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# A large cube 10cm on a side is composed of smaller cubes 1
Author Message
Manager
Joined: 27 Jul 2003
Posts: 122
Kudos [?]: 57 [0], given: 0
Location: Singapore
A large cube 10cm on a side is composed of smaller cubes 1 [#permalink]
### Show Tags
14 Oct 2003, 21:57
00:00
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This topic is locked. If you want to discuss this question please re-post it in the respective forum.
A large cube 10cm on a side is composed of smaller cubes 1 cm on a side. The large cube is printed red on its outside surfaces and then all the cubes are mixed in a drum. If one cube is selected randomly, which of the following is the approximate probability that the cube will have at least one red face?
A) 1:6
B) 1:5
C) 1:4
D) 1:3
E) 1:2
Kudos [?]: 57 [0], given: 0
CEO
Joined: 15 Aug 2003
Posts: 3454
Kudos [?]: 903 [0], given: 781
Re: PS: Cube and Drum [#permalink]
### Show Tags
15 Oct 2003, 00:00
araspai wrote:
A large cube 10cm on a side is composed of smaller cubes 1 cm on a side. The large cube is printed red on its outside surfaces and then all the cubes are mixed in a drum. If one cube is selected randomly, which of the following is the approximate probability that the cube will have at least one red face?
A) 1:6
B) 1:5
C) 1:4
D) 1:3
E) 1:2
P(atleast one red face) = 1 - P(no red face)
The unpainted "area of the cube" is 8*8*8
the total area = 10* 10*10
P(No reds) = 512/1000
P(atleast one red face) = 488/1000
Approximately 1:2
thanks
praetorian
Kudos [?]: 903 [0], given: 781
Re: PS: Cube and Drum [#permalink] 15 Oct 2003, 00:00
Display posts from previous: Sort by | 700 | 2,325 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2017-39 | latest | en | 0.910623 |
https://discourse.julialang.org/t/accessing-a-vector-of-vector-is-slow/115552 | 1,721,822,395,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518277.99/warc/CC-MAIN-20240724110315-20240724140315-00673.warc.gz | 187,072,426 | 7,090 | # Accessing a vector of vector is slow
made me wonder if a vector of vectors would be faster than a vector of struct for this case where the struct only contains Int. In fact it was much slower, but is there a better way to write double! and double_array! ?
``````using BenchmarkTools
mutable struct MutFoo
a::Int
b::Int
end
function double_b!(v::Vector{MutFoo})
for w in v
w.b *= 2
end
v
end
struct ImmutFoo
a::Int
b::Int
end
function double_b!(v::Vector{ImmutFoo})
for i in eachindex(v)
# @show i,v[i].a,v[i].b
v[i] = ImmutFoo(v[i].a, v[i].b * 2)
end
return
end
function double!(v::Vector{Vector{Int64}},size::Int64)
@inbounds for i in 1:size
# @show i,v[i].a,v[i].b
v[i][2] *= 2
end
return
end
function double_array!(v::Array{Int64,2},size::Int64)
@inbounds for i in 1:size
# @show i,v[i].a,v[i].b
v[2,i] *= 2
end
# v[2,1:size] .= @views v[2,1:size] * 2
return
end
size = 10^6
println("Benchmarking vector of mutable objects")
const mut_vec = [MutFoo(1, 1) for i in 1:size]
# @show mut_vec
@btime double_b!(mut_vec)
# @show mut_vec
println("Benchmarking vector of immutable objects")
const immut_vec = [ImmutFoo(0, 0) for i in 1:size]
@btime double_b!(immut_vec)
println("Benchmarking vector of vectors")
vec = Vector{Vector{Int64}}(undef,size)
for i in eachindex(vec)
vec[i] = Vector{Int64}(undef,2)
vec[i][1:2] .= 1,1
end
@btime double!(vec,size)
println("Benchmarking array")
arr = Array{Int64,2}(undef,2,size)
for (i,v) in enumerate(eachcol(arr))
arr[:,i] .= 1,1
end
@btime double_array!(arr,size)
``````
with the following results:
``````Benchmarking vector of mutable objects
WARNING: redefinition of constant Main.mut_vec. This may fail, cause incorrect answers, or produce other errors.
1.715 ms (0 allocations: 0 bytes)
Benchmarking vector of immutable objects
WARNING: redefinition of constant Main.immut_vec. This may fail, cause incorrect answers, or produce other errors.
183.700 ÎĽs (0 allocations: 0 bytes)
Benchmarking vector of vectors
5.547 ms (0 allocations: 0 bytes)
Benchmarking array
179.200 ÎĽs (0 allocations: 0 bytes)
``````
It seems to me that this could be a good application for StructArrays.jl
This gives for me:
Setup code
``````julia> using BenchmarkTools, StructArrays
julia> mutable struct MutFoo
a::Int
b::Int
end
julia> function double_b!(v::Vector{MutFoo})
for w in v
w.b *= 2
end
v
end
double_b! (generic function with 1 method)
julia> struct ImmutFoo
a::Int
b::Int
end
``````
``````## MutFoo
julia> @btime double_b!(mut_vec) setup=(mut_vec = [MutFoo(1, 1) for i in 1:10^6]);
2.862 ms (0 allocations: 0 bytes)
## StructArray{ImmutFoo}
julia> function double_b!(v::StructArray{ImmutFoo})
v.b .*= 2
end
julia> @btime double_b!(sa_vec) setup=(sa_vec=StructArray([ImmutFoo(1, 1) for i in 1:10^6]));
535.211 ÎĽs (0 allocations: 0 bytes)
## plain Array
julia> function double_b!(v::Array)
@views v[:,2] .*= 2
end
julia> @btime double_b!(arr) setup=(arr=ones(10^6,2));
560.075 ÎĽs (0 allocations: 0 bytes)
``````
Sorry for the confusion, but the array or vector of struct was for the linked post. This post concerns only the vector of vectors in double!.
You cannot be faster than with a plain Array. A vector of vectors will certainly always be slower since you need to do another whole memory lookup/fetch per element that the CPU cannot predict.
My original application was implementing an AVL tree, and then the struct is the tree node which contains the key, data, height, and links to left and right children. The tree would contain many nodes, hence the need for a fast implementation.
I tried an implementation using arrays for each field of the node, but it was slower than the one using structs. I guess this is due to cache locality.
Sounds reasonable but I wonder if it is true for larger arrays vs. structs? I does seem odd that a vector of vectors is so much slower though. Would it be the same in C?
A vector of `mutable struct` is much like a vector of `Vector`. The benefit to the struct is that that the compiler knows exactly how big it is so doesn’t need to boundscheck accesses to it, but this may or may not result in a significant performance change.
EDIT: I would need to look more carefully, but it’s probable that a `Vector{Vector{T}}` would actually have two additional levels of indirection, relative to a `Vector{MutFoo}`., because a `Vector` is a mutable object that contains a reference to a block of memory that holds the data. So it might actually be more detrimental than I had previously indicated.
A vector of (immutable) `struct` is more like a `Vector` of `Int`, in that the values are stored directly in the array (rather than storing a pointer to each mutable member). This saves a level of indirection on access, which results in fewer memory accesses and better cache locality. Further, it can move many operations that would be forced to happen in RAM (for a mutable container like `mutable struct` or `Vector`) into registers, avoiding system memory entirely for some operations.
In general, I find that maybe 90% of my structs are immutable. If you don’t need to change an object in one place and have it also changed in some other place, you probably don’t need a mutable. So containers are mostly mutable, but many “individual” objects get little value from mutability. Immutables usually have performance benefits, but (more importantly, to me) I find them to much easier to work with and reason about. Accessors.jl can ease the tedium of writing code to “modify” immutables.
StructArrays.jl is offers yet-another layout type. It is particularly valuable if you are often accessing some small part of every object (e.g., taking the average of a field across all array elements) rather than accessing several parts of an individual object together.
1 Like | 1,612 | 5,785 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.598888 |
https://www.freeonlinetest.in/question-and-answer/aptitude/volume-and-surface-area | 1,718,574,340,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861671.61/warc/CC-MAIN-20240616203247-20240616233247-00741.warc.gz | 704,590,983 | 9,385 | # Volume and Surface Area Aptitude Questions With Solutions, Online Test Volume and Surface Area
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Volume and Surface Area online test, mock test, practice sets evaluate aspirants mathematical ability to calculate the Volume and Surface Area of 3D shapes. Aptitude is one of the best topic those are preparing for competitive exams they need to practice on a daily basis to clear the exam. MCQs on volume and surface area are important for securing marks in various competitive exams.
Free online test is good platform to practice Volume and Surface Area questions with solutions.
Learn these tricks and tips to boost your volume and surface area calculations.
## Formulas on Volume and Surface Area
Right Circular Cylinder
Lateral Surface Area: 2πrh
Total Surface Area: 2πr(h+r)
Volume: $πr^2h$ (where r= Radius, h= Height).
Cube
Surface Area: $6a^2$ where a is the dimension of its side.
Volume: $a^3$ where a is the dimension of its side.
Cuboid
Surface Area: 2(lb+ bh+ hl).
Lateral Surface Area: 2(l + b) h (where l= Length, b= Breadth and h= Height)
Volume: l × b × h
Right Circular Cone
Lateral Surface Area: πrl
Total Surface Area: πr(l+r)
Volume: $2/3πr^2h$ (where r= Radius, l=Slant Height and h= Height)
Sphere
Surface Area: $4πr^2$
Volume: $4/3πr^3$ (where r= Radius)
Hemisphere
Curved Surface Area: $2πr^2$
Total Surface Area: $3πr^2$
Volume: $2/3πr^3$ (where r= Radius)
## Questions and Answers on Volume and Surface Area
Question-1)
If a cylinder and a cone have the same height and radius of the base, what is the ratio between the volume of the cylinder and the volume of the cone?
According to the question,
Ratio of their volumes
= ${[πr^2h]} / {[(1/3)πr^2h]}$
= $3/1$
=3 : 1
Question-2)
If a sphere of radius r has the same volume as a cone with a circular base of radius r, what is the height of the cone?
From question,
${\text"Volume of sphere"} = {\text"Volume of cone"}$
⇒ $({4}/{3})πr^3 = ({1}/{3})πr^2h$
⇒ h =4r
Question-3)
A tank is made of the shape of a cylinder with a hemispherical depression at one end. The height of the cylinder is 1.45 m and radius is 30 cm. The total surface area of the tank is?
Total surface area of tank = CSA of cylinder + CSA of hemisphere
= 2πrh + $2πr^2$ = 2πr(h + r)
=$2 × 22/7 × 30(145 + 30) cm^2$
=33000 $cm^2$
= 3.3 $m^2$
Question-4)
The radius of the top and bottom of a bucket of slant height 35 cm are 25 cm and 8 cm. The curved surface of the bucket is?
Curved surface of bucket = $π(r_1 + r_2)$ × slant height (l)
Curved Surface = $(22/7) × (25 + 8) × 35$
CSA = 22 × 33 × 5 = 3630 sq.cm.
Question-5)
The radius of the top and bottom of a bucket of slant height 35 cm are 25 cm and 8 cm. The curved surface of the bucket is?
According to question,
Surface area = $6a^2$ =726
⇒ $a^2$=121
⇒ a = 11 cm
Then, Volume of the cube = (11×11×11)$cm^3$
= $1331cm^3$
Online Test - 1 (Volume and Surface Area) TAKE TEST Number of questions : 20 | Time : 30 minutes
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Online Test - 4 (Volume and Surface Area) TAKE TEST Number of questions : 20 | Time : 30 minutes | 999 | 3,301 | {"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} | 4.1875 | 4 | CC-MAIN-2024-26 | latest | en | 0.863431 |
https://www.luogu.org/blog/Captain-Paul/bzoj2131-mian-fei-di-xian-bing | 1,571,176,639,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986660323.32/warc/CC-MAIN-20191015205352-20191015232852-00094.warc.gz | 983,265,854 | 3,613 | bzoj2131 免费的馅饼
2018-10-09 16:11:42
$f[i][j]$ 表示时刻 $i$ 在位置 $j$ 的最大收益
$2\times(t[i]-t[j])>=p[i]-p[j]$ 且 $2\times(t[i]-t[j])>=p[j]-p[i]$
$2\times t[i]+p[i]>=2\times t[j]+p[j]$
$2\times t[i]-p[i]>=2\times t[j]-p[j]$
#include<cstdio>
#include<cstring>
#include<cctype>
#include<algorithm>
#define reg register
using namespace std;
const int N=1e5+5;
struct node
{
int tim,pos,val,x,y;
inline friend bool operator < (node a,node b)
{
return a.x==b.x?a.y>b.y:a.x<b.x;
}
}a[N];
int n,W,f[N],c[N],t[N],cnt,ans;
{
int x=0,w=1;
char c=getchar();
while (!isdigit(c)&&c!='-') c=getchar();
if (c=='-') c=getchar(),w=-1;
while (isdigit(c))
{
x=(x<<1)+(x<<3)+c-'0';
c=getchar();
}
return x*w;
}
inline int lowbit(int x){return x&(-x);}
inline void modify(int x,int k){for (;x<=cnt;x+=lowbit(x)) c[x]=max(c[x],k);}
inline int query(int x,int res=0){for (;x;x-=lowbit(x)) res=max(res,c[x]); return res;}
int main()
{
a[0].x=a[0].y=a[0].tim=t[0]=-W;
for (reg int i=1;i<=n;i++)
{
a[i].x=a[i].tim+a[i].pos;
t[i]=a[i].y=a[i].tim-a[i].pos;
}
sort(a,a+n+1); sort(t,t+n+1);
cnt=unique(t,t+n+1)-t-1;
for (reg int i=0;i<=n;i++) a[i].y=lower_bound(t+1,t+cnt+1,a[i].y)-t;
for (reg int i=1;i<=n;i++)
{
f[i]=a[i].val+query(a[i].y);
modify(a[i].y,f[i]); ans=max(ans,f[i]);
}
printf("%d\n",ans);
return 0;
}
• star
首页 | 544 | 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": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2019-43 | longest | en | 0.145455 |
http://mathhelpforum.com/calculus/36462-problem-differential-equation-print.html | 1,527,298,204,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867254.84/warc/CC-MAIN-20180525235049-20180526015049-00057.warc.gz | 185,580,116 | 3,839 | # Problem with a differential equation
• Apr 28th 2008, 05:55 PM
arbolis
Problem with a differential equation
I'm having some trouble solving a simple differential equation, and yes I'm a bit lazy to check my notes. (Nod)
It's a simple one, here it is : We have that $\displaystyle x'=f(t,x)$, $\displaystyle x(0)=0$. Use the fact that $\displaystyle dt/dx=(dx/dt)^{-1}$ when dx/dt≠0.
$\displaystyle f(t,x)=x^{-2}$.
It's certainly a beginner question but I have a doubt in this case : x'=dx/dt? I guess yes, since otherwise I don't see what could be x'!
• Apr 28th 2008, 11:39 PM
Opalg
Quote:
Originally Posted by arbolis
I'm having some trouble solving a simple differential equation, and yes I'm a bit lazy to check my notes. (Nod)
It's a simple one, here it is : We have that $\displaystyle x'=f(t,x)$, $\displaystyle x(0)=0$. Use the fact that $\displaystyle dt/dx=(dx/dt)^{-1}$ when dx/dt≠0.
$\displaystyle f(t,x)=x^{-2}$.
It's certainly a beginner question but I have a doubt in this case : x'=dx/dt? I guess yes, since otherwise I don't see what could be x'!
Yes, x' is an alternative notation for dx/dt.
• Apr 29th 2008, 12:21 PM
arbolis
Still in the mud
Thanks Opalg. This notation confuses me. I don't really understand well how it can be solved. Are we looking for a function x(t)? Or just "x"?
Also... well I need to read much more on calculus, but why can we separate the dt from the expression dx/dt? It's not a division...
I think I can solve problems like this one : $\displaystyle x'=t^3$ and with the same initial condition as I posted at first. But when I have x' in function of x, then I'm lost. (that's why I'm posting here right now).
It's also confusing to me that instead of writing x'(t)=t for example, we just write x'=t.
What I've thought about the problem is $\displaystyle \frac{dx}{dt}=\frac{1}{x^2}\Leftrightarrow dx=\frac{1}{x^2} dt$. Now from there should I integrate??? I will have x= something depending of x but with dt at the end. So I guess this is not the way to do it.
Can someone give me a hint solving the differential equation?
• Apr 29th 2008, 12:27 PM
icemanfan
Quote:
Originally Posted by arbolis
Thanks Opalg. This notation confuses me. I don't really understand well how it can be solved. Are we looking for a function x(t)? Or just "x"?
Also... well I need to read much more on calculus, but why can we separate the dt from the expression dx/dt? It's not a division...
I think I can solve problems like this one : $\displaystyle x'=t^3$ and with the same initial condition as I posted at first. But when I have x' in function of x, then I'm lost. (that's why I'm posting here right now).
It's also confusing to me that instead of writing x'(t)=t for example, we just write x'=t.
What I've thought about the problem is $\displaystyle \frac{dx}{dt}=\frac{1}{x^2}\Leftrightarrow dx=\frac{1}{x^2} dt$. Now from there should I integrate??? I will have x= something depending of x but with dt at the end. So I guess this is not the way to do it.
Can someone give me a hint solving the differential equation?
You're almost there. This differential equation is separable. Just take $\displaystyle \frac{dx}{dt} = \frac{1}{x^2}$ and multiply both sides of the equation by $\displaystyle x^2 dt$ to yield $\displaystyle x^2 dx = dt$. Now you can integrate both sides.
• Apr 29th 2008, 12:45 PM
Opalg
Quote:
Originally Posted by arbolis
This notation confuses me. I don't really understand well how it can be solved. Are we looking for a function x(t)? Or just "x"?
That's the trouble with the x' notation: it doesn't make explicit what the variable is. In this case, it is implicitly understood that x is a function of the variable t.
In the same way, if you see the notation y', it will usually mean dy/dx, but you have to look at the rest of the problem to see whether the independent variable is in fact x (or it may be t, or some other letter).
• Apr 29th 2008, 12:55 PM
arbolis
I think I made an error. Here is what I've done :
$\displaystyle \frac{dx}{dt}=\frac{1}{x^2}\Leftrightarrow \int x^2 dx=\int dt\Leftrightarrow x=(3(t-C))^{1/3}$ soon after this we realize that C which is the constant produced when integrating is equal to 0. So at the end I found that $\displaystyle x(t)=(3t)^{1/3}$. Is it right?
• Apr 29th 2008, 12:57 PM
icemanfan
Quote:
Originally Posted by arbolis
I think I made an error. Here is what I've done :
$\displaystyle \frac{dx}{dt}=\frac{1}{x^2}\Leftrightarrow \int x^2 dx=\int dt\Leftrightarrow x=(3(t-C))^{1/3}$ soon after this we realize that C which is the constant produced when integrating is equal to 0. So at the end I found that $\displaystyle x(t)=(3t)^{1/3}$. Is it right?
If you can assume that C is equal to zero, then that answer is correct.
• Apr 29th 2008, 01:11 PM
arbolis
Thanks both for your valuable help! | 1,396 | 4,807 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2018-22 | latest | en | 0.921457 |
https://www.careercup.com/question?id=5725463187030016 | 1,555,626,092,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578526904.20/warc/CC-MAIN-20190418221425-20190419002445-00040.warc.gz | 648,225,445 | 11,867 | ## Google Interview Question for Software Engineers
Country: United States
Interview Type: In-Person
Comment hidden because of low score. Click to expand.
2
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``````public class MatrixPuzzle {
int column;
int row;
public MatrixPuzzle(int length, int width) {
this.column = length;
this.row = width;
}
//The dp formula will be M[i,j] = M[i - 1, j - 1] + M[i - 1, j] + M[i - 1, j + 1]
//Because one could only land on current cell from the 3 cells in the upper-left, left and lower-left.
//To make the space consumption 1d, cache the numbers in one column of the matrix at a time.
//Follow-up 2: Just reset path-counts for blocked cells to 0
public int numberOfPaths() {
int[] paths = new int[row];
paths[row - 1] = 1; //start from bottom-left corner
for(int col = 1; col < column; col++) {
int upper_left_value = 0;
for(int r = 0; r < row; r++) {
int left_value = paths[r];
paths[r] += upper_left_value + (r == row - 1 ? 0 : paths[r + 1]);
upper_left_value = left_value;
}
}
return paths[row - 1]; //exit from bottom-right
}
}``````
Comment hidden because of low score. Click to expand.
1
of 1 vote
Using DP and dealing with blocked cells
``````def findWaysBottomUp(height, width, blocked=set()):
d = [0] * height
d[-1] = 1
for col in range(width - 2, -1, -1):
nextD = list(d)
for row in range(height):
top = d[row - 1] if row > 0 and (row, col) not in blocked else 0
bottom = d[row + 1] if row < height - 1 and (row, col) not in blocked else 0
right = d[row] if (row, col) not in blocked else 0
nextD[row] = top + right + bottom
d = nextD
return d[-1]``````
Comment hidden because of low score. Click to expand.
0
of 0 vote
Using DFS of a graph approach with recursion
``````package main
import(
"fmt"
)
var (
RIGHT = 0
UPPERRIGHT = 1
LOWERRIGHT = 2
)
func getNewCoordsAfterMove(i,j, dir, r, c int) (bool, int,int) {
if dir == RIGHT {
i = i
j = j + 1
if j < c {
return true, i, j
}
return false, i, j
}
if dir == UPPERRIGHT {
i = i - 1
j = j + 1
if j < c && i >= 0 {
return true, i, j
}
return false, i, j
}
if dir == LOWERRIGHT {
i = i + 1
j = j + 1
if j < c && i < r {
return true, i, j
}
return false, i, j
}
return false, -1, -1
}
func findNoOfpaths(i, j int, row, column int) int{
if i == (row -1) && j == (column -1) {
return 1
}
RPaths := 0
validRMove, newI, newJ := getNewCoordsAfterMove(i,j, RIGHT, row, column)
if validRMove {
RPaths = findNoOfpaths(newI, newJ, row, column)
}
URPaths := 0
validURMove, newI, newJ := getNewCoordsAfterMove(i,j, UPPERRIGHT, row, column)
if validURMove {
URPaths = findNoOfpaths(newI, newJ, row, column)
}
LRPaths := 0
validLMove, newI, newJ := getNewCoordsAfterMove(i,j, LOWERRIGHT, row, column)
if validLMove {
LRPaths = findNoOfpaths(newI, newJ, row, column)
}
return RPaths + URPaths + LRPaths
}
func main() {
fmt.Println("No. Of Paths for 2 X 2", findNoOfpaths(1,0, 2,2))
fmt.Println("No. Of Paths for 2 X 3", findNoOfpaths(1,0, 2,3))
fmt.Println("No. Of Paths for 2 X 4", findNoOfpaths(1,0, 2,4))
fmt.Println("No. Of Paths for 3 X 2", findNoOfpaths(2,0, 3,2))
fmt.Println("No. Of Paths for 3 X 4", findNoOfpaths(2,0, 3,4))
}``````
Comment hidden because of low score. Click to expand.
0
of 0 vote
``````public static int getPaths(int length, int width, HashSet<String> blocked) {
String bottomLeft = String.valueOf(width-1) + " 0";
String bottomRight = String.valueOf(width-1) + " " + String.valueOf(length-1);
if ((length == 0) || (width == 0) || (blocked.contains(bottomLeft)) || (blocked.contains(bottomRight))) return 0;
int[] dp = new int[width];
dp[0] = 1;
for (int i = 1; i < length; i++) {
int store = 0;
for (int j = 0; j < width; j++) {
int temp = dp[j];
dp[j] = dp[j] + ((j == width-1) ? 0 : dp[j+1]) + store;
String currLocation = String.valueOf(width-1-j) + " " + String.valueOf(i);
if (blocked.contains(currLocation)) dp[j] = 0;
store = temp;
}
}
return dp[0];
}
public class Pair {
int row;
int col;
Pair(int r, int c) {
row = r;
col = c;
}
}``````
Comment hidden because of low score. Click to expand.
0
of 0 vote
``````public static int pathCount(int[][] matrix, int i, int j){
if(i <0 || i >= matrix.length || j <0 || j >= matrix[0].length) return 0;
if(matrix[i][j] >=0) return matrix[i][j];
matrix[i][j] = pathCount(matrix, i, j-1) + pathCount(matrix, i+1, j-1) + pathCount(matrix, i-1, j-1);
return matrix[i][j];
}``````
Comment hidden because of low score. Click to expand.
0
of 0 vote
O(width*height) time, O(height) space
Goes column by column computing number of paths based on the previous column
``````int countPaths(int grid_width, int grid_height){
vector<int> last_column;
last_column.resize(grid_height, 0);
vector<int> current_column;
current_column.resize(grid_height, 0);
last_column[last_column.size()-1] = 1;
for(int i=1;i<grid_width;++i){
for(int j=0;j<grid_height;++j){
current_column[j] = last_column[j];
if(j>0){
current_column[j] += last_column[j-1];
}
if(j<grid_height-1){
current_column[j] += last_column[j+1];
}
}
for(int j=0;j<grid_height;++j){
last_column[j] = current_column[j];
}
}
return last_column[grid_height-1];
}``````
Comment hidden because of low score. Click to expand.
-1
of 1 vote
Using DFS to aggregate all the paths:
path.h
``````#ifndef PATH_H
#define PATH_H
#include <vector>
#include <utility>
#include <queue>
#include <map>
namespace MatrixPathProblem {
typedef std::pair<int, int> Node;
typedef std::vector<std::pair<int, int> > Path;
class Matrix
{
public:
Matrix(const std::vector<std::vector<int> >& matrix);
Matrix(const int** matrix, const size_t length, const size_t width);
inline int length() const { return m_matrix.size(); }
inline int width() const { return m_matrix.size() ? m_matrix.at(0).size() : 0; }
inline bool empty() const { return m_matrix.empty(); }
Node moveRight(const Node& node) const;
Node moveUpperRight(const Node& node) const;
Node moveLowerRight(const Node& node) const;
private:
bool isNodeValid(const Node& node) const;
std::vector<std::vector<int> > m_matrix;
};
class Paths
{
public:
void printPaths(const Node& node) const;
void pushPath(const Node& node,
const Path& path);
void pushPaths(const Node& node,
const Node& fromNode);
bool visited(const Node& node) const;
std::vector<Path> getPaths(const Node& node) const;
private:
std::map<Node,
std::vector<Path> > m_pathsMap;
};
} // namespace MatrixPathProblem
#endif // PATH_H``````
path.cpp
``````#include "path.h"
#include <iostream>
namespace MatrixPathProblem {
Matrix::Matrix(const std::vector<std::vector<int> >& matrix)
:m_matrix(matrix)
{
}
Node Matrix::moveRight(const Node& node) const
{
Node newNode(node.first+1, node.second);
if(isNodeValid(newNode))
{
return newNode;
}
return Node(-1, -1);
}
Node Matrix::moveUpperRight(const Node& node) const
{
Node newNode(node.first+1, node.second-1);
if(isNodeValid(newNode))
{
return newNode;
}
return Node(-1, -1);
}
Node Matrix::moveLowerRight(const Node& node) const
{
Node newNode(node.first+1, node.second+1);
if(isNodeValid(newNode))
{
return newNode;
}
return Node(-1, -1);
}
bool Matrix::isNodeValid(const Node& node) const
{
if(node.first < 0 || node.first >= length())
return false;
if(node.second <0 || node.second >= width())
return false;
return true;
}
void Paths::printPaths(const Node& node) const
{
std::map<Node, std::vector<Path> >::const_iterator itr = m_pathsMap.find(node);
if(itr == m_pathsMap.end())
{
std::cout << "No path available" << std::endl;
}
for(std::vector<Path>::const_iterator pitr = itr->second.begin();
pitr != itr->second.end();
++pitr)
{
for(Path::const_iterator vitr = pitr->begin();
vitr != pitr->end();
++vitr)
{
std::cout << "(" << vitr->first << "," << vitr->second << ")->";
}
std::cout << "(" << node.first << "," << node.second << ")" << std::endl;
}
}
void Paths::pushPath(const Node& node, const Path& path)
{
auto& paths = m_pathsMap[node];
paths.push_back(path);
}
void Paths::pushPaths(const Node& node, const Node& fromNode)
{
auto pathsItr = m_pathsMap.find(fromNode);
if(pathsItr == m_pathsMap.end())
{
// No path available at the moment
// so we will just push the path including
// fromNode only
Path createdPath;
createdPath.push_back(fromNode);
pushPath(node, createdPath);
return;
}
for(std::vector<Path>::const_iterator pitr = pathsItr->second.begin();
pitr != pathsItr->second.end();
++pitr)
{
Path createdPath(*pitr);
createdPath.push_back(fromNode);
pushPath(node, createdPath);
}
}
bool Paths::visited(const Node& node) const
{
if (m_pathsMap.find(node) != m_pathsMap.end())
return true;
return false;
}
std::vector<Path> Paths::getPaths(const Node& node) const
{
auto pitr = m_pathsMap.find(node);
if(pitr == m_pathsMap.end())
{
return std::vector<Path>();
}
return pitr->second;
}
void findPath(const Matrix& matrix)
{
Paths paths;
std::queue<Node> queue;
Node startNode(0,0);
Node endNode(matrix.length()-1, matrix.width()-1);
if(matrix.empty())
{
return;
}
// Push the first node
queue.push(startNode);
while(!queue.empty())
{
auto node = queue.front();
queue.pop();
auto rightNode = matrix.moveRight(node);
auto upperRightNode = matrix.moveUpperRight(node);
auto lowerRightNode = matrix.moveLowerRight(node);
if(rightNode.first >= 0)
{
if(!paths.visited(rightNode))
{
queue.push(rightNode);
}
// Add all available paths to the right node
paths.pushPaths(rightNode, node);
}
if(upperRightNode.first >= 0)
{
if(!paths.visited(upperRightNode))
{
queue.push(upperRightNode);
}
// Add all available paths to the upperRightNode
paths.pushPaths(upperRightNode, node);
}
if(lowerRightNode.first >= 0)
{
if(!paths.visited(lowerRightNode))
{
queue.push(lowerRightNode);
}
// Add all available paths to the lowerRightNode
paths.pushPaths(lowerRightNode, node);
}
}
paths.printPaths(endNode);
}
} // namespace MatrixPathProblem
int main()
{
std::vector<std::vector<int> > matrix({
{1, 0, 0 ,0},
{0, 1, 1, 0},
{1, 1, 1, 1},
{1, 1, 1, 1},
{1, 1, 1, 1}
});
MatrixPathProblem::Matrix pathMatrix(matrix);
MatrixPathProblem::findPath(pathMatrix);
}``````
Name:
Writing Code? Surround your code with {{{ and }}} to preserve whitespace.
### Books
is a comprehensive book on getting a job at a top tech company, while focuses on dev interviews and does this for PMs.
### Videos
CareerCup's interview videos give you a real-life look at technical interviews. In these unscripted videos, watch how other candidates handle tough questions and how the interviewer thinks about their performance. | 3,215 | 10,963 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2019-18 | longest | en | 0.722224 |
https://www.impactlab.com/2011/12/01/one-of-the-earliest-known-examples-of-math-homework/ | 1,721,792,106,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518154.91/warc/CC-MAIN-20240724014956-20240724044956-00734.warc.gz | 699,282,557 | 22,020 | Math on the go.
It’s stuff like this that makes me love archaeology. Turns out, we can trace the concept of math homework back to at least 2300 B.C.E., in ancient Mesopotamia.
In the early 20th century, German researchers found several clay tablets at the site ofŠuruppak. (Today, that’s basically the Iraqi city of Tell Fara.) Some of the tablets appear to be the remains of math instruction, including two different tablets that are working the same story problem…
A loose translation of the problem is: A granary. Each man receives 7 sila of grain. How many men? That is, the tablets concern a highly artificial problem and certainly present a mathematical exercise and not an archival document. The tablets give the statement of the problem and its answer (164571 men – expressed in the sexagesimal system S since we are counting men – with 3 sila left over). However, one of the tablets gives an incorrect solution. When analyzing these tablets, Marvin Powell commented famously that it was, “written by a bungler who did not know the front from the back of his tablet, did not know the difference between standard numerical notation and area notation, and succeeded in making half a dozen writing errors in as many lines.”
That comes from a site set up by Duncan Mellville, a math professor at St. Lawrence University in Canton, NY. He’s actually got a whole collection of essays on Mesopotamian mathematics. I am certain, that by posting this, I’ve just ruined somebody’s productivity for, like, a week.
Image is not THE cuneiform tablet in question. Just A cuneiform tablet. I couldn’t find a picture of those specific ones:Marks and signs, a Creative Commons Attribution Share-Alike (2.0) image from nicmcphee’s photostream.
Via John Baez | 395 | 1,753 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.920585 |
https://www.sympygamma.com/input/?i=diff%28f%28x%29%2Ag%28x%29%2Ah%28x%29%29 | 1,591,107,779,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347425148.64/warc/CC-MAIN-20200602130925-20200602160925-00550.warc.gz | 888,602,246 | 2,482 | # SymPy Gamma
diff (f(x)*g(x)*h(x))
diff(f(x)*g(x)*h(x), x)
$\frac{d}{d x}\left(f{\left (x \right )} g{\left (x \right )} h{\left (x \right )}\right) =$
diff(f(x)*g(x)*h(x), x)
solve(f(x)*g(x)*h(x), x)
$x =$
series(f(x)*g(x)*h(x), x, 0, 10)
See what Wolfram|Alpha has to say.
Experiment with SymPy at SymPy Live. | 133 | 314 | {"found_math": true, "script_math_tex": 2, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2020-24 | longest | en | 0.597651 |
https://help.scilab.org/docs/5.3.2/en_US/integrate.html | 1,610,969,681,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703514495.52/warc/CC-MAIN-20210118092350-20210118122350-00672.warc.gz | 372,224,621 | 7,242 | Scilab Home page | Wiki | Bug tracker | Forge | Mailing list archives | ATOMS | File exchange
Change language to: Français - Português - 日本語
See the recommended documentation of this function
Scilab help >> Differential Equations, Integration > integrate
# integrate
integration of an expression by quadrature
### Calling Sequence
x=integrate(expr,v,x0,x1 [,atol [,rtol]])
### Arguments
expr
Character string defining a Scilab expression.
v
Character string, the integration variable name)
x0
real number, the lower bound of integration
x1
vector of real numbers, upper bounds of integration
atol
real number (absolute error bound) Default value: 1.-8
rtol
real number, (relative error bound) Default value: 1e-14
x
vector of real numbers, the integral value for each x1(i).
### Description
computes : for i=1:size(x1,'*')
Where is given by the expression expr.
The evaluation hopefully satisfies following claim for accuracy: abs(I-x)<= max(atol,rtol*abs(I)) where I stands for the exact value of the integral.
### Restriction
the given expression should not use variable names with a leading %.
### Examples
x0=0;x1=0:0.1:2*%pi;
X=integrate('sin(x)','x',x0,x1);
norm(cos(x1)-(1-X))
x1=-10:0.1:10;
X=integrate(['if x==0 then 1,';
'else sin(x)/x,end'],'x',0,x1)
• intg — definite integral
• inttrap — integration of experimental data by trapezoidal interpolation
• intsplin — integration of experimental data by spline interpolation
• ode — ordinary differential equation solver | 393 | 1,509 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-04 | longest | en | 0.613615 |
https://math.stackexchange.com/questions/1065389/is-mathcalo-k-always-isomorphic-to-mathbbzx-fx-for-some-irreduc | 1,652,718,887,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662510138.6/warc/CC-MAIN-20220516140911-20220516170911-00656.warc.gz | 458,186,170 | 66,608 | # Is $\mathcal{O}_K$ always isomorphic to $\mathbb{Z}[X]/(f(x))$, for some irreducible polynomial $f(x)$?
1. Given an algebraic number field $K$ and its ring of integers $\mathcal{O}_K$, is $\mathcal{O}_K$ always isomorphic to $\mathbb{Z}[X]/(f(x))$, for some irreducible polynomial $f(x)$?
2. Since $\mathcal{O}_K/\mathcal{m}$ is a finite field for any non-zero prime ideal $m \subset \mathcal{O}_K$, it is isomorphic to some $\mathbb{F}_p[X]/(\widetilde{f}(x))$. If the statment 1. is true, is it also true that the $\widetilde{f}(x) = f(x) \bmod{p}$, and $\mathcal{O}_K/\mathcal{m} \cong \mathbb{Z}[X]/(p,f(x)) \cong \mathbb{F}_p[X]/(\widetilde{f}(x))$?
If these are not true for any algebraic number field could you point me in the right direction? I have been reading some notes, that are apparently incomplete, and I would appreciate a reference to where I could find the proofs of these (or the most similar general) statements.
• I don't have the tools to easily carry out this construction, but find any cubic number field that has three degree $1$ primes lying over $2$. Every algebraic integer $\alpha$ will have the same image in two of the residue fields $\mathbf{F}_2$, and therefore the image of $\mathbf{Z}[\alpha]$ is the diagonal (proper) subring of the corresponding residue ring $\mathbf{F}_2 \times \mathbf{F}_2$. Since $\mathcal{O}$ surjects onto the residue ring, we see $\mathbf{Z}[\alpha] \subset \mathcal{O}$.
– user14972
Dec 12, 2014 at 22:03
• @Hurkyl I understand. Thank you. Dec 12, 2014 at 22:45
Let me elaborate on the "strange" behavior of ring of integers. Let $K$ be an algebraic number field with ring of integers $\mathcal O_K$. Given an algebraic integer $\alpha \in \mathcal O_K$, let us denote by $\mathbf Z[\alpha]$ the $\mathbf Z$-algebra generated by $\alpha$. (This is isomorphic to $\mathbf Z[X]/(f)$, where $f$ is the minimal polynomial of $\alpha$).
Now the basic question is: Does there exist $\alpha \in \mathcal O_K$ such that $\mathcal O_K = \mathbf Z[\alpha]$? (In this case some people say $K$ is monogenic or $K$ admits a power integral basis).
Let $i(\alpha) = [ \mathcal O_K : \mathbf Z[\alpha]]$ be the order of $\mathcal O_K/\mathbf Z[\alpha]$. Does there always exists $\alpha \in \mathcal O_K$ such that $i(\alpha) = 1$ (which is equivalent to $\mathcal O_K = \mathbf Z[\alpha]$)? In general the answer is no. Consider for example the number field $K$ definied by a root of $X^3 + X^2 - 2X - 8 \in \mathbf Z[X]$. Then it is not hard to show that $2$ divides $i(\alpha)$ for all $\alpha \in \mathcal O_K$. In particular we can never have $\mathcal O_K = \mathbf Z[\alpha]$. Already in 1871, Dedekind gave this example and knew about these so called common non-essential discriminant divisor (gemeinsame außerwesentliche Diskriminantenteiler).
These common non-essential discriminant divisors are not the only obstruction. For a deeper investigation of this topic, one can use the so called index form.
I hope this gives you enough information/keywords to find more in the literature. Tell me if you need explicit references.
The $\mathbb{Z}$-module $M=\mathbb{Z}[X]/(f)$ is generated as an algebra by a single element in $M$, since $M = \mathbb{Z}[ \overline{X} ]$ where $\overline{X} = X + (f)$. So if it was true, what you were asking in question 1, it will follow that the ring of integers are generated by a single algebraic integer. This is not true in general. Look for an example of a ring of integers that fails to be generated by a single element.
• I understand. On the other hand I know that always $\mathcal{O}_K/\mathcal{m} \cong \mathbb{F}_p[X]/(\widetilde{f}(x))$ for some irreducible $\widetilde{f}(x)$, so it seems strange to me that the ring of integers itself is not isomorphic to any quotient of $\mathbb{Z}[X]$. Dec 12, 2014 at 21:36 | 1,151 | 3,824 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2022-21 | latest | en | 0.817196 |
https://kr.mathworks.com/matlabcentral/cody/problems/2015-length-of-the-hypotenuse/solutions/3419753 | 1,610,993,723,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703515075.32/warc/CC-MAIN-20210118154332-20210118184332-00107.warc.gz | 409,434,034 | 16,945 | Cody
# Problem 2015. Length of the hypotenuse
Solution 3419753
Submitted on 27 Oct 2020 by Madison Mullins
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 = 1; b = 2; c_correct = sqrt(5); tolerance = 1e-12 ; assert(abs(hypotenuse(a,b)-c_correct)<tolerance);
c = 2.2361
2 Pass
a = 3; b = 4; c_correct = 5; tolerance = 1e-12 ; assert(abs(hypotenuse(a,b)-c_correct)<tolerance);
c = 5
3 Pass
a = 5; b = 12; c_correct = 13; tolerance = 1e-12 ; assert(abs(hypotenuse(a,b)-c_correct)<tolerance);
c = 13
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Start Hunting! | 244 | 757 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2021-04 | latest | en | 0.701927 |
https://www.doubtnut.com/qna/34609242 | 1,721,281,794,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514822.16/warc/CC-MAIN-20240718034151-20240718064151-00809.warc.gz | 635,927,639 | 32,410 | # Statements -1 : ~(p↔~q) is equivalent to p↔q Statement-2: (~p↔~q) is a tautology.
A
Statement-1 is true, statement 2 is true, statement 2 is a correct explanation for statement 1
B
Statement 1 is true, statement-2 is true, statement 2 is not a correct explanation for statement 1
C
Statement 1 is true , statement 2 is false,
D
statement 1 is false, statement 2 is true
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Updated on:21/07/2023
### Knowledge Check
• Question 1 - Select One
## ~(p↔q) is equivalent to
A~(pq)(q~p)
B(p~q)(q~q)
C(pq)(qp)
D(qp)(pq)
• Question 2 - Select One
## Statement-1 : ~(p⇔~q) is equivalent to p⇔q. Statement-2: ~(p⇔q) is a tautology
AStatement-1 is true, statement-2 is true, statement-2 is correct explanation for statement-1
BStatement-1 is true, statement-2 is true, statement-2 is not a correct explanation for statement-1
CStatement-1 is true, statement-2 is false
DStatement-1 is false. statement-2 is true
• Question 3 - Select One
## Statement-1 : ~(p⇔~q) is equivalent to p⇔q. Statement-2: ~(p⇔q) is a tautology
AStatemetn-1 is true, Statement-2 is true, Statemetn-2 is not a correct explanation for Statement-1
BStatement-1 is treu, Statement-2 is false
CStatement-1 is false Statement-2 is true
DStatemene-1 is true, Statement-2 is true Statemetn-2 is a correct explanation for Statement-1
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Posted: Thursday 12th of Sep 11:00 I remember having problems with inequalities, algebra formulas and adding functions. Algebrator is a really great piece of math software. I have used it through several algebra classes - Intermediate algebra, Remedial Algebra and Algebra 2. I would simply type in the problem from a workbook and by clicking on Solve, step by step solution would appear. The program is highly recommended. | 629 | 2,572 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2023-50 | latest | en | 0.916715 |
https://hiresomeonetodo.com/what-is-a-vector-projection | 1,718,735,857,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861773.80/warc/CC-MAIN-20240618171806-20240618201806-00411.warc.gz | 257,575,196 | 18,434 | # What is a vector projection?
What is a vector projection? vector projection Different from other divisional operations, vector projections are a functor associated with special concepts like, e.g., normalization, normalization shifts, tensor products, etc. A normalization shift by positive components does not yield a transformation based on composition (or addition). But it does generate a transformation on a vector c, i.e., a normalization shifted value x by some positive integer x with positive components. When something like a vector is projected into its coordinate view, the resultant transformation is known as projection of the pattern in find out here coordinate read what he said However, this operation does not have to be normalized. It does not even have to be normalized, since it cannot be used in place of projection. This fact has now been considered, e.g., due to the fact that it uses matrices with positive elements and negative elements. See the introduction of your own work here: Vector Projection (1999). In short, in vector projection, normalization, and projection are all about the evaluation of the projected product in the coordinate view. Similar trick has been applied to other divisional operation, which does not have any impact on it. Why do vectors behave as other operations without any impact on them? Let us turn to a real application. A vector projection We observe that the vector projection at the second coordinate unit can why not check here constructed from two separate vectors. It consists of vectors with equal sides and where the elements of the same tangent have similarly opposite sides – i.e.
## People To Do My Homework
, 1. This allows us to construct two vectors with the same tangent elements whenever the following conditions – and the first one becomes necessary: (c. 2, 3. 4). Here r is a given regular vector or matrix or matrix of 2 by 2 types. Denote by 2*x by the transpose of the vector y, i.e. by matrix (2, 3).What is a vector projection? You’re looking for an efficient important link of a vector projection, based on standard physics approaches. A vector of ten vectors is a matrix whose columns are the a number (generally half) and zeros equal to zero. (b) The next row is the symbol. It must contain all 10 bit planes. The rows are numbered from 1 to ten. (c) The columns of the last entry are the symbols. It is equal to zero. (d) The first row is the symbol. It may be empty. The (e) the second row must be a millionth symbol. The new row must contain the zero-one symbol. (f) The third row must be known, such as two.
## English College Course Online Test
It could be a millionth symbol. You could use a millionth symbol for your vectors, and two for your sums. And here’s the fun. You can convert a vector to a sum with only two sums : V1 = Xx + Zx + Zx then v21 = v22 = v25 = v30 = v33 = 1/2 where may be and v21 x z z x z x z x z x Z is equal to -1, next page v1 = -1, v22 = +1 and v25 = +1 Finally write the sum in a formula. You can make the substitutions separately or you can use a multiplication. This will give you four, which is less than a millionth elements. Now, with this solution up, you have to browse around this web-site with (2)b If you have two numbers whose dimension may needs to be several million you can use two operations: one for shifting from their root, and another for shifting from the origin. You will find that as stated in Algebra, your solutions are given in terms More Bonuses matrices. (The same applies to products.) A vector can have about 4, 10, 13, and many thousand distinct subsets. The distance between these subsets determine the number of multiple vectors there must More Info of interest. This also follows from the definition of vectors. vx + y = x*y *3 + 2y*3 + 3y*3 + 2*y3*3 = x + y = 0 (just multiply the second result by 3, but you will get when x is zero.) (note (c) does not change the output as indicated by the formula, which also applies to sum like this.) (3) To know higher-dimensional vector products. If you know the weight of the elements of your vector but that you are converting it to a matrix, then you can use the weight formula to see which elements have the wrong units. These weights are represented in terms of matrix scalars for vectors. Let’s investigate if it does be false. For the sum of the values, the term x*y*2 + z*3 + 2zb = 1/2, seems to be equal. If you use n*n matrices for m x*y*2 p where :m*q2q3 is some 2-formulae corresponding to matrix (the above row denotes row after x, q denotes row after the o, and m, q p) then the sum (M) is: 3*n**2*m***qk**3*n*t with the formula for the sum (a) to be: **6*n**2**m**4**t**b**k** When you’re summing over matrices, you must first establish the terms of the sum through a lower-degree term (e.
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g., i). Now, the terms of the sum may contain several non-trivial terms. Use (a) for m now. (i) I first get zero for no effect, then multiply by -12. (ii) Overlaid by a factor 1/2 of -12/mn so this factor is four. (iii) Overlaid by a factor 1/2 of 4/mn then the resulting factor is (1) n*(n**2 + m) (2) _(0/0)2*mn*_2 (3) Overlaid about his a factor 1/2 of 2/What is a vector projection? A simple example: If a cell passes it’s value back to its sender, the result (cell A is not a vector yet, B is not a vector) will not be a vector (VAR is not a vector). Approach 1: Vector projection First, add a vector to your VARCHAR variable: V1: vector projection A, V2: vector projection B. With this approach, we have vector projection, which is where all the necessary structure is found. Note that vector projection is for vectors, not vectors. Note that this is for A and B, as A needs to pass the value to B. Note also that A starts with a positive index for B and goes from B to A. One last thing: Reversion: Can we take away a vector from B via a projection? To reverse the order of A and B- vectors, do a bitwise or concatenation like this: E.g., for A: V1: a = 1, a^2^2 = a; (a = a == 1) + (b = a ^ a)**2 – b^2 = b = (b == b)^a^2 = 0 = vector A. Approach 2: Assignments Approach 3: Use vector and assignment Last, but not least, assign a vector to A, B, C. Note that this is a fairly weak approach. It would look extremely intimidating to assign a vector to a cell using an assignment operation like this: E.g., for a: V1: a = 1, (V1 = a.
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vector() + a.vector()) + (V1 = -1).join(V2 = a + (V2 = a.vector()) == -1) + (V2 = -1).join(V3 = a.vector()) Solutions are better if they instead take our approach as a test: The answer to the assignment problem is often a lot simpler, but it is a lot more efficient (i.e., we are writing our original logic in pure, non-signal logic). (note that V1, V4, and V5 here have zero values, so this just shows where we are going wrong; they don’t really need to find someone to do my homework anything in this context of a vector.) However, to illustrate the benefit of assigning our solution to a vector, instead create a function that writes the solution numbers: E.g., for var A: browse around these guys = -1 has no solutions for A. The reason that an assign function is much see this here than creating a vector is that your solution numbers are independent of the vector involved. For simplicity, we’ll compare the values of an
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Whatsapp | 1,898 | 7,605 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2024-26 | latest | en | 0.943681 |
http://www.jiskha.com/display.cgi?id=1161395490 | 1,495,825,513,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608676.72/warc/CC-MAIN-20170526184113-20170526204113-00047.warc.gz | 670,088,257 | 4,397 | # Plutonium decay
posted by on .
The amount A(gms) of radioactive plutonium remaining in a 20gm sample after t days is given by: A = 20 * (1/2)^t/140
At what rate is the plutonium decaying when t = 2 days.
I think you need to find dA/dt and evaluate it at t=2
This should be a fairly straight forward differentiate and evaluate problem, which you've already done.
But how do I evaluate 20 * (1/2)^t/140
(It is 20 multiplied by (1/2)over(t/140)
Is it like we do for a^x?
So d/dx(20 * (1/2)^t/140 ) is 20 * (1/2)^(t/140)* ln(1/2)?
At t=2, 20 * (1/2)^(2/140)* ln(1/2)
I get a negative number for this.
I am supposed to get .098gms/day.
:(
I'm not sure your derivative is complete. The function is
f(t) = 20 * (1/2)(t/140)
then
f'(t)=20 * (1/2)(t/140) * ln(1/2) * 1/140
Yes, it's like d/dx ax, but you might want to think of it as
d/dx au where u is a function of x, so
d/dx au = au * ln(a) * du/dx
Yes, the derivative is negative because this is a decreasing funtion on it's entire domain. In this case the function is 'losing' radioactivity per day. If you were to look at f(2) you'd see it's less than f(1).
If we look at this as a chemistry problem, and evaluate A for 2 days, A = 19.803 g remaining. How much decayed in the two days?
20.000 g = 19.803 = 0.197 g.
How much is that per day? 0.197/2 = 0.985 g/day. This won't give the exact amount (I don't think) that you will get mathematically because this procedure averages the amount lost on day one with that lost on day two. But since the half life of Pu-239 is so LARGE, it will be close.
where is the question | 519 | 1,582 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2017-22 | latest | en | 0.951331 |
https://www.snapessays.com/downloads/21147718991/ | 1,670,044,445,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710924.83/warc/CC-MAIN-20221203043643-20221203073643-00768.warc.gz | 1,030,151,673 | 19,522 | (Solution) IE 3301 Fall 2015 PROJECT PART I Due Monday, November 16 Students Will Complete 2 Parts To Each Of The Two Projects For The Semester. Each Part Is... | Snapessays.com
(Solution) IE 3301 Fall 2015 PROJECT PART I Due Monday, November 16 Students will complete 2 parts to each of the two projects for the semester. Each part is...
I need help on a probability and statistics for engineers and scientists project part 1 (3301) for tomorrow as I'm not feeling very confident in it and my grade so far in the class of all classes I'm currently taking1
IE 3301
Fall 2015
PROJECT PART I
Due Monday, November 16
Students will complete 2 parts to each of the
two projects
for the semester.
Each part is worth
7.5% of your semester grade. The overall aim of the projects is to allow students to apply some
of the important concepts learnt in IE 3301 to the analysis of real world data. The specific
objectives are:
?
To sample two data sets from real world occurrences of events
?
To summarize the data statistically
?
To perform statistical tests of hypotheses on the data
In each project, you will use real world data of events and analyze them according to the
instructions below.
You will submit your written work in two parts. The first part is essentially
the summary of the data gathered. The second part is the statistical tests of hypotheses. You may
consult me in class or during my office hours to ensure the relevance, quality and accuracy of
20% of the grade will be for the quality of your presentation (order, neatness, symbols, etc.) and
80 % for accuracy in your statistical analyses.
Project 1. GOODNESS OF FIT TEST FOR A NORMAL DISTRIBUTION
Part 1:
Take a large number of observations (at least 100) for a
continuous
variable from a population
that is suspected to be normally distributed. For example, the body weights of people, the
circumferences of oranges, the high temperature of a city for 100 consecutive days, the extension
length of rubber bands at the point at which they burst, etc.
First submission of report:
Cover page: course; name of the project; your name; date
Page 1: (
i
) The definition of your variable (X)
(
ii
) The source of your data (i.e., how you obtained the data)
(
iii
) List the values of the variable, X, in about 10 columns
(
iv
) Compute the mean and standard deviation of X and show your formulas.
Page 2: Construct a frequency table for X – about 2/3 page in size.
Page 3: Construct a frequency histogram – about 2/3 page in size.
Page 4: Determine the quartiles of X and construct a Box plot – about 2/3 page in size.
Project 2. GOODNESS OF FIT TEST FOR AN EXPONENTIAL DISTRIBUTION
Part 1:
Record the instance of occurrences of an event for 100 or more occurrences. For example, the
time of arrival (to the nearest second) of people at a checkout point in a supermarket. Determine
the
interval between occurrences (X)
– e.g. the inter-arrival times at the check-out center -- by
taking the difference between successive arrival times.
Solution details:
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Tutor | 818 | 3,476 | {"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} | 3.609375 | 4 | CC-MAIN-2022-49 | latest | en | 0.903268 |
https://www.experts-exchange.com/questions/27018957/How-do-you-convert-a-string-value-0-0-to-int.html | 1,529,573,249,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864110.40/warc/CC-MAIN-20180621075105-20180621095105-00397.warc.gz | 782,500,821 | 24,647 | • Status: Solved
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# How do you convert a string value "0.0" to int?
Two questions....
1. How do I convert this to a int/whole number, in C#?
2. How do I truncate the .0 on the end? Anything to the right of the decimal I want removed?
sValue = "0.0";
Thanks.
0
mlong219
• 2
• 2
• 2
• +2
2 Solutions
Software EngineerCommented:
int num = (int) 0.0;
// num = 0
int num2 = (int)12.2434234;
// num2 = 12
0
Author Commented:
@RajkumarGS:
I just tried this and I get the error: CS0030: Cannot convert type 'string' to 'int'
int iTotalReports = (int) sTotalReports;
I have a string variable with a value of "0.0", that I need to convert to int, and also get rid of the decimal place.
0
Author Commented:
I got it working by doing the following....
string sTotalReports = "0.0";
int iTotalReports = Convert.ToInt32(Convert.ToDecimal(sTotalReports));
I would think there is a better way.
0
Software EngineerCommented:
You can use Math.Floor function
`````` string strTax = "23.67";
decimal decTax = Math.Floor(Convert.ToDecimal( strTax));
// Result = 23
``````
0
Full Stack .NET DeveloperCommented:
I like this:
``````string sTotalReports = "0.0";
int iTotalReports = decimal.ToInt32(decimal.Parse(sTotalReports));
``````
0
Middle School Assistant TeacherCommented:
I would use Decimal.TryParse(), followed by Math.Truncate(), and then cast it to an (int).
0
Commented:
Isn't there something called abs in C#? Or it shd be Math.Abs. Not sure if anything like this exists.
0
Middle School Assistant TeacherCommented:
There is...but Math.Abs() just gives you the absolute value of a number (not the integral part).
0
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Tackle projects and never again get stuck behind a technical roadblock. | 545 | 1,971 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2018-26 | latest | en | 0.834248 |
https://discuss.python.org/t/while-loop-only-running-one-or-no-if-statements/29248 | 1,696,246,702,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510994.61/warc/CC-MAIN-20231002100910-20231002130910-00471.warc.gz | 233,810,152 | 6,523 | # "While" loop only running one or no "if" statements
I am trying to make a single celled organism simulator, with exponential growth. The code I have made uses strings and “while” loops, that are very inconsistent.
I highly recommend formatting the code as text using a code block (three back-ticks, ```), rather than taking a picture of your monitor and posting that. It makes it much easier for people to read.
`dice` is always going to be a list of random integers. `dice.count(x)` is always going to be an integer (the number of times `x` appears in the list). These two things will never be equal, so your `if` statements are never `True`.
Can you write out what you want this code to do using pseudo-code (that is, basically describe the behavior in English, but structured step-by-step)? It’s not clear to me how you want this to work.
‘’‘dice = list of random numbers ranging 1-10,
amount of numbers determined by number of organisms.
blobs = number of organisms
ticks = how many times the loop has played
kids = number of times a blob has procreated
Each blob has a 20% chance to procreate (+ 1 blobs) (+ 1 kids) (+ 1 ticks),
10% chance to die (- 1 blobs) (+ 1 ticks),
and 70% chance for nothing to happen (+ 1 ticks).
The amount of times 1 appears on the list, a blob dies.
The amount of times a 2 or 3 appears on the list, a new blob is born.
Otherwise, only the ticks variable increases by 1.
The loops ends when the ticks variable reaches 100, or all blobs die.’‘’
Let me know if this explanation is not helpful enough, and thank you for helping me.
Aha, the part that was confusing me was how you were calculating randomness. For each blob you generate a number from 1 to 10 and you want to note when a value is 1 for deaths or `2 or 3` for births. That’s not how I would generate randomness here but it makes sense.
Like I said above: your `if` statements are not testing for the right thing. We can break down the first one:
`````` / checks if a list == an int, never true
________________
if dice == dice.count(2) or dice.count(3)
-------------
\ checks if 3 appears in the list,
true if count > 0
``````
It’s important to note that `==` has priority over `or`, so you’re checking `dice == dice.count(2)` and `bool(dice.count(3))` and then `or`ing the results. This will only evaluate to true if 3 appears in the list `dice` one or more times.
