled. The Art of Scientific Computing second ed. All articles with unsourced statements Articles with unsourced statements dholesky February Articles with unsourced statements from June Articles with fatorisation statements from October Articles with French-language external links. The input A must be nonsparse. Usage notes and limitations: The Cholesky algorithmused to calculate the decomposition matrix Lis a modified version of Gaussian elimination.
Translated by Mouseover text to see original. Select the China site in Chinese or English for best site performance.
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How to implement advanced trading strategies using time series analysis, machine learning and Bayesian statistics with R and Python. Fundamentals of Matrix Computations. Choose a web site to get translated content where available and see local events and offers. How to find new trading strategy ideas and objectively assess them for your portfolio using a custom-built backtesting engine in Python. It may also happen that matrix A comes from an energy functional, which must be cholexky from physical considerations; this happens frequently in the numerical solution of partial differential equations.
If it is not, chol uses the complex conjugate transpose of the upper triangle as the lower triangle.
There are various methods for chloesky the Cholesky decomposition. However, the decomposition need not be unique when A is positive semidefinite. Now, suppose that the Cholesky decomposition is applicable. | 1,066 | 5,438 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-17 | latest | en | 0.802438 |
https://ar5iv.labs.arxiv.org/html/1205.6113 | 1,713,645,710,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817674.12/warc/CC-MAIN-20240420184033-20240420214033-00893.warc.gz | 85,998,771 | 116,487 | # Combined Preconditioning with Applications in Reservoir Simulation
Xiaozhe Hu Department of Engineering Mechanics, Kunming University of Science and Technology, China; Department of Mathematics, The Pennsylvania State University, University Park, PA 16802. Email: hu_x@math.psu.edu Shuhong Wu RIPED, PetroChina Company Limited, China. Email: wush@petrochina.com.cn Xiao-Hui Wu ExxonMobil Upstream Research Company, USA. Email: xiao-hui.wu@exxonmobil.com Jinchao Xu Department of Mathematical Sciences, The Pennsylvania State University, University Park, PA 16802. Email: xu@math.psu.edu. The research of this author was supported in part by NSF Grant DMS-0915153, DOE Grant DE-SC0006903, LLNL Subcontract B596235, and NSFC Grant 91130011/A0117. Chen-Song Zhang NCMIS & LSEC, Academy of Mathematics and System Sciences, Beijing, China 100190. Email: zhangcs@lsec.cc.ac.cn. This author is supported by the Dean Startup Fund of NCMIS, Academy of Mathematics and System Sciences. Shiquan Zhang Department of Mathematics, Sichuan University, China. Email: shiquanz3@gmail.com Ludmil Zikatanov Department of Mathematics, The Pennsylvania State University, University Park, PA 16802. Email: ltz@math.psu.edu
###### Abstract
We develop a simple algorithmic framework to solve large-scale symmetric positive definite linear systems. At its core, the framework relies on two components: (1) a norm-convergent iterative method (i.e. smoother) and (2) a preconditioner. The resulting preconditioner, which we refer to as a combined preconditioner, is much more robust and efficient than the iterative method and preconditioner when used in Krylov subspace methods. We prove that the combined preconditioner is positive definite and show estimates on the condition number of the preconditioned system. We combine an algebraic multigrid method and an incomplete factorization preconditioner to test the proposed framework on problems in petroleum reservoir simulation. Our numerical experiments demonstrate noticeable speed-up when we compare our combined method with the standalone algebraic multigrid method or the incomplete factorization preconditioner.
###### keywords:
Combined preconditioning, Multigrid method, Incomplete LU factorization, Reservoir Simulation
###### AMS:
65F08, 65F10, 65N55, 65Z05
## 1 Introduction
We consider a sparse linear system of equations that arises from the discretizations of the elliptic partial differential equations (PDEs) used to simulate the physical processes in petroleum reservoirs. The Petroleum Reservoir Simulation (PRS) is a tool for predicting hydrocarbon reservoir performance under various operating regimes. PRS helps engineers to obtain information pertaining to the processes that take place within oil reservoirs – information that can be used to maximize recovery and minimize environmental damage. The crucial aspect of PRS is its ability to solve large-scale discretized PDEs, which are strongly coupled, indefinite and often non-symmetric. Solving these linear systems consumes most of the computational time in all modern reservoir simulators – more than 75% of the computational time in general. Furthermore, the demand for more accurate simulations has led to larger discrete reservoir models. And, this increase in model size results in larger linear systems, that are more difficult, or even impossible to solve in an acceptable amount of time using standard direct or standard iterative solvers.
Over the last 30 years, incomplete LU factorization (ILU) has become one of the most commonly used methods for solving large sparse linear equations arising in PRS. First developed in the 1960s [6, 27], ILU methods provide an approximation of the exact LU factorization (computed via Gaussian elimination) by specifying the sparsity of the $L$ and $U$ factors. These methods need not be convergent when used in linear iterative procedures, but they do provide preconditioners that are used in Krylov subspace methods. Thought to have introduced the word preconditioning, Evans [11] may also have been the first to use ILU as a preconditioner. As noted, due to their simplicity, ILU methods are of particular interest to researchers in the field of reservoir simulation. However, when ILU preconditioners are applied to problems in PRS, their performance usually deteriorates as the number of grid-blocks increases.
To solve discretized scalar elliptic PDEs (Poisson-like PDEs), multigrid (MG) methods are efficient and provide a solution in optimal time and memory complexity in cases that allow standard coarse spaces to be used (see [13, 31, 2, 25] and references therein for details). However, the multigrid methods with standard coarse spaces make extensive use of the analytic information from the discretized equation and geometric information explicitly related to the discretization. This makes such methods difficult to use, such that in practice more sophisticated methods, such as algebraic multigrid (AMG) methods are preferred. There are many different types of AMG methods – the classical AMG ([22, 4]), smoothed aggregation AMG ([26, 5]), AMGe ([16, 18]), etc. However, despite their differences, they generally do not require geometric information from the grids. Interested readers can refer to [20, 29, 3, 8, 12, 28] and references therein for details on AMG methods. Due to their efficiency, scalability, and applicability, AMG methods have become increasingly popular in practice (see e.g., [24]) including in modern petroleum reservoir simulations [23, 9, 10, 15, 14]. However, the performance and efficiency of MG methods may degenerate as the physical and geometric properties of the problems become more complex. In such circumstances, there is an increase in AMG-setup time (the time needed to construct coarse spaces and operators) and superfluous fill-in in the coarse grid operators, which makes applying relaxation more expensive on coarser levels.
In this paper, we provide a simple and transparent framework for combining preconditioners like ILU or Jacobi or additive Schwarz preconditioners with AMG or another norm-convergent iterative method for large-scale symmetric positive definite linear system of equaitons. In the combined preconditioners we propose, the component provided by the norm-convergent iterative method need not be very effective when used alone. However, as we shall see, when it is paired with an existing preconditioner (such as an ILU preconditioner), the result is an efficient method. We show that the combined preconditioner is positive definite (SPD) under some mild assumptions. This guarantees the convergence of the conjugate gradient (CG) method for the preconditioned system. We apply the combined preconditioner to petroleum reservoir simulations that involve highly heterogeneous media and unstructured grids in three spatial dimensions (with faults and pinch-outs). The numerical results show an improved performance that justifies using the combined preconditioners in PRS.
The rest of the paper is organized as follows. The algorithmic framework is introduced and analyzed in Section 2. As an example in Section 3, we formally present the combined preconditioner by describing its essential components, e.g., an ILU preconditioner and an AMG method. In Section 4, we recall the mixed-hybrid finite element method used to provide a discretization of an important component of reservoir simulation – an elliptic PDE with heterogenous coefficients. Numerical results showing the improvement in performance are presented in Section 5, and some concluding remarks are given in Section 6.
## 2 Definition of the combined preconditioner
Consider the linear system
(1) $Au=f,$
with an SPD coefficient matrix $A$. Let $(\cdot,\cdot)$ be an inner product on a finite-dimensional Hilbert space $V$; its induced norm is $\|\cdot\|$. The adjoint of $A$ with respect to $(\cdot,\cdot)$, denoted by $A^{T}$, is defined by $(Au,v)=(u,A^{T}v)$ for all $u,v\in V$. $A$ is SPD if $A^{T}=A$ and $(Av,v)>0$ for all $v\in V\backslash\{0\}$. As $A$ is SPD with respect to $(\cdot,\cdot)$, the bilinear form $(A\cdot,\cdot)$ defines an inner product on $V$, denoted by $(\cdot,\cdot)_{A}$, and the induced norm of $A$ is denoted by $\|\cdot\|_{A}$.
In what follows, we also need an operator $S$, referred to as a smoother. In some of the results, we distinguish two cases: (1) norm-convergent smoother, i.e., $(I-SA)$ is a contraction in $\|\cdot\|_{A}$-norm; and (2) more generally, a non-expansive smoother, i.e., $(I-SA)$ is a non-expansive operator in $\|\cdot\|_{A}$ norm. Note that norm-convergent implies non-expansive.
Next, in Algorithm 2.1, we combine a non-expansive iterative method $S$ with an existing SPD preconditioner $B$, and as a result we get a preconditioner $B_{\text{co}}$.
It is easy to see that Algorithm 2.1 defines a linear operator $B_{\text{co}}$ and that
(2) $I-B_{\text{co}}A=(I-S^{T}A)(I-BA)(I-SA).$
From (2), it follows that
$\displaystyle B_{\text{co}}$ $\displaystyle=\widetilde{S}+(B-S^{T}AB-BAS+S^{T}ABAS)$ (3) $\displaystyle=\widetilde{S}+(I-S^{T}A)B(I-AS).$
Here $\widetilde{S}$ is the symmetrization of $S$, defined as
$I-\widetilde{S}A=(I-S^{T}A)(I-SA),\quad\quad\widetilde{S}=S+S^{T}-S^{T}AS.$
We now explain the properties pertaining to the notions of norm-convergent and non-expansive smoothers, which we refer to in the subsequent sections.
• Norm-convergent smoother. In this case, $(I-SA)$ is a contraction in $\|\cdot\|_{A}$-norm; i.e., there exists $\rho\in[0,1)$, such that for all $v\in V$ we have
(4) $\|(I-SA)v\|^{2}_{A}\leq\rho\|v\|^{2}_{A}.$
Note that under this assumption, $S$ is invertible. Indeed, we have,
$\|SAv\|_{A}=\|(I-(I-SA))v\|_{A}\geq\|v\|_{A}-\|(I-SA)v\|_{A}\geq(1-\sqrt{\rho})\|v\|_{A},$
which means $SA$ is coercive and implies that $S$ is invertible. Recall also that (4) holds if and only if the symmetrization $\widetilde{S}$ is SPD; i.e.,
$\|I-SA\|_{A}<\rho,\quad\mbox{if and only if}\quad(\widetilde{S}v,v)>0,\quad\mbox{for all}\quad v\in V\setminus\{0\}.$
• Non-expansive smoother. This is a more general case in which we assume that
(5) $\|(I-SA)v\|^{2}_{A}\leq\|v\|^{2}_{A}\quad\forall v\in V.$
One important difference between these two cases is this: with the non-expansive case, $\widetilde{S}$ can be positive semi-definite, whereas in the norm-convergent case, $\widetilde{S}$ is positive definite.
The following result shows that the operator $B_{\text{co}}$ is SPD and, therefore, can be employed as a preconditioner for the preconditioned conjugate gradient (PCG) method. A preliminary version of the following theorem can be found in [33].
###### Theorem 1.
Assume that $S:V\rightarrow V$ is a non-expansive smoother. Moreover, the operator $B:V\rightarrow V$ is SPD. Then, $B_{\text{co}}$ defined in (2) is SPD.
###### Proof.
For any $x\in V$, $x\neq 0$, we define $\tilde{x}:=(I-SA)x$. Thus, we obtain
$\displaystyle((I-B_{\text{co}}A)x,x)_{A}$ $\displaystyle=$ $\displaystyle((I-BA)(I-SA)x,(I-SA)x)_{A}$ $\displaystyle=$ $\displaystyle((I-BA)\tilde{x},\tilde{x})_{A}=(\tilde{x},\tilde{x})_{A}-(A\tilde{x},A\tilde{x})_{B}$
where $(\cdot,\cdot)_{B}$ is an inner product on $V$ defined as $(\cdot,\cdot)_{B}:=(B\cdot,\cdot)$ since $B$ is SPD.
Further, if $\tilde{x}=0$, then $((I-B_{\text{co}}A)x,x)_{A}=0<\|x\|_{A}^{2}$. On the other hand, for $\tilde{x}\neq 0$, we obtain from $(A\tilde{x},A\tilde{x})_{B}>0$ that
$\|x\|_{A}^{2}-(B_{\text{co}}Ax,x)_{A}=((I-B_{\text{co}}A)x,x)_{A}<\|\tilde{x}\|_{A}^{2}\leq\|x\|_{A}^{2},\quad\Longrightarrow\quad(B_{\text{co}}Ax,Ax)>0.$
As $x\in V$ is an arbitrary nonzero element, and $A$ is non-singular, we conclude that $B_{\text{co}}$ is SPD. Finally, in view of (3) and because both $A$ and $B$ are symmetric, it is easy to see that $B_{\text{co}}$ is symmetric. This completes the proof. ∎
###### Remark 2.2.
Note that the assumption in the theorem is that the error transfer for the convergent iteration $I-SA$ is non-expansive in the $A$ norm and does not require $I-SA$ to be a contraction.
###### Remark 2.3.
Theorem 1 shows that the operator $B_{\text{co}}$ is SPD. Hence, we can use the PCG method to solve (1) with $B_{\text{co}}$ as a preconditioner, and in exact arithmetic, PCG method will always be convergent. Algorithm 2.1 is relatively simple, but the order in which $S$ and $B$ are applied is crucial to the positive definiteness property of $B_{\text{co}}$. For example, consider the operator $\overline{B}$ defined by
(6) $I-\overline{B}A=(I-BA)(I-\widetilde{S}A)(I-BA).$
In this case, it is not true in general that $\overline{B}$ is positive definite. For example, if we use the ILU method to define a preconditioner $B$, finding the right scaling to ensure the positive definiteness of $\overline{B}$ could be a difficult task, and with indefinite $\overline{B}$ the PCG will not converge.
###### Remark 2.4.
Note that directly using Algorithm 2.1 as an iterative method may not result in a convergent method. This is because though we assume that $B$ is only SPD, we do not assume that $I-BA$ is a contraction. The contraction property of $I-BA$ (and also of $I-B_{\text{co}}A$) is that any eigenvalue of $BA$ satisfies $0<\lambda(BA)\leq\omega<2$. However, we do not assume that such a contraction property holds for $B$.
### 2.1 Other derivations of the combined preconditioner
Here, we present two distinct points of view that can lead to algorithms such as Algorithm 2.1. One is from the point of view of the fictitious or auxiliary space techniques, developed in [19] and [32], and another is from the point of view of the block factorization techniques’s, developed in [28, Chapter 5]. Below, we derive Algorithm 2.1 using these techniques. Of course, all three derivations lead to one and the same method and they are equivalent to each other. However, they present different points of view and help us understand different aspects of the method.
#### 2.1.1 Derivation via the auxiliary space method
An additive version of the Algorithm 2.1 was derived via fictitious or auxiliary space techniques developed in [19] and [32]. Equivalently, Algorithm 2.1 can be viewed as a successive or multiplicative version of the auxiliary space additive preconditioner described below. Here, we use only one auxiliary space, which will turn out to be the same as $V$, but with a different inner product. We define
(7) $\overline{V}=V\times W_{1},$
where $W_{1}$ is an auxiliary (Hilbert) space with an inner product
$\overline{a}_{1}(\cdot,\cdot)=(\cdot,\cdot)_{A_{1}}.$
We also introduce an operator $\Pi_{1}:W_{1}\mapsto V$ in order to define the additive preconditioner, and define the latter as follows:
(8) $\widehat{B}=\widetilde{S}+\Pi_{1}A_{1}^{-1}\Pi_{1}^{T}.$
A distinctive feature of the auxiliary space method is the presence of $V$ in (7) as a component of $\overline{V}$ and the presence of the symmetric positive definite operator $\widetilde{S}:V\mapsto V$. $\widetilde{S}$ is assumed to be SPD in order to guarantee that $\widehat{B}$ is SPD and can be applied as a preconditioner for the PCG method. It can be proved that the condition number of the preconditioned system can be bounded as follows (see [32]):
(9) $\kappa(\widehat{B}A)\leq c_{0}^{2}(c_{s}^{2}+c_{1}^{2}),$
if
$\|\Pi_{1}w_{1}\|^{2}_{A}\leq c_{1}^{2}\|w_{1}\|^{2}_{A_{1}},\qquad w_{1}\in W_{1},$
and
$(\widetilde{S}Av,v)_{A}\leq c_{s}^{2}(v,v)_{A},\qquad\forall v\in V.$
Moreover, for each $v\in V$, there exists $w_{1}\in W_{1}$ such that
$v=v_{0}+\Pi_{1}w_{1},\quad v_{0}=v-\Pi_{1}w_{1},\quad\text{and}\quad(\widetilde{S}Av_{0},v_{0})_{A}+(w_{1},w_{1})_{A_{1}}\leq c_{0}^{2}(v,v)_{A}.$
Taking $W_{1}=V$ but with the inner product defined by $B$, $A_{1}^{-1}=B$, and $\Pi=I$, gives the additive version $\widehat{B}$ of the preconditioner $B_{\text{co}}$. Note that when $W_{1}=V$, $B$ is SPD and $\widetilde{S}$ is the symmetrization of the smoother $S$, the preconditioner $\widehat{B}$ is SPD for both the norm-convergent smoother and the non-expansive smoother.
