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https://space.stackexchange.com/questions/39746/optimal-shape-for-a-spinning-ship-station-downsides-of-spheres-radiation-shi | 1,709,188,214,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474784.33/warc/CC-MAIN-20240229035411-20240229065411-00199.warc.gz | 529,545,725 | 41,064 | # Optimal shape for a spinning ship/station - downsides of spheres? (Radiation shielding)
What is the optimum shape for a spacecraft?
So I've already read this link, and a few other similar ones, and the main issues I am already aware of... but yet it still seems like a large majority of the designs for rotating (for 'gravity') ships and stations use a torus, or perhaps a cylinder.
Why is this?
The main issue I have, is the radiation shielding. It seems that if you're using water (for example), you would need 2 tonnes of water for every square meter of surface area. So if you use a torus, that's a huge amount of extra weight. A sphere is (by a fair margin) a shape with a better volume/surface area ratio. And saving those 2 tonnes per square meter would make your fuel consumption significantly better.
Is there something I'm missing? I suspect there is. In my head a torus just seemed like a much more 'streamlined' shape (not aerodynamically, just much less waste of volume). But it seems like the sphere gives you much more volume for much less mass, mostly due to the shielding requirements?
## 1 Answer
A spherical spinning station faces a number of problems. First, different latitudes will have different amounts of centrifugal "gravity" (due to different radii to the central axis), which is fine under some scenarios but not at all helpful if you only care about the parts that are at your target acceleration and maybe a bit at free-fall in the middle (which is how most spinning stations seem to be designed). In other words, a cylinder or toroid trades away a lot of total interior volume : mass ratio to maximize the interior volume at desired radius : mass ratio. Additionally, a spherical spinning station is going to try to become an oblate spheroid (as the earth itself is) due to the same centrifugal force pushing outward hardest along the equator, which poses some design difficulties and requires more mass to compensate (the earth's solution - fill the middle with solid or molten rock and metal to keep the bulge under control - is obviously not viable here).
Finally, you're talking about "a large majority of the designs" without mentioning if you're considering any other filter, which - for anything that has never yet been built and would be very expensive to build - means you're implicitly getting a filter for the minimum viable product (MVP). The MVP for a spinning station isn't trying to maximize the interior volume : mass ratio, it's trying to get something that has any usable amount of interior volume at the desired acceleration, for the minimum cost. To use some made-up numbers, a station with 20x the interior volume for merely 2x the cost just might not be worth it, if you don't need that much volume.
• "To use some made-up numbers, a station with 20x the interior volume for merely 2x the cost just might not be worth it, if you don't need that much volume." I agree with this, and this was essentially the logic I was using when I was originally considering a torus-style design. However that only worked up to the point where I realised the torus needs a huuuuge tonnage of radiation shielding because of the increased surface area. In theory, some of that tonnage could be used to turn the station into a sphere instead, ending up with essentially 'free' extra volume. Nov 5, 2019 at 22:43
• In the vast majority of orbits, you don't really need radiation shielding on any part not directly exposed to the sun. If you're measuring total surface area, you're going to get a vastly inflated number from a torus. Depending on orientation, you could restrict shielding to only the outer edge, or to only the top (or bottom) faces, and you wouldn't need to shield most of the middle of the torus either. Nov 6, 2019 at 0:06
• This isn't being thought of as a LEO design. It would be for other parts of the solar system, so no protection from the magnetosphere. Mars, for instance, won't protect it much. Unless you stay shielded behind the planet, but then you can't use solar panels at all. Edit: Also I thought that cosmic radiation was at least as dangerous as solar radiation? So you'd need shielding on all surfaces. Nov 6, 2019 at 0:13
• Here are 2 factors to consider. First, the shielding doesn't have to rotate. The centrifuge can be inside the shell of shielding. That reduces the structural integrity all that mass would need if it were rotating. It also reduces the gyroscopic effect a lot, allowing easier attitude changes. Nov 8, 2019 at 3:17
• The other factor is how the weight ratio improves with size. If a half meter of ice is needed for shielding, that would be a huge portion of the weight for a habitat that is 3 or 4 meters across. However as the overall size increases, the thickness of the shield stays the same. If the habitat were 50 meters across or more, the shielding would be a much smaller portion of the total mass. Nov 8, 2019 at 3:23 | 1,107 | 4,930 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2024-10 | latest | en | 0.953167 |
www.dessieblog.com | 1,670,421,010,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711162.52/warc/CC-MAIN-20221207121241-20221207151241-00657.warc.gz | 58,008,241 | 8,365 | # Just passed 100 posts…
… in case anyone cares. I just thought it was interesting – the picture is number 100.
# True math….
From Scizzy’s site, shamelessly stolen, b/c it’s oh so funny:
From a strictly mathematical viewpoint it goes like this:
What Makes 100%? What does it mean to give MORE than 100%? Ever wonder about those people who say they are giving more than 100%? We have all been to those meetings where someone wants you to give over 100%. How about achieving 103%? What makes up 100% in life?
If:
A B C D E F G H I J K L M N O P Q R S T U V W X Y Z is represented as:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26.
Then:
H-A-R-D-W-O-R-K
8+1+18+4+23+15+18+11 = 98%
and
K-N-O-W-L-E-D-G-E
11+14+15+23+12+5+4+7+5 = 96%
But,
A-T-T-I-T-U-D-E
1+20+20+9+20+21+4+5 = 100%
And,
B-U-L-L-S-H-I-T
2+21+12+12+19+8+9+20 = 103%
AND, look how far ass kissing will take you.
A-S-S-K-I-S-S-I-N-G
1+19+19+11+9+19+19+9+14+7 = 118%
So, one can conclude with mathematical certainty that while Hard Work and Knowledge will get you close, and Attitude will get you there, it’s the Bullshit and Ass Kissing that will put you over the top!
# Pic from Convention
Who says it’s all work? This is my favourite picture from convention a few weeks back.
# Hmm. It’s Springtime Now.
So I thought it was time for a new, spring-ish look for ye olde blog. (translation: I got bored with the green. LOL)
Time has flown by. I am swamped as heck at work, since I am pretty much working 6 days a week until we get someone new hired for the graveyard shift on Saturday night/Sunday morning. Das no fun, but the money is a good thing with the wedding fast approaching. (Current countdown at the time of this typing: 12wks 2days 4hrs 18mins 34secs)
To do in the next couple of weeks: find some bridesmaid’s dresses we all can live with that don’t break the bank, and finalise invitations. Anyone know where I can find plain old boring cream coloured invitation card stock w/ matching envelopes? It’s much more complicated than I would have thought to find. BOOOOO!!!!!!!!!!
We have decided on a honeymoon location (well, two in our case). We are going to New York City for a week, then for a 7 day cruise to the Caribbean. It’s going to be the longest vacation I have taken in like 5 years, no longer. The last one I took was only a week. ROFL, its gonna be my longest ever.
Other than that, I have nothing interesting going on in my life. Depressing, isn’t it? | 740 | 2,463 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | latest | en | 0.919249 |
https://flyerlqwx.web.app/hellriegel82111kowu/calculating-value-based-on-noi-and-cap-rate-272.html | 1,632,668,519,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057882.56/warc/CC-MAIN-20210926144658-20210926174658-00584.warc.gz | 297,508,993 | 6,358 | ## Calculating value based on noi and cap rate
Jun 4, 2019 Understand How Cap Rate Plays Into Your Investing Decisions. is calculated by dividing the property's annual net operating income (NOI), before returns may be harder to achieve, which can lead to a lower property value. In order to calculate the overall return on their investment, investors will have Mar 25, 2019 Cap rates let the lender determine their LTV (Loan to Value ratio) on a bought at too low a cap rate based on a rosy view of increased NOI
Capitalization Rate, or Cap Rate, is a calculation tool used to value real estate, mostly commercial and multi-family properties. It is the NOI, Net Operating Income of the property divided by the current market value or purchase price. NOI equals all revenue from the property minus all necessary operating expenses. The capitalization rate (cap rate) is simply the ratio of an income property's Net Operating Income (NOI) to the value of its property asset. For example, if a property has recently sold for \$1,500,000 and its NOI was \$120,000, then the cap rate is \$120,000/\$1,500,000, or 8%. The cap rate formula is NOI / property value x 100. Let’s take a look at a quick example of how to calculate NOI. Your gross rental income is \$60,000, your occupancy rate is 85 percent and your operating expenses are \$15,000. To calculate the cap rate of a property, you simply divide the NOI by the value of the property. This calculation will give you a percentage that indicates the annual return on your investment. Although the basic structure of the calculation is straightforward, there are a lot of factors that may affect the cap rate of a property. It assigns a property value equal to the net operating income divided by the cap rate. For example, a small rental property in San Francisco with a net operating income of \$100,000 and a cap rate of 7 percent is valued at \$1,428,571. The same property with a 10 percent cap rate would have a value of \$1 million. Basically, the cap rate is the ratio of net operating income (NOI) to property value or sales price. cap rate = net operating income / property value In other words, this ratio is a straightforward way to measure the relationship between the return generated by the property and the price of it.
## Aug 8, 2019 The capitalization rate is determined by two methods; the net operating income of a property divided by its value or purchase price or by a formula
Capitalization Rate, or Cap Rate, is a calculation tool used to value real estate, mostly commercial and multi-family properties. It is the NOI, Net Operating Jun 29, 2018 It assigns a property value equal to the net operating income divided by the cap rate. For example, a small rental property in San Francisco with a Jan 15, 2020 To calculate the cap rate of a property, you simply divide the NOI by the value of the property. Don't guess the value of your investment, calculate it. You will be able to quickly and efficiently compare properties based on a As the name suggests, it calculates the cap rate based on the value of the Basically, the cap rate is the ratio of net operating income (NOI) to property value or A cap rate is calculated by dividing the Net Operating Income (NOI) of a property by the purchase price (for new purchases) or the value (for refinances). Cap Rate Oct 13, 2019 This measure is computed based on the net income which the property is In the most popular formula, the capitalization rate of a real estate investment is Capitalization Rate = Net Operating Income / Current Market Value.
### Jan 15, 2020 To calculate the cap rate of a property, you simply divide the NOI by the value of the property. Don't guess the value of your investment, calculate it. You will be able to quickly and efficiently compare properties based on a
Oct 1, 2018 Our free cap rate calculator generates a property's net operating income and cap rate based on inputs including property value, gross income The cap rate is a convenient and quick method to determine if the value or the value of a property based on the first year's stabilized net operating income. Feb 5, 2019 Capitalization Rate (Cap Rate): The cap rate is the ratio between the first year Net Operating Income (NOI) and the purchase price of the property. Thus, the NCREIF cap rates are based on historical accounting NOI. the denominator of the cap rate calculation is the value reported by the member for the Jun 3, 2019 The income capitalization formula is as follows: Market Value = Net Operating Income (NOI) / Capitalization Rate. After calculating a property's
### A high cap rate means a lower property value. Conversely, the lower your cap rate, the higher your income property's value. Divide your net operating income (NOI) by the cap rate to calculate your property value. This reflects your income stream, which is an important consideration in determining a sale price for your self-storage facility.
Jul 24, 2018 I'll also share examples of how to use the cap rate formula in a very property by looking at its net operating income compared to its value. into four classes (A, B , C, and D) based on their location and building condition. Apr 25, 2016 The calculation is based on the Net Operating Income the property generates divided by the Purchase Price. Lower cap rates (3-5%) generally The capitalization rate is one method used to determine the value of income an NOI of \$130,000, then it would be said to have a cap rate of 6.5% (or 6.5 cap). market value of this property would be \$1.86 million based on a 7% cap rate.
## Jun 29, 2018 It assigns a property value equal to the net operating income divided by the cap rate. For example, a small rental property in San Francisco with a
The capitalization rate is one method used to determine the value of income an NOI of \$130,000, then it would be said to have a cap rate of 6.5% (or 6.5 cap). market value of this property would be \$1.86 million based on a 7% cap rate. A property's capitalization rate represents its rate of return, based on the expected income The interrelationship of NOI, cap rate and property value means that a in the NOI calculation can likewise affect a comparable property's cap rate. Oct 1, 2018 Our free cap rate calculator generates a property's net operating income and cap rate based on inputs including property value, gross income
The formula for Cap Rate is equal to Net Operating Income (NOI) divided by the current market value of the asset. Where: Net operating income is the annual income Annual Income Annual income is the total value of income earned during a fiscal year. | 1,477 | 6,639 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2021-39 | latest | en | 0.919779 |
https://math.stackexchange.com/questions/844471/integral-int-0-infty-tan-1-left-frac2axx2c2-right-sinbx-d | 1,721,483,420,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763515164.46/warc/CC-MAIN-20240720113754-20240720143754-00784.warc.gz | 349,029,454 | 41,696 | # Integral: $\int_0^\infty \tan^{-1}\left(\frac{2ax}{x^2+c^2} \right)\sin(bx) \; dx$
$$\displaystyle \int_0^\infty \tan^{-1}\left(\frac{2ax}{x^2+c^2} \right)\sin(bx) \; dx=\frac{\pi}{b}e^{-b\sqrt{a^2+c^2}}\sinh (ab)$$
I found this integral from here: http://integralsandseries.prophpbb.com/post2652.html?sid=d6641d4d4a3726f1b27bbb4b98ca840a and the solution uses contour integration. I am wondering if there is a way to solve it without using contour integration. I tried differentiating wrt $a$ and $c$ but in both cases, the resulting expression was dirty which made me reluctant to proceed further. I am out of ideas for this one.
Any help is appreciated. Thanks!
• Since the integrand has odd parity w.r.t. $b$, I experimented with the difference $I(a,b,c)-I(a,-b,c)=2I(a,b,c)=-\frac{2\pi}{b}\sinh{\left(b\sqrt{a^2+c^2}\right)}\sinh{(ab)}$, with the hope of exploiting hyperbolic trig identities for more insight into the symmetries of the problem. But then I realized that taking the sum of $I(b)$ and $I(-b)$ instead of their difference leads to a clear paradox. So I'm guessing there are probably unmentioned assumptions on the domains of the parameters that need to be specified for the internal consistency of the problem... Commented Jun 23, 2014 at 17:16
• @DavidH: Yes, that looks strange. The poster in the linked thread didn't mention anything about the domain so maybe we are already supposed to work within the domain where the result is true? But then, I am not sure what should be the correct domain, any ideas? Commented Jun 23, 2014 at 17:22
• Well, for each of the parameters $a,b,c$, if the domain includes $1$, then the domain also includes any positive real value because of scaling arguments (//waves hands vigorously//). So at the very least the domain includes the positive reals. Commented Jun 23, 2014 at 17:36
• Moving on to ideas for solving the integral itself, my gut tells me the methods for solving this integral math.stackexchange.com/questions/9402/… should be generalizable to the problem at hand. Commented Jun 23, 2014 at 18:43
In order to prove the final result I will need to state a lemma that will be used later.
Lemma$\require{autoload-all}$ $1$:
$$\int_0^\infty \! \frac{\cos(bx)}{x^2+\alpha} \mathrm{d}x = \frac{\pi e^{-b\sqrt{\alpha}}}{2b\sqrt{\alpha}}\tag{1}$$
Proof here.
Consider
$$I = \int_0^\infty\!\! \tan^{-1}\left(\frac{2ax}{x^2+c^2} \right)\sin(bx) \; \mathrm{d}x$$
Integrate by parts
$$I = \int_0^\infty \!\!\frac{2 a \left(c^2-x^2\right) \cos (b x)}{x^4 +(4 a^2+2 c^2) x^2+c^4} \; \mathrm{d}x$$
Decompose this function by partial fractions $$\frac{2 a \left(c^2-x^2\right) }{x^4 +(4 a^2+2 c^2) x^2+c^4} = \frac{a_-}{x^2+x_0} + \frac{a_+}{x^2+x_1}$$
It so happens that
$$x_0 = 2 a^2+2 a\sqrt{a^2+c^2}+c^2,\quad x_1 = 2 a^2-2a \sqrt{a^2+ c^2}+c^2$$
$$a_- =\frac{2a(c^2+x_0)}{x_1-x_0}, \quad a_+ = \frac{2a(c^2+x_1)}{x_0-x_1}$$
Note that both $x_0$ and $x_1$ are greater than $0$.
Re-write the integral
\begin{align} I &= a_-\int_0^{\infty} \!\! \frac{\cos(bx)}{x^2+x_0} \, \mathrm{d}x + a_+\int_0^{\infty} \!\! \frac{\cos(bx)}{x^2+x_1} \, \mathrm{d}x\\[.3cm] &= \frac{2a(c^2+x_0)}{x_1-x_0}\int_0^{\infty} \!\! \frac{\cos(bx)}{x^2+x_0} \, \mathrm{d}x +\frac{2a(c^2+x_1)}{x_0-x_1}\int_0^{\infty} \!\! \frac{\cos(bx)}{x^2+x_1} \, \mathrm{d}x\end{align}
Using $(1)$:
\begin{align} I &= \frac{2a(c^2+x_0)}{x_1-x_0}\cdot\frac{\pi e^{-b\sqrt{x_0}}}{2b\sqrt{x_0}} - \frac{2a(c^2+x_1)}{x_1-x_0}\cdot\frac{\pi e^{-b\sqrt{x_1}}}{2b\sqrt{x_1}}\\[.3cm] &= \left(\frac{a\pi}{b(x_1-x_0)}\right)\left(\frac{(c^2+x_0)e^{-\sqrt{x_0}}}{\sqrt{x_0}} - \frac{(c^2+x_1)e^{-\sqrt{x_1}}}{\sqrt{x_1}}\right)\end{align}
I will digress here to state (without proof but easily verified) that $$\frac{c^2+x_0}{\sqrt{x_0}}= \frac{c^2+x_1}{\sqrt{x_1}} = \frac{x_1-x_0}{2a}.$$ This allows us a tremendous simplification so that we can write
\begin{align} I = \left(\frac{\pi}{2b}\right)\left(e^{-b\sqrt{x_1}}-e^{-b\sqrt{x_0}} \right). \end{align}
It can also be shown that $$\sqrt{x_1} = -a+\sqrt{a^2+c^2}$$ $$\sqrt{x_0} = a+\sqrt{a^2+c^2}$$
Simply square each side to find the desired equality.
We can now complete the proof:
\begin{align} I &= \frac{\pi}{2b}\left(e^{-b\sqrt{x_1}}-e^{-b\sqrt{x_0}} \right) \\[.2cm] &= \frac{\pi}{2b}\left(\exp\left(ab-b\sqrt{a^2+c^2}\right)-\exp\left(-ab-b\sqrt{a^2+c^2}\right) \right) \\[.2cm] &= \frac{\pi}{b}\exp\left(-b\sqrt{a^2+c^2}\right)\frac12(\exp(ab)-\exp(ab)) \\[.2cm] &= \dfrac{\pi}{b}\exp\left(-b\sqrt{a^2+c^2}\right)\sinh{ab}\end{align}
If you are interested in working through the simplifications that I did not prove, I recommend that you begin by squaring each side after verifying that each side shares the same sign.
• I think partial fractions is the only way. Can you please complete your solution so that I can check it off as the answer or shall I write down an answer? Thank you! :) Commented Jun 25, 2014 at 5:25
• I will finish it tonight when I am more available.
Commented Jun 25, 2014 at 13:50
• @PranavArora Just a ping to say that I have completed my answer. Feel free to ask if you need anything clarified.
Commented Jun 27, 2014 at 7:02
• Excellent work! Thanks a lot. :) Commented Jun 27, 2014 at 8:06
$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{\infty}\arctan\pars{2ax \over x^{2} + c^{2}}\sin\pars{bx}\,\dd x ={\pi \over \verts{b}} \expo{-\verts{b}\root{\vphantom{\Large A}a^{2} + c^{2}}}\sinh\pars{ab}: \ {\large ?}}$
\begin{align}&\color{#c00000}{\int_{0}^{\infty}\arctan\pars{2ax \over x^{2} + c^{2}}\sin\pars{bx}\,\dd x} =\half\,\sgn\pars{ab}\Im\int_{-\infty}^{\infty} \arctan\pars{2\verts{a}x \over x^{2} + c^{2}}\expo{\ic\verts{b}x}\,\dd x \\[3mm]&=\half\,\sgn\pars{ab}\,\Im\int_{-\infty}^{\infty}{\ic \over 2}\,\ln\pars{% 1 - 2\verts{a}x\ic/\bracks{x^{2} + c^{2}}\over 1 + 2\verts{a}x\ic/\bracks{x^{2} + c^{2}}}\expo{\ic\verts{b}x}\,\dd x \\[3mm]&={1 \over 4}\,\sgn\pars{ab}\,\Re\int_{-\infty}^{\infty}\ln\pars{% x^{2}- 2\verts{a}\ic x + c^{2} \over x^{2} + 2\verts{a}\ic x + c^{2}} \ \underbrace{\expo{\ic\verts{b}x}\,\dd x} _{\ds{\dd\pars{\expo{\ic\verts{b}x} \over \ic\verts{b}}}} \end{align} $$\begin{array}{|c|}\hline \\ \quad\mbox{Here, we used the identity}\quad \arctan\pars{x} = {\ic \over 2}\,\ln\pars{1 - x\ic \over 1 + x\ic}\quad \\ \\ \hline \end{array}$$
Integrating by parts: \begin{align}&\color{#c00000}{% \int_{0}^{\infty}\arctan\pars{2ax \over x^{2} + c^{2}}\sin\pars{bx}\,\dd x} \\[3mm]&=-\,{1 \over 4}\,\sgn\pars{ab}\,\Re\int_{-\infty}^{\infty} \pars{{2x - 2\verts{a}\ic \over x^{2} - 2\verts{a}\ic x + c^{2}}- {2x + 2\verts{a}\ic \over x^{2} + 2\verts{a}\ic x + c^{2}}} \expo{\ic\verts{b}x}\,{\dd x \over \ic\verts{b}} \\[3mm]&=-\,{\sgn\pars{a} \over 2b}\,\Im\int_{-\infty}^{\infty} \pars{{x - \verts{a}\ic \over x^{2} - 2\verts{a}\ic x + c^{2}}- {x + \verts{a}\ic\over x^{2} + 2\verts{a}\ic x + c^{2}}} \expo{\ic\verts{b}x}\,\dd x \end{align}
\begin{align} &\mbox{Zeros of}\quad x^{2} - 2\verts{a}\ic x + c^{2} =0 \quad\mbox{are given by}\quad \left\lbrace\begin{array}{rcl} \phantom{-}x_{1} & = &\pars{\verts{a} + \root{a^{2} + c^{2}}}\ic \\[2mm] \phantom{-}x_{2} & = & \pars{\verts{a} - \root{a^{2} + c^{2}}}\ic \end{array}\right. \\[3mm]&\mbox{Zeros of}\quad x^{2} + 2\verts{a}\ic x + c^{2} =0 \quad\mbox{are given by}\quad \left\lbrace\begin{array}{rcl} -x_{1} & = &\pars{-\verts{a} - \root{a^{2} + c^{2}}}\ic \\[2mm] -x_{2} & = & \pars{-\verts{a} + \root{a^{2} + c^{2}}}\ic \end{array}\right. \end{align} Note that $\ds{\Im\pars{x_{1}} > 0}$ and $\ds{\Im\pars{x_{2}} < 0}$.
Therefore, \begin{align} &\color{#c00000}{% \int_{0}^{\infty}\arctan\pars{2ax \over x^{2} + c^{2}}\sin\pars{bx}\,\dd x} \\[3mm]&=-\,{\sgn\pars{a} \over 2b}\,\Im\pars{% 2\pi\ic\,{\pars{x_{1} - \verts{a}\ic}\expo{\ic\verts{b}x_{1}}\over 2x_{1} - 2\verts{a}\ic} - 2\pi\ic\,{\pars{-x_{2} + \verts{a}\ic}\expo{-\ic\verts{b}x_{2}}\over -2x_{2} + 2\verts{a}\ic}} \\[3mm]&=-\,{\pi\sgn\pars{a} \over 2b}\braces{% \exp\pars{-\verts{b}\bracks{\verts{a} + \root{a^{2} + c^{2}}}} - \exp\pars{\verts{b}\bracks{\verts{a} - \root{a^{2} + c^{2}}}}} \\[3mm]&=-\,{\pi\sgn\pars{a} \over 2b}\bracks{% \expo{-\verts{b}\root{\vphantom{\Large A}a^{2} + c^{2}}} \pars{\expo{-\verts{ab}} - \expo{\verts{ab}}}} \\[3mm]&={\pi\sgn\pars{a} \over b} \expo{-\verts{b}\root{\vphantom{\Large A}a^{2} + c^{2}}}\sinh\pars{\verts{ab}} ={\pi \over \verts{b}} \expo{-\verts{b}\root{\vphantom{\Large A}a^{2} + c^{2}}}\sinh\pars{ab} \end{align}
\begin{align} &\color{#66f}{\large% \int_{0}^{\infty}\arctan\pars{2ax \over x^{2} + c^{2}}\sin\pars{bx}\,\dd x ={\pi \over \verts{b}} \expo{-\verts{b}\root{\vphantom{\Large A}a^{2} + c^{2}}}\sinh\pars{ab}} \end{align}
• Thank you Sir Felix but I forgot to state that I am looking for elementary methods. +1. :) Commented Jun 25, 2014 at 5:26
• @PranavArora I think the $\large\sin\left(bx\right)$ factor avoids we use any other methods. However, it's still interesting your point of view and I believe we have to think a little bit more about that. Thanks. Commented Jun 25, 2014 at 5:34
• Brad's solution might work, I am working on it to see if it gives the result. Commented Jun 25, 2014 at 5:35
When we see an arctan, sometimes parts is a good start.
Let $$\displaystyle u=\tan^{-1}\left(\frac{2ax}{x^{2}+c^{2}}\right), \;\ dv=\sinh(bx)dx, \;\ du=\frac{-2a(x+c)(x-c)}{x^{4}+2(2a^{2}+c^{2})x^{2}+c^{4}}dx, \;\ v=\frac{-1}{b}\cos(bx)$$
The uv part goes to 0 and we have remaining due to it being an even function:
$$\frac{2a}{b}\int_{-\infty}^{\infty}\frac{(x-c)(x+c)\cos(bx)}{x^{4}+2(2a^{2}+c^{2})x^{2}+c^{4}}dx$$
Use a semicircle in the UHP and consider:
$$f=\frac{2a}{b}\int_{C}\frac{(z+c)(z-c)e^{ibz}}{z^{4}+2(2a^{2}+c^{2})z^{2}+c^{4}}$$
The portion around the arc tends to 0 as $$R\to \infty$$.
It has poles at:
$$z=(\sqrt{a^{2}+c^{2}}+a)i, \;\ z=(\sqrt{a^{2}+c^{2}}-a)i$$
$$2\pi i \cdot Res(f, \;\ (\sqrt{a^{2}+c^{2}}+a)i)=\frac{\pi }{b}e^{-ab}e^{-b\sqrt{a^{2}+c^{2}}}$$
$$2\pi i \cdot Res(f, \;\ (\sqrt{a^{2}+c^{2}}-a)i)=\frac{\pi}{b}e^{ab}e^{-b\sqrt{a^{2}+c^{2}}}$$
sum the residues in the UHP and note the exponential identity for sinh(ab). Thus, obtaining:
$$=\frac{\pi}{b}e^{-b\sqrt{a^{2}+c^{2}}}\sinh(ab)$$ | 4,781 | 11,364 | {"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": 8, "wp-katex-eq": 0, "align": 9, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2024-30 | latest | en | 0.867025 |
https://www.zoho.com/finance/essential-business-guides/inventory/what-is-carrying-cost.html | 1,716,970,914,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059221.97/warc/CC-MAIN-20240529072748-20240529102748-00631.warc.gz | 914,622,276 | 13,680 | # What is Carrying Cost?
Carrying cost is the amount that a business spends on holding inventory over a period of time. It is the cost of owning, storing, and keeping the items in stock.
## Significance of carrying cost
Carrying cost includes the cost of renting the warehouse where the stock is kept, operating the warehouse, paying the salaries of the employees working at the warehouse, any loss of inventory due to theft and damage, and insuring the inventory. Carrying costs are usually 15% to 30% of the value of a company’s inventory. This is a significant figure as it tells the company how long they can keep their inventory before they start losing money over unsalable items. Additionally, it shows how much they need to sell and buy in order to maintain appropriate inventory levels. Calculating carrying cost and knowing how to minimize it can help a company reclaim money tied up in inventory and increase its profits.
## Components of carrying cost
The four main components of carrying cost are:
1. Capital cost
2. Inventory service cost
3. Inventory risk cost
4. Storage space cost
### Capital cost
Capital cost is the largest component of carrying cost incurred by businesses. It includes the interests added and the cost of money invested in the inventory. Capital cost is always expressed as a percentage of the total value of the inventory being held. For example, if a company reports that its capital cost is 30% of its total inventory costs, and the total inventory is worth \$8,000, then the company’s capital cost is \$2,400.
### Inventory service cost
Inventory service cost includes IT hardware, applications, tax, and insurance. The company’s insurance costs are dependent on the type of goods in inventory and the level of inventory. The level of inventory is the amount of inventory the company keeps on hand to fulfill its orders—a high level of inventory makes it easier to meet the customer demand. High levels of inventory attract higher insurance premiums and taxes, raising the total inventory service cost.
### Inventory risk cost
Carrying inventory comes with risk. Inventory risk costs include the shrinkage of inventory (which refers to the loss of products because of factors other than sale), theft, and administrative errors (such as misplaced goods, errors in shipping, or late system updates). Another risk factor is product value depletion: if items are stored for too long in the inventory, their value can drop to a fraction of what they were originally worth.
### Storage space cost
Storage space cost includes the rent paid to warehouse your products, air conditioning and heating, lighting, transportation, and other costs associated with the physical warehouse. This cost has a fixed component and a variable component. The rent is a fixed cost, whereas the costs of handling the materials will vary constantly based on demand and the number of products stocked.
## How to calculate carrying cost
Calculating your carrying cost percentage is important for calculating the profit you’re making on your inventory.Carrying costs are always expressed as a percentage of the total value of inventory. Carrying costs are always expressed as a percentage of the total value of inventory. They’re equal to the inventory holding sum divided by the total value of inventory, then multiplied by 100.
Carrying cost (%) = Inventory holding sum / Total value of inventory x 100
The inventory holding sum is simply the total of all four components of carrying cost.
Inventory holding sum = Inventory service cost + Inventory risk cost + Capital cost + Storage cost
1. Calculate the value of each of your inventory cost components (inventory service cost, inventory risk cost, capital cost, and storage cost).
2. Add the inventory cost components to get the inventory holding sum.
3. Determine the total value of your inventory.
4. Divide the inventory holding sum by the total value of inventory and multiply by 100.
For example, let’s look at the carrying cost for a
Motorcycle retailer who carries inventory for all his bike models. The total value of his inventory is \$50,000. His inventory holding sum is \$10,000 (which includes the inventory service cost, risk cost, capital cost and storage cost). His inventory carrying cost, expressed as a percentage, is:
Carrying cost (%) = Inventory holding sum / Total value of inventory x 100
= 10,000 / 50,000 x 100
= 0.2 x 100
= 20%
The carrying cost incurred by the motorcycle retailer is 20% of his total inventory value.
Therefore, carrying costs enables you to find out your profit against incurred against the inventory you are holding. This cost ensures that you do not run into grave losses by holding inventory over a long period of time. Always the carrying cost should only be in limits of 20% – 30% of your total inventory value.
### Related Posts
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1. ###### Ageze Aklilu 4 years ago
Calculate cost carrying A/R if all costomer don’t take the discount
Hi Ageze, | 1,128 | 5,078 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.878546 |
https://www.convertunits.com/from/kwan/to/denier | 1,686,235,694,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224655027.51/warc/CC-MAIN-20230608135911-20230608165911-00211.warc.gz | 761,226,971 | 12,539 | ## Convert kwan [Japan] to denier [France]
kwan denier
How many kwan in 1 denier? The answer is 0.00034. We assume you are converting between kwan [Japan] and denier [France]. You can view more details on each measurement unit: kwan or denier The SI base unit for mass is the kilogram. 1 kilogram is equal to 0.26666666666667 kwan, or 784.3137254902 denier. Note that rounding errors may occur, so always check the results. Use this page to learn how to convert between kwan [Japan] and denier [France]. Type in your own numbers in the form to convert the units!
## Quick conversion chart of kwan to denier
1 kwan to denier = 2941.17647 denier
2 kwan to denier = 5882.35294 denier
3 kwan to denier = 8823.52941 denier
4 kwan to denier = 11764.70588 denier
5 kwan to denier = 14705.88235 denier
6 kwan to denier = 17647.05882 denier
7 kwan to denier = 20588.23529 denier
8 kwan to denier = 23529.41176 denier
9 kwan to denier = 26470.58824 denier
10 kwan to denier = 29411.76471 denier
## Want other units?
You can do the reverse unit conversion from denier to kwan, or enter any two units below:
## Enter two units to convert
From: To:
## 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! | 477 | 1,637 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-23 | latest | en | 0.78076 |
https://nrich.maths.org/public/leg.php?code=-256&cl=4&cldcmpid=1402 | 1,511,108,012,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805687.20/warc/CC-MAIN-20171119153219-20171119173219-00785.warc.gz | 671,525,438 | 4,776 | # Search by Topic
#### Resources tagged with Harmonic mean similar to About Pythagorean Golden Means:
Filter by: Content type:
Stage:
Challenge level:
### There are 3 results
##### Stage: 5
What is the relationship between the arithmetic, geometric and harmonic means of two numbers, the sides of a right angled triangle and the Golden Ratio?
### Pythagorean Golden Means
##### Stage: 5 Challenge Level:
Show that the arithmetic mean, geometric mean and harmonic mean of a and b can be the lengths of the sides of a right-angles triangle if and only if a = bx^3, where x is the Golden Ratio.
### Classical Means
##### Stage: 5 Challenge Level:
Use the diagram to investigate the classical Pythagorean means. | 158 | 718 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2017-47 | latest | en | 0.82861 |
https://brilliant.org/problems/can-euler-still-help/ | 1,607,117,221,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141743438.76/warc/CC-MAIN-20201204193220-20201204223220-00271.warc.gz | 217,122,741 | 8,410 | # Can Euler still help?
Algebra Level 4
If the expression $i\pi + e + 1$ can be expressed in the form $\ln(k)$ where $k$ is a real number, determine $k$ to three decimal places.
× | 55 | 182 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 4, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2020-50 | latest | en | 0.680361 |
https://www.studypool.com/discuss/205016/help-please-25?free | 1,480,868,301,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541324.73/warc/CC-MAIN-20161202170901-00109-ip-10-31-129-80.ec2.internal.warc.gz | 1,020,248,481 | 14,001 | Mathematics Tutor: None Selected Time limit: 1 Day
Aug 6th, 2014
note that by SSS triangle ABE is congruent to ADE
This means that angle AEB is 90 degrees and angle AED is 90 degrees
since we know that abd is 70, and abe=abd, we see that bae=20 degrees (because it is a triangle and 180-90-70=20)
Using similar reasoning and knowing the triangles are congruent we know that ead is 20
True.
Aug 6th, 2014
...
Aug 6th, 2014
...
Aug 6th, 2014
Dec 4th, 2016
check_circle | 158 | 474 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-50 | longest | en | 0.949236 |
https://www.javaexercise.com/python/convert-python-list-to-dictionary | 1,660,215,743,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571284.54/warc/CC-MAIN-20220811103305-20220811133305-00344.warc.gz | 752,224,488 | 11,119 | # Python List to Dictionary : How to convert a list into dictionary
Python List is a linear data structure that is used to store data. In this topic, we will create a dictionary by using the list. Dictionary also a data structure that is used to store data in key-value pair.
Here we have some examples to explain conversion between list and dictionary data structure.
Lets start with a simple example in which we have a list of numeric items and converting it to the dictionary. We are checking type of data after converting as well. Look at the following example.
## Python Example: Convert List to Dictionary
``````# Take a list
listItems = [2,3,6,8,11,20,3,40]
# Print the list
print("List items:\n",listItems, type(listItems))
# converting list to dictionary
dic = dict.fromkeys(listItems)
print("Dictionary Elements:")
for element in dic :
print(element, end = ' ')
print( type(dic))``````
Output:
List items:
[2, 3, 6, 8, 11, 20, 40] <class 'list'>
Dictionary Elements:
2 3 6 8 11 20 40 <class 'dict'>
Notice: The newly created dictionary contains only unique items from the list. It is because dictionary contains unique keys only but allows duplicate values.
If we iterate the dictionary in key-value pair then it will show all the value by default are set to none.
## Iterating Dictionary
``````# Take a list
listItems = [2,3,6,8,11,20,3,40]
# Print the list
print("List items:\n",listItems, type(listItems))
# converting list to dictionary
dic = dict.fromkeys(listItems)
print("Dictionary Elements:")
print(dic, type(dic))``````
Output:
List items:
[2, 3, 6, 8, 11, 20, 3, 40] <class 'list'>
Dictionary Elements:
{2: None, 3: None, 6: None, 8: None, 11: None, 20: None, 40: None} <class 'dict'>
## Convert List to Dictionary Using Zip() function
In a scenario, where we have two lists and want to convert them into a dictionary then zip() function is suitable. Python zip() is a function that is used to combine two iterable objects and returns a list of tuples.
Here, we have passed two lists in zip() function to get a list of tuples, where each tuple contains an entry from both the lists. After that we created a dictionary object from this list of tuples.
``````# Take a list
list1 = ["One","Two","Three","Four"]
# Print the list
print("List1 items:\n",list1, type(list1))
# Take another list
list2 = [1,2,3,4]
print("List2 items:\n",list2, type(list2))
# converting list to dictionary
ziplists = zip(list1,list2)
dic = dict(ziplists)
print("Dictionary Elements:")
print(dic, type(dic))``````
Output:
List1 items:
['One', 'Two', 'Three', 'Four'] <class 'list'>
List2 items:
[1, 2, 3, 4] <class 'list'>
Dictionary Elements:
{'One': 1, 'Two': 2, 'Three': 3, 'Four': 4} <class 'dict'>
## Converting List of tuples into Dictionary
If we already have a list of tuples then we don't need to use zip() function that we did in above example. We can directly pass this list into the dict() function and get the dictionary. See the below example.
``````# Take a list
listOfTuples = [(1,"One"),(2,"Two"),(3,"Three"),(4,"Four")]
# Print the list
print("List items:\n",listOfTuples, type(listOfTuples))
# converting list to dictionary
dic = dict(listOfTuples)
print("Dictionary Elements:")
print(dic, type(dic))``````
Output:
List items:
[(1, 'One'), (2, 'Two'), (3, 'Three'), (4, 'Four')] <class 'list'>
Dictionary Elements:
{1: 'One', 2: 'Two', 3: 'Three', 4: 'Four'} <class 'dict'>
### Conclusion
Well, in this topic, we learnt to convert python list to dictionary. We used single list and multiple lists to convert into dictionary. In an example, we used zip() function to combine two lists and then convert to dictionary. We have used several example to explain the conversion process.
Trending | 1,028 | 3,728 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2022-33 | latest | en | 0.717879 |
https://hassamuddin.com/blog/kosaraju/ | 1,713,438,536,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817206.28/warc/CC-MAIN-20240418093630-20240418123630-00421.warc.gz | 267,670,652 | 9,592 | Recently I’ve been retaking some classes that I had taken a few years ago, now with much more experience and a better mindset to learn. Specifically, I’m retaking Stanford’s Algorithms: Design and Analysis class on Coursera.
In doing so, I’ve gained a significant amount of appreciation for the elegant techniques and methods that algorithm developers have found. Most recently, I learned about Kosaraju’s algorithm for finding strongly connected components (SCCs) in a directed graph, and I thought it was ridiculously elegant and clean.
You’re never going to use Kosaraju’s Algorithm in real life. In fact, there’s a faster solution to this problem using Tarjan’s Algorithm. That’s not the point of studying algorithms. I’m still making my way through Stanford’s Algorithms course, but I’m not just learning about algorithms and how to analyze them. What I’m really learning is how to apply these problem-solving principles that are the foundation of all clever algorithm design.
Strongly Connected Components ¶
In an undirected graph, it’s clear to see what a “connected” component is. If two nodes have a path between them, they are connected, and the connected components are the chunks of nodes that aren’t isolated.
Above, the nodes 1, 2, and 3 are connected as one group, 4 and 5, as well as 6 and 7, are each a group as well. It is super clear what the different components in this graph are, and determining connected components in an undirected graph is a piece of cake.
So what happens when we start talking about directed graphs?
Well, just looking at it like we did earlier, there is only one group of connected nodes. So there’s only one connected component, right?
Trick question. To answer that question, we need to define connected components for directed graphs.
Definition ¶
A more theoretical definition of SCCs would talk about equivalence relations and subgraphs, but we don’t care about that. A Strongly Connected Component is the smallest section of a graph in which you can reach, from one vertex, any other vertex that is also inside that section. In slightly more theoretical terms, an SCC is a strongly connected subgraph of some larger graph G.
So that graph above has four SCCs. Every single node is its own SCC. If I go to node 2, I can never go to any other node, and then back to node 2. I’m going to leave it as an exercise to the reader to prove, or at least think about, the idea that for any given directed acyclic graph (DAG) with n nodes, there are exactly n SCCs.
Why? ¶
Okay, cool, we can define and reason about SCCs in graphs. So what, why do we care?
I mean honestly, it’s just cool. It’s a nice way to reason about and visualize complicated graphs, and you can use SCCs to analyze and think about larger graphs like social media networks or even the internet. SCCs can also be used to simplify the 2-SAT problem, but that’s beyond our scope.
It’s just cool, we can learn and understand a lot about large networks. Pinpoint weak points, make pretty diagrams, whatever you desire. But what’s even cooler is some of the ingenuity behind the algorithms that one might use to find SCCs.
Finding normal connected components ¶
Before we dive into Kosaraju’s Algorithm, let’s discuss how we’d calculate the connected components in an undirected graph. If we do a DFS (or BFS), on a given node, we’ll find all the connected nodes. If we iterate over every single node and DFS, whenever we iterate over a node that hasn’t been seen, it’s a connected component. The (python-esque) pseudocode might look something like:
``````def dfs_loop(G):
n = 0
for node in G:
if not node.visited:
n++
dfs(G, node)
def dfs(G, v):
v.visited = True
for edge:(v, w) in G:
if not w.visited:
dfs(G, w)
``````
and now n will count the number of components in an undirected graph. What if we tried this on a directed graph?
Counting SCCs ¶
There are 4 strongly connected components in this graph G: {1, 2, 3}, {4}, {5, 6, 7, 8}, {9, 10, 11}. Let’s try that same method on this example graph. Say we start at node 10, we’ll hit 9 and 10, and only those three nodes. Okay, that was easy. What if we start at node 3? We’ll hit 1, 2, 4, 5…
So our method works, sometimes. What do we do?
Kosaraju’s Algorithm ¶
Let’s apply some intuition. If we “compressed” this graph, converting each set of SCCs into one supernode, we’d get a simple set of 4 nodes. In fact, we get that same graph that we saw earlier. This graph is a DAG, it ends. If we start in one of the “sink” SCCs, we will only find the things in that SCC. We can keep working our way backward, and every time we call DFS, we will only see the nodes in that SCC.
Kosaraju’s Algorithm applies this idea to find all the SCCs in a given graph in linear time. We do one DFS pass trying to find the correct order of nodes, and then we do another pass going in that order.
How do we find that correct order? Ideally, we want to be starting in one of the sink SCCs. If we do a normal post-order DFS, we can be certain that the last node is not going to be inside of a sink SCC. The first node may be inside a sink SCC, but we can’t be sure. However, if we do a post-order DFS on the reversed graph, we know that the last node will be inside of a sink SCC. This reversed order is the correct order. I urge you to validate this yourself, as it might not be instantly clear, but you’ll find that this is indeed true.
Using this knowledge, we just need to do two passes of DFS. One on the reversed graph, and one in the newfound order.
At the suggestion of aneksteind, if you’re interested in an actual implementation of Kosaraju’s Algorithm, take a look at their gist. Or, just in some psuedocode:
``````s = None
order = []
def dfs_loop(G):
for node in G:
if not node.explored:
s = node
dfs(G, node)
def dfs(G, v):
v.explored = True | 1,412 | 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-2024-18 | longest | en | 0.959387 |
https://www.roofingproslewisvilletx.com/what-is-sports-spread-betting-how-do-spread-betting-works/ | 1,713,596,866,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817491.77/warc/CC-MAIN-20240420060257-20240420090257-00401.warc.gz | 865,674,422 | 129,301 | The negative or minus sign means you’re betting on the favored team, and the number after the negative sign is the amount you have to place to win \$100. There are three different types of odds used in soccer betting. Odds are the probability of a certain outcome to happen in a soccer match.
## How Do Parlay Odds Work?
Based on this, there has a movement to attract more people to horse racing by making it more accessible to the average punter. Ten years ago, if you were going to Cheltenham, all the odds would be displayed https://al-muammal.org/lacing-your-man-one-womans-story/ as fractional odds. Betting odds allow you to calculate how much money you will win if you make a bet. Let’s use the same examples as before, with the same replacement of numbers for letters, i.e. 4/1 becomes A/B. Quite simply, for every value of B that you bet, you will win A, plus the return of your stake.
## Calculating Implied Probability With Decimal Odds
If the game lands on three, you’d lose half your bet (Under 2.5) and refunded the other half . Soccer is the most widely-bet sport worldwide, and there are a number of different ways to bet an individual game. There are also all sorts of competitions with varying rules, so it’s always good to be informed before placing a bet. This article will help both novice and experienced bettors get familiar with betting on the most popular sport in the world. There are a variety of ways to bet a side and total in soccer matches, including three-way moneylines or backing the double chance.