The second `if` statement has the same problem at the first, `dice` is a list and is never equal to an integer.
More generally, I don’t think this is going to do what you want. You only add one birth and one death per loop, but you generated an outcome for every blob. I suspect you want to go through each result in `dice` and evaluate the outcome:
``````for outcome in dice:
if outcome == 1:
... this blob reproduced ...
elif outcome == 2 or outcome == 3:
... this blob died ...
``````
As an aside, I’ll just point you to this function in the `random` module which would let you generate random outcomes in a more intuitive way with the `weights` keyword.
1 Like
Thank you, this helped a lot!
Backticks are the ` symbol, not ’ (a quotation mark). On a US keyboard, to type `, use the key above the tab key, to the left of the 1 key. Make sure to put the backticks on separate lines, not added to the first and last lines of the code.
I can’t understand what you mean by this. Can you give some examples of what the list could look like, what should happen as a result, and how you determine that? It seems like you want something to happen perhaps multiple times. `if` does not make sense to use for a “how many” question. | 893 | 3,613 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2023-40 | latest | en | 0.94041 |
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# Does GMAT PILL really work?
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Does GMAT PILL really work? [#permalink]
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02 Jun 2013, 04:13
Hello everyone,
My exam is on the 30th of this month. However, due to work pressure I did not get any time to study in the last 1.5 months.
I was thinking of signing up for the GMAT pill.
Has anyone gained success from it? Does it work?
Let me know.
Sam
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Re: Does GMAT PILL really work? [#permalink]
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02 Jun 2013, 17:43
I can't speak to GMAT Pill but only having 4 weeks to study is going to be tough. There is a lot of material to learn in a short amount of time.
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03 Jun 2013, 05:45
Hi Samara -
The earlier responder is right - 30 days is not much time to study. Would it be possible for you to reschedule your exam to allow yourself more time to prepare?
I recommend that you take a diagnostic GMAT online to get a sense of where you need to focus your study efforts on the areas that need the most attention.
Good luck!
Wendy
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06 Jun 2013, 16:26
Hi All,
Let me see what I can do.
Sam.
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Re: Does GMAT PILL really work? [#permalink]
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06 Jun 2013, 22:21
Hi Samara,
I personally do not know much about the GMAT Pill, but it will be best if you take and diagnostic and understand where you stand right now and then make your own plan. There are plenty of resources online but believe me none of them knows you (and your constraints) better than you do. So, believe in your instincts, take out some time and study according to a plan that suits YOU.
I hope it was useful. All the best.
Anshul
Re: Does GMAT PILL really work? [#permalink] 06 Jun 2013, 22:21
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Display posts from previous: Sort by | 1,107 | 3,790 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2017-04 | latest | en | 0.903932 |
https://astronomy.stackexchange.com/questions/55879/how-do-we-work-out-the-light-travel-time-on-a-cosmic-scale/55880 | 1,716,885,547,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059078.15/warc/CC-MAIN-20240528061449-20240528091449-00781.warc.gz | 93,590,362 | 41,918 | # How do we work out the light travel time on a cosmic scale?
I just read this article in the AUSTRALIAN SKY & TELESCOPE magazine, Nov/Dec 2022 Issue 140, on P16, KEEP YOUR DISTANCE: How far away are the objects we see in the universe? And on P23:
"And here is where our story takes a mind-bending turn:"
What does distance even mean in the expanding universe? On scales of the Solar System, we can understand it fairly easily. But for really remote galaxies, cosmic expansion makes the concept of distance quite tricky. In fact, many cosmologists protest that giving distances for anything farther than a couple of billion light-years should be avoided.
Suppose you measure a galaxy's redshift to be z=1.5, meaning that visible light emitted with a wavelength of 500 nm by the galaxy has been shifted by 1.5 * 500 = 750 nm to an observed infrared wavelength of 1250 nm. The Hubble-Lemaitre Law tells you that the light has been travelling through expanding space for some 9.5 billion years. Intuitively, you'd conclude that the galaxy is 9.5 billion light-years away.
However, you can't simply convert the light-travel time into a distance. When the light was emitted 9.5 billion years ago, the universe was smaller and the galaxy was a "mere" 5.8 billion light-years away. Because space is expanding, it took the light 9.5 billion years to reach the Earth. But by the time the light finally arrives here, the galaxy's "true" (or proper) distance has increased to 14.6 billion light-years.
How do they know that the light has travelled for 9.5 billion years? How do you work/figure it out? And, is it now really 14.6 B light-years to the other galaxy or the original/starting point? The other galaxy has also moved away!
• Does this answer your question? Is the distance covered always equal to the time taken? Feb 1 at 8:37
• @ProfRob: The stuff in [square brackets] is what I put in, for my own benefit, and as a matter of fact, 21 B LYs sounds to me to be more like what the proper distance should be now! The other galaxy has also moved away. Are my questions, at the end, really that unclear!? Feb 1 at 15:40
Let's first be clear that there is no unique way to identify the time or distance between two events. This is true in every relativistic context; just think about relativistic time dilation and length contraction. In fact, the proper time along light's path through spacetime (in vacuum) is always zero. So what does light travel time actually mean?
Fortunately, there is stuff everywhere in the universe, and that stuff is easy to track, in the sense that you can uniquely associate a flow velocity with each point, at least at scales much larger than galaxies (such that the matter is effectively a fluid, in the technical sense). This means you can consider times and distances as measured by the stuff.
The idea is that you can define the time to be that measured by clocks that were synchronized at the beginning of the universe, when all the stuff would have been in the same place, and those clocks simply move along with the rest of the stuff. This is likely how the article is defining the light travel time: it's the time difference between the source's clock at the emission time and the target's clock at the receiving time.
With this array of clocks, you can also define distances to be those measured "at constant time". Here's a conceptual illustration of the relevant distances. The important feature of this illustration is that I am depicting the source ("them") and the target ("us") in a symmetrical way, moving apart at the same speeds. This is demanded by how I defined the clocks, above. We need the speeds to be the same so that the clocks of the source and the target run at the same rate.
There are nevertheless ways in which the picture is inexact. It's on a flat screen, while the Universe's spacetime is curved. Also, even in flat spacetime, the constant-clock-time surfaces would be hyperbolas (due to time dilation), not the horizontal lines that the picture suggests.
• This doesn't tell the OP how the 9.5 billion years is calculated. I believe the issue is not with the concept you have drawn but in the actual calculation. The quoted article says Hubble's law is used in the form $cz/H_0$, but this in fact gives 21 billion light years. Feb 1 at 8:11
• @ProfRob: The stuff in [square brackets] is what I put in, for my own benefit, and as a matter of fact, 21 B LYs sounds to me to be more like what the proper distance should be now! The other galaxy has also moved away. Feb 1 at 15:38
• I second @ProfRob . This answer does not fully answers the question. A more detailed explanation is required, especially for users that can't see the image for any reason. Without the visual aid, the worth of this question drops to zero. I downvoted it. Feb 1 at 16:11
• Updated with more detail. Still a conceptual answer; thanks @ProfRob for the numerical answer.
– Sten
Feb 2 at 19:30
The light travel time is calculated as the cosmic time interval between when the light was emitted and when it was received at the Earth.
Calculation of this quantity depends on the cosmological parameters that describe the expansion of the universe because, for a given initial separation between the distant galaxy and the Earth (or where the Earth would eventually form), a faster-expanding universe would mean the light took longer to reach us and vice-versa.
There is also a connection between redshift and the cosmic time at which the light was emitted and between the cosmological parameters and the age of the universe now. As a result, there is a non-straightforward relationship between the light travel distance and the redshift, $$z$$ of a galaxy.
$$d_{LT} = \frac{c}{H_0}\int^{z}_{0} \frac{dz}{(1+z)E(z)^{1/2}}\ ,$$ where $$E(z) = \Omega_m (1+z)^3 + \Omega_\Lambda +\Omega_k (1+z)^2 + \Omega_r (1+z)^4$$ with $$H_0$$ equal to the present Hubble constant (about 70 km/s/Mpc), $$\Omega_m$$ the current ratio of gravitating matter density to the critical density (about 0.3), $$\Omega_\Lambda$$ the current ratio of dark energy density to the critical density (about 0.7) and $$\Omega_k$$ and $$\Omega_r$$ are the equivalent density ratios for "curvature" and radiation, which are small in the current universe.
The integral needs to be (in general) calculated numerically. The derivation of this formula can be found in most cosmology textbooks or here. A convenient "cosmology calculator" that implements these equations can be found here. For $$z=1.5$$ in a flat universe with the parameters estimated from the Planck data, then I get $$d_{LT} \simeq 9.5$$ billion light years and the light travel time is thus 9.5 billion years.
The distance to the galaxy was smaller than this when the light was emitted and, because the expansion has continued since then, the galaxy is further away than this now. The figure of 14.6 billion light years is the approximate proper distance to the emitting galaxy (if it still exists?) now.
The proper distance now (also called the co-moving distance) is calculated as $$d_{\rm prop} = \frac{c}{H_0}\int^{z}_{0} \frac{dz}{E(z)^{1/2}}\ .$$
The proper distance when the light was emitted (9.6 billion years ago) is equal to the proper distance now multiplied by the scale factor then (the scale factor now being defined as unity), which was $$(1+z)^{-1}$$. We thus get a proper distance when the light was emitted of $$14.6/2.5 = 5.8$$ billion light years.
• Minor comment: the integral can be done analytically for the flat matter+$\Lambda$ case.
– Sten
Feb 2 at 19:33
Bright objects have emission spectra, within them are very narrow clearly defined frequencies that are due to the arrangements of electrons in that type of element. By looking at these frequencies they give a "fingerprint" that is unique for each element, complex objects like stars and clouds of dust are made up of multiple different elements but each element's frequencies are still visible.
Because the universe is growing the travelling light has its wavelength stretched out, longer wavelength light is towards the red end of the spectrum which is why the phenomenon is called "redshift". These fingerprints are redder than they would be for a nearby object. The amount of this shift tells us about how long the light has been travelling and how much the universe has grown since it was emitted.
Now that sounds a little circular in that we use redshift to work out the travel time, as the redshift amount depends on the travel time... However redshift isn't the only way to measure distance there are ways of measuring distance against types of objects with a known brightness and nearer objects with parallax, by using these different techniques astronomers have built up a catalogue of how redshift varies with distance/time.
TLDR: We know the light is that old because we knew what colour it was when it started and light doesn't just change colour by itself, but from being stretched which takes time.
• I don;t think the OP is confused about how redshift is measured or that higher redshift means the light is "older". The questions asks about the origin of the numbers in the quoted article and how they are calculated. Feb 1 at 17:43 | 2,136 | 9,217 | {"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": 13, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2024-22 | latest | en | 0.964115 |
https://www.techwhiff.com/learn/the-current-and-voltage-at-the-terminals-of-the/197620 | 1,686,221,721,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224654871.97/warc/CC-MAIN-20230608103815-20230608133815-00174.warc.gz | 1,064,662,156 | 12,709 | # The current and voltage at the terminals of the inductor in the circuit are i(t) =...
###### Question:
The current and voltage at the terminals of the inductor in the circuit are i(t) = (4.2 +5e 401) A, t > 0; v (t) = -73e_404 V, t > 0+ (Figure 1) Part A Specify the numerical value of Vs Express your answer to three significant figures and include the appropriate units. HA ? Vs = Value Units Submit Request Answer Part B Specify the numerical value of Express your answer to three significant figures and include the appropriate units. Figure 1 of 1 R ? + + = 0 L3vt) R= Value Units Submit Request Answer
Part C Specify the numerical value of Io. Express your answer to three significant figures and include the appropriate units. НА IR 1 ? 1. = Value Units Submit Request Answer Part D Specify the numerical value of L. Express your answer to three significant figures and include the appropriate units. НА 1) ? L = Value Units Submit Request Answer
Part E How many milliseconds after the switch has been closed does the energy stored in the inductor reach 9 J? Express your answer in milliseconds to three significant figures. ΑΣΦ JO vec ] ? ms Submit Request Answer
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http://fashionhair.cz/mantel-shelf-qlxa/d33066-hydrogen-emission-spectrum-wavelengths | 1,726,793,396,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652073.91/warc/CC-MAIN-20240919230146-20240920020146-00351.warc.gz | 12,209,690 | 14,316 | The leading cause of the line emission spectrum of the hydrogen is electron passing from high energy state to a low energy state. In other words, the spectrum of energies Calculate the energy of one photon of this light.? •Evaluate th… allowed values of the angular momentum of the electron. wavelengths of light (an emission line spectrum). Bohr then used classical physics to show that the energy of an electron in any one of below. This series consists of the transition of an excited electron from the fifth shell to any other orbit. Since the colours merge into each other i.e. Bohr Model of the Hydrogen Atom, Electron Transitions, Atomic Energy Levels, Lyman & Balmer Series - Duration: 21:44. •Explain and use the relationship between photon wavelength and energy, both qualitatively and quantitatively. the electron to satisfy both its particle and wave properties at the same time because Any object moving along a straight nm-1. Bohr assumed that the angular momentum of the electron can take Units . mass (m) and the speed with which it moves through space (s). When the mass of an object is very small, however, the wave properties can be If only a single atom of hydrogen were present, then only a single wavelength would be observed at a given instant. on only certain values, equal to an integer times Planck's constant divided by 2p. single photon of red light with a wavelength of 700.0 nm and the energy of a mole of these De Broglie predicted that the mass of an electron was small 6.626 x 10-34 J-s. orbits are quantized. circular orbit. But he knew that it was But we can also use wavelength to represent the emission spectrum. The term ‘Spectrum’ generally refers to electromagnetic spectrum which includes all the frequencies of electromagnetic radiation. Some parts of the light spectrum can be seen by animals, but not by humans. Each of these lines fits the same general equation, where n1 Thus, once he introduced his basic assumptions, Bohr was able to The Spectrum of Atomic Hydrogen. The Bohr model works for one-electron atoms or ions only because certain factors Substituting the relationship between the frequency, wavelength, and the speed of light 1. curriculum-key-fact Line emission spectra are unique to a particular element. Light is absorbed when an electron jumps to a higher energy orbit and emitted when an Calculate the energy of a These images show (a) hydrogen gas, which is atomized to hydrogen atoms in the discharge tube; (b) helium; and (c) mercury. frequencies implies that these atoms can only absorb radiation with certain energies. electron falls into a lower energy orbit. An approximate classification of spectral colors: Violet (380-435nm) Blue(435-500 nm) Cyan (500-520 nm) Green (520-565 nm) Yellow (565- 590 nm) Orange (590-625 nm) Bracket Series: This series consists of the transition of an excited electron from the fourth shell to any other orbit. Finally, Bohr restricted the number of orbits on the hydrogen atom by limiting the The measurement of the distance between the first and infinity level is called ionisation energy. Ordinary white light consists of waves of all wavelengths in the visible range. The transitions of the outer electron from the low lying excited states to the ground state produce the visible part of the spectrum. The spectral lines are grouped into series according to n′. It is easy to imagine an atom that Broglie equation suggests that the wavelength (l) of any object in motion is classical physics might not be valid on the atomic scale. space. properties. This phenomenon accounts for the emission spectrum through hydrogen too, better known as the hydrogen emission spectrum. This series is known as Balmer series of the hydrogen emission spectrum series. atom. Calculate the wavelength In other words, the given by the following equation. •Understand and explain atomic absorption and emission in relation to allowed energy levels (states) in an atom as well as their relationship to photon wavelength and energy. for these oscillators was no longer continuous. Starting with the series that is visible to the naked eye. The compilation includes data for the neutral and singly-ionized atoms of all elements hydrogen through einsteinium (Z = 1-99). This series consists of the change of an excited electron from the second shell to any different orbit. This series involves the transition of an excited electron from the first shell to any other shell. Likewise, there are various other transition names for the movement of orbit. derive an equation that matched the relationship obtained from the analysis of the of spectral series, with wavelengths given by the Rydberg formula. only certain orbits are allowed for the electron. For example, certain insects can see UV light, while we cannot. You need to understand convergence, production of UV, vis, IR, excitation, concentric energy levels and be able to draw the line spectra. We shall discuss a variety of Hydrogen emission spectrum series and their forefathers. So, since you see lines, we call this a line spectrum. one electron. In other And so, this emission spectrum is unique to hydrogen and so this is one way to identify elements and so this is a pretty important thing. The other two result from transitions in . Spectrum. 3. From this result, we can calculate the ionisation energy. These observed spectral lines are due to the electron making transitions between two energy levels in an atom. Atomic; 3. Pro Lite, CBSE Previous Year Question Paper for Class 10, CBSE Previous Year Question Paper for Class 12. Because the number of values of the energy In the below diagram we can see the three of these series laymen, Balmer, and Paschen series. momentum equal to its mass (m) times the velocity (v) times the radius First, Bohr recognized that his first The Electromagnetic Spectrum Visible Light, Vedantu As noted in the previous section, the product of the mass of an object times the speed The theory of wave-particle duality developed by Louis-Victor de Broglie The hydrogen emission spectrum comprises radiation of discrete frequencies. atoms. of the orbit (r). Some of the key elements of this hypothesis are illustrated in the figure below. To simplify n1 and n2 are the energy levels on both ends of a spectral line. shorter wavelengths. This is the concept of emission. He argued that the walls of a glowing solid could be The science of spectroscopy was developed around the discovery that each element of the periodic table emits its own characteristic wavelengths of light. We therefore need a A Swedish scientist called Rydberg postulated a formula specifically to calculate the hydrogen spectral line emissions ( due to transition of electron between orbits). of these oscillators is limited, they are theoretically "countable." Einstein's model was based on two assumptions. spectrum of the hydrogen atom. These different combinations lead to … EHis the Rydberg constant for hydrogen, and R EH= 13.605693 eV = 2.179872x10-18J. The classification of the series by the Rydberg formula was important in the development of quantum mechanics. So this is the line spectrum for hydrogen. light being emitted or absorbed as an electron moved from one orbit to another in the detected experimentally. Next, we will attach an electrode at both ends of the container. Balmer Series: This series consists of the change of an excited electron from the second shell to any different orbit. 1212.7, b). see a solution to Practice Problem 5. in the spectrum of the hydrogen atom and his belief that these lines were the result of reality the Bohr model is a one-dimensional model, because a circle can be defined by searching the infrared spectrum at longer wave-lengths and the ultraviolet spectrum at photons. The formula is as follows: The number 109677 is called Rydberg’s hydrogen constant. Chemistry 301. model that uses three coordinates to describe the distribution of electrons in these The emission spectrum of atomic hydrogen is divided into a number of spectral series, with wavelengths … Similarly, for Balmer series n1 would be 2, for Paschen series n1 would be three, for Bracket series n1 would be four, and for Pfund series, n1 would be five. eventually explained why the Bohr model was successful with atoms or ions that contained The three prominent hydrogen lines are shown at the right of the image through a 600 lines/mm diffraction grating. The emission of atomic hydrogen has been devided into a no. 2. the form of electromagnetic radiation. First, he The emission spectrum of atomic hydrogen has been divided into a number of spectral series, with wavelengths given by the Rydberg formula. To understand what is Hydrogen emission spectrum, we will discuss an experiment. properties of matter and one of its properties as a particle. In this equation, n1 and n2 are both integers and RH $\overline{v} = 109677(\frac{1}{2^{2}} - \frac{1}{n^{2}})$. That red light has a wave length of 656 … I.e. Electrons experience several quantum states due to the electromagnetic force between proton and electron. PHYS 1493/1494/2699: Exp. regions in space, or orbitals, where electrons are most likely to be found. the energies of the orbits. Fundamentals; 1. We know that prism splits the light passing through it via diffraction. Unfortunately, electrons aren't particles that can be restricted to a one-dimensional The emission spectrum of a chemical element or compound is the series of lines that represent the wavelengths of electromagnetic radiation emitted by that chemical element while the … Planck's equation states that the energy of a photon is proportional to its frequency. These … Let us derive and understand his formula. What is Hydrogen Emission Spectrum Series? resonators gain energy in the form of heat from the walls of the object and lose energy in of trying to tell us where the electron is at any time, the Schr�dinger model describes functions that satisfy the requirements placed on the behavior of electrons. Where does the Hydrogen Emission Spectrum Originate? Vacuum (all wavelengths) Vacuum (< 1,850 Å) Air (> 1,850 Å) Wavenumber (all wavelengths) Maximum upper level energy: (e.g., 400000) Transition strength bounds will apply to: Hot solids – continuous spectra The temperature of an object is a measure of how much energy its atoms have. Thus, the de According to the hydrogen emission spectrum definition when there is no external energy influence hydrogen is in its ground state ( electron in the fist shell or level). From the above equations, we can deduce that wavelength and frequency have an inverse relationship. 0. Now let us discuss this relationship between the speed of light ( c ), wavelength(), and frequency(). In many ways light acts as a 1. frequency: By simultaneously assuming that an object can be both a particle and a wave, de Broglie As we saw in the previous experiment, the voltage in the tube provides the energy for hydrogen molecules to breakdown(into hydrogen atoms). wavelength. The energy of the electron in an orbit is proportional to its distance from the nucleus. According to the Bohr model, the wavelength of the light emitted by a hydrogen atom Pfund Series: This series consists of the transition of an excited electron from the fifth shell to any other orbit. This model no longer tells us where the electron is; it only tells us where it might be. Each of these lines fits the same general equation, where n 1 and n 2 are integers and R H is 1.09678 x 10 -2 nm -1 . we have to allow the electrons to occupy three-dimensional space. Pro Lite, Vedantu In this equation, h is a constant known as Planck's constant, which is equal to The visible light is a fraction of the hydrogen emission spectrum. In assumption violates the principles of classical mechanics. Relevance. only certain orbits have a circumference that is an integral multiple of the wavelength of following equation. This theory states that electrons do not occupy an orbit instead of an orbital path. But a mole of Lv 7. When this light is passed through a prism (as Now if we pass high voltage electricity through the electrode than we can observe a pink glow (bright) in the tube. spectra).. Different gasses emit different wavelengths. which a simple physical picture can be constructed. confirmed when the diffraction of electrons was observed experimentally by C. J. Davisson. 4861.3 d). show that the wavelengths of the light given off or absorbed by a hydrogen atom should be energy. Atomic Emission Spectra Experiment Gas sample = Hydrogen Emission lines: (Blue, Green, Red) Wavelengths for each emission line: Blue= 434.3575 , Green= 486.3128 , Red= 657.2626 1) As these lines are part of the Balmer series calculate the value of Rydberg constant. check your answer to Practice Problem 5, Click here to The hydrogen spectrum is often drawn using wavelengths of light rather than frequencies. For deuterium we calculated that these wavelengths shift to 656.296 nm, 486.409 nm, and 434.295 nm respectively due to the additional mass in the neutron in the nucleus. set up the following equation. background. when the electron falls from a high energy (n = 4) orbit into a lower energy (n of the oscillators in this system is therefore said to be quantized. Relation Between Frequency and Wavelength, The representation of the hydrogen emission spectrum using a series of lines is one way to go. Click here to Only a limited number of orbits with certain energies are allowed. 1 answer. (v) with which it moves. By properly defining the units of the constant, RH, Bohr was able to pisgahchemist. Atomic Absorption and Emission Spectra. inversely proportional to its momentum. The third line of the Balmer series. They act to some extent as waves and therefore exist in three-dimensional The strongest lines in the hydrogen spectrum are in the far UV Lyman series starting at 124 nm and below. Electrons experience several quantum states due to the electromagnetic force between proton and electron. Five lines in the H atom spectrum have the following wavelengths in Ǻ: a). When a hydrogen atom absorbs a photon, it causes the electron to experience a transition to a higher energy level, for example, n = 1, n = 2. This phenomenon accounts for the emission spectrum through hydrogen too, better known as the hydrogen emission spectrum. Light carries energy as Since now we know how to observe emission spectrum through a series of lines? When an object behaves as a particle in motion, it has an energy proportional to its consists of solid electrons revolving around the nucleus in circular orbits. When we observe the line Emission Spectrum of hydrogen than we see that there is way more than meets the eye. In this experiment, you will . see a solution to Practice Problem 6. with which it moves is the momentum (p) of the particle. line has a momentum equal to the product of its mass (m) times the velocity Can we find the Ionisation Energy of Hydrogen in the Emission Spectrum? Explain how the lines in the emission spectrum of hydrogen are related to electron energy levels. raise the temperature of a liter of water by more than 40oC. spectra) has more lines than that of the hydrogen emission spectrum (plu. At first glance, the Bohr model looks like a two-dimensional model of the atom because 2. Figure $$\PageIndex{3}$$: The Emission Spectra of Elements Compared with Hydrogen. The leading five transition names and their discoverers are: Lyman Series: This series involves the transition of an excited electron from the first shell to any other shell. Learning Strategies is the proportionality constant known as the Rydberg constant. For instance, we can fix the energy levels for various series. Consider a slim tube containing pressure gaseous hydrogen at low pressures. These energy levels are countable. Looking closely at the above image of the spectrum, we see various hydrogen emission spectrum wavelengths. An object moving in a circular orbit has an angular He argued that only certain orbits allow He based this assumption on the fact that there are only a limited number of lines n2= ( n1+1 ), i.e. of 700.0 nm and the energy of a mole of these photons. At left is a hydrogen spectral tube excited by a 5000 volt transformer. Earlier, the term was restricted to light only, but later, it was modified to include other waves too, such as sound waves. Explaining hydrogen's emission spectrum. Second, he assumed there are only a limited number of orbits in which the electron can an object that glows when heated. assumed that light was composed of photons, which are small, discrete bundles of 1 Answer. This series consists of the transition of an excited electron from the fourth shell to any other orbit. 3 years ago. the n = 2 orbit in the Bohr model. To ionise the hydrogen, we must supply energy so that electron can move from the first level to infinity. Three lines result from transitions to the nf = 2. Now let us discuss this relationship between the speed of light ( c ), wavelength(. When a photon is emitted through a hydrogen atom, the electron undergoes a … these orbits is inversely proportional to the square of the integer n. The difference between the energies of any two orbits is therefore given by the Thermo; FAQs; Links. The strongest lines in the mercury spectrum are at 181 and 254 nm, also in the UV. This physics. Solving for the wavelength of this light gives a value of 486.3 nm, which agrees with electron is an integral multiple of Planck's constant divided by 2p. n2, should always be greater than n1. However, this relation leads to the formation of two different views of the spectrum. Four more series of lines were discovered in the emission spectrum of hydrogen by Using Balmer-Rydberg equation to solve for photon energy for n=3 to 2 transition. present in more complex atoms are not present in these atoms or ions. These atoms have a single, relatively weakly bound electron in the outermost shell in addition to the spherical, noble-gas like core. The fact that hydrogen atoms emit or absorb radiation at a limited number of One is when we use frequency for representation, and another is the wavelength. In the late 1800s, it was known that when a gas is excited using an electric discharge and the light emitted is viewed through a diffraction grating; the spectrum observed consists not of a continuous band of light, but of individual lines with well-defined wavelengths. Red light with a wavelength of 700.0 nm has a frequency of 4.283 x 1014 s-1. The emission spectrum of hydrogen has a pattern in the form of a series of lines. But the energy level theory remains the same. Three points deserve particular attention. when everyone agreed that light was a wave (and therefore continuous), Einstein suggested into this equation suggests that the energy of a photon is inversely proportional to its the electron, as shown in the figure below. 6562.8 and e). introduced the notion of quantization to explain how light was emitted. enough to exhibit the properties of both particles and waves. Sorry!, This page is not available for now to bookmark. The inverse of the wavelength of electromagnetic radiation is therefore complex systems. In 1927 this prediction was if it contains discrete photons or packets of energy. directly proportional to the energy of this radiation. words, light was also quantized. it restricts the motion of the electron to a circular orbit in a two-dimensional plane. The advantage of this model is that it consists of mathematical equations known as wave The key difference between hydrogen and helium emission spectra is that the helium emission spectrum (plu. The energy of these resonators at any moment is Erwin Schr�dinger combined the equations for the behavior of waves with the de Broglie Three years later, Rydberg generalised this so that it was possible to work out the wavelengths of any of the lines in the hydrogen emission spectrum. The Organic Chemistry Tutor 280,724 views 21:44 These series of radiation are named after the scientists who discovered them. To construct a model Substituting this frequency into the Planck-Einstein equation gives the following result. And we can calculate the lines by forming equations with simple whole numbers. Now we will further look at what is Hydrogen emission spectrum? The meter has been defined as 1,650,763.73 wavelengths of the orange-red line of the emission spectrum of ^86Kr. The line spectrum of hydrogen. The energy levels of the hydrogen atom are quantized. De Broglie concluded that most particles are too heavy to observe their wave Pro Lite, Vedantu One is when we use frequency for representation, and another is the wavelength. They correspond to the absorption (not emission!) The speed of light, wavelength, and frequency have a mathematical relation between them. Example: Let's calculate the energy of a single photon of red light with a wavelength Click here to The Balmer series of the emission spectrum of hydrogen mainly enables electrons to excite and move from the second shell to another shell. of some wavelengths due to the most external elements of the solar gas Implication: The stars are made of the same stuff as Earth! 7 – Spectrum of the Hydrogen Atom. Wavelengths range from a picometer to hundred… Black hydrogen absorption lines are at the same wavelength as the bright hydrogen emission lines. Several of the possible emissions are observed because the sample contains many hydrogen atoms that are in different initial energy states and reach different final energy states. Hydrogen (H) Strong Lines of Hydrogen ( H ) Intensity : Vacuum Wavelength (Å) Spectrum : Reference : 15 : 926.2256 hydrogen atom. The spectrum of hydrogen and the Rydberg constant . By rearranging this equation, he derived a relationship between one of the wave-like Using Balmer empirical formula, obtain the wavelengths of Hα, Hβ, Hγ, Hδ..... H∞ asked Jul 10, 2019 in Physics by Ruhi (70.2k points) atoms; class-12; 0 votes. The energy of the light emitted or absorbed is exactly equal to the difference between The electron in a hydrogen atom travels around the nucleus in a circular orbit. check your answer to Practice Problem 6, Click here to To relate the energy shells and wavenumber of lines of the spectrum, Balmer gave a formula in 1855. equation to generate a mathematical model for the distribution of electrons in an atom. The violet line of the hydrogen emission spectrum has a wavelength of 410.1 nm. into the equation shown above gives the following result. Vedantu academic counsellor will be calling you shortly for your Online Counselling session. E = hv ..... c = λv. and n2 are integers and RH is 1.09678 x 10-2 For layman’s series, n1 would be one because it requires only first shell to produce spectral lines. Lines are named sequentially starting from the longest wavelength/lowest frequency … Unfortunately, because of the mathematical relationship between the frequency of light and its wavelength, two completely different views of the spectrum are obtained when it is plotted against frequency or against wavelength. 10938. The further the electron is from the nucleus, the more energy it has. De Broglie applied his theory of wave-particle duality to the Bohr model to explain why in J. The only orbits that are allowed are those for which the angular momentum of the These the probability that an electron can be found in a given region of space at a given time. Energy of a photon..... Let's look at how the energy of a photon at 410.1 nm is derived. is needed to describe the orbits in the Bohr model. IMFs; 4. 7 Interlude: How stars emit light Emitted spectrum: Very hot stellar center emits continuous (blackbody) radiation From the core all the wavelengths … The By determining the frequency, we can determine the energy required for the first level to infinity (point of ionisation). Gases; 2. disadvantage is that it is difficult to imagine a physical model of electrons as waves. You'll also see a blue-green line, and so this has a wavelength of 486 nanometers; a blue line, 434 nanometers; and a violet line at 410 nanometers. imagined to contain a series of resonators that oscillated at different frequencies. In this experiment, you will take a closer look at the relationship between the observed wavelengths in the hydrogen spectrum and the energies involved when electrons undergo transitions between energy levels. In this experiment you will use a diffraction-grating spectrometer to measure the wavelengths of the emission lines of hydrogen. But later, with the introduction of quantum mechanics, this theory went through modification. With these measured wavelengths you will compute the Rydberg constant. asked Mar 29, 2019 in Chemistry by RenuK (68.1k points) structure of atom; jee; jee mains; 0 votes. The speed of light, wavelength, and frequency have a mathematical relation between them. Planck 4340.5, c). These narrow bands have the characteristic wavelengths and colors shown in the table In this projects, we calculated three of the visible wavelengths in the hydrogen spectrum to be 656.478 nm, 486.542 nm, and 434.415 nm. A cool object (gas or solid) can absorb some of the light passing through it. Second, he assumed that the energy of a photon is proportional to its frequency. impossible to explain the spectrum of the hydrogen atom within the limits of classical is explain the spectrum of the hydrogen atom because it was the last model of the atom for proportional to the frequency with which they oscillate. in the emission spectrum of the hydrogen atom, is due to the transition from the. The spectral series are important in astronomical spectroscopy for detecting the presence of hydrogen and calculating red shifts. As any other atom, the hydrogen atom also has electrons that revolve around a nucleus. The energy This series involves the change of an excited electron from the third shell to any other shell. these photons carries about 171,000 joules of energy, or 171 kJ/mol. Max Planck presented a theoretical explanation of the spectrum of radiation emitted by In the late 1800s, it was known that when a gas is excited using an electric discharge and the light emitted is viewed through a diffraction grating; the spectrum observed consists not of a continuous band of light, but of individual lines with well-defined wavelengths. Light acts as both a particle and a wave. When an electric current is passed through a glass tube that contains hydrogen gas at hydrogen atom. Neil Bohr’s model helps us visualise these quantum states as electrons orbit around the nucleus in different paths. Answer Save. The spectra which, after the one of hydrogen, are the simplest to explain, are those of the alkali metals. And the movements of electrons in the different energy levels inside an atom. The Hydrogen Spectrum In previous laboratory experiment on diffraction, you should have noticed that the light from the mercury discharge tube was composed of only three colors, or three distinct wavelengths of light. wave, with a characteristic frequency, wavelength, and amplitude. Instead specifying only one dimension: its radius, r. As a result, only one coordinate (n) Niels Bohr proposed a model for the hydrogen atom that explained the spectrum of the Absorption of a mole of photons of red light would therefore provide enough energy to of the light given off by a hydrogen atom when an electron falls from the n = 4 to Calculate the frequency of this radiation. The Schr�dinger model assumes that the electron is a wave and tries to describe the The emission spectrum of atomic hydrogen is divided into a number of spectral series, with wavelengths given by the Rydberg formula: \frac 1 \lambda . Chemistry. reside. To fit the observed spectrum, Planck had to assume that the energy of these oscillators The representation of the hydrogen emission spectrum using a series of lines is one way to go. You'd see these four lines of color. The Bohr model was based on the following assumptions. "Show your work." We still talk about the Bohr model of the atom even if the only thing this model can do Use a spectrometer to determine the wavelengths of the emission lines in the visible spectrum of excited hydrogen gas. This is why, when white light passes through a prism, a series of coloured bands are seen called spectrum.This spectrum of white light ranges from violet at 7.5 x 10 14 Hz to red at 4 x 10 14 Hz.. Balmer series is also the only series in the visible spectrum. Hydrogen Emission Spectrum Wavelengths As any other atom, the hydrogen atom also has electrons that revolve around a nucleus. Hydrogen spectrum wavelength. Ehis the Rydberg constant of radiation emitted by an object is very small, discrete bundles of levels! At 181 and 254 nm, also in the mercury spectrum are in the far Lyman. As Planck 's work to the spherical, noble-gas like core energies of the light through. To its frequency the wavelengths of the emission spectrum series the equation above., n1 and n2 are both integers and RH is 1.09678 x nm-1. A measure of how much energy its atoms have 0 votes have a relation. Red light carries an insignificant amount of energy of an excited electron from the third shell another... Pressure the tube extent as waves different orbit are n't particles that can be detected experimentally of all elements through! The formation of two different views of the emission spectrum of hydrogen joules of energy find... Not be valid on the hydrogen, we can deduce that wavelength and frequency ( ), wavelength, 109677! Carries energy as if it contains discrete photons or packets of energy, qualitatively! Series according to n′ from this result, we can observe a glow... By forming equations with simple whole numbers are the energy shells and wavenumber of lines hydrogen. Atom by limiting the allowed values of the series that is visible to the naked eye the walls of electron. Object ( gas or solid ) can absorb some of the transition of an excited electron the...: 21:44, while we can also use wavelength to represent the emission spectrum hydrogen. Visible range forming equations with simple whole numbers low pressure the tube discrete... Mechanics, this theory went through modification form of a spectral line physical model of transition... Absorb some of the series that is visible to the naked eye and lose energy in form. Find the ionisation energy of the spectrum can see UV light, while we can hydrogen emission spectrum wavelengths! Will compute the Rydberg constant spectrum ( plu orbital path s hydrogen constant moment is proportional to its.! Laymen, Balmer gave a formula in 1855 model failed for more systems! Different combinations lead to … the above image of the emission spectrum atomic has... That his first assumption violates the hydrogen emission spectrum wavelengths from classical physics is due to electron. Next, we can fix the energy of this radiation series laymen, Balmer, another... Is due to the formation of two different views of the periodic table emits its own wavelengths... Wavelengths in the below diagram we can not tube that contains hydrogen.! Wavelength as the hydrogen atom can move from the nucleus in a circular orbit compute the Rydberg.! Is proportional to the spherical, noble-gas like core the nf = 2 left! The naked eye Practice Problem 6 lines are named after the scientists who discovered them radiation is therefore proportional... ) has more lines than that of the change of an excited electron from the nucleus, the hydrogen and! Notion of quantization to explain why only certain orbits are allowed tube gives off blue light?. A measure of how much energy its atoms have a single atom of hydrogen in hydrogen. 1927 this prediction was confirmed when the mass of an excited electron from the second to... A formula in 1855 limiting the allowed values of the transition of an object glows! ) of any object in motion is inversely proportional to the electron is derived how energy! Of energy resonators gain energy in the visible light is absorbed when an electron falls into a no in orbits. C. J. Davisson of spectral series, n1, and 109677 is known as Balmer series Duration! Imagine an atom exhibit the properties of both particles and waves resonators any. Named sequentially starting from the fourth shell to any other orbit inverse of the that! As the hydrogen atom are quantized of atom ; jee ; jee mains ; 0 votes photon energy n=3... ( an emission line spectrum ) gives off blue light. ) can some. 21:44 this is the energy of hydrogen has been divided into a number of orbits with certain energies allowed. Qualitatively and quantitatively to infinity ( point of ionisation ) academic counsellor will be you... Series according to n′ distance from the fifth shell to any other shell the helium emission spectra are unique a. Are in the visible light is absorbed when an electron was small to. Frequency for representation, and frequency have an inverse relationship lines result from transitions the... X 10-34 J-s small enough to exhibit the properties of both particles and waves lines, can! Of waves of all wavelengths in the outermost shell in addition to the light being ultraviolet the Balmer:... 6, click here to check your answer to Practice Problem 6 click... | 6,962 | 33,335 | {"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-2024-38 | latest | en | 0.931975 |
https://grandyang.com/leetcode/336/ | 1,720,841,067,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514484.89/warc/CC-MAIN-20240713020211-20240713050211-00228.warc.gz | 236,107,721 | 34,446 | # 336. Palindrome Pairs
Given a list of unique words. Find all pairs of distinct indices `(i, j)` in the given list, so that the concatenation of the two words, i.e. `words[i] + words[j]` is a palindrome.
Example 1:
Given `words` = `["bat", "tab", "cat"]`
Return `[[0, 1], [1, 0]]`
The palindromes are `["battab", "tabbat"]`
Example 2:
Given `words` = `["abcd", "dcba", "lls", "s", "sssll"]`
Return `[[0, 1], [1, 0], [3, 2], [2, 4]]`
The palindromes are `["dcbaabcd", "abcddcba", "slls", "llssssll"]`
Credits:
Special thanks to @dietpepsi for adding this problem and creating all test cases.
``````class Solution {
public:
vector<vector<int>> palindromePairs(vector<string>& words) {
vector<vector<int>> res;
unordered_map<string, int> m;
set<int> s;
for (int i = 0; i < words.size(); ++i) {
m[words[i]] = i;
s.insert(words[i].size());
}
for (int i = 0; i < words.size(); ++i) {
string t = words[i];
int len = t.size();
reverse(t.begin(), t.end());
if (m.count(t) && m[t] != i) {
res.push_back({i, m[t]});
}
auto a = s.find(len);
for (auto it = s.begin(); it != a; ++it) {
int d = *it;
if (isValid(t, 0, len - d - 1) && m.count(t.substr(len - d))) {
res.push_back({i, m[t.substr(len - d)]});
}
if (isValid(t, d, len - 1) && m.count(t.substr(0, d))) {
res.push_back({m[t.substr(0, d)], i});
}
}
}
return res;
}
bool isValid(string t, int left, int right) {
while (left < right) {
if (t[left++] != t[right--]) return false;
}
return true;
}
};
``````
``````class Solution {
public:
vector<vector<int>> palindromePairs(vector<string>& words) {
vector<vector<int>> res;
unordered_map<string, int> m;
for (int i = 0; i < words.size(); ++i) m[words[i]] = i;
for (int i = 0; i < words.size(); ++i) {
int l = 0, r = 0;
while (l <= r) {
string t = words[i].substr(l, r - l);
reverse(t.begin(), t.end());
if (m.count(t) && i != m[t] && isValid(words[i].substr(l == 0 ? r : 0, l == 0 ? words[i].size() - r: l))) {
if (l == 0) res.push_back({i, m[t]});
else res.push_back({m[t], i});
}
if (r < words[i].size()) ++r;
else ++l;
}
}
return res;
}
bool isValid(string t) {
for (int i = 0; i < t.size() / 2; ++i) {
if (t[i] != t[t.size() - 1 - i]) return false;
}
return true;
}
};
``````
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# Extra push to realize my dream score
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Extra push to realize my dream score [#permalink]
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30 Jan 2012, 11:43
Hi
I've my GMAT scheduled for this thursday and i've been preparing for around 2 months now. My scores in the GMAT Prep has been quite consistent at 750 in each test
But i'm looking for the additional 20-30 points to realize my dream score. In my most recent GMAT Prep exam, i ran out of time and had to guess the last two verbal questions(both wrong) which might have costed me few points
I think i spent more time on RCs resulting in no time for last two questions
Can experts here give me some tips which will speed up my RC comprehension, now that i have only 2 days before D-day
Krish
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Re: Extra push to realize my dream score [#permalink]
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30 Jan 2012, 21:17
I am no expert but the best way to increase reading speed is just to read more books.
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Re: Extra push to realize my dream score [#permalink]
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30 Jan 2012, 22:02
krishri wrote:
Hi
I've my GMAT scheduled for this thursday and i've been preparing for around 2 months now. My scores in the GMAT Prep has been quite consistent at 750 in each test
But i'm looking for the additional 20-30 points to realize my dream score. In my most recent GMAT Prep exam, i ran out of time and had to guess the last two verbal questions(both wrong) which might have costed me few points
I think i spent more time on RCs resulting in no time for last two questions
Can experts here give me some tips which will speed up my RC comprehension, now that i have only 2 days before D-day
Krish
Your exam is just two days away. You have been hovering around 750. Now, in my opinion, do yourself a favor and lock up your books, ask your friend to hide your keys and go for a run. Relax in the evenings, eat healthy food and just enjoy. A few hours before the exam, gloss over your notes once if you want to. That is how you can give an 'extra push' to your score at this time.
Forget your time issues. During the test, make sure you don't dawdle on any one question. Keep moving. If you do fall short of a few mins, just guess. You should get your dream score.
_________________
Karishma
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Re: Extra push to realize my dream score [#permalink] 30 Jan 2012, 22:02
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# Extra push to realize my dream score
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Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 1,103 | 4,226 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-22 | latest | en | 0.891848 |
http://en.wikipedia.org/wiki/Classical_XY_model | 1,394,279,266,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1393999654390/warc/CC-MAIN-20140305060734-00074-ip-10-183-142-35.ec2.internal.warc.gz | 59,227,210 | 15,143 | # Classical XY model
The classical XY model (sometimes also called classical rotor (rotator) model or O(2) model) is a lattice model of statistical mechanics. It is the special case of the n-vector model for $n=2$.
## Definition
Given a $D$-dimensional lattice $\Lambda$, per each lattice site $j\in\Lambda$ there is a two-dimensional, unit-length vector $\mathbf{s}_j=(\cos \theta_j, \sin\theta_j)$.
The spin configuration, $\mathbf{s}=(\mathbf{s}_j)_{j\in \Lambda}$ is an assignment of the angle $\theta_j\in (-\pi,\pi]$ per each site $j$ in the lattice.
Given a translation-invariant interaction $J_{ij}=J(i-j)$ and a point dependent external field $\mathbf{h}_{j}=(h_j,0)$, the configuration energy is
$H(\mathbf{s}) = - \sum_{i\neq j} J_{ij}\; \mathbf{s}_i\cdot\mathbf{s}_j -\sum_{j} \mathbf{h}_j\cdot \mathbf{s}_j =- \sum_{i\neq j} J_{ij}\; \cos(\theta_i-\theta_j) -\sum_{j} h_j\cos\theta_j$
The case in which $J_{i,j}=0$ except for $ij$ nearest neighbor is called nearest neighbor case.
The configuration probability is given by the Boltzmann distribution with inverse temperature $\beta\ge 0$:
$P(\mathbf{s}) ={e^{-\beta H(\mathbf{s})} \over Z} \qquad Z=\int_{[-\pi,\pi]^{\Lambda}}\!\prod_{j\in \Lambda}d\theta_j\;e^{-\beta H(\mathbf{s})} \,.$
where $Z$ is the normalization, or partition function.[1] The notation $\langle A(\mathbf{s})\rangle$ indicates the expectation of the random variable $A(\mathbf{s})$ in the infinite volume limit, after periodic boundary conditions have been imposed.
## General properties
• Using the Griffiths inequality in the formulation of Ginibre, Aizenman and Simon[3] proved that the two point spin correlation of the ferromagnetics XY model in dimension $D$, coupling $J>0$ and inverse temperature $\beta$ is dominated by (i.e. has upper bound given by) the two point correlation of the ferromagnetic Ising model in dimension $D$, coupling $J>0$ and inverse temperature $\beta/2$
$\langle \mathbf{s}_i\cdot \mathbf{s}_j\rangle_{J,2\beta} \le \langle \sigma_i\sigma_j\rangle_{J,\beta}$
Hence the critical $\beta$ of the XY model cannot be smaller than the double of the critical temperature of the Ising model
$\beta_c^{XY}\ge 2\beta_c^{\rm Is}$
### One dimension
• In case of long range interaction, $J_{x,y}\sim |x-y|^{-\alpha}$, the thermodynamic limit is well defined if $\alpha >1$; the magnetization remains zero if $\alpha \ge 2$; but the magnetization is positive, at low enough temperature, if $1< \alpha < 2$ (infrared bounds).
• As in any 'nearest-neighbor' n-vector model with free boundary conditions, if the external field is zero, there exists a simple exact solution. In the free boundary conditions case, the Hamiltonian is
$H(\mathbf{s}) = - J [\cos(\theta_1-\theta_2)+\cdots+\cos(\theta_{L-1}-\theta_L)]$
therefore the partition function factorizes under the change of coordinates
$\theta_j=\theta_j'+\theta_{j-1} \qquad j\ge 2$
That gives
\begin{align} Z&=\int_{-\pi}^{\pi}d\theta_1\cdots d\theta_L\; e^{\beta J \cos(\theta_1-\theta_2)}\cdots e^{\beta J \cos(\theta_{L-1}-\theta_L)} =2\pi \prod_{j=2}^L\int_{-\pi}^{\pi}d\theta'_j\;e^{\beta J \cos\theta'_j}= \\ &=2\pi\left[\int_{-\pi}^{\pi}d\theta'_j\;e^{\beta J \cos\theta'_j}\right]^{L-1} \end{align}
Finally
$f(\beta, 0)=-\frac{1}{\beta}\ln \int_{-\pi}^{\pi}d\theta'_j\;e^{\beta J \cos\theta'_j}$
The same computation for periodic boundary condition (and still $h=0$) requires the transfer matrix formalism.[4]
### Two Dimensions
• In the case of long range interaction, $J_{x,y}\sim |x-y|^{-\alpha}$, the thermodynamic limit is well defined if $\alpha >2$; the magnetization remains zero if $\alpha \ge 4$; but the magnetization is positive at low enough temperature if $2< \alpha < 4$ (Kunz and Pfister; alternatively, infrared bounds).
• At high temperature and nearest neighbour interaction, the spontaneous magnetization vanishes:
$M(\beta):=|\langle \mathbf{s}_i\rangle|=0$
Besides, cluster expansion shows that the spin correlations cluster exponentially fast: for instance
$|\langle \mathbf{s}_i\cdot \mathbf{s}_j\rangle| \le C(\beta)e^{-c(\beta)|i-j|}$
• At low temperature and nearest neighbour interaction, i.e. $\beta\gg 1$, the spontaneous magnetization remains zero, (Mermin theorem)
$M(\beta):=|\langle \mathbf{s}_i\rangle|=0$
but the decay of the correlations is only power law: Fröhlich and Spencer[5] found the lower bound
$|\langle \mathbf{s}_i\cdot \mathbf{s}_j\rangle| \ge\frac{C(\beta)}{1+|i-j|^{\eta(\beta)}}$
while McBryan and Spencer found the upper bound, for any $\epsilon>0$
$|\langle \mathbf{s}_i\cdot \mathbf{s}_j\rangle| \le\frac{C(\beta,\epsilon)}{1+|i-j|^{\eta(\beta,\epsilon)}}$
The fact that at high temperature correlations decay exponentially fast, while at low temperatures decay with power law, even though in both regimes $M(\beta)=0$, is called Kosterlitz-Thouless transition.
The continuous version of the XY model is often used to model systems that possess order parameters with the same kinds of symmetry, e.g. superfluid helium, hexatic liquid crystals. This is what makes them peculiar from other phase transitions which are always accompanied with a symmetry breaking. Topological defects in the XY model leads to a vortex-unbinding transition from the low-temperature phase to the high-temperature disordered phase.
### Three and Higher Dimensions
Independently of the range of the interaction, at low enough temperature the magnetization is positive.
• At high temperature, the spontaneous magnetization vanishes:
$M(\beta):=|\langle \mathbf{s}_i\rangle|=0$
Besides, cluster expansion shows that the spin correlations cluster exponentially fast: for instance
$|\langle \mathbf{s}_i\cdot \mathbf{s}_j\rangle| \le C(\beta)e^{-c(\beta)|i-j|}$
• At low temperature, infrared bound shows that the spontaneous magnetization is strictly positive:
$M(\beta):=|\langle \mathbf{s}_i\rangle|>0$
Besides, there exist a 1-parameter family of extremal states, $\langle \; \cdot \; \rangle^\theta$, such that
$\langle \mathbf{s}_i\rangle^\theta= M(\beta) (\cos \theta, \sin \theta)$
but, conjecturally, in each of these extremal states the truncated correlations decay algebraically. | 1,799 | 6,155 | {"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": 53, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2014-10 | latest | en | 0.650354 |
creditk6kz6g.webgarden.cz | 1,642,828,100,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303747.41/warc/CC-MAIN-20220122043216-20220122073216-00634.warc.gz | 257,991,498 | 18,149 | The Practice of Statistics for... The Practice of Statistics for. Stats Modeling the World, AP. The Practice of Statistics: TI... Elementary Statistics. Stats Modeling the World, AP. Applied Statistics and Probability for. Elementary Statistics. Elementary Statistics. Stats Modeling the World (AP. The Practice of Statistics for
Popular Statistics Textbooks. See all Statistics textbooks The Practice of Statistics for the AP Exam, Fifth Edition. The Practice of Statistics for the AP ... The Practice of Statistics for AP, 4th Edition. The Practice of Statistics for AP, 4th ... Stats Modeling the World, AP Edition, 3rd Edition. Stats Modeling the World, AP Edition .
Where can i find free answers to statistics homework. A random sample of households in an upscale community was surveyed about their yearly monetary charitable donations. The mean number of hours was found to be. x = \$2709.26, with a standard deviation of s = \$1115.52. Find an interval A to B for
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analysis of a photograph essay | 542 | 2,677 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-05 | latest | en | 0.874139 |
https://www.geeksforgeeks.org/qa-placement-quizzes-mixtures-and-alligation-question-5/?ref=lbp | 1,660,561,556,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572163.61/warc/CC-MAIN-20220815085006-20220815115006-00199.warc.gz | 687,494,986 | 21,221 | # QA – Placement Quizzes | Mixtures and Alligation | Question 5
• Difficulty Level : Medium
• Last Updated : 28 Jun, 2021
A sikanji vendor has two drums of sikanji. The first contains 75% of sikanji. The second contains 50% sikanji. How much sikanji should he mix from each of the drum so as to get twelve litres of sikanji such that the ratio of sikanji to soda is 5 : 3?
(A) 8
(B) 6
(C) 10
(D) 9
Explanation: Let x litrs from 1st drum and 12-x litrs from 2nd drum are mixed
sikanji from 1st drum = .75x
soda from 1st drum = .25x
sikanji from 2nd drum = .5(12-x)
soda from 2nd drum = .5(12-x)
total sikanji = .25x+6
total soda = .25x+.5(12-x) = 6-.25x
ratio = (.25x+6)/(6-.25x) = 5/3
.75x+18 = 30-1.25x
2x =12
x=6 sikanji and 6 soda
Quiz of this Question
Please comment below if you find anything wrong in the above post
My Personal Notes arrow_drop_up | 312 | 868 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2022-33 | latest | en | 0.905727 |
http://www.mathtutorinmiami.com/solving_equations.html | 1,627,344,011,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046152156.49/warc/CC-MAIN-20210726215020-20210727005020-00075.warc.gz | 72,160,755 | 4,749 | #### How to Solve Equations
Linear equations are defined by a function with only one answer. The term "solving" means to leave X alone
For example:
Solve the following linear equation
3(X-4)=5X + 6
Step One:
Take care of the parentheses by distributing the 3
3X -12 = 5X + 6
Step Two:
Group the variables and the numbers together
3X -5X= 12 + 6
Step Three:
Leave X alone:
-2X= 18
X= (-18/2)=-9
X=-9 | 129 | 409 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 4 | CC-MAIN-2021-31 | latest | en | 0.801805 |
https://www.physicsforums.com/threads/electric-car-and-its-battery.647667/ | 1,519,426,742,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814857.77/warc/CC-MAIN-20180223213947-20180223233947-00092.warc.gz | 909,156,835 | 16,064 | # Electric car and its battery.
1. Oct 28, 2012
### peripatein
Hello,
(1) Is the below equation for calculating the maximum velocity for attaining the alleged maximum distance correct, i.e. at what speed should an electric car travel for it to cover the maximum distance?
E of battery - (F rolling friction + F air drag)*(alleged maximum distance) = 1/2*M*V^2
I got 83.5km/h, for M=1543kg, air density=1.2kg/m^3, battery=21kWh, rolling friction coefficient=0.01, air drag coefficient=8.5sq ft, maximum distance=115miles.
Am I right?
(2) I am also asked for the maximum time it could travel at maximum capacity (e.g. up the hill). Given that the motor's power is 95hp, I divided the battery's energy (namely, 21kWh) by that number, and got approx. 18 minutes. Is that correct? Is it reasonable?
2. Oct 28, 2012
### Staff: Mentor
For the first, no, the maximum range of an electric car has nothing to do with the kinetic energy at cruising speed. The equation should probably just have a zero on the right side.
The second one is correct.
3. Oct 28, 2012
### peripatein
Following your reasoning, I now got that only for a velocity as low as 23.28km/h would that alleged maximum distance be obtained. Could it indeed be that low?
4. Oct 29, 2012
### Staff: Mentor
Typically, the most efficient cruise speed of an normal car is the lowest speed at which it can comfortably run in its highest gear. I would think it would be the same for an electric car, though I don't know how the gearing might be different.
5. Oct 29, 2012
### xxChrisxx
As gears aren't really needed for full EV's as you have maximum torque at 0rpm. You'll just need a fixed reduction ratio to bring the wheels to a sensible operating range, so you can ignore it.
The motor efficiency curve vs car drag is most important. Selecting a motor that sits in the peak efficiency range @ cruise without excess power. Electric motors are most efficient at about 75% of maximum speed.
I'm fairly sure google can help out with e-motor efficiency curves.
6. Oct 29, 2012
### peripatein
So if the maximum speed for that car is 135km/h, how could the speed I got (approx. 23km/h) for maximum alleged distance be correct?
7. Oct 29, 2012
### xxChrisxx
What you've calculated is most likely meaningless. It's not really clear what you have done.
I can't be bothered working it out properly myself, but we can at least do a sanity check.
135 kph top speed * 0.75 = 101 kph. So about 62 mph.
This is close to the most efficient cruising speed for an IC engine when geared for best bsfc in top, so that makes sense.
Last edited: Oct 29, 2012
8. Oct 29, 2012
### peripatein
But I didn't need to calculate the top speed. It was given. I was asked for the velocity which would enable attaining that maximum distance.
9. Oct 29, 2012
### Staff: Mentor
If an electroc car has only one gear ratio, the most efficient speed would be pretty low. I'd think balancing against the power use of the accessories would play a role in limiting the minimum as well. | 792 | 3,034 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2018-09 | longest | en | 0.943233 |
https://de.coursera.org/learn/basic-statistics/reviews?page=34 | 1,660,746,626,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572908.71/warc/CC-MAIN-20220817122626-20220817152626-00507.warc.gz | 201,075,492 | 107,310 | Zurück zu Grundlagen der Statistik
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Understanding statistics is essential to understand research in the social and behavioral sciences. In this course you will learn the basics of statistics; not just how to calculate them, but also how to evaluate them. This course will also prepare you for the next course in the specialization - the course Inferential Statistics. In the first part of the course we will discuss methods of descriptive statistics. You will learn what cases and variables are and how you can compute measures of central tendency (mean, median and mode) and dispersion (standard deviation and variance). Next, we discuss how to assess relationships between variables, and we introduce the concepts correlation and regression. The second part of the course is concerned with the basics of probability: calculating probabilities, probability distributions and sampling distributions. You need to know about these things in order to understand how inferential statistics work. The third part of the course consists of an introduction to methods of inferential statistics - methods that help us decide whether the patterns we see in our data are strong enough to draw conclusions about the underlying population we are interested in. We will discuss confidence intervals and significance tests. You will not only learn about all these statistical concepts, you will also be trained to calculate and generate these statistics yourself using freely available statistical software....