#### 2.1.2 Derivation via block factorization
In this section, we present a derivation following the lines in [28, Chapter 5]. Let us introduce the operator $\accentset{\bm{=}}{B}:V\times V\mapsto V\times V$ in block factored form:
$\accentset{\bm{=}}{B}=\left(\begin{matrix}I&-S^{T}A\\ 0&I\end{matrix}\right)\left(\begin{matrix}\widetilde{S}&0\\ 0&B\end{matrix}\right)\left(\begin{matrix}I&0\\ -AS&I\end{matrix}\right).$
We then set
(10) $\accentset{\bm{\approx}}{B}=\begin{matrix}&\\ (I,&I)\\ &\end{matrix}\accentset{\bm{=}}{B}\left(\begin{matrix}I\\ I\end{matrix}\right),\qquad\accentset{\bm{\approx}}{B}:V\mapsto V.$
A straightforward calculation shows that $B_{\text{co}}$ and $\accentset{\bm{\approx}}{B}$ are the same. Regarding the positive definiteness of $\accentset{\bm{\approx}}{B}$, note that for the norm-convergent smoother, it is immediately evident that $\accentset{\bm{\approx}}{B}$ is SPD, because both $B$ and $\widetilde{S}$ are SPD. For the non-expansive smoother, the positive-definiteness of $\accentset{\bm{\approx}}{B}$ does not immediately follow from the form of the preconditioner given in (10). The arguments of Theorem 1 (or similar) are needed to conclude that $\accentset{\bm{\approx}}{B}$ is SPD.
### 2.2 Effectiveness of $B_{\text{co}}$
In this section, we show that the combined preconditioner, under suitable scaling assumptions performs no worse than its components. As the numerical tests show (see Section 5), the combination of ILU (for $B$) and AMG (for $S$) performs significantly better than its components. Let us set
(11) $m_{1}=\lambda_{\max}(BA)\quad\mbox{and}\quad m_{0}=\lambda_{\min}(BA).$
Without loss of generality (with proper scaling), we can assume that the preconditioner $B$ is such that the following inequalities hold:
(12) $m_{1}>1\geq m_{0}>0.$
We now prove a result that compares $\kappa(B_{\text{co}}A)$ with $\kappa(\widetilde{S}A)$ and $\kappa(BA)$ under the assumptions that $B$ and $S$ are such that both (12) and (4) are satisfied.
###### Theorem 2.5.
If $S$ is a norm-convergent smoother and $B_{\text{co}}$ is defined as in (2), then
(13) $\kappa(B_{\text{co}}A)\leq\frac{(1-m_{1})(1-\rho)+m_{1}}{(1-m_{0})(1-\rho)+m_{0}},$
and
(14) $\kappa(B_{\text{co}}A)<\kappa(BA).$
Furthermore, if $S$ is such that (4) holds with $\rho\geq 1-\frac{m_{0}}{m_{1}-1}$, then
(15) $\kappa(B_{\text{co}}A)\leq\kappa(\widetilde{S}A).$
###### Proof 2.6.
From the assumption stated in (12), we immediately conclude that $B$ is SPD and $\kappa(BA)=m_{1}/m_{0}$. By the definition of $\widetilde{S}$, we have
$0\leq((I-\widetilde{S}A)w,w)_{A}=((I-SA)w,(I-SA)w)_{A}=\|(I-SA)w\|^{2}_{A}\leq\rho\|w\|^{2}_{A},$
where we have used the assumption of the convergence of $S$ in the last inequality. By choosing $v=Aw$, we can obtain
(16) $(1-\rho)(A^{-1}v,v)\leq(\widetilde{S}v,v)\leq(A^{-1}v,v).$
On the other hand, as $m_{1}=\lambda_{\max}(BA)$ and $m_{0}=\lambda_{\min}(BA)$, we have
$m_{0}(A^{-1}v,v)\leq(Bv,v)\leq m_{1}(A^{-1}v,v).$
By the definition of $B_{\text{co}}$ (3), we have
$\displaystyle(B_{\text{co}}v,v)$ $\displaystyle=(\widetilde{S}v,v)+(B(I-AS)v,(I-AS)v)$ $\displaystyle\leq(\widetilde{S}v,v)+m_{1}(A^{-1}(I-AS)v,(I-AS)v)$ $\displaystyle=(1-m_{1})(\widetilde{S}v,v)+m_{1}(A^{-1}v,v).$
Similarly, we can derive that
$\displaystyle(B_{\text{co}}v,v)$ $\displaystyle=(\widetilde{S}v,v)+(B(I-AS)v,(I-AS)v)$ $\displaystyle\geq(\widetilde{S}v,v)+m_{0}(A^{-1}(I-AS)v,(I-AS)v)$ $\displaystyle=(1-m_{0})(\widetilde{S}v,v)+m_{0}(A^{-1}v,v).$
As $m_{1}>1\geq m_{0}$, we have
$[(1-m_{0})(1-\rho)+m_{0}](A^{-1}v,v)\leq(B_{\text{co}}v,v)\leq[(1-m_{1})(1-\rho)+m_{1}](A^{-1}v,v);$
then the condition number of $B_{\text{co}}A$ is bounded by
$\kappa(B_{\text{co}}A)\leq\frac{(1-m_{1})(1-\rho)+m_{1}}{(1-m_{0})(1-\rho)+m_{0}}.$
If $m_{1}>1$, then
$(1-m_{1})(1-\rho)+m_{1}
and if $1\geq m_{0}>0$, then
$(1-m_{0})(1-\rho)+m_{0}\geq m_{0}.$
Note that
$\kappa(B_{\text{co}}A)\leq\frac{(1-m_{1})(1-\rho)+m_{1}}{(1-m_{0})(1-\rho)+m_{0}}<\frac{m_{1}}{m_{0}}=\kappa(BA).$
Hence, the inequality (14) holds if $m_{1}>1\geq m_{0}>0$.
On the other hand, (15) follows from
(17) $\frac{(1-m_{1})(1-\rho)+m_{1}}{(1-m_{0})(1-\rho)+m_{0}}\leq\frac{1}{1-\rho}=\kappa(\widetilde{S}A),$
where the last equality comes from (16). Note that
$\frac{(1-m_{1})(1-\rho)+m_{1}}{(1-m_{0})(1-\rho)+m_{0}}=\frac{1+\rho(m_{1}-1)}{(1-\rho)+\rho m_{0}}.$
Note that $\frac{a+c}{b+d}\leq\frac{a}{b}$ if $\frac{c}{d}\leq\frac{a}{b}$ and $a,b,c,d>0$. Therefore,(17) holds if $\frac{m_{1}-1}{m_{0}}\leq\frac{1}{1-\rho}$, i.e., $\rho\geq 1-\frac{m_{0}}{m_{1}-1}$.
###### Remark 2.7.
If $\frac{m_{1}-1}{m_{0}}<\frac{1}{1-\rho}$, then (15) becomes $\kappa(B_{\text{co}}A)<\kappa(\widetilde{S}A)$.
###### Remark 2.8.
If either $\widetilde{S}$ or $B$ works well as a preconditioner itself, there is no need to use the more complicated $B_{\text{co}}$. However, we are interested in cases in which neither $\widetilde{S}$ nor $B$ works effectively alone. In these cases, we use $B_{\text{co}}$ to combine these two, and we use Theorem 2.5 to guarantees that $B_{\text{co}}$ will be a better preconditioner than either $B$ or $\widetilde{S}$ under reasonable conditions. The condition
$\rho\geq 1-m_{0}/(m_{1}-1)$
means that both $\widetilde{S}$ and $B$ yield slow convergence as
$\rho\geq 1-m_{0}/(m_{1}-1)\approx 1-\kappa(BA)^{-1}\quad\text{if}\ m_{1}>>1.$
Therefore, if $B$ is not a good preconditioner, i.e., $\kappa(BA)$ is large, then $\rho\approx 1$, which implies that $\widetilde{S}$ does not work well either. The new preconditioner $\widetilde{B}$ is no worse than $B$ or $\widetilde{S}$ alone. In fact, the new preconditioner may be capable of performing much better than either $B$ or $\widetilde{S}$ based on our numerical experiments (see Section 5).
## 3 A simple example: ILU+AMG
Algorithm 2.1 provides an approach to combining ILU and AMG methods. In the combined use of ILU and AMG for the linear system (1), AMG serves as $S$ and ILU serves as $B$. Next, we specify our choice of ILU and AMG for problems specific to reservoir simulation. However, we would like to emphasize that the combined preconditioner works for a wide range of iterative methods and preconditioners as long as they satisfy the assumptions in Theorem 1.
### 3.1 Incomplete LU factorization
Incomplete LU factorizations compute a sparse lower triangular matrix $L$ and a sparse upper triangular matrix $U$ so that the residual matrix $R=A-LU$ satisfies certain conditions. A general algorithm of ILU can be obtained by Gaussian elimination and dropping some of the elements in the off-diagonal positions. There are many variants of ILU preconditioners. They differ in terms of the rules that govern the prescribed fill-in of the factors during the ILU factorization procedure. For example, discarding the fill-in based on position gives the ILU($k$) method; discarding the fill-in based on the values of the corresponding entries in the $L$ or $U$ factors gives the threshold ILU method. There are also ILU methods for which the fill-in is managed based on a combination of positions and values or based on other dropping strategies. For details about the different ILU methods, we refer to monographs [1, 7] and Saad [21]. Here we use the notation and terminology from Saad [21].
We consider ILU($k$) (Algorithm 3.1), which to our knowledge was originally introduced for reservoir simulations in [30]. The ILU is based on the level of fill to determine the off-diagonal positions in the $L$ and $U$ factors where the entries fill-in will not be introduced (or as is often said, off-diagonal positions for which the fill-in entries are dropped). Next, we define level of fill, and the detailed algorithm is given in Algorithm 3.1.
###### Definition 3.9(Level of fill).
The initial level of fill of the elements of a sparse matrix $A$ defined as
$\mathcal{L}_{ij}=0\ \text{if}\ a_{ij}\neq 0\ \text{or}\ i=j,\ \text{otherwise}\ \mathcal{L}_{ij}=\infty.$
When an entry $a_{ij}$ is updated in the factorization procedure $a_{ij}:=a_{ij}-a_{ik}*a_{kj}$, its level of fill is also updated by
$\mathcal{L}_{ij}=\min\{\mathcal{L}_{ij},\,\mathcal{L}_{ik}+\mathcal{L}_{kj}+1\}.$
###### Remark 3.10.
Although the level of fill depends on the location of the element, the rationale is that the level of fill should indicate the magnitude of the element. The higher the level of fill, the smaller the element. For example, when the level of fill of $a_{ij}$ is $k$ this means that the size is $|a_{ij}|=O(\epsilon^{k})$ for some $\epsilon<1$.
### 3.2 Algebraic multigrid methods
As an algebraic variant of MG methods, AMG methods are widely applicable and the focus of current intensive development. AMG methods have mesh-independent convergence rates and optimal computational complexity for a wide range of problems.
Any AMG method consists of two phases: the SETUP phase and the SOLVE phase. In the SETUP phase, the intergrid operator $P_{l}$ is constructed, and the coarse grid matrix is defined as
$A_{l}=P_{l}^{T}A_{l+1}P_{l},\quad l=L-1,\ldots,0,$
and $A_{L}=A$. In the SOLVE phase, the smoother $S_{l}$ and coarse-grid correction are applied recursively as shown in the following general V-cycle MG Algorithm 3.2.
In AMG methods, $P_{l}$ must be constructed based on algebraic principles, which presents certain challenges. The key to fast convergence is the complementary nature of operators $S_{l}$ and $P_{l}$. That is, errors not reduced by $S_{l}$ must be interpolated well by $P_{l}$. We choose the classical AMG as the iterative method $S$ in our Algorithm 2.1. This method constructs the intergrid operator $P_{l}$ in two steps. First, the classical C-F splitting method is used, and then the operator $P_{l}$ is constructed by classical Ruge-Stüben interpolation; see [20] for details.
### 3.3 Application of Algorithm 2.1
In this paper, we consider SPD coefficient matrices $A$ only. In this case, ILU($k$) is replaced by an Incomplete Cholesky (IC) factorization. It is easy to see that IC($k$) is SPD, which satisfies the assumption on $B$ in Theorem 1. For AMG methods, we have $A_{l}=P_{l}^{T}A_{l+1}P_{l}$, $l=L-1,\ldots,0$. $A$ is SPD and $P_{l}$ is constructed in the classical AMG method; therefore, it is guaranteed that $A_{l}$ is also SPD for $l=0,1,\ldots,L$. Hence, it is easy to see that using the standard Gauss-Seidel (GS) smoother on each level gives a convergent iterative method. According to Algorithm 3.2, $B_{l}$, which stands for V-cycle MG on level $l$ ($l=1,2,\ldots,L$), can be defined recursively as
$I-B_{l}A_{l}=(I-S_{l}^{T}A_{l})(I-P_{l-1}B_{l-1}P_{l-1}^{T}A_{l})(I-S_{l}A_{l}),$
with $B_{0}=A_{0}^{-1}$. As
$\|I-S_{l}A_{l}\|_{A_{l}}<1\ \text{and}\ \|I-P_{l-1}B_{l-1}P_{l-1}^{T}A_{l}\|_{A_{l}}\leq 1,$
by induction assumption, we can show that
$\|I-B_{l}A_{l}\|_{A_{l}}<1.$
Thus the classical AMG is convergent for SPD problems by mathematical induction. Hence, ILU($k$) and the classical AMG can be combined using Algorithm 2.1 to yield the following corollary:
###### Corollary 3.11(Symmetric Positive Definiteness of ILU-AMG).
If we choose $S$ from the classical AMG methods and $B$ from ILU($k$) in Algorithm 2.1, then the operator $B_{\text{co}}$ defined in (2) is SPD.
## 4 A model problem in reservoir simulation
Petroleum reservoir simulation provides information about processes that take place within oil reservoirs. It is, therefore, of great assistance in efforts to achieve optimal recovery. Modern reservoir simulation faces increasingly complex physical models and uses highly unstructured grids. This results in Jacobian systems that are more difficult at each step. In this section, we describe a model problem in PRS and a discretization method that has great promise for efficient resource recovery, which is used for numerical tests in the next section.
Let $\Omega$ be a bounded domain in $\mathbb{R}^{3}$ in our consideration of the following second-order elliptic problem:
(18) $-\nabla\cdot(a\nabla p)+cp=f\quad\text{in }\Omega.$
In reservoir simulation, the model problem (18) usually occurs after temporal semi-discretization of the mathematical models that describe the multiphase flow in porous media. Such scalar problems also occur in more sophisticated preconditioning techniques for coupled systems of PDEs (see [33] for further details). The unknown function $p$ in (18) is the pressure; $a\in[L^{\infty}(\Omega)]^{3\times 3}$ is the diffusion tensor and it usually depends on the permeability and viscosity, etc.; $c\in L^{\infty}(\Omega)$ is nonnegative and proportional to the inverse of the time step size in an implicit temporal scheme (such as the Backward Euler method). We assume that $\partial\Omega$ has two non-overlapping parts, $\Gamma_{D}$ and $\Gamma_{N}$, such that $\overline{\Gamma_{D}\cup\Gamma_{N}}=\partial\Omega$. The model problem (18) is completed by the boundary conditions $p=g_{D}$ on $\Gamma_{D}$ and $(a\nabla p)\cdot\boldsymbol{n}=g_{N}$ on $\Gamma_{N}$, where $\boldsymbol{n}$ is the outward unit normal vector of $\partial\Omega$, and where $g_{D}$ and $g_{N}$ are given.
The difficulties that arises from (18) in regard to solving the linear system (1) are mainly due to the complex geometry and complicated physical properties of the reservoir, which often result in a heterogeneous diffusion tensor with large jumps, and in distorted, degenerated, and/or non-matching meshes with faults and pinch-outs. Figure 1 shows an example of the computational domain and the permeability of this reservoir in one horizontal layer.
We use the mixed-hybrid finite element method to discretize (18). Due to its local conservation property and intrinsic and accurate approximation of the flux $\boldsymbol{u}:=-a\nabla p$, the mixed-hybrid method is preferred in reservoir simulations. Let $\Omega_{h}=\{T\}$ be a triangulation of $\Omega$, and let $\Gamma_{h}$ be the set of boundaries of $T$ in $\Omega_{h}$ with the decomposition
$\Gamma_{h}^{\partial}=\{\Gamma\in\Gamma_{h}:\Gamma\in\partial\Omega\},\quad\Gamma_{h}^{0}=\Gamma_{h}\setminus\Gamma_{h}^{\partial}.$
We define the finite dimensional spaces as
$\displaystyle V_{h}$ $\displaystyle:=\{\boldsymbol{v}\in[L^{2}(\Omega)]^{d}:\boldsymbol{v}|_{T}\in V_{h}(T),\;\forall\,T\in\Omega_{h}\}$ $\displaystyle Q_{h}$ $\displaystyle:=\{q\in L^{2}(\Omega):q|_{T}\in P_{0}(T),\;\forall\,T\in\Omega_{h}\}$ $\displaystyle\Lambda_{h}$ $\displaystyle:=\{\mu\in L^{2}(\Gamma_{h}):\mu|_{\Gamma}\in P_{0}(\Gamma),\;\forall\,\Gamma\in\Gamma_{h}^{0};\;\mu|_{\Gamma}=0,\;\forall\,\Gamma\in\Gamma_{h}^{\partial}\},$
where $V_{h}(T)$ is the Kuznetsov-Repin element on the polyhedral elements [17]. This element is based on the Raviart-Thomas element on the local conforming tetrahedral partitioning of $T$. $P_{k}(T)$ denotes the set of polynomials of a degree not greater than $k$, $k\geq 0$. $V_{h}$, $Q_{h}$, and $\Lambda_{h}$ are used to approximate the flux $\boldsymbol{u}$; the pressure $p$ associated with elements in $\Omega_{h}$; and the Lagrange multipliers $\lambda_{h}$ are associated with faces in $\Omega_{h}$.
The mixed-hybrid finite element formulation for (18) can be written as follows: Find $(\boldsymbol{u}_{h},p_{h},\lambda_{h})\in V_{h}\times Q_{h}\times\Lambda_{h}$, $\boldsymbol{u}_{h}\cdot\boldsymbol{n}=-g_{N,h}$ on $\Gamma_{h}$, where $g_{N,h}$ is an appropriate approximation of $g_{N}$, such that
$\displaystyle(a^{-1}\boldsymbol{u}_{h},\boldsymbol{v})-\sum_{T}\{(p_{h},\text{div}\ \boldsymbol{v})_{T}-(\lambda_{h},\boldsymbol{v}\cdot\boldsymbol{n}_{T})_{\partial T}\}$ $\displaystyle=-\int_{\Gamma_{D}}g_{D}(\boldsymbol{v}\cdot\boldsymbol{n})\mathrm{d}s,$ $\displaystyle\forall\boldsymbol{v}\in V_{h}$ $\displaystyle-\sum_{T}(\text{div}\ \boldsymbol{u}_{h},q)_{T}-(cp_{h},q)$ $\displaystyle=-\int_{\Omega}fq\mathrm{d}x,$ $\displaystyle\forall q\in Q_{h}$ $\displaystyle\sum_{T}(\mu,\boldsymbol{u}_{h}\cdot\boldsymbol{n}_{T})_{\partial T}$ $\displaystyle=0,$ $\displaystyle\forall\mu\in\Lambda_{h}$
where $\boldsymbol{n}_{T}$ denotes the outward unit normal to $\partial T$.