All things being equal, spreads would be centred around the current market price with a point in either direction built in to account for the broker’s commission. Financial spread betting is not so much an investment as a speculation, calling a particular direction in the movement of a market to profit directly from the broker. Partly for this reason, spread betting is regulated by the Financial Conduct Authority rather than the Gambling Commission, although all profits remain tax free as with traditional gambling. Spread betting is a form of wagering that rewards – or penalises – the accuracy of a prediction, as opposed to traditional fixed odds betting that usually relies on a simple win/lose dynamic.
## Games Handicap Betting Or Spread Betting
Once all the handicaps have been applied, the final league table in the eyes of the bookmaker can be worked out and you will discover whether your bet was successful. Again, it is essential to keep in mind that in handicap betting, the handicap only applies to the selection you are betting on. Always bear in mind that in handicap betting, the handicap only applies to the selection you are betting on. Spreadex are offering a 1st England Wicket spread of 28-32, and you choose to sell £2 at 28. In classic England fashion, they collapse quickly, losing a wicket after just 5 runs, leave you with a profit of £46 ((28 – 5) x £2). Let’s say England were slightly quicker off the mark, bowling out an Aussie batsman after 60 runs total.
There is a wide selection of special bitcoin basketball betting offers as well. Totals BettingIn our hypothetical game between the Knicks and the Celtics, the over/under number might be 188. Again, if the combined score is exactly 188 points, the bet is considered a push, or a tie, and no money changes hands. Of the four major sports, basketball is one of the easiest to bet on. Along with football, basketball uses the point spread for wagering on sides , as well as an over/under number. | 760 | 3,525 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | latest | en | 0.949384 |
https://math.answers.com/other-math/What_is_378472_rounded_to_the_nearest_thousand | 1,685,956,850,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224651815.80/warc/CC-MAIN-20230605085657-20230605115657-00354.warc.gz | 437,416,668 | 49,576 | 0
# What is 378472 rounded to the nearest thousand?
Wiki User
2014-06-04 17:54:56
378000
Vella Little
Lvl 9
2022-03-04 13:11:22
Study guides
20 cards
## A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials
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Wiki User
2014-06-04 17:54:56
Rounded to the nearest thousand, it is 378,000. | 136 | 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} | 2.578125 | 3 | CC-MAIN-2023-23 | longest | en | 0.813891 |
https://books.google.com.jm/books?qtid=f37e7869&dq=editions:UOM39015063898350&lr=&id=nicAAAAAYAAJ&sa=N&start=90 | 1,713,996,592,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296819971.86/warc/CC-MAIN-20240424205851-20240424235851-00555.warc.gz | 130,194,295 | 5,911 | Books Books
The perimeters of two regular polygons of the same number of sides, are to each other as their homologous sides, and their areas are to each other as the squares of those sides (Prop.
A Text-book of Geometry - Page 211
by George Albert Wentworth - 1888 - 386 pages
## Plane and Solid Geometry
Arthur Schultze, Frank Louis Sevenoak - Geometry - 1902 - 394 pages
...= OD : O'D' = AO: A'O'. (Why ?) But P:P' = AB:A'B' = AD:A'D'. (Why?) 407. COR. The areas of regular polygons of the same number of sides are to each other as the squares of their radii or apothems. « Ex. 948. The lines joining the midpoints of the radii of a regular...
## Examinations Papers
1902 - 762 pages
...from three given points and from a given straight line. When is it impossible to do so ? Q. Prove that the perimeters of two regular polygons of the same number of sides are to one another as the radii of their circumscribing circles. Prove that in a given circle the perimeter...
## Plane Geometry by the Suggestive Method
John Alton Avery - Geometry, Modern - 1903 - 136 pages
...to any vertex of a regular polygon bisects the angle at the vertex. 143. The perimeters of regular polygons of the same number of sides are to each other as any two homologous sides. 144. Find the area of a square inscribed in a circle whose radius is 6. 145....
## Plane and Solid Geometry
Fletcher Durell - Geometry - 1911 - 553 pages
...Hence .Kand Kf are similar. Art. 321. QED 268 BOOK V. PLANE GEOMETRY PROPOSITION VI. THEOREM 434. I. The perimeters of two regular polygons of the same...other as the radii of their circumscribed circles, or as the radii of their inscribed circles; II. Their areas are to each other as the squares of these...
## Plane Geometry
Fletcher Durell - Geometry, Plane - 1904 - 382 pages
...homologous sides proportional. Hence K and K' are similar. Art. 321. QED PROPOSITION VI. THEOKEM 434. I. The perimeters of two regular polygons of the same...other as the radii of their circumscribed circles, or as the radii of their inscribed circles; II. Their areas are to each other as the squares of these...
## Plane and Solid Geometry
George Albert Wentworth - Geometry - 1904 - 496 pages
...O'A' = OM : O'M'. § 445 §364 § 431 § 436 Also, . § 357 §351 § 361 Ax. 1 QED 448. COR. The areas of two regular polygons of the same number of sides are to each other as the squares of the radii of the circumscribed circles, and of the inscribed circles. § 413 PROPOSITION...
## Plane Geometry
Fletcher Durell - Geometry, Plane - 1904 - 382 pages
...PLANE GEOMETRY PROPOSITION VI. THEOREM 484. I. The perimeters of two regular polygons of the stone number of sides are to each other as the radii of their circumscribed circles, or as the radii of their inscribed circles; II. Their areas are to each other as the squares of these...
## Catalogue ...
Yale University. Sheffield Scientific School - 1905 - 1074 pages
...equivalent to a given square and having the difference of its base and altitude equal to a given line. 4. The perimeters of two regular polygons' of the same number of sides are in the same ratio as the radii of the inscribed or circumseribed circles. 5. To construct a circle... | 851 | 3,206 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2024-18 | latest | en | 0.864892 |
https://www.dataunitconverter.com/petabyte-to-zettabit | 1,708,470,487,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473347.0/warc/CC-MAIN-20240220211055-20240221001055-00167.warc.gz | 749,122,451 | 16,982 | # PB to Zbit → CONVERT Petabytes to Zettabits
expand_more
info 1 PB is equal to 0.000008 Zbit
Input Petabyte (PB) - and press Enter.
## Petabyte (PB) Versus Zettabit (Zbit) - Comparison
Petabytes and Zettabits are units of digital information used to measure storage capacity and data transfer rate.
Both Petabytes and Zettabits are the "decimal" units. One Petabyte is equal to 1000^5 bytes. One Zettabit is equal to 1000^7 bits. There are 125,000 Petabyte in one Zettabit. Find more details on below table.
Unit Name Petabyte Zettabit
Unit Symbol PB Zb or Zbit
Standard decimal decimal
Defined Value 10^15 or 1000^5 Bytes 10^21 or 1000^7 Bits
Value in Bits 8,000,000,000,000,000 1,000,000,000,000,000,000,000
Value in Bytes 1,000,000,000,000,000 125,000,000,000,000,000,000
## Petabyte (PB) to Zettabit (Zbit) Conversion - Formula & Steps
The PB to Zbit Calculator Tool provides a convenient solution for effortlessly converting data units from Petabyte (PB) to Zettabit (Zbit). Let's delve into a thorough analysis of the formula and steps involved.
Outlined below is a comprehensive overview of the key attributes associated with both the source (Petabyte) and target (Zettabit) data units.
Source Data Unit Target Data Unit
Equal to 1000^5 bytes
(Decimal Unit)
Equal to 1000^7 bits
(Decimal Unit)
The conversion diagram provided below offers a visual representation to help you better grasp the steps involved in calculating Petabyte to Zettabit in a simplified manner.
÷ 1000
÷ 1000
÷ 1000
x 1000
x 1000
x 1000
Based on the provided diagram and steps outlined earlier, the formula for converting the Petabyte (PB) to Zettabit (Zbit) can be expressed as follows:
diamond CONVERSION FORMULA Zbit = PB x 8 ÷ 10002
Now, let's apply the aforementioned formula and explore the manual conversion process from Petabyte (PB) to Zettabit (Zbit). To streamline the calculation further, we can simplify the formula for added convenience.
FORMULA
Zettabits = Petabytes x 8 ÷ 10002
STEP 1
Zettabits = Petabytes x 8 ÷ (1000x1000)
STEP 2
Zettabits = Petabytes x 8 ÷ 1000000
STEP 3
Zettabits = Petabytes x 0.000008
Example : By applying the previously mentioned formula and steps, the conversion from 1 Petabyte (PB) to Zettabit (Zbit) can be processed as outlined below.
1. = 1 x 8 ÷ 10002
2. = 1 x 8 ÷ (1000x1000)
3. = 1 x 8 ÷ 1000000
4. = 1 x 0.000008
5. = 0.000008
6. i.e. 1 PB is equal to 0.000008 Zbit.
Note : Result rounded off to 40 decimal positions.
You can employ the formula and steps mentioned above to convert Petabytes to Zettabits using any of the programming language such as Java, Python, or Powershell.
### Unit Definitions
#### What is Petabyte ?
A Petabyte (PB) is a decimal unit of digital information that is equal to 1,000,000,000,000,000 bytes (or 8,000,000,000,000,000 bits) and commonly used to measure the storage capacity of enterprise storage arrays and data centers. It is also used to express data transfer speeds and in the context of data storage and memory, the binary-based unit of Pebibyte (PiB) is used instead.
arrow_downward
#### What is Zettabit ?
A Zettabit (Zb or Zbit) is a decimal unit of measurement for digital information transfer rate. It is equal to 1,000,000,000,000,000,000,000 (one sextillion) bits. It is used to measure the speed of extremely high-speed data transfer over communication networks, such as high-speed internet backbones and advanced computer networks. The zettabit is part of the International System of Units (SI) and the prefix zetta indicates multiplication by the seventh power of 1000.
## Excel Formula to convert from Petabyte (PB) to Zettabit (Zbit)
Apply the formula as shown below to convert from 1 Petabyte (PB) to Zettabit (Zbit).
A B C
1 Petabyte (PB) Zettabit (Zbit)
2 1 =A2 * 0.000008
3
If you want to perform bulk conversion locally in your system, then download and make use of above Excel template.
## Python Code for Petabyte (PB) to Zettabit (Zbit) Conversion
You can use below code to convert any value in Petabyte (PB) to Petabyte (PB) in Python.
petabytes = int(input("Enter Petabytes: "))
zettabits = petabytes * 8 / (1000*1000)
print("{} Petabytes = {} Zettabits".format(petabytes,zettabits))
The first line of code will prompt the user to enter the Petabyte (PB) as an input. The value of Zettabit (Zbit) is calculated on the next line, and the code in third line will display the result.
## Frequently Asked Questions - FAQs
#### How many Zettabits(Zbit) are there in a Petabyte(PB)?expand_more
There are 0.000008 Zettabits in a Petabyte.
#### What is the formula to convert Petabyte(PB) to Zettabit(Zbit)?expand_more
Use the formula Zbit = PB x 8 / 10002 to convert Petabyte to Zettabit.
#### How many Petabytes(PB) are there in a Zettabit(Zbit)?expand_more
There are 125000 Petabytes in a Zettabit.
#### What is the formula to convert Zettabit(Zbit) to Petabyte(PB)?expand_more
Use the formula PB = Zbit x 10002 / 8 to convert Zettabit to Petabyte.
#### Which is bigger, Zettabit(Zbit) or Petabyte(PB)?expand_more
Zettabit is bigger than Petabyte. One Zettabit contains 125000 Petabytes.
## Similar Conversions & Calculators
All below conversions basically referring to the same calculation. | 1,505 | 5,231 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2024-10 | latest | en | 0.768117 |
https://testbook.com/question-answer/a-car-covers-200-meter-in-20-seconds-after-startin--615da675602f1a010dd7e663 | 1,642,563,742,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301263.50/warc/CC-MAIN-20220119033421-20220119063421-00194.warc.gz | 561,166,973 | 30,598 | # A car covers 200 meter in 20 seconds after starting from rest with a constant acceleration. Find the acceleration.
1. 2 m/s2
2. 4 m/s2
3. 0.5 m/s2
4. 1 m/s2
## Answer (Detailed Solution Below)
Option 4 : 1 m/s2
## Detailed Solution
Concept:
Acceleration: The rate of change of velocity is called acceleration.
Equations of Motion
The equations of motion establish the relationship between acceleration, time, distance, initial speed, and final speed for a body moving in a straight line with uniform acceleration.
The equations are
v = u + at - (1)
$$s = ut + \frac{1}{2}at^2$$ - (2)
v2 = u2 + 2as - (3)
v is final velocity, u is initial velocity, t is time, a is acceleration, s is the distance travelled.
Calculation:
Initial speed u = 0 (Started from rest)
distance travelled s = 200 m
time taken t = 20 sec
acceleration a = ?
Using the second equation
$$s = ut + \frac{1}{2}at^2$$
$$\implies 200 = (0) t + \frac{1}{2}a(20)^2$$
a = 400 / 400 = 1 m s- 2. | 304 | 981 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2022-05 | latest | en | 0.85469 |
https://www.traditionaloven.com/building/masonry/concrete/convert-dekagram-dkg-dag-concrete-to-pound-lb-of-concrete.html | 1,596,633,249,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735958.84/warc/CC-MAIN-20200805124104-20200805154104-00422.warc.gz | 874,208,267 | 11,844 | Concrete 1 dekagram mass to pounds converter
# concrete conversion
## Amount: 1 dekagram (dag - dkg) of mass Equals: 0.022 pounds (lb) in mass
Converting dekagram to pounds value in the concrete units scale.
TOGGLE : from pounds into dekagrams in the other way around.
## concrete from dekagram to pound Conversion Results:
### Enter a New dekagram Amount of concrete to Convert From
* Whole numbers, decimals or fractions (ie: 6, 5.33, 17 3/8)
* Precision is how many numbers after decimal point (1 - 9)
Enter Amount :
Decimal Precision :
CONVERT : between other concrete measuring units - complete list.
Conversion calculator for webmasters.
## Concrete
This general purpose concrete formulation, called also concrete-aggregate (4:1 - sand/gravel aggregate : cement - mixing ratio w/ water) conversion tool is based on the concrete mass density of 2400 kg/m3 - 150 lbs/ft3 after curing (rounded). Unit mass per cubic centimeter, concrete has density 2.41g/cm3. The main concrete calculator page.
The 4:1 strength concrete mixing formula uses the measuring portions in volume sense (e.g. 4 buckets of concrete aggregate with 1 bucket of water.) In order not to end up with a too wet concrete, add water gradually as the mixing progresses. If mixing concrete manually by hand; mix dry matter portions first and only then add water. This concrete type is commonly reinforced with metal rebars or mesh.
Convert concrete measuring units between dekagram (dag - dkg) and pounds (lb) but in the other reverse direction from pounds into dekagrams.
conversion result for concrete: From Symbol Result To Symbol 1 dekagram dag - dkg = 0.022 pounds lb
# Converter type: concrete measurements
This online concrete from dag - dkg into lb converter is a handy tool not just for certified or experienced professionals.
First unit: dekagram (dag - dkg) is used for measuring mass.
Second: pound (lb) is unit of mass.
## concrete per 0.022 lb is equivalent to 1 what?
The pounds amount 0.022 lb converts into 1 dag - dkg, one dekagram. It is the EQUAL concrete mass value of 1 dekagram but in the pounds mass unit alternative.
How to convert 2 dekagrams (dag - dkg) of concrete into pounds (lb)? Is there a calculation formula?
First divide the two units variables. Then multiply the result by 2 - for example:
0.022046226218488 * 2 (or divide it by / 0.5)
QUESTION:
1 dag - dkg of concrete = ? lb
1 dag - dkg = 0.022 lb of concrete
## Other applications for concrete units calculator ...
With the above mentioned two-units calculating service it provides, this concrete converter proved to be useful also as an online tool for:
1. practicing dekagrams and pounds of concrete ( dag - dkg vs. lb ) measuring values exchange.
2. concrete amounts conversion factors - between numerous unit pairs.
3. working with - how heavy is concrete - values and properties.
International unit symbols for these two concrete measurements are:
Abbreviation or prefix ( abbr. short brevis ), unit symbol, for dekagram is:
dag - dkg
Abbreviation or prefix ( abbr. ) brevis - short unit symbol for pound is:
lb
### One dekagram of concrete converted to pound equals to 0.022 lb
How many pounds of concrete are in 1 dekagram? The answer is: The change of 1 dag - dkg ( dekagram ) unit of concrete measure equals = to 0.022 lb ( pound ) as the equivalent measure for the same concrete type.
In principle with any measuring task, switched on professional people always ensure, and their success depends on, they get the most precise conversion results everywhere and every-time. Not only whenever possible, it's always so. Often having only a good idea ( or more ideas ) might not be perfect nor good enough solution. If there is an exact known measure in dag - dkg - dekagrams for concrete amount, the rule is that the dekagram number gets converted into lb - pounds or any other concrete unit absolutely exactly.
Conversion for how many pounds ( lb ) of concrete are contained in a dekagram ( 1 dag - dkg ). Or, how much in pounds of concrete is in 1 dekagram? To link to this concrete dekagram to pounds online converter simply cut and paste the following.
The link to this tool will appear as: concrete from dekagram (dag - dkg) to pounds (lb) conversion.
I've done my best to build this site for you- Please send feedback to let me know how you enjoyed visiting. | 1,011 | 4,368 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2020-34 | latest | en | 0.828316 |
http://engineeronadisk.com/V3/engineeronadisk-93.html | 1,679,846,516,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945473.69/warc/CC-MAIN-20230326142035-20230326172035-00589.warc.gz | 15,864,230 | 4,372 | ## 1. Economic Justification
• When considering the economic value of a decision, one method is the payback period.
• Simple estimates for the initial investment and yearly savings are,
• There are clearly more factors than can be considered, including,
changes in material use
opportunity cost
setup times
change in inventory size
material handling change
• The simple models ignore the conversion between present value and future value. (ie, money now is worth more than the same amount of money later)
• Quite often a Rate of Return (ROR) will be specified by management. This is used in place of interest rates, and can include a companies value for the money. This will always be higher than the typical prime interest rate.
• So far we haven’t considered the effects of taxes. Basically corporate taxes are applied to profits. Therefore we attempt to distribute expenses evenly across the life of a project (even though the majority of the money has been spent in the first year). This distribution is known as depreciation.
• Methods for depreciation are specified in the tax laws. One method is straight line depreciation.
1.1 Problems
Problem 1.1 Consider an assembly line that is currently in use, and the system proposed to replace it. The product line is expected to last 5 years, and then be sold off. The corporate tax rate is 50% and the company policy is to require a 17% rate of return. Should we keep the old line, or install the new one?
Current Manual Line:
- used 2000 hrs/yr with 10 workers at \$20/hr each
- maintenance is \$20,000/yr
- the current equipment is worth \$20,000 used
Proposed Line:
- the equipment will cost \$100,000 and the expected salvage value at the end of the project is \$10,000
- 2 workers are required for 1000 hours year at \$40/hr each
- yearly maintenance will be \$40,000 | 407 | 1,845 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2023-14 | latest | en | 0.928542 |
https://education.ti.com/en-au/australiancurriculumnspired/aus-nz/detail?id=EEE3A96A15D24F5294FA7C4EA725B5C2&t=B712BC19306D415784592BEFAA62BCCF | 1,701,907,756,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100626.1/warc/CC-MAIN-20231206230347-20231207020347-00401.warc.gz | 264,876,715 | 14,592 | Education Technology
• ##### Device
• TI-Nspire™ CAS
• ##### Software
TI-Nspire™ CAS
ACMNA213
ACMNA233
ACMNA239
ACMSP246
3.2
# Year 10: Probable Factors
by Peter Fox
#### Objectives
• Apply the distributive law to the expansion of algebraic expressions, including binomials, and collect like terms where appropriate (ACMNA213)
• Expand binomial products and factorise monic quadratic expressions using a variety of strategies (ACMNA233)
• Explore the connection between algebraic and graphical representations of relations such as simple quadratics, circles and exponentials using digital technology as appropriate (ACMNA239)
• Describe the results of two- and three-step chance experiments, both with and without replacements, assign probabilities to outcomes and determine probabilities of events. Investigate the concept of independence (ACMSP246)
#### Vocabulary
• Quadratic Equations,
• Distributivity,
• Monomials,
• Binomials,
• Probability,
• Experiments
#### About the Lesson
Students create a simulation using TI-nspire that produces four dice rolls simultaneously. The outcome from each is used to form a single quadratic equation. TI-nspire is then used to determine if the quadratic factorises, a simulation that is repeated by students to obtain an approximate answer to the questions "What is the probability that a randomly generated quadratic will factorise?" Students are then guided through a series of scaffolded questions to determine the theoretical probability. | 322 | 1,496 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2023-50 | latest | en | 0.849187 |
https://studyres.com/doc/287075/quick-review-of-parallel--perpendicular--and-skew | 1,618,669,104,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038460648.48/warc/CC-MAIN-20210417132441-20210417162441-00316.warc.gz | 648,755,496 | 6,610 | Survey
* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project
Document related concepts
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Transcript
```Quick Review of Parallel, Perpendicular, and Skew
Parallel Lines: Do not intersect, coplanar
Skew Lines: Do not intersect, not coplanar
Perpendicular Lines: Intersect at right angle
Sep 165:40 AM
DO NOW:
Aug 249:55 PM
1
What is a Transversal?
A line that intersects two or more coplanar lines at different points. For a transversal passing through two lines, how many distinct angles are formed? Sep 208:06 AM
Angles Formed by Transversals (No. 1)
Two angles are corresponding angles if they have corresponding positions.
Sep 208:06 AM
2
Angles Formed by Transversals (No. 2)
Two angles are alternate interior angles if they lie between the two lines and on opposite sides of the transversal.
Sep 208:06 AM
Angles Formed by Transversals (No. 3)
Two angles are alternate exterior angles if they lie outside the two lines and on opposite sides of the transversal.
Sep 208:06 AM
3
Angles Formed by Transversals (No. 4)
Two angles are consecutive interior angles if they lie between the two lines and on the same side of the transversal.
Sep 208:06 AM
Exploring Parallel Lines
Sep 208:06 AM
4
Angle Relationships
Corresponding Angles
Congruent
Congruent
Consecutive Interior Angles
Supp.
Alternate Interior Angles
Alternate Exterior Angles
Congruent
Sep 208:06 AM
Find x!!!!!!
Sep 208:06 AM
5
Let's Try Another
Sep 208:06 AM
One More :)
Sep 208:06 AM
6
Practice: 3.1 WKBK pg 44 #11‐16
3.2 WKBK pg 47 #11‐22
Sep 811:43 PM
7
```
Related documents | 482 | 1,652 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2021-17 | latest | en | 0.847485 |
https://www.cs.cmu.edu/~motionplanning/student_gallery/2007/Hyungsoo/finalProject/Motion_Planning_Project.htm | 1,623,900,210,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487626465.55/warc/CC-MAIN-20210617011001-20210617041001-00488.warc.gz | 649,453,556 | 2,911 | Motion Planning : Final Project
Due : Dec/10/2007
Hyun Soo Park, Iacopo Gentilini
Goal
Find path from start position to goal position using D* algorithm and update its position using sonar sensor and vision.
Environment
Robot : Nomad scout robot (equiped with 16 sonar sensors, camera and differential wheel)
Workspace : NSH, A floor at CMU
Map building
Map has been built using sonar sensors and it is discretized in order to generate grid map.
(red x : sonar output)
Main Algorithm
• LOAD map
• Path planning with D*
• WHILE Mission completed || Can't find path
• READ sonar data
• READ current position(encorder output)
• IF (current position == way point)
• UPDATE map using sonar data
• IF (Good region)
• Image Processing
• UPDATE current position
• ELSE
• Kalman filtering using sonar data
• ENDIF
• D*
• ENDIF
• MOVE to way point
• IF (current position == goal position)
• Mission completed = true
• ENDIF
• ENDWHILE
Project Detail
1. D* algorithm
Map of workspace is updated every moment when robot reaches to the way point. When robot reaches to the way point, it senses obstacles. If the obstacle is not in before map, then it addes new obstacles. Since map is updated, the cost of each neighbor is changed so it has to generate new path. This is D* algorithm and it is called whenever cost is changed.
2. Kalman filtering using sonar sensor
Since our map is primarily composed of long corridor, we decided to use Kalman filter in order to calibrate its right/left position. Since encorder output is quite reliable, Kalman filter is useful when the orientation of robot is not accurate.
3. Image Processing
This algorithm also generates whenever it reaches to way point. The FindLANdamrk (FLAN) recognizes the presence of landmarks (pink boxes) within the images captured by the on-board camera. Then, if the robot lies within a region defined by three landmarks, the algorithm computates the robot global position and orientation just using as input the global coordinates of the three closest landmarks and the three angles measured in the image between robot and landmarks. If the robot does not lie within the triangle defined by the three pik boxes, then the position error could be very high. However, if it lies in the ˇ°good regionˇ± the error of the algorithm is about three inches.
Demo
VIDEO
Result
Presentation Slides
Paper presentation : Numerical Potential Field Techniques for Robot Path Planning (Sept, 19, 2007)
Project Proposal (Oct, 3, 2007)
Project 1st Progress (Oct, 22, 2007)
Project 2nd Progress (Nov, 13, 2007)
Project Final (Dec, 10, 2007)
Special Thanks
Professor Howie Choset, TA Hyungpil Moon, Biorobotics lab family
Reference
[1] H. Choset, K. M. Lynch, S. Hutchinson, G. Kantor, W. Burgard, L. E. Kavraki and S. Thrun, Principles of Robot Motion: Theory, Algorithms, and Implementations,
MIT Press, Boston, 2005. | 705 | 2,885 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2021-25 | latest | en | 0.846066 |
http://www.hindawi.com/journals/jam/2012/720192/ | 1,472,501,919,000,000,000 | application/xhtml+xml | crawl-data/CC-MAIN-2016-36/segments/1471982965886.67/warc/CC-MAIN-20160823200925-00072-ip-10-153-172-175.ec2.internal.warc.gz | 487,937,193 | 71,492 | `Journal of Applied MathematicsVolume 2012 (2012), Article ID 720192, 14 pageshttp://dx.doi.org/10.1155/2012/720192`
Research Article
## Implicit and Explicit Iterations with Meir-Keeler-Type Contraction for a Finite Family of Nonexpansive Semigroups in Banach Spaces
1School of Control and Computer Engineering, North China Electric Power University, Baoding 071003, China
2School of Mathematics and Physics, North China Electric Power University, Baoding 071003, China
3Department of Mathematics Education and the RINS, Gyeongsang National University, Jinju 660-701, Republic of Korea
Received 31 December 2011; Accepted 28 January 2012
Copyright © 2012 Jiancai Huang et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
#### Abstract
We introduce an implicit and explicit iterative schemes for a finite family of nonexpansive semigroups with the Meir-Keeler-type contraction in a Banach space. Then we prove the strong convergence for the implicit and explicit iterative schemes. Our results extend and improve some recent ones in literatures.
#### 1. Introduction
Let be a nonempty subset of a Banach space and be a mapping. We call nonexpansive if for all . The set of all fixed points of is denoted by , that is, .
One parameter family is said to a semigroup of nonexpansive mappings or nonexpansive semigroup on if the following conditions are satisfied: (1) for all ; (2) for all ; (3)for each , for all ; (4)for each , the mapping from , where denotes the set of all nonnegative reals, into is continuous.
We denote by the set of all common fixed points of semigroup , that is, and by the set of natural numbers.
Now, we recall some recent work on nonexpansive semigroup in literatures. In [1], Shioji and Takahashi introduced the following implicit iteration for a nonexpansive semigroup in a Hilbert space: where and . Under the certain conditions on and , they proved that the sequence defined by (1.1) converges strongly to an element in .
In [2], Suzuki introduced the following implicit iteration for a nonexpansive semigroup in a Hilbert space: where and . Under the conditions that , he proved that defined by (1.2) converges strongly to an element of . Later on, Xu [3] extended the iteration (1.2) to a uniformly convex Banach space that admits a weakly sequentially continuous duality mapping. Song and Xu [4] also extended the iteration (1.2) to a reflexive and strictly convex Banach space.
In 2007, Chen and He [5] studied the following implicit and explicit viscosity approximation processes for a nonexpansive semigroup in a reflexive Banach space admitting a weakly sequentially continuous duality mapping: where is a contraction, and . They proved the strong convergence for the above iterations under some certain conditions on the control sequences.
Recently, Chen et al. [6] introduced the following implicit and explicit iterations for nonexpansive semigroups in a reflexive Banach space admitting a weakly sequentially continuous duality mapping: where is a contraction, and . They proved that defined by (1.4) and (1.5) converges strongly to an element of , which is the unique solution of the following variation inequality problem:
For more convergence theorems on implicit and explicit iterations for nonexpansive semigroups, refer to [713].
In this paper, we introduce an implicit and explicit iterative process by a generalized contraction for a finite family of nonexpansive semigroups in a Banach space. Then we prove the strong convergence for the iterations and our results extend the corresponding ones of Suzuki [2], Xu [3], Chen and He [5], and Chen et al. [6].
#### 2. Preliminaries
Let be a Banach space and the duality space of . We denote the normalized mapping from to by defined by where denotes the generalized duality pairing. For any with and , it is well known that the following inequality holds:
The dual mapping is called weakly sequentially continuous if is single valued, and , where denotes the weak convergence, then weakly star converges to [1416]. A Banach space is called to satisfy Opial’s condition [17] if for any sequence in , , It is known that if admits a weakly sequentially continuous duality mapping , then is smooth and satisfies Opial’s condition [14].
A function is said to be an -function if , for any , and for every and , there exists such that , for all . This implies that for all .
Let be a mapping. is said to be a -contraction if there exists a -function such that for all with . Obviously, if for all , where , then is a contraction. is called a Meir-Keeler-type mapping if for each , there exists such that for all , if , then .
In this paper, we always assume that is continuous, strictly increasing and , where , is strictly increasing and onto.
The following lemmas will be used in next section.
Lemma 2.1 (see [18]). Let be a metric space and be a mapping. The following assertions are equivalent: (i) is a Meir-Keeler-type mapping;(ii)there exists an -function such that is a -contraction.
Lemma 2.2 (see [19]). Let be a Banach space and be a convex subset of . Let be a nonexpansive mapping and be a -contraction. Then the following assertions hold: (i) is a -contraction on and has a unique fixed point in ;(ii)for each , the mapping is of Meir-Keeler-type and it has a unique fixed point in .
Lemma 2.3 (see [20]). Let be a Banach space and be a convex subset of . Let be a Meir-Keeler-type contraction. Then for each there exists such that, for each with .
Lemma 2.4 (see [21]). Let C be a closed convex subset of a strictly convex Banach space E. Let be a nonexpansive mapping for each , where is some integer. Suppose that is nonempty. Let be a sequence of positive numbers with . Then the mapping defined by is well defined, nonexpansive and holds.
Lemma 2.5 (see [22]). Assume that is a sequence of nonnegative real numbers such that where is a sequence in and is a sequence in such that(i);(ii);(iii) or .Then .
#### 3. Main Results
In this section, by a generalized contraction mapping we mean a Meir-Keeler-type mapping or - contraction. In the rest of the paper we suppose that from the definition of the -contraction is continuous, strictly increasing and is strictly increasing and onto, where , for all . As a consequence, we have the is a bijection on .
Theorem 3.1. Let be a nonempty closed convex subset of a reflexive Banach space which admits a weakly sequentially continuous duality mapping from into . For every , let be a semigroup of nonexpansive mappings on such that and be a generalized contraction on . Let and be the sequences satisfying and . Let be a sequence generated by Then converges strongly to a point , which is the unique solution to the following variational inequality:
Proof. First, we show that the sequence generated by (3.1) is well defined. For every and , let and define by where . Since is nonexpansive, is nonexpansive. By Lemma 2.2 we see that is a Meir-Keeler-type contraction for each . Hence, each has a unique fixed point, denoted as , which uniquely solves the fixed point equation (3.3). Hence generated by (3.1) is well defined.
Now we prove that generated by (3.1) is bounded. For any , we have Using (3.4), we get and hence which implies that Hence This shows that is bounded, and so are , and .
Since is reflexivity and is bounded, there exists a subsequence such that for some as . Now we prove that . For any fixed , we have By hypothesis on , , we have Further, from (3.9) we get Since admits a weakly sequentially duality mapping, we see that satisfies Opial’s condition. Thus if , we have This contradicts (3.11). So .
In (3.5), replacing with and with , we see that which implies that Now we prove that is relatively sequentially compact. Since is weakly sequentially continuous, we have which implies that If , then is relatively sequentially compact. If , we have . Since is continuous, . By the definition of , we conclude that , which implies that is relatively sequentially compact.
Next, we prove that is the solution to (3.2). Indeed, for any , we have Therefore, Since and is weakly sequentially continuous, we have This shows that is the solution of the variational inequality (3.2).
Finally, we prove that is the unique solution of the variational inequality (3.2). Assume that with is another solution of (3.2). Then there exists such that . By Lemma 2.3 there exists such that . Since both and are the solution of (3.2), we have Adding the above inequalities, we get which is a contradiction. Therefore, we must have , which implies that is the unique solution of (3.2).
In a similar way it can be shown that each cluster point of sequence is equal to . Therefore, the entire sequence converges strongly to . This completes the proof.
If letting for all in Theorem 3.1, then we get the following.
Corollary 3.2. Let be a nonempty closed convex subset of a reflexive Banach space which admits a weakly sequentially continuous duality mapping from into . For every ( ), let be a semigroup of nonexpansive mappings on such that and be a generalized contraction on . Let and be sequences satisfying . Let be a sequence generated by Then converges strongly to a point , which is the unique solution to the following variational inequality:
Theorem 3.3. Let be a nonempty closed convex subset of a reflexive and strictly convex Banach space which admits a weakly sequentially continuous duality mapping from into . For every , let be a semigroup of nonexpansive mappings on such that and be a generalized contraction on . Let and be the sequences satisfying . Let be a sequence generated Then converges strongly to a point , which is the unique solution of variational inequality (3.2).
Proof. Let and . Now we show by induction that It is obvious that (3.25) holds for . Suppose that (3.25) holds for some , where . Observe that Now, by using (3.24) and (3.26), we have By induction we conclude that (3.25) holds for all . Therefore, is bounded and so are , , .
For each and , define the mapping , where . Then we rewrite the sequence (3.24) to Obviously, each is nonexpansive. Since is bounded and is reflexive, we may assume that some subsequence of converges weakly to . Next we show that . Put , , and for each . Fix . By (3.28) we have So, for all , we have
Since has a weakly sequentially continuous duality mapping satisfying Opials condition, this implies . By Lemma 2.4, we have for each . Therefore, . In view of the variational inequality (3.2) and the assumption that duality mapping is weakly sequentially continuous, we conclude that
Finally, we prove that as . Suppose that . Then there exists and subsequence of such that for all . Put , and . By Lemma 2.3 one has for all . Now, from (2.2) and (3.28) we have It follows that where is a constant.
Let and . It follows from (3.33) that It is easy to see that , and (noting (3.28)) Using Lemma 2.5, we conclude that as . It is a contradiction. Therefore, as . This completes the proof.
If letting for all in Theorem 3.3, then we get the following.
Corollary 3.4. Let be a nonempty closed convex subset of a reflexive and strictly convex Banach space which admits a weakly sequentially continuous duality mapping from into . For every , let be a semigroup of nonexpansive mappings on such that and be a generalized contraction on . Let and be sequences satisfying . Let be a sequence generated Then converges strongly to a point , which is the unique solution of variational inequality (3.2).
Remark 3.5. Theorem 3.1 and Corollary 3.2 extend the corresponding ones of Suzuki [2], Xu [3], and Chen and He [5] from one nonexpansive semigroup to a finite family of nonexpansive semigroups. But Theorem 3.3 and Corollary 3.4 are not the extension of Theorem 3.2 of Chen and He [5] since Banach space in Theorem 3.3 and Corollary 3.4 is required to be strictly convex. But if letting in Theorem 3.3 and Corollary 3.4, we can remove the restriction on strict convexity and hence they extend Theorem 3.2 of Chen and He [5] from a contraction to a generalized contraction.
Remark 3.6. Our Theorem 3.1 extends and improves Theorems 3.2 and 4.2 of Song and Xu [4] from a nonexpansive semigroup to a finite family of nonexpansive semigroups and a contraction to a generalized contraction. Our conditions on the control sequences are different with ones of Song and Xu [4].
#### Acknowledgment
This work was supported by the Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science, and Technology (Grant Number: 2011-0021821).
#### References
1. N. Shioji and W. Takahashi, “Strong convergence theorems for asymptotically nonexpansive semigroups in Hilbert spaces,” Nonlinear Analysis, vol. 34, no. 1, pp. 87–99, 1998.
2. T. Suzuki, “On strong convergence to common fixed poitns of nonexpansive semigroups in HIlbert spaces,” Proceedings of the American Mathematical Society, vol. 131, pp. 2133–2136, 2002.
3. H. K. Xu, “A strong convergence theorem for contraction semigroups in banach spaces,” Bulletin of the Australian Mathematical Society, vol. 72, no. 3, pp. 371–379, 2005.
4. Y. Song and S. Xu, “Strong convergence theorems for nonexpansive semigroup in Banach spaces,” Journal of Mathematical Analysis and Applications, vol. 338, no. 1, pp. 152–161, 2008.
5. R. D. Chen and H. M. He, “Viscosity approximation of common fixed points of nonexpansive semigroups in Banach space,” Applied Mathematics Letters, vol. 20, no. 7, pp. 751–757, 2007.
6. R. D. Chen, H. M. He, and M. A. Noor, “Modified mann iterations for nonexpansive semigroups in Banach space,” Acta Mathematica Sinica, English Series, vol. 26, no. 1, pp. 193–202, 2010.
7. I. K. Argyros, Y. J. Cho, and X. Qin, “On the implicit iterative process for strictly pseudo-contractive mappings in Banach spaces,” Journal of Computational and Applied Mathematics, vol. 233, no. 2, pp. 208–216, 2009.
8. S. S. Chang, Y. J. Cho, H. W. Joseph Lee, and C. K. Chan, “Strong convergence theorems for Lipschitzian demi-contraction semigroups in Banach spaces,” Fixed Point Theory and Applications, vol. 2011, Article ID 583423, 10 pages, 2011.
9. Y. J. Cho, L. B. Ćirić, and S. H. Wang, “Convergence theorems for nonexpansive semigroups in CAT (0) spaces,” Nonlinear Analysis, vol. 74, no. 17, pp. 6050–6059, 2011.
10. Y. J. Cho, S. M. Kang, and X. Qin, “Strong convergence of an implicit iterative process for an infinite family of strict pseudocontractions,” Bulletin of the Korean Mathematical Society, vol. 47, no. 6, pp. 1259–1268, 2010.
11. W. Guo and Y. J. Cho, “On the strong convergence of the implicit iterative processes with errors for a finite family of asymptotically nonexpansive mappings,” Applied Mathematics Letters, vol. 21, no. 10, pp. 1046–1052, 2008.
12. H. He, S. Liu, and Y. J. Cho, “An explicit method for systems of equilibrium problems and fixed points of infinite family of nonexpansive mappings,” Journal of Computational and Applied Mathematics, vol. 235, no. 14, pp. 4128–4139, 2011.
13. X. Qin, Y. J. Cho, and M. Shang, “Convergence analysis of implicit iterative algorithms for asymptotically nonexpansive mappings,” Applied Mathematics and Computation, vol. 210, no. 2, pp. 542–550, 2009.
14. J. P. Gossez and E. L. Dozo, “Some geometric properties related to the fixed point theoryfor nonexpansive mappings,” Pacific Journal of Mathematics, vol. 40, pp. 565–573, 1972.
15. J. S. Jung, “Iterative approaches to common fixed points of nonexpansive mappings in Banach spaces,” Journal of Mathematical Analysis and Applications, vol. 302, no. 2, pp. 509–520, 2005.
16. J. Schu, “Approximation of fixed points of asymptotically nonexpansive mappings,” Proceedings of the American Mathematical Society, vol. 112, pp. 143–151, 1991.
17. Z. Opial, “Weak convergence of the sequence of successive approximations for nonexpansive mappings,” Bulletin of the American Mathematical Society, vol. 73, pp. 591–597, 1967.
18. T. C. Lim, “On characterizations of Meir-Keeler contractive maps,” Nonlinear Analysis, vol. 46, no. 1, pp. 113–120, 2001.
19. A. Petrusel and J. C. Yao, “Viscosity approximation to common fixed points of families of nonexpansive mappings with generalized contractions mappings,” Nonlinear Analysis, vol. 69, no. 4, pp. 1100–1111, 2008.
20. T. Suzuki, “Moudafi's viscosity approximations with Meir-Keeler contractions,” Journal of Mathematical Analysis and Applications, vol. 325, no. 1, pp. 342–352, 2007.
21. R. E. Bruck, “Properties of fixed-point sets of nonexpansive mappings in Banach spaces,” Transactions of the American Mathematical Society, vol. 179, pp. 251–262, 1973.
22. H. K. Xu, “An iterative approach to quadratic optimization,” Journal of Optimization Theory and Applications, vol. 116, no. 3, pp. 659–678, 2003. | 4,331 | 16,931 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2016-36 | latest | en | 0.834483 |
https://mathematica.stackexchange.com/questions/101995/using-select-on-a-2d-table-with-a-criteria-that-accepts-3-parameters | 1,709,077,488,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474688.78/warc/CC-MAIN-20240227220707-20240228010707-00320.warc.gz | 389,177,555 | 40,000 | # Using Select on a 2d table with a criteria that accepts 3 parameters
I have a table
sets := Table[{i, j, k}, {i, 1, 10}, {j, 1, 10}, {k, 1, 10}]
and I want to use a function that accepts three parameters and returns a boolean, as a criteria in Select. The function looks like this:
isPytagorean[{a_, b_, c_}] := a^2 == b^2 + c^2
I tried a couple of things but none of them worked. This one returns an empty list:
Select[sets, isPytagorean]
How should I use Select to get a new table of tables for which the criteria isPytagorean is met?
• You'll want to Flatten[] sets first before trying Select[], for starters. But for your actual problem, FindInstance[] might be more useful. Dec 13, 2015 at 17:11
• Cases[sets, _?(isPytagorean), -1] Dec 13, 2015 at 18:26
sol = Position[Map[isPytagorean, sets, {3}], True]
{{5, 3, 4}, {5, 4, 3}, {10, 6, 8}, {10, 8, 6}}
rev = Reverse /@ Union[Sort /@ sol]
{{5, 4, 3}, {10, 8, 6}}
isPytagorean /@ rev
{True, True}
Update
The following is much faster:
Cases[PowersRepresentations[#, 2, 2] & /@ Range@100, {{0, _}, _}]
{{{0, 5}, {3, 4}}, {{0, 10}, {6, 8}}}
{a, b, c} /. Solve[
a^2 == b^2 + c^2 && 1 <= a <= 10 && 1 <= b <= 10 && 1 <= c <= 10 && {a, b, c} ∈ Integers,
{a, b, c}
]
tup = Tuples[Range[10], 3];
can = Pick[tup, Sort[#].DiagonalMatrix[{1, 1, -1}].Sort[#] & /@ tup, 0]
Union[Sort /@ can]
yields: {{3, 4, 5}, {6, 8, 10}} | 535 | 1,395 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2024-10 | latest | en | 0.716532 |
https://911weknow.com/m2m-day-88-how-to-fairly-scramble-a-rubiks-cube | 1,718,335,716,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861520.44/warc/CC-MAIN-20240614012527-20240614042527-00372.warc.gz | 67,959,206 | 10,852 | # M2M Day 88: How to fairly scramble a Rubik’s Cube
This post is part of Month to Master, a 12-month accelerated learning project. For January, my goal is to solve a Rubik?s Cube in under 20 seconds.
In the video where I solve the Rubik?s Cube in 17 seconds, before I start solving, I scramble the cube according to a series of 20 moves.
After sharing the video, many people asked where this scramble came from, if it?s random, if I?ve practice it before, etc. So, I thought I?d address that today?
I use an app called ChaoTimer to automatically generate new, random scrambles on every solve.
Each scramble is expressed using Basic Notation, which is depicted below.
Under the Basic Notation scheme, R means ?turn the right face of the cube clockwise?. R? (R prime), means ?turn the right face of the cube counterclockwise?. R2 means ?turn the right face of the cube 180 degrees?. And so on, with F = front; B = back; L = Left; R = Right; U = Up; D = Down.
You can see in the screen shot above that I?ve only solved the Rubik?s Cube 853 times this month while using the app. There are 43,252,003,274,489,856,000 possible Rubik?s Cube scrambles, all of which can be generated in 20 moves.
Thus, I?m nearly certain that I?ve never had the same scramble twice.
Anyway, if you?re interested in speed cubing, I?d highly recommend downloading the ChaoTimer app. Not only does it generate scrambles, but it also keeps time and crunches a bunch of statistics about your solves. | 376 | 1,478 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2024-26 | latest | en | 0.928556 |
https://math.stackexchange.com/questions/3294631/how-to-show-that-a-finite-dimensional-linear-system-of-odes-cannot-be-chaotic | 1,627,232,673,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151699.95/warc/CC-MAIN-20210725143345-20210725173345-00688.warc.gz | 410,387,259 | 38,793 | # How to show that a finite dimensional linear system of ODEs cannot be chaotic?
Consider the general linear system $$\dot{x}=Ax$$ where $$A\in\mathbb{R}^n\times\mathbb{R}^n$$, and $$x\in\mathbb{R}^n$$. Many sources assert that such a system cannot be chotic for any $$n$$, for example
https://en.wikipedia.org/wiki/Chaos_theory
How to know whether an Ordinary Differential Equation is Chaotic?
I'm looking for a simple reasoning, either by a proof, or by geometrical considerations, that will show that such systems can never be chaotic.