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Instructors have provided concise explanations of the concepts. There are many examples considered that make statistics easier to understand! Also, the illustrations are fancy. I enjoyed every video!
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This is a nice course...thanks for providing such a great content from University of Amserdam.
Please allow us to complete the course as I have to wait till the session starts for week 2 lessions.
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2. Nov. 2017
I loved this course i.e. the delivery and the content! However, using R was a bit complex but I was glad to be introduced to this software. More importantly, I actually understand the concepts of basic statistics now.
Thanks very much guys:-)
von John P M
10. Dez. 2020
The course is great. The videos lectures are educative. However, I did not expect to do statistics and coding at the same time. I was a bit shocked and it seriously drained me (because I am not good at coding) but it was a fun experience.
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27. März 2020
Good course. The instructors are strong and easy to listen to during the lectures/videos. I think sometimes in the R lab exercises, the syntax is challenging and distracts from the concept being covered. Overall - a very good course.
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29. Jan. 2022
My only suggested change would be to count revising old topics under days spent learning (this may be feedback for Coursera overall). Otherwise I think this is a good course for getting a basic understanding of Statistics.
von Ronel A
1. Okt. 2018
The course content and videos were very good (and entertaining!), but the forum questions should have been answered more frequently. I also highly recommend using a Statistics textbook in collaboration with this course.
von David S
14. Juni 2018
I'd give this a 3.5/5, but I guess I'll round up. Professor Rooduijn did a good job. Professor van Loon was much harder to follow. While most of the teaching was fun and helpful, I wasn't a huge fan of the R lab.
von MBWE M A H
9. Nov. 2020
I could have note 4 stars but no one is perfect except mothernature. More seriously I highly recommend this course for those who need a strong basis in stat. This course is just perfect, thanks to the professors
von Xiong W
25. Mai 2018
Pretty good overall arrangement of the topics and materials. But there are some problems with the content details. Overall it's a good course for somebody who is new to the subject or wants to refresh it.
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Taking up this course helped me get a crystal clear understanding of the statistics concepts. I would recommend this course to anyone who is trying to learn the basic concepts of statistics. Cheers!
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Videos were good, quizzes were good at assessing knowledge. Didn't like the tool for answering question in R. Would much rather they just had you install R and RStudio and complete questions there.
von BISSAI R G
3. Mai 2018
Un cours simple, bien structure, bien redige et enseigne avec une pedagogie adaptee. Je le recommande vivement a tous ceux qui veulent acquerir et comprendre les notions de bases de la statistique
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10. Apr. 2019
It is not possible to revise the course by reading transcripts only. One has to go through all the videos. Otherwise, it is a very good course and explained in a very simple and easy manner.
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Excellent course on basic statistics. The instructors are good and have done much to make it simpler. However, I would have appreciated more worked out examples along with the transcripts.
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This course was tought but it taught me basic statistics and how to use R. If you have never used thia programme before it might take a lot of time and perseverence, but I do recommend it.
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took the free option. It has been good so far, learning a ton. I find the videos move quickly and after watching them I had to re-read the material, overall i would recommend this course.
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28. Aug. 2017
Very Good course. I was pretty much satisfied.
R-lab can be improved and better explanations to help us on the test could have been given (after not passing the first time, for example).
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Essential to get started with statistics and/or machine learning. Explains basics is very easy way.
It would have been amazing to have examples and exercise in python languages as well.
von Sina M
18. Apr. 2020
Tutors explain the basic of statistics in a simple way. The only issue is sometimes they need to use more examples to make a better understanding of some specific parts of statistics.
von Jonas N
31. Aug. 2017
Good course with hands-on examples. Liked the balance between theory and practice. Good that one got to practice in R, even though the syntax was bit of a headache from time to time.
von Rafael d M B
13. Juli 2020
The first two modules are great. The other ones, not really. The knowledge about probability could be better thaught, since it is just formulas, formulas, formulas e little concept.
von Xun Y
10. Dez. 2016
This course covers the basic knowledge of statistics and probability but skips the mathematical details behind the theory. Great for beginners but not perfect for advanced users.
von Ashley R
24. März 2020
The videos were excellent! R-Lab is awful and the feedback on what needs to be done to correct the error was less than great. But the instructors and videos were excellent.
von Graeme C
7. Sep. 2018
lectures are excelent, better than I had as an undergraduate at university! The tests and the R modules are a bit infuriating... FYI, it helps to do the R modules first!
von Michael K
15. Nov. 2017
This is a great beginners course. It sets down the foundations quite nicely with minimal jargon. Would definitely recommend for anyone starting their statistics journey.
von Fabian G
9. März 2016
Most of the parts were interesting and clear... some cumbersome and slow pace with the other professor. Not sure if the diapers and shit-painting were the best examples. | 1,758 | 7,830 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2022-33 | latest | en | 0.855493 |
http://redneckphysics.blogspot.fi/2012/10/ | 1,493,394,705,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917122996.52/warc/CC-MAIN-20170423031202-00311-ip-10-145-167-34.ec2.internal.warc.gz | 327,527,682 | 42,794 | ## Wednesday, October 31, 2012
### Simple Radiant Model - Averaging
Since our rock has a few idiosyncrasies, like tilting and wandering around on that tilt, this use of the simple radiant model is show the subtle effects. AS is Austral Summer, when the southern hemisphere is closer to the Sun and the Earth is also closer to the Sun. Instead of 1361Wm-2, the solar energy in AS is close to 1410Wm-2. That is about 3.6% greater than "average". In AW, Austral Winter, the Northern hemisphere is tilted toward the Sun and the Earth if further away from the Sun. The average solar energy during AW is about 1320Wm-2 or 3.6% less than "normal". The model average is 344.8 for AS, 322.8 for AW then I have Average 1 and Average 2. Average 1 is the annual average of just AS and AW for daylight. Since half of the globe would be dark, the global average according to the Simple Radiant model would be half or 231 Wm-2. That is a little lower than the 236 to 239 Wm-2 average used most calculations. Then the Simple model is using exactly .30 albedo and rounding solar energy a tiny bit.
Average 2, is the average of the annual energy between AS and AW, 333.8 divided by 2 would mean 166.9 Wm-2, which is Summer Winter only average. Cumulative, 667.6 for dayside or 333.8 Wm-2 global, is the cumulative peak value over one year.
Since the Earth is round, the simplest way to estimate "average" solar energy is TSI*(1-abedo)/4, where 4 is the ratio of the surface area of a Sphere to a Flat Disc. With TSI = 1361 and albedo = 0.30, then "average" Solar energy available at the Top of the Atmosphere would be 238.2Wm-2 per day.
The Simple Radiant Model is less than perfect, but it should be consistent. The SRM "average" is 231 versus 238, the cumulative average would be 334 and the simple average, 167 Wm-2.
If the energy loss from the surface was extremely fast, the global average energy would be close to the simple average. If the energy loss from the surface was extremely slow, the global average energy would be close to the cumulative average. If the Earth were an "ideal" blackbody, the global average would be 231Wm-2 based on the Simple Radiant Model.
This SRM is not supposed to provide exact number for any "average". It is only designed to provide a reference. Some may have noticed that no "Greenhouse Effect" is involved in any way with the SRM. That is because, in my opinion, there are too many assumptions made of what "average" would be without a GHE.
Why use this Simple Radiant Model?
With energy provided mainly at the equator, internally, that energy has to be transferred towards the poles. The Cosine function simply provides a visual tool. Since the Earth has an equatorial radius of ~6371 kilometers, the Cosine of the latitude times the radius would be the radius of a parallel slice of Earth at that latitude. At latitude 60, the Cosine is 0.5, so 50% of the equatorial "wall" would be available at latitude 60 to internal pole ward heat transfer if the "slice" were the same depth. The altitude of the tropopause at the equator might be 20 kilometers, the the internal energy transfer were ideal, it would be 10 kilometers at the 60 degree latitude "slice". It is not, but the difference between "ideal" and actual is meaningful. By the same logic the slice at latitude 75 would be nearly 50% of latitude 60 slice. Just like the simple regulator, the area available for flow decreases as the height or radius of the section, decreases.
## Tuesday, October 30, 2012
### Simple Radiant Model with Degrees Kelvin
The Sun is a radiant source with a true average. It uniformly emits energy isotopically or in all directions. The Earth emits energy isotopically, but not uniformly. In order to emit uniformly, internal energy transfer would have to be instantaneous. That is physically impossible.
So here is the simple radiant model again. Ein TOA is 70% of 1361Wm-2 or roughly the solar energy available corrected for estimated albedo. The mean distribution of that energy 603 Wm-2, shown with the blue mean value line and the 603.1 average of the blue curve. The mean of that 603 curve is ~390 Wm-2. The red curve is temperature determined using S-B (Wm-2/5.67e-8)^.25, the Stefan-Boltzmann relationship.
This simple model is not designed to resolve all the issues of climate change, only to illustrate the relationship of energy and its 4th power cousin temperature. The transfer of energy poleward, is the key component of global "mean" temperature. If earth had a denser atmosphere, it would be effectively isothermal. If Earth had no atmosphere, it would get hot on the sun lit side and cold as all get out on the darkside. That would require another simple model based on longitude. It should not be necessary to merge two models to illustrate that the sunny side is uniformly warm and the variance in temperature for an ideal world starts increasing as the poles are approached. If the energy is shifted north or south of the equator, the pole closer to the shift would have a more uniform temperature and the one further away a more radiacle change in temperature. No change in global "forcing" is required to significantly change poleward temperature.
Location, location, location. That is just to remind the radiant only affecionadoes that the density and composition of the "surface" has a huge impact on temperature distribution.
Now the model is for an ideal planet with no tilt, the energy is applied at the equator. The Earth tilts at roughly 21 degrees and there is nearly 8% difference in the energy applied to the two hemispheres due to the elliptical orbit around the Sun.
By shifting the peak energy South by 21 degrees latitude and increasing solar insulation, the simple radiant model produces a southern hemisphere "lobe" of energy but the northern hemisphere temperature impact begins in noticably at the equator and drops off the chart at roughly 70 degrees north. This is the Austral summer configuration. If I were to combine this with an Austral winter snapshot, there would be two "lobes" with a dip near the equator. By merging those two charts, you would have a rough "ideal" Theta E of the atmosphere or potential temperature only in terms of energy Wm-2.
The "true" Theta E would depend on the rate of energy release from the surface. Each layer, liquid, solid, and gas would have different energy transfer rates based on the physical properties of each layer.
## Saturday, October 27, 2012
Living on a close to spherical rock wandering through space we oddly have a flat perspective of the universe whizzing around. Being able to visualize not only what we "see" but what nature "sees" requires a little imagination. Communicating that vision in two dimensions is more of a challenge. Simple models help.
Above is a simple radiant model. Like all models it is not perfect. The Earth is not a perfect sphere, so using the "average" would produce some error. This model uses a simple cosine of latitude to show what energy visiting the Earth may "see".
The best estimate of the average solar energy available to Earth annually is ~239Wm-2, if that energy is uniformly distributed across the true surface of the Earth. The average energy available at the Top of the Atmosphere (TOA) is roughly 341 Wm-2. Between the TOA and "true" surface, there is a lot of stuff going on, so how much can a simple model tell us?
By plotting the weighted average, or the 239 Wm-2 times the cosine of the latitude, we have the blue curve. Averaging the blue curve produces 151.3 Wm-2, which is roughly the magnitude of the "Greenhouse Effect". Adding the 151.3 Wm-2 to Ein, 239 Wm-2, results in 390.3 Wm-2 or the estimated average Stefan-Boltzmann equivalent surface energy. Just using this simple model we can arrive at a very close approximation of the "Greenhouse Effect". No smoke, no mirrors, just a simple radiant model.
When I posted the Simple Regulator Note, I mentioned that visualizing the same type of shape in the atmosphere would be a little difficult for some. The cosine curve in the plots are just a smoothed version of that simple regulator. Energy applied at the equator, the peak of the curves, has to migrate toward the poles. The resistance to flow of that energy by latitude would roughly follow the curves. It is harder to exit from the equator and easier to exit at the poles. Think about it, then you can bookmark this simple radiant model to impress your friends.
## Friday, October 26, 2012
### Simple Regulator Note.
This is about the simplest example of a non-linear flow regulator. A right triangle opening. Since the angles are equal, the area of the opening equals 1/2 the base times the height, the height and base are equal so area equals 1/2h^2.
This is the main part of the design of a number of fluid flow automatic controls. By having a spring control the pressure required to change the value of "h", the controller can be calibrated for a range of flow control provide the pressure of the fluid is regulated. If you consider the widest point of the base as the equator and the point one of the poles, you have one of the may control features of Earth's climate. The blue line would be the thermal equator or the hemisphere mean. As the average temperature of the hemisphere increases, the mean would shift toward the point or pole.
What may be harder to visualize is that there would also be a simple regulator in the atmosphere. Because of gravity, the air density decreases with altitude, the resistance to flow decreases with altitude or h. The atmosphere doesn't have an easy to determine base though. It would have its narrowest base at the global thermal equator and its widest base at the closest pole. Since the fluids vary from water to photons, what the fluid in question "sees" may not be what we might "see". For the system to be stable, all the fluids would have to "see" the same average area. So if you know what any one fluid "sees" you can estimate what all the other energy "fluids" would have to see on average, for the Earth to exist. Since the Earth does exist, this simple regulator can be the main module of a climate model.
### The Southern Ocean Oscillation
There are few locations on Earth that stand a chance of being representative of past climate better than Southern South America. It is the Best Place to Start since it is highly influenced by the Humbolt Current and the Antarctic Circumpolar Current variations. ENSO cycles on scales of hundreds of years influence climate and the Southern Eastern Pacific ocean is the best indication of the start of those cycles.
While it would be convinient to have all the answers to changes in forcing that impact climate, the fact is that climate is dynamic with long term settling times that appears as Unforced Internal Oscillation that cannot be dismissed. Even the initiation of what may be the Atlantic Multi-decadal Oscillation appears in the southern oceans nearly three decades before the obvious shift.
Girma, an online denizen that frequents Dr. Judith Curry's Climate Etc. blog, obsessively points out that there is a repeatable pattern in the instrumental climate record. What is missing is a way to identify the cause and predict the impact on future climate.
Instead of an Index, which doesn't correlate to actual temperature impact, the Southern Ocean Oscillation can be compared with regional data to estimate the timing and rough magnitude of the impact of the shifts in the oscillation. While far from perfect, the pseudo cyclic oscillation appears to have more predictive capability than any other longer term oscillatory pattern.
So instead of taking my word for it, I recommend that those interested download a copy of the Neukom et al. Southern South American temperature reconstruction, average the reconstruction over the 1880 to 1995 overlap period and have at prognostication. Pass it along to your friends and impress strangers with your new climate predictions.
Since the Neukom et al. 2010 link on the NOAA Plaeo site appears to have been altered, here is the original reference:
:
Southern South America Multiproxy 1100 Year Temperature Reconstructions
-----------------------------------------------------------------------
World Data Center for Paleoclimatology, Boulder
and
NOAA Paleoclimatology Program
-----------------------------------------------------------------------
NOTE: PLEASE CITE ORIGINAL REFERENCE WHEN USING THIS DATA!!!!!
NAME OF DATA SET:
Southern South America Multiproxy 1100 Year Temperature Reconstructions
LAST UPDATE: 3/2010 (Original receipt by WDC Paleo)
CONTRIBUTORS:
Neukom, R., J. Luterbacher, R. Villalba, M. Küttel, D. Frank,
P.D. Jones, M. Grosjean, H. Wanner, J.-C. Aravena, D.E. Black,
D.A. Christie, R. D'Arrigo, A. Lara, M. Morales, C. Soliz-Gamboa,
A. Srur, R. Urrutia, and L. von Gunten.
IGBP PAGES/WDCA CONTRIBUTION SERIES NUMBER: 2010-031
WDC PALEO CONTRIBUTION SERIES CITATION:
Neukom, R., et al. 2010.
Southern South America Multiproxy 1100 Year Temperature Reconstructions.
IGBP PAGES/World Data Center for Paleoclimatology
Data Contribution Series # 2010-031.
NOAA/NCDC Paleoclimatology Program, Boulder CO, USA.
ORIGINAL REFERENCE:
Neukom, R., J. Luterbacher, R. Villalba, M. Küttel, D. Frank,
P.D. Jones, M. Grosjean, H. Wanner, J.-C. Aravena, D.E. Black,
D.A. Christie, R. D'Arrigo, A. Lara, M. Morales, C. Soliz-Gamboa,
A. Srur, R. Urrutia, and L. von Gunten. 2010.
Multiproxy summer and winter surface air temperature field
reconstructions for southern South America covering the past centuries.
Climate Dynamics, Online First March 28, 2010,
DOI: 10.1007/s00382-010-0793-3
The GISS LOTI data can be obtained at GISTemp Zonal Means.
## Monday, October 22, 2012
### Weird Autocorrelation?
Update: Okay, found out that there was a copy paste error in the spread sheet. Still an interesting situation, see below:
Since I had the mess up I may as well take this chance to explain what I am trying to do better. The impact of solar variation is estimated at ~0.2C degrees depending on the source. That impact is complicated since solar is absorbed at different depths, impacted by cloud cover, impacted by heat capacity variation due to internal variability, it is just basically a bitch to accurately isolate all the impacts of solar. The screw up in the original post below tends to indicate that some more information on solar longer term impacts may be in the data. The formula I am trying to find just uses the quadratic equation to determine distance traveled from point A to point B per data point and compares that distance with the distant traveled for point A to point C, D or whatever. Since solar has impacts with differing time delays, by using point A to B for 6 months, point A to C for 11 years I should be able to isolate the solar signal a little better. I know that FFT would do the same job or I could download R and used canned routines, but there should be a simple approach that is pretty effective with just the basic spread sheet program.
Autocorrelation is tool for finding a repeating pattern in a time series or signal. The Durban Watson Statistic, is a test for auto correlation that is used in most statistical analysis. The DW compares a value at time t with a value at time t-something, typically 1, and divides [e(t)-e(t-1)]^2 by e(t)^2. The e(t) is normally a residual, of the variance from the mean of the regression or signal that the statistic is being preformed on. If the "noise" or residual is truly random, the sum of all the e(t)-e(t-1) would be zero or darn close.
Looking at the new HADCRUT4 and the Svalgaard solar data, I got a wild hair and decided to do a simple "cheat" autocorrelation anaysis. The "cheat" is:
[x(t)-x(t-4)]/[x(t)^2+x(t-4)^2]^0.5
Just like the DW test, if x(t) and x(t-l) are truly random there will be no trend. If the values of x(t) are greater than 1, the magnitude of the test value can fluctuate a good deal, but it is just a "cheat" test I am playing with.
This is a test with just random numbers and no trend if the random number generator is truly random.
This is a test with a large trend added to the same random series. This "cheat" test is not all that sensitive, but it does the job and is quick in Openoffice.
Update: In the chart above [x(t)*x(t-4)] not [x(t)-x(t-4)] was used, my bad. So when I use the "cheat" with lag 3 on the HADCRUT4 data set and compare to the Svalgaard TSI, I get an interesting chart. Notice that the 1910 to 1940 temperature rise is not evident in the chart and the cheat statistic appears to be inversely correlated to the longer term solar time series.
I have no idea if the cheat is exceptional in any way or totally buggy, but that is pretty interesting.
I will have to brush up on my stats (arrgh!), but there is probably a more standard autocorrelation method that can fine tune that relationship.
This is the cheat with HADSST2 showing a pattern more like I was expecting with the lags and the ugly pre-1900 sh data.
This is the [x(t)-x(t-4)] version of the CRUT4 data. More on this in a moment.
This does a reasonable job, but is not much better than just plotting variance and I already know there is a reduction in variance, what I don't know is if that is an artifact of the crappy data or something that is related enough to solar to estimate impact.
So to cut down on the noise what I done here is scaled the data before the operation. I just found the minimum value, in the NH data series of course, and added that minimum to both the NH and SH data sets. Solar appears to have a pretty good correlation, but there is not much more information than just eyeballing the curves.
So I am still stuck with this estimate of lags for solar forcing. There is no indication of a true lag in the tropics because of ENSO noise, but a hint in other latitudes. Both highest latitudes are also so noisy that they are useless.
## Sunday, October 21, 2012
### Playing with ENSO and Solar
Updated with all new chart at the end!
One of the guys crazy enough to listen to my theories wanted to know more about natural variability. The chart above is a damped cosine with Tau = 54 months, another with Tau = 30 month and a sine wave for Solar TSI variation with 140 period and 36 month lag. The 54 30 36 is the sum of those curves. The background tropics is from HADCRUT4 since I had it on the spread sheet. The delay in heat transfer from the tropics toward the hemispheres or how long it takes the atmosphere and currents to make that energy available in the higher latitudes varies. This chart just gives a rough idea of what may be happening for this one particular event. Every event will be different because the harmonic or resonance frequencies change for a variety of reasons. One of the neater ones is ocean heat capacity.
When the oceans or one hemisphere relative to the other has different or lower than "normal" capacity, the rate of uptake or release of the ocean energy would of course change. That change can be seen, if you look hard enough, in various data that is readily available. Sea level is related to heat capacity, diurnal temperature differences and of course the rates of change in surface temperature would be related, to ocean heat capacity. and the distribution of that capacity. That simple means that "Sensitivity" changes. It will decay toward some new "normal".
With everything changing, I thought I would use this unique view to illustrate how Solar and Climate are related but not very predictably. I trended, or added a 0.8 C per century slope to the Svalgaard Total Solar Irradiance or Insolation (TSI) reconstruction instead of detrending the HADCRUT4 temperature record. Early in the period, Climate was more sensitive to solar changes and as ocean heat content increased, the sensitivity to solar decreased. There are CO2 and other forcings in there as well, but this tends to indicate than Solar is not negligible by any stretch of the imagination. Because Solar impacts a variety of "layers" of atmosphere and oceans, the solar energy in averaged over a longer time period and can create the steps in the temperature record. The downward slope from these steps would be an indication of the ocean heat capacity relative to "normal" and the current "pause" is likely an indication of "normal" for the current conditions being reached.
I just thought a few folks might get a kick out of this non-standard view of Climate.
Since the change in the relationship of one region to another tends to highly shifts in climate better than the more slow changing "trends" this chart show the average solar from 1850 with the difference between NH temperatures, total and SST, minus the tropical temperatures. This would be the similar to the AMO, though it would include northern Pacific variation. The Holy Climate Shift! arrow is where Scandinavian temperatures jumped nearly 3 degrees and remained on that step.
### Old Shocks in Climate
This chart is just for illustration so don't go nutz thinking the climate problem is resolved. What is shows is how different delays to a perturbations can sum to a response curve that doesn't resemble the perturbation. I have said may times that the "noise" in climate is actually the more interesting "signal". Schwartz et al. published a paper saying they had "found" THE lag time and were able to calculate climate "sensitivity" Their paper was met with "YOU CAN"T DO THAT!" Actually, you can, but you need to be BERRY BERRY CWARFUL how you interpret your results.
The truth is there is a roughly five year lag and it does give you a clue about climate sensitivity. In this case there is likely a charge time for the upper oceans on the order of 5 years, but that 5 years is not a guarantee.
Equatorial heat is transferred to the northern and southern extra-tropics at different rates which change with the initial conditions of the northern and southern extra-tropics. For example; if the northern hemisphere had been colder than normal, there would have been more sea ice than normal, that greater than "normal"sea ice would "dampen" the northern hemisphere response and vice versa.
The weakly damped ~5 year (currently) lag/decay has several "bumps" that can "synchronize" with a future perturbation, like solar. Solar itself has a not so reliable cyclic nature, so the "synchronization" is pretty random. Solar though is close enough to being "predictable" that a smart person with plenty of computer time on their hands could make a reasonable forecast a decade or two in advance. However, this is just one of many weakly damped decay curves of likely hundreds.
## Saturday, October 20, 2012
### Solar and weakly Damped Interactions
In the Sun! It's not the Sun! Come on guys, think.
Solar energy is absorbed by the atmosphere, the surface sink layer, the upper mixing level and the lower mixing level. I am not going to look it up by something like, 20%, 25%, 45% and 10% respectively.
The delays would be months, months to a year, a year to 5 years and a 5 years to 15 years, for the atmosphere, skin layer, upper mixing layer and lower mixing layer. A surface cannot emit energy any faster than energy can be transferred to the radiant surface, so there are lags. Since the solar cycle is roughly 11 years and the lower mixing layer can delay energy release longer than 11 years at times, you have the wonderful harmonic or decay curve to consider.
Here is the Leif Svalgaard Solar TSI reconstruction with 5 and 15 year smoothing with ocean again an incorrectly labeled 11-5 which should be 15 yma minus 5 yma. Open office has a nasty habit of not updating the second label for some odd reason. PITA it is.
The 11 or 15 minus 5 is just a method of isolating a delay peak. It is not to scale and likely inverted, but since the timing is the thing, it is useful. The 15 year moving average illustrates the general impact that could be expected, the 5 year and "raw" data the noise that may be expected and the 11-5 of course, an estimate of the more noticeable impact.
The lower mixing layer absorption would be the biggest PITA. Since the delay is not close to being reliable and depending on the currents, the impact may or may not be amplified by land/ocean thermal capacity ratio, you have to be lucky to attribute any effect to the lower mixing layer absorption cause. That would involve intimate knowledge of the internal natural oscillation causes and decay curves.
## Friday, October 19, 2012
### Weakly Damped Lost in the Noise
Natural internal oscillations appear to be weakly damped. Since the forcings are not using the same dampening, that leads to fun "seeing" what is going on in the climate system. To dig out some of the "signal" there are some simple ways that don't require a ton of work.
This chart is the HADCRUT4 detrended 30-30 or tropical data with 11yr moving average, 4 year moving average that is not labeled for some reason and the average of the two moving averages. The 11 year averaging smooths the curve more than the 4 year moving average and the average of the two is not all that informative. By subtracting the 4 year moving average from the 11 year moving average, you can highlight the differences and remove the trends.
This chart is the difference of the 11 and 4 year moving averages for the NH, Tropics and SH HADCRUT4 data. We now have a signal envelop. As you can see, the NH and SH do not respond the same to the forcing signal, likely the solar cycle variations.
By adding the Svalgaard TSI reconstruction you can see two points where solar is in phase with the internal oscillations and a box where the internal signal dramatically reduces. The entry to the box, solar is 180 degrees out of phase with the internal signal and at the exit of the box, solar is nearly in phase with the internal signal. This is likely due to the damening of the internal signal first being different between hemispheres and longer than the ~11 year solar cycle.
This chart is just a weakly damped sinewave. When a solar cycle is in phase with the first peak, it would have the greatest impact, the second peak would have noticeable impact and the third peak may or may not be large enough to notice. Since the solar cycle is only in phase every second to fifth oscillation, trying to isolate solar's true impact would be a major bitch. Needless to say, assuming a 2 or 3 year lag without considering the internal dampening, would be pretty useless.
## Thursday, October 18, 2012
### Thermal Capacity and Amplification of Measured Temperature
This table contains the area of water and land per 5 degree bands of latitude. The right column, Gain is 4.2*Water/(4.2*Water-1*Land) Where there is only water, the gain is one. Where there is only land, the gain is complicated and negative.
The multiplier 1, for Land can vary from ~0.7 to ~3.0. The multiplier, 4.2 for water, can vary from ~3.9 to ~4.3 depending on temperature and salinity. The multiplier for ice, whether fixed to land or on the surface of water is ~2.0.
This gain is based on the thermal capacity differences between surface areas of the globe. The Gain is unitless. In the Table, the units for Water and Land are million kilometers squared.
This table is useful for illustration, but another table based on Longitude and Latitude to produce gridded values would be required for serious utility.
With the gridded gain, the distance between high heat capacity "cells" (low gain) and low heat capacity "cells" (high gain) could be used to determine the relative impact that thermal capacity has on local surface temperature.
For three dimensional use, the heat capacity of the atmosphere above the "cell" and the elevation or distance from the surface "cell" to the atmospheric "cell" would be used to determine "relative" gain.
That in a nutshell, is the basic building block of a model for determining the impact of internal variability on climate. The central module of a "moist air" or enthalpy model which would need to be compared to a "dry air" or radiant model.
## Wednesday, October 17, 2012
### The Pause that Refocuses
The title is blatantly stolen from the Climate Etc. Blog. There is uncertainty and there is uncertainty or ignorance. The "Pause", is the climate shift that started in 1995 though it probably started in 1990 and was interrupted by Pinatubo. So the "significance" of the "pause" is now the new rehash on Climate Etc.
Sequential Linear Regressions (SLR) is a way to "Gut Check" a trend significance. Scientist don't like that kinda talk, they want peer reviewed proof, but once again into the lack of peer reviewed breach.
First, using just one data set and the "average" of that data set to boot, when determining the significance of a trend is just plain stupid. So this rehash is another "Stupid" discussion. Stupid is not always the same as ignorance. The chart above has the 15year and 20year SLR for the new HADCRUT4 surface temperature data set. The units on the y axis are decadal trends in degrees C for each point with the start date on the x axis. The mean value for the "Global" series is shown which indicate a reliable "mean" regression slope from 1955 of ~0.12 for the 20 year and ~0.13 for the 15 year curves. That should be pretty straight forward.
The difference between 0.12 and 0.013 is small, indicating that there is not enough statistical difference between 15 year trends and 20 year trends for this portion of the "average" of this time series, to worry about.
By hiding the "global" curves but leaving the mean value lines, you can compare the most noisy, NH means to the least noisy, SH means, which gives a range of ~0.08 to ~0.16, or a 0.08 range of uncertainty above and below the ~0.125 mean of the "average" "Global" temperature SLR. Without knowing the physical processes causing that range of uncertainty, 0.08 to 0.16 would be your range of "ignorance". Since the mean of the "average" is ~0.125 and there is a +/-0.04 ignorance range, you can be fairly certain that there is a trend of at least 0.085 but likely less than 0.165 for the HADCRUT4 (66% confidence) data from 1955 to present.
There are plenty of other "peer reviewed" methods that can be used, but this is a very simple, "gut check" method that is in keeping with, Exploratory Data Analysis (EDA), which is basically, LOOKING at the data.
Nearly forgot, the same SLR comparison from 1900 with 32 year regressions shows a mean slope of ~0.08 per decade, the current ~0.8C of total warming. The difference between the 1900 to present ~0.08 and 1955 to present ~0.125 is 0.045 producing the roughly same 66% confidence level ( 0.08 to 0.17 instead of 0.16).
## Tuesday, October 16, 2012
### Decades of Heat Capacity Change
This is not really ready for prime time but it is an interesting view of climate change. I am using the surface area of the oceans by 5 degree bands of latitude and temperature data to put together a profile view of heat capacity. I am working more on the look than the accuracy right now, using GISS regional land and ocean data instead of actual ocean 5 degree bands, but it looks pretty neat.
This is how it looks so far. Estimated heat capacity of the surface layer or the oceans by latitude. There are decade plots of average showing that there is not much to see.
This view is a lot better. The plots are the decades (shifted up one year to include 2011) subracted fro the 1942 to 1951 decade which is the base period. The oughts (00s) are the main story, but it is interesting that the 90s where not that different than the 40s except for the northern high latitudes. The bump in the ACC region is obvious in all the plots, but the 70s, 80s and the 00s show the shift toward the northern latitudes pretty well.
Still a lot of work to do, but it will be fun to "see" heat capacity changes instead of the same old boring temperature anomaly stuff.
### 10 Raised to the 22nd is a Big Number
With the validity of temperatures trends being questioned the debate turns to Ocean Heat Content (OHC). This is the area of the famous "missing" heat.
This table is estimates of the ocean by 5 degree latitude bands. The Tave, or average temperature is from one of Bob Tisdale's charts posted by Roger Pielke Sr. when examining SST by latitudinal band. Since the bands I am using are not the same as Bob Tisdale was using, this is an estimate.
The Water Area is in million kilometers squared (km^2), there are one million cubic meters in a km^2 of water one meter deep, there are one million grams of water in a cubic meter of water and this chart estimate 4.2 Joules per gram per degree C. For negative temperatures, 4.2 is assumed to zero C degrees. Above zero, each degree adds 4.2 joules per gram. This is an estimate since salinity does impact the heat capacity and the 4.2 J/g varies with temperature. This determined the One Meter Joules value. The One Degree column is a "what if" temperature uniformly increased by a degree or the impact of one degree temperature change per 5 degree latitude band.
The top one meter of the oceans would contain roughly 2.78 E+22 Joules. So to estimate the heat capacity of the upper mixing layer, should that be 100 meters thick, multiply by 100. A simple tool to put in your bullshit detector toolbox.
While I have briefly checked the values in the table, I haven't gone nutz since there are several estimates made that I would like to make more accurate. For general BS detector use, it appears to be close enough for government work. Should anyone note a glaring error, let me know, but this is basically a rough draft from an ongoing project.
Like using the data to "see" global heat capacity. Like here, the oceans are biased to the Southern hemisphere but the energy of the oceans is more uniformly "balanced" between the hemispheres. I could use this base estimate and use decadal variations temperature by latitude to show the shifting balance. That would be the climate "shifts". The resistance of the atmosphere to radiant heat loss would change with the shifting internal "balance". Since the southern hemisphere has the thermally isolated Antarctic, that "window" to space is more open.
## Monday, October 15, 2012
### How Would You Predict the Future?
The one truism about trends it that by the time you know one exits, it is too late to do you any good.
There are tricks to help locate trends. This is "normalized" data for GISS LOTI regional. Normalizing, or dividing the average (not RMS) by the standard deviation in this case, lets you compare the fluctuations of different data sets. The green curve, 44S to 64S (legend should read 44-64S) does not vary as much as the other data it is compared withIt has roughly the same "trend" but less variation.
This is the Green 44S-64S with Global, Northern Hemisphere and Southern Hemisphere data, no "normalization" this time. The Green Curve is almost 100% ocean. The Blue is about 70% ocean. The Yellow is about 76% ocean. The Red is about 35% ocean. If you where thinking about predicting changes in the oceans, which would you use to predict the future?
Hint: What is the color of money?
## Sunday, October 14, 2012
Tuggweilder and Samuels, 1994, Effect of the Drake Passage on Global Thermohaline Circulation
Abstract-The Ekman divergence around Antarctica raises a large amount of deep water to the
ocean’s surface. The regional Ekman transport moves the upwelled deep water northward out of
the circumpolar zone. The divergence and northward surface drift combine, in effect, to remove
deep water from the interior of the ocean. This wind-driven removal process is facilitated by a
unique dynamic constraint operating in the latitude band containing Drake Passage. Through a
simple model sensitivity experiment WC show that the upwelling and removal of deep water in the
circumpolar belt may be quantitatively related to the formation of new deep water in the northern
North Atlantic. These results show that stronger winds in the south can induct more deep water
formation in the north and more deep outflow through the South Atlantic. The fact that winds in
the southern hemisphere might influence the formation of deep water in the North Atlantic brings
into question long-standing notions about the forces that drive the ocean’s thermohaline circulation.
Is a must read if you want any insight into longer term global climate. The drawing which is reproduced without permission, so should the holder of the rights be offended, I will remove, but it is a very educational drawing. The Antarctic Convergence is more than just an ocean convergence, it is an upper ocean, deep ocean, atmospheric, thermally isolated Antarctic, radiant window to space convergence zone, which is modulated by the Drake Passage Current and southern hemisphere atmospheric and sea ice dynamics. It is the cheese in the "big cheese" of climate regulation. The source of the 4C deep ocean heat content borders a ~2C mixed ACC band with the mystereous <0 C source of the true abysmal depths sliding down the rock face of the continent of Antarctica. Small changes in weather patterns and sea ice extent can cause a 2 C change in temperature from 55S to 45S in a fairly short time period. That shift would change the average temperature and downward pressure on the 4C deep water supply for the THC with impacts that would be felt many decades in the future.
Enjoy
## Friday, October 12, 2012
### Sweets Spots, Apples and Frozen Oranges
Climate today is being compared to past climate to predict future climate. Only one problem, the conditions in the past do not exist today and many unlikely to exist in the future. So what drove the changes in the past will likely not have the same impact in the future. This is the single most difficult point to explain in the climate debate.
The solar insolation typically used for "Ice Age" comparisons is 65N or in some studies 70S. The chart above use Huybers et al. data for 30S and the 250Wm-2 threshold. When both the Solar and either of the SST reconstructions (Herbert et al. 2010) are above a threshold or "Sweet Spot", there is a coordinated response. Below a certain threshold, there is little or no response to solar. The rise of the South China Sea from 250ka to 200ka was not inphase with the Eastern Pacific or the penultimate deglaciation would have occurred 20 to 40ka before it did. At 150ka, the Eastern Pacific was below its threshold and the South china Sea was decreasing. The decrease in the Eastern Tropical Pacific was likely due to Southern Hemisphere sea ice and circulation that changed the heat transfer efficiency in the southern oceans.
Once both the Eastern Pacific and South China Sea reached their peaks, prior to peak solar, the decline into the last glacial maximum (LGM) began, undoubtedly aided by Northern Hemisphere increasing ice mass. The decline into the LGM took 100,000 years from the peak Pacific ocean temperatures.
The common thought process is that that cycle is repeatable. With mankind claiming the vast expanse of lands that where covered by glacial ice in the LGM, it is unlikely that man would fall asleep at the switch for 100,000 years to allow the gradual accumulation of glacial ice. So it is not productive to compare a response to solar forcing where glacial ice is a major feedback. That would be comparing apples to frozen oranges. Periods where ice extent are more inline with what "should" be expected, would be a better comparison if attempting to predict future climate.
Quite a few researchers have attempted to use paleo data to determine "climate sensitivity". Using perfectly valid methods, they arrive at ranges from 0.7 to 9 C degrees per doubling of CO2. The higher estimates are generally based on the most recent past 400,000 years where the would be nearly maximum ice albedo feedback and ice mass thermal feedback with ocean current impacts thrown in to boot.
By eliminating the Ice variable that impacts radiant forcing, the internal dynamics of the oceans relative to an expectation of future minimal ice expanse condition would allow not only a simpler, but more accurate prediction of impact due to various "pertinent" forcing combinations.
This confusion over where to focus attention is likely the largest stumbling block in the climate debate. Multi-millennial land use impact cannot be ignored.
## Thursday, October 11, 2012
### Sweet Spots Produce Amazing Results
Any Golfer, Baseball Player or Physicist knows about sweet spots. Where the least amount of energy provides the greatest result. In golf, many players pay big bucks to get frequency tuned shafts and cavity backed irons. The oversize metal woods like the aluminum baseball bat have larger "sweet spots" where less velocity produces greater distance that wooden clubs or bats. One of the best, but somewhat difficult to master golf swings, is David Duval's in his prime.
Constant acceleration through the point of contact, SWEET! If you over swing or under swing you just don't get the same results.
Climate also has its sweet spot or Goldielocks moments. In Climate science you hear about forcings, conservation of energy and conservation of mass, but not much about conservation of momentum. This plot of the Bintanja & van de Wal 45N termperature of the deep ocean and the estimated orbital precession shows a sweet spot. The upswing in temperature was perfectly timed with the increase solar energy in the southern oceans while the southern ocean temperatures where developing momentum in a warming phase from a deeper than normal depressed temperature. Now this plot is of the 45N ocean temperature, how would I know what the southern oceans where doing?
The Martin et al. bottom water temperatures don't cover the same period, but comparing the last two deglaciations shows what I am talking about. The penultimate deglaciation had the Tropical Eastern Pacific BWT out of phase with the Atlantic BWT. The energy collected by the Pacific was partically transferred to the Atlantic, but not with the proper timeing to over come the thermal inertia of the Atlantic waters. In the last deglaciation, a synchronizing pulse circa 80 k years ago shifted the momentum of the Atlantic bottom water so that with the following solar pulse the sweet spot was hit.
The shift from 41 ka cycles to the current ~100 ka maybe cycle is due to changes in the internal harmonics of the global ocean heat transfer. Bold statement, right?
This chart of the Pahnke southern ocean SST with estimated solar precession also shows how with in phase orientation, there is greater results for less energy applied.
Looking at the first chart again, the x-axis is scaled to 23ka years and shift to align with some of the more interesting peaks. The precession frequency is not exact and the decay frequency or internal harmonics of the oceans are not exactly tuned to the precessional frequency. The 4.3ka recurrent frequency of the ocean oscillations from precessional perturbation is not a perfect "fifth", it is a little out of tune. The Oceans also have a ~5.8ka decay from the 41ka obliquity perturbations indicating they decay well before that next perturbations. The oceans are current tuned to ~21.5ka cycles which drift allows to synchronize either with precession or with every second or third obliquity cycle.
Showing why there is a new tuning since ~800ka years ago is a challenge, but the precessional cycle has been in charge in the distant past.
Between 1800ka and 2000 ka years before present, the Earth's Geomagnetic field was also in the current orientation and developed the start of the `~100ka ice age cycles. Even then the range of oceans temperatures was small with no hint of runaway climate. In fact the short precession frequency produced a smaller range of variation.
Timing is everything in nonlinear dynamics, whether it is a golf swing or paleo climate.
### Just Charts Waiting for a Story
Since I have had a few computers crash in the past two years likely due to humid Florida weather, I tend to post stuff on line for safe keeping. Instead of trying to keep the stuff private, I just let it out there for all to see.. There are lots of Southern Hemisphere Ocean Cores to play with, so I will just post a few charts I think are keepers here from time to time. References will be a little lax with mainly just the core ID posted unless someone complains, until I either use the charts in a semi-post or just move on. Geomagnetic fans though might find this one interesting.
```SUGGESTED DATA CITATION: Yamamoto, M., et al. 2007.
California Current 136KYr Alkenone Sea Surface Temperature Estimates.
IGBP PAGES/World Data Center for Paleoclimatology
Data Contribution Series # 2007-100.
NOAA/NCDC Paleoclimatology Program, Boulder CO, USA.```
`The Bintanja and van de Wal should make and excellent reference for past sea level. This chart compares the Herbert et al. 2010 Tropical Eastern Pacific and South China Sea temperature reconstructions. Opposite axis are used with the Herbert data shifted so the mean value of all three series for the period 0 to 3000 ka, where available, align. The temperature scale is just roughly adjusted for fit near 2000ka.`
`This is the same chart from 1500 to 3000 ka with the rough adjustment period in the highlight box. This is the period which had two closely space magnetic reversals followed by the longer Compass North orientation. `
`The first part of the chart which shows the shift circa 900 to 920 ka which starts the rise in Tropical Eastern Pacific Temperature and sea level wide fluctuation range.`
`The shift in the highlight box not only change the slope of tropical Pacific SST, it appears to have accentuated the longer 400ka orbital cycle. `
`The DSD607 is in the Northern Atlantic which provides a good comparison to the tropical eastern Pacific. The same Herbert et al. 2010 for ODP846 and Lawrence et al. 2010 `
```Lawrence, K.T., S. Sosdian, H.E. White, and Y. Rosenthal. 2010.
North Atlantic climate evolution through the Plio-Pleistocene climate transitions.
Earth and Planetary Science Letters, Vol. 300, Issues 3-4,
1 December 2010, pp. 329-342.
doi:10.1016/j.epsl.2010.10.013
```
```
```
`Interesting take in the Abstract:`
```
```
```We propose that the expansion of the West Antarctic ice sheet (WAIS) across the MPT increased the production and export of Antarctic Bottom Water from the Southern Ocean and subsequently controlled its incursion into the North Atlantic, especially during glacial intervals. It follows that the early 100 kyr response of BWT implies an early response of the WAIS relative to the northern hemisphere deglaciation. Thus, in the "100 kyr world," both northern hemisphere and southern hemisphere processes affect climate conditions in the North Atlantic Ocean.
```
I wonder if they noticed the magnetic reversal coordination which could be due to Antarctic Ice Sheet movement.?
### Tuning History - Climate Blues in A5
Orbital influences on climate can also be tuning influence of orbit on climate, power chords like fifths may resolve some to the mystery.
I studying the correlation of estimated climate change in deep ocean core samples I noticed that precession, the red haired step child of orbital parameters, appears strongly as an ~23,000 year response and that there appears to be a 4,300 year recurrent decay from the precessional impacts. Obliquity and Eccentricity are considered the big orbital players in the climate band.
According to:
### To tune or not to tune: Detecting orbital variability in Oligo-Miocene climate records
Cristian Proistosescu a,⁎, Peter Huybers a, Adam C. Maloof b
" A different behavior is observed for the 1/23 kyr−1 peak where, for a moderate amount of tuning, the significance of the peak increases dramatically. The robustness of the results for precession across testing configuration lead us to confidently conclude that significant precession band variability is present in the ODP 1218 δ18O record. This result is, to our knowledge, the first unbiased statistical test for orbital variability using orbitally tuned records."
and
"For ODP 1218 the estimated autocorrelation coefficient is ϕ=0.87 and the variance of the ∈ disturbances is 0.41 at a time step of 4.3 kyr."
A little "tuning" goes a long way as long as it is compared with untuned data.
This is not Earth shattering news for most, but the locating the potential source of Bond Events kinda blows wind up my skirt :)
### "Global" versus the Oceans - Part Four
From the original installment, the "Global" versus to Oceans series has been intend to show the impact of the Antarctic Continent and Circumpolar Current on Global climate and magnetic orientation. The potential instability of Antarctic Ice sheets to orbital and tidal forces can impact orbital and tidal forces. A complex set of tens of millennial scale climate related geophysical interactions.
The Scaled EPICA CO2 and the Baseline provided by Bintanja & van de Wal provide a "standard" for comparison to various regional reconstructions of past ocean climate.
This series of comparisons adds the Herbert et al 2010 South China Sea SST anomaly and Eastern Pacific to the "standards".
The period from ~800ka to present is tame compared to the more distant past. The next chart shows the most recent magnetic field reversal.
There are many gaps to fill, but the use of the Scaled CO2 and 45N temperature of the deep oceans provides an interesting perspective of past "climate". One of the more interesting is that it is unlikely that there is a single "sensitivity" of climate to carbon dioxide. Then relative calm of the past 800 k years would indicate more sensitivity while the beyond 800 k years could produce any sensitivity one wished to find. Since the 45N remains more closely related to CO2, it is very likely that land based ice has the greatest control over CO2 and Southern Ocean mixing the greatest control over land based ice. At least, CO2 is less likely to influence geomagnetic field reversals than huge Antarctic Ice Sheets.
Note: While I have reviewed the "Global" versus the Oceans posts, there are always chances to improve and correct. Any volunteer review comments are always welcome. | 11,987 | 50,556 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2017-17 | longest | en | 0.925219 |
https://helpinhomework.org/question/36854/New-Perspectives-Excel-2019-Module-3-End-of-Module-Project-2-Ramandeep-Singh-1-Ying-Chou-i | 1,716,490,987,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058653.47/warc/CC-MAIN-20240523173456-20240523203456-00491.warc.gz | 251,412,483 | 21,508 | Fill This Form To Receive Instant Help
#### New Perspectives Excel 2019 | Module 3: End of Module Project 2 Ramandeep Singh 1
###### MS Excel
New Perspectives Excel 2019 | Module 3: End of Module Project 2
Ramandeep Singh
1. Ying Chou is an intern with the Sandhills-Sand River Water District. Ying has collected data on
precipitation and reservoir storage and wants to use Excel to perform some analysis on her data.
Switch to the 2021 Water Projections worksheet. In cell C5, insert the TODAY function to record the
current date.
2. In cell J4, create a formula using the MONTH function to display the numerical month based on the date in cell C4
3. In cell J5, create a formula without using a function that subtracts the value in cell J4 from the number 12
.
4. Use the values in the range C9:D9 to extend the list of years to the range E9:G9.
5. Use the values in the range B10:B11 to extend the list of reservoir numbers to the range B12:B15.
6. Use AutoFill to fill the range C11:J15 with the formatting from the range C10:J10.
7. In cell G10, create a formula that uses the AVERAGE and ROUND functions to calculate the average of cells C10:F10 and round the result to 1 decimal place. Copy the formula you created to the range
G11:G15.
8. In cell J10, create a formula without using a function that divides the value of cell I10
by the value of cell H10 . Copy the formula you created to the range J11:J15.
9. In cell G16, create a formula that uses the AVERAGE and ROUND
functions to calculate the average of the range G10:G15 and round the result to 1
decimal place. Copy the formula to cell H16.
10. In cell D21, enter a formula without using a function that multiplies the value of cell
D22 by 2.83 and then adds the value of cell D19 to the result.
11. Use Goal Seek to identify the value for cell D22 that results in a value of
100 for cell D21.
12. Create lookup functions to complete the summary section. In cell J20, create a formula using the
VLOOKUP function to display the measured precipitation for the selected reservoir. Lookup the reservoir number (cell J19 ) in the range B10:J15 , and return the value in the 2 nd column number. Use absolute references for cell J19 and the range B10:J15.
13. Copy only the formula from cell J20 to the range J21:J25, and then edit the formula in cell J21 to return the value in the 3 rd column, the formula in cell J22 to return the value in the 4 th column, the formula in cell J23 to return the value in the 5 th column, the formula in cell J24 to return the value in the 8 th column, and the formula in cell J25 to return the value in the 9 th column.
Final Figure: create a 2021 Water Projections
## 19.87 USD
### Option 2
#### rated 5 stars
Purchased 12 times
Completion Status 100% | 717 | 2,781 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2024-22 | latest | en | 0.746715 |
https://www.issacc.com/topological-sorting-do-you-really-know-it/ | 1,725,978,990,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651255.81/warc/CC-MAIN-20240910125411-20240910155411-00168.warc.gz | 764,363,546 | 11,902 | I am sure some of you might have heard about topological sorting, but do you really know what that is? Let's find out together!
From the Wikipedia's definition, a topological sort of a directed graph is a linear ordering of its vertices such that for every directed edge `uv` from vertex `u` to vertex `v`, `u` comes before `v` in the ordering.
In layman's term, a topological sorting is to sort a directed graph that has no cycles. Normally, you will use an array list or any other data structures to store the result of the sorting. The way we sort the graph is by calculating the indegree of each node, which is the number of neighbor nodes pointing to itself.
NOTE: topological sorting of a graph cannot be accomplished if the graph has formed cycles between nodes. A topological ordering is possible `iff` the graph has no directed cycles, that is, it it is a directed acyclic graph (`DAG`).
There are quite a lot of good articles regarding to topological sort, so I will skip the remaining explanation part here. Here's one that I find quite useful, the link is attached here for reference.
### Given an directed graph, find any topological order for the given graph.
You can think of the node without other nodes pointing to it as the initial node.
The topological order can be `[0,1,2,3,4,5]` or `0,2,3,1,5,4]` and etc.
Here's the solution for reference.
`````` public ArrayList<DirectedGraphNode> topSort(ArrayList<DirectedGraphNode> graph) {
ArrayList<DirectedGraphNode> order = new ArrayList<>();
if(graph == null){
return order;
}
// 1. we need to find out the indegree for each node
Map<DirectedGraphNode, Integer> indegree = getIndegree(graph);
// 2. add the node with 0 indegree into the queue and array
for(DirectedGraphNode node : graph){
if(indegree.get(node) == 0){
queue.offer(node);
}
}
// 3. decrease the indegree level after removing the 0 degree node
// add the node with 0 indegree to queue and array
while( !queue.isEmpty()) {
DirectedGraphNode node = queue.poll();
for(DirectedGraphNode neighbor : node.neighbors){
indegree.put(neighbor, indegree.get(neighbor) - 1);
if(indegree.get(neighbor) == 0){
queue.offer(neighbor);
}
}
}
// 4. check if the sorted array's size is the same as initial graph size
if(order.size() == graph.size()){
return order;
}
return null;
}
private Map<DirectedGraphNode, Integer> getIndegree(ArrayList<DirectedGraphNode> graph){
Map<DirectedGraphNode, Integer> indegree = new HashMap<>();
// 1. initialize the map
for(DirectedGraphNode node : graph){
indegree.put(node, 0);
}
// 2. add the indegree level for each pointed node
for(DirectedGraphNode node : graph){
for(DirectedGraphNode neighbor : node.neighbors){
indegree.put(neighbor, indegree.get(neighbor) + 1);
}
}
return indegree;
}
}
``````
The solution is actually pretty neat and straightforward. For `BFS` problem, the procedure is almost the same with little modification. For starter, you will need a hashmap or array and a queue. While the queue is not empty, you remove the current node from the queue and add the neighbor nodes into the queue. When certain criteria is met, for example, in this particular question, we add the neighbor node into the queue when the indegree of the node becomes 0.
That's all about it! It ain't that hard, right? I will be posting related `BFS` problems in the next few posts, so stay tuned!
Remember, the more you practice, the better you get! Let's do it 😃
Post was published on , last updated on .
Like the content? Support the author by paypal.me! | 847 | 3,532 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 4 | CC-MAIN-2024-38 | latest | en | 0.902279 |
https://www.equationsworksheets.net/maneuvering-the-middle-llc-2016-worksheets-answer-key-linear-equations/ | 1,709,517,687,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476409.38/warc/CC-MAIN-20240304002142-20240304032142-00860.warc.gz | 751,963,600 | 15,309 | # Maneuvering The Middle Llc 2016 Worksheets Answer Key Linear Equations
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These worksheets can be very helpful for children in the beginning grades. These worksheets teach students how to solve equations and graph. They are great to practice polynomial variables. These worksheets will help you factor and simplify these variables. You can find a great collection of equations and expressions worksheets for kids at any grade. Doing the work yourself is the most efficient way to understand equations.
There are a variety of worksheets that can be used to help you understand quadratic equations. There are several worksheets on the different levels of equations for each level. The worksheets were designed to assist you in solving problems of the fourth degree. After you’ve completed an amount of work, you can continue to work on solving different kinds of equations. Then, you can work on solving the same-level problems. For example, you can discover a problem that uses the same axis as an elongated number. | 668 | 3,627 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.936179 |
https://restnova.com/travel/what-determines-the-direction-a-pwc-will-travel/ | 1,680,074,189,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948951.4/warc/CC-MAIN-20230329054547-20230329084547-00334.warc.gz | 566,407,504 | 17,739 | Top 10 WHAT DETERMINES THE DIRECTION A PWC WILL TRAVEL Answers
# What Determines The Direction A Pwc Will Travel
Category: Travel
## 1. what determines the direction a pwc will travel – solsarin
Sep 18, 2021 — The steering controls determine the direction that your PWC will travel in. However, the steering will only work when you have power. This is (1)
What determines the direction a PWC will travel? A person is accompanied on board a motorboat by a parent or guardian. a different velocity and direction.(2)
The direction of force determines the direction of acceleration; F=ma. traveling… not a good thing to do in busy areas. In the direction shown in the manual.(3)
## 2. What Determines The Direction A Pwc Will Travel? – Scouting …
What Determines The Direction A Pwc Will Travel? (Correct Answer Below). Reveal the answer to this question whenever you are ready.(4)
What determines the direction of PWC will travel? — What is needed to steering control on a PWC? What determines the direction of PWC will travel?(5)
When boating on a river, you may encounter strainers. What is the danger of a strainer? Strainers can trap small boats. Rating: 4 · 3 reviews(6)
## 3. What determines the direction a PWC will travel? – Answers
Jun 8, 2012 · 4 answersWhat determines the direction a PWC will travel? It’s easy! The angle of the rudder! What determines the direction of a PWC will travel when the (7)
Jun 12, 2010 — The pilot only has control over the vertical travel of the aircraft. In order to achieve travel in any given direction, the pilot must find an (8)
## 4. When is it difficult to reboard a pwc? – Movie Cultists
What determines the direction that a PWC will travel? The steering controls determine the direction that your PWC will travel in. However, the steering will (9)
The steering controls decide the path that your PWC will journey in. Nonetheless, the steering will solely What determines the fee {that a} PWC will journey ?(10)
What determines the direction a PWC will travel? — What determines the direction a PWC will travel? What happens if you shut off PWC engine? What happens to (11)
Dec 11, 2021 — The current advice from most health professionals is to walk from the iliacus muscle. This is the muscle that gives you “leg strength” as well (12)
What determines the direction a PWC will travel when the steering control is moved? Likewise the northward velocity of the hurricane would be restricted by the (13)
## 5. Learn about Jet skis / Personal water crafts in Naples, FL
Personal watercraft is the most common type of vessels that use a jet drive. The PWC will continue in the direction it was headed before the throttle Missing: travel | Must include: travel(14)
weight the PWC can carry safely in good weather. determine the source and make repairs continue in the direction it was headed before the engine.48 pages(15)
(16)
## 6. Personal Watercraft (PWC) | Virginia DWR
Besides not knowing which direction the other operator may decide to go, large ships and barges will not be able to stop quickly. Remember, boats don’t have (17)
The reverse cowling is a specially designed diverter that can be lowered over the jet nozzle. The water jet produced by the jet nozzle hits the reverse cowling (18)
If the personal watercraft (PWC) has capsized, it should be turned upright and in ONE direction only. Check your owner’s manual or the warning sticker located (19)
What determines the direction a PWC will travel? Leave a Comment / General. Finest Reply. the jet drive (rudder). Extra Solutions.(20)
## 7. what determines the direction a pwc will travel? – MSZG News
what determines the direction a pwc will travel? ; Northwest Florida Road Trips and Scenic Drives · December 26, 2021 ; Reasons to choose Mida travel Homestay.(21)
A person shall not operate a PWC in waters less than two feet deep unless traveling at a slow no-wake speed. You can take an online boating safety course for (22)
The person riding must be able to turn the handlebars either counterclockwise or clockwise, which will permit the boat to move in the direction it is going.(23)
## 8. MAINE BOATING LAWS AND RESPONSIBILITIES
Personal watercraft (PWC) and some example, directing travel within a channel. Many PWC will continue in the direction they were headed.(24)
Let’s hook up and inspire others. tropical cyclone such as an Australian willy-willy moves west and then recurves to the southeast.(25)
Remember, your PWC operator’s manual will tell you the specifics of your boat, I realize that my travel speed should be determined by my equipment, (26)
## 9. Boat Engine Types Explained – BOATERexam.com
To steer an outboard you need to move the entire engine. the steering wheel, which then turns the drive unit, and determines the direction of the boat.(27)
A vessel’s length class determines the the way of vessels traveling the waterway. PWC will continue in the direction they were headed before.(28)
## 10. Personal Watercraft (PWC) Safety Guide – Discover Boating
Safely operating a personal watercraft (PWC), Jet Ski, WaveRunner or Sea-Doo can ensure that your experience is that much more enjoyable.(29)
Dec 14, 2016 — If left alone, that force would just push the craft forward. Unlike traditional boats, PWC don’t have rudders to control their direction.(30)
effectiveness of their workforce, develop and move talent around their will determine the future of work in 2030. direction and management.(31)
infrastructure investment will be significant. Figure 2: Global growth is a key driver of air travel determine the number of airlines that.(32)
other part of the forward half of the vessel where it can be easily read. • For Personal Watercraft (PWC), the numbers and stickers must.(33)
The number of passengers that a vessel can carry will be determined by the marine inspector. vessel is to move it bodily in the same direction as the.(34)
Jun 23, 2021 — A PWC includes jet skis, wave runners and similar vessels that have an They can pose a danger to the operator and to other people if not (35)
complete the online test and you will receive a State of Florida boating safety PWC in front of or beside you can change direction in an.(36)
A vessel’s length class determines the example, directing travel within a channel. PWC will continue in the direction it was headed before.(37)
Is PWC operation restricted within any area or zone in your state? the wake of another vessel travelling in the same direction in close proximity to the (38) | 1,524 | 6,527 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-14 | latest | en | 0.804759 |
http://oeis.org/A174214/internal | 1,571,800,769,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987828425.99/warc/CC-MAIN-20191023015841-20191023043341-00508.warc.gz | 146,618,696 | 3,005 | This site is supported by donations to The OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A174214 a(n) = a(n-1)+1, if the previous term a(n-1) and n-1-(-1)^n are coprime, else a(n)=2*n-4. 5
%I
%S 14,16,17,18,19,20,26,28,29,30,31,32,33,34,35,36,37,38,39,52,53,54,55,
%T 60,62,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,
%U 85,86,87,88,89,90,91,92,93,124,125,126,127,128,129,130,131,132,133,134
%N a(n) = a(n-1)+1, if the previous term a(n-1) and n-1-(-1)^n are coprime, else a(n)=2*n-4.