## 5 Numerical Experiments
It is easy to see that the resulting discrete linear system of the mixed-hybrid finite element method in the previous section has the following matrix form:
$\begin{pmatrix}\mathsf{D}&\mathsf{B}^{T}&\mathsf{C}^{T}\\ \mathsf{B}&-\mathsf{M}&0\\ \mathsf{C}&0&0\end{pmatrix}\begin{pmatrix}\widetilde{u}\\ \widetilde{p}\\ \widetilde{\lambda}\end{pmatrix}=\begin{pmatrix}\widetilde{f_{1}}\\ \widetilde{f_{2}}\\ 0\end{pmatrix}.$
Notice that the matrix $\mathsf{D}$ is block diagonal with each block corresponding to the flux unknowns in one element. Hence, we can easily invert $\mathsf{D}$ and obtain (1) with
(19) $A=\begin{pmatrix}\mathsf{BD^{-1}B}^{T}+\mathsf{M}&\mathsf{BD^{-1}C}^{T}\\ \mathsf{CD^{-1}B}^{T}&\mathsf{CD^{-1}C}^{T}\end{pmatrix},\quad u=\begin{pmatrix}\widetilde{p}\\ \widetilde{\lambda}\end{pmatrix},\quad\text{and}\ f=\begin{pmatrix}\mathsf{BD^{-1}}\widetilde{f_{1}}-\widetilde{f_{2}}\\ \mathsf{CD^{-1}}\widetilde{f_{1}}\end{pmatrix},$
where it is well-known that $A$ is SPD.
In the numerical experiments, performed on a Dell Precision desktop computer, we solve the above linear system by PCG with different preconditioners. We pick eight problems from the real petroleum reservoir data. Table 1 gives the degrees of freedom (DOFs) and the number of non-zeros (NNZ) for the test problems. Here, the difference between Model 1 (3) and 2 (4) is that the permeabilities in Model 1 (3) are homogeneous whereas the permeabilities in Model 2 (4) have large jumps. Figure 1 shows the computational domain of Models 3 and 4, and the highly heterogenous permeabilities used in Model 4. Models 5–8 are test problems that are relatively large in size. The computational domain of Models 5 and 6 are the same, but the physical properties, such as porosity and permeability, are different. Models 7 and 8 share the same computational domain, but the permeabilities used in Model 7 are from real reservoir data and the permeabilities in Model 8 are artificially adjusted in order to make the resulting linear system very difficult to solve. | 10,236 | 34,320 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 365, "math_alttext": 3, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2024-18 | latest | en | 0.807038 |
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Use this math bundle to teach, reinforce and practice the Distributive Property of Multiplication.. The animated and interactive PowerpPoint explicitly explains this property. Then use the center and game for follow-up and review. Included in the
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Use this math bundle to teach, practice and reinforce area and perimeter. This package includes a PowerPoint, a game, and a center. Get the students immersed in learning about perimeter and area. What's Included Perimeter and Area
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I have been teaching for over 30 years, teaching grades 1 - 4. I have also taught adults in graduate school courses as well as been a leader in staff development. I am a certificated bilingual teacher, currently in a Sheltered Immersion Class (English Language Learners) teaching 3rd grade. I have also served as: • bilingual resource teacher • literacy coach • mentor teacher • conference presenter I have never "stood still" with my teaching and am always learning and applying new ways and strategies. Check out my newest resources for Google Classroom! I'm now in year three of teaching in a 1:1 Chromebook classroom. I use the Chromebooks daily with my students. I also am fortunate to have 8 iPads that I use for small groups such as, Guided Reading.
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Single dad raising two boys. Sometimes I feel school never stops!. My colleagues are also my friends (many of which I've had their children in my class). One of my kids is attending the school I teach at, while the other one just started Middle School. Visit MY BLOG | 1,040 | 4,183 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2017-26 | latest | en | 0.894105 |
https://oeis.org/A343588 | 1,653,473,591,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662584398.89/warc/CC-MAIN-20220525085552-20220525115552-00283.warc.gz | 483,597,803 | 3,659 | The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A343588 Sums of two odd fourth powers. 2
2, 82, 162, 626, 706, 1250, 2402, 2482, 3026, 4802, 6562, 6642, 7186, 8962, 13122, 14642, 14722, 15266, 17042, 21202, 28562, 28642, 29186, 29282, 30962, 35122, 43202, 50626, 50706, 51250, 53026, 57122, 57186, 65266, 79186, 83522, 83602, 84146, 85922, 90082, 98162 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 LINKS Jinyuan Wang, Table of n, a(n) for n = 1..2000 EXAMPLE 2 is in the sequence since 2 = 1^4 + 1^4. 82 is in the sequence since 82 = 1^4 + 3^4. MATHEMATICA Take[Union[Total /@ Tuples[(2 Range[0, 70] - 1)^4, 2]], 70] CROSSREFS Cf. A342832 (sums of two distinct odd fourth powers). Sequence in context: A020955 A230398 A120826 * A259308 A202965 A307583 Adjacent sequences: A343585 A343586 A343587 * A343589 A343590 A343591 KEYWORD nonn AUTHOR Wesley Ivan Hurt, Apr 20 2021 STATUS approved
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Last modified May 25 05:58 EDT 2022. Contains 354048 sequences. (Running on oeis4.) | 494 | 1,345 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-21 | latest | en | 0.631155 |
https://aiida.readthedocs.io/projects/aiida-core/en/v1.6.1/topics/calculations/concepts.html | 1,669,866,519,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710789.95/warc/CC-MAIN-20221201021257-20221201051257-00836.warc.gz | 123,792,027 | 10,581 | # Concepts¶
A calculation is a process (see the process section for details) that creates new data. Currently, there are two ways of implementing a calculation process:
The first one is the simplest of the two and is basically a python function that is magically transformed into a process. This is ideal for calculations that are not very computationally intensive and can be easily implemented in a python function. For more taxing calculations, typically performed by external codes that are optionally run on remote computing clusters, the calculation job is the better alternative.
In the following sections, both concepts will be explained but without going too much into detail on how to implement or run them. For a more detailed exposé, please refer to the respective advanced sections on calculation functions and calculation jobs.
## Calculation functions¶
Consider the following computational task at hand:
Given three integers, add the first two and then multiply the sum by the third.
In plain python code, the solution would look something like the following:
# -*- coding: utf-8 -*-
return x + y
def multiply(x, y):
return x * y
This simple code snippet will achieve the goal of getting the desired result, however, the provenance is lost. There is no connection between the output of the functions and their inputs. The remedy to this problem is the calcfunction(). The calcfunction in AiiDA is a function decorator that transforms a regular python function in a calculation process, which automatically stores the provenance of its output in the provenance graph when executed. Updating the previous snippet with calcfunction decorators yields:
# -*- coding: utf-8 -*-
from aiida.engine import calcfunction
@calcfunction
return x + y
@calcfunction
def multiply(x, y):
return x * y
The only thing we had to do to decorate the two functions was to add the line @calcfunction just before the function definition. Adding the decorator tells AiiDA that the provenance for this function should be stored in the provenance graph when it is executed. This means linking up the inputs and the outputs for a calculation node, which represents the function that was executed. The final change that has to be performed to make this possible, is to make the inputs and the outputs storable. In the previous snippet, the inputs are plain python integer types, which cannot be automatically stored in the provenance graph as nodes. To solve this, one only has to wrap them in the Int node sub class, which makes them storable in the database:
# -*- coding: utf-8 -*-
from aiida.engine import calcfunction
from aiida.orm import Int
@calcfunction
return x + y
@calcfunction
def multiply(x, y):
return x * y
The only difference with the previous snippet is that all inputs have been wrapped in the Int class. The result that is returned by the function, is now also an Int node that can be stored in the provenance graph, and contains the result of the computation.
Note
Since x and y inside the add and multiply functions are already Int instances the sum will also be one. This is true because all arithmetic operators also work on the base AiiDA classes (Int, Float, etc.) as they would on the equivalent python types. It is important to realize though that only Node instances, or sub classes thereof can be stored. For more information on how to return results from process functions, refer to the advanced section.
With these trivial changes, the full provenance of the result produced by running the function is maintained and looks like the following:
Fig. 13 The provenance generated by the calcfunction example
The example above already shows how a calcfunction can be run: simply by calling it. The value that is returned is the result returned by the definition of the function. However, sometimes one would also like to have a reference to the calculation node that represents the execution of the function in the provenance graph. The following example shows two additional launch functions that will return a tuple, which in addition to the results, also return the pk or the node associated with the process
# -*- coding: utf-8 -*-
from aiida.engine import calcfunction, run, run_get_node, run_get_pk
from aiida.orm import Int
@calcfunction
return x + y
x = Int(1)
y = Int(2)
result, node = run_get_node(add, x, y)
result, pk = run_get_pk(add, x, y)
This was a very short and limited description of calculation functions. For a more detailed description of launching them, please refer to the section on launching processes. If you want more details on implementing calculation functions and best practices, refer to the section on working with calculation functions.
## Calculation jobs¶
In the previous section on calculation functions, we showed how a simple python function can be transformed into a process, such that when it is launched, its execution is recorded automatically in the provenance graph. However, not all computations are well suited to be implemented as a python function, but rather are implemented as a separate code, external to AiiDA. To interface an external code with the engine of AiiDA, the CalcJob process class was introduced. A detailed explanation of how to implement it, the interface and best practices, can be found in a later section. Here, instead, we will focus on the big picture and explain in broad lines how a calculation job models the execution of an external code and what tasks it performs when launched.
To illustrate how a calculation job operates, we need an external code. Let’s imagine an external code that consists of a bash script that reads an input file containing two integers, sums them and prints the result in the standard output using echo, for example:
#!/bin/bash
# Read two integers from file 'aiida.in' and echo their sum
x=$(cat aiida.in | awk '{print$1}')
y=$(cat aiida.in | awk '{print$2}')
echo $(($x + \$y ))
When run, this script reads the contents of a file called aiida.in and expects that it contains two integers. It will parse these into the variables x and y and then print their sum. When you want to run this ‘code’ through AiiDA, you need to tell how AiiDA should run it. The ArithmeticAddCalculation is a calculation job implementation that forms an interface to accomplish exactly that for the example bash script. A CalcJob implementation for a specific code, often referred to as a calculation plugin, essentially instructs the engine how it should be run. This includes how the necessary input files should be created based on the inputs that it receives, how the code executable should be called and what files should be retrieved when the calculation is complete. Note the files should be ‘retrieved’ because calculation jobs can be run not just on the localhost, but on any computer that is configured in AiiDA, including remote machines accessible over for example SSH.
Since a CalcJob is a process just like the calculation functions described before, they can be run in an identical way.
# -*- coding: utf-8 -*-
from aiida.engine import run
from aiida.plugins import CalculationFactory
inputs = {
'x': Int(1),
'y': Int(2),
}
the provenance generated by running the calculation job will look something like this:
Fig. 14 The provenance generated by the calculation job example
The execution of the calculation job is represented in the provenance graph by a process node, i.e. the pink square labeled C:sub:1 in Fig. 14. The integer data nodes x and y that were passed as inputs are linked to the calculation job as such, as well as the third input code. This input is required for all calculation jobs as it represents the external code that is actually executed. These code nodes are instances of the Code class, which is a sub-class of Data. This means that code instances are a sort of data node. Its function is to record the path to the executable and some other code related attributes defined during the code setup.
The calculation job produced two outputs, an integer node, containing the sum of x and y and a FolderData node, containing the output files that were retrieved. Note that all outputs of calculation jobs (except for the retrieved node) are technically not created by the calculation job itself, but rather by an implementation of the Parser class. In principle, this step is optional, and so a calculation job is therefore not required to produce any outputs, except for the retrieved folder data node, which will always be there. How the parser fits into the concept of calculation jobs will be addressed in this section.
To arrive at the provenance graph shown above in Fig. 14, the engine performed quite some tasks. When a calculation job is launched, the engine will take it roughly through the following steps:
• Upload: the calculation job implementation is used to transform the input nodes into the required input files, which are uploaded to a ‘working’ directory on the target machine
• Submit: to execute the calculation, a job is submitted to the scheduler of the computer on which the input code is configured.
• Update: the engine will query the scheduler to check for the status of the calculation job
• Retrieve: once the job has finished, the engine will retrieve the output files, specified by the calculation plugin and store them in a node attached as an output node to the calculation
All of these tasks require the engine to interact with the computer, or machine, that will actually run the external code. Since the Code that is used as an input for the calculation job, which is configured for a specific Computer, the engine knows exactly how to execute all these tasks. The CalcJob implementation itself then is completely independent of the machine the code will be run on. To run the calculation job on a different machine, all you have to do is change the code input to one that is configured for that machine. If the machine is not the localhost, the engine will need a way to connect to the remote machine in order to perform each of the four tasks listed above. The mechanism that allows the engine to connect to the remote machine is called a transport and therefore the tasks it performs using this transport are called transport tasks.
### Exponential backoff mechanism¶
In the case of calculation jobs being executed on a remote machine, the engine will have to connect to the machine for each of the transport tasks. In connecting to the remote, a whole host of potential problems may occur that would cause the calculation job to fail. For example, the remote machine may be down and as a result unreachable, or the engine itself may lose its internet connection. However, these problems are often temporary. To prevent the calculation job from excepting and it being lost forever, an exponential backoff mechanism has been implemented. Whenever the engine performs a transport task but encounters an exception, instead of letting the calculation job fail, it will reschedule the same task to be executed again at a later time. The task will be automatically rescheduled until it finishes successfully, where the interval between tries increases exponentially. If after 5 consecutive tries, the task still fails, instead of rescheduling it, the engine will simply pause the calculation job. The output of verdi process list will give more information on why the task failed:
PK Created State Process label Process status
---- ---------- ------------ -------------------------- ---------------------------------------------------------------------------------------
151 1h ago ⏸ Waiting ArithmeticAddCalculation Pausing after failed transport task: retrieve_calculation failed 5 times consecutively
Total results: 1
When there are calculation jobs that have been paused because the transport tasks have failed multiple times, the user has the time to investigate the problem. If the problem is determined to be temporary and it has been resolved, one can use verdi process play to resume the paused processes. The engine will then automatically reschedule the task that failed last and the calculation job will continue where it left off.
This exponential backoff mechanism makes the engine very robust with respect to calculation jobs, reducing the loss of computational resources due to temporary problems to an absolute minimum.
Note
The parameters, such as the delays between retries and the maximum number of retries, are currently not configurable, but they might be in the future.
### Parsers¶
The previous section explained how the CalcJob class functions as an interface between AiiDA’s engine and an external piece of code. The calculation job plugin will instruct the engine how the transport tasks should be accomplished. However, as mentioned before, those tasks stop after the output files have been retrieved, which the engine will attach as a FolderData node with the label 'retrieved' to the calculation job node. As far as the calculation job goes that is all that is absolutely required. However, often one wants to parse those output files into some specific outputs that should be represented as individual nodes in the provenance graph. This can be accomplished by implementing the Parser class and specifying it in the inputs of the calculation job. In that case, the engine will call the parser after the output files created by the job have been successfully retrieved. In the parser implementation, the retrieved files can then be parsed and converted into output nodes. For technical details on how to implement a parser for a calculation job and how to specify it in the inputs, please refer to the detailed parser section, | 2,734 | 13,745 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2022-49 | latest | en | 0.905643 |
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# ds:geometry
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10 Apr 2006, 05:56
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In the xy-plane, does the line with equation y=3x+2 contain the point (r,s)?
1, (3r+2-s)(4r + 9 -s) = 0
2, (4r -6 -s )(3r + 2 - s) = 0
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10 Apr 2006, 07:22
In the xy-plane, does the line with equation y=3x+2 contain the point (r,s)?
1, (3r+2-s)(4r + 9 -s) = 0
2, (4r -6 -s )(3r + 2 - s) = 0
Lets look at the question first. We have an equation giving relationship between x & y. And the question asks us if (r,s) is contained in y=3x+2
Usually when we want to see if if something is contained in a given line equation we have to sub those values and see if the equation holds good.
Ah so we should be looking for values of r & s in statements 1 and 2
Good now armed with this information lets look at statement 1
(3r+2-s)(4r + 9 -s) = 0
Great they already factored it for us so
3r+2-s = 0 --> 3r-s = -2 ------------Eqn(1)
4r+9-s = 0 ----> 4r-s = -9 ----------Eqn(2)
Solving Eqn (1) and Eqn(2) we get r= -7 and s= -19
Sub (r,s) in the question stem y=3x+2 ---> -19 = 3(-7) +2--> -19 = -19
Great Statement 1 is sufficient
Statement(2)
(4r -6 -s )(3r + 2 - s) = 0
This one is factored as well.
(4r -6 -s ) = 0 -------> 4r-s = 6 ------Eqn(3)
(3r+2-s) = 0 ------> 3r-s = -2 ------Eqn(4)
Solving Eqn (3) and Eqn (4) we get r = 8 , s= 26
Substitute in question stem
y= 3x + 2 --> 26 = 3(8) + 2 ---> 26 = 26
Statement 2 is sufficient as well.
Hence
Heman & the master of the universe
Last edited by heman on 16 Sep 2006, 17:21, edited 2 times in total.
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10 Apr 2006, 07:44
heman,
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10 Apr 2006, 07:45
macca wrote:
In the xy-plane, does the line with equation y=3x+2 contain the point (r,s)?
1, (3r+2-s)(4r + 9 -s) = 0
2, (4r -6 -s )(3r + 2 - s) = 0
s= 3r +2
from 1
3r + 2 = s
s = 4r + 9
From these two statements we can deduce that (r,s) may or may not be on line
From 2
s = 4r - 6
s = 3r+2
Again (r,s) may or may not be on the line.
