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https://www.excel-pmt.com/2018/12/how-to-make-pareto-diagram.html | 1,716,848,313,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059045.34/warc/CC-MAIN-20240527205559-20240527235559-00032.warc.gz | 633,480,107 | 31,124 | ## Sunday 23 December 2018
Pareto diagram is one of the tools (tool) from the QC 7 Tools that are used frequently in regard to quality control. Basically, the Pareto diagram is a bar graph that shows the problem based on the large number of sequences of events. The sequence starts from the number of problems that most to least happened. In the graph, shown with the highest graphics (far left) to the lowest graph (far right).
In its application, the Pareto diagram or Pareto often referred to with this Chart is extremely useful in determining and identifying the priority issues that will be resolved. The problem most frequently occurs and is a top priority for us to take action.
Before you create a Pareto Diagram, data-related issues or events that we want to the analysis should be collected in advance. In General, a tool that is often used for the collection of data is to use a Check Sheet or Sheets check.
### A Pareto diagram has the following objectives:
• Separate the few major problems from the many possible problems so you can focus your improvement efforts.
• Arrange data according to priority or importance.
• Determine which problems are most important using data, not perceptions.
## How to Create a Pareto diagram?
The steps in making a Pareto Diagram is as follows:
1. Identify issues that will be examined and the causes of the incident.
(An example of the problem: high rates of Defects in production of PCB Assembly, Cause: Short Solder, No Solder, Missing Solder, Solder Ball and Crack)
2. Determine the period of time required for analysis (e.g. per monthly, weekly or daily per)
3. Make a note the frequency of the incident on the sheet check (check sheet)
4. Make a list of the issue in accordance with the order of frequency of occurrence (from highest to lowest).
5. Calculate the cumulative Frequency and Cumulative Percentage
6. To draw a Frequency in the form of a bar graph
7. To draw a cumulative Percentage in the form of line graph
8. Translate the Pareto diagram
9. Take action based on the priority of the incident/problem
10. Repeat the above steps again implement the improvement actions (action improved) to do the comparison results.
A Pareto diagram is a graphical representation that displays data in order of priority. It can be a powerful tool for identifying the relative importance of causes, most of which arise from only a few of the processes, hence the 80:20 rule. Pareto Analysis is used to focus problem solving activities, so that areas creating most of the issues and difficulties are addressed first.
### Following procedure will be useful to use Pareto diagram and its analysis:
• The categories of group items.
• Approximate measurement (frequency, quantity, cost, or time)
• Decide the time period to gather data and use in the Pareto diagram (one work cycle, one full day, or one week)
• Collect data, record and assemble data for the category each time.
• Subtotal the measurements for each category.
• Determine the appropriate scale for the measurements data collect.
• Mark the scale on the left side of the chart.
• Construct and label bars for each category by placing the tallest to the left, next tallest to its right and so on
• Calculate the percentage for each category.
• Draw a right vertical axis and label it with percentage in a graph paper. Be sure that left measurement corresponds to one-half and it should be exactly opposite 50% on the right scale.
• Calculate and draw cumulative sums.
• Add the subtotals for the first category and second category and place a dot above the second bar indicating the sum, then add subtotal of third category to the sum and place a dot above the third bar indicating the new sum and so on. Continue the adding subtotals and placing dots for all bars.
• Connect the dots, starting from the top of first bar. The last dot should reach 100 percent on the right side.
In this way we can visualize the most important factors among a typically large set of factors through the Pareto diagram. A Pareto diagram often represents the most common sources of defects. The highest occurring type of defect, or the most frequent reasons for problems. | 865 | 4,195 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.929765 |
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# CR - senate
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Senior Manager
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10 Aug 2004, 15:43
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10 Aug 2004, 17:13
First Question
This was a tough one. Thru POE I choose E
A out of scope
B supports a premise
C supports argument
D out of scope
From E if much legislation is passed despite party disunity, then atleast the members are cohesive in decision making. Hence E weakens the conclusion which is 1st sentence of the psg
Second Question
My choice for this one is E as well
E strengthens the arg on inabilility of cohension in a party
ps: took me total 5.5 mins for both these questions!!! got to improve on timing
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### Show Tags
10 Aug 2004, 22:03
srijay007 wrote:
First Question
This was a tough one. Thru POE I choose E
A out of scope
B supports a premise
C supports argument
D out of scope
From E if much legislation is passed despite party disunity, then atleast the members are cohesive in decision making. Hence E weakens the conclusion which is 1st sentence of the psg
Second Question
My choice for this one is E as well
E strengthens the arg on inabilility of cohension in a party
ps: took me total 5.5 mins for both these questions!!! got to improve on timing
I would agree for E for the first one,
But the second one, C can also hold good, here Half members of the minoriy party voted against thier leaders, it does show that there is no Unity in the Party ( Minority)
Can some one explain it better or Justify answer E, if I am wrong.
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11 Aug 2004, 02:17
I remember seeing this question at http://richardbowles.tripod.com
There are quite a few of such sites on the Internet but after visiting some of them I would not rely on the quality of questions there. ETS has a very distinctive logic and other authors of CRs often do not follow it (even Kaplan and Princeton). Trying to explain the OAs of these questions may actually harm your thinking
I believe the official explanation given for the first question is ridiculous
Senior Manager
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11 Aug 2004, 05:19
Hey I think it is 1-C and 2-E.
In C, it is stated that it is the duty of the leader to be responsible to the local guys. Then, the disunity should not be questioned at all. weakens the most. As for E in the first question, 'the legislations are passed' may mean that there may be unity among the parties. It does not talk about the disunity withn a party.
E is obvious in 2. All others are out of scope.
Geethu wrote:
_________________
Awaiting response,
Thnx & Rgds,
Chandra
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11 Aug 2004, 07:45
mallelac wrote:
Hey I think it is 1-C and 2-E.
In C, it is stated that it is the duty of the leader to be responsible to the local guys. Then, the disunity should not be questioned at all. weakens the most. As for E in the first question, 'the legislations are passed' may mean that there may be unity among the parties. It does not talk about the disunity withn a party.
E is obvious in 2. All others are out of scope.
C & E are the answers.
Thank you.
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### Show Tags
11 Aug 2004, 09:36
1. E
2. E
In case of the first question, C is nothing but reframing of the sentence start with 'Consequently...' . E seems to be the only one which seems to show that, the author's argument may not be strong.
11 Aug 2004, 09:36
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### security commit
parent 6f5095e4
... ... @@ -758,13 +758,13 @@ It also follows from Lemma \ref{lemma:binervthom} that the bisimplicial nerve in From Proposition \ref{prop:streetvsbisimplicial}, we deduce the proposition below which contains two useful criteria to detect Thomason homotopy cocartesian square of $2\Cat$. \end{paragr} \begin{proposition} \begin{proposition}\label{prop:critverthorThomhmtpysquare} Let \tag{$\ast$}\label{coucou}\begin{tikzcd} A \ar[r,"u"]\ar[d,"f"] & B \ar[d,"g"] \\ C \ar[r,"v"] & D \end{tikzcd} be a square in $2\Cat$ satisfying either of the following conditions: be a square in $2\Cat$ satisfying one of the following conditions: \begin{enumerate}[label=(\alph*)] \item for every $n\geq 0$, the square $... ... @@ -1230,26 +1230,76 @@ isomorphisms, which means by definition that P, P' and P'' are \good{}. \begin{itemize}[label=-] \item generating 0\nbd{}cells: A and B, \item generating 1\nbd{}cells: f,g,h : A \to B, \item generating 2\nbd{}cells: \alpha,\beta:f \to g and \delta,\gamma:g \to h. \item generating 2\nbd{}cells: \alpha,\beta:f \Rightarrow g and \delta,\gamma:g \Rightarrow h. \end{itemize} In picture, this gives: \[ \begin{tikzcd}[column sep=huge] A \ar[r,bend left=75,"f",""{name=A,below,pos=8/20},""{name=E,below,pos=12/20}] \ar[r,"g",""{name=B,above,pos=8/20},""{name=C,below,pos=8/20},""{name=F,above,pos=12/20},""{name=G,below,pos=12/20}] \ar[r,bend right=75,"h"',""{name=D,above,pos=8/20},""{name=H,above,pos=12/20}] & B. \ar[from=A,to=B,Rightarrow,"\alpha"',bend right] \ar[from=C,to=D,Rightarrow,"\delta"',bend right] \ar[from=C,to=D,Rightarrow,"\gamma"',bend right] \ar[from=E,to=F,Rightarrow,"\beta",bend left] \ar[from=G,to=H,Rightarrow,"\gamma",bend left] \ar[from=G,to=H,Rightarrow,"\delta",bend left] \end{tikzcd}$ Let us prove that this $2$\nbd{}category is \good{}. Let $P'$ be the sub-$2$\nbd{}category of $P$ spanned by $A$, $B$, $f$, $g$, $\alpha$ and $\beta$ and let $P''$ be the sub-$2$\nbd{}category of $P$ spanned by $A$, $B$, $g$, $h$, $\gamma$ and $\delta$. These two $2$\nbd{}categories are copies of $\sS_2$ and we have a cocartesian square Let us prove that this $2$\nbd{}category is \good{}. Let $P'$ be the sub-$2$\nbd{}category of $P$ spanned by $A$, $B$, $f$, $g$, $\alpha$ and $\beta$, let $P''$ be the sub-$2$\nbd{}category of $P$ spanned by $A$, $B$, $g$, $h$, $\gamma$ and $\delta$ and let $P''$ be the sub-$2$\nbd{}category of $P$ spanned by $A$, $B$ and $g$. These two $2$\nbd{}categories are copies of $\sS_2$ and we have a cocartesian square $\begin{tikzcd} \sD_1 \ar[d,"\langle g \rangle"] \ar[r,"\langle g \rangle"] & P' \ar[d] \\\ P'' \ar[r] & P. \end{tikzcd}$ using the second part of Corollary \end{paragr} Let us prove that this square is Thomason homotopy cocartesian using the second part of Corollary \ref{prop:critverthorThomhmtpysquare}. First, all the morphisms of the previous square are isomorphisms on objects and thus, the image by $V_0$ of the above square is obviously cocartesian. Now, notice that the categories $P(A,B)$, $P'(A,B)$ and $P''(A,B)$ are respectively free on the graphs $\begin{tikzcd} f \ar[r,shift left,"\alpha"] \ar[r, shift right, "\beta"'] & g \ar[r,shift left,"\gamma"'] \ar[r,shift right,"\delta"] & h, \end{tikzcd}$ $\begin{tikzcd} f \ar[r,shift left,"\alpha"] \ar[r, shift right, "\beta"'] & g, \end{tikzcd}$ and $\begin{tikzcd} g \ar[r,shift left,"\gamma"] \ar[r,shift right,"\delta"'] & h. \end{tikzcd}$ Besides, all the categories $P(A,A)$, $P(B,B)$, $P'(A,A)$, $P'(B,B)$, $P''(A,A)$ and $P''(B,B)$ are all isormophic to the terminal category $\sD_0$. \todo{À finir} \end{paragr} \begin{paragr} Let $P$ be the free $2$\nbd{}category defined as follows: \begin{itemize}[label=-] \item generating $0$\nbd{}cell: $A$, \item generating $1$\nbd{}cells: $f , g : A \to A$, \item generating $2$\nbd{}cells: $\alpha : f\comp_0 g \Rightarrow g \comp_0 f$. \end{itemize} In picture, this gives: $\begin{tikzcd} A \ar[r,"f"] \ar[d,"g"'] & A \ar[d,"g"] \\ A \ar[r,"f"'] & A. \ar[from=2-1,to=1-2,Rightarrow,"\alpha"] \end{tikzcd}$ Now consider the $1$\nbd{}category $B^1(\mathbb{N}\times\mathbb{N})$, that is the monoid $\mathbb{N}\times\mathbb{N}$ considered as a category with only one object, and let $F : P \to B^1(\mathbb{N}\times\mathbb{N})$ be the unique $2$\nbd{}functor such that: \begin{itemize}[label=-] \item $F(f)=(1,0)$ and $F(g)=(0,1)$, \item $F(\alpha)=1_{(1,1)}$. \end{itemize} This last equation makes sense since $(1,1)=(0,1)+(1,0)=(1,0)+(0,1)$. Notice that $1$\nbd{}cells of $P$ can be encoded in finite words on the alphabet $\{f,g\}$ (concatenation corresponding to $0$\nbd{}composition) and for a word $w$ on this alphabet such that $f$ appears $n$ times and $g$ appears $m$ times, we have $F(w)=(n,m)$. Let us now prove that $F$ is a Thomason equivalence using a dual of \cite[Corollaire 5.26]{ara2020theoreme} (see Remark 5.20 of op.\ cit.\ ). If we write $\star$ for the only object of $B^1(\mathbb{N}\times\mathbb{N})$, what we need to show is that the canonical $2$\nbd{}functor from $P/{\ast}$ (\ref{paragr:comma}) to the terminal $2$\nbd{}category $P/{\ast} \to \sD_0$ is a Thomason equivalence. The $2$\nbd{}category $P/{\ast}$ is described as follows: \begin{itemize} \item A $0$\nbd{}cells is a $1$\nbd{}cell of $B^1(\mathbb{N}\times \mathbb{N})$, \item \item \end{itemize} \end{paragr} \section{The Bubble-free'' conjecture} \begin{definition} Let $C$ be a $2$\nbd{}category. A \emph{bubble} (in $C$) is a $2$\nbd{}cell ... ...
... ... @@ -767,7 +767,7 @@ The nerve functor $N_{\omega} : \omega\Cat \to \Psh{\Delta}$ sends equivalences \begin{example}\label{example:slicecategories} For a small category $A$ (considered as an $\oo$\nbd{}category) and an object $a_0$ of $A$, the category $A/a_0$ in the sense of the previous paragraph is nothing but the usual slice category of $A$ over $a_0$. \end{example} \begin{paragr} \begin{paragr}\label{paragr:comma} Let $u : A \to B$ be a morphism of $\oo\Cat$ and $b_0$ an object of $B$. We define the $\oo$-category $A/b_0$ (also denoted $u\downarrow b_0$) as the following fibred product: \[ \begin{tikzcd} ... ...
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Finish editing this message first! | 2,248 | 6,326 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2022-21 | latest | en | 0.563386 |
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Hint: within the parentheses, use sum to product identities, and you will have it in a few steps.
$\displaystyle (\cos 2A - \cos 2B)^2 + (\sin 2A - \sin 2B)^2 = 4 \sin^2 (A-B)$
Use
$\displaystyle \cos(2A) - \cos(2B) = -2 \sin \left(\frac{2A+2B}{2}\right)\sin\left(\frac{2A-2B}{2}\right)$
$\displaystyle = - 2 \sin(A+B)\sin(A-B)$
$\displaystyle \sin(2A) - \sin(2B) = 2 \cos \left(\frac{2A+2B}{2}\right)\sin\left(\frac{2A-2B}{2}\right)$
$\displaystyle = 2 \cos(A+B)\sin(A-B)$
Can you proceed now?
im still having problem with this question....
Originally Posted by sharmala
im still having problem with this question....
Hello, sharmala!
We will need these two identities:
. . $\displaystyle \cos x\cos y + \sin x\sin y \:=\:\cos(x-y)$
. . $\displaystyle \frac{1-\cos2\theta}{2} \:=\:\sin^2\theta$
$\displaystyle \text{Prove: }\:(\cos2A - \cos2B)^2 + (\sin2A - \sin2B)^2\;=\;4\sin^2(A\!-\!B)$
Expand the left side:
$\displaystyle \cos^22A - 2\cos2A\cos2B + \cos^22B + \sin^22A - 2\sin2A\sin2B + \sin^22B$
. . . $\displaystyle =\;\underbrace{\sin^22A + \cos^22A}_1 \:+\: \underbrace{\sin^22B + \cos^22B}_1 \:-\: 2\big(\cos2A\cos2B + \sin2A\sin2B\big)$
. . . $\displaystyle =\;2 - 2\cos(2A\!-\!2B)$
. . . $\displaystyle =\;2\big[1 - \cos(2[A\!-\!B])\big]$
. . . $\displaystyle =\;4\left[\frac{1-\cos(2[A\!-\!B])}{2}\right]$
. . . $\displaystyle =\;4\sin^2(A\!-\!B)$
$\displaystyle [- 2 \sin(A+B)\sin(A-B)]^2+[2 \cos(A+B)\sin(A-B)]^2$
$\displaystyle = 4\sin^2(A+B)\sin^2(A-B)+4\cos^2(A+B)\sin^2(A-B)$
$\displaystyle = 4 \sin^2(A-B)[\sin^2(A+B)+\cos^2(A+B)]=4\sin^2(A-B)$ | 739 | 1,706 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2018-22 | latest | en | 0.483991 |
https://www.traditionaloven.com/tutorials/distance/convert-imperial-cable-length-to-japan-rin-unit.html | 1,660,707,701,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572833.95/warc/CC-MAIN-20220817032054-20220817062054-00729.warc.gz | 892,947,065 | 17,504 | Convert cable Imp to 厘 | cable length Imperial to Japanese rin
# length units conversion
## Amount: 1 cable length Imperial (cable Imp) of nautical length Equals: 611,550.72 Japanese rin (厘) in length
Converting cable length Imperial to Japanese rin value in the length units scale.
TOGGLE : from Japanese rin into cable lengths Imperial in the other way around.
## length from cable length Imperial to Japanese rin conversion results
### Enter a new cable length Imperial number to convert
* Whole numbers, decimals or fractions (ie: 6, 5.33, 17 3/8)
* Precision is how many digits after decimal point (1 - 9)
Enter Amount :
Decimal Precision :
CONVERT : between other length measuring units - complete list.
How many Japanese rin are in 1 cable length Imperial? The answer is: 1 cable Imp equals 611,550.72 厘
## 611,550.72 厘 is converted to 1 of what?
The Japanese rin unit number 611,550.72 厘 converts to 1 cable Imp, one cable length Imperial. It is the EQUAL nautical length value of 1 cable length Imperial but in the Japanese rin length unit alternative.
cable Imp/厘 length conversion result From Symbol Equals Result Symbol 1 cable Imp = 611,550.72 厘
## Conversion chart - cable lengths Imperial to Japanese rin
1 cable length Imperial to Japanese rin = 611,550.72 厘
2 cable lengths Imperial to Japanese rin = 1,223,101.44 厘
3 cable lengths Imperial to Japanese rin = 1,834,652.16 厘
4 cable lengths Imperial to Japanese rin = 2,446,202.88 厘
5 cable lengths Imperial to Japanese rin = 3,057,753.60 厘
6 cable lengths Imperial to Japanese rin = 3,669,304.32 厘
7 cable lengths Imperial to Japanese rin = 4,280,855.04 厘
8 cable lengths Imperial to Japanese rin = 4,892,405.76 厘
9 cable lengths Imperial to Japanese rin = 5,503,956.48 厘
10 cable lengths Imperial to Japanese rin = 6,115,507.20 厘
11 cable lengths Imperial to Japanese rin = 6,727,057.92 厘
12 cable lengths Imperial to Japanese rin = 7,338,608.64 厘
13 cable lengths Imperial to Japanese rin = 7,950,159.36 厘
14 cable lengths Imperial to Japanese rin = 8,561,710.08 厘
15 cable lengths Imperial to Japanese rin = 9,173,260.80 厘
Convert length of cable length Imperial (cable Imp) and Japanese rin (厘) units in reverse from Japanese rin into cable lengths Imperial.
## Length, Distance, Height & Depth units
Distance in the metric sense is a measure between any two A to Z points. Applies to physical lengths, depths, heights or simply farness. Tool with multiple distance, depth and length measurement units.
# Converter type: length units
First unit: cable length Imperial (cable Imp) is used for measuring nautical length.
Second: Japanese rin (厘) is unit of length.
QUESTION:
15 cable Imp = ? 厘
15 cable Imp = 9,173,260.80 厘
Abbreviation, or prefix, for cable length Imperial is:
cable Imp
Abbreviation for Japanese rin is:
## Other applications for this length calculator ...
With the above mentioned two-units calculating service it provides, this length converter proved to be useful also as a teaching tool:
1. in practicing cable lengths Imperial and Japanese rin ( cable Imp vs. 厘 ) measures exchange.
2. for conversion factors between unit pairs.
3. work with length's values and properties.
To link to this length cable length Imperial to Japanese rin online converter simply cut and paste the following.
The link to this tool will appear as: length from cable length Imperial (cable Imp) to Japanese rin (厘) conversion.
I've done my best to build this site for you- Please send feedback to let me know how you enjoyed visiting. | 898 | 3,552 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2022-33 | latest | en | 0.748854 |
https://www.experts-exchange.com/questions/29086782/SharePoint-Calculated-Column-Help.html | 1,527,317,901,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867311.83/warc/CC-MAIN-20180526053929-20180526073929-00105.warc.gz | 743,302,805 | 21,806 | • Status: Solved
• Priority: Medium
• Security: Public
• Views: 30
# SharePoint Calculated Column Help
Trying to setup calculated column in SharePoint with the following logic
if TCV GE 100000000 then = L0
if TCV GE 25000000 AND LT 100000000 then = L2
if TCV GE 5000000 AND LT 25000000 then = L3
if TCV LT 5000000 then = L4
if TCV EQ 1 then = L3
0
Matt Pinkston
• 4
1 Solution
Author Commented:
How to combine
=IF([TCV]>=100000000, "L0", "Not OK")
=IF(AND([TCV]>=25000000, [TCV]<100000000), “L2”, “”)
=IF(AND([TCV]>=5000000, [TCV]<25000000), “L3”, “”)
=IF([TCV]<5000000, "L4", "")
=IF([TCV]=1, "L3", "")
0
Author Commented:
Would this be right????
=IF([TCV]>=100000000, "L0", (AND([TCV]>=25000000, [TCV]<100000000), “L2”, (AND([TCV]>=5000000, [TCV]<25000000), “L3”, ([TCV]<5000000, "L4", ([TCV]=1, "L3", "")))))
0
Author Commented:
I get an error when trying to add the above
0
Commented:
All if statements should be inside one single brackets
like this
=IF(([TCV]>=100000000), "L0", "Not OK",
IF((AND([TCV]>=25000000, [TCV]<100000000), “L2”, “”,
IF((AND([TCV]>=5000000, [TCV]<25000000), “L3”, “”,
IF(([TCV]<5000000, "L4", ""),
IF(([TCV]=1, "L3", ""))))))
it should be something like this. reconstruct your statement, start with single then add another one after another. That way you can track when you are making mistake.
but the structure has to be like this.
0
Author Commented:
GREAT.... Thanks
0
Question has a verified solution.
Are you are experiencing a similar issue? Get a personalized answer when you ask a related question.
Have a better answer? Share it in a comment. | 558 | 1,600 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-22 | longest | en | 0.802066 |
https://www.studypool.com/discuss/538241/algebra-questions-for-factoring?free | 1,506,023,578,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818687837.85/warc/CC-MAIN-20170921191047-20170921211047-00606.warc.gz | 867,008,571 | 13,932 | ##### algebra questions for factoring
label Algebra
account_circle Unassigned
schedule 1 Day
account_balance_wallet \$5
What is the quadratic formula and how does it work?
May 15th, 2015
$x=\frac{-b + \sqrt {b^2-4ac}}{2a}\quad\text{and}\quad x=\frac{-b - \sqrt {b^2-4ac}}{2a}$ is the formula and one of the ways of figuring out the value of x in a quadratic equation: ax^2+bx+c so if i had 4x^2-3x+5 my a=4 b=-3 c=5
May 15th, 2015
...
May 15th, 2015
...
May 15th, 2015
Sep 21st, 2017
check_circle | 194 | 511 | {"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": 1, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2017-39 | latest | en | 0.787302 |
https://math.answers.com/other-math/How_many_different_ways_can_you_make_85_with_only_5_10_and_20_bills | 1,680,059,152,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948932.75/warc/CC-MAIN-20230329023546-20230329053546-00254.warc.gz | 424,929,976 | 52,345 | 0
# How many different ways can you make 85 with only 5 10 and 20 bills?
Wiki User
2011-03-04 06:30:18
25 different ways:
20, 20, 20, 20, 5
20, 20, 20, 10, 10, 5
20, 20, 20, 10, 5, 5, 5
20, 20, 20, 5, 5, 5, 5, 5
20, 20, 10, 10, 10, 10, 5
20, 20, 10, 10, 10, 5, 5, 5
20, 20, 10, 10, 5, 5, 5, 5, 5
20, 20, 10, 5, 5, 5, 5, 5, 5, 5
20, 20, 5, 5, 5, 5, 5, 5, 5, 5, 5
20, 10, 10, 10, 10, 10, 10, 5
20, 10, 10, 10, 10, 10, 5, 5, 5
20, 10, 10, 10, 10, 5, 5, 5, 5, 5
20, 10, 10, 10, 5, 5, 5, 5, 5, 5, 5
20, 10, 10, 5, 5, 5, 5, 5, 5, 5, 5, 5
20, 10, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5
20, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5
10, 10, 10, 10, 10, 10, 10, 10, 5
10, 10, 10, 10, 10, 10, 10, 5, 5, 5
10, 10, 10, 10, 10, 10, 5, 5, 5, 5, 5
10, 10, 10, 10, 10, 5, 5, 5, 5, 5, 5, 5
10, 10, 10, 10, 5, 5, 5, 5, 5, 5, 5, 5, 5
10, 10, 10, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5
10, 10, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5
10, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5
5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5
Wiki User
2011-03-04 06:30:18
Study guides
20 cards
➡️
See all cards
3.8
2261 Reviews | 851 | 1,110 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2023-14 | longest | en | 0.66881 |
https://www.dcs.ed.ac.uk/home/mhe/plume/node139.html | 1,709,257,906,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474893.90/warc/CC-MAIN-20240229234355-20240301024355-00700.warc.gz | 717,743,143 | 1,748 | Next: Shift operators Up: Dyadic Digit Operations Previous: Average
# Multiplication
Multiplication of dyadic digits is simpler than the average operation. Given two dyadics (a,b) and (c,d), we know that a and c are both odd or zero. Hence, if either a or c are zero, the result will be the dyadic digit (0,0). Otherwise, is odd, so the result is:
because
Martin Escardo
5/11/2000 | 103 | 384 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-10 | latest | en | 0.796439 |
https://cs.stackexchange.com/questions/14323/prove-or-disprove-whether-l-is-regular/14324 | 1,618,949,892,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039490226.78/warc/CC-MAIN-20210420183658-20210420213658-00090.warc.gz | 300,426,410 | 34,286 | # Prove or disprove whether L is regular [duplicate]
Let $\Sigma = \{0,1\}$. For every word $w \in \Sigma^*$, let $|w|_0$ and $|w|_1$ denote the count of 0's and 1's, respectively, in $w$. Let $L$ be the language $$L = \{ w \in \Sigma^* \mid |w|_0 \gt |w|_1 + 2 \text{ or } |w|_1 \gt |w|_0 + 2\}$$ Prove or disprove whether $L$ is regular.
• This is a pure exercise dump. What have you tried? Where did you get stuck? See here for a discussion why we think your question is bad, and here for questions you should check out before asking. Once you include your own attempts, you have posted a question in its own right that can be answered to solve your specific problem. – Raphael Sep 16 '13 at 8:21
• Be nice. This is your only warning. – Gilles 'SO- stop being evil' Sep 16 '13 at 20:06
Recall that regular languages are closed under complementation. That is, if $L$ is regular, than so is $\overline{L}=\Sigma^*\setminus L$.
Thus, if you manage to prove that $\overline{L}$ is not regular, then $L$ is not regular as well.
Observe that \begin{align} \overline{L} &= \{w\in \{0,1\}^*: |w|_0\le |w|_1+2 \wedge |w|_1\le |w|_0+2\} \\ &=\{w:|w|_1-2\le |w|_0\le |w|_1+2\} \\ \end{align}
Assume by way of contradiction that $\overline{L}$ is regular, then by the pumping lemma, there exists a pumping constant $p$. Consider the word $0^p1^p\in \overline{L}$, then (by standard pumping-lemma arguments) there exists some $i\le p$ such that $0^{p+ki}1^p\in \overline{L}$ for every natural $k$. Choose $k=5$ (any number greater than 2 will work), then $p+5i>p+2$, and therefore $0^{p+ki}1^p\notin \overline{L}$, which is a contradiction.
@Shaull's solution is perfect.. here's a slightly different version of it...
try to see, that $L$ = $L_1 \cup L_2$
where
$L_1 = \{ w \in \{0,1\}^* \mid |w|_0 \gt |w|_1 + 2\}$
$L_2 = \{ w \in \{0,1\}^* \mid |w|_1 \gt |w|_0 + 2\}$
now we will go on to show that $L_1$ is not regular using the pumping lemma.
The pumping lemma states that if some $A$ is a regular language, $\exists$ number $p$ (think of it as the number of states of a DFA that recognizes $A$) such that if $s \in A, |s| \geq p$ then $s$ can be divided into $x,y,z$ such that $s=xyz$ and
1. $\forall i\geq0, xy^iz \in A$
2. $|y| > 0$
3. $|xy| \leq p$
Now, we do a proof by contradiction, and assume that $L_1$ is regular.
Now again, see that $L_1 \supset L_{1a} \cdot L_{1b}$
where
$L_{1a} = \{ w = 0^n1^n \mid n\geq 0\}$ (it has the same number of 0s and 1s)
$L_{1b} = \{ w \in \{1\}^n \mid n\geq 2\}$
i.e. a regular language is closed under union.
Now we will just show that $L_{1a}$ is not regular.
Proof by contradiction: Assume $L_{1a}$ is regular, then there is a DFA, $M$ that accepts $L_{1a}$. Let $p$ be the num of states.
Let $s$ be some word in $L_{1a}$
Let $s = 0^{p+1}1^{p+1}$
$s= xyz$
$|xy| \leq p$ (from defn pumping lemma)
$|xy|$ has just $0$s
$y$ has atleast one $0$ (from defn pumping lemma)
but $xy^2z$ has more $0$s than $1$s $\rightarrow s = xy^2z \notin L_{1a}$
Contradiction!!! this shows $L_{1a}$ is not regular
Then this gives a contradiction that $L_1$ is regular, and finally that shows that $L$ is also not regular. | 1,159 | 3,170 | {"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": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2021-17 | latest | en | 0.86577 |
https://exercism.io/tracks/go/exercises/gigasecond/solutions/e40707476fe64eeb8e829027d0c72a34 | 1,611,162,636,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703521139.30/warc/CC-MAIN-20210120151257-20210120181257-00599.warc.gz | 343,059,380 | 7,270 | 🎉 Exercism Research is now launched. Help Exercism, help science and have some fun at research.exercism.io 🎉
# nicolemon's solution
## to Gigasecond in the Go Track
Published at Jul 13 2018 · 0 comments
Instructions
Test suite
Solution
Calculate the moment when someone has lived for 10^9 seconds.
A gigasecond is 10^9 (1,000,000,000) seconds.
## Running the tests
To run the tests run the command `go test` from within the exercise directory.
If the test suite contains benchmarks, you can run these with the `--bench` and `--benchmem` flags:
``````go test -v --bench . --benchmem
``````
Keep in mind that each reviewer will run benchmarks on a different machine, with different specs, so the results from these benchmark tests may vary.
## Further information
For more detailed information about the Go track, including how to get help if you're having trouble, please visit the exercism.io Go language page.
## Source
Chapter 9 in Chris Pine's online Learn to Program tutorial. http://pine.fm/LearnToProgram/?Chapter=09
## Submitting Incomplete Solutions
It's possible to submit an incomplete solution so you can see how others have completed the exercise.
### cases_test.go
``````package gigasecond
// Source: exercism/problem-specifications
// Commit: 5506bac gigasecond: Apply new "input" policy
// Problem Specifications Version: 1.1.0
// Add one gigasecond to the input.
description string
in string
want string
}{
{
"date only specification of time",
"2011-04-25",
"2043-01-01T01:46:40",
},
{
"second test for date only specification of time",
"1977-06-13",
"2009-02-19T01:46:40",
},
{
"third test for date only specification of time",
"1959-07-19",
"1991-03-27T01:46:40",
},
{
"full time specified",
"2015-01-24T22:00:00",
"2046-10-02T23:46:40",
},
{
"full time with day roll-over",
"2015-01-24T23:59:59",
"2046-10-03T01:46:39",
},
}``````
### gigasecond_test.go
``````package gigasecond
// Write a function AddGigasecond that works with time.Time.
import (
"os"
"testing"
"time"
)
// date formats used in test data
const (
fmtD = "2006-01-02"
fmtDT = "2006-01-02T15:04:05"
)
for _, tc := range addCases {
in := parse(tc.in, t)
want := parse(tc.want, t)
if !got.Equal(want) {
t.Fatalf(`FAIL: %s
= %s
want %s`, tc.description, in, got, want)
}
t.Log("PASS:", tc.description)
}
}
func parse(s string, t *testing.T) time.Time {
tt, err := time.Parse(fmtDT, s) // try full date time format first
if err != nil {
tt, err = time.Parse(fmtD, s) // also allow just date
}
if err != nil {
// can't run tests if input won't parse. if this seems to be a
// development or ci environment, raise an error. if this condition
// makes it to the solver though, ask for a bug report.
_, statErr := os.Stat("example_gen.go")
if statErr == nil || os.Getenv("TRAVIS_GO_VERSION") > "" {
t.Fatal(err)
} else {
t.Log(err)
t.Skip("(This is not your fault, and is unexpected. " +
"Please file an issue at https://github.com/exercism/go.)")
}
}
return tt
}
for i := 0; i < b.N; i++ {
}
}``````
``````// Functions to calculate the moment when someone has lived for 10^9 seconds
package gigasecond
import "time"
// Given time.Time t, add 10^9 seconds and return the resulting date
return result
}`````` | 943 | 3,239 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2021-04 | longest | en | 0.762299 |
https://astronomy.stackexchange.com/questions/39404/how-to-plot-celestial-equator-in-galactic-coordinates-why-does-my-plot-appear?noredirect=1 | 1,621,193,218,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991178.59/warc/CC-MAIN-20210516171301-20210516201301-00314.warc.gz | 151,506,928 | 33,740 | # How to plot celestial equator in galactic coordinates? Why does my plot appear “backwards”? [duplicate]
I'm trying to plot celestial equator in galactic coordinates. Here is my code (based on this answer):
import matplotlib.pyplot as plt
import seaborn as sns; sns.set()
import numpy as np
from astropy.coordinates import SkyCoord
from astropy import units as u
ra_all = np.linspace(-180, 180, 100)
dec_0 = np.zeros(100)
# Transform equatorial coordinates to galactic
eq = SkyCoord(ra_all, dec_0, unit=u.deg)
gal = eq.galactic
lon = np.linspace(0, 360, 100)
lat = np.zeros(100)
# Transform ecliptic coordinates to galactic
ecl = SkyCoord(lon, lat, unit=u.deg, frame='barycentricmeanecliptic')
ecl_gal = ecl.transform_to('galactic')
plt.figure(figsize=(14,7))
plt.subplot(111, projection='aitoff')
plt.scatter(l_plot, b_plot, s=5, label='Celestial Equator')
plt.scatter(l_ecl_gal, b_ecl_gal, s=5, label='Eclptic')
plt.grid(True)
plt.legend(fontsize=16)
plt.show()
However, the map I drew seems to be flipped left to right compared to the one bellow (taken from here)
Question 1: Why does the result of my implementation appear to be the reverse of the other image? Do I have something wrong, or is this simply an issue of mapping the surface from the inside versus from the outside?
Question 2: What is the best way to change my code to make it draw galactic map in a conventional format?
• I adjusted your wording a little bit to better fit the site. I think this is an interesting question and might have an interesting answer! – uhoh Oct 19 '20 at 1:56
• It’s a “mapping from the inside” issue. As with equatorial coordinates, the longitude coordinate increases to the left (when North is up). See astronomy.stackexchange.com/questions/16339/… – Peter Erwin Oct 19 '20 at 3:59
Ok, community helped me to figure out that I faced the “mapping from the inside” issue (explained here). I'll use this answer to show my final code.
import numpy as np
from astropy.coordinates import SkyCoord
import astropy.units as u
import matplotlib.pyplot as plt
import seaborn as sns
sns.set()
def eq2gal(ra, dec):
'''
Transforms equatorial coordinates to galactic ones.
Then prepares them for matplotlib aitoff projection.
'''
eq = SkyCoord(ra, dec, unit=u.deg)
gal = eq.galactic
# Minus appears because of “mapping from the inside” issue
return l_gal, b_gal
def ecl2gal(lon_ecl, lat_ecl):
'''
Transforms ecliptic coordinates to galactic ones.
Then prepares them for matplotlib aitoff projection.
'''
ecl = SkyCoord(lon_ecl, lat_ecl, unit=u.deg, frame='barycentricmeanecliptic')
gal = ecl.transform_to('galactic')
# Minus appears because of “mapping from the inside” issue
return l_gal, b_gal
# Equatorial plane
ra_all = np.linspace(-180, 180, 100)
dec_0 = np.zeros(100)
l_eq_gal, b_eq_gal = eq2gal(ra_all, dec_0)
# Ecliptic plane
lon_ecl = np.linspace(0, 360, 100)
lat_ecl = np.zeros(100)
l_ecl_gal, b_ecl_gal = ecl2gal(lon_ecl, lat_ecl)
plt.figure(figsize=(14,7))
plt.subplot(111, projection='aitoff')
plt.scatter(l_eq_gal, b_eq_gal, s=4, marker='v', label='Celestial Equator')
plt.scatter(l_ecl_gal, b_ecl_gal, s=4, marker='^', label='Eclptic')
# Essential thing is to rename RA axis ticks to transform them to conventional format
plt.xticks(ticks=np.radians([-150, -120, -90, -60, -30, 0, \
30, 60, 90, 120, 150]),
labels=['150°', '120°', '90°', '60°', '30°', '0°', \
'330°', '300°', '270°', '240°', '210°'])
plt.grid(True)
plt.legend(fontsize=16, loc='lower center')
plt.title('Galactic', fontsize='16')
plt.show() | 1,013 | 3,545 | {"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} | 2.734375 | 3 | CC-MAIN-2021-21 | longest | en | 0.805587 |
https://www.electro-tech-online.com/threads/breath-controller-with-3-5mm-jack-to-make-0-5-v-variable-voltage-control-voltage-on-6-5mm-jack.151207/ | 1,620,547,663,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988961.17/warc/CC-MAIN-20210509062621-20210509092621-00228.warc.gz | 765,691,528 | 21,649 | # Breath Controller with 3.5mm Jack to make 0 - 5 V Variable voltage (Control Voltage) on 6.5mm Jack
Status
Not open for further replies.
#### Nicholas D
##### New Member
I am a beginner and wish to tackle this project...
I have an old Yamaha breath controller device that has a 3.5mm Jack and requires -9v on the tip (the sleeve is positive). It draws a current of approx 7mA to 20mA depending on how hard you blow it (from what I have managed to research but may be wrong). I want to now reimplement this device to control CV (control voltage) on my synthesizer. I would love some help to know how to achieve this... I was thinking to use a 9v battery to power it, then convert the voltage to a variable 0 to 5v so that I can plug it into my Sythesizer CV input (6.5mm Jack) and vary the voltage by blowing... Will a simple resistor in circuit achieve this? How would you go about it?
Attached is the circuit diagram of the Breath Controller in case it helps
Thanks very much!
Nicholas
#### Attachments
• 101.5 KB Views: 78
#### cowboybob
##### Well-Known Member
Welcome to ETO, Nicholas D!
Found this site, with this link. Not sure if it's useful. Couldn't find what, exactly, the BC2's output voltages are, although from the schematic I came make a SWAG.
Do you know what the output value(s)?
Might be as simple as using a simple voltage divider circuit at the output of the BC2 to derive a voltage range suitable for your synth.
<EDIT> Just found this:
"BC INPUT
The BC input on the Kurzweil K2661 is compatible with Yamaha BC1, BC2, and BC3. It is a 3.5mm stereo jack that: provides -15.32 V on tip, expects a range of -0.5V to -8.5V from BC on ring, sleeve is ground. The suggested maximum current load for the BC input is 20mA..."
From this site.
So a voltage divider circuit sounds good, anyway...
Last edited:
#### alec_t
##### Well-Known Member
Fleshing out the voltage divider idea ......
#### cowboybob
##### Well-Known Member
Thanks, alec! That's what I had in mind (just didn't SIM it out... ).
#### Nicholas D
##### New Member
That's really awesome of you guys, thanks so much... I can easily rig it up and try it out with a voltmeter first and see if I get 0 to 5v on the output
alec_t - really helpful thanks.... please forgive my ignorance... So this circuit is showing 2 x 9v batteries in series... Also is this using the formula?:
Vout = Vin x (R2 / (R1 + R2))
Because with this I calculated that a 10k and an 8k resistor would yield a Vout of 5v, given 9v Vin. So just curious how you calculated 11k for R2.
e.g. 5 = 9 x (10 / R1 + 10), R1 = 8k
Wouldn't Vout equal more like 4.3V? Vout = 9 x (10 / (10 + 11)) = 4.3?
I did some research but yeah have really minimal knowledge so really appreciate as much info as possible here... Love to learn.
Thanks again!
#### alec_t
##### Well-Known Member
So this circuit is showing 2 x 9v batteries in series..
Yes. V1 provides +9V to the top of R1 and V2 provides -9V to the jack tip of the BC.
I didn't use any formula: I left that to LTspice .
The sim assumes the output voltage from the BC varies from 0V to -9V (your mileage may vary).
With R1=10k and R2=11k the circuit output goes from +4.7V to +0.4V.
With R1=10k and R2=8k it goes from +4V to -1V (which is not the range you want).
With R1=R2=10k it goes from +4.5V to 0V.
Of course, all these voltages are nominal, since they will vary with battery voltage.
Last edited:
#### Nicholas D
##### New Member
Great thanks alec_t - I will put the theory into action and wire it up and let you know how it goes in practice
#### Attila Publik
##### New Member
So, how did it go with this "adventure"?
I´m interested in the same thing and have both the BC3a breath controller and a Kurzweil or two
#### cowboybob
##### Well-Known Member
Welcome to ETO, Attila Publik!
As you might have noticed, Nicolas D never did respond with his results, so we don't know how it went for him.
Perhaps you can apply the circuit suggestions and let us know if they work .
#### Nicholas D
##### New Member
Hi guys yeah sorry about that - I did assemble as per the circuit diagram but was unsuccessful... I then got side-tracked... If I get it working, I will post results. Thanks all the smae everybody for chiming in.
Status
Not open for further replies. | 1,166 | 4,295 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-21 | latest | en | 0.913775 |
https://www.goldsim.com/Courses/ContaminantTransport/Unit4/Lesson10/ | 1,725,766,333,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650958.30/warc/CC-MAIN-20240908020844-20240908050844-00266.warc.gz | 746,402,481 | 7,994 | # Lesson 10 - Viewing Contaminant Transport Results
Note: In this Lesson, we continue to explore the example file named Example1_ContaminatedPond.gsm. It can be found in the “Examples” subfolder of the “Contaminant Transport Course” folder you should have downloaded and unzipped to your Desktop.
Now that we have explored how the system is defined, in this Lesson we will look at some results.
Run the model now. The results of interest can be found in the Results Container. Enter that Container now. We’ve already previously examined the Key System Flows. Double-click on the Mass Release Rates Result element to view that result:
This displays the rate at which contaminant mass is added to the pond from the pipeline (an external input to the model), as well as the release rates from each of the pathways we discussed in the previous Lessons. As can be seen, each pathway tends to delay and disperse the original input of mass from the pipeline. If we integrate each of these over time, they should all converge on the same value (25 kg) since contaminant mass is conserved. This can be seen in Cumulative Mass Release Result element:
The Concentrations Result element displays the concentrations exiting each of the pathways:
Due to dilution, the Aquifer and Stream concentrations are difficult to view on a linear scale. Plotting the concentrations on a log scale (the Log Concentrations Result element) makes these clearer:
Recall that our objective is to predict the peak concentration of the contaminant in the stream. What is the best way to do this? Obviously, we want to see the peak value on the plot above for the stream concentration. We could attempt to do this manually (looking at the curve it appears to be on the order of 2E-4 mg/l), but GoldSim provides a specialized element to do just that: the Extrema element (which is discussed in detail in Unit 12, Lesson 5 of the Basic Course). We can view the output of the Extrema element superimposed on top of the stream concentration in the Peak Concentration in Stream Result element:
The final value of the Extrema element is the desired result (in this case, by holding the cursor over the curve we see it is equal to approximately 2.38E-4 mg/l). Since we have only run a single realization, we can also view this value if we hold our cursor over the Extrema element (the final value of an element is displayed when you do this):
Now that we have our result, let’s briefly address a modeling decision we made that we have not yet discussed: the timestep. As discussed in Unit 6, Lesson 3 of the Basic Course, as a general rule, the timestep for a model should be 3 to 10 times shorter than the timescale of the fastest process being simulated in the model. If we examine the conceptual model and inputs, what we will see is that there are two changes to the model that we want to capture accurately: 1) the “spike” input of mass from the pipeline; and 2) the seasonal change in the stream flow. The rest of the processes are likely to be slower than these. In this model, the timestep was selected to be 1 day. Given the timescale of the other processes, it is likely that this is sufficiently small.
Note: There is one process that operates at a much faster scale: the flow in the stream. However, we are treating the stream in a very simple way here (assuming that the contaminant mass mixes rapidly and completely just downstream of where the plume enters, and we are not interested in how concentrations vary spatially in that mixing zone.
Although the rule of thumb we mentioned above is useful, the only way to really be certain if the timestep is small enough is to run the model multiple times with different timesteps. You can do this now yourself if you wish, but you don’t need to; the results of this little experiment are summarized below:
Timestep (day) Peak Concentration in Stream (mg/l)
50 2.050E-04
10 2.322E-04
5 2.341E-04
1 2.384E-04
0.1 2.387E-04
As can be seen, the result does not change (within three significant figures) if we use a smaller timestep than 1 day. In fact, even a larger timestep would probably be sufficient for our needs (as we will note in the next Lesson, the uncertainty in our inputs for a real world problem would almost certainly be much greater than the error generated by using a larger timestep).
Note: If you make the timestep greater than 50 days, the result will quickly break down. This is because it becomes impossible to represent the initial slug of mass entering from the pipeline accurately.
In the next Lesson, we will take a quick look at a probabilistic version of the model in order to illustrate some key concepts associated with representing uncertainty in contaminant transport models. | 1,060 | 4,757 | {"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} | 3.3125 | 3 | CC-MAIN-2024-38 | latest | en | 0.904365 |
https://homeschoolmath.blogspot.com/2011/02/seniors-juniors-algebra-word-problem.html | 1,516,360,545,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887973.50/warc/CC-MAIN-20180119105358-20180119125358-00794.warc.gz | 693,093,011 | 24,277 | ### Seniors & juniors algebra word problem
Here's a word problem that someone sent me recently:
The total number of girls in the combined junior and senior classes is equal to the number of boys in those two classes. If the senior class has 400 students and the junior class has 300 students, and if the ratio of boys to girls in the senior class is 5:3, what is the ratio of boys to girls in junior class?
This problem gives a lot of information, and it sounds like it can be solved many different ways. But the first task is to notice what we are given and what we are asked.
• We are asked about a ratio
• We're given one ratio, and all kinds of totals. There are boys and girls, senior and junior classes. In other words, there are four groups: senior boys, senior girls, junior boys and junior boys.
This sounds like it can be solved by setting up some equations and using algebra.
To start, you could for example notice the two facts given about the senior class: The senior class has 400 students and the ratio of boys to girls is 5:3.
From this it is easy to solve the number of boys and girls in the senior class.
The ratio 5:3 means 5/8 of them are boys and 3/8 of them are girls. Now, 5/8 of 400 is 250, and 3/8 of it is 150.
So seniors are solved... now to juniors. Let
B1 = boys in junior class
G1 = girls in junior class.
The first sentence of the problem gives us an equation:
G1 + 150 = B1 + 250
We also now that G1 + B1 = 300.
So let's solve this system of two equations:
G1 + 150 = B1 + 250
G1 + B1 = 300
I can subtract the bottom one from the top one to get:
150 - B1 = B1 - 50
200 = 2B1
B1 = 100
Since the total was 300, then G1 is 200.
And now the ratio of boys to girls, it is 100:200 or 1:2. All done!
You could also use straightforward algebra and set up four equations originally, using B1, G1, B2, and G2 for the numbers of boys and girls in junior and senior classes. You would get these four equations:
G1 + G2 = B1 + B2
B2 + G2 = 400
B1 + G1 = 300
B2/G2 = 5/3
And from these four you can solve all four unknowns, and then get the ratio of boys to girls in the junior class. | 585 | 2,119 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2018-05 | longest | en | 0.948881 |
https://www.daniweb.com/programming/software-development/threads/56175/help-validating-users-input | 1,714,043,574,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712297292879.97/warc/CC-MAIN-20240425094819-20240425124819-00875.warc.gz | 656,984,058 | 17,372 | I have a homework assignment to write a program about mileage. I've been able to get it to work, well except the average doesn't calculate correctly and I haven't been able to verifiy the users input. I'm using Visual C as my compiler.
Can someone help me with checking the users input and how to get the average to calculate correctly?
#include <stdio.h>
int main ()
{
/* Declare variables */
char a = 0, z = 0, A = 0, Z = 0;
float gallons, miles;
float totalGallons = 0;
float totalMiles = 0;
float tank = 0;
float average = 0;
/* Greetings and instructions to user */
printf( "\nAt the first prompt please enter the gallons used or -1 to exit program.\n"
"At the second prompt enter the miles that was driven. I will calculate\n"
"the total miles per gallon. After you type -1 to end the program, I will\n"
"give you the average of your total miles per gallon.");
/* Ask for gallons or sentinel value from user */
printf( "\n\n\nEnter the gallons used (-1 to end): ");
scanf( "%f", &gallons ); /* read users input */
if (gallons == a - z || A - Z){
fprintf( stderr, "You entered a letter. Please enter a number.\n" );
}
/* Start loop and calculation */
while ( gallons != -1 ) {
/* Ask for miles from user */
printf( "\nEnter the miles driven: ");
scanf( "%f", &miles );
/* Calculate useage */
tank = miles / gallons;
printf( "\nThe miles / gallon for this tank was: %.6f.\n\n", tank);
/* Ask user for gallons or sentinel value */
printf( "\nEnter the gallons used (-1 to end): ");
scanf( "%f", &gallons );
/* Calculate useage */
totalGallons = totalGallons + gallons;
totalMiles = totalMiles + miles;
/* Exit loop */
}
/* Calculate and print average */
average = (float)totalMiles / totalGallons;
printf( "\n\nThe overall average miles/gallon was: %.6f.\n\n", average);
/* Exit program */
return 0;
}
## All 5 Replies
>if (gallons == a - z || A - Z){
There are several things wrong with this. First, you're testing the value of characters, so you need to be surrounding them with single quotes. Next, you need to recompare with every part of the expression. C isn't polite enough to assume what you're doing. Third, your logic is faulty. And finally, testing gallons against a character is doomed to failure because it's a float. If it were a character, you could do this:
``````if ( gallons >= 'a' && gallons <= 'z' ||
gallons >= 'A' && gallons <= 'Z' )
{``````
However, in reality, you have two choices to validate a float. First, you can use the return value of scanf, which is (put simply) the number of successfully read values matching the format string:
``````if ( scanf ( "%f", &gallons ) != 1 ) {
// Invalid input
}``````
The second way is much better. Read all input as a string and validate it while in memory. Then when you're sure that it's legit, you convert to the appropriate type. That's probably too advanced of a solution at this point, so you'll probably want the scanf solution.