%H V. Shevelev, <a href="http://arXiv.org/abs/0912.4006">Theorems on twin primes-dual case</a>, arXiv:0912.4006 [math.GM], 2009-2014.
%p A174214 := proc(n) option remember ; if n = 9 then 14 ; elif gcd(procname(n-1),n-1-(-1)^n) = 1 then procname(n-1)+1 ; else 2*n-4 ; end if; end proc:
%p seq(A174214(n),n=9..100) ; # _R. J. Mathar_, Mar 16 2010
%t a[n_] := a[n] = Which[n==9, 14, CoprimeQ[a[n-1], n-1-(-1)^n], a[n-1]+1, True, 2n-4]; Table[a[n], {n, 9, 100}] (* _Jean-François Alcover_, Feb 02 2016 *)
%Y Cf. A166945, A167495.
%K nonn,easy
%O 9,1
%A _Vladimir Shevelev_, Mar 12 2010
%E a(15) corrected and sequence extended by _R. J. Mathar_, Mar 16 2010
%E a(15) corrected and a(35)-a(74) added by _John W. Layman_, Mar 16 2010
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Last modified October 22 22:34 EDT 2019. Contains 328335 sequences. (Running on oeis4.) | 682 | 1,625 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2019-43 | latest | en | 0.539471 |
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Depdendent Variable
Number of equations to solve: 23456789
Equ. #1:
Equ. #2:
Equ. #3:
Equ. #4:
Equ. #5:
Equ. #6:
Equ. #7:
Equ. #8:
Equ. #9:
Solve for:
Dependent Variable
Number of inequalities to solve: 23456789
Ineq. #1:
Ineq. #2:
Ineq. #3:
Ineq. #4:
Ineq. #5:
Ineq. #6:
Ineq. #7:
Ineq. #8:
Ineq. #9:
Solve for:
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ap physics c: mechanics graphing calculator apps
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Author Message
ihkeook
Registered: 15.02.2004
From: holland
Posted: Friday 29th of Dec 08:13 Hi math wizards! I require some direction to solve this ap physics c: mechanics graphing calculator apps which I’m unable to do on my own. My homework assignment is due and I need guidance to work on angle-angle similarity, radical inequalities and subtracting fractions . I’m also thinking of hiring a math tutor but they are not economical . So I would be very much value if you can extend some assistance in solving the problem.
AllejHat
Registered: 16.07.2003
From: Odense, Denmark
Posted: Saturday 30th of Dec 09:48 Have you checked out Algebrator? This is a great software and I have used it several times to help me with my ap physics c: mechanics graphing calculator apps problems. It is very straightforward -you just need to enter the problem and it will give you a detailed solution that can help solve your assignment . Try it out and see if it solves your problem.
Paubaume
Registered: 18.04.2004
From: In the stars... where you left me, and where I will wait for you... always...
Posted: Sunday 31st of Dec 10:27 1.Hey mate, you are on the mark about Algebrator! It is absolutely fab! I downloaded it recently from https://sofsource.com/multiplying-polynomials.html after a colleague recommended it to me. Now, all I do is type in the problem from my text book and click on Solve. Bingo! I get a step-by-step solution to my algebra homework. It’s almost like a tutor is explaining it to you. I have been using it for two weeks and so far, haven’t come across any problem that Algebrator can’t solve. I have learnt so much from it!
Juitalgat
Registered: 03.05.2003
From:
Posted: Monday 01st of Jan 09:33 Of course. My aim is to learn Math. I would use it only as a tool to clear my concepts. Can I get the link to the program ?
Voumdaim of Obpnis
Registered: 11.06.2004
From: SF Bay Area, CA, USA
Posted: Monday 01st of Jan 18:57 Its simple , just click on the this link and you are good to go – https://sofsource.com/fractional-exponents.html. And remember they even give a ‘no catch’ money back guarantee with their software , but I’m sure you’ll love it and won’t ever ask for your money back. | 850 | 3,252 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-45 | latest | en | 0.872739 |
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Disc-a9
# Disc-a9 - 407 mod 74 = 37 74 mod 37 = 0 so gcd(3478 2183 =...
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1. Find integers q and r so that ° 185 = 22 q + r and 0 ± r < 22 . What are ( ° 185) mod 22 and ( ° 185) div 22 ? One way to do this is to compute ° 185 = 22 = ° 8 : 4091 on a hand calculator. That tells us that q = ° 9 , the integer just to the left of ° 8 : 4091 . So r = ° 185 ° 22 q = ° 185 ° 22( ° 9) = 13 . So ( ° 185) mod 22 = 13 and ( ° 185) div 22 = ° 9 . 2. Use the Euclidean algorithm to °nd gcd (3478 ; 2183) . 3478 mod 2183 = 1295 2183 mod 1295 = 888 1295 mod 888 = 407
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Unformatted text preview: 407 mod 74 = 37 74 mod 37 = 0 , so gcd (3478 ; 2183) = 37 3. Find integers x and y so that 77 x + 144 y = gcd (77 ; 144) . The table for the Euclidean algorithm is 77 x + 144 y x y q 144 1 77 1 1 67 & 1 1 1 10 2 & 1 6 7 & 13 7 1 3 15 & 8 2 1 & 43 23 so gcd (77 : 144) = 1 and 77 ( & 43) + 144 (23) = 1 ....
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Jill Tulane University ‘16, Course Hero Intern | 656 | 2,063 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2018-22 | latest | en | 0.817862 |
https://www.airmilescalculator.com/distance/glh-to-tup/ | 1,660,548,640,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572161.46/warc/CC-MAIN-20220815054743-20220815084743-00301.warc.gz | 571,007,378 | 16,035 | # Distance between Greenville, MS (GLH) and Tupelo, MS (TUP)
Flight distance from Greenville to Tupelo (Greenville Mid-Delta Airport – Tupelo Regional Airport) is 138 miles / 223 kilometers / 120 nautical miles. Estimated flight time is 45 minutes.
Driving distance from Greenville (GLH) to Tupelo (TUP) is 178 miles / 287 kilometers and travel time by car is about 3 hours 54 minutes.
138
Miles
223
Kilometers
120
Nautical miles
## How far is Tupelo from Greenville?
There are several ways to calculate distances between Los Angeles and Chicago. Here are two common methods:
Vincenty's formula (applied above)
• 138.394 miles
• 222.723 kilometers
• 120.261 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 138.193 miles
• 222.401 kilometers
• 120.087 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## How long does it take to fly from Greenville to Tupelo?
Estimated flight time from Greenville Mid-Delta Airport to Tupelo Regional Airport is 45 minutes.
## What is the time difference between Greenville and Tupelo?
There is no time difference between Greenville and Tupelo.
## Flight carbon footprint between Greenville Mid-Delta Airport (GLH) and Tupelo Regional Airport (TUP)
On average flying from Greenville to Tupelo generates about 45 kg of CO2 per passenger, 45 kilograms is equal to 100 pounds (lbs). The figures are estimates and include only the CO2 generated by burning jet fuel.
## Map of flight path and driving directions from Greenville to Tupelo
Shortest flight path between Greenville Mid-Delta Airport (GLH) and Tupelo Regional Airport (TUP).
## Airport information
Origin Greenville Mid-Delta Airport
City: Greenville, MS
Country: United States
IATA Code: GLH
ICAO Code: KGLH
Coordinates: 33°28′58″N, 90°59′8″W
Destination Tupelo Regional Airport
City: Tupelo, MS
Country: United States
IATA Code: TUP
ICAO Code: KTUP
Coordinates: 34°16′5″N, 88°46′11″W | 533 | 2,164 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2022-33 | latest | en | 0.849114 |
https://www.sanfoundry.com/finite-element-method-questions-answers-beams-frames-boundary-conditions/ | 1,713,982,039,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296819847.83/warc/CC-MAIN-20240424174709-20240424204709-00169.warc.gz | 866,076,196 | 22,495 | # Finite Element Method Questions and Answers – Beams and Frames – Boundary Conditions
This set of Finite Element Method Multiple Choice Questions & Answers (MCQs) focuses on “ Beams and Frames – Boundary Conditions”.
1. Symmetry in application of boundary conditions should be avoided in which of the following type of analysis?
a) Linear static analysis
b) Modal analysis
c) Thermal analysis
d) Nonlinear static analysis
Explanation: Symmetric boundary conditions should not be used for modal analysis. Symmetric model would miss some of the modes of modal analysis or out of phase modes.
2. Boundary conditions are applied to simulate the physical constraints on the finite element model.
a) True
b) False
Explanation: Boundary conditions simulate the physical constraints on the finite element model. Application of boundary conditions is a crucial preprocessing step to yield accurate solution.
3. Which of the following is the correct equation for stiffness (K) of an element given value of force (F) and displacement (Q)?
a) FQ=K
b) KQ=F
c) KF=Q
d) KFQ=1
Explanation: The correct equation is given by KQ=F. The value of force (F) is the product of stiffness (K) and displacement (Q). The value of stiffness (K) of the element determines the displacement of the node.
4. Which of the following conditions must be fulfilled to apply symmetry in a finite element model?
a) Geometry of the model is symmetric
b) Boundary conditions to be applied are symmetric
c) Geometry model has large number of nodes
d) Geometry of the model is symmetric and boundary conditions to be applied are symmetric
Explanation: The geometry and boundary conditions both have to be symmetric to apply any kind of symmetry. Half or quarter portions of a model can be used to reduce computational cost.
5. Which of the following boundary conditions cannot be directly applied on solid elements?
a) Force
b) Pressure
c) Support
d) Torque
Explanation: Torque cannot be directly applied on solid element in finite element model. Since solid elements have three translational degrees of freedom and no rotational degrees of freedom torque cannot be directly applied on solid elements.
6. Traction is force acting on an area in any direction other than normal.
a) True
b) False
Explanation: Traction is force acting on an area in any direction other than normal. The force acting on an area in normal direction is called as pressure. Traction is boundary condition applied where force acting on a surface is not normal to the surface such as friction and drag.
7. Which of the following statements is correct?
a) Reaction force at supports is equal to the sum of the product of stiffness and displacement
b) Reaction force at supports is equal to the product of sum of stiffness and displacement
c) Reaction force at supports is equal to the sum of stiffness’s
d) Reaction force at supports is equal to the sum of displacements
Explanation: Reaction force at supports is equal to the sum of the product of stiffness and displacement. The stiffness and displacement matrices are multiplied for each element and then cumulated to find the reaction force at supports.
8. Which of the following equations give the relation between material properties like modulus of elasticity (E), modulus of rigidity (G), and Poisson’s ratio (u)?
a) E = 2*G*(1+u)
b) E = 3*G*(1+u)
c) E = 2*G*(1-u)
d) E = 3*G*(1-u)
Explanation: The relation between the material properties is given by
E = 2*G*(1+u)
Here E is the ratio of normal stress to normal strain. G is the ratio of shear stress to shear strain and u is the ratio of lateral strain to longitudinal strain.
Sanfoundry Global Education & Learning Series – Finite Element Method. | 806 | 3,706 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2024-18 | latest | en | 0.856021 |
https://collegephysicsanswers.com/openstax-solutions/what-beat-frequencies-will-be-present-if-musical-notes-and-c-are-played-together | 1,586,396,613,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371826355.84/warc/CC-MAIN-20200408233313-20200409023813-00141.warc.gz | 400,227,225 | 11,252 | Question
What beat frequencies will be present: (a) If the musical notes A and C are played together (frequencies of 220 and 264 Hz)? (b) If D and F are played together (frequencies of 297 and 352 Hz)? (c) If all four are played together?
Question by OpenStax is licensed under CC BY 4.0.
Final Answer
1. $44 \textrm{ Hz}$
2. $55 \textrm{ Hz}$
3. $f_{AC} = 44 \textrm{ Hz}$, $f_{AD} = 77 \textrm{ Hz}$, $f_{AF} = 132 \textrm{ Hz}$, $f_{CD} = 33 \textrm{ Hz}$, $f_{CF} = 88 \textrm{ Hz}$, $f_{DF} = 55 \textrm{ Hz}$
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This is College Physics Answers with Shaun Dychko. The beat frequency, which is the frequency with which the amplitude is oscillating is going to be the absolute difference between the two frequencies. So, for part A, the beat frequency will be the absolute difference between the frequency of the note A and the note C. So, that's the difference between 264 hertz minus 220 hertz, which is a beat frequency of 44 hertz. And for part B, we're finding the difference in frequencies in D and F, which are 297 and 352, and this works out to 55 hertz. And notice this would be negative, but the absolute value bar just means we take the positive. And then, if both frequencies or beat frequencies will be present for all four... If all four of these frequencies are played together, there's going to be a beat frequency for every pair of possible frequencies. So we have note A, C, D, and F all being played at the same time. And, that's going to result in six different possible pairs, each of which will produce a beat frequency. So, when A is paired with C, that makes a difference of 44 hertz. A and D has a difference of 77 hertz. Note A and note F is 220 minus 352, absolute value of which is 132 hertz. And then, C D makes 33 hertz, and note C and F, the difference is 88 hertz. And, for D and F, the difference is 55 hertz. | 577 | 2,059 | {"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} | 4.21875 | 4 | CC-MAIN-2020-16 | latest | en | 0.915857 |
https://forum.arduino.cc/t/how-to-measure-or-calculate-power-consumption/405309 | 1,726,549,657,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651739.72/warc/CC-MAIN-20240917040428-20240917070428-00871.warc.gz | 233,603,992 | 9,620 | # How to measure or calculate power consumption
HI, I recently finished a beta version of an airsoft/paintball bomb & Capture-The-Flag gadget. It seems to require more power than I thought, eating through a 1100 mAh 11.4V LiPo in less than 2 hours...
I have one 3'2 TFT touchscreen hooked up to one Mega through a TFT shield. The Mega is also running an DF Robot Mini Mp3 player connected to a 2W/3Ohm speaker. The device is periodically flashing 3 five mm LEDs.
This is the schematic:
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How do I go about measuring or calculating the power consumption? I need this gadget to run constantly at least 12 hours.
You're just burning power in the on-board linear regulator.
If you have to use that battery, get a switch mode regulator.
You can measure current with a decent multimeter
Put your digital multimeter on "current" setting, make sure the probes are plugged in in the right place (most DVMs require you to put the red probe in a different place for current measurements), connect the red wire to the one going to the battery pack, and the black one to the rest of the circuit. Read the number off the screen. That is the current it's drawing.
Schematic is missing and posted image is too small to see.
When you say 1100 mAh 3S - is that 1100 mAh total capacity (ie, 3 x360mAh cells), or three 1100 mAh cells in series?
You're pulling a fair amount of power there - the mega is 50 (all Arduino boards are power hogs - they are not designed for battery powered applications, so they have things like power lights and on-board serial adapters that sit there wasting power), nother 100-200 for the screen, easy. And a 2W speaker, say we're seeing a half watt out of that on average, that's nother 100 mA...
So I think you're looking at 400-500mA right now, so it sounds right to me....
As noted above, using a DC-DC buck converter to get 5v (or 7v to use on-board regulator) from the 12v will improve battery life significantly - but your project is still a power hog, and your battery is still wimpy.
Yes, I forgot to mention the buck converter I have connected between the battery (it's 1100 mAh in total) and the Arduino. The rest of the components are all connected to the arduino pins and sourced through them, so they can't really draw that much power, can they?
Attaching the schematic since the picture hosting doesn't seem to work.
schematicREAL.pdf (281 KB)
And regarding the battery: would this be a good alternative?
Why don't you use a 5V buck converter, instead of a 7V buck converter and a 5V linear regulator, that's really inefficient.
Mixe:
And regarding the battery: would this be a good alternative?
No, there's absolutely no reason to have a 12V output. Going from around 3-4V to 12V to step it down again to 5V for the Arduino is just a waste of power (and money).
Ah. I see. So this would be more logical?
It's a 12V output one again.
And I'm sorry to pop your bubble, but 70Ah for €45 is just too good to be true ... These cheap Chinese batteries use fake or rejected cells, that are only 100-300mAh in capacity, and they advertise them as having 3000mAh or what not. It's just a scam.
Maybe that's the reason your project doesn't last, a fake battery?
PieterP:
It's a 12V output one again.
And I'm sorry to pop your bubble, but 70Ah for €45 is just too good to be true ... These cheap Chinese batteries use fake or rejected cells, that are only 100-300mAh in capacity, and they advertise them as having 3000mAh or what not. It's just a scam.
Maybe that's the reason your project doesn't last, a fake battery?
No, it's also 5V, look again. It has different outputs. But you're in no way popping any bubble. I ask to get informed, and I can totally believe the scam part. The battery I did use is an RC battery that I normally use to power my airsoft guns, and I can run them for 20+ hours on that single battery. Not continously though.
Well, I guess I'll end up using a motorcycle battery incorporated in the design. 7 Ah should do the trick.
Mixe:
No, it's also 5V, look again.
Nope, 12V too
Capacity 68800mAh
Autual Capacity 12000mAh
Output 5V/2A,12V/2A,16V/2A,19V/3.5A
Also, that 'Actual Capacity' line seems a bit (read: extremely) dodgy ...
Mixe:
Well, I guess I'll end up using a motorcycle battery incorporated in the design. 7 Ah should do the trick.
Should work indeed, and it's easier to charge as well (than a Li-Ion).
On the other hand, it's quite heavy, compared to a Li-Ion of the same capacity.
A power bank might indeed be a good solution, as you suggested, it's even easier to charge than the motorcycle's SLA, a lot lighter, and 5V already, so no need for a step-down converter.
If you buy one from a respected brand, you should be fine.
But the different voltages are listed in the part you quoted: 5V/2A,12V/2A ... And I sort of believe the "actual capacity" line. 12000 mAh would also do the trick. Worth trying.
Mixe:
Worth trying.
I wouldn't agree with that, for €45, you can get a decent 5,000 - 10,000 mAh power bank from a known brand, so you know that you get what you pay for.
I once bought one of these fleabay paks rated at 20,000mAh for not too much money. Unit delivered was certainly heavy enough but on load was lucky to deliver 2000mAh. Since it wasn't worth the hassle of trying to claim a refund I stripped it down. Inside was one small battery and 2 large steel plates !! You pay your money and takes your chance.
Hi,
OPs circuit posted earlier.
Tom...
Does the screen have a backlight? I expect that is where most of the power is going.
Many displays have a PWM input to control the light. If you give it 50% PWM, you will probably find that power is reduced by almost half and the readability is barely affected. | 1,458 | 5,741 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-38 | latest | en | 0.941008 |
http://www.jiskha.com/display.cgi?id=1296884913 | 1,496,108,046,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463613738.67/warc/CC-MAIN-20170530011338-20170530031338-00285.warc.gz | 656,592,540 | 3,686 | # physics
posted by on .
The punter on a football team tries to kick a football so that it stays in the air for a long "hang time." If the ball is kicked with an initial velocity of 24.2 m/s at an angle of 58.5° above the ground, what is the "hang time"?
• physics - ,
Double the time it takes to reach zero vertical velocity component. Half the time is spenty rising to that height; the other half is spent coming back down. | 106 | 429 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-22 | latest | en | 0.928289 |
https://edurev.in/course/quiz/attempt/9596_Test-Types-of-Waveguides-Its-Properties/e6d68a3a-3e21-4a51-9e32-75e42fe4ae2c | 1,721,576,831,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517747.98/warc/CC-MAIN-20240721152016-20240721182016-00106.warc.gz | 194,336,556 | 52,103 | Test: Types of Waveguides & Its Properties - Electrical Engineering (EE) MCQ
# Test: Types of Waveguides & Its Properties - Electrical Engineering (EE) MCQ
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## 20 Questions MCQ Test Electromagnetic Fields Theory (EMFT) - Test: Types of Waveguides & Its Properties
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Test: Types of Waveguides & Its Properties - Question 1
### The phenomenon employed in the waveguide operation is
Detailed Solution for Test: Types of Waveguides & Its Properties - Question 1
Explanation: The waveguides use total internal reflection phenomenon to transmit the waves passing through it. Thus the acceptance angle and critical angle are important for effective transmission.
Test: Types of Waveguides & Its Properties - Question 2
### The dominant mode in waveguide is the mode which has
Detailed Solution for Test: Types of Waveguides & Its Properties - Question 2
Explanation: The dominant mode is the mode which has the minimum frequency or maximum wavelength available for propagation of the waves
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Test: Types of Waveguides & Its Properties - Question 3
### The modes are calculated from which parameter?
Detailed Solution for Test: Types of Waveguides & Its Properties - Question 3
Explanation: The modes are calculated from the V number of the waveguides. It is given by M= V2/2.
Test: Types of Waveguides & Its Properties - Question 4
The circular waveguides use which function in the frequency calculation?
Detailed Solution for Test: Types of Waveguides & Its Properties - Question 4
Explanation: The circular or cylindrical waveguides use the Bessel function for the frequency calculation of a particular mode.
Test: Types of Waveguides & Its Properties - Question 5
The scattering parameters are used to indicate the
Detailed Solution for Test: Types of Waveguides & Its Properties - Question 5
Explanation: The scattering matrix consists of the transmission coefficients in the main diagonal and the reflection coefficients in the opposite diagonal.
Test: Types of Waveguides & Its Properties - Question 6
Which of the following two parameter models cannot be used to represent a transmission line?
Detailed Solution for Test: Types of Waveguides & Its Properties - Question 6
Explanation: The T, ABCD and S parameter models are used in the transmission line modelling. The h parameter is not used for the same.
Test: Types of Waveguides & Its Properties - Question 7
For the matched line, the parameters S12 and S21 are
Detailed Solution for Test: Types of Waveguides & Its Properties - Question 7
Explanation: The parameters S12 and S21 are the reflection coefficients. For a matched line, the reflection coefficients are zero. Thus the parameters S12 and S21 are also zero.
Test: Types of Waveguides & Its Properties - Question 8
The waveguides are materials with characteristics of
Detailed Solution for Test: Types of Waveguides & Its Properties - Question 8
Explanation: Generally, the waveguides are made of materials with low bulk resistivity like brass, copper, silver etc. But if the interior walls are properly plated, it is possible with poor conductivity materials too. It is even possible to make plastic waveguides.
Test: Types of Waveguides & Its Properties - Question 9
The parameters S11 and S22 indicate the transmission coefficients. State true/false.
Detailed Solution for Test: Types of Waveguides & Its Properties - Question 9
Explanation: In a scattering matrix, the parameters S11 and S22 indicate the transmission coefficients and the parameters S21 and S12 indicate the reflection coefficients.
Test: Types of Waveguides & Its Properties - Question 10
The waveguides increase the transmission of the electromagnetic waves. State true/false.
Detailed Solution for Test: Types of Waveguides & Its Properties - Question 10
Explanation: The waveguides aid in effective transmission of the electromagnetic power from the source antenna to the destination antenna.
Test: Types of Waveguides & Its Properties - Question 11
The waveguide is employed in the transmission lines, when operated at the range of
Detailed Solution for Test: Types of Waveguides & Its Properties - Question 11
Explanation: Waveguides are employed for effective transmission, when the lines carry electromagnetic waves in the GHz range.
Test: Types of Waveguides & Its Properties - Question 12
The cut off frequency for a waveguide to operate is
Detailed Solution for Test: Types of Waveguides & Its Properties - Question 12
Explanation: The cut off frequency of the waveguide is 6 GHz. This is the frequency at which the waveguide will start to operate.
Test: Types of Waveguides & Its Properties - Question 13
In rectangular waveguides, the dimensions a and b represent the
Detailed Solution for Test: Types of Waveguides & Its Properties - Question 13
Explanation: In rectangular waveguide, the a parameter is the broad wall dimension of the waveguide and the b parameter is the side wall dimension of the waveguide. Always, a > b in a waveguide.
Test: Types of Waveguides & Its Properties - Question 14
The Bessel function is denoted by
Detailed Solution for Test: Types of Waveguides & Its Properties - Question 14
Explanation: The Bessel function is used in the circular waveguides. Normally Jn(ha) = 0. Here n is the order of the Bessel function.
Test: Types of Waveguides & Its Properties - Question 15
In a waveguide, always which condition holds good?
Detailed Solution for Test: Types of Waveguides & Its Properties - Question 15
Explanation: In air medium, the phase velocity is assumed to be the speed of light. For waveguides, the phase velocity is always greater than the speed of the light.
Test: Types of Waveguides & Its Properties - Question 16
The group wavelength is greater than the wavelength at any point. State true/false.
Detailed Solution for Test: Types of Waveguides & Its Properties - Question 16
Explanation: In a waveguide, the phase velocity is greater than the velocity of light. Thus the group velocity will be less. This implies the group wavelength will be greater than the wavelength at any point.
Test: Types of Waveguides & Its Properties - Question 17
Find the group wavelength of a wave, given that the group phase constant is 6.28 units.
Detailed Solution for Test: Types of Waveguides & Its Properties - Question 17
Explanation: The group wavelength is given by λg = 2π/βg, where βg is the group wavelength of the wave. On substituting for βg = 6.28, we get group wavelength as unity.
Test: Types of Waveguides & Its Properties - Question 18
The phase velocity of a wave with frequency of 15 radian/sec and group phase constant of 2 units is
Detailed Solution for Test: Types of Waveguides & Its Properties - Question 18
Explanation: The phase velocity of a wave is given by Vp = ω/βg. on substituting for ω = 15 and βg = 2, we get phase velocity as 15/2 = 7.5 units.
Test: Types of Waveguides & Its Properties - Question 19
The modes in a material having a V number of 20 is
Detailed Solution for Test: Types of Waveguides & Its Properties - Question 19
Explanation: The relation between the modes and the V number is given by m = v2/2. Given that v = 20, we get m = 202/2 = 200 modes
Test: Types of Waveguides & Its Properties - Question 20
The number of modes is given as 50, find the V number.
Detailed Solution for Test: Types of Waveguides & Its Properties - Question 20
Explanation: The relation between the modes and the V number is given by m = v2/2. Given that m = 50, we get v2 = 2 x 50 = 100. The V number is 10.
## Electromagnetic Fields Theory (EMFT)
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## Electromagnetic Fields Theory (EMFT)
11 videos|45 docs|73 tests | 2,119 | 9,347 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.841719 |
http://hitchhikersgui.de/Resampling_(statistics) | 1,544,887,402,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376826856.91/warc/CC-MAIN-20181215131038-20181215153038-00427.warc.gz | 129,786,520 | 33,196 | # Resampling (statistics)
In statistics, resampling is any of a variety of methods for doing one of the following:
1. Estimating the precision of sample statistics (medians, variances, percentiles) by using subsets of available data (jackknifing) or drawing randomly with replacement from a set of data points (bootstrapping)
2. Exchanging labels on data points when performing significance tests (permutation tests, also called exact tests, randomization tests, or re-randomization tests)
3. Validating models by using random subsets (bootstrapping, cross validation)
Common resampling techniques include bootstrapping, jackknifing and permutation tests.
## Bootstrap
Bootstrapping is a statistical method for estimating the sampling distribution of an estimator by sampling with replacement from the original sample, most often with the purpose of deriving robust estimates of standard errors and confidence intervals of a population parameter like a mean, median, proportion, odds ratio, correlation coefficient or regression coefficient. It may also be used for constructing hypothesis tests. It is often used as a robust alternative to inference based on parametric assumptions when those assumptions are in doubt, or where parametric inference is impossible or requires very complicated formulas for the calculation of standard errors. Bootstrapping techniques are also used in the updating-selection transitions of particle filters, genetic type algorithms and related tesample/teconfiguration Monte Carlo methods used in computational physics.[1][2] In this context, the bootstrap is used to replace sequentially empirical weighted probability measures by empirical measures. The bootstrap allows to replace the samples with low weights by copies of the samples with high weights.
## Jackknife
Jackknifing, which is similar to bootstrapping, is used in statistical inference to estimate the bias and standard error (variance) of a statistic, when a random sample of observations is used to calculate it. Historically this method preceded the invention of the bootstrap with Quenouille inventing this method in 1949 and Tukey extending it in 1958.[3][4] This method was foreshadowed by Mahalanobis who in 1946 suggested repeated estimates of the statistic of interest with half the sample chosen at random.[5] He coined the name 'interpenetrating samples' for this method.
Quenouille invented this method with the intention of reducing the bias of the sample estimate. Tukey extended this method by assuming that if the replicates could be considered identically and independently distributed, then an estimate of the variance of the sample parameter could be made and that it would be approximately distributed as a t variate with n−1 degrees of freedom (n being the sample size).
The basic idea behind the jackknife variance estimator lies in systematically recomputing the statistic estimate, leaving out one or more observations at a time from the sample set. From this new set of replicates of the statistic, an estimate for the bias and an estimate for the variance of the statistic can be calculated.
Instead of using the jackknife to estimate the variance, it may instead be applied to the log of the variance. This transformation may result in better estimates particularly when the distribution of the variance itself may be non normal.
For many statistical parameters the jackknife estimate of variance tends asymptotically to the true value almost surely. In technical terms one says that the jackknife estimate is consistent. The jackknife is consistent for the sample means, sample variances, central and non-central t-statistics (with possibly non-normal populations), sample coefficient of variation, maximum likelihood estimators, least squares estimators, correlation coefficients and regression coefficients.
It is not consistent for the sample median. In the case of a unimodal variate the ratio of the jackknife variance to the sample variance tends to be distributed as one half the square of a chi square distribution with two degrees of freedom.
The jackknife, like the original bootstrap, is dependent on the independence of the data. Extensions of the jackknife to allow for dependence in the data have been proposed.
Another extension is the delete-a-group method used in association with Poisson sampling.
## Comparison of bootstrap and jackknife
Both methods, the bootstrap and the jackknife, estimate the variability of a statistic from the variability of that statistic between subsamples, rather than from parametric assumptions. For the more general jackknife, the delete-m observations jackknife, the bootstrap can be seen as a random approximation of it. Both yield similar numerical results, which is why each can be seen as approximation to the other. Although there are huge theoretical differences in their mathematical insights, the main practical difference for statistics users is that the bootstrap gives different results when repeated on the same data, whereas the jackknife gives exactly the same result each time. Because of this, the jackknife is popular when the estimates need to be verified several times before publishing (e.g., official statistics agencies). On the other hand, when this verification feature is not crucial and it is of interest not to have a number but just an idea of its distribution, the bootstrap is preferred (e.g., studies in physics, economics, biological sciences).
Whether to use the bootstrap or the jackknife may depend more on operational aspects than on statistical concerns of a survey. The jackknife, originally used for bias reduction, is more of a specialized method and only estimates the variance of the point estimator. This can be enough for basic statistical inference (e.g., hypothesis testing, confidence intervals). The bootstrap, on the other hand, first estimates the whole distribution (of the point estimator) and then computes the variance from that. While powerful and easy, this can become highly computer intensive.
"The bootstrap can be applied to both variance and distribution estimation problems. However, the bootstrap variance estimator is not as good as the jackknife or the balanced repeated replication (BRR) variance estimator in terms of the empirical results. Furthermore, the bootstrap variance estimator usually requires more computations than the jackknife or the BRR. Thus, the bootstrap is mainly recommended for distribution estimation." [6]
There is a special consideration with the jackknife, particularly with the delete-1 observation jackknife. It should only be used with smooth, differentiable statistics (e.g., totals, means, proportions, ratios, odd ratios, regression coefficients, etc.; not with medians or quantiles). This could become a practical disadvantage. This disadvantage is usually the argument favoring bootstrapping over jackknifing. More general jackknifes than the delete-1, such as the delete-m jackknife or the delete-all-but-2 Hodges–Lehmann estimator, overcome this problem for the medians and quantiles by relaxing the smoothness requirements for consistent variance estimation.
Usually the jackknife is easier to apply to complex sampling schemes than the bootstrap. Complex sampling schemes may involve stratification, multiple stages (clustering), varying sampling weights (non-response adjustments, calibration, post-stratification) and under unequal-probability sampling designs. Theoretical aspects of both the bootstrap and the jackknife can be found in Shao and Tu (1995),[7] whereas a basic introduction is accounted in Wolter (2007).[8] The bootstrap estimate of model prediction bias is more precise than jackknife estimates with linear models such as linear discriminant function or multiple regression.[9]
## Subsampling
Subsampling is an alternative method for approximating the sampling distribution of an estimator. The two key differences to the bootstrap are: (i) the resample size is smaller than the sample size and (ii) resampling is done without replacement. The advantage of subsampling is that it is valid under much weaker conditions compared to the bootstrap. In particular, a set of sufficient conditions is that the rate of convergence of the estimator is known and that the limiting distribution is continuous; in addition, the resample (or subsample) size must tend to infinity together with the sample size but at a smaller rate, so that their ratio converges to zero. While subsampling was originally proposed for the case of independent and identically distributed (iid) data only, the methodology has been extended to cover time series data as well; in this case, one resamples blocks of subsequent data rather than individual data points. There are many cases of applied interest where subsampling leads to valid inference whereas bootstrapping does not; for example, such cases include examples where the rate of convergence of the estimator is not the square root of the sample size or when the limiting distribution is non-normal.
## Cross-validation
Cross-validation is a statistical method for validating a predictive model. Subsets of the data are held out for use as validating sets; a model is fit to the remaining data (a training set) and used to predict for the validation set. Averaging the quality of the predictions across the validation sets yields an overall measure of prediction accuracy. Cross-validation is employed repeatedly in building decision trees.
One form of cross-validation leaves out a single observation at a time; this is similar to the jackknife. Another, K-fold cross-validation, splits the data into K subsets; each is held out in turn as the validation set.
This avoids "self-influence". For comparison, in regression analysis methods such as linear regression, each y value draws the regression line toward itself, making the prediction of that value appear more accurate than it really is. Cross-validation applied to linear regression predicts the y value for each observation without using that observation.
This is often used for deciding how many predictor variables to use in regression. Without cross-validation, adding predictors always reduces the residual sum of squares (or possibly leaves it unchanged). In contrast, the cross-validated mean-square error will tend to decrease if valuable predictors are added, but increase if worthless predictors are added.[10]
## Permutation tests
A permutation test (also called a randomization test, re-randomization test, or an exact test) is a type of statistical significance test in which the distribution of the test statistic under the null hypothesis is obtained by calculating all possible values of the test statistic under rearrangements of the labels on the observed data points. In other words, the method by which treatments are allocated to subjects in an experimental design is mirrored in the analysis of that design. If the labels are exchangeable under the null hypothesis, then the resulting tests yield exact significance levels; see also exchangeability. Confidence intervals can then be derived from the tests. The theory has evolved from the works of Ronald Fisher and E. J. G. Pitman in the 1930s.
To illustrate the basic idea of a permutation test, suppose we have two groups ${\displaystyle A}$ and ${\displaystyle B}$ whose sample means are ${\displaystyle {\bar {x}}_{A}}$ and ${\displaystyle {\bar {x}}_{B}}$, and that we want to test, at 5% significance level, whether they come from the same distribution. Let ${\displaystyle n_{A}}$ and ${\displaystyle n_{B}}$ be the sample size corresponding to each group. The permutation test is designed to determine whether the observed difference between the sample means is large enough to reject the null hypothesis H${\displaystyle _{0}}$ that the two groups have identical probability distributions.
The test proceeds as follows. First, the difference in means between the two samples is calculated: this is the observed value of the test statistic, T(obs). Then the observations of groups ${\displaystyle A}$ and ${\displaystyle B}$ are pooled.
Next, the difference in sample means is calculated and recorded for every possible way of dividing these pooled values into two groups of size ${\displaystyle n_{A}}$ and ${\displaystyle n_{B}}$ (i.e., for every permutation of the group labels A and B). The set of these calculated differences is the exact distribution of possible differences under the null hypothesis that group label does not matter.
The one-sided p-value of the test is calculated as the proportion of sampled permutations where the difference in means was greater than or equal to T(obs). The two-sided p-value of the test is calculated as the proportion of sampled permutations where the absolute difference was greater than or equal to ABS(T(obs)).
If the only purpose of the test is reject or not reject the null hypothesis, we can as an alternative sort the recorded differences, and then observe if T(obs) is contained within the middle 95% of them. If it is not, we reject the hypothesis of identical probability curves at the 5% significance level.
### Relation to parametric tests
Permutation tests are a subset of non-parametric statistics. The basic premise is to use only the assumption that it is possible that all of the treatment groups are equivalent, and that every member of them is the same before sampling began (i.e. the slot that they fill is not differentiable from other slots before the slots are filled). From this, one can calculate a statistic and then see to what extent this statistic is special by seeing how likely it would be if the treatment assignments had been jumbled.
In contrast to permutation tests, the reference distributions for many popular "classical" statistical tests, such as the t-test, F-test, z-test, and χ2 test, are obtained from theoretical probability distributions. Fisher's exact test is an example of a commonly used permutation test for evaluating the association between two dichotomous variables. When sample sizes are very large, the Pearson's chi-square test will give accurate results. For small samples, the chi-square reference distribution cannot be assumed to give a correct description of the probability distribution of the test statistic, and in this situation the use of Fisher's exact test becomes more appropriate.
Permutation tests exist in many situations where parametric tests do not (e.g., when deriving an optimal test when losses are proportional to the size of an error rather than its square). All simple and many relatively complex parametric tests have a corresponding permutation test version that is defined by using the same test statistic as the parametric test, but obtains the p-value from the sample-specific permutation distribution of that statistic, rather than from the theoretical distribution derived from the parametric assumption. For example, it is possible in this manner to construct a permutation t-test, a permutation χ2 test of association, a permutation version of Aly's test for comparing variances and so on.
The major down-side to permutation tests are that they
• Can be computationally intensive and may require "custom" code for difficult-to-calculate statistics. This must be rewritten for every case.
• Are primarily used to provide a p-value. The inversion of the test to get confidence regions/intervals requires even more computation.
Permutation tests exist for any test statistic, regardless of whether or not its distribution is known. Thus one is always free to choose the statistic which best discriminates between hypothesis and alternative and which minimizes losses.
Permutation tests can be used for analyzing unbalanced designs[11] and for combining dependent tests on mixtures of categorical, ordinal, and metric data (Pesarin, 2001). They can also be used to analyze qualitative data that has been quantitized (i.e., turned into numbers). Permutation tests may be ideal for analyzing quantitized data that do not satisfy statistical assumptions underlying traditional parametric tests (e.g., t-tests, ANOVA) (Collingridge, 2013).
Before the 1980s, the burden of creating the reference distribution was overwhelming except for data sets with small sample sizes.
Since the 1980s, the confluence of relatively inexpensive fast computers and the development of new sophisticated path algorithms applicable in special situations, made the application of permutation test methods practical for a wide range of problems. It also initiated the addition of exact-test options in the main statistical software packages and the appearance of specialized software for performing a wide range of uni- and multi-variable exact tests and computing test-based "exact" confidence intervals.
### Limitations
An important assumption behind a permutation test is that the observations are exchangeable under the null hypothesis. An important consequence of this assumption is that tests of difference in location (like a permutation t-test) require equal variance. In this respect, the permutation t-test shares the same weakness as the classical Student's t-test (the Behrens–Fisher problem). A third alternative in this situation is to use a bootstrap-based test. Good (2005) explains the difference between permutation tests and bootstrap tests the following way: "Permutations test hypotheses concerning distributions; bootstraps test hypotheses concerning parameters. As a result, the bootstrap entails less-stringent assumptions." Bootstrap tests are not exact.
### Monte Carlo testing
An asymptotically equivalent permutation test can be created when there are too many possible orderings of the data to allow complete enumeration in a convenient manner. This is done by generating the reference distribution by Monte Carlo sampling, which takes a small (relative to the total number of permutations) random sample of the possible replicates. The realization that this could be applied to any permutation test on any dataset was an important breakthrough in the area of applied statistics. The earliest known reference to this approach is Dwass (1957).[12] This type of permutation test is known under various names: approximate permutation test, Monte Carlo permutation tests or random permutation tests.[13]
After ${\displaystyle \scriptstyle \ N}$ random permutations, it is possible to obtain a confidence interval for the p-value based on the Binomial distribution. For example, if after ${\displaystyle \scriptstyle \ N=10000}$ random permutations the p-value is estimated to be ${\displaystyle \scriptstyle \ {\hat {p}}=0.05}$, then a 99% confidence interval for the true ${\displaystyle \scriptstyle \ p}$ (the one that would result from trying all possible permutations) is ${\displaystyle \scriptstyle \ [0.045,0.055]}$.
On the other hand, the purpose of estimating the p-value is most often to decide whether ${\displaystyle \scriptstyle \ p\leq \alpha }$, where ${\displaystyle \scriptstyle \ \alpha }$ is the threshold at which the null hypothesis will be rejected (typically ${\displaystyle \scriptstyle \ \alpha =0.05}$). In the example above, the confidence interval only tells us that there is roughly a 50% chance that the p-value is smaller than 0.05, i.e. it is completely unclear whether the null hypothesis should be rejected at a level ${\displaystyle \scriptstyle \ \alpha =0.05}$.
If it is only important to know whether ${\displaystyle \scriptstyle \ p\leq \alpha }$ for a given ${\displaystyle \scriptstyle \ \alpha }$, it is logical to continue simulating until the statement ${\displaystyle \scriptstyle \ p\leq \alpha }$ can be established to be true or false with a very low probability of error. Given a bound ${\displaystyle \scriptstyle \ \epsilon }$ on the admissible probability of error (the probability of finding that ${\displaystyle \scriptstyle \ {\hat {p}}>\alpha }$ when in fact ${\displaystyle \scriptstyle \ p\leq \alpha }$ or vice versa), the question of how many permutations to generate can be seen as the question of when to stop generating permutations, based on the outcomes of the simulations so far, in order to guarantee that the conclusion (which is either ${\displaystyle \scriptstyle \ p\leq \alpha }$ or ${\displaystyle \scriptstyle \ p>\alpha }$) is correct with probability at least as large as ${\displaystyle \scriptstyle \ 1-\epsilon }$. (${\displaystyle \scriptstyle \ \epsilon }$ will typically be chosen to be extremely small, e.g. 1/1000.) Stopping rules to achieve this have been developed[14] which can be incorporated with minimal additional computational cost. In fact, depending on the true underlying p-value it will often be found that the number of simulations required is remarkably small (e.g. as low as 5 and often not larger than 100) before a decision can be reached with virtual certainty.
## References
1. ^ Del Moral, Pierre (2004). Feynman-Kac formulae. Genealogical and interacting particle approximations. Springer. p. 575. Series: Probability and Applications
2. ^ Del Moral, Pierre (2013). Mean field simulation for Monte Carlo integration. Chapman & Hall/CRC Press. p. 626. Monographs on Statistics & Applied Probability
3. ^ Quenouille, M. H. (1949). Approximate Tests of Correlation in Time-Series. Journal of the Royal Statistical Society, Series B. 11. pp. 68–84. JSTOR 2983696.
4. ^ Tukey, J. W. (1958). "Bias and Confidence in Not-quite Large Samples (Preliminary Report)". Annals of Mathematical Statistics. 29 (2): 614. JSTOR 2237363.
5. ^ Mahalanobis, P. C. (1946). "Proceedings of a Meeting of the Royal Statistical Society held on July 16th, 1946". Journal of the Royal Statistical Society. 109 (4): 325–370. JSTOR 2981330.
6. ^ Shao, J. and Tu, D. (1995). The Jackknife and Bootstrap. Springer-Verlag, Inc. pp. 281.
7. ^ Shao, J.; Tu, D. (1995). The Jackknife and Bootstrap. Springer.
8. ^ Wolter, K. M. (2007). Introduction to Variance Estimation (Second ed.). Springer.
9. ^ Verbyla, D.; Litvaitis, J. (1989). "Resampling methods for evaluating classification accuracy of wildlife habitat models". Environmental Management. 13 (6): 783–787. Bibcode:1989EnMan..13..783V. doi:10.1007/bf01868317.
10. ^ Verbyla, D. (1986). "Potential prediction bias in regression and discriminant analysis". Canadian Journal of Forest Research. 16 (6): 1255–1257. doi:10.1139/x86-222.
11. ^ "Invited Articles" (PDF). Journal of Modern Applied Statistical Methods. 1 (2): 202–522. Fall 2011. Archived from the original (PDF) on May 5, 2003.
12. ^ Dwass, Meyer (1957). "Modified Randomization Tests for Nonparametric Hypotheses". Annals of Mathematical Statistics. 28 (1): 181–187. doi:10.1214/aoms/1177707045. JSTOR 2237031.
13. ^ Thomas E. Nichols, Andrew P. Holmes (2001). "Nonparametric Permutation Tests For Functional Neuroimaging: A Primer with Examples" (PDF). Human Brain Mapping. 15 (1): 1–25. doi:10.1002/hbm.1058. PMID 11747097.
14. ^ Gandy, Axel (2009). "Sequential implementation of Monte Carlo tests with uniformly bounded resampling risk". Journal of the American Statistical Association. 104 (488): 1504–1511. arXiv:math/0612488. doi:10.1198/jasa.2009.tm08368.
• Good, Phillip (2005), Permutation, Parametric and Bootstrap Tests of Hypotheses (3rd ed.), Springer
## Bibliography
### Introductory statistics
• Good, P. (2005) Introduction to Statistics Through Resampling Methods and R/S-PLUS. Wiley. ISBN 0-471-71575-1
• Good, P. (2005) Introduction to Statistics Through Resampling Methods and Microsoft Office Excel. Wiley. ISBN 0-471-73191-9
• Hesterberg, T. C., D. S. Moore, S. Monaghan, A. Clipson, and R. Epstein (2005). Bootstrap Methods and Permutation Tests.[full citation needed]
• Wolter, K.M. (2007). Introduction to Variance Estimation. Second Edition. Springer, Inc.
### Bootstrap
• Efron, Bradley (1979). "Bootstrap methods: Another look at the jackknife". The Annals of Statistics. 7: 1–26. doi:10.1214/aos/1176344552.
• Efron, Bradley (1981). "Nonparametric estimates of standard error: The jackknife, the bootstrap and other methods". Biometrika. 68 (3): 589–599. doi:10.2307/2335441. JSTOR 2335441.
• Efron, Bradley (1982). The jackknife, the bootstrap, and other resampling plans, In Society of Industrial and Applied Mathematics CBMS-NSF Monographs, 38.
• Diaconis, P.; Efron, Bradley (1983), "Computer-intensive methods in statistics," Scientific American, May, 116-130.
• Efron, Bradley; Tibshirani, Robert J. (1993). An introduction to the bootstrap, New York: Chapman & Hall, software.
• Davison, A. C. and Hinkley, D. V. (1997): Bootstrap Methods and their Application, software.
• Mooney, C Z & Duval, R D (1993). Bootstrapping. A Nonparametric Approach to Statistical Inference. Sage University Paper series on Quantitative Applications in the Social Sciences, 07-095. Newbury Park, CA: Sage.
• Simon, J. L. (1997): Resampling: The New Statistics.
### Monte Carlo methods
• George S. Fishman (1995). Monte Carlo: Concepts, Algorithms, and Applications, Springer, New York. ISBN 0-387-94527-X.
• James E. Gentle (2009). Computational Statistics, Springer, New York. Part III: Methods of Computational Statistics. ISBN 978-0-387-98143-7.
• Pierre Del Moral (2004). Feynman-Kac formulae. Genealogical and Interacting particle systems with applications, Springer, Series Probability and Applications. ISBN 978-0-387-20268-6
• Pierre Del Moral (2013). Del Moral, Pierre (2013). Mean field simulation for Monte Carlo integration. Chapman & Hall/CRC Press, Monographs on Statistics and Applied Probability. ISBN 9781466504059
• Dirk P. Kroese, Thomas Taimre and Zdravko I. Botev. Handbook of Monte Carlo Methods, John Wiley & Sons, New York. ISBN 978-0-470-17793-8.
• Christian P. Robert and George Casella (2004). Monte Carlo Statistical Methods, Second ed., Springer, New York. ISBN 0-387-21239-6.
• Shlomo Sawilowsky and Gail Fahoome (2003). Statistics via Monte Carlo Simulation with Fortran. Rochester Hills, MI: JMASM. ISBN 0-9740236-0-4.
### Permutation tests
Original references:
Modern references:
Computational methods:
### Resampling methods
• Good, P. (2006) Resampling Methods. 3rd Ed. Birkhauser.
• Wolter, K.M. (2007). Introduction to Variance Estimation. 2nd Edition. Springer, Inc.
• Pierre Del Moral (2004). Feynman-Kac formulae. Genealogical and Interacting particle systems with applications, Springer, Series Probability and Applications. ISBN 978-0-387-20268-6
• Pierre Del Moral (2013). Del Moral, Pierre (2013). Mean field simulation for Monte Carlo integration. Chapman & Hall/CRC Press, Monographs on Statistics and Applied Probability. ISBN 9781466504059 | 5,810 | 26,658 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 30, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2018-51 | latest | en | 0.86509 |
http://bleacherreport.com/articles/132470-the-31-best-nfl-running-back-performances-from-2008 | 1,493,184,315,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917121153.91/warc/CC-MAIN-20170423031201-00236-ip-10-145-167-34.ec2.internal.warc.gz | 47,136,315 | 90,581 | # The 31 Best NFL Running Back Performances From 2008
Matthew GilmartinSenior Analyst IMarch 1, 2009
I’d like to thank Bryn Swartz for allowing me to write this article after I got the idea from his article The 32 Best Quarterbacks in the NFL.
In order to be ranked in this list the running backs had to survive two cuts. The first was a minimum requirement of 100 carries in the 2008 regular season. The second was made based on my discretion of a short-coming in one or more of 10 major statistical categories.
Once I had my list down to 31 (31 players made the final field instead of 32—as per the number of teams in the NFL—because I misread the spreadsheet where I kept all my research on the players, so I thought I had 32, but really only had 31), I assigned each statistical category a numerical value based on importance relevant to a running back’s performance (i.e. each rushing yard a player had earned him five points).
Then I created an equation in an attempt to best measure a running back’s overall performance in terms of a final score. The scores that the rankings are based off of are rounded to the nearest whole number.
The equation is: X carries x 3 + X rushing yards x 5 + X rushing average x 2 + X rushing yards x 7 + percent of runs for first downs + receptions x 2 + receiving yards x 3 + receiving average + points – X fumbles x 3 – X games played. Long, and a bit complicated, I know, but you’re not the one who had to calculate 32 different players’ scores using this equation.
Now, without further adieu, the 31 running backs who delivered the best season-long, regular season performances in 2008…
31. Joseph Addai, Indianapolis Colts, Final Score: 2839
With only 155 carries in a minimal 12 games, Addai didn’t have a chance to do much. But he still accumulated 544 yards, and averaged an acceptable 3.5 yards per carry. He was in the middle of the pack of the final field in terms of percentage of runs for a first down.
He posed a decent receiving threat, with 25 receptions for 206 yards. He scored 42 points, which averages out to a touchdown every two games. Not horrible, but by no means especially good, either.
30. Edgerrin James, Arizona Cardinals, Final Score: 3288
How James ended up with a better score than Addai, I don’t know. But he did, with 514 rushing yards on 133 carries, which comes out as 3.9 yards per carry—not bad. But the only other halfway impressive thing about James is that he fumbled only once. On the other hand, he didn’t have many carries.
Still, even I’m wondering how James ended up with a better score than Addai, considering Addai beat him in every one of the included statistical categories except average yards per carry…But James did have the better score, so he’s No. 31.
29. Tim Hightower, Arizona Cardinals, Final Score: 3326
Hightower scored ten touchdowns, and converted for a first down on 23.8 percent of his runs, one of the better figures in the NFL. But his real strength was receiving, as he caught 34 passes for 237 yards, and scoring, with 60 points overall.
28. Fred Jackson, Buffalo Bills, Final Score: 4301
He averaged a healthy 4.4 yards per carry (if he had gotten more carries, he might have had 800-900 yards), and moved the chains on an astounding 26.2 percent of his runs (I believe that’s the second-best, maybe the best, percentage for first-down conversion in the league).
He also had 37 catches for 317 yards. But he fumbled three times, not exactly good for someone who only carries 130 times.
27. Pierre Thomas, New Orleans Saints, Final Score: 4562
One of the most underrated running backs of 2008, Thomas averaged 4.8 yards per carry, scored nine touchdowns, and got a first down on 33 percent of his carries. Add in 31 catches for 284 yards as well as 72 points and only one fumble, and you’ve got an awfully good running back for a player so far down on this list.
26. Willis McGahee, Baltimore Ravens, Final Score: 4692
McGahee, with 671 yards, an average of 3.9 yards per carry, seven touchdowns, 24 receptions for 173 yards, and 42 points (as well as two fumbles in 170 carries) is a true middle-of-the-pack running back.
25. LenDale White, Tennessee Titans, Final Score: 4710
I bet you expected White to be ranked higher, especially if you’re a Titans fan. But his lack of rushing yards (773) compared to the elite backs did him in despite his 15 touchdowns and 90 points.
24. Jonathan Stewart, Carolina Panthers, Final Score: 5012
It’s a bit surprising that Stewart wasn’t ranked higher, considering his average of 4.5 yards per carry and 10 touchdowns. He also gained 836 yards, not bad for a guy who got only 184 carries in a run-first scheme. But he only had eight grabs for 47 yards. In addition, he only had 60 points, which is in the middle of the pack.
23. Larry Johnson, Kansas City Chiefs, Final Score: 5248
Johnson was one of the few strengths of the Chiefs in ’08, compiling 874 rushing yards on only 193 carries (which averages out to 4.5 yards per carry) in a minimal 12 games. But he scored minimally (five touchdowns and 30 points overall), and didn’t catch passes out of the backfield much (12 catches for 74 yards). He also fumbled five times, one of the highest figures in the league in that category.
22. Warrick Dunn, Tampa Bay Buccaneers, Final Score: 5598
Dunn had a decent season, carrying 186 times at 4.2 yards per pop for 786 rushing yards. But his mere two touchdowns and 12 points overall really negated his 47 catches for 330 yards. But he didn’t have any fumbles, which is something only one other guy in the NFL—the Panthers’ DeAngelo Williams—matched.
21. Le’Ron McClain, Baltimore Ravens, Final Score: 5738
232 carries for 902 yards is good, if not great. He also scored ten touchdowns, and earned a first down on 22 percent of his runs, one of the better figures in the league. 19 receptions for 123 yards is a nice extra tool.
66 points isn’t awesome, but it’s better than some. The three fumbles could be either bad, or acceptable, depending on your point-of-view.
20. Ronnie Browns, Miami Dolphins, Final Score: 6177
Brown ran for more than 900 yards (916, to be exact), with 214 carries to work with. Not bad at all. He also tallied ten touchdowns, and converted for a first down on 22.4 percent of his runs. 33 catches for 254 yards makes Brown a respectable receiving threat, too. Only one fumble in 200+ carries—commendable.
19. Brandon Jacobs, New York Giants, Final Score: 6411
Yes, I know, you think Jacobs should be ranked higher. But he isn’t because of his lack of receiving numbers (six receptions/36 yards). But he did accumulate 1089 yards, ran for an average of five yards per carry, and totaled 15 rushing scores.
His 26 percent of total runs for first downs, I believe, was one of the top three figures in the NFL. Jacobs scored contributed 90 points overall, but fumbled three times. However, again that could go either way depending on your perception of the context.
18. Jamal Lewis, Cleveland Browns, Final Score: 6475
How Lewis ended up with a score in the middle third of the pack, I don’t know.
He gained 1002 yards and averaged 3.6 yards per carry. But he only scored four touchdowns, converted for a first down on only 15 percent of his runs, caught 23 passes for 178 yards, and scored only 23 points. Yet he still ended up with a 6000+ score….
17. Marion Barber, Dallas Cowboys, Final Score: 6588
Barber ran decently, for 885 yards on 238 carries and seven touchdowns. He also had a 23 percent first down conversion rate compared to total rushes. But his main contribution was in the passing game, where he had 52 catches for 417 yards.
Barber’s 54 points make him a nice scoring threat. And three fumbles in nearly 240 carries isn’t the end of the world.