Combining 1 and 2
There is only one equaltion on which (r,s) lies
hence C
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10 Apr 2006, 07:50
You need both the Choices to arrive at one solution. Hence C.
- Vipin
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10 Apr 2006, 08:27
gmat_crack Can you tell me what is the flaw in my solution.?
Heman
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10 Apr 2006, 10:55
In the xy-plane, does the line with equation y=3x+2 contain the point (r,s)?
1, (3r+2-s)(4r + 9 -s) = 0
2, (4r -6 -s )(3r + 2 - s) = 0
Q, y=3x+2, intersects y axis at 2 and slope of 3 (imagine)
s=3r+2 (same line equation, definetely intersects), may or may not intersect
s=4r+9 (intersects y axis at 9 and more slope) so it has to intersect line in Q
From 2,
s = 4r-6(intersects y axis on -6 and slope of 4), may or may not intersect
s=3r+2(same line equation) definetely intersects
Does this make sense? Am i thinking in right direction?
Both ways we cannot say. I think answer is E.
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10 Apr 2006, 12:05
Hope we can get some help from HongHu!
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10 Apr 2006, 22:21
My guess is B.
For the points (r,s) to lie on the line y=3x+2, the slope of r,s should be equal to the slope of y=3x+2 and since 2 provids us with this (s=3r+2) my guess is B.
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10 Apr 2006, 22:30
in the two equations substitute s = 3r+2
if the LHS == 0 then the equation holds true.
U can see its true for both 1 and 2.
Hence D
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11 Apr 2006, 00:53
pseudo wrote:
My guess is B.
For the points (r,s) to lie on the line y=3x+2, the slope of r,s should be equal to the slope of y=3x+2 and since 2 provids us with this (s=3r+2) my guess is B.
hmm, but the point (r,s) could be on a line parallell to y=3x+2?
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11 Apr 2006, 00:55
macca wrote:
pseudo wrote:
My guess is B.
For the points (r,s) to lie on the line y=3x+2, the slope of r,s should be equal to the slope of y=3x+2 and since 2 provids us with this (s=3r+2) my guess is B.
hmm, but the point (r,s) could be on a line parallell to y=3x+2?
I am not sure whether you can just use the slope!
You have to prove that there are/is atleast 1 point where they intersect!
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11 Apr 2006, 12:25
1, (3r+2-s)(4r + 9 -s) = 0
2, (4r -6 -s )(3r + 2 - s) = 0
It is C.
You cannot simply plug in s = 3r +2. You have to determine whether the equations given have a root that is common with s = 3r + 2.
Statement 1:
s = 4r + 9 OR 3r + 2. Therefore insufficient.
Statement 2:
s = 4r - 6 OR 3r + 2. Also insufficient.
Statements 1 and 2: Taking both equations, we know that s MUST equal 3r + 2, since it is a root to both equations. Therefore C.
If you simply substitute s = 3r + 2, of course you'll get both sides equal to zero, because it is one of the two roots of each equation. This does not necessarily mean that both statements are insufficient by themselves.
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11 Apr 2006, 17:01
jcgoodchild wrote:
You cannot simply plug in s = 3r +2. You have to determine whether the equations given have a root that is common with s = 3r + 2.
Well it depends. You can plug in s = 3r +2. In essense what you are doing is illustrating that at that point the same values of s,r makes sense for both the equation. Though, you cannot say that either equation 1 or 2 is equivalent to 0.
However, if it is equivalent to 0, you can show that the two equations are valid for the same values of s,r! I.e. the two curve/lines intersect.
Another way to explain the above would be to draw the graph of the two equations!
If they intersect, they will have the same values of s,r satisfying the two equations!
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12 Apr 2006, 01:50
macca wrote:
In the xy-plane, does the line with equation y=3x+2 contain the point (r,s)?
1, (3r+2-s)(4r + 9 -s) = 0
2, (4r -6 -s )(3r + 2 - s) = 0
A) (3r+2-s)(4r + 9 -s) = 0
=> (3r+2-s) = 0 or (4r + 9 -s)=0
so any pair of (r,s) which satisfy one of above will solve Eq-A
In sufficient
B) Also in sufficient with same logic
A+B to gather means only 3r+2-s=0 (s=3r+2) is valid option Hence C is the answer.
-------------------------------------------------
Other way to visualize this problem is to consider r and s as variables, this means 3r+2-s=0 is a Line(line1), so does 4r+9-s=0(line2), so any value on line 1 or line 2 will satifsy eq. one so we really dont know which exact line we are talking about.
Same logic will apply to B
But if we see A, B simultaniously only points on Line 3r+2-s=0 will satisfy both equation, hence C is the answer.
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12 Apr 2006, 06:27
OA is B!
Don't understand how to solve this problem.
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12 Apr 2006, 08:00
macca wrote:
OA is B!
Don't understand how to solve this problem.
What is source of your question?
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12 Apr 2006, 08:23
The source is gmatprep
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12 Apr 2006, 13:22
The official answer is wrong and I say this full authority.
In order for a point with coordinates (r,s) to lie on a line y=3x+2, the slope between the point (r,s) and another point lying on line y-3x+2 should be 3.
Lets say x =1 then y= 5
So slope between (R,S) and (1,5) should be 3.
Now lets take B
(4r -6 -s )(3r + 2 - s) = 0
Which means that 4r-s-s =0 and also 3r+2-s=0
Simply those so that
r= (s+6)/4 …. (1)
s= s = 3r+2…. (2)
now take the slope between (R,S) and (1,5)
slope = s-5 / r-1
substiture r and s from 1 and 2 and you will get:
slope = 12(r-1) / (s+2)
substitute s from 2
slope = 12(r-1) / 3r+4
you are stuck and you cannot find the slope. So B can not be the answer.
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12 Apr 2006, 21:00
Do they have any explanation? This sounds weird!
12 Apr 2006, 21:00
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https://www.amazingtelescopes.com/interesting-about-telescopes/how-to-determine-telescope-resolution.html | 1,652,706,482,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662510117.12/warc/CC-MAIN-20220516104933-20220516134933-00751.warc.gz | 716,048,640 | 13,199 | How To Determine Telescope Resolution?
Due to the fact that 1 radian equals 57.3 degrees and 1 degree equals 3600 arcseconds, the resolution is equal to 0.000000069 radians multiplied by (57.3 degrees/radian) multiplied by (3600 arcseconds/1 degree) equals 0.014 degrees. One thousand milliarcseconds equals one arcsecond, hence the resolution is 0.014 arcsecond x (1000 milliarcsecond / arcsecond) = 14 milliarcseconds, or 0.014 arcsecond x (1000 milliarcsecond / arcsecond).
• This formula calculates the resolving power in arc seconds of a telescope with a diameter D in meters that is collecting light with a wavelength in nanometers equal to the resolving power in arc seconds of a telescope with a diameter D in meters that is collecting light with a wavelength in nanometers equal to The resolving power of a telescope may be calculated using the following formula: = 2.06×105(D) = 2.06 x 10 5 (D), where is the resolving power. The wavelength of light is represented by the symbol
How do you find theoretical resolution?
How to figure out what the resolution of a microscope should be.
1. NA is equal to n x sin. For example, n is the refractive index of the imaging medium, and is half of the angular aperture of the objective. For example, d = 2/NA A specimen is imaged using a wavelength of light that is equal to 2 NA2. R= 1.22 NAobj+NAcond.
2. D= 2 NA2
3. R= 1.22 NAobj+NAcond.
You might be interested: How Far Will The James Webb Telescope See? (Solution found)
What is the resolution limit of a telescope?
These equipment, which are made up of a collection of telescopes and known as astronomical interferometers, may attain angular resolutions as small as one-hundredth of an arcsecond at optical wavelengths, and much more precision at x-ray wavelengths.
What is Rayleigh resolution?
Specifically, according to Rayleigh’s criteria, the resolution of an optical microscope is defined as the shortest distance between two point sources at which their existence is discernible in the picture (1).
How do you find the angular resolution of a telescope?
The angular resolution is related to the ratio of the wavelength, l, of the radiation divided by the diameter of the telescope: q = l/D. The wavelength, l, of the radiation is proportional to the diameter of the telescope. If you have 20/20 vision, your visual acuity is restricted by the diameter of your pupil, and glasses will not be able to improve your vision any more.
On what factor the limit of resolution of telescope depends?
The wavelength of light used by a telescope determines its resolving capability.
How much more light does an 8 meter telescope gather than a 2 meter telescope?
A) The light-collecting area of the 8-meter telescope is 16 times greater than that of the 2-meter telescope.
How is Rayleigh resolution calculated?
This is known as the Rayleigh criteria, and it is represented by the equation D=1.22D = 1.22 D. The Rayleigh criterion yields the lowest feasible angle between point sources, or the highest possible resolution. Once this angle has been determined, the distance between stars may be estimated using the information provided about their distance from us.
You might be interested: Which Is Better A Sky Watcher Or Orion Dob. Telescope? (Solved)
What is B in Rayleigh criterion?
When determining the smallest resolvable angle, the Rayleigh criteria is used. Entering known values results in a result. Solution for (b) The distance between two objects that are separated by an angle and that are a distance apart from each other is
Is higher or lower angular resolution better?
“Higher” resolution does not necessarily imply superior performance; rather, it implies that only larger objects can be resolved; yet, we want to be able to resolve tiny objects, such as the angular distance between a pair of double stars. As a result, in this circumstance, tiny is preferable.
What is the formula for angular resolution?
When viewed from a distance, the two patterns appear to be similar; but, when you go closer to them, you will notice that the lines are scarcely distinguishable and that you will be unable to distinguish between the two pictures. Based on this distance L, you may determine the angular resolution of your eyes as follows: angular resolution = (2 mm)/L. (in radians).
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http://automation.isa.org/2017/03/electromagnetic-flowmeters-simple-operation-high-accuracy/ | 1,506,235,498,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818689897.78/warc/CC-MAIN-20170924062956-20170924082956-00381.warc.gz | 33,166,724 | 26,553 | # Primer: Electromagnetic Flowmeters Simple to Operate, High Accuracy
This post was written by Scott Stewart, senior product consultant with Yokogawa Corporation.
Electromagnetic flowmeters, or as they are known by their more common name, magmeters, have become one of the most common flow metering technologies over the past 50 or more years. They are simple to operate, have no moving parts, cause no flow obstruction, and can provide a high level of accuracy and turndown ratio. The only hard-and-fast rule of application is the process liquid must be electrically conductive.
### Simple working concept
High school physics tell us that a conductor passing through a magnetic field will generate a voltage that is proportional to the speed with which the conductor is moving. You might recognize it as Faraday’s law of magnetic induction.
In the case of a magmeter, the liquid itself is the conductor, and the magnetic field is created by coils placed around the pipe. Two electrodes placed on opposite sides of the pipe perpendicular to the liquid flow and the magnetic poles measure the induced voltage. The voltage generated is proportional to the velocity, which is then converted to a flow rate. The liquid flow, magnetic field, and line between the electrodes form the x, y, and z axes.
Figure 1 goes into more detail about the actual construction. Magmeter users appreciate how the interior surface of the device can match the upstream and downstream pipe diameter, and how there are no obstructions into the flow stream. The flowmeter itself must have an insulating lining, which can be made from rubber, Teflon, ceramic, or other materials capable of standing up to erosive or corrosive media.
Figure 1. Magmeter construction is very straightforward. The magnetic flux field (B) is generated by a coil that is mounted perpendicular to what are typically two electrodes. They will pick up the induced electromotive force (voltage). The magnetic field generates an electromotive force proportional to the magnetic flux density (B), the velocity of the conductor (V), and the diameter of the pipe (D).
Magmeters are highly scalable, running from 0.1 inch to 104 inches or more in diameter. They work well for clean fluids, such as water, acids, and caustics—or in heavy particulate applications, such as paper stock, pulp, and lime slurries. As mentioned earlier, the main requirement is the process media must have some level of conductivity, with a typical minimum range of 1 micro-Siemen (µS) to 5 µS. As long as the process meets this requirement, the meter should perform satisfactorily in the application. Most oil-based liquids are not conductive, which knocks magmeters out of many refinery applications.
### The exciting part
Since their introduction in the 1950s, magmeters have evolved enormously from a technology standpoint. A typical meter from the first decade would have used AC to excite the coil, so it used the available electrical line frequency of 50 Hz or 60 Hz. Users soon found the high sampling rate this frequency offered to be particularly well suited for noisy slurry applications, and it offered fast response to changes in flow rate. However, users also noticed these AC-powered magmeters did not have a stable zero point when there was no flow. If nothing was moving in the line, the devices tended to wander. They also used a large amount of energy, with many units drawing as much as 300 W.
As magmeter technology advanced, the next generation excited the coils with pulsed square-wave DC operating around 6.25 Hz to 11 Hz. This worked well enough in most situations and delivered the sought-after zero stability, but lower excitation frequencies could not handle noise caused by a high solids content. The low sampling rates also made them sluggish when responding to rapidly changing flow rates.
As new power supply circuits developed, instrumentation designers had more options available to try different excitation frequencies. This led to adjustable sensors where a user could choose the best frequency for the application. If the process is noisy with slurries or flow rates change rapidly, the user chooses a high frequency. For an intermittent process, such as a batch application with periods when flow stops, the user chooses the low frequency.
This approach works, but as transmitters got smarter, it was possible to incorporate circuitry capable of using both frequencies simultaneously. Dual-frequency excitation allows the transmitter to superimpose two frequencies on top of each other at the same time (figure 2). A low-frequency component of 6.25 Hz and a high-frequency component of 75 Hz work together for higher performance than either one functioning alone (figure 3).
Figure 2. Creating a waveform with characteristics of both AC and DC allows a magmeter to use the advantages of both.
Figure 3. Dual-frequency sensors deliver performance advantages without requiring a user to choose one or the other or to switch at various points.
### Reducing power consumption
Early magmeters needed a powerful magnetic field to create a signal strong enough to be measured accurately and linearized enough to provide a reliable and accurate flow reading. This is why early units drew so much power. With improvements in transmitter designs, it became possible to scale everything down, even while improving performance.
Units that once drew 300 W can now operate with 10 W to 15 W. Naturally, the physical size of a sensor has a major influence on power consumption; a 36-inch sensor requires more power to maintain a magnetic field than a 4-inch sensor. Some designs go even further with power reduction, to the point where two-wire, loop-powered designs are available in sensor sizes up to 8 inches. Bear in mind, when only 0.3 W is available (0.1 percent of what earlier units often needed), there is some compromise in performance.
Nonetheless, being able to replace other loop-powered flowmeters with a magmeter offers compelling advantages. Differential pressure and mechanical flowmeters are still very common, but obstructions in the flow path can cause clogging and pressure loss. Having the free flow of a magmeter without needing to update wiring is very attractive.
These low-power units have some limitations. Check with your supplier, but they typically require relatively high conductivity (10 µS to 20 µS) for the process liquid and a slight reduction in accuracy. Four-wire magmeters are usually capable of accuracies of ±0.2 percent to ±0.5 percent of flow rate. Loop-powered magmeters usually start at ±0.5 percent of flow. Moreover, four-wire units have a higher effective turndown ratio, handling liquid velocities ranging from 0.33 feet per second (fps) to 33 fps.
### Simple installation requirements
Magmeters have few installation constraints, but observing them goes a long way toward ensuring the best possible performance. First and foremost is keeping the pipe full of liquid. If long horizontal runs accumulate slugs of air, this will cause inaccurate readings. Keep the sensor at a low point in the piping, or better yet, install it in a vertical section with flow going up.
Second, like most flowmeter designs, having straight pipe upstream and downstream from the sensor reduces turbulence and provides more accurate readings. The flow profile through a pipe is never entirely uniform, but reducing turbulence keeps it more predictable. Having at least five diameters of straight pipe upstream and at least two diameters downstream will be a major help. Having longer straight runs is even better.
No single flowmeter technology is truly universal. Every flow instrument has its limitations, so every application has to be evaluated carefully to ensure the appropriate process requirements are met. Magmeters are particularly well suited to situations with difficult liquids or where solids tend to cause clogging, so make sure they are in your application toolbox.
Scott Stewart has been a senior product consultant with Yokogawa Corporation for 11 years, part of a 25-year career in industrial automation. Prior experience includes 10 years with Magnetrol International as a level and flow specialist in the southeast U.S. He graduated from Texas State University with a degree in marketing. Stewart is currently training for the Ride the Rockies Bicycle Tour in Colorado and likes to tinker with classic automobiles.
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Medium difficulty brain teasers and puzzles
Same Number of Handshakes
In some cultures, it is typical for guests at an event to shake hands when they meet other guests (when there isn’t a pandemic). Some might be more social and shake hands with a lot of other guests, while others may be less social and shake hands with few or no other guests.
Can you prove that, regardless of the number of guests at the event, there must be at least two guests with the same number of handshakes at the event? In other words, can you prove that it is impossible for every guest to have shaken a different number of hands?
Assume guests can’t shake hands with themselves.
Continue reading “Same Number of Handshakes”
Middle Four Letters
The middle four letters of these 8-letter words are shown. Can you figure out the full 8-letter words? There may be multiple suitable answers for some of the words.
1. _ _ EQUA _ _
2. _ _ DUST _ _
3. _ _ CUBA _ _
4. _ _ COLA _ _
5. _ _ DIRE _ _
6. _ _ MESA _ _
7. _ _ TACO _ _
8. _ _ OPIC _ _
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Inscribed Squares
Above are two identical isosceles right triangles containing two inscribed squares.
In one, a perfect square has been inscribed such that two sides line up with the two legs of the right triangle. In the other, a perfect square has been inscribed such that one side lines up with the hypotenuse of the right triangle.
Is the inscribed square on the left larger or the one on the right?
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Card Flipping Endgame
There are n playing cards lined up face-down in a row. Every turn, a pair of adjacent cards with the left card face-down is randomly selected (i.e., a pair of cards has no chance of being selected if the left card is face-up, otherwise all pairs are equally likely to be selected). Both cards are then flipped over (face-down to face-up or face-up to face-down).
Prove that after enough turns, it will eventually be impossible to select a pair of cards with the left card face-down.
When you reach this card flipping endgame, will the rightmost card be face-up or face-down?