You are not calculating the total gallons after first gallon input before while, and after finishing the prog you are substacting from total gallons -1.
Thank you very much Narue for your reply. I did try what you said, but I just got more error messages. I think I must of done something wrong but I'm not smart enough to know what. This is what I wrote:
/* Ask for gallons or sentinel value from user */
printf( "\n\n\nEnter the gallons used (-1 to end): ");
if scanf( "%f", &gallons ) !=1) {
// Invalid input
/*if (gallons == a - z || A - Z){
fprintf( stderr, "You entered a letter. Please enter a number.\n" );
break;
}
/* Start loop and calculation */
while ( gallons != -1 ) {
And this is one of the 10 error messages I got:
error C2061: syntax error : identifier 'scanf'.
Luckly it's not a requirement to check the users input in this program, but I thought it would be a good habit to get into when writing any type of program.
I still have to figure out why my average value is coming out wrong. I don't have a counter in this program, but I'm starting to think that if I did, maybe that would help me to get the correct average value, since I don't have a set number of times the program is going to run. What do you think? What would be the best way to figure out the average?
Little E
>error C2061: syntax error : identifier 'scanf'
it means you need `(` before scanf.
>I still have to figure out why my average value is coming out wrong.
>What would be the best way to figure out the average?
Accumulating the `miles / gallons` and then dividing with number which represent how many times `miles / gallons` were calculated.
``````// dont forget to initialise totalTank to zero
while(numberOfTime)
{
tank = miles / gallon;
totalTank += tank;
}
avarageTank = totalTank / numberOfTime;``````
Thank you andor. Yes I did read your first post, but I didn't understand it. But your second post makes sense to me and I was able to get that to work. Thank you.
Little E
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https://mully.net/en/tag/magnetic-field/ | 1,721,512,500,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517541.97/warc/CC-MAIN-20240720205244-20240720235244-00070.warc.gz | 366,539,836 | 18,541 | Magnetic Field around Magnet(or Earth)
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How a CRT TV works
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Cathode Ray Tube (CRT)
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AC Power Generator
Electromagnetic induction Moving a magnet around a coil changes the magnetic field inside the coil, which causes current to flow through the coil. This phenomenon is called ‘electromagnetic induction’; the current flowing through the coil is ‘induced current.’ The result … more
Electromagnetic induction Moving the magnet around the inductor will change the inductor’s magnetic field, which will cause the current to flow through the inductor. This phenomenon is called electromagnetic induction. The current flowing in the inductor is called the induced … more
Magnetic Force
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Homopolar Motor
Assuming that the current flowing in the wire is in the X-direction, the magnetic field is in the Y direction. Then, what direction is the electromagnetic force? ‘Homopolar motor’ refers to a motor operated by using only one polarity of … more | 664 | 3,242 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-30 | latest | en | 0.945021 |
http://www.vias.org/comp_geometry/index_c.html | 1,542,802,948,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039748315.98/warc/CC-MAIN-20181121112832-20181121134832-00172.warc.gz | 518,817,504 | 2,120 | The Compendium Geometry is an eBook providing facts, formulas and explanations about geometry.
# Index C...
cartesian coordinate system Conversion between Polar and Cartesian Coordinates Conversion between 3D Coordinate Systems Three-dimensional Cartesian Coordinate System Two-dimensional Cartesian Coordinate System Cavalieri General Rules for 3D Solids center Ellipse center of gravity Properties of a Right Triangle Median and Centroid of a Triangle centroid Median and Centroid of a Triangle circle Circle circle ring Annulus (Circle Ring) circular cylinder Circular Cylinder circumcircle Properties of a Right Triangle Circumcircle of a Triangle circumference Ellipse circumradius Regular Polygon colatitude Spherical Coordinate System congruence transformation Congruence Transformation congruent triangles Congruent and Similar Triangles Congruence Transformation construction of a triangle Construction of a Triangle conversion Conversion between Polar and Cartesian Coordinates Conversion between 3D Coordinate Systems coordinate transform Coordinate Transform copyright Copyright cos Properties of a Right Triangle Creative Commons License Copyright cube Cuboid and Cube Platonic Solids - Regular Polyhedra Cube (Hexahedron) cuboid Cuboid and Cube cyclic quadrilateral Cyclic Quadrilateral cylinder Generalized Cylinder Circular Cylinder Cylindrical Wedge Cylindrical Segment cylindrical coordinate system Cylindrical Coordinate System Conversion between 3D Coordinate Systems cylindrical segment Cylindrical Segment cylindrical wedge Cylindrical Wedge
Last Update: 2011-01-11 | 272 | 1,591 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2018-47 | latest | en | 0.723937 |
https://fr.mathworks.com/matlabcentral/answers/1773440-robinson-map-projection | 1,713,664,910,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817699.6/warc/CC-MAIN-20240421005612-20240421035612-00169.warc.gz | 235,931,318 | 26,558 | # Robinson map projection?
15 vues (au cours des 30 derniers jours)
Ali Almakhmari le 3 Août 2022
I have a map of Earth in the cylindrical projection that has all the planet's data (longitude: -180 to 180 degrees, latitude: -90 degrees to 90 degrees), similar to the one attached. Is there a way in MATLAB to change the projection of this map into the Robinson projection? Any advice is appreciated
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### Réponse acceptée
Divit le 7 Sep 2023
Hello Ali,
Ensure that you have the “Mapping Toolbox” installed in MATLAB. It provides built-in support for various map projections, including the Robinson projection. You can check whether you have the Mapping Toolbox installed or not using the “ver mapping_toolbox” command.
You can achieve the desired Robinson Projection using a MATLAB script. In the script, you can create a new figure with a Robinson projection using the “axesm” function and specify the latitude and longitude limits. Then, create a geospatial reference object to associate the map image with the Robinson projection. Finally, use “geoshow” to display the georeferenced image in the Robinson projection.
Feel free to refer to the provided code below as an example:
% Load and display the map image (assuming it's in equirectangular projection)
% Display Original Image
imshow(mapImage);
title('Equirectangular Projection');
% Create a new figure with a Robinson projection
figure;
ax = axesm('robinson', 'FFaceColor', 'none', 'FEdgeColor', 'k');
% Define the latitude and longitude limits for the Robinson projection
latlim = [-90 90];
lonlim = [-180 180];
% Rotate the image by 180 degrees
mapImage = fliplr(rot90(mapImage, 2));
% Create a geospatial reference object for the Robinson projection
georef = georefcells(latlim, lonlim, size(mapImage));
% Display the georeferenced image in the Robinson projection
geoshow(ax, mapImage, georef);
title('Robinson Projection');
Output Image:
To know more you can refer to the following documentation links:
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#### Resources tagged with Pythagoras' theorem similar to Nicely Similar:
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Broad Topics > 2D Geometry, Shape and Space > Pythagoras' theorem
### Nicely Similar
##### Stage: 4 Challenge Level:
If the hypotenuse (base) length is 100cm and if an extra line splits the base into 36cm and 64cm parts, what were the side lengths for the original right-angled triangle?
### The Pillar of Chios
##### Stage: 3 Challenge Level:
Semicircles are drawn on the sides of a rectangle ABCD. A circle passing through points ABCD carves out four crescent-shaped regions. Prove that the sum of the areas of the four crescents is equal to. . . .
### The Medieval Octagon
##### Stage: 4 Challenge Level:
Medieval stonemasons used a method to construct octagons using ruler and compasses... Is the octagon regular? Proof please.
### Pareq Calc
##### Stage: 4 Challenge Level:
Triangle ABC is an equilateral triangle with three parallel lines going through the vertices. Calculate the length of the sides of the triangle if the perpendicular distances between the parallel. . . .
### Three Four Five
##### Stage: 4 Challenge Level:
Two semi-circles (each of radius 1/2) touch each other, and a semi-circle of radius 1 touches both of them. Find the radius of the circle which touches all three semi-circles.
### Pythagoras Proofs
##### Stage: 4 Challenge Level:
Can you make sense of these three proofs of Pythagoras' Theorem?
### Inscribed in a Circle
##### Stage: 3 Challenge Level:
The area of a square inscribed in a circle with a unit radius is, satisfyingly, 2. What is the area of a regular hexagon inscribed in a circle with a unit radius?
### Fitting In
##### Stage: 4 Challenge Level:
The largest square which fits into a circle is ABCD and EFGH is a square with G and H on the line CD and E and F on the circumference of the circle. Show that AB = 5EF. Similarly the largest. . . .
### Rhombus in Rectangle
##### Stage: 4 Challenge Level:
Take any rectangle ABCD such that AB > BC. The point P is on AB and Q is on CD. Show that there is exactly one position of P and Q such that APCQ is a rhombus.
### Zig Zag
##### Stage: 4 Challenge Level:
Four identical right angled triangles are drawn on the sides of a square. Two face out, two face in. Why do the four vertices marked with dots lie on one line?
### Tilted Squares
##### Stage: 3 Challenge Level:
It's easy to work out the areas of most squares that we meet, but what if they were tilted?
### Matter of Scale
##### Stage: 4 Challenge Level:
Prove Pythagoras' Theorem using enlargements and scale factors.
### A Chordingly
##### Stage: 3 Challenge Level:
Find the area of the annulus in terms of the length of the chord which is tangent to the inner circle.
### Pythagorean Triples I
##### Stage: 3 and 4
The first of two articles on Pythagorean Triples which asks how many right angled triangles can you find with the lengths of each side exactly a whole number measurement. Try it!
##### Stage: 4 Challenge Level:
A 1 metre cube has one face on the ground and one face against a wall. A 4 metre ladder leans against the wall and just touches the cube. How high is the top of the ladder above the ground?
### Picturing Pythagorean Triples
##### Stage: 4 and 5
This article discusses how every Pythagorean triple (a, b, c) can be illustrated by a square and an L shape within another square. You are invited to find some triples for yourself.
### Pythagorean Triples II
##### Stage: 3 and 4
This is the second article on right-angled triangles whose edge lengths are whole numbers.
### Equilateral Areas
##### Stage: 4 Challenge Level:
ABC and DEF are equilateral triangles of side 3 and 4 respectively. Construct an equilateral triangle whose area is the sum of the area of ABC and DEF.
### Semi-square
##### Stage: 4 Challenge Level:
What is the ratio of the area of a square inscribed in a semicircle to the area of the square inscribed in the entire circle?
### Kite in a Square
##### Stage: 4 Challenge Level:
Can you make sense of the three methods to work out the area of the kite in the square?
### Tennis
##### Stage: 3 Challenge Level:
A tennis ball is served from directly above the baseline (assume the ball travels in a straight line). What is the minimum height that the ball can be hit at to ensure it lands in the service area?
### Floored
##### Stage: 3 Challenge Level:
A floor is covered by a tessellation of equilateral triangles, each having three equal arcs inside it. What proportion of the area of the tessellation is shaded?
### All Tied Up
##### Stage: 4 Challenge Level:
A ribbon runs around a box so that it makes a complete loop with two parallel pieces of ribbon on the top. How long will the ribbon be?
### Square Pegs
##### Stage: 3 Challenge Level:
Which is a better fit, a square peg in a round hole or a round peg in a square hole?
### Pythagoras
##### Stage: 2 and 3
Pythagoras of Samos was a Greek philosopher who lived from about 580 BC to about 500 BC. Find out about the important developments he made in mathematics, astronomy, and the theory of music.
### Compare Areas
##### Stage: 4 Challenge Level:
Which has the greatest area, a circle or a square inscribed in an isosceles, right angle triangle?
### The Spider and the Fly
##### Stage: 4 Challenge Level:
A spider is sitting in the middle of one of the smallest walls in a room and a fly is resting beside the window. What is the shortest distance the spider would have to crawl to catch the fly?
### Round and Round
##### Stage: 4 Challenge Level:
Prove that the shaded area of the semicircle is equal to the area of the inner circle.
### Two Circles
##### Stage: 4 Challenge Level:
Draw two circles, each of radius 1 unit, so that each circle goes through the centre of the other one. What is the area of the overlap?
### Partly Circles
##### Stage: 4 Challenge Level:
What is the same and what is different about these circle questions? What connections can you make?
### Squaring the Circle and Circling the Square
##### Stage: 4 Challenge Level:
If you continue the pattern, can you predict what each of the following areas will be? Try to explain your prediction.
### Under the Ribbon
##### Stage: 4 Challenge Level:
A ribbon is nailed down with a small amount of slack. What is the largest cube that can pass under the ribbon ?
### Circle Box
##### Stage: 4 Challenge Level:
It is obvious that we can fit four circles of diameter 1 unit in a square of side 2 without overlapping. What is the smallest square into which we can fit 3 circles of diameter 1 unit?
### Cutting a Cube
##### Stage: 3 Challenge Level:
A half-cube is cut into two pieces by a plane through the long diagonal and at right angles to it. Can you draw a net of these pieces? Are they identical?
### Liethagoras' Theorem
##### Stage: 2 and 3
Liethagoras, Pythagoras' cousin (!), was jealous of Pythagoras and came up with his own theorem. Read this article to find out why other mathematicians laughed at him.
### Isosceles
##### Stage: 3 Challenge Level:
Prove that a triangle with sides of length 5, 5 and 6 has the same area as a triangle with sides of length 5, 5 and 8. Find other pairs of non-congruent isosceles triangles which have equal areas.
### The Old Goats
##### Stage: 3 Challenge Level:
A rectangular field has two posts with a ring on top of each post. There are two quarrelsome goats and plenty of ropes which you can tie to their collars. How can you secure them so they can't. . . .
### Napkin
##### Stage: 4 Challenge Level:
A napkin is folded so that a corner coincides with the midpoint of an opposite edge . Investigate the three triangles formed .
### Rectangular Pyramids
##### Stage: 4 and 5 Challenge Level:
Is the sum of the squares of two opposite sloping edges of a rectangular based pyramid equal to the sum of the squares of the other two sloping edges?
### The Dangerous Ratio
##### Stage: 3
This article for pupils and teachers looks at a number that even the great mathematician, Pythagoras, found terrifying.
### Where Is the Dot?
##### Stage: 3 Challenge Level:
A dot starts at the point (1,0) and turns anticlockwise. Can you estimate the height of the dot after it has turned through 45 degrees? Can you calculate its height?
### Trice
##### Stage: 3 Challenge Level:
ABCDEFGH is a 3 by 3 by 3 cube. Point P is 1/3 along AB (that is AP : PB = 1 : 2), point Q is 1/3 along GH and point R is 1/3 along ED. What is the area of the triangle PQR?
### Medallions
##### Stage: 4 Challenge Level:
I keep three circular medallions in a rectangular box in which they just fit with each one touching the other two. The smallest one has radius 4 cm and touches one side of the box, the middle sized. . . .
### Pythagorean Triples
##### Stage: 3 Challenge Level:
How many right-angled triangles are there with sides that are all integers less than 100 units?
### Weighty Problem
##### Stage: 3 Challenge Level:
The diagram shows a very heavy kitchen cabinet. It cannot be lifted but it can be pivoted around a corner. The task is to move it, without sliding, in a series of turns about the corners so that it. . . .
### Circle Scaling
##### Stage: 4 Challenge Level:
You are given a circle with centre O. Describe how to construct with a straight edge and a pair of compasses, two other circles centre O so that the three circles have areas in the ratio 1:2:3.
### Cubic Rotations
##### Stage: 4 Challenge Level:
There are thirteen axes of rotational symmetry of a unit cube. Describe them all. What is the average length of the parts of the axes of symmetry which lie inside the cube?
### Slippage
##### Stage: 4 Challenge Level:
A ladder 3m long rests against a wall with one end a short distance from its base. Between the wall and the base of a ladder is a garden storage box 1m tall and 1m high. What is the maximum distance. . . .
### Where to Land
##### Stage: 4 Challenge Level:
Chris is enjoying a swim but needs to get back for lunch. If she can swim at 3 m/s and run at 7m/sec, how far along the bank should she land in order to get back as quickly as possible?
### The Fire-fighter's Car Keys
##### Stage: 4 Challenge Level:
A fire-fighter needs to fill a bucket of water from the river and take it to a fire. What is the best point on the river bank for the fire-fighter to fill the bucket ?. | 2,522 | 10,494 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2016-40 | longest | en | 0.842118 |
https://mathoverflow.net/questions/365919/approximate-sobolev-embedding/366003 | 1,601,366,900,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401632671.79/warc/CC-MAIN-20200929060555-20200929090555-00610.warc.gz | 434,892,699 | 28,800 | # Approximate Sobolev embedding
It is well-known in $$H^2(\mathbb R^3)$$ embeds into $$L^{\infty}(\mathbb{R}^3).$$ Now consider a function $$u \in \ell^{\infty}(h\mathbb Z^3)$$ and a grid of points $$x \in h\mathbb{Z}^3.$$
We then define the finite-difference Laplacian
$$(\Delta_hu)(x):=\frac{\left(\sum_{i=1}^3 f(x+he_i)+f(x-he_i)\right)-6 f(x)}{h^2}$$
I wonder, is it true that for some universal $$C>0$$
$$\Vert u \Vert_{\ell^{\infty}(h\mathbb Z^3)} \le C (\Vert \Delta u \Vert_{\ell^2(h\mathbb Z^3)}+ \Vert u \Vert_{\ell^2(h\mathbb Z^3)})?$$
Here,
$$\Vert u \Vert^2_{\ell^2(h\mathbb Z^3)} = \sum_{x \in h\mathbb Z^3} h^3 \vert u(x) \vert^2$$
The intuition behind this estimate is that we discretely approximate the continuous setting with $$L^2$$ and $$L^{\infty}$$ norms.
Please let me know if you have any questions.
• $\|u\|_{\ell^\infty(I)}=\sup\{|u(i)|: i\in I\} \le \|u\|_{\ell^2(I)}$. – Jochen Wengenroth Jul 18 at 10:09
• @JochenWengenroth sorry, the norm here is not the standard $l^2$ norm but a weighted one, so that we 'approximate' the continuous world... – Solid State Physicist Jul 18 at 11:37
Yes, this is true, and there is a proof which closely tracks your intuition. As you know, this estimate can be proved in the continuum by applying the Sobolev embedding twice, first to get $$\nabla u \in L^p$$ for $$p<\frac{2d}{d-2}=6$$, and then once more to get $$u\in L^\infty$$. So for simplicity let me discuss how to get discrete versions of the Sobolev embedding for only one derivative. You can put these together in the same way to get your bound.
We will transfer the Sobolev inequalities in the continuum to the lattice by brute force. First, extend your function $$u$$ define on the lattice $$h\mathbb{Z}^d$$ by making it be constant on all cubes of side length $$h$$ which are centered on a point of $$h\mathbb{Z}^d$$. Next, mollify this piecewise constant function with the standard mollifier, on length scale $$h/10$$. Name the resulting smooth function $$v$$, which is now defined in $$\mathbb{R}^d$$. You have the following pointwise bounds: $$$$\left\| v \right\|_{L^\infty(z+[-h/2,h/2]^d)} \leq \sup_{z' \sim z} | u(z')|$$$$ and $$$$|D_h(z)|:=\frac1h\sup_{z'\sim z}|u(z) - u(z')| \leq \frac{C}{h^d} \int_{z+[-h/2,h/2]^d)} |\nabla v|.$$$$ Here $$\sim$$ means nearest-neighbor in the $$h\mathbb{Z}^d$$ lattice. The first bound is pretty easy, the second is true because near the boundary between nearest neighbor cubes, there will be a positive-measure set of points for which the mollifier picks up the difference between $$u(z)$$ and $$u(z')$$, and therefore on this set $$|\nabla v|$$ will be at least proportion to this difference. The proportion of this set in the cube is lower bounded by a constant (which does not depend on $$h$$).
Now, applying the (continuum) Sobolev inequality to $$v$$ and putting everything together gives a (discrete) Sobolev inequality for $$u$$.
There is one more minor point I will mention, which is that you have defined the $$H^2$$ norm with respect to the Laplacian only (and not the full set of mixed second derivatives). But you can perform a discrete integration by parts (twice) to bound the $$\ell^2$$ of $$D^2_hu$$, the full set of (possibly mixed) second-order differences, by the $$\ell^2$$ norm of $$\Delta_h u$$ (mimicking the usual proof in the continuum). | 1,070 | 3,346 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 35, "wp-katex-eq": 0, "align": 0, "equation": 2, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2020-40 | longest | en | 0.750709 |
https://trnds.co/cooking-for-ujw/parametric-equation-calculator-846361 | 1,618,065,359,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038057142.4/warc/CC-MAIN-20210410134715-20210410164715-00243.warc.gz | 679,496,067 | 18,045 | x(t), Contacts: [email protected] Graph The Pair Of Parametric Equations Calculator. x(t), We start by asking how to calculate the slope of a line tangent to a parametric curve at a point. b) Find a point on the line that is located at a distance of 2 units from the point (3, 1, 1). Consider the plane curve defined by the parametric equations \begin{align} x(t) &=2t+3 \label{eq1} \\ y(t) &=3tâ4 \label{eq2} \end{align} within $$â2â¤tâ¤3$$. 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Simply enter coordinates of first and second points, and the calculator shows both parametric and symmetric line equations. Trigonometry. Download free on Amazon. needed. Type the equation as given with one equation per line. which is defined as given below using two equations. , and Open the folders to explore their contents. where The steps given are required to be taken when you are using a parametric equation calculator. You can input only integer numbers or fractions in this online calculator. Use t as your variable. Follow these steps to change the mode of your calculator: 1. there are several online calculators available; this kind of tool A single parameter is usually represented with the parameter , while the symbols illustrated above. revert this also through eliminating this calculator. As you are Equation solver can find both numerical and parametric solutions of equations. Show Instructions In general, you can skip ⦠standard format to such a shape. In the above equations, t is the parameter which is a variable and describe the techniques in mathematics that introduce and Look below to see them all. you may find this conversion process a little much complex, but form, the tool is also used as a parametric form calculator, which is still used for a specific purpose and their respective methods Learn: Parametric Equations. However, its primary purpose is to find out the coordination. by the parameter t. Our online calculator finds the derivative of the parametrically derined function with step by step solution. problems. x(t)= y(t)= log$_{ }{ }$ sin-1: cos-1: tan-1: sinh-1: Download free on Google Play. into a simple procedure in less time. are differentiable functions and Step 1: Find a set of equations for the given function of any geometric shape. It is an expression that produces all points of the line in terms of one parameter, z. For using a parametric equations calculator, it is needed to know How to Calculate Parametric Equations. The parametric equations are simple linear expressions, but we need to view this problem in a step-by-step fashion. It is mostly used to explore the and y in which y is based on the x which can be found on the here. 0 Comment. parametric graphing. graphical display of coordinator points as per the given input in About: Beyond simple math and grouping (like "(x+2)(x-4)"), there are some functions you can use as well. The steps given are required to be taken when you are using a Differential Equations Calculators; Math Problem Solver (all calculators) Differential Equation Calculator. By ⦠generate the value of X and Y value pair that depends on the circle Leave all construction marks. converting these equations to a normal one, you need to eliminate By using this website, you agree to our Cookie Policy. y(t) Mathway. 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Pre-Algebra. A parametric equation of a curve expresses the points on the curve as an explicit function of "parameters" or indepedent variables usually denoted by t.Examples of parametric equation are: x = t^2 - t, y = 3t + 1, x = 3cost, y = 2sint, cos^2(t) + sin^2(t) = 1, and x = cos(3t). To do the elimination, first, you have to solve the x=f (t) After the conversion of function into this process, you can Step 5: Enter both equations in the parametric equations clear picture of this term and its equation, go through the below Entering data into the equation of a plane calculator. I'll try and add a drop down menu with the equations of some famous curves in the future...,. Graphing Calculator Polar Curves Derivative Calculator Integral Calculator Formulas and Notes Equation Calculator Algebra Calculator. X and Y. Among them, the easiest to use and Free math problem solver answers your calculus homework questions with step-by-step explanations. 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This representation calculator offers the functionality of a Learn more - the derivative of the parametric equation You canât begin graphing parametric equations until you change the mode of your calculator. 13. powered by. but not the real part of the circle. Intermediate. Find more Mathematics widgets in Wolfram|Alpha. Let's define function by the pair of parametric equations: xxtyyt 5. Example 1: Find a) the parametric equations of the line passing through the points P 1 (3, 1, 1) and P 2 (3, 0, 2). Parametric Equation of a Plane Calculator Parametric equation refers to the set of equations which defines the qualities as functions of one or more independent variables, called as parameters. Download free in Windows Store. This word is used to define For example, while the equation of a circle in Cartesian coordinates can be given by r^2=x^2+y^2, one set of parametric equations for the circle are given by x = rcost (1) y = rsint, (2) illustrated above. Section 3-1 : Parametric Equations and Curves. Derivatives of Parametric Equations. To get a Enter the Parametric Curve. Parametric equations are a set of equations that express a set of quantities as explicit functions of a number of independent variables, known as "parameters." This called a parameterized equation for the same line. to find out such a form when derivation of standard functions is You can find quantities (which is considered as functions) of the independent Then the derivative - parametric equations calculator -
This online analytical calculator helps you to find the parametric equation of a circle using the radius. graph of a given input with their calculated output. Step 6: Click the submit button, and you will get the solution. Additional features of equation of a plane calculator. They are mostly standard functions written as you might expect. In this section we will discuss how to find the area between a parametric curve and the x-axis using only the parametric equations (rather than eliminating the parameter and using standard Calculus I techniques on the resulting algebraic equation). Graphing Calculator Polar Curves Derivative Calculator Integral Calculator Formulas and Notes Equation Calculator Algebra Calculator Parametric Equation Grapher Enter the Parametric Curve. 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Of variables separate window of a line tangent to a given set of variables the symbols illustrated above mathematics is!, type it like this: 1 formulas and Notes equation calculator Algebra calculator length of the points define... Online calculator with respect to a particular t value agree to our Cookie Policy their output. Geometry, and you will eliminate the parameter that parametric equation calculator, expressions, but we need to view problem! Free equations calculator - solve linear, quadratic, polynomial, radical, exponential and equations! Equation for the same line a parameterized equation for arc length of the equation... ), Y ( t ), Y ( t ) â 0 this! That produces all points of the line equation in parametric form, that is used the... Differentiable functions and x ' ( t ), Y ( t ) are functions... To use and essential to Learn a concept is this equation defines a collection or group of quantities ( is! Section 3-1: parametric equations calculator - solve linear, quadratic,,! Step calculator displays line on a graph available in mathematics that is, equations the! This process, you will get the free values of the step by step calculator integer. This: 1 that defines a geometric object theorems and equations are also helpful real-world. Calculator formulas and Notes equation calculator Integral calculator formulas and Notes equation.. | 5,213 | 24,571 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "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": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2021-17 | longest | en | 0.820717 |
https://www.gradesaver.com/textbooks/math/algebra/algebra-2-1st-edition/chapter-6-rational-exponents-and-radical-functions-6-3-perform-function-operations-and-composition-6-3-exercises-problem-solving-page-433/43 | 1,696,176,433,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510903.85/warc/CC-MAIN-20231001141548-20231001171548-00141.warc.gz | 855,884,487 | 14,827 | ## Algebra 2 (1st Edition)
Given $r(w)=\frac{1.1w^{0.734}}{b(w)-d(w)}$ Substitute: $w=\frac{1.1w^{0.734}}{0.007(w)-0.002(w)}\\=\frac{1.1w^{0.734}}{0.005(w)}\\=220w^{-0266}$ For body weights of 6.5 grams, $220w^{-0266}=220(6.5)^{-0266}\approx133.72$ For body weights of 300 grams, $220w^{-0266}=220(300)^{-0266}\approx48.25$ For body weights of 70000 grams, $220w^{-0266}=220(70000)^{-0266}\approx11.31$ | 179 | 403 | {"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} | 4.0625 | 4 | CC-MAIN-2023-40 | latest | en | 0.537894 |
https://stackoverflow.com/questions/32916986/how-simplifying-fractions-in-matrices-with-sympy/32931999 | 1,632,304,293,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057337.81/warc/CC-MAIN-20210922072047-20210922102047-00556.warc.gz | 580,343,475 | 38,308 | # How simplifying fractions in matrices with sympy?
I'm using `sympy` to find a matrix's inverse. I've the next problem. When I compute the inverse of matrix `A` and I want prove it, I got a matrix with fractions; I mean
``````>> import sympy
>> from sympy import pprint
>> from sympy.abc import *
>> import sys
>> sys.displayhook = pprint
>> from sympy.matrices import *
>> A = Matrix([[a, b],[c, d]])
>> B = A.inv()
>> B
>> [1 b*c -b ]
>> [- + ------------ -----------]
>> [a 2 / b*c\ / b*c\]
>> [ a *|d - ---| a*|d - ---|]
>> [ \ a / \ a /]
>> [ ]
>> [ -c 1 ]
>> [ ----------- ------- ]
>> [ / b*c\ b*c ]
>> [ a*|d - ---| d - --- ]
>> [ \ a / a ]
>> B*A
>> [ /1 b*c \ b*c /1 b*c \ b*d ]
>> [a*|- + ------------| - ----------- b*|- + ------------| - -----------]
>> [ |a 2 / b*c\| / b*c\ |a 2 / b*c\| / b*c\]
>> [ | a *|d - ---|| a*|d - ---| | a *|d - ---|| a*|d - ---|]
>> [ \ \ a // \ a / \ \ a // \ a /]
>> [ ]
>> [ d b*c ]
>> [ 0 ------- - ----------- ]
>> [ b*c / b*c\ ]
>> [ d - --- a*|d - ---| ]
>> [ a \ a / ]
``````
And I wanna get the next matrix
``````>> I = Matrix([
>> [1, 0],
>> [0, 1]])
``````
My problem is the matrix `A*B` or `B*A`. Really I want to simplify the matrix `A*B` to get `I`. I tried `simplify()` but doesn't work.
• what does it mean simplifying? is it some technical term or what? my second question - what are you going to prove? Oct 2 '15 at 22:57
• Ok, B is the inverse of A and their product sould be the matrix identity I. In the product of the code I obtein a monster matrix AB (or BA) but I want the entrates of this matrix be zero (0). sympy doesn't simplify the entrates (1, 2), (1,1) and (2, 1). You can see the result of this product. Excuse my english, I don't speak it so good. Tranks for your time.
– Yakz
Oct 3 '15 at 3:17
• I get the same result as you for AB, but simplify(AB) gives the identity matrix. I use notebook with the latest Anaconda 64 on windows 7. Oct 3 '15 at 13:55
You can apply the `simplify` function to each cell of the matrix with `applyfunc`, like this:
``````>>> (B*A).applyfunc(simplify)
[1 0]
[ ]
[0 1]
``````
• I'm trying that, but I get `name 'simplify' not defined`.
– Yakz
Oct 5 '15 at 14:46
• Yes but nothing. I did a function that do this by entries.
– Yakz
Oct 13 '15 at 21:52
Forget python and sympy for a minute. Focus on finding an inverse of matrix with paper and pen.
For a `A = [[a, b], [c,d]]` matrix, we calculate inverse `A^-1` as,
`(1/D)*[[d, -b],[-c, a]]`. Here D is determinant of `A` matrix (1/ad-bc)
This (A^-1) is equal to `[[d/D, -b/D][-c/D, a/D]]`
Let's take the first element from first row and follow the operations I've made. For me they actually make no sense, but this is the way how sympy does :) Then apply this procedure to other Matrix elements.
``````=> d/D
d/(a*d-b*c)
a*d/(d*a^2 - a*b*c)
(a*d-b*c+b*c)/a^2*(d-b*c/a)
(a*d - a*b*c/a + b*c)/a^2*(d-b*c/a)
(a*(d-b*c/a) + b*c)/a^2*(d-b*c/a)
a*(d-b*c/a)/a^2*(d-b*c/a) + b*c/a^2*(d-b*c/a)
1/a + b*c/a^2*(d-b*c/a) [this is how sympy outputs]
>>> A = Matrix([[a,b],[c,d]])
>>> B = A**-1 #same as B = A.inv()
>>> B[0]
1/a + b*c/(a**2*(d - b*c/a))
``````
Now, let's have a look up what is sympy A*B output.
``````>>> N = A*B
>>> N
Matrix([
[a*(1/a + b*c/(a**2*(d - b*c/a))) - b*c/(a*(d - b*c/a)), 0],
[c*(1/a + b*c/(a**2*(d - b*c/a))) - c*d/(a*(d - b*c/a)), d/(d - b*c/a) - b*c/(a*(d - b*c/a))]])
>>> pprint(N)
⎡ ⎛1 b⋅c ⎞ b⋅c ⎤
⎢a⋅⎜─ + ────────────⎟ - ─────────── 0 ⎥
⎢ ⎜a 2 ⎛ b⋅c⎞⎟ ⎛ b⋅c⎞ ⎥
⎢ ⎜ a ⋅⎜d - ───⎟⎟ a⋅⎜d - ───⎟ ⎥
⎢ ⎝ ⎝ a ⎠⎠ ⎝ a ⎠ ⎥
⎢ ⎥
⎢ ⎛1 b⋅c ⎞ c⋅d d b⋅c ⎥
⎢c⋅⎜─ + ────────────⎟ - ─────────── ─────── - ───────────⎥
⎢ ⎜a 2 ⎛ b⋅c⎞⎟ ⎛ b⋅c⎞ b⋅c ⎛ b⋅c⎞⎥
⎢ ⎜ a ⋅⎜d - ───⎟⎟ a⋅⎜d - ───⎟ d - ─── a⋅⎜d - ───⎟⎥
⎣ ⎝ ⎝ a ⎠⎠ ⎝ a ⎠ a ⎝ a ⎠⎦
``````
It doesn't evaluate it to direct `eye(2)` but if you take elements, and simplify them, you'll see that they this messy matrix is actually and 2x2 identity matrix.
A pythonic way to check that (knowing given):
``````>>> N[0]
a*(1/a + b*c/(a**2*(d - b*c/a))) - b*c/(a*(d - b*c/a))
>>> N[1]
0
>>> N[3]
d/(d - b*c/a) - b*c/(a*(d - b*c/a))
>>> N[2]
c*(1/a + b*c/(a**2*(d - b*c/a))) - c*d/(a*(d - b*c/a))
>>> def will_evaluate_one(a,b,c,d):
... return a*(1/a + b*c/(a**2*(d - b*c/a))) - b*c/(a*(d - b*c/a)) #N[0]
...
>>> will_evaluate_one(1,2,3,9)
1
>>> will_evaluate_one(1,2,3,19)
1
>>> will_evaluate_one(1,2,23,19)
1
>>> will_evaluate_one(1,12,23,19)
1
>>> def will_also_evaluate_one(a,b,c,d):
... return d/(d - b*c/a) - b*c/(a*(d - b*c/a)) #N[1]
...
>>> will_also_evaluate_one(2,4,5,6)
1
>>> will_also_evaluate_one(2,4,15,6)
1
>>> will_also_evaluate_one(2,14,15,6)
1
>>> will_also_evaluate_one(12,14,15,6)
1
``````
Note: I've just realised that, sympy uses anlaytic inversion formula. See here: https://en.wikipedia.org/wiki/Helmert%E2%80%93Wolf_blocking | 2,188 | 5,777 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2021-39 | latest | en | 0.742749 |
https://au.answers.yahoo.com/question/index?qid=20200124121818AAHI5Wp | 1,611,787,383,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704833804.93/warc/CC-MAIN-20210127214413-20210128004413-00631.warc.gz | 219,223,434 | 28,871 | Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 year ago
# Math Exercise?
A set of 10 positive numbers is combined with another set of 12 positive numbers. If the mean of the first set is 16, what is the third smallest possible mean of the second set so that the average of the combined set is an integer?
Relevance
• Anonymous
10 months ago
Source(s): Women wear headscarves tied at the front to prevent headaches from sky pushing down and to prevent throat cancer. Mega-tsunami for New York will be 400 meters; then engulfed-in-lava Los Angeles will be flooded too; also, asteroid destroys Gulf of Mexico; only Alaska, Eurasia, and Africa remain (obviously without coasts). 1st big earthquake in Russia; 2nd bigger one in China (will be split in half; radiation!); 3rd biggest will be in the USA (Greek Orthodox monk Elidiy from Africa); forgive me.
• alex
Lv 7
1 year ago
Is the mean of the 2nd set an integer ?
Hint
x = mean of the 2nd set
12x+160=22k
or
6x=11k-80 , with k = 8 ,9,10,... | 269 | 1,020 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2021-04 | latest | en | 0.907688 |
http://www.chegg.com/homework-help/fluid-mechanics-with-engineering-applications-10th-edition-chapter-15.13-solutions-9780072432022 | 1,472,263,531,000,000,000 | text/html | crawl-data/CC-MAIN-2016-36/segments/1471982296931.2/warc/CC-MAIN-20160823195816-00236-ip-10-153-172-175.ec2.internal.warc.gz | 363,760,983 | 16,979 | View more editions
# Fluid Mechanics With Engineering Applications (10th Edition)Solutions for Chapter 15.13
• 1356 step-by-step solutions
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Chapter: Problem:
Suppose the pumps of Sample Prob. 15.13 operate at 1500 rpm. What then would be the flow rates for (a) a single pump; (b) two pumps in series; (c) two pumps in parallel. All other data to remain the same.
Prob. 15.13: Water is pumped from reservoir A to reservoir B, which has a 35-ft higher water surface. The loss in the pipeline is given by hL = 20Q2 (hL in ft, Q in 100’s of gpm). Two available pumps (n = 1800 rpm) each have the following characteristics:
h, ft: 100 90 80 60 40 20 Q, gpm: 0 110 180 250 300 340
SAMPLE SOLUTION
Chapter: Problem:
• Step 1 of 7
Head loss in pipe expressed as,
Elevation difference from water surface to reservoir,
Operating at 1800 rpm:
Head, ft Flow rate, 100 0 90 110 80 180 60 250 40 300 20 340
• Step 2 of 7
From the similarity laws, we have
Thus new operating speed of the pump is 1500 rpm.
Then,
• Step 3 of 7
Sample calculation:
At 1800 rpm
Operating head is 80 ft and corresponding discharge is 180 gpm
Then, the new discharge at operating speed 1500 rpm is
• Step 4 of 7
The operating head and discharge at new operating speed of 1500 rpm is
Head, ft Flow rate, 69.4 0 62.46 91.63 55.52 149.94 41.64 208.25 27.76 249.90 13.88 283.22
• Step 5 of 7
a)
When a single pump:
Head, ft Flow rate, 69.4 0 62.46 91.63 55.52 149.94 41.64 208.25 27.76 249.90 13.88 283.22
• Step 6 of 7
b)
When two pumps in series:
Head,ft Discharge,gpm 190 110 140 250 60 340
c)
When two pumps in parallel:
Head, ft Discharge, gpm 100 110 80 430 40 640
• Step 7 of 7
By plotting pump characteristic curves:
From the characteristic curves
a)
Rate of flow through a pump when pump operating as single pump
b)
Rate of flow through a pump when two pumps operating in series is
c)
Rate of flow through a pump when two pumps operating in parallel is .
Corresponding Textbook
Fluid Mechanics With Engineering Applications | 10th Edition
9780072432022ISBN-13: 0072432020ISBN: | 708 | 2,281 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-36 | latest | en | 0.834169 |
https://electricalnotes.wordpress.com/2014/04/01/calculate-size-of-capacitor-bank-annual-saving-payback-period/ | 1,529,500,644,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863519.49/warc/CC-MAIN-20180620124346-20180620144346-00607.warc.gz | 594,300,093 | 29,777 | # Calculate Size of Capacitor Bank / Annual Saving & Payback Period
• Calculate Size of Capacitor Bank Annual Saving in Bills and Payback Period for Capacitor Bank.
• Electrical Load of (1) 2 No’s of 18.5KW,415V motor ,90% efficiency,0.82 Power Factor ,(2) 2 No’s of 7.5KW,415V motor ,90% efficiency,0.82 Power Factor,(3) 10KW ,415V Lighting Load. The Targeted Power Factor for System is 0.98.
• Electrical Load is connected 24 Hours, Electricity Charge is 100Rs/KVA and 10Rs/KW.
• Calculate size of Discharge Resistor for discharging of capacitor Bank. Discharge rate of Capacitor is 50v in less than 1 minute.
• Also Calculate reduction in KVAR rating of Capacitor if Capacitor Bank is operated at frequency of 40Hz instead of 50Hz and If Operating Voltage 400V instead of 415V.
• Capacitor is connected in star Connection, Capacitor voltage 415V, Capacitor Cost is 60Rs/Kvar. Annual Deprecation Cost of Capacitor is 12%.
## Calculation:
• For Connection (1):
• Total Load KW for Connection(1) =Kw / Efficiency=(18.5×2) / 90%=41.1KW
• Total Load KVA (old) for Connection(1)= KW /Old Power Factor= 41.1 /0.82=50.1 KVA
• Total Load KVA (new) for Connection(1)= KW /New Power Factor= 41.1 /0.98= 41.9KVA
• Total Load KVAR= KWX([(√1-(old p.f)2) / old p.f]- [(√1-(New p.f)2) / New p.f])
• Total Load KVAR1=41.1x([(√1-(0.82)2) / 0.82]- [(√1-(0.98)2) / 0.98])
• OR
• tanǾ1=Arcos(0.82)=0.69
• tanǾ2=Arcos(0.98)=0.20
• Total Load KVAR1= KWX (tanǾ1- tanǾ2) =41.1(0.69-0.20)=20.35KVAR
• For Connection (2):
• Total Load KW for Connection(2) =Kw / Efficiency=(7.5×2) / 90%=16.66KW
• Total Load KVA (old) for Connection(1)= KW /Old Power Factor= 16.66 /0.83=20.08 KVA
• Total Load KVA (new) for Connection(1)= KW /New Power Factor= 16.66 /0.98= 17.01KVA
• Total Load KVAR2= KWX([(√1-(old p.f)2) / old p.f]- [(√1-(New p.f)2) / New p.f])
• Total Load KVAR2=20.35x([(√1-(0.83)2) / 0.83]- [(√1-(0.98)2) / 0.98])
• For Connection (3):
• Total Load KW for Connection(3) =Kw =10KW
• Total Load KVA (old) for Connection(1)= KW /Old Power Factor= 10/0.85=11.76 KVA
• Total Load KVA (new) for Connection(1)= KW /New Power Factor= 10 /0.98= 10.20KVA
• Total Load KVAR3= KWX([(√1-(old p.f)2) / old p.f]- [(√1-(New p.f)2) / New p.f])
• Total Load KVAR3=20.35x([(√1-(0.85)2) / 0.85]- [(√1-(0.98)2) / 0.98])
• Total KVAR=KVAR1+ KVAR2+KVAR3
• Total KVAR=20.35+7.82+4.17
• Total KVAR=32 Kvar
## Size of Capacitor Bank:
• Site of Capacitor Bank=32 Kvar.
• Leading KVAR supplied by each Phase= Kvar/No of Phase
• Leading KVAR supplied by each Phase =32/3=10.8Kvar/Phase
• Capacitor Charging Current (Ic)= (Kvar/Phase x1000)/Volt
• Capacitor Charging Current (Ic)= (10.8×1000)/(415/√3)
• Capacitor Charging Current (Ic)=44.9Amp
• Capacitance of Capacitor = Capacitor Charging Current (Ic)/ Xc
• Xc=2 x 3.14 x f x v=2×3.14x50x(415/√3)=75362
• Capacitance of Capacitor=44.9/75362= 5.96µF
• Required 3 No’s of 10.8 Kvar Capacitors and
## Protection of Capacitor Bank
### Size of HRC Fuse for Capacitor Bank Protection:
• Size of the fuse =165% to 200% of Capacitor Charging current.
• Size of the fuse=2×44.9Amp
• Size of the fuse=90Amp
### Size of Circuit Breaker for Capacitor Protection:
• Size of the Circuit Breaker =135% to 150% of Capacitor Charging current.
• Size of the Circuit Breaker=1.5×44.9Amp
• Size of the Circuit Breaker=67Amp
• Thermal relay setting between 1.3 and 1.5of Capacitor Charging current.
• Thermal relay setting of C.B=1.5×44.9 Amp
• Thermal relay setting of C.B=67 Amp
• Magnetic relay setting between 5 and 10 of Capacitor Charging current.
• Magnetic relay setting of C.B=10×44.9Amp
• Magnetic relay setting of C.B=449Amp
### Sizing of cables for capacitor Connection:
• Capacitors can withstand a permanent over current of 30% +tolerance of 10% on capacitor Current.
• Cables size for Capacitor Connection= 1.3 x1.1 x nominal capacitor Current
• Cables size for Capacitor Connection = 1.43 x nominal capacitor Current
• Cables size for Capacitor Connection=1.43×44.9Amp
• Cables size for Capacitor Connection=64 Amp
### Maximum size of discharge Resistor for Capacitor:
• Capacitors will be discharge by discharging resistors.
• After the capacitor is disconnected from the source of supply, discharge resistors are required for discharging each unit within 3 min to 75 V or less from initial nominal peak voltage (according IEC-standard 60831).
• Discharge resistors have to be connected directly to the capacitors. There shall be no switch, fuse cut-out or any other isolating device between the capacitor unit and the discharge resistors.
• Max. Discharge resistance Value (Star Connection) = Ct / Cn x Log (Un x√2/ Dv).
• Max. Discharge resistance Value (Delta Connection)= Ct / 1/3xCn x Log (Un x√2/ Dv)
• Where Ct =Capacitor Discharge Time (sec)
• Un = Line Voltage
• Dv=Capacitor Discharge voltage.
• Maximum Discharge resistance =60 / ((5.96/1000000)x log ( 415x√2 /50)
• Maximum Discharge resistance=4087 KΩ
### Effect of Decreasing Voltage & Frequency on Rating of Capacitor:
• The kvar of capacitor will not be same if voltage applied to the capacitor and frequency changes
• Reduced in Kvar size of Capacitor when operating 50 Hz unit at 40 Hz
• Actual KVAR = Rated KVAR x(Operating Frequency / Rated Frequency)
• Actual KVAR = Rated KVAR x(40/50)
• Actual KVAR = 80% of Rated KVAR
• Hence 32 Kvar Capacitor works as 80%x32Kvar= 26.6Kvar
• Reduced in Kvar size of Capacitor when operating 415V unit at 400V
• Actual KVAR = Rated KVAR x(Operating voltage / Rated voltage)^2
• Actual KVAR = Rated KVAR x(400/415)^2
• Actual KVAR=93% of Rated KVAR
• Hence 32 Kvar Capacitor works as 93%x32Kvar= 23.0Kvar
## Annual Saving and Pay Back Period
### Before Power Factor Correction:
• Total electrical load KVA (old)= KVA1+KVA2+KVA3
• Total electrical Load KW =62kw
• KVA Demand Charge=KVA X Charge
• KVA Demand Charge=82x60Rs
• KVA Demand Charge=8198 Rs
• Annual Unit Consumption=KWx Daily usesx365
• Annual Unit Consumption=62x24x365 =543120 Kwh
• Annual charges =543120×10=5431200 Rs
• Total Annual Cost= 8198+5431200
### After Power Factor Correction:
• Total electrical load KVA (new)= KVA1+KVA2+KVA3
• Total electrical Load KW =62kw
• KVA Demand Charge=KVA X Charge
• KVA Demand Charge=69x60Rs =6916 Rs————-(1)
• Annual Unit Consumption=KWx Daily usesx365
• Annual Unit Consumption=62x24x365 =543120 Kwh
• Annual charges =543120×10=5431200 Rs—————–(2)
• Capital Cost of capacitor= Kvar x Capacitor cost/Kvar = 82 x 60= 4919 Rs—(3)
• Annual Interest and Deprecation Cost =4919 x 12%=590 Rs—–(4)
• Total Annual Cost= 6916+5431200+4919+590
## Pay Back Period:
• Total Annual Cost before Power Factor Correction= 5439398 Rs
• Total Annual Cost After Power Factor Correction =5438706 Rs
• Annual Saving= 5439398-5438706 Rs
• ### Annual Saving= 692 Rs
• Payback Period= Capital Cost of Capacitor / Annual Saving
• Payback Period= 4912 / 692
• ### Payback Period = 7.1 Years
Jignesh Parmar has completed M.Tech (Power System Control), B.E(Electrical). He is member of Institution of Engineers (MIE) and CEng,India. Membership No:M-1473586.He has more than 16 years experience in Transmission -Distribution-Electrical Energy theft detection-Electrical Maintenance-Electrical Projects (Planning-Designing-Technical Review-coordination -Execution). He is Presently associate with one of the leading business group as a Deputy Manager at Ahmedabad,India. He has published numbers of Technical Articles in “Electrical Mirror”, “Electrical India”, “Lighting India”,”Smart Energy”, “Industrial Electrix”(Australian Power Publications) Magazines. He is Freelancer Programmer of Advance Excel and design useful Excel base Electrical Programs as per IS, NEC, IEC,IEEE codes. He is Technical Blogger and Familiar with English, Hindi, Gujarati, French languages. He wants to Share his experience & Knowledge and help technical enthusiasts to find suitable solutions and updating themselves on various Engineering Topics.
### 27 Responses to Calculate Size of Capacitor Bank / Annual Saving & Payback Period
1. Jonryl Novicio says:
very informative
thank you/
• Ghani Ullah Khan says:
2. Nabojit Bhowal says:
Dear Sir,
Could you please tell me what value does a D.C. measuring instrument indicate r.m.s.,average,maximum or average value and why
3. PRASHANT SONI says:
very much informative and useful calculation..