It is easy to show that such systems are subject to the superposition principle, i.e., a linear combination of two solutions is again a solution: If $$\dot{x} = Ax$$ and $$\dot{y} = Ay$$, and $$z=αx+βy$$ is our linear combination, then $$\dot{z} = \frac{\mathrm{d}}{\mathrm{d}t} (αx+βy) = α\dot{x} + β\dot{y} = αAx + βAy = A(αx+βy) = Az .$$ Suppose the dynamics is sensitive to initial conditions, i.e., any small perturbation will blow up (until it reaches the size of the attractor). Now, we can decompose the perturbed solution into the unperturbed solution and the perturbation, which must be again a solution due to the superposition principle. In particular we can also scale down any solution arbitrarily to use it as a perturbation. As this perturbation must blow up to extent, so must the solution, which is thus unbounded.
More formally, sensitivity to initial conditions means that for some $$δ>0$$ (at most the diameter of the attractor) and any $$ε>0$$, any trajectories starting with a distance of $$ε$$ will become $$δ$$ apart after a sufficiently large time: $$∃δ>0 : ∀ε>0: ∀x,y,|x(0)-y(0)|≤ε : ∃t: |x(t)-y(t)|≥δ,$$ where $$x$$ and $$y$$ are solutions of your system. Now choose $$y=x+\frac{ε}{|x(0)|} x$$, which is a solution due to the superposition principle. Inserting this into the above yields: $$∃δ>0 : ∀ε>0: ∃t: ε \frac{|x(t)|}{|x(0)|}≥δ.$$ Thus $$x$$ is unbounded.
• sensitive to initial conditions but unbounded. An example for this is the solution to $$\dot{x} = x$$.
• bounded but not sensitive to initial conditions, namely converging to a fixed point, limit cycle (periodic dynamics), or limit torus (quasi-periodic dynamics). An example for this is the solution to $$\dot{x} = -x$$. | 645 | 2,257 | {"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": 20, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2021-31 | latest | en | 0.882423 |
https://www.infogalactic.com/info/Boundary_layer | 1,632,252,959,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057227.73/warc/CC-MAIN-20210921191451-20210921221451-00304.warc.gz | 838,288,991 | 25,176 | # Boundary layer
Boundary layer visualization, showing transition from laminar to turbulent condition
In physics and fluid mechanics, a boundary layer is the layer of fluid in the immediate vicinity of a bounding surface where the effects of viscosity are significant. In the Earth's atmosphere, the atmospheric boundary layer is the air layer near the ground affected by diurnal heat, moisture or momentum transfer to or from the surface. On an aircraft wing the boundary layer is the part of the flow close to the wing, where viscous forces distort the surrounding non-viscous flow. See Reynolds number.
Laminar boundary layers can be loosely classified according to their structure and the circumstances under which they are created. The thin shear layer which develops on an oscillating body is an example of a Stokes boundary layer, while the Blasius boundary layer refers to the well-known similarity solution near an attached flat plate held in an oncoming unidirectional flow. When a fluid rotates and viscous forces are balanced by the Coriolis effect (rather than convective inertia), an Ekman layer forms. In the theory of heat transfer, a thermal boundary layer occurs. A surface can have multiple types of boundary layer simultaneously.
The viscous nature of airflow reduces the local velocities on a surface and is responsible for skin friction. The layer of air over the wing's surface that is slowed down or stopped by viscosity, is the boundary layer. There are two different types of boundary layer flow: laminar and turbulent.[1]
Laminar Boundary Layer Flow
The laminar boundary is a very smooth flow, while the turbulent boundary layer contains swirls or "eddies." The laminar flow creates less skin friction drag than the turbulent flow, but is less stable. Boundary layer flow over a wing surface begins as a smooth laminar flow. As the flow continues back from the leading edge, the laminar boundary layer increases in thickness.
Turbulent Boundary Layer Flow
At some distance back from the leading edge, the smooth laminar flow breaks down and transitions to a turbulent flow. From a drag standpoint, it is advisable to have the transition from laminar to turbulent flow as far aft on the wing as possible, or have a large amount of the wing surface within the laminar portion of the boundary layer. The low energy laminar flow, however, tends to break down more suddenly than the turbulent layer.
## Aerodynamics
Ludwig Prandtl
Laminar boundary layer velocity profile
The aerodynamic boundary layer was first defined by Ludwig Prandtl in a paper presented on August 12, 1904 at the third International Congress of Mathematicians in Heidelberg, Germany. It simplifies the equations of fluid flow by dividing the flow field into two areas: one inside the boundary layer, dominated by viscosity and creating the majority of drag experienced by the boundary body; and one outside the boundary layer, where viscosity can be neglected without significant effects on the solution. This allows a closed-form solution for the flow in both areas, a significant simplification of the full Navier–Stokes equations. The majority of the heat transfer to and from a body also takes place within the boundary layer, again allowing the equations to be simplified in the flow field outside the boundary layer. The pressure distribution throughout the boundary layer in the direction normal to the surface (such as an airfoil) remains constant throughout the boundary layer, and is the same as on the surface itself.
The thickness of the velocity boundary layer is normally defined as the distance from the solid body at which the viscous flow velocity is 99% of the freestream velocity (the surface velocity of an inviscid flow). Displacement Thickness is an alternative definition stating that the boundary layer represents a deficit in mass flow compared to inviscid flow with slip at the wall. It is the distance by which the wall would have to be displaced in the inviscid case to give the same total mass flow as the viscous case. The no-slip condition requires the flow velocity at the surface of a solid object be zero and the fluid temperature be equal to the temperature of the surface. The flow velocity will then increase rapidly within the boundary layer, governed by the boundary layer equations, below.
The thermal boundary layer thickness is similarly the distance from the body at which the temperature is 99% of the temperature found from an inviscid solution. The ratio of the two thicknesses is governed by the Prandtl number. If the Prandtl number is 1, the two boundary layers are the same thickness. If the Prandtl number is greater than 1, the thermal boundary layer is thinner than the velocity boundary layer. If the Prandtl number is less than 1, which is the case for air at standard conditions, the thermal boundary layer is thicker than the velocity boundary layer.
In high-performance designs, such as gliders and commercial aircraft, much attention is paid to controlling the behavior of the boundary layer to minimize drag. Two effects have to be considered. First, the boundary layer adds to the effective thickness of the body, through the displacement thickness, hence increasing the pressure drag. Secondly, the shear forces at the surface of the wing create skin friction drag.
At high Reynolds numbers, typical of full-sized aircraft, it is desirable to have a laminar boundary layer. This results in a lower skin friction due to the characteristic velocity profile of laminar flow. However, the boundary layer inevitably thickens and becomes less stable as the flow develops along the body, and eventually becomes turbulent, the process known as boundary layer transition. One way of dealing with this problem is to suck the boundary layer away through a porous surface (see Boundary layer suction). This can reduce drag, but is usually impractical due to its mechanical complexity and the power required to move the air and dispose of it. Natural laminar flow techniques push the boundary layer transition aft by reshaping the aerofoil or fuselage so that its thickest point is more aft and less thick. This reduces the velocities in the leading part and the same Reynolds number is achieved with a greater length.
At lower Reynolds numbers, such as those seen with model aircraft, it is relatively easy to maintain laminar flow. This gives low skin friction, which is desirable. However, the same velocity profile which gives the laminar boundary layer its low skin friction also causes it to be badly affected by adverse pressure gradients. As the pressure begins to recover over the rear part of the wing chord, a laminar boundary layer will tend to separate from the surface. Such flow separation causes a large increase in the pressure drag, since it greatly increases the effective size of the wing section. In these cases, it can be advantageous to deliberately trip the boundary layer into turbulence at a point prior to the location of laminar separation, using a turbulator. The fuller velocity profile of the turbulent boundary layer allows it to sustain the adverse pressure gradient without separating. Thus, although the skin friction is increased, overall drag is decreased. This is the principle behind the dimpling on golf balls, as well as vortex generators on aircraft. Special wing sections have also been designed which tailor the pressure recovery so laminar separation is reduced or even eliminated. This represents an optimum compromise between the pressure drag from flow separation and skin friction from induced turbulence.
When using half-models in wind tunnels, a peniche is sometimes used to reduce or eliminate the effect of the boundary layer.
## Boundary layer equations
The deduction of the boundary layer equations was one of the most important advances in fluid dynamics (Anderson, 2005). Using an order of magnitude analysis, the well-known governing Navier–Stokes equations of viscous fluid flow can be greatly simplified within the boundary layer. Notably, the characteristic of the partial differential equations (PDE) becomes parabolic, rather than the elliptical form of the full Navier–Stokes equations. This greatly simplifies the solution of the equations. By making the boundary layer approximation, the flow is divided into an inviscid portion (which is easy to solve by a number of methods) and the boundary layer, which is governed by an easier to solve PDE. The continuity and Navier–Stokes equations for a two-dimensional steady incompressible flow in Cartesian coordinates are given by
${\partial u\over\partial x}+{\partial \upsilon\over\partial y}=0$
$u{\partial u \over \partial x}+\upsilon{\partial u \over \partial y}=-{1\over \rho} {\partial p \over \partial x}+{\nu}\left({\partial^2 u\over \partial x^2}+{\partial^2 u\over \partial y^2}\right)$
$u{\partial \upsilon \over \partial x}+\upsilon{\partial \upsilon \over \partial y}=-{1\over \rho} {\partial p \over \partial y}+{\nu}\left({\partial^2 \upsilon\over \partial x^2}+{\partial^2 \upsilon\over \partial y^2}\right)$
where $u$ and $\upsilon$ are the velocity components, $\rho$ is the density, $p$ is the pressure, and $\nu$ is the kinematic viscosity of the fluid at a point.
The approximation states that, for a sufficiently high Reynolds number the flow over a surface can be divided into an outer region of inviscid flow unaffected by viscosity (the majority of the flow), and a region close to the surface where viscosity is important (the boundary layer). Let $u$ and $\upsilon$ be streamwise and transverse (wall normal) velocities respectively inside the boundary layer. Using scale analysis, it can be shown that the above equations of motion reduce within the boundary layer to become
$u{\partial u \over \partial x}+\upsilon{\partial u \over \partial y}=-{1\over \rho} {\partial p \over \partial x}+{\nu}{\partial^2 u\over \partial y^2}$
${1\over \rho} {\partial p \over \partial y}=0$
and if the fluid is incompressible (as liquids are under standard conditions):
${\partial u\over\partial x}+{\partial \upsilon\over\partial y}=0$
The order of magnitude analysis assumes the streamwise length scale significantly larger than the transverse length scale inside the boundary layer. It follows that variations in properties in the streamwise direction are generally much lower than those in the wall normal direction. Apply this to the continuity equation shows that $\upsilon$, the wall normal velocity, is small compared with $u$ the streamwise velocity.
Since the static pressure $p$ is independent of $y$, then pressure at the edge of the boundary layer is the pressure throughout the boundary layer at a given streamwise position. The external pressure may be obtained through an application of Bernoulli's equation. Let $u_0$ be the fluid velocity outside the boundary layer, where $u$ and $u_0$ are both parallel. This gives upon substituting for $p$ the following result
$u{\partial u \over \partial x}+\upsilon{\partial u \over \partial y}=u_0{\partial u_0 \over \partial x}+{\nu}{\partial^2 u\over \partial y^2}$
For a flow in which the static pressure $p$ also does not change in the direction of the flow then
${\partial p\over\partial x}=0$
so $u_0$ remains constant.
Therefore, the equation of motion simplifies to become
$u{\partial u \over \partial x}+\upsilon{\partial u \over \partial y}={\nu}{\partial^2 u\over \partial y^2}$
These approximations are used in a variety of practical flow problems of scientific and engineering interest. The above analysis is for any instantaneous laminar or turbulent boundary layer, but is used mainly in laminar flow studies since the mean flow is also the instantaneous flow because there are no velocity fluctuations present. This simplified equations is parabolic PDE and can be solved using a similarity solution often referred to as the Blasius boundary layer
## Turbulent boundary layers
The treatment of turbulent boundary layers is far more difficult due to the time-dependent variation of the flow properties. One of the most widely used techniques in which turbulent flows are tackled is to apply Reynolds decomposition. Here the instantaneous flow properties are decomposed into a mean and fluctuating component. Applying this technique to the boundary layer equations gives the full turbulent boundary layer equations not often given in literature:
${\partial \overline{u}\over\partial x}+{\partial \overline{v}\over\partial y}=0$
$\overline{u}{\partial \overline{u} \over \partial x}+\overline{v}{\partial \overline{u} \over \partial y}=-{1\over \rho} {\partial \overline{p} \over \partial x}+ \nu \left({\partial^2 \overline{u}\over \partial x^2}+{\partial^2 \overline{u}\over \partial y^2}\right)-\frac{\partial}{\partial y}(\overline{u'v'})-\frac{\partial}{\partial x}(\overline{u'^2})$
$\overline{u}{\partial \overline{v} \over \partial x}+\overline{v}{\partial \overline{v} \over \partial y}=-{1\over \rho} {\partial \overline{p} \over \partial y}+\nu \left({\partial^2 \overline{v}\over \partial x^2}+{\partial^2 \overline{v}\over \partial y^2}\right)-\frac{\partial}{\partial x}(\overline{u'v'})-\frac{\partial}{\partial y}(\overline{v'^2})$
Using a similar order-of-magnitude analysis, the above equations can be reduced to leading order terms. By choosing length scales $\delta$ for changes in the transverse-direction, and $L$ for changes in the streamwise-direction, with $\delta<, the x-momentum equation simplifies to:
$\overline{u}{\partial \overline{u} \over \partial x}+\overline{v}{\partial \overline{u} \over \partial y}=-{1\over \rho} {\partial \overline{p} \over \partial x}-\frac{\partial}{\partial y}(\overline{u'v'})$.
This equation does not satisfy the no-slip condition at the wall. Like Prandtl did for his boundary layer equations, a new, smaller length scale must be used to allow the viscous term to become leading order in the momentum equation. By choosing $\eta<<\delta$ as the y-scale, the leading order momentum equation for this "inner boundary layer" is given by:
$0=-{1\over \rho} {\partial \overline{p} \over \partial x}+{\nu}{\partial^2 \overline{u}\over \partial y^2}-\frac{\partial}{\partial y}(\overline{u'v'})$.
In the limit of infinite Reynolds number, the pressure gradient term can be show to have no effect on the inner region of the turbulent boundary layer. The new "inner length scale" $\eta$ is a viscous length scale, and is of order $\frac{\nu}{u_*}$, with $u_*$ being the velocity scale of the turbulent fluctuations, in this case a friction velocity.
Unlike the laminar boundary layer equations, the presence of two regimes governed by different sets of flow scales (i.e. the inner and outer scaling) has made finding a universal similarity solution for the turbulent boundary layer difficult and controversial. To find a similarity solution that spans both regions of the flow, it is necessary to asymptotically match the solutions from both regions of the flow. Such analysis will yield either the so-called log-law or power-law.
The additional term $\overline{u'v'}$ in the turbulent boundary layer equations is known as the Reynolds shear stress and is unknown a priori. The solution of the turbulent boundary layer equations therefore necessitates the use of a turbulence model, which aims to express the Reynolds shear stress in terms of known flow variables or derivatives. The lack of accuracy and generality of such models is a major obstacle in the successful prediction of turbulent flow properties in modern fluid dynamics.
A constant stress layer exists in the near wall region. Due to the damping of the vertical velocity fluctuations near the wall, the Reynolds stress term will become negligible and we find that a linear velocity profile exists. This is only true for the very near wall region.
## Heat and mass transfer
In 1928, the French engineer André Lévêque observed that convective heat transfer in a flowing fluid is affected only by the velocity values very close to the surface.[2][3] For flows of large Prandtl number, the temperature/mass transition from surface to freestream temperature takes place across a very thin region close to the surface. Therefore, the most important fluid velocities are those inside this very thin region in which the change in velocity can be considered linear with normal distance from the surface. In this way, for
$u(y) = u_0 \left[ 1 - \frac{(y - h)^2}{h^2} \right] = u_0 \frac{y}{h} \left[ 2 - \frac{y}{h} \right] \;,$
when $y \rightarrow 0$, then
$u(y) \approx 2 u_0 \frac{y}{h} = \theta y$,
where θ is the tangent of the Poiseuille parabola intersecting the wall. Although Lévêque's solution was specific to heat transfer into a Poiseuille flow, his insight helped lead other scientists to an exact solution of the thermal boundary-layer problem.[4] Schuh observed that in a boundary-layer, u is again a linear function of y, but that in this case, the wall tangent is a function of x.[5] He expressed this with a modified version of Lévêque's profile,
$u(y) = \theta(x) y$.
This results in a very good approximation, even for low $Pr$ numbers, so that only liquid metals with $Pr$ much less than 1 cannot be treated this way.[4] In 1962, Kestin and Persen published a paper describing solutions for heat transfer when the thermal boundary layer is contained entirely within the momentum layer and for various wall temperature distributions.[6] For the problem of a flat plate with a temperature jump at $x = x_0$, they propose a substitution that reduces the parabolic thermal boundary-layer equation to an ordinary differential equation. The solution to this equation, the temperature at any point in the fluid, can be expressed as an incomplete gamma function.[3] Schlichting proposed an equivalent substitution that reduces the thermal boundary-layer equation to an ordinary differential equation whose solution is the same incomplete gamma function.[7]
## Convective transfer constants from boundary layer analysis
Paul Richard Heinrich Blasius derived an exact solution to the above laminar boundary layer equations.[8] The thickness of the boundary layer $\delta$ is a function of the Reynolds number for laminar flow.
$\delta \approx {5.0 \times x \over \sqrt {Re}}$
$\delta$ = the thickness of the boundary layer: the region of flow where the velocity is less than 99% of the far field velocity $v_\infty$; $x$ is position along the semi-infinite plate, and $Re$ is the Reynolds Number given by $\rho v_\infty x / \mu$ ($\rho =$ density and $\mu =$ dynamic viscosity).
The Blasius solution uses boundary conditions in a dimensionless form:
${v_x - v_S \over v_\infty - v_S} = {v_x \over v_\infty} = {v_y\over v_\infty}= 0$ at $y=0$
${v_x - v_S \over v_\infty - v_S} = {v_x \over v_\infty} = 1$ at $y=\infty$ and $x=0$
Velocity Boundary Layer (Top,orange) and Temperature Boundary Layer (Bottom, green) share a functional form due to similarity in the Momentum/Energy Balances and boundary conditions.
Note that in many cases, the no-slip boundary condition holds that $v_S$, the fluid velocity at the surface of the plate equals the velocity of the plate at all locations. If the plate is not moving, then $v_S = 0$. A much more complicated derivation is required if fluid slip is allowed.[9]
In fact, the Blasius solution for laminar velocity profile in the boundary layer above a semi-infinite plate can be easily extended to describe Thermal and Concentration boundary layers for heat and mass transfer respectively. Rather than the differential x-momentum balance (equation of motion), this uses a similarly derived Energy and Mass balance:
Energy: $v_x {\partial T \over \partial x} + v_y {\partial T \over \partial y} = {k \over \rho Cp}{\partial^2 T \over \partial y^2}$
Mass: $v_x {\partial c_A \over \partial x} + v_y {\partial c_A \over \partial y} = D_{AB}{\partial^2 c_A \over \partial y^2}$
For the momentum balance, kinematic viscosity $\nu$ can be considered to be the momentum diffusivity. In the energy balance this is replaced by thermal diffusivity $\alpha = {k / \rho C_P}$, and by mass diffusivity $D_{AB}$ in the mass balance. In thermal diffusivity of a substance, $k$ is its thermal conductivity, $\rho$ is its density and $C_P$ is its heat capacity. Subscript AB denotes diffusivity of species A diffusing into species B.
Under the assumption that $\alpha = D_{AB} = \nu$, these equations become equivalent to the momentum balance. Thus, for Prandtl number $Pr = \nu/\alpha = 1$ and Schmidt number $Sc = \nu/D_{AB} = 1$ the Blasius solution applies directly.
Accordingly, this derivation uses a related form of the boundary conditions, replacing $v$ with $T$ or $c_A$ (absolute temperature or concentration of species A). The subscript S denotes a surface condition.
${v_x - v_S \over v_\infty - v_S} = {T - T_S \over T_\infty - T_S} = {c_A - c_{AS} \over c_{A\infty} - c_{AS}}= 0$ at $y=0$
${v_x - v_S \over v_\infty - v_S} = {T - T_S \over T_\infty - T_S} = {c_A - c_{AS} \over c_{A\infty} - c_{AS}} = 1$ at $y=\infty$ and $x=0$
Using the streamline function Blasius obtained the following solution for the shear stress at the surface of the plate.
$\tau_0 = \left( {\partial v_x \over \partial y} \right) _{y=0}=0.332 {v_\infty \over x} Re^{1/2}$
And via the boundary conditions, it is known that
${v_x - v_S \over v_\infty - v_S} = {T - T_S \over T_\infty - T_S} = {c_A - c_{AS} \over c_{A\infty} - c_{AS}}$
We are given the following relations for heat/mass flux out of the surface of the plate
$\left( {\partial T \over \partial y} \right) _{y=0}=0.332 {T_\infty - T_S \over x} Re^{1/2}$
$\left( {\partial c_A \over \partial y} \right) _{y=0}=0.332 {c_{A\infty} - c_{AS} \over x} Re^{1/2}$
So for $Pr=Sc=1$
$\delta =\delta _T= \delta _c= {5.0*x\over\sqrt{Re}}$
Where $\delta_T,\delta_c$ are the regions of flow where $T$ and $c_A$ are less than 99% of their far field values.[10]
Because the Prandtl number of a particular fluid is not often unity, German engineer E. Polhausen who worked with Ludwig Prandtl attempted to empirically extend these equations to apply for $Pr\ne 1$. His results can be applied to $Sc$ as well.[11] He found that for Prandtl number greater than 0.6, the thermal boundary layer thickness was approximately given by:
Plot showing the relative thickness in the Thermal boundary layer versus the Velocity boundary layer (in red) for various Prandtl Numbers. For $Pr=1$, the two are equal.
${\delta \over \delta_T} = Pr^{1/3}$ and therefore ${\delta \over \delta_c} = Sc^{1/3}$
From this solution, it is possible to characterize the convective heat/mass transfer constants based on the region of boundary layer flow. Fourier’s law of conduction and Newton’s Law of Cooling are combined with the flux term derived above and the boundary layer thickness.
${q\over A} = -k \left({\partial T \over \partial y} \right)_{y=0} = h_x(T_S-T_\infty)$
$h_x = 0.332{k \over x} Re^{1/2}_x Pr^{1/3}$
This gives the local convective constant $h_x$ at one point on the semi-infinite plane. Integrating over the length of the plate gives an average
$h_L = 0.664{k \over x} Re^{1/2}_L Pr^{1/3}$
Following the derivation with mass transfer terms ($k$ = convective mass transfer constant, $D_{AB}$ = diffusivity of species A into species B, $Sc = \nu / D_{AB}$ ), the following solutions are obtained:
$k'_x = 0.332{D_{AB} \over x} Re^{1/2}_x Sc^{1/3}$
$k'_L = 0.664{D_{AB} \over x} Re^{1/2}_L Sc^{1/3}$
These solutions apply for laminar flow with a Prandtl/Schmidt number greater than 0.6.[10]
## Naval architecture
Many of the principles that apply to aircraft also apply to ships, submarines, and offshore platforms.
For ships, unlike aircraft, one deals with incompressible flows, where change in water density is negligible (a pressure rise close to 1000kPa leads to a change of only 2–3 kg/m3). This field of fluid dynamics is called hydrodynamics. A ship engineer designs for hydrodynamics first, and for strength only later. The boundary layer development, breakdown, and separation become critical because the high viscosity of water produces high shear stresses. Another consequence of high viscosity is the slip stream effect, in which the ship moves like a spear tearing through a sponge at high velocity.
## Boundary layer turbine
This effect was exploited in the Tesla turbine, patented by Nikola Tesla in 1913. It is referred to as a bladeless turbine because it uses the boundary layer effect and not a fluid impinging upon the blades as in a conventional turbine. Boundary layer turbines are also known as cohesion-type turbine, bladeless turbine, and Prandtl layer turbine (after Ludwig Prandtl).
## Predicting Transient Boundary Layer Thickness in a Cylinder Using Dimensional Analysis
By using the transient and viscous force equations for a cylindrical flow you can predict the transient boundary layer thickness by finding the Womersley Number ($N_w$).
Transient Force = $\rho v w$
Viscous Force = ${\mu v\over\delta_1}$
Setting them equal to each other gives:
$\rho v w={\mu v\over\delta_1}$
Solving for delta gives:
$\delta_1=\sqrt{\mu\over\rho w}=\sqrt{\ v\over\ w}$
In dimensionless form:
${L\over\delta_1}={L\sqrt{w\over\ v}}=N_w$
Where $N_w$ = Womersley Number; $\rho$ = density; $v$ = velocity; $\delta_1$ = length of transient boundary layer; $\mu$ = viscosity; $L$ = characteristic length.
## Predicting Convective Flow Conditions at the Boundary Layer in a Cylinder Using Dimensional Analysis
By using the convective and viscous force equations at the boundary layer for a cylindrical flow you can predict the convective flow conditions at the boundary layer by finding the dimensionless Reynolds Number ($N_R$).
Convective Force = $\rho v^2\over\ L$
Viscous Force = ${\mu v\over\delta_2^2}$
Setting them equal to each other gives:
${\rho v^2\over\ L}={\mu v\over\delta_2^2}$
Solving for delta gives:
$\delta_2=\sqrt{\mu L\over\rho v}$
In dimensionless form:
${L\over\delta_2}={\sqrt{\rho v L\over\mu}}=\sqrt{N_R}$
Where $N_R$ = Reynolds Number; $\rho$ = density; $v$ = velocity; $\delta_2$ = length of convective boundary layer; $\mu$ = viscosity; $L$ = characteristic length.
## References
1. http://www.faa.gov/regulations_policies/handbooks_manuals/aviation/pilot_handbook/. Missing or empty |title= (help)<templatestyles src="Module:Citation/CS1/styles.css"></templatestyles>
2. Lévêque, A. (1928). "Les lois de la transmission de chaleur par convection". Annales des Mines ou Recueil de Mémoires sur l'Exploitation des Mines et sur les Sciences et les Arts qui s'y Rattachent, Mémoires (in French). XIII (13): 201–239.CS1 maint: unrecognized language (link)<templatestyles src="Module:Citation/CS1/styles.css"></templatestyles>
3. Niall McMahon. "André Lévêque p285, a review of his velocity profile approximation".<templatestyles src="Module:Citation/CS1/styles.css"></templatestyles>
4. Martin, H. (2002). "The generalized Lévêque equation and its practical use for the prediction of heat and mass transfer rates from pressure drop". Chemical Engineering Science. 57 (16). pp. 3217–3223. doi:10.1016/S0009-2509(02)00194-X.<templatestyles src="Module:Citation/CS1/styles.css"></templatestyles>
5. Schuh, H. (1953). "On asymptotic solutions for the heat transfer at varying wall temperatures in a laminar boundary layer with Hartree's velocity profiles". Jour. Aero. Sci. 20 (2). pp. 146–147.<templatestyles src="Module:Citation/CS1/styles.css"></templatestyles>
6. Kestin, J. and Persen, L.N. (1962). "The transfer of heat across a turbulent boundary layer at very high prandtl numbers". Int. J. Heat Mass Transfer. 5: 355–371. doi:10.1016/0017-9310(62)90026-1.CS1 maint: multiple names: authors list (link)<templatestyles src="Module:Citation/CS1/styles.css"></templatestyles>
7. Schlichting, H. (1979). Boundary-Layer Theory (7 ed.). New York (USA): McGraw-Hill.<templatestyles src="Module:Citation/CS1/styles.css"></templatestyles>
8. Blasius, H. (1908). "Grenzschichten in Flüssigkeiten mit kleiner Reibung". Z. Math. Phys. 56: 1–37. (English translation)
9. Martin, Michael J. Blasius boundary layer solution with slip flow conditions. AIP conference proceedings 585.1 2001: 518-523. American Institute of Physics. 24 Apr 2013.
10. Geankoplis, Christie J. Transport Processes and Separation Process Principles: (includes Unit Operations). Fourth ed. Upper Saddle River, NJ: Prentice Hall Professional Technical Reference, 2003. Print.
11. Pohlhausen, E. (1921), Der Wärmeaustausch zwischen festen Körpern und Flüssigkeiten mit kleiner reibung und kleiner Wärmeleitung. Z. angew. Math. Mech., 1: 115–121. doi: 10.1002/zamm.19210010205 | 7,094 | 28,772 | {"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": 128, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.892338 |
https://dave.thehorners.com/aboutme/80-random-thoughts/563-pi-314unicode-character-u03c0 | 1,653,349,995,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662562106.58/warc/CC-MAIN-20220523224456-20220524014456-00790.warc.gz | 243,435,513 | 11,312 | Dave Horner's Website - Yet another perspective on things...
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Rough Hits : 4908956
how did u find my site?
which seems more true?
"If you want to build a ship, don't drum up people to collect wood and don't assign them tasks and work, but rather teach them to long for the endless immensity of the sea."
- Antoine de Saint Exupéry
# pi 3.14...Unicode Character “?” (U+03C0).
Saturday, 10 October 2020 16:01
# celebrate and think about π π pie
pi has a infinite structure. a simple shape which is irrational and transcendental. found in the strangest and commonest of places. tagged with our name / symbol. ?????C0 8; C0; C0 π (𝚷)
From the Greek letter ?. a product over a set of terms. 𝝥
Unicode Character “?” (U+03C0). Name: Greek Small Letter Pi.
Unicode Character “????” (U+1D70B) Name: Mathematical Italic Small Pi.
Proof that ?() is irrational
Lindemann–Weierstrass theorem - (transcendental proof stuff) a symmetric polynomial whose arguments are all conjugates of one another gives a rational number.
missed half the point?; circles carve out a space; it finds a voice in random noise.
the mathematical constant — the ratio of a circle's circumference to its diameter.
* hough * hough * hough *
you can learn a lot from a transcendental infinite constant.
? | greek small letter pi (U+03C0) @ Graphemica
In 1873, British mathematician William Shanks stunned the world of mathematics when he claimed to have calculated pi to 707 digits. In 1945, D.F. Ferguson, using pen and paper, identified an error |?????C0 8; in Shanks' work in the 527th place. Ferguson then carried the computation to 808 places.
In 1948, Wrench and Smith verified Ferguson's work—and continued the calculation to 1,000 places using a gear-driven calculator. In 1961, Wrench and Daniel Shanks (unrelated to William Shanks) used an IBM 7090 to calculate pi to 100,265 digits. The printout was formally presented to the Smithsonian Institution—and their feat earned a place in the Guinness Book of Records. Pi Computation Wizard John Wrench Dies at Age 97 | Mathematical Association of America - Few today remember what mathematician John William Wrench Jr. did 60 years ago, just before the advent of the electronic computer. He and fellow mathematician Levi Smith calculated the value of pi to more than 1,000 decimal places. (a lil more flare and he might have submitted 2 pow 10, 2^10, 1024!)
pi is found everywhere; an elusive infinite number. pi is a number which works|sings sweetly with however much precision you require; time, space, circles, and random noise.
zoom out and rotate your head however you like to see the spiral. we all follow implicitly; its in there.
it has a shape. a diameter and circumference. it has a name. it teases it's universality. concrete infinite. circle of pie.
just select a circle Hough Transform (CHT) local maxima in an accumulator matrix.
we may not all see the same color <span style="color:blue">; yours and mine being equal in complex polar coordinates or otherwise. circles, have a smell | poetry. we all see the pie piece and imagine the pie hole.
find pi in an infinite converging series with leibniz; https://youtu.be/uH4trBNn540
find pi with mandelbrot set and imaginary numbers / complex planes. sqr(-1); and real;
Mandelbrot Set
find pi in random points inside and outside a circle. - https://www.youtube.com/watch?v=5cNnf_7e92Q
# celebrate pi day
Emma Haruka Iwao Sets Guinness World Record For Most Accurate Value Of Pi : NPR - the most accurate value of pi on 2019/March. calculated 31,415,926,535,897 digits of pi
find pi in colliding-balls and bouncing boxes elastic idealized frictionless worlds with mass, velocity, and collision detection.
pi sometimes feels too good to be true. just square the circle. or was it, circle the square? pls provide the ghost solution to the penrose tiling?
Roger Penrose wins 2020 Nobel Prize in Physics for discovery about black holes | University of Cambridge - the most important contribution to the general theory of relativity since Einstein.
cheers for Professor Sir Roger Penrose, his son!!! (MC Escher) and the tribar
my son Gabe 10/10/2020 asked me today; what's the biggest number you know of. best I could do was a googleplex for a name. when asked what it was; all I could say was a lot of 9's plus 1.
A googolplex is the number 10googol, or equivalently, 10(10100). Written out in ordinary decimal notation, it is 1 followed by 10100 zeroes, that is, a 1 followed by a googol zeroes.
This still is the largest number I can name off the top of my head. correct and accurate big numbers are hard. if I looked; I bet there's at least + 1 bigger yet. i still have to ask why? what's important about the next 1024 precise digits of pie? who cares?
2019/March we celebrated the most accurate value of pi calculated 31,415,926,535,897 digits of pi.
when do we get a googleplex of pi digits? does it/is matter? who cares if they calculated pi &pi π digits of π. 3.14 is enough most times. c. 250 BCE Archimedes bound pie up (?223?71 < ? < ?22?7) || (223⁄71 < π < ?22⁄7) how many places/digits of π does your app need? what's the resolution and precision needed for your confidence level at what goal? when does our π constant error in the 527th place begin to matter||show in our solution?
finding πpieπ with precision takes time and resources; writing known digits of pi is easy to write; calculating digits of pi and verification is NP-Hard. When does the 50,000,000,000,000th place of pi matter?;
according to wikipedia: book can be printed with 106 zeros (around 400 pages with 50 lines per page and 50 zeros per line) therefore, it requires 10^(94) such books to print all the zeros of a googolplex (that is, printing a googol zeros). If each book had a mass of 100 grams; the mass of all such books required to write out a googolplex would be vastly greater than the masses of the Milky Way and the Andromeda galaxies combined (by a factor of roughly 2.0 x 1050), and greater than the mass of the observable universe by a factor of roughly 7 x 1039.. that's a lot of zeros. but who carries around books these days?! why print and write such a number to paper? what's that for dude?
I guess my son will have to forgive me for the 9's and 1 joke. Did I mention he just turned 9? It's the best explanation I had at the time.
```want is 1+1 11 or 2 but it kan be 4 bekus my name is poop on xbox you can add me on xbox
```
my son; would like you to know thats his joke, he made it.
Chronology of computation of ?
radon transform tangent; complex penrose transform local maxima in an accumulator matrix; modified.
pi in complex penrose. what would they see.
to circle back to my boy in the car yesterday: I didn't realize at the time; it is 1 + (10100-1) 9's. a one zero goo gol. a 10googol;
I wonder how long it will take to get the goo gle of pi. Jan-2020 Gabe and I still working on phonics and we've only got 13 9's pi + 1 * 50 today; that took us almost a year to compute we(a|e)k computers (303 days!). wonder what places they'll see 314 days in 2020. someone's working on it; from many perspectives.
nine year olds usually don't get order of operations and parentheses! it takes time. we all have our tricks. tell/ask me 1+1; I just need to know what's 1 * 1 first?
29 January 2020 using y-cruncher v0.7.7 303 days 50,000,000,000,000 decimal places.
The last 100 decimal digits of the latest world record computation are:
```1151172718 2444229740 0412605840 3026105553 7774728936 : 49,999,999,999,950
8888086663 6658909667 9659924528 1042319124 0640849268 : 50,000,000,000,000
```
how pretty but pointless patterns in polar plots of primes prompt pretty important ponderings on properties of those primes. distribution of primes numbers together with rational approx. for pi. polar coordinates of pretty pointless patterns. +2pi doesn't change the location. (r,0)=(p,p) prime p;
these pointless Archimedean spirals of primes with properties! π π
https://youtu.be/EK32jo7i5LQ
some people use computers to calculate millions and billions of digits of pi. It seemed so useless to me. so what; kinda 3.14 and 2 and about 3 qtr. fuzzy e fizz buzz.
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Last Updated on Monday, 15 March 2021 18:01 | 2,157 | 8,236 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-21 | latest | en | 0.89025 |
https://www.ato.gov.au/law/view/document?LocID=%22REG%2F20150038%2F21%22 | 1,713,240,679,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817043.36/warc/CC-MAIN-20240416031446-20240416061446-00219.warc.gz | 605,350,484 | 32,590 | EXCISE REGULATION 2015
PART 4 - TOBACCO, FUEL, AND DELIVERY OF CERTAIN GOODS WITHOUT ENTRY
Division 2 - Matters relating to tobacco
SECTION 21 AMOUNT OF DUTY TO DETERMINE THE AMOUNT OF A PENALTY
21(1)
This section sets out how to work out the amount of duty that is to be used to determine the amount of a penalty relating to a quantity of tobacco leaf.
Note:
This section is made for paragraph (b) of the penalty in each of subsections 28(1) , 30(1) , 31(1) , 33(1) , 35(1) , 39K(2) and (3) , 39M(2) , 44(4) and (8) , 117C(1) , 117D(1) , 117F(1) and 117H(1) of the Act.
21(2)
If the tobacco leaf was seized in a bale, the amount of duty is worked out using the formula:
(Weight − 2) × Rate
where:
rate
means the rate of excise duty that applies, on the penalty day, to a kilogram of tobacco leaf that is manufactured into excisable goods and entered for home consumption.
weight − 2
means the gross weight of the bale and the tobacco leaf, in kilograms, minus 2 kilograms.
21(3)
If:
(a) the tobacco leaf was not seized; and
(b) there is sufficient evidence to show that the quantity of tobacco leaf was in a bale;
the amount of duty is worked out using the formula:
Weight × Rate
where:
rate
means the rate of excise duty that applies, on the penalty day, to a kilogram of tobacco leaf that is manufactured into excisable goods and entered for home consumption.
weight
means 100 kilograms.
21(4)
If the tobacco leaf was seized in unbaled form, the amount of duty is worked out using the formula:
Weight × Rate
where:
rate
means the rate of excise duty that applies, on the penalty day, to a kilogram of tobacco leaf that is manufactured into excisable goods and entered for home consumption.
weight
means the weight of the tobacco leaf, in kilograms.
21(5)
If:
(a) the tobacco leaf was not seized; and
(b) there is sufficient evidence to show that the quantity of tobacco leaf was in unbaled form;
the amount of duty is worked out using the formula:
Weight × Rate
where:
rate
means the rate of excise duty that applies, on the penalty day, to a kilogram of tobacco leaf that is manufactured into excisable goods and entered for home consumption.
weight
means the weight of the tobacco leaf, in kilograms, as shown by the evidence.
This information is provided by CCH Australia Limited Link opens in new window. View the disclaimer and notice of copyright. | 624 | 2,391 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-18 | latest | en | 0.933954 |
http://isabelle.in.tum.de/repos/isabelle/file/9eede540a5e8/src/HOL/Typedef.thy | 1,563,658,144,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526670.1/warc/CC-MAIN-20190720194009-20190720220009-00446.warc.gz | 78,153,447 | 4,801 | src/HOL/Typedef.thy
author berghofe Wed May 07 10:56:50 2008 +0200 (2008-05-07) changeset 26802 9eede540a5e8 parent 26151 4a9b8f15ce7f child 27295 cfe5244301dd permissions -rw-r--r--
Deleted instantiation "set :: (type) itself".
``` 1 (* Title: HOL/Typedef.thy
```
``` 2 ID: \$Id\$
```
``` 3 Author: Markus Wenzel, TU Munich
```
``` 4 *)
```
``` 5
```
``` 6 header {* HOL type definitions *}
```
``` 7
```
``` 8 theory Typedef
```
``` 9 imports Set
```
``` 10 uses
```
``` 11 ("Tools/typedef_package.ML")
```
``` 12 ("Tools/typecopy_package.ML")
```
``` 13 ("Tools/typedef_codegen.ML")
```
``` 14 begin
```
``` 15
```
``` 16 ML {*
```
``` 17 structure HOL = struct val thy = theory "HOL" end;
```
``` 18 *} -- "belongs to theory HOL"
```
``` 19
```
``` 20 locale type_definition =
```
``` 21 fixes Rep and Abs and A
```
``` 22 assumes Rep: "Rep x \<in> A"
```
``` 23 and Rep_inverse: "Abs (Rep x) = x"
```
``` 24 and Abs_inverse: "y \<in> A ==> Rep (Abs y) = y"
```
``` 25 -- {* This will be axiomatized for each typedef! *}
```
``` 26 begin
```
``` 27
```
``` 28 lemma Rep_inject:
```
``` 29 "(Rep x = Rep y) = (x = y)"
```
``` 30 proof
```
``` 31 assume "Rep x = Rep y"
```
``` 32 then have "Abs (Rep x) = Abs (Rep y)" by (simp only:)
```
``` 33 moreover have "Abs (Rep x) = x" by (rule Rep_inverse)
```
``` 34 moreover have "Abs (Rep y) = y" by (rule Rep_inverse)
```
``` 35 ultimately show "x = y" by simp
```
``` 36 next
```
``` 37 assume "x = y"
```
``` 38 thus "Rep x = Rep y" by (simp only:)
```
``` 39 qed
```
``` 40
```
``` 41 lemma Abs_inject:
```
``` 42 assumes x: "x \<in> A" and y: "y \<in> A"
```
``` 43 shows "(Abs x = Abs y) = (x = y)"
```
``` 44 proof
```
``` 45 assume "Abs x = Abs y"
```
``` 46 then have "Rep (Abs x) = Rep (Abs y)" by (simp only:)
```
``` 47 moreover from x have "Rep (Abs x) = x" by (rule Abs_inverse)
```
``` 48 moreover from y have "Rep (Abs y) = y" by (rule Abs_inverse)
```
``` 49 ultimately show "x = y" by simp
```
``` 50 next
```
``` 51 assume "x = y"
```
``` 52 thus "Abs x = Abs y" by (simp only:)
```
``` 53 qed
```
``` 54
```
``` 55 lemma Rep_cases [cases set]:
```
``` 56 assumes y: "y \<in> A"
```
``` 57 and hyp: "!!x. y = Rep x ==> P"
```
``` 58 shows P
```
``` 59 proof (rule hyp)
```
``` 60 from y have "Rep (Abs y) = y" by (rule Abs_inverse)
```
``` 61 thus "y = Rep (Abs y)" ..
```
``` 62 qed
```
``` 63
```
``` 64 lemma Abs_cases [cases type]:
```
``` 65 assumes r: "!!y. x = Abs y ==> y \<in> A ==> P"
```
``` 66 shows P
```
``` 67 proof (rule r)
```
``` 68 have "Abs (Rep x) = x" by (rule Rep_inverse)
```
``` 69 thus "x = Abs (Rep x)" ..
```
``` 70 show "Rep x \<in> A" by (rule Rep)
```
``` 71 qed
```
``` 72
```
``` 73 lemma Rep_induct [induct set]:
```
``` 74 assumes y: "y \<in> A"
```
``` 75 and hyp: "!!x. P (Rep x)"
```
``` 76 shows "P y"
```
``` 77 proof -
```
``` 78 have "P (Rep (Abs y))" by (rule hyp)
```
``` 79 moreover from y have "Rep (Abs y) = y" by (rule Abs_inverse)
```
``` 80 ultimately show "P y" by simp
```
``` 81 qed
```
``` 82
```
``` 83 lemma Abs_induct [induct type]:
```
``` 84 assumes r: "!!y. y \<in> A ==> P (Abs y)"
```
``` 85 shows "P x"
```
``` 86 proof -
```
``` 87 have "Rep x \<in> A" by (rule Rep)
```
``` 88 then have "P (Abs (Rep x))" by (rule r)
```
``` 89 moreover have "Abs (Rep x) = x" by (rule Rep_inverse)
```
``` 90 ultimately show "P x" by simp
```
``` 91 qed
```
``` 92
```
``` 93 lemma Rep_range:
```
``` 94 shows "range Rep = A"
```
``` 95 proof
```
``` 96 show "range Rep <= A" using Rep by (auto simp add: image_def)
```
``` 97 show "A <= range Rep"
```
``` 98 proof
```
``` 99 fix x assume "x : A"
```
``` 100 hence "x = Rep (Abs x)" by (rule Abs_inverse [symmetric])
```
``` 101 thus "x : range Rep" by (rule range_eqI)
```
``` 102 qed
```
``` 103 qed
```
``` 104
```
``` 105 end
```
``` 106
```
``` 107 use "Tools/typedef_package.ML"
```
``` 108 use "Tools/typecopy_package.ML"
```
``` 109 use "Tools/typedef_codegen.ML"
```
``` 110
```
``` 111 setup {*
```
``` 112 TypedefPackage.setup
```
``` 113 #> TypecopyPackage.setup
```
``` 114 #> TypedefCodegen.setup
```
``` 115 *}
```
``` 116
```
``` 117 text {* This class is just a workaround for classes without parameters;
```
``` 118 it shall disappear as soon as possible. *}
```
``` 119
```
``` 120 class itself = type +
```
``` 121 fixes itself :: "'a itself"
```
``` 122
```
``` 123 setup {*
```
``` 124 let fun add_itself tyco thy =
```
``` 125 let
```
``` 126 val vs = Name.names Name.context "'a"
```
``` 127 (replicate (Sign.arity_number thy tyco) @{sort type});
```
``` 128 val ty = Type (tyco, map TFree vs);
```
``` 129 val lhs = Const (@{const_name itself}, Term.itselfT ty);
```
``` 130 val rhs = Logic.mk_type ty;
```
``` 131 val eq = HOLogic.mk_Trueprop (HOLogic.mk_eq (lhs, rhs));
```
``` 132 in
```
``` 133 thy
```
``` 134 |> TheoryTarget.instantiation ([tyco], vs, @{sort itself})
```
``` 135 |> `(fn lthy => Syntax.check_term lthy eq)
```
``` 136 |-> (fn eq => Specification.definition (NONE, (("", []), eq)))
```
``` 137 |> snd
```
``` 138 |> Class.prove_instantiation_instance (K (Class.intro_classes_tac []))
```
``` 139 |> LocalTheory.exit
```
``` 140 |> ProofContext.theory_of
```
``` 141 end
```
``` 142 in TypedefPackage.interpretation add_itself end
```
``` 143 *}
```
``` 144
```
``` 145 instantiation bool :: itself
```
``` 146 begin
```
``` 147
```
``` 148 definition "itself = TYPE(bool)"
```
``` 149
```
``` 150 instance ..