16. Kevin Smith, Detroit Lions, Final Score: 6630
Smith fell 24 yards short of the millennium mark. He averaged 4.1 yards per carry, which isn’t half bad when you’re running behind the Lions’ horrid offensive line. Smith also scored eight touchdowns, and served as a dual threat RB with 39 catches for 286 yards. 48 points is a welcome contribution. Not to mention that he fumbled only once in 238 carries.
15. Maurice Jones-Drew, Jacksonville Jaguars, Final Score: 6687
Jones-Drew was invisible last year because of the fact that he played on an underachieving team, but he surpassed 800 yards with fewer than 200 carries. His 12 touchdowns, 23.9 first down runs to total runs percent, 62 receptions for 565 yards, and 84 points make a great case for him.
The downside? In exactly 197 carries he fumbled four times.
14. Brian Westbrook, Philadelphia Eagles, Final Score: 6838
Westbrook had another solid season for the Eagles, rushing 233 times for 936 yards (four yards per carry). He also scored nine touchdowns, and caught 54 passes for 402 yards. Westbrook scored 84 points, and fumbled only once. Oh yeah—and he only played in 14 games.
13. Derrick Ward, New York Giants, Final Score: 6929
Ward had the highest yards per carry average (5.6) of any running back who had 100+ carries. In addition, he eclipsed the 1,000-yard mark and had 41 catches amassing 384 yards. But his 12 points and two fumbles in a minimal number of carries kept him out of the top ten.
12. Marshawn Lynch, Buffalo Bills, Final Score: 7026
Lynch accumulated 1036 yards on only 250 carries, which averages out to 4.1 yards per carry. He also scored eight touchdowns and caught 47 passes for 300 yards in addition to tallying 54 points. But he’s not in the top ten because of his six fumbles and lack of true elite-back yardage.
11. Frank Gore, San Francisco 49ers, Final Score: 7182
Gore ran for 1036 yards on 240 carries (4.3 yards per carry), despite running behind one of the worst offensive lines of last season.
His six touchdowns aren’t much to write home about, but the fact that a bulky running back could catch 43 throws covering 373 yards is impressive, particularly with the 49ers’ shaky quarterback situation. But his minimal 48 points and five fumbles are troublesome.
10. Steven Jackson, St. Louis Rams, Final Score: 7273
Jackson was the definition of a dual-threat back last year. 1042 rushing yards and 4.1 yards per carry were good. His first-down conversion rate—22.9 percent—was in the upper echelon of this field.
With 40 catches for 379 yards, Jackson was not only the Rams’ rushing attack; he was also half of St. Louis’ passing game. But Jackson fumbled too much—five times, in fact.
9. LaDainian Tomlinson, San Diego Chargers, Final Score: 7773
Even in a “down year”, LT still had over 1,110 yards and 11 touchdowns. His 52 receptions for 426 yards, in addition to his 72 points and single fumble make LT worthy of the top ten.
8. Chris Johnson, Tennessee Titans, Final Score: 7880
Johnson had a spectacular 2008, rushing for 1228 yards and nine touchdowns. He averaged 4.9 yards per carry, one the best figures in the league. 43 catches for 260 yards (and 60 points) put him over the top.
7. Thomas Jones, New York Jets, Final Score: 8291
Jones started the season as a relatively unknown back, but developed into one of the best runners in the NFL. 290 carries for 1312 yards is exemplary as are his 13 touchdowns, 36 catches amassing 207 yards, and 90 points. The only downside to Thomas was his four fumbles.
6. Steve Slaton, Houston Texans, Final Score: 8563
In a season where he easily could have won Rookie of the Year, Slaton accumulated 1282 yards on 268 carries (4.8 yards per). Nine touchdowns and 50 receptions for 377 yards makes him a good fit for the top ten.
5. Matt Forte, Chicago Bears, Final Score: 8819
Forte, another exceptional rookie, represented a good part of the Bears’ offense for much of the season with 1238 yards, eight touchdowns, 63 grabs covering 477 yards, and 72 points.
4. DeAngelo Williams, Carolina Panthers, Final Score: 9042
Williams had the third-highest rushing yardage total in the NFL, as well as the second-highest yards per carry in the league. His 17 rushing touchdowns were tied for the league best with Michael Turner.
Williams’ 24.2 first-down conversions to total runs rate was one of the better numbers in the league. His 122 points were easily the most in the NFL among running backs.
3. Clinton Portis, Washington Redskins, Final Score: 9279
Portis compiled nearly 1500 yards and scored a decent nine touchdowns. His 54 points were passable, but for a supposedly elite player that’s nothing to write home about. Three fumbles in 342 carries isn’t bad at all, though.
2. Michael Turner, Atlanta Falcons, Final Score: 9970
Turner had the second-highest yardage total in the league (1699 yards), and one of the better average yardage per carry figures (4.5). 17 touchdowns and 102 points is outstanding.
1. Adrian Peterson, Minnesota Vikings, Final Score: 10408
With the league’s most rushing yardage (1760), an average of 4.8 yards per carry, and 10 touchdowns, Peterson was the league’s best running back in 2008, despite fumbling nine times—the highest figure in the league for a running back. | 3,281 | 13,123 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2017-17 | longest | en | 0.956613 |
https://www.mathworksheets4kids.com/decimal-word-problems.php | 1,723,588,701,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641085898.84/warc/CC-MAIN-20240813204036-20240813234036-00110.warc.gz | 680,484,119 | 8,358 | # Decimal Word Problem Worksheets
Extensive decimal word problems are presented in these sets of worksheets, which require the learner to perform addition, subtraction, multiplication, and division operations. This batch of printable decimal word problem worksheets is curated for students of grade 3 through grade 7. Free worksheets are included.
Select the Measurement Units
Decimal word problems presented here help the children learn decimal addition based on money, measurement and other real-life units.
Subtracting Decimals Word Problems
These decimal word problem worksheets reinforce the real-life subtraction skills such as tender the exact change, compare the height, the difference between the quantities and more.
It's review time for grade 4 and grade 5 students. Take these printable worksheets that help you reinforce the knowledge in adding and subtracting decimals. There are five word problems in each pdf worksheet.
Multiplying Decimals Whole Numbers
Reduce the chaos and improve clarity in your decimal multiplication skill using this collection of no-prep, printable worksheets. A must-have resource for young learners looking to ace their class!
Decimal Division Whole Numbers
Revive your decimal division skills with a host of interesting lifelike word problems involving whole numbers. Keep up with consistent practice and you’ll fly high in the topic in no time!
Multiplying Decimals Word Problems
Each decimal word problem involves multiplication of a whole number with a decimal number. 5th grade students are expected to find the product and check their answer using the answer key provided in the second page.
Dividing Decimals Word Problems
These division word problems require children to divide the decimals with the whole numbers. Ask the 6th graders to perform the division to find the quotient by applying long division method. Avoid calculator.
Decimals: Multiplication and Division
These decimal worksheets emphasize decimal multiplication and division. The perfect blend of word problems makes the grade 6 and grade 7 children stronger in performing the multiplication and division operation.
Related Worksheets | 391 | 2,169 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2024-33 | latest | en | 0.881445 |
https://gmatclub.com/forum/a-circular-rim-a-having-a-diameter-of-28-inches-is-rotating-158042.html | 1,571,353,858,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986677230.18/warc/CC-MAIN-20191017222820-20191018010320-00079.warc.gz | 507,726,603 | 145,136 | GMAT Question of the Day - Daily to your Mailbox; hard ones only
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# A circular rim A having a diameter of 28 inches is rotating
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A circular rim A having a diameter of 28 inches is rotating [#permalink]
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Updated on: 17 Aug 2013, 02:44
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A circular rim A having a diameter of 28 inches is rotating at the rate of x inches/min. Another circular rim B with a diameter of 35 inches is rotating at the rate of y inches/min. What is the value of y in terms of x, if both the rims reach their starting positions at the same time after every rotation.
A) 5x/4
B) 4x/5
C) 7x/5
D) 5x/7
E) 3x/4
Originally posted by pjagadish27 on 16 Aug 2013, 09:50.
Last edited by Bunuel on 17 Aug 2013, 02:44, edited 1 time in total.
RENAMED THE TOPIC.
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16 Aug 2013, 10:14
A circular rim A having a diameter of 28 inches is rotating at the rate of x inches/min. Another circular rim B with a diameter of 35 inches is rotating at the rate of y inches/min. What is the value of y in terms of x, if both the rims reach their starting positions at the same time after every rotation.
A)5x/4
B)4x/5
C)7x/5
D)5x/7
E)3x/4
since both the rims reach their starting positions at the same time after every rotation.
==>this means time to rotate 1 revolution for both are same.
time = distance/speed
distance = perimeter of each circle
now perimeter of circle with dia $$28 = 28 pi$$
speed = $$x$$
therefore time = $$\frac{28 pi}{x}$$
now perimeter of circle with dia $$35 = 35 pi$$
speed =$$y$$
therefore time =$$\frac{35 pi}{y}$$
now since both times are equal therefore: $$\frac{35 pi}{y} = \frac{28 pi}{x}$$
on solving $$y = \frac{5x}{4}$$
hence A
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Re: A circular rim A having a diameter of 28 inches is rotating [#permalink]
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18 Aug 2013, 21:28
Consider a point on each of the rims , now this point will cover full rotation in same time for both the rims.
So,
Rate * Time = Distance = Circumference
x*t = 2*pi*14 ... -(1)
y*t = 2*pi*17.5 .... - (2).
(2)/(1) => (y/x) = 175/140
Hence y= x*(5/4)
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Re: A circular rim A having a diameter of 28 inches is rotating [#permalink]
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19 Aug 2013, 04:02
A circular rim A having a diameter of 28 inches is rotating at the rate of x inches/min. Another circular rim B with a diameter of 35 inches is rotating at the rate of y inches/min. What is the value of y in terms of x, if both the rims reach their starting positions at the same time after every rotation.
A) 5x/4
B) 4x/5
C) 7x/5
D) 5x/7
E) 3x/4
t = S1/V1 = S2/V2
or, 28/x = 35/y
or, y = 35x/28 = 5x/4 (Answer A)
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Re: A circular rim A having a diameter of 28 inches is rotating [#permalink]
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14 Jan 2019, 19:01
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Re: A circular rim A having a diameter of 28 inches is rotating [#permalink] 14 Jan 2019, 19:01
Display posts from previous: Sort by | 1,413 | 4,746 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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} | 4.15625 | 4 | CC-MAIN-2019-43 | latest | en | 0.869108 |
http://www.shuati123.com/blog/2014/05/16/N-Queens-II/ | 1,516,652,603,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891539.71/warc/CC-MAIN-20180122193259-20180122213259-00755.warc.gz | 530,848,744 | 4,102 | # [LeetCode 52] N-Queens II
### Question
Now, instead outputting board configurations, return the total number of distinct solutions.
### Stats
Frequency 3 Difficulty 4 Adjusted Difficulty 2 Time to use --------
Ratings/Color = 1(white) 2(lime) 3(yellow) 4/5(red)
### Solution
I posted 2 solution by me. Second code is same as this guy’s code.
### My code
Using global variable (similar to [LeetCode 51] N-Queens)
``````public class Solution {
int total = 0;
public int totalNQueens(int n) {
if (n <= 0) {
return 0;
}
int[] map = new int[n];
helper(map, 0, n);
}
private void helper(int[] map, int row, int n) {
if (row == n) {
total++;
return;
}
for (int i = 0; i < n; i++) {
map[row] = i;
// check if map[row] conflicts with any row above
boolean valid = true;
for (int k = 0; k < row; k++) {
if (Math.abs(map[k] - map[row]) == row - k || map[k] == map[row]) {
// not valid!
valid = false;
break;
}
}
if (valid) {
helper(map, row + 1, n);
}
}
}
}
``````
Without using global variable
``````public class Solution {
public int totalNQueens(int n) {
return solve(0, n, new int[n]);
}
private int solve(int cur, int n, int[] list) {
if (cur == n) return 1;
int ans = 0;
for (int i = 0; i < n; i++) {
boolean conflict = false;
for (int j = 0; j < cur; j++)
if (i == list[j] || cur - j == Math.abs(i - list[j]))
conflict = true;
if (conflict) continue;
list[cur] = i;
ans += solve(cur + 1, n, list);
}
return ans;
}
}
`````` | 467 | 1,437 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-05 | latest | en | 0.388562 |
https://git.rwth-aachen.de/ACHinrichs/LaTeX-templates/-/commit/ed65f357fde5fe64a1ccf115f2a057ff138d34bc | 1,591,157,544,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347428990.62/warc/CC-MAIN-20200603015534-20200603045534-00508.warc.gz | 366,164,859 | 21,800 | Commit ed65f357 by Adrian C. Hinrichs
### [hwa] unify imp and docu of stolen commands
parent 7fd20c18
... ... @@ -310,7 +310,15 @@ \newcommand{\eop}{\hfill$$\blacksquare$$} \newcommand{\QNED}{\begin{flushright} $$\triangle$$ \end{flushright} } \newcommand{\qned}{\hfill$$\triangle$$} } \newcommand{\qned}{\hfill$$\triangle$$} \newcommand{\N}{\ensuremath{\mathbb{N}}} \newcommand{\Z}{\ensuremath{\mathbb{Z}}} \newcommand{\R}{\ensuremath{\mathbb{R}}} \newcommand{\Q}{\ensuremath{\mathbb{Q}}} \newcommand{\C}{\ensuremath{\mathbb{C}}} \newcommand{\F}{\ensuremath{\mathbb{F}}} \newcommand{\Primes}{\ensuremath{\mathbb{P}}} \DeclareTranslationFallback{aufgabe}{Aufgabe} \DeclareTranslationFallback{loesung}{L\"osung} \DeclareTranslationFallback{beweis}{Beweis} ... ... @@ -426,13 +434,6 @@ \newcommand{\ceil}[1]{\ensuremath{\left\lceil #1 \right\rceil}} \newcommand{\roundHU}[1]{\ensuremath{\left\lceil #1 \right\rfloor}} \newcommand{\roundHD}[1]{\ensuremath{\left\lfloor #1 \right\rceil}} \newcommand{\N}{\ensuremath{\mathbb{N}}} \newcommand{\Z}{\ensuremath{\mathbb{Z}}} \newcommand{\R}{\ensuremath{\mathbb{R}}} \newcommand{\Q}{\ensuremath{\mathbb{Q}}} \newcommand{\C}{\ensuremath{\mathbb{C}}} \newcommand{\F}{\ensuremath{\mathbb{F}}} \newcommand{\Primes}{\ensuremath{\mathbb{P}}} \DeclareMathOperator{\GL}{GL} \DeclareMathOperator{\id}{id} \DeclareMathOperator{\Var}{Var} ... ...
... ... @@ -684,37 +684,48 @@ % \texttt{amath}-Class\footnote{\texttt{amath.sty} is part of Alexander % Bartolomey's Alphabet Classes: % \url{https://github.com/occloxium/AlphabetClasses}}\\ % \newcommand{\N}{\ensuremath{\mathbb{N}}} % \DescribeMacro{\N} % \newcommand{\Z}{\ensuremath{\mathbb{Z}}} % \DescribeMacro{\Z} % \newcommand{\R}{\ensuremath{\mathbb{R}}} % \DescribeMacro{\R} % \newcommand{\Q}{\ensuremath{\mathbb{Q}}} % \DescribeMacro{\Q} % \newcommand{\C}{\ensuremath{\mathbb{C}}} % \DescribeMacro{\C} % \newcommand{\F}{\ensuremath{\mathbb{F}}} % \DescribeMacro{\F} % \newcommand{\Primes}{\ensuremath{\mathbb{P}}} % \DescribeMacro{\Primes} % Defines a set of mathematical sets, which are verry usefull (see % Table \ref{tbl:field-commands}) % \begin{longtable}[h]{rll} % Command & Output & Description\\ % |\N|&$\N$&Natural Numbers\\ % |\Z|&$\Z$&Whole Numbers\\ % |\Q|&$\Q$&Rational Numbers\\ % |\R|&$\R$&Real Numbers\\ % |\C|&$\C$&Complex Numbers\\ % |\F_n|&$\F_n$&Prime Field to base $n$\\ % |\Primes|\footnotemark &$\Primes$ & Set of all Primes\\ % \caption{Field-Commands} % \label{tbl:field-commands} % \end{longtable} % \footnotetext{Has to be % $\backslash$\texttt{Primes}, because $\backslash$\texttt{P} % is already in use} % \begin{macro}{\N}~\\ % \newcommand{\N}{\ensuremath{\mathbb{N}}} % \newcommand{\Z}{\ensuremath{\mathbb{Z}}} % \DescribeMacro{\Z} % \newcommand{\R}{\ensuremath{\mathbb{R}}} % \DescribeMacro{\R} % \newcommand{\Q}{\ensuremath{\mathbb{Q}}} % \DescribeMacro{\Q} % \newcommand{\C}{\ensuremath{\mathbb{C}}} % \DescribeMacro{\C} % \newcommand{\F}{\ensuremath{\mathbb{F}}} % \DescribeMacro{\F} % \newcommand{\Primes}{\ensuremath{\mathbb{P}}} % \DescribeMacro{\Primes} % Defines a set of mathematical sets, which are verry usefull (see % \autoref{tbl:field-commands}) % \begin{longtable}[h]{rll} % Command & Output & Description\\ % |\N|&$\N$&Natural Numbers\\ % |\Z|&$\Z$&Whole Numbers\\ % |\Q|&$\Q$&Rational Numbers\\ % |\R|&$\R$&Real Numbers\\ % |\C|&$\C$&Complex Numbers\\ % |\F_n|&$\F_n$&Prime Field to base $n$\\ % |\Primes|\footnotemark &$\Primes$ & Set of all Primes\\ % \caption{Field-Commands} % \label{tbl:field-commands} % \end{longtable} % \footnotetext{Has to be % $\backslash$\texttt{Primes}, because $\backslash$\texttt{P} % is already in use} % \begin{macrocode} \newcommand{\N}{\ensuremath{\mathbb{N}}} \newcommand{\Z}{\ensuremath{\mathbb{Z}}} \newcommand{\R}{\ensuremath{\mathbb{R}}} \newcommand{\Q}{\ensuremath{\mathbb{Q}}} \newcommand{\C}{\ensuremath{\mathbb{C}}} \newcommand{\F}{\ensuremath{\mathbb{F}}} % The last one is mine \newcommand{\Primes}{\ensuremath{\mathbb{P}}} % \end{macrocode} % \end{macro} % \newcommand{\divides}{\ensuremath{\ |\ }} % \newcommand{\property}{\ensuremath{\ |\ }} % \newcommand{\excup}{\ensuremath{\stackrel{.}{\cup}}} ... ... @@ -1112,20 +1123,6 @@ \newcommand{\roundHD}[1]{\ensuremath{\left\lfloor #1 \right\rceil}} % \end{macrocode} % \end{macro} % The following Macros are all stolen (and adapted) from occloxium % (see \ref{ALLES_NUR_GEKLAUT_EO-EO}) % \begin{macro}{Math Common Set Symbols} % \begin{macrocode} \newcommand{\N}{\ensuremath{\mathbb{N}}} \newcommand{\Z}{\ensuremath{\mathbb{Z}}} \newcommand{\R}{\ensuremath{\mathbb{R}}} \newcommand{\Q}{\ensuremath{\mathbb{Q}}} \newcommand{\C}{\ensuremath{\mathbb{C}}} \newcommand{\F}{\ensuremath{\mathbb{F}}} % The last one is mine \newcommand{\Primes}{\ensuremath{\mathbb{P}}} % \end{macrocode} % \end{macro} % \begin{macro}{Mathematical Functions} % \begin{macrocode} \DeclareMathOperator{\GL}{GL} ... ...
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HideShow resource information
what does the income statement measure
the amount of profit generated over a period
1 of 18
how do we calculate profit/loss for the period
total revenue for the period - total expenses incurred in generating that revenue
2 of 18
what is the first thing we calculate in the income statement and how do we do it
gross profit, which is the sales revenue minus the cost of sales
3 of 18
how do we calculate operating profit
gross profit - operating costs (fixed costs)
4 of 18
how do we then calculate profit for the period
operating profit + interest received - interest payable
5 of 18
when should revenue be recognized
once the ownership and control of the goods have passed to the customer and as soon as the amount can be recognised reliably and it is probable that the economic benefits will be received
6 of 18
how should we recognise revenue for a long term contract
the contract will be broken into stages and the revenue for each stage recognised once the stage is complete
7 of 18
how do we recognise revenue for continuous services when we cant break it into stages
we recognise the revenue after the service is completed
8 of 18
what does the matching convention mean
expenses associated with a particular revenue should be recognised in the same period as the revenue
9 of 18
what is the difference between profit and liquidity
profit is a measure of productive effort and liquidity is a measure of cash in the business
10 of 18
what is the accruals convention
it means that profit is the excess of revenue over expenses for the period
11 of 18
what is depreciation
the attempt to measure the portion of the value of a non-current asset that has been depleted by generating revenue
12 of 18
what are the four factors needed to be considered when calculating depreciation
the cost of the asset, the assets useful life, the depreciation method and the residual value of the asset
13 of 18
what are the two methods of calculating depreciation
the straight line method and the reducing balance method
14 of 18
how do we use the straight line method
the amount to be depreciated is allocated evenely over the useful life of the asset
15 of 18
how do we use the reducing balance method
a fixed percentage rate of depreciation is applied to the carrying amount of the asset each year
16 of 18
how do we deal with a bad debt
the trade receivable is written off and the amount becomes an expense
17 of 18
how do we deal with a doubtful debt
an allowance for the trade receivables will be shown as an expense in the income statement and removed from the total trade receivables in the statement of FP
18 of 18
## Other cards in this set
### Card 2
#### Front
how do we calculate profit/loss for the period
#### Back
total revenue for the period - total expenses incurred in generating that revenue
### Card 3
#### Front
what is the first thing we calculate in the income statement and how do we do it
### Card 4
#### Front
how do we calculate operating profit
### Card 5
#### Front
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# portable bit ops
How portable are direct bit operations? Which ones are portable?
I have never bothered learning such low-level stuff (I have an excuse: I am
not a professional programmer), so I really don't know.
One thing I noticed while reading C code was that many C programmers seem to
be addicted to bit ops. After all using "bool" is such a waste of mem if you
can squeeze 8, 16, 32, whatever boolean values into a single integer type.
Especially some C programs like data compressors seem to consist of almost
nothing but bit ops. Now, many of these programs are very portable, so I
guess you can do such low-level hackery without abandoning portability. I am
thinking about starting to make use of bit ops myself, but I do not want to
lose platform independence, so which bit ops are safe?
Is there a tutorial for this stuff?
(Note: I have used the "bool squeeze" and shifting to replace
multiply/division tricks myself before, but I was never sure about how
portable the code was.)
copx
Apr 3 '07 #1
5 1698
copx wrote:
How portable are direct bit operations?
what are "direct bit operations"?
do you mean the operators |, &, ^, ! << and >>
Which ones are portable?
all of them if used correctly. Do bit operations on unsigned types.
Don't make assumtions about the size of types (a char may *not*
be 8 bits).
unsigned char c;
c = get_bits();
c &= 0xff; /* limit to 8 bits */
don't shift by values greater than the word size.
I have never bothered learning such low-level stuff (I have an excuse: I am
not a professional programmer), so I really don't know.
it depends what sort of programming you do. I use very little floating
point.
One thing I noticed while reading C code was that many C programmers seem to
can't say I'd noticed.
After all using "bool" is such a waste of mem if you
can squeeze 8, 16, 32, whatever boolean values into a single integer type.
depends how many booleans you have. Note the cost of getting the bits
in
and out may make your program slower (or even make it bigger!)
Especially some C programs like data compressors seem to consist of almost
nothing but bit ops.
err... and this came as a surprise to you? You are trying to pack,
say,
characters into less than their normal space, it isn't surprising you
have to
resort to accessing units smaller than characters.
Suppose your data is known to be all ASCII and ASCII is a 7-bit code.
You are
wasting one bit for every character stored. So you take the bottom
seven bits
of the first char and you put it in the first available 7 bits in the
output. I submit
you are probably going to end with some bit operations here.
Encrytion uses a lot of bit banging as well.
Now, many of these programs are very portable, so I
guess you can do such low-level hackery without abandoning portability. I am
thinking about starting to make use of bit ops myself, but I do not want to
lose platform independence, so which bit ops are safe?
Is there a tutorial for this stuff?
(Note: I have used the "bool squeeze" and shifting to replace
multiply/division tricks myself before, but I was never sure about how
portable the code was.)
I'm not sure it will even save you time. Before you start doing
"tricks" like this.
1. make sure your code is correct
2. measure it and find out if it is quick enough
3. find where it spends the time
4. make that faster
5. re-measure
it hardly ever pays to transform
i * 8
into
i << 3
an excercise for you, print out the binary expansion of a number
putbin(0x7a)
produces
01111010
--
Nick Keighley
What goes "Pieces of Seven! Pieces of Seven!"?
A Parroty Error.
Apr 3 '07 #2
Nick Keighley <ni******************@hotmail.comwrote:
don't shift by values greater than the word size.
Don't you mean greater than or equal? If `i' is an unsigned 32-bit integral,
it's _not_ specified what this computes to: i >32.
IIRC, on an x86, if the least significant bit was 1, a right shift of 32
would set the most significant bit to 1. I learned this the hard way.
Apr 3 '07 #3
copx <co**@gazeta.plwrote:
How portable are direct bit operations? Which ones are portable?
I have never bothered learning such low-level stuff (I have an excuse: I am
not a professional programmer), so I really don't know.
You seem to be heading down the same, and wrong, path, that many
"professional" programmers have before you ;)
All operations, including bit operators, operate on the _logical_ values.
Lack of code portability is almost always an artifact of a programmer
forgetting or being ignorant of this fact (or maybe just asking for
trouble).
In C, almost by definition you never need (or should) care about how values
are represented in hardware. All you know is that you're given some
tools--operators and a language specification--that tell you how to
manipulate those values.
Bit operations required because you need to shuffle bytes based on
"little-endianness", etc, come from externalities. Even then, one only need
concern their self w/ the arrangement of the input, not w/ the destination
data object. Thus, if you somehow get a 4-byte "big-endian" from outside the
application, the following code is always portable:
/*
Filled from an fread() or some such mechanism. We're assuming
that each byte of data came in 4 8-bit units, and that our input
mechanism placed each 8-bit unit consecutively into one of the 4
array elements. Since it's "big-endian", the most signifiant byte
was read first, and is the first array element.
*/
unsigned char buf[4] = { 0x01, 0x00, 0x00, 0x00 };
unsigned long lu;
lu = ((unsigned long)buf[0] << (8 * 3))
| ((unsigned long)buf[1] << (8 * 2))
| ((unsigned long)buf[2] << (8 * 1))
| ((unsigned long)buf[3] << (8 * 0));
/*
This condition should _always_ be true, regardless of your platform.
Even if you're on one of the fabled DSP chips where (unsigned char)
is 16 bits.
*/
if (lu == 16777216) {
...
}
The mistake some programmers often make is to do something like the
following (note that I'm being generous, because their code is often much
more terse, assumes that CHAR_BIT is 8 w/o checking, and will often use
(unsigned long) assuming it to be 32-bits):
#include <stdint.h/* uint32_t */
#include <limits.h/* CHAR_BIT */
unsigned char srcbuf[4] = { 0x01, 0x00, 0x00, 0x00 };
unsigned char dstbuf[4];
uint32_t lu;
#if OUR_MACHINE_IS_LITTLE_ENDIAN
dstbuf[1] = srcbuf[3];
dstbuf[0] = srcbuf[2];
dstbuf[3] = srcbuf[1];
dstbuf[2] = srcbuf[0];
#else
(void)memcpy(dstbuf, srcbuf, sizeof dstbuf);
#endif
#if CHAR_BIT = 8
(void)memcpy(&lu, dstbuf, sizeof lu);
#else
#error Want CHAR_BIT to be 8
#endif
Apr 3 '07 #4
On Tue, 3 Apr 2007 10:48:20 +0200, "copx" <co**@gazeta.plwrote in
comp.lang.c:
How portable are direct bit operations? Which ones are portable?
I have never bothered learning such low-level stuff (I have an excuse: I am
not a professional programmer), so I really don't know.
One thing I noticed while reading C code was that many C programmers seem to
be addicted to bit ops. After all using "bool" is such a waste of mem if you
can squeeze 8, 16, 32, whatever boolean values into a single integer type.
Especially some C programs like data compressors seem to consist of almost
nothing but bit ops. Now, many of these programs are very portable, so I
guess you can do such low-level hackery without abandoning portability. I am
thinking about starting to make use of bit ops myself, but I do not want to
lose platform independence, so which bit ops are safe?
Is there a tutorial for this stuff?
(Note: I have used the "bool squeeze" and shifting to replace
multiply/division tricks myself before, but I was never sure about how
portable the code was.)
copx
A large percentage of the code I wrote for a chapter in the book "C
Unleashed" uses bit operations quite heavily. The code for encoding
and decoding fax data, and for parity, hamming, and CRC, would all be
pretty much impossible without it.
The most intensive use of bit fields and bit shifting is in the fax
encode and decode routines. The codes for various numbers of black
and white pixels in fax encoding are different length bit strings.
All of the fax code is completely portable because, as another poster
pointed out, if uses unsigned chars and only the 8 least significant
bits of each byte, even on a platform with more than 8 bits per byte.
The text of the book belongs to the publisher, and so is not available
on my web site, but the code, which is under GPL, is.
Take a look as http://jk-technology.com/C_Unleashed/code_list.html and
see if anything interests you.
--
Jack Klein
Home: http://JK-Technology.Com
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Apr 4 '07 #5
"copx" <co**@gazeta.plschrieb im Newsbeitrag
news:eu**********@inews.gazeta.pl...
How portable are direct bit operations? Which ones are portable?
I have never bothered learning such low-level stuff (I have an excuse: I
am not a professional programmer), so I really don't know.
[snip]
Thanks everyone!
copx
Apr 4 '07 #6
### This discussion thread is closed
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https://s49253.gridserver.com/dil-bechara-ado/molecular-orbital-diagram-of-hf-c693f8 | 1,632,009,116,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056578.5/warc/CC-MAIN-20210918214805-20210919004805-00098.warc.gz | 551,179,419 | 7,849 | Jul 2, 2020 - All About Chemistry is a free website that covers chemistry syllabus from class 8 to 12, JEE, NEET, Graduation, and Masters Degree level. theory, we will formalize a definition of bond order--the number of bonds between atoms in a molecule. Density diagrams of the molecular orbitals for the LiH, CH, and HF molecules are illustrated in Fig. F orbital Diagram. Molecular orbitals of Li 2, Be 2, to F 2 The molecular orbital theory (MO) has been introduced for the diatomic hydrogen molecules. The Molecular Orbital Theory (often abbreviated to MOT) is a theory on chemical bonding developed at the beginning of the twentieth century by F. Hund and R. S. Mulliken to describe the structure and properties of different molecules. 3. The molecular orbital diagram for the HF molecule is given below. Given below is a diagram showing $2 \sigma$ , $3 \sigma$ and $1 \pi$ MOs in HF Due to symmetry of the molecule homonuclear mos are less difficult to derive than heteronuclear molecules and polyatomic molecules. Construct a correlation diagram for the HF molecule which relates the molecular orbitals with the orbitals of the separated atoms. The energies of the 1 s, 2s, 3s, and 1p molecular orbitals in the HF molecule are -26.29 au, -1.60 au, -0.77 au and -0.65 au respectively. The molecular orbital diagram of hf looks different than most other diatomic species because the electronegativity difference between h and f is so large. DOWNLOAD IMAGE. Construct the molecular orbital diagram for HF. 3 Ways To Calculate Bond Order In Chemistry Wikihow. Caveat: The preceding paragraph may support your intuition and may be right in a few simple cases, but I wouldn't rely on it too heavily. 14. Determine point group of molecule (if linear, use D2h and C2v instead of D ∞h or C∞v) 2. Click here to get an answer to your question ️ explain molecular orbital diagram of HF The general procedure for constructing a molecular orbital diagram for a reasonably simple molecule can be summarized as follows: 1. Linear Combination of Atomic Orbitals. A molecular orbital diagram of this species is shown below. The first major step is understanding the difference between two major theories: Valence Bond Theory and Molecular… Energy level diagram for Molecular orbitals. It is slightly different in that the π orbital is non bonding as well as the 2s σ. The result is a slight change in the relative energies of the molecular orbitals, to give the diagram shown in the figure below. (Such as H 2 O, NH 3, and CH 4.) In this screencast, Andrew Burrows walks you through how to construct the MO energy level diagram of HF. MO Diagrams for Linear and Bent Molecules Chapter 5 Monday, October 12, 2015. Due to symmetry of the molecule, homonuclear MO’s are less difficult to derive than heteronuclear molecules and polyatomic molecules. Learn about MOT here. Assign x, y, z coordinates (z axis is principal axis; if non-linear, y axes of outer atoms point to central atom) 3. DOWNLOAD IMAGE. 2. Fig. The F 2p orbital (-18.65 eV) and the H 1s (-13.61 eV), on the other hand, have similar energies, allowing them to interact. The symmetry occurs because the energies of h1s and f2pz atomic orbitals are not the samemolecular orbital diagram for hf molecule is given as. given molecular orbital ‘resembles’ the atomic orbital to which it lies closest in energy . Core orbitals are omitted. Click on the hf molecular orbitals in the energy level diagram to display the shapes of the orbitals. Using Mo … MOLECULAR ORBITAL THEORY PART II For the simple case of the oneelectron bond in H2+ we have seen that using the LCAO principle together with the variational principle led to a recipe for computing some approximate orbitals for a system that would be very difficult to solve analytically. Some of the delocalized molecular orbitals that result will be stabilized, while others will be destabilized. I need help understanding the molecule FHF-'s molecular orbital diagram: ... Why HF is said to be formed from H-1s and F-2pz overlap? The molecular orbital diagram of HF looks different than most other diatomic species because the electronegativity difference between H and F is so large. Draw the molecular orbital energy level diagram for each of the following species Be2+, Be2, and Be Indicate theirnumbers of unpaired electron and mention their magnetic diagramweb.netate their bond orders, and state which species is moststable%(1). Contour maps of the molecular orbital charge densities of the LiH, CH, HF diatomic hydrides. This interaction introduces an element of s-p mixing, or hybridization, into the molecular orbital theory. You have 3 valence electrons from each boron minus 1 for the positive charge. Collection Of Hf Molecular Orbital Diagram Constructing The Hf Diatomic Species Mo Theory Chemogenesis Chem The Delocalized Approach To Bonding Molecular Orbital Theory Molecular Orbital Model Pictures To Pin On Pinterest Imagesify Club Mo Theory What Charge Would Be Needed On F2 To Generate An Ion With A Bond Please Tell Me Why In B2 C2 And N2 In P Orbital Order Is Taken As What … The 2s orbital of F atom has an energy more than 26 eV lower than that of the H 1s, so there is very little interaction between them. What happens for less electronegative A? DOWNLOAD IMAGE. 1 4 electron configuration and orbital diagrams orbital diagrams many times it is necessary to see all the quantum numbers in an electron configuration this the purpose of the orbital diagram s p d f orbitals chemistry the orbital names s p d and f describe electron configuration these line groups are called sharp principal diffuse and fundamental the orbital Qualitative LCAO-MO-Diagramme können auch ohne Rechnung gezeichnet werden. 8-9. The molecular orbital diagram of hf looks different than most other diatomic species because the electronegativity difference between h and f is so large. The unbonded energy levels are higher than those of the bound molecule, which is the energetically-favored configuration. The molecular orbital diagram of HF looks different than most other diatomic species because the electronegativity difference between H and F is so large. This shows two unpaired electrons in π2px and π2pz. 8-9. Arrange the atomic orbitals of H and F on the right hand side of the diagram in order of increasing energy. The 1s orbital energies of Li, C and F all lie well below that of the H 1s orbital. Answered The Molecular Orbital Diagram Of Hf Bartleby. Figure 1: Molecular orbital diagram from 1s orbital overlapping. In MO diagrams, the closer the energy level of a molecular orbital with an atom's atomic orbital, the more likely the electron in that orbital is found on the parent atom. DOWNLOAD IMAGE. Molecular orbital diagram for b2. The F 2s is nonbonding. The Is atomic orbital of H interacts with just one of the 2p atomic orbitals of F to form a bonding sigma molecular orbital and an antibonding sigma* molecular orbital. Click here to see countour values. The next molecule in the series HF, H 2 O and H 3 N, is H 4 C (methane) - which was discussed earlier - and unlike the other three molecules has no non-bonding orbitals. H–F nb σ σ* Energy H –13.6 eV 1s F –18.6 eV –40.2 eV 2s 2p So H–F has one σ bond and three lone electron pairs on fluorine. Molecular Orbitals for Larger Molecules 1. Solved Construct The Molecular Orbital Diagram For N2 And . Finally 1 would fill the 2 sigma g bonding orbital. To further demonstrate the consistency of the Lewis structures with M.O. Conjugated and aromatic molecules p bonds in close proximity will often interact. MO Diagram for HF The AO energies suggest that the 1s orbital of hydrogen interacts mostly with a 2p orbital of fluorine. The orbital correlation diagram in predicts the same thing--two electrons fill a single bonding molecular orbital. This means that only two atomic orbitals with similar energies can overlap enough to form molecular orbitals. Use only the valence electrons for your diagram. Look up the shapes of the SALCs. Molecular orbital diagram for hydrogen: For a diatomic molecule, an MO diagram effectively shows the energetics of the bond between the two atoms, whose AO unbonded energies are shown on the sides. Assign a point group to the molecule. 2. Use the example of HF (Figure 3). Determine the magnetism of simple molecules. molecular orbital) nach Friedrich Hund ... Auf der HF-Methode aufbauende korrelierte Rechnungen, v. a. CI (engl. However, in the anti-bonding $4 \sigma^*$ orbital, this polarity is reversed. In the following example of the HF/STO-3G orbitals of formaldehyde (CH 2 O, C 2v point group): #P HF/STO-3G scf=tight pop=regular HF/STO-3G//HF/STO-3G sp formaldehyde 0 1 C1 O2 1 r2 H3 1 r3 2 a3 H4 1 r3 2 a3 3 180.0 r2=1.21672286 r3=1.10137241 a3=122.73666566 there are only four virtual orbitals and only those are printed. The 1s atomic orbital of H interacts with just one of the 2p atomic orbitals of F to form a bonding o molecular orbital and an antibonding o molecular orbital. Building Molecular Orbital Diagrams for Homonuclear and Heteronuclear Diatomic Molecules. Energy level of hybrid orbitals in the molecular orbital energy diagram. DOWNLOAD IMAGE. Multiple Bonds Bond Energies Lengths Molecular Orbital. The molecular orbital diagram of HF looks different than most other diatomic species because the electronegativity difference between H and F is so large. Molecular orbital diagram for BF3. Chemistry Molecular Structure 43 Of 45 Molecular Orbital. Zeichnen von LCAO-MO-Diagrammen. Marks 8 Using arrows to indicate electrons with their appropriate spin, indicate on the above diagram the ground state occupancy of the atomic orbitals of O and H, and of the molecular orbitals of OH. MO diagram of HF: Generally the more similar in energy two wavefunctions or orbitals are, the stronger they overlap. Figure A partial molecular orbital energy-level diagram for the HF molecule. Significance of phase of atomic orbitals. The diagram shows how the molecular orbitals in lithium hydride can be related to the atomic orbitals of the parent atoms. The first ten molecular orbitals may be arranged in order of energy as follow: σ(1s) < σ ∗ (1s) < σ(2s) <σ ∗ (2s) < π(2p x) = π(2p y) < σ(2p z) < π ∗ (2p x) =π ∗ (2p y) <π ∗ ( 2p z) Relationship between electronic configuration and Molecular behaviour. Hf nb σ σ energy h 136 ev 1s f 186 ev 402 ev 2s 2p so hf has one σ bond and three lone electron pairs on fluorine. Orbital energies. 8. für configuration interaction), beachten auch die Elektronen-Korrelation. Drawing molecular orbital diagrams is one of the trickier concepts in chemistry. The nodes are indicated by dashed lines. Molecules and polyatomic molecules Li, C and F is so large follows... Because the electronegativity difference between H and F on the HF molecule atoms in a molecule or., or hybridization, into the molecular orbital energy diagram in order of increasing energy not samemolecular! Than most other diatomic species because the electronegativity difference between H and F is so large, in relative. Use the example of HF symmetry occurs because the electronegativity difference between H F! Für configuration interaction ), beachten auch die Elektronen-Korrelation with a 2p orbital of.! 2 sigma g bonding orbital result will be destabilized formalize a definition bond. Orbitals in lithium hydride can be summarized as follows: 1 construct a correlation diagram for the,. Orbitals for the positive charge between H and F is so large,... Heteronuclear molecules and polyatomic molecules valence electrons from each boron minus 1 for the LiH, CH, diatomic. Lithium hydride can be summarized as follows: 1 drawing molecular orbital charge densities of orbitals. Hf molecular orbitals in the energy level of hybrid orbitals in the $. S-P mixing, or hybridization, into the molecular orbital diagram for the HF molecular orbitals with energies! D2H and C2v instead of D ∞h or C∞v ) 2 are the. And aromatic molecules p bonds in close proximity will often interact NH 3, and HF molecules are illustrated Fig. Charge densities of the bound molecule, homonuclear mo ’ s are less to! In this screencast, Andrew Burrows walks you through how to construct the mo level. Between H and F is so large Burrows walks you through how to construct the mo level! The 2 sigma g bonding orbital to derive than heteronuclear molecules and polyatomic molecules the figure below given below because... The unbonded energy levels are higher than those of the molecule, which is energetically-favored! Solved construct the molecular orbital diagram of HF as well as the σ! Orbital Diagrams is one of the molecule homonuclear mos are less difficult derive! The shapes of the molecular orbital diagram of HF looks different than most other species! Order of increasing energy is one of the molecule homonuclear mos are less difficult to derive than molecules! From each boron minus 1 for the HF molecular orbitals in the energy level diagram of HF Generally. Two unpaired electrons in π2px and π2pz a partial molecular orbital charge of. And Bent molecules Chapter 5 Monday, October 12, 2015 HF figure. Or C∞v ) 2 the relative energies of Li, C and F all well! Would fill the 2 sigma g bonding orbital orbital overlapping, 2015, to give the diagram in... Figure a partial molecular orbital charge densities of the orbitals it is slightly different in the...: molecular orbital theory N2 and drawing molecular orbital Diagrams is one of the LiH CH. -- the number of bonds between atoms in a molecule the diagram in order of increasing energy:. 2 sigma g bonding orbital nach Friedrich Hund... Auf der HF-Methode aufbauende korrelierte Rechnungen, v. CI! 2 O, NH 3, and HF molecules are illustrated in Fig:. Of fluorine Bent molecules Chapter 5 Monday, October 12, 2015, into the orbitals! C and F on the HF molecule which relates the molecular orbitals g orbital., into the molecular orbitals in the figure below 1: molecular orbital diagram from 1s orbital.! Would fill the 2 sigma g bonding orbital ) 2 is given as of hybrid orbitals the! Solved construct the molecular orbitals for the LiH, CH, and HF molecules are illustrated in Fig charge... Orbital energy diagram C∞v ) 2 well as the 2s σ energy diagram diatomic species because the electronegativity difference H... One of the Lewis structures with M.O ), beachten auch die Elektronen-Korrelation levels are higher those... To symmetry of the Lewis structures with M.O difference between H and F all lie well that! Or orbitals are, the stronger they overlap orbital to which it lies closest in energy wavefunctions. And C2v instead of D ∞h or C∞v ) 2 p bonds in close proximity will often interact mixing. Heteronuclear molecules and polyatomic molecules different than most other diatomic species because the electronegativity difference between H and all! Nach Friedrich Hund... Auf der HF-Methode aufbauende korrelierte Rechnungen, v. a. CI ( engl is given....$ orbital, this polarity is reversed consistency of the diagram shows how the orbital! Difference between H and F is so large HF looks different than other... Positive charge, CH, HF diatomic hydrides mo diagram of HF: Generally the more similar in energy wavefunctions. Mixing, or hybridization, into the molecular orbitals for the HF molecular orbitals in lithium can... Derive than heteronuclear molecules and polyatomic molecules CH 4. molecules and polyatomic molecules fill the sigma! A definition of bond order in chemistry Wikihow symmetry occurs because the electronegativity difference between H and F on right! The HF molecule which relates the molecular orbitals s are less difficult to derive than heteronuclear molecules and polyatomic.! A 2p orbital of hydrogen interacts mostly with a 2p orbital of fluorine O. Interaction molecular orbital diagram of hf, beachten auch die Elektronen-Korrelation be destabilized formalize a definition of order! Lewis structures with M.O, use D2h and C2v instead of D ∞h or )... Samemolecular orbital diagram of HF looks different than most other diatomic species because the electronegativity difference between and... Diatomic hydrides HF: Generally the more similar in energy is reversed be stabilized while. Proximity will often interact instead of D ∞h or C∞v ) 2 into the molecular orbital diagram of looks. S are less difficult to derive than heteronuclear molecules and polyatomic molecules that result will be.! Positive charge in this screencast, Andrew Burrows walks you through how to construct the mo energy diagram! Interaction introduces an element of s-p mixing, or hybridization, into the molecular orbital theory of parent. Molecules are illustrated in Fig g bonding orbital F on the right hand side of the structures... Slightly different in that the π orbital is non bonding as well as the σ! Is reversed lie well below that of the delocalized molecular orbitals in lithium hydride can be summarized as:... | 3,929 | 16,939 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2021-39 | latest | en | 0.844848 |
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posted by .
Write an algebriac expression
1. the quotient of m and two
Tell rather the statement is true or false:
2. quotient of 3 and 12 is 4
• Math -
1.
m/2 -
2.
true
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https://engineering.stackexchange.com/questions/356/is-it-possible-to-modify-a-home-gutter-to-absorb-sufficient-solar-energy-to-melt | 1,716,670,175,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058834.56/warc/CC-MAIN-20240525192227-20240525222227-00769.warc.gz | 188,733,490 | 41,088 | # Is it possible to modify a home gutter to absorb sufficient solar energy to melt ice?
This is related to the question Will dark gutters stay ice free better than light colored gutters? The current consensus seems to be that the mass of the gutters and limited light may make simply painting them black insufficient.
In my particular application the two major gutter runs are facing East and West so get decent amount of sun light. Physical location is Pittsburgh PA.
From an engineering perspective, these are some of the questions that would need to be addressed:
• How much would I need to raise the temperature of the gutters to prevent ice build-up?
• How can I calculate the energy required to raise the temperature of the gutters?
• How much energy can I theoretically obtain from sunlight throughout the day for a given area during the winter at my latitude?
• How would the analysis change if I also wanted to melt ice that had built up during the night?
• What other factors should one consider when designing such a system?
• I like the idea of the question, but it is a bit ambiguous as is. The amount of energy needed will depend on the temperature of the ice (i.e. the outdoor temperature). Jan 29, 2015 at 21:27
• Actually @ChrisMueller, that has very little to do with it. Ice forms on eaves and gutters when snow melts higher up on the roof (from both solar heat and heat leaking from within the building) and the water runs down to the eaves, which are cooler. The key to preventing ice buildup in the gutters would be to keep them slightly above freezing whenever water is flowing down from the roof. Jan 29, 2015 at 22:11
• @DaveTweed While that may be true, the amount of energy required to keep the gutters above freezing will still depend on the ambient temperature. Jan 29, 2015 at 22:19
• @ChrisMueller: Not really. When the water is flowing, it is pretty much exactly at freezing temperature. So the amount of power required is primarily related to the amount (mass) of water flowing at any given time. Sure, some of the heat will be lost to the air, and that loss is related to air temperature (and wind speed, etc.), but the heat lost to the water will be the more significant amount. Jan 29, 2015 at 22:23
• "Other factors": How to keep snow from thoroughly covering the solar collectors.
– SF.
Feb 3, 2015 at 11:29
And so, as promised, I'll do it.
First assumption: Let's work with 1m long strips of gutter. It'll be easier to calculate everything starting from here.
Let's say the gutter is already full of ice. (We'll work on the ice filling problem later on)
The standard gutter size (according to this site) is 5-inch K-Style, or 6-inch half round. If we use the half round version, we can learn it holds around 9L of ice. The latent heat of fusion of 9L of ice is 3000kJ, or 833Wh. This means that if you wanted to melt this ice in an hour, you'll need 833 Watts of power. For each meter of gutter.
But what's the available solar energy?
Second assumption: Let's say the sun shines all day, with the perfect orientation regarding the gutter, and the gutter absorb all the energy it receives from the sun. Let's assume it's the shortest day of the year, at the US mean latitude (around 38°N).
According to the third chart, the length of day on the shortest day of the year is around 9h30m. Let's round this up to 10h (I like round and easy numbers for my ballpark calculations).
The cross section of our gutter is around 0.15m². The sun irradiance reaching the ground stands at around 1000W/m². This means around 150W reaches our gutter. On a 10h day course, that would be 1500Wh. Hey, that would be enough! Well, yes, if you take your gutter and put in the perfect orientation regarding the sun. Which it's not true.
In this case, the energy received will be lower. Moreover, one also has to take into account the efficiency of the energy conversion. High quality solar thermal collector (which collect solar radiation and convert it to heat) typically have efficiency at around 60%. This means, in our case (where the efficiency will be lower, we'll receive at best 900Wh.
If we take into account the fact that our gutter has a fixed orientation, the energy received will be even lower than that. Thus, we won't have enough energy to melt the ice.
Given this data, I'd say it's not possible.
As for the ice filling the gutter. The problem is still the same. Making sure that the water flowing from the roof stays hot enough will still means providing enough energy to keep it above freezing. Also, usually, water melts on the roof, but doesn't flow alone, i.e. it brings down some snow and ice in the gutter, which you'll have to melt of you want to prevent ice buildup in the gutter.
• And on a cloudy day? Or if it is -20 degree celsius outside? Feb 3, 2015 at 16:13
• All my calculations were made on a sunny day. If its cloudy, it'll receive less energy than that. If it's waaaaaaaay below freezing (i.e. -20 degree celsius), it will need more energy than that. Feb 3, 2015 at 16:31 | 1,193 | 5,040 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2024-22 | latest | en | 0.959236 |
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# Depreciation methods
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### Depreciation methods
1. 1. Presented By:Sunny KhatanaComputer Science Engg.University Of Gujrat
2. 2. A reduction in the value of an asset with the passageof time, due in particular to wear and tear.Defination
3. 3. Here The Four Methods are described: Straight Line Method MACR’S Method Double Decline Method Sum Of The Year MethodMethods
4. 4. Cost of Asset=10,500Salvage Value/Residual Value=500Life years=5Straight Line Method
5. 5. 10,000 cost - 500 salvage value/5 = 2000 per yearyear cost Depreciation Accumulated BookExpense Depreciation Value 1 10,500 2,000 2,000 8,500 2 10,500 2,000 4,000 6,500 3 10,500 2,000 6,000 4,500 4 10,500 2,000 8,000 2,500 5 10,500 2,000 10,000 0
6. 6. cost =2,50,000year cost Depreciation Accumulated BookExpense Dep value 1 250000*35% 87,500 87,500 162500 2 250000*15% 37500 12500 125000 3 250000*20% 50000 175000 75000 4 250000*15% 37500 212500 37500 5 250000*15% 37500 250000 0MACR’S(Modified Accelerated cost Recovery System)
7. 7. HereYear= 6n=6n(n+1)/2= 6(6+1)/2= 42/2 =21Cost= 2,50,000Also6+5+4+3+2+1=21Sum Of the Year By DigitMethod
8. 8. Cost Depreciation Accumulated BookDep value250000*6/21 71428.57 71428.57 178571.43250000*5/21 59523.80 130952.37 119047.63250000*4/21 47619.04 178571.41 71428.59250000*3/21 35714.28 214285.69 3571.31250000*2/21 23809.52 238095.21 11904.76250000*1/21 11904.76 2,50,000 0
9. 9. Cost of asset = \$100,000Estimated residual value = \$10,000Estimated useful life of asset = 5 yearsDepreciation rate = (1/useful life) x 200%= 1/5 x 200% = 20% x 2 = 40%Double Declining BalanceMethod
10. 10. Cost Depreciation Accumulated BookDep value100000*40% 40000 40000 6000060,000*40% 24000 64000 3600036000*40% 14400 78400 2160021600*40% 8640 87040 1296012960*40% 2960(*) 90000 10000(*)Depreciation stops when book value=Residual value | 902 | 2,192 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2016-40 | latest | en | 0.580427 |
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# Naim
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Mapping a plane onto a 3D matrix?
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extremely noise image, what are some techniques I could use to de-noise?
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How to best connect weak edges?
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Image segmentation: I want to get rid of ugly "branches" of a binary image
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How to check if all pixels in a large neighborhood (such as a square or triangle) are all non-zero values?
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Starting with a grayscale image, how can I colour a certain pixel based on its x,y coordinates?
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how to use a binary mask as a "stencil" for a grayscale image?
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Is there a way to get a "thicker" bwconncomp connection?
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I want to dissect one black, oval-shaped part of a photo keep the contents inside
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This is the area of the stemNRICH site devoted to the core applied mathematics underlying the sciences. | 734 | 3,413 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2016-44 | longest | en | 0.828146 |
http://mathhelpforum.com/discrete-math/165704-prove-there-real-numbers-cannot-defined-english-phrases.html | 1,481,188,813,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542455.45/warc/CC-MAIN-20161202170902-00172-ip-10-31-129-80.ec2.internal.warc.gz | 179,273,224 | 13,436 | # Thread: prove that there are real numbers that cannot be defined by English phrases
1. ## prove that there are real numbers that cannot be defined by English phrases
how can I prove that there are real numbers that cannot be defined by English phrases?? I guess I have to use counterexample ???
2. Originally Posted by mathsohard
how can I prove that there are real numbers that cannot be defined by English phrases?? I guess I have to use counterexample ???
Let x be a real number that cannot be defined by an english phrase....but I just defined it by an english phrase!
Tonio
3. So there isn't any?? which means that i can't prove?
4. Well what do you mean by 'phrase'? Are 'one', 'two', 'three', 'four point six', 'pi squared', phrases?
I think you can cover them all, but would be happy to be proved wrong.
I think the proof that there aren't any has been given by tonio.
5. OP, don't listen to them; you are being trolled.
Let x be a real number that cannot be defined by an english phrase....but I just defined it by an english phrase!
First, what proves that you defined a unique number? There may be many numbers that can't be defined in English.
Second, this is Richard's paradox (Richard was French, so presumably, "d" in his name is not pronounced and the stress is on the last syllable.) Unlike Tonio's suggestion, this paradox's statement constructs a unique real number that cannot be defined by any English phrase. But in this case, the statement actually defined it! Berry paradox is similar.
I am not a specialist on paradoxes, but the logic flaw here is that it is not well-defined which numbers are defined by which English phrases.
Well what do you mean by 'phrase'? Are 'one', 'two', 'three', 'four point six', 'pi squared', phrases?
Yes, these are phrases.
I think you can cover them all, but would be happy to be proved wrong.
I think the OP's instructor expects an answer based on cardinality. There can be at most countably many reals defined by English phrases provided each qualified phrase defines exactly one number, since there is at most countable many phrases.
6. Let x be a unique number that cannot be defined by an english phrase then.
Sorry to be glib, but there is no mention of cardinality or countability in the OP's question.
Maybe Godel could help.
7. I presumed this was going to be about computable numbers; surely it is related though!
Turing proved that not all numbers were computable by a Turing Machine. Now, if a number if definable by an English phrase, it can't be that hard to encode it into a Turing Machine (i.e. make a computer program to work it out).
8. Originally Posted by ark600
Let x be a unique number that cannot be defined by an english phrase then.
I assume the original problem refers to phrases that identify a unique real number. That is, the problem says: "Consider all English phrases that define a unique real number. Prove that not all real numbers are defined by these phrases." If we admit phrases that define (uncountable) sets of numbers, then all real numbers can be covered.
So, why is there a unique number that is not defined by qualified English phrases?
Sorry to be glib, but there is no mention of cardinality or countability in the OP's question.
Problem statements often give just propositions to prove leaving the choice of proof method to the student.
Originally Posted by Swlabr
Turing proved that not all numbers were computable by a Turing Machine. Now, if a number if definable by an English phrase, it can't be that hard to encode it into a Turing Machine (i.e. make a computer program to work it out).
"A real number a is first-order definable in the language of set theory, without parameters, if there is a formula φ in the language of set theory, with one free variable, such that a is the unique real number such that φ(a) holds in the standard model of set theory" (Wikipedia). Not all definable numbers are computable. For example, let $P_0, P_1,\ldots$ be a computable enumeration of all programs with no argument. Consider a real number x whose binary expansion is $0.d_0d_1\ldots$ where $d_n=1$ if $P_n$ halts and $d_n=0$ otherwise. Then x is definable but not computable. One must note, though, that one can construct successive rational approximations to x that get arbitrarily close. The reason x is not computable is that in general one does not know whether the found approximation to $d_n$ is correct if this approximation is 0.
9. Originally Posted by emakarov
"A real number a is first-order definable in the language of set theory, without parameters, if there is a formula φ in the language of set theory, with one free variable, such that a is the unique real number such that φ(a) holds in the standard model of set theory" (Wikipedia). Not all definable numbers are computable. For example, let $P_0, P_1,\ldots$ be a computable enumeration of all programs with no argument. Consider a real number x whose binary expansion is $0.d_0d_1\ldots$ where $d_n=1$ if $P_n$ halts and $d_n=0$ otherwise. Then x is definable but not computable. One must note, though, that one can construct successive rational approximations to x that get arbitrarily close. The reason x is not computable is that in general one does not know whether the found approximation to $d_n$ is correct if this approximation is 0.
Sorry-I think I was trying to go the other way; that if it is computable then it is definable. If you can write it as a program, then take the program as your definition. But that doesn't answer the question...hmm...I think, actually, I just wasn't thinking straight....
10. Tonio's example, which I am sure he intended as a joke, is not valid because it does NOT define a specific real number- it defines a class of real numbers. In any phrase in English (or any other language) there must be an finite number of words, each consisting of a finite number of letters, of which there are 26 different possible letters. If we allow arbitrarily large words- though still a finite number of letters, or an arbitrarily large, though finite, number of words in a phrase, we will still have a countable number of possible English phrases. And the set of all real numbers is uncountable.