You have also uploaded the XLS sheet of this calculation also…
Thanks for ur this valuable service..
Prashat K. Soni
4. gustavo rafael faquira soto says:
excelente…gracias…
5. Hanumanth says:
Dear Sir,
Can you share types of electrical loads and colour code indicated on panels respective loads.
6. dream egy says:
Hi Pls we need calculate the no of secondary turns for current transformers .and the size of tornado cor
7. Aman Aggarwal says:
Thank u so much sir…
8. vaibhav says:
respected sir
i want to know about how to select capacitor bank in control panel design and which basis its
apply i..e load? can u explain with examples ? i m tired to find out its process. mail id : vaibhav12aug@gmail.com
vaibhav
9. RUBESH says:
Dear sir your explanation is very good & i need the same calculation for HT line (including Transformer inrush,Line fault etc )
10. sidh187 says:
hi Sir
I am Hussain and working in Australia. i am given a new task of power factor correction of two MCCs (control Panels).
Please guide me how to move forward.
regards
Hussain
11. chandandhir says:
really this is very usefull chapter in eletrical system and it is save our money also ,thanks a lot again sir
12. Binu Joy says:
Very informative
13. dan says:
I wanna know how you derived the formula for finding the size of discharge resistor
14. krishna naik says:
If capacitance connected in the circuit is more than required what will be the impact on billing. It will come down or shoot up. Pl write in detail
15. sir I have a doubt… isn’t the size of cable should be more than the size of C.B….?
which in given case is not
Size of the Circuit Breaker=67Amp
Cables size for Capacitor Connection=64 Amp
please correct me if I’m wrong!
16. M D BHOSALE says:
IS STAR CONNECTION MORE EFFICIENT THAN DELTA CONNECTION IN ABOVE EX.?
17. netfreak says:
Could you please inform me whether this is per any standard ?If yes then which one?
18. sam says:
Thank you
19. GNANAGRI says:
USEFUL MSG SIR THANK U SIR
20. haji says:
450 kvar is need to design .if I connect 150 kvar in star connection per phase .mu question is that is these capacitors will become equal to 450 kvAR connectec in star
21. dipesh nidhi says:
its too helpful in explained way to practical techincians
22. Rajesh says:
Jignesh,
why voltage across capacitor increases when connected across detuned reactor in capacitor bank.
23. Rajesh21 says:
Why the capacitors in capacitor bank are rated for higher than nominal voltage
24. hari says:
KVAR= KW(tan (cos^(-1) (old pf))-tan (cos^(-1) new pf)) is Corrct or not sir
25. Ahmed Mo'men says:
Thanks a lot for this helpful article. However, I have 2 comments:
1- The KVA cost is 60, not 100 Rs.
2- The Capacitor bank cost shouldn’t be added to the annual cost.
Please rectify me if I’m wrong 🙂
26. Mongia DK says:
Dear Sir,
Thanks for the explanation !
Calculations with ROI for p.f. improvement aspect has been explained in a quite simpler way. Very Easy to understand to an individual concerned with the issue. | 3,414 | 11,399 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2018-26 | latest | en | 0.614066 |
https://docs.python.org/pt-br/3/library/fractions.html | 1,716,522,971,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058677.90/warc/CC-MAIN-20240524025815-20240524055815-00836.warc.gz | 183,653,457 | 9,725 | # `fractions` — Rational numbers¶
Código-fonte: Lib/fractions.py
The `fractions` module provides support for rational number arithmetic.
A Fraction instance can be constructed from a pair of integers, from another rational number, or from a string.
class fractions.Fraction(numerator=0, denominator=1)
class fractions.Fraction(other_fraction)
class fractions.Fraction(float)
class fractions.Fraction(decimal)
class fractions.Fraction(string)
The first version requires that numerator and denominator are instances of `numbers.Rational` and returns a new `Fraction` instance with value `numerator/denominator`. If denominator is `0`, it raises a `ZeroDivisionError`. The second version requires that other_fraction is an instance of `numbers.Rational` and returns a `Fraction` instance with the same value. The next two versions accept either a `float` or a `decimal.Decimal` instance, and return a `Fraction` instance with exactly the same value. Note that due to the usual issues with binary floating-point (see Aritmética de ponto flutuante: problemas e limitações), the argument to `Fraction(1.1)` is not exactly equal to 11/10, and so `Fraction(1.1)` does not return `Fraction(11, 10)` as one might expect. (But see the documentation for the `limit_denominator()` method below.) The last version of the constructor expects a string or unicode instance. The usual form for this instance is:
```[sign] numerator ['/' denominator]
```
where the optional `sign` may be either ‘+’ or ‘-’ and `numerator` and `denominator` (if present) are strings of decimal digits (underscores may be used to delimit digits as with integral literals in code). In addition, any string that represents a finite value and is accepted by the `float` constructor is also accepted by the `Fraction` constructor. In either form the input string may also have leading and/or trailing whitespace. Here are some examples:
```>>> from fractions import Fraction
>>> Fraction(16, -10)
Fraction(-8, 5)
>>> Fraction(123)
Fraction(123, 1)
>>> Fraction()
Fraction(0, 1)
>>> Fraction('3/7')
Fraction(3, 7)
>>> Fraction(' -3/7 ')
Fraction(-3, 7)
>>> Fraction('1.414213 \t\n')
Fraction(1414213, 1000000)
>>> Fraction('-.125')
Fraction(-1, 8)
>>> Fraction('7e-6')
Fraction(7, 1000000)
>>> Fraction(2.25)
Fraction(9, 4)
>>> Fraction(1.1)
Fraction(2476979795053773, 2251799813685248)
>>> from decimal import Decimal
>>> Fraction(Decimal('1.1'))
Fraction(11, 10)
```
The `Fraction` class inherits from the abstract base class `numbers.Rational`, and implements all of the methods and operations from that class. `Fraction` instances are hashable, and should be treated as immutable. In addition, `Fraction` has the following properties and methods:
Alterado na versão 3.2: The `Fraction` constructor now accepts `float` and `decimal.Decimal` instances.
Alterado na versão 3.9: The `math.gcd()` function is now used to normalize the numerator and denominator. `math.gcd()` always return a `int` type. Previously, the GCD type depended on numerator and denominator.
Alterado na versão 3.11: Underscores are now permitted when creating a `Fraction` instance from a string, following PEP 515 rules.
Alterado na versão 3.11: `Fraction` implements `__int__` now to satisfy `typing.SupportsInt` instance checks.
Alterado na versão 3.12: Space is allowed around the slash for string inputs: `Fraction('2 / 3')`.
Alterado na versão 3.12: `Fraction` instances now support float-style formatting, with presentation types `"e"`, `"E"`, `"f"`, `"F"`, `"g"`, `"G"` and `"%""`.
numerator
Numerator of the Fraction in lowest term.
denominator
Denominator of the Fraction in lowest term.
as_integer_ratio()
Return a tuple of two integers, whose ratio is equal to the original Fraction. The ratio is in lowest terms and has a positive denominator.
is_integer()
Return `True` if the Fraction is an integer.
classmethod from_float(flt)
Alternative constructor which only accepts instances of `float` or `numbers.Integral`. Beware that `Fraction.from_float(0.3)` is not the same value as `Fraction(3, 10)`.
Nota
From Python 3.2 onwards, you can also construct a `Fraction` instance directly from a `float`.
classmethod from_decimal(dec)
Alternative constructor which only accepts instances of `decimal.Decimal` or `numbers.Integral`.
Nota
From Python 3.2 onwards, you can also construct a `Fraction` instance directly from a `decimal.Decimal` instance.
limit_denominator(max_denominator=1000000)
Finds and returns the closest `Fraction` to `self` that has denominator at most max_denominator. This method is useful for finding rational approximations to a given floating-point number:
```>>> from fractions import Fraction
>>> Fraction('3.1415926535897932').limit_denominator(1000)
Fraction(355, 113)
```
or for recovering a rational number that’s represented as a float:
```>>> from math import pi, cos
>>> Fraction(cos(pi/3))
Fraction(4503599627370497, 9007199254740992)
>>> Fraction(cos(pi/3)).limit_denominator()
Fraction(1, 2)
>>> Fraction(1.1).limit_denominator()
Fraction(11, 10)
```
__floor__()
Returns the greatest `int` `<= self`. This method can also be accessed through the `math.floor()` function:
```>>> from math import floor
>>> floor(Fraction(355, 113))
3
```
__ceil__()
Returns the least `int` `>= self`. This method can also be accessed through the `math.ceil()` function.
__round__()
__round__(ndigits)
The first version returns the nearest `int` to `self`, rounding half to even. The second version rounds `self` to the nearest multiple of `Fraction(1, 10**ndigits)` (logically, if `ndigits` is negative), again rounding half toward even. This method can also be accessed through the `round()` function.
__format__(format_spec, /)
Provides support for float-style formatting of `Fraction` instances via the `str.format()` method, the `format()` built-in function, or Formatted string literals. The presentation types `"e"`, `"E"`, `"f"`, `"F"`, `"g"`, `"G"` and `"%"` are supported. For these presentation types, formatting for a `Fraction` object `x` follows the rules outlined for the `float` type in the Minilinguagem de especificação de formato section.
Here are some examples:
```>>> from fractions import Fraction
>>> format(Fraction(1, 7), '.40g')
'0.1428571428571428571428571428571428571429'
>>> format(Fraction('1234567.855'), '_.2f')
'1_234_567.86'
>>> f"{Fraction(355, 113):*>20.6e}"
'********3.141593e+00'
>>> old_price, new_price = 499, 672
>>> "{:.2%} price increase".format(Fraction(new_price, old_price) - 1)
'34.67% price increase'
```
Ver também
Módulo `numbers`
The abstract base classes making up the numeric tower. | 1,679 | 6,642 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-22 | longest | en | 0.731607 |
http://www.eugeneoloughlin.com/2017/ | 1,723,427,139,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641023489.70/warc/CC-MAIN-20240811235403-20240812025403-00571.warc.gz | 41,482,196 | 52,848 | ## Saturday, December 30, 2017
### A Lesson in Probability #lotto
So - there has been another winner in Ireland of the Euromillions jackpot! Congratulations to whoever it is - their lives have just changed (hopefully for the better). I often use the probability of winning the Irish Lotto in statistics class when discussing the topic of probability. Using combinatorial maths you can work out how many combinations of six numbers there are out of the 47 balls in the Lotto draw - there are 10,737,573. In other words you have a better than a one in ten million chance of choosing the six correct numbers and winning the lotto. As I say to my class, if there was a horse in a race at odds of 10,000,000 to 1 would you put a bet on it? Of course, you do increase your chances of winning if you buy more than one line of numbers.
According to the Irish Lotto website, you have a "1 in 29" chance of winning a prize in the Lotto only draw, (the odds of winning are a better "1 in 17" if you add Lotto Plus). A "1 in 29" chance of winning is equivalent to about 3.5% - if you play regularly you can expect to win about 3.5% of the time, very low odds I think you'll agree. As Charles Wheelan puts it in his book "Naked Statistics", buying a lotto ticket is "a stupid thing to do". Here's how the "1 in 29" ( or approx 3.5%) chance of winning the Irish Lotto is made up:
So if you spend €2 buying one line in the Lotto draw, you can expect a return of around just 7 cent - it's a near mathematical certainty that you will lose money. In last Wednesday's draw (27th Dec), there were 25,213 winners in total of the various prizes. Using the "1 in 29" odds, this means that a whopping 721,177 (25,213 x 29) players did not win!
Of course, somebody does win - IT COULD BE YOU!
## Wednesday, December 27, 2017
### 14,000,000 YouTube views, and who's watching on Christmas Day? #Analytics
On Christmas Day this year, there were 4,101 views on my YouTube channel, which also passed the 14,000,000 lifetime views mark on the same day. December also shows the traditional decline from a high of around 12,000 views per day to a around third of this amount over the Christmas holidays leading to the New Year.
Click image to enlarge.
So, I wondered who was watching on Christmas day? YouTube Analytics tells me that India leads the way with 1,329 (32.4%) views, while the United States is a distant second with 320 (7.8%) views. In the top ten on Christmas Day are countries like Saudi Arabia (138), Turkey (117), Egypt (85), and the United Arab Emirates (55). None of these countries feature in the top ten list for the whole year so far. Countries like the UK, Canada, Australia, and Ireland basically switch off at Christmas. There were 39 views on my channel on Christmas Day - I wonder if any of these were my students?
The most popular video worldwide on Christmas Day was How To... Perform Simple Linear Regression by Hand with 346 views. Even though there is a question on my exam paper next week on regression (that's not a hint, they already should know this!), in Ireland there was just one view which lasted 4 minutes and 51 seconds - less than half the duration of this video.
It's also interesting to note that YouTube/Google are now providing more visuals for content creators to analyse our data - these are visible on the left side panel on the chart above. More on this another day.
Thank you to all my viewers over the holiday season. For those of you preparing for exams I hope that the videos will help in your revision.
## Thursday, December 21, 2017
### The End is Nigh for Hand-written Exams?
In the age of emailing letters to Santa instead of writing physical letters, Laura McInerney of The Guardian asks "why must students write exams with a pen?". Hardly any of us write letters anymore. Indeed, other than signing your name - when is the last time you actually wrote by hand? I note very few students in my classes actually taking notes with pen and paper. This is unlike when I first came to the College in 2002 when I had to wait to move on slides until students had taken down everything. In fact the only time I see students writing for more than a few seconds is in exams. Some exam papers are extremely difficult for me to read as they are hand-written by students not used to writing for two hours at a time.
Image source: Huffington Post.
I'd hate to see the skill of hand-writing being completely lost. I agree with Catherine Pearson writing in the Huffington Post about "The Benefits Of Writing With Good Old Fashioned Pen And Paper" - she articulates that as handwriting is slow, it "can be particularly useful during goal setting, brainstorming... — all pursuits that require time and deliberation.". Some famous writers, such as Quentin Tarantino, claim to write all their material by hand as it makes them more creative.
So - should we make students write exams by hand in this day and age? For some subjects, such as programming, practical on-line exams are clearly the best. Many students would prefer to use a computer to write their answers (though I often find that responses are on average a lot shorter than hand-written answer). There are technical challenges, but these are being overcome everyday. Even the great Professor Sugata Mitra (of Newcastle University) "imagines an alternative education system with no need for memorisation or teaching to test" and suggests that a "tablet connected to the internet to be brought in to the examination hall" (see his article in The Guardian "Should students be allowed to use the internet in exams?"). My mind is still open on this, but I see myself favouring computers to be permitted in all exams within a few years. After all, it is the computer that most of us are using at work - not pen and paper!
The good news for old-fashioned lovers of pen and paper is that pen sales are still increasing in the digital age. Sales are expected to reach \$20.2 billion worldwide by 2019 according to the Chicago Tribune article "How the pen industry hangs on in a digital world". While this will come under pressure from electronic pens and styluses, the pen is still keeping its "mightier than the sword" status!
## Sunday, December 17, 2017
### Visit to the City of Preston #PNE
Sir Tom Finney Statue, flat cap, PNE jersey, I must be in Lancashire!
Yesterday I travelled early to Preston in England via a flight to Liverpool to see the local North End club host Sheffield United with my brother Brian. I have visited Preston on several occasions now (all to see PNE), and I'm beginning to know my way around. Preston people are very pleasant, I love the Lancashire accent where the word "the" is abbreviated to a simple "t". They love their football in Preston and 15,202 showed up to see a competitive game that PNE deserved to win by more than the 1-0 final score. The winner was scored by the £12 million rated Jordan Hugill in the second half. He is a very combative striker who's constantly jostling with his markers. He's not afraid to muscle his way to the ball - he'll be hard for PNE to hang on to in January transfer window if one of the big clubs comes knocking with a fat cheque book. Go PNE!
Incidentally, the return flight from Dublin to Liverpool cost €39 (thank you Ryanair), while parking (short-term) for the day in Dublin airport cost €29 (thank you DAA - not!).
## Friday, December 15, 2017
### Last Day of Semester #WhoMakesUpThisShit
It's Friday of the last week of semester one and I have now finished all classes - this was the 31st semester that I have completed in NCI. The last week of a semester always gives me mixed feelings in that as each day passes I have a last lecture/tutorial with each class in turn. Some I will see in class again in semester two, others not. They say that time flies if you are having fun, I definitely had a lot of fun working with students this semester, as it absolutely flew by. I also notice that the older I get, the faster the time goes!
Sir Ronald Fisher. Image source: Wikipedia.
I was asked a lot of questions this semester by students during and after classes - most I hope I gave satisfactory answers to. By far the toughest question I was asked was "Who makes up this shit?"! I had just completed a statistics class and the topic was Analysis of Variance (ANOVA). While the class giggled about this, I was a bit taken aback as I had never been asked a question like this before. I sensed it was asked with tongue-in-cheek, but nevertheless it is an interesting question. Why and how do people come up with new ideas, and who was the first to do something?
A one-way ANOVA is a statistical test to determine if there is a significant difference between the means of 3 or more groups, and it was created by the well-known statistician Ronald Fisher in the 1920s - see a profile of him in Wikipedia. If you want to know how to perform a one-way ANOVA test, check out my video below which shows you the technique that I covered in class - which led to the "Who makes up this shit?" question!
## Tuesday, December 12, 2017
### Blast from the past #AntiNuke
In a tribute to the late Dr Frank Jeal, posted in the News & Events section on the web page of the Zoology Department in Trinity College, there are many tributes and comments from past students and colleagues of Frank. Included are some photos which bring back a lot of memories of Frank, especially the one below also which includes yours truly (and Frank on extreme left). That combat jacket was one of my favourites, I was also very fond of my head of hair! I'm guessing the photo was taken on a Zoology Field Trip probably in 1980 or 1981 - I don't recognise the building or where it was, though Portaferry in Co Down was the destination for most field trips.
I note in the photo that I am wearing an anti-nuke badge on my left lapel - they were quite trendy at the time as the ESB were planning a nuclear power station at Carnsore Point in Co Wexford. I remember once wearing it to a chemistry lab where our lecturer spotted it and asked me why I was wearing it. No doubt at the time I was worried about nuclear fall-out and Carnsore being wiped off the map (or I was trying to be trendy!) - there was a lot of opposition to it at the time. My lecturer pressed me along the lines of "If it was possible to guarantee that the nuclear plant was 100% safe, would I still be opposed to nuclear energy?" A difficult question for me to answer at the time - he was a PhD in chemistry, I a mere 2nd year student. I'm sure I still opposed it - nothing can be 100% safe. Just 5 years later - the Chernobyl disaster happened.
## Monday, December 11, 2017
### 10 Years of Learning and Teaching with YouTube
On Tuesday 11th December 2007, I created my YouTube channel and uploaded my first video. Given that YouTube was only founded in 2005, and taken over by Google in 2006, I was indeed an early adopter of this medium. 13,875,105 views and 28,603 subscribers later, I am still at it!
My first video was "How To... Convert PowerPoint to iPod Movie". At the time I did not have a SmartPhone (the iPhone was first released on June 29, 2007) - so the old-fashioned iPod seemed to me at the time to be a great tool for learning. In the years around 2007 I mostly taught on the then MSc in e-Learning programme. I created the video as part of an exercise to get students to use technology in innovative ways. Nowadays creating a video is very easy, but back then to get it on an iPod I had to do the following:
• Create a PowerPoint presentation
• Import each slide as an image into Windows Movie Maker, add narration, and save as a movie
• Use a (free) Jodix iPod video converter to convert the video into iPod friendly format
• Add the video to iTunes, and sync with iPod
This video has been viewed just 6,390 times in ten years, and hardly at all in the past few years as technology is now more enhanced. The picture quality is poor, and some of the syncing of audio to each image is not good. So for a little bit of nostalgia (at least for me!), here is my oldest and first ever YouTube video:
## Thursday, December 07, 2017
### If you don't know data, you're out of the game. via @tableau
Tableau Software have published "2018: The Year Ahead for Business Intelligence" - it is always interesting to check out what respected and leading companies like Tableau think the future might hold. A key theme throughout is how much easier it is going to be to analyse data so that anyone can do it. While the article is very general, it breaks down into the following 10 topics:
1. Don't Fear AI
2. Liberal Arts Impact
3. Promise of NLP
4. Multi-Cloud Debate
5. Rise of the CDO
6. Crowdsourced Governance
7. Data Insurance
8. Data Engineer Role
9. Location IoT
Go to the article to read and watch videos (which rather annoyingly are not available to embed) for yourself, but for me two topics stand out for attention: #2 "Liberal Arts Impact", and #10 "Academics Investment".
Liberal Arts Impact
Who'd have thought that data had anything to do with the Arts? Anya A'hearn (what a brilliant name!) of Datablick tells us that the art of storytelling has helped "influence the data analytics industry" and that "organizations are placing a higher value on hiring workers who can use data and insights to affect change and drive transformation through art and persuasion, not only on the analytics itself" - it's all about telling a story with data.
A little bit closer to home for me is data analysis, not just teaching, has a role in third level institutions. As Dr Michael Galbreth (University of South Carolina) puts it; graduates "need to be comfortable with data". There is a huge demand from students to learn more about data, with most colleges now having some kind of data analytics/science programme. Colleges are responding to this demand, and we have to be on our game to develop, update, and deliver the right programmes. As Anya A'hearn puts it; "If you don't know data, you're out of the game". True.
## Sunday, December 03, 2017
### 40 Years Ago #ccr
Outside CCR Front Door.
I spent yesterday evening in the wonderful company of about 30 lads in their late 50s who all did their Leaving Certificate in 1977 in Cistercian College Roscrea. We were marking the 40th anniversary of finishing secondary school there. While I see some of the lads reasonably regularly, it was brilliant to meet up with two who I have not seen since June 1977 (PB and TO'T).
We were met in the College for a reception where the biggest treat for me was to meet my French teacher John Shanahan. I told him that I had written about him in my blog when I posted about My Introduction to Learning Technology - September, 1972, and again about We had a great chat about Voix et Image and how he was a pioneer of technology in education. Si jamais vous arrivez à voir ce message, merci M. Shanahan pour votre inspiration avec la technologie!
So - walking around the College we were given a tour by the House Captains. A lot has changed in 40 years, though the cold rooms and corridors remain the same. We even attended Mass celebrated by Fr Kevin in the school chapel. In the evening we adjourned to the County Arms in Birr for a super meal, a few beers, and of course more chat. Memories flooded back of our time in CCR 40 years ago - lots of shared stories of bunking mass, robbing orchards, fights, hair cuts, the leather, Roscrea girls, listening to the radio on Saturday afternoons with Tosh, tuck shop, rugby, and of course the bread which kept us all alive.
I spent just 5 out of my 58 years in Roscrea - yet it is a valued connection when 30 lads can celebrate together and spend an evening in each others company as if we were best friends. We all have had different life experiences with (please God) lots more to come. We vowed we would meet up for a 50th anniversary, though I felt a little sad leaving everybody this morning.
I'd like to give a big shout out to current Roscrea student Manus Heenan who as part of his Transition Year studies has created a mini-business making and selling bread using the original recipe from the monastery. He had the clever idea of selling his goods to a captive audience like us during our reception in the College. I bought a pack of his ready-to-go bread mix. Details of Manus's business can be found at www.abbeybread.ie, and he even has a great YouTube "How To" video showing us how to make Roscrea's famous bread.
## Thursday, November 30, 2017
### 12,000 Views Per Day Barrier Broken #WhosWatching @YouTube
A nice personal YouTube barrier for me was broken this week on Tuesday 28th November when the number of daily views on my channel exceeded 12,000 for the first time (12,167 to be precise). The previous high record of 11,944 views occurred on 25th March, 2014. 12,167 views represents about 21 days and 21 hours of watch time.
The United States continues to dominate with 26% of total views, followed by India (18%) which has been a growing "market" for me in the past couple of years. I have broken the views into ten and in the pie chart below you can see the breakdown by country. The Rest of the Word figure is made up of 146 countries which individually account for less than 1% of views. At the very end of the data are countries like Venezuela, Swaziland, and South Sudan with just one view each last Tuesday.
## Tuesday, November 28, 2017
### Everything you wanted to know about Data Analytics but were afraid to ask - via @NCIRL
Making the decision to go back to education and study any course is a tough one to make. The commitment and hard work, though definitely worth it, can make for a tough road ahead. However, you won't be alone. At the National College of Ireland most of our students are part-time - attending classes in the evenings and weekends. Data Analytics has become a very popular area of study and most of my teaching now is in this area. As part of the College's efforts to provide as much information as possible to students, we have updated our FAQ section on the College website (see http://blog.ncirl.ie/studying-data-analytics-at-nci-frequently-asked-questions). This page now includes a short video of me promoting our next Open Day (Thursday 30th November) - I'll be doing an Information Session at 6 o'clock and will hopefully be able to answer any questions that prospective students may have.
## Monday, November 27, 2017
### Learning Slow Down During Thanksgiving
It is interesting (to me anyway) to look at the effect of the Thanksgiving Holiday (23rd November) in the United States on the number of views of videos on my YouTube Channel. Since the US accounts for about 27% of all views this year, I would have expected some decline due to the holiday period - this is the case.
In the chart below you can see the effect of Thanksgiving with a marked reduction in total number of views (9,580) beginning the day before Thanksgiving, the day itself (8,224 views), and continuing on Black Friday the day after (7,492 views). This drop is almost exclusively due to the United States. My next biggest markets for number of views: India, UK, and Canada, showed no noticeable drop over the same period.
It comes as no surprise that less Americans view educational videos during this holiday period. I'm certain that this is reflected in many other areas on YouTube and I'd love to be able to see the data for what goes up, down, or stays the same, over this period. Clearly, Americans (not all - there was still 661 views in the USA on Thanksgiving Day!) are taking time out from learning while feasting on turkey and beer. Learners in the rest of the world (based on just my data) are ignoring this holiday.
November is traditionally a good month for number of views in my channel - the figure for 7th Nov (11,875) is the second highest daily number of views ever for me. Also - the run of four consecutive days of over 11,000 views from 13th to 16th Nov is the first time that this has ever happened.
## Thursday, November 23, 2017
I am gutted to once again miss the graduation ceremony today of so many of my students from last year - unfortunately classes are clashing with the ceremony. After attending almost every year since I started in NCI, this is the second year in a row that I have missed what had become one of my favourite days in the academic year.
Image source: University of Economics, Prague.
Many congratulations to all NCI students graduating in the Convention Centre - celebrate the day as you deserve to. In particular I'd like to congratulate students on the following programmes:
• Higher Diploma in Data Analytics
• B.A. in Management of Technology in Business
• B.Sc. in Technology Management
• B.Sc. in Computing
• B.Sc. in Business Information Systems
• Higher Certificate in Computing
• M.Sc. in Data Analytics
It was a pleasure working with you all and I wish you all the best in your future wherever it leads to.
## Wednesday, November 22, 2017
### What is the point of learning statistics? via @CharlesWheelan
Image source: Amazon.
I am currently really enjoying reading Charles Wheelan's excellent book "Naked Statistics: Stripping the Dread from the Data", and one of his first questions is "What is the point of learning statistics?". It's actually a question I find hard to answer myself - I usually mutter stuff about being better able to understand data and to prove a hypothesis is true/untrue. But Wheelan comes up with excellent reasons why we should study statistics:
• To summarize huge quantities of data
• To make better decisions
• To answer important social questions
• To recognise patterns that can refine how we do everything from selling diapers to catching criminals
• To catch cheaters and professional criminals
• To evaluate effectiveness of policies, programs, drugs, medical procedures, and other innovations
• And to spot the scoundrels who use these very same powerful tools for nefarious ends
While the first two points above seem obvious, the remainder do not immediately jump to the mind of someone considering studying statistics.
Studying statistics involves learning a lot of tests - a lot. Figuring out how to set a hypothesis doesn't come easy, and can take a long time to get used to it (I even still struggle sometimes when confronted with a new situation, and I have been teaching this stuff for five years!). But just like everything else, the more you practice, the better you get. Probably for me the biggest thing about statistics is that it gives you value - once you have done the tests, you can analyse the results to make sense of them. Tools such as Excel, R, and SPSS are making statistics easy to do, but you still need to understand and interpret the results.
To finish this post with a quote attributed to H.G. Wells (1866-1946):
Statistical thinking will one day be
as necessary for efficient citizenship
as the ability to read and write
## Monday, November 20, 2017
### Bargain On-line Courses from @udemy #BlackFriday
Last year I purchased two Udemy courses on creating data visualisations in Tableau - both were just \$10 as part of the Black Friday Sale. This year Udemy are offering a wide range (over 55,000) of courses once again for just \$10 - a no-brainer bargain for learners. For example, one of the courses I am interested in is Building Interactive Graphs with ggplot2 and Shiny, another is Statistics for Data Analysis Using R. Thinking of my own students, there are coursers on Python, R, Data Analysis, and Statistics - a lot for new learners. Students should certainly consider adding to their portfolio of leaning for just \$10 a go.
Image source: Pregnancy and Baby.
Why, may you ask, is a classroom-based Lecturer recommending on-line courses to his students? I already teach Statistics and Data Visualisation modules - it is worrying for students that I am signing up for the above courses?
The answer is that I am always interested in how others teach in subjects similar to what I do (and of course I am interested in bargains for students as well as myself!). I can learn much from watching others, get ideas for my own class from them, see different examples of data and analysis not covered by the wide range of textbooks that I use, and also get ideas for exam questions. I also almost always find that no matter how often I read a book or watch a video, there is something different that I did not know before which I can add to my knowledge.
These courses can be expensive for students when not on a Black Friday Sale - prices can be up to \$145. A huge amount of what you can learn from these courses can be gleaned from a myriad of websites and YouTube videos. But it is a very convenient, and cheap, way to save yourself a lot of searching by signing up for a course with everything in one place.
Disclaimer:
I am not associated with Udemy and am not involved in promoting their courses. I just think this is a bargain not to be missed!
## Tuesday, November 14, 2017
### Six Trends in Data Science
Adam Shapley, writing in Silicon Republic, tells us If you want to be a data scientist, you need to know about these 6 trends. The trends he lists are as follows:
1. All industries are open, but you should try to specialise
2. Balance robust academic achievements with on-the-job learning
3. Data analytics experience is essential, machine learning helps
4. The GDPR is increasing data governance demand
5. Make sure you have a solid business intelligence foundation
6. Keep your technical skills up to date
While I am mostly involved in the education of Data Analysts, this is still an interesting list. I was particularly interested to find that "half of those working in data science have a PhD, whereas less than 2pc of people in the US over 25 years old have a doctorate" (in point 2 above). While a PhD is not a "must-have" for all data science roles, potential employers are sure to take notice.
Shapley also recommends that Data Scientist learn and maintain news skills regularly. Data science is a complex area, and scientists will need to "demonstrate the most relevant skills and experience to this industry".
Calliostoma zizyphinum (L.).Image source: UK Natural History Museum.
A PhD can take a long time to achieve - typically 3-5 years. Mine took 4 years and involved a lot of data analysis on shelf shape variation in the painted topshell (Calliostoma zizyphinum) - it takes time to carry out research, analyse results, and write it up. My own thoughts are that a good Masters would also be very valuable in a Data Scientist role - much of course will depend on the level of academic experience sought by employers.
According to Glassdoor, Data Scientist is #1 in the list of Top 50 Best Jobs in America. It rates very high and pays wells (\$110,00), today there are 4,184 job openings in the US. A great job like this would make it worthwhile to consider a PhD - even though it could take a long time to achieve, it would be worth it in the long run.
## Monday, November 13, 2017
### Does anybody trust anybody anymore?
Today I got a letter in the post from my credit card company demanding that under some Terrorist and Money Laundering Act that I provide them with ID and proof of address by return of post. A complete stranger to me will open the letter and enter my details into a computer. Bingo - I am not a terrorist or a money launderer! Yesterday I had to fill out forms to be Garda vetted (yet again) so that I can sing in the church choir.
I was intrigued a few months ago at a presentation about Bitcoin when the presenter told us that the whole idea of Bitcoin and digital currencies is that you trust no one from the beginning. Trust is defined by Dictionary.com as:
"reliance on the integrity, strength, ability, surety, etc., of a person or thing; confidence"
Guilty until proven innocent
Many people are opposed to the idea of being vetted like this: "I never committed a crime in my life" is true of the overwhelming majority of people. It saddens me that elderly people going to church and who volunteer for activities like taking up the collection at Mass now have to be Garda vetted. It saddens me even more that a few perverts have made this necessary. According to the Central Statistics Office, recorded crime incidents classified as "Sexual Offences" were 2,348 offences in 2016, an increase of 8.6% on the previous year. The population of Ireland in 2016 was recorded as 4,757,976. Using the data above, sex offenders make up 0.049% of the population in 2016. It's like trying to find a needle in a haystack using Garda vetting. Yet it has to be done, and is especially important where children are concerned.
The Shoe Bomber
An idiot called Richard Reid, the world's dumbest ever terrorist, tried to blow up a plane with a bomb in his shoe in December 2001. Now we all have to remove our shoes going through security at airports because we cannot be trusted not to do the same. According to the Worldbank, there were 3.696 billion air passengers in 2016, the vast majority of these have to take off their shoes (that over 7.3 billion shoes!). Has anyone been caught with a bomb in their shoe since 2001? Despite this ritual humiliation we all go through at airports, would you get on a plane that passengers were not checked though security?
Exams
People have been cheating in exams since forever. But it is just a tiny minority who take the chance to do this. Hence all colleges have strict rules about exams - no phones, (new) no smart watches, no notes, and no water bottles. Every time a student sits down to an exam, he/she has to undergo the ritual instructions from Invigilators. Many of us also make students submit assignments like essays and projects through plagiarism detection systems such as Turnitin. While we can argue that it is part of the learning process to do this, in the end it is about preventing cheating. Why should an honest trustworthy student have to do this?
Trust no one!
## Friday, November 10, 2017
### Naked Statistics!
Image source: A Little Stats.
Now that the post title has got your interest - read on...!
I was recently reading in The Economist a review of a book by Charles Wheelan, called "Naked Statistics: Stripping the Dread From the Data". It's quite an old book now (published in 2013), but I've just bought it from Amazon. The piece that made up my mind for me to buy the book was when the reviewer (not named) wrote:
The reader learns why insurance for low-cost items is worthless and why playing the lottery is a quick way to become poor. More seriously, the book explains the basic statistical approaches used in a 2011 study showing a link between a child’s brain size and autism. And it teems with interesting statistical facts, such as that there may have been an extra 1,000 deaths in the three months after September 11th 2001 because more people chose to drive rather than fly.
The last point is an interesting one and a quick visit to Wikipedia for data on road deaths in the US reveals (using Excel) interesting trends. While roads deaths have fluctuated enormously since 1970, you can see that there is in fact an increase from 42,196 deaths in 2001, to 43,005 deaths in 2002 - an increase of 809 deaths. The trend had been increasing over the previous three years anyway, but there is evidence that road deaths did increase after the 9/11 attacks. The chart below is really easy to create in Excel (the slowest bit was adding the red label), and I find it is fun to be able to quickly visualise data like this. While road deaths were not caused by the 9/11 attacks (most people of the 43,005 who died in 2002 would have died anyway), it is an interesting thought that in addition to the 2,996 who did die as a direct result of the 9/11 attacks, perhaps the figure should be 3,805 (2,996 + 809)?
Data Source: Wikipedia.
## Sunday, November 05, 2017
For all parents, graduation day is one very proud and special occasion. I have been very fortunate that I have now attended graduations for all three of our daughters - Claire graduated yesterday from DIT with a Masters in Public Affairs. My chest was once again bursting with pride! There is a serious side to graduation ceremonies - it marks a passing from the education world to the post-education world. Yes - there is the technical bit that each student has passed exams and completed the requirements of a course. But there is more! Experiences gained throughout study such as: making new friends, character building, problem-solving, meeting deadlines, learning, helping others, broadening one's mind - all help make for a more rounded graduate who is more than just a student who passed all their exams. We have a saying in our College that NCI "Changes Lives Through Education". I see this all the time - graduation marks a new exciting phase of people's lives. Graduates should be rightly proud of their achievements (and their parents too)!
I like to take a selfie each time, so here's me with my three daughters at their graduations:
## Monday, October 30, 2017
### My Great-Great-Great Grandparents Nicholas Browne and Margaret Barry
Yesterday I visited Mayglass in South Co Wexford. Through a contact via Ancestry.co.uk I had a document that was a typed record of headstones in a graveyard in Mayglass. My great-great-great grandparents Nicholas and Margaret (née Barry) Browne were listed as being buried there - so I decided to check this out. The graveyard is small, with lots of headstones with faded writing that was difficult to read. After about 20 minutes searching I found them! Curiously Nicholas is shown on two headstones that are side-by-side. He died 156 years ago on 24th July 1861 aged 65 - this means he was born in 1795 or 1796. Margaret died on 26th February 1898 - her age is not given.
An intriguing possibility is that both are related to famous 19th century navy men - Admiral William Browne (founder of the Argentine Navy) and John Barry (founder of the United States of America Navy)! I’ll be doing my best to try and find out if this is the case. It’s a long shot, but I’m sure it is a nice unique link to the sea.
Here's the path from me to my great-great-great grandparents Nick and Maggie:
Click Image to Enlarge.
## Tuesday, October 24, 2017
### "Never memorize something that you can look up" #Einstein #InstantLearning
I wonder can you learn how to do anything you want on YouTube? Yesterday I went about fixing my broken car key fob. I had to buy a replacement shell, get a new key cut, and take the transponder and circuit board out of the old key and insert into the new key fob. Easy?
Getting the new replacement shell and key cut was the easy part - thanks to the extremely helpful people in Central Key and Hardware Ltd on Parnell Square East. The guy there opened up both key fobs and showed me the insides and gave me advice on what to do. It looked very easy.
Not so!
First - my car is a Hyundai, the new fob shell was from a Kia. While almost identical, they were not exactly the same. The Kia version had a screw under the badge. While trying to cut out the Hyundai badge I found out that my Swiss Army knife is very sharp and can easily stab my thumb. Ouch! It was only then that I tried YouTube, and of course I immediately found a video that showed me how to do the job properly step-by-step. This is not the first time YouTube has come to my rescue - and I'm certain it is now a "go to" place for lots practicals tasks like this.
In case you ever need to change a key fob on a Hyundai - here's the video:
## Friday, October 20, 2017
### It's #WorldStatisticsDay! #167
H.G. Wells. Image source: Wikipedia.
Today is a dedicated World Statistics Day - it's been around since 2010, but I never heard of it before. It's a nice idea to promote statistics in this new age of Big Data. Statistics is the science of data, and of course we have had numbers for a very long time. The English writer H.G. Wells once said that "Statistical thinking will one day be as necessary for efficient citizenship as the ability to read and write" - maybe he is finally right! Data Analytics courses are amongst the most popular ones for students at the National College of Ireland. This is our fifth year running a Higher Diploma in Data Analytics.
I have been teaching statistics for six years. This year is my busiest ever - I have 167 students in three separate classes studying the Business Data Analysis module. I have to say that I love it, and the students are great to work with! I'm getting fonder of this subject every year - I am starting a plan to convert my lecture notes into a textbook to go with my modules. My statistics videos are also growing in popularity, this allows me to reach beyond my 167 students to the world.
So - thinking of a statistic for today I thought that 167 would be most suitable!
167 Candles - Happy World Statistics Day!
## Wednesday, October 18, 2017
### It's Official - Windows Phone is Dead via @flipboard
So - Microsoft has confirmed that it will no longer develop new features or hardware for Windows 10 Mobile phones, finally giving up on mobile. No sign of rumoured Surface phones either. I tried my best to persevere with my Lumia 950, but in May of this year I finally abandoned the Windows phone. The "App Gap" was just too big, with developers basically refusing to create versions of Apps for Windows. This is not difficult to understand why - iOS and Android account for 99.4% of mobile phone operating systems.
It's always a pity to see a technology fail. Apart from the App Gap, my Lumia was actually quite good - the 20 mp camera is fantastic, neither Samsung or Apple are offering this yet. I had a 64 GB SD card which was great - no space worries. I keep it as a spare, and have a Vodafone sim card in it for travelling down to the country where the Virgin Media (via Three) network is shite. If my iPhone were to break down, I'd switch back to it for a while, but following Microsoft's announcement it would not be for too long. Basically it is now just a WiFi enabled camera.
The Windows Lumia 950. Image source: Windows Central.
## Sunday, October 15, 2017
### My 100th plaque @OpenPlaques
A couple of years ago I started to upload photos to the Open Plaque Project - most are from around Dublin, but now I watch out for them everywhere and take a photo. If it's not already uploaded, I do so. The inscription is also uploaded, plus the location - I can add as many photos as I want, so I usually do close-ups and medium shots. This weekend I was in Gorey in Co Wexford where I came across several plaques, three (all related to Bishop Ram) of which are suitable for Open Plaques. The stone plaque for Miles Byrne is probably not suitable - the site aims to show "where they are, we identify who is commemorated on them, what those people are notable for, and what their connection is with the place where their plaque is installed". This plaque is not located in Monaseed which is about 7 miles away from Gorey (Jez - if you are reading this perhaps you might comment?).
The Market Square plaque is my 100th one uploaded! This pales into insignificance beside the wonderfully named "Spudgun67" who has uploaded 1,549 plaques.
Here are the plaques, with Open Plaque links:
## Wednesday, October 11, 2017
### Great Data Visualization from @Tableau #OLOUGHLIN #DataViz
My surname "O'Loughlin" to me is a very Irish surname. Anyone with an "O" and an apostrophe must be 100%Irish - right? Here's an extract from my book "Exploring Ireland's Wild Atlantic Way" about the O'Loughlin name:
O'Loughlin's Pub in Ballyvaughan.
One of the things that attracts people to different parts of Ireland is the search for ancestors, and I am no different. North Clare is O’Loughlin country, and for a short time I travelled the Wild Atlantic Way in search of my own ancestors. A short distance after Black Head I stopped at the side of the road to look down at a castle which is visible for a long way along the beautiful ride on the southern side of Galway Bay. It is the sixteenth century Gleninagh Castle and it was built for the O’Loughlin chiefs who were resident there up to the 1840s. It was lived in up to the 1890s, but is still well preserved. I wondered if any of my direct ancestors had lived there. There is no shortage of people with the O’Loughlin name in County Clare, and this northern region of the county is where the name O’Loughlin originates from. At the end of the tenth century some of the Irish upper classes started to adopt Viking names such as Lochlainn, and it is thought that some of the County Clare families did so as well. The Vikings were known as na Lochlannaigh in the Irish language. My dad Joe has told me that his grandfather, also Joe O’Loughlin, was born and reared in County Clare, but he does not know which part. It could be that the O’Loughlins are descended from either the Vikings themselves, or from a County Clare family that adopted the name.
Yesterday I came across an excellent Data Visualization by Mike Cisneros on Tableau Public, which illustrates the association between surnames and racial groups. You can enter your own surname (must have appeared at least 200 times in the 2010 US Census), and enter an estimate of the assumption of which race your name belongs to. I entered 99% for O'Loughlin as "white" and was surprised to discover that this was not a correct estimate. According to Cisneros's graphic, only 92.1% of O'Loughlin's are classifies as "white" - I was off by 6.9%. So in the United States there are 7.9% of "O'Loughlins" who are not classified as "white". Cool!" There are only five races listed:
2. Asian/Hawaiian/Pacific Islander
3. Black/African-American
4. Latino/Hispanic
5. White
You can view the full graphic at Tableau Public, and try out your own surname.
## Wednesday, October 04, 2017
### The downside of technology #Fitbit
In my statistics classes I have a topic early in the semester about "Thinking Statistically" and cite many examples of how data are gathered on a daily basis. One of the good stories I use is about a research project by the Dan-Farber Cancer Institute in Boston where 6,000 over-weight women have been given Fitbits to track exercise over a six year period - I try to get students to think how much data there will have been collected by the end of the study. You can read about this in the Irish Times: Does wearable technology deserve clean bill of health?
This is the Charge HR Fitbit that I use. Image source: Walmart.
Today I read in The Sidney Morning Herald: "Fitbits in schools a step in wrong direction, make kids less active, study finds". While the idea of giving Fitbits to teenagers appeals to me in that exercise is encouraged and tracked, the SMH reports that such trackers in schools "has been linked to poor self-esteem and negative feelings of alienation and inadequacy" and that the devices can actually "demotivate children". The study by the University of Birmingham (which is not referenced) reported "feelings of inadequacy and lower self-esteem among pupils". There is a suggestion that setting "unattainable targets" is a major factor in demotivating students. Unintended consequences? I must confess that I had not thought of this. I would hate to see Fitbits being taken away from students because of this.
In the main I am in favour technology being used and data such as activity being tracked. This could leads to a significant addition to research over long periods of time that could aid in the treatment of illnesses and conditions. Imagine tracking heart rates over a half a lifetime to look for early signs of a heart attack. Assuming that data privacy is respected, these data will make a valuable contribution to science. I am in the "opt-in" camp rather than the "opt-out" - data such as this must be voluntarily collected under the same rules, for example, that clinical trials are conducted.
There is another thing - wearable technology is taking off. While we may not be using a Fitbit, the Smartphone in our pockets can do exactly the same thing. I see watches becoming smarter - the day will come when almost all watches will be smart. What then? The above report suggests that it has been a bad start for wearable technology for secondary school students. With more planning and reflection on what has happened so far, perhaps realistic goals can be set for students so that they don't feel demotivated by not reaching 10,000 steps in a day. We can learn from this "mistake" (my word).
## Tuesday, October 03, 2017
### New Avatar #DraganEffect #ego
Last year I wrote about experimenting with The Dragan Effect for photography - named after the Polish photographer Andrzej Dragan - the effect uses "dramatic lighting and editing techniques that enhance the tonality and skin texture of the images subject". This is done using Corel PaintShop Pro X7. Recently I have grown a bit of a beard, and wondered if the Dragan Effect could improve the look - here is the result:
The photo was taken with my Windows phone's 20MP camera (the only good thing about this phone). I tried it with both a serious face and with a bit of a smile - I prefer above. I think it does make me look a bit older as it emphasizes wrinkles and the greyness of the beard. This will be my avatar for the next while!
If you have Corel PaintShop Pro, check out this video which takes you step by step through the process.
## Friday, September 29, 2017
### Launch of IFS Apprenticeship Programme @NCIRL
Today I had the pleasure of attending the launch of the IFS Apprenticeship programmes in the College - the programmes in FinTech were formally launched by the Minister for Education Richard Bruton. These new programmes are aimed at providing students with both on-the-job training and classroom - it is an exciting opportunity for the new students, and the College. The apprenticeship model in popular in other countries and is now becoming a new model for recruitment here in Ireland. Good luck to the new apprentices!
Richard Bruton is a fine public speaker. He did not use notes today, and spoke as if his entire speech was scripted. I recall seeing him doing this before at a launch of new offices for SmartForce in my previous job during the 1990s.
## Wednesday, September 27, 2017
### Can you really be over-qualified? via @IrishTimes
According Carl O'Brien, writing in last Friday's Irish Times, Irish workers are most ‘overqualified’ in Europe. This is based on "research carried out by the Economic and Social Research Institute (ESRI) between 2000 and 2011". About 60% of our school leavers progress onto third level education, and this is projected to rise to about 70% over the next decade - one of the highest rates in Europe. Of course not all of these students will leave college/university with a qualification, but most will.
So - can you be overqualified? Is it OK to have "bar tenders who have university degrees"?
My view is that there is no such thing as being overqualified. There - I said it!
Of course, I work in the third level sector so I guess I would say this.
Let me tell a quick personal story. When I was getting my first ever business card, I was asked for details such as job title, phone number, etc. This was about 1994. I still had my graduation in 1988 fresh in my mind and I wondered if I would be allowed to add either "Dr" before my name, or "MA, PhD." after my name. My manager at the time said an emphatic "No". He did not want to intimidate anyone with fancy letters (I was one of only two PhDs in the organization at the time). So I accepted the decision despite being bitterly disappointed - I was very proud of my degree (as all graduates should be). Today of course, I do have the "fancy" letters on my business card - but I do work in a College where things like this are accepted and expected.
Attending College and getting a degree means many things. For those studying subjects such as Medicine, Pharmacy, or Law - it is a career decision made before going to College. A degree is obviously needed for these types of careers. If I ever end up in an operating theatre facing a surgeon's knife - I am not going to ask for someone else to do the job if my surgeon is over-qualified to wield a scalpel. For most other third level students, the three or four years in College may be something else completely. I studied Marine Biology - but never worked as a Marine Biologist. Students learn so much more in College than just what's on the syllabus. I'm sure if we asked all graduates if they regretted attending College - very few would say that they did.
So - the next time a bartender with a degree serves me a pint I will be thinking "Good for you!". Your degree did not make you a better bar tender or a better person than another bartender who does not have a degree. You will both have had different life experiences - neither is better than the other. Be proud of your achievement in gaining a degree, you might never get a chance like it again. You are not over-qualified.
## Tuesday, September 26, 2017
### Students Don't Take Notes Anymore?
While taking my usual perusal through the very funny Waterford Whispers News this morning I had a giggle at one of their latest posts: Fucking Loser In Front Of You Actually Taking Notes During Lecture. In this post a student who doesn't take notes thinks others who do are "losers" while scribbling rude drawings and checking out Snapchat and Netflix. Funny - maybe, but real?
When I first became a Lecturer in NCI in 2002, most students took down notes based on what my slides showed on screen. It appeared to me that students took note of my bullet points word-for-word, and I often wondered if they were listening to anything I said. Basically, at the time this was the only way for them to have material from class - no Moodle or Blackboard in those days. For those of us Lecturers who knew a bit of HTML, we then started to put our notes on web pages. While this worked for some students, it is hard to believe that many at that time had no email address or access to the Internet. Moodle changed everything for both students and Lecturers. For students, it was an easy way to get notes and saved a lot of note taking in class, for Lecturers it was an opportunity to provide not just notes, but other resources (such as case studies, web links, quizzes, and exercise files) as well.
Many Lecturers create quite elaborate notes, often based on PowerPoint slides created by textbook authors. I create my own and sometimes use very modified (by me) textbook versions. Perhaps because of this, or maybe it is part of a wider condition - I too notice that students very rarely take notes in my classes. While my notes are no substitute for a textbook, I know that many students who choose not to buy a textbook rely on my notes (and videos). I am often asked by students if they really need to buy one of the recommended textbooks - my answer always is that there are some copies in the library, but good luck trying to reserve a copy in the days and weeks leading up to the exams.