```
``` 151
```
``` 152 end
```
``` 153
```
``` 154 instantiation "fun" :: ("type", "type") itself
```
``` 155 begin
```
``` 156
```
``` 157 definition "itself = TYPE('a \<Rightarrow> 'b)"
```
``` 158
```
``` 159 instance ..
```
``` 160
```
``` 161 end
```
``` 162
```
``` 163 hide (open) const itself
```
``` 164
```
``` 165 end
``` | 2,362 | 6,420 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2019-30 | longest | en | 0.835577 |
https://www.brightstorm.com/math/calculus/applications-of-the-derivative/optimization-using-the-second-derivative-test/ | 1,679,788,541,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945376.29/warc/CC-MAIN-20230325222822-20230326012822-00185.warc.gz | 764,363,102 | 27,521 | ###### Norm Prokup
Cornell University
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
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# Optimization Using the Second Derivative Test - Concept
Norm Prokup
###### Norm Prokup
Cornell University
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
Share
Some optimization problems can be solved by use of the second derivative test. If the second derivative is always positive, the function will have a relative minimum somewhere. If it is always negative, the function will have a relative maximum somewhere. Other ways of solving optimization problems include using the closed interval method or the first derivative test.
We have one more optimization method to talk about and it's the second derivative test for absolute max and min. Let me show you how it works, suppose you have a function of y equals f of x and that function you know has a second derivative which is always positive everywhere on it's domain. Well if it has a critical point and that critical point will be an absolute minimum it's pretty much guaranteed. So that tells us how second derivative test basically works, you start with the continuous function on some interval i that's its domain and suppose f prime of c is 0 that is one critical point.
If the second derivative is always positive on domain then f will have an absolute minimum so think second derivative is positive it'll be shaped like this, there will be a minimum at x=c and if the second derivative is always negative on the interval it'll have an absolute maximum at x=c that's the second derivative test. Remember the second derivative has to be always positive or always negative. | 389 | 1,915 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-14 | latest | en | 0.947707 |
https://grandpaperwriters.com/answered-write-a-matlab-algorithm-and-scripts-to-solve-the-following-questionsuggested-variable-names-in-brackets-you-are-required-to-utilize-a-switch-statement-in-at-least-one-part-of-y/ | 1,701,638,148,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100508.53/warc/CC-MAIN-20231203193127-20231203223127-00261.warc.gz | 328,741,596 | 14,262 | Answered! Write a MATLAB algorithm and scripts to solve the following question(suggested variable names in brackets). You are required to utilize a switch statement in at least one part of your code.:…
Write a MATLAB algorithm and scripts to solve the following question(suggested variable names in brackets). You are required to utilize a switch statement in at least one part of your code.:
My daughter in grade 4 is actively learning math and needs an “app” to help he check her homework. She is learning addition, subtraction, multiplication and two types of division. She wants this app to
Don't use plagiarized sources. Get Your Custom Essay on
Answered! Write a MATLAB algorithm and scripts to solve the following question(suggested variable names in brackets). You are required to utilize a switch statement in at least one part of your code.:…
GET AN ESSAY WRITTEN FOR YOU FROM AS LOW AS \$13/PAGE
(a) Ask her the first number (a)
(b) Ask her the second number (b)
(c) present her with options of +, -, *, ÷ and ÷ with remainder and ask her to choose one by entering 1, 2, 3, 4, or 5 (op).
(d) Perform the request operation a (op) b and output the entire equation (see Assignment 3 solutions as good example). Keep in mind that in Grade 4 they don’t have decimals or negative numbers yet so:
(a) subtraction is only allowed if the result is not negative, otherwise you should print out an informative message (understandable by the intended user)
(b) division is only allowed if its not “bad” but not only does this mean the second number can not be zero, but that it must also perfectly divide. Each case should have its own descriptive error message
(c) divide with remainder does allow non-perfect division, but must indicate how many times b divided a and what the remainder is (example: 5 ÷ 2 is 2 with a reminder of 1).
prompt = ‘Please enter first number : ‘;
x = input(prompt)
prompt1 = ‘Please enter second input : ‘;
y = input(prompt1)
disp(‘2. Subtraction’);
disp(‘3. Multiplication’);
disp(‘4. Division’);
disp(‘5. Modulus’);
if ch == 1
fprintf(‘Addition of x and y is : %d n’, x+y)
elseif ch == 2
if((x > 0 && y > 0) && (x > y))
fprintf(‘Subtraction of x and y is : %d n’, x-y)
elseif(x < y)
disp(‘1st number is smaller than second number’);
else
disp(‘Negative numbers are not allowed for subtraction’);
end
elseif ch == 3
fprintf(‘Multiplication of x and y is : %d n’, x*y)
elseif ch == 4
if(x > 0 && y > 0)
fprintf(‘Division of x and y is : %d n’, fix(x/y))
else
disp(‘Cannot Perform division with zero’);
end
elseif ch == 5
fprintf(‘Modulus of x and y is : %d n’, mod(x,y))
else
disp(‘Invalid Choice’);
end | 704 | 2,646 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2023-50 | latest | en | 0.868005 |
http://slideplayer.com/slide/3614050/ | 1,527,180,337,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794866511.32/warc/CC-MAIN-20180524151157-20180524171157-00451.warc.gz | 278,290,345 | 21,553 | # Sampling Distribution of a Sample Proportion Lecture 25 Sections 8.1 – 8.2 Fri, Feb 29, 2008.
## Presentation on theme: "Sampling Distribution of a Sample Proportion Lecture 25 Sections 8.1 – 8.2 Fri, Feb 29, 2008."— Presentation transcript:
Sampling Distribution of a Sample Proportion Lecture 25 Sections 8.1 – 8.2 Fri, Feb 29, 2008
Sampling Distributions Sampling Distribution of a Statistic
The Sample Proportion The letter p represents the population proportion. The symbol p ^ (“p-hat”) represents the sample proportion. p ^ is a random variable. The sampling distribution of p ^ is the probability distribution of all the possible values of p ^.
Example Suppose that 2/3 of all males wash their hands after using a public restroom. Suppose that we take a sample of 1 male. Find the sampling distribution of p ^.
Example W N 2/3 1/3 P(W) = 2/3 P(N) = 1/3
Example Let x be the sample number of males who wash. The probability distribution of x is xP(x)P(x) 01/3 12/3
Example Let p ^ be the sample proportion of males who wash. (p ^ = x/n.) The sampling distribution of p ^ is p^p^ P(p^)P(p^) 01/3 12/3
Example Now we take a sample of 2 males, sampling with replacement. Find the sampling distribution of p ^.
Example W N W N W N 2/3 1/3 2/3 1/3 2/3 1/3 P(WW) = 4/9 P(WN) = 2/9 P(NW) = 2/9 P(NN) = 1/9
Example Let x be the sample number of males who wash. The probability distribution of x is xP(x)P(x) 01/9 14/9 2
Example Let p ^ be the sample proportion of males who wash. (p ^ = x/n.) The sampling distribution of p ^ is p^p^ P(p^)P(p^) 01/9 1/24/9 1
Samples of Size n = 3 If we sample 3 males, then the sample proportion of males who wash has the following distribution. p^p^ P(p^)P(p^) 01/27 =.03 1/36/27 =.22 2/312/27 =.44 18/27 =.30
Samples of Size n = 4 If we sample 4 males, then the sample proportion of males who wash has the following distribution. p^p^ P(p^)P(p^) 01/81 =.01 1/48/81 =.10 2/424/81 =.30 3/432/81 =.40 116/81 =.20
Samples of Size n = 5 If we sample 5 males, then the sample proportion of males who wash has the following distribution. p^p^ P(p^)P(p^) 01/243 =.004 1/510/243 =.041 2/540/243 =.165 3/580/243 =.329 4/580/243 =.329 132/243 =.132
Our Experiment In our experiment, we had 80 samples of size 5. Based on the sampling distribution when n = 5, we would expect the following Value of p ^ 0.00.20.40.60.81.0 Actual Predicted0.33.313.226.3 10.5
The pdf when n = 1 01
The pdf when n = 2 011/2
The pdf when n = 3 011/32/3
The pdf when n = 4 011/42/43/4
The pdf when n = 5 011/52/53/5 4/5
1 8/10 The pdf when n = 10 02/104/106/10
Observations and Conclusions Observation: The values of p ^ are clustered around p. Conclusion: p ^ is close to p most of the time.
Observations and Conclusions Observation: As the sample size increases, the clustering becomes tighter. Conclusion: Larger samples give better estimates. Conclusion: We can make the estimates of p as good as we want, provided we make the sample size large enough.
Observations and Conclusions Observation: The distribution of p ^ appears to be approximately normal. Conclusion: We can use the normal distribution to calculate just how close to p we can expect p ^ to be.
One More Observation However, we must know the values of and for the distribution of p ^. That is, we have to quantify the sampling distribution of p ^.
The Central Limit Theorem for Proportions It turns out that the sampling distribution of p ^ is approximately normal with the following parameters.
The Central Limit Theorem for Proportions The approximation to the normal distribution is excellent if
Example If we gather a sample of 100 males, how likely is it that between 60 and 70 of them, inclusive, wash their hands after using a public restroom? This is the same as asking the likelihood that 0.60 p ^ 0.70.
Example Use p = 0.66. Check that np = 100(0.66) = 66 > 5, n(1 – p) = 100(0.34) = 34 > 5. Then p ^ has a normal distribution with
Example So P(0.60 p ^ 0.70) = normalcdf(.60,.70,.66,.04737) = 0.6981.
Why Surveys Work Suppose that we are trying to estimate the proportion of the male population who wash their hands after using a public restroom. Suppose the true proportion is 66%. If we survey a random sample of 1000 people, how likely is it that our error will be no greater than 5%?
Why Surveys Work Now we have
Why Surveys Work Now find the probability that p^ is between 0.61 and 0.71: normalcdf(.61,.71,.66,.01498) = 0.9992. It is virtually certain that our estimate will be within 5% of 66%.
Why Surveys Work What if we had decided to save money and surveyed only 100 people? If it is important to be within 5% of the correct value, is it worth it to survey 1000 people instead of only 100 people?
Quality Control A company will accept a shipment of components if there is no strong evidence that more than 5% of them are defective. H 0 : 5% of the parts are defective. H 1 : More than 5% of the parts are defective.
Quality Control They will take a random sample of 100 parts and test them. If no more than 10 of them are defective, they will accept the shipment. What is ? What is ?
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http://mathhelpforum.com/calculus/23256-tough-definite-integral.html | 1,481,054,847,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541995.74/warc/CC-MAIN-20161202170901-00109-ip-10-31-129-80.ec2.internal.warc.gz | 173,005,504 | 10,064 | 1. ## Tough definite integral
$dx/(2x+7)$
limits are from 2 to 0.
The only one on my homework I cannot get...any help would be appreciated.
2. Originally Posted by calculuslyfrustrated
$dx/(2x+7)$
limits are from 2 to 0.
What's the matter?
Note that
$\int_2^0 {\frac{1}
{{2x + 7}}\,dx} = \frac{1}
{2}\int_2^0 {\frac{{(2x + 7)'}}
{{2x + 7}}\,dx} = \frac{1}
{2}\left( {\ln \left| {2x + 7} \right|\Big|_2^0} \right).$
You may go on from there. | 181 | 448 | {"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": 3, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2016-50 | longest | en | 0.834169 |
https://community.wolfram.com/groups/-/m/t/497083 | 1,685,625,042,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224647810.28/warc/CC-MAIN-20230601110845-20230601140845-00622.warc.gz | 208,857,190 | 22,422 | # Can I use Manipulate[expr ] to plot AND calculate simultaneously?
Posted 8 years ago
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Hi,I have multiple functions of $x$ that depend on specific parameters, say $f(x,a,b,c)$, $g(x,a,b,d)$, $h(x,b,c,d)$. I can plot these functions simultaneously and use Manipulate[] to set the range for the parameters $a,b,c,d,..$ etc., and look at how the functions behave simultaneously under a change in parameters (using the slide-bar). That's very neat. Then, I was interested in seeing the difference between, say, $g$ and $h$, so I plotted Abs[g[x,a,b,d]-h[x,b,c,d]]. Let's call that function $y(x,a,b,c,d)$.All these functions are plotted on a certain domain ( $x$ from $0$ to $\pi$) and I would like to calculate $\int_{0}^{\pi}y(x,a,b,c,d) dx$. Given a set of parameters that I fix (say $a=1,b=1,c=1,d=1$), I should get a scalar output. But I would like to see this output at the same time I vary each slidebar in the Manipulate output. In other words, I want to be able to play with the parameters and see the plot change accordingly, and at the same time, see how the result of my definite integral varies. Is there a way to do this with Manipulate? If so, how?I have tried to look in the documentation but there doesn't seem to be any way of manipulating multiple "tasks" at the same time. Thanks in advance! | 361 | 1,340 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2023-23 | longest | en | 0.870397 |
https://www.convertunits.com/from/bushel+%5BUS,+dry%5D/to/barrel | 1,628,223,680,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046152112.54/warc/CC-MAIN-20210806020121-20210806050121-00259.warc.gz | 720,369,613 | 16,906 | ## ››Convert bushel [US, dry] to barrel [US, liquid]
bushel [US, dry] barrel
Did you mean to convert bushel [US, dry] bushel [UK] to barrel [US, liquid] barrel [US, beer] barrel [US, dry] barrel [US, petroleum] barrel [UK] barrel [UK, wine]
## ››More information from the unit converter
How many bushel [US, dry] in 1 barrel? The answer is 3.3837575834577.
We assume you are converting between bushel [US, dry] and barrel [US, liquid].
You can view more details on each measurement unit:
bushel [US, dry] or barrel
The SI derived unit for volume is the cubic meter.
1 cubic meter is equal to 28.377593070673 bushel [US, dry], or 8.3864143251288 barrel.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between bushels and barrels.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of bushel [US, dry] to barrel
1 bushel [US, dry] to barrel = 0.29553 barrel
5 bushel [US, dry] to barrel = 1.47765 barrel
10 bushel [US, dry] to barrel = 2.95529 barrel
20 bushel [US, dry] to barrel = 5.91059 barrel
30 bushel [US, dry] to barrel = 8.86588 barrel
40 bushel [US, dry] to barrel = 11.82118 barrel
50 bushel [US, dry] to barrel = 14.77647 barrel
75 bushel [US, dry] to barrel = 22.16471 barrel
100 bushel [US, dry] to barrel = 29.55294 barrel
## ››Want other units?
You can do the reverse unit conversion from barrel to bushel [US, dry], or enter any two units below:
## Enter two units to convert
From: To:
## ››Definition: Barrel
The definition of a US non-beer liquid barrel is 31½ US gallons (119.2 litres), or half a hogshead. This is different from a barrel of US beer, which is defined by tax law.
## ››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! | 628 | 2,198 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2021-31 | latest | en | 0.868935 |
https://number.academy/2008939 | 1,716,580,619,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058736.10/warc/CC-MAIN-20240524183358-20240524213358-00389.warc.gz | 366,025,065 | 11,131 | # Number 2008939 facts
The odd number 2,008,939 is spelled 🔊, and written in words: two million, eight thousand, nine hundred and thirty-nine, approximately 2.0 million. The ordinal number 2008939th is said 🔊 and written as: two million, eight thousand, nine hundred and thirty-ninth. The meaning of the number 2008939 in Maths: Is it Prime? Factorization and prime factors tree. The square root and cube root of 2008939. What is 2008939 in computer science, numerology, codes and images, writing and naming in other languages
## What is 2,008,939 in other units
The decimal (Arabic) number 2008939 converted to a Roman number is (M)(M)(V)MMMCMXXXIX. Roman and decimal number conversions.
#### Time conversion
(hours, minutes, seconds, days, weeks)
2008939 seconds equals to 3 weeks, 2 days, 6 hours, 2 minutes, 19 seconds
2008939 minutes equals to 4 years, 1 month, 3 weeks, 2 days, 2 hours, 19 minutes
### Codes and images of the number 2008939
Number 2008939 morse code: ..--- ----- ----- ---.. ----. ...-- ----.
Sign language for number 2008939:
Number 2008939 in braille:
QR code Bar code, type 39
Images of the number Image (1) of the number Image (2) of the number More images, other sizes, codes and colors ...
## Share in social networks
#### Is Prime?
The number 2008939 is a prime number.
#### Factorization and factors (dividers)
The prime factors of 2008939
Prime numbers have no prime factors smaller than themselves.
The factors of 2008939 are 1, 2008939.
Total factors 2.
Sum of factors 2008940 (1).
#### Prime factor tree
2008939 is a prime number.
#### Powers
The second power of 20089392 is 4.035.835.905.721.
The third power of 20089393 is 8.107.748.148.603.240.448.
#### Roots
The square root √2008939 is 1417,370453.
The cube root of 32008939 is 126,179533.
#### Logarithms
The natural logarithm of No. ln 2008939 = loge 2008939 = 14,513117.
The logarithm to base 10 of No. log10 2008939 = 6,302967.
The Napierian logarithm of No. log1/e 2008939 = -14,513117.
### Trigonometric functions
The cosine of 2008939 is -0,8988.
The sine of 2008939 is -0,438359.
The tangent of 2008939 is 0,487716.
## Number 2008939 in Computer Science
Code typeCode value
2008939 Number of bytes1.9MB
Unix timeUnix time 2008939 is equal to Saturday Jan. 24, 1970, 6:02:19 a.m. GMT
IPv4, IPv6Number 2008939 internet address in dotted format v4 0.30.167.107, v6 ::1e:a76b
2008939 Decimal = 111101010011101101011 Binary
2008939 Decimal = 10210001202011 Ternary
2008939 Decimal = 7523553 Octal
2008939 Decimal = 1EA76B Hexadecimal (0x1ea76b hex)
2008939 BASE64MjAwODkzOQ==
2008939 MD57b26e5ccb0cc0433ecf27945a26b44d2
2008939 SHA1d5cc3e2f67b4b5c930866880536dc2c99316366f
2008939 SHA2245fdfb99de4bcbcc49214ee29bdfe091e9c7d02a4079ef70a8ff44bd3
2008939 SHA2564d3e71662e6570106dbc7fccef85b3da922e588f2894c0acf8f39c4258d565ca
2008939 SHA384fc9e86de2cd9f0596fe3070ee2fbaf70380f5f1ae31439acc3b571eb7c628bc4197af5f5a58970405eb7793b11905b24
More SHA codes related to the number 2008939 ...
If you know something interesting about the 2008939 number that you did not find on this page, do not hesitate to write us here.
## Numerology 2008939
### Character frequency in the number 2008939
Character (importance) frequency for numerology.
Character: Frequency: 2 1 0 2 8 1 9 2 3 1
### Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 2008939, the numbers 2+0+0+8+9+3+9 = 3+1 = 4 are added and the meaning of the number 4 is sought.
## № 2,008,939 in other languages
How to say or write the number two million, eight thousand, nine hundred and thirty-nine in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 2.008.939) dos millones ocho mil novecientos treinta y nueve German: 🔊 (Nummer 2.008.939) zwei Millionen achttausendneunhundertneununddreißig French: 🔊 (nombre 2 008 939) deux millions huit mille neuf cent trente-neuf Portuguese: 🔊 (número 2 008 939) dois milhões e oito mil, novecentos e trinta e nove Hindi: 🔊 (संख्या 2 008 939) बीस लाख, आठ हज़ार, नौ सौ, उनतालीस Chinese: 🔊 (数 2 008 939) 二百万八千九百三十九 Arabian: 🔊 (عدد 2,008,939) مليونان و ثمانية آلاف و تسعمائة و تسعة و ثلاثون Czech: 🔊 (číslo 2 008 939) dva miliony osm tisíc devětset třicet devět Korean: 🔊 (번호 2,008,939) 이백만 팔천구백삼십구 Danish: 🔊 (nummer 2 008 939) to millioner ottetusinde og nihundrede og niogtredive Hebrew: (מספר 2,008,939) שני מיליון ושמונת אלפים תשע מאות שלושים ותשע Dutch: 🔊 (nummer 2 008 939) twee miljoen achtduizendnegenhonderdnegenendertig Japanese: 🔊 (数 2,008,939) 二百万八千九百三十九 Indonesian: 🔊 (jumlah 2.008.939) dua juta delapan ribu sembilan ratus tiga puluh sembilan Italian: 🔊 (numero 2 008 939) due milioni e ottomilanovecentotrentanove Norwegian: 🔊 (nummer 2 008 939) to million åtte tusen ni hundre og trettini Polish: 🔊 (liczba 2 008 939) dwa miliony osiem tysięcy dziewięćset trzydzieści dziewięć Russian: 🔊 (номер 2 008 939) два миллиона восемь тысяч девятьсот тридцать девять Turkish: 🔊 (numara 2,008,939) ikimilyonsekizbindokuzyüzotuzdokuz Thai: 🔊 (จำนวน 2 008 939) สองล้านแปดพันเก้าร้อยสามสิบเก้า Ukrainian: 🔊 (номер 2 008 939) два мільйони вісім тисяч дев'ятсот тридцять дев'ять Vietnamese: 🔊 (con số 2.008.939) hai triệu tám nghìn chín trăm ba mươi chín Other languages ...
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```
Two cells A and B of e.m.f 1.3 V and 1.5 V resp are arranged as shown in the figure. The voltmeter reads 1.45 V. The voltmeter is assumed to be ideal. Then show that r 1 = 3 r2 .
```
7 years ago
Share
``` Let I be current in circuit and E1=1.3V and E2=1.5V and E = 1.45V
by kirchoffs laws
(E2-E1)-I(R1+R2) = 0
and E = E2 - IR2
so , 0.05 = IR2 .........(1)
0.20=I(R1+R2) .........(2)
on solvine above 2 eqns, we have R1 = 3R2
--
regards
Ramesh
```
7 years ago
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The cieling of long hall is 25 m high What is the maximum horizontal distance that a ball thrown with a speed of 40 ms can go without hitting the ceiling of wall? Plz explain with a DIAGRAM
Let the angle of throwing be Q with respect to the ground So, the range is R = 40*cosQ*t where t = time taken reach ground. Now, we know the time taken to reach the top of its flight is half...
Tapas Khanda 5 months ago
Let the angle of throwing be Q with respect to the ground So, the range is R = 30*cosQ*t where t = time taken reach ground. Now, we know the time taken to reach the top of its flight is half...
Tapas Khanda 5 months ago
Diagram is not required to solve this question. You can directly use the formula to find range : R = (v^2/g)*sin2Q Maximum height reached, H = v^2*sin^2Q/2g So, 25 = 40^2*sin^2Q/(2*9.8) So,...
Tapas Khanda 5 months ago
What harmfull gases are produced when we burn plastics? And can we burn plastics?
I`m doing a project, can someone help me, How to reduce those toxic contents when plantic are burnt Like sulphur, nitrogen etc
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More Questions On Electric Current | 1,292 | 4,792 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2017-04 | longest | en | 0.753952 |
https://www.globalguideline.com/interview_questions/Answer.php?a=How_many_moles_of_HCl_are_present_in_70_L_of_a_33_M_HCl_solution | 1,547,973,142,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583705091.62/warc/CC-MAIN-20190120082608-20190120104608-00187.warc.gz | 821,404,286 | 8,223 | How many moles of HCl are present in .70 L of a .33 M HCl solution?
► First, remember definition of M (moles), M = moles of species / L.
0.33 M = 0.33 moles HCl / L
► Then, multiple your volume by the molar concentration:
0.33 moles HCl / L x 0.70 L = 0.231 moles HCl
It is helpful to carry the units with your calculations. That way you can check that numerators and denominators cancel to give you the units of your answer. | 128 | 430 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2019-04 | longest | en | 0.902781 |
https://www.physicsforums.com/threads/free-fall-acceleration-problem.430343/ | 1,508,570,860,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824618.72/warc/CC-MAIN-20171021062002-20171021082002-00159.warc.gz | 990,741,243 | 14,209 | # Free fall acceleration problem
1. Sep 19, 2010
### pikwa10
1. The problem statement, all variables and given/known data
Hello. Can someone please give me a clue as to what I'm doing wrong and why I'm doing it wrong? Thank you. The problem is:
A steel ball is dropped from a building's roof and passes a window, taking 0.13 s to fall from the top to the bottom of the window, a distance of 1.20 m. It then falls to a sidewalk and bounces back past the window, moving from bottom to top in 0.13 s. Assume that the upward flight is an exact reverse of the fall. The time the ball spends below the bottom of the window is 2.23 s. How tall is the building?
2. Relevant equations
V= Vo + at
X - Xo= Vo t + 1/2 a t^2
3. The attempt at a solution
V= -1.20 m/ 0.13 s = - 9.23 m/s (velocity of ball when it just passed the window)
-9.23 m/s= 0 m/s + (-9.8 m/s^2) t ---> t= 0.94 s to reach - 9.23 m/s
0.94 s to reach the window + 0.13 s to pass the window + 2.23/2 (1.12 s) to hit the ground= t total= 2.19 s to reach sidewalk
delta x= 0 m/s (2.19 s) + 1/2 (-0.9 m/s^2) (2.19 s) ^2
delta x= -24 m ---> 24 m is roughly the height of the building
2. Sep 19, 2010
### SiYuan
V= Vo + at
V= -1.20 m/ 0.13 s = - 9.23 m/s (velocity of ball when it just passed the window) | 435 | 1,272 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2017-43 | longest | en | 0.825472 |
https://www.2classnotes.com/cbse-question-papers/8th-science-periodic-test-ii-2017-18-dps/ | 1,652,703,295,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662510117.12/warc/CC-MAIN-20220516104933-20220516134933-00362.warc.gz | 720,884,815 | 13,683 | Monday , May 16 2022
# 8th Science Periodic Test II (2017-18)
## 8th Science Periodic Test II (2017-18)
School Name: Delhi Public School, Sector 24, Rohini, Delhi 110085 India
Time: 1 hour
M.M. 30 marks
Date: 26/07/2017
Class: VIII
#### Question: 1. (1×5 = 5)
1. A ball of dough is rolled into a flat chapatti. Name the force exerted to change the shape of the dough.
2. Name a non metal which is a good conductor of electricity.
3. A girl is pushing a box towards east direction. In which direction should her friend push the box so that it moves faster in the same direction?
4. A non-metal X in gaseous form is greenish yellow in colour and is used in water purification. Identify the non metal.
5. Define the term Malleability.
#### Question: Choose the correct option: (1×3 = 3) (I). Two forces A and B act on an object in opposite directions. A is bigger than B. The net force on the object is
1. A+B acting in the direction of A
2. A-B acting in the direction of B
3. A+B acting in the direction of B
4. A-B acting in the direction of A
#### (II). Which of the following does not change when a force acts on an object?
1. Direction of motion
2. State of rest
3. Shape
4. mass
1. Ca(SO3)2
2. CaSO3
3. CaSO4
4. Ca(SO4)2
#### Question: 5. (2 marks)
1. What is meant by atmospheric pressure?
2. We do not get crushed the huge atmospheric pressure. Why?
#### Question: 6. Give reasons: (2 marks)
1. The wheels of military tanks are covered with a broad rubber strip.
2. Electrical wires are made up of metals like aluminium and copper.
#### Question: 7. (3 marks)
1. Define pressure.
2. A man standing on a wooden plank 100 cm long and 40 cm wide exerts a force of 60 N. Calculate the pressure on the ground.
#### Question: 8. (3 marks)
1. Define corrosion
2. State the conditions necessary for corrosion of copper to take place.
3. Write the chemical equation for correction of copper.
#### Question: 9. Complete and balance the following equations: (3 marks)
1. Al + H2SO4 →
2. Ag + H2S →
#### Question: 10. (a). A small piece of sodium metal was introduced into a beaker containing cold water. A vigorous reaction with evolution of fumes was observed in the beaker. After the reaction, the solution in the beaker was tested with a red litmus paper. (5 marks)
1. Which gas was given out during the reaction?
2. What change would be observed in the red litmus paper?
3. Write the balanced chemical equation for the reaction of sodium with water.
#### (b). Rohan prepared a blue colored solution of copper sulphate in beaker A and placed an iron nail in it. Swati prepared a pale green solution of ferrous sulphate in breaker B and placed a copper wire in it.
1. What changes will they observe in the two beakers after an hour? Give reasons for you answer.
2. Write chemical equation for the reactions taking place.
3. What are such reactions called?
## 11th Class Mathematics Term 2 Question Paper 2021-22
School Name: Himalaya Public School, Sector 13, Rohini, Delhi 110085 India Class: 11th Standard (CBSE) Subject: Mathematics Time … | 827 | 3,066 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2022-21 | longest | en | 0.85592 |
https://mathoverflow.net/questions/282432/does-determinacy-in-l-mathbbr-implies-projective-determinacy-in-v | 1,675,664,343,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500304.90/warc/CC-MAIN-20230206051215-20230206081215-00142.warc.gz | 398,126,766 | 25,707 | # Does determinacy in $L(\mathbb{R})$ implies projective determinacy (in $V$)?
Does $AD^{L(\mathbb{R})}$ directly implies projective determinacy? At least it certainely implies $PD$'s consistency.
• Yes, of course it does. All projective sets are obviously in $L (\mathbb R)$ so they have winning strategies (in $L (\mathbb R)$ and therefore in $V$; note that any possible set of moves for either player is coded by a real). Oct 1, 2017 at 12:11
• let A be projective, f be I's winning strategy of A in L(R). assume there is a play according to f and II wins, assume the result is a sequence of integer s. now consider this strategy for II: if II is facing an initial segment of s, choose the next term of s, else choose 0. this strategy beats f and it is in L(R), contradicts the assumption that f is a winning strategy. Apr 21, 2022 at 16:42
Suppose $M$ is an inner model (of $\mathsf{ZF}$) with the same reals as $V$, and let $A\subseteq \mathbb R$ be a set of reals in $M$. Suppose further that $A$ is determined in $M$. Under these assumptions, $A$ is also determined in $V$. The point is that since winning strategies are coded by reals, and any possible run of the game for $A$ is coded by a real, then any run of the game in $V$ is already in $M$.
Similarly, since $M$ and $V$ share the same reals, they agree on what sets are first-order definable over the reals, that is, they agree on what sets of reals are projective. It follows that if projective determinacy holds in $M$ then it holds in $V$. In fact, much more is true in the sense that $M$ and $V$ agree on many more definitions over the reals than just first-order (for instance, $L(\mathbb R)^M=L(\mathbb R)$).
It follows immediately from these two observations (taking $M=L(\mathbb R)$) that if determinacy holds in $L(\mathbb R)$, then in $V$ all sets of reals in $L(\mathbb R)$ are determined. :-) That is, $\mathsf{AD}^{L(\mathbb R)}$ implies $L(\mathbb R)$-determinacy. In particular, projective determinacy holds in $V$.
Naturally, $V$ may have more sets of reals than $M$, some of which may not be determined. Also, in general, there may be notions of definability that do not coincide in $M$ and $V$ (for instance, by forcing we may easily change what sets of reals are in $\mathsf{HOD}$). It may be that some definition $\phi$ gives in $M$ a set $A$ that is determined in $M$. As long as we do not change the reals, that set $A$ will be determined in $V$, but running the same definition $\phi$ in $V$ may result in a different set $\hat A$ that may very well fail to be determined. | 723 | 2,563 | {"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} | 3.1875 | 3 | CC-MAIN-2023-06 | latest | en | 0.935627 |
https://www.transum.org/Software/SW/Starter_of_the_day/students/Know_Your_Place.asp?Level=3 | 1,656,697,907,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103943339.53/warc/CC-MAIN-20220701155803-20220701185803-00057.warc.gz | 1,095,534,906 | 12,780 | ## Perform some calculations without a calculator which require a knowledge of place value.
##### Place ValueLevel 1Level 2Level 3Level 4DescriptionHelpMore Place Value
This is level 3: find the missing numbers in the calculations. You will be awarded a Transum trophy if you get at least 15 of the 18 questions correct and you do this activity online.
8 x ? = 320 600 x ? = 42000 40 x ? = 240000 8 x ? = 4 0.6 x ? = 0.48 0.7 x ? = 0.035 28 ÷ ? = 0.7 6 ÷ ? = 20 42 ÷ ? = 600 1.6 ÷ ? = 0.02 0.35 ÷ ? = 0.007 0.003 x ? = 0.000003 2 × ? + 900 = 2500 600 × ? + 0.06 = 1800000.06 0.005 × ? + 4000 = 4003.5 0.6 + ? × 300 = 120000.6 800 + ? × 0.06 = 1160 0.006 + ? × 40 = 32000.006
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Monday, November 10, 2014
"In the decimal number system the value of a digit depends on its position in the number. Each position has a value ten times the position to its right. This is unlike other number systems such as Roman numerals. It is this place value that makes arithmetic easier to perform. Make sure you know the position of the digits that represent thousands, hundreds, tens, units, tenths, hundredths and thousandths at least."
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## Help
In the decimal number system the value of a digit depends on its position in the number. Each position has a value ten times the position to its right. This is unlike other number systems such as Roman numerals. It is this place value that makes arithmetic easier to perform. Make sure you know the position of the digits that represent thousands, hundreds, tens, units, tenths, hundredths and thousandths at least.
Here is an interactive Place Value Chart to help with you calculations.
## Examples
60 x 70 = 6 x 10 x 7 x 10 = 42 x 100 = 4200
20000 ÷ 500 = 20 x 1000 ÷ 5 ÷ 100 = 4 x 10 = 40
Or 20000 ÷ 500 = 200 ÷ 5 = 40
Don't wait until you have finished the exercise before you click on the 'Check' button. Click it often as you work through the questions to see if you are answering them correctly. You can double-click the 'Check' button to make it float at the bottom of your screen.
Answers to this exercise are available lower down this page when you are logged in to your Transum account. If you don’t yet have a Transum subscription one can be very quickly set up if you are a teacher, tutor or parent.
Close | 1,712 | 7,463 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.902882 |
http://betterlesson.com/lesson/resource/2513672/getting-her-supplies | 1,487,663,234,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501170696.61/warc/CC-MAIN-20170219104610-00514-ip-10-171-10-108.ec2.internal.warc.gz | 31,022,392 | 21,033 | ## Getting her supplies - Section 3: Independent Practice
Getting her supplies
# Addition Fluency: Adding with 0, 1, 2
Unit 3: Addition Strategies
Lesson 2 of 9
## Big Idea: No solving required! Children who understand why a number remains the same when added to zero will not need to memorize each combination individually. Also, fluency can increase by practicing adding 1 or 2 to a number.
Print Lesson
5 teachers like this lesson
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43 minutes
### Jennifer Moon
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Environment: Suburban | 351 | 1,481 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2017-09 | latest | en | 0.862626 |
https://oeis.org/A112710 | 1,571,616,868,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987750110.78/warc/CC-MAIN-20191020233245-20191021020745-00284.warc.gz | 618,284,814 | 3,802 | This site is supported by donations to The OEIS Foundation.
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A112710 Partial sums of Catalan numbers A000108 multiplied by powers of -3. 2
1, -2, 16, -119, 1015, -9191, 87037, -851186, 8531044, -87167702, 904619302, -9509144240, 101036961052, -1083385595648, 11708378016712, -127402051267703, 1394629704432367, -15347719608268403, 169699299111055897, -1884322608673443833, 21003350078068124587 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,2 COMMENTS Fourth column (m=3) of triangle A112707 (partial sums of Catalan numbers multiplied by powers of nonpositive numbers). LINKS FORMULA a(n)= A112707(n+3, 3), n>=0. G.f.: c(-3*x)/(1-x), where c(x):=(1-sqrt(1-4*x))/(2*x) is the o.g.f. of Catalan numbers A000108. G.f. for unsigned sequence: c(3*x)/(1+x), hence |a(n)| is the convolution of the sequences {C(n)*3^n} and {(-1)^n}, where C(n):=A000108(n). CROSSREFS Cf. A064306 (unsigned m=2 case). Sequence in context: A159324 A088755 A136782 * A046105 A086854 A258683 Adjacent sequences: A112707 A112708 A112709 * A112711 A112712 A112713 KEYWORD sign,easy AUTHOR Wolfdieter Lang, Oct 31 2005 STATUS approved
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Last modified October 20 20:13 EDT 2019. Contains 328272 sequences. (Running on oeis4.) | 497 | 1,534 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2019-43 | latest | en | 0.626529 |
https://math.stackexchange.com/questions/2609180/constraints-to-the-hamiltonian-path-can-one-tell-if-a-path-is-hamiltonian-by-lo | 1,561,382,993,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999539.60/warc/CC-MAIN-20190624130856-20190624152856-00333.warc.gz | 509,905,062 | 37,811 | # constraints to the hamiltonian path: can one tell if a path is hamiltonian by looking at it?
A few days ago, the channel Numberphile released this video on the square-sum problem. The show later released this follow-up video on how they proved that every number from 25 to 91 can have themselves and all the preceding numbers be ordered so the sum of any two adjacent numbers add up to a square. (confused? check out the video!) Anyway, I wanted to see if I could get, say $100$ somehow, but the lines linking the numbers on a graph are way too complex for me to tell where a line goes. So I'm doing it the hard way...
First, I made a table from $1-100$ dictating what numbers would be added to said number to get a square number, like $100$ or $4$. To keep this question short, I have the first 12 rows...
$$\begin{matrix} \text{number}&4&9&16&25&36&49&64&81&100&\cdots\\ -&-&-&-&-&-&-&-&-&-&-\\ 1&3&8&15&24&35&48&63&80&99&\cdots\\ 2&\varnothing&7&14&23&34&47&62&79&98&\cdots\\ 3&\color{red}{1}&6&13&22&33&46&61&78&97&\cdots\\ 4&\varnothing&5&12&21&32&45&60&77&96&\cdots\\ 5&\varnothing&\color{red}{4}&11&20&31&44&59&76&95&\cdots\\ 6&\varnothing&\color{red}{3}&10&19&30&43&58&75&94&\cdots\\ 7&\varnothing&\color{red}{2}&9&18&29&42&57&74&93&\cdots\\ 8&\varnothing&\color{red}{1}&\varnothing&17&28&41&56&73&92&\cdots\\ 9&\varnothing&\varnothing&\color{red}{7}&16&27&40&55&72&91&\cdots\\ 10&\varnothing&\varnothing&\color{red}{6}&15&26&39&54&71&90&\cdots\\ 11&\varnothing&\varnothing&\color{red}{5}&14&25&38&53&70&89&\cdots\\ 12&\varnothing&\varnothing&\color{red}{4}&13&24&37&52&69&88&\cdots\\ \vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\ddots\\ \end{matrix}$$
Anyway, i did some more math and calculated the number of times that each number from $1-100$ can be added to another number between $1-100$ and made this list. (if i used this method on this question it'd be more user-friendly...)
$$\begin{matrix} \text{number of times}&\text{numbers}\\ 1&\varnothing\\ 2&\varnothing\\ 3&\varnothing\\ 4&64&65&66&67&68&81&82&83&84&85&86&87&88&89&90&91&92&93&94&95&\\ 5&37&38&39&40&41&42&43&49&50&51&52&53&54&55&56&57&58&59&60&61&62&63&69&70&71&72&73&74&75&76&77&78&79&80\\ 6&16&17&18&19&20&26&27&28&29&30&31&32&33&34&35&36&44&45&46&47&48\\ 7&9&10&11&12&13&14&15&21&22&23&24&25\\ 8&2&4&5&6&7&8\\ 9&1&3\\ \end{matrix}$$ (this is only logging the first 100 numbers) Because of how long this is taking to write, i think someone has already done this. but here's where it leads to hamiltonian: a "cross every edge" graph has one requirement: have either two or zero odd vertex. but i don't know any hamiltonian graph requirements: is there a way to tell if a graph is or isn't hamiltonian by looking at it? does anyone know if $100$ would work? Thanks in advance.
Hamiltonian paths are hard. That is why they are interesting.
You mention "cross every edge" graphs (the formal name is Eulerian) have the requirement that there are only two or zero odd degree vertices (the degree of a vertex is the number of edges incident to a vertex, what you're calling the "number of times" a number can be added to another to sum to a square). This requirement is actually sufficient as well: every graph with only two or zero odd degree vertices will be Eulerian. Even more, one of the proofs of this fact gives an algorithm for finding the walk that crosses every edge in any such graph, and this algorithm is relatively fast.
But for Hamiltonian paths, no simple characterization (like some condition on the degrees) is known. In fact, any algorithm that you could come up with to find Hamiltonian paths would be pretty slow (technically, I mean that the problem of determining if a general graph has a Hamiltonian path is NP-complete). For this reason, mathematicians have turned to necessary conditions and sufficient conditions for Hamiltonian paths. These are what might allow you to determine if a graph has a Hamiltonian path "by looking at it":
Necessary Conditions: these are conditions that all graphs that have Hamiltonian paths satisfy. For instance, connectedness (there has to be a way to get from any vertex to any other vertex) is a necessary condition. You can think of these conditions as barriers: if a graph doesn't satisfy this condition, then it definitely doesn't have a Hamiltonian path. These will give you quick negative answers.
Sufficient Conditions: these are conditions that, when satisfied, guarantee the graph to have a Hamiltonian cycle. For instance, a famous theorem of Dirac says that if a graph has $n$ vertices and the degree of each vertex is at least $\frac{n}{2}$, then the graph is Hamiltonain. These are usually relatively strong conditions, so there will definitely be graphs that have Hamiltonian cycles that do not satisfy these conditions. However, when satisfied, these conditions will give you quick positive answers.
The holy grail would be to find a condition that is both necessary and sufficient, but as that is unlikely, it is (more or less) the subject of research to close the gap between these two types of conditions.
Side note: there are a whole bunch of structures like Hamiltonian paths that researchers are interested in necessary and sufficient conditions for. Most notably hard problems are Hamiltonian cycles, clique tilings, cycle covers, and so on. The conditions and their proofs start to get much more complicated, so there's not much more I can say without going into more gory details.
So what about this square-sum problem? It doesn't seem like anyone has a good idea. These graphs seem like they have too few edges to be able to use the various sufficient conditions we have, but there is also no obvious reason they don't violate the necessary conditions we have, so we are kind of stuck in the middle. We would have to use more information about how the graph is constructed, i.e. number theory, to solve this problem. I know the problem has been verified up to 299, and my suspicion is that for large enough $n$ (number of vertices), these graphs do have Hamiltonian cycles. (Proofs might only work for very large $n$ though.)
TL;DR: if you want to find a Hamiltonian path in the graph you have, there are some slow algorithms out there, or you could just look for it yourself and get lucky (since it seems these graphs have several Hamiltonian paths). If you're looking for conditions about the graph that would help, you're more than likely out of luck!
• I should've seen this coming, there being a tag for Hamiltonian paths. – Alexander Day Jan 17 '18 at 20:31
• You could've also seen it coming because someone hasn't already answered it before. But that's what makes it exiting! Any contribution helps. – Bob Krueger Jan 17 '18 at 20:49 | 1,919 | 6,742 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2019-26 | longest | en | 0.70462 |
https://polytope.miraheze.org/wiki/Octagonal-truncated_dodecahedral_duoprism | 1,723,584,029,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641085898.84/warc/CC-MAIN-20240813204036-20240813234036-00521.warc.gz | 348,140,505 | 12,255 | # Octagonal-truncated dodecahedral duoprism
Octagonal-truncated dodecahedral duoprism
Rank5
TypeUniform
Notation
Bowers style acronymOtid
Coxeter diagramx8o x5x3o ()
Elements
Tera20 triangular-octagonal duoprisms, 12 octagonal-decagonal duoprisms, 8 truncated dodecahedral prisms
Cells160 triangular prisms, 30+60 octagonal prisms, 96 decagonal prisms, 8 truncated dodecahedra
Faces160 triangles, 240+480 squares, 60 octagons, 96 decagons
Edges240+480+480
Vertices480
Vertex figureDigonal disphenoidal pyramid, edge lengths 1, (5+5)/2, (5+5)/2 (base triangle), 2+2 (top), 2 (side edges)
Measures (edge length 1)
Circumradius${\displaystyle {\sqrt {\frac {45+4{\sqrt {2}}+15{\sqrt {5}}}{8}}}\approx 3.24418}$
Hypervolume${\displaystyle 5{\frac {99+99{\sqrt {2}}+47{\sqrt {5}}+47{\sqrt {10}}}{6}}\approx 410.60782}$
Diteral anglesTodip–op–odedip: ${\displaystyle \arccos \left(-{\sqrt {\frac {5+2{\sqrt {5}}}{15}}}\right)\approx 142.62263^{\circ }}$
Tiddip–tid–tiddip: 135°
Odedip–op–odedip: ${\displaystyle \arccos \left(-{\frac {\sqrt {5}}{5}}\right)\approx 116.56505^{\circ }}$
Todip–trip–tiddip: 90°
Odedip–dip–tiddip: 90°
Central density1
Number of external pieces40
Level of complexity30
Related polytopes
ArmyOtid
RegimentOtid
DualOctagonal-triakis icosahedral duotegum
ConjugatesOctagrammic-truncated dodecahedral duoprism, Octagonal-quasitruncated great stellated dodecahedral duoprism, Octagrammic-quasitruncated great stellated dodecahedral duoprism
Abstract & topological properties
Euler characteristic2
OrientableYes
Properties
SymmetryH3×I2(8), order 1920
ConvexYes
NatureTame
The octagonal-truncated dodecahedral duoprism or otid is a convex uniform duoprism that consists of 8 truncated dodecahedral prisms, 12 octagonal-decagonal duoprisms and 20 triangular-octagonal duoprisms. Each vertex joins 2 truncated dodecahedral prisms, 1 triangular-octagonal duoprism, and 2 octagonal-decagonal duoprisms.