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Calculator Error Riddle
A kid is adding consecutive integers on a calculator, one at a time, starting with 1 + 2 + 3 + … and so on. At one point you notice the sum is now 100, but that shouldn’t be possible if the kid was adding this way. The kid tells you that he made an error and subtracted exactly one of the numbers he was supposed to add.
What is the number he subtracted?
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Unique Number of Letters
Out of all whole numbers between one and five thousand, there is only one number that has a unique number of letters when you spell it out. What number is it?
Include spaces and dashes. For example, “twenty-four” and “two hundred” both have 11 letters.
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Further Examples of Camera Design ... Web Page
"Camera Design Using Locus of Unit Monochromats," paper for CIC 14 in pdf form
Jim Worthey Home Page
Foveon X3 Sensors, With Prefilter
Locus of Unit Monochromats for Foveon X3 Sensors With Prefilter Locus Derived by "Fit First" Method
Locus of Unit Monochromats for Foveon X3 Sensors With Prefilter Locus Derived by "Fit First" Method
Usage note: To see some preset views, click 'view' arrows below the 3D picture.
In the virtual reality picture above, the Locus of Unit Monochromats for the human 2° observer is drawn in the usual way, as the edge of the multicolored surface. The LUM of the camera is indicated by spheres. The short arrows indicate the transition from the camera's LUM itself to the best fit, so the arrow tips are the best fit curve. If the spheres lay right along the human LUM, that would mean that the camera fulfills Maxwell-Ives. In that case, the camera's LUM would be the best fit, and the arrows would have zero length.
The camera LUM is derived by the "fit first" steps:
• Find a best fit to human LUM by a linear combination of the camera's sensor functions. The best fit functions will look more or less like achromatic, red-green, blue-yellow, but will not be an orthnormal set.
• Preserving that sequence, orthonormalize the functions by the Gram-Schmidt method .
• Combining those 3 functions into a 3-dimensional graph gives the camera LUM, indicated by spheres.
• Yes, the best fit found as an intermediate step is the same best fit indicated by the tips of the short arrows.
The steps just specified all result in adding and subtracting of the camera sensor functions. We can then ask if the red-green function is really made by subtracting the camera's green function from its red function, and so forth. Such questions are answered by the transformation matrix relating the camera's orthonormal basis to the sensor functions. Let the camera sensors be the columns of array rgbSens, and CamOmega be the camera's orthonormal basis. Then
CamOmega = rgbSens*Y, where
Y = inv(CamOmega'*rgbSens) .
The tiny apostrophe, ', denotes matrix transpose. The transform Y can be found for any camera, and indeed for the eye itself.
2° observer himself
Foveon X3, no filter
Prefiltered Foveon X3
Y = inv(OrthoBasis'*rgbbar) =
0.0725 0.267 0.0376 0.0447 -0.31 -0.0543 0 0 0.138
Y = inv(CamOmega'*rgbSens) =
0.962 5.56 1.919 2.41 -6.98 -5.72 -0.783 1.61 5.84
Y = inv(CamOmega'*rgbSens) =
-0.544 9.4 2.67 5.61 -13.1 -7.86 -1.72 3.56 7.99
Column amplitudes =
0.0852 0.409 0.153
Column amplitudes =
2.71 9.06 8.40
Column amplitudes =
5.89 16.5 11.5
Foveon X3 Filtered: LUM by FF Method, Projected into v2-v1 and v2-v3 planes
In the graphs below, the thicker functions on the left are the basis for the camera's LUM, shown above by spheres. The thicker functions on the right are the basis for the best fit curve, shown above as the tips of arrows.
Foveon X3 Filtered: Orthonormal Basis and Fit to 2º Observer
Links:
Further Examples of Camera Design ... Web Page
"Camera Design Using Locus of Unit Monochromats," paper for CIC 14 in pdf form
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https://affairscloud.com/quants-quiz-data-interpretation-set-13/ | 1,696,333,643,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511075.63/warc/CC-MAIN-20231003092549-20231003122549-00064.warc.gz | 105,483,909 | 38,823 | # Quants Questions : Data Interpretation Set 13
Hello Aspirants. Welcome to Online Quantitative Aptitude section in AffairsCloud.com. Here we are creating question sample From Data Intrepretation that is important for IBPS,SBI exam,SSC exam, Railways Exam,FCI exam and other competitive exams. We have included Some questions that are repeatedly asked in exams !!
I.Study the information carefully to answer the given questions
In a college out of 600 students, boys and girls are in the ratio 8: 7. Each students went to either physics lab or chemical lab or both, 40% of the boys attend only physics lab, total 20% children went to both the labs out of which 25% are boys, 45% of the girls went to only chemistry lab.
1. Total how many boys went to only physics lab ?
A)175
B)158
C)156
D)134
E)None of these
B)158
Explanation :
8:7 = 320(boys):280(girls)
20% both = 20*600/100 = 120
B – (phy ) = 40*320/100 = 128
B – (both) = 25*120/100 = 30
G – (chem) = 45*280/100 = 126
Total no of boys(phy) = 128+30 = 158
2. How many girls went to only physics lab ?
A)62
B)46
C)64
D)60
E) None of these
C) 64
Explanation :No of girls (phy) = (280-126 – 90) = 64
3. Total number of students went to physics lab is what percentage of the total number of students in the college
A)22%
B)32%
C)33%
D)36%
E) None of these
B)32%
Explanation :
Total no of students went to only phy = 128+64 = 192
% = 192*100/600 = 32%
4. What is the total number of girls went to chemistry lab ?
A)216
B)210
C)267
D)261
E) None of these
A)216
Explanation :Total no of Girls(chem) = 126+90 = 216
5. What is the total number of students went to only chemistry lab ?
A)244
B)248
C)284
D)288
E) None of these
D)288
Explanation :Total no of students(chem) = (320-128-30)B + 126G = 162 + 126 = 288
II. Study the information carefully to answer the given questions
No of 5 types of books sold and returned(due to missing of pages) by a publisher over the years(data in 10000)
Book B1 B2 B3 B4 B5 Year Sold Return Sold Return Sold Return Sold Return Sold Return 2001 22 4 52 5 17 2.5 82 7 62 6 2002 37 5 47 4 22 2.55 77 6 60 6 2003 17 2.5 42 4.5 19 2.75 60 4 64 5.5 2004 27 2.25 44 4.3 27 3.5 67 5 42 3.5 2005 32 3.5 50 4.5 32 4 70 5 47 4 2006 29 3.5 43 4.1 28 2.75 74 2.5 52 4.5
1. In the case of B2 type of books, in which year was the ratio of return to sold highest among the five years ?
A)2001
B)2002
C)2003
D)2004
E)2005
C)2003
Explanation :
2001 = 5/52 =0.0962
2002 = 4/47 = 0.0851
2003 = 4.5/42 = 0.1071
2004 =4.3/44 = 0.0977
2005 = 4.5/50 = 0.09
2006 = 4.1/43 = 0.0953
2. In the year 2002 , ratio of book return to sold was below 8.5% in which type of books ?
A)B1
B)B2
C)B3
D)B4
E)B5
D)B4
Explanation :
B1 =(5*100)/37 = 13.51%
B2 =(4*100)/47 = 8.51%
B3 =(2.55*100)/22 = 11.59%
B4 =(6*100)/77 = 7.79%
B5 =(6*100)/60 = 10%
3. In the case of B5 type of books, in which year was the return to sold rate is lowest among the 6 years ?
A)2006
B)2002
C)2003
D)2005
E)2004
E)2004
Explanation :
2001 = 6/62 = 0.0968
2002 = 6/60 = 0.10
2003 = 5.5/64 = 0.0859
2004 = 3.5/42 = 0.0833
2005 = 4/47 = 0.0851
2006 = 4.5/52 = 0.0865
4. The acceptable B5 type of books in 2003 was what percent of that of that in 2002 ?
A)108.33%
B)103.88%
C)108.44%
D)103.54%
E)None of these
A)108.33%
Explanation :
(64-5.5)*10,000 = 585000………….2003
(60-6)*10,000 = 540000……………..2002
% = (585000*100)/540000 = 108.33%
5. The number of B3 type of books returned in 2002 is approximately what % more or less than B4 type of books rejected in 2003 ?
A)36.2%
B)32.65%
C)36.25%
D)15.65%
E)None of these
C)36.25%
Explanation :
2002 = 2.55*10000 = 25500
2003 = 4*10000 = 40000
% = [(40000-25500)*100]/ 40000 = 36.25%
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Author Message
TerryE
Super User
Joined: 16 Jul 2006
Posts: 550
Location: UK
Posted: Mon Jul 24, 2006 7:32 pm Post subject: How To: paste a formula to a Column in Calc
Quite often I've wanted to add a formula to a column in Excel. This is easy once you have mastered the trick of RC notation. Even if you haven't you just type the formula into the first field, toggle into the debugger immediate pane (yup, the VBA IDE has one of those) and type ?ActiveCell.FormulaR1C1 and up it comes to paste into your code. So then you do something like
Code: '***** VBA EXCEL Version **** nRows = vbaSheet.Cells.SpecialCells(xlCellTypeLastCell).Row vbaSheet.Range(“D2).Resize(nRows-1).FormulaR1C1 = _ “=R[-4]C[-3]+R[-3]C[-3]+R[-2]C[-3]”
I know this looks yukky, but you don't have to compose it this form. The real advantage of R1C1 notation in that it is cell relative so the formula is exactly the same for the whole row.
Now in OOo Basic:
• We don't have a cell relative format so we have to apply the formula to the first cell
• We can then fill the entire range. Note that the Calc Fill function does automatically relocate formulae.
• If you are doing large ranges then it is well worth having enableAutomaticCalculation turned off and do a manual calculate
• Sometimes the first thing that you want to do after doing this is to anchor the fields by replacing the formula with the values. I realise that if you are going to do this then you could just as easily do this with a programmatic For ... Next loop, but my response to this is Time it!. Using the Calc calculation engine is faster and is easier to code.
So here is the OOo Basic Code
Code: ***** OOB UNO Calc Version **** PastFormulaToColumn(sFormula, oSheet, nColumn, nStartRow, nEndRow, bPasteValues)
That's even easier than VBA! Well apart from this canned routine below that you will need to put in your code or library.
Code: ***** OOB UNO Calc Version **** Function PastFormulaToColumn(sFormula, oSheet, nColumn, nStartRow, _ Optional ByVal nEndRow, Optional ByVal bPasteValues) ' ' Routine to Paste Array into a specified column on a worksheet in thisComponent ' Dim oCursor, oRange, vData On Error Goto PasteHandler ' Apply default values for last two arguements If IsMissing(nEndRow) Then nEndRow = -1 If IsMissing(bPasteValues) Then bPasteValues = False ' If nEndRow is negative then -1 denote last row, -2 last but one etc. If nEndRow < 0 Then oCursor = oSheet.CreateCursorByRange(oSheet.getCellByPosition(0, 0)) oCursor.GotoEndOfUsedArea(False) nEndRow = oCursor.RangeAddress.endRow + 1 + nEndRow End If oSheet.getCellByPosition(nColumn, 1).Formula = sFormula oRange = oSheet.getCellRangeByPosition(nColumn, nStartRow, _ nColumn, nEndRow) oRange.fillSeries(com.sun.star.sheet.FillDirection.TO_BOTTOM, _ com.sun.star.sheet.FillMode.SIMPLE,0,0,0) ThisComponent.calculate() If bPasteValues Then vData = oRange.DataArray ' The intermediate variant is needed oRange.DataArray = vData ' Otherwise this is a No-Op !! End If PastFormulaToColumn = True Exit Function PasteHandler: PastFormulaToColumn = False End Function
Hope that you fnd this is useful. Comments gratefully received and if you think that I am having a pop at Calc compared to Excel, then remember that Calc has some nice features like RegExp pattern matching, the ability to call Basic functions as worksheet functions and the fact its free //Terry
_________________
Terry
WinXPSP3, OOo 2.4.1, Ubunto 8.04 for development
Also try the Official OOo Community Forum where I mainly post now.
noranthon
Super User
Joined: 07 Jul 2005
Posts: 3318
Posted: Mon Jul 24, 2006 8:28 pm Post subject:
Another member (Pitounet, I think it was) suggested the method .fillAuto The following is an example of a routine which inserts a formula into the top cell of a column range, copies it into the other cells of the range and, finally, converts the formulas to values:
Code: 'Calculate move. =IF((E2-E3)=0;"";MROUND(((E3-E2)/E2)*100;0.01)) oCell2 = oSheet2.getCellByPosition ( 6, 2 ) oTarget = oSheet2.getCellRangeByPosition( 6, 2, 6, lFoot ) sString = "=IF((E2-E3)=0;" & Chr\$( 34 ) & Chr\$( 34 ) & _ ";MROUND(((E3-E2)/E2)*100;0.01))" oCell2.setFormula( sString ) : oTarget.fillAuto( 0, 1 ) : oDoc.calculate() oTarget.setDataArray( oTarget.getDataArray )
Column E contains daily closing prices of stock trades, extracted from a .csv file. Column G is filled with a calculation of price movement from day to day.
_________________
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TerryE
Super User
Joined: 16 Jul 2006
Posts: 550
Location: UK
Posted: Tue Jul 25, 2006 3:16 am Post subject: ``
• fillAuto and fillSeries are alternative methods within the same service, with fillSeries just being specfic about what fill opoeration you want to do.
• Since I was giving a book example I used the correct constants, though I agree that hard coding then as (0,0,0,0,0) is briefer -- or in you case (1,0)
• As I said in my code the reason that I don't recommend oRange.DataArray = oRange.DataArray or the syntactically equivalent functional form that you oTarget.setDataArray( oTarget.getDataArray ) is that it doesn't work. At least in test on my OOo (V 2.0.3) it does the equivalent of a Paste and not a Paste Value, and this kind of defeats the purpose, eh?
• On a programming practice point of view I can't understand why people use & Chr(34 & when the language syntax gives an efficient alternative to use the repeated ". I would have written your formula:
Code: . . . , "=IF((E2-E3)=0;"""";MROUND(((E3-E2)/E2)*100;0.01))" , . . .
My view is that like your example I would tend to go for brevity for an inline code fragment, but that for an out of line utility routine, I would always go for flexibility, clarity and run-time efficiency.
Since I had a comment about coding practice, I'll mention another pet hate of mine that I sometimes see in coding examples: coding constructs like
Code: s= "Select a, b " s = s & "from someTable " s = s & "where c='user22'"
to enhance readability and avoid long constants that run over the editor window. I would alway use split line concatination (and lay it out so it is truly readible. For example
Code: s= "SELECT a, b " & _ "FROM someTable " & _ "WHERE c='user22'"
as the second for is easier to read and since OOo Basic folds constants at complie time actually runs a lot faster.
_________________
Terry
WinXPSP3, OOo 2.4.1, Ubunto 8.04 for development
Also try the Official OOo Community Forum where I mainly post now.
noranthon
Super User
Joined: 07 Jul 2005
Posts: 3318
Posted: Tue Jul 25, 2006 4:50 am Post subject: I still have version 2.0.2 and Target.setDataArray(Source.getDataArray) enters the results of formulas, which appears to be according to the API. I must admit my use of Chr\$ was in imitation of those who preceded me along the path. If your usage is acceptable, it seems to be merely a matter of taste._________________ search forum by month
TerryE
Super User
Joined: 16 Jul 2006
Posts: 550
Location: UK
Posted: Tue Jul 25, 2006 6:13 am Post subject:
Agreed largely about the taste issue, but there is a runtime speed hit as well because of this constant folding issue. The s = s + "next bit" or "bit" & var & "bit" format actualy runs about 10 times slower for three chunks -- not relevant set up code but maybe material in an inner loop.
As to the x.setDataArray(x.getDataArray) working in 2.0.2 thanks for confirming that. I looked at my test case again and realised that I'd missed setting the PasteValue flag = true for that case. Sorry. The simple test which demonstrates this "Feature by Design" as MS would say is
Code: Sub Main rB2 = ThisComponent.Sheets(0).getCellByPosition(1, 1) rB2a = rB2 rB2b = ThisComponent.Sheets(0).getCellByPosition(1, 1) rB2.Formula = "=Row()" : rB2.DataArray = rB2.DataArray : Print rB2.Formula rB2.Formula = "=Row()" : rB2a.DataArray = rB2.DataArray : Print rB2.Formula rB2.Formula = "=Row()" : rB2b.DataArray = rB2.DataArray : Print rB2.Formula rB2.Formula = "=Row()" : t=rB2.DataArray : rB2.DataArray=t : Print rB2.Formula rB2.Formula = "=Row()" : rB2.setDataArray(rB2.getDataArray): Print rB2.Formula
This prints ROW() for the first and second cases and 2 for the rest. I thought that it might have been the compiler optimising away a statement of the form x=x even if it's not a noop, but that wouldn't explain the second case. I think its a runtime issue. I'll post a bug report if it isn't already logged. //Terry
Good to have the feedback and debate. //Terry
_________________
Terry
WinXPSP3, OOo 2.4.1, Ubunto 8.04 for development
Also try the Official OOo Community Forum where I mainly post now.
TerryE
Super User
Joined: 16 Jul 2006
Posts: 550
Location: UK
Posted: Sun Aug 06, 2006 4:00 am Post subject: My apologies to Noranthon about my snipe on using Chr\$(34) instead of "" inside a string to represent embedded double quotes. This is sencond nature to me because I come from the VBA world. I've only recently realised that this syntax has only been properly supported in OOo Basic since OOo version 2.0. It just didn't work in OOo 1. Given the very low key announcement of the improvements to Basic in OOo 2.0, I can well understand why experienced OOo Basic users haven't yet picked up this practice. For those interested, they can see a readable version of this announcement here ab02vba announcement_________________Terry WinXPSP3, OOo 2.4.1, Ubunto 8.04 for development Also try the Official OOo Community Forum where I mainly post now.
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You cannot vote in polls in this forum | 2,812 | 10,385 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-22 | longest | en | 0.882309 |
https://stacks.math.columbia.edu/tag/0CUK | 1,718,481,199,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861606.63/warc/CC-MAIN-20240615190624-20240615220624-00520.warc.gz | 485,473,860 | 6,604 | Lemma 70.5.4. Let $S$ be a scheme. Let $X = \mathop{\mathrm{lim}}\nolimits _{i \in I} X_ i$ be the limit of a directed inverse system of algebraic spaces over $S$ with affine transition morphisms (Lemma 70.4.1). Let $x \in |X|$ with images $x_ i \in |X_ i|$. If each $X_ i$ is decent, then $\overline{\{ x\} } = \mathop{\mathrm{lim}}\nolimits _ i \overline{\{ x_ i\} }$ as sets and as algebraic spaces if endowed with reduced induced scheme structure.