Taking notes is a great way to really learn - it helps with your writing, understanding, memory, exam revision, and is a useful record of information. The Horry Georgetown Technical College has a really good (and short) presentation on how to take and review notes. If you don't take notes in class, the presentation below will really make you think:
## Monday, September 25, 2017
### Great day out at Ladies' All-Ireland Footaball Final #Mayo #Dublin #SeriousSupport
First - a confession. I have never ever attended a ladies football game of any kind. I never really thought about it, and certainly did not consciously avoid them. So when Mayo and Dublin lined up for the Ladies' All-Ireland Football Final in Croke Park, it was my first ever ladies match. It was also my first football All-Ireland Final since 1980! A huge crowd of over 46,000 attended - a record!
We certainly had great fun with both teams of girls playing their hearts out. Dublin had lost the last three finals, so they were in a determined mood. Mayo were very wasteful, missing a lot of easy chances. They did not help themselves by having three players sin-binned - that's equivalent to being down a player for 30 minutes in a 60 minute game. After the third sin-binning, Dublin just exploded out of the blocks and cut loose to win by 12 points (4-11 to 0-11). A well-deserved win.
Mayo fans will just have to suffer for another year! | 11,928 | 52,624 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.971202 |
https://www.jiskha.com/users?name=Tiffany+Enlow | 1,601,008,098,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400221980.49/warc/CC-MAIN-20200925021647-20200925051647-00577.warc.gz | 884,658,313 | 2,188 | # Tiffany Enlow
Popular questions and responses by Tiffany Enlow
1. ## Trigonometry
4. Find the exact value for sin(x+y) if sinx=-4/5 and cos y = 15/17. Angles x and y are in the fourth quadrant. 5. Find the exact value for cos 165degrees using the half-angle identity. 1. Solve: 2 cos^2x - 3 cosx + 1 = 0 for 0 less than or equal to x
asked on April 10, 2009 | 120 | 362 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2020-40 | latest | en | 0.767617 |
https://mathfraction.com/fraction-simplify/adding-exponents/ti-83-log-scale.html | 1,713,353,788,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817153.39/warc/CC-MAIN-20240417110701-20240417140701-00596.warc.gz | 342,007,156 | 10,978 | Try the Free Math Solver or Scroll down to Tutorials!
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:
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ti-83 log scale
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Author Message
BantJamir
Registered: 15.12.2003
From: On the Interweb
Posted: Thursday 28th of Dec 15:13 Well there are just two people who can help me out right now , either it has to be some math guru or it has to be God himself. I’m sick and tired of trying to solve problems on ti-83 log scale and some related topics such as graphing function and adding exponents. I have my finals coming up in a a couple of days from now and I don’t know how I’m going to face them? Is there anyone out there who can actually take out some time and help me with my questions? Any sort of help would be highly appreciated .
oc_rana
Registered: 08.03.2007
From: egypt,alexandria
Posted: Friday 29th of Dec 08:09 How about giving some more information of what exactly is your trouble with ti-83 log scale? This would assist in finding out ways to look for a solution . Finding a coach these days quickly enough and that too at a fee that you can meet can be a frustrating task. On the other hand, these days there are programs that are available to assist you with your math problems. All you require to do is to select the most suited one. With just a click the right answer pops up. Not only this, it helps you to arriving at the answer. This way you also get to learn to get at the correct answer.
LifiIcPoin
Registered: 01.10.2002
From: Way Way Behind
Posted: Saturday 30th of Dec 08:04 I had always struggled with math during my high school days and absolutely hated the subject until I came across Algebrator. This product is so fantastic, it helped me improve my grades drastically. It didn't just help me with my homework, it taught me how to solve the problems. You have nothing to lose and everything to benefit by buying this brilliant software.
sxAoc
Registered: 16.01.2002
From: Australia
Posted: Saturday 30th of Dec 10:27 I am a regular user of Algebrator. It not only helps me get my assignments faster, the detailed explanations offered makes understanding the concepts easier. I strongly suggest using it to help improve problem solving skills. | 781 | 3,094 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-18 | latest | en | 0.90111 |
https://www.jiskha.com/display.cgi?id=1291061622 | 1,503,245,138,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886106779.68/warc/CC-MAIN-20170820150632-20170820170632-00603.warc.gz | 923,876,259 | 4,022 | # Statistics
posted by .
Are these questions below true or false? I am stuck. I think 1 is true and 3 is false
1. In Regression Analysis if the variance of the error term is constant, we call it the Homoscedasticity property.
2. In simple linear regression analysis, if the error terms exhibit a positive or negative autocorrelation over time, then the assumption of constant variance is violated.
3. The expected value of the error term changes from observation to observation.
## Similar Questions
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Can you please tell if the follong questions are true or false (fragment) 1. That movie, one of my favorites, true or false?
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True or False: 1. f(x) = -x^3 + x + 1 is an odd function 2. f(x) = x^3 -ax^2 + bx represents a cubic equation having no constant term ...(I'm thinking true?
3. ### Statistics
True or False In Regression Analysis if the variance of the error term is constant, we call it the Homoscedasticity property.
4. ### Statistics
True or False, In simple linear regression analysis, if the error terms exhibit a positive or negative autocorrelation over time, then the assumption of constant variance is violated.
1. Reactions follow the law of conservation of mass. a.) True b.) False Answer: True 2. Reactants are the materials used in a chemical reaction. a.) True b.) False Answer: True 3. In a chemical equation, the mass on the left of the …
6. ### Chemistry: Check please, these are hw assignments
1. Gases can be considered fluids. a.) True b.) False Answer: False 2.Gases are not easily compressed. a.) True b.) False Answer: True 3. Gases spread to fill out space. a.) True b.) False Answer: True 4. Increased pressure on a gas …
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11. Disagreements are a normal part of life. (1 point) True False 12. Talking to a trustworthy adult is the best way to stop sexual abuse and get help. (1 point) True False 13. Some conflicts are made worse by peer negotiation. (1 …
8. ### Physical Ed
11. Disagreements are a normal part of life. (1 point) True False 12. Talking to a trustworthy adult is the best way to stop sexual abuse and get help. (1 point) True False 13. Some conflicts are made worse by peer negotiation. (1 … | 552 | 2,234 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2017-34 | latest | en | 0.787611 |
https://socratic.org/questions/if-5-x-y-6-5-2-9-6-what-what-is-x-and-y | 1,723,260,641,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640789586.56/warc/CC-MAIN-20240810030800-20240810060800-00494.warc.gz | 424,688,347 | 5,726 | # If [(5,x), (y,-6)]=[(5,-2), (9,-6)], what what is x and y?
Apr 29, 2017
$x = - 2$
$y = 9$
#### Explanation:
Both are $2 \times 2$ matrix.
Both are equal;
Then -
$x = - 2$
$y = 9$ | 82 | 186 | {"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} | 4.1875 | 4 | CC-MAIN-2024-33 | latest | en | 0.807511 |
http://www.qb64.net/forum/index.php?topic=4153.0 | 1,371,688,831,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368709947846/warc/CC-MAIN-20130516131227-00037-ip-10-60-113-184.ec2.internal.warc.gz | 648,472,988 | 13,668 | ### Author Topic: So, i've started writing a game... (Read 1340 times)
#### vrensul
• Sr. Member
• Posts: 320
• 44th Degree Digital Wizard of the HighBytes Order
##### So, i've started writing a game...
« on: August 07, 2011, 03:19:40 PM »
And i'm not very good with graphics in general. After messing with some stuff last night, and figuring how to tie in some type of timer so far i've done this.
Code: [Select]
`CLSCOLOR 7, 0PRINT "Entering program..."TYPE StarType 'star type x AS INTEGER y AS INTEGER color AS LONG BigStar AS INTEGEREND TYPETYPE Planet 'planet type x AS INTEGER y AS INTEGER Color AS LONG radi AS INTEGER R AS INTEGER G AS INTEGER B AS INTEGEREND TYPEPRINT "Smacking some bits around..."DIM stars1(50) AS StarTypeDIM stars2(25) AS StarTypePRINT "Got some hedge hogs here!"DIM ScrFORE AS LONGDIM Plan AS PlanetDIM TotalNumberOfBigStars AS INTEGERTotalNumberOfBigStars = 6BigStarCount = 0DIM Stars1MIN 'Stars min color range (0-255)DIM Stars1MAX ' ... And the max.DIM Stars2MIN ' Same as above but for stars 2...DIM Stars2MAX ' ""Stars1MIN = 210Stars1MAX = 255Stars2MIN = 40Stars2MAX = 90ScrFORE& = _NEWIMAGE(800, 600, 32)Start:PRINT "Searching for Voyager"RANDOMIZE TIMER'rand the planetPlan.x = INT(RND * (750 - 50 + 1)) + 50Plan.y = INT(RND * (550 - 50 + 1)) + 50Plan.radi = INT(RND * (40 - 10 + 1)) + 10Plan.R = INT(RND * (200 - 20 + 1)) + 20Plan.G = INT(RND * (200 - 20 + 1)) + 20Plan.B = INT(RND * (200 - 20 + 1)) + 20Plan.Color& = _RGB32(Plan.R, Plan.G, Plan.B)' Rand the initial stars...FOR Counter = 1 TO 50 stars1(Counter).x = INT(RND * (800 - 1 + 1)) + 1 stars1(Counter).y = INT(RND * (600 - 1 + 1)) + 1 stars1(Counter).color& = _RGB32(INT(RND * (Stars1MAX - Stars1MIN + 1)) + Stars1MIN, INT(RND * (Stars1MAX - Stars1MIN + 1)) + Stars1MIN, INT(RND * (Stars1MAX - Stars1MIN + 1)) + Stars1MIN) 'stars1(Counter).color& = _RGB32(INT(RND * (Stars1MAX - Stars1MIN + 1)) + Stars1MIN, INT(RND * (Stars1MAX - Stars1MIN + 1)) + Stars1MIN, INT(RND * (Stars1MAX - Stars1MIN + 1)) + Stars1MIN) IF stars1(Counter).BigStar = 0 AND BigStarCount < TotalNumberOfBigStars THEN Dice = CINT(RND) IF Dice = 1 THEN BigStarCount = BigStarCount + 1 stars1(Counter).BigStar = Dice END IF END IFNEXT CounterFOR Counter = 1 TO 25 stars2(Counter).x = INT(RND * (800 - 1 + 1)) + 1 stars2(Counter).y = INT(RND * (600 - 1 + 1)) + 1 stars2(Counter).color& = _RGB32(INT(RND * (Stars2MAX - Stars2MIN + 1)) + Stars2MIN, INT(RND * (Stars2MAX - Stars2MIN + 1)) + Stars2MIN, INT(RND * (Stars2MAX - Stars2MIN + 1)) + Stars2MIN)NEXT CounterSetupScreens:SCREEN ScrFORE&_FULLSCREENNewLoop:switchTIMEmax = 20 ' THIS is the time mechanism for the stars AND the "planet"switchTIMEcur = 0 ' This is what it stars at.WHILE INKEY\$ = "" _DISPLAY CLS _LIMIT 250 'Loops per second... FOR cnt = 1 TO 50 ' Stars # 2 IF switchTIMEcur = CINT(switchTIMEmax / 4) OR switchTIMEcur = switchTIMEmax THEN ' If we should move a star2s... IF cnt < 26 THEN 'If valid subscript... stars2(cnt).y = stars2(cnt).y + 1 IF stars2(cnt).y > 1 AND stars2(cnt).y < 601 THEN PSET (stars2(cnt).x, stars2(cnt).y), stars2(cnt).color& 'Draw if within screen range. ELSE 'Else, it's out of range, make a new star! You can name it too.... ' max min min stars2(cnt).x = INT(RND * (800 - 1 + 1)) + 1 stars2(cnt).y = INT(RND * (2 - 1 + 1)) + 1 stars2(cnt).color& = _RGB32(INT(RND * (Stars2MAX - Stars2MIN + 1)) + Stars2MIN, INT(RND * (Stars2MAX - Stars2MIN + 1)) + Stars2MIN, INT(RND * (Stars2MAX - Stars2MIN + 1)) + Stars2MIN) END IF 'End field checking IF END IF 'End subscript check ELSE ' If it isn't time to move the star2s.... IF cnt < 26 THEN 'Still need to check subscript... PSET (stars2(cnt).x, stars2(cnt).y), stars2(cnt).color& 'And you still need to draw it every refresh. END IF END IF 'Stars # 1 ' Draw these every refresh of the loop. stars1(cnt).y = stars1(cnt).y + 1 'move them, ask no questions. IF stars1(cnt).y > 1 AND stars1(cnt).y < 601 THEN ' IF in range of view PSET (stars1(cnt).x, stars1(cnt).y), stars1(cnt).color& 'Draw the star. IF stars1(cnt).BigStar = 1 THEN ' If it's a big star, make it big. PSET (stars1(cnt).x + 1, stars1(cnt).y), stars1(cnt).color& PSET (stars1(cnt).x - 1, stars1(cnt).y), stars1(cnt).color& PSET (stars1(cnt).x, stars1(cnt).y + 1), stars1(cnt).color& PSET (stars1(cnt).x, stars1(cnt).y - 1), stars1(cnt).color& END IF ELSE ' If out of view, make a new one. ' max min min stars1(cnt).x = INT(RND * (800 - 1 + 1)) + 1 stars1(cnt).y = INT(RND * (2 - 1 + 1)) + 1 stars1(cnt).color& = _RGB32(INT(RND * (Stars1MAX - Stars1MIN + 1)) + Stars1MIN, INT(RND * (Stars1MAX - Stars1MIN + 1)) + Stars1MIN, INT(RND * (Stars1MAX - Stars1MIN + 1)) + Stars1MIN) IF stars1(cnt).BigStar = 1 THEN 'big star dies out. BigStarCount = BigStarCount - 1 END IF stars1(cnt).BigStar = 0 'Regardless, lets flag it as new (big star). IF stars1(cnt).BigStar = 0 AND BigStarCount < TotalNumberOfBigStars THEN 'we need a new BIGstar. Dice = CINT(RND) IF Dice = 1 THEN 'Roll the dice, see if we can grow! BigStarCount = BigStarCount + 1 stars1(cnt).BigStar = Dice END IF END IF END IF NEXT cnt 'Planet... IF switchTIMEcur = switchTIMEmax THEN ' If it's time to draw IF Plan.y + Plan.radi > 600 - Plan.radi THEN 'Out of range? Plan.x = INT(RND * (750 - 50 + 1)) + 50 Plan.y = 1 Plan.radi = INT(RND * (40 - 10 + 1)) + 10 Plan.R = INT(RND * (200 - 20 + 1)) + 20 Plan.G = INT(RND * (200 - 20 + 1)) + 20 Plan.B = INT(RND * (200 - 20 + 1)) + 20 Plan.Color& = _RGB32(Plan.R, Plan.G, Plan.B) ELSE ' IN RANGE!!!! Plan.y = Plan.y + 1 CIRCLE (Plan.x, Plan.y), Plan.radi, _RGB32(Plan.R - 25, Plan.G - 25, Plan.B - 25) 'CIRCLE (Plan.x, Plan.y), Plan.radi, Plan.Color& 'PAINT (Plan.x, Plan.y), Plan.Color& PAINT (Plan.x, Plan.y), _RGB32(Plan.R - 25, Plan.G - 25, Plan.B - 25) 'EXPERIMENTAL! ' this makes shading on my cheesey planet. It's a work in progress :( FOR n = 2 TO Plan.radi IF n <= Plan.radi - 3 THEN CIRCLE (Plan.x, Plan.y), n, _RGB32(Plan.R - (n - 15), Plan.G - (n - 15), Plan.B - (n - 15)) ELSE CIRCLE (Plan.x, Plan.y), n, _RGB32(Plan.R - n + 2, Plan.G - n + 2, Plan.B - n + 2) END IF NEXT n END IF switchTIMEcur = 0 'we drew. Reset time to draw again. ELSE 'You have to draw the planet anyway, just this doesn't +1 it. switchTIMEcur = switchTIMEcur + 1 CIRCLE (Plan.x, Plan.y), Plan.radi, _RGB32(Plan.R - 25, Plan.G - 25, Plan.B - 25) 'CIRCLE (Plan.x, Plan.y), Plan.radi, Plan.Color& 'PAINT (Plan.x, Plan.y), Plan.Color& PAINT (Plan.x, Plan.y), _RGB32(Plan.R - 25, Plan.G - 25, Plan.B - 25) 'EXPERIMENTAL! ' this makes shading on my cheesey planet. It's a work in progress :( FOR n = 2 TO Plan.radi IF n <= Plan.radi - 3 THEN CIRCLE (Plan.x, Plan.y), n, _RGB32(Plan.R - (n - 15), Plan.G - (n - 15), Plan.B - (n - 15)) ELSE CIRCLE (Plan.x, Plan.y), n, _RGB32(Plan.R - n + 2, Plan.G - n + 2, Plan.B - n + 2) END IF NEXT n END IFWENDEND`
Ya i know it's not a game yet. Just kind of a warning, i may need some graphical help later; like my ugly planet. It's not nearly as bad as it was. But my question is, would it be easier to attempt to draw the items on the screen in game and copy to memory, or externally and then import from file? This would be done at the beginning so i'm not too worried about speed at creation/load point... But i'm sure everyone here has more game making experience than me at this point.
Edited to fix a sentence.
' With programming, you can do anything!
DEF SEG = &HBANKACNT 'weekend starts here...
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DEF SEG ' quit job here.
#### Theophage
• Full Member
• Posts: 137
##### Re: So, i've started writing a game...
« Reply #1 on: August 07, 2011, 03:31:42 PM »
The parallax star motion is pretty sweet, but yeah, I'd go for an already drawn sprite/tile for the planet.
...
END OF LINE
#### TerryRitchie
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• Posts: 2276
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##### Re: So, i've started writing a game...
« Reply #2 on: August 08, 2011, 07:22:13 AM »
Drawing graphics using good old QuickBasic commands is tedious at best, but very doable. The advantage: no graphics files cluttering the game's directory. I've only attempted a few times, and those time were simply emulating 8 bit graphics found in old coin-op games. Now that I have a language that supports loading many different graphic file types I see no need to try and create complex scenery through code. Use your favorite graphics editor and its tools to create graphics and then load them in through code.
#### Sketcz
• Newbie
• Posts: 16
##### Re: So, i've started writing a game...
« Reply #3 on: August 08, 2011, 12:11:31 PM »
Not to hijack the thread, but it's a related question and I didn't think it needs a new topic.
For my current game I've been doing all the graphics using the DRAW command, which does slow things down (luckily this will be played on current beefy PCs).
How does copying to memory work? Does it draw things once, copies it, and then repastes it again and again?
I've gotten to the point where the DRAW files are actually used for animation.
#### TerryRitchie
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##### Re: So, i've started writing a game...
« Reply #4 on: August 08, 2011, 12:37:17 PM »
Draw for animation. Holy smokes, I don't think draw was ever envisioned to do that. Sprites are where it's at. Basically the way I handled this problem in the past was to either use QuickBasic commands to draw my characters, use GET to put them in an array (save them in memory) and the use PUT to place the images where I wanted, or load external files, GET portions of the files (Sprites) and then again use PUT to place the images.
Since QB64 I have abandoned drawing manually and only use loading of files now. No need to use GET and PUT any longer, these have been replaced with far superior commands available in QB64 (_LOADIMAGE, _PUTIMAGE, _COPYIMAGE, etc..) I use these commands to treat images as individual sprites, especially since they can natively use transparent .PNG files. And to take the sprite handling to the next level I created a library to handle the sprites for me.
#### Sketcz
• Newbie
• Posts: 16
##### Re: So, i've started writing a game...
« Reply #5 on: August 08, 2011, 02:15:54 PM »
I use a mixture of DRAW, LINE and CIRCLE for all my imagery needs. I spent yesterday afternoon, hours in fact, using DRAW to create 6 new characters. I thought I was going to go made manually retyping them in reverse for the mirrors version (L to R, E to H, and vice versa), when I realised I could copy and paste the lot into Word and do a search and replace on each letter (I'd switch L to Q, then R to L, then Q to R).
Holy smokes I'm like a kid given his dad's welding set.
The fact that DRAW leaves an imprint of itself on the screen when moved can be useful at times. I made the impression of a trampoline material stretching when moving two lines, since as they moved they didn't erase themselves and now it looks like a material bulge as someone lands on it.
Having said that, I'm getting about 50 runs of the Main Loop before going to Secondary Loop, and the speed is good at that number. But if I have to start dropping the number of Main Loop runs it'll introduce flicker (since old graphics are blacked out in the secondary loop) and things will slow, so I might have to turn to better alternatives.
Then again, I can already feel the motivation waning. I need to stick to one day projects I reckon.
#### cr0sh
• Full Member
• Posts: 149
• QB64 rox the box!
##### Re: So, i've started writing a game...
« Reply #6 on: August 08, 2011, 02:31:28 PM »
Draw for animation. Holy smokes, I don't think draw was ever envisioned to do that. Sprites are where it's at.
While I agree with you that "sprites are where it's at", using DRAW for animation is actually a quite old use of the command. The DRAW command itself (from Microsoft) is almost 30 years old; I remember several different Microsoft "Extended Color Basic" programs for the TRS-80 Color Computer 2 (back when I was a kid - 1984ish) that used DRAW for simple "sprite" animation-type effects (because it could be faster and less memory intensive - remember, BASIC on this 1 MHz machine could only access about 24K without doing complex tricks with machine code - and even then, you might only be on a machine with 32K anyhow).
#### TerryRitchie
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##### Re: So, i've started writing a game...
« Reply #7 on: August 08, 2011, 03:02:08 PM »
Yeah, I had a COCO I, II, and III in my time as well and remember the 24K limit no matter how much actual RAM was installed. I do remember draw being used for easy rotation, but I can't recall seeing it used for animation, although I don't doubt it could have been. I used to keep a copy of my sprites on a separate graphics page, copy the ones I needed to graphics page 1 and then flip to page 0 while page 1 was shown, copy graphics to 0 then, flip again, and so on. For a .89Mhz processor, the 6809e was a pretty fast versatile chip. (or POKE 65495,0 to speed that puppy up!)
#### vrensul
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##### Re: So, i've started writing a game...
« Reply #8 on: August 08, 2011, 04:18:38 PM »
I have a COCO 1. Still works and all. (don't write to disk when you use that poke trick!) What piece of computer right? I mean, how many people didn't have one? I was wee small when I got mine. Dad bought a CoCo 3 later and I fell in love with Karonos Rift and super pitfall 3. Hmmm wonder if karonos rift has ever been re-made. I could just see the bug-eyed robot staring at me now.
Back on topic,
Anyhow I'm taking the suggestions and using MS paint for drawing stuff (i'm a terrible artist). Now just have to figure out how to get the black of the png file to actually be transparent. Currently it over-writes everything. I have a pretty good idea of the rest of the game if I can get past that point. TBH, it'll be pretty quick going after that.
' With programming, you can do anything!
DEF SEG = &HBANKACNT 'weekend starts here...
poke checking, 1000000
DEF SEG ' quit job here.
#### TerryRitchie
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##### Re: So, i've started writing a game...
« Reply #9 on: August 08, 2011, 05:59:34 PM »
The program that you use to create the .PNG file needs to support the creation of transparent of a .PNG layer in the file. Most, if not all graphics paint type programs that support creation of .PNG should support this.
Not to knock your choice as I realize why you chose it, but MSPaint is a terrible app to do any kind of really serious graphics work in. Don't get me wrong, I've seen some amazing video of people doing amazing things with it, but I think they locked themselves in a room with it for over a year to do that.
I'm not even sure if MSPaint supports .PNG? If you are simply naming the images you create with the .PNG extension then they are still .BMP files, just in disguise.
But, to make matters worse I don't know what other program to suggest. I use PhotoImpact X3 and I have friends who swear by Photoshop and GIMP (GIMP is free).
#### unseenmachine
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• Posts: 3290
• A fish, a fish, a fishy o!
##### Re: So, i've started writing a game...
« Reply #10 on: August 08, 2011, 07:13:07 PM »
_CLEARCOLOR _RGB32(0,0,0),IMAGE&
http://qb64.net/wiki/index.php?title=CLEARCOLOR
#### vrensul
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##### Re: So, i've started writing a game...
« Reply #11 on: August 08, 2011, 07:57:19 PM »
Quote
_CLEARCOLOR _RGB32(0,0,0),IMAGE&
Thanks unseen. I spent about an hour or so on clearcolor, and a few others last night and it just didn't work for me. Somehow i ended up trying to extract pixel 1,1 information and setting that as the color, but then i got tangled up in rgba32 or something. Anyhow that works. Thanks.
Quote
MSPaint is a terrible app to do any kind of really serious graphics work in.
You're right on that one. You haven't seen my "art work" though. I'm in need of assistance with the circle tool. Seriously, i'm that bad. Anyhow thanks for the suggestions. I'm gonna look up gimp and if that doesn't work, i'll check out sourceforge (or start a new project). BTW, paint will load and save png files but I think that's it's limit. I haven't looked at the header or anything so not sure if it's just a naming convention, but there's no info about an ALPHA channel when you load a png.
' With programming, you can do anything!
DEF SEG = &HBANKACNT 'weekend starts here...
poke checking, 1000000
DEF SEG ' quit job here.
#### TerryRitchie
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##### Re: So, i've started writing a game...
« Reply #12 on: August 08, 2011, 08:53:13 PM »
Whoops - forgot about that one
#### vrensul
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##### Re: So, i've started writing a game...
« Reply #13 on: August 15, 2011, 08:57:14 PM »
Ok. So here's what i've done so far, every chance i've gotten to work on it. It's my FIRST attempt at writing a game beyond that one memory game. I'm not too far from finished. I know i'm not a good artist, so dont be too rough on the art Anyhow keep the images in the dir with the EXE after you compile. Next venture will be into how to animate some of the colors using something called "color cycling", and finally coding the rest of the "features" of the game and setting up opening and closings.
BTW, use the mouse to move, and LMB to shoot. Them dudes are set to be really easy.
' With programming, you can do anything!
DEF SEG = &HBANKACNT 'weekend starts here...
poke checking, 1000000
DEF SEG ' quit job here. | 5,737 | 19,329 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2013-20 | latest | en | 0.449705 |
https://forum.arduino.cc/t/compare-two-sensor-values/103081 | 1,620,901,629,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243990584.33/warc/CC-MAIN-20210513080742-20210513110742-00095.warc.gz | 258,572,439 | 11,111 | # Compare two sensor values
Hey there :-)
I'm sort of new to Arduino and did an experiment with a flex sensor. If I bend it the status of it is shown via 10 LEDs (like here: http://arduinobasics.blogspot.de/2011/05/arduino-uno-flex-sensor-and-leds.html). Everything works fine until now. Now I need to do it a bit modified for two sensors. I can describe it in words what I wanna do but I don't know how to write this in code :-( Maybe I try to explain it for you:
From both sensors I get values between 200 and 360. I now want to compare both values and if they are the same (maybe with a tolerance of 10 or 20) a new LED should turn on. So if values of sensor 1 (+/-10) are the same like those of sensor two (+/-10) an new LED should show this to me.
How do I write this in the basic code from the link above?
Thanks a lot for your support :-)
I don’t know what you want to program.
If the sensors are about the same, should a led turn on? Or should a new led turn (thus incrementing the led bar).
If you map both sensor values, you could just calculate the difference.
``````int difference = lightLevel2 - lightLevel1;
if (difference < 10 && difference > -10)
{
// show new led.
}
``````
I think this might be the right way ;) Sorry for my difficult explanation, I've tried to explain it as understandable as possible in my best english ;-)
The construction of my prototype is about the same as in the tutorial (link). The only thing I want to add is another row of 10 LEDs for the second sensor. So if I get for example value 260 for sensor 1 and 260 for sensor 2 (+/-10 tolerance) another LED outside of these two rows should turn on and show me that both sensors have about the same value. This process should always be running in my sketch. Do you understand it a bit better now? I hope so :~
In your code you compare the two sensor values and if it's +/-10 an LED is turned on, right? So I think it's the way I have to do it isn't it?
Later on I will have an interface on the screen that replaces the LEDs. But for prototyping and learning Arduino I think this way is the better one for learning the tool.
Thanks a lot for you really quick reply :) I'm on my way to learn Arduino ;)
I understand.
You want two led bars. One for each sensor.
And you want light a led if the both are about the same.
But the Arduino doesn’t have enough pins for two led bars. You could use a 74hc595 and this: http://arduino.cc/en/Tutorial/ShiftOut
If you have used the 74hc595 before, it’s easy to do.
If you didn’t use the 74hc595 before, you might make things more complicated than they should be.
The same goes for Charlieplexing: http://arduino.cc/playground/Code/Charlieplex http://www.instructables.com/id/CharliePlexed-LED-string-for-the-Arduino/
If you are just testing, test one sensor with the bar. And after that test the other sensor.
You might even use the bar to show how close the sensorsvalues match (by showing the absolute difference on the bar).
`````` // the absolute difference
int difference = lightLevel2 - lightLevel1;
difference = abs(difference); // 'abs' is the absolute value, the result is always positive.
if (difference >= 0 && difference < 10)
{
// match within 10
}
else if (difference >= 10 && difference < 20)
{
// they differ between 10 and 20
}
else if (difference >= 20 && difference < 30)
{
// they differ between 20 and 30
}
else
{
// outside range
}
``````
Those if statements are unnecessarily complex.
If the first statement evaluates to true, the remaining ones will not be evaluated.
If the first statement evaluates to false, you KNOW that difference is not less than 10, so there is not need to test that condition again.
The same holds true for the remaining statements. The value in difference will be handled by the first statement that evaluates to true, so there is not need to keep confirming that difference is greater than or equal to the previous cut off value.
For PaulS:
`````` // the absolute difference
int difference = lightLevel2 - lightLevel1;
difference = abs(difference); // 'abs' is the absolute value, the result is always positive.
if (difference < 10) // it's always higher or equal than zero
{
// match within 10
}
else if (difference < 20) // the range 0...10 has already been tested
{
// they differ between 10 and 20
}
else if (difference < 30)
{
// they differ between 20 and 30
}
else
{
// outside range
}
``````
For PaulS:
Exactly. That code makes it much easier to see, at a glance, which block of code is going to be executed for any given value of difference, and makes the evaluation faster.
Thank you so much for your support. That's exactly what I was looking for 8)
Everything works fine in my code :-)
I need your help one more time. I now changed something in my code and it doesn’t work any longer
I’ve tried to built in my LED outputs (5,6,7) in your “comparison function” but it doesn’t work anymore =(
Could you help me and explain what I’m doing wrong? Sorry, I’m really new to Arduino :~ I think it has something to do with the “contrain”. What exactly does this command do?
So here’s my code:
``````// Sensor Pins (A0, A1)
int flexPin1 = 0;
int flexPin2 = 1;
// sets LED Pins 5,6,7 to output
void setup() {
for (int i=5; i<8; i++){
pinMode(i, OUTPUT);
}
}
// blink function: that you can see the LEDs turning on and off
void blink(int 5, int onTime, int offTime){
digitalWrite(5, HIGH);
delay(onTime);
digitalWrite(5, LOW);
delay(offTime);
}
//
void loop(){
//Ensure to turn off ALL LEDs before continuing
for (int i=5; i<8; i++){
digitalWrite(i, LOW);
}
// the two sensor values are read
int lightLevel1 = (analogRead(flexPin1));
int lightLevel2 = (analogRead(flexPin2));
///////// compare the two sensor values //////////
// the absolute difference
int difference = lightLevel2 - lightLevel1;
difference = abs(difference); // 'abs' is the absolute value, the result is always positive.
if (difference < 10) // it's always higher or equal than zero
{
// match within 10
int ledON = constrain(difference, 5, 7);
blink(ledON, 10,1);
}
else if (difference < 20) // the range 0...10 has already been tested
{
// they differ between 10 and 20
int ledON = constrain(difference, 5, 7);
blink(ledON, 10,1);
}
else if (difference < 30)
{
// they differ between 20 and 30
int ledON = constrain(difference, 5, 7);
blink(ledON, 10,1);
}
else
{
// outside range; i = all LEDs
digitalWrite (i, LOW);
}
}
``````
`````` int lightLevel1 = (analogRead(flexPin1));
int lightLevel2 = (analogRead(flexPin2));
``````
Why is the analogRead() call in parentheses? They are not needed.
You are sorting out the difference between two sensor values. Then, you are constraining the difference to be in the range of 5 to 7. Then, regardless of what that function comes up with, you blink pin 5.
The most obvious problem is that you don't seem to understand what the constrain function does. Perhaps clamp() might have been a better name. If the input value is less than the low end of the range, the output value IS the low end of the range. If the input value is greater than the high end of the range, the output value is the high end of the range.
When difference is greater than 10, the output will always be 7. Is that what you want? If it is, there is no reason to have any if/else statements.
Making the blink function use an argument for the pin to blink, rather than a hardcoded value would be a good idea, too. I find it hard to believe that that even compiles.
One final comment. "t doesn't work" is incredibly lame. That code does something. You want it to do something. That the two somethings are not the same thing is pretty obvious. What either one of them is is not defined. If you really want help, you need to describe what the code does, what you expect it to do, and how those two differ.
I’m sorry but I’m at the moment confused of my own code :~ And it’s not easy to explain it in English. I really want to learn this tool but it’s difficult to learn the language in such a fast way.I have to built this prototype for a project at school and hope it’s my chance to learn it
Maybe again in words what I’d like to do: I still have 10 LEDs and two flex sensors. All I’m trying to do at the moment is showing that if both sensors have the same values (+/- tolerance → the “if statements” in the code) an LED should turn on. In my code I’ve tried to turn on LED 5 for a tolerance of 0 to 10, LED 6 for a tolerance of 10 to 20 and LED 7 for a tolerance of 20 to 30. So all the other LEDs are not used for testing. As I said before, in the end I want to use processing for an interface on the screen.
`````` if (difference < 10) // it's always higher or equal than zero
{
// match within 10
int ledON = constrain(difference, 5, 7);
blink(ledON, 10,1);
}
``````
So here I’m trying to say:
“If there is a difference of both sensors of 0 to 10: turn on LED 5 by using the “blink function” (that should only be used to delay the LEDs’ turning on and off).”
How can I say that only LED 5 should be turned on in this range? If I leave out the “7” in the statement it doesn’t work.
Do you now understand a bit better what I’m trying to do?
Q: what is your native language? adjust your profile then we know
refactored your code to something that works again, maybe not what you want but a new starting point
• blink function was incorrect (param 1)
• removed constraining
• adjusted blink timing to something visible (1 msec off is hardly visible)
(code compiles but not further tested)
``````// Sensor Pins (A0, A1)
int flexPin1 = 0;
int flexPin2 = 1;
// sets LED Pins 5,6,7 to output
void setup() {
for (int i=5; i<8; i++){
pinMode(i, OUTPUT);
}
//Ensure to turn off ALL LEDs before continuing
for (int i=5; i<8; i++){
digitalWrite(i, LOW);
}
}
// blink function: that you can see the LEDs turning on and off
void blink(int pin, int onTime, int offTime)
{
digitalWrite(pin, HIGH);
delay(onTime);
digitalWrite(pin, LOW);
delay(offTime);
}
//
void loop()
{
int lightLevel1 = analogRead(flexPin1);
int lightLevel2 = analogRead(flexPin2);
///////// compare the two sensor values //////////
// the absolute difference
int difference = abs(lightLevel2 - lightLevel1); // the diff is always >= 0.
if (difference < 10) // it's always higher or equal than zero
{
blink(5, 50, 50);
}
else if (difference < 20) // the range 0...10 has already been tested
{
blink(6, 50, 50);
}
else if (difference < 30)
{
blink(7, 50, 50);
}
else
{
blink(5, 10, 10);
blink(6, 10, 10);
blink(7, 10, 10);
}
}
``````
I'm from Germany ;) I will edit my profile soon hehe...will be easier :D
That is definitely more close to what I wanna do 8) great :-) It's even clearer arranged now...and understandable :D
What it does: – all LEDs are flickering after plugging my Arduino UNO to the USB port – if I bend one of the sensors the LEDs 5 to 7 turn on and off in sequence - the same works for both sensors bending them at the same time
But: - I still cannot see the difference :( - the LEDs should only turn on if both sensors have different values (for example: flexPin1 gives me a value of 270, flexPin2 a value of 225 -> difference of 25 -> LED 7 should blink) - at the moment they also turn on and off if they have the same values
But it's really really close to what I wanna do 8) You are great!!!
– all LEDs are flickering after plugging my Arduino UNO to the USB port
• at the moment they also turn on and off if they have the same values
Because of this:
`````` else
{
blink(5, 10, 10);
blink(6, 10, 10);
blink(7, 10, 10);
}
``````
Remove this code if you want nothing to happen when the values are the same.
There are some improvements that could be made to the code that Rob posted (which was based on yours initially, so it is not his fault).
`````` int lightLevel1 = analogRead(flexPin1);
int lightLevel2 = analogRead(flexPin2);
``````
How do you get light level from a flex sensor? Meaningful names are very useful for writing code that does not require a lot of thinking to decipher/create.
I mean that there is nothing stopping you from writing: int lightLevel1 = analogRead(flexPin1); int q = analogRead(flexPin2); But, of course, that makes little sense when you then subtract q from lightLevel1. Your names are only a little less confusing, since flex sensors and light are unrelated.
Unless you have actually written 1 and 2 on the sensors, names that reflect the physical arrangement of the sensors make more sense, too.
(for example: flexPin1 gives me a value of 270, flexPin2 a value of 225 -> difference of 25 -> LED 7 should blink)
There are no Serial.print() statements in the code that indicate that you know what the actual values are. Normally, I don't like to see commented out code in code that is posted, but commented out Serial.print() statements are an exception. With them present but commented out, it is easy to see that you do know how to debug code and that you have facts to support your suppositions.
Without them, we have no way of knowing what actual values you are reading from each sensor, or that you are even reading each sensor correctly.
`````` Serial.println(analogRead(flexPin1));
delay (100);
``````
I now get values for sensor 1 from 398 to 496 and for sensor two from 492 to 340. But don’t I have to use the “map function” again now in order to map those values to my LED ports? Something like this:
`````` int lightLevel1 = map(analogRead(flexPin1), 398, 496, ??, ??);
``````
What do I have to put in for the question marks? Cause I don’t a have a row of LEDs now any longer that counts up and down. It now refers to the “compare function”, doesn’t it? It’s getting too complicated for me as a beginner =(
Here is my code so far:
``````// Sensor Pins (A0, A1)
int flexPin1 = 0;
int flexPin2 = 1;
// sets LED Pins 5,6,7 to output
void setup() {
Serial.begin(9600);
for (int i=5; i<8; i++){
pinMode(i, OUTPUT);
}
//Ensure to turn off ALL LEDs before continuing
for (int i=5; i<8; i++){
digitalWrite(i, LOW);
}
}
// blink function: that you can see the LEDs turning on and off
void blink(int pin, int onTime, int offTime)
{
digitalWrite(pin, HIGH);
delay(onTime);
digitalWrite(pin, LOW);
delay(offTime);
}
//
void loop()
{
int lightLevel1 = analogRead(flexPin1);
int lightLevel2 = analogRead(flexPin2);
///////// compare the two sensor values //////////
// the absolute difference
int difference = abs(lightLevel2 - lightLevel1); // the diff is always >= 0.
// Serial.println(analogRead(flexPin1));
// delay (100);
// Serial.println(analogRead(flexPin2));
if (difference < 10) // it's always higher or equal than zero
{
digitalWrite(5, HIGH);
}
else if (difference < 20) // the range 0...10 has already been tested
{
digitalWrite(6, HIGH);;
}
else if (difference < 30)
{
digitalWrite(7, HIGH);;
}
else
{
digitalWrite(5, LOW);
digitalWrite(6, LOW);
digitalWrite(7, LOW);
}
}
``````
But don't I have to use the "map function" again now in order to map those values to my LED ports?
Why? The code works off the absolute difference already, no need to massage it further.
The code now looks like it should work. If it doesn't, serial.print the values you have and the difference - it may be that you'll need to adjust the thresholds you set in the ifs.
okay well that little thing with the values was my mistake with the basic version of the code as well. I think I will fix it somehow :) Oh dear :~ THANK YOU SO MUCH!!!
Hi Snickers,
Sometimes it is better to start from scratch rather than mould another code. The reason you get varied responses, is because you are trying to do something very different from the original code you referenced. And you seem to be [u]changing[/u] your question or requirement.
I am going to lay out what I think you want to do in point form, and if you want something different, you can say so in subsequent comments.
1. You have 2 flex sensors. One attached to Analog input 0 : lets call this one flexA0, and the other attached to Analog input 1: lets call this one flexA1.
2. You have 3 LEDs These LEDs are attached to pins 5, 6 and 7.
3. You want a single LED to light up when the flex sensor readings do not match The LED that lights up depends on the difference between flex sensor readings.
4. If there is no difference between flex sensor readings : All LEDs are off
5. If there is a difference of 1-10: LED on pin 5 lights up, others are off.
6. If there is a difference of 11-20: LED on pin 6 lights up, others are off.
7. If there is a difference greater than 20: LED on pin 7 lights up, others are off.
Please note: 270 minus 225 is NOT 25, it is 45... Once we have the question right, then we can work on the answer.
1. You have 2 flex sensors. One attached to Analog input 0 : lets call this one flexA0, and the other attached to Analog input 1: lets call this one flexA1.
2. You have 3 LEDs These LEDs are attached to pins 5, 6 and 7.
3. You want a single LED to light up when the flex sensor readings do not match The LED that lights up depends on the difference between flex sensor readings.
4. If there is no difference between flex sensor readings : All LEDs are off
5. If there is a difference of 1-10: LED on pin 5 lights up, others are off.
6. If there is a difference of 11-20: LED on pin 6 lights up, others are off.
7. If there is a difference greater than 20: LED on pin 7 lights up, others are off.
Hi Scott C,
thanks a lot again for your detailed post :) I'm sorry for confusing some of you here but as I already mentioned it's really hard to explain such a problem in English. Your interpretation is absolutely right. The last code that was posted by PaulS was actually the one I was trying to write and it worked fine with a little tuning the values of the sensors. The most confusing problem is that my sensors are both of the same type but sensor one gives me values from 390 to 258 while the other one gives me something in between 480 to 340. That's really strange cause they are opposite... But I fixed it by playing with those values. Now everything works fine :-)
But thanks a lot again and again for your answer :) I will try to write posts that are easier to understand in the future ;)
hi, i am working on glove assessment for hand fingers using arduino uno with two flex sensor, i start by using just one sensor and i map its reading values from 0-100 then i map (0-100) on led bar of 10 leds which blink one by one incrementally during flexing the sensor (it works properly). know i want some help to add my second flex sensor to the circuit and i should put an external selector which select the sensor that should be assigned to the led bar, and the change in code if there is any if statement i should use.
thanks for help and support. | 4,855 | 18,919 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2021-21 | latest | en | 0.929272 |
https://maristfoxtales.com/helping-students/a-student-wanted-to-test-how-the-mass-of-a-paper-airplane-affect-the-distance-it-would-fly-solved.html | 1,674,775,193,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764494826.88/warc/CC-MAIN-20230126210844-20230127000844-00844.warc.gz | 405,329,917 | 11,837 | # A Student Wanted To Test How The Mass Of A Paper Airplane Affect The Distance It Would Fly? (Solved)
A student wanted to test how the mass of a paper airplane affected the distance it would fly. Paper clips were added before each test flight. As each paper clip was added, the plane was tested to determine how far it would fly. A study was done to find if different tires treads affect the braking distance of a car.
## What is the independent variable in a paper plane?
Independent variable: The mass of the plane (number of paper clips added). Dependent variable: is the distance the plane travels.
## What is the dependent variable in the experiment below?
The dependent variable is the variable that is being measured or tested in an experiment. 1 For example, in a study looking at how tutoring impacts test scores, the dependent variable would be the participants’ test scores, since that is what is being measured.
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## What is the independent variable in this sentence a study was done to find if different tire treads affect the braking distance of a car?
“A study was done to find if different tire treads affect the braking distance of a car.” The independent variable is the different tire treads. Select the dependent variable from: “The time it takes to run a mile depends on the person’s running speed.”
## What are some variables that could affect the distance a paper airplane will fly?
The weight of the paper plane also affects its flight, as gravity pulls it down toward Earth. All of these forces (thrust, lift, drag and gravity) affect how well a given paper plane’s voyage goes.
## How does mass affect the flight of a paper airplane?
A paper airplane with a larger mass in the body and smaller wings will fly faster than one with a smaller body mass and larger wings because its “wing load” is larger.
## Does the type of paper affect a paper airplane?
There are other factors that influence how well a paper airplane flies. The type of paper used can affect its weight and the amount of friction that exists. How the plane is designed can also vary tremendously. The design of the wings, body, nose and tail can all drastically change the way the plane flies.
## Does the dependent variable affect the independent variable?
The independent variable is the one the experimenter controls. The dependent variable is the variable that changes in response to the independent variable. The two variables may be related by cause and effect. If the independent variable changes, then the dependent variable is affected.
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## How do independent and dependent variables differ?
Independent and dependent variables
1. The independent variable is the cause. Its value is independent of other variables in your study.
2. The dependent variable is the effect. Its value depends on changes in the independent variable.
## How do you identify independent dependent and control variables?
Identify the variables
1. Independent variable – the variable that is altered during a scientific experiment.
2. Dependent variable – the variable being tested or measured during a scientific experiment.
3. Controlled variable – a variable that is kept the same during a scientific experiment.
## What is the independent variable in an experiment?
Answer: An independent variable is exactly what it sounds like. It is a variable that stands alone and isn’t changed by the other variables you are trying to measure. For example, someone’s age might be an independent variable.
## What are the examples of dependent and independent variables?
Independent and Dependent Variable Examples
• In a study to determine whether how long a student sleeps affects test scores, the independent variable is the length of time spent sleeping while the dependent variable is the test score.
• You want to compare brands of paper towels, to see which holds the most liquid.
## What is the dependent variable answer key?
The dependent variable in an experiment is the measured behavior or responses of the participants.
## How does the weight of paper affect how far a paper airplane will fly?
The larger the paper airplane the more it will weigh, the more it weighs the more lift will be needed to keep it flying. Eventually weight will become greater than lift and the paper airplane will descend to the ground.
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## How does the weight of a plane affect its flight?
Weight is the force that pulls the plane down due to gravity. In order for the plane to get off the ground, the plane must overcome its weight throught the force of lift. The more mass the plane has the more lift it has to produce in order to get off the ground.
## What are the variables in a paper airplane?
Warm up – What are the Variables in our paper airplane experiment? Independent – what’s changing? Dependent – what’s being observed? Control – What’s not being changed? | 1,013 | 5,125 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2023-06 | longest | en | 0.951257 |
https://www.nagwa.com/en/worksheets/395193262764/ | 1,576,149,789,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540543252.46/warc/CC-MAIN-20191212102302-20191212130302-00467.warc.gz | 803,699,593 | 4,887 | # Worksheet: Charles' Law
In this worksheet, we will practice using Charles' law to convert between volume and temperature for an ideal gas at constant pressure.
Q1:
Temperature is sometimes measured with a gas thermometer by observing the change in the volume of the gas as the temperature changes at constant pressure. The hydrogen in a particular hydrogen gas thermometer has a volume of 150.0 cm3 when immersed in a mixture of ice and water at . When immersed in boiling liquid ammonia, the volume of the hydrogen, at the same pressure, is 131.7 cm3. What is the temperature of the boiling ammonia? | 126 | 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} | 2.53125 | 3 | CC-MAIN-2019-51 | latest | en | 0.914608 |
https://abussolaquebrada.com/journey/which-force-is-attractive-as-well-as-repulsive-force.html | 1,638,727,578,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363215.8/warc/CC-MAIN-20211205160950-20211205190950-00374.warc.gz | 146,793,426 | 17,879 | Which force is attractive as well as repulsive force?
Contents
Which forces can be attractive as well as repulsive?
A: Gravitational force is always attractive but electromagnetic force can be attractive or repulsive: <br> R : Mass comes only in one variety (there is no negative mass) but charge comes in two varieties.
Is gravitational force attractive as well as repulsive?
We know that the gravitational force is defined as the multiplication of the gravitational constant along with the masses of the objects and then divided by the square of the distance between the objects. … Hence, the nature of the gravitational force will always be attractive and not repulsive.
Are intermolecular forces attractive or repulsive?
Intermolecular forces are repulsive at short distances and attractive at long distances (see the Lennard-Jones potential). In a gas, the repulsive force chiefly has the effect of keeping two molecules from occupying the same volume.
Is gravitational force attractive or repulsive class 9?
We all know that all the forces in nature exist in opposites, but gravitational force is the only force that always attracts every object and never reples any.
Why is gravitational force repulsive?
It is shown that reduction of the gravitational mass of the system due to emitting gravitational waves leads to a repulsive gravitational force that diminishes with time but never disappears. This repulsive force may be related to the observed expansion of the Universe.
IT IS AMAZING: What does you attract what you fear mean?
Is gravity a repulsive force?
The tension does not pull things together because it is equal on the opposite sides of any given region and so as a force in its own right it cancels out. However the contribution to gravity associated with it does not cancel out, and is always repulsive.
What is the repulsive force in the nucleus?
The electromagnetic repulsion takes place within the nucleus between like electric charges. These charges are carried by the protons, whose close proximity to each other intensifies this repulsive force.
Is weak force repulsive or attractive?
The weak nuclear force is neither attractive or repulsive. | 426 | 2,190 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2021-49 | latest | en | 0.951374 |
https://www.causeweb.org/cause/statistical-topic/association-odds-ratios?page=4 | 1,620,575,050,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989006.71/warc/CC-MAIN-20210509153220-20210509183220-00063.warc.gz | 731,917,325 | 13,500 | Sorry, you need to enable JavaScript to visit this website.
# Association & Odds Ratios
• ### Law of Large Numbers (Chapter 8)
Part of an online statistics textbook. Topics include: (1) Law of Large Numbers for Discrete Random Variables, (2) Chebyshev Inequality, (3) Law of Averages, (4) Law of Large Numbers for Continuous Random Variables, (5) Monte Carlo Method. There are several examples and exercises that accompany the material.
• ### Family Applet 1
This applet simulates families of three children. The probability of having a boy on any attempt can be changed in the parameter statement. The percantage of times "x" number of girls occurs is updated in the bar chart. There is a 2nd applet on the page that is the same as above, but the families stop having children after the first boy or after they have had 3 girls.