## Vertex coordinates
The vertices of an octagonal-truncated dodecahedral duoprism of edge length 1 are given by all even permutations of the last three coordinates of:
• ${\displaystyle \left(\pm {\frac {1}{2}},\,\pm {\frac {1+{\sqrt {2}}}{2}},\,0,\,\pm {\frac {1}{2}},\,\pm {\frac {5+3{\sqrt {5}}}{4}}\right),}$
• ${\displaystyle \left(\pm {\frac {1}{2}},\,\pm {\frac {1+{\sqrt {2}}}{2}},\,\pm {\frac {1}{2}},\,\pm {\frac {3+{\sqrt {5}}}{4}},\,\pm {\frac {3+{\sqrt {5}}}{2}}\right),}$
• ${\displaystyle \left(\pm {\frac {1}{2}},\,\pm {\frac {1+{\sqrt {2}}}{2}},\,\pm {\frac {3+{\sqrt {5}}}{4}},\,\pm {\frac {1+{\sqrt {5}}}{2}},\,\pm {\frac {2+{\sqrt {5}}}{2}}\right),}$
• ${\displaystyle \left(\pm {\frac {1+{\sqrt {2}}}{2}},\,\pm {\frac {1}{2}},\,0,\,\pm {\frac {1}{2}},\,\pm {\frac {5+3{\sqrt {5}}}{4}}\right),}$
• ${\displaystyle \left(\pm {\frac {1+{\sqrt {2}}}{2}},\,\pm {\frac {1}{2}},\,\pm {\frac {1}{2}},\,\pm {\frac {3+{\sqrt {5}}}{4}},\,\pm {\frac {3+{\sqrt {5}}}{2}}\right),}$
• ${\displaystyle \left(\pm {\frac {1+{\sqrt {2}}}{2}},\,\pm {\frac {1}{2}},\,\pm {\frac {3+{\sqrt {5}}}{4}},\,\pm {\frac {1+{\sqrt {5}}}{2}},\,\pm {\frac {2+{\sqrt {5}}}{2}}\right).}$
## Representations
An octagonal-truncated dodecahedral duoprism has the following Coxeter diagrams:
• x8o x5x3o (full symmetry)
• x4x x5x3o () (octagons as ditetragons) | 1,231 | 3,267 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 10, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2024-33 | latest | en | 0.662783 |
http://mathhelpforum.com/discrete-math/191639-graphs-dijkstra-s-algorithm.html | 1,508,586,123,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824733.32/warc/CC-MAIN-20171021095939-20171021115939-00237.warc.gz | 217,731,732 | 11,136 | 1. ## Graphs, Dijkstra's algorithm
Hello,
I've got some problems to solve and would appreciate any advice.
a) Give an example of graph which involves negative edge path costs and nevertheless Dijkstra's algorithm finds shortest path from any vertex (but not from all of them).
b) Give an example of graph which involves negative edge path costs and nevertheless Dijkstra's algorithm finds shortest path from all vertices.
c) Give an example of graph and its vertex such that output of Dijstra's algorithm woun't be shortest paths from given vertex (compare shortest paths from algorithm with real shortest path from given vertex).
d) Determine all $n \in N$ for which the following applies: removing any two edges of the graph $K_n$ there is no more than (n - 3)-connected graph.
Thanks
Token
2. ## Re: Graphs, Dijkstra's algorithm
a)
b)
c)
Starts from A
d) $n=3$ When we remove 2 edges from complete graph $K_3$ we will get disconnected graph which is 0-connected. (correct me if i am wrong please).
What do you think?
3. ## Re: Graphs, Dijkstra's algorithm
Well, i hope my results are correct as anybody didn't reply.
Now I got another Problem to solve:
Consider the following algorithm for finding shortest paths in undirected final graph that can involve negatively weighted edges. Let's say the smallest value of edge will be $h$. Then we add $\left | h \right |$ to all edges. Now all of our edges are positively weighted. Choose vertex (let's say $v$) and run Dijkstra's algorithm on this vertex. Prove or disprove that with this way we get the shortest paths from vertex $v$ in previous given graph (from length of shortest path to arbitary vertex found by Dijkstra's algorithm we should subtract $k.\left | h \right |$, where $k$ is length of path).
Firstly I thought that this doesn't work, but I found THIS , where somebody wrote that is should work. And also found something like Johnson's algorithm that seems to be similar. So i am now little confused. If you could help me with this at least.
thanks
token
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# Dijkstra algorithm solved examples in maths
Click on a term to search for related topics. | 507 | 2,173 | {"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": 10, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2017-43 | longest | en | 0.93018 |
http://tfresource.org/index.php?title=Destination_Choice:_Theoretical_Foundations&oldid=7943 | 1,579,797,587,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250611127.53/warc/CC-MAIN-20200123160903-20200123185903-00016.warc.gz | 156,762,544 | 12,771 | This page is open for editing because it is part of the Incubator. Have something to add? Please register so you can contribute. Have an option you would like to share? Please click on the 'Talk' button to enter the dialogue. The TF Resource Volunteers appreciate your feedback and interest.
Three primary theoretical starting points for developing destination choice models dominate current practice:
• Gravity models.
• Entropy maximization (also known as information minimization) models.
• Random utility models.
These three modeling approaches are, under appropriate assumptions, mathematically equivalent, and so are special cases of what can be generally called spatial interaction models. All these models attempt to address the same problem, as illustrated in Figure 1, in which spatial interactions (usually trips) between locations in space (typically traffic zones) are to be predicted, given limited, more macro, information concerning these interactions, such as the number of trips originating in each zone and/or the number of trips destined to each zone.
Figure 1: The Spatial Interaction Problem
## Gravity Models
Gravity models have been in use by geographers, market researchers, transportation modelers and many others for well over a hundred years. The starting point for these models, as the name implies, is Newton’s Law of Gravity:
$$F_{ij}={Gm_i}{m_j}{{d_{ij}}^{-2}}$$
where $$F_{ij}$$ is the gravitation force between bodies i and j, $$m_x$$ is the mass of body x, $$d_{ij}$$ is the distance between bodies i and j, and G is the gravitational constant. That is, the gravitational force (or interaction) between the bodies is proportional to their masses (size) and inversely related to the distance between them: bigger bodies closer together have a greater interaction.
Gravity models of human spatial interaction adopt the same assumption: the amount of interaction between two locations (usually represented by trips, but could also be flows of money, information, etc.) is proportional to the “size” (“attractiveness”) of the two locations and the extent of their physical separation (measured in distance or travel time). That is, gravity models assume:
$$T_{ij}∝{X_i}$$$$T_{ij}∝{X_j}$$$$T_{ij}∝{f_{ij}}$$ or, $$T_{ij}=k{X_i}{X_j}{f_{ij}}$$
where Tij is the number of trips between origin i and destination j, Xi is some measure of the “size” of location i, fij is a measure of the “impedance” (difficulty of travel) between locations i and j, and k is a constant of proportionality. Fij is an inverse function of distance or travel time; that is, for example if t_ij is the travel time from i to j, then: (df_ij)/(dt_ij ) <0.
The constant k is chosen so that (2) satisfies known constraints on the interactions being predicted. If, for example, Xi = Oi, the number of shopping trips predicted by a trip generation model to originate in zone i, then it is reasonable to impose the constraint on our model that:
∑_j'▒T_ij' = O_i ∀ i (3)
Substituting (2) into (3) and solving for k yields:
k =1 / ∑_j'▒X_j' f_ij' or T_ij=O_i X_j f_ij / ∑_j'▒X_j' f_ij' (4)
Equation (4) is called a singly-constrained gravity model, since only a single constraint (equation (3)) has been imposed on the model. Very many examples of singly-constrained destination choice models exist, in a variety of applications. In particular, note that instead of constraining the predicted trips to match predetermined trip origin totals (an origin-constrained model), it would have been possible to constrain the trips to sum to predetermined trip destination totals instead (a destination constrained model).
In some cases, such as predicting work locations, it may make sense to constrain the predicted origin-destination (O-D) trips to equal both trip origin and trip destination totals (as typically determined by trip generation models). In this case, Xj is usually defined as Dj, the predetermined number of trips destined to location j, and, in addition to constraint (3), a second constraint is added that equation (2) must also satisfy:
∑_i'▒T_i'j = D_j ∀ j (5)
Imposing both constraints (3) and (5) on the model results in a doubly-constrained gravity model, which can be expressed as:
T_ij=O_i D_j^* f_ij / ∑_j'▒D_j'^* f_ij' (6)
where D*j is a modified attraction term that is iteratively defined so that constraint (5) is satisfied for all destinations j.
## Entropy Maximization Models
While it is intuitively plausible that should trips go to “bigger” (more attractive) destinations as well as to destinations that are closer to, rather than farther from, the trip origin, gravity models have always been criticized for their apparently ad hoc derivation: why should human interactions necessarily follow the same “law” as gravitational bodies? Beginning with Alan Wilson’s seminal paper in 1967,[1] a sound statistical theory underlying gravity models was developed. Wilson showed that the statistically most likely trip matrix, T, is given by maximizing the entropy function:
max┬T∑_i▒∑_j▒〖T_ij ln(T_ij 〗) (7)
subject to known constraints. In the case of a doubly-constrained model, at a minimum, these are constraints (3) and (5), plus typically a third constraint which often takes the form:
(∑_i▒〖∑_j▒T_ij t_ij 〗)/T= t ̅ ∀ i,j (8)
where T is the (known) total number of trips in the system, and t ̅ is the observed average travel time. In other words, equation (8) states that a feasible predicted trip matrix is one in which the predicted average travel time (the left-hand side of (8) equals the observed average travel time).
Solving this mathematical program yields the following trip distribution model:
T_ij = A_i B_j O_i D_j e^(βt_ij ) (9)
where ( < 0) is an estimated parameter and Ai and Bj are “balancing factors” that are iteratively defined so that constraints (3) and (5) are both satisfied:
A_i =1 /∑_(j^')▒B_(j^' ) D_(j^' ) e^(βt_ij' ) (10.1)
B_j =1 /∑_(i^')▒A_(ij^' ) O_(ij^' ) e^(βt_i'j ) (10.2)
It can be shown that BjDj = D*j, the “modified attraction term” in the gravity model formulation, equation (6). Noting this and substituting (10.1) into (9) yields:
T_ij=O_i D_j^* e^(βt_ij ) / ∑_j'▒D_j'^* e^(βt_ij' ) (11)
This is exactly the doubly-constrained gravity model (equation (6)) with the specific impedance function f_ij = e^(βt_ij ). In other words, the “ad hoc” gravity model, “properly specified” is the statistically most likely model of a trip O-D matrix, given known constraints. This provides very strong theoretical support for “gravity-like” spatial interaction models. Other important points to note include: A specific entropy model specification is determined by the choice of constraints imposed on the model. The general procedure for specifying an entropy model is defined below. Arbitrarily complex specifications can be generated, providing that an appropriate constraint set can be specified. In particular, the impedance functional form derives from the constraint(s) written concerning transportation level-of-service variables. In the example above, imposing the constraint that the predicted system-wide average travel should equal the observed average time in the base data yields a negative exponential impedance function. If instead, one wrote a constraint in which the predicted average of ln(Tij) equals the observed average value, then the resulting impedance function would take the form of a negative power function (tij)-b.
## Random Utility Models
By far the most common type of destination choice model used in practice is some form of random utility model, usually a multinomial logit model or a nested logit model (e.g., a nested destination-mode choice model). Random utility (discrete choice) models are used throughout travel demand modeling given their strong theoretical foundations in microeconomic theory and their practical and efficient analytical function forms.
A typical logit destination choice model for the probability that destination j is chosen given trip origin i (Pj|i) is often generically expressed as: P_(j|i) =e^(V_(j|i) )/(∑_j'▒e^(V_(j^' |i) ) ) (12)
where Vj|i is the systematic utility of destination j given origin i. Logit destination choice models are widely used for a variety of reasons including: Flexibility in specifying the utility function (any relevant variable can be readily included). Readily available parameter estimation software. Familiarity with the method. Computational efficiency. Support for both disaggregate (person-level) and aggregate (trip flows) formulations.
Detailed discussion of the specification and use of logit destination choice models is provided on many other pages throughout this wiki.
Mathematical Equivalence of Gravity, Entropy and Logit Models
It is commonplace in the literature to state that “destination choice” (i.e., disaggregate logit) models are superior in performance to “gravity models”. This, however, is a somewhat misleading statement in that it reflects the common practice in terms of how “gravity” and “logit” models are typically implemented, rather than fundamental differences in the mathematics of the two approaches. In practice, “gravity” models are often aggregate (based on O-D flows instead of individual trips) and very simply specified in terms of both attraction/size variables and impendence functions (including sometimes the use of distance rather than travel times). “Logit” models, on the other hand, are usually disaggregate (based on individual trips) and can have an extensive set of explanatory attraction variables in the utility function. Given this typically more extensive set of explanatory variables, it is not surprising that such “logit” models outperform the more simply specified “gravity” models.
But, as Anas (1983)[2] first observed, “gravity” models as derived through entropy maximization can be formulated at the disaggregate (individual trip) level as well, and can incorporate any number of explanatory variables. In particular, any linear-in-the-parameters utility function typically used in logit destination choice models can be replicated in an entropy model. Further, if consistently defined at the same level of aggregation, the same set of explanatory variables and the same base data are used for parameter estimation, then it can be shown that the estimated parameters for the two models will be identical. Thus, logit and entropy (gravity) models are, in fact, not different models but are mathematically the same model.
As a simple illustration of this, equation (4) can be rearranged to yield:
T_ij= O_i {X_j f_ij/∑_j'▒〖X_j' f_ij' 〗} = O_i P_(j|) where: P_(j|i) =(X_j f_ij)/(∑_(j^')▒〖X_(j^' ) f_(ij^' ) 〗)=e^(ln(X_j )+ln(f_ij ) )/(∑_j'▒e^(ln(X_j' )+ln(f_ij' ) ) ) (13)
If we assume that: f_ij= e^(βt_ij ) then equation (13) becomes: P_(j|i) =e^(ln(X_j )+ βt_ij )/(∑_j'▒e^(ln(X_j' )+ βt_ij' ) ) (14)
Equation (14) is a simple logit destination choice model.
This mathematical equivalency with entropy models only holds for multinomial logit models, not for random utility models in general. The ability to theoretically derive logit models from two very different starting points, one behavioral (people choose alternatives so as to maximize their personal utility) and one statistical (deriving most likely choice probabilities given known constraints on these probabilities), however, is striking and arguably reinforces the case for use of logit models in applications where the underlying assumptions of the model (e.g., statistical independence of the alternatives) holds.
## Other Destination Choice Model Formulations
Historically, other approaches to destination choice models have been developed, including intervening opportunities models and competing opportunities models. In general, these approaches tend to be computationally more intensive without generating improved fits to observed data than more conventional methods and so are rarely used in current practice. Brief descriptions of these methods are provided here for historical documentation.
### Intervening Opportunities Models
Eric Miller to provide.
### Competing Opportunities Models
Eric Miller to provide.
## Doubly-Constrained Gravity/Entropy Models
### Gravity Formulation
Given observed or predicted trip origins, Oi and destinations, Dj, for a set of zones, the spatial interaction modeling problem is to predict the origin-destination trips, Tij, between every origin zone i and destination zone j. The predicted trips must satisfy the following two constraints:
∑_j'▒T_ij' = O_i ∀ i (15) ∑_i'▒T_i'j = D_j ∀ j (16)
If we assume that these trips are proportional to a zone-to-zone impedance function, fij, then the following gravity model will always satisfy constraint (15);
T_ij=O_i D_j f_ij/∑_(j^')▒D_(j^' ) f_(ij^' ) (17)
Equation (17), however, will not satisfy constraint (16) except, perhaps, in the most trivial cases. The only way that a spatial interaction model can satisfy both the “row constraints” (15) and the “column constraints” (16) is through an iterative solution procedure in which the actual “attraction term” (Dj) is replaced by a “modified attraction term” Dj*,k :
D_j^(*,k+1) = D_j^(*,k) (D_j/∑_i▒T_ij^k ) (18.1) D_j^(*,0)= D_j (18.2)
T_ij^(k+1)=O_i D_j^(*,k) f_(ij )/ ∑_(j^')▒D_(j^')^(*,k) f_(ij^' ) (19)
Where T_ij^k is the predicted trips from zone i to zone j in the kth iteration and the procedure iterates until: (D_j/∑_i▒T_ij^k ) ≤ ϵ ∀ j (20)
### Entropy Formulation
The equivalent entropy formulation involves solving the following mathematical program:
max┬T∑_i▒∑_j▒〖T_ij ln(T_ij 〗) (21) Subject to: Constraints (15) and (16)
(∑_i▒〖∑_j▒T_ij t_ij 〗)/T= t ̅ ∀ i,j (22)
Using the method of Lagrange to maximize (21) subject to the equality constraints (15), (16) and (22) eventually yields the solution:
T_ij = A_i B_j O_i D_j e^(βt_ij ) (23)
where ( < 0) is an estimated parameter and Ai and Bj are “balancing factors” that are iteratively defined so that constraints (3) and (5) are both satisfied:
A_i =1 /∑_(j^')▒B_(j^' ) D_(j^' ) e^(βt_ij' ) (10.1)
B_j =1 /∑_(i^')▒A_(ij^' ) O_(ij^' ) e^(βt_i'j ) (10.2)
To numerically compute the balancing factors, the Bj terms can be initialized to 1.0, the Ai terms can be computed given these Bj’s, and the algorithm iterates until the factors converge to stable values.
A common criticism of doubly-constrained gravity models is the supposedly ad hoc nature of the balancing procedure using the “modified attraction terms” described above. As shown by the entropy model formulation, however, a doubly-constrained matrix can only be computed iteratively: no analytical closed-form solution is possible. Further, with manipulation, it can be shown that BjDj = Dj* , the “modified attraction term” in the gravity model formulation, equation (19). Hence the gravity model balancing procedure is not, in fact “ad hoc” but rather the correct method for computing a doubly-constrained matrix.
## Estimating Gravity/Entropy Model Parameters
Various ad hoc procedures are sometimes used to estimate gravity model parameters. Given the entropy interpretation of the gravity model, however, the method for parameter estimation is unambiguous: parameters must be chosen so that the underlying constraints of the model hold. For example, for the simple doubly-constrained model given by equation (23) the parameter must be chosen so that associated constraint generating this parameter, equation (22) holds. This problem can be solved using the standard Newton-Raphson root-finding method to find the value of that satisfies equation (22).
Further, in the case of singly-constrained models gravity/entropy models the set of constraints generating the set of parameters to be estimated are exactly the set of equations defining the first-order conditions for maximizing the long-likelihood function for the corresponding multinomial logit model. Thus, standard logit model parameter estimation procedures can be used.
## Developing an Entropy Model
Explanatory variables are entered into an entropy model by writing a constraint for each variable. In the sections above, travel time was entered into the model by writing a constraint involving it (predicted average time = observed average time). This can be repeated for as many variables as desired. As an example, consider a singly- (origin-) constrained shopping destination choice model in which one wants the following explanatory variables to enter the impedance (utility) function: ln(Fj) (where Fj is the amount of retail floorspace in zone j) A “dummy variable” CBDj (where CBDj = 1 if zone j is in the city’s central business district).
The full mathematical program to solve to generate this model is:
max┬T∑_i▒∑_j▒〖T_ij ln(T_ij 〗) (25)
Subject to: ∑_j'▒T_ij' = O_i ∀ i (26) ∑_i▒∑_j▒〖T_ij 〖CBD〗_j 〗=¯CBD =Total trips destined for the CBD (27)
∑_j▒〖T_ij ln(F_j )= ¯F〗 =Total weighted ln(floorspace)visited (28)
(∑_i▒〖∑_j▒T_ij t_ij 〗)/T= t ̅ ∀ i,j (29)
To solve this, construct the Lagrangian equation:
〖max┬T S=〗∑_i▒∑_j▒〖T_ij ln(T_ij 〗)- ∑_i▒〖λ_i [∑_(j^')▒T_(ij^' ) - O_i ] 〗-μ[∑_i▒∑_j▒〖T_ij 〖CBD〗_j 〗-¯CBD]- γ[∑_j▒〖T_ij ln(F_j )- ¯F〗]- β[∑_i▒∑_j▒〖T_ij t_ij-T¯t〗]
To maximize, solve the first-order optimality conditions:
∂S/(∂T_ij )=0=ln(T_ij )-1- λ_i- μ〖CBD〗_j- γln(F_j )- βt_ij which yields: T_ij=e^(1+λ_i + μ〖CBD〗_j + γln(F_j )+βt_ij ) (30)
Substituting (30) into (26) and solving for 1+i yields:
e^(1+λ_i )=O_i / ∑_j'▒e^(μ〖CBD〗_j' + γln(F_j' )+βt_ij' ) (31)
Substituting (31) into (30) yields:
T_ij= O_i e^(μ〖CBD〗_j + γln(F_j )+βt_ij ) / ∑_j'▒e^(μ〖CBD〗_j' + γln(F_j' )+βt_ij' ) (32)
Equation (32) is the desired singly-constrained entropy trip destination model. Note that it can be rewritten as:
T_ij= O_i F_j^γ e^(μ〖CBD〗_j + βt_ij ) / ∑_j'▒〖F_j'^γ e^(μ〖CBD〗_j' + βt_ij' ) 〗 (33)
which is often the format used for “gravity” models.
Equation (32) also defines the destination probability choice model:
P_(j|i)= e^(μ〖CBD〗_j + γln(F_j )+βt_ij ) / ∑_j'▒e^(μ〖CBD〗_j' + γln(F_j' )+βt_ij' ) (34)
As previously discussed, equation (33) is a multinomial logit model, which can be estimated using standard logit estimation software.
# References
1. Wilson, A.G., “A Statistical Theory of Spatial Distribution Models”, Chapter 3 in R. Quandt (ed) The Demand for Travel: Theory and Measurement, Lexington, Mass: Lexington Books, 1970, . 55-82.
2. Anas, A., “Discrete choice theory, information theory, and the multinomial logit and gravity models”, Transportation Research B 17, 1983, 13-23. | 4,850 | 18,540 | {"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": 2, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2020-05 | latest | en | 0.928249 |
https://getgoodcollegegrades.com/paper/which-of-the-following-is-a-vector-quantity | 1,695,373,966,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506339.10/warc/CC-MAIN-20230922070214-20230922100214-00357.warc.gz | 313,908,274 | 8,822 | # Which of the following is a vector quantity?
Discipline: Physics
Paper Format: APA
Pages: 1 Words: 275
Question
Which of the following is a vector quantity? speed time force displacement temperature momentum velocity acceleration The sum of two vectors is the largest when the two vectors are. pointing in the same direction pointing in opposite directions perpendicular to each other positioned at a 45 degree angle
The scalars are the quantities that are explained only by the magnitude.
Hence, the incorrect options are,
• Speed
• Time
• Temperature
The speed, time and temperature have no direction. They have only magnitude. Hence, they are scalar quantities.
Vectors are the quantities that are explained by both magnitude and direction.
The force, momentum, displacement, velocity, and acceleration have both magnitude and direction. Hence, the correct options are,
• Force
• Displacement
• Momentum
• Velocity
• Acceleration
Force, displacement, momentum, velocity, and acceleration are vector quantities.
The scalars are the quantities that are explained only by the magnitude. The velocity is a vector quantity. The momentum and acceleration depend on the velocity. Hence, they are also vector quantities.
The force is directly proportional to acceleration. Hence, the force is a vector quantity.
The incorrect options are,
• Pointing in opposite direction.
• Perpendicular to each other.
• Positioned at a angle.
The expression for resultant of vector is given below:
When the two vectors are in opposite direction substitute for .
…… (1)
If the vectors are in opposite direction, then the resultant vector is the difference in magnitude of the two vectors.
If the two vectors are perpendicular substitute for .
…… (2)
If the two vectors are positioned at angle substitute for .
…… (3)
When the two vectors are in same direction substitute for .
…… (4)
If the two vectors are parallel then the resultant vector is the sum of two vectors. Thus, the sum of the two vectors is largest when the vectors are in same direction.
From the above equations (1), (2), (3), and (4) we get the correct option,
• Pointing in the same direction.
The sum of the two vectors is largest when the vectors are pointing in the same direction.
When the vectors are pointing in same direction, the angle between them is zero.
For the two vectors pointing in the opposite direction the angle between them is | 489 | 2,437 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2023-40 | longest | en | 0.891487 |
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May 6, 2016
# Homework Help: physics
Posted by Anonymous on Friday, March 23, 2012 at 2:04pm.
A 59.1 kg ice skater, moving at 13.3 m/s, crashes into a stationary skater of equal mass. After the collision, the two skaters move as a unit at 6.65 m/s. Suppose the average force a skater experience without breaking a bone is 4656 N. If the impact time is 0.166 s, what is the magnitude of the average force each skater experiences? | 128 | 438 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2016-18 | longest | en | 0.917989 |
http://www.pinoybix.org/2015/02/answers-in-physics-part10.html | 1,511,149,216,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805911.18/warc/CC-MAIN-20171120032907-20171120052907-00560.warc.gz | 489,647,429 | 92,949 | # MCQs in Physics Part X - Answers
Answers key for the Compiled MCQs in Physics Part 10 of the series as one topic in General Engineering and Applied Sciences (GEAS) in the ECE Board Exam.
Below are the answers key for the Multiple Choice Questions in Physics - MCQs Part 10.
451. Acceleration
452. Accuracy
453. Angular Acceleration
454. Area
455. Angular Velocity
456. Beam
457. Beam Bridge
458. none of the above (bend)
459. Biomechanics
460. Brace
461. Brittle
462. Buckle
463. Cable
464. Cantilever
465. Chemical Equilibrium
466. Compression
467. Core
468. deform
469. Distance
470. modulus of elasticity
471. Engineering
472. Equilibrium
473. Force
474. Gravitation
475. gravitational constant
476. Height
477. Hydraulics
478. Inertia
479. Instantaneous
480. invariant mass
481. Irreversible
482. joint
483. kilogram
484. length
485. Lever
487. magnitude
488. mass
489. matter
490. mechanics
491. measurement
492. momentum
493. Newton’s Law of Motion
494. Pile
495. physical body
496. quantity
497. rigid
498. science
499. stress
500. shear stress
### Online Questions and Answers in Physics Series
Following is the list of practice exam test questions in this brand new series:
College Physics MCQs
PART 1: MCQs from Number 1 – 50 Answer key: PART I
PART 2: MCQs from Number 51 – 100 Answer key: PART II
PART 3: MCQs from Number 101 – 150 Answer key: PART III
PART 4: MCQs from Number 151 – 200 Answer key: PART IV
PART 5: MCQs from Number 201 – 250 Answer key: PART V
PART 6: MCQs from Number 251 – 300 Answer key: PART VI
PART 7: MCQs from Number 301 – 350 Answer key: PART VII
PART 8: MCQs from Number 351 – 400 Answer key: PART VIII
PART 9: MCQs from Number 401 – 450 Answer key: PART IX
PART 10: MCQs from Number 451 – 500 Answer key: PART X
### Complete List of MCQs in General Engineering and Applied Sciences per topic
#### Search! Type it and Hit Enter
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# Trying to find the lowest value per Dimension, possibly aggr()?
Hello,
I have a chart and I need to identify the lower Incentive value for each %ADAcctKey.
The Incentive value is a calculated value. In the example below 445 would flag 10% and for 2544 the lowest flagged would be 0%.
%ADAcctKey MetricDesc QA Adherence Revenue Saves Level Target Incentive 445 QA 87.50% Target 80% >= 10% 445 Adherence 88.04% Target 88% >= 10% 2544 Save Rate 26.09% Stretch 22% >= 20% 2544 QA 81.68% Target 80% >= 15% 2544 Adherence 74.34% Below 85% < 0% 2544 Revenue \$335.40 Below \$750 < 0%
Thank you for you help.
• ###### Re: Trying to find the lowest value per Dimension, possibly aggr()?
Use the following expression:
• ###### Re: Trying to find the lowest value per Dimension, possibly aggr()?
That's the first thing I tried but I get blanks for all values. I am wondering if it's because of the Incentive column and how or when it's calculated.
With the below formula populating the Incentive column, would the values exist at the run time of the formula Aggr(Min(Incentive), %ADAcctKey)
Incentive =
if(QA=0,0,
if(Productivity=0,0,
if([Revenue]=0,0,
if(Saves=0,0,
AchievedIncentive)))))
• ###### Re: Trying to find the lowest value per Dimension, possibly aggr()?
Maybe like this
Aggr(
if(QA=0,0,
if(Productivity=0,0,
if([Revenue]=0,0,
if(Saves=0,0,
AchievedIncentive)))))
)
• ###### Re: Trying to find the lowest value per Dimension, possibly aggr()?
Swuehl thanks,
Still getting blanks. I think I may need to do some remodeling to simplify some things.
• ###### Re: Trying to find the lowest value per Dimension, possibly aggr()?
Ok, could you describe how your data model is built, and which dimensions and expressions you're using in the chart.
It would be very helpful if you could upload a small sample QVW that demonstrates your data and chart.
• ###### Re: Trying to find the lowest value per Dimension, possibly aggr()?
It's working but doing something very wrong.
So I have changed my model and it is now allowing me to calculate using the above formula but it's giving an odd result.
I am using the formula Aggr(Min(Achieved), %ADAcctKey) To Identify the Achieved Level by the agent.
Notice the Minimum Score to the right which shows that value of 0 as the lowest achieved and this is correct. It is just on the incorrect line. When I create a second chart to provide a list of agents lowest achieved level with only that row I now get incorrect rows.
%ADAcctKey Category IncentiveLevel Achieved Score Minimum Score 37 Procuctivity Stretch 0.2 0.888 37 QA Target 0.15 91.74 0 37 Revenue Threshold 0.05 1223.98 37 Adherence Below 0 0.8454
The consistent part is that the value is placed on the first portion of our load script as seen below. Notice how the first Metric loaded below in the Resident load is QA. All the lowest values in the chart are displaying there. When I moved QA to the bottom of the script, all scores then showed in the row for Adherence, the new first loaded.
/***[ Initial load where each agents score for all categories is in a single row]***/
AgentName,
RoleShortDesc,
RoleID,
Procuctivity,
Saves,
QA,
Revenue
FROM
[Customer Care Incentive Extract.xls]
(biff, embedded labels, table is Sheet1\$);
/***[ Each metric is then split out into a single column and later joined to the KPI range]***/
tmpIncentiveScores:
RoleID & 2 as VariableKey,
AgentName,
RoleShortDesc,
RoleID,
QA as IncentiveScore,
'QA' as IncentiveCategory
Concatenate(tmpIncentiveScores)
RoleID & 2 as VariableKey,
AgentName,
RoleShortDesc,
RoleID,
• ###### Re: Trying to find the lowest value per Dimension, possibly aggr()?
have you tried using no distinct?
Aggr( nodistinct Min(Achieved), %ADAcctKey)
also why not use min(total <%ADAcctKey > Min(Achieved)) ? this tends to give you the same results that by using aggr but has better performance
• ###### Re: Trying to find the lowest value per Dimension, possibly aggr()?
Ramon,
Thank you for the help.
The formula Aggr( nodistinct Min(Achieved), %ADAcctKey) places the lowest value in all rows for the agent which will work but does not allow us to only display the single row.
With the formula min(total <%ADAcctKey > Min(Achieved)) I get all null values. I am curious as to the need for the '<>' around <%ADAcctKey >.
Thanks again.
• ###### Re: Trying to find the lowest value per Dimension, possibly aggr()?
sorry it should be min(total <%ADAcctKey > (Achieved))
the <> after a total does a grouping by that field
• ###### Re: Trying to find the lowest value per Dimension, possibly aggr()?
You probably have some code after the script snippet you've posted above, but just to be sure:
Are you dropping table load_IncentiveScores after transforming the columns to rows? If not, you will create a synthetic table that will link the columns with same name. In your scenario, I believe that's not what you want.
Do you have a synthetic table in your data model?
Again, if you want us to help you, you need to post more detailed information about your data model, best by creating a small sample QVW that demonstrates this issue.
• ###### Re: Trying to find the lowest value per Dimension, possibly aggr()?
App showing the Aggr() value on the wrong row.
I also attempted this with Rank() during my process but that is additional work on the load script.
Look at the table 'Using Aggr() - Minimum Score' with EmpID 31480 if it's not selected. You will see using Aggr( Min(AchievedIncentive), EmpID) places the lowest score in 15% when it's really 10%. This works correctly on the majority of the values.
Thank you everyone for your help. I have been able to create what I needed and part of the problem was how the app that feeds these numbers was built. I have uploaded an app that takes the raw scores and builds what I need. I will work further to have the back in the source app but this is working for now. | 1,566 | 6,031 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-30 | latest | en | 0.846967 |
http://hi.gher.space/forum/viewtopic.php?f=3&t=2077&p=23803 | 1,553,507,887,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203865.15/warc/CC-MAIN-20190325092147-20190325114147-00334.warc.gz | 86,933,572 | 5,879 | ## x4oW(1/2)o and x4oW(3/2)o
Higher-dimensional geometry (previously "Polyshapes").
### x4oW(1/2)o and x4oW(3/2)o
I've been mainly looking after people, and my polytope thoughts have been with these pair of interesting polyhedra. They both have squares as faces, the dihedral or margin-angle being 60 degrees and 120 degrees respectively. [These are regular figures of infinite densities].
What makes these pair interesting, is that they correspond to an infinite cover of squares from the triangular and hexagonal prism, respectively. This means that if you put six x4oW(1/2)o around an edge, you get the same ring of squares as an x4oW(3/2)o. If we suppose that Proffessor Coxeter is right, and that a polytope whose faces are centre-inversion, also is centre-inversion, it means that the 1/2 form forms the squares of a hexagram prism, and that three of these stacked on opposite squares, will form the opposite squares of 3/2 form.
What is even more interesting is that x4oW(1/2)x = x4oW(3/2)o. One can see from the bridging constants for these groups, that the nodes of the first are at the ratios of x, q, x, and of the second, x, q, h. This means that there are no further subgroups of these figures. I thought one could expand the edges of the first out to the sphere of the second, but this would not fit well either, because it would imply that x12o is a subgroup, when it clearly isn't.
Both of these are of course, infinitely dense, as neither of xW(1/2)o or xW(3/2)o close. The vertex figure, for an edge of root-2, of these polygons are q:q:1. and q:q:h respectively. I calculated the van Oss polyhedra for these as W(20/7) and W(16/5) respectively, these are fairly near the hexagon, the first slightly smaller, the second slightly larger.
Later i was able to calculate the Schläfli Index for these two groups (3 and 1 respectively), which leads to the correct vertex-diameters for both of these figures D2 = 7/3, and D2 = 5 respectively).
But there are some interesting things i am still unsure of, as i have not as yet walked the surface of x4oW(1/2)o far enough to verify a direct connection to imply central inversion, or that at least three edges cross in right angles, that one can state catergorically that central inversion holds. It is certainly true that the number system known for this system does not support three equal edges crossing at right angles (although they do so in pairs), because the third axis is at ratio of h to the first two.
The dream you dream alone is only a dream
the dream we dream together is reality.
\(Latex\) at https://greasyfork.org/en/users/188714-wendy-krieger
wendy
Pentonian
Posts: 1874
Joined: Tue Jan 18, 2005 12:42 pm
Location: Brisbane, Australia | 713 | 2,720 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-13 | latest | en | 0.967514 |
https://education.blurtit.com/4521742/how-to-work-out-x-is-an-angle-at-a-point-between-two-tangents-the-tangents-join-as-a | 1,669,761,363,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710711.7/warc/CC-MAIN-20221129200438-20221129230438-00466.warc.gz | 241,964,202 | 10,309 | # How to work out? X is an angle at a point between two tangents. The tangents join as a chord, which is the base of another triangle, top angle 43 (given)
I used the alternate segment theorem to determine that an angle in the triangle with x on was 43 degrees.
This was an isosceles triangle because tangents from a point are equal in length. Base angles in an isosceles triangle are equal so that meant that:
x = 180 - (2 x 43)
x = 180 - 86
x = 94 degrees
Because angles in a triangle sum to 180 degrees.
thanked the writer. | 140 | 533 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2022-49 | latest | en | 0.954144 |
http://www.blog44.ca/evelync/2020/03/12/a-post-about-maximum-surface-area/ | 1,618,734,302,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038469494.59/warc/CC-MAIN-20210418073623-20210418103623-00259.warc.gz | 115,446,549 | 14,314 | ## Evelyn Posts Stuff
### The stuff that Evelyn posts
Hello!
This post marks the end of another Scimatics project. This one was all about surface area, volume, and 3D shapes. One of my favourite things to do in math.
If you couldn’t tell, I was not trying to be sarcastic. I actually really enjoy calculating the surface area and volume of random prisms. I’m a weirdo, but whatever. We can discuss my strange interests another time.
The premise of this project was to design an inanimate object on a 3D program called Tinkercad. We either had to design the object so that it had as much surface area or volume as possible. At first, I didn’t really understand the project. If you’re trying to get maximum volume, why not make the object the size of Vancouver? The size of Earth? It turns out that it had to be smaller that 10cm squared. That made a bit more sense.
If you read this blog often, or, in fact, any of my peers’ blogs, you may have guessed that we had a driving question for this project. You might also be a PLP teacher. That is likely the case.
How can we maximize the surface area or volume of an object?
So, basically what I just explained.
Given Mr. Gross’s desire for short blog posts, let’s get right to the point (she said, after writing a 4 paragraph introduction).
The standard 3 Curricular Competencies:
• Reasoning and Analyzing
• Applying and Innovating
• Communicating and Representing
Cool. Let’s begin.
I decided at the beginning of this project that I wanted to create a slide, with maximum surface area. I was randomly fiddling around with Tinkercad when I had that stroke of inspiration.
Side note: before this project, I had never used Tinkercad, or any sort of 3D modelling software for that matter. I had seen my sister working with it, though, so I used my limited knowledge of the program to make random objects. That’s how I figured out about the slide.
After working on the slide some more, I realized that I definitely did not had the required 10 shapes. I added one of those spinny merry-go-round things to make a full-on playground.
Here is my final creation:
I think it turned out well. After this screenshot was taken, I added a base so that the pieces would stick together when we 3D printed them.
Next I had to calculate the surface area and volume of my creation. It took me so long to calculate it all, but eventually I was finished and left with this chaotic mess:
I later made a more legible good copy to hand in.
Finally, I calculated the surface area-to-volume ratio. Drumroll, please…
It didn’t work out exactly as planned, but what can you do? Whatever kid is going to play on it won’t care, anyways.
Now I suppose I should be getting on with the Curricular Competencies.
Reasoning and Analyzing: Model mathematics in contextualized experiences.
I used Tinkercad to make a playground 3D model, which included 17 shapes. I attempted to optimize my design for surface area, but didn’t exactly succeed. I could have made the pieces thinner and longer for more surface area. Oh well, at least I know where I went wrong.
Applying and Innovating: Contribute to care for self, others, community, and world through personal or collaborative approaches.
For this competency, I think I did pretty well. Better than usual, in fact, because I did not feel the need to take repeated breaks, like in some other projects. I also complete all the required workbook pages relatively accurately.
Communicating and Representing: Explain and justify mathematical ideas and decisions.
I calculated the surface area and volume of all my shapes precisely and accurately. My ratio isn’t what I hoped it would be, but at least it’s correct! My Keynote presentation includes images of my design, clearly shows the calculations that I made to determine the surface area and volume, and the ration was displayed visibly and prominently at the end.
So I guess that’s it!
I am in the process of writing 2 other posts before spring break. Hopefully they will be out before Friday, since that’s when they’re due. A busy time for this blog!
Cheers,
Evelyn 👩🏽
## 1 Comment
1. Great job! I like some of the examples about using field experience such as their vacation and random acts of their personal lives that they can share to put in their backpack. | 964 | 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} | 3.359375 | 3 | CC-MAIN-2021-17 | longest | en | 0.961707 |
https://onews.info/what-everyday-objects-are-7-inches-long/ | 1,709,296,376,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475238.84/warc/CC-MAIN-20240301093751-20240301123751-00670.warc.gz | 428,580,534 | 15,666 | # What everyday objects are 7 inches long?
## What everyday objects are 7 inches long?
So you can line up 9 pennies to get around 7 inches in length. A US dollar bill measures almost 7 inches so this is another easy way to visualize this length….The larger blades are just over 7 inches long.
• 10 Dimes.
• 4 Golf Balls or 2 Softballs. …
• 5 Toothpicks. …
• 2 Credit Cards. …
Nov 19, 2021
## What object is equivalent to an inch?
The most common object used to estimate inches is the adult thumb, which is about 1 inch wide. Other options may include a water bottle cap, a detachable pencil eraser, the width of a standard rubber eraser, the length of a paperclip, and the length of a standard small sewing pin.
## What objects are 5 inches?
6 Things that are 5 Inches Long
• 5 US Quarters or a Dollar Bill. A US quarter coin measures around 1 inch in diameter, thus it equals one-fifth of 5 inches. …
• 2 Adult Thumbs. An adult thumb measures about an inch in length starting from the tip of the thumb down to the top knuckle.
Nov 19, 2021
## What is the measurement of 8 inches?
What part of a foot is 8 inches?
Inch Decimal of a Foot
6 inches 0.500
7 inches 0.583
8 inches 0.667
9 inches 0.750
Jan 8, 2022
## How long is a pencil?
A standard, hexagonal, "#2 pencil" is cut to a hexagonal height of 1⁄4-inch (6 mm), but the outer diameter is slightly larger (about 9⁄32-inch (7 mm)) A standard, #2, hexagonal pencil is 19 cm (7.5 in) long.
## How much is an inch visually?
One inch (2.5 cm) is roughly the measurement from the top knuckle on your thumb to your thumb tip.
## How can I measure 6 inches without a ruler?
0:342:04Measuring stuff without a ruler. A Mere Mortals woodworking quick tip.YouTube
## How long is an inch visually?
One inch (2.5 cm) is roughly the measurement from the top knuckle on your thumb to your thumb tip.
## How can I measure 8 inches without a ruler?
0:342:04Measuring stuff without a ruler. A Mere Mortals woodworking quick tip.YouTube
## How big is an inch on your finger?
One inch (2.5 cm) is roughly the measurement from the top knuckle on your thumb to your thumb tip.
## How long is a eraser?
The size is approximately 1.8 inches (4.5 cm) X 0.7 inches (1.8 cm) X 0.5 inches (1.3 cm).
## What’s the length of a paperclip?
A #1 regular paperclip measures approximately 1.375 inches (in).
## How can I measure 7 inches without a ruler?
0:342:04Measuring stuff without a ruler. A Mere Mortals woodworking quick tip.YouTube
## Is a dollar bill 6 inches?
See our other measuring without a tape measure or ruler. Just a note on accuracy. Since a dollar bill is 6.14″ and not 6″ exactly the marks will be off by a fraction of an inch each.
## How long is a piece of chalk?
Each standard chalk size measures 3 1/4" long by 3/8" inch in diameter.
## What is the length of the pencil?
Knowledge of pencils The length is 172mm or more in JIS standard. The first person to decide the length closest to this is Germany Luther Farber. It proposes to “7 inches (17.78 cm)” around 1840. This length is said to have taken the length from the place of the adult hand to the tip of the middle finger.
## How long is an inch on your finger?
The length between your thumb tip and the top knuckle of your thumb is roughly one inch.
## Can u eat chalk?
Chalk is considered non-toxic in small amounts. If large amounts are eaten, it can be irritating to the stomach and cause vomiting. Chalk can be a choking hazard for very young children. CAUTION: Eating pool or billiard chalk can be different than school and blackboard chalk because it may also contain lead.
## What is school chalk made of?
chalk, soft, fine-grained, easily pulverized, white-to-grayish variety of limestone. Chalk is composed of the shells of such minute marine organisms as foraminifera, coccoliths, and rhabdoliths. The purest varieties contain up to 99 percent calcium carbonate in the form of the mineral calcite.
## What is the length of notebook?
The standard everyday book is 110 x 210 mms (4.33 x 8.26 inches) and can be useful for both horizontal and vertical layouts. The size was popularized by the Traveler's Company TN (Traveler's Notebook,) and has been carried all over the globe to document travel.
## Is your pinky finger an inch?
It turns out that many of us humans have pinky tips that are all around the same length, and handily, that length is about 1 inch.
## Why do I crave dirt?
You crave and eat clay or dirt They draw blood to determine if you lack nutrients such as iron, zinc, and vitamins. In most cases, supplying your body with the nutrients it craves through a healthy diet and supplements resolves your pica.
## Can you eat paper?
Paper is mostly composed of cellulose, a harmless organic compound found in plants. But humans lack the enzymes necessary to properly digest it, which is to say Milbank's column “will come through his GI tract in much the same form it came in,” Staller said.
## What does chalk taste like?
It varies for different people but it's the chalky taste that edible chalk and clay lovers adore. Edible chalk has a very clean fresh taste and always remains monolithic. Some chalks are crunchy and some chalks are soft depending on the type.
## Can you tell the size of a man by his thumb?
Sorry boys, but it does seem the size of your fingers is a pretty good guide to your manhood. Long regarded as a myth and source of much amusement, scientists have now confirmed that the length of a man's index finger relative to his ring finger can reveal his penis size.
## Can I eat chalk?
Chalk is considered non-toxic in small amounts. If large amounts are eaten, it can be irritating to the stomach and cause vomiting. Chalk can be a choking hazard for very young children. CAUTION: Eating pool or billiard chalk can be different than school and blackboard chalk because it may also contain lead.
## Why do I suddenly like mustard?
It turns out that this craving for variety is hardwired into us as humans. So when we experience something new like spicy mustard or sweet honey-mustard dressing, our brain releases dopamine which creates pleasurable feelings and makes us want more. Our ancestors enjoyed mustard, and you can too!
## Can you eat your own poop?
According to the Illinois Poison Center, eating poop is “minimally toxic.” However, poop naturally contains the bacteria commonly found in the intestines. While these bacteria don't harm you when they're in your intestines, they're not meant to be ingested in your mouth.