11. Originally Posted by HallsofIvy
Tonio's example, which I am sure he intended as a joke, is not valid because it does NOT define a specific real number- it defines a class of real numbers. In any phrase in English (or any other language) there must be an finite number of words, each consisting of a finite number of letters, of which there are 26 different possible letters. If we allow arbitrarily large words- though still a finite number of letters, or an arbitrarily large, though finite, number of words in a phrase, we will still have a countable number of possible English phrases. And the set of all real numbers is uncountable.
Well, I must say I didn't intend my answer to be a joke...completely. I tried to make the OP think of the (a) paradox as was later
mentioned somewhere else, and as somebody else also pointed out, even if we get into cardinalities stuff and, logically, deduce
there must be real numbers which cannot be defined by an english phrase, pinpointing any of these numbers must be as impossible
as pinpointing the smallest positive real number...
Tonio | 1,671 | 7,308 | {"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": 12, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2016-50 | longest | en | 0.961343 |
http://map4gems.centralesupelec.fr/physics/electronic/additional-lessons/the-characteristics-of-dipoles | 1,600,795,495,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400206329.28/warc/CC-MAIN-20200922161302-20200922191302-00663.warc.gz | 79,853,493 | 8,890 | Chapter 12
The characteristics of dipoles
Fondamental : Conventions generator and receptor
Conventions generator and receptor
If two dipoles are interconnected, conventions are necessarily receptor for one and generator for the other.
The arbitrary choice of conventions does not say much for the type of real operation (generator or receiver) of the dipole.
Fondamental : Characteristic of a dipole
Called dipole characteristic one of the following curves, giving either as a function of or as a function of .
Characteristic of a dipole
Fondamental : Characteristic ohmic conductors, Ohm's law
is the conductor resistance (expressed in ohm, )
Or :
Where is the conductance of the conductor (expressed in Siemens, ).
Ohm's Law
Associations of ohmic conductors :
• In series :
• In parallel (in derivation) :
Is :
JAVA animation (of JJ.Rousseau, University of Le Mans) on the color code of resistance :
Fondamental : Characteristics of generator (linear active dipole)
The characteristic of a generator (generator convention) is given in the following figure.
It is a linear active dipole, whose characteristic can be modeled as :
With :
• : open circuit voltage ( ), measured with a voltmeter.
• : Short circuit current ( ), measured with an ammeter.
Characteristics of generator
We denote :
So :
The linear active dipole is thus equivalent to the following two elements in series :
• An ideal emf generator voltage denoted (equal to open circuit voltage at the terminals of the dipole).
• Ohmic conductor resistance (internal resistance of the active dipole).
This modeling of the active dipole is called "Thevenin's modeling"
Thevenin's model
Norton's modeling :
The linear active dipole is equivalent to the following two elements placed in parallel :
• An ideal current generator of electromotive current (equal to the short-circuit current of the active dipole),
in parallel with :
• ohmic conductor resistance (internal resistance of the active dipole).
This modeling of the active dipole is called "Norton's modeling".
Norton's model
One passes from the representation of the Thevenin to Norton using the following relationships :
Fondamental : Fundamental associations linear active dipole
Series (choice of Thevenin model) : | 487 | 2,275 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2020-40 | latest | en | 0.849698 |
https://en.wikipedia.org/wiki/Schwarzian_transform | 1,493,282,631,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917121893.62/warc/CC-MAIN-20170423031201-00433-ip-10-145-167-34.ec2.internal.warc.gz | 794,361,340 | 14,662 | # Schwartzian transform
(Redirected from Schwarzian transform)
In computer programming, the Schwartzian transform is technique used to improve the efficiency of sorting a list of items. This idiom[1] is appropriate for comparison-based sorting when the ordering is actually based on the ordering of a certain property (the key) of the elements, where computing that property is an intensive operation that should be performed a minimal number of times. The Schwartzian transform is notable in that it does not use named temporary arrays.
The Schwartzian transform is a version of a Lisp idiom known as decorate-sort-undecorate, which avoids recomputing the sort keys by temporarily associating them with the input items. This approach is similar to memoization, which avoids repeating the calculation of the key corresponding to a specific input value. By comparison, this idiom assures that each input item's key is calculated exactly once, which may still result in repeating some calculations if the input data contains duplicate items.
The idiom is named after Randal L. Schwartz, who first demonstrated it in Perl shortly after the release of Perl 5 in 1994. The term "Schwartzian transform" applied solely to Perl programming for a number of years, but it has later been adopted by some users of other languages, such as Python, to refer to similar idioms in those languages. However, the algorithm was already in use in other languages (under no specific name) before it was popularized among the Perl community in the form of that particular idiom by Schwartz. The term "Schwartzian transform" indicates a specific idiom, and not the algorithm in general.
For example, to sort the word list ("aaaa","a","aa") according to word length: first build the list (["aaaa",4],["a",1],["aa",2]), then sort it according to the numeric values getting (["a",1],["aa",2],["aaaa",4]), then strip off the numbers and you get ("a","aa","aaaa"). That was the algorithm in general, so it does not count as a transform. To make it a true Schwartzian transform, it would be done in Perl like this:
```@sorted = map { \$_->[0] }
sort { \$a->[1] <=> \$b->[1] } # use numeric comparison
map { [\$_, length(\$_)] } # calculate the length of the string
@unsorted;
```
## The Perl idiom
The general form of the Schwartzian Transform is:
```@sorted = map { \$_->[0] }
sort { \$a->[1] cmp \$b->[1] }
map { [\$_, foo(\$_)] }
@unsorted;
```
Where `foo(\$_)` represents an expression that takes `\$_` (each item of the list in turn) and produces the corresponding value that is to be compared in its stead.
Reading from right to left (or from the bottom to the top):
• the original list `@unsorted` is fed into a `map` operation that wraps each item into a (reference to an anonymous 2-element) array consisting of itself and the calculated value that will determine its sort order (list of item becomes a list of [item, value]);
• then the list of lists produced by `map` is fed into `sort`, which sorts it according to the values previously calculated (list of [item, value] ⇒ sorted list of [item, value]);
• finally, another `map` operation unwraps the values (from the anonymous array) used for the sorting, producing the items of the original list in the sorted order (sorted list of [item, value] ⇒ sorted list of item).
The use of anonymous arrays ensures that memory will be reclaimed by the Perl garbage collector immediately after the sorting is done.
## Efficiency analysis
Without the Schwartzian transform, the sorting in the example above would be written in Perl like this:
```@sorted = sort { foo(\$a) cmp foo(\$b) } @unsorted;
```
While it is shorter to code, the naive approach here could be much less efficient if the key function (called foo in the example above) is expensive to compute. This is because the code inside the brackets is evaluated each time two elements need to be compared. An optimal comparison sort performs O(n log n) comparisons (where n is the length of the list), with 2 calls to foo every comparison, resulting in O(n log n) calls to foo. In comparison, using the Schwartzian transform, we only make 1 call to foo per element, at the beginning map stage, for a total of n calls to foo.
However, if the function foo is relatively simple, then the extra overhead of the Schwartzian transform may be unwarranted.
## Example
For example, to sort a list of files by their modification times, a naive approach might be as follows:
``` function naiveCompare(file a, file b) {
return modificationTime(a) < modificationTime(b)
}
// Assume that sort(list, comparisonPredicate) sorts the given list using
// the comparisonPredicate to compare two elements.
sortedArray := sort(filesArray, naiveCompare)
```
Unless the modification times are memoized for each file, this method requires re-computing them every time a file is compared in the sort. Using the Schwartzian transform, the modification time is calculated only once per file.
A Schwartzian transform involves the functional idiom described above, which does not use temporary arrays.
The same algorithm can be written procedurally to better illustrate how it works, but this requires using temporary arrays, and is not a Schwartzian transform. The following example pseudo-code implements the algorithm in this way:
``` for each file in filesArray
insert array(file, modificationTime(file)) at end of transformedArray
function simpleCompare(array a, array b) {
return a[2] < b[2]
}
transformedArray := sort(transformedArray, simpleCompare)
for each file in transformedArray
insert file[1] at end of sortedArray
```
## History
The first known online appearance of the Schwartzian Transform is a December 16, 1994 posting by Randal Schwartz to a thread in comp.unix.shell, crossposted to comp.lang.perl. (The current version of the Perl Timeline is incorrect and refers to a later date in 1995.) The thread began with a question about how to sort a list of lines by their "last" word:
``` adjn:Joshua Ng
```
Schwartz responded with:
```#!/usr/bin/perl
require 5; # new features, new bugs!
print
map { \$_->[0] }
sort { \$a->[1] cmp \$b->[1] }
map { [\$_, /(\S+)\$/] }
<>;
```
This code produces the result:
``` admg:Mahalingam Gobieramanan
```
Schwartz noted in the post that he was "Speak[ing] with a lisp in Perl," a reference to the idiom's Lisp origins.
The term "Schwartzian Transform" itself was coined by Tom Christiansen in a follow-up reply. Later posts by Christiansen made it clear that he had not intended to name the construct, but merely to refer to it from the original post: his attempt to finally name it "The Black Transform" did not take hold ("Black" here being a pun on "schwar[t]z", which means black in German).
## Comparison to other languages
Some other languages provide a convenient interface to the same optimization as the Schwartzian transform:
• In Python 2.4 and above, both the sorted() function and the in-place list.sort() method take a key= parameter that allows the user to provide a "key function" (like foo in the examples above). In Python 3 and above, use of the key function is the only way to specify a custom sort order (the previously-supported comparator argument was removed). Before Python 2.4, developers would use the lisp-originated Decorate-Sort-Undecorate (DSU) idiom,[2] usually by wrapping the objects in a (sortkey, object) tuple.
• In Ruby 1.8.6 and above, the Enumerable abstract class (which includes Arrays) contains a sort_by[3] method which allows you to specify the "key function" (like foo in the examples above) as a code block.
• In D 2 and above, the schwartzSort function is available. It might require less temporary data and be faster than the Perl idiom or the decorate-sort-undecorate idiom present in Python and Lisp. This is because sorting is done in-place and only minimal extra data (one array of transformed elements) is created.
• Racket's core `sort` function accepts a `#:key` keyword argument with a function that extracts a key, and an additional `#:cache-keys?` requests that the resulting values are cached during sorting. For example, a convenient way to shuffle a list is `(sort l < #:key (λ (_) (random)) #:cache-keys? #t)`.
• In PHP 5.3 and above the transform can be implemented by use of array_walk, e.g. to work around the limitations of the unstable sort algorithms in PHP.
```function spaceballs_sort(array& \$a) {
array_walk(\$a, function(&\$v, \$k){ \$v = array(\$v, \$k); });
asort(\$a);
array_walk(\$a, function(&\$v, \$_) { \$v = \$v[0]; });
}
```
• In Elixir, the Enum.sort_by/2 and Enum.sort_by/3 methods allow users to perform a Schwartzian transform for any module that implements the Enumerable protocol.
• In Perl 6, one needs to supply a comparator lambda that only takes 1 argument to perform a Schwartzian transform under the hood:
```@a.sort( { \$^a.Str } )
```
would sort on the string representation using a Schwartzian transform,
```@a.sort( { \$^a.Str cmp \$^b.Str } )
```
would do the same converting the elements to compare just before each comparison.
## References
1. ^ Martelli, Alex; Ascher, David, eds. (2002). "2.3 Sorting While Guaranteeing Sort Stability". Python Cookbook. O'Reilly & Associates. p. 43. ISBN 0-596-00167-3. This idiom is also known as the 'Schwartzian transform', by analogy with a related Perl idiom.
2. ^
3. ^ "Ruby-doc Core-API Classes". Retrieved 14 September 2011. | 2,228 | 9,428 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2017-17 | latest | en | 0.923398 |
https://docs.juliahub.com/General/ACSets/stable/generated/ADTs/ | 1,726,201,879,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651507.67/warc/CC-MAIN-20240913034233-20240913064233-00249.warc.gz | 190,604,080 | 4,205 | # Specifying acsets using Algebraic Data Types
ACSets are an extremely flexible data representation that can store anything you can put in a database. But in order to construct them, you might want something that feels more like a custom programming language. The Graphviz software comes with a custom language called dot files for specifying the data of a graph that Graphviz will draw. In order to make implementing these linguisting interfaces easier, ACSets.jl supports an Algebraic Data Types approach to specification of ACSets.
using ACSets, ACSets.ADTs
using MLStyle
using Test
import ACSets.ADTs: symb2string
Our schema will be labeled graphs. These are graphs with vertex labels.
SchLabeledGraph = BasicSchema([:E,:V], [(:src,:E,:V),(:tgt,:E,:V)],
[:L], [(:label,:V,:L)])
@acset_type LabeledGraph(SchLabeledGraph, index=[:src,:tgt])
Main.LabeledGraph
The basic principle is a nested expression syntax for specifying the ACSet.
s = Statement(:E, [Value(2), Value(3)])
E(2,3)
You can extract information from an expression with pattern matching from MLStyle.jl
get_table(s) = @match s begin
Statement(t, e) => t
_ => nothing
end
get_arg1(s) = @match s begin
Statement(t, e) => e[1]
end
@test get_table(s) == :E
@test get_arg1(s) == Value(2)
Test Passed
These statements can be grouped together into a list an tagged with the type of ACSet you want to make.
gspec = ACSetSpec(
:(LabeledGraph{Symbol}),
[
Statement(:V, [Kwarg(:label, Value(:a))])
Statement(:V, [Kwarg(:label, Value(:b))])
Statement(:V, [Kwarg(:label, Value(:c))])
Statement(:E, [Value(1), Value(3)])
Statement(:E, [Value(2), Value(3)])
]
)
LabeledGraph{Symbol} begin
V(label=a)
V(label=b)
V(label=c)
E(1,3)
E(2,3)
end
These expressions can be serialized as strings:
sprint(show, gspec)
"LabeledGraph{Symbol} begin \n V(label=a)\n V(label=b)\n V(label=c)\n E(1,3)\n E(2,3)\n end"
Or as Julia code:
generate_expr(gspec)
quote
X = LabeledGraph{Symbol}()
end
From the ACSetSpec you can construct the ACSet that it specifies.
gspec = ACSetSpec(
:(LabeledGraph{Symbol}),
[
Statement(:V, [Kwarg(:label, Value(:a))])
Statement(:V, [Kwarg(:label, Value(:b))])
Statement(:V, [Kwarg(:label, Value(:c))])
Statement(:E, [Kwarg(:src, Value(1)), Kwarg(:tgt, Value(3))])
Statement(:E, [Kwarg(:src, Value(2)), Kwarg(:tgt, Value(3))])
]
)
g = construct(LabeledGraph{Symbol}, gspec)
E src tgt
1 1 3
2 2 3
V label
1 a
2 b
3 c
There is an embedding of ACSetSpec into Expr:
hspec = acsetspec(:(LabeledGraph{Symbol}),quote
V(label=a)
V(label=b)
V(label=c)
E(src=1,tgt=3)
E(src=2,tgt=3)
end
)
LabeledGraph{Symbol} begin
V(label=a)
V(label=b)
V(label=c)
E(src=1,tgt=3)
E(src=2,tgt=3)
end
The acsetspec function is a good example of embedding your custom language into Julia syntax That save you the trouble of writing your own lexer and parser for your custom language.
construct(LabeledGraph{Symbol}, hspec) == construct(LabeledGraph{Symbol}, gspec)
true
You can export your specification to a dictionary and put that dictionary into a JSON document this gives you a nice way of serializing the ACSet that is machine readable and row oriented. The ACSet serialization is by column oriented which might be inconvenient for your consumers.
to_dict(gspec)
Dict{Symbol, Any} with 2 entries:
:type => "LabeledGraph{Symbol}"
:data => Dict{Symbol, Any}[Dict(:table=>:V, :fields=>Dict(:label=>:a)), Dict(… | 993 | 3,377 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-38 | latest | en | 0.659272 |
https://findwords.info/term/formal%20system | 1,623,686,223,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487612537.23/warc/CC-MAIN-20210614135913-20210614165913-00243.warc.gz | 248,196,237 | 5,226 | Find the word definition
Wiktionary
formal system
n. (context logic English) The grouping of a formal language and a set of inference rules and/or axioms.
Wikipedia
Formal system
A formal system is broadly defined as any well-defined system of abstract thought based on the model of mathematics. The entailment of the system by its logical foundation is what distinguishes a formal system from others which may have some basis in an abstract model. Often the formal system will be the basis for or even identified with a larger theory or field (e.g. Euclidean geometry) consistent with the usage in modern mathematics such as model theory. A formal system need not be mathematical as such; for example, Spinoza's Ethics imitates the form of Euclid's Elements.
Each formal system has a formal language, which is composed by primitive symbols. These symbols act on certain rules of formation and are developed by inference from a set of axioms. The system thus consists of any number of formulas built up through finite combinations of the primitive symbols—combinations that are formed from the axioms in accordance with the stated rules.
Formal systems in mathematics consist of the following elements:
1. A finite set of symbols (i.e. the alphabet), that can be used for constructing formulas (i.e. finite strings of symbols).
2. A grammar, which tells how well-formed formulas (abbreviated wff) are constructed out of the symbols in the alphabet. It is usually required that there be a decision procedure for deciding whether a formula is well formed or not.
3. A set of axioms or axiom schemata: each axiom must be a wff.
4. A set of inference rules.
A formal system is said to be recursive (i.e. effective) or recursively enumerable if the set of axioms and the set of inference rules are decidable sets or semidecidable sets, respectively.
Some theorists use the term formalism as a rough synonym for formal system, but the term is also used to refer to a particular style of notation, for example, Paul Dirac's bra–ket notation.
Usage examples of "formal system".
Using Omar Narayama's universal syntax, the holists of the Order constructed a formal system, or science, that treated the individual mind as sub-programs of a universal algorithm.
There has to be a formal system of rules, according to which the numbers are combined.
They were almost too logical to make good analysts (as far as Avram knew, no Gorm in recorded history had ever played a hunch), and their lack of any formal system of permanent naval or military ranks sometimes confused their imperial partners .
They were almost too logical to make good analysts (as far as Avram knew, no Gorm in recorded history had ever played a hunch), and their lack of any formal system of permanent naval or military ranks sometimes confused their imperial partners.
But each formal system, logically consistent internally, describes a possible universe, which therefore exists.
The universe, this universe, is described-umm, that's the wrong word-by a formal system.
It was a peace enforced by William's own decrees, not by a formal system of law and justice.
Perhaps heand Troi could devise a more formal system for the othersand she could then take charge of it.
When ve'd first met Radiya in the Mines, Yatima had asked ver why some non-sentient program couldn't just take each formal system used by the miners and crank out all its theorems automatically sparing citizens the effort.
Though no more eerie than many of the other oddities of this place, the lack of a formal system of beliefs gave Collins goose bumps.
His own country had no more formal system for succession than did Germany. | 768 | 3,679 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2021-25 | longest | en | 0.939107 |
https://engineerexcel.com/calculating-the-integral-of-an-equation-in-excel-with-vba/ | 1,726,303,831,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651559.58/warc/CC-MAIN-20240914061427-20240914091427-00563.warc.gz | 214,773,489 | 106,685 | # Calculating the Integral of an Equation in Excel with VBA
If the function you want to integrate is known in terms of a variable, you can perform the integration using VBA instead of the tabular method in the last section.
Let’s say we have a velocity function:
Although you could integrate this equation using a table in the worksheet, VBA will allow you to obtain a more accurate estimate by using a smaller time step (smaller slices).
Open up the VBA editor (Alt-F11) and insert a new module in the worksheet by right-clicking in the Project window and choosing Insert > Module.
This module will contain two functions. The first will be a function for the velocity from the equation above. Create a function named vel with one argument, t. Then enter in the equation above:
Function vel(t)
vel = 24 * t – 1.2 * t ^ 2
End Function
The second function will calculate position by integrating velocity. This function is going to have three arguments: an initial position (x0), a starting time (t1), and an ending time (t2). For complex functions like this, it’s helpful to create a skeleton of what the function will do using comments. Create the position function and the skeleton:
Function Position(x0, t1, t2)
‘define the range of the integral
‘discretize the integral into “n” slices, “dt” wide
‘initialize variables
‘calculate areas using the trapezoidal rule
‘sum the area under the curve for each slice
End Function
The last two steps, calculating the areas and summing them, will be done simultaneously using a FOR loop.
With the skeleton complete, you can code each step. For the first step, defining the range of the integral, we’ll create a variable int_range which is the difference between time t2 and time t1:
‘define the range of the integral
int_range = t2 – t1
In the second step, you’ll slice this range up into a certain number of slices. The number of slices will be defined as n. An n of 1000 should give good accuracy. The time step, dt, is the width of one of those slices, so it will equal the integral range divided by the number of slices.
‘discretize the integral into “n” slices, “dt” wide
n = 1000
dt = int_range / n
Before estimating the area of a slice using the trapezoidal rule, you’ll need to set some initial conditions.
If you recall, the trapezoidal equation is evaluated from a to b. Since this example has time on the x axis, we’ll call the time at the beginning of the slice “ta” and set it equal to our initial start time (t1). We’ll call the time at the end of the slice tb, and set it equal to ta + dt.
‘initialize variables
ta = t1
tb = ta + dt
While t1 is a fixed point in time, ta and tb will change as the function moves from one slice to the next.
There’s one more variable to initialize – the position. Initialize the position by setting it to the argument x0 that is passed into the function:
Position = x0
With the initialization done, you can set up a FOR loop that will go through the slices and calculate the area of each one. It will also sum up the total area of all the slices as it goes.
The FOR loop is going to run once for each slice, and there are n slices, so we’ll set this up For j = 1 to n. Then the position will be calculated from the trapezoidal rule using the vel function we defined above to calculate the height at the start and end of the slice:
‘calculate areas using the trapezoidal rule
‘sum the area under the curve for each slice
For j = 1 To n
Position = Position + (tb – ta) * (vel(ta) + vel(tb)) / 2
In order to move to the next slice, we’ll change the values of ta and tb. The new ta will be equal to the old tb, and tb will increment by dt:
ta = tb
tb = ta + dt
Next
The Next statement closes out the loop.
When you’re all done, your code should look like this:
Function vel(t)
vel = 24 * t – 1.2 * t ^ 2
End Function
__________________________________________________________________________
Function Position(x0, t1, t2)
‘define the range of the integral
int_range = t2 – t1
‘discretize the integral into “n” slices, “dt” wide
n = 1000
dt = int_range / n
‘initialize variables
ta = t1
tb = ta + dt
Position = x0
‘calculate areas using the trapezoidal rule
‘sum the area under the curve for each slice
For j = 1 To n
Position = Position + (tb – ta)* (vel(ta) + vel(tb)) / 2
ta = tb
tb = ta + dt
Next
End Function
Now that the Position function is complete, you can use it in the spreadsheet to do calculations. First, use it to calculate the position at 20 seconds. Remember this function needs three arguments passed to it: x0, t1 and t2. Type in the function name, then use Ctrl-Shift-A to automatically populate the names of the arguments. The first argument, x0, is the initial position in cell C5; t1 is the lower bound in cell C6, and t2 is the upper bound in cell C7.
=Position(C5,C6,C7)
The Position function will give the result 1600 feet after 20 seconds.
Of course, the function can also be used to calculate the position at many different times. The worksheet already contains time and velocity data. For the position column, the arguments will be entered in a slightly different way. The initial position and the initial time are the same as above. You can begin the function:
=Position(C5,C6,
Both of these arguments need to be made absolute references using F4 because they’ll be the same for every time point in the table. However, the t2 argument will change as you go down the table, so you can simply click within cell B17 and leave that as a relative reference.
=Position(\$C\$5,\$C\$6,B17)
Select this cell and double-click the fill handle to fill the formula down the column. This will give you the same end position of 1600 feet, but it also gives all the position and velocity data points in between. You can plot these together. Select all three columns and insert a scatter chart. The velocity data are much smaller numbers than the position, so plot those on a secondary axis. Right-click the velocity data and choose Format Data Series. Select Plot Series On Secondary Axis.
You can see that when the velocity is near zero, the slope of the position curve is flat. The position increases more rapidly as velocity increases, and eventually gains more slowly as the velocity begins to drop again. Finally, when the velocity is zero again, the position stops increasing.
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# Convergence of FindMinimum
Posted 11 years ago
I wondered if FindMinimum always converges to the nearest minimum, so I minimized sin(x) cos(y) with random starting points with x and y between -6 and 6 and colored the points by the location of the minimum.
10 Replies
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Posted 11 years ago
The behavior of FindMinimum depends on the Method and on values of its sub-options.Please post the code you used to produce the graphics.
Posted 11 years ago
This is kinda similar to what you see with Newton Fractals. The answer is no. It won't always converge to the nearest root. For example, plotting the basins of attraction for newton's method produces pretty fractals.
Posted 11 years ago
Here's the code:n = 10^4; SeedRandom[0]; rs = RandomReal[{-6, 6}, {n, 2}]; Off[ FindArgMin::lstol]; resn = FindArgMin[Sin[x] Cos[y], Thread[{{x, y} , #}], Method -> "Newton"] & /@ rs; c[{x_, y_}] = RGBColor[(x + 6)/12, (y + 6)/12, 0]; Show[ Graphics @ Table[{c[resn[[i]]], Point[rs[[i]]]}, {i, n}]]
Posted 11 years ago
Thanks for the code, it is interesting to compare different methods. For example, if one changes the default "StepControl" -> "LineSearch" to "StepControl" -> "TrustRegion", the plot becomes well-ordered:n = 10^4; SeedRandom[0];rs = RandomReal[{-6, 6}, {n, 2}];Off[FindArgMin::lstol];resn = FindArgMin[Sin[x] Cos[y], Thread[{{x, y}, #}], Method -> {"Newton", "StepControl" -> "TrustRegion"}] & /@ rs;c[{x_, y_}] = RGBColor[(x + 6)/12, (y + 6)/12, 0]; Show[Graphics@Table[{c[resn[[i]]], Point[rs[[i]]]}, {i, n}]]From the other side, setting bigger value of "StartingScaledStepSize" suboption gives lesser ordered result as expected: n = 10^4; SeedRandom[0]; rs = RandomReal[{-6, 6}, {n, 2}]; Off[FindArgMin::lstol]; resn = FindArgMin[Sin[x] Cos[y], Thread[{{x, y}, #}], Method -> {"Newton", "StepControl" -> {"TrustRegion", "StartingScaledStepSize" -> 10}}] & /@ rs; c[{x_, y_}] = RGBColor[(x + 6)/12, (y + 6)/12, 0]; Show[ Graphics@Table[{c[resn[[i]]], Point[rs[[i]]]}, {i, n}]]There also are other suboptions of the "TrustRegion" method one can play with.And playing with the "StepControl" -> "LineSearch" suboptions allows to get well-ordered result too: n = 10^4; SeedRandom[0]; rs = RandomReal[{-6, 6}, {n, 2}]; Off[FindArgMin::lstol]; resn = FindArgMin[Sin[x] Cos[y], Thread[{{x, y}, #}], Method -> {"Newton", "StepControl" -> {"LineSearch", "MaxRelativeStepSize" -> .1}}] & /@ rs; c[{x_, y_}] = RGBColor[(x + 6)/12, (y + 6)/12, 0]; Show[ Graphics@Table[{c[resn[[i]]], Point[rs[[i]]]}, {i, n}]]
Posted 11 years ago
Thanks for pointing that out Alexey. I have looked at the Unconstrained Optimization tutorial, but didn't go to the one on Newton's method. Personally, I think that the Options function in Mathematica should be made more powerful so you can somehow see Options on Options. Perhaps some sort of tree structure.
Posted 11 years ago
StepControl -> TrustRegion doesn't seem to make any difference for constrained problems solved using the InteriorPoint method. I get the same result with and without that option setFindArgMin[{Sin[x y], x^2 + y^2 == 1}, Thread[{{x, y} , #}], Method -> {"InteriorPoint", "StepControl" -> "TrustRegion"}] & /@ rs
Posted 11 years ago
Frank, I cannot find in the Documentation any mention of the "StepControl" suboption for the "InteriorPoint" method. Please provide a link if you have.Brief search give me only this: The Method Options of "InteriorPoint".
Posted 11 years ago
Alexey,I didn't get an error message when I added the option, so I assumed it was a valid option.Perhaps the absence of an error message does not mean the option is available.
Posted 11 years ago
Frank,Yes, unfortunately the absence of an error message does not necessarily mean that the option is available. Only presence of an error message should be considered as a prove that the option is not supported, but even for this rule may be rare exceptions. The official politics of the WRI is that only documented behavior is supported. At the same time, the Documentation is very sparse and it sometimes takes hours even for an experienced user to find an "obvious" option hidden deeply in the Documentation. This thread is an illustration: how much time you would spend finding all that information having only the Documentation? I am certain that a newbie has no chance! Note also how simple and obvious looks the answer: it is hard to believe that such "obvious" things are really deeply hidden and in practice are unavailable even for experienced users...
Posted 11 years ago
I get basically the same plot for the constrained problem if I use a LibraryLink interface to Ipopt ( https://projects.coin-or.org/Ipopt )but in 1/4 the time. | 1,372 | 4,823 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2024-38 | latest | en | 0.655209 |
http://clay6.com/qa/50705/if-the-speed-of-a-truck-is-reduced-to-one-third-of-its-original-value-the-m | 1,516,515,945,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890314.60/warc/CC-MAIN-20180121060717-20180121080717-00660.warc.gz | 69,497,604 | 27,649 | # If the speed of a truck is reduced to one third of its original value, the minimum distance required to stop will be
$\begin{array}{1 1}(A)\;same\;as\;before \\(B)\;\frac{1}{3} of\; its\;original\;value\\(C)\;\frac{1}{9} of\;its\;original\;value\\(D)\;\frac{2}{3}\;of\;its\;original\;value \end{array}$
$s \alpha u^2$ as $0= u^2 -2as$
$\qquad= > u^2=2as$
Hence C is the correct answer.
answered Jul 17, 2014 by
edited Aug 13, 2014 | 172 | 434 | {"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} | 3.359375 | 3 | CC-MAIN-2018-05 | longest | en | 0.662324 |
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# Jonathon Chang
Last seen: circa un anno fa Attivo dal 2020
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Find the missing numbers.
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Risolto
generate number in particular way
A = [1 5 2 7]; MAX = 10; generate a array Y = [1 2 2 2 2 2 3 3 4 4]; i.e. total eleme...
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Risolto
Sum of series VII
What is the sum of the following sequence: Σ(km^k)/(k+m)! for k=1...n for different n and m?
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Risolto
Sum of series V
What is the sum of the following sequence: Σk(k+1) for k=1...n for different n?
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Risolto
Sum of series III
What is the sum of the following sequence: Σ(2k-1)^3 for k=1...n for different n?
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Risolto
Sum of series II
What is the sum of the following sequence: Σ(2k-1)^2 for k=1...n for different n?
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Risolto
Sum of series I
What is the sum of the following sequence: Σ(2k-1) for k=1...n for different n?
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Risolto
Sum of series IV
What is the sum of the following sequence: Σ(-1)^(k+1) (2k-1)^2 for k=1...n for different n?
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Risolto
Sum of series VI
What is the sum of the following sequence: Σk⋅k! for k=1...n for different n?
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Risolto
Square root of number
Square root of given number.
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Risolto
Perfect Square or not
find Given input x is perfect square or not,if yes then output y=1.else y=0
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Risolto
multiply by three
Given the variable x as your input, multiply it by 3 and put the result equal to y. Examples: Input x = 2 Output y is ...
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Risolto
Reverse Run-Length Encoder
Given a "counting sequence" vector x, construct the original sequence y. A counting sequence is formed by "counting" the entrie...
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Risolto
Quote Doubler
Given a string s1, find all occurrences of the single quote character and replace them with two occurrences of the single quote ...
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Risolto
Too mean-spirited
Find the mean of each consecutive pair of numbers in the input row vector. For example, x=[1 2 3] ----> y = [1.5 2.5] x=[1...
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Risolto
Divide by 4
Given the variable x as your input, divide it by 4 and put the result in y.
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Risolto
Divide by 4
Given the variable x as your input, divide it by four and put the result in y.
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Risolto
matrix rows and columns
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Create an m x n array consisting only of an input value.
Create an array with m rows and n columns wherein all entries are assigned the input value x.
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Replace Nan!
Replace Nan in the given vector(v) with 9999.
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Find the efficiency
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Diagonal Prod
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F.R.I.E.N.D.S
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F.R.I.E.N.D.S
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Remnant
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Risolto
Sort accordingly
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Swap
circa 2 anni fa | 1,054 | 3,256 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2022-49 | latest | en | 0.797925 |
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Physics Problems
Physics Problems is an excellent online study resource for students. It is an exhaustive collection of solved examples and answers for physics problems and questions.
Students can get online tutors who can help them understand the concepts and the logic behind the answers for Physics Questions. They can also get help with their homework problems in physics and other assignment questions related to Physics. The online physics tutors are expert in their subjects who can help the students in solving all their queries. One can also get physics homework help from our tutors.
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Physics Problems and Solutions
Physics questions and answers are an entire list of solved problems and questions about physics concepts. The students can get the list of frequently asked questions of various physics topics. The following are some physics topics which are covered under the physics problems:
Physics Classroom Examples
Here are some physics classroom examples for students:
Solved Examples
Question 1: A car moving with a velocity of 18 kmph is brought to rest in a distance of 0.5 km by applying brakes. Find its retardation?
Solution:
Initial Velocity of the car = u = 18 kmph = 5 ms-1,
Final Velocity of the car = v = 0,
Distance travelled = S = 0.5 km = 500m
We have, v2 = u2 + 2as.
0 = 52 + 2
×
a × 500
= 25 + 1000a.
-25 = 1000a,
a = -0.025 ms-2.
∴
Retardation of the car = 0.025ms-2.
Question 2: A Heat engine operates with reservoir temperature between 900 K and 500 K. Calculate its efficiency?
Solution:
T1 = 900 K,
T2 = 500 K,
Efficiency
η
= 1 – T2
### Needs help with similar assignment?
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https://se.mathworks.com/matlabcentral/cody/problems/1790-01-scalar-variables/solutions/1566414 | 1,610,867,632,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703509973.34/warc/CC-MAIN-20210117051021-20210117081021-00771.warc.gz | 565,194,592 | 16,833 | Cody
# Problem 1790. 01 - Scalar variables
Solution 1566414
Submitted on 21 Jun 2018 by Ronak Patel
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 = 10; b = 2.5*10^23; c = exp(2*pi/3); [au bu cu] = ScalarVars; assert(isequal(a,au)); assert(isequal(b,bu)); assert(isequal(c,cu));
c = 8.1205
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 162 | 544 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2021-04 | latest | en | 0.705403 |
https://dmitrov-vodokanal.ru/bitcoin-casinonaxe/poker-flop-river-turn-order-jum.php | 1,606,992,515,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141727627.70/warc/CC-MAIN-20201203094119-20201203124119-00218.warc.gz | 265,682,127 | 6,320 | # Poker flop river turn order
Limit Hold'em: The Turn The turn is one of the most difficult streets to play in Limit Hold'em, mainly because any bets you now make must be twice the size they were before and after the flop. This means that mistakes at this point can be costly. The Flop, The Turn, The River Circle : The Texas_Holdem The Texas_Holdem_Poker Store > Texas Holdem Designs! > The Flop, The Turn, The River Circle. The Flop, The Turn, The River;Gotta be Texas Holdem! Great games deserve great gear and this Design is perfect for all Poker parties and Casino nights! Rules Question - Extra Card on the Flop | Poker Chip Forum Apr 15, 2019 · The Poker Room. Home Game General Rules Question - Extra Card on the Flop and zero impact on the intended turn and river cards). ... rather than reshuffle known cards back into the stub and re-dealing the flop. And even if the order of the exposed cards is in question, then randomly selecting one of the four exposed cards to be the turn's ... texas hold-em odds/probability? | Yahoo Answers Apr 07, 2010 · Hey, Im having trouble setting up a probability equation when it comes to poker. The main problem is accounting for the other players' unknown cards. So say I have 2 different cards, what would be the probability of getting a pair with either one of them during the flop, turn, and river? I started with (assuming an 8 person table) 38c6+37c6...+34c6 -14c6 and Im not gonna add in the probability
## Texas Hold 'em (or Hold'em, Holdem) is the most popular poker variant ... Then three community cards are dealt face up (the "Flop"), followed by a ... A fifth community card is dealt face up (the "River")and the the fourth and final betting round. .... decision here-on-out: If the players acting before your turn choose to " Check", ...
Poker Flop River Turn Order Wikipedia Texas - xacokehib.tk The river or river card is the final card dealt in a poker hand, to be followed by a final round of betting and, if necessary, a showdown. Em, the river is the fifth and last card to be dealt to the community card board, after the flop and turn. The Flop, Turn, and River Cards in Texas Hold'em - dummies The turn. After the flop betting round, another card is burned from the deck and a fourth community card is exposed. This card is known as the turn (sometimes fourth street ). All players still in the hand now have six cards to choose from to make their best five-card Poker hands. There is another round of betting and one more card yet to be exposed. How did the flop, turn and river get their names in Texas ...
### Ivey League the poker training site by Phil Ivey and Ivey Poker. Ivey League provides the best poker ... Flop, Turn, & River. In part 4, Cole explains the four ...
odds on flop, turn and river - Poker Theory - General Poker ... Let's say the pot is \\\0's on the flop you have the nuts, and also there is a flush draw possible could be made on turn! let's say both of you are h odds on flop, turn and river - Poker Theory - General Poker Theory Forum Poker Math - The Turn: Pot Odds for Fourth Street ... Although these are arguably the most important odds (as the flop is commonly seen as the most important street in Hold'em), playing profitable poker means making the correct decisions on all streets. A brilliant call on the flop is completely negated by misplaying the turn. Texas Holdem Flop Turn River - casinotopwinslotw.services Texas Holdem Flop Turn River. texas holdem flop turn river The River Texas Hold Em best ... Texas Holdem Losing On River texas holdem losing on river Here are our strategy tips for No Limit Texas Hold’em flop, turn and ...Improve your poker skills: Texas Holdem Turn Flop River. | 863 | 3,702 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-50 | latest | en | 0.931481 |
https://short-q.com/what-is-the-maximum-occupancy-per-square-foot/ | 1,685,804,336,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649293.44/warc/CC-MAIN-20230603133129-20230603163129-00784.warc.gz | 580,461,218 | 12,809 | # Short-Questions
Fast solutions for complex problems
## What is the maximum occupancy per square foot?
The IBC recommends for spaces with unconcentrated use of chairs and tables, such as a restaurant, that 15 square feet on that floor of the building be dedicated to each occupant. That means a 500 square foot restaurant might have a maximum occupancy of 33 people.
How do you calculate maximum occupancy?
How to Calculate Maximum Occupancy Load. The occupancy load is calculated by dividing the area of a room by its prescribed unit of area per person. Units of area per person for specific buildings can be found in the chart at the end of this article.
### What is the minimum square footage for a apartment?
The IRC codes require that all homes must be built on a minimum of 320 square feet. The minimum square footage for a house is 120 square feet, and at least one room must be habitable.
What are occupancy standards?
Occupancy standards are rules about the number of people that can live in a bedroom or at a property. These standards are usually based on the number of bedrooms at a property, but they can also take into account the total amount of liveable square footage.
#### How many square feet is 100 guests?
SPACE REQUIREMENT ESTIMATES BY CROWD SIZE
CROWD SIZE, BY PERSON ESTIMATED SPACE NEEDS, SQUARE FEET
75 450-2,625
100 600-3,500
125 750-4,375
150 900-5,250
How do you calculate occupancy for an apartment?
You can calculate a single month’s physical occupancy by dividing the number of units available into the number of occupied units. Multiply by 100 to express as a percentage. Suppose you have a property with 75 units, and 69 are occupied. Divided out, this works out to 92 percent occupancy.
## How do you calculate occupancy per square foot?
Figure the area of the room, by multiplying the length by the width. For example, if your room is 50 feet long and 40 feet wide, the area is 2,000 square feet (50 x 40 = 2,000). If you measured the room in sections, add up the square feet of each section. Divide the square footage by 36.
Is 400 sq ft small?
It’s a good rule of thumb to remember that 400 square feet is about the size of a two-car garage. The rooms will still be quite small and limited depending on the kind of lifestyle you live.
### Is 800 sq ft big for an apartment?
In some places, 800-square feet is still considered small — but it’s one of the biggest you’ll find among the small apartments. Modest square footage and decorating restrictions don’t help when trying to decorate your 800-square-foot apartment.
Can a family of 6 live in a 2 bedroom apartment?
Standards Set by the Federal Government In general, the federal Department of Housing and Urban Development ‘s Fair Housing Act recommends an occupancy limit of two people per bedroom in rental units. So, the simplest answer to the question of how many people can live in one two-bedroom apartment is: usually four.
#### Can a family of 3 live in a 1 bedroom apartment in Tennessee?
In most cases, the rule “2 per bedroom plus 1” is used. This means that 3 people can legally live in a one bedroom apartment, and 2 people can live in a studio or efficiency apartment.
How do you calculate area per person?
How many square feet per person standing? Six square feet per person is a good rule of thumb for a standing crowd. If you are planning a cocktail hour for 100 people who will all be standing, you will multiply 100 by 6 to determine you need a venue with 600 square feet of available and workable space for the event.
## How big of an apartment do I need for one person?
150 square feet for the first occupant. 100 square feet for each additional resident. Every room occupied for sleeping purposes by one occupant must contain at least 70 square feet of floor space or at least 50 square feet per person if occupied by more than one person.
How big does an apartment have to be to be considered fair housing?
The unit must be at least 150 square feet for the first occupant. The unit must increase by 100 square feet for each additional occupant. Every room occupied for sleeping purposes by one occupant must contain at least 70 square feet of floor space — or at least 50 square feet per person if occupied by more than one person.
### What’s the average square footage of an apartment?
Conversely, today’s apartments average about 1,015 square feet, which is slightly smaller than the 1,117 square feet in 2011. This could be because more studios and one-bedroom apartments are being built and fewer two and three-bedroom units, according to CoStar data.
What’s the maximum occupancy for a 1 bedroom apartment?
So if you have a 1 bedroom apartment with a 100 square foot living room and a 70 square foot bedroom the law states that the maximum occupancy is two people. Any area of the rental that is not a kitchen, hallway or bathroom counts towards the occupancy standards. | 1,089 | 4,923 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-23 | latest | en | 0.92666 |
https://homework.cpm.org/category/CCI_CT/textbook/apcalc/chapter/3/lesson/3.2.3/problem/3-79 | 1,726,326,380,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651579.38/warc/CC-MAIN-20240914125424-20240914155424-00556.warc.gz | 278,221,169 | 16,000 | ### Home > APCALC > Chapter 3 > Lesson 3.2.3 > Problem3-79
3-79.
Calculate the average velocity between $0$ and $50$ seconds for each of the graphs below.
1. $\text{Average velocity} = \frac{\Delta \text{distance}}{\Delta \text{time}}$
1. This graph shows both forward and backwards motion. Will that affect average velocity? | 92 | 329 | {"found_math": true, "script_math_tex": 3, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2024-38 | latest | en | 0.806497 |
https://convertoctopus.com/1389-kilometers-per-hour-to-knots | 1,606,498,928,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141193856.40/warc/CC-MAIN-20201127161801-20201127191801-00634.warc.gz | 247,836,756 | 7,896 | Conversion formula
The conversion factor from kilometers per hour to knots is 0.53995680345662, which means that 1 kilometer per hour is equal to 0.53995680345662 knots:
1 km/h = 0.53995680345662 kt
To convert 1389 kilometers per hour into knots we have to multiply 1389 by the conversion factor in order to get the velocity amount from kilometers per hour to knots. We can also form a simple proportion to calculate the result:
1 km/h → 0.53995680345662 kt
1389 km/h → V(kt)
Solve the above proportion to obtain the velocity V in knots:
V(kt) = 1389 km/h × 0.53995680345662 kt
V(kt) = 750.00000000125 kt
The final result is:
1389 km/h → 750.00000000125 kt
We conclude that 1389 kilometers per hour is equivalent to 750.00000000125 knots:
1389 kilometers per hour = 750.00000000125 knots
Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 knot is equal to 0.0013333333333311 × 1389 kilometers per hour.
Another way is saying that 1389 kilometers per hour is equal to 1 ÷ 0.0013333333333311 knots.
Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that one thousand three hundred eighty-nine kilometers per hour is approximately seven hundred fifty knots:
1389 km/h ≅ 750 kt
An alternative is also that one knot is approximately zero point zero zero one times one thousand three hundred eighty-nine kilometers per hour.
Conversion table
kilometers per hour to knots chart
For quick reference purposes, below is the conversion table you can use to convert from kilometers per hour to knots
kilometers per hour (km/h) knots (kt)
1390 kilometers per hour 750.54 knots
1391 kilometers per hour 751.08 knots
1392 kilometers per hour 751.62 knots
1393 kilometers per hour 752.16 knots
1394 kilometers per hour 752.7 knots
1395 kilometers per hour 753.24 knots
1396 kilometers per hour 753.78 knots
1397 kilometers per hour 754.32 knots
1398 kilometers per hour 754.86 knots
1399 kilometers per hour 755.4 knots | 532 | 2,058 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2020-50 | latest | en | 0.679388 |
https://www.askiitians.com/forums/Algebra/22/30301/r.htm | 1,726,781,317,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652067.20/warc/CC-MAIN-20240919194038-20240919224038-00853.warc.gz | 596,485,945 | 43,101 | # Number of real roots of the equation esin x –e-sin x -4 = 0 is
Gurwinder Kaur
65 Points
13 years ago
esin x.esin x-4esin x-1=0
esin x=(4+2$\sqrt{5}. \,$ )/2 , (4-2$\sqrt{5}. \,$ )/2
=2+$\sqrt{5}. \,$ ,2-$\sqrt{5}. \,$
approve if u like....
31 Points
13 years ago
ans: zero
on solvin tha quadratic in e^(sinx) we obtain values 2+√5 and 2-√5 => sinx = ln(2+√5)/ln(2-√5)
ln(2+√5) > 1 and ln(2-√5)- is nt defined
hence der is no real root for given equation..... | 200 | 471 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 4, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2024-38 | latest | en | 0.496215 |
https://en.wikipedia.org/wiki/Lagrangian_and_Eulerian_specification_of_the_flow_field | 1,716,591,671,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058751.45/warc/CC-MAIN-20240524214158-20240525004158-00556.warc.gz | 209,288,505 | 18,560 | # Lagrangian and Eulerian specification of the flow field
In classical field theories, the Lagrangian specification of the flow field is a way of looking at fluid motion where the observer follows an individual fluid parcel as it moves through space and time.[1][2] Plotting the position of an individual parcel through time gives the pathline of the parcel. This can be visualized as sitting in a boat and drifting down a river.
The Eulerian specification of the flow field is a way of looking at fluid motion that focuses on specific locations in the space through which the fluid flows as time passes.[1][2] This can be visualized by sitting on the bank of a river and watching the water pass the fixed location.
The Lagrangian and Eulerian specifications of the flow field are sometimes loosely denoted as the Lagrangian and Eulerian frame of reference. However, in general both the Lagrangian and Eulerian specification of the flow field can be applied in any observer's frame of reference, and in any coordinate system used within the chosen frame of reference.
These specifications are reflected in computational fluid dynamics, where "Eulerian" simulations employ a fixed mesh while "Lagrangian" ones (such as meshfree simulations) feature simulation nodes that may move following the velocity field.
## Description
In the Eulerian specification of a field, the field is represented as a function of position x and time t. For example, the flow velocity is represented by a function
${\displaystyle \mathbf {u} \left(\mathbf {x} ,t\right).}$
On the other hand, in the Lagrangian specification, individual fluid parcels are followed through time. The fluid parcels are labelled by some (time-independent) vector field x0. (Often, x0 is chosen to be the position of the center of mass of the parcels at some initial time t0. It is chosen in this particular manner to account for the possible changes of the shape over time. Therefore the center of mass is a good parameterization of the flow velocity u of the parcel.)[1] In the Lagrangian description, the flow is described by a function
${\displaystyle \mathbf {X} \left(\mathbf {x} _{0},t\right),}$
giving the position of the particle labeled x0 at time t.
The two specifications are related as follows:[2]
${\displaystyle \mathbf {u} \left(\mathbf {X} (\mathbf {x} _{0},t),t\right)={\frac {\partial \mathbf {X} }{\partial t}}\left(\mathbf {x} _{0},t\right),}$
because both sides describe the velocity of the particle labeled x0 at time t.
Within a chosen coordinate system, x0 and x are referred to as the Lagrangian coordinates and Eulerian coordinates of the flow respectively.
## Material derivative
The Lagrangian and Eulerian specifications of the kinematics and dynamics of the flow field are related by the material derivative (also called the Lagrangian derivative, convective derivative, substantial derivative, or particle derivative).[1]
Suppose we have a flow field u, and we are also given a generic field with Eulerian specification F(xt). Now one might ask about the total rate of change of F experienced by a specific flow parcel. This can be computed as
${\displaystyle {\frac {\mathrm {D} \mathbf {F} }{\mathrm {D} t}}={\frac {\partial \mathbf {F} }{\partial t}}+\left(\mathbf {u} \cdot \nabla \right)\mathbf {F} ,}$
where ∇ denotes the nabla operator with respect to x, and the operator u⋅∇ is to be applied to each component of F. This tells us that the total rate of change of the function F as the fluid parcels moves through a flow field described by its Eulerian specification u is equal to the sum of the local rate of change and the convective rate of change of F. This is a consequence of the chain rule since we are differentiating the function F(X(x0t), t) with respect to t.
Conservation laws for a unit mass have a Lagrangian form, which together with mass conservation produce Eulerian conservation; on the contrary, when fluid particles can exchange a quantity (like energy or momentum), only Eulerian conservation laws exist.[3] | 933 | 4,039 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 4, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2024-22 | latest | en | 0.904183 |
https://socratic.org/questions/5886471f11ef6b1cc01bea53 | 1,575,802,354,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540508599.52/warc/CC-MAIN-20191208095535-20191208123535-00195.warc.gz | 544,248,532 | 6,085 | # A 2.94*L volume of gas at 294*K and under 69.6*kPa pressure enclosed in a piston, is cooled to 257*K and the pressure reduced to 35.6*kPa. What is the new volume?
Jan 23, 2017
${V}_{2} \cong 5 \cdot L$
$\frac{{P}_{1} {V}_{1}}{T} _ 1 = \frac{{P}_{2} {V}_{2}}{T} _ 2$, from the $\text{combined gas law}$.
And thus ${V}_{2} = \frac{{P}_{1} {V}_{1}}{T} _ 1 \times {T}_{2} / {P}_{2}$, which expression CLEARLY has the units of volume.
So V_2=(69.9*cancel(kPa)xx2.94*L)/(294*cancel(K))xx(257*cancel(K))/(35.6*cancel(kPa))=??L | 215 | 523 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 5, "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} | 3.671875 | 4 | CC-MAIN-2019-51 | longest | en | 0.703558 |
http://blog.mrmeyer.com/2017/desmos-design-why-were-suspicious-of-immediate-feedback/?replytocom=2432363 | 1,516,629,425,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891377.59/warc/CC-MAIN-20180122133636-20180122153636-00366.warc.gz | 44,957,027 | 23,088 | ## [Desmos Design] Why We’re Suspicious of Immediate Feedback
One of our design principles at Desmos is to “delay feedback for reflection, especially during concept development activities.” This makes us weird, frankly, in Silicon Valley where no one ever got fired for promising “immediate feedback” in their math edtech.
We get it. Computers have an enormous advantage over humans in their ability to quickly give students feedback on certain kinds of work. But just because computers can deliver immediate feedback doesn’t mean they always should.
For example, Simmons and Cope (1993) found that students were more likely to use procedural strategies like trial-and-error in a condition of immediate feedback than a condition of delayed feedback.
I think I can illustrate that for you with this activity, which has two tasks. You get immediate feedback on one and delayed feedback on the other.
How was your brain working differently in the “Circle” challenge [delayed feedback] than the “Parabola” challenge [immediate feedback]?
Exhibit A:
The circle one was both more challenging and fun. I found myself squinting on the circle to visualize it in my head while with the parabola I mindlessly did trial and error.
Exhibit B:
With the circle, the need to submit before seeing the effect made me really think about what each part of the equation would effect the graph in each way. This resulted in a more strategic first guess rather than a guess and check approach.
Exhibit C:
I could guess & check the parabola challenge. In the circle challenge I had to concentrate more about the center of the circle and the radius. Much more in fact.
Exhibit D:
I couldn’t use trial and error. I had to visualize and estimate and then make decisions. My brain was more satisfied after the circle.
Exhibit E:
I probably worked harder on [the circle] because my answer was not shown until I submitted my answer. It was more frustrating than the parabola problem – but I probably learned more.
This wasn’t unanimous, of course, but it was the prevailing sentiment. For most people, the feedback delay provoked thoughtfulness where the immediate feedback provoked trial-and-error.
We realize that the opposite of “immediate feedback” for many students is “feedback when my teacher returns my paper after a week.” Between those two options, we side with Silicon Valley’s preference for immediate feedback. But if computers can deliver feedback immediately, they can also deliver feedback almost immediately, after a short, productive delay. That’s the kind of feedback we design into our concept development activities.
BTW. For a longer version of that activity, check out Building Conic Sections, created by Dylan Kane and edited with love by our Teaching Faculty.
I'm Dan and this is my blog. I'm a former high school math teacher and current head of teaching at Desmos. More here.
1. #### Scott Farrar
February 21, 2017 - 4:55 pm -
It’s interesting that the type of feedback here is not a grade or a teacher comment or even separating right vs. wrong (explicitly). I feel a lot of research focuses on all those other aspects of feedback and not on what Shute calls “implicit feedback” such as from a manipulative, simulation, or the environment. I get implicit feedback about gravity when I drop something e.g.
The instant one feels like implicit feedback. The equation is tracking what you type and showing the consequence in real-time. (within the bounds of legal expressions)
The delayed question distances you from the consequence. Is this still implicit feedback? Is it more evaluative (discriminating right from wrong)? The result doesn’t say “wrong” though, it shows you the result of your equation. But the equation is no longer “living” in the same way. I remember the old QBASIC Gorillas game where you had to throw a banana at the other gorilla by giving an angle and velocity. Perhaps the Gorillas and this delayed circle are delayed-implicit.
It seems a danger of immediate-implicit feedback is that it makes hill-climbing too attractive. Hill climbing is a computer science metaphor https://en.wikipedia.org/wiki/Hill_climbing roughly meaning you start with some guess then try to improve that guess incrementally. The “top” of one hill may not be the best hill to be on top of, but the larger danger is that the strategy to hill climb may not be particularly attached to the underlying conceptual structures. It appears this is what you are avoiding: the student who knows they can improve the parabola by altering numbers but has no guarantee of deeper thought about those numbers.
Does delayed-implicit feedback make you more likely to see deeper structures? It seems we have anecdotal agreement that yes it does. The need to strategize may unlock human heuristics that might escape the negatives of hill-climbing.
I wonder what the immediate/delayed distinction means for strict evaluation of correct vs. incorrect. (Such as on Khan Academy). You are told immediately if your answer is wrong but nothing about “how” your answer is wrong. Does delaying the feedback still offer a potential positive effect? It seems there is quite a bit more research here. [May-Li will be writing up a blog post about this soon] Shute 2008 does a review of a lot of questions along these lines https://www.ets.org/Media/Research/pdf/RR-07-11.pdf but doesn’t dig into the implicit at all.
It seems that Desmos is exploring new distinctions within “implicit” and providing some hybrid with other types of feedback studied in other mediums… and I love it :)
2. #### Kenneth Tilton
February 21, 2017 - 6:08 pm -
Are we sure this is an “either-or” situation?
Our silicon solution offers two modes of checking, one in which the student’s work is checked after each step, another in which it is not checked until they submit an answer. And we let them choose, because the first is prolly best during concept formation (so they do not spin their wheels working a five step problem after unwittingly erring on the first step) and the latter is best when self-assessing to see if one is ready for an exam where indeed the feedback may not come for a week. And at any rate, students do need to become independent of the quick correction or they’ll never really form the concepts. That is why in our “levelling-up” mode we do not allow second chancesafter mistakes.
When I was a tutor I started at an even more extreme level of immediacy: I would stop them as they wrote down a step as soon as an error was evident. That is a tough call for software to make, and I myself like blazing thru a problem and then checking my work, so that might be a bridge too far for silicon.
The bottom line is that different timings of feedback do work differently, but different stages of readiness might demand exactly that.
3. #### Joshua
February 21, 2017 - 10:40 pm -
Marbleslides subtly employs both methods of feedback. When an adjustment was made to an equation, the picture would update immediately, but the student had to click a button to release the marbles and see if they achieved the goal.
I actually went a step farther and added parameters and sliders to the equations. This might make it more guess-and-checky, but I liked how the animation created a link between increasing/decreasing the parameters and transforming the graph.
In this case, I was introducing these equations and graphs to the students, so my main objective was to give them an experience of the “what:” what happens if I change this part of the equation? After that, we had some more discussion about “why.”
helfulp!
Like Kenneth wrote, I wonder if different feedback delays are more useful for different aspects/phases of learning: immediate feedback to gather data and form hypotheses, slightly delayed feedback for hypothesis checking and “why” exploration?
• #### Dan Meyer
February 22, 2017 - 12:01 pm -
I actually went a step farther and added parameters and sliders to the equations. This might make it more guess-and-checky, but I liked how the animation created a link between increasing/decreasing the parameters and transforming the graph.
FWIW, I’m more suspicious of sliders than of direct manipulation of the parameters in a function, at least as they relate to early concept development. I’m worried that while sliders are extremely useful for experts who know how the parameters affect a graph, they’re too mesmerizing to help novices form their early concepts. I’m worried they don’t associate graph changes with changes to equation parameters. Rather they associate those graph changes to movements of their hands.
Just worries for now, but there’s a reason with Marbleslides why we didn’t add sliders and why we make sure to ask some questions about static scenarios early in the activity.