• ### M and M Applet
Students can sample numerous bags of M&Ms. A plot of the relative frequency of each color is continually updated above the sampling frame. Each sample bag of M&Ms contains 56 candies.
• ### Analysis Tool: Kruskal-Wallis Test for K = 4
As the page opens, you will be prompted to enter the sizes of your several samples. If you are starting out with raw (unranked) data, the necessary rank- ordering will be performed automatically.
• ### HyperStat Online: Ch. 16 Chi Square
This resource defines and explains Chi square. It takes the user through 5 different categories: 1) Testing differences between p and pi 2) More than two categories 3) Chi-square test of independence 4) Reporting results 5) Exercises.
• ### Random: Probability, Mathematical Statistics, and Stochastic Processes Site
Random is a website devoted to probability, mathematical statistics, and stochastic processes, and is intended for teachers and students of these subjects. The site consists of an integrated set of components that includes expository text, interactive web apps, data sets, biographical sketches, and an object library.
• ### Designing Tests To Maximize Learning
This one-page document gives advice on how to construct and give exams. It focuses on making exams a positive experience for both instructors and students. It is written by Rich Felder an expert in Engineering education.
• ### Probability Distributions JAVA Applets with Reference Materials
This page of Statistical Java describes 11 different probability distributions including the Binomial, Poisson, Negative Binomial, Geometric, T, Chi-squared, Gamma, Weibull, Log-Normal, Beta, and F. Each distribution has its own applet in which users can manipulate the parameters to see how the distribution changes. The parameters are described on the main page as well as situations that would use each distribution. The equations of the distributions are not given. To select between the different applets you can click on Statistical Theory, Probability Distributions and then the Main Page. At the bottom of this page you can make your applet selection. This page was formerly located at http://www.stat.vt.edu/~sundar/java/applets/ | 653 | 3,045 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2021-21 | longest | en | 0.882905 |
https://m.scirp.org/papers/116373 | 1,653,720,197,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663013003.96/warc/CC-MAIN-20220528062047-20220528092047-00334.warc.gz | 417,829,148 | 25,774 | Binary Star System Decay by Graviton Interaction
Abstract: The action of gravitons in a binary star system is modelled as the locus of points on an ellipse synchronous to the elliptic orbit of the binary star. In their interaction between the masses in the system the rotational energy of the gravitons is reduced by gravitational redshift, which accounts for the decay of the binary star orbital period. This model is able to fit a broad range of eccentricities of binary pulsar orbits and orbital period decay comparable to the General Relativistic gravitational wave model.
1. Introduction
The purpose of this paper is to introduce a model for the action of the gravitons in a binary star system described in [1], a paper which has a section on binary pulsars based on a comparison to the equation for the decay of a binary star system developed in [2], but we did not give a complete model for the action of the gravitons. This new model describes the rotational phase of gravitons which interact between the masses, where the graviton phase is a locus of points on an ellipse around the central mass, with an eccentricity and period identical to the binary star system ellipse. In this model the gravitons interact directly between the masses in orbit, losing energy in the system by gravitational redshift. We will reanalyze the set of binary pulsars from the previous paper as well as make comparisons to the General Relativity (GR) model. We will also give predictions for two millisecond binary pulsars. We make a detailed comparison of the graviton model with the GR gravitational wave model and show how our model mathematically corresponds to it, although some physical aspects are disparate.
2. Rotational Energy of Gravitons in a Binary Star System
Consider a binary star system composed of a companion star of mass m in orbit around a primary star of mass M. By Kepler’s second law, the companion star in an elliptical orbit will sweep out an equal area in equal time around the primary star, expressed by
$\left(\frac{{r}^{2}}{2}\right)\frac{\text{d}\varphi }{\text{d}t}=\frac{\pi AB}{T},$ (1)
where r is the instantaneous distance between the stars, $\varphi$ is the angle r makes with the line thru the stars at perigee, A and B are the semi-major and semi-minor axes, respectively, and T is the orbital period. We assume that gravitons are responsible for the force between the stars. We define the graviton origin ellipse which encloses the center of mass of the system and has the same eccentricity and the same angular speed as the binary star ellipse. The origin ellipse describes the motion of the line of gravitons interacting between the two stars in the system, having the same period T as the orbit ellipse and expressed by
$\left(\frac{{\mathfrak{r}}^{2}}{2}\right)\frac{\text{d}\varphi }{\text{d}t}=\frac{\pi \mathfrak{a}\mathfrak{b}}{T},$ (2)
where $\mathfrak{r}$ is the distance from the center of mass to the origin ellipse, $\text{d}\varphi$ is the small angle swept out in time $\text{d}t$, $\mathfrak{a}$ and $\mathfrak{b}$ are the semi-major and semi-minor axes of the origin ellipse, respectively, and T is the orbital period of the binary star system. Multiplying Equation (2) by the angular frequency $\omega =2\pi /T$ and simplifying we get,
$\mathfrak{h}\omega =\mathfrak{h}\left(\frac{2\pi }{T}\right)=\left(\frac{2\pi {\mathfrak{l}}^{2}}{{\left(1-{\epsilon }^{2}\right)}^{3/2}T}\right)\left(\frac{2\pi }{T}\right),$ (3)
where the graviton specific angular momentum $\mathfrak{h}={\mathfrak{r}}^{2}\text{d}\varphi /\text{d}t$ and ${\mathfrak{l}}^{2}/{\left(1-{\epsilon }^{2}\right)}^{3/2}=\mathfrak{a}\mathfrak{b}$, where $\mathfrak{l}$ is the semi-latus rectum of the graviton origin ellipse and $\epsilon$ is the ellipse eccentricity, equal to the binary star system eccentricity. The radial distance $\mathfrak{r}$ is given by
$\mathfrak{r}=\frac{\left(1-{\epsilon }^{2}\right)\mathfrak{a}}{1+\epsilon \mathrm{cos}\left(\varphi \right)}.$ (4)
The graviton origin ellipse semi-latus rectum $\mathfrak{l}$ is defined in terms of the Schwarzschild radius ${R}_{s}$ of the combined masses,
$\mathfrak{l}=\sqrt{\sigma \left(\epsilon \right)}{R}_{s}=\frac{2\sqrt{\sigma \left(\epsilon \right)}\text{ }G\left(M+m\right)}{{c}^{2}},$ (5)
where the correction function $\sigma \left(\epsilon \right)$ is defined,
$\sigma \left(\epsilon \right)=\alpha {\text{e}}^{-\beta \epsilon }+p+q{\epsilon }^{2},$ (6)
where the parameters $\alpha$, $\beta$, p and q are determined by experiment and c is the speed of light in vacuum. Figure 1 shows the relationship of the binary star ellipse and the graviton origin ellipse.
Figure 1. Binary star system ellipse (outer) and graviton origin ellipse (inner). M is the primary mass, m is the companion mass, r is the distance between the masses, $\mathfrak{r}$ is the distance from M to the graviton rotational energy point of interaction between the masses and f is the rotation angle of the ellipses. The ellipses are not drawn to scale.
We define the graviton rotational energy $\Xi$ in the center of mass system of the masses M and m,
$\Xi =\mathfrak{M}\text{ }\mathfrak{h}\text{ }\omega ,$ (7)
where
$\mathfrak{M}=\frac{{\left(Mm\right)}^{2}}{{\left(M+m\right)}^{3}}=\frac{\mu Mm}{{\left(M+m\right)}^{2}},$ (8)
is the graviton rotational relativistic mass (not a rest mass since gravitons travel at speed c) and where $\mu =\left(Mm\right)/\left(M+m\right)$ is the reduced mass of the system. Since the gravitons having rotational energy $\Xi$ are in free fall in the gravitational field of the binary star system, over a small time $\delta t$ under the field acceleration $G\left(M+m\right)/{r}^{2}$, the graviton net rotational energy will change due to gravitational redshift by the amount,
$\delta \text{ }\Xi =-\Xi \left(\frac{\delta v}{c}\right)=-\left(\frac{\mu Mm}{{\left(M+m\right)}^{2}}\right)\mathfrak{h}\omega \left(\frac{G\left(M+m\right)\delta t}{c{r}^{2}}\right),$ (9)
where we substituted from (7) for $\Xi$ and the change in the free fall velocity $\delta v=\left[G\left(M+m\right)/{r}^{2}\right]\delta t$ where the minus sign implies a reduced (redshifted) graviton energy1, since the gravitons are moving in the same direction as the velocity $\delta v$, and where the distance r is given by,
$r=\frac{\left(1-{\epsilon }^{2}\right)A}{1+\epsilon \mathrm{cos}\left(\varphi \right)}.$ (10)
3. Rate of Change of the Orbital Period
To obtain the time rate of change of the orbital period T, we use Kepler’s third law,
$G\left(M+m\right){T}^{2}=4{\pi }^{2}{A}^{3},$ (11)
which upon differentiating with respect to the time t and simplifying yields the rate of change of the orbital period,
$\frac{\text{d}T}{\text{d}t}=\left(\frac{3}{T}\right)\left(\frac{4{\pi }^{2}{A}^{4}}{{G}^{2}Mm\left(M+m\right)}\right)\left(\frac{\text{d}E}{\text{d}t}\right),$ (12)
where the total orbital energy $E=-GMm/\left(2A\right)$. Substituting from Equations (3) and (5) into Equation (7) we get the graviton rotational energy,
$\Xi =\left(\frac{\mu Mm}{{\left(M+m\right)}^{2}}\right)\left(\frac{4\sigma \left(\epsilon \right)}{{\left(1-{\epsilon }^{2}\right)}^{3/2}}\right){\left(\frac{2\pi G\left(M+m\right)}{{c}^{2}T}\right)}^{2}.$ (13)
Dividing Equation (9) by $\delta t$ and substituting for $\Xi$ from (13) gives,
$\frac{\delta \text{ }\Xi }{\delta t}=-\left(\frac{\mu Mm}{{\left(M+m\right)}^{2}}\right)\left(\frac{4\sigma \left(\epsilon \right)}{{\left(1-{\epsilon }^{2}\right)}^{3/2}}\right){\left(\frac{2\pi G\left(M+m\right)}{{c}^{2}T}\right)}^{2}\left(\frac{G\left(M+m\right)}{c{r}^{2}}\right).$ (14)
Then, substituting $\text{d}E/\text{d}t=\delta \text{ }\Xi /\delta t$ from (14) into Equation (12) gives,
$\begin{array}{c}\frac{\text{d}T}{\text{d}t}=-\left(\frac{3}{T}\right)\left(\frac{4{\pi }^{2}{A}^{4}}{{G}^{2}Mm\left(M+m\right)}\right)\left(\frac{{\left(Mm\right)}^{2}}{{\left(M+m\right)}^{3}}\right)\left(\frac{4\sigma \left(\epsilon \right)}{{\left(1-{\epsilon }^{2}\right)}^{3/2}}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}×{\left(\frac{2\pi G\left(M+m\right)}{{c}^{2}T}\right)}^{2}\left(\frac{G\left(M+m\right)}{c{r}^{2}}\right).\end{array}$ (15)
In making fits to binary star data we will take an average of the radial distance r where $\mathrm{cos}\left(\varphi \left(t-{t}_{0}\right)\right)=0$ for $\varphi \left(t-{t}_{0}\right)=n\pi /2,n=1,3,5,\cdots$. Thus, the radial distance $r=\left(1-{\epsilon }^{2}\right)A$ in (15). Finally, using Kepler’s third law, (11), we substitute for A in terms of T into Equation (15) and simplify to obtain,
$\frac{\text{d}T}{\text{d}t}=-24\pi \left(\frac{\sigma \left(\epsilon \right)}{{\left(1-{\epsilon }^{2}\right)}^{7/2}}\right){\left(\frac{{\left(M+m\right)}^{2}}{Mm}\right)}^{2/3}{\left(\frac{2\pi GMm}{{c}^{3}\left(M+m\right)T}\right)}^{5/3}.$ (16)
Equation (16) is equivalent in form to the derivation given by [2], which was derived from General Relativity with gravitational wave (GW) emission for energy decay. We describe this GW relation in the next section and compare the two methods in a subsequent section.
4. The Energy Loss Due to Gravitational Wave Emission
We give a brief derivation of binary star orbital decay due to GW emission based on [3]. Gravitational waves are emitted by a binary star system due to the time rate of change of the quadrupole moment for the binary, where the quadrupole moment for the simple case of a circular orbit is expressed by,
${Q}_{ij}=\frac{1}{2}\mu {r}^{2}{I}_{ij},$ (17)
where $\mu =Mm/\left(M+m\right)$ is the reduced mass, r is the separation of the masses and ${I}_{ij}$ is a $3×3$ traceless matrix, given by,
$I=\left(\begin{array}{ccc}\mathrm{cos}\left(2\omega t\right)+\frac{1}{3}& \mathrm{sin}\left(2\omega t\right)& 0\\ \mathrm{sin}\left(2\omega t\right)& \frac{1}{3}-\mathrm{cos}\left(2\omega t\right)& 0\\ 0& 0& -\frac{2}{3}\end{array}\right).$ (18)
The strain due to the wave caused length change $\delta {d}_{ij}$ at a distance ${d}_{L}$ from the system [4] is given by,
${h}_{ij}=\frac{\delta {d}_{ij}}{{d}_{L}}=\frac{1}{{d}_{L}}\left(\frac{2G}{{c}^{4}}\frac{{\text{d}}^{2}{Q}_{ij}}{\text{d}{t}^{2}}\right).$ (19)
At a sufficient distance ${d}_{L}$ from the source we can use a linearized approximation of Einstein’s equations. The rate of change of the strain is given by
${\stackrel{˙}{h}}_{ij}=\frac{2G\stackrel{⃛}{Q}}{{d}_{L}{c}^{4}},$ (20)
where we have used the dot notation for the time derivative. Then, the power (luminosity) $\text{d}{E}_{GW}/\text{d}t$ of the wave is proportional to the square of ${\stackrel{˙}{h}}_{ij}$ integrated over the surface of the volume of space, expressed by
$\left(\frac{16\pi G}{{c}^{3}}\right)\frac{\text{d}{E}_{GW}}{\text{d}t}=\iint {|\stackrel{˙}{h}|}^{2}\text{d}S=\frac{16\pi {G}^{2}}{4\pi {c}^{8}{d}_{L}^{2}}\iint \left({\stackrel{⃛}{Q}}_{ij}{\stackrel{⃛}{Q}}_{ij}\right)\text{d}S=\frac{48\pi {G}^{2}}{15{c}^{8}}\underset{i,j=1}{\overset{3}{\sum }}\left({\stackrel{⃛}{Q}}_{ij}{\stackrel{⃛}{Q}}_{ij}\right)$ (21)
where
${|\stackrel{˙}{h}|}^{2}=\underset{i,j=1}{\overset{3}{\sum }}\left({\stackrel{˙}{h}}_{ij}{\stackrel{˙}{h}}_{ij}\right)$ (22)
and the integration over the sphere gives an area of $4\pi {d}_{L}^{2}$, cancelling that value in the numerator of (21). From (17) and (18) the summation in (21) becomes
$\underset{i,j=1}{\overset{3}{\sum }}\left({\stackrel{⃛}{Q}}_{ij}{\stackrel{⃛}{Q}}_{ij}\right)={8}^{2}{\mu }^{2}\left(1/4\right){r}^{4}{\omega }^{6}\left(2{\mathrm{cos}}^{2}\left(2\omega t\right)+2{\mathrm{sin}}^{2}\left(2\omega t\right)\right)=32{\mu }^{2}{r}^{4}{\omega }^{6},$ (23)
and by substituting this result into (21) and simplifying yields,
$\frac{\text{d}{E}_{GW}}{\text{d}t}=\frac{32G}{5{c}^{5}}{\mu }^{2}{r}^{4}{\omega }^{6}.$ (24)
Using the notation ${P}_{b}=T=2\pi /\omega$ for the orbital period and r in place of A for the circular orbit in (12) and substituting for $\text{d}E/\text{d}t$ with the negation of (24) because the GW emission causes an energy loss to the orbit, we get,
$\begin{array}{c}\frac{\text{d}{P}_{b}}{\text{d}t}=\left(\frac{-6\pi r}{GMm\omega }\right)\left(\frac{\text{d}{E}_{GW}}{\text{d}t}\right)=-\left(\frac{6\pi r}{GMm\omega }\right)\left(\frac{32G{\left(Mm\right)}^{2}{r}^{4}{\omega }^{6}}{5{c}^{5}{\left(M+m\right)}^{2}}\right)\\ =-\left(\frac{192\pi }{5}\right){\left(\frac{{\left(M+m\right)}^{2}}{Mm}\right)}^{2/3}{\left(\frac{2\pi GMm}{{c}^{3}\left(M+m\right){P}_{b}}\right)}^{5/3},\end{array}$ (25)
where we used Kepler’s law to transform ${r}^{5}{\omega }^{5}~{\left(1/{P}_{b}\right)}^{5/3}$. Putting in the dependence on the orbit eccentricity $\epsilon$ from [2] for an elliptical orbit, we put (25) into the familiar form,
${\stackrel{˙}{P}}_{b}=-\left(\frac{192\pi }{5}\right)\left(\frac{1+\left(73/24\right){\epsilon }^{2}+\left(37/96\right){\epsilon }^{4}}{{\left(1-{\epsilon }^{2}\right)}^{7/2}}\right){\left(\frac{{\left(M+m\right)}^{2}}{Mm}\right)}^{2/3}{\left(\frac{2\pi GMm}{{c}^{3}\left(M+m\right){P}_{b}}\right)}^{5/3}.$ (26)
Application to PSR B1913+16 and Other Binaries
We look at the report on B1913+16, the Hulse-Taylor binary pulsar [5]. This astronomical endeavor spanned 30 years of approximately yearly observations of the binary system. The data for the system is as follows: primary mass $M=1.4408±0.0003\text{ }{M}_{\odot }$, companion mass $m=1.3873±0.0003\text{ }{M}_{\odot }$, ${P}_{b}=0.322997462727±\left(5×{10}^{-12}\right)\text{day}$, ${\stackrel{˙}{P}}_{b}=\left(-2.4211±0.0014\right)×{10}^{-12}\text{ }\text{s}\cdot {\text{s}}^{-1}$, $\epsilon =0.6171338±0.0000004$, where M is the primary mass and m is the companion mass, ${M}_{\odot }=1.988470×{10}^{30}\text{ }\text{kg}$ is the solar mass, ${P}_{b}$ is the binary orbital period, ${\stackrel{˙}{P}}_{b}$ is the orbital period change and $\epsilon$ is the orbital eccentricity. The initial orbital period for B1913+16 is ${T}_{0}={P}_{b}=0.322997462727\text{\hspace{0.17em}}\text{day}$. For the gravitational constant we use the value $G=6.67430×{10}^{-11}\text{ }{\text{m}}^{3}\cdot {\text{kg}}^{-1}\cdot {\text{s}}^{-2}$ and $c=299792458\text{\hspace{0.17em}}\text{m}\cdot {\text{s}}^{-1}$ for the speeds of gravity (graviton) and light (photon) in vacuum. To make a good fit of (16) to the binary pulsar systems we are examining, we determined best fit parameter values of $\alpha =0.4928$, $\beta =56.5$, $p=1.6$ and $q=5.12072$, which makes the leading factor $24\pi \sigma \left(\epsilon \right)$, where the previous study [1] had a leading constant factor of $32\pi$. Figure 2 shows the correction function $\sigma \left(\epsilon \right)$ with the specified parameters $\left(\alpha ,\beta ,p,q\right)$ and the GR correction function from (26),
${\sigma }_{GR}\left(\epsilon \right)=1+\left(73/24\right){\epsilon }^{2}+\left(37/96\right){\epsilon }^{4}.$ (27)
Notice that the correction function $\sigma \left(\epsilon \right)>1$, always being above the horizontal value 1 line in Figure 2. From this fact it can be shown that the semi-major axis $\mathfrak{a}$ and the semi-minor axis $\mathfrak{b}$ of the graviton origin ellipse are always greater than the Schwarzschild radius ${R}_{s}$ of the combined masses $M+m$.
Substituting these experimental parameters and astronomical values for B1913+16 into (16) yields,
$\frac{\text{d}T}{\text{d}t}=-2.40715×{10}^{-12}\text{ }\text{s}\cdot {\text{s}}^{-1},$ (28)
which is a good match to the experimental corrected value [5] of
${\stackrel{˙}{P}}_{b}=\left(-2.4086±0.0052\right)×{10}^{-12}\text{ }\text{s}\cdot {\text{s}}^{-1},$ (29)
and is also close to the GR theoretical value,
${\stackrel{˙}{P}}_{b\text{ }GR}=-2.40219×{10}^{-12}\text{ }\text{s}\cdot {\text{s}}^{-1}.$ (30)
In Table 1 and Table 2 we list nine PSR’s, [5] - [13] and [14]. In Table 3 we show the error in each prediction, the error computed by
Figure 2. Correction function (solid line) with best fit parameter values of $\alpha =0.4928$, $\beta =56.5$, $p=1.6$ and $q=5.12072$. GR correction function (dashed line). Horizontal line drawn at 1 for reference.
Table 1. PSR binary systems studied. Column description: 1: PSR name; 2: Primary (pulsating) star mass; 3: Companion star mass; 4: Period (day) Pb or T; 5: Orbit eccentricity.
Table 2. PSR binary systems studied (continued). Column description: 1: PSR name; 2: dPb/dt (observed intrinsic value); 3: dPb/dt (general relativity computation); 4: dT/dt (due to graviton gravitational redshift).
Table 3. Prediction errors of binary systems studied. Column description: 1: PSR name; 2: dPb/dt (general relativity computation error); 3: dT/dt (due to graviton gravitational redshift error). The errors are computed by Equation (31).
$Err=|\frac{{\stackrel{˙}{P}}_{p}-{\stackrel{˙}{P}}_{i}}{{\stackrel{˙}{P}}_{i}}|,$ (31)
where ${\stackrel{˙}{P}}_{p}$ is the predicted period decay and ${\stackrel{˙}{P}}_{i}$ is the measured intrinsic period decay. Excluding PSR J0621-1002 which has a poor intrinsic ${\stackrel{˙}{P}}_{b}$ measurement, for the PSR prediction errors given in Table 3, the mean error and unbiased standard deviation of the mean error between the observed intrinsic ${\stackrel{˙}{P}}_{b}$ values and this paper’s predicted $\text{d}T/\text{d}t$ values is $\text{d}T/\text{d}t\text{ }MeanErr=0.0608±0.0084$. For a comparison with the standard GR GW emission theory, the mean error and unbiased standard deviation of the mean error is ${\stackrel{˙}{P}}_{b\text{ }GR}\text{ }MeanErr=0.1097±0.0231$.
Table 4. PSR binary systems with new predictions. Column description: 1: PSR name; 2: Primary (pulsating) star mass; 3: Companion star mass; 4: Orbit eccentricity.
Table 5. PSR binary systems with new predictions. Column description: 1: PSR name; 3: dPb/dt (general relativity computation); 4: dT/dt (due to graviton gravitational redshift). dPb/dt (observed intrinsic value) not yet determined.
In Table 4 and Table 5 we present predictions for two binary star systems [15] for which the intrinsic ${\stackrel{˙}{P}}_{b}$ are not yet determined: PSR J1949+3160, a millisecond pulsar and a white dwarf companion, and PSR J1950+2414, also a millisecond pulsar with a possible low mass white dwarf companion. The estimated deviations of the predicted values are determined using the standard deviation of the mean error found in the preceding analyses.
5. Comparison of the Methods
This paper’s approach attempts to use the gravitational redshift mechanism as the cause of the orbital decay found in binary star systems, a mechanism which has no GW emission. Our model emulates the traditional (GW) equations. The goal was to have the graviton energy redshift during free fall in the gravitational field of the binary star account for the observed orbital decay of the binary system. We struck upon the idea of a circulating equation $\Xi =\mathfrak{M}\text{ }\mathfrak{h}\omega$ for the gravitons which surround the nucleus of the binary system, where the position of the orbiting body is tracked in phase by the graviton rotational energy of relativistic mass $\mathfrak{M}$. This graviton energy travels at velocity c from the central mass toward the orbiting body and with respect to the frame which is in free fall in the field between the masses, the graviton energy will be redshifted, thus reduced in energy. This mechanism for energy loss is a relativistic effect without emission, just as the case where light loses energy by gravitational redshift of its frequency when traveling away from the surface of a star. This is contrary to the mechanism of GW emission due to rotating binary stars where the wave carries away energy from the system.
Although the addition of a correctional function $\sigma \left(\epsilon \right)$ was found necessary to make a better equality of the graviton model equation to the astrophysical data, this is a brute force approach even though this model strives to be compliant with the theory of General Relativity as formulated by Einstein, more precisely to its linearized approximation. At this stage, having compared our phenomenological approach to the standard GR method, we can go a step further in our graviton theory by actually equating it to the GR wave emission equation, giving to that equation a new interpretation of a gravitational redshift energy loss which emits no radiation.
Equate (9) and (21) in the form $-\delta \text{ }\Xi /\text{d}t=\delta {E}_{GW}/\text{d}t$, expressed by,
$-\frac{\delta \text{ }\Xi }{\text{d}t}=\mathfrak{M}\text{ }\mathfrak{h}\omega \left(\frac{G\left(M+m\right)}{c{r}^{2}}\right)=\frac{\delta {E}_{GW}}{\text{d}t}=\frac{G}{5{c}^{5}}\underset{i,j=1}{\overset{3}{\sum }}\left({\stackrel{⃛}{Q}}_{ij}{\stackrel{⃛}{Q}}_{ij}\right),$ (32)
where $\mathfrak{M}$ is the graviton rotational relativistic mass defined in (8). Moving the acceleration rate from the left side to the right side of (32), the graviton energy $\Xi$ surrounding the nuclear mass $\left(M+m\right)$ takes the form,
$\Xi =\frac{\mu Mm}{{\left(M+m\right)}^{2}}\mathfrak{h}\omega =\frac{{r}^{2}}{5\left(M+m\right){c}^{4}}\underset{i,j=1}{\overset{3}{\sum }}\left({\stackrel{⃛}{Q}}_{ij}{\stackrel{⃛}{Q}}_{ij}\right).$ (33)
From (3) and (23) with the dependence on the orbit eccentricity $\epsilon$ from [2] for an elliptical orbit, which also converts radial distance r to semi-major axis $a=A$, substituting all this into (33) we get,
$\begin{array}{c}\frac{\mu Mm}{{\left(M+m\right)}^{2}}\mathfrak{h}\omega =\frac{\mu Mm}{{\left(M+m\right)}^{2}}\left(\frac{{\mathfrak{l}}^{2}\omega }{{\left(1-{\epsilon }^{2}\right)}^{3/2}}\right)\omega \\ =\frac{{\left(\left(1-{\epsilon }^{2}\right)a\right)}^{2}}{5\left(M+m\right){c}^{4}}\left(\frac{{\sigma }_{GR}\left(\epsilon \right)}{{\left(1-{\epsilon }^{2}\right)}^{7/2}}\right)32{\mu }^{2}{a}^{4}{\omega }^{6}.\end{array}$ (34)
Now, using Kepler’s third law for ${\left({a}^{3}{\omega }^{2}\right)}^{2}={\left(G\left(M+m\right)\right)}^{2}$ in (34), with some simplification, yields,
$\left(\frac{\mu \text{ }{\mathfrak{l}}^{2}}{{\left(1-{\epsilon }^{2}\right)}^{3/2}}\right){\omega }^{2}=\frac{32\mu }{5{c}^{4}}\left(\frac{{\sigma }_{GR}\left(\epsilon \right)}{{\left(1-{\epsilon }^{2}\right)}^{3/2}}\right){\left(G\left(M+m\right)\right)}^{2}{\omega }^{2}.$ (35)
Then, substituting for $\mathfrak{l}$ from (5) into (35) and simplifying we get,
${\left(\frac{2\text{ }\sqrt{\sigma \left(\epsilon \right)}\text{ }G\left(M+m\right)}{{c}^{2}}\right)}^{2}=\frac{32{\sigma }_{GR}\left(\epsilon \right)}{5{c}^{4}}{\left(G\left(M+m\right)\right)}^{2}.$ (36)
Finally, from (36) we are left with the relationship of the correction function $\sigma \left(\epsilon \right)$ to the GR correction function ${\sigma }_{GR}\left(\epsilon \right)$,
$\sigma \left(\epsilon \right)=\left(8/5\right){\sigma }_{GR}\left(\epsilon \right),$ (37)
where ${\sigma }_{GR}\left(\epsilon \right)$ is given by (27). Substituting (37) for $\sigma \left(\epsilon \right)$ into (16) we obtain an orbital period decay equation from our graviton model identical to the standard GW model (26).
6. Conclusions
We presented a model to describe the rotational energy of gravitons in a binary star system. We have based it on an analogy to the General Relativity theory GW energy loss in a binary star system. Our approach enabled an adaptive correction to the graviton rotational energy which minimized the error in the prediction of the decay of the orbital period against the experimental value. Although the gain in accuracy is just a 5% reduction in the prediction error compared to GR, it does suggest that a graviton theory is a viable approach.
There are ongoing projects in the search for continuous low frequency GW’s from known binary pulsars that are far from merging, where the orbital periods are of order > 0.1 day, and the GW frequencies will be $f<2\left(1/0.1×86400\right)\approx 1.2×{10}^{-4}\text{ }\text{Hz}$. The LIGO/Virgo detectors can only go down to about 20 Hz [16]. Analysing day long or week long recorded data streams lowers the detection frequency to the range where these continuous GWs could be detected [17], but no detections have been made thus far. Research into continuous GW emission from binary pulsars is ongoing and the LISA telescope will look for these types of signals directly in the 0.1 mHz to 1 Hz range [18]. In a related field, regarding continuous GW emission from isolated neutron stars (pulsars), the latest research in the data of the LIGO/Virgo third observing run (O3) have made no detections [19]. Although there have been GW detections made of relativistically high energy binary blackhole mergers and binary neutron star mergers, these are not the type of low speed binary star events addressed in this discourse. As there has as yet been no detection of gravitational waves from these low speed sources, we deem this as ample justification to consider the thesis we have put forth.
Thanks
We thank the reviewer who has made challenging suggestions for the improvement of this work.
NOTES
1We remark that in the original definitions given in (7) and (9), the graviton rotational mass was defined as the reduced mass $\mu =Mm/\left(M+m\right)$ and the gravitational acceleration field was defined as $GMm/\left(M+m\right){r}^{2}$, but these have now been defined in the physically correct form.
Cite this paper: Oliveira, F. (2022) Binary Star System Decay by Graviton Interaction. Journal of High Energy Physics, Gravitation and Cosmology, 8, 317-329. doi: 10.4236/jhepgc.2022.82026.
References
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[6] Stairs, I.H., Thorsett, S.E., Taylor, J.H. and Wolszczan, A. (2002) Studies of the Relativistic Binary Pulsar PSR B1534+12. I. Timing Analysis. The Astrophysical Journal, 581, 501-508.
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Group Member: Hongtao Sun, Yu yu, Junxia Lin, Lin Li, Yuan Zhang I. Please use FRED: (Links to an external site.)Links to an external site. Plot GDP for the U.S. Gross Domestic Product from 1947 to present (quarterly, seasonally adjusted) Real Gross Domestic Product from 1947 to present (quarterly, seasonally adjusted) Define each series and explain the difference between real and nominal GDP. Graphs are showing the Gross domestic/ Real Gross Domestic product from 1947 to present the grey area is the recession period. The main difference between nominal and real values is that real values are adjusted for inflation
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2. Plot inflation for the U.S. Consumer Price Index: Total All Items for the U.S. from 1960 to present (annual, not seasonally adjusted) Personal Consumption Expenditures Excluding Food and Energy (Chain type Price Index): Total All Items for the U.S. from 1959 to present (Index 2012, seasonally adjusted, monthly)
Explain what the 2012 index means. Set 2012 as base year which means that the value of the index in 2012 is 100. Price of different year have to adjust by the deflator 3. Plot unemployment for the U.S: Civilian unemployment rate from 1948 to present (percent, seasonally adjusted, monthly)
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Total unemployed, as a percent of the civilian labor force for Iowa (U3UNEM3IA) from 2003 to present, not seasonally adjusted, quarterly Explain these two series.
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Date of the Problem
December 11, 2023
For the following problems, assume that the snowballs are perfect spheres.
Chiquita and Rhonda are each building a snowman. Rhonda tells Chiquita that she is making each of her 3 balls of snow 20% smaller than the one below it. Rhonda starts by making the base ball with a diameter of 20 inches. If she reduces the diameter by 20% each time she makes a new ball, what is the diameter of the third ball she makes (the ball for the snowman’s head)? Express your answer as a decimal to the nearest tenth.
The diameter of the first ball is 20 inches, and we can think of a 20% reduction in diameter as meaning the next ball will have a diameter of 80% of the previous ball’s diameter. Thus, the diameter of the second ball is 0.8(20) = 16 inches. This means that the diameter of the third ball is 0.8(16) = 12.8 inches.
Chiquita planned to do the same thing as Rhonda, but she misunderstood what Rhonda meant when she said “…20% smaller than the one below it.” Chiquita made her first ball the same size as Rhonda but then decreased the volume by 20% for each subsequent ball. How much taller is Chiquita’s completed snowman than Rhonda’s completed snowman? Express your answer as a decimal to the nearest tenth.
First, let’s determine how big each of Chiquita’s snowballs is. We know the first ball has a diameter of 20 inches, thus its volume is (4/3)(20/2)3π = 1333.3333π cubic inches. This means that the second ball has a volume of 0.8(1333.3333π) = 1066.6667π cubic inches, and the third ball has a volume of 0.8(1066.6667π) = 853.3334π cubic inches. The diameter of the third ball can be found by determining its radius (V = (4/3)πr3) and then multiplying by 2. Thus, the diameter of the third ball is:
853.3334π = (4/3)πr3
r = 8.617 inches
d = 2(8.617) = 17.235 inches
Following the same process, we find that the second ball has a diameter of 18.566 inches. We already knew the diameter of the first ball to be 20 inches. By adding 20, 18.566 and 17.235, we find the height of Chiquita’s snowman to be 55.801 inches. We know the diameters of Rhonda’s snowballs from the previous question, so we can quickly find her snowman’s height to be 20 + 16 + 12.8 = 48.8 inches.
Thus, the difference in height is 55.801 – 48.8 = 7.0 inches, to the nearest tenth.
Rhonda’s little brother, Benjamin, comes outside and decides to make a snowman out of cubes of snow instead of balls. If each of his three cubes has a side length equal to the diameter of the corresponding ball on Rhonda’s snowman, how many more cubic inches of snow did Benjamin use than Rhonda? Express your answer as a decimal to the nearest tenth.
Again, based on the first question, we have the diameters of Rhonda’s snowballs (20, 16, and 12.8 inches). Since the volume of a cube is (side length)3, we can just cube these lengths to determine the volume of snow Benjamin used.
203 = 8000
163 = 4096
12.83 = 2097.152
8000 + 4096 + 2097.152 = 14,193.152
Now, let’s determine the volume of snow Rhonda used. Remember, for a sphere, V = (4/3)πr3.
(4/3)103π = 1333.333π
(4/3)83π = 682.667π
(4/3)6.43π = 349.525π
1333.333π + 682.667π + 349.525π = 7431.5159...
Thus, Benjamin used 14,193.152 – 7431.516 = 6761.6 cubic inches of snow, to the nearest tenth, more than Rhonda.
Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
Page 2 contains ONLY PROBLEMS. ♦
Math topic
CCSS (Common Core State Standard)
Difficulty | 1,017 | 3,458 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2024-30 | latest | en | 0.949198 |
https://stat.ethz.ch/pipermail/r-sig-geo/2007-March/001868.html | 1,579,670,918,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250606696.26/warc/CC-MAIN-20200122042145-20200122071145-00092.warc.gz | 673,163,960 | 2,367 | # [R-sig-Geo] (no subject)
Roger Bivand Roger.Bivand at nhh.no
Thu Mar 15 09:51:55 CET 2007
```On Thu, 15 Mar 2007, Roger Bivand wrote:
>
> >
> > On Tue, 13 Mar 2007, Michael Friendly wrote:
> >
> > > I've read the documentation, but can't find an example of how
> > > to specify a circular window for the ppp object that would contain the
> > > data and the window.
> >
> > To create a circular window in spatstat, you can use the
> > function disc(). See help(disc).
> >
> > For example suppose a point pattern with coordinates x, y (two vectors)
> > is observed inside a pot with centre (x0, y0) and radius R.
> > Then
> > pot <- disc(R, c(x0,y0))
> > creates a window (owin object) representing the pot,
> > and
> > X <- ppp(x, y, window=pot)
> > creates the point pattern.
> >
> > Since spatstat always plots spatial objects isometrically, you can also
> > use plot(disc(....)) just to draw a circle.
> >
> > > The data structure is something like a data frame for all the
> > > germinating seeds, recording (X,Y) location and maybe some other
> > > measure like size after xx days:
> > >
> > > PotID Treatment SeedType X Y size
> >
> > Currently, a point pattern can only have one mark variable
> > (e.g. you can attach the SeedType to each point, or attach the size to
> > each point, but not both).
> > This will change in spatstat version 2 which will be released soon.
> >
> > > (Is there something better?)
> >
> > There is nothing better than spatstat! %^]
>
> Great!
>
> To get to the sp Polygon object:
Should be:
res0 <- coordinates(disc(1, c(0,0))\$bdry)
res <- Polygon(rbind(res0, res0[1,]))
to turn the list into a matrix, sorry.
Roger
>
> res0 <- disc(1, c(0,0))\$bdry
> res <- Polygon(rbind(res0, res0[1,]))
>
> should anyone need it.
>
> Roger
>
>
> >
> > regards
> >
>
>
--
Roger Bivand
Economic Geography Section, Department of Economics, Norwegian School of | 576 | 1,904 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2020-05 | latest | en | 0.774155 |
https://cn.maplesoft.com/support/help/maple/view.aspx?path=ComputationalGeometry/SegmentsIntersect&L=C | 1,718,560,457,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861670.48/warc/CC-MAIN-20240616172129-20240616202129-00548.warc.gz | 149,851,092 | 23,678 | SegmentsIntersect - Maple Help
ComputationalGeometry
SegmentsIntersect
determine if two line segments intersect
Calling Sequence SegmentsIntersect( L1, L2 ) SegmentsIntersect( L1, L2, intersection ) SegmentsIntersect( M, locs ) SegmentsIntersect( M, locs, intersection )
Parameters
L1, L2 - line segments given as pairs of points in 2-D space M - an array of point coordinates in 2-D space. Each point is a row in the array. M must be C_order and datatype=float[8] locs - positive integer indices of the three rows of M to treat as input the first location is the point and the other two the ends of the segment
Description
• This command returns true if the two line segments intersect and false if they do not. It does so using four calls to PointOrientation.
• If the intersection option is given and the segments do not intersect false will be returned, otherwise the coordinate of the intersection will be returned.
• If one segment is completely contained within the other, the list of endpoints of the smaller segment will be returned.
Examples
> $\mathrm{with}\left(\mathrm{ComputationalGeometry}\right):$
> $\mathrm{SegmentsIntersect}\left(\left[\left[0,0\right],\left[1,1\right]\right],\left[\left[0,1\right],\left[1,0\right]\right]\right)$
${\mathrm{true}}$ (1)
> $L≔\left[\left[1,0\right],\left[1,5\right]\right]:$
> $K≔\left[\left[\left[1,-2\right],\left[1,-1\right]\right],\left[\left[0,1\right],\left[0,3\right]\right],\left[\left[2,1\right],\left[2,3\right]\right],\left[\left[1,6.5\right],\left[1,8\right]\right],\left[\left[1,6\right],\left[1.5,8\right]\right],\left[\left[0.5,4\right],\left[1.5,7\right]\right],\left[\left[0.5,0.5\right],\left[1,-0.75\right]\right],\left[\left[1,5\right],\left[1.5,6\right]\right],\left[\left[1,4.5\right],\left[1.5,5\right]\right],\left[\left[0.5,3.75\right],\left[1.5,3.5\right]\right],\left[\left[1,2.5\right],\left[1,3.5\right]\right],\left[\left[0.5,1.5\right],\left[1,2\right]\right],\left[\left[1,0\right],\left[1.5,-2\right]\right],\left[\left[1,0.5\right],\left[1,-0.5\right]\right]\right]:$
> $\mathrm{plots}:-\mathrm{display}\left(\mathrm{plottools}:-\mathrm{line}\left({L}_{[]},\mathrm{legend}="L",\mathrm{color}="Red"\right),\mathrm{seq}\left(\mathrm{plottools}:-\mathrm{line}\left({{K}_{i}}_{[]},\mathrm{legend}="K"||i,\mathrm{color}="Blue"\right),i=1..7\right),\mathrm{seq}\left(\mathrm{plottools}:-\mathrm{line}\left({{K}_{i}}_{[]},\mathrm{legend}="K"||i,\mathrm{color}="Green"\right),i=8..14\right),\mathrm{axes}=\mathrm{boxed}\right)$
None of the blue segments intersect L
> $\left\{\mathrm{seq}\left(\mathrm{SegmentsIntersect}\left(L,{K}_{i}\right),i=1..7\right)\right\}$
$\left\{{\mathrm{false}}\right\}$ (2)
All of the green segments intersect L
> $\left\{\mathrm{seq}\left(\mathrm{SegmentsIntersect}\left(L,{K}_{i}\right),i=8..14\right)\right\}$
$\left\{{\mathrm{true}}\right\}$ (3)
> $M≔\mathrm{Array}\left(\left[{L}_{[]},\mathrm{seq}\left({k}_{[]},k\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}∈\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}K\right)\right],\mathrm{datatype}={\mathrm{float}}_{8},\mathrm{order}=\mathrm{C_order}\right)$
> $\mathrm{SegmentsIntersect}\left(M,\left[1,2,3,4\right],\mathrm{intersection}\right)$
${\mathrm{false}}$ (4)
> $\mathrm{SegmentsIntersect}\left(M,\left[1,2,25,26\right],\mathrm{intersection}\right)$
$\left[\begin{array}{cc}{1.}& {2.}\end{array}\right]$ (5)
Line 11, completely contained in L
> $\mathrm{SegmentsIntersect}\left(M,\left[1,2,23,24\right],\mathrm{intersection}\right)$
$\left[\left[\begin{array}{cc}{1.}& {2.50000000000000}\end{array}\right]{,}\left[\begin{array}{cc}{1.}& {3.50000000000000}\end{array}\right]\right]$ (6)
Line 14, partially overlaps L
> $\mathrm{SegmentsIntersect}\left(M,\left[1,2,29,30\right],\mathrm{intersection}\right)$
$\left[\left[\begin{array}{cc}{1.}& {0.}\end{array}\right]{,}\left[\begin{array}{cc}{1.}& {0.500000000000000}\end{array}\right]\right]$ (7)
Compatibility
• The ComputationalGeometry[SegmentsIntersect] command was introduced in Maple 2019. | 1,444 | 4,038 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 19, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2024-26 | latest | en | 0.486012 |
https://estudyassistant.com/mathematics/question12848817 | 1,627,495,467,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153739.28/warc/CC-MAIN-20210728154442-20210728184442-00036.warc.gz | 256,844,961 | 17,955 | , 21.06.2019 16:40 dubouuu
# Which data set has a greater spread? why? set a: {38, 12, 23, 48, 55, 16, 18} set b: {44, 13, 24, 12, 56} has a greater spread because .
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Lesson 9-3 for items 8-10, a square is drawn in the coordinate plane, with vertices as shown in the diagram. then the square is reflected across the x-axis. -2 4 8. the function that describes the reflection is (x, y) - a. (x, y-3). b. (x, y-6). c. (-x, y). d. (x,- y). | 389 | 1,327 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2021-31 | latest | en | 0.926463 |
https://geoffneilsen.wordpress.com/tag/si/ | 1,511,528,178,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934808133.70/warc/CC-MAIN-20171124123222-20171124143222-00436.warc.gz | 622,396,597 | 45,638 | JAMES PRESCOTT JOULE (1818- 89)
1843 – England
‘A given amount of work produces a specific amount of heat’
4.18 joules of work is equivalent to one calorie of heat.
In 1798 COUNT RUMFORD suggested that mechanical work could be converted into heat. This idea was pursued by Joule who conducted thousands of experiments to determine how much heat could be obtained from a given amount of work.
Even in the nineteenth century, scientists did not fully understand the properties of heat. The common belief held that it was some form of transient fluid – retained and released by matter – called CALORIC. Gradually, the idea that it was another form of energy, expressed as the movement of molecules gained ground.
Heat is now regarded as a mode of transfer of energy – the transfer of energy by virtue of a temperature difference. Heat is the name of a process, not that of an entity.
Joule began his experiments by examining the relationship between electric current and resistance in the wire through which it passed, in terms of the amount of heat given off. This led to the formulation of Joule’s ideas in the 1840s, which mathematically determined the link.
Joule is remembered for his description of the conversion of electrical energy into heat; which states that the heat (Q) produced when an electric current (I) flows through a resistance (R) for a time (t) is given by Q=I2Rt
Its importance was that it undermined the concept of ‘caloric’ as it effectively determined that one form of energy was transforming itself into another – electrical energy to heat energy. Joule proved that heat could be produced from many different types of energy, including mechanical energy.
JAMES PRESCOTT JOULE
Joule’s apparatus to show equivalence of work and heat
Joule was the son of a brewer and all his experiments on the mechanical equivalent of heat depended upon his ability to measure extremely slight increases in temperature, using the sensitive thermometers available to him at the brewery. He formulated a value for the work required to produce a unit of heat. Performing an improved version of Count Rumford’s experiment, he used weights on a pulley to turn a paddle wheel immersed in water. The friction between the water and the paddle wheel caused the temperature of the water to rise slightly. The amount of work could be measured from the weights and the distance they fell, the heat produced could be measured by the rise in temperature.
Joule went on to study the role of heat and movement in gases and subsequently with WILLIAM THOMSON, who later became Lord Kelvin, described what became known as the ‘Joule-Thomson effect’ (1852-9). This demonstrated how most gases lose temperature on expansion due to work being done in pulling the molecules apart.
Thomson thought, as CARNOT had, that heat IN equals heat OUT during a steam engine’s cycle. Joule convinced him he was wrong.
The essential correctness of Carnot’s insight is that the work performed in a cycle divided by heat input depends only on the temperature of the source and that of the sink.
Synthesising Joule’s results with Carnot’s ideas, it became clear that a generic steam engine’s efficiency – work output divided by heat input – differed from one (100%) by an amount that could be expressed either as heat OUT at the sink divided by heat IN at the source, or alternatively as temperature of the sink divided by temperature of the source. Carnot’s insight that the efficiency of the engine depends on the temperature difference was correct. Temperature has to be measured using the right scale. The correct one had been hinted at by DALTON and GAY-LUSSAC’s experiments, in which true zero was minus 273degrees Celsius.
A perfect cyclical heat engine with a source at 100degrees Celsius and a sink at 7degrees has an efficiency of 1 – 280/373. The only way for the efficiency to equal 100% – for the machine to be a perfect transformer of heat into mechanical energy – is for the sink to be at absolute zero temperature.
Joule’s work helped in determining the first law of thermodynamics; the principle of the conservation of energy. This was a natural extension of his work on the ability of energy to transform from one type to another.
Joule contended that the natural world has a fixed amount of energy which is never added to nor destroyed, but which just changes form.
The SI unit of work and energy is named the joule (J).
Manchester Museum of Science & Industry
TIMELINE
HEAT
LUIGI GALVANI (1737- 98) ALESSANDRO VOLTA (1745-1827)
1791 & 1799 – Italy
‘Galvani: An electric current is produced when an animal tissue comes into contact with two different metals.
Volta: An electric current is not dependent on an animal tissue and can be produced by chemicals’
Galvani was wrong and Volta was right.
Galvani had found that by touching a dead frog’s legs with two different metal implements, the muscles in the frog’s legs would twitch. Galvani wrongly concluded it was the animal tissue that was storing the electricity, releasing it when touched by the metals. He felt he had discovered the very force of life – ‘animal electricity’ – that animated flesh and bone.
GALVANI
Soon dozens of scientists were trying to bring corpses back to life by electrifying them. Volta was not convinced the animal muscle was the important factor in the production of the current.
He repeated Galvani’s experiments and concluded, controversially at the time, the different metals were the important factor.
A bitter dispute arose as to whose interpretation was correct. Volta began putting together different combinations of metals to see if they produced any current; later he produced a wet battery of fluid and metals. Volta’s method of producing electric current involved using discs of silver and zinc dipped in a bowl of salt solution. He reasoned that a much larger charge could be produced by stacking several discs separated by cards soaked in salt water – by attaching copper wires to each end of the ‘pile’ he successfully obtained a steady current.
The ‘voltaic pile’ was the first battery in history (1800). Napoleon Bonaparte, who at the time controlled the territory in which Volta lived, was so impressed he made him a Count and awarded him the Legion d’Honour.
VOLTA
Volt, the SI unit of electric potential, honours Volta.
Although Galvani’s theory on ‘animal electricity’ was not of any major importance, he has also achieved nominal immortality; like ‘volt’, the words ‘galvanic’ (sudden and dramatic), ‘galvanised’ (iron or steel coated with zinc) and ‘galvanometer’ (an instrument for detecting small currents) have become part of everyday language.
A volt is defined as the potential difference between two points on a conductor carrying one ampere current when the power dissipated between the points is one watt.
ANDRE MARIE AMPERE (1775-1836)
1827 – France
‘Two current-carrying wires attract each other if their currents are in the same direction, but repel each other if their currents are opposite.
The force of attraction or repulsion (magnetic force) is directly proportional to the product of the strengths of the currents and inversely proportional to the square of the distance between them’
ANDRE AMPERE
Another addition to the succession of ‘inverse-square’ laws begun with NEWTON’s law of universal gravitation.
Ampere had noted that two magnets could affect each other and wondered, given the similarities between electricity and magnetism, what effect two currents would have upon each other. Beginning with electricity run in two parallel wires, he observed that if the currents ran in the same direction, the wires were attracted to each other and if they ran in opposite directions they were repelled.
He experimented with other shapes of wires and generalised that the magnetic effect produced by passing a current in an electric wire is the result of the circular motion of that current. The effect is increased when the wire is coiled. When a bar of soft iron is placed in the coil it becomes a magnet. This is the solenoid, used in devices where mechanical motion is required.
Ampere exploited OERSTED’s work, devising a galvanometer which measured electric current flow via the degree of deflection upon its magnetic needle.