## Can you eat grass?
More than 400 types of grasses can be eaten worldwide. Grasses are known for being edible and healthy eating because of their proteins and chlorophyll. Magnesium, phosphorus, iron, calcium, potassium, and zinc are commonly found in grasses. Grasses show up in your every-day foods, too.
## What happens if I eat slate pencil?
Slate pencil eating side effects are detrimental to the stomach and gut. Damages the inner lining of the mouth or Canker sores – Side effects of eating slate pencils involve the mouth as well. Slate pencils are entirely insoluble in water, and chewing them damages the inner lining of your mouth. | 1,739 | 7,183 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2024-10 | latest | en | 0.876597 |
https://socratic.org/questions/what-is-the-slope-if-the-secant-line-of-the-function-y-4x-2-2x-1-between-x-3-and#465439 | 1,670,297,128,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711069.79/warc/CC-MAIN-20221206024911-20221206054911-00176.warc.gz | 557,077,930 | 5,946 | # What is the slope if the secant line of the function y = 4x^2 – 2x + 1 between x = 3 and x = 6?
Aug 18, 2017
$\text{slope } = 34$
#### Explanation:
$\text{the slope of the secant line is}$
•color(white)(x)m=(f(6)-f(3))/(6-3)
$\text{that is "" difference in y"/"difference in x"" between the 2 points}$
$f \left(6\right) = 4 {\left(6\right)}^{2} - 2 \left(6\right) + 1$
$\textcolor{w h i t e}{\times x} = 144 - 12 + 1$
$\textcolor{w h i t e}{\times x} = 133$
$f \left(3\right) = 4 {\left(3\right)}^{2} - 2 \left(3\right) + 1$
$\textcolor{w h i t e}{\times x} = 36 - 6 + 1$
$\textcolor{w h i t e}{\times x} = 31$
$\Rightarrow m = \frac{133 - 31}{3} = \frac{102}{3} = 34$ | 294 | 683 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 11, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2022-49 | latest | en | 0.592283 |
https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems&diff=111716 | 1,576,461,812,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541315293.87/warc/CC-MAIN-20191216013805-20191216041805-00295.warc.gz | 278,909,537 | 15,704 | # Difference between revisions of "2019 AMC 8 Problems"
## Problem 1
Ike and Mike go into a sandwich shop with a total of $30.00$ to spend. Sandwiches cost $4.50$ each and soft drinks cost $1.00$ each. Ike and Mike plan to buy as many sandwiches as they can and use the remaining money to buy soft drinks. Counting both soft drinks and sandwiches, how many items will they buy?
$\textbf{(A) }6\qquad\textbf{(B) }7\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad\textbf{(E) }10$
## Problem 2
$[asy] draw((0,0)--(3,0)); draw((0,0)--(0,2)); draw((0,2)--(3,2)); draw((3,2)--(3,0)); dot((0,0)); dot((0,2)); dot((3,0)); dot((3,2)); draw((2,0)--(2,2)); draw((0,1)--(2,1)); label("A",(0,0),S); label("B",(3,0),S); label("C",(3,2),N); label("D",(0,2),N); [/asy]$ Three identical rectangles are put together to form rectangle , as shown in the figure below. Given that the length of the shorter side of each of the smaller rectangles is $5$ feet, what is the area in square feet of rectangle $ABCD$?
$\textbf{(A) }45\qquad\textbf{(B) }75\qquad\textbf{(C) }100\qquad\textbf{(D) }125\qquad\textbf{(E) }150$
## Problem 3
3. Which of the following is the correct order of the fractions $\frac{15}{11}$, $\frac{19}{15}$, and $\frac{17}{13}$, from least to greatest?
$\textbf{(A) }\frac{15}{11} < \frac{17}{13} < \frac{19}{15}\qquad\textbf{(B) }\frac{15}{11} < \frac{19}{15} < \frac{17}{13}\qquad\textbf{(C) }\frac{17}{13} < \frac{19}{15} < \frac{15}{11}\qquad\textbf{(D) }\frac{19}{15} < \frac{15}{11} < \frac{17}{13}\qquad\textbf{(E) }\frac{19}{15} < \frac{17}{13} < \frac{15}{11}$
## Problem 4
Quadrilateral $ABCD$ is a rhombus with perimeter $52$ meters. The length of diagonal $\overline{AC}$ is $24$ meters. What is the area in square meters of rhombus $ABCD$?
$[asy] draw((-13,0)--(0,5)); draw((0,5)--(13,0)); draw((13,0)--(0,-5)); draw((0,-5)--(-13,0)); dot((-13,0)); dot((0,5)); dot((13,0)); dot((0,-5)); label("A",(-13,0),W); label("B",(0,5),N); label("C",(13,0),E); label("D",(0,-5),S); [/asy]$
$\textbf{(A) }60\qquad\textbf{(B) }90\qquad\textbf{(C) }105\qquad\textbf{(D) }120\qquad\textbf{(E) }144$
## Problem 5
A tortoise challenges a hare to a race. The hare eagerly agrees and quickly runs ahead, leaving the slow-moving tortoise behind. Confident that he will win, the hare stops to take a nap. Meanwhile, the tortoise walks at a slow steady pace for the entire race. The hare awakes and runs to the finish line, only to find the tortoise already there. Which of the following graphs matches the description of the race, showing the distance $d$ traveled by the two animals over time $t$ from start to finish? [img]https://latex.artofproblemsolving.com/e/f/9/ef92362133ad95b0788184ff802df2094337d377.png[/img] [img width="65" height="5"]https://latex.artofproblemsolving.com/6/8/2/682b63fdba98e33c13055463223d6c8ea409cf23.png[/img]
## Problem 6
There are $81$ grid points (uniformly spaced) in the square shown in the diagram below, including the points on the edges. Point $P$ is in the center of the square. Given that point $Q$ is randomly chosen among the other 80 points, what is the probability that the line $PQ$ is a line of symmetry for the square?
$[asy] draw((0,0)--(0,8)); draw((0,8)--(8,8)); draw((8,8)--(8,0)); draw((8,0)--(0,0)); dot((0,0)); dot((0,1)); dot((0,2)); dot((0,3)); dot((0,4)); dot((0,5)); dot((0,6)); dot((0,7)); dot((0,8)); dot((1,0)); dot((1,1)); dot((1,2)); dot((1,3)); dot((1,4)); dot((1,5)); dot((1,6)); dot((1,7)); dot((1,8)); dot((2,0)); dot((2,1)); dot((2,2)); dot((2,3)); dot((2,4)); dot((2,5)); dot((2,6)); dot((2,7)); dot((2,8)); dot((3,0)); dot((3,1)); dot((3,2)); dot((3,3)); dot((3,4)); dot((3,5)); dot((3,6)); dot((3,7)); dot((3,8)); dot((4,0)); dot((4,1)); dot((4,2)); dot((4,3)); dot((4,4)); dot((4,5)); dot((4,6)); dot((4,7)); dot((4,8)); dot((5,0)); dot((5,1)); dot((5,2)); dot((5,3)); dot((5,4)); dot((5,5)); dot((5,6)); dot((5,7)); dot((5,8)); dot((6,0)); dot((6,1)); dot((6,2)); dot((6,3)); dot((6,4)); dot((6,5)); dot((6,6)); dot((6,7)); dot((6,8)); dot((7,0)); dot((7,1)); dot((7,2)); dot((7,3)); dot((7,4)); dot((7,5)); dot((7,6)); dot((7,7)); dot((7,8)); dot((8,0)); dot((8,1)); dot((8,2)); dot((8,3)); dot((8,4)); dot((8,5)); dot((8,6)); dot((8,7)); dot((8,8)); label("P",(4,4),NE); [/asy]$
$\textbf{(A) }\frac{1}{5}\qquad\textbf{(B) }\frac{1}{4} \qquad\textbf{(C) }\frac{2}{5} \qquad\textbf{(D) }\frac{9}{20} \qquad\textbf{(E) }\frac{1}{2}$
## Problem 7
Shauna takes five tests, each worth a maximum of $100$ points. Her scores on the first three tests are $76$, $94$, and $87$ . In order to average $81$ for all five tests, what is the lowest score she could earn on one of the other two tests?
$\textbf{(A) }48\qquad\textbf{(B) }52\qquad\textbf{(C) }66\qquad\textbf{(D) }70\qquad\textbf{(E) }74$
## Problem 8
Gilda has a bag of marbles. She gives 20% of them to her friend Pedro. Then Gilda gives 10% of what is left to another friend, Ebony. Finally, Gilda gives 25% of what is now left in the bag to her brother Jimmy. What percentage of her original bag of marbles does Gilda have left for herself?
$\textbf{(A) }20\qquad\textbf{(B) }33\frac{1}{3}\qquad\textbf{(C) }38\qquad\textbf{(D) }45\qquad\textbf{(E) }54$
## Problem 9
Alex and Felicia each have cats as pets. Alex buys cat food in cylindrical cans that are $6$ cm in diameter and $12$ cm high. Felicia buys cat food in cylindrical cans that are $12$ cm in diameter and $6$ cm high. What is the ratio of the volume one of Alex's cans to the volume one of Felicia's cans?
$\textbf{(A) }1:4\qquad\textbf{(B) }1:2\qquad\textbf{(C) }1:1\qquad\textbf{(D) }2:1\qquad\textbf{(E) }4:1$
## Problem 11
The eighth grade class at Lincoln Middle School has $93$ students. Each student takes a math class or a foreign language class or both. There are $70$ eigth graders taking a math class, and there are $54$ eight graders taking a foreign language class. How many eigth graders take $\textit{only}$ a math class and $\textit{not}$ a foreign language class?
$\textbf{(A) }16\qquad\textbf{(B) }23\qquad\textbf{(C) }31\qquad\textbf{(D) }39\qquad\textbf{(E) }70$
## Problem 13
A $\textit{palindrome}$ is a number that has the same value when read from left to right or from right to left. (For example 12321 is a palindrome.) Let $N$ be the least three-digit integer which is not a palindrome but which is the sum of three distinct two-digit palindromes. What is the sum of the digits of $N$?
$\textbf{(A) }2\qquad\textbf{(B) }3\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad\textbf{(E) }6$
## Problem 14
Isabella has $6$ coupons that can be redeemed for free ice cream cones at Pete's Sweet Treats. In order to make the coupons last, she decides that she will redeem one every $10$ days until she has used them all. She knows that Pete's is closed on Sundays, but as she circles the $6$ dates on her calendar, she realizes that no circled date falls on a Sunday. On what day of the week does Isabella redeem her first coupon?
$\textbf{(A) }$Monday$\qquad\textbf{(B) }$Tuesday$\qquad\textbf{(C) }$Wednesday$\qquad\textbf{(D) }$Thursday$\qquad\textbf{(E) }$Friday
## Problem 15
On a beach $50$ people are wearing sunglasses and $35$ people are wearing caps. Some people are wearing both sunglasses and caps. If one of the people wearing a cap is selected at random, the probability that this person is is also wearing sunglasses is $\frac{2}{5}$. If instead, someone wearing sunglasses is selected at random, what is the probability that this person is also wearing a cap? $\textbf{(A) }\frac{14}{85}\qquad\textbf{(B) }\frac{7}{25}\qquad\textbf{(C) }\frac{2}{5}\qquad\textbf{(D) }\frac{4}{7}\qquad\textbf{(E) }\frac{7}{10}$
## Problem 16
Qiang drives $15$ miles at an average speed of $30$ miles per hour. How many additional miles will he have to drive at $55$ miles per hour to average $50$ miles per hour for the entire trip?
$\textbf{(A) }45\qquad\textbf{(B) }62\qquad\textbf{(C) }90\qquad\textbf{(D) }110\qquad\textbf{(E) }135$
## Problem 25
Alice has $24$ apples. In how many ways can she share them with Becky and Chris so that each of the people has at least $2$ apples?
$\textbf{(A) }105\qquad\textbf{(B) }114\qquad\textbf{(C) }190\qquad\textbf{(D) }210\qquad\textbf{(E) }380$ | 2,943 | 8,228 | {"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": 69, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2019-51 | longest | en | 0.625367 |
http://nrich.maths.org/9623 | 1,502,935,249,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886102819.58/warc/CC-MAIN-20170817013033-20170817033033-00641.warc.gz | 326,188,040 | 5,356 | # Recording Mathematics
## Recording Mathematics
This collection of activities offers an opportunity for you to focus on children's own creative representations and recordings. Read the article to find out why we have selected these particular tasks. Click the 'Submit a solution' link on any live problem to find out the closing date.
### Primary Children's Mathematical Recording
##### Stage: Early years, 1 and 2
This article for teachers outlines different types of recording, depending on the purpose and audience.
### What Was in the Box?
##### Stage: 1 Challenge Level:
This big box adds something to any number that goes into it. If you know the numbers that come out, what addition might be going on in the box?
##### Stage: 1 Challenge Level:
There are three baskets, a brown one, a red one and a pink one, holding a total of 10 eggs. Can you use the information given to find out how many eggs are in each basket?
##### Stage: 1 Challenge Level:
If you put three beads onto a tens/ones abacus you could make the numbers 3, 30, 12 or 21. What numbers can be made with six beads?
### The Amazing Splitting Plant
##### Stage: 1 Challenge Level:
Can you work out how many flowers there will be on the Amazing Splitting Plant after it has been growing for six weeks?
### School Fair Necklaces
##### Stage: 1 and 2 Challenge Level:
How many possible necklaces can you find? And how do you know you've found them all?
### What's in the Box?
##### Stage: 2 Challenge Level:
This big box multiplies anything that goes inside it by the same number. If you know the numbers that come out, what multiplication might be going on in the box?
### Money Bags
##### Stage: 2 Challenge Level:
Ram divided 15 pennies among four small bags. He could then pay any sum of money from 1p to 15p without opening any bag. How many pennies did Ram put in each bag?
### Dice in a Corner
##### Stage: 2 Challenge Level:
How could you arrange at least two dice in a stack so that the total of the visible spots is 18?
### Domino Square
##### Stage: 2, 3 and 4 Challenge Level:
Use the 'double-3 down' dominoes to make a square so that each side has eight dots. | 500 | 2,171 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2017-34 | latest | en | 0.919161 |
https://openstax.org/books/college-physics-2e/pages/27-glossary | 1,716,425,938,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058575.96/warc/CC-MAIN-20240522224707-20240523014707-00611.warc.gz | 384,422,042 | 97,018 | Skip to ContentGo to accessibility pageKeyboard shortcuts menu
College Physics 2e
# Glossary
axis of a polarizing filter
the direction along which the filter passes the electric field of an EM wave
birefringent
crystals that split an unpolarized beam of light into two beams
Brewster’s angle
$θb=tan−1n2n1,θb=tan−1n2n1,$ where $n2n2$ is the index of refraction of the medium from which the light is reflected and $n1n1$ is the index of refraction of the medium in which the reflected light travels
Brewster’s law
$tanθb=n2n1tanθb=n2n1$, where $n1n1$ is the medium in which the incident and reflected light travel and $n2n2$ is the index of refraction of the medium that forms the interface that reflects the light
coherent
waves are in phase or have a definite phase relationship
confocal microscopes
microscopes that use the extended focal region to obtain three-dimensional images rather than two-dimensional images
constructive interference for a diffraction grating
occurs when the condition $dsinθ=mλ(for m=0,1,–1,2,–2,…)dsinθ=mλ(for m=0,1,–1,2,–2,…)$ is satisfied, where $dd$ is the distance between slits in the grating, $λλ$ is the wavelength of light, and $mm$ is the order of the maximum
constructive interference for a double slit
the path length difference must be an integral multiple of the wavelength
contrast
the difference in intensity between objects and the background on which they are observed
destructive interference for a double slit
the path length difference must be a half-integral multiple of the wavelength
destructive interference for a single slit
occurs when $Dsinθ=mλ,(form=1,–1,2,–2,3,…)Dsinθ=mλ,(form=1,–1,2,–2,3,…)$, where $DD$ is the slit width, $λλ$ is the light’s wavelength, $θθ$ is the angle relative to the original direction of the light, and $mm$ is the order of the minimum
diffraction
the bending of a wave around the edges of an opening or an obstacle
diffraction grating
a large number of evenly spaced parallel slits
direction of polarization
the direction parallel to the electric field for EM waves
horizontally polarized
the oscillations are in a horizontal plane
Huygens’s principle
every point on a wavefront is a source of wavelets that spread out in the forward direction at the same speed as the wave itself. The new wavefront is a line tangent to all of the wavelets
incoherent
waves have random phase relationships
interference microscopes
microscopes that enhance contrast between objects and background by superimposing a reference beam of light upon the light emerging from the sample
optically active
substances that rotate the plane of polarization of light passing through them
order
the integer $mm$ used in the equations for constructive and destructive interference for a double slit
phase-contrast microscope
microscope utilizing wave interference and differences in phases to enhance contrast
polarization
the attribute that wave oscillations have a definite direction relative to the direction of propagation of the wave
polarization microscope
microscope that enhances contrast by utilizing a wave characteristic of light, useful for objects that are optically active
polarized
waves having the electric and magnetic field oscillations in a definite direction
Rayleigh criterion
two images are just resolvable when the center of the diffraction pattern of one is directly over the first minimum of the diffraction pattern of the other
reflected light that is completely polarized
light reflected at the angle of reflection $θbθb$, known as Brewster’s angle
thin film interference
interference between light reflected from different surfaces of a thin film
ultraviolet (UV) microscopes
microscopes constructed with special lenses that transmit UV rays and utilize photographic or electronic techniques to record images
unpolarized
waves that are randomly polarized
vertically polarized
the oscillations are in a vertical plane
wavelength in a medium
$λ n=λ/nλ n=λ/n$, where $λλ$ is the wavelength in vacuum, and $nn$ is the index of refraction of the medium
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© Jan 19, 2024 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University. | 1,129 | 4,846 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 20, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2024-22 | latest | en | 0.871417 |
http://www.chegg.com/homework-help/questions-and-answers/forceplatform-tool-used-analyze-performance-athletes-bymeasuring-vertical-force-athlete-ex-q211549 | 1,474,799,750,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738660181.83/warc/CC-MAIN-20160924173740-00190-ip-10-143-35-109.ec2.internal.warc.gz | 373,070,134 | 13,848 | A forceplatform is a tool used to analyze the performance of athletes bymeasuring the vertical force that the athlete exerts on the groundas a function of time. Starting from rest, a 65kg athletejumps down onto the platform from a height of .006m. Whileshe is in contact with the platform durning the time interval0<t<.8s, the force she exerts on it is described as thefunction:
F=(9200N/s)t - (11500 N/s^2)t^2
A) Whatimpulse did the athlete recieve from the platform?
B) Withwhat speed did she reach the platform
C) Withwhat speed did she leave it?
D) Towhat height did she jump upon leaving the platform? | 158 | 608 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-40 | latest | en | 0.939756 |
http://conversion.org/mass/milligram/bag-portland-cement | 1,660,451,053,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571993.68/warc/CC-MAIN-20220814022847-20220814052847-00296.warc.gz | 9,707,813 | 6,909 | # Milligram to bag (Portland cement) conversion
Conversion number between Milligram [mg] and bag (Portland cement) is 2.3453432147327 × 10-8. This means, that Milligram is smaller unit than bag (Portland cement).
### Contents [show][hide]
Switch to reverse conversion:
from bag (Portland cement) to Milligram conversion
### Enter the number in Milligram:
Decimal Fraction Exponential Expression
[mg]
eg.: 10.12345 or 1.123e5
Result in bag (Portland cement)
?
precision 0 1 2 3 4 5 6 7 8 9 [info] Decimal: Exponential:
### Calculation process of conversion value
• 1 Milligram = (exactly) (0.000001) / (42.63768278) = 2.3453432147327 × 10-8 bag (Portland cement)
• 1 bag (Portland cement) = (exactly) (42.63768278) / (0.000001) = 42637682.78 Milligram
• ? Milligram × (0.000001 ("kg"/"Milligram")) / (42.63768278 ("kg"/"bag (Portland cement)")) = ? bag (Portland cement)
### High precision conversion
If conversion between Milligram to kilogram and kilogram to bag (Portland cement) is exactly definied, high precision conversion from Milligram to bag (Portland cement) is enabled.
Decimal places: (0-800)
Milligram
Result in bag (Portland cement):
?
### Milligram to bag (Portland cement) conversion chart
Start value: [Milligram] Step size [Milligram] How many lines? (max 100)
visual:
Milligrambag (Portland cement)
00
102.3453432147327 × 10-7
204.6906864294655 × 10-7
307.0360296441982 × 10-7
409.381372858931 × 10-7
501.1726716073664 × 10-6
601.4072059288396 × 10-6
701.6417402503129 × 10-6
801.8762745717862 × 10-6
902.1108088932595 × 10-6
1002.3453432147327 × 10-6
1102.579877536206 × 10-6
Copy to Excel
## Multiple conversion
Enter numbers in Milligram and click convert button.
One number per line.
Converted numbers in bag (Portland cement):
Click to select all
## Details about Milligram and bag (Portland cement) units:
Convert Milligram to other unit:
### Milligram
Definition of Milligram unit: =1x10-6 kg.
Convert Bag (Portland cement) to other unit:
### bag (Portland cement)
Definition of bag (Portland cement) unit: ≡ 94 lb av . = 94 x 0.45359237 kg
← Back to Mass units | 650 | 2,119 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2022-33 | latest | en | 0.662553 |
https://math.answers.com/Q/6_is_what_percent_of_50 | 1,638,821,487,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363312.79/warc/CC-MAIN-20211206194128-20211206224128-00159.warc.gz | 452,246,016 | 71,232 | 0
# 6 is what percent of 50?
Wiki User
2009-11-29 22:55:38
6 is 12% of 50.
Wiki User
2009-11-29 22:55:38
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Q: 6 is what percent of 50? | 127 | 268 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2021-49 | latest | en | 0.787848 |
https://www.aqua-calc.com/calculate/food-calories/substance/coffee-blank-creamer-coma-and-blank-upc-column--blank-050000787005 | 1,566,623,351,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027319724.97/warc/CC-MAIN-20190824041053-20190824063053-00443.warc.gz | 704,159,298 | 7,699 | # Calories of COFFEE CREAMER, UPC: 050000787005
## food calories and nutrients calculator
### compute calories and nutrients of generic and branded foods
#### See how many calories in0.1 kg (0.22 lbs) ofCOFFEE CREAMER, UPC:050000787005
From kilocalories(kcal) kilojoule(kJ) Carbohydrate 0 0 Fat 0 0 Protein 0 0 Other 233 974.87 Total 233 974.87
Nutrient (find foodsrich in nutrients) Unit Value /100 g BasicAdvancedAll Proximates Energy kcal 233 Protein g 0 Total lipid (fat) g 10 Carbohydrate,bydifference g 33.33 Sugars, total g 33.33 Minerals Sodium, Na mg 33 Lipids Fatty acids,totalsaturated g 0 Fatty acids,totalmonounsaturated g 6.67 Fatty acids,totalpolyunsaturated g 0 Cholesterol mg 0
#### Weight
gram 100 ounce 3.53 kilogram 0.1 pound 0.22 milligram 100 000
• About COFFEE CREAMER, UPC: 050000787005
• COFFEE CREAMER, UPC: 050000787005 contain(s) 233 calories per 100 grams or ≈3.527 ounces [ price ]
• Ingredients: WATER, SUGAR, VEGETABLE OIL (HIGH OLEIC SOYBEAN AND/OR HIGH OLEIC CANOLA), AND LESS THAN 2% OF SODIUM CASEINATE (A MILK DERIVATIVE), NATURAL AND ARTIFICIAL FLAVORS, MONO- AND DIGLYCERIDES, DIPOTASSIUM PHOSPHATE, CELLULOSE GEL, CELLULOSE GUM, CARRAGEENAN.
• The calories and nutrients calculator answers questions like these: How many nutrients (amino acids, lipids, minerals, proximates and vitamins) in a selected food per given weight? How many total calories (calories from carbohydrates, fats and proteins) in a selected food per given weight? The total number of calories and amount of nutrients are calculated based on the selected food and its given weight, and using the USDA Food Composition Databases. Visit our food calculations forum for more details.
• A few foods with a name containing, like or similar to COFFEE CREAMER, UPC: 050000787005:
• COFFEE-MATE, LITE COFFEE CREAMER, THE ORIGINAL, UPC: 050000301324 weigh(s) 101.44 gram per (metric cup) or 3.39 ounce per (US cup), and contain(s) 500 calories per 100 grams or ≈3.527 ounces [ weight to volume | volume to weight | price | density ]
• COFFEE-MATE, SUGAR FREE COFFEE CREAMER, FRENCH VANILLA, UPC: 050000848119 contain(s) 100 calories per 100 grams or ≈3.527 ounces [ price ]
• Beverages, coffee, instant, vanilla, sweetened, decaffeinated, with non dairy creamer contain(s) 465 calories per 100 grams or ≈3.527 ounces [ price ]
• COFFEE CREAMER, UPC: 688267086861 weigh(s) 88.76 gram per (metric cup) or 2.96 ounce per (US cup), and contain(s) 571 calories per 100 grams or ≈3.527 ounces [ weight to volume | volume to weight | price | density ]
• NONDAIRY COFFEE CREAMER, UPC: 035826064943 weigh(s) 84.54 gram per (metric cup) or 2.82 ounce per (US cup), and contain(s) 600 calories per 100 grams or ≈3.527 ounces [ weight to volume | volume to weight | price | density ]
• Reference (ID: 144519)
• USDA National Nutrient Database for Standard Reference; National Agricultural Library; United States Department of Agriculture (USDA); 1400 Independence Ave., S.W.; Washington, DC 20250 USA.
#### Foods, Nutrients and Calories
PLANTAIN CHIPS, UPC: 019061190612 contain(s) 500 calories per 100 grams or ≈3.527 ounces [ price ]
TOASTED HEMP SEEDS, UPC: 697658983108 weigh(s) 169.07 gram per (metric cup) or 5.64 ounce per (US cup), and contain(s) 567 calories per 100 grams or ≈3.527 ounces [ weight to volume | volume to weight | price | density ]
Foods high in Methionine
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CaribSea, Freshwater, Super Naturals, Peace River weighs 1 489.72 kg/m³ (93.00018 lb/ft³) with specific gravity of 1.48972 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Balsa, Light weighs 92.11 kg/m³ (5.75024 lb/ft³) [ weight to volume | volume to weight | price | density ]
Volume to weightweight to volume and cost conversions for Refrigerant R-124, liquid (R124) with temperature in the range of -40°C (-40°F) to 82.23°C (180.014°F)
#### Weights and Measurements
A meter per minute squared (m/min²) is a derived metric SI (System International) measurement unit of acceleration
The fuel consumption or fuel economy measurement is used to estimate gas mileage and associated fuel cost for a specific vehicle.
Convert stone per metric tablespoon to tonne per US fluid ounce or convert between all units of density measurement
#### Calculators
Ellipse calculator, equations, area, vertices and circumference | 1,285 | 4,530 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2019-35 | latest | en | 0.648689 |
https://plainmath.net/71007/h-s-frac-s-s-s-the-inverse | 1,653,450,684,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662578939.73/warc/CC-MAIN-20220525023952-20220525053952-00022.warc.gz | 484,426,246 | 14,341 | # H(s)=\frac{s-5}{(s-3)(s-1)} The inverse Laplace transform of H(s) is equal to f*g
$H\left(s\right)=\frac{s-5}{\left(s-3\right)\left(s-1\right)}$
The inverse Laplace transform of H(s) is equal to
$f\cdot g$
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Note that we can write
$F\left(s\right)=\frac{s-5}{s-3}=1-\frac{2}{s-3}$
Then, $f\left(t\right)=\delta \left(t\right)-2{e}^{3t}u\left(t\right)$ and
$h\left(t\right)={\int }_{0}^{t}{e}^{t-{t}^{\prime }}\left(\delta \left({t}^{\prime }\right)-2{e}^{3{t}^{\prime }}\right){dt}^{\prime }$
$={e}^{t}-{e}^{t}\left({e}^{2t}-1\right)$
$=2{e}^{t}-{e}^{3t}$
Note that partial fraction expansion makes things easier. We simply write
$H\left(s\right)=\frac{2}{s-1}-\frac{1}{s-3}$ | 365 | 947 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 26, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2022-21 | latest | en | 0.670798 |
https://discourse.automationgame.com/t/guide-suspension-crash-course/2908 | 1,656,122,112,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103033925.2/warc/CC-MAIN-20220625004242-20220625034242-00311.warc.gz | 251,848,605 | 10,586 | # Guide: Suspension Crash Course
[size=200]Suspension Crash Course[/size]
Hey guys,
I thought I might put together a short suspension crash course for those who might not have too much knowledge about that part of car design. The full “Beginner’s Guide to Car Design” still is quite a long way off (that will come after the Car Designer is feature complete), so this can shorten the wait. I hope you enjoy!
What I want to explain:
[ul]]which tests are simulated/:m]
]what the different graphs say/:m]
]what the settings in the suspension tab do/:m][/ul]
I hope it helps with better understanding of what is happening and maybe even with finding bugs. Feel free to ask here if something remains unclear!
[size=150]Tests and Graphs[/size]
Steady state circle test - Yaw rate graph
This tests simulates a car on a skidpad which is driving in a circle with a radius of 50 meters. It starts very slowly and gets faster step by step. Depending on how the driver has to change the steering wheel angle at the various speeds, it is possible to determine if a car is understeering, oversteering or neutral. If the driver needs to steer more, the car is understeering, if he has to steer less, the car is oversteering.
Neutral is represented by the yellow line in the yaw rate graph. Everything below is "under"steering, everything above is "over"steering.
Out of this test you get the maximum lateral acceleration (g-Forces) and, after future updates, an influence on how sporty or tame your car is (depending on over- or understeer).
Oversteer:
abload.de/thumb/overtvjzt.png
Understeer:
abload.de/thumb/underhrkj1.png
Body roll test
The body roll test is similar to the moose or fishhook test. The maximum lateral acceleration out of the circle test is taken into account: The car is driving a corner with that acceleration in one direction and pulling to the other direction after a short time. The focus is on the roll angle of the car body. This will first reach positive numbers and then negative ones, levelling off at a certain value.
Out of this test you get the maximum absolute roll angle and if that becomes too big and things like track width, height of center of gravity, roll inertia and suspension roll centres are coming together in a unfortunate way, the car can roll over. Currently, this is quite difficult to achieve, as the fake total weight slider moves down the center of gravity quite a bit.
Big maximum roll angles will be rated not tame and not sporty.
abload.de/thumb/rollzajva.png
Body bump test
In this test, the car is driving over a short, 2 cm high obstacle. This bump goes from the road through the wheel and the suspension into the body. Both parallel wheel travel and opposite wheel travel on every axle are simulated (so to stiff sway bars will make the suspension utterly hard for the opposite wheel travel test), but only parallel wheel travel is shown in the graphs.
The movement of front and rear tires is shown with the red and green lines, the body movement at the respective axle with the blue and yellow lines.
Depending on how much of the bump is going into the body and how long it takes to stop moving again, the comfort rating will be good or bad in the future. The stress on the tires (depending on how much the tire moves up and down) will go into tameness. Higher profile tires will be softer and absorb more of the bump than low profile tires.
abload.de/thumb/bumpl5jbo.png
[size=150]Suspension Settings[/size]
Camber °]
Negative camber can increase the potential side forces of the tires of one axle but will make the tires wear faster. If you have an understeering car and want to make it more neutral, try giving it more negative camber at the front. If you have an oversteering car, give it more negative camber at the rear.
The effect of this can only be seen in the yaw rate graph and the maximum lateral acceleration.
Springs [N/m]
Harder springs make the car roll less but less comfortable in the bump test. Also, softer springs at the front axle make it understeer less and softer at the rear make it oversteer less.
The effect of this can be seen in all of the three mentioned tests and graphs.
Dampers [Ns/m]
Dampers try to “dampen” fast body roll and fast wheel travel. So harder dampers will cause less body roll and less stress on the tires in the bump test, but in the car you will notice more of the bump because the suspension gets stiffer.
Effects can be seen in the roll and the bump test, not in the circle/yaw rate test, as this is a static maneuvre.
Sway Bars [Nm/rad]
Sway bars make your car roll less by forcing the wheel on the one side of the respective axle to go into the same direction as the wheel on the other side of the axle. This means your suspension will get stiffer at the opposite wheel travel (body bump) test and therefore less comfortable. In addition to that you can influence over- or understeer: Stiffer sway bars will weaken the axle where they are installed. So making the front sway bar harder will make the car understeer, making the rear sway bar harder will make the car oversteer.
Effects can be seen in all three of the mentioned tests.
Ride Height [mm]
Ride height (together with suspension geometry -> roll centres) has an influence on how much the body rolls. So more ride height will give higher maximum roll angles. In the circle test you might sometimes want a bit more roll angle: The more the body rolls, the more the outside wheels will travel upwards and the more negative camber they can build up (this only is right for certain suspension types, others can behave the opposite way!).
Effects of this can be seen in the roll test, the circle test (and also in acceleration and braking tests, as it influences dynamic weight distribution.
Total Weight [kg]
Higher total weight will make the car more comfortable as the body barely moves when the wheels travel up and down. But it might roll more and will become worse at cornering. The car will probably even need better, stiffer tires to carry the load.
Effects of this can be seen in all three suspension tests as well as in acceleration, braking and fuel economy.
Weight Distribution
Weight distribution has a major affect on over- and understeer, it’s nearly as important as the tire parameters. A front-heavy car will barely be able to oversteer (in static maneuvres as the circle test) and vice versa. Also it determines how your spring stiffnesses have to be distributed between front and rear for a smooth ride. It can affect the body roll, too, if different suspensions are fitted to the front and rear axle.
Effects of this can be seen in all three suspension tests as well as in acceleration and braking (dynamic weight distribution).
1 Like
Perfect! Thanks for this. Had a fair idea of what does what from the youtube video but this certainly clarifies things
Good post! Just made me wiser again with this game!
LOL I thought that if I made the dampening stiffness higher the person would feel less road imperfections
How do i make the supsension so the driver feels less road imperfections
1. Read the guide
2. Understand the guide
3. ???
4. Profit!
Also: make springs and dampers softer.
Thank you soo much! This has really helped me get smooth riding cars. I do have some questions that I didn’t really see answered in this (maybe they’re too advanced?).
My questions:
1. Can I get a reference as to the body bump? Say 0 is Rolls Royce and 4 is a Scania or Peterbilt?
2. What would be an appropriate city car roll angle? Sports cars?
3. What would be an average cornering rating for a city car? I usually aim for 0.85-0.9 g (using hard, long life or medium compound tires).
1. The tires go over a 2 cm bump and then you see how mouch the body moves at the front and the rear. The less it moves the more comfy it is. It’s difficult to give you and absolute value because it depends a lot on body weight, unsprung mass and tire profile. Perfect would be 0 body movement, but that’s impossible and would ruin other car stats such as cornering. With the next update you will get a comfort rating which includes the results from the bump test. Then you can just optimise the comfort rating.
2. For both cases: The less, the better. If you want to get it close to 0, you will very probably ruin your comfort rating (hard springs, dampers and anti roll bars). So the right compromise has to be found. Up to 5 degrees should be quite fine.
3. 0.8 to 1.0 g is fine. In the next update tires will have more grip so that common high profile tires are better in corners than now. The difference between high profile and low profile tires will be reduced a bit, too.
I’m really looking forward to this next update. Will we be getting this car here (if you don’t mind me asking)?
It’s currently not even in closed beta, so I guess not. But maybe Andrew finds the time to work on it to bring it back. There are other more important things to do though.
Ah that sucks, guess I’ll have to wait on that one. There’s supposed to be a lot of cars in this update, right? I’m pretty sure I’ll find one in there that’ll be just as nice, if not better. This update will also help me get accurate weight ratings down because as it stands I keep building 2500 lb (give or take 300 lbs) cars.
Technically, you do - up to a point - then it’s downhill all the way from there…
Great guide, many thanks!
This guide is very helpful. I finally built a car that makes 1.00g on the skidpad. Now to make a floppy comfy couch car.
Even with this guide i’m having great difficulty trying to make an understeering car oversteer. Generally RWD cars would tend to oversteer, and yet all those I build understeer no matter how I set the suspension. Perhaps I’ll try a lighter engine for now.
To do this, make sure of a couple things.
1. Make sure your rear tires aren’t a ton wider than your fronts. (This will make the most difference)
2. Decrease your rear camber/increase your front camber
3. Increase your rear stiffness
4. Don’t have too much downforce on the rear.
I know this works, oversteer was one of the big problems I had in the rally championship.
[quote=“gt1cooper”]
To do this, make sure of a couple things.
1. Make sure your rear tires aren’t a ton wider than your fronts. (This will make the most difference)
2. Decrease your rear camber/increase your front camber
3. Increase your rear stiffness
4. Don’t have too much downforce on the rear.
I know this works, oversteer was one of the big problems I had in the rally championship.[/quote]
Thanks dude, I’ll give it a try. I did manage to make one before I read this: a 595hp, 500+ torque, 6.9 V8 car with 285mm wide rears haha.
The yaw rate graph is measured without drive torque, i.e. off-power. You won’t notice a difference between FWD, RWD and AWD (except secondary effects coming from a different weight distribution for example). Front heavy cannot be made oversteer very easily. Mid- and rear-engined cars will tend to oversteer a lot more.
Nice! this helped me making cars that understeer a little less, understeer was my weak point in the BRC…
Thanks | 2,559 | 11,104 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2022-27 | latest | en | 0.918384 |
https://amezomovies.com/how-to-find-outliers-with-iqr-using-python-built-in/ | 1,680,422,553,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296950422.77/warc/CC-MAIN-20230402074255-20230402104255-00613.warc.gz | 132,323,581 | 16,315 | # How to Find Outliers With IQR Using Python – Built In
Every data set has issues, or points that don’t make sense. These points, referred to as outliers, can either show issues in the data collection process or real phenomena that are not representative of what typically happens. Here are a few examples of outliers that I’ve seen in real data sets:
Including outliers in your data analysis skews your data set and negatively impacts the results of your analysis. Therefore it’s important to make sure your data set excludes all outliers, and only uses the realistic data.
Let’s talk about how to do that using IQR (interquartile ranges).
Before talking through the details of how to write Python code removing outliers, it’s important to mention that removing outliers is more of an art than a science. You need to carefully determine what is an outlier and what is not based on the context of your project. Here are a few examples using the outliers described above:
More From Peter GrantHow to Use Float in Python (With Sample Code!)
With that word of caution in mind, one common way of identifying outliers is based on analyzing the statistical spread of the data set. In this method you identify the range of the data you want to use and exclude the rest. To do so you:
Here’s a Python-based example using NumPy to exclude the highest and lowest five percent of all data points from a data set.
That code yields the following outputs:
The first line in the above code imports the NumPy package for use in the analysis process. If you want to do data science with Python I recommend getting very, very familiar with NumPy.
The second line sets a seed for NumPy’s randomization code, telling it to return the same quasi-random numbers every time.
The third line creates a new variable, `random_data`, which is an array of 100 random values between `0` and `1`
The fourth line is where the magic starts. That line calls the NumPy percentile function to identify the value of the data at the ninety-fifth percentile and the fifth percentile, then stores those values in the `p95` and `p5` variables. These values set the bound which will later be used to limit the data set.
The next two lines are print statements showing what’s happening. The first prints `p95`and `p5`so that you can see the values at those percentiles. You can see they’re quite close to 95 percent and five percent of the upper range of the data set which, in a non-normal data set, is what we expect. The second prints the length of `random_data`, showing that it still contains the 100 values that were originally entered.
The following two lines both reduce the data set based on the bounds specified above. The first line modifies `random_data` to only include the values that are less than `p95`, and the following line adjusts it again to include only values that are greater than `p5`. In this way, the data set is reduced to include only values within the bounds set by the fifth and ninety fifth percentiles of the data set. This reduces the data set to 90 percent of the total values, and is equivalent to stating the largest and smallest five percent are all outliers.
The final line prints the length of `random_data` after modification, and we can see that it’s now reduced to 90 data points as expected.
A Deeper Dive Into OutliersHow to Find Outliers (With Examples)
In order to limit the data set based on the percentiles you must first decide what range of the data set you want to keep. One way to examine the data is to limit it based on the IQR. The IQR is a statistical concept describing the spread of all data points within one quartile of the average, or the middle 50 percent range. The IQR is commonly used when people want to examine what the middle group of a population is doing. For instance, we often see IQR used to understand a school’s SAT or state standardized test scores.
When using the IQR to remove outliers you remove all points that lie outside the range defined by the quartiles `+/- 1.5 * IQR`. For example, consider the following calculations.
The following code shows an example of using IQR to identify and remove outliers.
Which returns the following outputs:
The code rejecting outliers using IQR has is different from the prior example code in the following ways:
The outputs show that the code follows the same processes with the new requirements. It prints that the third quartile is at approximately`0.68`, and the first quartile is at approximately `-0.67`. Given that NumPy’s `standard_normal` function uses a standard deviation of `1`, these numbers are almost exactly as expected. The code then prints that the total data set holds 100,000 points. The IQR is identified at `1.34`, which leads to upper and lower bounds of `2.69` and `-2.68`. Filtering the code to only values within those two thresholds yields a data set of 99,249 points, indicating that 751 were outside of that range and removed.
And that’s how you do it! You can now think about how to identify outliers through both a practical and statistical approach, how to write generic code to remove outliers and how to evaluate your data set using the common interquartile range metric.
Built In’s expert contributor network publishes thoughtful, solutions-oriented stories written by innovative tech professionals. It is the tech industry’s definitive destination for sharing compelling, first-person accounts of problem-solving on the road to innovation.
source | 1,156 | 5,462 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-14 | latest | en | 0.896759 |
https://la.mathworks.com/matlabcentral/answers/756789-horizontal-line-in-plot-up-to-a-graph | 1,618,884,092,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038921860.72/warc/CC-MAIN-20210419235235-20210420025235-00621.warc.gz | 457,859,195 | 23,559 | # Horizontal line in plot up to a graph
24 views (last 30 days)
Mepe on 26 Feb 2021
Commented: Mepe on 26 Feb 2021
How can I generate a horizontal line starting at x = 0 up to a graph (see figure, red line)? The x, y coordinates are of course available for the graph.
KSSV on 26 Feb 2021
If you have the xlimits (x0,x1) and the y-value..the required equations is y = k.
x = 1:10 ;
k = 3 ;
y = k*ones(*Size(x)) ;
plot(x,y)
Also have a look on line.
##### 2 CommentsShowHide 1 older comment
Mepe on 26 Feb 2021
The value x1 should be freely chosen. From the point of intersection with the function (x, y) the line should run horizontally to x = 0. | 202 | 647 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2021-17 | latest | en | 0.827733 |
http://acrophobic.me/post/oscillators-circuit-diagrams-archives | 1,566,474,162,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027317113.27/warc/CC-MAIN-20190822110215-20190822132215-00367.warc.gz | 12,502,120 | 8,810 | # oscillators circuit diagrams archives
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Oscillators Circuit Diagrams Archives Circuit Diagrams Oscillators Circuit Diagrams Archives Circuit Diagramz is a website for electronics and schematics circuit diagrams. Oscillators Circuit Diagrams Archives Page 2 of 7 ... Oscillators Circuit Diagrams Archives Page 2 of 7 Circuit Diagramz is a website for electronics and schematics circuit diagrams. Transistors Tutorial Free Electric Circuits Textbooks Transistors Tutorial Part 7: "Oscillators" ... Figure 1 is a simple block diagram showing an amplifier and a block representing the many oscillator ... Because of the inherently high Q or frequency selectivity of LC networks or resonant tank circuits, LC oscillators produce more precise sinewave outputs even when the loop gain of the circuit ... Oscillators – delabs Schematics – Electronic Circuits Electronic Designs and Schematic Diagrams. Oscillators. Square and Triangle Opamp Oscillator. 2018 12 17. By: delabs. On: December 17, 2018. In: Opamp Circuits. Tagged: Oscillators. Here is the circuit of a Op Amp based Square Wave Generator. One of the main application of this is in a Simple PWM circuit and Triangle Opamp Oscillator ... Schematics of delabs Circuit Diagrams: Crystal ... The circuit above is a parallel resonant oscillator circuit. The Crystal works by the piezoelectric principle, piezo means pressure. The electric field causes the impedance of the crystal to change. The LP Record Player needle is the reverse of this, the bumps on the spiral groove of the record applies pressure to needle which generates ... Oscillator Circuits – Page 10 – Circuit Wiring Diagrams Oscillator and signal generator circuits diagram. Tuned Collector Oscillator. Posted by Circuit Diagram in Oscillator Circuits. The basic circuit of a tuned collector oscillator is shown in figure. It is called the tuned collector oscillator, because the tuned circuit is connected to the collector. ... To know the basics of oscillators, its ... Positive Feedback Amplifiers (Oscillators) LC and Crystal ... Tuned Oscillators (Radio Frequency Oscillators) 17 Tuned oscillator is a circuit that generates a radio frequency output by using LC tuned (resonant) circuit. Because of high frequencies, small inductance can be used for the radio frequency of oscillation. Tuned input and tuned output Oscillator tuned output tuned input feedback coupling ci RF ... Oscillator Circuits – Page 2 – Circuit Wiring Diagrams When we use the value of parts as schematic diagram of wien bridge oscillator circuit will have frequencies about 20 Hz – 22.5 Khz by adjust P1 and maximum harmonic distortion approximately 2 %. This circuit use dual voltage positive and negative so we use the voltage converter circuit as Figure 2 input is 9V to 4.5V, Ground, 4.5V. Oscillator Circuits Oregon State University C. Tuned Oscillator Circuits Tuned Oscillators use a parallel LC resonant circuit (LC tank) to provide the oscillations. There are two common types: • Colpitts – The resonant circuit is an inductor and two capacitors. •Hartley– The resonant circuit is a tapped inductor or two inductors and one capacitor. Folkscanomy Electronics: Books on Electronics, Circuits ... Folksonomy: A system of classification derived from the practice and method of collaboratively creating and managing tags to annotate and categorize content; this practice is also known as collaborative tagging, social classification, social indexing, and social tagging. Coined by Thomas Vander...