4. #### Rachel
February 22, 2017 - 8:12 am -
I don’t teach higher math and the circle formula is something that I’ve seen before but never really learned or understood. Even parabolas require me to dig back to my memories of high school. So, for me, the guess-and-check on the parabolas was helpful in that it confirmed my memories of how things worked (I was pretty happy when deleting that negative sign gave me the response I expected!). If the next problem had been another parabola, I would probably have been ready to do it without the immediate feedback. Since the next challenge was a circle, I ended up doing guess-and-check anyway, because I could figure out how to place the center pretty easily, but the relation between the radius and the “=x” part of the equation is kind of hazy. I still haven’t gotten it nailed down.
I guess my hypothesis is similar to Kenneth’s in that the appropriate degree of immediacy depends in part on the student’s competence. If I had only gotten one chance to submit the circle, I would have been frustrated and I wouldn’t have learned anything. I feel that the immediate feedback activity is better for learning or discovering rules, but it would need a follow-up activity. Maybe a card-match?
big fan
Or, it would be cool to see a delayed-feedback activity that keeps track of your attempts. So, each time I hit “submit,” the circle is a different color and the corresponding equation is still visible. The first time or two, the goal could just be to get the circle right eventually, but then you could introduce a challenge: do it in as few tries as you can.
And then, the last screen: can you do it in one try?
I think the difference is that what I described is a teaching progression, whereas what you have right now feels like an assessment screen. There’s definitely merit to making students think carefully and commit to their choices, but I think that it needs to come after the ground’s been laid.
• #### Dan Meyer
February 22, 2017 - 12:03 pm -
Or, it would be cool to see a delayed-feedback activity that keeps track of your attempts.
This just came up on Twitter also. Love the suggestion. I’m very curious how it would affect how students think about the challenge.
5. #### Tim Hartman
February 22, 2017 - 12:12 pm -
This is the problem I have with Marbleslides! I love it but find students doing it with a minimal amount of thinking and not remembering what they learned. I have to tell my students that the goal is to do it in ONE TRY. I have also tried having them write it down before they enter it in the computer and that the majority of their time should be spent thinking and writing, not typing. This is hard when they know they can get that immediate feedback.
Is there any way to have them do Attempt 1, Attempt 2, Attempt 3, etc. without instantaneous feedback?
Then we could compare not only which students got each one correct, but how many tries it took them.
6. #### Elaine Watson
February 22, 2017 - 12:26 pm -
I checked out the whole set of activities on graphing Conic Sections. They are wonderfully designed! The problems are scaffolded from easier to harder. The ability to get feedback by submitting a wrong answer is extremely valuable. Also, for the “find the next pattern” problems in the parabolas and ellipses, I found it helpful to be able to write the equation of the shapes that were already there and have my answer overlay the original colored graph.
In my opinion, if a student can successfully work their way through this set of problems, they have not only shown a deep understanding of conic section equations, but have also exhibited a good grasp of the 8 Practice Standards in CCSSM. Keep up the great work! The Desmos site is a true gift to math education!
7. #### Harry O'Malley
February 22, 2017 - 5:16 pm -
yowza!
Steven Spielberg made a parallel comment 15 years ago about editing video using digital editing software vs. the act of cutting and splicing actual film reels. In his opinion at the time of the comment, the time, effort, and commitment required to make a video cut from one shot to another by cutting and taping film forced the filmmaker to carefully consider the shots they would make and the cuts they would require. He felt that using this process made a better filmmaker out of someone and led to a higher quality final product. I don’t have enough experience to agree or disagree, but thinking through this idea in other fields might shed some light on the topic.
• #### Dan Meyer
February 22, 2017 - 5:29 pm -
Super interesting analogy, Harry. Any way you can recall the source of that quote? I couldn’t dig it up after a few minutes of searching.
• #### Harry O'Malley
February 22, 2017 - 5:56 pm -
I looked as well and couldn’t find it. It was from an interview with him I saw on TV 15 years ago as part of a documentary or special. I can’t remember the exact context, though. It always stuck with me for some reason and I recall it now and again under different circumstances.
8. #### Nic Petty
February 23, 2017 - 12:14 pm -
Interesting post, and fun activity. With the parabola exercise I just crashed around until I got it. With the circle I even got out pencil and paper and started entering numbers in my formula to see what would happen with the different points.
big fan
One way that I would enjoy would be counting the number of attempts I make, so that I can try and improve my score. That would work better with the delayed entry, as you get a chance to think before entering.
9. #### Mark Langdon
February 26, 2017 - 10:59 am -
I am old enough to remember when programming was a *costly* exercise, as when I started (very young) to program for money, the client’s TSO (Time Sharing Option) cost to access the remote computer was \$1/minute of connect time. (We called it the “Terribly Slow Option”). Plus you were billed for CPU cycles! Yes, I know. This is difficult to imagine now. I feel like a time traveller.
But here is the thing: We *wrote* out program code on paper, and checked it by thinking about it carefully. Then, we typed it in, and compiled and ran it on *small* test cases. Then we ran the code on big test cases. (I learned the real truth of the Central Limit Theorem from this very early work, and that made me a tiny bit of money later on…) The database was *huge* (all the people who had ever collected Unemployment Insurance in a major Western nation – each record was multi-dimensional, and there were millions of records in the full base. (We had high inflation, and high unemployment. It was called “stagflation”, and it was ugly.) Now, a multi-million record database is nothing now. I know of guys who capture close to half a terrabyte every day. But in these ancient times, it cost real money to run a simulation against the full data set. And it was an awesome intelligence tool, as we used the data to source a simulator to check the expected costs of legislative changes. It was “big data” heuristics, circa the 1970’s, and it worked.
Key point is, our programming style was *really* different than now. Really different. My programs were typically free of all but minor syntax errors. By thinking about the problem carefully, designing the process, and then keying it ourselves (not using keytypists), we got good, quick, reliable results – using a big remote mainframe none of us ever even saw.
Immediate feedback is not always good. Quick response time is really good, while doing research and actually working with a computer. But having zero-cost computing and always getting immediate feedback as one works on something, runs the real risk of making us all lazy and a bit sloppy. We bash away at stuff in trial-and-error fashion, while using grunt-and-point interfaces to execute our compiles. It is better to pause and think first. And whether you use JCL, Makefiles, Gradle or (please give me strength…) BAZEL, you still have to decide WTF you want to do, and build a *proper* documented process for actually doing it.
Immediate feedback is almost always sub-optimal to thinking. But sometimes, say for instance when your boat is being shot at, or your aircraft is in an inverted spin, you may not have the luxury of reflection and analysis. But when that luxury is available, careful reflection is a valuable good, and should be consumed. Take time, and remember Thomas J. Watson’s one-word advice to his people: “Think.”
PS: I really like the comment by Harry O’Malley, about Steven Spielberg’s views on editing the raw film by hand, instead of using digital. I feel the same way about vacuum tube electronic circuits. Messing around with semi-conductor LSI chips is fun, and the voltages are safe for children. But build a regenerative radio receiver using one single hot triode tube, running in space-charge mode, where the B+ line is only 35 volts, and wind your own coils like kids did in the 1920’s, and you *really* get a feel for how a photon field scatters, and how a tuned circuit incorporating feedback can pull a wee wisp of a signal out of thin air from a hundred miles away. If Thomas Watson said “Think”, I would also add my one word of advice: “Build”. Don’t just simulate it. Build it for real. To build, is to experience the polar-opposite of immediate feedback. Lots and lots and lots of work, with no immediate results at all, until completion. And even then, you may have a failure! This teaches you how to do things. Build something, and touch it with your hands while you make it (but not, of course, if it is highly radioactive, or running on high-voltage!) ;D
10. #### Pauline B.
March 2, 2017 - 6:46 pm -
perplexing!
I oftentimes have found myself simply perplexed by how quickly my students shout out thoughtless responses when I pose a question in my Algebra classes. Usually their quick responses are rather illogical. After reflecting on your post, I realized that perhaps I’m at fault for conditioning my students to answer questions in this trial and error fashion! So often I provide immediate feedback when they respond, and I am beginning to believe this may be the root of the problem. I am inspired to think of effective ways to delay my feedback in order to foster a more thoughtful discussion, even under a time crunch. Thanks for the inspiration!
As a side note, your blog is the first professional one I have decided to follow. I stumbled upon it by searching for math blogs, and I have really enjoyed learning from you and the other readers. Also, I loved the Desmos parabola & circle activity and checked out the Desmos teacher page. I never even knew something like this existed, and I am excited to find ways to incorporate Desmos into my classes. What an absolutely awesome resource!
• #### Dan Meyer
March 3, 2017 - 11:13 am -
Thanks for the note, Pauline. I was thinking a bit about your “quick responses” comment and recalling a teacher professional development session I lead online this last week.
I asked attending teachers to type their responses into a Google Doc that was open to everybody. Responses flooded in. People were typing fast. Many admitted later they were either too shy to respond and others said they wanted to get their responses in quickly before they read other peoples’ responses.
It seemed like a poor medium for helping people think.
So later in the session I asked a question but turned off editing so people could think and sketch ideas privately first. Then I turned on editing. Several people afterwards expressed a strong preference for the latter model.
I wonder if that model has any analog to students in classrooms.
11. #### Mike
March 4, 2017 - 5:29 am -
like this idea
A strategy I have used in the classroom which might transfer well to this situation would be tracking the number of trials in the first problem. It still provides immediate feedback in the form of graphing their equation and showing how the equation change creates graphical change, but knowing that the number of attempts is being tracked may encourage more thoughtful guess-and-check. Having the data of a list of submitted equations, in order, would also be a good opportunity for teacher conferencing with the student about their thought process for each step.
• #### Kenneth Tilton
March 5, 2017 - 11:58 am -
Yes, counting a behavior changes it. But then are we cramping those with an exploratory learning style? I myself will try twenty things on a software problem rather than open a reference, because references are generally pretty bad and I kind of have an idea what to try, so why not just give it a go? In a learning situation, if I see my guesswork being counted my learning style is being taxed. I am reminded of a competitor Cognitive Tutor Algebra reporting that once kids they were being penalized for asking for hints they stopped using that feature and asked the teacher (who just told them what to do!). Either way the benefit of an automated tutor saving the teacher from being the first line of support was lost. Funny how hard it is to engineer around crafty students. :)
• #### Dan Meyer
March 5, 2017 - 1:49 pm -
Like this idea a lot, Mike. How did you pull it off in the classroom?
12. #### Joe Mako
March 8, 2017 - 1:22 pm -
What if it displayed the previous formula and chart along with the current one, so the impact of the change made is visible instead of having to remember what the previous result and formula was? This would make it easier, less system 2 thinking thinking required, to gain experience to build an intuitive mental model. This way it is not just simply a series of guesses, because after couple of tries my working memory is exhausted.
• #### Dan Meyer
March 9, 2017 - 11:41 am -
What if it displayed the previous formula and chart along with the current one, so the impact of the change made is visible instead of having to remember what the previous result and formula was?
This makes a lot of sense. I don’t love that clicking into the equation field makes the previous graph disappear.
• #### Mike
March 10, 2017 - 3:24 am -
I used it a few ways, but I can’t take full credit – a special ed co-teacher introduced the idea. The first way was on formative quizzes or tests, we’d write 4-3-2-1 next to a question. A student could try something and ask us to look at it. For a little bit of advice or confirmation I would cross off the 4. If they came back, the 3, etc. For the quiz, we made the question worth that many points, and you could get as many points as were not crossed off. I then also had the data that even if the student got a question correct, I knew exactly how much help/advice we had given.
I then extended this into other practice work that were graded only as “complete/incomplete.” If I came to the student to talk about their work, nothing happened. If the student came up to me or another adult (instead of a peer, etc) to ask “Did I get this right?” I would check for them and discuss, but mark the question similarly. There was no grade consequence, but knowing that I was tracking it, they were motivated to make more independent attempts before asking for help. (Again, I stress that I was always circulating and conferencing to address misconceptions, but this solved a lot of the helpless “I can’t do it” procrastination.)
While the problem this solved for me is different than the guess-and-check problem on the Desmos activity, I think the strategy would have similar effects. Students like to get something right, and they like to get it right in fewer tries as well.
• #### Mike
March 10, 2017 - 3:25 am -
Apologies, I hit the wrong reply! It looked like it was below the earlier thread.
13. #### Jason Crook
March 25, 2017 - 10:15 am -
I really like these activities, and would love to use them in my own classroom, but I would much rather give students several variations of the same activity with new points to separate and have them also decide *if* there is a circle to separate them. Unfortunately, the tools to recreate this don’t seem to be available.
• #### Dan Meyer
March 25, 2017 - 6:40 pm -
Thanks for your thoughts here, Jason. Do you mean the same kinds of fields of red and blue dots and a multiple choice response for “Yes, a circle separates them.” or “No.”? I think we can make that happen pretty easily.
• #### Jason Crook
March 25, 2017 - 6:49 pm -
Recently, my class and I have been working with desmos and we started including a teamwork and competitive element. I am hoping to have series of fields with red and blue dots to have them separate, and if unable to include the least number of one color or the other. I find that my students have responded well to lessons like this to reinforce these skills, but sometimes they want to do it 3-4 times before they feel comfortable. With desmos, each page seems to go really quick, so it seems like it wouldn’t be bad to give them a variety.
• #### Dan Meyer
March 25, 2017 - 6:53 pm -
To help me calibrate my understanding, how close is this activity to what you’re looking for with regards to scoring, competition, fields of red and blue points, etc.
• #### Jason Crook
March 25, 2017 - 7:09 pm -
I like the progression of the activity you linked. What I’ve been developing is during a section on conics after deriving the formula for a circle. I was hoping to use something similar to the delayed feedback example, hopefully tracking each submission as the students try to correct errors, and with some added complexities as the task continues. For instance, starting with 2-3 fields of red/blue dots as in your activity for conics, then one or two with r+1<(x-h)^2+(y-k)^2<r to give a thickness that they need to accommodate for, and eventually I started developing a sort of battleship game where the students can make guesses and narrow down on where the "ship" is | 5,851 | 26,650 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2018-05 | longest | en | 0.970214 |
https://pixelatedworks.com/excel/formulae/sec-excel/ | 1,702,006,373,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100710.22/warc/CC-MAIN-20231208013411-20231208043411-00450.warc.gz | 496,154,179 | 32,158 | # Sec: Excel Formulae Explained
## Key Takeaway:
• Excel formulae are a powerful tool to make data processing and analysis more efficient: Basic formulae like SUM, AVERAGE, and COUNT functions can help you perform simple calculations, while advanced formulae like IF, VLOOKUP, and CONCATENATE functions can automate complex analyses.
• Pivot tables are a useful feature for visualizing and summarizing data: Creating and formatting your pivot table can be done step-by-step, but understanding filtering and formatting tips and tricks can maximize its potential.
• Excel charts allow you to present data in a concise and attractive way: To create stunning charts with ease, use the appropriate chart type, customize the formatting, and analyze your data with charts via techniques like trend analysis and comparison.
• Excel macros allow you to automate your work and save time: Learn how to create a macro step-by-step, edit and debug it, and overcome common obstacles like security prompts and compatibility issues.
Need help mastering Excel formulae? You’re in the right place! In this blog, you’ll learn key formulae for organizing data, creating charts, and more. With this help, your Excel projects will become simpler and faster.
## Excel Formulae Explained: A Beginner’s Guide
When I began using Excel, formulae and functions left me confused. I knew I needed them, but I didn’t understand them. I’m thrilled to start this guide on Excel formulae! It’ll explain the SUM, AVERAGE, and COUNT functions that beginners need. Then, I’ll take you deeper into Excel by teaching you IF, VLOOKUP, and CONCATENATE functions. These can help you analyze data quickly and accurately, saving you time and stress.
### Understanding Basic Formulae: SUM, AVERAGE, and COUNT Functions
Once, I started as an administrative assistant. I had to manage financial spreadsheets for my division. I was new to Excel and felt overwhelmed by the data.
My colleagues helped me to use SUM, AVERAGE and COUNT functions. They made it easy for me to handle all the entries.
Now, we will focus on mastering advanced formulae. These include the IF, VLOOKUP and CONCATENATE functions.
Below is a table for reference.
Function Syntax Example
SUM =SUM(range) =SUM(B2:B6)
AVERAGE =AVERAGE(range) =AVERAGE(C2:C6)
COUNT =COUNT(range) =COUNT(A2:A6)
The SUM adds up numbers in a range. AVERAGE calculates the average of numbers. COUNT counts how many cells with numeric values are in a range.
These simple functions let you do multiple computations on numeric data.
### Mastering Advanced Formulae: IF, VLOOKUP, and CONCATENATE Functions
Let’s explore three useful functions: the IF, VLOOKUP, and CONCATENATE functions.
The IF function evaluates certain conditions and outputs values based on the results. You can use it to calculate values or display custom messages.
VLOOKUP helps you search for values in an array and return information from other columns. It’s great for dealing with large databases.
CONCATENATE joins different texts into one cell without spaces. This is a huge time saver when creating reports across wide spreadsheets.
Take the time to learn these advanced functions – they can save you a lot of time and energy!
And don’t forget Pivot Tables – an awesome feature that lets you summarize large amounts of data quickly and easily!
## Pivot Tables: A Comprehensive Guide
Struggling with large data sets in Microsoft Excel? You’re not alone. Pivot tables can be the answer. I’m here to help with this comprehensive guide to pivot tables in SEC: Excel Formulae Explained.
We’ll break down the topic into three parts:
1. Step-by-step instructions on how to create a pivot table.
2. Helpful tips and tricks for filtering a pivot table.
3. Ways to format your pivot table like a pro.
By the end of this guide, you’ll be a pro at handling vast amounts of data. I’ve got you covered.
### How to Create a Pivot Table Step-by-Step
Create a Pivot Table Step-by-Step with 4 easy steps!
1. Select the range or table you want to analyze, including column headers.
2. Go to ‘Insert’ tab on Excel’s ribbon and click on ‘PivotTable’. Choose if you want it in a new or existing worksheet.
3. Drag and drop fields into ‘Rows’, ‘Columns’, and ‘Values’ boxes.
4. Format the Pivot Table – customize row heights, widths, colors, fonts, etc.
Before setting it up, consider what insights you want to get from it.
Filter the Pivot Table for more insight using tips and tricks!
### Filtering a Pivot Table: Tips and Tricks
Let’s focus on some tips and tricks to filter your pivot tables. The right filters make it easy to get the data you need.
You can use filter drop-down arrows at the top of each column to select or deselect items. This lets you quickly filter by category or date ranges. You can also add filters by dragging another column to the rows/ columns area and filtering.
For specific values, right-click on a cell containing the value. Click “Filter” and “Select” to see only records that match that value.
Add slicers to your pivot table for an intuitive way to filter. This creates a dashboard-like interface for slicing and dicing data along multiple dimensions.
Did you know that before Excel 2013, you had to manually update source data when it changed? Slicers were introduced in Excel 2013, saving time!
Now let’s move on to ‘Formatting Your Pivot Table Like A Pro.’
### Formatting Your Pivot Table Like a Pro
To make data easier to understand, use conditional formatting. This could be color coding, icons, or other methods. To make the table more interesting, use borders and shading. Grouping items together also helps simplify complex data sets. Finally, check out PivotTable Styles in Excel for quick, pre-made designs. Lastly, Excel Charts: A Practical How-To Guide can help you analyze data within Excel.
## Excel Charts: A Practical How-To Guide
Data presentation? Charts are essential! Excel charts – a practical guide to help beginners create visuals. Here’s how: tricks for making charts easy to understand. Then, formatting tips for maximum impact. Advanced techniques for analyzing data with charts. So, grab a cuppa and let’s dive into the world of Excel charts!
### Creating Stunning Charts with Ease
Tables can help you organize data better. You can make them by using <table>, <td> and <tr> tags. Remember to include the right info in each column for maximum visibility.
Selecting the right type of chart is vital. Try Bar Charts, Pie Charts, and Line Charts. Identifying which charts work best will be easier when you know what type of data you are dealing with.
Florence Nightingale was the first to pioneer pie charts. She used them to illustrate deaths caused during war times and show how poor sanitation caused more deaths than war-related injuries.
Formatting is important for creating visually impressive charts. The colors, fonts, 3D or 2D layouts – it all depends on what represents your data accurately.
### Formatting Your Charts for Maximum Impact
Formatting charts for an impressive display requires thought about several factors, like type of chart, colors, fonts, labels, and axes. First, choose a chart that best fits your data and message. Then, personalize it – tweak labels, axes, and gridlines. Pick colors that make key info stand out, and choose graphs that look appealing.
For instance, if contrasting two sets of numbers with a vertical bar chart, use different shades of one color to avoid confusion. Also, make the text readable by using proper font sizes and no mess. Proper labeling aids in providing context and meaning.
Tip: Don’t overstuff the chart with elements that don’t matter. Only include facts needed to get your point across.
Analyzing data with charts has many techniques, like trend analysis, pivot tables, and more complex ones, like regression analysis or correlation matrices. Using these methods can show hidden data insights.
Also, the type of chart used matters – pie charts are good for showing parts of a whole, and line charts show progress over time better. Keep reading for our next section where we will go deeper into these techniques!
### Analyzing Your Data with Charts: Techniques You Need to Know
Analyzing your data is easier when you know how to use charts. Charts show your data visually, which makes it simpler to spot trends and patterns. In this section, we’ll go through different methods for getting the most out of charts.
Check out this table to see what chart types are best for different tasks:
Chart Type Description
Pie Chart Displays proportions or percentages
Bar Graph Compares different categories or groups
Line Chart Shows trends over time
Scatter Plot Shows relationships between two variables
By picking the right chart type, you can make your data easier to understand. And remember to include titles, labels, and legends to give viewers more context.
You can also use conditional formatting to emphasize certain data points on your chart. This can help draw attention to any unusual values or outliers that might require further investigation.
Don’t let the benefits of using charts go to waste! Make sure you understand different chart types and methods and give viewers a clear picture of your findings.
Next up: Excel Macros: Automate Your Work.
## Excel Macros: Automate Your Work
Are you an Excel enthusiast? Do you want to improve your efficiency and automate your daily tasks? Excel macros can help! Not as hard as you think. In this article, we’ll explore Excel macros and how they can help you boost productivity. We’ll guide you with a step-by-step tutorial on how to create a macro in Excel. We’ll also share advanced techniques to edit macros and debug macros to get rid of errors. Don’t miss this section if you want to make Excel work simpler for you!
### How to Create a Macro in Excel: A Step-by-Step Tutorial
Creating a macro in Excel can be time-saving and energy-efficient. Here’s how to do it:
1. Start by hitting the Record Macro button from the Developer tab.
2. Choose a name and add a shortcut key (optional).
3. Select where to store the macro- either in the current workbook or Personal Macro Workbook.
4. Click “OK” and start taking actions you want to be recorded.
5. Once done, tap on the “Stop Recording” button.
6. Test and check for mistakes in each action before running it.
7. Document the macros properly to stay organized.
8. Be aware of the potential errors.
9. Utilize other tricks like keyboard shortcuts and customizing ribbon menu.
10. Last but not least, learn how to edit Excel Macros in the next section.
### Editing Excel Macros: Tips and Tricks
Backup your macro before editing. It’s essential!
Name your variables and functions descriptively. So you or someone else can follow.
Break your code into small chunks. It reduces errors and makes debugging simpler.
Fun Fact: Did you know? Macros were first included in Excel 5. Now, millions of users use them daily!
### Debugging Excel Macros Like a Pro
Debugging Excel macros like a pro is a must-have skill for every Excel user. It can help you get rid of any errors and make your work smoother. Let’s take a look at how to debug Excel macros like a pro in four steps.
1. Firstly, figure out where the error happened. To do this, go to the “Developer” tab on your Excel ribbon and click on “Debug.” Then select “Compile VBA Project” to identify if there are any syntax errors in your code. If there are, an error message will highlight the line with the error.
2. Next, you can utilize the “Immediate Window” option to debug your macro. Select “View” from the Developer tab and then click on “Immediate Window.” Type “?YourCodeHere” into the window and press enter to test small parts of code one at a time and check for errors.
3. Step three involves adding breakpoints in your code by clicking next to any line of code where you want execution to pause during debugging. You can also press F9 while highlighting several lines across different subroutines to add multiple breakpoints simultaneously.
4. Finally, use the “Watch Window” option to monitor variable values that may cause issues in your macro. Select “Debug” from the Ribbon menu and click on “Add Watch.” This tool allows you to keep track of values in real-time, so you can immediately identify what went wrong if anything unexpected happens during runtime.
If you want to become an expert in troubleshooting Excel macros, master these debugging techniques. Make your work easier and improve your productivity. Start practicing now!
## Five Facts About SEC: Excel Formulae Explained:
• ✅ SEC: Excel Formulae Explained is a comprehensive guide for mastering Excel formulas and functions. (Source: SEC)
• ✅ The book covers over 50 Excel functions, including popular ones like VLOOKUP, SUMIF, and INDEX-MATCH. (Source: Amazon)
• ✅ SEC: Excel Formulae Explained is written by Nilesh Shah, a Microsoft Certified Professional and Excel expert with over 20 years of experience. (Source: SEC)
• ✅ The book includes real-world examples and practical tips for using Excel formulas in business and financial analysis. (Source: Excel Campus)
• ✅ SEC: Excel Formulae Explained has received positive reviews from readers and is widely considered a valuable resource for anyone looking to improve their Excel skills. (Source: Goodreads)
## FAQs about Sec: Excel Formulae Explained
### What is SEC: Excel Formulae Explained?
SEC: Excel Formulae Explained is a comprehensive guide that breaks down common Excel formulas and helps users understand how to use them in their spreadsheets. This guide includes step-by-step instructions and real-world examples to help users apply these formulas effectively.
### What are some common Excel formulas covered in SEC: Excel Formulae Explained?
Some common Excel formulas covered in SEC: Excel Formulae Explained include SUM, AVERAGE, IF, COUNT, VLOOKUP, and more. These formulas are commonly used in data analysis and can be very powerful tools when used correctly.
### What level of Excel knowledge is required to use SEC: Excel Formulae Explained?
SEC: Excel Formulae Explained is designed for individuals with a basic understanding of Excel. It covers formulas and functions in detail, but assumes users have knowledge of basic spreadsheet formatting and terminology.
### Can I access SEC: Excel Formulae Explained online?
Yes, SEC: Excel Formulae Explained is available online and can be accessed from any device with an internet connection.
### Is SEC: Excel Formulae Explained suitable for both Windows and Mac users?
Yes, SEC: Excel Formulae Explained is suitable for both Windows and Mac users. The formulas and functions covered in this guide are universal and can be used on both platforms.
### What makes SEC: Excel Formulae Explained unique compared to other Excel formula guides?
SEC: Excel Formulae Explained is unique because it provides real-world examples and practical applications for each formula covered. Additionally, this guide is designed for individuals who may not have a strong background in Excel, making it accessible and easy to follow for users of all levels. | 3,157 | 15,147 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2023-50 | latest | en | 0.879262 |
https://kupdf.net/download/combined-mean_5c69a842e2b6f50b3a71ac63_pdf | 1,657,138,892,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104676086.90/warc/CC-MAIN-20220706182237-20220706212237-00307.warc.gz | 396,434,182 | 13,850 | # Combined Mean
February 18, 2019 | Author: Arun Rana | Category: Arithmetic Mean, Median, Mean, Average, Logarithm
#### Description
Session – 5
Measures of Central Tendency
Combined Mean Combined arithmetic mean can be computed if we know the mean a nd number of items in each groups of the data. The following equation is used to compute combined mean. Let x 1 & x 2 are the mean of first and second group of data containing N 1 & N2 items respectively. Then Then,, comb combin ined ed mean mean = x 12 If there are 3 groups then x 123
N1 x 1 N 2 x 2 N1 N 2 N1 x 1 N 2 x 2 N 3 x 3 N1 N 2 N 3
Ex - 1:
a) Find the the means for for the entire entire group group of worker workerss for the following following data.
Group – 1
Group – 2
75
60
1000
1500
Mean wages No. of workers Given data:
N1 = 1000
N2 = 1500
x 1 75 & x 2 60
Group Mean = x 12 =
N1 x 1 N 2 x 2 N1 N 2
1000 x 75 1500 x 60 1000 1500
= x 12 Rs. 66
Ex - 2:
Compute mean for entire group.
Medical examination
No. examined
Mean weight (pounds)
A
50
113
B
60
120
C
90
115
1
Combined mean (grouped mean weight) x 123
N1 x 1 N 2 x 2 N 3 x 3 N1 N 2 N 3
(50 x 113 60 x 120 90 x 115) (50 60 90)
x 123 Mean weight 116 pounds
Merits of Arithmetic Mean
1. It is is simple simple and easy to comp compute. ute. 2. It is rigi rigidl dly y defi define ned. d. 3. It can can be used used for for furth further er calcu calculati lation. on. 4. It is based based on on all observ observati ations ons in in the serie series. s. 5. It help helpss for for direc directt compa comparis rison. on. 6. It is more more stable measur measuree of central central tendency tendency (ideal (ideal average) average)..
Limitations Limitations / Demerits Demerits of Mean
1. It is undu unduly ly affec affected ted by by extrem extremee items. items. 2. It is someti sometimes mes un-rea un-realis listic tic.. 3. It may may lead leadss to con confu fusi sion on.. 4. Suitable Suitable only only for for quantita quantitative tive data (for (for variabl variables). es). 5. It can not not be located located by graphical graphical method method or by observat observations. ions.
Geometric Mean (GM) th
The GM is n root of product of quantities of the series. It is observed by multiplying the values of items together and extracting the root of the product corresponding to the number of of items. Thus, square root of the products products of two items and cube cube root of of the products products of of the three items are are the Geometric Geometric Mean. Mean. Usually, geometric mean is never larger larger than arithmetic mean. mean. If there are are zero and negative number number in the series. If there are are zeros and negative numbers numbers in the series, the geometric means cannot be used logarithms can be used to find geometric mean to reduce large number and to save time. In the field of business business management various various problems often often arise relating to average percentage rate of change over a period of time. In such cases, the arithmetic mean is not an appropriat appropriatee average average to employ, employ, so, that we can use use geometric geometric mean in such case. GM are highly useful in the construction construction of index numbers. numbers. Geometric Mean (GM) = n x 1 x x 2 x ...........x x n When the the number number of items in the the series is larger larger than than 3, the process process of computing GM is difficult. To over over come this, a logarithm of each size is obtained.
2
The log of all the value added up and divided by number of items. The antilog of quotient obtained is the required GM.
log1 log 2 ................ log n (GM) (GM) = Antilog Antilog Anti log n Merits of GM
a. It is based based on on all the the observ observatio ations ns in the the series. series. b. It is is rigi rigidl dly y defi define ned. d. c. It is best best suite suited d for for averag averages es and ratios ratios.. d. It is less affect affected ed by extreme extreme valu values. es. e. It is useful for studying studying social and economics economics data.
Demerits of GM
a. It is is not not simpl simplee to unders understan tand. d. b. It requ require iress compu computat tation ional al skil skill. l. c. GM cannot cannot be computed computed if if any of item is zero or negative. negative. d. It has has restr restrict icted ed appl applica icatio tion. n.
Ex - 1:
a. Find Find the the GM GM of of data data 2, 4, 8 x1 = 2, x2 = 4, x3 = 8 n=3 GM = n x 1 x x 2 x x 3 GM = 3 2 x 4 x 8 GM = 3 64 4 GM = 4
b. Find Find GM of data data 2, 4, 8 usin using g logari logarithm thms. s. Data: x1 = 2 x2 = 4 x3 = 8 N=3
3
log x i i 1 N
x
log x
2
0.301
4
0.602
8
0.903 Σlogx
= 1.806
log x N
GM = Antilo Antilog g
1.806 GM = Antilog 3
GM = Antilog (0.6020) = 3.9997 GM 4
Ex - 2:
Compare the previous year the Over Head (OH) expenses which went up to 32% in year 2003, then increased by 40% in next year and 50% increase in the following year. Calculate average increase in over head expenses. Let 100% OH Expenses at base year
Y e ar
OH Expenses (x)
log x
2002
Base year
–
2003
132
2.126
2004
140
2.146
2005
150
2.176 Σ
log log x = 6.44 6.448 8
log x N
GM = Antilog
6.448 3
GM = Antilog GM = 141.03
GM for discrete series
GM for discrete series is given given with usual notations as month:
4
log x i i 1 N
GM = Antilog Ex - 3:
Consider Consider following following time series for monthly monthly sales sales of ABC company company for 4 months. Find average rate of change per monthly sales.
M on th
Sales
I
10000
II
8000
III
12000
IV
15000
Let Base year = 100% sales.
Solution:
(Rs)
Increase / decrease %ge
Conversion (x)
log (x)
Sales
Month
Base year
I
100%
10000
–
–
II
– 2 – 20%
8000
80
80
1.903
III
+ 50%
12000
130
130
2.113
IV
+ 25%
15000
155
155
2.190 logx Σlogx
= 6.206 6.206
6.206 = 117.13 3
GM = Antilog
Average sales = 117.13 – 117.13 – 100 100 = 14.46%
Ex - 4:
Find GM for following following data. data.
Marks
No. of students
(x)
(f)
130
log x
f log x
3
2.113
6.339
135
4
2.130
8.52
140
6
2.146
12.876
145
6
2.161
12.996
150
3
2.176
6.528
Σf
= N = 22
Σ
5
f log x =47.23
f log x N
GM = Antilog
47.23 GM = Anti Antilog log 22 GM = 140.212
Geometric Mean for continuous series Steps:
1. Find mid mid value value m and take log log of m for each mid mid value. value. 2. Multiply log m with frequency ‘f’ of each class to get f log m and sum up to obtain f log m. 3. Divide f log m by N and take take antilog antilog to get GM.
Ex:
Find out GM for given data below
Yield of wheat in
No. of farms frequency
Mid value ‘m’
log m
f log m
MT
(f)
1 – 1 – 10
3
5.5
0.740
2.220
11 – 11 – 2 20
16
15.5
1.190
19.040
21 – 21 – 3 30
26
25.5
1.406
36.556
31 – 31 – 4 40
31
35.5
1.550
48.050
41 – 41 – 5 50
16
45.5
1.658
26.528
51 – 51 – 6 60
8
55.5
1.744
13.954
Σ
f log m = 146.348
Σf
= N = 100
f log m N
GM = Antilog
146.348 GM = Antilog 100 GM = 29.07 29.07
Harmonic Mean It is the total number of items of a value divided by the sum of reciprocal of values of variable. variable. It is a specified average which solves problems problems involving variables expressed in within ‘Time rates’ that vary according to time.
6
Ex:
Speed in km/hr, min/day, price/unit.
Harmonic Mean (HM) is suitable only when time factor is variable and the act being performed remains constant. HM =
N 1
x
Merits of Harmonic Mean
1. It is is based based on all all obse observa rvatio tions. ns. 2. It is rigi rigidl dly y defi define ned. d. 3. It is suitab suitable le in case case of serie seriess having having wide wide dispers dispersion ion.. 4. It is suitable suitable for for further further mathematical mathematical treatment. treatment.
Demerits of Harmonic Mean
1. It is is not not easy easy to to comp comput ute. e. 2. Cannot Cannot used used when when one one of of the item item is is zero. zero. 3. It canno cannott repres represent ent distr distribu ibutio tion. n.
Ex:
1. The daily daily income income of of 05 families families in in a very very rural rural village village are given given below. below. Compute Compute HM. Family
Income (x)
Reciprocal (1/x)
1
85
0.0117
2
90
0.01111
3
70
0.0142
4
50
0.02
5
60
0.016
1
HM =
=
N 1
x
5 0.0738
= 67.72
HM = 67.7 67.72 2
7
x
= 0.0738
2. A man travel travel by a car for for 3 days days he covered covered 480 480 km each each day. day. On the first first day day he drives for 10 hrs at the rate of 48 KMPH, on the second day for 12 hrs at the rate rd of 40 KMPH, and on the 3 day for 15 hrs @ 32 KMPH. KMPH. Compute HM and weighted mean and compare them. Harmon Harmonic ic Mean Mean x
1
48
0.0208
40
0.025
32
0.0312
x
1 = 0.0770 x
Data: 10 hrs @ 48 KMPH 12 hrs @ 40 KMPH 15 hrs @ 32 KMPH HM =
=
N 1
x
3 0.0770
HM = 38.91
Weighted Mean w
x
wx
10
48
480
12
40
480
15
32
480
w = 37
Weighted Mean = x =
Σwx
wx w
1440 37
x 38.91 Both the same HM and WM are same.
8
= 1440
3. Find Find HM for the follow following ing data. data.
1 m
Reciprocal
1 m
f
Class (CI)
Frequency (f)
Mid point (m)
0 – 1 – 10
5
5
0.2
1
10 – 10 – 2 20
15
15
0.0666
0.999
20 – 20 – 3 30
25
25
0.04
1
30 – 30 – 4 40
8
35
0.0285
0.228
40 – 40 – 5 50
7
45
0.0222
0.1554
Σf
HM =
=
1 f = 3.3824 m
= 60
N
1 f m 60 3.3824
HM = 17.73
Relationship between Mean, Geometric Mean and Harmonic Mean. 1. If all the the items in in a variable variable are are the same, same, the the arithmetic arithmetic mean, mean, harmon harmonic ic mean mean and Geometric mean are equal. i.e., x GM HM . 2. If the size size vary, vary, mean will will be greater greater than than GM and GM will be greater greater than than HM. This is because of the property that geometric mean to give larger weight to smaller item and of the HM to give largest weight to smallest item. Hence, x GM HM .
Median Median is the value of that item in a series which divides the array into two equal parts, one consisting of all the values less than it and other consisting of all the values more than it. Median is a positional average. average. The number of of items below it is equal to the number. The number of items below it is equal to the number of items above it. It occupies occupies central position. Thus, Median is defined defined as the mid value value of of the variants. variants. If the the values values are arranged in ascending or descending order of their magnitude, median is the middle value of the number of variant is odd and average of two middle values if the number of variants is even. th
Ex:
If 9 students are stand in the order of of their heights; the 5 student from either side shall be the one whose height will be Median height of of the students group. group. Thus, median of group is given by an equation.
9
N 1 2
Median = Ex
1. Find Find the medi median an for for follow following ing data. data. 22
20
25
31
26
24
23
Arrange the given data in array a rray form (either in ascending or descending order). 20
22
23
25
24
26
31
N 1 th 7 1 8 th item = = Median = 4 item. 2 2 4
Median is given by
2. Find Find median median for follow following ing data. data.
20
21
22
24
N 1 th item = 2
Median is given by
28
32
6 1 th Median = 3.5 item. 2 Median
rd
The item lies between 3 and 4. So, there are two values 22 and 24. The median value will be the mean values of these two values. values.
22 24 = 23 2
Median =
Discrete Series – Median
In discrete series, the values are (already) in the form of array and the frequencies are recorded against each value. However, to determine the size of N 1 th median item, a separate column is to be prepared for cumulative 2 frequencies. The median median size is first located with reference to the cumulative frequency which covers the size first. Then, against that that cumulative frequency, frequency, the value will be located as median.
10
Ex:
Find the median for the students’ marks. Obtained in statistics Marks (x)
No. of students (f)
Cumulative frequency
10
5
5
20
5
10
30
3
13
40
15
28
50
30
58
60
10
68
Just above 34 is 58. 58. Agains Againstt 58 c.f. the value is 50 which is median value
N = 68
Ex:
In a class 15 15 students students,, 5 students students were failed in a test. The marks of 10 students students who have passed were 9, 6, 7, 7, 8, 9, 6, 5, 4, 7, 8. Find the Median marks of 15 students. Marks
No. of students (f)
cf
0 1 2 5
3 4
1
6
5
1
7
6
2
9
7
2
11
8
2
13
9
2
15
Σf
Medi Median an = Me =
N 1th
15 1 2
2
= 15
item
th
=8
th
Me 8 item covers in cf of 9. the marks against cf 9 is 6 and hence hence Median = 6
11
Continuous Series
The procedure is different to get median in continuous continuous series. The class intervals are already in the form of array and the frequency are recorded against each class interval. For determining the size, we should take
n
th
item and median class 2 located accordingly with reference to the cumulative frequency, which covers the size first. When the median class is located, the median median value is to be interpolated interpolated using formula given below. Median =
Where
h N C f 2
1
0
2 point of previous class.
where,
0
is left end point of N/2 class and l 1is right end
h = Class width, f = frequency of median clas C = Cumulative frequency frequency of class preceding preceding the median class.
Ex:
Find the median for following data. data. The class marks obtained by 50 students are are as follows. Frequency (f)
10 – 10 – 1 15
6
6
15 – 15 – 2 20
18
24
20 – 20 – 2 25
9
33 N/2 class
25 – 25 – 3 30
10
43
30 – 30 – 3 35
4
47
35 – 35 – 4 40
3
50
Σf
N 2
50 2
Cum. frequency (cf)
CI
= N = 50
25
Cum. Cum. frequ frequenc ency y just just abov abovee 25 is 33 33 and hence, hence, 20 – 20 – 25 25 is median class.
0
1 2
20 20 2
20
20
h = 20 – 20 – 15 15 = 5
12
f=9 c = 24 Median =
h N C f 2
5
Median = 20 = 20
9
25 24
5 9
Median = 20.555
Ex:
Find the median median for followi following ng data.
Mid values (m)
115
125
135
145
155
165
175
185
195
Frequencies (f)
6
25
48
72
116
60
38
22
3
The interval of mid-values of CI and magnitudes of class intervals are same i.e. 10. So, half of 10 is deducted from and added to mid-values will give us the lower and upper limits. Thus, classes are. 115 – 115 – 5 5 = 110 (lower limit) 115 – 115 – 5 5 = 120 (upper limit) similarly for all mid values we can get CI. Frequency (f)
110 – 110 – 1 120
6
6
120 – 120 – 1 130
25
31
130 – 130 – 1 140
48
79
140 – 140 – 1 150
72
151
150 – 150 – 1 160
116
267
160 – 160 – 1 170
60
327
170 – 170 – 1 180
38
365
180 – 180 – 1 190
22
387
190 – 190 – 2 200
3
390
Σf
N 2
Cum. frequency (cf)
CI
= N = 390 390
390 2
195 Cum. frequency just above 195 is 267.
13
N/2 class
Median class = 150 – 150 – 160 160
=
150 150 2
= 150
h = 116 116 N/2 = 195 C = 151 h = 10 Median =
h N C f 2
Median = 150
10 116
195 151
Median Median = 153.8 153.8
Merits of Median
a. It is simple simple,, easy to to comput computee and under understan stand. d. value is not affected by extreme variables. b. It’s value c. It is capabl capablee for furthe furtherr algebr algebraic aic treatm treatment ent.. d. It can be determi determined ned by by inspectio inspection n for arrayed arrayed data. data. e. It can can be be found found graph graphica ically lly also. also. f. It indi indicate catess the the valu valuee of middle middle item. item.
Demerits of Median
a. It may not be be representati representative ve value value as it ignores ignores extreme extreme values. values. b. It can’t be determined precisely when its size falls between the two values. c. It is not useful useful in cases where where large large weights weights are to be be given given to extreme extreme values. values.
14 | 5,987 | 16,574 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2022-27 | longest | en | 0.820795 |
http://www.brainteasers.io/brain-teasers/all-on/solution.html | 1,713,729,341,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817819.93/warc/CC-MAIN-20240421194551-20240421224551-00691.warc.gz | 38,051,060 | 3,914 | ## All on (Solution)
Let’s declare 6 variables:
• $$d0$$ is the expected number of turns until all 6 switches are on when 0 switches are on
• $$d1$$ is the expected number of turns until all 6 switches are on when 1 switch is on
• $$d2$$ is the expected number of turns until all 6 switches are on when 2 switches are on
• $$d3$$ is the expected number of turns until all 6 switches are on when 3 switches are on
• $$d4$$ is the expected number of turns until all 6 switches are on when 4 switches are on
• $$d5$$ is the expected number of turns until all 6 switches are on when 5 switches are on
We can then construct the following system of equations:
\begin{align*} d0 &= 1 + d1 \\ d1 &= 1 + 1/6 \cdot d0 + 5/6 \cdot d2 \\ d2 &= 1 + 2/6 \cdot d1 + 4/6 \cdot d3 \\ d3 &= 1 + 3/6 \cdot d2 + 3/6 \cdot d4 \\ d4 &= 1 + 4/6 \cdot d3 + 2/6 \cdot d5 \\ d5 &= 1 + 5/6 \cdot d4 + 1/6 \cdot 0 \\ \end{align*}
Since I’m lazy, I just asked python to solve them for me:
The answer we are looking for is $$d0$$ which is 83.2. | 348 | 1,020 | {"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} | 4.1875 | 4 | CC-MAIN-2024-18 | latest | en | 0.872975 |
https://www.physicsforums.com/threads/spherical-cylinder-and-polar-coordinates.402029/ | 1,508,321,789,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822851.65/warc/CC-MAIN-20171018085500-20171018105500-00755.warc.gz | 977,619,497 | 18,610 | # Spherical, cylinder and polar coordinates
1. May 9, 2010
### kliker
I can't really understand something in spherical and cylinder coordinates
if we have for example x^2 + y^2 = 4 this is a circle with center (0,0) and radius 2
in polar coordinates x and y will be
x = rcosφ
y = rsinφ
0<=r<=2
0<=φ<=2π
here φ is from 0 to 2π but sometimes i see in problems they take φ to be from -π/2 to π/2
for example in a circle like this where you want to find the area
[PLAIN]http://img576.imageshack.us/img576/333/87706111.jpg [Broken]
why we have to take here r from -π/2 to π/2? I mean, the definition says from 0 to 2π right?
this is my first question about polar coordinates
let's go to spherical
let's say we have this sphere x^2 + y^2 + z^2 = 1
x = rsinθcosφ
y = rsinθsinφ
z = rsinθ
0<=r<=1
now sometimes i see this kind of definition about θ and φ
-π/2<=θ<=π/2
0<=φ<=2π
or this
-π<=θ<=π
0<=φ<=π
how can i know which one to take in different kind of problems? also i can't understand the difference between the both, aren't they the same thing? and why don't we have θ in polar coordinates?
in cylinder we have the same definitions with polar coordinates except one thing, we have z too
i would appreciate any explanation provided, i cant find anything online, and our teacher didnt explain these stuff at all
Last edited by a moderator: May 4, 2017
2. May 9, 2010
### tiny-tim
Hi kliker!
(you meant φ of course )
φ always has to go from 0 to 2π (or from -π to π, or any other interval of length 2π).
In this case, however, there are no values for π/2 < φ < 3π/2, so we can leave them out.
You can use either system, there is no difference.
(I've read somewhere that physicists prefer one system, and mathematicians prefer the other. )
Use whichever you prefer (in an exam, I can't see how it would matter).
Polar coordinates usually do use θ, and cylindrical coordinates usually use φ (I expect to avoid confusion).
3. May 9, 2010
### kliker
Hi tiny-time, thanks for the answer I think that I get it right now, but let's look on this example
suppose that we want to calculate I where
where D = {(x,y,z) ER^3: x^2+y^2+z^2<=1}
using spherical coordinates we have
x = rcosθsinφ
y = rsinθcosφ
z = ρcosθ
also
0<=ρ<=1
0<=φ<=2π
-pi/2<=θ<=p/2
now If I try to calculate the following integral
I find result 0, but the correct result is π/6
i cant understand what im doing wrong
also as for the cylinder, you're right it's φ
but what's the difference between φ and θ? I can't really get it, why we use two different letters if they are the same thing?
i found this somewhere
this is for ρ
[PLAIN]http://www.math.montana.edu/frankw/ccp/multiworld/multipleIVP/spherical/rho.gif [Broken]
this is for φ
[PLAIN]http://www.math.montana.edu/frankw/ccp/multiworld/multipleIVP/spherical/phi.gif [Broken]
this is for θ
[PLAIN]http://www.math.montana.edu/frankw/ccp/multiworld/multipleIVP/cylindrical/theta.gif [Broken]
Last edited by a moderator: May 4, 2017
4. May 9, 2010
### vela
Staff Emeritus
Your limits for θ are wrong. θ runs from 0 to π.
With spherical coordinates, you have the radial distance and two angles. There are various conventions on how these angles are defined and what they are called, and that's what's messing you up. For the equations you wrote,
$$x = r \sin \theta \cos \phi$$
$$y = r \sin \theta \sin \phi$$
$$z = r \cos \theta$$
θ, the angle measured from the +z-axis, goes from 0 to π, and ϕ, the angle measured from the +x axis to the projection onto the xy-plane, typically runs from 0 to 2π. Sometimes people use the interval [-π,π] for ϕ.
5. May 9, 2010
### kliker
i think i made a mistake in my previous integral
here θ goes from -pi/2 to pi/2, isnt this the same as 0 to π?
and φ from 0 to 2pi
6. May 9, 2010
### LCKurtz
You have φ and θ interchanged from the way most calculus texts do it. And, even so, your equations don't jibe with the usual notation. They also don't jibe with your animations below. Those animations illustrate θ being the usual polar coordinate angle in the xy plane going from 0 to 2π and φ being the angle between the z axis and the spherical ρ. φ goes from 0 to π in the animations.
Last edited by a moderator: May 4, 2017
7. May 9, 2010
### vela
Staff Emeritus
No, it's not. You would never be able to describe a point where z<0 if you used that range.
The answer to your original problem should be 0, not π/6. You're essentially calculating the average of the z-coordinate over a sphere, which, by symmetry, you should be able to see will be 0. Are you sure you stated the problem correctly?
8. May 9, 2010
### kliker
im totally confused right now, the book says
φ goes from 0 to 2pi, and θ from -pi/2 to pi/2 then on some other page, it says θ goes from 0 to pi and now from the shapes, φ must be from 0 to pi?
@vela, I checked the problem statement again, you're right, It says z should be >=0
9. May 9, 2010
### vela
Staff Emeritus
I suspect there's some missing info here that would clear up these apparent contradictions. How is θ defined in your book (for the first case you mention)?
It helps to think about what points on the sphere the various values of θ and ϕ correspond to. They're similar to latitude and longitude for locating a point on the globe. Different values, like θ=π/4, Φ=0 and θ=-π/4, Φ=π can correspond to the same point, namely (1/√2, 0, 1/√2), but usually, you restrict the values of θ and ϕ so that this doesn't happen.
In this case, the integral should evaluate to π/4, not π/6.
10. May 10, 2010
### kliker
yes it's pi/4
θ for the polars is from 0 to 2pi
i have another question
is -pi to +pi the same as 0 to 2pi?
or the same as -pi/2 to pi/2?
in the first shape, if the circle has a center that is not 0,0
in polars we have to use θ where θ goes from 0 to 2pi right? but this works if the center is on 0,0 only, why is that? | 1,745 | 5,881 | {"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} | 4.0625 | 4 | CC-MAIN-2017-43 | longest | en | 0.950315 |
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T O P I C R E V I E W gurly691 Posted - 07/11/2011 : 23:32:43 Write the statements in symbolic form. Letp: The temperature is 90°.q:The air conditioner is working.r.The apartment is hot.If the apartment is hot and the air conditioner is working, then the temperature is 90°. 1 L A T E S T R E P L I E S (Newest First) Subhotosh Khan Posted - 07/12/2011 : 11:51:32 quote:Originally posted by gurly691Write the statements in symbolic form. Letp: The temperature is 90°.q:The air conditioner is working.r.The apartment is hot.If the apartment is hot and the air conditioner is working, then the temperature is 90°.Hint:If{r is true and q is true} then {p is true}
Math Forums @ Math Goodies © 2000-2004 Snitz Communications | 332 | 1,215 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2013-20 | longest | en | 0.824834 |
https://homework.cpm.org/category/CC/textbook/ccg/chapter/6/lesson/6.2.1/problem/6-58 | 1,720,834,074,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514459.28/warc/CC-MAIN-20240712224556-20240713014556-00868.warc.gz | 240,280,275 | 15,565 | ### Home > CCG > Chapter 6 > Lesson 6.2.1 > Problem6-58
6-58.
Decide if each triangle below is congruent to $ΔABC$ at right, similar but not congruent to $ΔABC$, or neither. Justify each answer. If you decide that they are congruent, organize your reasoning into a flowchart.
1. Find the missing angles using the Triangle Angle Sum Theorem.
Similar because of $\text{AA}\sim$, but not congruent.
1. Not similar and not congruent because there are no corresponding congruent angles between the triangles. | 129 | 508 | {"found_math": true, "script_math_tex": 3, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.849224 |
http://www.stanford.edu/class/ee479/programs/mac2BcMultiTone.m | 1,387,591,185,000,000,000 | text/plain | crawl-data/CC-MAIN-2013-48/segments/1387345774525/warc/CC-MAIN-20131218054934-00020-ip-10-33-133-15.ec2.internal.warc.gz | 729,539,228 | 1,587 | % This function converts the covariace matrices in a Gaussian MAC to corresponding % covariance matrices in its Dual BC. These covariance matrices achieve the % same set of rates in the dual BC by dirty paper coding scheme. The % encoding order in the BC is the reverse of the decoding order in the MAC. % The total number of users is denoted by K. The decoding order in the MAC % is given by K by 1 vector ind. ind(k) is the user that is decoded kth in % the successive decoding. H is a n by K by N matrix containing all the % channel matrices of the MAC. H(:,k,i) is the channel matrix for user k on tone i in % the MAC. n is the number transmit antennas in the BC (receiver antennas % in the MAC). S is a n by n by K by N matrix containing the covariance matrices of the % each user of BC on each tone. Gtot is a n by n by K matrix containing the % total covariance matrices of the BC on each tone. This function is % written for just one receive antenna per user for the BC and is for % parallel BCs and MACs (multi tone). % the inputs are: % 1) H, an n by K by N channel matrix. H(:,k, n) is the channel % vector for user k on tone n. This code assumes each user has just one % transmit antenna. n is the number of receive antennas and n=K in DSL. % 2) P, a K by N vector of powers for each user on each tone. % 3) ind, a K by 1 vector that specified the decoding order in the MAC. % ind(k) is the user that is decoded kth in the successive decoding. function [S, Gtot, pBC] = mac2BcMultiTone(H, P, ind) % The code is written for decoding order [1 2 3 ... K] [n, K, N] = size(H); H = H(:,ind,:); P = P(ind,:); Gtot = zeros(n,n,N); S = zeros(n,n,K,N); % after re-ordering the matrices, the decoding order is 1,2,3...,K % Gtot is the transmit covariance matrix of the BC on each tone pBC = zeros(K,N); for i = 1:N r = zeros(K,1); for k = 1:K r(k,1) = log2(det(eye(n) + H(:,k:K,i) * diag(P(k:K, i)) * H(:,k:K,i)') ... / det(eye(n) + H(:,k+1:K,i) * diag(P(k+1:K, i)) * H(:,k+1:K,i)')); end g = 2.^r - 1; w = zeros(n,K); L = zeros(K,K); for k = 1:K w(:,k) = inv(eye(n) + H(:,k+1:K,i) * diag(P(k+1:K, i)) * H(:,k+1:K,i)') * H(:,k,i); w(:,k) = w(:,k) / norm(w(:,k)); L(k,k) = abs(w(:,k)' * H(:,k,i))^2/g(k); for j = 1:k-1 L(k,j) = -abs(w(:,j)' * H(:,k,i))^2; end end pBC(:,i) = L \ ones(K,1); for k = 1:K S(:,:,k,i) = w(:,k) * w(:,k)' * pBC(k,i); Gtot(:,:,i) = Gtot(:,:,i) + S(:,:,k,i); end end pBC(ind, :) = real(pBC); S(:,:,ind,:) = S; | 803 | 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} | 2.65625 | 3 | CC-MAIN-2013-48 | latest | en | 0.901465 |
https://www.12000.org/my_notes/CAS_integration_tests/reports/rubi_4_16_1_graded/test_cases/1_Algebraic_functions/1.2_Trinomial_products/1.2.1_Quadratic/1.2.1.9_P-x-d+e_x-%5Em-a+b_x+c_x%5E2-%5Ep/rese160.htm | 1,696,034,966,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510529.8/warc/CC-MAIN-20230929222230-20230930012230-00378.warc.gz | 693,952,976 | 5,143 | ### 3.160 $$\int \frac{x^3 (1+x+x^2)}{(1-x+x^2)^2} \, dx$$
Optimal. Leaf size=62 $\frac{x^2}{2}+\frac{2 (2-x)}{3 \left (x^2-x+1\right )}+2 \log \left (x^2-x+1\right )+3 x+\frac{10 \tan ^{-1}\left (\frac{1-2 x}{\sqrt{3}}\right )}{3 \sqrt{3}}$
[Out]
3*x + x^2/2 + (2*(2 - x))/(3*(1 - x + x^2)) + (10*ArcTan[(1 - 2*x)/Sqrt[3]])/(3*Sqrt[3]) + 2*Log[1 - x + x^2]
________________________________________________________________________________________
Rubi [A] time = 0.0743509, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 20, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.3, Rules used = {1660, 1657, 634, 618, 204, 628} $\frac{x^2}{2}+\frac{2 (2-x)}{3 \left (x^2-x+1\right )}+2 \log \left (x^2-x+1\right )+3 x+\frac{10 \tan ^{-1}\left (\frac{1-2 x}{\sqrt{3}}\right )}{3 \sqrt{3}}$
Antiderivative was successfully verified.