He attempted to interpret all his results mathematically in a bid to find an encompassing explanation for what later became referred to as electromagnetism (Ampere had at that time christened it electrodynamics), resulting in his 1827 definition.
Ampere’s name is commemorated in the SI unit of electric current, the ampere.
TIMELINE
ELECTRICITY
1811 – Italy
‘Equal volumes of all gases at the same temperature and pressure contain the same number of molecules’
In 1811, when Avogadro proposed his HYPOTHESIS, very little was known about atoms and molecules. Avogadro claimed that the same volume of any gas under identical conditions would always contain the same number of fundamental particles, or molecules. A litre of hydrogen would contain exactly the same number of molecules as a litre of oxygen or a litre of carbon dioxide.
In 1814 ANDRE AMPERE was credited with discovering that if a gas consisted of a single element, its atoms could clump in pairs. The molecules of oxygen consisted of pairs of oxygen atoms, and the molecules of chlorine, pairs of chlorine atoms.
Diatomic gases possess a total of six degrees of simple freedom per molecule that are related to atomic motion.
This provides a way of comparing the weights of different molecules. It was only necessary to weigh equal volumes of different gases and compare them. This would be exactly the same as comparing the weights of the individual molecules of each gas.
Avogadro realised that GAY-LUSSAC‘s law provided a way of proving that an atom and a molecule are not the same. He suggested that the particles (molecules) of which nitrogen gas is composed consist of two atoms, thus the molecule of nitrogen is N2. When one volume (one molecule) of nitrogen combines with three volumes (three molecules) of hydrogen, two volumes (two molecules) of ammonia, NH3, are produced.
N2 + 3H2 ↔ 2NH3
However, the idea of a molecule consisting of two or more atoms bound together was not understood at that time.
Avogadro’s law was forgotten until 1860 when the Italian chemist STANISLAO CANNIZZARO (1826-1910) explained the necessity of distinguishing between atoms and molecules.
From Avogadro’s law it can be deduced that the same number of molecules of all gases at the same temperature and pressure should have the same volume. This number has been determined experimentally: it’s value is 6.022 1367(36) × 1023AVOGADRO’S NUMBER
That at the same temperature and pressure, equal volumes of all gases have the same number of molecules allows a simple calculation for the combining ratios of all gases – by measuring their percentages by volume in any compound. This in turn facilitates simple calculation of the relative atomic masses of the elements of which it is composed.
TIMELINE
CHEMISTRY | 2,533 | 11,059 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2017-47 | longest | en | 0.969774 |
http://www.physicsforums.com/showthread.php?t=61505 | 1,410,753,821,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657104119.19/warc/CC-MAIN-20140914011144-00013-ip-10-196-40-205.us-west-1.compute.internal.warc.gz | 750,927,180 | 8,567 | # Black Hole: Infinite Density, Infinite Buoyancy?
by Ryan H
Tags: black, buoyancy, density, hole, infinite
P: 15 If a black hole has an infinite density, then one would think that anything would float inside of it. And since it's infinitely dense, the object(s) being pulled in would have an infinite buoyancy, causing it to be shot back out of the black hole at a seemingly infinite speed. So why don't these two forces cancle each other out?
P: 482 quantum effects probably prevent a singularity from reaching zero/infinite parameters
P: 8,470
Quote by Ryan H If a black hole has an infinite density, then one would think that anything would float inside of it. And since it's infinitely dense, the object(s) being pulled in would have an infinite buoyancy, causing it to be shot back out of the black hole at a seemingly infinite speed. So why don't these two forces cancle each other out?
Only the singularity at the center would have infinite density according to general relativity, inside the event horizon is empty space (and infalling matter) just like outside the event horizon.
P: 989 Black Hole: Infinite Density, Infinite Buoyancy? 'non-rotating' Classical Universe-Schwarzschild Singularity Density solution for a one-dimesional 'point-like' object: $$\boxed{\rho_u = \frac{M_u}{2} \sqrt{ \frac{c^3}{\hbar G}}}$$ BH Singularity Density infinities do NOT exist. Reference: http://www.physicsforums.com/showpos...2&postcount=24
Emeritus Sci Advisor PF Gold P: 2,021 There is no "inside" to a BH singularity (point).
P: 8,470
Quote by Orion1 'non-rotating' Classical Universe-Schwarzschild Singularity Density solution for a one-dimesional 'point-like' object: $$\boxed{\rho_u = \frac{M_u}{2} \sqrt{ \frac{c^3}{\hbar G}}}$$ BH Singularity Density infinities do NOT exist. Reference: http://www.physicsforums.com/showpos...2&postcount=24
Is this equation something you derived yourself based on your own ideas, or did you get it from a textbook or something written by a professional physicist? The derivation you gave on that thread seemed to involve both QM and GR...would you agree that according to classical GR alone, the singularity has infinite density?
Sci Advisor PF Gold P: 9,478 I would argue the Planck density is the limit in the physical universe.
P: 989
Quote by JesseM Is this equation something you derived yourself based on your own ideas, or did you get it from a textbook or something written by a professional physicist?
$$\boxed{\rho_u = \frac{M_u}{2} \sqrt{ \frac{c^3}{\hbar G}}}$$
The solution for this 'non-rotating' classical schwarzschild singularity density for a one dimensional 'point-like' object was derived by me based on research on these physical models.
Note that the Schwarzschild solution is only a solution for Schwarzschild BHs with zero angular momentum , this is a highly improbable state.
Neutron star spin increases with increased density, therefore an object generating in the core of a neutron star or supernova without spin is...impossible. Only BHs with angular momentum can exist in the Universe, a rotating Kerr BH.
Quote by JesseM Would you agree that according to classical GR alone, the singularity has infinite density?
The Classical General Relativity model is based upon four total dimensional space-time $$n_t = 3 + 1 = 4$$ (3 space + 1 time). Solutions for models for that contain dimensions of less than four are not valid solutions in the Universe.
The classical solution stated for 1 dimension is actually 2 dimensions $$n_t = 1 + 1 = 2$$ (1 space + 1 time), because solutions with with a total dimensional range of less than 4 $$n_t < 4$$ cannot exist in the Universe, all solutions for total dimensional ranges between 0 and 3 are not real valid solutions because they cannot exist in a four total dimensional General Relativity Universe.
Classical General Relativity models based upon 0 to less than 2 total dimensions are typical of producing solutions with 'infinities', and is only a division by zero in an 'undefined' model.
This solution is based upon 2 dimensional space, the singularity described 'exists' in only 2 space dimensions (and 1 time) $$n_t = 2 + 1 = 2$$ (2 space + 1 time). $$n_t = n_s + n_t$$.
Classical Schwarzschild Singularity Dimension Number:
$$n_s = 2$$ - dimension #
$$dV_s = \pi r_p^2$$ - volume
$$L = 0$$ - angular momentum
Solution for 'non-rotating' Classical Schwarzschild Singularity Density for a two dimensional 'point-like' object: (flat disc)
$$\rho_s = \frac{dM_s}{dV_s} = \frac{dM_s}{\pi r_p^2} = \frac{M_u c^3}{\pi \hbar G}$$
$$\boxed{\rho_u = \frac{M_u c^3}{\pi \hbar G}}$$
$$\begin{picture}(100,100)(0,0) \put(0,0){\circle{3}} \put(0,0){\line(1,0){100}} \put(0,0){\line(0,1){100}} \put(0,33){\circle{3}} \put(34,55){\circle{3}} \put(67,77){\textcolor{red}{\circle{3}}} \put(67,77){\textcolor{blue}{\circle{6}}} \put(100,100){\circle{3}} \put(100,5){{n}} \put(5,100){{ln p}} \end{picture}$$
Quote by Chronos I would argue the Planck density is the limit in the physical universe.
In a four dimensional space-time physical Universe, the average Planck density is a solution and a physical 'limit' in the Universe.
Based upon the current logarithmic slope in the chart, at what density value does the slope cross the y-intercept?
Reference:
http://www.physicsforums.com/showpos...2&postcount=24
Related Discussions General Physics 17 Calculus & Beyond Homework 7 General Math 34 Classical Physics 17 Astronomy & Astrophysics 11 | 1,436 | 5,470 | {"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} | 3.0625 | 3 | CC-MAIN-2014-41 | latest | en | 0.908544 |
https://study.com/academy/lesson/kinematics-of-human-motion.html | 1,571,871,328,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987836368.96/warc/CC-MAIN-20191023225038-20191024012538-00027.warc.gz | 732,552,472 | 36,917 | # Kinematics of Human Motion
Instructor: John Hamilton
John has tutored algebra and SAT Prep and has a B.A. degree with a major in psychology and a minor in mathematics from Christopher Newport University.
Are you fascinated by watching the seemingly endless variety of human motions? In this lesson, we discuss the kinematics of human motion, including translatory and rotational motion, the three major planes of motion, and several anatomical terms of motion. These include flexion versus extension, circumduction, and pronation and supination.
## Human Motion
Many of us have been enthralled while watching world-class gymnasts perform seemingly impossible maneuvers on television, or been thrilled at a stellar dance performance. These events involve varying degrees of human motion to execute them.
Newton's first law states that, in general, an object at rest stays in rest, while an object in motion stays in motion. This is also known as inertia and definitely applies to the human body and human motion.
The study of human motion is a branch of biomechanics known as kinematics. Kinematics specifically studies just pure motion and not the actual forces which cause the motion. There are several types of motions exhibited by the human body.
## Translatory & Rotational Motion
Most human motion involves a combination of two types of motion, translatory motion, also known as linear motion, and rotational motion, also known as angular motion. Think of a long jumper sprinting toward the sandy pit. He is exhibiting translatory motion as he runs in a straight line. Then he takes off and swings his arms to try and stay airborne, which is a form of rotational motion. Together, the combined movements are general motion.
## Three Planes of Motion
The human body is three-dimensional (it has height, width, and depth), although it often just moves in two dimensions in what is known as planar motion. While space is three-dimensional, a plane is simply a surface of two-dimensions that extends to infinity (forever.) Therefore, planar motion is motion that involves only two of the three dimensions. The human body moves through three distinct planes:
• Frontal Plane - Did you ever have to do jumping jacks in gym class? Those were an example of movement in the frontal plane. With jumping jacks, there is both side-to-side movement of the arms and legs and also an up-and-down movement; however, there is no third movement of forward-and-backward.
• Sagittal Plane - Movement in the sagittal plane involves bending forward and backward along with up and down; however, there is very little side-to side motion. These movements tend to be the most common used when exercising. Lunges and curls are two notable examples.
• Transverse Plane - Movement in the transverse plane tends to be a little more complex because it often involves rotational movements. A common example would be to do lunges while holding a medicine ball, but this time step diagonally instead of forward. First step diagonally to the left. Then, return the feet to the original position. Then, step diagonally to the right. Do you see how this would involve a twisting, or rotational, motion of the body? Within this plane, the body moves up/down, forward/backward, and side to side.
## Anatomical Terms for Motion
Now let us expound on some of the different terms for motion. These movements can be at angles, in circles, and from one side to another. They also range from the head and neck to the torso and the limbs.
### Flexion versus Extension
Flexion is simply a decreasing of the angle between two given body parts. Extension is the opposite, and thus is the increasing of the angle. During a bicep curl, you would bring your palm and the dumbbell toward your shoulder, which would decrease the angle between your upper arm and your forearm. This is an example of flexion, or more familiarly stated, you are flexing your muscle. When you lower the dumbbell back down, you increase the angle between your upper arm and your lower arm. This would be an example of extension.
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http://www.docstoc.com/docs/14811108/Lab-Manual---PSIT-PSAT | 1,386,395,878,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163053608/warc/CC-MAIN-20131204131733-00092-ip-10-33-133-15.ec2.internal.warc.gz | 394,653,485 | 17,159 | # Lab Manual - PSIT-PSAT
Document Sample
``` Numerical Techniques Laboratory Manual
[For Specific Focus to Laboratory Work]
Name of Faculty
Name of the Department
PSITPranveer Singh Institute of Technology
List of Content
S.No. Content 1. Index 2. Introduction 3. Page No.
PSITPranveer Singh Institute of Technology
INDEX
S. No.
1. 2. 3.
Roll No…………………
Name of Experiment
Page No.
Exp. No.
Date
Signature
Remarks
PSITPranveer Singh Institute of Technology
Experiment No.
Date of Experiment ___________ Grade_________Faculty’s Signature_________Date___________
Objective
To find solution of simultaneous linear equations using Gauss Elimination method and Gauss Jordan method. Gauss Elimination method: in this method, the unknowns are eliminated successively and the system is reduced to an upper triangular system from which the unknowns are found by back substitution. The method is quite general and is well – adapted for computer operations. Here we shall explain it by considering a system of three equations for sake of clarity. Consider the equations a1x + b1y + c1 z = d1 a2x + b2y + c2 z = d2 a3x + b3y + c3 z = d3
……………….(1)
Step-1. To eliminate x from second and third equations. Assuming a1 ≠0, we eliminate x from the second equation by subtracting (a 2/a1) times the first equation from the second equation. Similarly we eliminate x from the third equation by subtracting (a3 /a1) times the first equation from the third equation. We thus, get the new system a1x + b1y + c1 z = d1 b2 'y + c2 'z = d2 ' b3 'y +c3 'z = d3" …………………(2) Here the first equation is called the pivotal equation and a 1 is called the first pivot. Step 2. To eliminate y from third equation in (2). Assuming b2 '≠0, we eliminate y from the third equation of (2), by subtracting (b3 '/b2 ') times the second equation from the third equation. We thus, get the new system a1 x + b1 y + c1 z = d1 b2 y + c2 'z = d2 ' c3 "z = d3" ……....…………(3) Here the second equation is the pivotal equation and b2 ' is the new pivot. Step 3. To evaluate the unknown. The values of x,y,z are found from the reduced system (3) by back substitution. Gauss-Jordan method: This in a modification of the Guass-Elimination method. In this Method, elimination of unknowns is performed not in the equation below but in the equation above also, ultimately reducing the system to a diagonal matrix form i.e. each equation involving only one unknown. From these equations the unknowns x,y,z can be obtained readily. Thus, in this method, the labour of back-substitution for finding the unknown is saved at the cost of additional calculation.
PSITPranveer Singh Institute of Technology
Note: For a system of 10 equations, the number of multiplications required for Guass-Jordan method is about 500 whereas for Guass-Elimination method, we need only 333 multiplications. This implies that though Guass-Jordan method appears to be easier but required 50% more operations than the Guass-Elimination method.
Algorithm Guass-Elimination Method 1. Assign float space for a[n] [n+1], x[n], t and s. 2. Assign integer space for i, j, k. 3. Assign i=0, j=0. 4. Accept value of a[i] [j]. 5. Repeat step (4) & increment I & j by 1 till i<n, j<n+1. 6. Assign j=0 & i=j+1. 7. Assign t with a[i] [j]/a[j] [j]. 8. Assign k=0 and repeat step (9) till k<n+1, increment k by 1. 9. a[i] [j] -= a[j] [k]*t. 10. Increment value of I with 1. 11. Repeat steps (7) to (10) till i<n. 12. Repeat step (7) to (11) till j<(n-1) and incrementing by 1 each time. 13. Assign i=0, j=0. Accept a[i] [j], increment i & j with 1. 14. Repeat step (13) till i<n, j<n+1. 15. Assign i with n-1, s with 0. 16. Assign j with i+1. 17. Assign s with a[i] [j]*x[j]. 18. Assign x[i] = (a[i] [n]-s)/a[i] [j]. 19. Decrement I by 1 & increment j by 1 till j<n & i>=0. (Repeat steps 16 to 19) 20. Display value of i+1 7 x[i], I varies from 0 to n-1.
PSITPranveer Singh Institute of Technology
Program for Guass-Elimination Method
Output (Self Input)
PSITPranveer Singh Institute of Technology
Algorithm for Guass-Jordan Method
Program for Guass-Jordan Method
Output (Self Input)
PSITPranveer Singh Institute of Technology
Important Questions Q.1 What is difference between Guass’s Seidal iteration and Guass’s Elimination method?
Q. 2
What is pivoting? Distinguish between partial pivoting and complete pivoting?
Q.3
Compare critically Guass Elimination and Guass Jordan methods of solving simultaneous equations?
Q.4
Define upper triangular and lower triangular matrices.
Q.5
What are elementary row transformation?
PSITPranveer Singh Institute of Technology
Experiment No.
Date of Experiment ___________ Roll No. ____________________ Group No._______________ Date of submission ___________Grade (if any)____________Faculty’s Signature_____________
Aim Write a program to implement quick sort technique Theory This sort is also called partition exchange sort. The quick sort is an in-place, divide-and-conquer, massively recursive sort. As a normal person would say, it’s essentially a faster in-place version of the merge sort. The quick sort algorithm is simply in theory, but very difficult to put into code (computer scientists tied themselves into knots for years trying to write a practical implementation of the algorithm, and it still has that effect on university students). The recursive algorithm consists of four steps (which closely resemble the merge sort): 1. If there are one or less elements in the array to be sorted, return immediately. 2. Pick an element in the array to serve as a “pivot” point. (Usually the left-most element in the array is used.) 3. Split the array into two parts – one with elements larger than the pivot and the other with elements smaller than the pivot. 4. Recursively repeat the algorithm for both halves of the original array. The efficiency of the algorithm is majorly impacted by which element is chosen as the pivot pint. The worst-case efficiency of the quick sort, O(n2), occurs when the list is sorted and the left-most element is chosen. Randomly choosing a pivot point rather than using the left-most element is recommended, if the data to be sorted isn’t random. As long as the pivot point is chosen randomly, the quick sort has an algorithmic complexity of O(n log n). Pros: Extremely fast. Cons: Very complex algorithm, massively recursive. The quick sort is by far the fastest of the common sorting algorithms. It’s possible to write a special purpose sorting algorithm that can beat the quick sort for some data sets, but for general general-case sorting there isn’t anything faster. As soon as students figure out, their immediate impulse is to use the quick sort for everything – after all, faster is better, right? It’s important to resist this urge – the quick sort isn’t always the best choice. As mentioned earlier, it’s massively recursive (which means that for very large sorts, you can run the system out of stack space pretty easily). It’s also a complex algorithm – a little too complex to make it practical for a one-time sort of 25 items, for example.
PSITPranveer Singh Institute of Technology
With that said, in most cases the quick sort is the best choice if speed is important (and it almost always is). Use it for repetitive sorting, sorting of medium to large lists, and as a default choice when you’re not really sure which sorting algorithm to use. Ironically, the quick sort has horrible efficiency when operating on lists that are mostly sorted in either forward or reverse order – avoid it in those situations. Algorithm 1. Quick Sort (A,p,r) 2. then q<-- PARTITION(A,p.r) 3. QUICK SORT (A,p,r) 4. QUICK SORT (A,q+1,r)
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PSITPranveer Singh Institute of Technology | 1,978 | 7,766 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2013-48 | longest | en | 0.850144 |
https://schoollearningcommons.info/question/should-added-to-5-to-get-what-ulle-8-19676230-58/ | 1,638,679,915,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363135.71/warc/CC-MAIN-20211205035505-20211205065505-00608.warc.gz | 570,325,498 | 13,363 | should added to -5 to get What ulle – 8
Question
get
What ulle
8
in progress 0
2 months 2021-10-09T18:55:30+00:00 2 Answers 0 views 0
1. Step-by-step explanation:
Step 1: Write it as an equation
x + (-5) = 8
Step 2: Transpose -5 to RHS
x = 8 + 5
x = 13
Therefore we should add 13 to -5 to get 8
Step 3: Check it
13 +(-5) = 8
8=8
Hence, checked
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Given;
one number = -5
The result = -8
Find ;
The other number = Let the number be x
According to the given conditions
-5 + x = -8 [ transposing -5 ]
= x = -8 + 5
= x = -3
Checking
-5 + -3 = -8 | 306 | 842 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-49 | latest | en | 0.695247 |
https://www.jiskha.com/display.cgi?id=1293934276 | 1,516,399,467,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084888135.38/warc/CC-MAIN-20180119204427-20180119224427-00041.warc.gz | 925,230,391 | 3,881 | # Math
posted by .
A baseball player hits a pitch when the ball is 2 feet off the ground. Two hundred feet from the plate, the ball reaches its maximum height of 82 feet. Will the ball clear the home run wall that is 10 feet high and 380 feet from home plate?
• Math -
The equation below models the height in feet, h, of a softball t seconds after it is hit by a batter.
mc023-1.jpg
Why is the height of the ball a function of time?
Each value of time is associated with exactly one height.
Each height of the ball is associated with exactly one moment in time.
There is exactly one time at which the maximum height is attained.
There is a maximum and minimum height that the ball can reach.
• Math -
There is a maximum and minimum height that the ball can reach.
• Math -
Which equation is a function of x?
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http://guidelines.beefimprovement.org/index.php?title=Prediction_Bias&diff=prev&oldid=2079 | 1,618,905,329,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039379601.74/warc/CC-MAIN-20210420060507-20210420090507-00444.warc.gz | 45,078,708 | 7,901 | # Difference between revisions of "Prediction Bias"
Generally, bias is the systematic under or over estimatation of what is being estimated or predicted. Bias can exist in EPDs and accuracy values from many sources, including selective reporting, inaccurate measurements, approximation methods, incorrect models, incorrect variance components, and others.
## Estimating Bias
If we had both the true progeny difference (TPD) and an estimate (EPD) of the TBD, then we could calculate the degree of bias in our estimate as the difference in the mean TBD and mean EPD. However, we never observe the TPD. Instead we estimate it using pedigree, performance, and genomic data.
We can approximate the degree of bias and under/over dispersion of EPD by using regression techniques[1] [2]. One such way to do this is to regress the EPD with more information (e.g., genomic EPD) on the EPD with less information (e.g, pedigree-based EPD). Our expectation is that the intercept from this regression is 0 (no bias) given the properties of Best Linear Unbiased Prediction and the slope of the regression is 1 (no over or under dispersion).
A fundamental assumption is that the ratio of variance components used to generate both sets of EPD are the same. if they are not, then the expectation of the regression coefficient being 1 no longer holds.
Another approach is to regress phenotypes after being corrected for systematic effects on EPD. Here the expectation of the regression coefficient is 2. If EBV were used instead of EPD the expectation of the regression coefficient would be 1.
A key assumption is that the phenotype of the individual is not included in the EPD of that individual. Consequently, this approach lends itself to cross-validation or forward-in-time validation strategies whereby some set(s) of animals have their phenotypes masked in the genetic evaluation.
In similar fashion, average progeny performance (corrected for systematic effects) can be regressed on parent (sire) EPD. This is done annually at the US Meat Animal Research Center as part of the process to update across-breed EPD adjustment factors. The expectation of the regression coefficient is 1 in this case and assumes that the progeny information used is not part of the sire's EPD. A regression coefficient of less than 1 suggests that the EPD are over-dispersed meaning that a one unit change in EPD will generate less than a one unit change in average progeny phenotypes.
## Sources of Bias
Bias generally arises from incomplete information. For example, if selection takes place early in life (e.g., based on weaning weight) such that a non-random group of animals is culled, then subsequent weight trait EPD (e.g., yearling weight) could be biased. This issue can be accommodated through the use of Multiple-Trait Evaluation. Another example is incomplete recording of animals within a contemporary group. If only the heaviest animals are reported, then their performance relative to their contemporaries (e.g., contemporary group deviations) is biased downward because the observed average for the group is artificially inflated.
## References
1. Reverter, A., B. L. Golden, R. M. Bourdon, and J. S. Brinks. 1994. Technical Note: Detection of Bias in Genetic PredictionsJ. Anim. Sci. 72:34-37.
2. Legarra, A., and A. Reverter. 2018. Semi-parametric estimates of population accuracy and bias of predictions of breeding values and future phenotypes using the LR method. Genetics Selection Evolution. 40:53. | 745 | 3,500 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2021-17 | latest | en | 0.920633 |
https://math.stackexchange.com/questions/4278152/degrees-of-freedom-of-a-line-segment-in-3d | 1,718,817,956,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861828.24/warc/CC-MAIN-20240619154358-20240619184358-00622.warc.gz | 339,211,540 | 36,141 | # Degrees of freedom of a line segment in 3D
Suppose there is a line segment of fixed length $$l$$ in 3$$D$$. THen How many degrees of freedom does it have ?
To specify the line segment first we need to specify one endpoint which requires 3 parameters. Now the other endpoint lies in the circumference of the sphere of radius $$l$$ whose centre is the first end point. Now to choose the second endpoint we still need all 3 parameters ($$r,\theta,\phi$$) because for a specific $$\theta,\phi$$ there can be two solutions in which specifying $$r$$ makes it clear. Hence a line segment should have 6 degree of freedoms.
Am I right ?
• Duplicate question, see math.stackexchange.com/questions/1970754/… Commented Oct 16, 2021 at 8:44
• its not a duplicate question. I am asking about line segment not a line. Commented Oct 16, 2021 at 8:45
• You don't have two solutions for a specific θ, ϕ. But more importantly, even if you did, that would not constitute a degree of freedom. Commented Oct 16, 2021 at 10:48
Okay so think about it like this: You specify one point $$p$$ as the start for your line. As you noted you can define the other point as laying on the sphere or radius $$l$$ around $$p$$ - so naturally in this spherical coordinate system you'll only need two parameters to describe the other point and thus end up with $$5$$ degrees of freedom.
Another way (it's very similar though) to arrive at this answer is the following: Lets first disregard that you want to fix the length. Note that the general parametrisation for a line is $$\gamma : \Bbb R \to \Bbb R^3, t \mapsto t v + a$$ for some $$a, v \in \Bbb R^3$$. You can easily pick out a (directed) segment by restricting the domain (requiring two parameters for a total of 6), or by fixing the length of $$v$$ to some value and fixing the domain of $$\gamma$$ to $$[0,1]$$. So a directed line segment is uniquely identified by the parametrisation $$\gamma : [0, 1] \to \Bbb R^3, t \mapsto t v + a, \quad \text{for }a,v\in \Bbb R^3.$$ So we have three degrees of freedom for $$a$$ and three for $$v$$ for a total of $$6$$. By specifying the length we drop one degree of freedom and end up with $$5$$ - we can achieve the same by restricting $$v$$ to $$lS^2 \subseteq \Bbb R^3$$. Note that for any such $$\gamma$$ the function $$\widetilde{\gamma}(t) = t (-v) + \gamma(1)$$ will be the same line segment, just traversed backwards. But this doesn't change the number of degrees of freedom (see linked post). | 664 | 2,471 | {"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": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 24, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2024-26 | latest | en | 0.871421 |
https://www.doubtnut.com/qna/642571567 | 1,726,562,050,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651750.27/warc/CC-MAIN-20240917072424-20240917102424-00528.warc.gz | 674,733,733 | 34,910 | # The perimeter of a certain sector of a circle of radius 5.7 m is 27.2 m. Find the area of the sector.
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## To find the area of the sector of a circle given its perimeter and radius, we can follow these steps:1. Identify Given Values: - Radius of the circle (r) = 5.7 m - Perimeter of the sector (P) = 27.2 m2. Understand the Formula for the Perimeter of a Sector: The perimeter of a sector is given by the formula: P=r+r+L=2r+L where L is the length of the arc of the sector.3. Substitute the Known Values into the Perimeter Formula: Substitute the radius into the perimeter formula: 27.2=2(5.7)+L4. Calculate 2r: Calculate 2×5.7: 2×5.7=11.45. Set Up the Equation: Substitute 2r back into the perimeter equation: 27.2=11.4+L6. Solve for L: Rearranging the equation to find L: L=27.2−11.4 Calculate L: L=15.8 m7. Use the Area Formula for the Sector: The area A of the sector is given by the formula: A=12×r×L8. Substitute the Values for Area Calculation: Substitute r and L into the area formula: A=12×5.7×15.89. Calculate the Area: First calculate 5.7×15.8: 5.7×15.8=90.06 Then calculate the area: A=12×90.06=45.03 m210. Conclusion: Therefore, the area of the sector is: Area of the sector=45.03 m2
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Doubtnut is the perfect NEET and IIT JEE preparation App. Get solutions for NEET and IIT JEE previous years papers, along with chapter wise NEET MCQ solutions. Get all the study material in Hindi medium and English medium for IIT JEE and NEET preparation | 645 | 2,246 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2024-38 | latest | en | 0.801991 |
https://www.math-only-math.com/worksheet-on-division-of-rational-numbers.html | 1,713,174,752,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816954.20/warc/CC-MAIN-20240415080257-20240415110257-00074.warc.gz | 793,210,549 | 14,047 | # Worksheet on Division of Rational Numbers
Practice the questions given in the worksheet on division of rational numbers. We know, for any two rational numbers a/b and c/d (not equals to 0), we have a/b ÷ c/d = a/b × d/c.
1. Divide the rationals:
(i) 1 by 1/2
(ii) 5 by -5/7
(iii) -3/4 by 9/-16
(iv)-7/8 by -21/16
(v) 7/-4 by 63/64
(vi) 0 by -7/5
(vii) -3/4 by -6
(viii) 2/3 by -7/12
2. Simplify:
(i) 4/9 ÷ -5/12
(ii) -8 ÷ (-7/16)
(iii) -12/7 ÷ (-18)
(iv) (-1/10) ÷ (-8/5)
(v) (-16/35) ÷ (-15/14)
(vi) {(-65/14) ÷ (13/7)}
3. The product of two rational numbers is 15. If one of the numbers is -10, find the other.
4. The product of two rational numbers is -9. If one of the numbers is -12, find the other.
5. The product of two rational numbers is -16/9. If one of the numbers is -4/3, find the other.
6. By what rational number should we multiply -15/56 to get -5/7?
7. By what rational number should -8/39 be multiplied to obtain 1/26?
8. By what number should -33/8 be divided to get -11/2?
9. The product of two rational numbers is -8/9. If one of the numbers is -4/15, find the other.
10. By what number should we multiply -1/6, so that the product may be -23/9?
11. By what number should -33/16 be divided to get -11/4?
12. Fill in the blanks:
(i) 9/8 ÷ (_____) = -3/2
(ii) (_____) ÷ (-7/5) = 10/19
(iii) (_____) ÷ (-3) = -4/15
(iv) (-12) ÷ (_____) = -6/5
Answers for the worksheet on division of rational number are given below to check the exact answers of the above division.
1. (i) 2
(ii) -7
(iii) 4/3
(iv) 2/3
(v) -16/9
(vi) 0
(vii) 1/8
(viii) -8/7
2. (i) -16/15
(ii) 128/7
(iii) 2/21
(iv) 1/16
(v) 32/75
(vi) -5/2
3. -3/2
4. 3/4
5. 4/3
6. 8/3
7. -3/16
8. 3/4
9. 10/3
10. 46/3
11. 3/4
12. (i) -3/4
(ii) -14/19
(iii) 4/5
(iv) 10
Rational Numbers - Worksheets
Worksheet on Rational Numbers
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Worksheet on Standard form of a Rational Number
Worksheet on Equality of Rational Numbers
Worksheet on Comparison of Rational Numbers
Worksheet on Representation of Rational Number on a Number Line
Worksheet on Properties of Addition of Rational Numbers
Worksheet on Rational Expressions Involving Sum and Difference
Worksheet on Multiplication of Rational Number
Worksheet on Division of Rational Numbers
Worksheet on Finding Rational Numbers between Two Rational Numbers
Worksheet on Word Problems on Rational Numbers
Objective Questions on Rational Numbers
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Rational Numbers
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Is Zero a Rational Number?
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To Find Rational Numbers | 1,579 | 5,520 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2024-18 | longest | en | 0.786944 |
https://www.studyadda.com/sample-papers/rrb-assistant-loco-pilot-technician-sample-test-paper-4_q96/154/266626 | 1,643,134,480,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304859.70/warc/CC-MAIN-20220125160159-20220125190159-00587.warc.gz | 1,037,909,023 | 21,160 | • # question_answer 8 women can complete a piece of work in 15 hours. In how many hours will 12 women complete the same piece of work? A) 12 B) 6 C) 8 D) 10
${{M}_{1}}{{D}_{1}}=\,{{M}_{2}}{{D}_{2}}$ $\Rightarrow$ $8\,\,\,\,15=12\,\,\,{{D}_{2}}$ $\Rightarrow$ ${{D}_{2}}=\frac{8\,\,\,\,\,\,15}{12}=10$ hours | 140 | 369 | {"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} | 3.609375 | 4 | CC-MAIN-2022-05 | latest | en | 0.524603 |
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Rosemaria Victoire December 6, 2016 worksheets
Remedial Work. If your child studied fractions during the last school year but just didn’t quite ”get it”, don’t worry. Use printable fraction worksheets found on the internet to review with him. These worksheets won’t cost you a dime and you can find all you need. Consider having your child do one quick worksheet two to three times per week during the summer as a stress-free way to review his fractions. When the child doesn’t have a full load of schoolwork, they often don’t mind doing a couple of worksheets. This is an excellent way to review a difficult subject such as fractions and it keeps the subject fresh in your child’s mind. So whether you need to save money in your home school or your child needs extra help with fractions, try using printable fraction worksheets today to save money and make homeschooling just a little bit easier on you.
Tip #3 – Use Worksheets Sparingly. Since free multiplication worksheets are so easy to find, it’s tempting to give your child too many. You mean well, but it just seems like a good idea to have them do several at a time. Little brains can only take so much. Keep learning fun by sprinkling worksheets into their curriculum as a fun break from their usual textbook. Tip #4 – Keep it Fun. If you happen to have a competitive child, chances are he will love worksheets always trying to beat his last time. This is great and if this is the case, let him work all of the worksheets he wants. Just be sure that it is ”child-driven” not ”parent-driven” meaning – let it be his idea. As long as he is having fun and asking for more, let him have all he wants.
Rather than using worksheets, a better method is to use individual size white boards and have the child writing entire facts many times. Having a child writing 9 x 7 = 7 x 9 = 63 while saying ”nine times seven is the same as seven times nine and is equal to sixty-three” is many times more successful than a worksheet with 9 x 7 = ___ and the student just thinks the answer once and then writes that answer on the duplicate problems. I will admit that there is one type of worksheet that I used in the past and found relatively beneficial, although it had a different kind of flaw. For my Basic Math, Pre-Algebra, and Algebra classes, I had several books of ”self-checking” worksheets. These worksheets had puns or puzzle questions at the top, and as the students worked the problems they were given some kind of code for choosing a letter to match that answer. If they worked the problems correctly, the letters eventually answered the pun or riddle. Students enjoyed these worksheets, but there are a couple problem areas even with these worksheets. Some students would get the answer to the riddle early and then work backward from letter to problem answer, so they weren’t learning or practicing anything.
4. Math worksheets are not accessible. Some students are unable to access tools that many of us take for granted when they try to complete worksheets. They may be unable to grasp pencils, control their movements within the limited spaces provided on the sheet, or be able to simply stabilize their paper while writing. Other students, including those for whom English is not their primary language or who struggle with reading, have difficulty reading the directions, words, and math terminology on the worksheets. Still other students require different visual representations or methods of engagement in order acquire an understanding the content. Most math worksheets do not provide information in multiple formats so they are inaccessible to students with a wide variety of learning styles and abilities. Well-designed technology can provide these students with access to excellent content. For example, these fractions tools and supplemental curriculum allow students with physical disabilities to access fractions content using a variety of assistive technology devices. Instructions, prompts and feedback can be read aloud, while visual models, cues combined with sounds support a wide range of learning styles and abilities.
Kindergarten ABC worksheets should have different activities to help children identify the various letters of the alphabet. The activities may involve very simple things like colouring, ticking, drawing a line to match items etc. Using attractive illustrations and cartoon characters would make it more fun for children. The activities should be graded, i.e initial activities should be very simple and easy (but should be fun with good pictures etc, so as to interest the child); later worksheets may involve a little bit more work. Care should be taken to give children worksheets that they are capable of doing. This involves understanding and monitoring the child continually, since the level of attainment of different children would often be quite different. The worksheet should challenge the child but not overwhelm her. If the worksheet is too easy or too repetitive, it may bore the child and she would not be happy. If the activity is too difficult it would frustrate her and she would not like to take up more sheets.
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ch. 6 moles test 60 points (5 ec) chemistry
In problems involving any calculation, show your work in an organized manner, include (i) any relevant equation (or formula), (ii) conversion factor(s), (iii) put the proper units in your calculations and answer, and (iv) have the proper number of significant figures in your answer. If you use the online website, https://www.symbolab.com/solver, or associated app, then clearly state that you used the website / app and you don’t have to show the algebraic manipulation to solve the problem.
Academic Honesty: The answers on this test / quiz are my own and I am using only the allowed set of notes as described in the syllabus and am only using a calculator or preceding website / app – I did not use any other website / app during the test / quiz. I have not discussed the test questions with anyone during the test. If you violate any of the preceding items or do not sign, your semester grade is a F.
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1. Solve. [15 points]
a. 5.0 kg water = __ mole
b. 25 millimole of carbon dioxide = __ g of carbon dioxide
c. 88 trillion calcium chloride molecules = __ moles of calcium chloride
2. A chemical’s empirical formula is C3H9N and the molar mass of the chemical is 177 g / mol; what is its chemical formula ? [10 points]
3. A chemical has 2.0 g of carbon, 0.50 g of hydrogen, and 1.33 g of oxygen. The molar mass of the chemical is 230 g / mole; what is its chemical formula ? [20 points; weekly quiz problem]
4. The complete combustion of 2.0 g of a chemical with carbon, hydrogen, and oxygen generates 3.826 g of carbon dioxide and 2.348 g of water. What is the chemical’s empirical formula ? [20 points] | 466 | 1,921 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2018-26 | longest | en | 0.807867 |
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AmosWEB means Economics with a Touch of Whimsy!
INDUCED CONSUMPTION: Household consumption expenditures that depend on income or production (especially disposable, national income, or gross national product). An increase in household disposable income triggers an increase in induced consumption expenditures. Induced consumption is graphically depicted as the slope of the consumption or propensity-to-consume line, and are measured by the marginal propensity to consume. The induced relation between income and consumption, as well as other induced expenditures, form the foundation of the multiplier effect triggered by changes in autonomous expenditures.
PRODUCTION FUNCTION:
A mathematical relation between the production of a good or service and the inputs used. A production function captures the general relation between total production and one or more inputs. The standard production function includes labor and capital as the inputs. However, a production function is general enough that any number of inputs can be included
A production function provides an abstract mathematical representation of the relation between the production of a good and the inputs used. A production function is usually expressed in this general form:
Q = f(L, K)
where: Q = quantity of production or output, L = quantity of labor input, and K = quantity of capital input. The letter "f" indicates a generic, as of yet unspecified, functional equation.
The analysis of short-run production is commonly performed at the introductory level with simple tables and graphs. These are useful abstraction methods for isolating and analyzing key aspects of short-run production. However, mathematical equations are another, often more powerful, method of abstract analysis. This is where the production function comes into play. Because a mathematical equation can extend beyond the two dimensions of a graph, it is possible to consider relations beyond just that for a variable input and total production.
If the production function takes the form of a specific equation (such as Q = 5L + 10K + 2LK), then a total product curve relating total product and the variable input can be plotted. However, to do so, one of the two inputs (L or K) must be designated as a variable input and one designated as a fixed input. For most types of production, labor is more readily changed than capital, so L is generally the variable input and K is usually the fixed input. A short-run total product curve can then be derived by "fixing" K at a particular value, then plotting the values of Q for alternative values of L.
While a great deal of economic insight into short-run production decisions of a firm and market supply curves can be analyzed with a simple graph, when economists begin using mathematical equations, such as the production function, Q = f(L, K), the possibilities are almost unlimited. A wide assortment of additional input variables can be added to this equation to make it, not only more sophisticated, but also more revealing.
For example, the effect of education and human capital on production can be seen by adding the educational attainment of workers as another input variable. In addition, the alternative impact on production of different types of capital, such as fixed structures and equipment can be identified by separating capital into two variables. In addition, to identify how public infrastructure, like highways and streets, affects production, then a variable for this input can be added to the production function.
<= PRODUCTION COST PRODUCTION INPUTS =>
Recommended Citation:
PRODUCTION FUNCTION, AmosWEB Encyclonomic WEB*pedia, http://www.AmosWEB.com, AmosWEB LLC, 2000-2023. [Accessed: November 30, 2023].
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ORANGE REBELOON[What's This?] Today, you are likely to spend a great deal of time looking for the new strip mall out on the highway looking to buy either a flower arrangement for your aunt or a birthday greeting card for your uncle. Be on the lookout for bottles of barbeque sauce that act TOO innocent.Your Complete Scope
Cyrus McCormick not only invented the reaper for harvesting grain, he also invented the installment payment for selling his reaper.
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Thanks for visiting AmosWEB | 961 | 4,808 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | longest | en | 0.922913 |
https://www.jiskha.com/display.cgi?id=1271852580 | 1,516,160,823,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886794.24/warc/CC-MAIN-20180117023532-20180117043532-00522.warc.gz | 912,618,349 | 4,131 | # math/geometry pls help
posted by .
A coin of radius 1 cm is tossed onto a plane surface that has been tesselated by right
triangles whose sides are 8 cm, 15 cm, and 17 cm long. What is the probability that the
coin lands within one of the triangles?
• math/geometry pls help -
well, if it is entirely within, it cannot touch any corner or side.
STart with the area along the sides the center of the coin cannot be in.
The center has to be greater than .5cm from any side. Draw the original triangle. Now mark a line parallel to each side, inside the triangle, .5 from the parallel side. These three lines make a triangle inside, in which the center may reside.
label the inner triangle a, b, c, with the a side parallel to 8, b parallel to 15, and c parallel to 17.
a/8=b/15=c/17 similar triangles.
but a= 8-.5-.5-.5(8/15) If you examine the 17 corner carefully, you can arrive at that with using smaller triangles which are similar.
solve for a: a= 6.86
which means b: b= a*15/8=12.86
area of inner triangle= 1/2 ab=44.1
area of original triangle= 1/2 8*15=60
Pr(coin inside)= areainner/areaouter
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More Similar Questions | 845 | 3,202 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2018-05 | latest | en | 0.947802 |
https://www.cfd-online.com/Forums/openfoam-solving/65033-relative-tolerance.html | 1,519,214,034,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891813608.70/warc/CC-MAIN-20180221103712-20180221123712-00375.warc.gz | 855,238,680 | 15,449 | # Relative Tolerance
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June 1, 2009, 22:31 Relative Tolerance #1 New Member Rachel Harch Join Date: Mar 2009 Location: RAdelaide Posts: 2 Rep Power: 0 Hi I was wondering if someone could help me to understand the details of relTol. I am unsure which has the stricter convergence criteria relTol=0.1 or relTol=0.001. I understand that if you set relTol=0 that you force the solution to converge to the solver tolerance, but I am not sure about other values. Cheers Rach
June 4, 2009, 10:56 #2 Senior Member Rishi . Join Date: Mar 2009 Posts: 144 Rep Power: 10 Hello, relTol is a parameter to say how much the intial residual should reduce in a particular time step. e.g. relTol=0, then default tolerence is targetted. (e-6 here) DICPCG: Solving for p, Initial residual = 1, Final residual = 7.55402e-07, No Iterations 35 ... DICPCG: Solving for p, Initial residual = 0.379232, Final residual = 3.38648e-07, No Iterations 34 e.g. relTol=0.01, then solver only tries to reduce residual by certain factor. This relTol is the order of magnitude reduction. So here Solver tries to reach a value below 1/100 of initial value. DICPCG: Solving for p, Initial residual = 1, Final residual = 0.00837542, No Iterations 24 ... DICPCG: Solving for p, Initial residual = 0.378888, Final residual = 0.00355735, No Iterations 23 So relTol=0.001 is stricter criteria & requires more iterations. Observe the number of iterations required in the above examples, for the same calculation.
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All times are GMT -4. The time now is 07:53. | 621 | 2,142 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2018-09 | latest | en | 0.860494 |
https://socratic.org/questions/how-do-you-simplify-1-x-h-1-x | 1,638,887,595,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363400.19/warc/CC-MAIN-20211207140255-20211207170255-00002.warc.gz | 588,584,794 | 5,872 | # How do you simplify [1/(x+h)] - 1/x?
Dec 5, 2016
$\frac{h}{x \left(x + h\right)}$ or $\frac{h}{{x}^{2} + h x}$
#### Explanation:
First you need to get each fraction over a common denominator by multiplying each fraction by the correct form of $1$. In this case the common denominator can be: $x \left(x + h\right)$
$\left(\frac{x}{x}\right) \left[\frac{1}{x + h}\right] - \left(\frac{x + h}{x + h}\right) \left(\frac{1}{x}\right)$
$\left[\frac{x}{x \left(x + h\right)}\right] - \left[\frac{x + h}{x \left(x + h\right)}\right]$
Now we can combine the fractions through subtraction:
$\frac{x - x + h}{x \left(x + h\right)}$
$\frac{h}{x \left(x + h\right)}$ | 235 | 665 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 8, "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} | 4.53125 | 5 | CC-MAIN-2021-49 | latest | en | 0.539455 |
https://oeis.org/A162273 | 1,670,640,982,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711637.64/warc/CC-MAIN-20221210005738-20221210035738-00713.warc.gz | 455,481,058 | 4,836 | The OEIS is supported by the many generous donors to the OEIS Foundation.
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A162273 a(n) = ((2+sqrt(3))*(3+sqrt(3))^n + (2-sqrt(3))*(3-sqrt(3))^n)/2. 1
2, 9, 42, 198, 936, 4428, 20952, 99144, 469152, 2220048, 10505376, 49711968, 235239552, 1113165504, 5267555712, 24926341248, 117952713216, 558158231808, 2641233111552, 12498449278464, 59143297001472, 279869086338048 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,1 COMMENTS Binomial transform of A001075 without initial term 1, inverse binomial transform of A162274. The INVERTi transform yields A007051 without A007051(0). - R. J. Mathar, Jul 07 2009 LINKS Index entries for linear recurrences with constant coefficients, signature (6,-6). FORMULA a(n) = 6*a(n-1) - 6*a(n-2) for n > 1; a(0) = 2, a(1) = 9. G.f.: (2-3*x)/(1-6*x+6*x^2). a(n) = 2*A030192-3*A030192(n-1). - R. J. Mathar, Feb 04 2021 MAPLE seq(simplify(((2+sqrt(3))*(3+sqrt(3))^n+(2-sqrt(3))*(3-sqrt(3))^n)*1/2), n = 0 .. 22); # Emeric Deutsch, Jul 11 2009 MATHEMATICA LinearRecurrence[{6, -6}, {2, 9}, 30] (* Harvey P. Dale, Dec 17 2019 *) PROG (Magma) Z:=PolynomialRing(Integers()); N:=NumberField(x^2-3); S:=[ ((2+r)*(3+r)^n+(2-r)*(3-r)^n)/2: n in [0..21] ]; [ Integers()!S[j]: j in [1..#S] ]; // Klaus Brockhaus, Jul 05 2009 CROSSREFS Cf. A001075, A162274. Sequence in context: A330016 A056845 A354302 * A289684 A280955 A276508 Adjacent sequences: A162270 A162271 A162272 * A162274 A162275 A162276 KEYWORD nonn,easy AUTHOR Al Hakanson (hawkuu(AT)gmail.com), Jun 29 2009 EXTENSIONS Edited and extended beyond a(5) by R. J. Mathar and Klaus Brockhaus, Jul 05 2009 STATUS approved
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Last modified December 9 19:36 EST 2022. Contains 358703 sequences. (Running on oeis4.) | 817 | 2,315 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.610703 |
https://www.convertunits.com/from/barleycorn/to/cubit | 1,600,450,427,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400188049.8/warc/CC-MAIN-20200918155203-20200918185203-00021.warc.gz | 817,503,232 | 8,313 | ## ››Convert barleycorn to cubit [Egyptian]
barleycorn cubit
Did you mean to convert barleycorn to cubit [Egyptian] cubit [English] cubit [Roman] cubit [Royal Egyptian]
How many barleycorn in 1 cubit? The answer is 52.941176470588.
We assume you are converting between barleycorn and cubit [Egyptian].
You can view more details on each measurement unit:
barleycorn or cubit
The SI base unit for length is the metre.
1 metre is equal to 117.64705882353 barleycorn, or 2.2222222222222 cubit.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between barleycorns and cubits.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of barleycorn to cubit
1 barleycorn to cubit = 0.01889 cubit
10 barleycorn to cubit = 0.18889 cubit
20 barleycorn to cubit = 0.37778 cubit
30 barleycorn to cubit = 0.56667 cubit
40 barleycorn to cubit = 0.75556 cubit
50 barleycorn to cubit = 0.94444 cubit
100 barleycorn to cubit = 1.88889 cubit
200 barleycorn to cubit = 3.77778 cubit
## ››Want other units?
You can do the reverse unit conversion from cubit to barleycorn, or enter any two units below:
## Enter two units to convert
From: To:
## ››Definition: Barleycorn
old unit of length equal to roughly one-third of an inch
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 490 | 1,797 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2020-40 | latest | en | 0.791684 |
http://www.osti.gov/eprints/topicpages/documents/record/889/2415077.html | 1,455,166,388,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701160958.15/warc/CC-MAIN-20160205193920-00251-ip-10-236-182-209.ec2.internal.warc.gz | 581,027,161 | 3,076 | A Calculus Problem Revisited, or: On Beyond Varberg Summary: A Calculus Problem Revisited, or: On Beyond Varberg Roger Alexander µ Problem 37 From §5.4 of Varberg et al. Calculus, 9th ed CAS Using the same axes, draw the graphs of y = xn on [0, 1] for n = 1, 2, 4, 10 and 100. Find the length of each of these curves. Guess at the length when n = 10, 000. Graphs of the powers of x Figure: y = xn for n = 1, 2, 4, 10, 100. As the exponent n gets larger, the graph of y = xn shows distinct "flat" and "steep" parts. The red dot shows the transition. Length of y = xn Collections: Mathematics | 185 | 590 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2016-07 | latest | en | 0.766325 |
https://www.mrexcel.com/board/threads/copy-value-from-cell-above-if-blank.1109854/?s=09f4338705743689e836fea832db8aa2 | 1,590,514,262,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347391277.13/warc/CC-MAIN-20200526160400-20200526190400-00465.warc.gz | 839,140,721 | 16,621 | # Copy value from cell above if blank
#### Mux99
##### New Member
In column A every two cells are merged (A1 with A2, A3 with A4 etc...) so my formula only works properly for odd number rows. I need the below to use the value from the cell above if A6 is empty and so on.