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# Lamp A flashes every 6 seconds, lamp B flashes every 8 seconds, lamp
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Lamp A flashes every 6 seconds, lamp B flashes every 8 seconds, lamp [#permalink]
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05 Jan 2016, 12:46
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Lamp A flashes every 6 seconds, Lamp B flashes every 8 seconds, Lamp C flashes every 10 seconds. At a certain instant of time all three lamps flash simultaneously. During the period of 2 minutes after that how many times will exactly two lamps flash?
A) 5
B) 6
C) 9
D) 12
E) 15
I've found different answer than an official one. What do you think, guys? What would be your answer?
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Re: Lamp A flashes every 6 seconds, lamp B flashes every 8 seconds, lamp [#permalink]
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05 Jan 2016, 18:41
2
lamp A flashes every 6 secs,
b flashes every 8 secs,
c flashes every 10 secs.
At certain time three flashes simultaneously.
after that calculate when the three lamps flashes simultaneously in 2 minutes.
take a L.C.M of three 6,8,10 sec then it is 120 secs.
so exactly after 2 minutes it flashes simultaneoulsy.
question stem asks exactly two lamps falshes .
take lcm of a and b .6 and 8 is 24 secs. so two lamp flashes 5times in 2minutes.
lcm of b and c . 8 and 10 secs is 40 secs. so both flashes 3times in 2minutes.
lcm of c and a ,6 and 10 secs is 30 secs . pair flashes 4times..
all three pairs flashes simultaenoulsy at 2 minutes. we have to minus one from all three pairs.
then 12-3=9.
so option C is correct.
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Re: Lamp A flashes every 6 seconds, lamp B flashes every 8 seconds, lamp [#permalink]
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05 Jan 2016, 20:55
4
TeymurHajiyev wrote:
Lamp A flashes every 6 seconds, Lamp B flashes every 8 seconds, Lamp C flashes every 10 seconds. At a certain instant of time all three lamps flash simultaneously. During the period of 2 minutes after that how many times will exactly two lamps flash?
A) 5
B) 6
C) 9
D) 12
E) 15
I've found different answer than an official one. What do you think, guys? What would be your answer?
You can easily enumerate this since 120 secs is a small time frame.
Exactly two lamps can flash in three ways:
Lamps A and B but not C: LCM of 6 and 8 is 24. It shouldn't be divisible by 10
24, 48, 72, 96, 120 (ignore - divisible by 10)
Lamps B and C but not A: LCM of 8 and 10 is 40. It shouldn't be divisible by 6
40, 80, 120 (ignore - divisible by 6)
Lamps A and C but not B: LCM of 6 and 10 is 30. It shouldn't be divisible by 8
30, 60, 90, 120 (ignore - divisible by 8)
There are total 9 cases.
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Re: Lamp A flashes every 6 seconds, lamp B flashes every 8 seconds, lamp [#permalink]
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06 Jan 2016, 21:03
TeymurHajiyev wrote:
Lamp A flashes every 6 seconds, Lamp B flashes every 8 seconds, Lamp C flashes every 10 seconds. At a certain instant of time all three lamps flash simultaneously. During the period of 2 minutes after that how many times will exactly two lamps flash?
A) 5
B) 6
C) 9
D) 12
E) 15
I've found different answer than an official one. What do you think, guys? What would be your answer?
Hi TeymurHajiyev,
Since you mentioned coming across at least one DIFFERENT answer, what was THAT answer? Did it include an explanation?
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Posts: 8296
Re: Lamp A flashes every 6 seconds, lamp B flashes every 8 seconds, lamp [#permalink]
### Show Tags
06 Jan 2016, 21:59
3
TeymurHajiyev wrote:
Lamp A flashes every 6 seconds, Lamp B flashes every 8 seconds, Lamp C flashes every 10 seconds. At a certain instant of time all three lamps flash simultaneously. During the period of 2 minutes after that how many times will exactly two lamps flash?
A) 5
B) 6
C) 9
D) 12
E) 15
I've found different answer than an official one. What do you think, guys? What would be your answer?
Hi,
since 120 secs is too small a time, one can work out each time two lights are on...
lets see the way/procedure incase the time frame is much more..
we will solve this Q by that procedure..
we have three different time at which the bulbs are on.. 6,8,10..
the solution is based on LCM concept..
1) we will get the entire time in same units, that is minutes..
2) take LCM of all three, this along with its multiples till the total time, will be the time all three will be on simulatenously..
to find how many times all three are on simultaneously, divide the TOTAL TIME by LCM, the INTEGER value will give us the number, say A..
3) Next would be to take LCM of each pair of time and divide total time by each of these three LCMs. Again the integer value will give us the time two blink together..
4) Since we are asked exactly 2 of them blink together, we subtract A, all three blinking together, from each of the three answers we get from (3) above.
5) Add all three new values and we have the answer....
lets use this in this Q..
1) total time=120 secs..
2) LCM of 6,8,10=120.. times=120/120=1... so all three will switch on together only once , A=1..
3) three pairs..
6,8.. LCM=24...times=120/24=5
6,10.. LCM=30..times=120/30=4
8,10.. LCM=40..times=120/40=3
4) three sets of two blinking together
5-1=4
4-1=3
3-1=2
5) total times= 4+3+2=9
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Re: Lamp A flashes every 6 seconds, lamp B flashes every 8 seconds, lamp [#permalink]
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07 Jan 2016, 00:58
Thank you guys: Psiva00734, VeritasPrepKarishma and chetan2u, now it's much clearer for me.
Dear Rich, I had answer 5, which was based on wrong LCM of 20, 15 and 12 (60 times). I messed up with "seconds" and "times", but now I feel better
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Lamp A flashes every 6 seconds, lamp B flashes every 8 seconds, lamp [#permalink]
### Show Tags
05 Feb 2020, 19:22
TeymurHajiyev wrote:
Lamp A flashes every 6 seconds, Lamp B flashes every 8 seconds, Lamp C flashes every 10 seconds. At a certain instant of time all three lamps flash simultaneously. During the period of 2 minutes after that how many times will exactly two lamps flash?
A) 5
B) 6
C) 9
D) 12
E) 15
I've found different answer than an official one. What do you think, guys? What would be your answer?
This 3 lamps will flash simultaneously flash every 2 minutes. Lamp A and B will flash together at every 24 seconds, Lamp B and C will flash together at every 40 seconds, and Lamp A and C will do at every 30 seconds. So, total number of times during the period of 2 minutes when 2 lamps will flash together is (120/24) + (120/30)+(120/40) - 3 (at 120 second 3 lamps will flash simultaneously, as this is counted in each of the 3 cases, we have to deduct it) = 5 + 4 +3 - 3= 9.
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Re: Lamp A flashes every 6 seconds, lamp B flashes every 8 seconds, lamp [#permalink]
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05 Feb 2020, 20:39
TeymurHajiyev wrote:
Lamp A flashes every 6 seconds, Lamp B flashes every 8 seconds, Lamp C flashes every 10 seconds. At a certain instant of time all three lamps flash simultaneously. During the period of 2 minutes after that how many times will exactly two lamps flash?
A) 5
B) 6
C) 9
D) 12
E) 15
I've found different answer than an official one. What do you think, guys? What would be your answer?
What is the source of this question? It does not seem like a GMAT official question.
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Re: Lamp A flashes every 6 seconds, lamp B flashes every 8 seconds, lamp [#permalink]
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09 Feb 2020, 05:20
1
TeymurHajiyev wrote:
Lamp A flashes every 6 seconds, Lamp B flashes every 8 seconds, Lamp C flashes every 10 seconds. At a certain instant of time all three lamps flash simultaneously. During the period of 2 minutes after that how many times will exactly two lamps flash?
A) 5
B) 6
C) 9
D) 12
E) 15
I've found different answer than an official one. What do you think, guys? What would be your answer?
Since the LCM of 6, 8, and 10 is 120, the lamps will flash at the same time every 120 seconds, or every 2 minutes. Furthermore, the LCM of 6 and 8 is 24, that of 8 and 10 is 40, and that of 6 and 10 is 30. Therefore, lamps A and B will flash every 24 seconds, B and C every 40 seconds, and A and C every 30 seconds.
In the 120-second, or 2-minute, period, A and B will flash 5 times, at 24, 48, 72, 96, and 120 seconds; B and C will flash 3 times, at 40, 80, and 120 seconds; and A and C will flash 4 times, at 30, 60, 90, and 120 seconds.
Thus, in the 2-minute period, two lamps will flash 5 + 3 + 4 = 12 times, but we cannot count the flashing of all three lights at 120 seconds, as this does not satisfy the event “EXACTLY TWO lamps flash.” Since this event was counted 3 times, our final answer must be 12 - 3 = 9.
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Re: Lamp A flashes every 6 seconds, lamp B flashes every 8 seconds, lamp [#permalink] 09 Feb 2020, 05:20
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# Lamp A flashes every 6 seconds, lamp B flashes every 8 seconds, lamp
new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne | 3,266 | 11,294 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2020-16 | latest | en | 0.860994 |
https://www.airmilescalculator.com/distance/hsv-to-lws/ | 1,632,030,009,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056711.62/warc/CC-MAIN-20210919035453-20210919065453-00636.warc.gz | 669,966,097 | 81,646 | # Distance between Huntsville, AL (HSV) and Lewiston, ID (LWS)
Flight distance from Huntsville to Lewiston (Huntsville International Airport – Lewiston–Nez Perce County Airport) is 1772 miles / 2851 kilometers / 1540 nautical miles. Estimated flight time is 3 hours 51 minutes.
Driving distance from Huntsville (HSV) to Lewiston (LWS) is 2309 miles / 3716 kilometers and travel time by car is about 39 hours 4 minutes.
## Map of flight path and driving directions from Huntsville to Lewiston.
Shortest flight path between Huntsville International Airport (HSV) and Lewiston–Nez Perce County Airport (LWS).
## How far is Lewiston from Huntsville?
There are several ways to calculate distances between Huntsville and Lewiston. Here are two common methods:
Vincenty's formula (applied above)
• 1771.702 miles
• 2851.277 kilometers
• 1539.567 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 1768.671 miles
• 2846.401 kilometers
• 1536.934 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## Airport information
A Huntsville International Airport
City: Huntsville, AL
Country: United States
IATA Code: HSV
ICAO Code: KHSV
Coordinates: 34°38′13″N, 86°46′30″W
B Lewiston–Nez Perce County Airport
City: Lewiston, ID
Country: United States
IATA Code: LWS
ICAO Code: KLWS
Coordinates: 46°22′28″N, 117°0′53″W
## Time difference and current local times
The time difference between Huntsville and Lewiston is 2 hours. Lewiston is 2 hours behind Huntsville.
CDT
PDT
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 198 kg (436 pounds).
## Frequent Flyer Miles Calculator
Huntsville (HSV) → Lewiston (LWS).
Distance:
1772
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
1772
Round trip? | 529 | 2,023 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2021-39 | latest | en | 0.824881 |
https://www.visithowto.com/how-to-measure-amperage/ | 1,656,442,346,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103573995.30/warc/CC-MAIN-20220628173131-20220628203131-00018.warc.gz | 1,136,027,355 | 89,224 | # How To Measure Amperage
If you’re working on an electrical project, you may need to check the amperage or how much electricity is flowing through the circuit.If you want to know if something is pulling more energy than it should, you might need to measure the Amps.If you’re trying to determine if a component in your vehicle is draining the battery, measuring the amperage can be useful.If you use safety around electrical components and have a multimeter, it’s easy to measure Amps.
### Step 1: If you want to know the maximum power of your battery or breaker, you need to check it.
Before attaching your multimeter to the circuit, you need to make sure it is rated for the number of Amps traveling through that circuit.You can find the maximum Amp on the back of the device or in the instruction manual for most power sources.Do not attempt to test more currents than the highest dial setting if you want to check how high the dial goes.The maximum current is also known as the maximum Amps.
### Step 2: If your multimeter isn’t rated high enough for the circuit, use a plug-in clamp.
The accessory can extend the range.Plug the leads into the multimeter and attach the other end to the circuit in the way you would attach a multi meter.The hot or live wire is usually black, red, blue, or some other color except white or green.When you are using a multimeter, it doesn’t become part of the circuit.
### Step 3: Push the probe into the sockets on the multimeter.
The probes that came with the meter should have a red and black one.One end of the probe has a plug in it.The black probe should always be plugged into the COM sockets.If the port isn’t marked with “COM”, you may see a negative symbol.When you measure the current, you’ll have to hold your leads in place.Attach them to the circuit, freeing up your hands.Both types of probes will connect to the meter in the same way.
### Step 4: The red probe should be placed in the sockets labeled “A.”
Depending on the functions of your meter, you may have several places where you can plug in a red probe.The port is marked with the letter A.There are two sockets with an A, one labeled as 10A and the other labeled mA, both of which are designed to measure current up to 10 Amp.To make sure you don’t overload the meter, select the higher A or 10A setting if you’re not sure which one to use.You could see ports labeled V for voltage or for ohms.You can ignore these for the test.
### Step 5: The meter has either AC or DC current on it.
Unless your meter is designed to only be used on AC or DC circuits, you need to choose which one you’re testing.If you’re not sure, check your power source again for that information.The voltage should be listed as well.AC, or alternating current, is used in items like household appliances and electric motors, while direct current is utilized in battery-powered motors and devices.Unless there’s a transformer that converts electricity to DC, the power in a residential setting is going to be AC.
### Step 6: The dial should be turned to an amplifier setting that’s higher than what you’re measuring.
If you want to test the maximum currents, you should look for a dial on your meter that is higher than the number.If you want to be safe, you can turn the dial all the way to the maximum, but you may not get a reading if the current is too low.You need to turn the dial down if that happens.If the current is stronger than you thought, setting your meter to handle more Amps will help protect against blowing the fuses.You could destroy the meter if the current is too high.You don’t have to manually adjust a dial if the dial is auto-ranging.If this is the case, you won’t see a dial with the settings, and the meter will be labeled as auto-ranging, or you can see it on the display.
### Step 7: The power to the circuit needs to be turned off.
If your circuit is powered by a battery, it’s a good idea to take the negative lead out.If you have to turn off the power at a breaker, you should turn the switch off.Attach the meter to the circuit.
### Step 8: The power supply has a red wire.
To complete a circuit, you need to test the amount of current flowing through it.The first thing to do is to shut off the power to the circuit and then remove the positive wire from the source.You may have to cut the wire with wire clippers to break the circuit.If you see a cap where the power source meets the wire going to the device you’re testing, you can remove it and untangle the wires from around each other.You can connect the wires by clips.The black wire doesn’t need to be plugged in.The black is negative in an alternating current circuit, while the hot wire is positive.
### Step 9: The ends of the wire should be removed.
You’ll need to wrap a small piece of wire around the multimeter prongs or expose enough wire to make them secure.If the wire is insulated all the way to the end, you can squeeze your wire clippers just enough to cut into the rubber insulation.Pull the clippers away from you to remove the insulation.If you accidentally cut into the wire, you can try again.You need to remove the end of the wire from the power source and the device you’re testing.
### Step 10: Wrap the positive wire around the multimeter.
Wrap the exposed end of the red wire that’s coming away from the power source around the multimeter probe or use alligator clips to connect it to the wire.Attach the wire securely to make sure you get an accurate reading.The meter just needs to complete the circuit if you connect the positive probe to the power source or device.It’s fine if you can attach the wires the other way.Attaching the positive wire first will help prevent a short if the negative wire touches a ground.If the reading has a negative sign in front of it, it means you put the leads on backwards.Fix it by reversing the leads.
### Step 11: Turn on the circuit if you connect the multimeter probe to the remaining wire.
Attach the positive wire from the electrical component you’re trying to test to the multimeter probe.When you touch the black probe to the wire, the power will be restored.Turn the power back on if you turned it off.This is the end of the wire you disconnected from the power supply.If you’re testing in a car, don’t start the car or try to turn on any fans, lights or anything else, as you could overload the meter.
### Step 12: As you read the meter, leave the probes in place.
You should see a number on the digital display when the meter is in place.This is the amount of current you have.For the most precise measurement, leave the probes on the circuit for at least 60 seconds to make sure the current is stable.If the reading is less than the sensitive setting, the meter should be disconnected and the red probe moved to a different area.The current of the circuit you’re testing will be shown in this reading.That’s the amount of electricity that can flow through that current. | 1,494 | 6,942 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.92217 |
https://oeis.org/A244760/internal | 1,582,720,555,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146342.41/warc/CC-MAIN-20200226115522-20200226145522-00277.warc.gz | 477,217,497 | 3,302 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A244760 a(n) = Sum_{k=0..n} C(n,k) * (1 + 3^k)^(n-k) * 2^k. 3
%I
%S 1,4,24,232,3840,111904,5785344,529662592,85449338880,24204383609344,
%T 11986829259362304,10361640102119729152,15589910824599107174400,
%U 40815393380277274447519744,185575767151388880816233447424,1465910356757779350231777997914112
%N a(n) = Sum_{k=0..n} C(n,k) * (1 + 3^k)^(n-k) * 2^k.
%H G. C. Greubel, <a href="/A244760/b244760.txt">Table of n, a(n) for n = 0..89</a>
%F E.g.f.: Sum_{n>=0} exp((1+3^n)*x) * (2*x)^n/n!.
%F O.g.f.: Sum_{n>=0} (2*x)^n/(1 - (1+3^n)*x)^(n+1).
%F a(n) ~ c * 3^(n^2/4) * 2^((3*n+1)/2) / sqrt(Pi*n), where c = sum_{k=-inf..+inf} 1/(3^(k^2) * 2^k) = 1.88621563508001862566062... if n is even, and c = sum_{k=-inf..+inf} 1/(3^((k+1/2)^2) * 2^(k+1/2)) = 1.88659407336643412717014... if n is odd. - _Vaclav Kotesovec_, Jan 25 2015
%e E.g.f.: A(x) = 1 + 4*x + 24*x^2/2! + 232*x^3/3! + 3840*x^4/4! + 111904*x^5/5! +...
%e ILLUSTRATION OF INITIAL TERMS:
%e a(1) = (1+3^0)^1 + (1+3^1)^0*2 = 4;
%e a(2) = (1+3^0)^2 + 2*(1+3^1)^1*2 + (1+3^2)^0*2^2 = 24;
%e a(3) = (1+3^0)^3 + 3*(1+3^1)^2*2 + 3*(1+3^2)^1*2^2 + (1+3^3)^0*2^3 = 232;
%e a(4) = (1+3^0)^4 + 4*(1+3^1)^3*2 + 6*(1+3^2)^2*2^2 + 4*(1+3^3)^1*2^3 + (1+3^4)^0*2^4 = 3480; ...
%t Table[Sum[Binomial[n,k] * (1 + 3^k)^(n-k) * 2^k,{k,0,n}],{n,0,20}] (* _Vaclav Kotesovec_, Jan 25 2015 *)
%o (PARI) {a(n) = sum(k=0,n,binomial(n,k) * (1 + 3^k)^(n-k)*2^k )}
%o for(n=0,25,print1(a(n),", "))
%o (PARI) /* E.g.f. Sum_{n>=0} exp((1+3^n)*x)*(2*x)^n/n! */
%o {a(n)=n!*polcoeff(sum(k=0, n, exp((1+3^k)*x +x*O(x^n))*(2*x)^k/k!), n)}
%o for(n=0,25,print1(a(n),", "))
%o (PARI) /* O.g.f. Sum_{n>=0} (2*x)^n/(1 - (1+3^n)*x)^(n+1): */
%o {a(n)=polcoeff(sum(k=0, n, (2*x)^k/(1-(1+3^k)*x +x*O(x^n))^(k+1)), n)}
%o for(n=0,25,print1(a(n),", "))
%Y Cf. A244755, A244756, A244754, A243918.
%K nonn
%O 0,2
%A _Paul D. Hanna_, Jul 05 2014
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Last modified February 26 07:17 EST 2020. Contains 332277 sequences. (Running on oeis4.) | 1,189 | 2,402 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2020-10 | latest | en | 0.378951 |
https://www.gradesaver.com/textbooks/math/calculus/calculus-with-applications-10th-edition/chapter-4-calculating-the-derivative-4-2-derivatives-of-products-and-quotients-4-2-exercises-page-216/6 | 1,563,858,359,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195528869.90/warc/CC-MAIN-20190723043719-20190723065719-00388.warc.gz | 694,193,275 | 12,346 | ## Calculus with Applications (10th Edition)
${g^,}\,\left( t \right) = 36{t^3} + 24t$
$\begin{gathered} g\,\left( t \right) = \,{\left( {3{t^2} + 2} \right)^2} \hfill \\ Write\,\,as\,a\,\,product\,\, \hfill \\ g\,\left( t \right) = \left( {3{t^2} + 2} \right)\left( {3{t^2} + 2} \right) \hfill \\ Use\,\,the\,\,product\,\,rule\,\,to\,\,find\,\,{g^,}\,\left( t \right) \hfill \\ {g^,}\,\left( t \right) = \,\left( {3{t^2} + 2} \right){\left( {3{t^2} + 2} \right)^,} + \left( {3{t^2} + 2} \right){\left( {3{t^2} + 2} \right)^,} \hfill \\ {g^,}\,\left( t \right) = 2\left( {3{t^2} + 2} \right){\left( {3{t^2} + 2} \right)^,} \hfill \\ Then \hfill \\ {g^,}\,\left( t \right) = 2\left( {3{t^2} + 2} \right)\,\left( {6t} \right) \hfill \\ Multiplying \hfill \\ {g^,}\,\left( t \right) = 12t\left( {3{t^2} + 2} \right) \hfill \\ {g^,}\,\left( t \right) = 36{t^3} + 24t \hfill \\ \end{gathered}$ | 456 | 889 | {"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} | 4.46875 | 4 | CC-MAIN-2019-30 | latest | en | 0.487755 |
https://www.thestudentroom.co.uk/showthread.php?t=2004507 | 1,493,557,946,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917125532.90/warc/CC-MAIN-20170423031205-00487-ip-10-145-167-34.ec2.internal.warc.gz | 950,161,617 | 37,727 | You are Here: Home >< Maths
# Quick trig question
1. how does the above =
tan2x = 1
2. Divide both sides by and you obtain
edit: when division by isn't allowed, i.e. when , it is clear that and
3. (Original post by IShouldBeRevising_)
how does the above =
tan2x = 1
It doesn't.
4. (Original post by hassi94)
It doesn't.
tan 2x = -1 *
5. (Original post by IShouldBeRevising_)
tan 2x = -1 *
Divide both sides by cos2x.
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Top tips from students who have already aced their exams | 355 | 1,322 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2017-17 | longest | en | 0.933275 |
harris666.flex-driver.com | 1,642,987,551,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304345.92/warc/CC-MAIN-20220123232910-20220124022910-00000.warc.gz | 329,277,165 | 11,245 | # How to Interpret Scores for Predictions
How do we calculate a score for a prediction? The score is a logarithm of the probability. For example, if 80% of the population believes that an event will occur, the predicted outcome would have a score of -0.22 while 20% would have a score of -1.6. The objective is to maximize the score, which is a function of the number of possible outcomes. However, there are some caveats that should be kept in mind when interpreting a score.
When viewing the results, typically the logarithm of the particular probability of your occasion occurring is used. A positive log-odds report indicates that typically the event is even more likely than not. In other words, a good log-odds score indicates that the celebration is more probable than not. A higher score is a good sign. A low score is furthermore not necessarily bad, but is a lot more suitable for comparison purposes. A low quality score is not necessarily indicative of a bad model, but is more appropriate to comparing designs.
The log-odds score is an easy and convenient way to compare different foretelling of methods. It will be a score that represents the logarithm of the likelihood of an event, and compares it to be able to a null design. A high rating indicates that the event is even more likely than typically the null model. The low score, nevertheless, does not necessarily mean that the outcome is a great one. It’s just more precise than a null model.
In certain situations, a lower score does not really necessarily mean which a model is poor. It just indicates that the result is not really a good fit for your test. This is better to utilize a higher-quality score to compare different models. It is a indication of an insufficient model. If this isn’t, it’s most likely not. Then, again, a low score does not always mean a bad result.
Inside a statistical type, the scoring may be the numeric values in the results of a new statistical model. In time series versions, scoring is the numerical value of the particular observed data. Within a regression model, it could be the probability of an event. In typically the case of time series models, a new score can label the outcome of a test. That can refer to a numeric value or a probability. With regard to example, it might be the predicted associated with a great event.
In a time series model, the particular score refers in order to the probability regarding an event happening. In a category model, scoring pertains to the course or outcome associated with the test. In a new graphical model, credit scoring is the excess weight or value given to a info set as a new result of evaluation. In addition, it refers to the outcome regarding an event. The prediction of a specific analyze is based on the axis from the distribution.
The log-odds report is the quantity of anticipated events, divided by simply the number of variables. The log-odds scores are a logarithm from the probability regarding an event. Typically the axis from the log-odds is defined because the number of times the score will happen. The scores are produced from the values of the data points and therefore are correlated. Typically the predicted score is a way of measuring the probability in the occurrence of a specific event.
The log-odds score is usually a measure of the particular likelihood of a great event. The log-odds score is the logarithm of typically the probability of a good event. The higher typically the log-odds, the more likely a good event is to occur. The likelihood of the event is a factor of its magnitude, therefore it should be obtained into account. A good log-odds score signifies a positive relationship.
To maximize the expected reward, the actual probability of an occasion must be reported. Or else, some other probability may give a lower anticipated score. The logarithmic score is the only real type of conjecture that may be suited to this purpose. The above methods will help you determine which estimations have the best quality and are the majority of reliable. The top quality of a model will certainly affect the amount of predictions it could make. As well as the accuracy of the type, the reliability from the prediction should be high.
To be able to use a equipment learning model regarding predicting events, you should consider the type of data as well as file format. A good example may be the type associated with data you could have. A person can use a LUIS model in order to predict the possibility of an occasion occurring. Using the LUIS model can predict events by simply looking at the particular data. This 파라오 슬롯 is a useful technique for determining the likelihood of a celebration. A successful classification formula can identify numerous potential incidents, like a disaster. | 979 | 4,790 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2022-05 | longest | en | 0.934611 |
http://www.mathworks.com/help/symbolic/mupad_ref/graph-createrandomgraph.html?requestedDomain=www.mathworks.com&nocookie=true | 1,511,418,016,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806736.55/warc/CC-MAIN-20171123050243-20171123070243-00222.warc.gz | 420,757,632 | 13,527 | # Documentation
### This is machine translation
Translated by
Mouseover text to see original. Click the button below to return to the English verison of the page.
# `Graph`::`createRandomGraph`
Generates a random graph.
MATLAB live scripts support most MuPAD functionality, though there are some differences. For more information, see Convert MuPAD Notebooks to MATLAB Live Scripts.
## Syntax
```Graph::createRandomGraph(`VertexNr`, `EdgeNr`, <`Directed | Undirected`>)
```
## Description
`Graph::createRandomGraph` generates a random graph.
`Graph::createRandomGraph(VertexNr, EdgeNr)` generates a random graph with VertexNr vertices and EdgeNr edges.
### Note
If the number EdgeNr is too great (i.e. ), a complete graph will be created.
`Graph::createRandomGraph(VertexNr, EdgeNr, Undirected)` generates a random graph with VertexNr vertices and 2 EdgeNr edges is created. This is due to the fact that no odd number of undirected edges could be created otherwise.
## Examples
### Example 1
The following graph was created `randomly`, meaning that your results will most probably differ:
```G := Graph::createRandomGraph(5,6): Graph::printGraphInformation(G)```
```Vertices: [1, 2, 3, 4, 5] Edges: [[1, 4], [2, 1], [2, 4], [2, 5], [3, 1], [3, 4]] Vertex weights: no vertex weights. Edge descriptions: no edge descriptions. Edge weights: no edge weights. Edge costs: no edge costs. Adjacency list (out): 1 = [4], 2 = [1, 4, 5], 3 = [1, 4], 4 = [], 5 = [] Adjacency list (in): 1 = [2, 3], 2 = [], 3 = [], 4 = [1, 2, 3], 5 = [2] Graph is directed. ```
### Example 2
The same number of vertices, but this time the edges are undirected (and therefore the number of Edges is (2 EdgeNr)). As you can clearly see, the edges differ from the edges created above:
```G := Graph::createRandomGraph(5, 6, Undirected): Graph::printGraphInformation(G)```
```Vertices: [1, 2, 3, 4, 5] Edges: [[1, 4], [1, 5], [2, 3], [2, 4], [2, 5], [3, 2], [4, 1], [4, 2], [4\ , 5], [5, 1], [5, 2], [5, 4]] Vertex weights: no vertex weights. Edge descriptions: no edge descriptions. Edge weights: no edge weights. Edge costs: no edge costs. Adjacency list (out): 1 = [4, 5], 2 = [3, 4, 5], 3 = [2], 4 = [1, 2, 5], 5\ = [1, 2, 4] Adjacency list (in): 1 = [4, 5], 2 = [3, 4, 5], 3 = [2], 4 = [1, 2, 5], 5 \ = [1, 2, 4] Graph is undirected. ```
### Example 3
If the number of edges to be created extends the possible limit (), a complete graph will be returned:
```G := Graph::createRandomGraph(3, 6, Undirected): Graph::printGraphInformation(G)```
```Warning: Unable to produce the required number of edges. Creating a complete graph instead. [Graph::createRandomGraph] ```
```Vertices: [1, 2, 3] Edges: [[1, 2], [1, 3], [2, 1], [2, 3], [3, 1], [3, 2]] Vertex weights: no vertex weights. Edge descriptions: no edge descriptions. Edge weights: no edge weights. Edge costs: no edge costs. Adjacency list (out): 1 = [2, 3], 2 = [1, 3], 3 = [1, 2] Adjacency list (in): 1 = [2, 3], 2 = [1, 3], 3 = [1, 2] Graph is undirected. ```
## Parameters
`VertexNr` Positive integer `EdgeNr` Positive integer
## Options
`Directed` If `Directed` is stated, a directed Graph is created `Default` `Undirected` If `Undirected` is stated, an undirected Graph is created.
`Graph` | 1,053 | 3,254 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2017-47 | latest | en | 0.789294 |
http://www.statmodel.com/discussion/messages/12/171.html?1314230863 | 1,500,849,055,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424623.68/warc/CC-MAIN-20170723222438-20170724002438-00649.warc.gz | 558,125,027 | 14,007 | 3-level analysis by HLM and Mplus
Message/Author
bmuthen posted on Tuesday, March 26, 2002 - 10:39 am
A frequent question is how 3-level modeling of growth of students within schools compares to HLM-type analysis. Here are some answers.
Actually, these analyses are one and the same. The 3-level HLM formulation describes across-time variation at level 1 involving random intercept and slope coefficients, level 2 studies the across-student variation of those coefficients, while level 3 studies their across-school variation. In the latent variable context, level 1 and level 2 are combined into the "Within" part of the model, i.e. the part describing variation across students. The "Between" part describes across-school variation and corresponds to level 3 of HLM. In this way, the 3-level HLM model is turned into a 2-level latent variable model. The fact that level 1 and level 2 are both considered in the Within part of the latent variable model is due to viewing the level 1 across-time variation as a multivariate observation vector rather than as a univariate repeated observation. Level 1 of the Within part is the latent variable measurement model part and level 2 of the Within part is the structural part.
So in summary, Mplus estimates a random coefficient model and the Between component intercept and growth coefficients are representing variation across clusters in the coefficients defined for Within (students), just as in HLM analysis. If you use the ML estimator you get the same estimates.
aboabdulmalik posted on Thursday, September 29, 2005 - 10:05 pm
Hi,
I am examining the mediational effect of (COLL) on the relationship between CL and TS. I have the predictor CL at the school level, the mediator COLL at the school level, and the outcome variable, TS, at the individual level. Thus, it is 2 > 2 > 1 (i.e., CL > COLL > TS; with a direct path from the predictor to the outcome as well).
I am only familiar with the software HLM, through my independent reading. I wanted to examine the mediational effect but i was not able to regress the mediational variable on the predictor variable. Regressing the mediator on the predictor is one essential equation to examine mediational models. Can i examine this equation using OLS through SPSS regression. The rationale is that both the mediator and the predictor variables are both at the school level. While for other equations, i am using HLM to regress the outcome on the predictor and the outcome on the mediator.
If i use SPSS regression to regress the mediator on the predictor, how comparable the coefficients produced by SPSS and HLM. Is there any implimintation required while running the regression. Or, is there any function within the software HLM to run a mediational modeling. I know that MPlus enables to examine the mediational modeling, but my limited time and limited statistics will not help me. The best i can use is the simple 2-level HLM.
I appreciate your help.
Another question is related to examining the assumptions before using HLM. I checked my data in terms of outliers, normality, and linearity through SPSS. I followed Tabachnick and Fidell's 2001) discussion of regression assumptions. My question, then, is there any examinations that are specific to HLM and how to run these examination. Can you refer me to any published work with this regard.
Thank you
bmuthen posted on Friday, September 30, 2005 - 9:15 am
I think the easiest way to get the correct estimates and particularly the correct standard errors is to use Mplus for this modeling. That would make teh analysis very straightforward.
Marc Reis posted on Monday, November 21, 2005 - 12:36 pm
Hello,
I would like to use Mplus to estimate the effect of the group-level variable organizational climate on an individual-level outcome. Organizational climate is measured with multiple individual-level (continous) indicators, so there are two scources of error variance for the aggregated organizational climate score for each organizational unit: the variation among the items (due to the fact that they are not perfectly reliable) and among the members of each organizational unit (assuming that there is a "true score" for each group). Since three-level-modeling is not yet available, are there any other ways to estimate the model?
Many thanks for any suggestions!
Linda K. Muthen posted on Monday, November 21, 2005 - 6:39 pm
Multiple indicators do not count as a level in Mplus because it takes a multivariate approach to multilevel modeling. In Mplus this would be a two-level model if I am understanding you correctly.
Marc Reis posted on Tuesday, November 22, 2005 - 2:46 am
My problem is that organizational climate is defined as a latent group-level-variable that influences a latent individual-level variable. The idea was to specify the following model:
%within%
DV_within BY i1-i6;
%between%
ORG_CLIMATE BY i7-i12;
DV_between BY i1-i6;
DV_between ON ORG_CLIMATE;
I'am unsure about how Mplus treats i7-i12 in this case? Would ORG_CLIMATE be a "correct measure" of the latent group-level variable? (e.g. would it be reasonable to save a factor score for each group?)
Linda K. Muthen posted on Tuesday, November 22, 2005 - 7:30 am
You would specify the model as you have done above. You would also have BETWEEN = i7-i12; in the VARIABLE command.
Marc Reis posted on Tuesday, November 22, 2005 - 1:15 pm
I tried to specify BETWEEN = i7-i12; in the VARIABLE command, but Mplus doesn´t allow group-level variables to have within-group variation. Note that i7-i12 are individual level indicators that I would like to aggegrate to the group level. So maybe there's another way or I made a mistake.
bmuthen posted on Tuesday, November 22, 2005 - 4:36 pm
If i7-i12 are scores on individuals, then you don't put those variables on the Between = list. And, you might want to use a within-level factor, say
%Within%
wORG by i7-i12;
%Between%
bORG by i7-i12;
This assumes that you are interested in the within structure of these variables as well as the between structure (and it may not be the same).
Another alternative is to simply aggregate each variable to the between level, i.e. creating cluster means, and then treat these as Between = variables with only between-level variation - and then specify the between-level factor model you have.
Anonymous posted on Wednesday, November 23, 2005 - 1:39 am
Hello,
I have an idea about the above discussion, maybe this helps. In general there might be two reasons to specify a model with latent variables: The indicators are no perfect measures (--> error variance) and they usually do not measure the factor to the same extent (--> different factor loadings). So requesting factor scores means actually weighting the indicators, doesn´t it?
The question is whether it is necessary to weight the members of a group to obtain a better estimate of the group-level variable. Assuming that there is no special sampling procedure, I don´t see a rationale to weight one group member more than others. So from this point of view, it would be reasonable to simply compute the cluster mean as Bengt suggested, maybe based on the individual-level factor scores. But I am unsure whether to compute the factor scores based on the original covariance matrix or the dissaggregated within covariance matrix.
Maybe someone would like to comment...
bmuthen posted on Wednesday, November 23, 2005 - 6:37 pm
Comments are invited regarding this. Concerning whether to compute factor scores based on the original cov matrix or within cov matrix, I would say that if factor scores are needed for a multilevel setting, you are better off getting factor scores from a multilevel factor analysis model.
chantanee posted on Thursday, August 10, 2006 - 2:19 am
The purpose of my study was to find out the relationships among multilevel variables, student variables, classroom variables, and school variables, effected on student science learning achievement of Thai upper secondary school students. The study consisted of 3 sub objectives: (1) to identify student variables directly effected on science learning achievement (2) to identify the direct influences of classroom variables and cross level interaction between classroom variables and student variables effected on student science learning achievement, and (3) To identify the direct influences of school variables and cross level interaction between school variables and classroom variables or between school variables and student variables effected on student science learning achievement.
The sample of the study employed multi-stage random sampling. It consisted of 3 groups: (1) 132 school administrators: principals, assistant principals and heads of science department from 44 public upper secondary schools in Thailand, (2) 132 science teacher who taught Chemistry, Biology and Physics in the classroom, from 88 classroom samples ( 2 classrooms per school), and (3) 2,488 Grade 11 science students.
If it’s possible, could you kindly give me some advice for these question?
1. Are the results effectiveness to report?
2. Are two classrooms per school powerful enough for employing HLM?
Bengt O. Muthen posted on Thursday, August 10, 2006 - 6:36 pm
This sounds like multilevel modeling would work well - you have enough schools and classrooms. 2 classrooms per school is a bare minimum which does not allow you to study many classroom variables. Mplus does not currently handle this 3-level model.
Jeff Cookston posted on Thursday, October 19, 2006 - 2:23 pm
We have a dataset that includes three waves of data collected on a number of family context constructs. LGM has been our preferred method of analysis, but we're now planning to collect 2 days of cortisol samples from our participants with multiple cortisol samples each day. Most prior studies that use cortisol data tend to model the daily hormone patterns using HLM. I'm confused how we can use our LGMs of the family context constructs to predict the HLM based hormone patterns. Could the cortisol data be modeled in LGM and then used as a dual process predicted by family context?
Linda K. Muthen posted on Thursday, October 19, 2006 - 4:27 pm
Are you referring to HLM the program or hierarchical linear modeling in general?
Jeff Cookston posted on Thursday, October 19, 2006 - 8:33 pm
Sorry, hierarchical linear modeling in general.
Linda K. Muthen posted on Friday, October 20, 2006 - 8:02 am
The SEM and HLM growth models differ in two basic ways. One is the treatment of time scores. In SEM, they are treated as parameters in the model. In HLM, they are treated as data. The second is the treatment of time-varying covariates. The regression coefficients are fixed in SEM and random in HLM. Mplus can have time scores as parameters or data and can have fixed or random coefficients for time-varying covariates. So I think you should be okay.
Frank Gallo posted on Sunday, August 09, 2009 - 8:02 pm
Dear Dr. Muthen
I am a beginner with Mplus. I am using Mplus Version 5.21. I have stratified data: police arrests (n =3,300) within police departments (n = 16) that serve community population levels (n = 4). The DV police force is continuous. I have a mixture (nominal, ordinal, ratio) of 21 covariates at level 1 and none at levels 2 and 3. Community levels are fixed effects. Would the multilevel modeling features of Mplus handle these data? Thank you.
Best regards,
Frank
Linda K. Muthen posted on Monday, August 10, 2009 - 6:44 am
It sounds like you have a two-level cross-sectional model which can be estimated in Mplus. The problem I see is that you have only 16 police departments. It is usually recommended to have a minimum of 30 clusters.
Utkun Ozdil posted on Thursday, December 16, 2010 - 11:28 am
Hi,,
I collected my data from a university's three faculties (Faculty of Education, Faculty of Engineering, and Faculty of Arts and Sciences). In each of these faculties were involved second, third, and fourth grade undergraduates. So, I have students nested within grade levels within faculties. This led me to analyze a three-level model.
Does MPlus handle such data analysis as mine or is the HLM program more appropriate?
Thanks...
Utkun
Linda K. Muthen posted on Thursday, December 16, 2010 - 2:49 pm
Mplus does not currently have three-level cross-sectional models. HLM does. Your data, however, are not suitable for multilevel modeling given that faculty and grade cannot be considered random modes.
Jing Zhang posted on Wednesday, August 24, 2011 - 3:35 pm
Dear Dr. Muthen,
In your post dated on March 26, 2002, you talked about how to deal with 3-level modeling of growth of students within schools. You said that level 1 and level 2 are combined into the "Within" part of the model by viewing the level 1 across-time variation as a multivariate observation vector rather than as a univariate repeated observation. My question is that:
Does this mean that the data in long format will not work, and the data has to be changed to wide format if it is in long format?
Thanks,
Jing
Linda K. Muthen posted on Wednesday, August 24, 2011 - 5:07 pm
Yes.
Andrea M Reina Tamayo posted on Monday, September 22, 2014 - 8:12 am
I am working on a three-level CFA model (unbalanced data). The method I used was ESM to examine the variability of a continues variable. Because the method is so intense, I used three items to capture the construct.
Person gave ratings on these items three times a day, for five days. Therefore, moments were nested within days, and days within people. These are the 3 levels.
I gave a unique ID to every person, and this unique ID appears in the data 15 times per person.
Example below for one person a bit of a second person. M stands for moment.
ID Day M X1 X2 X3
1234 1 1 1 6 7
1234 1 2 2 4 6
1234 1 3 3 . .
1234 2 1
1234 2 2
1234 2 3
1234 3 1
1234 3 2
1234 3 3
1234 4 1
1234 4 2
1234 4 3
1234 5 1
1234 5 2
1234 5 3
1567 1 1
1567 1 2
1567 1 3
1567 2 1
When I run my analysis MPlus gives me a warning message:
*** WARNING
Clusters for DAY with the same IDs have been found in different clusters
for RESP_NR. These clusters are assumed to be different because clusters for
DAY are not allowed to appear in more than one cluster for RESP_NR.
I want to know what does this warning message mean? Can I trust my output with it, or is it affecting my results?
I would appreciate your help!
Linda K. Muthen posted on Monday, September 22, 2014 - 9:39 am
Please send the full output and your license number to support@statmodel.com.
Luo Wenshu posted on Tuesday, March 24, 2015 - 7:42 pm
Hi Dr. Muthen,
I am using MPlus to run a 2-level HLM model and have the following questions.
1) Is there a default setting for centering of predictors, grandmean or groupmean?
2) For random slopes, if we find they are not statistically significant, does this mean we do not need to build level 2 model with predictors for these random slopes and just turn to fixed model for slopes?
3) Do we need to allow random intercepts and slopes correlated at Level 2?
Thank you very much.
Bengt O. Muthen posted on Wednesday, March 25, 2015 - 7:36 am
1) The default is no centering.
2) You may still find significant influence of level-2 predictors on the random slopes.
3) Yes.
Luo Wenshu posted on Thursday, March 26, 2015 - 1:25 am
Thank you very much Dr. Muthen,
For the correlations among random intercepts and slopes at Level 2, what is the default setting in MPlus? It seems that the corrlelations are fixed to be zeros by default.
Bengt O. Muthen posted on Thursday, March 26, 2015 - 8:23 am
It depends on the analysis setting. You see in the output what is done in each case. If the covariance is not there, add it.
Melody Kung posted on Thursday, August 06, 2015 - 12:28 pm
Hi Drs. Muthen,
I am running a 2-level model with 2 independent, latent, between-level variables.
Using example of 9.12 in the manual as a guide, I specified "within" and "between" variable names in the VARIABLES section and then defined the latent variables and their indicators in the MODEL section, followed by the "%WITHIN%" and "%BETWEEN%" statements. The example does not include latent variables, whereas my model does.
Can I include latent variables in the %BETWEEN% statements? When I try to do so, the error message that shows up states that the two latent variables in the BETWEEN option are unknown, even though I specified the latent variables in the MODEL section.
I hope my question makes sense. Thanks for any help you can provide.
Linda K. Muthen posted on Thursday, August 06, 2015 - 1:48 pm
Please send the output and your license number to support@statmodel.com.
Cynthia Yuen posted on Tuesday, February 09, 2016 - 1:56 pm
Hello,
I have a longitudinal study where teens completed daily diaries for 14 days each year for 3 years. Thus, days are nested within years which are nested within individuals. I am interested in whether an individual level variable (e.g., sex) moderates certain slopes at the daily level (e.g., conflict --> distress within a day). I'm having trouble figuring out the appropriate analysis to run -- would this be a three-level model such as in example 9.20 of the version 7.6 guidebook?
Thank you!
Bengt O. Muthen posted on Tuesday, February 09, 2016 - 6:21 pm
Having only 3 years on Level-2 is too few for a 3-level analysis. Perhaps you could do a 2-level analysis of days within subject and let year be represented by dummy covariates.
Cynthia Yuen posted on Thursday, February 11, 2016 - 7:41 am
Thanks for the quick response! Would it be better to do something like 9.12 or 9.13 instead and model growth within a two-level model? Two of our main questions are whether the daily relations between events (e.g., conflict --> distress) change as teens age, and whether some individual-level characteristics like ethnicity predict how/whether these slopes change over time. Do you have any advice on how to appropriately model this?
Bengt O. Muthen posted on Friday, February 12, 2016 - 5:26 pm
I don't hear that you have a growth model situation but a regression of distress on conflict - where that regression may change over year (I assume, not over the days). If so, I would do a 2-level regression where level 1 is time (the 14 days) and level 2 is subject. Year can be level-2 and can predict the DV distress and perhaps the slope by creating Year*conflict and letting that influence distress. But it is a research question which I don't have enough background in your study to really answer.