Proof. Set $Z = \overline{\{ x\} } \subset |X|$ and $Z_ i = \overline{\{ x_ i\} } \subset |X_ i|$. Since $|X| \to |X_ i|$ is continuous we see that $Z$ maps into $Z_ i$ for each $i$. Hence we obtain an injective map $Z \to \mathop{\mathrm{lim}}\nolimits Z_ i$ because $|X| = \mathop{\mathrm{lim}}\nolimits |X_ i|$ as sets (Lemma 70.5.1). Suppose that $x' \in |X|$ is not in $Z$. Then there is an open subset $U \subset |X|$ with $x' \in U$ and $x \not\in U$. Since $|X| = \mathop{\mathrm{lim}}\nolimits |X_ i|$ as topological spaces (Lemma 70.5.2) we can write $U = \bigcup _{j \in J} f_ j^{-1}(U_ j)$ for some subset $J \subset I$ and opens $U_ j \subset |X_ j|$, see Topology, Lemma 5.14.2. Then we see that for some $j \in J$ we have $f_ j(x') \in U_ j$ and $f_ j(x) \not\in U_ j$. In other words, we see that $f_ j(x') \not\in Z_ j$. Thus $Z = \mathop{\mathrm{lim}}\nolimits Z_ i$ as sets.
Next, endow $Z$ and $Z_ i$ with their reduced induced scheme structures, see Properties of Spaces, Definition 66.12.5. The transition morphisms $X_{i'} \to X_ i$ induce affine morphisms $Z_{i'} \to Z_ i$ and the projections $X \to X_ i$ induce compatible morphisms $Z \to Z_ i$. Hence we obtain morphisms $Z \to \mathop{\mathrm{lim}}\nolimits Z_ i \to X$ of algebraic spaces. By Lemma 70.4.3 we see that $\mathop{\mathrm{lim}}\nolimits Z_ i \to X$ is a closed immersion. By Lemma 70.4.4 the algebraic space $\mathop{\mathrm{lim}}\nolimits Z_ i$ is reduced. By the above $Z \to \mathop{\mathrm{lim}}\nolimits Z_ i$ is bijective on points. By uniqueness of the reduced induced closed subscheme structure we find that this morphism is an isomorphism of algebraic spaces. $\square$
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### help virial theorem
Posted: Tue Aug 18, 2009 2:43 am
Can anyone give me a simple explanation for the virial theorem and an example. I understand it's something like <T>=-(1/2)U where T=total kinetic energy and U=total potential energy. Now I saw a solution on physicsGRE.net for 0177 problem 3 where somebody says U=3/2T. I may understand 3 as being something like the number of degrees of freedom for translational motion but I don't get how the 2 comes down instead of up.
Thanks a lot
### Re: help virial theorem
Posted: Tue Aug 18, 2009 8:45 am
why do you need virial thm to answer this question.. simply equate the centrifugal with the gravitational force (non inertial frame)? in physicsgre.net there are many redundant NEC's
### Re: help virial theorem
Posted: Tue Aug 18, 2009 9:46 am
I will explain briefly, not providing a full proof. Consider a collection of particles whose position vectors r_a and momenta p_a are both bounded (i.e. they remain finite for all values of the time). The virial theorem is -in a more general case- an expression about the average kinetic energy of the system of the forementioned particles. Specifically it is:
<T> = -(1/2) * <Σ F_a*r_a> (1)
where
T = total kinetic energy
Σ = summation over a = the number of particles of the system
F_a = the force on the a-th particle
r_a = the position of the a-th particle
where the right hand side was named by Clausius "virial". Notice also the factor -1/2 on the right hand side (RHS). If the forces on each particle are derivable from some potential U_a, then (1) transforms to the following form:
<T> = (1/2) <Σ r_a * grad(U_a)> (2), since in this case F_a = - grad(U_a).
Of particular interest is the case of 2 particles that interact via some central force-> F analogous to (r^n) , where n = some exponent.
In this case the potential energy of interaction takes the form U = k*r^(n+1), where k = proportionality constant. Thus, for this particular case it is (r = relative position of the 2 particles, U = potential energy of interaction), in spherical coordinates (see the RHS side of (2)):
r*grad(U) = r*dU/dr = k(n+1)*r^(n+1) = (n+1)*U (3), since U = k*r^(n+1).
Combining (3) and (2) (for the case of 2 particles always) we find that
<T> = (n+1)/2 * <U> (4).
If the mutual interaction is gravitational then n = -2 (see the force dependence on r above, how it was defined) and from (4) it comes out that
<T> = -(1/2) * <U> .
This is a useful relation for calculations concerning the planetary motion.
In the alternative answer of GR0177 #3, the argumentation is correct but the virial theorem does not seem to be employed correctly. In general, think carefully before reading the solutions given on this site. I have met several errors in the quoted answers. Think before you "digest" the quoted answer.
Physics_auth
### Re: help virial theorem
Posted: Tue Aug 18, 2009 10:34 am
Thank you physics_auth. I was scared that all I know suddenly became wrong... | 796 | 2,973 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2020-50 | latest | en | 0.890062 |
https://dmoj.ca/problem/lgc1p7/editorial | 1,675,703,179,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500356.92/warc/CC-MAIN-20230206145603-20230206175603-00191.warc.gz | 215,170,483 | 8,333 | ## Editorial for Lyndon's Golf Contest 1 P7 - Fun Factoring
Remember to use this editorial only when stuck, and not to copy-paste code from it. Please be respectful to the problem author and editorialist.
Submitting an official solution before solving the problem yourself is a bannable offence.
Author: Dingledooper
#### 83 bytes
Since can be as large as , an solution is insufficient. We can optimize it to by observing that for every factor of such that , there exists a corresponding factor . An example implementation is given below:
n=int(input())
i=1
while i*i<=n:
if n%i<1:
print(i)
if i*i-n:print(n//i)
i+=1
#### 64 bytes
In our -byte solution, an if statement is used to ensure that no duplicates are printed. Note also that print is repeated twice. We can simplify the logic by instead storing a set of {i,n//i}, and outputting the set with each number on its own line, which can be done with the sep argument. This yields a -byte solution:
n=int(input())
i=1
while i*i<=n:
if n%i<1:print(*{i,n//i},sep='\n')
i+=1
We can further shorten this to bytes by making use of Python's short-circuiting operators. With or, we can cause print to execute only when n%i is falsy, which then allows us to move everything onto one line:
n=int(input())
i=1
while i*i<=n:n%i or print(*{i,n//i},sep='\n');i+=1
There is still more to optimize here. In particular, another -byte save is possible by outputting via map:
n=int(input())
i=1
while i*i<=n:n%i<1in map(print,{i,n//i});i+=1
The mechanics are left as an exercise to the reader (Hint: look up "chained comparison").
#### 54 bytes
Recall back to the naive solution of iterating from to to check if it is a divisor. A natural question is, can we optimize this method by pruning specific values that don't need to be checked? Yes in fact, we can!
For the sake of example, take . Obviously, we know that divides once, and divides twice. But what about all the numbers in between? It's obvious that none of the numbers in the range need to be checked, since divided by any of these numbers would result in a value greater than but less than , which cannot be a whole number. The same argument can be applied starting from . We know that divides twice, and divides thrice, but since every number in the range divides by a fractional amount between and , those values can be pruned. These observations prompt the following strategy:
Initialize a variable , representing the current divisor to check for. We would like to skip all "unnecessary" values, which can formally be defined as all values , such that . Therefore, the next "useful" value is the largest such that , which we can calculate to be . Implementing this algorithm leads us to the -byte solution:
n=i=int(input())
while i:n%i or print(i);i=n//(n//i+1)
Now, let's try to prove that it has a time complexity of . Notice that the number of iterations is equivalent to asking for the number of distinct values of , for .
If , then there can obviously be at most distinct values of . If , then it implies that , which also can take on at most values. Therefore, the total complexity is . For reference, the exact number of iterations for any given is . | 766 | 3,181 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2023-06 | latest | en | 0.887545 |
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Concept: Significant Figure Rules with Multiplication and Division.
# Problem: You may want to reference (Pages 24 - 28) Section 1.6 while completing this problem.Carry out the following operations, and express the answers with the appropriate number of significant figures.(3.30×104)(0.2502)
###### FREE Expert Solution
Recall that when multiplying, we round the product or quotient so that it has the same number of significant figures as the least-precise measurement—the measurement with the fewest significant figures.
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###### Problem Details
You may want to reference (Pages 24 - 28) Section 1.6 while completing this problem.
Carry out the following operations, and express the answers with the appropriate number of significant figures.
(3.30×104)(0.2502) | 184 | 839 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2021-17 | latest | en | 0.823947 |
https://www.gradesaver.com/textbooks/math/trigonometry/CLONE-68cac39a-c5ec-4c26-8565-a44738e90952/chapter-5-trigonometric-identities-section-5-3-sum-and-difference-identities-for-cosine-5-3-exercises-page-221/82 | 1,721,043,797,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514696.4/warc/CC-MAIN-20240715102030-20240715132030-00688.warc.gz | 717,365,639 | 12,661 | ## Trigonometry (11th Edition) Clone
The reduction formula is $$\cot\theta$$
*Summary of the method: For a formula $f(Q\pm\theta)$ 1) See that $Q$ terminates on the $x$ or $y$ axis. If it terminates on the $x$ axis, go for Case 1. If it terminates on the $y$ axis, go for Case 2. 2) Case 1: - For a small positive value of $\theta$, determinate $Q\pm\theta$ lies in which quadrant. - If $f\gt0$, use a $+$ sign. If $f\lt0$, use a $-$ sign. - The reduced form will have that sign, $f$ the function and $\theta$ the angle. 3) Case 2: - For a small positive value of $\theta$, determinate $Q\pm\theta$ lies in which quadrant. - If $f\gt0$, use a $+$ sign. If $f\lt0$, use a $-$ sign. - The reduced form will have that sign, cofunction of$f$ as the function and $\theta$ the angle. $$\tan(270^\circ-\theta)$$ 1) $270^\circ$ terminates on the $y$ axis. We go for Case 2. 2) As $\theta$ is a very small positive value, which means $\theta\gt0$, $$270^\circ\gt(270^\circ-\theta)\gt180^\circ$$ So $270^\circ-\theta$ lies in quadrant III. 3) Tangent is positive in quadrant III. So we use a $+$ sign. 4) In case 2, we use the cofunction of the given formula as the function. That means the given formula is tangent, so the reduced form would have cotangent as the function, combined with the positive sign proved above. Overall, the reduced form would be $$\cot\theta$$ | 421 | 1,361 | {"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} | 4.375 | 4 | CC-MAIN-2024-30 | latest | en | 0.782987 |
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2-101.
The region bounded by $y = − x + 6$ and the coordinate axes is rotated about the $y$-axis. Calculate the volume of the resulting solid.
Sketch the region so you can visualize the flag.
The axis of rotation (the pole) is the $y$-axis.
The resulting solid will be a cone.
Which axis contains the radius ($r$) and which contains the height ($h$)? | 113 | 410 | {"found_math": true, "script_math_tex": 5, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-38 | latest | en | 0.812688 |
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Course name: Python for Data Science
#### These are the answers for course Python for Data Science NPTEL Assignment 3 Week 3
Q1. Choose the appropriate command(s) to filter those booking details whose reservation_status are a No-show?
a. data_hotel_ns datahotel. loc[data_hotel.reservation_status=’No-Show’]
b. data_hotel_ns = data_hotel[ data _hotel.reservation_status = “No-Show’]
c. data hotel_ns = data_hotel.reservation_status.loc[data_hotel.isin([‘No-Show’])]
d. data_hotel_ns = data_hotel.loc [data hotel.reservation_status.isin([ No-Show’])]
Q2. From the same data, find how many bookings were not cancelled in the year 2017?
a. 9064
b. 6231
c. 9046
d. None of the above
Q3. From the total bookings that were made in 2017 and not cancelled, which month had the highest number of repeated guests?
a. July
b. February
c. January
d. None of the above
These are the answers for course Python for Data Science NPTEL Assignment 3 Week 3
Q4. Which of the following commands can be used to create a variable Flag, and set the values as Premium when the rating is equal to or greater than 3.25, and otherwise as Regular?
a. dt_cocoa[°Flag’] = [“Premium” if x 3.25 else “Regular” for x in dt_cocoa[‘Rating’ ]]
b. dt_cocoa[“Flag’] = [“Premium” if x 3.25 else “Regular” for x in dt_cocoa[ _` Rating ‘]]
c. dt_cocoa[“Flag”] = np.where(dt_cocoa[ “Rating”] < 3.25, “Regular”, “Premium”)
d. None of the above
Q5. Which instruction can be used to impute the missing values in the column Review Data from the dataframe dt_cocoa by grouping the records company–wise?
Q6. After checking the data summary, which feature requires a data conversion considering the data values held?
a. Rating
b. Review Date
c. Company
d. None of the above
These are the answers for course Python for Data Science NPTEL Assignment 3 Week 3
Q7What is the maximum average rating for the cocoa companies based out of Guatemala?
a. 43.
b. 53.
c. 42.
d. None of the above
Q8. Which pandas function is used to stack the dataframes vertically?
a. pd.merge()
b. pd.concat()
c. join()
d. None of the above
These are the answers for course Python for Data Science NPTEL Assignment 3 Week 3
Q9. Of the following set of statements, which of them can be used to extract the column Direction as a separate dataframe?
a. df_weather[[_`Direction ‘ ]]
b. df_weather.iloc[:,0]
c. df_weather.loc[:.[ ‘Direction ‘]]
d. None of the above
These are the answers for course Python for Data Science NPTEL Assignment 3 Week 3
Q10. Which one of these students’ average score across all subjects was the lowest? Which subject has the highest average score across students?
a. Harini, Maths
b. Sathi, Maths
c. Harini, Physics
d. Rekha, Maths
These are the answers for course Python for Data Science NPTEL Assignment 3 Week 3
Python for Data Science NPTEL All weeks: https://progies.in/answers/nptel/python-for-data-science
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# Problems in parallel computations of tree functions
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### Problems in parallel computations of tree functions
1. 1. Algorithms Parallel Algorithms 1
2. 2. 2 An overview of lecture • The Euler tour technique • Computation of different tree functions • Tree contraction • Evaluation of arithmetic expressions
3. 3. 3 Problems in parallel computations of tree functions • Computations of tree functions are important for designing many algorithms for trees and graphs. • Some of these computations include preorder, postorder, inorder numbering of the nodes of a tree, number of descendants of each vertex, level of each vertex etc.
4. 4. 4 Problems in parallel computations of tree functions • Most sequential algorithms for these problems use depth-first search for solving these problems. • However, depth-first search seems to be inherently sequential in some sense.
5. 5. 5 Parallel depth-first search • It is difficult to do depth-first search in parallel. • We cannot assign depth-first numbering to the node n unless we have assigned depth- first numbering to all the nodes in the subtree A.
6. 6. 6 Parallel depth-first search • There is a definite order of visiting the nodes in depth-first search. • We can introduce additional edges to the tree to get this order. • The Euler tour technique converts a tree into a list by adding additional edges.
7. 7. 7 Parallel depth-first search • The red (or, magenta ) arrows are followed when we visit a node for the first (or, second) time. • If the tree has n nodes, we can construct a list with 2n - 2 nodes, where each arrow (directed edge) is a node of the list.
8. 8. 8 Euler tour technique • For a node v T, p(v) is the parent of v. • Each red node in the list represents an edge of the nature < p(v) , v >. • We can determine the preorder numbering of a node of the tree by counting the red nodes in the list.
9. 9. 9 Euler tour technique • Let T = (V, E) be a given tree and let T’ = (V, E’ ) be a directed graph obtained from T. • Each edge (u, v) E is replaced by two edges < u, v > and < v, u >. • Both the indegree and outdegree of an internal node of the tree are now same. • The indegree and outdegree of a leaf is 1 each. • Hence T’ is an Eulerian graph.
10. 10. 10 Euler tour technique • An Euler circuit of a graph is an edge-disjoint circuit which traverses all the nodes. • A graph permits an Euler circuit if and only if each vertex has equal indegree and outdegree. • An Euler circuit can be used for optimal parallel computation of many tree functions. • To construct an Euler circuit, we have to specify the successor edge for each edge.
11. 11. 11 Constructing an Euler tour • Each edge on an Euler circuit has a unique successor edge. • For each vertex v V we fix an ordering of the vertices adjacent to v. • If d is the degree of vertex v, the vertices adjacent to v are: adj(v) = < u0, u1, …, ud -1 > • The successor of edge < ui, v > is: s(< ui, v >) = < v, u(i + 1) mod d >, 0 i (d - 1)
12. 12. 12 Constructing an Euler tour
13. 13. 13 Correctness of Euler tour • Consider the graph T’ = (V, E’ ) , where E’ is obtained by replacing each e E by two directed edges of opposite directions. Lemma: The successor function s defines only one cycle and not a set of edge-disjoint cycles in T’. Proof: We have already shown that the graph is Eulerian. • We prove the lemma through induction.
14. 14. 14 Correctness of Euler tour basis: When the tree has 2 nodes, there is only one edge and one cycle with two edges. Suppose, the claim is true for n nodes. We should show that it is true when there are n + 1 nodes.
15. 15. 15 Correctness of Euler tour • We can introduce an extra node by introducing a leaf to an existing tree, like the leaf v. • Initially, adj(u) = <…, v’, v’’, …> . Hence, s(< v’, u >) = < u, v’’ >.
16. 16. 16 Correctness of Euler tour • After the introduction of v, adj(u) = <…, v’, v, v’’, …> • s(< v’, u >) = < u, v > and s(< v, u >) = < u, v’’ > • Hence, there is only one cycle after v is introduced.
17. 17. 17 Construction of Euler tour in parallel
18. 18. 18 Construction of Euler tour in parallel • We assume that the tree is given as a set of adjacency lists for the nodes. The adjacency list L[v] for v is given in an array. • Consider a node v and a node ui adjacent to v. • We need: – The successor < v, u(i + 1) mod d > for < ui, v >. This is done by making the list circular. – < ui, v >. This is done by keeping a direct pointer from ui in L[v] to v in L[ui].