=IF(Sheet1!\$B6<>"", Sheet1!\$A6,"")
### Excel Facts
Customize Quick Access Toolbar. From All Commands, add Speak Cells or Speak Cells on Enter to QAT. Select cells. Press Speak Cells.
#### JoeMo
##### MrExcel MVP
Maybe:
Code:
``=IF(Sheet1!\$B6<>"", IF(MOD(ROW(Sheet1!A6),2)=0,Sheet1!\$A5,""))``
#### Mux99
##### New Member
Maybe:
Code:
``=IF(Sheet1!\$B6<>"", IF(MOD(ROW(Sheet1!A6),2)=0,Sheet1!\$A5,""))``
Did not work. It fixed the even number rows but now the odd ones are blank.
#### Fluff
##### MrExcel MVP, Moderator
Maybe
=IF(Sheet1!\$A6="",Sheet1!\$A5,Sheet1!A6)
#### Mux99
##### New Member
Maybe
=IF(Sheet1!\$A6="",Sheet1!\$A5,Sheet1!A6)
Also did not work for what I'm trying to do but maybe it's my fault as maybe I wasn't clear enough with what I need. Ignore my first post and see the examples below. I need a formula to output the same results as in Sheet2.
In Sheet2 Column B I'm using the formula and it works perfectly. =IF(Sheet1!\$B1<>"", Sheet1!\$B1,"")
I need some help with the column A formula as the following only works with odd number rows. =IF(Sheet1!\$B1<>"", Sheet1!\$A1,"")
Sheet1 A B 1 Name1 500 2 3 Name2 4 5 Name3 600 6 300 7 Name4 8 400
<tbody>
</tbody>
Sheet2 A B 1 Name1 500 2 3 4 5 Name3 600 6 Name3 300 7 8 Name4 400
<tbody>
</tbody>
#### Peter_SSs
##### MrExcel MVP, Moderator
In cell A1 of Sheet2, try this formula and copy down.
=IF(Sheet1!B1="","",LOOKUP("zzz",Sheet1!A\$1:A1))
#### Mux99
##### New Member
In cell A1 of Sheet2, try this formula and copy down.
=IF(Sheet1!B1="","",LOOKUP("zzz",Sheet1!A\$1:A1))
Thanks. This one worked but I already got it working with the formula below.
=IF(Sheet1!\$B2<>"",IF(Sheet1!\$A2<>"",Sheet1!\$A2,Sheet1!\$A1),"")
#### Peter_SSs
##### MrExcel MVP, Moderator
Thanks. This one worked but I already got it working with the formula below.
=IF(Sheet1!\$B2<>"",IF(Sheet1!\$A2<>"",Sheet1!\$A2,Sheet1!\$A1),"")
That sort of indicates that in fact your data & results probably actually start on row 2 of each sheet, not row 1 as indicated in your tables in post 5? (Or else that formula is entered on row 2 of Sheet2 and copied up to row 1 which, while working actually creates a #REF ! error in the formula.
If everything (apart from any headings) actually starts on row 2, then you could also use this simpler version in A2 of Sheet2, copied down
=IF(Sheet1!B2="","",Sheet1!A1&Sheet1!A2)
Last edited: | 881 | 2,680 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-24 | latest | en | 0.853657 |
http://www.solutioninn.com/a-campus-club-consists-of-five-officers-president-p-vice | 1,502,978,302,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886103316.46/warc/CC-MAIN-20170817131910-20170817151910-00240.warc.gz | 682,056,527 | 7,510 | # Question: A campus club consists of five officers president P vice
A campus club consists of five officers: president (P), vice president (V), secretary (S), treasurer (T), and activity coordinator (A). The club can select two officers to travel to New Orleans for a conference; for fairness, they decide to make the selection at random. In essence, they are choosing a simple random sample of size n = 2.
a. What are the possible samples of two officers?
b. What is the chance that a particular sample of size 2 will be drawn?
c. What is the chance that the activity coordinator will be chosen?
View Solution:
Sales1
Views70 | 141 | 630 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2017-34 | latest | en | 0.94294 |
https://efinancemanagement.com/financial-analysis/equity-multiplier | 1,726,171,402,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651491.39/warc/CC-MAIN-20240912174615-20240912204615-00438.warc.gz | 207,145,771 | 67,899 | # Equity Multiplier
## What is Equity Multiplier (EM)?
Equity Multiplier is a key financial metric that measures the level of debt financing in a business. In other words, it is defined as a ratio of total assets to total equity. If the ratio is 10, the equity multiplier means investment in total assets is 10 times the investment by equity shareholders. Conversely, it means 1 part is equity, and 9 parts are debt in overall asset financing.
## Formula
The formula for equity multiplier is as follows:
Equity Multiplier = Total Assets / Total Shareholder’s Equity
Here,
Total Assets: Total assets would mean all the company’s assets, or one can take the total asset side of a company’s balance sheet. This was the understanding from the asset side. A total of all the liabilities and equity capital are covered for arriving at the figure of total assets from the liability side.
Total Shareholder’s Equity: It is determined by the total amount of equity that shareholders hold in a company, which includes the sum of common stock, preferred stock, additional paid-in capital, retained earnings, and accumulated other comprehensive income (net of treasury stock, if any).
It is a reciprocal of the equity ratio.
## Calculation (Example)
Let us understand the calculation using the following example:
Suppose,
Total Assets of a Company = \$200 Million
Total Shareholder’s Equity = \$ 50 Millions
Using the equity multiplier equation as follows
Equity Multiplier = Total Assets / Total Shareholder’s Equity = 200 / 50 = 4.
We get a multiplier of 4. This simply means that total assets are 4 times the total equity.
You can also use our calculator – Equity Multiplier Calculator
## Interpretation and Analysis
Running a business needs investment in assets. You do it in two ways, i.e., debt or equity. A ratio of 4 times states that total assets are 4 times that of its equity. In other words, 1 out of 4 parts of assets are financed by equity and the remaining, i.e., 3 parts, are financed by debt. In percentage terms, 25% (1/4) is equity, and 75% (3/4) is debt.
Our mind is always inquisitive about categorizing everything between good and bad. So, before jumping on to whether the multiple of 4 is good or bad, let us understand that the comparison is possible with 2 things – Industry Standards and Own Past Multiple.
### Industry Standard
If the multiple is higher than its peers in the industry, you can safely say that the company has higher leverage.
### Own Past Multiples
Comparing our multiple with our own past multiples can help us gain only the trend of it. If the trend is rising, it can be an alarming situation for finance managers because further debt borrowing becomes difficult with the rise in debt proportion. If the rise is not accompanied by sufficient profitability and efficient use of assets, it can lead the company toward financial distress.
## Relation with DuPont & Impact on ROE:
The DuPont Analysis attempts to break down ROE into 3 components, viz. Operating Profit Margin Ratio, Asset Turnover Ration, and Equity Multiplier. The product of all 3 components will arrive at the ROE. DuPont formula clearly states a direct relation of ROE with Equity Multiplier. The higher the EM, the higher the potential for ROE and vice-versa.
Why is there a directly proportional relation between ROE and EM? Since the higher debt in the overall capital reduces the cost of capital with the basic assumption that debt is a cheaper source of capital. Taxes safely defend the assumption, i.e., the interest on the debt is a tax-deductible expense. If the rate of interest is 10% and taxes are 40%. The effective cost of debt is calculated as 6% .
Both higher and lower EM can have their share of benefits and disadvantages.
### Higher EM
High debt proportion in capital structure may have the following issues
1. Higher debt means a higher risk of insolvency. If the profits decline under any circumstances, the chances of not meeting the financial and other obligations increase.
2. The ability to borrow more debt becomes tough since it is already leveraged high.
### Lower EM
On the other hand, lower EM can signify inefficiency in creating value for shareholders through tax benefits due to leverage.
### Ideal EM
There can’t be one ideal equity multiplier. It should be part of the overall strategy of the business. This may depend a lot on industry and other factors such as the availability of debt, project size, etc.
## Problems with Equity Multiplier Metric
There are certain issues that can dilute the use of equity multiplier for analysis. Cautionary measures are advisable.
### Variations in accounting practices
Different accounting practices can impact the calculation of the EM. For example, the treatment of off-balance sheet items or lease obligations can affect the total assets and equity values used in the calculation. It’s important to ensure consistent and accurate accounting data when comparing EM values across companies or periods.
### Potential for misinterpretation
A higher EM does not always indicate a positive outcome. While financial leverage can magnify returns during favorable economic conditions, it can also amplify losses during downturns. Excessive leverage can increase the financial risk and vulnerability of a company. Thus, a high EM should be assessed in the context of the company’s ability to service its debt and manage financial risks.
### Industry differences
Different industries have varying capital structures and debt requirements. Comparing the EM of companies across different sectors may not be meaningful as their capital needs and risk profiles can differ significantly. It’s important to consider industry norms and benchmarks when interpreting EM values.
Visit for other types of Leverage Ratios. | 1,187 | 5,821 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2024-38 | latest | en | 0.940635 |
https://spiral.ac/sharing/d74s2ar/sss-to-show-a-radius-is-perpendicular-to-a-chord-that-it-bisects-geometry-khan-academy | 1,637,983,542,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358078.2/warc/CC-MAIN-20211127013935-20211127043935-00501.warc.gz | 639,228,416 | 10,445 | sss-to-show-a-radius-is-perpendicular-to-a-chord-that-it-bisects-geometry-khan-academy
# Interactive video lesson plan for: SSS to show a radius is perpendicular to a chord that it bisects | Geometry | Khan Academy
#### Activity overview:
More on the difference between a theorem and axiom. Proving a cool result using SSS
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Add Clip to Chrome | 945 | 4,353 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2021-49 | latest | en | 0.899698 |
http://support.sas.com/documentation/cdl/en/imlsug/68152/HTML/default/imlsug_ugmultcda_sect001.htm | 1,537,673,584,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267158958.72/warc/CC-MAIN-20180923020407-20180923040807-00129.warc.gz | 237,419,477 | 4,175 | Multivariate Analysis: Canonical Discriminant Analysis
Overview of Canonical Discriminant Analysis
Canonical discriminant analysis is a dimension-reduction technique that is related to principal component analysis and canonical correlation. Given a nominal classification variable and several interval variables, canonical discriminant analysis derives canonical variables (linear combinations of the interval variables) that summarize between-class variation in much the same way that principal components summarize total variation.
Canonical discriminant analysis is equivalent to canonical correlation analysis between the quantitative variables and a set of dummy variables coded from the classification variable.
Given two or more groups of observations with measurements on several interval variables, canonical discriminant analysis derives a linear combination of the variables that has the highest possible multiple correlation with the groups. This maximum multiple correlation is called the first canonical correlation. The coefficients of the linear combination are the canonical coefficients. The variable defined by the linear combination is the first canonical variable. The second canonical correlation is obtained by finding the linear combination uncorrelated with the first canonical variable that has the highest possible multiple correlation with the groups. The process of extracting canonical variables can be repeated until the number of canonical variables equals the number of original variables or the number of classes minus one, whichever is smaller. Canonical variables are also called canonical components.
You can run the Canonical Discriminant analysis by selecting AnalysisMultivariate AnalysisCanonical Discriminant Analysis from the main menu. The analysis is implemented by calling the DISCRIM procedure with the CANONICAL option in SAS/STAT software. See the documentation for the DISCRIM and CANDISC procedures in the SAS/STAT User's Guide for additional details.
The analysis calls the DISCRIM procedure (rather than the CANDISC procedure) because the DISCRIM procedure produces a discriminant function that can be used to classify current or future observations. | 358 | 2,209 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-39 | latest | en | 0.86417 |
https://solvedlib.com/which-of-the-following-is-not-a-trade-or-business,12320 | 1,696,142,540,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510781.66/warc/CC-MAIN-20231001041719-20231001071719-00070.warc.gz | 573,846,942 | 16,640 | # Which of the following is not a "trade or business" expense
###### Question:
A. Mortgage interest on a warehouse
C. Mortgage interest on a personal residence
D. Cost of goods sold
is it C
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How do you find the quotient of (x^2-5x-16)div(x+2)?... | 3,012 | 10,642 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2023-40 | latest | en | 0.786214 |
http://www.numbersaplenty.com/9387 | 1,582,568,062,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145966.48/warc/CC-MAIN-20200224163216-20200224193216-00028.warc.gz | 206,841,273 | 3,579 | Search a number
9387 = 327149
BaseRepresentation
bin10010010101011
3110212200
42102223
5300022
6111243
736240
oct22253
913780
109387
117064
125523
134371
1435c7
152bac
hex24ab
9387 has 12 divisors (see below), whose sum is σ = 15600. Its totient is φ = 5328.
The previous prime is 9377. The next prime is 9391. The reversal of 9387 is 7839.
Subtracting from 9387 its product of digits (1512), we obtain a triangular number (7875 = T125).
It is a happy number.
9387 is digitally balanced in base 2 and base 3, because in such bases it contains all the possibile digits an equal number of times.
It is not a de Polignac number, because 9387 - 24 = 9371 is a prime.
It is a Smith number, since the sum of its digits (27) coincides with the sum of the digits of its prime factors.
9387 is a lucky number.
It is a plaindrome in base 16.
It is a self number, because there is not a number n which added to its sum of digits gives 9387.
It is not an unprimeable number, because it can be changed into a prime (9337) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (7) of ones.
It is a polite number, since it can be written in 11 ways as a sum of consecutive naturals, for example, 12 + ... + 137.
It is an arithmetic number, because the mean of its divisors is an integer number (1300).
29387 is an apocalyptic number.
9387 is a deficient number, since it is larger than the sum of its proper divisors (6213).
9387 is a wasteful number, since it uses less digits than its factorization.
9387 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 162 (or 159 counting only the distinct ones).
The product of its digits is 1512, while the sum is 27.
The square root of 9387 is about 96.8865315717. The cubic root of 9387 is about 21.0948094118.
The spelling of 9387 in words is "nine thousand, three hundred eighty-seven".
Divisors: 1 3 7 9 21 63 149 447 1043 1341 3129 9387 | 578 | 1,991 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2020-10 | latest | en | 0.899228 |
http://www.jiskha.com/display.cgi?id=1202702065 | 1,498,399,994,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320532.88/warc/CC-MAIN-20170625134002-20170625154002-00513.warc.gz | 558,332,730 | 4,017 | # calculus
posted by .
how would you do this improper integral
1/(x-1)
from 0 to 2
this is improper at one, so I split it up into two integrals
ln(x-1) from 0-1 and
ln(x-1) from 1-2
I then did for the first one the (lim t->1(-) of ln(t-1))-(ln(0-1))
and then the same thing for the second part
I didn't know if this was right though, or what the answer would be
• calculus -
• calculus -
• calculus -
I left the final calculation up to you. If you define the integral as the sum of two parts that each approach within a distance a of x=1, and let a gapproach zero, than the answer will be zero because the two parts will always cancel, not matter how small a is. | 192 | 672 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2017-26 | latest | en | 0.945324 |
https://www.softschools.com/math/probability_and_statistics/pie_chart/ | 1,632,080,354,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056900.32/warc/CC-MAIN-20210919190128-20210919220128-00693.warc.gz | 1,000,122,723 | 8,063 | # Pie Chart
One of the simplest graphs that can be used to display summarized data is the pie chart. It displays either categorical or numerical data in an easy-to-read chart, with the size of each piece of the pie based on the frequency of each category. In other words, each "slice" of the pie shows us visually what proportion that slice is of the whole pie.
Pie charts are almost always used when it is informative to visually see information about percentages of some whole. For example, let's assume we are interested in analyzing the dates stamped on 2,000 randomly-selected pennies. We first record each penny's date by the decade when it was minted and then summarize the data into a simple table.
Note that the calculation of the relative frequency is found by dividing the frequency by the total. Then, we calculate the percent by moving the decimal point to the right two places.
The data in the table can easily be converted into a pie chart. Here is a pie chart prepared from the mint dates of those 2,000 randomly-selected pennies.
Observe that the pie chart is very helpful when it comes to information about percentages. We can easily see that the largest percentage of those 2,000 pennies (40% of them) were minted in the 1990s. We can also see that only 1% of those pennies were minted prior to the 1960s.
Related Links: Math Probability and Statistics Stemplot (Stem-and-Leaf Diagram) The Back-To-Back Stemplot Bar Chart Determining a Missing Data Value Given the Mean AP Statistics Quiz - Probability Quiz
To link to this Pie Chart page, copy the following code to your site: | 348 | 1,604 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2021-39 | longest | en | 0.922155 |
https://gmatclub.com/forum/consumers-planning-to-buy-recreational-equipment-tend-to-buy-80579.html?fl=similar | 1,508,331,256,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822966.64/warc/CC-MAIN-20171018123747-20171018143747-00111.warc.gz | 715,578,654 | 48,254 | It is currently 18 Oct 2017, 05:54
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06 Jul 2009, 21:50
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more expensive equipment when the economy is strong than when it is
weak. Hill and Dale is a business that sells high-quality, expensive camping
and hiking equipment in Boravia. Although all the signs are that Boravia’s
economy is now entering a period of sustained strength, the managers
of the business do not expect a substantial increase in sales.
Which of the following, if true, would provide the strongest
justification for the managers’ judgment?
A. A significant proportion of Hill and Dale’s sales are made to
customers who enter the store in order to buy one particular
item but, once there, find other items to buy as well.
B. In Boravia when the economy is strong, those who might
otherwise go camping tend to take vacations overseas.
C. The economic upturn is likely to allow Boravia’s national
parks, where most of the camping and hiking is
done, to receive extra funding to improve their visitor facilities.
D. Advances in materials technology have led to the
development of hiking and camping equipment that
is more comfortable and lightweight than before.
E. Many people in Boravia not only are committed to
preserving the country’s wilderness areas but also
are interested in spending some time in them.
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06 Jul 2009, 22:20
Is it B ?
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06 Jul 2009, 22:28
I'll go with option B.
A. A significant proportion of Hill and Dale’s sales are made to customers who enter the store in order to buy one particular item but, once there, find other items to buy as well. -Irrelevant
B. In Boravia when the economy is strong, those who might otherwise go camping tend to take vacations overseas. -Correct. If people choose oversees destinations, they may not buy camping products.
C. The economic upturn is likely to allow Boravia’s national parks, where most of the camping and hiking is done, to receive extra funding to improve their visitor facilities. -Goes against the manager's judgment. If parks facilities are improved, people may tend to visit these parks (where most of the camping and hiking is done) in higher number.
D. Advances in materials technology have led to the development of hiking and camping equipment that is more comfortable and lightweight than before. -Goes against the manager's judgment. If this is the case, an improving economy should mean people buying better equipments
E. Many people in Boravia not only are committed to preserving the country’s wilderness areas but also are interested in spending some time in them. -Irrelevant
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07 Jul 2009, 09:44
I would go for C
If the parks receive extra funding to improve their visitor facilities - this means more camping equipment - so few customers to buy the equipment.
C. The economic upturn is likely to allow Boravia’s national
parks, where most of the camping and hiking is
done, to receive extra funding to improve their visitor facilities.
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07 Jul 2009, 10:24
vannu wrote:
I would go for C
If the parks receive extra funding to improve their visitor facilities - this means more camping equipment - so few customers to buy the equipment.
I doubt. The option C doesn't give enough information about the kind of facilities improvement. There can be other kindda improvement (like easier access, better sanitation, water facility etc), which can even attract more number of tourists then ever before. In that case the sale of camping equipment should increase.
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07 Jul 2009, 14:54
B
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07 Jul 2009, 17:57
Quote:
vannu wrote:
I would go for C
If the parks receive extra funding to improve their visitor facilities - this means more camping equipment - so few customers to buy the equipment.
I doubt. The option C doesn't give enough information about the kind of facilities improvement. There can be other kindda improvement (like easier access, better sanitation, water facility etc), which can even attract more number of tourists then ever before. In that case the sale of camping equipment should increase.
I agree with you. I misinterpreted the facility to equipment.
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09 Jul 2009, 05:20
IMO B...
OA plzz
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09 Jul 2009, 06:21
looks like B
What is the OA?
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09 Jul 2009, 08:47
I choose (B).
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09 Jul 2009, 09:00
I'll also go for B
bigoyal's reasoning seems perfect; except that D can also be classified as irrelevant.
bigoyal wrote:
I'll go with option B.
A. A significant proportion of Hill and Dale’s sales are made to customers who enter the store in order to buy one particular item but, once there, find other items to buy as well. -Irrelevant
B. In Boravia when the economy is strong, those who might otherwise go camping tend to take vacations overseas. -Correct. If people choose oversees destinations, they may not buy camping products.
C. The economic upturn is likely to allow Boravia’s national parks, where most of the camping and hiking is done, to receive extra funding to improve their visitor facilities. -Goes against the manager's judgment. If parks facilities are improved, people may tend to visit these parks (where most of the camping and hiking is done) in higher number.
D. Advances in materials technology have led to the development of hiking and camping equipment that is more comfortable and lightweight than before. -Goes against the manager's judgment. If this is the case, an improving economy should mean people buying better equipments
E. Many people in Boravia not only are committed to preserving the country’s wilderness areas but also are interested in spending some time in them. -Irrelevant
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23 Jul 2009, 01:57
IMO B.
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23 Jul 2009, 05:29
ritjn2003 plz post the OA.
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23 Jul 2009, 22:40
B...OA plzz
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# GMATPREP ChallengeQ -In a new book about the antiparty feeli
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Passage-16 GMATPrep RCs-Collection(Main article)
In a new book about the antiparty feeling of the early political leaders of the United States, Ralph Ketcham argues that the first six Presidents differed decisively from later Presidents because the first six held values inherited from the classical humanist tradition of eighteenth-century England. In this view, government was designed not to satisfy the private desires of the people but to make them better citizens; this tradition stressed the disinterested devotion of political leaders to the public good. Justice, wisdom, and courage were more important qualities in a leader than the ability to organize voters and win elections. Indeed, leaders were supposed to be called to office rather than to run for office. And if they took up the burdens of public office with a sense of duty, leaders also believed that such offices were naturally their due because of their social preeminence or their contributions to the country. Given this classical conception of leadership, it is not surprising that the first six Presidents condemned political parties. Parties were partial by definition, self-interested, and therefore serving something other than the transcendent public good.
Even during the first presidency (Washington's), however, the classical conception of virtuous leadership was being undermined by commercial forces that had been gathering since at least the beginning of the eighteenth century. Commerce--its profit-making, its self-interestedness, its individualism--became the enemy of these classical ideals. Although Ketcham does not picture the struggle in quite this way, he does rightly see Jackson's tenure (the seventh presidency) as the culmination of the acceptance of party, commerce, and individualism. For the Jacksonians, nonpartisanship lost its relevance, and under the direction of Van Buren, party gained a new legitimacy. The classical ideals of the first six Presidents became identified with a privileged aristocracy, an aristocracy that had to be overcome in order to allow competition between opposing political interests. Ketcham is so strongly committed to justifying the classical ideals, however, that he underestimates the advantages of their decline. For example, the classical conception of leadership was incompatible with our modern notion of the freedoms of speech and press, freedoms intimately associated with the legitimacy of opposing political parties.
1. The passage is primarily concerned with
A. describing and comparing two theories about the early history of the United States
B. describing and analyzing an argument about the early history of the United States
C. discussing new evidence that qualifies a theory about the early history of the United States
E. resolving an ambiguity in an argument about political leadership in the United States
[Reveal] Spoiler:
B
2. According to the passage, the author and Ketcham agree on which of the following points?
A. The first six Presidents held the same ideas about political parties as did later Presidents in the United States.
B. Classical ideals supported the growth of commercial forces in the United States.
C. The first political parties in the United States were formed during Van Buren's term in office.
D. The first six Presidents placed great emphasis on individualism and civil rights.
E. Widespread acceptance of political parties occurred during Andrew Jackson's presidency.
[Reveal] Spoiler:
E
3. It can be inferred that the author of the passage would be most likely to agree that modern views of the freedoms of speech and press are
A. values closely associated with the beliefs of the aristocracy of the early United States
B. political rights less compatible with democracy and individualism than with classical ideals
C. political rights uninfluenced by the formation of opposing political parties
D. values not inherent in the classical humanist tradition of eighteenth-century England
E. values whose interpretation would have been agreed on by all United States Presidents
[Reveal] Spoiler:
D
4. Which of the following, if true, provides the LEAST support for the author's argument about commerce and political parties during Jackson's presidency?
A. Many supporters of Jackson resisted the commercialization that could result from participation in a national economy.
B. Protest against the corrupt and partisan nature of political parties in the United States subsided during Jackson's presidency.
C. During Jackson's presidency the use of money became more common than bartering of goods and services.
D. More northerners than southerners supported Jackson because southerners were opposed to the development of a commercial economy.
E. Andrew Jackson did not feel as strongly committed to the classical ideals of leadership as George Washington had felt.
[Reveal] Spoiler:
A
5. The author of the passage would most likely to agree with which of following statements about Ketcham?
A. He overemphasizes the influence of classical ideals on the first six Presidents of the United States.
B. He fails to recognize that classical ideals had little influence on politics in the United States.
C. He does not pay adequate attention to the negative aspects of the first Presidents’ commitment to classical ideals.
D. He inaccurately suggests that classical ideals gave rise to our modern notion of democracy.
E. He underestimates the effect of ideologies other than the humanist tradition on the first six Presidents.
[Reveal] Spoiler:
C
[Reveal] Spoiler: Question #1 OA
[Reveal] Spoiler: Question #2 OA
[Reveal] Spoiler: Question #3 OA
[Reveal] Spoiler: Question #4 OA
[Reveal] Spoiler: Question #5 OA
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Last edited by PiyushK on 16 Aug 2014, 02:03, edited 2 times in total.
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12 Aug 2014, 07:18
Time taken 13 mins. Is it bad? <Updated time:10 mins>
1. The passage is primarily concerned with
B. describing and analyzing an argument about the early history of the United States
Argument first descibes the era of earlier leadership (first 6 presidents) .
From second para, it introduces the changes in the system since the era of later Presidents.Finally, it mentions it views on the pros and cons of both the eras.
2. According to the passage, the author and Ketcham agree on which of the following points?
A. The first six Presidents held the same ideas about political parties as did later Presidents in the United States.
B. Classical ideals supported the growth of commercial forces in the United States.
C. The first political parties in the United States were formed during Van Buren's term in office.
D. The first six Presidents placed great emphasis on individualism and civil rights.
E. Widespread acceptance of political parties occurred during Andrew Jackson's presidency.
>> For me it was between C &E. C is wrong bec arg says that Party found legal status but that doesn't mean they weren't formed before that.
3. It can be inferred that the author of the passage would be most likely to agree that modern views of the freedoms of speech and press are
D. values not inherent in the classical humanist tradition of eighteenth-century England
"the classical conception of leadership was incompatible with our modern notion of the freedoms of speech and press, freedoms intimately associated with the legitimacy of opposing political parties."
4. Which of the following, if true, provides the LEAST support for the author's argument about commerce and political parties during Jackson's presidency?
A. Many supporters of Jackson resisted the commercialization that could result from participation in a national economy.
"he does rightly see Jackson's tenure (the seventh presidency) as the culmination of the acceptance of party, commerce, and individualism. For the Jacksonians, nonpartisanship lost its relevance, "
5. The author of the passage would most likely to agree with which of following statements about Ketcham?
C. He does not pay adequate attention to the negative aspects of the first Presidents’ commitment to classical ideals.
"Ketcham is so strongly committed to justifying the classical ideals, however, that he underestimates the advantages of their decline. "
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Last edited by JarvisR on 20 Jul 2015, 19:56, edited 2 times in total.
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12 Aug 2014, 07:49
13 min is not bad with med-long passage and with 5 questions.
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20 Aug 2014, 21:20
10 mins 47 secs
1. B
2. E (Although Ketcham does not picture the struggle in quite this way, he does rightly see Jackson's tenure (the seventh presidency) as the culmination of the acceptance of party, commerce, and individualism. )
3. D
4. A
5. C
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Re: GMATPREP ChallengeQ -In a new book about the antiparty feeli [#permalink]
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21 Aug 2014, 02:24
maggie27 wrote:
10 mins 47 secs
1. B
2. E (Although Ketcham does not picture the struggle in quite this way, he does rightly see Jackson's tenure (the seventh presidency) as the culmination of the acceptance of party, commerce, and individualism. )
3. D
4. A
5. C
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Re: GMATPREP ChallengeQ -In a new book about the antiparty feeli [#permalink]
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19 Sep 2014, 06:58
8:41. All correct! thanks for the boost!
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Re: GMATPREP ChallengeQ -In a new book about the antiparty feeli [#permalink]
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20 Sep 2014, 23:09
10 min 41 secs All Correct!!!!
How will one rate this passage in terms Difficulty ?
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Re: GMATPREP ChallengeQ -In a new book about the antiparty feeli [#permalink]
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04 Nov 2014, 23:51
Could anyone please explain why option A is only chosen for Question 4 ( Which of the following provides LEAST Support)
My take from a given line in the passage: Since Commerce starting coming up under Jackson's rule, Commercialization was considered as the enemy of the classical ideals.........Although Ketcham does not cover this STRUGGLE.
So basically what is implied is that commerce had to struggle to come its way up. Then WHY is option A which in a way states that struggle(many protests against commercialization) considered as the one that provides LEAST support?
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26 Nov 2014, 22:59
time taken: 10:00
4th question incorrect.
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Re: GMATPREP ChallengeQ -In a new book about the antiparty feeli [#permalink]
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14 Apr 2015, 06:20
dream21 wrote:
10 min 41 secs All Correct!!!!
How will one rate this passage in terms Difficulty ?
9:30 and one mistake.
It's an easy passage so I will evaluate it about a 500 pts passage but i'm not a specialist.
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Re: GMATPREP ChallengeQ -In a new book about the antiparty feeli [#permalink]
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26 Jun 2015, 07:10
1B 2E 3D 4A 5C All Correct
5 minutes 26 seconds
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22 Dec 2015, 15:43
could someone explain why 3 is D and why 5 isn't E?
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Re: GMATPREP ChallengeQ -In a new book about the antiparty feeli [#permalink]
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11 May 2016, 06:56
Hi,
Thank you for posting such good question however I am unable to understand why 1B is correct , if someone can please help me understand . Would really appreciate. As per my reasoning E should be the answer of Q1. Please help.
Regards
Megha
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Re: GMATPREP ChallengeQ -In a new book about the antiparty feeli [#permalink]
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25 Jul 2016, 21:22
Time Taken=9 min
Got 4th question wrong...
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Re: GMATPREP ChallengeQ -In a new book about the antiparty feeli [#permalink]
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26 Jul 2016, 01:54
9 min,40 seconds.
All correct
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Re: GMATPREP ChallengeQ -In a new book about the antiparty feeli [#permalink]
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03 Aug 2016, 11:50
13 mins
all correct
got stuck on the 4th question ,wasted too much time.
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16 Aug 2016, 23:56
Interesting passage , took a lot of time to solve 4th question , still got it incorrect
10 mins , including 3 mins to read
- antiparty feeling of the early political leaders of the United States - classical conception of leadership
- commercial forces - profit-making, its self-interestedness, its individualism--became the enemy of these classical ideals
- classical conception of leadership was incompatible with our modern notion of the freedoms of speech and press
1.
B. describing and analyzing an argument about the early history of the United States
2 . " Although Ketcham does not picture the struggle in quite this way, he does rightly see Jackson's tenure (the seventh presidency) as the culmination of the acceptance of party, commerce, and individualism."
3. " the classical conception of leadership was incompatible with our modern notion of the freedoms of speech and press, freedoms intimately associated with the legitimacy of opposing political parties."
5. Ketcham is so strongly committed to justifying the classical ideals, however, that he underestimates the advantages of their decline.
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17 Aug 2016, 00:15
Quote:
4. Which of the following, if true, provides the LEAST support for the author's argument about commerce and political parties during Jackson's presidency ?
A. Many supporters of Jackson resisted the commercialization that could result from participation in a national economy.
B. Protest against the corrupt and partisan nature of political parties in the United States subsided during Jackson's presidency.
C. During Jackson's presidency the use of money became more common than bartering of goods and services.
D. More northerners than southerners supported Jackson because southerners were opposed to the development of a commercial economy.
E. Andrew Jackson did not feel as strongly committed to the classical ideals of leadership as George Washington had felt.
The below is Ron's answer to the above question - https://www.manhattanprep.com/gmat/foru ... 11468.html
first, you have to figure out what the highlighted thing means. (in general, if a question prompt contains vague words, you must clarify the vague language before answering the question.)
the highlighted thing, from the passage, means:
"he does rightly see Jackson's tenure (the seventh presidency) as the culmination of the acceptance of party, commerce, and individualism. "
(B) they started to accept parties, i.e., part of the highlighted thing.
(C) they started to accept modern commerce, i.e., part of the highlighted thing.
(D) jackson's opponents opposed him because he supported commerce, i.e., part of the highlighted thing.
(E) the classical ideals were anti-parties, so jackson's drifting away from those ideals supports the highlighted thing.
here (a) is the clear favorite, since it's the only answer choice that directly contradicts the highlighted thing.
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Re: GMATPREP ChallengeQ -In a new book about the antiparty feeli [#permalink]
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22 Aug 2016, 00:26
1
This post was
BOOKMARKED
Another question from this passage:
The author of the passage would be most likely to agree with which of the following statements about Ketcham?
a) He overemphasizes the influence of classical ideals on the first six presidents of the US.
b) He fails to recognize that classical ideals has little influence on politics in the US.
c) He does not pay adequate attention to the negative aspects of the first six presidents' commitment to classical ideals.
d) He inaccurately suggests that classical ideals gave rise to our modern notion of democracy.
e) He underestimates the effect of ideology other than the humanist tradition on the first six president.
OA: C
I narrowed it down to A & C but ended up choosing A which was incorrect. I picked it because the passage says "Even during the first presidency (Washington's), however, the classical conception of virtuous leadership was being undermined by commercial forces that had been gathering since at least the beginning of the eighteenth century." So that's why thought that Ketcham had "overemphasized the influence of classical ideals" even since the first president.
In hindsight, I understand why C is correct but not clear enough for me to make the distinction on the actual test. Can you better explain why C is correct and how I can avoid picking an answer like A?
Thanks!
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Re: GMATPREP ChallengeQ -In a new book about the antiparty feeli [#permalink]
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28 Oct 2016, 23:47
Time Taken : 10 mins. All correct. Easy passage. But a lengthy one.
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Re: GMATPREP ChallengeQ -In a new book about the antiparty feeli [#permalink] 28 Oct 2016, 23:47
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Display posts from previous: Sort by | 5,912 | 23,764 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2017-04 | latest | en | 0.825242 |
https://www.doorsteptutor.com/Exams/NSTSE/Class-5/Questions/Topic-Mathematics-0/Subtopic-Factors-and-Multiples-1/Part-3.html | 1,524,640,300,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947705.94/warc/CC-MAIN-20180425061347-20180425081347-00624.warc.gz | 781,292,958 | 10,661 | # Mathematics-Factors and Multiples (NSTSE (National Science Talent Search Exam- Unified Council) Class 5): Questions 17 - 24 of 84
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## Question number: 17
» Mathematics » Factors and Multiples
MCQ▾
### Question
How many prime numbers are there from 1 to 50?
### Choices
Choice (4) Response
a.
17
b.
20
c.
15
d.
10
## Question number: 18
» Mathematics » Factors and Multiples
MCQ▾
### Question
The figure given shows 5 numbers. (Model Test Paper)
Which of the 5 numbers are the prime factors of 130?
### Choices
Choice (4) Response
a.
3 and 10
b.
2 and 3
c.
10 and 13
d.
2 and 13
## Question number: 19
» Mathematics » Factors and Multiples
MCQ▾
### Question
Identify an example for twin primes.
### Choices
Choice (4) Response
a.
5, 9
b.
c.
21, 25
d.
13, 15
## Question number: 20
» Mathematics » Factors and Multiples
MCQ▾
### Question
Which of the following numbers which have no common factor except 1?
### Choices
Choice (4) Response
a.
Twin primes
b.
Co-primes
c.
Composite numbers
d.
None of the above
## Question number: 21
» Mathematics » Factors and Multiples
MCQ▾
### Question
How many prime numbers are there from 1 to 100?
### Choices
Choice (4) Response
a.
27
b.
20
c.
15
d.
25
## Question number: 22
» Mathematics » Factors and Multiples
MCQ▾
### Question
Study the numbers given in the box.
Identify the number that is not a divisor of the given numbers.
### Choices
Choice (4) Response
a.
39
b.
14
c.
26
d.
13
## Question number: 23
» Mathematics » Factors and Multiples
MCQ▾
### Question
How many times the H. C. F. of and 77 is their L. C. M. ?
### Choices
Choice (4) Response
a.
44
b.
3630
c.
11
d.
24
## Question number: 24
» Mathematics » Factors and Multiples
MCQ▾
### Question
Which of the following are multiples of 2?
### Choices
Choice (4) Response
a.
Composite number
b.
Odd number
c.
Prime number
d.
Even number
f Page | 636 | 2,217 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-17 | latest | en | 0.664308 |
https://rebmon.com/combination-of-addition-and-subtraction/ | 1,679,512,684,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296944452.74/warc/CC-MAIN-20230322180852-20230322210852-00185.warc.gz | 548,289,098 | 28,577 | # Combination of Addition and Subtraction
Right here we are going to speak in regards to the mixture of addition and subtraction.
As an instance issues involving each addition and subtraction, we first group all numbers with ‘+’ and ‘-‘ indicators. After discovering the sum of the numbers with a ‘+’ signal and equally the sum of all of the numbers with a ‘-‘ signal, we discover the distinction. If there isn’t a signal earlier than a quantity, it means a ‘+’ signal.
Listed below are the principles that can be utilized to resolve for addition (+) and subtraction (-) sums collectively:
I: Gather first, then take out.
II: Resolve the numbers in parentheses and remedy later.
To unravel questions that contain each addition and subtraction, we first add the required quantity after which subtract it from the whole.
one. Let’s take 11 + 26 – 12.
Step I: Add 11 and 26 first. The sum of 11 and 26 is 37. Step II: Now, subtract 12 from 37. 37 – 12 = 25
Reply: 11 + 26 – 12 = 25
2. Let’s give one other instance. A shopkeeper has 90 oranges. He offered 51 oranges on Monday and 25 on Tuesday. What number of oranges are left with you?
WITH Oranges offered on Monday 5 1 Oranges offered on Tuesday + 2 5 Complete oranges offered 7 6
Remaining Oranges = Complete Oranges – Oranges Offered
WITH whole oranges 9 0 oranges on the market – 7 6 Variety of Oranges Remaining 1 4
Solved Instance with Addition (+) and Subtraction (-) Collectively:
1. Resolve 3252 – 1802 + 2610
3 2 5 2
+ 2 6 1 0
5 8 6 2
Now
5 8 6 2
– 1 8 0 2
4 0 6 0
2. Discover the answer of (7259 – 5369) + 2156
6 11 15
7 2 5 9
5 3 6 9
1 8 9 0
Now
eleventh
1 8 9 0
+ 2 1 5 6
4 0 4 6
3. Subtract 4926 from the sum of 3051 and 4257
Sum of 3051 and 4257
one
3 0 5 1
+ 4 2 5 7
7 3 0 8
Now subtract 4926 from 7308
6 12 10
7 3 0 8
4 9 2 6
2 3 8 2
4. Resolve 481 + 356 – 168.
Step I: Add the numbers with the ‘+’ signal.
Step II: Subtract the ‘-‘ signal from the sum of the ‘+’ signal numbers.
Due to this fact, 481 + 356 – 168 = 669
5. 4235 – 4732 + 3543
pearl H T HE 4 2 3 5 + 3 5 4 3 7 7 7 8 Step I: First add 4235 + 3543
pearl H T HE 7 7 7 8 – 4 7 3 2 3 4 4 6 Step II: Subtract 4732 from the results of step I ie 7778
To recollect:
At all times settle for the + signal if there isn’t a +/- signal earlier than numbers.
Add the quantity first, then subtract.
Subtract numbers with a ‘-‘ signal.
Phrase Issues Associated to Addition and Subtraction:
6. Daniel has 478 chocolate chip cookies and 245 chocolate chip cookies. He distributed 323 cookies amongst his pals. What number of cookies are left with you?
Step I: Further
Complete variety of cookies = 478 + 245 = 753
Step II: Extraction
Cookies left with Daniel = 753 – 323 = 430
Questions and Solutions on Addition and Subtraction:
1. Fill within the blanks –
(i) 5555 – 3213 + 4235
pearl H T HE 5 5 5 5 + 4 2 3 5 pearl H T HE __ _ _ – 3 2 1 3
(ii) 5362 – 4141 + 2222
pearl H T HE 5 3 6 2 + 2 2 2 2 pearl H T HE __ _ _ –
(iii) 6247 + 2431 – 5132
pearl H T HE 6 2 4 7 + 2 4 3 1 pearl H T HE __ _ _ – 5 1 3 2
(iv) 4321 + 1541 – 3323
pearl H T HE 4 3 2 1 + 1 5 4 1 pearl H T HE __ _ _ –
one. (I)
pearl H T HE 5 5 5 5 + 4 2 3 5 9 7 9 0 pearl H T HE 9 7 9 0 – 3 2 1 3 6 5 7 7
(ii)
pearl H T HE 5 3 6 2 + 2 2 2 2 7 5 8 4 pearl H T HE 7 5 8 4 – 4 1 4 1 3 4 4 3
(iii)
pearl H T HE 6 2 4 7 + 2 4 3 1 8 6 7 8 pearl H T HE 8 6 7 8 – 5 1 3 2 3 5 4 6
(iv)
pearl H T HE 4 3 2 1 + 1 5 4 1 5 8 6 2 pearl H T HE 5 8 6 2 – 3 3 2 3 2 5 3 9
2. Make the next sums in your pocket book –
(i) 2222 + 3333 – 1111
(ii) 3122 + 4212 – 2112
(iii) 4364 + 3213 – 6243
(iv) 5154 + 2343 – 4265
(v) 8546 – 6324 – 2112
(vi) 9798 – 7656 + 7554
(vii) 8746 – 5534 + 2465
(viii) 3778 – 2563 + 4672
(ix) 2432 + 1342 – 1121
(x) 4365 + 434 – 1213
(xi) 3794 + 1213 – 2505
(xii) 7811 – 1222 + 234
(xiii) 5234 + 3259 – 4172
(xiv) 3777 + 2111 – 4352
(xv) 2725 + 1234 – 786
(xvi) 2575 + 2430 – 1225
(xvii) 7250 – 5450 + 3550
(xviii) 2775 + 1465 – 1275
(xix) 2537 – 2425 + 1212
(xx) 4533 – 2322 + 289
2. (i) 4444
(ii) 5222
(iii) 1334
(iv) 3232
(v) 110
(vi) 9696
(vii) 5677
(viii) 5887
(ix) 2653
(x) 3586
(xi) 2502
(xii) 6823
(xiii) 4321
(xiv) 1536
(xv) 3173
(xvi) 3780
(xvii) 5350
(xviii) 2965
(xix) 1324
(xx) 2500 | 1,787 | 4,308 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 5.03125 | 5 | CC-MAIN-2023-14 | latest | en | 0.847014 |
https://kidsworksheetfun.com/area-model-multiplication-worksheets/ | 1,713,078,208,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816875.61/warc/CC-MAIN-20240414064633-20240414094633-00525.warc.gz | 315,271,308 | 25,460 | # Area Model Multiplication Worksheets
Area Model Multiplication Worksheets. (feel free to use a separate piece of paper to show your work.) partner b. Math in basic terms is frequently all.
Multiplication property (valentine’s day themed) math worksheets multiplication tables (st. In this case a= (16.2 mm) (2.3 mm) then, use the area model representation to find the answer. 3 times table by andrea727:
### Math In Basic Terms Is Frequently All.
Change 16.2 and 2.3 to whole numbers and break up the numbers according to place value. Students will get an opportunity to work with the distributive property in this worksheet. Worksheets are division, area model division work no remainders, array area model for division, area model division, division made easy, modeling multiplication and division of fractions, fraction multiplication and division models, grade 4 supplement.
### 16.2 → 162 → 100 + 60 + 22.3 → 23 → 20 + 3.
The questions in the worksheets have a gradual increase in the level of difficulty. Multiply by one digit by hasker: The area model is also known as the box model.
### Find The Products Of The Numbers On The Bottom Layer.
Multiply fractions with area models. 2 digit multiplication with an area model worksheet via : Multiplication skills is a huge and inportant subject.using these worksheets to practice multiplying 2 dg x 2dg by use the area model multiplication,children can develop their understanding of multiplication and build upon their numeracy skills.worksheets include:2 digits x 2.
### Addition Worksheets And Subtraction Worksheets Aren’t What Most Young Children Need To Be Carrying Out Throughout Their Working Day.
The worksheet requires students to work with a set of problems on rewriting the multiplication expressions using the area model. Our pdf area model multiplication worksheets will get the learners upbeat and coming back for more! The area model of solving multiplication problems is derived from the concept of finding the area of a rectangle.
### Ask The Class How Many Inches Long It Is.
At first the kids may well not take pleasure in being given additional research although the positive aspects they may gain from honing the noticed and practicing. Radical 4 math area model puzzles via : *click on open button to open and print to worksheet. | 478 | 2,319 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | latest | en | 0.850345 |
http://tomewbank.com/forum/z9lhg1s.php?page=knitting-for-charity-near-me-d44ecc | 1,620,427,434,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988828.76/warc/CC-MAIN-20210507211141-20210508001141-00111.warc.gz | 49,593,677 | 10,011 | # knitting for charity near me
Figure 2. Such telescopes can gather more light, since larger mirrors than lenses can be constructed. A current exciting development is a collaborative effort involving 17 countries to construct a Square Kilometre Array (SKA) of telescopes capable of covering from 80 MHz to 2 GHz. The mirrors are extremely smooth and consist of a glass ceramic base with a thin coating of metal (iridium). Although Galileo is often credited with inventing the telescope, he actually did not. In this equation, 16 cm is the standardized distance between the image-side focal point of the objective lens and the object-side focal point of the eyepiece, 25 cm is the normal near point distance, and are the focal distances for the objective lens and the eyepiece, respectively. Simple telescopes can be made with two lenses. (a) Galileo made telescopes with a convex objective and a concave eyepiece. X rays ricochet off 4 pairs of mirrors forming a barrelled pathway leading to the focus point. The magnification, M, of a two-lens system is equal the product of the magnifications of the individual lenses: M = M 1 M 2 = (- d i1 / d o1) (- d i2 / d o2) Object at Infinity Look through the lenses at a distant object. He constructed several early telescopes, was the first to study the heavens with them, and made monumental discoveries using them. (b) What distance between the lenses will allow the telescope … Although it may seem like a crude device, a simple telescope nicely illustrates the basic working principles of more powerful astronomical instruments. A Galilean telescope has an objective lens with f 1 = 20 cm and the eyepiece lens with f 2 = -5 cm. Also, use the principal ray through the center of each lens to derive the angular magnification of the telescope: M= - … The focal length of the objective is +2.25 m and the angular magnification is magnitude 14. The greater the angular magnification M, the larger an object will appear when viewed through a telescope, making more details visible. The distance between the lenses is just their sum qo + pe. i = distance from lens to image. The project will use cutting-edge technologies such as adaptive optics in which the lens or mirror is constructed from lots of carefully aligned tiny lenses and mirrors that can be manipulated using computers. A telescope, in its original configuration (refractor), consists of two lenses. A 7.5× binocular produces an angular magnification of −7.50, acting like a telescope. The telescope eyepiece (like the microscope eyepiece) magnifies this first image. The first image is thus produced at di = fo, as shown in the figure. Therefore. To prove this, note that M 11 = m θ = +4 is the angular magnification. What he did was more important. Telescope Calculator Results: Focal Length: The distance (usually expressed in millimeters) from a mirror or lens to the image that it forms. An artist’s impression of the Australian Square Kilometre Array Pathfinder in Western Australia is displayed. Stars are so unimaginably far away that the light we receive from them arrives in rays that are perfectly parallel. Calculate the matrix for this system and find m θ. To prove this, note that. Find the distance between the objective and eyepiece lenses in the telescope in the above problem needed to produce a final image very far from the observer, where vision is most relaxed. The angular magnification M for a telescope is given by $M=\frac{\theta^{\prime}}{\theta }=-\frac{{f}_{\text{o}}}{{f}_{\text{e}}}\\$, where. The third lens acts as a magnifier and keeps the image upright and in a location that is easy to view. The image in most telescopes is inverted, which is unimportant for observing the stars but a real problem for other applications, such as telescopes on ships or telescopic gun sights. Nosotros y nuestros socios almacenaremos y/o accederemos a la información de tu dispositivo mediante el uso de cookies y tecnologías similares, a fin de mostrar anuncios y contenido personalizados, evaluar anuncios y contenido, obtener datos sobre la audiencia y desarrollar el producto. Figure 1. The minus sign indicates the image is inverted. These produce an upright image and are used in spyglasses. The second lens, the eyepiece, catches the light as it … The most common two-lens telescope, like the simple microscope, uses two convex lenses and is shown in Figure 1b. Some telescopes use extra lenses and/or mirrors to create a long effective focal length in a short tube. A telescope, in its original configuration (refractor), consists of two lenses. A telescope can also be made with a concave mirror as its first element or objective, since a concave mirror acts like a convex lens as seen in Figure 3. Because $\frac{1}{\infty}=0\\$, this simplifies to $\frac{1}{d_{\text{i}}}=\frac{1}{f_{\text{o}}}\\$, which implies that di = fo, as claimed. (a) What distance between the two lenses will allow the telescope to focus on an infinitely distant object and produce an infinitely distant image? If you use a concave lens for the eyepiece, then the distance between lenses needs to be the difference of their focal lengths, F - f. This telescope forms an image in the same manner as the two-convex-lens telescope already discussed, but it does not suffer from chromatic aberrations. If an upright image is needed, Galileo’s arrangement in Figure 1a can be used. Figure 1a shows a telescope made of two lenses, the convex objective and the concave eyepiece, the same construction used by Galileo. Unless otherwise stated, the lens-to-retina distance is 2.00 cm. To obtain the greatest angular magnification, it is best to have a long focal length objective and a short focal length eyepiece. Simple telescopes can be made with two lenses. X rays, with much more energy and shorter wavelengths than RF and light, are mainly absorbed and not reflected when incident perpendicular to the medium. The eyepiece forms a case 2 final image that is magnified. The first two lenses are far enough apart that the second lens inverts the image of the first one more time. In case of an astronomical telescope, the distance between the objective lens and the eyepiece is equal to : (final image is at ∞) View Answer The focal lengths of objective and eye lens of an astronomical telescope are respectively 2 meter and 5 cm. Telescopes, like microscopes, can utilize a range of frequencies from the electromagnetic spectrum. Basic Telescope Optics. Distance between two lenses of a telescope? A large reflecting telescope has an objective mirror with a 10.0 m radius of curvature. Para obtener más información sobre cómo utilizamos tu información, consulta nuestra Política de privacidad y la Política de cookies. The objective forms a case 1 image that is the object for the eyepiece. The mirrors for the Chandra consist of a long barrelled pathway and 4 pairs of mirrors to focus the rays at a point 10 meters away from the entrance. Four pairs of precision manufactured mirrors are exquisitely shaped and aligned so that x rays ricochet off the mirrors like bullets off a wall, focusing on a spot. It can be shown that the angular magnification of a telescope is related to the focal lengths of the objective and eyepiece; and is given by, $\displaystyle{M}=\frac{\theta^{\prime}}{\theta}=-\frac{f_{\text{o}}}{f_{\text{e}}}\\$. Distance between two lenses of a telescope? But they can be reflected when incident at small glancing angles, much like a rock will skip on a lake if thrown at a small angle. To determine the image distance, the lens equation can be used. The simplest answer is that there’s none: a pair of binoculars is, in essence, a pair of refracting telescopes mounted in parallel. Figure 4. A two-element telescope composed of a mirror as the objective and a lens for the eyepiece is shown. Instruments to make the image of the Australian Square Kilometre Array Pathfinder in Australia! Microscopes, can utilize a range of frequencies from the electromagnetic spectrum is 2.00.... Than lenses can be minimized by deforming or tilting the tiny lenses and mirrors study heavens! Chandra Observatory, a simple telescope nicely illustrates the basic working principles more. The focal length of the electromagnetic spectrum two-lens telescope, like the microscope )! Than lenses can be constructed stated, the image upright and in a reflecting telescope does exactly the thing. 2.50 cm focal length of the project is the angular magnification m, the larger an object will appear viewed., distance from Earth 385000 km ) mirrors to create a long focal lengths f1 = +24.1 cm f2. Allowing dim objects to be observed with greater magnification and better resolution angle is subtended a... Of more powerful astronomical instruments principles of more powerful astronomical instruments off 4 pairs of mirrors forming barrelled... Image in the figure by many factors, including lens quality and atmospheric disturbance through a telescope is used view! Off 4 pairs of mirrors forming a barrelled pathway leading to the focus point used by Galileo telescope requires more... Two-Lens telescope, in its original configuration ( refractor ), consists of lenses... Magnification is possible Australian Square Kilometre Array Pathfinder in Western Australia ( see figure 5 ) of! { \prime } } { \theta } \\ [ /latex ] −7.50, like. Angular magnification of a telescope is negative., uses two convex lenses and mirrors tus. [ /latex ] telescope that has a concave eyepiece larger an object will appear when viewed through telescope! Very long focal length of the first one, the larger an object will appear when through... Telescopes the focal length objective is +2.25 m and the concave eyepiece Australia telescope Compact Array, which six. M and the concave eyepiece, the same thing. +5.5 cm factors, including lens quality and disturbance! Has lenses with focal lengths of the two lenses s arrangement in figure 1a can be minimized by deforming tilting... | 2,197 | 9,979 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2021-21 | longest | en | 0.917041 |
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# CYCLES IN CASUALTY: - PowerPoint PPT Presentation
CYCLES IN CASUALTY:. Balancing Loops in the Insurance Industry Kawika Pierson MIT Sloan PhD Candidate. Presentation Outline. The Insurance Industry Past Research Economics Control Theory System Dynamics The Model Boundary Causal Loop Diagram Important Structures PID Control
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### CYCLES IN CASUALTY:
Balancing Loops in the Insurance Industry
Kawika Pierson
MIT Sloan PhD Candidate
Presentation Outline
• The Insurance Industry
• Past Research
• Economics
• Control Theory
• System Dynamics
• The Model
• Boundary
• Causal Loop Diagram
• Important Structures
• PID Control
• Behavior
• How You Can Help
The Insurance Industry
• Basic Idea
• Two Sides to the Business
• Insurance
• Investing
• Insurance Cycle – What is Cycling?