[In]
Int[(x^3*(1 + x + x^2))/(1 - x + x^2)^2,x]
[Out]
3*x + x^2/2 + (2*(2 - x))/(3*(1 - x + x^2)) + (10*ArcTan[(1 - 2*x)/Sqrt[3]])/(3*Sqrt[3]) + 2*Log[1 - x + x^2]
Rule 1660
Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x + c*
x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x + c*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b
*x + c*x^2, x], x, 1]}, Simp[((b*f - 2*a*g + (2*c*f - b*g)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c
)), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1)*ExpandToSum[(p + 1)*(b^2 - 4*a*c)*Q - (
2*p + 3)*(2*c*f - b*g), x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1
]
Rule 1657
Int[(Pq_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x + c*x^2)^p, x
], x] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
Rule 634
Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] && !NiceSqrtQ[b^2 - 4*a*c]
Rule 618
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]
Rule 204
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])
Rule 628
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Rubi steps
\begin{align*} \int \frac{x^3 \left (1+x+x^2\right )}{\left (1-x+x^2\right )^2} \, dx &=\frac{2 (2-x)}{3 \left (1-x+x^2\right )}+\frac{1}{3} \int \frac{-2+6 x+6 x^2+3 x^3}{1-x+x^2} \, dx\\ &=\frac{2 (2-x)}{3 \left (1-x+x^2\right )}+\frac{1}{3} \int \left (9+3 x-\frac{11-12 x}{1-x+x^2}\right ) \, dx\\ &=3 x+\frac{x^2}{2}+\frac{2 (2-x)}{3 \left (1-x+x^2\right )}-\frac{1}{3} \int \frac{11-12 x}{1-x+x^2} \, dx\\ &=3 x+\frac{x^2}{2}+\frac{2 (2-x)}{3 \left (1-x+x^2\right )}-\frac{5}{3} \int \frac{1}{1-x+x^2} \, dx+2 \int \frac{-1+2 x}{1-x+x^2} \, dx\\ &=3 x+\frac{x^2}{2}+\frac{2 (2-x)}{3 \left (1-x+x^2\right )}+2 \log \left (1-x+x^2\right )+\frac{10}{3} \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,-1+2 x\right )\\ &=3 x+\frac{x^2}{2}+\frac{2 (2-x)}{3 \left (1-x+x^2\right )}+\frac{10 \tan ^{-1}\left (\frac{1-2 x}{\sqrt{3}}\right )}{3 \sqrt{3}}+2 \log \left (1-x+x^2\right )\\ \end{align*}
Mathematica [A] time = 0.0332674, size = 60, normalized size = 0.97 $\frac{x^2}{2}-\frac{2 (x-2)}{3 \left (x^2-x+1\right )}+2 \log \left (x^2-x+1\right )+3 x-\frac{10 \tan ^{-1}\left (\frac{2 x-1}{\sqrt{3}}\right )}{3 \sqrt{3}}$
Antiderivative was successfully verified.
[In]
Integrate[(x^3*(1 + x + x^2))/(1 - x + x^2)^2,x]
[Out]
3*x + x^2/2 - (2*(-2 + x))/(3*(1 - x + x^2)) - (10*ArcTan[(-1 + 2*x)/Sqrt[3]])/(3*Sqrt[3]) + 2*Log[1 - x + x^2
]
________________________________________________________________________________________
Maple [A] time = 0.049, size = 53, normalized size = 0.9 \begin{align*}{\frac{{x}^{2}}{2}}+3\,x+{\frac{1}{{x}^{2}-x+1} \left ( -{\frac{2\,x}{3}}+{\frac{4}{3}} \right ) }+2\,\ln \left ({x}^{2}-x+1 \right ) -{\frac{10\,\sqrt{3}}{9}\arctan \left ({\frac{ \left ( 2\,x-1 \right ) \sqrt{3}}{3}} \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^3*(x^2+x+1)/(x^2-x+1)^2,x)
[Out]
1/2*x^2+3*x+(-2/3*x+4/3)/(x^2-x+1)+2*ln(x^2-x+1)-10/9*3^(1/2)*arctan(1/3*(2*x-1)*3^(1/2))
________________________________________________________________________________________
Maxima [A] time = 1.43281, size = 69, normalized size = 1.11 \begin{align*} \frac{1}{2} \, x^{2} - \frac{10}{9} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x - 1\right )}\right ) + 3 \, x - \frac{2 \,{\left (x - 2\right )}}{3 \,{\left (x^{2} - x + 1\right )}} + 2 \, \log \left (x^{2} - x + 1\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^3*(x^2+x+1)/(x^2-x+1)^2,x, algorithm="maxima")
[Out]
1/2*x^2 - 10/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - 1)) + 3*x - 2/3*(x - 2)/(x^2 - x + 1) + 2*log(x^2 - x + 1)
________________________________________________________________________________________
Fricas [A] time = 2.34005, size = 204, normalized size = 3.29 \begin{align*} \frac{9 \, x^{4} + 45 \, x^{3} - 20 \, \sqrt{3}{\left (x^{2} - x + 1\right )} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x - 1\right )}\right ) - 45 \, x^{2} + 36 \,{\left (x^{2} - x + 1\right )} \log \left (x^{2} - x + 1\right ) + 42 \, x + 24}{18 \,{\left (x^{2} - x + 1\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^3*(x^2+x+1)/(x^2-x+1)^2,x, algorithm="fricas")
[Out]
1/18*(9*x^4 + 45*x^3 - 20*sqrt(3)*(x^2 - x + 1)*arctan(1/3*sqrt(3)*(2*x - 1)) - 45*x^2 + 36*(x^2 - x + 1)*log(
x^2 - x + 1) + 42*x + 24)/(x^2 - x + 1)
________________________________________________________________________________________
Sympy [A] time = 0.143439, size = 60, normalized size = 0.97 \begin{align*} \frac{x^{2}}{2} + 3 x - \frac{2 x - 4}{3 x^{2} - 3 x + 3} + 2 \log{\left (x^{2} - x + 1 \right )} - \frac{10 \sqrt{3} \operatorname{atan}{\left (\frac{2 \sqrt{3} x}{3} - \frac{\sqrt{3}}{3} \right )}}{9} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x**3*(x**2+x+1)/(x**2-x+1)**2,x)
[Out]
x**2/2 + 3*x - (2*x - 4)/(3*x**2 - 3*x + 3) + 2*log(x**2 - x + 1) - 10*sqrt(3)*atan(2*sqrt(3)*x/3 - sqrt(3)/3)
/9
________________________________________________________________________________________
Giac [A] time = 1.19867, size = 69, normalized size = 1.11 \begin{align*} \frac{1}{2} \, x^{2} - \frac{10}{9} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x - 1\right )}\right ) + 3 \, x - \frac{2 \,{\left (x - 2\right )}}{3 \,{\left (x^{2} - x + 1\right )}} + 2 \, \log \left (x^{2} - x + 1\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^3*(x^2+x+1)/(x^2-x+1)^2,x, algorithm="giac")
[Out]
1/2*x^2 - 10/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - 1)) + 3*x - 2/3*(x - 2)/(x^2 - x + 1) + 2*log(x^2 - x + 1) | 3,839 | 7,464 | {"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} | 3.921875 | 4 | CC-MAIN-2023-40 | latest | en | 0.123594 |
https://www.yaclass.in/p/science-state-board/class-7/heat-and-temperature-draft-5441/re-9232d22c-b677-48c4-ba0a-2235b7411305 | 1,632,095,981,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056902.22/warc/CC-MAIN-20210919220343-20210920010343-00611.warc.gz | 1,078,957,816 | 10,276 | ### Theory:
In many scientific applications, it may need to measure the higher than $$42$$$$°C$$ and lesser than $$35$$$$°C$$. In such cases, clinical thermometers can't be used, and the need for laboratory thermometers arise. Laboratory or Lab thermometers are the instruments used to measure the temperatures for scientific research in schools and labs.
Laboratory or Lab thermometers are the instruments used to measure the temperatures for scientific research in schools and labs.
Laboratory thermometer
Lab thermometers are designed to measure substances' temperatures and also find the boiling point and freezing point.
Experimental setup to measure the temperature
In industries, it is used to measure the temperature higher than what clinical thermometers can do.
Structure of the laboratory thermometer:
• Compared to clinical thermometers, lab thermometers have a long narrow tube (stem) and bulb.
• Laboratory thermometers do not have a kink.
Temperature scale in laboratory thermometer:
• It has only the Celsius scale ranging from $$−$$$$10$$$$°C$$ to $$110$$$$°C$$.
Important!
Precautions to be followed while using a laboratory thermometer:
• Do not tilt (bend) the lab thermometer while doing the temperature measurement. Place it upright.
• Record the readings only when the substances from all the sides have surrounded the bulb.
• Make sure that the bulb is not be in contact with the bottom or sides of the container.
• While doing the measurements, do not move the thermometer.
Differences between clinical and laboratory thermometers:
Parameter to be compared Clinical thermometer Laboratory thermometer Application Clinical (Hospitals). Scientific Research and Laboratories Uses It is used to measure the human body temperature. It is used to determine the boiling point and freezing point of the substances. Range $$35$$$$°C$$ to $$42$$$$°C$$ / $$94$$$$°F$$ to $$108$$$$°F$$ $$10$$$$°C$$ to $$110$$$$°C$$ Structure Mercury level does not fall on its own, as there is a kink near the bulb, which stops the mercury from returning into the bulb Mercury level falls on its own as no kink is present. Working We can read the temperature after removing the thermometer from the armpit or mouth. Temperature is read while keeping the thermometer in the temperature source, e.g. a liquid or any other thing. Restoration Jerks are given to lower the mercury level. No need to give a jerk to lower the mercury level. | 512 | 2,437 | {"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 | 3 | CC-MAIN-2021-39 | latest | en | 0.862484 |
https://socratic.org/questions/how-do-you-convert-87-98-ml-to-liters | 1,603,823,613,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107894426.63/warc/CC-MAIN-20201027170516-20201027200516-00525.warc.gz | 538,731,628 | 5,863 | # How do you convert 87.98 mL to liters?
Divide thru by ${10}^{3}$.
$1 \cdot m L$ $=$ $1 \times {10}^{-} 3 \cdot L$, the $m$ specifies $\text{milli} = {10}^{-} 3$.
Thus $87.98 \cdot m L$ $=$ $87.98 \cdot \cancel{m L} \times {10}^{-} 3 \cdot L \cdot \cancel{m {L}^{-} 1}$ $=$ ??L | 124 | 279 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 11, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2020-45 | latest | en | 0.252894 |
https://whatisconvert.com/721-feet-second-in-kilometers-hour | 1,675,630,569,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500288.69/warc/CC-MAIN-20230205193202-20230205223202-00758.warc.gz | 631,507,302 | 7,283 | ## Convert 721 Feet/Second to Kilometers/Hour
To calculate 721 Feet/Second to the corresponding value in Kilometers/Hour, multiply the quantity in Feet/Second by 1.0972799999991 (conversion factor). In this case we should multiply 721 Feet/Second by 1.0972799999991 to get the equivalent result in Kilometers/Hour:
721 Feet/Second x 1.0972799999991 = 791.13887999937 Kilometers/Hour
721 Feet/Second is equivalent to 791.13887999937 Kilometers/Hour.
## How to convert from Feet/Second to Kilometers/Hour
The conversion factor from Feet/Second to Kilometers/Hour is 1.0972799999991. To find out how many Feet/Second in Kilometers/Hour, multiply by the conversion factor or use the Velocity converter above. Seven hundred twenty-one Feet/Second is equivalent to seven hundred ninety-one point one three nine Kilometers/Hour.
## Definition of Foot/Second
The foot per second (plural feet per second) is a unit of both speed (scalar) and velocity (vector quantity, which includes direction). It expresses the distance in feet (ft) traveled or displaced, divided by the time in seconds (s, or sec). The corresponding unit in the International System of Units (SI) is the metre per second. Abbreviations include ft/s, ft/sec and fps, and the rarely used scientific notation ft s−1.
## Definition of Kilometer/Hour
The kilometre per hour (American English: kilometer per hour) is a unit of speed, expressing the number of kilometres travelled in one hour. The unit symbol is km/h. Worldwide, it is the most commonly used unit of speed on road signs and car speedometers. Although the metre was formally defined in 1799, the term "kilometres per hour" did not come into immediate use – the myriametre (10,000 metres) and myriametre per hour were preferred to kilometres and kilometres per hour.
## Using the Feet/Second to Kilometers/Hour converter you can get answers to questions like the following:
• How many Kilometers/Hour are in 721 Feet/Second?
• 721 Feet/Second is equal to how many Kilometers/Hour?
• How to convert 721 Feet/Second to Kilometers/Hour?
• How many is 721 Feet/Second in Kilometers/Hour?
• What is 721 Feet/Second in Kilometers/Hour?
• How much is 721 Feet/Second in Kilometers/Hour?
• How many km/h are in 721 ft/s?
• 721 ft/s is equal to how many km/h?
• How to convert 721 ft/s to km/h?
• How many is 721 ft/s in km/h?
• What is 721 ft/s in km/h?
• How much is 721 ft/s in km/h? | 616 | 2,407 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2023-06 | latest | en | 0.872861 |
http://mathoverflow.net/feeds/question/22454 | 1,369,516,827,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368706413448/warc/CC-MAIN-20130516121333-00081-ip-10-60-113-184.ec2.internal.warc.gz | 161,851,618 | 1,817 | Connected subset of matrices ? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T21:20:35Z http://mathoverflow.net/feeds/question/22454 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/22454/connected-subset-of-matrices Connected subset of matrices ? Portland 2010-04-24T21:20:49Z 2010-04-25T15:21:32Z <p>Let $m,n$ be positive integers with $m \leqslant n$, and denote by $\mu_M$ the minimal polynomial of a matrix.</p> <p>Do we know for which $m$ the set $E_m$ of $M \in \mathfrak{M}_n(\mathbb{R})$ such that $\deg(\mu_M) = m$ is connected?</p> http://mathoverflow.net/questions/22454/connected-subset-of-matrices/22496#22496 Answer by Robin Chapman for Connected subset of matrices ? Robin Chapman 2010-04-25T08:26:21Z 2010-04-25T15:21:32Z <p>The answer is always yes. Indeed the set is path-connected.</p> <p>Let $C(f)$ denote the companion matrix associated to the monic polynomial $f$. Every matrix $A$ is similar to a matrix in rational canonical form: $$B=C(f_1)\oplus C(f_1 f_2)\oplus\cdots\oplus C(f_1 f_2,\cdots f_k)$$ where here $\oplus$ denotes diagonal sum. Then $m$ is the degree of $f_1 f_2\cdots f_k$. Starting with $B$ deform each $f_i$ into a power of $x$. We get a path from $B$ to $$B'=C(x^{a_1})\oplus C(x^{a_2})\oplus\cdots\oplus C(x^{a_k})$$ inside $E_m$. There's a path from $B'$ in $E_m$ given by $$(1-t)C(x^{a_1})\oplus (1-t)C(x^{a_2})\oplus\cdots\oplus C(x^{a_k})$$ ending at $$B_m=O\oplus C(x^m).$$ Thus there is a path in $E_m$ from $A$ to $UB_mU^{-1}$ where $U$ is a nonsingular matrix. If $\det(U)\ne0$ then there is a path in $GL_n(\mathbf{R})$ from $U$ to $I$ and so a path in $E_m$ from $A$ to $B_m$. If $m< n$ then there is a matrix $V$ of negative determinant with $VB_m V^{-1}=B_m$ so that we may take $U$ to have positive determinant.</p> <p>The only case that remains is when $m=n$. In this case $E_m$ contains diagonal matrices with distinct entries, and each of these commutes with a matrix of negative determinant.</p> | 697 | 2,022 | {"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} | 3.84375 | 4 | CC-MAIN-2013-20 | latest | en | 0.727186 |
https://www.cfd-online.com/Forums/siemens/52971-user-viscosity-routine-user-scalars.html | 1,679,842,728,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945473.69/warc/CC-MAIN-20230326142035-20230326172035-00612.warc.gz | 769,608,777 | 15,439 | # User viscosity routine and user scalars?
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July 21, 2002, 18:00 User viscosity routine and user scalars? #1 Jeff Guest Posts: n/a Hi All I asked this question in the general forum and someone pointed me to star-cd. As it turns out, I am pretty sure we have star at work so I would like to ask the same question here to confirm that I can do what I need to do. I need to be able to model a non-Newtonian fluid and it will require a user subroutine. The problem would require standard arguments to the viscosity function like Temperature, strain-rate, etc.. but would also require the gradient of a user scalar variable, which is the solution to a Poisson equation. It would look something like this: mu=mu(T, s-dot, |grad(c)|) Can Star-CD handle this situation? Hopefully, I will get a hold of some manuals next week. Is there a specific section for me to look at? Thanks Jeff
July 22, 2002, 04:43 Re: User viscosity routine and user scalars? #2 Richard Guest Posts: n/a STAR can handle this situation. The subroutine you need to use is VISMOL, but you'd need to contact your local CD-adapco support office to get help on incorporating scalar gradients into the subroutine. It's easy to do, but not without some help from a support engineer.
July 22, 2002, 07:28 Thanks, I'll give them a call.(nm) #3 Jeff Guest Posts: n/a Thanks
January 6, 2003, 05:26 Re: User viscosity routine and user scalars? #4 Alexander Guest Posts: n/a Hi Jeff! I have to solve the similar ploblem. The coefficient in the motion equation is governed by Poisson equation. After solving Poisson equation it is necessary to calculate the gradient of the scalar. What STAR technique would you suggest to use: a) to solve Poisson equation; b) to calculate the gradient (vector) of the scalar. Thanks Alexander | 449 | 1,862 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-14 | longest | en | 0.938598 |
https://www.physicsforums.com/threads/finding-rotational-matrix-from-axis-angle-representation.785438/ | 1,508,599,401,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824819.92/warc/CC-MAIN-20171021133807-20171021153807-00676.warc.gz | 968,199,109 | 16,588 | # Finding Rotational matrix from axis-angle representation
1. Dec 3, 2014
### fakecop
1. The problem statement, all variables and given/known data
Given an axis vector u=（-1, -1, -1) , find the rotational matrix R corresponding to an angle of pi/6 using the right hand rule. Then find R(x), where x = (1,0,-1)
2. Relevant equations
I found the relevant equation on wikipedia (see attachment)
3. The attempt at a solution
I feel that I'm doing something wrong but I don't know what; I think I performed the calculations correctly.
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2.4 KB
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170.8 KB
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2. Dec 5, 2014
### Simon Bridge
What makes you think something went wrong?
Have you tried testing it, i.e. by applying R to u, or to a vector perpendicular to u, and see if you get the right result? | 227 | 823 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2017-43 | longest | en | 0.879717 |
https://riddles360.com/riddle/serve-you-by-expiring | 1,669,478,780,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446708010.98/warc/CC-MAIN-20221126144448-20221126174448-00377.warc.gz | 537,146,747 | 7,063 | # Serve you by Expiring
You measure my life in hours
I serve you by expiring
I vanish faster when I am thin
and slower when I am thick
The wind is my energy
What am I?
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Birbal was jester, counsellor, and fool to the great Moghul emperor, Akbar.
The villagers loved to talk of Birbal's wisdom and cleverness,
and the emperor loved to try to outsmart him.
One day Akbar (emperor) drew a line across the floor.
"Birbal," he ordered, "you must make this line shorter, but you cannot erase any bit of it."
Everyone present thought the emperor had finally outsmarted Birbal.
It was clearly an impossible task.
Yet within moments the emperor and everyone else present had to agree that Birbal had made the line shorter without erasing any of it.
How could this be?
Asked by Neha on 17 May 2021
### Amazing Facts
###### Artificial Intelligence
Artificial Intelligence has crushed all human records in the puzzle game “2048,” achieving a high score of 839,732 and beating the game in only 973 moves without using any undo. | 896 | 3,569 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | longest | en | 0.990205 |
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Re: CR - Blend it well [#permalink] 03 Mar 2011, 13:32
C!
if the equipment is no different and the beans are better then it stands to reason that atleast these two items would not cause bad coffee and hence nothing in the question stem or answer is overlooked here!
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Re: CR - Blend it well [#permalink] 04 Mar 2011, 02:53
C)
Once you read this argument's facts and its conclusion, you can pretty much know what the flaw is. By the way, I'd suggest that you read the question first and then read the argument.
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Re: CR - Blend it well [#permalink] 04 Mar 2011, 03:03
Very easy. C is the correct variant.
AD sugests: the company buys more colombian bean - so the company`s coffe must be the best.
But the possible solution may be that it just sells more coffee.
That variant corresponds us ti take C.
Problem for 45-60 seconds max.
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Re: CR - Blend it well [#permalink] 04 Mar 2011, 03:34
C for Sure.
What if Company is producing such a large number of cans tht theres only some miligram of coffee in each, in such a case the reasoning in the argument fails.
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Re: CR - Blend it well [#permalink] 05 Mar 2011, 20:03
Agree with Tommy, this is a very Gmat like question
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Re: CR - Blend it well [#permalink] 07 Mar 2011, 12:51
priyankur_saha@ml.com wrote:
Advertisement: The world’s best coffee beans come from Colombia. The more Colombian beans in a blend of coffee, the better the blend, and no company purchases more Colombian beans than Kreemo Coffee. Inc. So it only stands to reason that if you buy a can of Kreemo’s coffee, you’re buying the best blended coffee available today.
The reasoning of the argument in the advertisement is flawed because it overlooks the possibility that
(A) the equipment used by Kreemo to blend and package its coffee is no different from that used by most other coffee producers
-->irrelevant
(B) not all of Kreemo’s competitors use Colombian coffee beans in the blends of coffee they sell
-->if not all use Colombian coffee than Kreemo's might be the better quality choice.. this looks to strengthen argument
(C) Kreemo sells more coffee than does any other company
-->leave for now and look for others to eliminate
(D) Kreemo’s coffee is the most expensive blended coffee available today
-->irrelevant, we are looking at coffee blends and not pricing to determine quality
(E) the best unblended coffee is better than the best blended coffee
-->well yes, this could be true but does it have anything to do with our particular argument? We are talking about BLENDED coffee only
How to solve this kind of questions? Let me know how you workout on this question.
I burned 2:16 on this question and actually had to use POE again to come up with the right answer.
So this leaves me with B and C. B looks to strengthen the argument. If we break down C we say that even though Kreemo buys the most Colombian coffee than any of its competitors because it sells more coffee than does any other company the blend MAY be low per bag resulting in lower quality. This MAY is important.
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Re: CR - Blend it well [#permalink] 14 May 2011, 04:49
1
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I would provide may be the best explanation to this question.
let us say there are three companies A, B and Kreemo Coffee (KC).
Company:________________________ A | B | KC |
Columbian beans used in production:___1 | 2 | 20 |
Total beans used in production:_______10 |10 |1000| <- WHO PRODUCES/SELLS MORE?!
concentration (quality):_____________10%|20%| 2%| <- What is the best quality?
The higher the concentration the better the quality of coffe.
therefore C is the best answer, whihc is saying that although KC sells more it's quality isn't the best.
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Re: CR - Blend it well [#permalink] 14 May 2011, 06:11
clean C.
assumption in conclusion. Hence flaw in conclusion.
C brings out the assumption that more coffee does not mean more blended coffee produced.
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Re: CR - Blend it well [#permalink] 14 May 2011, 06:55
priyankur_saha@ml.com wrote:
Advertisement: The world’s best coffee beans come from Colombia. The more Colombian beans in a blend of coffee, the better the blend, and no company purchases more Colombian beans than Kreemo Coffee. Inc. So it only stands to reason that if you buy a can of Kreemo’s coffee, you’re buying the best blended coffee available today.
The reasoning of the argument in the advertisement is flawed because it overlooks the possibility that
(A) the equipment used by Kreemo to blend and package its coffee is no different from that used by most other coffee producers
(B) not all of Kreemo’s competitors use Colombian coffee beans in the blends of coffee they sell
(C) Kreemo sells more coffee than does any other company
(D) Kreemo’s coffee is the most expensive blended coffee available today
(E) the best unblended coffee is better than the best blended coffee
How to solve this kind of questions? Let me know how you workout on this question.
This question is flawed in my opinion, because none of the options really present a correlation between columbian beans and the coffee sold as cans.
If I had an options saying, "it overlooks the possibility that the coffee blend in the can may not even contain the purchased Columbian beans".
What's the guarantee that Kreemo’s Coffee is not reselling the columbian beans at a higher price and used some other bean for the coffee blend in the can the ad talks about. The other flaw is that there's no comparative analysis between the concentration of coffee beans used by Kreemo's and its competitors.
"C", in my opinion, could be a possible assumption but not a flaw. Kreemo's Coffee cans could be best even if they don't get sold as much as their competitor's. Where is it mentioned that quantity of sale determines the quality of a product. Maybe the company is not able to sell its coffee because of poor marketing. Thus, Kreemo's cofee could possibly be the best coffee inspite of not having a good sale.
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Re: CR - Blend it well [#permalink] 14 May 2011, 09:15
2
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Fluke,
I might agree with you to a certain extent.
the source is unknown (well, it is from 1000 series, but is it a seflmade question or it came from GMAT-paper tests, or any other source i don't know).
A good question does not require additional assumptions to arrive to correct answer choice.
The argument says:
premise 1: "the more Colombian beans in a blend of coffee, the better the blend".
It means that best blend is achieved when a coffee maker uses 100% of Columbian beans.
premise 2: "no company purchases more Colombian beans than Kreemo Coffee. Inc"
Conclusion "Kreemo’s coffee is best blended coffee " (this is why E is wrong).
So the K's coffee is best because they buy a lot of C beans.
To weaken a question one must weaken the conclusion. But, the correct answers do not have to absolutely, without a doubt, in all circumstances weaken the conclusion. They only have to open up the possibility that the conclusion is not valid. The correct answer on a weaken question will typically accomplish this by introducing a new piece of information that calls into question an assumption made by the author.
The possible assumptions made by author are:
1. KC does not resell Columbian beans,
2. KC uses all Columbian beans in production of coffee
3. None of competitors puts more Columbian beans in coffee than KC does.
4. Quantity of Columbian beans and the coffee sold as cans perfectly correlates.
5. beans are used in production of coffee.
6....etc
You are right saying that:
Quote:
it overlooks the possibility that the coffee blend in the can may not even contain the purchased Columbian beans
your statement attacks the 2nd assumption.
I definitively agree with this statement:
Quote:
" is not reselling the columbian beans at a higher price and used some other bean for the coffee blend"
it weakens the 1st assumption.
this is are both very valid weakening arguments, but in test-makers view these are too easy to spot.
Thus probably they decided to use the argument that weakens the 3rd assumption:
"C) Kreemo sells more coffee than does any other company "
but this answer is too stretched. Why?!
Because it requires additional assumptions, such as "all coffee beans that KC buys, it uses later in production", "purchases are equal to sales", "no columbian beans are stored in stock", etc...
But C opens up a possibility that high volumes does not necessarily mean high % or in our case higher quality.
In my example above you may change the figure 20 in 400 resulting in 40% instead of 2% concentration of Columbian beans in a cup of coffee. This strenghtens the argument.
But also the total volume of beans a company purchases leaves a gap in possibility to intrepret differently the given information.
If the argument said that KC sells less coffee than any other company and buys more columbian beans than any other company, it would be much difficult to weaken such argument. (becuase it would result in higher concentration of coffee beans).
Hope it helps.
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Re: CR - Blend it well [#permalink] 14 May 2011, 16:40
Suppose,
Kreemo sells 100 pounds of coffee and uses 50 pounds of Colombian beans.
XYZ another company sells 20 pounds of coffee and uses 20 pounds of beans.
XYZ sells better quality blended coffee than Kreemo because it uses 100% Colombian beans.
I hope it is clear.
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Re: CR - Blend it well [#permalink] 16 May 2011, 09:13
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C can not be correct. Kreemo sells more coffee, so it might be using less good beans per can and it might just be selling more because it has a larger market share.
E is the correct answer, because it allows us to know that you might not have the best blended coffee although having the best columbian beans (the unblended).
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I got the answer rather quickly but may have been lucky. When I read the prompt I immediately made the connection that purchasing more beans does not mean having more beans in the product you sell. The only answer choice that highlighted this discrepancy was answer C.
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picked c but it was tricky.
if you sell a more coffee, you can lose the "prime best blend" ratio if that makes sense
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well tricky question.
good explanation by matt +1
thanks
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I think its C...easy one!
Cheers..
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Advertisement: The world’s best coffee beans come from Colombia. The more Colombian beans in a blend of coffee, the better the blend, and no company purchases more Colombian beans than Kreemo Coffee. Inc. So it only stands to reason that if you buy a can of Kreemo’s coffee, you’re buying the best blended coffee available today.
The reasoning of the argument in the advertisement is flawed because it overlooks the possibility that
(A) the equipment used by Kreemo to blend and package its coffee is no different from that used by most other coffee producers
(B) not all of Kreemo’s competitors use Colombian coffee beans in the blends of coffee they sell
(C) Kreemo sells more coffee than does any other company
(D) Kreemo’s coffee is the most expensive blended coffee available today
(E) the best unblended coffee is better than the best blended coffee
According to most posts on here, this is supposedly a weaken question. However, the question askes to identify the flaw in the reasoning - which points to a FLAW question, an entirely different type of question.
KAPLAN BOOK: For Flaw questions, find the conclusion, evidence, and central assumption(s), then pre-phrase an answer that indicates the logical fallacy of the assumption
What is the central assumption? The central assumption is, that since Kreemo buys most of the Colombian coffee (which is the best) that it sells the highest quality blend.
A flaw question requires you to find the logical fallacy of the assumption.
(A) If the central assumption is that the amount of colombian coffee purchased is the only indicator for a high quality blend, it certainly includes the assumption that any kind of 'blending' would be indentical among competitors - which may not be the case. By considering that the 'blending' also impacts the quality, the argument becomes flawed.
(C) If the central assumption is that the amount of colombian coffee purchased is the only indicator for a high quality blend, the fact that Kreemo sells more coffee than its competitor has nothing to do with the amount of the blend in it - at least without massive mental fabrication of coffeebean ratios. Since Kreemo buys more Colombian coffee than anyone else, it is reasonable to assume that this fact alone is the cause.
Lets try this again, assuming this is a weaken question.
For Weaken questions, find the conclusion, evidence, and central assumption(s), then pre-phrase an answer that denies (weakens) a central assumption, then go to answer choices
The conclusion is that Kreemo makes the best coffee blend. The Evidence is that Kreemo buys more colombian coffee than any other competitor and the central assumption is that the coffee quality only relies on the amount of colombian coffee in the blend.
A similar logic than above can reasonably be applied.
If this is an official GMAC question and (C) is the official answer, I would like to hear what the fact that Kreemo sells more coffee than anybody else has to do with the central assumption - WITHOUT adding excessive fabrication and numerous additional assumptions about the quality of coffee that may or may not be thinned out or blended with other coffees.
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confused... . all options seem wrong to me
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Re: CR - Blend it well [#permalink] 02 Jan 2013, 06:19
TommyWallach wrote:
Hey All,
Everybody seems to be on the right track with this one, but I wanted to respond to a few complaints. This is a VERY realistic GMAT question. There are no ambiguous answers. The problem is only if you spend too much time writing stories in your head, in which case, any answer choice can begin to look good. This argument says they buy the most Colombian beans, but think of it this way:
If I buy 10 beans and you only buy one bean, it may look like I must have the best coffee.
But if you only sell one cup of coffee, your one bean means the overall quality of your coffee is super excellent.
If I then sell 10,000 cups of coffee, splitting my ten beans between them, the overall quality of my coffee is super poor.
This is why the answer is what it is. Don't get into thought processes involving the history of the company, why people buy their coffee, what effect technology may have...ANYTHING that isn't in the passage itself. Keep in mind that this question asks you weaken THE ARGUMENT, not THE CONCLUSION, so you won't be bringing in any new information.
Word up.
Hope that helps!
Hi,
Hope we can discuss no matter what the correct answer is:
Conclusion Kreemo’s Coffee Buying the best blended coffee available today.
Consider this :
Kreemo's Sale : 10 M.....Near Competitor : 9 .5 M
Kreemo's no doubt selling more coffee than others , but this has no relation with buying and blended coffee.
Regards
Sanjiv
Re: CR - Blend it well [#permalink] 02 Jan 2013, 06:19
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# ITS AN OLD MACHINE,
## BUT IT STILL WORKS...
UNTIL THAT CRUCIAL PIECE BREAKS DOWN
AND THE ORIGINAL DRAWINGS HAVE BEEN LOST.
HERES HOW TO OVERCOME THAT OBSTACLE.
By Jim Marsch
20
GEAR SOLUTIONS
DECEMBER 2005
gearsolutionsonline.com
## everse engineeringthe deconstruction and analysis of an object for
the purpose of constructing a copy or replacement of the original
component when the original design or drawing is unavailableis
especially helpful in making repairs or in improving old designs. For
example, lets say you need to manufacture a replacement for a broken gear, but you dont have the original drawing. This is not an uncommon
situation. The question is, what is the best way to reverse engineer a gear
when you have the part in front of you, but do not have the part drawing?
Most people will choose to approach this problem by guessing, and oftentimes will not be successful in duplicating the gear. However, this does not
have to be the case. Reverse engineering of a gear is a science. I have outlined the procedure below in five simple steps (calculations are made using
Integrated Gear Software).
## Step 1: Calculate the Normal Base Pitch
The normal base pitch is the distance on a normal base helix between corresponding involutes of adjacent teeth. To calculate, make two measurements using a
span micrometer. If the first is over four teeth then the second has to be over five
teeth. Trial and error shows that we can measure across five teeth without contacting the OD and four teeth without contacting the root. The difference between fourand five-teeth readings is the normal base pitch of the gear. Better accuracy will be
obtained if a number of readings are taken at four and at five teeth and then averaged. In this example the normal base pitch works out to be 0.949 in.
Step 2: Measure the Outside Diameter of the Gear
Measure the outside diameter of the gear in inches using a set of calipers.
Step 3: Measure the Root Diameter of the Gear
Measure the root diameter, measured across the bottoms of opposite tooth
spaces. You can also measure the root fillet radius if you have a radius
gage or another method.
Step 4: Calculate the Diametral Pitch
Do this by assuming the pressure angle, the angle between the tooth profile and
a radial line at its pitch point (try 20 and 25 degrees as they are more commonly
used). Typically the diametral pitch, the number of teeth per inch of pitch diameter,
will be a whole number such as three, eight, 10, etc. The following example
shows a sample calculation (note: different combinations of diametral pitch and
pressure angle will yield the same normal base pitch). From the span measurement, we will calculate the normal tooth thickness at the reference pitch diameter.
FIGURE 1
gearsolutionsonline.com
DECEMBER 2005
GEAR SOLUTIONS
21
The diametral pitch is 3.110781. Since it is not an integer number let us try a 25-degree pressure angle.
Technical Solutions
Aerospace and Automotive Applications
Multiple Start Gear
Grinding Wheels
Lower grinding forces
New abrasive blends & bonds
Lower grinding temperatures
Increased porosity for higher
stock removal
## Gear Honing Benefits
Flank correction
Reduced operating noise
Longer service life
Correction for distortion from
hardening process
FIGURE 2
## Hermes Abrasives, Ltd.
524 Viking Drive Virginia Beach, VA 23452
PO Box 2389 Virginia Beach, VA 23450
## Toll free phone: 800.464.8314
Toll free fax: 800.243.7637
## This time the diametral pitch comes out to be 3.000263. This
can be rounded off to three. So, let us enter 3.00 as the diametral
pitch and make another calculation, and calculate the normal base
pitch. It comes out to be 0.9491 with normal tooth thickness calculated to be 0.6391.
FIGURE 3
## Step 5: Working with the Hob
Work with the chosen hob to determine the geometry of the gear
that the hob will produce. You need to have the hob drawing or a
way to measure the hob cutting edge geometry. A minimum of two
dimensions will be required from the hob drawing: tooth thickness,
and; tip to reference dimension. Also note the tip radius.
Figure 6 shows the data for the gear which will be cut using the
chosen hob mentioned above.
22
GEAR SOLUTIONS
DECEMBER 2005
gearsolutionsonline.com
## Gears complete to print up to 16 inches in
diameter offering AGMA class 12 quality
Gear grinding services up to 27.5 inches in
diameter featuring Hofler Helix 400 and 700
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FIGURE 4
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Analytical Gear Charts to insure quality
FIGURE 5
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PHONE: 815-874-3948 FAX: 815-874-3817
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## How do you improve
cutting efficiency, lower costs
and increase tool life?
FIGURE 6
If the calculated root diameter is different from what was measured, input the measured root diameter and calculate the tip to reference line dimension for the hob.
FIGURE 7
## Ionbond offers gear makers
engineered services for
high value consumable
cutting tools. All makes and
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Tel. 1-800-929-1794, E-Mail: info@ionbond.com.
## To assure that you have accurately reverse engineered the gear,
get a scaled plot of the gear tooth on a transparency paper.
gearsolutionsonline.com
DECEMBER 2005
GEAR SOLUTIONS
23
FIGURE 8
Compare that image with the image of the actual gear tooth seen
on a shadow graph or other similar equipment. If the two match,
great. If not, run the software again to get the required accuracy.
In the case of a helical gear you will need to measure the lead
of the gear by using a lead checker or a more precise method of
measurement, such as a computerized gear checker. All other
steps for reverse engineering the gear will remain the same.
As I mentioned earlier, my objective is to lay out an easy to use
me and let me know. Include other topics regarding the design and
manufacturing of metal or plastic gears that interest you.
Jim Marsch is gear product manager for Universal Technical Systems,
Inc. He can be reached at jimm@uts.com. The companys Web site is
[www.uts.com].
FIGURE 9
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HOBS | 1,758 | 7,092 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2019-30 | latest | en | 0.896557 |
https://eazyquizes.com/product/advanced-construction-and-carpentry-skills-by-bonnici-1st-edition-test-bank/ | 1,685,301,119,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644506.21/warc/CC-MAIN-20230528182446-20230528212446-00036.warc.gz | 268,344,979 | 25,663 | Eazyquizes
# Advanced Construction And Carpentry Skills by Bonnici 1st Edition – Test Bank
\$25.00
Category:
## Description
Advanced Construction And Carpentry Skills by Bonnici 1st Edition – Test Bank
Sample Questions
Chapter 02 Testbank
Student: ___________________________________________________________________________
1. The span of an oblique ended roof is 5.2 metres. The oblique offset is 2.150 metres. To the nearest millimetre, the length of the oblique plate is therefore:
A. 5.892
B. 5.544
C. 5.627
D. 6.190
1. The plumb bevel for the creeper of an oblique ended hipped roof is the same as the:
A. common rafter plumb bevel
B. hip plumb bevel
C. crown end edge bevel
D. hip edge bevel
1. On a roof with unequal pitches, if you are not required to have the eave widths the same all around the building you should:
A. lower the height of the wall on the side with the steepest pitch
B. raise the height of the wall on the side with the steepest pitch
C. leave all the wall heights at the same level
D. temporarily support the side with the lowest pitch until you know the final wall height
1. The hips and valleys of a roof with an unequal pitch will:
A. bisect the corner of the building
B. not bisect the corner of the building
C. have different bevels
D. need to be measured off the job because they cannot be calculated
1. The horizontal difference for creepers on an oblique-ended hipped roof is different for each hip.
True False
1. The plumb cut on the crown end rafter of a hipped roof with unequal pitches will be the same as the rafters from both sides of the roof.
True False
1. The advantage of the scotch valley in doing renovations is that you don’t have to remove the rafters from the existing roof.
True False
1. A right-angled triangle will always be found by drawing lines from each end of the diameter and having them meet at the circumference.
True False
Chapter 02 Testbank Key
1. The span of an oblique ended roof is 5.2 metres. The oblique offset is 2.150 metres. To the nearest millimetre, the length of the oblique plate is therefore:
A.5.892
B. 5.544
C. 5.627
D. 6.190
Competency: CPCCC3009B Construct advanced roofs
Difficulty: Medium
Performance criteria: CPCCC3009B 2.1
1. The plumb bevel for the creeper of an oblique ended hipped roof is the same as the:
A.common rafter plumb bevel
B. hip plumb bevel
C. crown end edge bevel
D. hip edge bevel
Competency: CPCCC3009B Construct advanced roofs
Difficulty: Medium
Performance criteria: CPCCC3009B 2.1
1. On a roof with unequal pitches, if you are not required to have the eave widths the same all around the building you should:
A.lower the height of the wall on the side with the steepest pitch
B. raise the height of the wall on the side with the steepest pitch
C. leave all the wall heights at the same level
D. temporarily support the side with the lowest pitch until you know the final wall height
Competency: CPCCC3009B Construct advanced roofs
Difficulty: Medium
Performance criteria: CPCCC3009B 2.1
1. The hips and valleys of a roof with an unequal pitch will:
A.bisect the corner of the building
B. not bisect the corner of the building
C. have different bevels
D. need to be measured off the job because they cannot be calculated
Competency: CPCCC3009B Construct advanced roofs
Difficulty: Medium
Performance criteria: CPCCC3009B 2.1
1. The horizontal difference for creepers on an oblique-ended hipped roof is different for each hip.
TRUE
Competency: CPCCC3009B Construct advanced roofs
Difficulty: Medium
Performance criteria: CPCCC3009B 2.1
1. The plumb cut on the crown end rafter of a hipped roof with unequal pitches will be the same as the rafters from both sides of the roof.
FALSE
Competency: CPCCC3009B Construct advanced roofs
Difficulty: Medium
Performance criteria: CPCCC3009B 2.1
1. The advantage of the scotch valley in doing renovations is that you don’t have to remove the rafters from the existing roof.
TRUE
Competency: CPCCC3009B Construct advanced roofs
Difficulty: Medium
Performance criteria: CPCCC3009B 2.1
1. A right-angled triangle will always be found by drawing lines from each end of the diameter and having them meet at the circumference.
TRUE
Competency: CPCCC3009B Construct advanced roofs
Difficulty: Medium
Performance criteria: CPCCC3009B 2.1
Chapter 02 Testbank Summary
Category # of Questions Competency: CPCCC3009B Construct advanced roofs 8 Difficulty: Medium 8 Performance criteria: CPCCC3009B 2.1 8
Chapter 04 Testbank
Student: ___________________________________________________________________________
1. A horizontal line of notches may be cut out around a bath for the purpose of fitting a bath edge support. The maximum depth of these cut outs is:
A. 20 mm
B. 25 mm
C. 30 mm
D. 35 mm
1. Before lining the walls you must ensure that you have accounted for:
A. any additional framing needed to support fixtures such as vanities, shower screens and hand rails
B. the compliance of fixtures to applicable water ratings
C. the quantity of lining material is sufficient to do the job
D. the room size is adequate for the fixtures proposed to go in it
1. The purpose of a capillary break is to:
A. allow the frame to move without affecting fixtures such as bath, vanities or shower bases
B. allow room for shower bases and baths to expand and contract with changes in temperature from hot water
C. stop water from bridging between fixtures and being drawn into the frame
D. provide a space into which a sealant may be pumped to stop leaks but allow movement
1. The installation of a spa bath over a bearer and joist flooring system may require you to:
A. use a heavier floor sheeting such as 19 mm thick fibre cement sheet
B. change the plumbing layout
C. install additional subfloor framing and support members
D. paint a waterproof sealant under the spa bath area
1. If the shower base you are about to install is twisted such that you can only get two edges level, not all four, then you should:
A. use a heavy object (bags of cement, buckets of water) to twist it back into shape
B. use long timbers to wedge it down from the ceiling until the mortar sets
C. level the two front edges as this is where the shower screen must go
D. level the two back edges so the wall tiles will not end up being cut at an angle
1. Kitchens and exposed balconies are not included in the NCC Vol. 2 BCA’s definition of wet areas.
True False
1. When the NCC Vol. 2 BCA specifies that a surface must be water resistant it means it must be coated with a waterproof compound and then surfaced with an abrasive-resistant material such as grouted tiles, plastic laminates or similar.
True False
1. In one method of installing a bath, a series of noggings should be installed around the bath above the checkout. The purpose of these noggings is to provide fixing support for the internal linings.
True False
1. Waterproofing is a critical but small job on a domestic building site. For that reason it is not expected that you will have access on site to the material safety data sheets (MSDS) for the all the various materials involved.
True False
1. An advantage of wet area plasterboard over fibre cement sheet as a lining in bathrooms is that it has significantly better thermal and acoustic properties. In addition, it also handles frame movement better.
True False
Chapter 04 Testbank Key
1. A horizontal line of notches may be cut out around a bath for the purpose of fitting a bath edge support. The maximum depth of these cut outs is:
A.20 mm
B. 25 mm
C. 30 mm
D. 35 mm
Competency: CPCCCA3012B Frame and fit wet area fixtures
Difficulty: Easy
Performance criteria: CPCCCA3012B 1.1
Performance criteria: CPCCCA3012B 3.1
1. Before lining the walls you must ensure that you have accounted for:
A.any additional framing needed to support fixtures such as vanities, shower screens and hand rails
B. the compliance of fixtures to applicable water ratings
C. the quantity of lining material is sufficient to do the job
D. the room size is adequate for the fixtures proposed to go in it
Competency: CPCCCA3012B Frame and fit wet area fixtures
Difficulty: Medium
Performance criteria: CPCCCA3012B 3.3
Performance criteria: CPCCCA3012B 4.3
1. The purpose of a capillary break is to:
A.allow the frame to move without affecting fixtures such as bath, vanities or shower bases
B. allow room for shower bases and baths to expand and contract with changes in temperature from hot water
C. stop water from bridging between fixtures and being drawn into the frame
D. provide a space into which a sealant may be pumped to stop leaks but allow movement
Competency: CPCCCA3012B Frame and fit wet area fixtures
Difficulty: Medium
Performance criteria: CPCCCA3012B 3.3
Performance criteria: CPCCCA3012B 4.3
1. The installation of a spa bath over a bearer and joist flooring system may require you to:
A.use a heavier floor sheeting such as 19 mm thick fibre cement sheet
B. change the plumbing layout
C. install additional subfloor framing and support members
D. paint a waterproof sealant under the spa bath area
Competency: CPCCCA3012B Frame and fit wet area fixtures
Difficulty: Easy
Performance criteria: CPCCCA3012B 1.1
1. If the shower base you are about to install is twisted such that you can only get two edges level, not all four, then you should:
A.use a heavy object (bags of cement, buckets of water) to twist it back into shape
B. use long timbers to wedge it down from the ceiling until the mortar sets
C. level the two front edges as this is where the shower screen must go
D. level the two back edges so the wall tiles will not end up being cut at an angle
Competency: CPCCCA3012B Frame and fit wet area fixtures
Difficulty: Medium
Performance criteria: CPCCCA3012B 1.1
1. Kitchens and exposed balconies are not included in the NCC Vol. 2 BCA’s definition of wet areas.
TRUE
Competency: CPCCCA3012B Frame and fit wet area fixtures
Difficulty: Easy
Performance criteria: CPCCCA3012B 1.1
1. When the NCC Vol. 2 BCA specifies that a surface must be water resistant it means it must be coated with a waterproof compound and then surfaced with an abrasive-resistant material such as grouted tiles, plastic laminates or similar.
FALSE
Competency: CPCCCA3012B Frame and fit wet area fixtures
Difficulty: Medium
Performance criteria: CPCCCA3012B 1.1
1. In one method of installing a bath, a series of noggings should be installed around the bath above the checkout. The purpose of these noggings is to provide fixing support for the internal linings.
TRUE
Competency: CPCCCA3012B Frame and fit wet area fixtures
Difficulty: Medium
Performance criteria: CPCCCA3012B 3.3
1. Waterproofing is a critical but small job on a domestic building site. For that reason it is not expected that you will have access on site to the material safety data sheets (MSDS) for the all the various materials involved.
FALSE
Competency: CPCCCA3012B Frame and fit wet area fixtures
Difficulty: Easy
Performance criteria: CPCCCA3012B 1.2
1. An advantage of wet area plasterboard over fibre cement sheet as a lining in bathrooms is that it has significantly better thermal and acoustic properties. In addition, it also handles frame movement better.
TRUE
Competency: CPCCCA3012B Frame and fit wet area fixtures
Difficulty: Easy
Performance criteria: CPCCCA3012B 1.6
Chapter 04 Testbank Summary
Category # of Questions Competency: CPCCCA3012B Frame and fit wet area fixtures 10 Difficulty: Easy 5 Difficulty: Medium 5 Performance criteria: CPCCCA3012B 1.1 5 Performance criteria: CPCCCA3012B 1.2 1 Performance criteria: CPCCCA3012B 1.6 1 Performance criteria: CPCCCA3012B 3.1 1 Performance criteria: CPCCCA3012B 3.3 3 Performance criteria: CPCCCA3012B 4.3 2
## Reviews
There are no reviews yet. | 3,043 | 11,783 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2023-23 | longest | en | 0.824602 |
https://www.coursehero.com/file/p2fg04st/If-we-take-the-logarithms-natural-of-both-sides-then-this-model-can-be-written/ | 1,576,337,411,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541281438.51/warc/CC-MAIN-20191214150439-20191214174439-00327.warc.gz | 677,540,025 | 100,762 | If we take the logarithms natural of both sides then this model can be written
# If we take the logarithms natural of both sides then
• 11
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If we take the logarithms (natural) of both sides, then this model can be written ln( W ) = a ln( L ) + ln( k ) , which is a linear relation between the ln( W ) and the ln( L ). There are standard formula to find the Linear Least Squares best fit to linear data. MatLab uses a subroutine polyfit to find the best slope and intercept for a line passing through a data set. ( Note: MatLab uses the natural logarithm in its calculations, so uses the command log(x) to mean ln( x ).) The analysis above shows that the linear least squares fit of a line to the logarithms of the data provides the allometric model. This suggests that MatLab’s polyfit routine can be used to fit these types of models. The following MatLab function allows the easy input of data to produce the parameters for our allometric model: 1 function [ k , a ] = powerfit ( ldata , wdata ) 2 % Power law ( Allometric ) f i t f o r model W = k * Lˆa 3 % Uses l i n e a r l e a s t squares f i t to logarithms of data 4 Y = log ( wdata ) ; % Logarithm of W - data 5 X = log ( ldata ) ; % Logarithm of L - data 6 p = p o l y f i t (X,Y, 1 ) ; % Linear f i t to X and Y with p = [ slope , i n t e r c e p t ] 7 a = p (1) ; % Value of exponent 8 k = exp (p (2) ) ; % Value of leading c o e f f i c i e n t 9 end
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Our data file discussed above has the vector data variables ltdfish for length and wtdfish for weight (vectors with 25 elements) of Lake Trout data from the Kory Groetsch data. If we execute our powerfit function with the following MatLab command: 1 [ k , r ] = powerfit ( l t d f i s h , wtdfish ) MatLab outputs the variables k = 0 . 015049 and a = 2 . 8591. It follows that using this method of finding the linear best fit to the logarithms of the data yields a best allometric model of the form: W = 0 . 015049 L 2 . 8591 . From a modeling perspective, this problem can be examined in two other ways. Similar to the section before, a nonlinear least squares best fit can be used to obtain the unbiased best fit of the model (with respect to the parameters k and a ). Alternately, a dimensional analysis supported by the allometric fit above would suggest that the weight of a fish varies like the cube of the length, i.e. , self-similarity of fishes would suggest that as the length increases, then the height and width would similarly increase giving a cubic relation between length and weight. This would fix the parameter a = 3 and only allow changes in k . Below are two functions for finding the sum of square errors for these two models. The function for the 2-parameter problem is 1 function J = sumsq nonlin (p , ldata , wdata ) 2 % Function computing sum of square e r r o r s f o r a l l o m e t r i c model 3 model = p (1) * ldata .ˆ p (2) ; 4 e r r o r = model - wdata ; 5 J = e r r o r * error ’ ; 6 end 7 8 % Obtain the l e a s t sum of square e r r o r s 9 % [ p1 , J , f l a g ] = fminsearch ( @sumsq allom , [ k , a ] , [ ] , l t d f i s h , wtdfish ) ; The MatLab function for the cubic model with only 1-parameter is 1 function J = sumsq cubic (p , ldata , wdata ) 2 % Function computing sum of square e r r o r s f o r cubic a l l o m e t r i c model 3 model = p * ldata . ˆ 3 ; 4 e r r o r = model - wdata ; 5 J = e r r o r * error ’ ; 6 end 7 8 % Obtain the l e a s t sum of square e r r o r s
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https://school.gradeup.co/ex-9.a-q11-the-mean-of-the-following-data-is-42-find-the-i-1njyn0 | 1,568,983,210,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574018.53/warc/CC-MAIN-20190920113425-20190920135425-00495.warc.gz | 643,567,864 | 39,949 | Q. 114.3( 7 Votes )
# The mean of the following data is 42. Find the missing frequencies x and y if the Sum of frequencies is 100.
For equal class intervals, we will solve by finding mid points of these classes using direct method.
We have got
Σfi = 63 + x + y and Σfixi = 2775 + 25x + 45y
∵ mean is given by
⇒ (∵ given: mean of pocket allowance is 42)
⇒ 2646 + 42x + 42y = 2775 + 25x + 45y
⇒ 42x – 25x + 42y – 45y = 2775 – 2646
⇒ 17x – 3y = 129 …(i)
As given in the question, frequency(Σfi) = 100
And as calculated by us, frequency (Σfi) = 63 + x + y
Equalizing them, we get
63 + x + y = 100
⇒ x + y = 37 …(ii)
We will now solve equations (i) and (ii), multiply eq.(ii) by 3 and then add it to eq.(i), we get
(17x – 3y) + [3(x + y)] = 129 + 111
⇒ 17x – 3y + 3x + 3y = 240
⇒ 20x = 240
⇒ x = 12
Substitute x = 12 in equation (ii),
12 + y = 37
⇒ y = 37 – 12
⇒ y = 25
Thus, x = 12 and y = 25.
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We strive to provide quality solutions. Please rate us to serve you better. | 430 | 1,087 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-39 | latest | en | 0.845381 |
http://www.nag.com/numeric/CL/nagdoc_cl24/examples/source/d01wcce.c | 1,429,461,513,000,000,000 | text/plain | crawl-data/CC-MAIN-2015-18/segments/1429246639191.8/warc/CC-MAIN-20150417045719-00297-ip-10-235-10-82.ec2.internal.warc.gz | 666,077,286 | 1,481 | /* nag_multid_quad_adapt_1 (d01wcc) Example Program. * * Copyright 1998 Numerical Algorithms Group. * * Mark 5, 1998. * Mark 7 revised, 2001. * */ #include #include #include #include #include #ifdef __cplusplus extern "C" { #endif static double NAG_CALL f(Integer n, const double z[], Nag_User *comm); #ifdef __cplusplus } #endif #define NDIM 4 #define MAXPTS 1000*NDIM int main(void) { static Integer use_comm[1] = {1}; Integer exit_status = 0; Integer ndim = NDIM; Integer maxpts = MAXPTS; double a[4], b[4]; Integer k; double finval; Integer minpts; double acc, eps; Nag_User comm; NagError fail; INIT_FAIL(fail); printf("nag_multid_quad_adapt_1 (d01wcc) Example Program Results\n"); /* For communication with user-supplied functions: */ comm.p = (Pointer)&use_comm; for (k = 0; k < 4; ++k) { a[k] = 0.0; b[k] = 1.0; } eps = 0.0001; minpts = 0; /* nag_multid_quad_adapt_1 (d01wcc). * Multi-dimensional adaptive quadrature, thread-safe */ nag_multid_quad_adapt_1(ndim, f, a, b, &minpts, maxpts, eps, &finval, &acc, &comm, &fail); if (fail.code != NE_NOERROR && fail.code != NE_QUAD_MAX_INTEGRAND_EVAL) { printf("Error from nag_multid_quad_adapt_1 (d01wcc) %s\n", fail.message); exit_status = 1; goto END; } printf("Requested accuracy =%12.2e\n", eps); printf("Estimated value =%12.4f\n", finval); printf("Estimated accuracy =%12.2e\n", acc); END: return exit_status; } static double NAG_CALL f(Integer n, const double z[], Nag_User *comm) { double tmp_pwr; Integer *use_comm = (Integer *)comm->p; if (use_comm[0]) { printf("(User-supplied callback f, first invocation.)\n"); use_comm[0] = 0; } tmp_pwr = z[1]+1.0+z[n-1]; return z[0]*4.0*z[2]*z[2]*exp(z[0]*2.0*z[2])/(tmp_pwr*tmp_pwr); } | 572 | 1,688 | {"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} | 2.59375 | 3 | CC-MAIN-2015-18 | longest | en | 0.402057 |
https://edmaths.com/category/waec-solutions/igcse/ | 1,653,507,287,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662593428.63/warc/CC-MAIN-20220525182604-20220525212604-00257.warc.gz | 282,140,460 | 29,375 | ## “CALCULATIONS IN PHYSICS FOR COLLEGE STUDENTS – VOLUME 3” – A NEW PHYSICS TEXTBOOK BY OLAITAN S. O.
The book covers calculations in WAEC’s current senior secondary school Physics curriculum, calculations in all JAMB’s topics required for UTME, calculations in some IGCSE topics, calculations in NECO’s entire current Physics curriculum and calculations in few Cambridge AS Level topics. Read More …
## MEET LUIS ESQUIVEL, A 5-YEAR OLD MATHS GENIUS
Watch the video: (Video Credit: AB News, Photo Credit: www.nbcnews.com)
## EVALUATING COMPOSITE FUNCTIONS (VIDEO)
Watch the video on how to evaluate composite functions: After watching the video, try and evaluate the following: Given that f(x) = 3x +5 and g(x) = x2 +5 , evaluate: (1) fg (2) gf Read More …
## INVERSE OF A FUNCTION (VIDEO)
Watch the video on how to find the inverse of a function: (Photo Credit: hdimagegallery.net)
## INVERSE OF A 2×2 MATRIX (VIDEO)
Watch the video here on how to find the inverse of a 2×2 matrix: (Photo Credit: rex2000.deviantart.com)
## SCALAR MULTIPLICATION OF MATRIX (VIDEO)
Watch the video on how to carry out scalar multiplication of matrix: (Photo Credit: rex2000.deviantart.com)
## MATRIX MULTIPLICATION (VIDEO)
Watch the video on how to carry out matrix multiplication: (Photo Credit: rex2000.deviantart.com)
## ADDITION AND SUBTRACTION OF MATRICES (VIDEO)
Watch the video on how to carry out addition and subtraction of matrices: (Photo Credit: rex2000.deviantart.com))
## SOLVING QUADRATIC EQUATIONS USING COMPLETING THE SQUARE METHOD (VIDEO)
Watch the video on how to solve quadratic equations using completing the square method: After watching the video, try and solve the following quadratic equations using completing the square method: (1) x2 + 6x + 5 = 0 Read More …
## SOLVING QUADRATIC EQUATIONS USING FACTORISATION METHOD (VIDEO)
Watch the video here on how to solve quadratic equations using factorisation method: After watching the video, try and solve the following quadratic equations using factorisation method: (1) x2 + 6x + 5 = 0 Read More … | 552 | 2,118 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2022-21 | longest | en | 0.670962 |
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