19. 19. 19 Construction of Euler tour in parallel • We can construct an Euler tour in O(1) time using O(n) processors. • One processor is assigned to each node of the adjacency list. • There is no need of concurrent reading, hence the EREW PRAM model is sufficient.
20. 20. 20 Rooting a tree • For doing any tree computation, we need to know the parent p(v) for each node v. • Hence, we need to root the tree at a vertex r. • We first construct an Euler tour and for the vertex r, set s(< ud -1, r >) = 0. • ud -1 is the last vertex adjacent to r. • In other words, we break the Euler tour at r.
21. 21. 21 Rooting a tree
22. 22. 22 Rooting a tree
23. 23. 23 Rooting a tree Input: The Euler tour of a tree and a special vertex r. Output: For each vertex v r, the parent p(v) of v in the tree rooted at r.
24. 24. 24 Rooting a tree begin 1. Set s(< u, r >) = 0, where u is the last vertex in the adjacency list of r. 2. Assign a weight 1 to each edge of the list and compute parallel prefix. 3. For each edge < x, y >, set x = p(y) whenever the prefix sum of < x, y > is smaller than the prefix sum of < y, x >. end
25. 25. 25 Rooting a tree
26. 26. 26 Computation of tree functions • Given a tree T, for many tree computations: – We first construct the Euler tour of T – Then we root the tree at a vertex • We can compute: – The postorder number of each vertex – The preorder number of each vertex – The inorder number of each vertex – The level of each vertex – The number of descendants of each vertex.
27. 27. 27 Tree Contraction • Some tree computations cannot be solved efficiently with the Euler tour technique alone. • An important problem is evaluation of an arithmetic expression given as a binary tree.
28. 28. 28 Tree Contraction • Each leaf holds a constant and each internal node holds an arithmetic operator like +, . • The goal is to compute the value of the expression at the root. • The tree contraction technique is a systematic way of shrinking a tree into a single vertex. • We successively apply the operation of merging a leaf with its parent or merging a degree-2 vertex with its parent.
29. 29. 29 The RAKE operation • Let T = (V, E) be a rooted binary tree and for each vertex v, p(v) is its parent. • sib(v) is the child of p(v). We consider only binary trees. • In the rake operation for a leaf u such that p(u) r. – Remove u and p(u) from T, and – Connect sib(u) to p(p(u)).
30. 30. 30 The RAKE operation • In our tree contraction algorithm, we apply the rake operation repeatedly to reduce the size of the binary tree. • We need to apply rake to many leaves in parallel in order to achieve a fast running time.
31. 31. 31 The RAKE operation • But we cannot apply rake operation to nodes whose parents are consecutive on the tree. • For example, rake operation cannot be applied to nodes 1 and 8 in parallel. • We need to apply the rake operation to non- consecutive leaves as they appear from left to right.
32. 32. 32 The RAKE operation • We first label the leaves consecutively from left to right. • In an Euler path for a rooted tree, the leaves appear from left to right. • We can assign a weight 1 to each edge of the kind (v, p(v)) where v is a leaf. • We exclude the leftmost and the rightmost leaves. These two leaves will be the two children of the root when the tree is contracted to a three-node tree. • We do a prefix sum on the resulting list and the leaves are numbered from left to right.
33. 33. 33 The RAKE operation • We now store all the n leaves in an array A. • Aodd is the subarray consisting of the odd- indexed elements of A. • Aeven is the subarray consisting of the even- indexed elements of A. • We can create the arrays Aodd and Aeven in O(1) time and O(n) work.
34. 34. 34 Tree contraction algorithm begin for iterations do 1. Apply the rake operation in parallel to all the elements of Aodd that are left children 2. Apply the rake operation in parallel to the rest of the elements in Aodd. 3. Set A := Aeven. end log( 1)n
35. 35. 35 Tree contraction algorithm
36. 36. 36 Correctness of tree contraction • Whenever the rake operation is applied in parallel to several leaves, the parents of any two such leaves are not adjacent. • The number of leaves reduces by half after each iteration of the loop. Hence the tree is contracted in O(log n) time. • Euler tour takes O(n) work. • The total number of operations for all the iterations of the loop is: ( ( )) 2i i n O O n
37. 37. 37 An example
38. 38. 38 An example
39. 39. 39 Evaluation of arithmetic expressions • If we evaluate an expression tree bottom-up, it will take O(n) time for a long and skinny tree. • Hence we apply tree contraction.
40. 40. 40 Evaluation of arithmetic expressions • We do not completely evaluate each internal node. We evaluate the internal nodes partially. • For each internal node v, we associate a label (av, bv). av and bv are constants. • The value of the expression at node is: (av X + bv), where X is an unknown value for the expression of the subtree rooted at v.
41. 41. 41 Keeping an invariant Invariant: • Let u be an internal node which holds the operation , . • Let v and w are the children of u with labels (av, bv) and (aw, bw). • Then the value at u is: val(u) = (avval(v) + bv) (awval(w) + bw)
42. 42. 42 Keeping an invariant
43. 43. 43 Applying the rake operation • The value at node u is: val(u) = (avcv + bv) (aw X + bw) • X is the unknown value at node w.
44. 44. 44 Applying the rake operation • The contribution of val(u) to the value of node p(u) is: au val(u) + bu = au[(avcv + bv) (aw X + bw)] + bu • We can adjust the labels of node w to (a’w , b’w) • a’w = au(avcv + bv) aw • b’w = au(avcv + bv) bw + bu
45. 45. 45 Complexity of expression evaluation • The correctness of the expression evaluation depends on correctly maintaining the invariants. • We start with a label (1, 0) for each leaf and correctly maintain the invariant at each rake operation. • We have already proved the correctness of the rake operation. • Hence, evaluation of an expression given as a binary tree takes O(n) work and O(log n) time.
46. 46. 46 An example
47. 47. 47 An example
48. 48. 48 An example | 2,935 | 11,001 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-30 | latest | en | 0.835658 |
https://www.physicsforums.com/threads/find-the-missing-number.205629/ | 1,481,376,433,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698543170.25/warc/CC-MAIN-20161202170903-00459-ip-10-31-129-80.ec2.internal.warc.gz | 988,016,633 | 14,940 | # Find The Missing Number
1. Dec 21, 2007
### Zetep
Shown below is a 4 by 4 matrix:
999 998 997 996
729 648 567 486
126 192 210 192
12 18 ? 18
So far, I've not come up with the 'missing' number, but in the process of attempting to develop some sort of 'magic' solution, I've found some interesting relationships with the other numbers.
For example, the average of a(1,2) and a(1,4), 998 and 996, is the a(1,3) number 997. Then, the average of a(2,2) and a(2,4), 648 and 486, is the C(2,3) number 567.
There seems to be other relationships, i.e., a(3,2) plus a(4,2) equals a(3,3). So does
a(3,4) plus a(4,4) equal a(3,3). Plus other similar associations.
Evidently, my once keen mind is not so keen these days with being able to solve this 'mysterious' matrix. Any counsel, advice and/or assistance would be seriously appreciated.
Zetep
2. Dec 22, 2007
### james1234567890
The missing number is 0. Each number in the matrix is the product of the digits of the number in the previous row but the same column of the matrix. So the missing number (4,3) is product of the digits of (3,3) or 210 = 2*1*0 = 0.
3. Dec 22, 2007
### Zetep
Found The Missing Number
Thanks for your reply ... very much appreciated. Obviously, the solution is straightforward when the method you described is taken. I'm curious to know how it was that you chose to apply your approach. Were there 'clues' given the matrix numbers that appeared visible to your keen eye?
Zetep
4. Dec 22, 2007
### james1234567890
Thanks a lot for your appreciation. This idea struck me when I looked at the first two numbers in the first column i.e. 999 and 729. I remembered that 729 is 9 cubed i.e. 9 * 9 * 9. To my surprise, the number below also follwed the same convention i.e. 126 = 7 * 2 * 9. Thereafter I tried to apply this rule to all numbers in the martrix and it turned out to be correct. | 567 | 1,872 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2016-50 | longest | en | 0.901308 |
https://aprove.informatik.rwth-aachen.de/eval/JAR06/JAR_TERM/TRS/SK90/4.33.trs.Thm26:POLO_FILTER:NO.html.lzma | 1,718,457,023,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861594.22/warc/CC-MAIN-20240615124455-20240615154455-00496.warc.gz | 73,602,519 | 1,608 | Term Rewriting System R:
[x]
a(b(a(x))) -> b(a(b(x)))
Termination of R to be shown.
` R`
` ↳Dependency Pair Analysis`
R contains the following Dependency Pairs:
A(b(a(x))) -> A(b(x))
Furthermore, R contains one SCC.
` R`
` ↳DPs`
` →DP Problem 1`
` ↳Argument Filtering and Ordering`
Dependency Pair:
A(b(a(x))) -> A(b(x))
Rule:
a(b(a(x))) -> b(a(b(x)))
The following dependency pair can be strictly oriented:
A(b(a(x))) -> A(b(x))
There are no usable rules w.r.t. to the AFS that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
POL(b(x1)) = x1 POL(a(x1)) = 1 + x1 POL(A(x1)) = 1 + x1
resulting in one new DP problem.
Used Argument Filtering System:
A(x1) -> A(x1)
b(x1) -> b(x1)
a(x1) -> a(x1)
` R`
` ↳DPs`
` →DP Problem 1`
` ↳AFS`
` →DP Problem 2`
` ↳Dependency Graph`
Dependency Pair:
Rule:
a(b(a(x))) -> b(a(b(x)))
Using the Dependency Graph resulted in no new DP problems.
Termination of R successfully shown.
Duration:
0:00 minutes | 322 | 1,067 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2024-26 | latest | en | 0.676534 |
http://www.gamedev.net/topic/643343-chasing/ | 1,480,870,074,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541324.73/warc/CC-MAIN-20161202170901-00399-ip-10-31-129-80.ec2.internal.warc.gz | 483,136,913 | 28,065 | • Create Account
## chasing
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9 replies to this topic
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Posted 20 May 2013 - 05:56 PM
Hello , here is an image show what I am gonna to do in my XNA game , I want to check if the player is inside zone range and if yes then attack the player , I tried to do this using checking ray but it's odd and give null results as player must be in front of enemy so he can be detected !
### #2David.M Members
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Posted 20 May 2013 - 06:39 PM
Since you didn't specify, I'm assuming 2D for this. I'm also guessing each enemy has their "zone" or view distance stored with the rest of their information. You can find the distance between the player and the enemy by using the following
float x = player.x - enemy.x; // Distance between x positions of player and enemy
float y = player.y - enemy.y; // Likewise, difference between y values of positions
float distance = Math.sqrt((x * x) - (y * y)); // Replace with your math library - this is a Java implementation
Then, if the distance calculated is less than than the enemy's view distance or zone, the enemy attacks. For making sure the enemy actually 'sees' the player, you can do a check where, if the enemy is facing left and the player's x position is left of (less than) the enemy's, attack. If the player's x position is right of (greater than) the enemy's, attack. You can adjust this to have the enemy have a view angle as well so he can't see floor to ceiling but has a more realistic feeling field of view.
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Posted 21 May 2013 - 04:52 AM
thanks David , sorry I am assuming 3d of course , I have bounding box surrounding my player and also for the enemy , now I can fire a ray to check collision in front of enemy then if I am collide with the ray enemy can attack else enemy can't , but this looks odd and very bad as I must be in front of enemy , so instead I want a zone ( circle ) which is dangerous circle so that when I am inside this circle enemy can attack
Note : anyone have another strategy pls share with some code if you can .
David , you said that I can do this by using view angle for each enemy instead , can you post some code if you can taking in mind the ray I can use for each model in the game .
Edited by mrmohadnan, 21 May 2013 - 05:03 AM.
### #4belfegor Members
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Posted 21 May 2013 - 05:50 AM
It doesn't matter if it is 2d or 3d, you could use the same math
float enemyConeAngle = 0.6f;
float enemyZoneRadius = 10.0f;
vec3 enemyPos = enemy.pos;
vec3 enemyDir = normalize(enemy.LookAt - enemyPos);
vec3 playerPos = player.pos;
vec3 toPlayerDir = normalize(playerPos - enemyPos);
float angle = acos(dot(enemyDir, toPlayerDir));
float distance = sqrt(dot(playerPos - enemyPos));
if(angle < enemyConeAngle && distance < enemyZoneRadius)
{
enemy.state = chase_player;
}
...
if(enemy.state == chase_player)
{
float moveSpeed = 3.0f;
enemy.pos += toPlayerDir * moveSpeed * deltaTime; // move towards player
}
On top of my head, might made mistake somewhere.
Edited by belfegor, 21 May 2013 - 05:57 AM.
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Posted 21 May 2013 - 05:55 AM
Oh again brother belfegor thaaaaaaaaaanks , I tried this code but the enemy move at his place ( no translation is done ! )not move , what do think belfegor ?
double distanceX = (double)(cci.CharacterController.Body.Position.X - dwarfChrachterController.Body.Position.X);
double distanceZ = (double)(cci.CharacterController.Body.Position.Z - dwarfChrachterController.Body.Position.Z);
if (Math.Sqrt( (distanceX * distanceX) + (distanceZ * distanceZ)) <=500)
{
RunController(dwarfAnimatior, dwarfwalk);
Vector3 direction = cci.CharacterController.Body.Position - dwarfChrachterController.Body.Position;
Vector3.Normalize(direction);
dwarfChrachterController.Body.Position += direction * 5 * gameTime.ElapsedGameTime.Seconds;
if (enemyRay.Intersects(cci.CharacterController.Body.CollisionInformation.BoundingBox) <= 50)
{
sound.playAh();
}
}
Edited by mrmohadnan, 21 May 2013 - 05:58 AM.
### #6belfegor Members
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Posted 21 May 2013 - 06:08 AM
You should learn to debug your code a little.For example put some breakpoints to see what is going on. You could put one in body of if(Math.Sqrt(...)) statement to see if code that is inside is executed.
In my engine 500 would be half of kilometer.
Edited by belfegor, 21 May 2013 - 06:15 AM.
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Posted 21 May 2013 - 06:54 AM
belfegor , even when I am in the range he move at his position not translated toward me ? ok I will brothers thanks again
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Posted 21 May 2013 - 03:37 PM
...........................
Edited by mrmohadnan, 21 May 2013 - 04:27 PM.
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Posted 21 May 2013 - 04:16 PM
yeah , thaaaanks belfegor , I got it working and now it's moving toward me , now just orient the enemy when he move toward me so that his body.front direction facing me ! can you refine your code a little bit as you do with steering ?
Ray enmyRay;
enmyRay.Position = dwarfChrachterController.Body.Position;
enmyRay.Direction = Vector3.Normalize(dwarfChrachterController.Body.OrientationMatrix.Forward);
double distanceX = (double)(cci.CharacterController.Body.Position.X - dwarfChrachterController.Body.Position.X);
double distanceZ = (double)(cci.CharacterController.Body.Position.Z - dwarfChrachterController.Body.Position.Z);
Vector3 velocity =(cci.CharacterController.Body.Position- dwarfChrachterController.Body.Position) ;
if (Math.Sqrt( (distanceX * distanceX) + (distanceZ * distanceZ)) <=500)
{
RunController(dwarfAnimatior, dwarfwalk);
Vector3.Normalize(velocity);
velocity *= 5.0f;
dwarfChrachterController.HorizontalMotionConstraint.MovementDirection += new Vector2(velocity.X,velocity.Z);
if (enmyRay.Intersects(cci.CharacterController.Body.CollisionInformation.BoundingBox) <= 50)
{
RunController(dwarfAnimatior, dwarfattak);
sound.playAh();
}
velocity = cci.CharacterController.Body.Position - dwarfChrachterController.Body.Position;
}
Edited by mrmohadnan, 21 May 2013 - 04:47 PM.
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Posted 21 May 2013 - 10:22 PM
I'm going to have to agree that you could/should be able to figure this out simply by stepping through your code. Debugging 101, sir.
Dave Mark - President and Lead Designer of Intrinsic Algorithm LLC
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Old topic!
Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic. | 1,849 | 7,071 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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} | 3.25 | 3 | CC-MAIN-2016-50 | latest | en | 0.932402 |
https://scienceoxygen.com/what-does-0-01-mean-in-chi-square/ | 1,675,194,978,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499890.39/warc/CC-MAIN-20230131190543-20230131220543-00205.warc.gz | 532,330,370 | 24,621 | # What does 0.01 mean in chi-square?
NS Table d – Chi-square. Critical values of the Chi-square (X2) distribution at p = 0.05, 0.01, & 0.001 for d = 1 – 20 degrees of freedom.
## What are the statistical tests in a level biology?
This resource summarises the four statistical tests required for A level biology (Standard Deviation, T-test, Spearman Rank, Chi-squared).
## What is the critical value a level biology?
The critical value is the value of chi squared that corresponds to a 0.05 (5%) level of probability that the difference between our observed and expected results is due to chance.
## What is a null hypothesis in a level biology?
Biology definition: A null hypothesis is an assumption or proposition where an observed difference between two samples of a statistical population is purely accidental and not due to systematic causes.
## What is a chi square test used for in biology?
• Chi-squared tests are used to determine whether the difference between an observed and expected frequency. distribution is statistically significant. It is possible to infer whether two genes are linked or unlinked by looking at the frequency distribution of potential phenotypes.
## What is the Student t-test a level biology?
The Student’s t-test is a statistical test that compares the mean and standard deviation of two samples to see if there is a significant difference between them.
## What is a running mean in biology?
The running mean is a simple technique that allows you to judge whether or not you have enough measurements or counts. A more statistically valid approach to determine the number of repeats required is to calculate the running mean.
## How do you use the Hardy Weinberg equation a level biology?
1. Step 1: Find q.
2. Step 2: Find p (the frequency of the dominant allele F). If q = 0.32, and p + q = 1.
3. Step 3: Find p2 (the frequency of homozygous dominant genotype)
4. Step 4: Find 2pq = 2 x (p) x (q)
5. Step 5: Check calculations by substituting the values for the three frequencies into the equation; they should add up to 1.