• Underwriting Loss Ratio or Combined Loss Ratio
• Expense Ratio – Expenses/Premiums
• Combined Ratio – Loss + Expense = (A + E) / P
The Insurance Industry
• Insurance Cycle – What Causes It?
• Industry View:
• “The next stage is precipitated by a catastrophe or similar significant loss, for example Hurricane Andrew or the attacks on the World Trade Center.” – “The Insurance Cycle” wikipedia
• “Using quarterly data from 1974 through 1990, we provide evidence of a long-run link between the general economy and the underwriting performance as measured by the combined ratio.” – Grace and Hotchkiss, 1995 J o Risk and Insurance
• “Fluctuations in the supply of property-liability insurance may be exacerbated by regulation.” Winter, 1991 Economic Inquiry
Past Research in Economics
• Early 1980’s through Mid 90’s
• Three Main Schools of Thought
• Cycle Caused by Interest Rate Fluctuations
• Doherty and Kang (1988) – Insurance Prices Change in Lagged Response to Interest Rates
• Grace and Hotchkiss (1995) – “External Impacts on the Property-Liability Insurance Cycle”
• Cycle Caused by Limits to the Supply of Insurance
• Winter (1988, 1991, 1994), Gron (1989, 1994)
• Cycle Caused by Feedback Processes
• Brockett and Witt (1982) – Loss expectations from the past inform current premiums, causing autocorrelation
Past Research in Control Theory
• If a Cycle Exists we Will Create a Lagged Negative Feedback Loop to Explain It
• Balzer and Benjamin 1980 – “Dynamic Response of Insurance Systems with Delayed Profit/Loss Sharing Feedback…” Journal of the Institute of Actuaries
• Zimbidis and Haberman 2001 – “The Combined Effect of Delay and Feedback on the Insurance Pricing Process: a Control Theory Approach” Insurance: Mathematics and Economics
Past Research in System Dynamics
• The Claims Game and Hanover Insurance
• “claims management, quality and costs”
• Quality = Claim Adjustment Quality
• Daniel H. Kim
• Learning Laboratories
• Peter Senge – “The Fifth Discipline”
• Moissis 1989 Masters Thesis (Sterman)
• Focuses on Determining Decision Rules
• Cavaleri and Sterman (1997) “Towards evaluation of systems thinking interventions: a case study”
• Improved Manager’s Mental Models
Past Research in System Dynamics
• Insurance Cycle…
• Are There Really no SD Articles on the Insurance Cycle?
• Thomas Beck
• Co-President of Swiss SD Society
• Works for Large Swiss Reinsurer
• No Published Articles on Insurance Cycle
The Model – Boundary
• Endogenous Variables
• Underwriting Quality (Risk)
• Claims
• Employees
• Exogenous Variables
• Desired Profit Margin
• Size of the Total Market
• Some Components of Administrative Costs
The Model – Boundary
• Many Feedbacks Excluded
• Size of the Insurance Market
• Investments and Interest Rates
• Free Capital’s Influence on Underwriting
• Effect of Time Pressure on Claim Settlement
• Competitive Effects on Profit Margins
• Random Claim Incidence
• Employee Productivity
• Is this Too Far Towards “Negative Loop w/ Delay”
The Model – PID Control
• Translating Equations to SD isn’t Always Easy
• Proportional Control = Standard Structure
• Integral Control = No Steady State Error
• Reasonable that People Use IC
• Derivative Control = Less Overshoot
• Less Likely that People Use DC
The Model – Behavior
• Displays Decaying Oscillation to Step Input
The Model – Behavior
• Instability A Function of Largest Source of Costs
The Model – Behavior
• Loop Gain Very Important
The Model – Potential Solutions
• Derivative Control of Premiums?
• Careful Tuning Is Necessary
• Managerial Implementation
• Industry Wide Application
• Why Do Quality Standards Change?
• Can This Loop Be Cut
• Life Insurance
• The Kalmanuclear Option?
• Optimal LINEAR Filter
• Just Build a Really Good Model Instead | 1,235 | 5,279 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2018-05 | latest | en | 0.793174 |
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Dear Readers, Exam Race for the Year 2019 has already started, To enrich your preparation here we are providing new series of Practice Questions on Quantitative Aptitude โ Section for CWC/FCI Exam. Aspirants, practice these questions on a regular basis to improve your score in aptitude section. Start your effective preparation from the right beginning to get success in upcoming CWC/FCI Exam.
CWC/FCI Prelims Quantitative Aptitude Questions (Day-32)
maximum of 10 points
Directions (1 – 5): What value should come in the place of question mark (?) in the following questions?
1) (4 ร 4)3 รท (512 รท 8)4 ร (32 ร 8)4 = (4)? + 4
a) 8
b) 5
c) 7
d) 6
e) None of these
2) 76 % of 1285 โ 35 % of 1256 =?
a) 537
b) 629
c) 451
d) 787
e) None of these
3) (2/7) of 5033 + 78 % of 650 = (?)2 + 181
a) 48
b) 46
c) 42
d) 39
e) None of these
4) 32 % of 480 + (5/7) of 1890 โ 28 % of 820 =?
a) 1336
b) 1178
c) 1452
d) 1274
e) None of these
5) ? *100/(140 ร 8 โ 680) = 330 + 452
a) 9857
b) 10362
c) 11238
d) 8564
e) None of these
Directions (6 โ 10): Study the following information carefully and answer the given questions:
Following table shows the total number items produced by 5 different companies (In thousands) and the percentage of items sold and the cost per items.
6) Total number of items sold by the company A is approximately what percentage of total number of items sold by the company C?
a) 100 %
b) 90 %
c) 80 %
d) 115 %
e) 65 %
7) Find the total cost earned by the company B and D together?
a) 1.728 Crore
b) 1.856 Crore
c) 1.642 Crore
d) 1.944 Crore
e) None of these
8) Find the ratio between the total number of items sold by the company B to that of company D?
a) 25: 36
b) 64: 95
c) 52: 77
d) 48: 69
e) None of these
9) Find the difference between the total cost earned by the company A to that of company E?
a) 0.351 Crore
b) 3.51 Crore
c) 0.0351 Crore
d) 35.1 lakhs
e) None of these
10) Total number of items produced by the company A and D together is approximately what percentage more/less than the total number of items produced by the company B and C together?
a) 25 % more
b) 20 % less
c) 15 % more
d) 15 % less
e) 20 % more
Direction (1-5) :
46 รท (512/8)4 ร (32 ร 8)4 = (4)x + 4
46 รท 644 ร 2564 = (4)x + 4
46 รท 412 ร 416 = 4x + 4
46 โ 12 + 16 = 4x + 4
410 = 4x + 4
x + 4 = 10
x = 6
(76/100)*1285 โ (35/100)*1256 = X
X = 976.6 โ 439.6
X = 537
(2/7)*5033 + (78/100)*650 = x2 + 181
1438 + 507 โ 181 = x2
X2 = 1764
X = 42
(32/100)*480 + (5/7)*1890 โ (28/100)*820 = x
X = 153.6 + 1350 โ 229.6
X = 1274
(x*100)/(1120 โ 680) = 330 + 2025
x/100)*(440) = 2355
x = 2355*(440/100)
x = 10362
Direction (6-10) :
Total number of items sold by the company A
= > 78*(60/100)
Total number of items sold by the company C
= > 82*(58/100)
Required % = {[78*(60/100)]/ [82*(58/100)]}*100 = 98 % = 100 %
The total cost earned by the company B and D together
= > 64*(72/100)*125 + 90*(76/100)*200
= > (5760 + 13680) thousands
= > 1.944 Crore
The total number of items sold by the company B
= > 64*(72/100)
The total number of items sold by the company D
= > 90*(76/100)
Required ratio = [64*(72/100)]: [90*(76/100)]
= > 64: 95
The total cost earned by the company A
= > 78*(60/100)*250 = 11700 thousand
= > 1.17 Crore
The total cost earned by the company E
= > 65*(84/100)*150 = 8190 thousand
= > 0.819 Crore
Required difference = 1.17 โ 0.819 = 0.351 Crore
Total number of items produced by the company A and D together
= > 78 + 90 = 168000
Total number of items produced by the company B and C together
= > 64 + 82 = 146000
Required % = [(168000 โ 146000)/146000]*100 = 15 % more
Online Mock Tests 2019:
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# X 1 2 3 4 5 6 P(X=x) k 2k 3k 4k 5k 6kThe probability distribution of discrete r.v. $$X$$ is Then $$P(X\le 4)$$ =
Solution
## Sum of the probabilities is equal to 1$$\therefore k+2k+3k+4k+5k+6k=1$$$$21k=1$$$$k=\dfrac{1}{21}$$$$P(X\leq 4)$$$$=P(1)+P(2)+P(3)+P(4)$$$$=k+2k+3k+4k$$$$=\dfrac{1}{21}+\dfrac{2}{21}+\dfrac{3}{21}+\dfrac{4}{21}$$$$=\dfrac{1+2+3+4}{21}$$$$=\dfrac{10}{21}$$$$\therefore P(X\leq 4)=0.476$$Applied Mathematics
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1. /**
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4. public class Rectangle
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10. Constructs a rectangle.
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13. */
14. public Rectangle(int aHeight, int aWidth)
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21. Constructs a rectangle that is a square.
22. @param aSide: the length of a side of the square
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24. public Rectangle(int aSide)
25. {
26. height = aSide;
27. width = aSide;
28. }
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30. /**
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32. @return the perimeter of the rectangle
33. */
34. public int perimeter()
35. {
36. return 2 * height + 2 * width;
37. }
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39. /**
40. Computes the area.
41. @return the area of the rectangle
42. */
43. public int area()
44. {
45. return height * width;
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8 2021 年以降の合計貢献数
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Contact Us: Have a question, suggestion, or concern? Visit our Help Center for assistance!Suggest-A-Calc: We are always looking for innovative, high-quality products. If you don't see your favorite calculator let us know and we'll notify you when it becomes available! Did you know: We have many calculator products that aren't yet listed on our website! CalculatorSource Home | Construction Calculators | Surveying Calculators | Real Estate Calculators | Electrical Calculators | Teacher Calculators | Student Calculators | Scientific Calculators | Graphing Calculators | Metric Calculators | Printing Calculators | Checkbook Calculators | Film and Video Calculators | Cooking Calculators | Boating Calculators | Gaming Calculators | Home Project Calculators | Accounting and Payroll Calculators | Scheduling and Billing Calculators | Aviation Calculators | Medical Calculators | Finance and Business Calculators | Discontinued Calculators | Calculator Accessories | Bookstore | Clearance Center | Custom Imprinting | Used Calculators | Sell Your Calculator Info and Our Policies | Help Center | Search | Product Index Copyright © 2002 intelliHoldings. | 751 | 3,386 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | longest | en | 0.816338 |
https://www.cloudtechtwitter.com/2022/05/question-28-find-leaders-in-array.html | 1,725,781,118,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650960.90/warc/CC-MAIN-20240908052321-20240908082321-00151.warc.gz | 684,630,208 | 20,388 | ## Wednesday, May 4, 2022
### Question 28 : Find leaders in an array.
We need to print all the leaders present in the array. Element is the leader if it is greater than right side of elements.
For example:
``````arr[]={14, 12, 70, 15, 99, 65, 21, 90}
Here 99 and 90 are leader elements``````
#### Solution 1:
Use two loops. Outer loop to iterate over array elements and inner loop to check for right elements of the array.
If the current element is greater than the right element then it is a leader.
Java code:
``````public static void findLeadersInAnArrayBruteForce(int arr[])
{
System.out.println("Finding leaders in an array using brute force : ");
for (int i = 0; i < arr.length; i++) {
for (int j = i+1; j < arr.length; j++) {
if(arr[i] <= arr[j])
{
break;
}
}
System.out.print(arr[i]+" ");
}
}``````
Time complexity : o(N^2)
#### Solution 2:
Let's find more optimized solution
We will use the property that the rightmost element is always a leader.
• We will start from the rightmost element and track max.
• Whenever we get the new max, that element is a leader.
Java code:
``````public static void findLeadersInAnArray(int arr[])
{
System.out.println("Finding leaders in an array : ");
int rightMax=arr[arr.length-1];
// Rightmost will always be a leader
System.out.print(rightMax+" ");
for (int i = arr.length-2; i>=0; i--) {
if(arr[i] > rightMax)
{
rightMax=arr[i];
System.out.print(" "+rightMax);
}
}
}``````
Time complexity : o(N)
### Java Program to find leaders in the array::
``````public class FindLeadersInArrayMain {
public static void main(String[] args) {
int arr[]={14, 12, 70, 15, 99, 65, 21, 90};
System.out.println("n==================");
}
{
System.out.println("Finding leaders in an array using brute force : ");
for (int i = 0; i < arr.length; i++) {
for (int j = i+1; j < arr.length; j++) {
if(arr[i] <= arr[j])
{
break;
}
}
System.out.print(arr[i]+" ");
}
}
{
System.out.println("Finding leaders in an array : ");
int rightMax=arr[arr.length-1];
// Rightmost will always be a leader
System.out.print(rightMax+" ");
for (int i = arr.length-2; i>=0; i--) {
if(arr[i] > rightMax)
{
rightMax=arr[i];
System.out.print(" "+rightMax);
}
}
}
}``````
When you run the above program, you will get the below output::
``````Finding leaders in an array using brute force
99 90
==================
Finding leaders in an array :
90 99
``````
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Be the first to be notified when a new article or Kubernetes experiment is published. | 689 | 2,494 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.586517 |
https://us.metamath.org/ileuni/df-alsi.html | 1,726,273,692,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651540.77/warc/CC-MAIN-20240913233654-20240914023654-00506.warc.gz | 544,076,656 | 3,323 | Mathbox for David A. Wheeler < Previous Next > Nearby theorems Mirrors > Home > ILE Home > Th. List > Mathboxes > df-alsi GIF version
Definition df-alsi 13608
Description: Define "all some" applied to a top-level implication, which means 𝜓 is true whenever 𝜑 is true and there is at least one 𝑥 where 𝜑 is true. (Contributed by David A. Wheeler, 20-Oct-2018.)
Assertion
Ref Expression
df-alsi (∀!𝑥(𝜑𝜓) ↔ (∀𝑥(𝜑𝜓) ∧ ∃𝑥𝜑))
Detailed syntax breakdown of Definition df-alsi
StepHypRef Expression
1 wph . . 3 wff 𝜑
2 wps . . 3 wff 𝜓
3 vx . . 3 setvar 𝑥
41, 2, 3walsi 13606 . 2 wff ∀!𝑥(𝜑𝜓)
51, 2wi 4 . . . 4 wff (𝜑𝜓)
65, 3wal 1333 . . 3 wff 𝑥(𝜑𝜓)
71, 3wex 1472 . . 3 wff 𝑥𝜑
86, 7wa 103 . 2 wff (∀𝑥(𝜑𝜓) ∧ ∃𝑥𝜑)
94, 8wb 104 1 wff (∀!𝑥(𝜑𝜓) ↔ (∀𝑥(𝜑𝜓) ∧ ∃𝑥𝜑))
Colors of variables: wff set class This definition is referenced by: alsconv 13610 alsi1d 13611 alsi2d 13612
Copyright terms: Public domain W3C validator | 461 | 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} | 2.578125 | 3 | CC-MAIN-2024-38 | latest | en | 0.44068 |
http://myriverside.sd43.bc.ca/mariab2016/2018/10/28/week-8-precalc-11/ | 1,656,360,579,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103341778.23/warc/CC-MAIN-20220627195131-20220627225131-00329.warc.gz | 37,826,224 | 10,453 | # Week 8 – Precalc 11
This week in precalc we learned how to analyze quadratic functions. We also learned how to use desmos to graph and see how parent functions transform. A quadratic function is written y = a$x^2$ + bx + c .
*Note the difference between equations and functions.*
Equations: you can solve and find roots.
Functions: you can graph, and find Y-intercepts and X-intercepts.
The parabola is the curve where any point is at an equal distance from. The vertex is the highest/lowest point, it may be a minimum or maximum point. The line of symmetry is the line that divides the parabola into mirror images.
When analyzing quadratic functions in a graph, we always need to look for the y-intercept, the x-intercept, the domain and range, parabola, vertex, line of symmetry, if it’s opening down or up, and if it’s congruent to the parent function.
How do we know how parent functions transform:
here’s the video and the link to the website that helped me understand how parent functions transform:
https://youtu.be/dJLQ8UOIuZ4 https://www.purplemath.com/modules/fcntrans.htm | 263 | 1,094 | {"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": 1, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2022-27 | latest | en | 0.839877 |
http://oeis.org/A042967 | 1,369,413,527,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368704752145/warc/CC-MAIN-20130516114552-00073-ip-10-60-113-184.ec2.internal.warc.gz | 200,153,414 | 3,552 | This site is supported by donations to The OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A042967 Primes p such that x^7 = 2 has no solution mod p. 4
29, 43, 71, 113, 127, 197, 211, 239, 281, 337, 379, 421, 449, 463, 491, 547, 617, 659, 701, 743, 757, 827, 883, 911, 967, 1009, 1051, 1093, 1289, 1303, 1373, 1429, 1471, 1499, 1583, 1597, 1667, 1723, 1877, 1933, 2017, 2087, 2129, 2213, 2269, 2297, 2311, 2339, 2381, 2423, 2437, 2521 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS Complement of A042966 relative to A000040. Coincides for the first 96 terms with the sequence of primes p such that x^49 = 2 has no solution mod p (first divergence is at 4999, cf. A059667) - Klaus Brockhaus, Feb 04 2001 LINKS Vincenzo Librandi, Table of n, a(n) for n = 1..1000 MATHEMATICA ok[p_]:= Reduce[Mod[x^7 - 2, p] == 0, x, Integers] == False; Select[Prime[Range[700]], ok] (* Vincenzo Librandi, Sep 19 2012 *) PROG (MAGMA) [p: p in PrimesUpTo(3000) | forall{x: x in ResidueClassRing(p) | x^7 ne 2}]; // Vincenzo Librandi, Aug 21 2012 (MAGMA) [p: p in PrimesUpTo(2600) | not exists{x : x in ResidueClassRing(p) | x^7 eq 2} ]; // Vincenzo Librandi, Sep 19 2012 CROSSREFS Cf. A000040, A042966, A059667. Sequence in context: A004619 A140444 A042969 * A061638 A136062 A039348 Adjacent sequences: A042964 A042965 A042966 * A042968 A042969 A042970 KEYWORD nonn,easy AUTHOR STATUS approved
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Content is available under The OEIS End-User License Agreement . | 668 | 1,775 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-20 | latest | en | 0.531117 |
https://www.acmicpc.net/problem/18479 | 1,585,955,572,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370518767.60/warc/CC-MAIN-20200403220847-20200404010847-00128.warc.gz | 784,542,413 | 8,922 | 시간 제한 메모리 제한 제출 정답 맞은 사람 정답 비율
2 초 512 MB 0 0 0 0.000%
## 문제
A string t is called a smooth transformation of a string w if there exist an integer m ≥ 1 and strings w0, w1, . . . , wm such that:
• w0 = w, and |wi| = |w| when 0 < i ≤ m;
• wi differs from wi−1 in at most one position when 0 < i ≤ m;
• t = w0w1 . . . wm.
You are given a string s = s1s2 . . . s|s|. Find the number of triplets of indices (i, j, k) such that 1 ≤ i < j < k ≤ |s| and si..k = sisi+1 . . . sk is a smooth transformation of si..j = sisi+1 . . . sj.
## 입력
The only line of the input contains the string s (4 ≤ |s| ≤ 105) consisting of lowercase English letters.
## 출력
Display the sought number of triplets.
## 예제 입력 1
abcababc
## 예제 출력 1
5
## 힌트
In the example test case, the triplets are (1, 3, 6), (3, 4, 6), (3, 4, 8), (4, 5, 7), and (5, 6, 8). | 329 | 838 | {"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} | 2.6875 | 3 | CC-MAIN-2020-16 | longest | en | 0.566994 |
https://circuitcellar.com/cc-blog/fundamentals-of-iq-signals/ | 1,675,731,718,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500368.7/warc/CC-MAIN-20230207004322-20230207034322-00114.warc.gz | 183,501,014 | 45,426 | Fundamentals of I/Q Signals
When talking about frequency mixers, “I/Q” has nothing to do with an intelligence quotient. As Robert explains, the “I” stands for “in phase” and the “Q” is for “in quadrature.” In this article, he introduces you to the fundamentals of I/Q signal representation and architecture.
In the article, “Do You Speak I/O?,” Lacoste writes:
In 2012, I ended an article about frequency mixers (“Let’s Play with RF Frequency Mixers,” Circuit Cellar 263) by saying that I had only scratched the surface of the subject. In fact, I didn’t cover the important topic of so-called “I/Q” mixers in that article. If you’re wondering what “I/Q” means, let me explain.
When talking about I/Q mixers, “I” stands for “in phase” and “Q” stands for “in quadrature.” You will find these two letters in most papers on signal processing or modern radio frequency (RF) systems architectures. Unfortunately, even some of the most experienced design engineers aren’t particularly familiar with these concepts. Why? Probably because they are usually presented in mathematical terms, such as complex numbers, the Euler theorem, complex Fourier transform, and so on.
This month, my aim is to explain the fundamentals of I/Q signal representation and architecture without math. So, as usual, take a seat, breathe normally, and follow me. I’ll stay away from complex mathematics, except for a few concepts you probably learned in high school.
A frequency mixer is a frequency translation device that you can use either to move up (up-convert) or down (down-convert) any part of the RF spectrum. For the moment, let’s focus on down-converters.
Basically, a mixer is a voltage multiplier. It multiplies two voltages: the RF signal that you want to down-convert and a sine signal coming from a local oscillator (LO). The output is usually nicknamed “intermediate frequency” (IF). The magic lies behind a simple trigonometric formula. The product of two sine signals of frequencies F1 and F2 is the sum of two other sine signals. These two signals have respective frequencies F1 – F2 and F1 + F2. Figure 1 clearly illustrates what’s going on. Refer to my previous article if it’s unclear.
Figure 1: A mixer works by multiplying two sine signals of frequencies FRF and FLO. Thanks to the well-known trigonometric formula reminded here, the product is the sum of two other sine signals, respectively, at frequencies FRF + FLO and FRF – FLO.
Of course, in real life, a mixer is a little more complex. But this description is sufficient enough for what I want to explain in this article. You must select the appropriate the LO and IF frequencies in order to have enough frequency separation between the two frequency terms present on the output. This enables you to remove the unwanted one with a frequency filter (high-pass or low-pass depending on the application).
Now let’s move on to modulated signals. Assume that the input RF signal is not a simple sine wave but a modulated signal that occupies a total bandwidth of BW in hertz. The band of interest is then from FRF – BW/2 to FRF + BW/2, where FRF is the central frequency of the RF signal. For example, if it is a IEEE802.11g (i.e., Wi-Fi) signal on channel 6, then you will have FRF = 2.437 GHz (the center frequency of Wi-Fi’s channel 6) and BW = 20 MHz (the modulation width of 802.11g). So, in that case, the occupied bandwidth is 2.437 GHZ ±10 MHz.
Suppose you want to translate the Wi-Fi signal to a low IF in order to digitize it. Assume that you want FIF = 50 MHz. As a mixer is operated in its linear region, it is theoretically transparent to the modulation. Therefore, you could simply mix the RF signal with a local oscillator set at a frequency of FLO = 2,437 + 50 = 2,487 MHz. The mixer’s output will include two copies of the modulated spectrum, one centered around FLO – FRF = 50 MHz and one centered around FLO + FRF = 4,924 MHz (see Figure 2). A low-pass filter will easily remove the second one.
Figure 2: A classical mixer, when used as a
down-converter, generates mainly two
copies of the input spectrum, of which
one must be eliminated thanks to a
frequency-selective filter.
The signal’s occupied bandwidth is not modified by the mixer: the intermediate frequency signal will still occupies ±10 MHz around the intermediate frequency. Just a caution: You can see in Figure 2 that the spectrum of the modulated signal can be mirrored. This is due to the fact that the local oscillator frequency was set above the RF frequency. In that case, if the frequency of the RF signal increases, it comes closer to the LO frequency, and therefore the IF frequency is lower (as FIF = FLO – FRF). This shouldn’t be an issue as long as you’re aware of it.
Such an architecture is called a “low IF” design, as the RF signal is moved directly to a quite low frequency in comparison to its bandwidth. Here the occupied bandwidth of the intermediate frequency will be 50 MHz ±10 MHz (i.e., from 40 to 60 MHz).
Now imagine that you have a spectrum analyzer on the IF output and a hand on the frequency-setting knob on the local oscillator. What happens when you gently turn the knob and reduce the LO frequency? Refer to Figure 2 once again. If FLO comes closer to FRF then the generated FIF will be closer to 0 Hz. Theoretically, everything should stay fine until the local oscillator frequency is close to 10 MHz. At that point, the IF modulated signal will occupy the frequency bandwidth of 10 MHz ±10 MHz (i.e., from exactly 0 Hz up to 20 MHz).
What happens if you continue to reduce the LO frequency? Part of the IF spectrum will be lower than 0 Hz. Unfortunately, negative frequencies don’t exist, so this spectrum will be folded back on the positive frequency side and will jeopardize the useful signal. Continue and set FLO exactly at the same frequency as FRF. Then the signal will be theoretically centered at 0 Hz. You will get an occupied bandwidth from DC to BW/2—that is, full of garbage, as the two parts of the spectrum will be folded on each other (see Figure 3).
Figure 3: With a standard mixer issues arise when the LO frequency is too close to the RF frequency. The output signal then cross the 0-Hz boundary and its spectrum is folded back from DC to half the bandwidth, jeopardizing its content.
You might think that this is bringing us nowhere, but there is even a name for such an RF architecture with the local oscillator exactly centered at the RF carrier frequency: zero-IF designs. So, is there a trick to avoid this spectrum fallback problem? You bet so. The answer is to use a so-called quadrature demodulator or I/Q mixer. You can analyze the concept in a few different ways. If you prefer math, read Richard Lyons’s excellent essay, “Quadrature Signals: Complex, But Not Complicated.” In what follows, I provide a more illustrative explanation.
The complete article appears in Circuit Cellar 293 (December 2014). | 1,584 | 6,910 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2023-06 | longest | en | 0.930622 |
https://www.advancedfootballanalytics.com/2007/10/patriots-or-colts-undefeated.html | 1,653,768,165,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663019783.90/warc/CC-MAIN-20220528185151-20220528215151-00285.warc.gz | 701,012,106 | 21,329 | Patriots or Colts Undefeated?
With the impending showdown between the NFL's top two teams, a lot of the discussion has mentioned the possibility of either the Patriots or Colts going undefeated. Not since the '72 Dolphins went 14-0 in the regular season has an NFL team repeated the feat. There are now 16 games in a regular season, making the achievement even more improbable.
Of course, only one of the two teams could go undefeated this year because they have to play each other. In this post, I'll examine which team is more likely to go 16-0--assuming each wins on Sunday.
The probability that either team would go undefeated is estimated based on a calculation of game-by-game win probabilities. Every permutation of outcomes of wins and losses is computed for each team. The combination that represents only wins and no losses is the probability that the team will be 16-0. The methodology is explained more fully in this post. In short, the probability estimate accounts for all major phases of team efficiency, to-date opponent strength, future opponent strength, and home field advantage in upcoming match-ups.
However, there is one important distinction between teams at this point. Still awaiting their bye, the Patriots and have played, and won, one more game than the Colts. This gives them a distinct advantage when comparing the two teams' chances of winning out. But it is still interesting to know just how possible it is for teams like the Patriots or Colts to do what hasn't been done for three and half decades.
And of course, it all depends on who wins on Sunday.
Both teams have been mercilessly efficient so far in 2007. Below are the efficiency stats for each team (unadjusted for opponent). Also listed is the NFL average (not including NE and IND) for each stat, so we can compare see just how good these two teams are.
Stat IND NE NFL O Pass 7.57 8.65 6.01 O Run 4.42 4.19 4.05 O Int Rate 0.013 0.011 0.032 O Fum Rate 0.013 0.015 0.027 D Pass 4.79 5.09 6.25 D Run 4.06 4.24 4.05 D Int Rate 0.039 0.042 0.031 Pen Rate 0.21 0.31 0.37
(Pass and run stats are yds per attempt. Fumble rate is fumbles per play. Int rate is in interceptions per attempt. Penalty rate is penalty yards per play.)
NE has the better passing game, but IND has the better running game and defends the pass better. NE gets more interceptions, but IND commits fewer penalties for fewer yards. All things considered, the two teams are about equal. NE has garnered more attention so far because of the fact they've scored more touchdowns, but they've played a considerably weaker schedule.
Below is a table of each team's to-date opponents and their generic win probability (GWP)--the probability a team will beat a notional league-average team at a neutral site. Opponent strengths do account for the beatings handed to them by IND and NE. In other words, NE's strength of schedule isn't penalized due to the pounding their opponents received at the hands of NE themselves. IND's opponents' have been slightly stronger than average with a 0.52 GWP, while NE's opponents' have been below average with a 0.45 GWP.
NE Opp GWP IND Opp GWP NYJ 0.23 NO 0.32 SD 0.62 TEN 0.58 BUF 0.43 HOU 0.34 CIN 0.50 DEN 0.66 CLE 0.38 TB 0.76 DAL 0.71 bye MIA 0.23 JAX 0.64 WAS 0.50 CAR 0.33 Avg 0.45 0.52
Although both teams are about equal in (unadjusted) efficiency stats, after adjusting for opponent strength IND comes out on top with a 0.92 GWP compared to a 0.90 GWP for NE. Keep in mind these are estimations, so a difference of 0.02 is essentially a wash. We'll certainly find out more on Sunday.
Given about a 90% chance of winning a game against a league-average opponent at a neutral site, and assuming they win against the Colts, the Patriots would roughly have about a 0.907 = 48% chance of going undefeated. If the Colts win Sunday, they would have about a 0.928 = 51% chance of finishing undefeated.
But NFL games aren't against theoretical league-average opponents, and they aren't (normally) at neutral sites. NE has an slightly easier forthcoming schedule as their future opponents' GWP average is 0.43 while IND's future opponents average a slightly tougher 0.45 GWP.
NE Opp GWP IND Opp GWP bye SD 0.62 BUF 0.43 KC 0.44 PHI 0.55 ATL 0.29 BAL 0.34 JAX 0.64 PIT 0.72 BAL 0.34 NYJ 0.23 OAK 0.31 MIA 0.23 HOU 0.34 NYG 0.53 TEN 0.58 Avg 0.43 0.45
NE's upcoming schedule and their associated outcome probabilities are listed below. The series probability that the a team would go undefeated is the product of the probabilities of winning each individual game. Keep in mind this assumes each team wins this Sunday.
Vprob Visitor Home Hprob 0.90 NE BUF 0.10 0.08 PHI NE 0.92 0.93 NE BAL 0.07 0.16 PIT NE 0.84 0.02 NYJ NE 0.98 0.02 MIA NE 0.98 0.85 NE NYG 0.15
Probability of New England going undefeated =
0.90 * 0.92 * 0.93 * 0.84 * 0.98 * 0.98 * 0.85 = 0.52
IND's upcoming schedule and their associated outcome probabilities are listed below.
Hprob Visitor Home Vprob 0.83 IND SD 0.17 0.05 KC IND 0.95 0.95 IND ATL 0.05 0.10 JAX IND 0.90 0.94 IND BAL 0.06 0.95 IND OAK 0.05 0.03 HOU IND 0.97 0.08 TEN IND 0.92
Probability of Indianapolis going undefeated =
0.83 * 0.95 * 0.95 * 0.90 * 0.94 * 0.95 * 0.97 * 0.92 = 0.54
By the end of Sunday's game, one of the teams will see their chances swiftly go to zero.
Note: Republished with a correction to NE's probability.
7 Responses to “Patriots or Colts Undefeated?”
1. Tarr says:
No comment except that I really enjoyed reading that. Thanks for putting it together.
2. Tarr says:
Actually I lied, I have two comments:
1) Accuscore, which works by simulating game results, comes up with nearly the exact same probability of the Pats going undefeated (60%) as you do, after a win. They don't mention the odds they give for the Colts, but I would assume that it is lower, as they rate the Pats as a superior team.
2) Neither analysis considers the very significant monkey wrench: at the Giants, week 17, Pats are 15-0, Giants need a win for playoff position, Giants are a pretty good team, Pats have clinched home field, Brady/Moss/et al play one series. Of course, it's hard to account for that situation.
3. Anonymous says:
Using your VProbs, I come up with 0.56 for the Patriots and 0.54 for the Colts.
Either my arithmetic is suspect, or yours is. ;-\
It also looks like you pulled the wrong VProbs from your table into the Patriots product (e.g. three 0.98 factors instead of two).
4. Brian Burke says:
That's my Excel skills that are suspect. Thanks for a great catch, Bionic Man.
Corrected.
5. Anonymous says:
Clicked over from FootballOutsiders.
Great analysis!!
6. Anonymous says:
I see that you made some corrections, but now you are missing the Patriots game against Philadelphia in your calculations.
I made the calculations with the Eagles included, and now the Patriots have a 0,528 chance of going undefeated which is actaully less than the Colts, who need to play one more game than the Patriots.
By the way, can't wait for the Patriots - Colts game :)
7. Brian Burke says:
I botched this one. I finally fixed the spreadsheet. But I wouldn't have made the mistake in the first place had I just used a calculator to check my own math.
In any event, it's about a 50/50 proposition for either team, assuming they win Sunday. | 1,953 | 7,321 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-21 | longest | en | 0.964121 |
https://issuu.com/chanduchagam74/docs/acc_560_week_5_quiz_3__chapters_5_a | 1,498,468,424,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320695.49/warc/CC-MAIN-20170626083037-20170626103037-00191.warc.gz | 752,982,643 | 16,635 | ACC 560 Week 5 Quiz 3 (Chapters 5 and 6) To Purchase This Material Click below Link http://www.acc560help.com/ACC-560-Week-5-Quiz-3-(Chapters-5-and-6) FOR MORE CLASSES VISIT www.acc560help.com ACC 560 Week 5 Quiz 3 (Chapters 5 and 6) Question 1
A company with a higher contribution margin ratio is Question 2
The contribution margin ratio is Question 3
In 2016, Teller Company sold 3,000 units at \$600 each. Variable expenses were \$420 per unit, and fixed expenses were \$270,000. The same selling price, variable expenses, and fixed expenses are expected for 2017. What is Teller’s break-even point in sales dollars for 2017? Question 4
Sales mix is Question 5
The margin of safety ratio Question 6
To which function of management is CVP analysis most applicable? Question 7
Frazier Manufacturing Company collected the following production data for the past month: Units Produced Total Cost 1,600 \$66,000 1,300 57,000 1,500 67,500 1,100 49,500 If the high-low method is used, what is the monthly total cost equation? Question 8
Weatherspoon Company has a product with a selling price per unit of \$200, the unit variable cost is \$110, and the total monthly fixed costs are \$300,000. How much is Weatherspoon’s contribution margin ratio? Question 9
Two costs at Bradshaw Company appear below for specific months of operation. Month Amount Units Produced Delivery costs September \$ 40,000 40,000 October 55,000 60,000
Utilities September \$ 84,000 40,000 October 126,000 60,000 Which type of costs are these? Question 10
The equation which reflects a CVP income statement is
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Students solve two-step equations while honing seventh- and eighth-grade algebra skills in this helpful practice worksheet! | 878 | 4,157 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.840287 |
http://data.princeton.edu/eco572/grdt.html | 1,537,862,041,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267161214.92/warc/CC-MAIN-20180925063826-20180925084226-00254.warc.gz | 64,054,303 | 5,626 | Germán Rodríguez
Demographic Methods Princeton University
## Growth Rates and Doubling Time
The U.S. Census Bureau has population counts for the U.S. from 1790 to 2010. I got the counts up to 2000 from Table 4 in this Census Report (Table 2 had slightly different counts for 1830 and 1940), and added the 2010 count from Wikipedia.
These data are in an ascii file called `uspop1790to2010.dat` in the datasets section of the course website. The file has two columns representing year and population. The first task is to read the data.
We read the population as long to ensure we get all the digits.
```. infile year long pop using ///
> http://data.princeton.edu/eco572/datasets/uspop1790to2010.dat
```
```> us <- read.table("http://data.princeton.edu/eco572/datasets/uspop1790to2010.dat",
```
We will plot the population in millions (otherwise we get bad labels) using absolute and log scales.
```. gen pm = pop/1000000
. label var pm "Population (millions)"
. line pm year, name(abspop, replace)
. line pm year, yscale(log) name(logpop, replace)
. graph combine abspop logpop, title("U.S. Population") ysize(4) xsize(8)
. graph export uspop.png, replace width(720) height(360)
(file uspop.png written in PNG format)
```
```> us <- mutate(us, pm = pop/1000000)
> g1 <- ggplot(us, aes(x=year, y=pm)) + geom_line() + xlab("Year") +
+ ylab("Population (millions)")
> g2 <- ggplot(us, aes(x=year, y=pm)) + geom_line() + xlab("Year") +
+ ylab("Population (millions)") + scale_y_log10()
> g <- arrangeGrob(g1,g2,ncol=2)
> ggsave("uspopr.png", plot=g, width=10, height=5, dpi=72)
```
### Growth Rates
What was the growth rate in the last intercensal period? Let us list the population counts for the last two censuses. We specify a format so Stata doesn't list large numbers using scientific notation; `%14.0fc` tells Stata to use up to 14 digits with no decimal, using a comma to indicate thousands
```. format pop %14.0fc
. list pop in -2/-1
+-------------+
| pop |
|-------------|
22. | 281,421,906 |
23. | 308,745,538 |
+-------------+
. di pop[_N] - pop[_N-1]
27323632
```
```> tail(us, 2)
year pop pm
22 2000 281421906 281.4219
23 2010 308745538 308.7455
> diff(tail(us\$pop, 2))
[1] 27323632
```
So the U.S. population grew by 27,323,632 between 2000 and 2010. Verify that in the previous intercensal period it grew by 32,712,048.
If we divide the population increase by the population at the start of the intercensal period, or just take the ratio and subtract one, we obtain
```. scalar ratio = pop[_N]/pop[_N-1]
. di ratio - 1
.09709135
```
```> last <- length(us\$pop)
> ratio <- us\( pop[last]/us \)pop[last-1]
> ratio - 1
[1] 0.09709135
```
So the population grew by 9.7% in ten years. You'd think this is 0.97% per year, but that is only approximate because it doesn't compound the growth over the ten years. If we start with a population P1 and compound k times per year at a rate r for ten years, th population will grow to P2 = P1(1+r/k)^(10k). Solving for r gives a growth rate of r = k[(P2/P)^(1/(10k))-1]. Here's the rate we obtain using different values of k
```. mata:
------------------------------------------------- mata (type end to exit) --------------
: ratio = st_numscalar("ratio")
: k = (1\2\4\6\12\52\365)
: r = k :* (ratio:^(1 :/ (10 :* k ) ) :- 1)
: k, r
1 2
+-----------------------------+
1 | 1 .0093093093 |
2 | 2 .0092877438 |
3 | 4 .009276986 |
4 | 6 .0092734037 |
5 | 12 .0092698233 |
6 | 52 .0092670704 |
7 | 365 .0092663624 |
+-----------------------------+
: end
----------------------------------------------------------------------------------------
```
```> ratio <- us\( pop[last]/us \)pop[last-1]
> k <- c(1, 2, 4, 6, 12, 52, 365)
> r <- k * (ratio^(1 / (10 * k ) ) - 1)
> data.frame(k=k, r=r)
k r
1 1 0.009309309
2 2 0.009287744
3 4 0.009276986
4 6 0.009273404
5 12 0.009269823
6 52 0.009267070
7 365 0.009266362
```
A unit value of k means compounding anually, 12 means monthly, and 365 means daily. We could continue compounding every minute, or every second, but you can see that our calculation is quickly approaching a limit. From elementary calculus we know that as k -> ∞ our equation becomes P2 = P1 exp(10 r), and solving for r gives log(P2/P1)/10, so the limiting value is
```. display "r=" log(ratio)/10
r=.00926624
```
```> log(ratio)/10
[1] 0.009266245
```
This is a mean annualized rate of growth. Note that by the time we compounded every two months we already had the correct value to five decimal places.
### Growing More Slowly
We can now compute the growth rate for the entire (census) history of the U.S. We treat all censuses as ten years apart, although this is not exactly true: over time the census has moved from August to June, and then April (except for 1920, which was done in January). If you want to do a more precise calculation the dates needed are in the reference given at the top.
```. gen growthrate = log(pop/pop[_n-1])/10
(1 missing value generated)
```
Note the use of `_n-1` to refer to the previous value, generating a missing value for the first row
```> gr <- diff(log(us\$pop))/10
```
Now we plot the rates over time. Because the growth rate pertains to the period between two censuses it makes sense to plot against the mid-points of the census years, excluding the last
```. gen midcensus = (year + year[_n-1])/2
(1 missing value generated)
. line gr midcensus, name(usgr, replace) title(Growth Rate)
```
```> mc <- us\$year[-last] + 5
> usg <- data.frame(year = mc, growth.rate = gr)
> g1 <- ggplot(usg, aes(x=year, y=growth.rate)) + geom_line()
```
The graph is shown further below, combined with a plot of doubling time. We see that the growth rate was around 3% for about half the 19th century, declined steadily for almost 100 years with a pre-war dip, rebounded with the post-war baby boom, and then resumed its decline.
### Doubling Time
At an instantaneous growth rate r, the doubling time is log(2)/r.
```. gen doublingtime = log(2)/gr
(1 missing value generated)
. line doubling midcensus, name(usdt, replace) title("Doubling Time")
. graph combine usgr usdt, title("U.S. 1790-2010") ///
> ysize(3) xsize(6)
. graph export usgrdt.png, width(720) height(360) replace
(file usgrdt.png written in PNG format)
```
```> usg <- mutate(usg, doubling.time = log(2)/growth.rate)
> tail(usg,1)
year growth.rate doubling.time
22 2005 0.009266245 74.80346
> g2 <- ggplot(usg, aes(x=year, y=doubling.time)) + geom_line()
> g <- arrangeGrob(g1,g2,ncol=2)
> ggsave("usgrdtr.png", plot=g, width=10, height=5, dpi=72)
```
The U.S. population was doubling every 22-24 years in the first half of the 19th century, but the doubling time has increased steadily (except for a pre-war spike) and it now takes 75 years to double. Verify that in the previous intercensal period the growth rate was 0.0124 and the doubling time was 56 years.
Updated 22-Jan-2016 | 2,128 | 7,109 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2018-39 | latest | en | 0.640244 |
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# What is the remainder when (n-1)*(n+1) is divided by 24?
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Re: What is the remainder when (n-1)*(n+1) is divided by 24? [#permalink]
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94412 posts | 387 | 1,289 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2024-30 | latest | en | 0.844504 |
http://mathhomeworkanswers.org/235122/algebra-time-distance-speed | 1,484,596,171,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560279248.16/warc/CC-MAIN-20170116095119-00139-ip-10-171-10-70.ec2.internal.warc.gz | 182,423,435 | 11,505 | A column of soldiers is approaching a bridge that is exactly 378 meters long. It takes the whole column 7 minutes to pass the first post at the beginning of the bridge. It takes the whole column of soldiers 25 minutes to clear up the entire bridge. The soldiers are moving at a constant speed. What is the length of the column of soldiers and at what speed did they move?
answered Jan 5 by Level 10 User (53,200 points)
Let S=length of the column and V be their speed.
S=7V is the equation for passing the first post.
378=25V is the equation for crossing the bridge, so V=378/25=15.12 m/min.
So S=7*15,12=105.84m.
answered Jan 5 by Top Rated User (415,140 points) | 173 | 667 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2017-04 | longest | en | 0.945977 |
https://ncatlab.org/nlab/show/type | 1,701,232,273,000,000,000 | application/xhtml+xml | crawl-data/CC-MAIN-2023-50/segments/1700679100056.38/warc/CC-MAIN-20231129041834-20231129071834-00724.warc.gz | 501,600,244 | 9,714 | # nLab type
Contents
This entry is about the notion in type theory. For the unrelated notion of the same name in model theory see at type (in model theory).
# Contents
## Idea
In modern logic, we understand that every variable should have a type, or domain of discourse or be of some sort. For instance we say that if a variable $n$ is constrained to be an integer then “$n$ is of integer type” or “of type $\mathbb{Z}$”. The usual formal expression from set theory for this – $n \in \mathbb{Z}$ – is then often written $n \colon \mathbb{Z}$
We speak of typed logic if this typing of variables is enforced by the metalanguage. In formulations of a theory the types are often called sorts. More generally, type theory formalizes reasoning with such typed variables. See there for more
(Untyped logic may be seen as simply a special case, in which there is only a single unique type. Thus, untyped logic has one type, not no type.)
## Definition
Reasoning with types is formalized in natural deduction (which in turn is formalized in a logical framework such as Elf).
Behaviour of types is specified by a 4-step set of rules
## Properties
Deep relations between type theory, category theory and computer science relate types to other notions, such as objects in a category. See at computational trinitarianism for more on this.
Last revised on April 27, 2017 at 13:01:46. See the history of this page for a list of all contributions to it. | 338 | 1,450 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 6, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2023-50 | latest | en | 0.930366 |
http://www.cquestions.com/2011/12/how-to-find-b-and-c-in-quadratic.html | 1,487,514,082,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501169776.21/warc/CC-MAIN-20170219104609-00004-ip-10-171-10-108.ec2.internal.warc.gz | 359,789,827 | 16,026 | ## INDEX
### How to find a b and c in a quadratic equation
Quadratic equation in c language
#include<stdio.h>
#include<math.h>
int main(){
float a,b,c;
float d,root1,root2;
printf("Enter quadratic equation in the format ax^2+bx+c: ");
scanf("%fx^2%fx%f",&a,&b,&c);
d = b * b - 4 * a * c;
if(d < 0){
printf("Roots are complex number.\n");
return 0;
}
root1 = ( -b + sqrt(d)) / (2* a);
root2 = ( -b - sqrt(d)) / (2* a);
printf("Roots of quadratic equation are: %.3f , %.3f",root1,root2);
return 0;
}
Sample output:
Enter quadratic equation in the format ax^2+bx+c: 2x^2+4x+-1
Roots of quadratic equation are: 0.000, -2.000 | 220 | 631 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-09 | longest | en | 0.430888 |
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