## How much data do you need to get to apply the chi-square test?
1. Degrees of freedom. That’s just the number of categories minus 1.
2. The alpha level(α). This is chosen by you, or the researcher. The usual alpha level is 0.05 (5%), but you could also have other levels like 0.01 or 0.10.
## What is the hypothesis for chi-square test?
The null hypothesis of the Chi-Square test is that no relationship exists on the categorical variables in the population; they are independent.
## How do you write a H0 and H1 hypothesis?
H0: defendant is innocent; • H1: defendant is guilty. H0 (innocent) is rejected if H1 (guilty) is supported by evidence beyond “reasonable doubt.” Failure to reject H0 (prove guilty) does not imply innocence, only that the evidence is insufficient to reject it.
## What is p-value in biology?
DEFINITION OF THE P-VALUE In statistical science, the p-value is the probability of obtaining a result at least as extreme as the one that was actually observed in the biological or clinical experiment or epidemiological study, given that the null hypothesis is true [4].
## How do you write a null and alternative hypothesis examples?
Null Hypothesis: On the average, the dosage sold under this brand is 50 mg (population mean dosage = 50 mg). Alternative Hypothesis: On the average, the dosage sold under this brand is not 50 mg (population mean dosage ≠ 50 mg). This is a two-sided alternative hypothesis.
## When can I use chi-square?
Market researchers use the Chi-Square test when they find themselves in one of the following situations: They need to estimate how closely an observed distribution matches an expected distribution. This is referred to as a “goodness-of-fit” test. They need to estimate whether two random variables are independent.
## Where do we use chi-square test?
Chi-square is most commonly used by researchers who are studying survey response data because it applies to categorical variables. Demography, consumer and marketing research, political science, and economics are all examples of this type of research.
## What are the 3 types of t tests?
Types of t-tests There are three t-tests to compare means: a one-sample t-test, a two-sample t-test and a paired t-test. The table below summarizes the characteristics of each and provides guidance on how to choose the correct test.
## Which t-test should I use?
If you are studying one group, use a paired t-test to compare the group mean over time or after an intervention, or use a one-sample t-test to compare the group mean to a standard value. If you are studying two groups, use a two-sample t-test. If you want to know only whether a difference exists, use a two-tailed test.
## Does at test give you ap value?
Every t-value has a p-value to go with it. A p-value from a t test is the probability that the results from your sample data occurred by chance. P-values are from 0% to 100% and are usually written as a decimal (for example, a p value of 5% is 0.05). Low p-values indicate your data did not occur by chance.
## What does p 0.05 mean in chi-square?
A statistically significant test result (P ≤ 0.05) means that the test hypothesis is false or should be rejected. A P value greater than 0.05 means that no effect was observed.
## What is p-value in chi-square?
P value. In a chi-square analysis, the p-value is the probability of obtaining a chi-square as large or larger than that in the current experiment and yet the data will still support the hypothesis. It is the probability of deviations from what was expected being due to mere chance.
## What is the chi-square critical value at a 0.05 level of significance?
The Chi-Square critical value for a significance level of 0.05 and degrees of freedom = 11 is 19.67514. Thus, if we’re conducting some type of Chi-Square test then we can compare the Chi-Square test statistic to 19.67514.
## What is a 5 year running mean?
The 5-yr running means of annual-mean surface temperature anomalies (K) averaged over the Sahara Desert (red), the entire tropics (blue), and tropical land (green) for (a) ERAI, (b) NCEP-2, (c) MERRA, and the (d) CRU TS3.
## What is Chi in biology?
Chi-square Test for Independence is a statistical test commonly used to determine if there is a significant association between two variables. For example, a biologist might want to determine if two species of organisms associate (are found together) in a community. Does Species A associate with Species B? Species A.
## What is a correlation coefficient in biology?
Coefficient of correlation (r) is the degree of relationship between two variables, i.e., x and y, whereas coefficient of determination (R2) shows percentage variation in y which is explained by all the x variables together. The value of “r” may vary from −1 to +1, whereas the value of “r2” lies between 0 and +1. | 1,576 | 6,928 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2023-06 | latest | en | 0.89248 |
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A284308 Number A(n,k) of singular vector tuples for a general k-dimensional {n}^k tensor; square array A(n,k), n>=1, k>=1, read by antidiagonals. 10
1, 1, 1, 1, 2, 1, 1, 6, 3, 1, 1, 24, 37, 4, 1, 1, 120, 997, 240, 5, 1, 1, 720, 44121, 51264, 1621, 6, 1, 1, 5040, 2882071, 23096640, 2940841, 11256, 7, 1, 1, 40320, 260415373, 18754813440, 14346274601, 180296088, 79717, 8, 1, 1, 362880, 31088448777, 24874143759360, 153480509680141, 9859397817600, 11559133741, 572928, 9, 1 (list; table; graph; refs; listen; history; text; internal format)
OFFSET 1,5 LINKS Alois P. Heinz, Antidiagonals n = 1..18, flattened Shalosh B. Ekhad and Doron Zeilberger, On the Number of Singular Vector Tuples of Hyper-Cubical Tensors, 2016; also arXiv preprint arXiv:1605.00172, 2016. Shmuel Friedland and Giorgio Ottaviani, The number of singular vector tuples and uniqueness of best rank-one approximation of tensors, Found. Comput. Math. 14 (2014), no. 6, 1209-1242. Bernd Sturmfels, Tensors and Their Eigenvalues, Notices AMS, 63 (No. 6, 2016), 606-606. EXAMPLE Square array A(n,k) begins: 1, 1, 1, 1, 1, 1, ... 1, 2, 6, 24, 120, 720, ... 1, 3, 37, 997, 44121, 2882071, ... 1, 4, 240, 51264, 23096640, 18754813440, ... 1, 5, 1621, 2940841, 14346274601, 153480509680141, ... 1, 6, 11256, 180296088, 9859397817600, 1435747717722810960, ... CROSSREFS Columns k=1-9 give: A000012, A000027, A271905, A272551, A283829, A283830, A287083, A287084, A287085. Rows n=1-3 give: A000012, A000142, A274308. Main diagonal gives A284309. Sequence in context: A144351 A213936 A142589 * A172400 A226691 A158389 Adjacent sequences: A284305 A284306 A284307 * A284309 A284310 A284311 KEYWORD nonn,tabl AUTHOR Alois P. Heinz, Mar 24 2017 STATUS approved
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Last modified July 24 20:26 EDT 2021. Contains 346273 sequences. (Running on oeis4.) | 898 | 2,379 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2021-31 | latest | en | 0.396924 |
https://lambdapage.org/25096/spatial-reasoning-worksheets-free/kent-11-free-non-verbal-reasoning-worksheets-paro-tuition-intended-for-spatial-reasoning-worksheets-free/ | 1,620,290,996,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988753.91/warc/CC-MAIN-20210506083716-20210506113716-00341.warc.gz | 381,210,954 | 7,691 | # kent 11 free non verbal reasoning worksheets paro tuition intended for spatial reasoning worksheets free
##### kent 11 free non verbal reasoning worksheets paro tuition intended for spatial reasoning worksheets free.
For the advanced exercise, if you think that your kids have already got their line traced tidily, then you also can make the other line examples with more complex shapes.
This is the very first step before your kid can write on a blank paper. Never push them to do so until they can manage to trace perfectly. You also need to make a summative evaluation by start counting the number together with kid.
Introducing the writing method for early kids is actually recommended. It can be done by using the worksheet for kindergarten that specific for the writing progress.
The worksheet contains of several letter with the word. This is a little bit challenging because you also need to teach about spelling the word. However, this activity can be fun and give a satisfying result if you make a proper learning process.
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### Mandala Worksheets
March 9, 2021
This also can be done by giving them an offer about the reward if they can finish the task. After done with the preparation, you can start instruct the kid to tracing the shapes. At this stage, you also can introduce the name of the shapes.
The next worksheet has the main purposes for introducing the number for kids. Like the alphabet, you can make a preparation with your kid by teaching them about the number from 0 to 9.
Take a look on the colored pictures, whether it’s tidy enough or not. Give them an instruction that next time they should put the color inside the area. It will make them understand about coloring the object in proper way.
It consists of several picture that your kid might love to trace. You can introduce the picture and even telling a story as the preparation. It helps the kid to get interested and start working on their tracing process.
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### Two By Two Multiplication Worksheets
March 3, 2021
You also can make a creative process by also singing the alphabet. It helps your kids to memorize about the letters of the alphabet. Of course, it also can be a fun activity to do with your kid. After finished singing together, instruct your kid to follow your writing.
This can be a creative solution that will make your kid enhance their imagination and creativity. This free activity is also good for redeem the boredom. During this session, you can play a cheerful music on the background. Ask your kid to dance together and they will enjoy for this activity.
If you think that spelling words and recognizing the number is too complex, then you can try a simple worksheet that has main purpose of tracing skill. It’s a line tracing worksheet that can be used to achieve this skill.
Here are several creative worksheets that can be easily print and use. Children will love to improve their skill by using these worksheets. Moreover, it also gives an efficient method for the teaching from parents.
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March 9, 2021 | 760 | 3,331 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2021-21 | latest | en | 0.952598 |
https://pt.scribd.com/document/150075336/Exp-5-Intro-to-AC-Phase-Control | 1,563,617,278,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526506.44/warc/CC-MAIN-20190720091347-20190720113347-00483.warc.gz | 503,677,357 | 64,308 | Você está na página 1de 8
# EEK 471 LAB 5
Universiti Sains Malaysia Electrical Engineering Department Advanced Power Electronic Laboratory EEK471
## Introduction to AC Phase Control
OBJECTIVE To become familiar with the principles of ac phase control. To observe the operation of full-wave and half-wave phase control circuits.
INTRODUCTION Control principles In ac phase control, one or more thyristors are used to control the current in an ac circuit. The thyristors may also rectify the ac source voltage to convert it into dc. A thyristor begins to conduct when a firing pulse is applied to the gate, providing that the thyristor is forward biased at that time. The firing pulses are timed with respect to the phase of the ac source voltage. In a singlephase circuit, the control circuit detects the beginning (0') of each cycle of the ac source voltage, that is, when the waveform passes through zero, going from negative to positive. Then, the control circuits ends firing pulses to the various thyristors. Each pulse is delayed a certain number of degrees, so the thyristors begin to conduct at precisely the desired point in the cycle. Full-wave phase control The circuit of Figure 1 can provide full-wave ac phase control. With this circuit, and the appropriate control circuitry, you can control the power PO consumed by the load. A common example of this type of circuit is the "dimmer" circuit used to control incandescent lamps.
## Figure 1: Full-wave ac phase control circuit.
Figure 2 shows the waveforms associated with this circuit, when the load is purely resistive. The control circuit sends a firing pulse to the gate of Q 1, after the beginning of the cycle so that this thyristor begins to conduct. lt conducts for the remainder of the first half-cycle. A second firing pulse is sent to Q1, 180 after the pulse to Q1, causing Q2 to conduct from that point to the end of the second half-cycle.
## EEK 471 LAB 5
Figure 2: Waveforms for a full-wave phase control circuit. Whenc = 0, the phase-control circuit conducts for the entire cycle and the load receives maximum power. When is greater than 0, the power is "off" for part of each cycle. The RMS voltage, current, and power transferred to the load are therefore reduced. When the load is purely resistive, the active power PO = EO x I dissipated by the load is always equal to, or less than, the apparent power PS = ES x I supplied by the source. When is greater than 0, the current undergoes a phase shift with respect to the source voltage ES, as shown in Figure 2. Because the current now lags with respect to the voltage, the ac source has to supply reactive power Q in addition to the active power P. The reactive power grows as the firing angle is increased. Half-wave phase control: controlled rectifier supplying a passive load Figure 3 shows a half-wave phase control circuit using a single thyristor. This is a controlled rectifies supplying a passive load. When the load is purely resistive, the waveforms of this circuit are shown in Figure 4.
## EEK 471 LAB 5
Figure 4: Waveforms for hall-wave phase control circuit (resistive load). When the load is inductive, the output voltage can be negative for part of the cycle, as shown in Figure 5. This is because an inductor stores energy in its magnetic field for part of the cycle, then gradually releases this energy. Current continues to flow, and the thyristors continue to conduct, until all the stored energy is released. Since this occurs some time after the ac source voltage has passed through zero, the output voltage becomes negative for part of the cycle.
## EEK 471 LAB 5
The negative part of the output voltage waveform reduces the average output voltage EO. To prevent the output voltage from going negative, a free-wheeling diode can be placed in the circuit as shown in Figure 6. When the output voltage begins to go negative, the free-wheeling diode begins to conduct. This maintains the output voltage at approximately zero while the energy stored in the inductor is being released. The output voltage waveform is the same as for a purely resistive load (Figure 4), and the average output voltage is therefore greater than it would be without the free-wheeling diode.
## Figure 6: Half-wave phase control circuit with free-wheeling diode.
EQUIPMENTS EMS 8821 Enclosure Power Supply EMS 8840-0A PE Power Supply EMS 8841-2A Power Thyristor Module EMS 9030-30 Thyristor Firing Unit EMS 8425 Lab-Volt AC Ammeter EMS 8426 Lab-Volt AC Voltmeter EMS 8412-05 Lab-Volt DC Voltmeter/Ammeter EMS 8311 Variable Resistance EMS 8325 Smoothing Inductor EMS 9056-15 and EMS 9056-05 Voltage/Current Isolator 24V AC Power Switch Textronic Oscilloscope Connection Leads
WARNINGS The voltages and currents that are used during this lab are larger and rated at 240VAC Line-toNeutral with current as high as 20 amps (or higher if circuits are improperly connected). Please take the proper precautions and use your head before touching any circuitry. NEVER change any circuit connections while the power supply is turned on. Ask the demonstrator to check your connections before turning on the switches. And follow the rating of voltmeters and ammeters given to prevent equipments from damaged.
## EEK 471 LAB 5
Experiment: Part I Full-wave phase control 1. In this part, you will observe the operation of a full-wave phase control circuit made with two thyristors connected back to back. Set up the circuit of Figure 1a.
## Figure 1a: Circuit for full-wave phase control.
2. Set the 24V ac power switch to the 1 (ON) position. The POWER ON LED on the Power THYRISTOR module should light up to indicate that the module is correctly powered. 3. Make the following settings on the Thyristor Firing Unit ANGLE CONTROL COMPLEMENT ANGLE CONTROL ARC COSINE DC SOURCE FIRING CONTROL MODE 0 0 Minimum 1~
4. Use the setting for parameters as in Table 1a. Set the main power switch to 1 (ON), and set the voltage control knob to only 50(%). Table 1a: Parameters setting i1 (A) E1 ac (V) e1 (V) 5 300 600 R1() 240
## Line Voltage (V) 0 - 240
I1 ac (A) 1.5
5. Vary the FIRING ANGLE on the Thyristor Firing Unit by adjusting the DC SOURCE control. Describe the effect on the output waveforms and on the power delivered to the load. 6. Set the voltage control knob to the 0 % position, set the main power switch to the 0 position and turn OFF the power supply. Explain the operation of this circuit.
## EEK 471 LAB 5
Experiment: Part II Half-wave phase control 1. In this part, you will observe the voltage across the thyristor and across the load in a halfwave control circuit. Set up the circuit of Figure 2a.
## Figure 2a: Circuit to observe the voltage across the thyristor.
2. Set the 24V ac power switch to the 1 (ON) position. The POWER ON LED on the Power THYRISTOR module should light up to indicate that the module is correctly powered. 3. Use the similar settings on the Thyristor Firing Unit ANGLE CONTROL COMPLEMENT ANGLE CONTROL ARC COSINE DC SOURCE FIRING CONTROL MODE 0 0 Minimum 1~
4. Use the setting for parameters as in Table 2a. Set the main power switch to 1 (ON), and set the voltage control knob to only 50(%).
## Table 2a: Parameters setting I1 dc (mA) e1 (V) 300 600
R1() 240
5. Vary the FIRING ANGLE on the Thyristor Firing Unit while observing the oscilloscope. Adjust the FIRING ANGLE to 120 and sketch the waveform in Figure 2b. Note that you need to adjust the FIRING ANGLE as closely as possible to the value given.
## Figure 2b: Voltage waveforms for thyrisor ( = 120).
6. Set the voltage control knob to the 0 % position, set the main power switch to the 0 position and turn OFF the power supply. Disconnect the Voltage Isolator from the circuit and connect it across the load as shown in Figure 2c.
Figure 2c: Circuit to observe the voltage across the load. 7. Set the main power switch to 1 (ON), and set the voltage control knob to only 50(%). Vary the FIRING ANGLE on the Thyristor Firing Unit while observing the oscilloscope. Adjust the FIRING ANGLE to 120 and sketch the waveform in Figure 2b. What is the conduction angle of the thyristor when the firing angle is 120? Conduction angle = ___________ 8. Set the voltage control knob to 0 % then set the main power switch to O (OFF).
## EEK 471 LAB 5
REVIEW QUESTIONS 1. What point in a single-phase ac waveform is used as a reference point for timing the thyristor gate pulses? 2. Why are the gate pulses in a thyristor circuit delayed with respect to the reference point. 3. What is the relationship of the gate pulses to the two thyristors in a full-wave ac phase control circuit? 4. Why does the source for a thyristor circuit sometimes have to supply reactive power? 5. What is the advantage of using a thyristor in a battery charging circuit? | 2,115 | 8,789 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-30 | latest | en | 0.881089 |
https://cracku.in/15-what-is-the-value-of-mmamx-ysy-xxay-xfor-x2-y3-x-cat-1996 | 1,716,480,689,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058642.50/warc/CC-MAIN-20240523142446-20240523172446-00561.warc.gz | 162,281,227 | 23,882 | Instructions
Answer the questions based on the following information. A, S, M and D are functions of x and y, and they are defined as follows.
$$A(x, y)=x + y$$
$$S(x, y)=x-y$$
$$M(x, y)=xy$$
$$D(x,y)=\frac{x}{y}$$. $$y\neq0$$
Question 15
# What is the value of $$M(M(A(M(x, y),S(y, x)),x),A(y, x))$$for $$x=2, y=3$$?
Solution
Given expression can be reduced to
M ( x(xy+y-x) , (y+x) )
Or x(x+y)(xy+y-x)
After putting value of x and y , expression will reduce to a value = 70. | 178 | 489 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2024-22 | latest | en | 0.818209 |
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