Luo Wenshu posted on Friday, April 15, 2016 - 4:46 pm
Dear Dr. Muthen,
In 2-level HLM analysis (student and class level), I see from the users guide that for level 2 predictors, we may use observed (mx) or latent(x). I know the observed level 2 can be obtained by aggregating level 1 scores at the class level.
1) How is the latent Level 2 predictor calculated?
2)Which one is preferred?
3)If I have a level 1 predictor as Rasch measure (i.e.,latent variable), do I still need to use latent variable of the predictor at Level 2?
Thank you very much!
Bengt O. Muthen posted on Friday, April 15, 2016 - 5:43 pm
1-2. See the paper on our website:
Lüdtke, O., Marsh, H.W., Robitzsch, A., Trautwein, U., Asparouhov, T., & Muthén, B. (2008). The multilevel latent covariate model: A new, more reliable approach to group-level effects in contextual studies. Psychological Methods, 13, 203-229.
3. If the Rasch indicators vary over level-2 units, you should express a latent variable model on this level also - the indicators are then the random intercepts of each observed indicator.
Luo Wenshu posted on Friday, April 15, 2016 - 6:11 pm
Thank you for the quick response, Dr. Muthen!
I used latent level 2 predictors in my 2 level random slopes analysis. Using the default estimator MLR, I got the following warning message. Does this mean the result is not trustworthy. Do I need to turn to MLF estimator as suggested.
WARNING: THE MODEL ESTIMATION HAS REACHED A SADDLE POINT OR A POINT WHERE THE
OBSERVED AND THE EXPECTED INFORMATION MATRICES DO NOT MATCH.
AN ADJUSTMENT TO THE ESTIMATION OF THE INFORMATION MATRIX HAS BEEN MADE.
THE CONDITION NUMBER IS -0.157D-01.
THE PROBLEM MAY ALSO BE RESOLVED BY DECREASING THE VALUE OF THE
MCONVERGENCE OR LOGCRITERION OPTIONS OR BY CHANGING THE STARTING VALUES
OR BY USING THE MLF ESTIMATOR.
In addition, I found level 2 predictors had big standard errors. I suspect multicollinearity issue at level 2 because correlations at level 2 are usually higher than correlations at level 1 between the same set of variables. How to solve the problem?
Thank you very much Again!
Bengt O. Muthen posted on Sunday, April 17, 2016 - 4:05 pm
Try a smaller mconvergence value than the one you see in the analysis Summary.
If this doesn't help, send output and license number to Support.
Katrina Jia Lin posted on Sunday, September 04, 2016 - 1:25 am
Dear Dr. Muthen,
I am conducting a three-level path analysis and encountered this error message.
*** ERROR in DEFINE command
The GROUPMEAN specification for TYPE=THREELEVEL must include the name of
the cluster variable for the cluster means. Problem with:
PAAM CSP
The lines for the centering I used is
DEFINE:
CENTER PAAM CSP(GROUPMEAN);
The cluster for level 2 is SCODE, and for level 3 is CCODE. I understand I will need to add SCODE to the line but I am not sure how. Can you please advise on this?
Thank you in advance.
Linda K. Muthen posted on Sunday, September 04, 2016 - 6:57 am
Please send the full output and your license number to support@statmodel.com. | 5,238 | 21,583 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-30 | latest | en | 0.928733 |
https://www.mrexcel.com/board/threads/tricky-index-match-function-please-help.640306/ | 1,696,452,531,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511406.34/warc/CC-MAIN-20231004184208-20231004214208-00159.warc.gz | 957,916,495 | 20,361 | #### cameronWFA
##### New Member
Hey Y'all,
Here is a simplified example for the spreadsheet I am trying to work with. I don't know how to post the sheet on the forum so it is just attached as an image.
Essentially, I want the cell that says "FORMULA CELL" to check the contents of the cell under the collumn "TOP DIVIDEND PAYING STOCK" (The cell under that collumn that is in its same row), and match it to the name that goes across row 3 (C3:G3). Then, after it has matched that, I want it to display the numerical value of that located collumn for the same row as "FORMULA CELL" and also the stock name it looked up under TOP DIVIDEND PAYING STOCK.
I hope that made sense, it is actually not all that complicated of a task I don't think. Here is the formula I am using now which I cant get to work correctly so that I can drag it up and it finds the value for each stock as it looks each one up:
Code:
``=INDEX(C4:G9, 9,MATCH(B9,\$C\$3:\$G\$3), 0)``
So basically, it is
INDEX(range, [row9], [match b9 to c3:g3])
And then I would ideally want it to output 32, the value in E9. I am hoping to get this thing working so I can drag the cell up and it would perform the same index/match function for row 8,7,6,5,etc... But currently I get the output "#VALUE"
Thank you all so much in advance for all your help. You guys rock!
Best,
Cam
### Excel Facts
Last used cell?
Press Ctrl+End to move to what Excel thinks is the last used cell.
I don't follow ,ainly because I didn't see any image, but this will not work:
=INDEX(C4:G9, 9,MATCH(B9,\$C\$3:\$G\$3), 0)
because C4:G9 is 6 rows tall and you're requesting row 9 from those 6 rows. Not going to work.
INDEX(range, [row9], [match b9 to c3:g3])
The most important thing to understand is that
[row9] part is not a Row# persay.
It's an Index #.
The nth Row within the referenced range.
So you referenced range C4:G9
Trying to go to the 9th row.
But there are only 6 rows (4 5 6 7 8 & 9) in C4:G9
So it's a Ref! Error.
INDEX(A1:A10,9) = A9 (the 9th row in A1:A10)
INDEX(A3:A17,9) = A11 (the 9th row in A3:A17)
Hope that helps.
Here's the picture! Sorry I forgot to attatch it.
Ah, I see. However even when I change that to the correct "index row" in this case 6, I still get #VALUE as an error.
Code:
``=INDEX(C4:G9, 6,MATCH(B9,\$C\$3:\$G\$3), 0)``
OK: Here is an image with the code in the top, and the output I am getting. I am now getting a value (32), however this value is incorrect. With the code I am using in that cell, I am trying/it should display cell F4, with the value 34. And then ideally I want to be able to drag it down, and the next one below it would say 34, a readout of Verizon (E5).
Im pullin my hair out here!! ;]
Best,
Cam
You're trying too hard:
=INDEX(C4:G4,MATCH(B4,\$C\$3:\$G\$3,0))
and fill down.
You're trying too hard:
=INDEX(C4:G4,MATCH(B4,\$C\$3:\$G\$3,0))
and fill down.
That worked great, Bob. Thanks a lot for the help.
Best,
Cam
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Go back | 1,197 | 3,998 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | latest | en | 0.94637 |
http://www.csci.csusb.edu/dick/maths/math_85_Statistics.txt | 1,540,289,547,000,000,000 | text/plain | crawl-data/CC-MAIN-2018-43/segments/1539583516123.97/warc/CC-MAIN-20181023090235-20181023111735-00211.warc.gz | 428,092,777 | 1,702 | .Open Statistics . Also See .See ./math_81_Probabillity.html . Online Statistics and Probability Calculators .See http://easycalculation.com/statistics/statistics.php .See http://statpages.org/ .See ../tools/stats.scm (Local Scheme functions) . Notes and definitions on statistics. STATISTICS::=following, .Net . Definition of a (finite) sample I :: Finite_Sets=given. `I` standards for index. Typically it is a range 1..`n` where `n` is the sample size. However in some languages/cultures `I` could be 0..`n`-1. In fact there is no reason to limit `I` to a range. Any finite set of indices will work. Sample::=I>->Real. Each index item has a measured value. Example. For example the list (1,2,3) has `I` = 1..3 and size=3. size::Real= |I|. n:=size. Local shorthand. For p:Real, p::Sample = I +> p. A coercion that converts a single value into a sample with the same value for each index. This turns out to be useful when we subtract the mean of a sample (a number) from every item in the sample. For x,y::\$Sample I will use `x` and `y` as the names of samples of data. . Statistics on one sample mean(x)::= +x/n. In \$STANDARD (+) is a serial operator that adds up all the items in its arguments. +(1,2,3) = (1+2+3) = 6. mean((1, 2, 3)) = +(1, 2, 3)/3 = 6/3 = 2. min, max, range, mode, histogram are to be done. .Hole stats1 ss(x)::=+(x*x). Sum of squares. ss((1, 2, 3)) = +(1*1, 2*2 , 3*3) = +(1,4,9) = 14. ms(x) ::= ss(x - mean(x) )/n. Mean squares about mean. (-1)|- ms(x) =( ss(x) - (+x)*mean(x))/n. Better for small hand calculations. ms((1,2,3)) = (14 - 6*2)/3 = 2/3. var(x)::= n * ms(x)/(n-1). Sample variance -- rescale to allow for estimating the mean. root_mean_square(x)::=sqrt(ms(x)). rms ::= root_mean_square. standard_deviation(x)::=sqrt( var (x) ). sd(x)::=standard_deviation(x). . Statistics on Two samples SP(x,y)::=+((x-mean(x))*(y-mean(y))). (-1)|- ss(x) = SP(x,x). MS(x,y)::=SP(x,y)/n. r(x,y)::= MS(x,y)/( sd(x)*sd(y)). Correlation coefficient -- Pearson. More... .Hole stats2 .Close.Net STATISTICS .Close Statistics in MATHS | 665 | 2,047 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2018-43 | latest | en | 0.720461 |
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## Course Information
Course: MATH 5121 · Topology 1 Web site: www.northeastern.edu/suciu/MATH5121/top1.fa15.html Instructor: Prof. Alex Suciu, < a.suciu@neu.edu > Time and Place: Th 2:30–4:00 pm, in 544 NI and Fr 11:00am–12:30pm in 555 NI Office Hours: Mon & Wed 4:40pm–5:40pm, in 435 LA, or by appointment Prerequisites: MATH 5101 (Analysis 1), MATH 5111 (Algebra 1) Textbooks: Topology (2nd Edition) by James R. Munkres, Prentice Hall, 2000 Algebraic Topology by Allen Hatcher, Cambridge University Press, 2002 (also here) Grade: Based on problem sets, class participation, and possibly a final exam
## Course Description
This course provides an introduction to the concepts and methods of Topology. It consists of three inter-connected parts.
1. Topological Spaces and Continuous Maps This part of the course serves as a quick introduction to General Topology. The objects of study are topological spaces and continuous maps between them. Key is the notion of homeomorphism, which leads to the study of topological invariants. The main properties that are studied are connectedness, path connectedness, and compactness, as well as their "local" versions. We also introduce several constructions of spaces, including identification spaces. 2. Fundamental Group and Covering Spaces This part of the course is a brief introduction to the methods of Algebraic and Geometric Topology. It starts with Poincaré's definition of the fundamental group of a space, and various methods to compute it, such as the Seifert-van Kampen theorem. It proceeds with the classification of surfaces, and a detailed study of covering spaces. Applications include the Brouwer fixed point theorem, the Borsuk-Ulam theorem, and the Nielsen-Schreier theorem. 3. Simplicial Complexes and Simplicial Homology Time permitting, this part of the course is a brief introduction to the methods of Combinatorial Topology and Homological Algebra. It starts with simplicial complexes and their realizations, and proceeds to simplicial homology groups, and ways to compute them. We will illustrate these techniques with concrete examples, and derive some applications.
For more information, including past exams and class projects, see these older syllabi, from 1998, 2001, 2003, 2005, 2007, 2008, 2009, 2011, and 2013. You may also want to look at some past qualifying exams in Topology, based in large part on the material covered in this course.
## Homework Assignments
Assignment Chapter Page Problems
Homework 1
Due Sept. 25
Munkres 2.16 92 5
Munkres 2.17 101-102 13, 19
Munkres 2.18 112 13
Munkres 2.20 126 3
Munkres 3.23 152 5
Homework 2
Due Oct. 8
Munkres 3.24 158 8, 9, 10
Munkres 3.25 162 5, 6, 7
Homework 3
Due Oct. 22
Munkres 2.22 145 3
Munkres 2.22 supplement 146 5
Munkres 3.26 170-172 1, 5, 8, 12
Homework 4
Due Oct. 30
Munkres 3.29 186 3, 5
Munkres 7.46 289 7
Munkres 9.51 330 3(b)(d)
Munkres 9.58 366 1&6, 8
Homework 5
Due Nov. 14
Munkres 9.52 335 6, 7
Munkres 9.53 341 4, 6(b)
Munkres 9.54 348 8
Munkres 9.58 367 9(a)-(d)
Homework 6
Due Nov. 30
Munkres 11.70 433 1, 2
Hatcher 1.2 53-54 4, 6, 8, 14
Homework 7
Due Dec. 11
Munkres 12.74 454 3, 6
Munkres 12.75 457 3
Munkres 13.79 483 5
Hatcher 1.3 79 9, 10
Department of Mathematics Office: 435 Lake Hall Messages: (617) 373-2450 Northeastern University Phone: (617) 373-3899 Fax: (617) 373-5658 Boston, MA, 02115 Email: a.suciu@neu.edu Directions | 1,126 | 3,453 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2017-39 | latest | en | 0.88978 |
http://gogeometry.com/school-college/p880-triangle-midpoint-perpendicular-hexagon-area-parallelogram.htm | 1,550,817,573,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247513661.77/warc/CC-MAIN-20190222054002-20190222080002-00495.warc.gz | 115,920,835 | 3,501 | # Online Geometry Problem 880: Triangle, Midpoints, Sides, Perpendiculars, Hexagon, Area. Level: High School, Honors Geometry, College, Mathematics Education
The area of a triangle ABC is 24. D, E, and F are the midpoints of BC, AC, and AB, respectively. Perpendiculars from E to AB and F to AC meet at G, perpendiculars from F to BC and D to AB meet at H, perpendiculars from D to AC and E to BC meet at M. Find the area of the hexagon DMEGFH. Geometry problem solving is one of the most challenging skills for students to learn. When a problem requires auxiliary construction, the difficulty of the problem increases drastically, perhaps because deciding which construction to make is an ill-structured problem. By “construction,” we mean adding geometric figures (points, lines, planes) to a problem figure that wasn’t mentioned as "given."
Home | Search | Geometry | Problems | All Problems | Open Problems | Visual Index | 10 Problems | Problems Art Gallery | Art | 871-880 | Triangle | Hexagon | Midpoint | Perpendicular | Parallelogram | Area | Triangle area | Solution / comment | By Antonio Gutierrez | 261 | 1,117 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-09 | longest | en | 0.897773 |
http://www.ncl.ucar.edu/Document/Functions/Built-in/uv2sfvpg.shtml | 1,441,254,415,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440645298781.72/warc/CC-MAIN-20150827031458-00225-ip-10-171-96-226.ec2.internal.warc.gz | 596,806,212 | 5,136 | NCL Home > Documentation > Functions > Spherical harmonic routines
# uv2sfvpg
Computes the stream function and velocity potential via spherical harmonics given u and v on a gaussian grid.
## Prototype
``` procedure uv2sfvpg (
u : numeric,
v : numeric,
sf : float or double,
vp : float or double
)
```
## Arguments
u
v
wind components (input, arrays with two or more dimensions, rightmost two dimensions must be nlat x nlon)
• input values must be in ascending latitude order
• input array must be on a global grid
sf
stream function (output, same dimensions as u and v, values will be in ascending latitude order)
vp
velocity potential (output, same dimensions as u and v, values will be in ascending latitude order)
## Description
Given wind components u and v, uv2sfvpg computes the stream function and the velocity potential and returns the results in the arrays sf and vp. uv2sfvpg operates on a gaussian grid.
This procedure does not handle missing values (defined by the _FillValue attribute). If any missing values are encountered in a particular 2D input grid, then all of the values in the corresponding output grids will be set to the missing value defined by the output grids' _FillValue attributes.
Note: For the arrays whose last two dimensions are nlat x nlon, the rest of the dimensions (if any) are collectively referred to as N. If the input/output arrays are just two dimensions, then N can either be considered equal to 1 or nothing at all.
Arrays which have dimensions N x nlat x nlon should not include the cyclic (wraparound) points when invoking the procedures and functions which use spherical harmonics (Spherepack).
If the input arrays u and v are on a fixed grid, uv2sfvpf should be used. Also, note that uv2sfvpg is the procedural version of uv2sfvpG.
## Examples
Example 1
Compute the stream function and velocity potential, given the u and v wind components on a gaussian grid.
```begin
nlat = 128 ; dimensions
mlon = 256
mlon1 = mlon+1
fbfile = "uv300.hs"
nrec = fbinnumrec(fbfile) ; total number of records in the file
ntim = nrec/2 ; number of time steps in dataset
uvmsg = 1e+36
sf = new ( (/nlat,mlon /), float, uvmsg ) ; stream function
vp = new ( (/nlat,mlon /), float, uvmsg ) ; velocity potential
do i = 0,nrec-1,2
month = 1 ; January
if (i .ge. 2) then
month = 7 ; July
end if
u = work(:,0:mlon-1)
v = work(:,0:mlon-1)
uv2sfvpg (u,v, sf,vp) ; u,v ==> stream function + velocity pot
end do
end
```
## Errors
If jer or ker is equal to:
1 : error in the specification of nlat
2 : error in the specification of nlon
4 : error in the specification of N (jer only) | 714 | 2,781 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2015-35 | longest | en | 0.704035 |
https://math.stackexchange.com/questions/2310721/how-to-calculate-sample-size-in-a-random-sampling | 1,561,398,054,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999620.99/warc/CC-MAIN-20190624171058-20190624193058-00032.warc.gz | 515,766,110 | 32,804 | # How to calculate sample size in a random sampling? [closed]
How one can calculate the sample size in any random sampling? Is it varies with sampling method or it is fixed for all methods? Explain it.
## closed as off-topic by NCh, Namaste, Claude Leibovici, Shailesh, C. FalconJun 8 '17 at 0:02
This question appears to be off-topic. The users who voted to close gave this specific reason:
• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – NCh, Namaste, Claude Leibovici, Shailesh, C. Falcon
If this question can be reworded to fit the rules in the help center, please edit the question.
• sample size($n$) is not fixed for a fixed population,it is given by the number of sample you have, so you can change it by taking more sample from population; but if there is a cost constraint, $n$ will have a upper bound. – MAN-MADE Jun 5 '17 at 15:36
The design of experiments includes planning to have a large enough sample to accomplish the task at hand. Here are two examples of frequently used statistical procedures, and how the sample size for them can be planned in advance.
CI with a desired margin of error. Perhaps the simplest case is to choose the $n$ that will make the margin of error in a confidence interval be of a desired size. For example, if you are sampling from a normal population with known standard deviation $\sigma$ and you want to use a 95% confidence interval (CI) to estimate the unknown sample mean $\mu$ within $\pm E.$
The CI is of the form $\bar X \pm 1.96\sigma/\sqrt{n}.$ So you can set $E = 1.96\sigma/\sqrt{n},$ and solve for $n = (1.96\sigma/E),$ where all the quantities on the right hand side are known. (If this $n$ is not an integer, it is customary to round up to the next larger integer.)
One-sample t test with a certain power. Suppose we plan to use normal data to test $H_0: \mu = 10$ vs $H_a: \mu > 10$ at the 5% level. If the true value of $\mu$ is actually as large as $\mu = 12,$ you would like to reject $H_0$ with probability 95%. In order to get an answer as to the size of $n$, you need to have an estimate of the population SD. (If you knew the exact value of $\sigma$, this would be a z test, not a t test.) Perhaps another similar experiment has been done before or perhaps you have a good idea of the precision of the equipment you will use. Such prior information can help to estimate $\sigma.$
This is a more complicated problem and is often solved using software. Here is an answer from Minitab statistical software. Notice that I guessed that $\sigma \approx 3.$ Then the procedure says I'd need about $n = 26$ observations for this experiment.
Power and Sample Size
1-Sample t Test
Testing mean = null (versus > null)
Calculating power for mean = null + difference
α = 0.05 Assumed standard deviation = 3
Sample Target
Difference Size Power Actual Power
2 26 0.95 0.951579
Your question is a very good one. In textbook statistics problems you are often given the results of an experiment or study, with little information or opportunity to think about what went before data collection. But experiments and studies need to be planned in advance. It can be a terrible waste of time and money to do an experiment that has too little data for the task at hand. Part of experimental design is to make reasonable assumptions and plan the sample size in advance. | 871 | 3,649 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2019-26 | latest | en | 0.938152 |
http://www.cfd-online.com/Forums/openfoam-programming-development/92188-implementation-schroedingers-equation-print.html | 1,438,321,435,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042988051.33/warc/CC-MAIN-20150728002308-00086-ip-10-236-191-2.ec2.internal.warc.gz | 360,003,678 | 2,529 | CFD Online Discussion Forums (http://www.cfd-online.com/Forums/)
- OpenFOAM Programming & Development (http://www.cfd-online.com/Forums/openfoam-programming-development/)
- - Implementation of Schroedinger's equation (http://www.cfd-online.com/Forums/openfoam-programming-development/92188-implementation-schroedingers-equation.html)
DiracRules September 5, 2011 15:17
Implementation of Schroedinger's equation
I'm quite new in OpenFoam, but I'd like to learn something more by writing some code.
I'd like to develop a solver for Schroedinger's equation.
I found that the diffusion equation can be modified to achieve this task.
But my question is:since Schroedinger's equation contains the imaginary unit, and I've not seen any solver using it, how can I describe the behaviour of the imaginary unit in openfoam?
I read this paper, but I'd like to develop something mine, or at least understand the mathematic at the basis of his use of the matrices in substitution to the imaginary unit.
Thank you all!!
mirko September 6, 2011 09:58
Quote:
Originally Posted by DiracRules (Post 322986) I'm quite new in OpenFoam, but I'd like to learn something more by writing some code. I'd like to develop a solver for Schroedinger's equation. I found that the diffusion equation can be modified to achieve this task. But my question is:since Schroedinger's equation contains the imaginary unit, and I've not seen any solver using it, how can I describe the behaviour of the imaginary unit in openfoam? I read this paper, but I'd like to develop something mine, or at least understand the mathematic at the basis of his use of the matrices in substitution to the imaginary unit. Thank you all!!
You will have to solve separately for real and imaginary parts just as the author is doing. There is a project on OpenFoam-extend to allow for more general types of unknowns (not just floating real numbers). But I am not familiar with its status regarding complex numbers.
Mirko
All times are GMT -4. The time now is 01:43. | 474 | 2,027 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2015-32 | longest | en | 0.928476 |
https://eduinput.com/introduction-to-algebraic-proofs/ | 1,701,763,581,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100550.40/warc/CC-MAIN-20231205073336-20231205103336-00894.warc.gz | 265,475,244 | 25,439 | Home | Math | Introduction to Algebraic Proofs
# Introduction to Algebraic Proofs
October 28, 2023
written by Rida Mirza
Algebraic proofs are like using algebra to show that something is true. They help us check if the math rules and statements in algebra are correct. So, they’re a big part of algebra, making sure we’re doing math right.
In this article, we will discuss introduction to algebraic proofs by covering some key concepts and strategies.
## Basic Principles of Algebraic Proofs
There are a few core principles that form the foundation of algebraic proofs:
• Start with known facts, definitions, axioms, or previously proven theorems. These are statements accepted as true, and can be used as the basis for making logical deductions.
• Apply rules of logic. Use standard methods of logical reasoning like induction, contradiction, contraposition, etc.
• Follow step-by-step derivations. Show your work and lay out each step in the proof clearly.
• End with the statement you set out to prove. This is your conclusion, which should follow logically from the previous steps.
## Rules of Division
Rules of division that are useful in algebraic proofs:
• Dividing both sides of an equation by the same non-zero number is valid. The resulting equation remains true.
• Dividing by a variable is permitted as long as you account for the possibility that the variable may equal 0.
• When dividing polynomials, you can only divide by a factor or monomial that divides evenly into the first term.
• The remainder theorem states that the remainder left over after dividing a polynomial by (x – a) is equal to the value of the polynomial evaluated at x = a.
## Solved Examples
### Example
Prove that if a/b = c/d, then ad = bc, where a, b, c, d are non-zero numbers.
Solution
Proof:
a/b = c/d
a = c/d * b (multiply both sides of the equation by b)
ad = (c/d * b)d (multiply both sides by d)
### Example
Prove that x2– 4 is divisible by (x+2).
Solution
Proof:
Let f(x) = x2 – 4
Substitute x = -2 into f(x):
f(-2) = (-2)2 – 4 = 4 – 4 = 0
Since f(-2) = 0, by the remainder theorem, x2 – 4 is divisible by (x+2).
Therefore, the statement is proven true.
## FAQs
### What are some common strategies for constructing algebraic proofs?
Induction, contradiction, contraposition, direct proof using axioms/theorems, examining cases, using mappings, division algorithms, and factoring polynomials are some common strategies.
### When do we need to start a proof with given facts or previously proven statements?
When proving purely algebraic statements, you can begin simply by writing the statement you want to prove. However, if you are proving a theorem relying on geometrical concepts or results from other branches of math, you would need to start by establishing relevant facts, axioms, definitions or previously proven theorems.
### Is mathematical induction applicable to algebraic proofs?
Yes, mathematical induction is an important technique used to prove many algebraic theorems and properties, especially those involving sequences, divisibility, and binomial coefficients.
### How can I get better at constructing algebraic proofs?
Practice! Try proving simple algebraic statements using different strategies. Work through examples from textbooks and review proofs of key theorems to become familiar with the structure and reasoning involved.
File Under: | 756 | 3,394 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.9375 | 5 | CC-MAIN-2023-50 | latest | en | 0.936789 |
http://mathhelpforum.com/algebra/28102-2x2-system-word-problem.html | 1,527,459,837,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794870470.67/warc/CC-MAIN-20180527205925-20180527225925-00444.warc.gz | 177,555,130 | 9,521 | # Thread: 2x2 System word problem?
1. ## 2x2 System word problem?
An underwater(but near the surface) explosion is detected by sonar on a ship 30s before it is heard on deck. If sound travels at 5,000 ft/s in water and 1,000 ft/s in air, how far is the ship from the explosion?
All I need is the equations and I'll be aright to go from their.
2. Let $\displaystyle t$ be equal to the time it takes to be heard by the sonar.
We then have that the time for it to be heard on deck is $\displaystyle t + 30$.
$\displaystyle v = \frac {d}{t}$
Therefore:
$\displaystyle d = vt$
And since the distances are equal:
$\displaystyle 1000(t + 30) = 5000t$
Now solve for $\displaystyle t$ to get:
$\displaystyle 4000t = 30000$
$\displaystyle t = 7.5$
Now just plug that into the equation $\displaystyle d = vt$ to get:
$\displaystyle d = 5000(7.5)$
$\displaystyle d = 37,500$ feet | 265 | 883 | {"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} | 4.25 | 4 | CC-MAIN-2018-22 | latest | en | 0.877297 |
https://la.mathworks.com/matlabcentral/cody/problems/44721-seperate-array-to-small-section-according-to-its-index-position/solutions/3779598 | 1,611,128,149,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703519923.26/warc/CC-MAIN-20210120054203-20210120084203-00606.warc.gz | 427,064,521 | 17,132 | Cody
# Problem 44721. Seperate array to small section according to its index position
Solution 3779598
Submitted on 22 Nov 2020
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Fail
for n = 3 : 12 y = odd_and_even_fun(n); while length(y) ~= 1 y = cellfun(@(x)x(:)',mat2cell(cell2mat(reshape(y,2,'')),2,repelem(numel(y{1}),length(y)/2)),'uni',0); end assert(isequal(y{1},0:2^n-1)) end
Error using mat2cell (line 89) Input arguments, D1 through D2, must sum to each dimension of the input matrix size, [2 8]. Error in Test1 (line 4) y = cellfun(@(x)x(:)',mat2cell(cell2mat(reshape(y,2,'')),2,repelem(numel(y{1}),length(y)/2)),'uni',0);
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Question
# Statement - 1 : if |A|=±1, then the matrix A is orthogonal matrix.Statement - 2 : If A is orthogonal matrix then if |A|=±1
A
Statement-1 is correct and Statement-2 is correct and Statement-2 is correct explanation of Statement-1
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B
Statement-1 is correct and Statement-2 is correct and Statement-2 is not the correct explanation of Statement-1
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C
Statement-1 is incorrect
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D
Statement-1=2 is incorrect
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Solution
## The correct option is C Statement-1 is incorrectIf A is orthogonal matrix , then AAT=IIf we apply det on both sides , we get |A|2=1⇒|A|=±1Therefore statement 2 is correctIf A is Involuntary matrix , then A=A−1If we apply det , we get |A|2=1⇒|A|=±1Therefore if |A|+± , then matrix A need not be orthogonalTherefore statement 1 is incorrectSo the correct option is C
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Join BYJU'S Learning Program | 369 | 1,283 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2024-22 | latest | en | 0.79811 |
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## #1 2013-02-27 05:05:01
jacks
Member
Registered: 2012-11-21
Posts: 132
### positive divisers of 10!
The no. of positive divisers of
which are is in the form of
where
Offline
## #2 2013-02-27 05:16:45
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
### Re: positive divisers of 10!
Hi;
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
## #3 2013-02-27 05:28:38
jacks
Member
Registered: 2012-11-21
Posts: 132
### Re: positive divisers of 10!
Yes Bobbym as usual you are always Right.
Would You like to explain it to me how can i tackle these type of Questions
Thanks
Offline
## #4 2013-02-27 05:35:48
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
### Re: positive divisers of 10!
We can put it into the form of the other question because it is easy to factor but what if the number were something that was not?
A math method from here I do not know and have not yet found in any references. I just computed the divisors and performed a mod 5 on them.
Last edited by bobbym (2013-02-27 05:36:05)
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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### Permutations and combinations
by merlyn (Sage)
on Jul 25, 2000 at 16:49 UTC Need Help??
Here's a couple of code snippets I keep handy when I want to generate all combinations (no replacement) and permutations (all possible ways of combining a list of varying things). Enjoy.
You can probably see that I speak Perl with a lisp sometimes.
```
print "permute:\n";
print "[", join(", ", @\$_), "]\n" for permute([1,2,3], [4,5,6], [7,8,9
+]);
print "combinations:\n";
print "[", join(", ", @\$_), "]\n" for combinations(1..5);
sub permute {
my \$last = pop @_;
unless (@_) {
return map [\$_], @\$last;
}
return map { my \$left = \$_; map [@\$left, \$_], @\$last } permute(@_);
}
sub combinations {
return [] unless @_;
my \$first = shift;
my @rest = combinations(@_);
return @rest, map { [\$first, @\$_] } @rest;
}
Replies are listed 'Best First'.
RE: Permutations and combinations
by AltBlue (Chaplain) on Aug 21, 2000 at 13:19 UTC
interesting approach that reminded me about some old project that needed some permutations generations stuff... so, i've checked up that code and here i come with a reviewed standalone version:
```#!/usr/bin/perl -w
# DESCRIPTION: Generate permutations in lexicographic order
# USAGE: ./permlex.pl <term1> <term2> <term3> .....
use strict;
die "bleah... nothing to permutate\n" if \$#ARGV<0;
my @terms = @ARGV; my \$n = \$#ARGV; my @a = (0..\$n);
genperm(); exit(0);
sub genperm {
print join(" ",@terms[@a]),"\n";
my (\$k,\$j) = (\$n-1,\$n);
\$k-- while (\$k>=0 and \$a[\$k]>\$a[\$k+1]);
return(0) if (\$k<0);
\$j-- while (\$a[\$k]>\$a[\$j]);
swap(\$j,\$k++); \$j=\$n;
swap(\$j--,\$k++) while (\$j>\$k);
genperm();
}
sub swap {
my (\$i,\$j) = @_; my \$t = \$a[\$i];
(\$a[\$i],\$a[\$j]) = (\$a[\$j],\$t);
}
as you may see, it's a pure lexicographic permutations generator algorithm, as in the books ;-)
oh, not to forget, just checked up on cpan and found out there is a Algorithm::Permute module. here is a lame example for module users ;-)
```#!/usr/bin/perl -w
use strict;
die "bleah... nothing to permutate\n" unless defined @ARGV;
use Algorithm::Permute qw(permute permute_ref);
print join(" ", @\$_), "\n" for permute(\@ARGV);
--
AltBlue.
your sub swap can be written without \$t:
```sub swap {
my (\$i,\$j) = @_;
@a[\$i,\$j] = @a[\$j,\$i];
}
lol, you bother to look upon this old piece of code :))
heh, of course that your snippet is good, but a swap routine could be written even simpler:
```sub swap { reverse @_ }
... heh, it's not the case for that code thou, as the array is a global one :) cheers.
--
AltBlue.
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# Capuchin monkeys often rub their bodies with a certain type
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Capuchin monkeys often rub their bodies with a certain type [#permalink]
### Show Tags
27 Feb 2013, 08:24
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Capuchin monkeys often rub their bodies with a certain type of millipede. Laboratory tests show that secretions from the bodies of these millipedes are rich in two chemicals that are potent mosquito repellents, and mosquitoes carry parasites that debilitate capuchins. Some scientists hypothesize that the monkeys rub their bodies with the millipedes because doing so helps protect them from mosquitoes.
Which of the following, if true, provides the most support for the scientists’ hypothesis?
A. A single millipede often gets passed around among several capuchins, all of whom rub their bodies with it.
B. The two chemicals that repel mosquitoes also repel several other varieties of insects.
C. The capuchins rarely rub their bodies with the millipedes except during the rainy season, when mosquito populations are at their peak.
D. Although the capuchins eat several species of insects, they do not eat the type of millipede they use to rub on their bodies.
E. The two insect-repelling chemicals in the secretions of the millipedes are carcinogenic for humans but do not appear to be carcinogenic for capuchins.
[Reveal] Spoiler: OA
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Re: Capuchin monkeys often rub their bodies with a certain type [#permalink]
### Show Tags
27 Feb 2013, 21:30
1
KUDOS
The argument tell us that the Capuchin monkeys rub their bodies with a certain type of millipede and that laboratory tests show that the secretions of these millipedes contain some chemicals that act like mosquito repellent. Based on this evidence, scientists believe that Capuchin monkeys rub these millipedes on their bodies in order to protect themselves from mosquitoes.
We are supposed to find a statement that supports this argument.
A. A single millipede often gets passed around among several capuchins, all of whom rub their bodies with it.
If a single millipede gets passed around several capuchins, the amount of secretions might reduce over time. So, monkeys could be rubbing these millipedes on their bodies due to some other reason. Also, this argument is quite irrelevant to the argument. WRONG
B. The two chemicals that repel mosquitoes also repel several other varieties of insects.
This statement tells us that these chemicals repel many other varieties of insects. However, this is irrelevant to the argument as it provides us no evidence that could support or refute the argument that Capuchin monkeys rub the millipedes on their bodies for their mosquito repellent properties.WRONG
C. The capuchins rarely rub their bodies with the millipedes except during the rainy season, when mosquito populations are at their peak.
This statement supports the argument. The fact that the Capuchin monkeys rub their bodies with these millipedes only during the rainy season, when mosquitoes are abundant, suggests that the monkeys could be doing so to protect themselves from the mosquitoes. Otherwise, they might have continued this behavior during the other seasons also. RIGHT
D. Although the capuchins eat several species of insects, they do not eat the type of millipede they use to rub on their bodies
The fact that the Capuchin monkeys do not eat the specific type of millipede provides no support to the argument. There could be multiple reasons for this behavior.
1. This could be a behavioral aspect
2. The millipedes do not taste good
3. The millipedes do not have as much nutrition as some other insects that are abundant in the area.
Hence, this statement is not the answer. WRONG
E. The two insect-repelling chemicals in the secretions of the millipedes are carcinogenic for humans but do not appear to be carcinogenic for capuchins.
The monkeys would not rub the insects on themselves just because they are non-carcinogenic. Some benefit would have to be reaped out of this rubbing. Hence, this statement gives us no evidence to support the argument and is WRONG.
Hope this helps! Let me know if I could clarify something for you.
fozzzy wrote:
Capuchin monkeys often rub their bodies with a certain type of millipede. Laboratory tests show that secretions from the bodies of these millipedes are rich in two chemicals that are potent mosquito repellents, and mosquitoes carry parasites that debilitate capuchins. Some scientists hypothesize that the monkeys rub their bodies with the millipedes because doing so helps protect them from mosquitoes.
Which of the following, if true, provides the most support for the scientists’ hypothesis?
A. A single millipede often gets passed around among several capuchins, all of whom rub their bodies with it.
B. The two chemicals that repel mosquitoes also repel several other varieties of insects.
C. The capuchins rarely rub their bodies with the millipedes except during the rainy season, when mosquito populations are at their peak.
D. Although the capuchins eat several species of insects, they do not eat the type of millipede they use to rub on their bodies.
E. The two insect-repelling chemicals in the secretions of the millipedes are carcinogenic for humans but do not appear to be carcinogenic for capuchins.
Re: Capuchin monkeys often rub their bodies with a certain type [#permalink] 27 Feb 2013, 21:30
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There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page. The difference of 2 positive numbers is 5 and the sum of their squares is , what are the numbers? I don't know who to vote best answer here.
Quadratic equations are very common in algebra. The graph of a quadratic equation has the shape of a parabola. Quadratic functions can be graphed on a coordinate plane. They are symmetrical around a single axis of symmetry. Therefore, when solving quadratic functions, one can find approximate values of the roots. For example, suppose the problem is to find two real numbers whose difference is 6 and whose product is If x equals one of the numbers, then x-6 will equal the other number.
Solving quadratic equations is an important skill in algebra. Some methods of solving quadratic equations include factoring, taking the square root of both sides, completing the square, and using the quadratic formula. There are many applications where quadratic equations can be used, such as motion, time, distance, and speed.
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# Mastering the Martingale: An In-depth Analysis of this Popular Roulette Strategy
Mastering the Martingale: An In-depth Analysis of this Popular Roulette Strategy | Casino Babes
# Mastering the Martingale: An In-depth Analysis of this Popular Roulette Strategy
By Walter Hemphill, Editor at CasinoBabes.net
Roulette has long been a favorite casino game, offering excitement and the potential for big wins. Among the various strategies used by players to increase their chances of success, the Martingale betting system stands out as one of the most popular.
## What is the Martingale Strategy?
The Martingale strategy is a simple betting system that originated in 18th-century France. Its core principle revolves around doubling your bet after every loss, with the aim of recovering all previous losses and generating a small profit.
Typically, this strategy is applied to even-money bets in roulette, such as Red/Black, Odd/Even, or 1-18/19-36. The logic behind the Martingale is that, eventually, you will win a bet and regain your initial losses plus a small profit, covering all previous losing bets.
## How does the Martingale Strategy Work?
Let’s consider a hypothetical scenario to demonstrate the Martingale strategy in action:
Starting with a basic bet of \$10 on Red in roulette, if you lose the first bet, you double your next bet to \$20. If you lose again, you double the bet once more to \$40. If you win on the third bet, you receive a \$40 payout, which covers your losses of \$10 and \$20, leaving you with a \$10 profit.
However, if you lose three times in a row, you would have lost a total of \$70 (\$10 + \$20 + \$40), and your next bet should be \$80. As you can see, the stakes increase rapidly when using the Martingale strategy, requiring a substantial bankroll to sustain potential losing streaks.
## The Pros and Cons of the Martingale Strategy
Despite its popularity, the Martingale strategy has both advantages and disadvantages that players should consider before implementing it in their roulette sessions.
### Pros:
• Simple and easy to understand.
• Offers potential for short-term profits.
• May help recover previous losses in small increments.
### Cons:
• Requires a large bankroll to sustain potential losses.
• No guaranteed long-term success. The casino always maintains an edge over players.
• Can hit table limits quickly, preventing further betting progression.
Players must be aware that the Martingale strategy does not alter the odds of the game itself. It is solely a betting system that aims to manipulate short-term outcomes with potential short bursts of profit.
### 1. Is the Martingale strategy suitable for all players?
No, the Martingale strategy is best suited for players with a large bankroll and a high tolerance for risk. It is not recommended for casual players or those with limited funds.
### 2. Can the Martingale strategy guarantee winnings?
No, the Martingale strategy does not guarantee winnings. While it can generate short-term profits, it cannot overcome the long-term house edge, which ensures that the casino always has an advantage in the game.
### 3. Are there any alternative strategies to the Martingale?
Yes, there are various strategies players can explore, such as the Fibonacci system, Labouchere system, or D’Alembert system. Each has its own unique approach to managing bets and minimizing risks.
### 4. Is the Martingale strategy legal?
Yes, the Martingale strategy is legal and widely used by players in many casino games, including roulette. However, its effectiveness and suitability depend on individual preferences and circumstances.
## Conclusion
The Martingale strategy is a popular betting system that offers potential short-term gains in roulette. It is a simple strategy to understand and implement, but it comes with risks and requires a substantial bankroll to mitigate potential losses. Players must remember that no strategy can guarantee long-term success, and the casino always maintains an advantage. As with any gambling strategy, responsible play is crucial.
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0 87 | 898 | 4,240 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2024-18 | latest | en | 0.901334 |
https://studylib.net/doc/9874171/4361t5_sampleproblems | 1,600,845,921,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400209999.57/warc/CC-MAIN-20200923050545-20200923080545-00439.warc.gz | 662,938,994 | 13,389 | # 4361t5_sampleproblems
```Test 5 practice problems
Problem 1
Mason, Inc. uses a standard costing system. Overhead costs are allocated based on
direct labor hours. The standard variable overhead and fixed overhead rates are \$1 and
\$5 per direct labor hour, respectively. Data relevant for the current period include:
Direct materials purchased
50,000 lbs. @ \$12 per lb.
Direct materials used
49,000 lbs.
Budgeted production
11,800 units
Actual production
12,000 units
Standard quantity of direct materials 3.75 lbs.
Direct materials standard price
\$12.50 per lb.
Direct labor costs incurred
75,000 hours @ \$12 per hour
Direct labor standard
6.4 hours
Standard direct labor cost
\$11 per hour
\$77,070
\$181,000
1.
2.
3.
4.
How much overhead was applied during the year?
Calculate all direct labor, direct materials, and overhead variances.
Assuming a 1.4% materiality threshold, identify which variances should be
investigated under management by exception.
5. Identify who is responsible for each variance.
Problem 2
Finish-It-Yourself Furniture Company manufactures replicas of antique oak filing
cabinets. Additional information is as follows:
Selling price
\$500 per filing cabinet
Variable production cost
\$170 per filing cabinet
Fixed production costs
\$8,000 per month
Variable selling and administration \$20 per filing cabinet
\$3,000 per month
a. In good form, prepare a variable costing income statement for a month in
which 100 filing cabinets are manufactured and 90 are sold, if the firm uses
variable costing. Assume no beginning inventory.
b. In good form, prepare an absorption costing income statement for a month
100 filing cabinets are manufactured and 90 are sold, if the firm uses
absorption costing and actual costing. Assume no beginning inventory.
c. What is the cost assigned to ending inventory under each of the above
costing methods?
d. Explain the differences between the two ending inventory valuations (do not
perform a computation for this answer).
e. Prepare a reconcile of operating incomes between variable costing and
absorption costing.
f. If the manager of Finish-It-Yourself Furniture Company is given a bonus
based on income, which type of costing income statement would you
recommend for evaluating manager performance? Justify your choice.
Problem 3
A company believes it can sell 2,000,000 units of its proposed new can opener at a
price of \$16.00 each.
A. If the company desires to make a markup of 30% on cost, how much is the target
cost per can opener?
B. If the company desires to make a markup of 30% on selling price, how much is
the target cost per can opener?
Problem 4
A company believes it can sell 500,000 of its proposed new laser mouse at a price of
\$11.00 each There will be \$800,000 in fixed costs associated with the mouse. If the
company desires to make a profit \$200,000, how much is the target variable cost per
mouse?
``` | 677 | 2,897 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-40 | latest | en | 0.869919 |
http://biology.stackexchange.com/questions/20364/calculation-of-the-bacterial-growth-rate-from-a-spectrophotomer-growth-curve | 1,469,821,877,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257831771.10/warc/CC-MAIN-20160723071031-00115-ip-10-185-27-174.ec2.internal.warc.gz | 24,271,646 | 17,682 | # Calculation of the bacterial growth rate from a spectrophotomer growth curve
Typically the microbial growth in liquid cultures is monitored by turbidity. Data is obtained with a spectrophotometer to measure optical density at 600nm. The slope of the bacterial kinetic curve in exponential phase is the growth rate.
But I have seen two ways of calculate the growth rate:
1. Growth rate = Maximum slope value of the Kinetic curve
2. $Log_{10}$ transformation of the growth data and then calculate the slope.
I think the second one is right if you want to calculate the generation time:
generation time = $\frac{ln(2)}{Growth rate}$.
What is the correct way to calculate the growth rate and then generation time?
-
After you have the data plotted, the exponential growth phase should appear as a line with positive slope. The logarithmic scale on the Y axis will automatically transform the exponential curve into a straight line. To determine growth rate in terms of generation per hour, you need to get the optical density at time 0, which is the beginning of the exponential phase, and time t, which is just some other time on the exponential phase. Use this equation to determine growth rate: $k =\frac{log(X_t) - log(X_0)}{0.301t}$
Once you know the growth rate, you can determine the generation time with $t_{gen} = \frac{1}{k}$, which gives time for 1 generation in hours. To convert to minutes, just multiply by 60. | 322 | 1,431 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2016-30 | latest | en | 0.910149 |
https://tw.tradingview.com/script/TSDzwIwk-No-Volume-SVAPO-LazyBear/ | 1,576,242,698,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540555616.2/warc/CC-MAIN-20191213122716-20191213150716-00417.warc.gz | 584,809,119 | 172,545 | # No Volume SVAPO [LazyBear]
5270瀏覽
5270
This is a no-volume version of Vervoort's SVAPO. The original version is @ Since it doesn't include volume in its calculations, you can use this on FX.
http://stocata.org/ta_en/proprietary.htm...
BTW, this is my 150th script. Plenty more to come, my to-publish queue is still full of new stuff :)
For a complete list of my other indicators, do check out the links below:
- Chart:
```//
// @author LazyBear
// List of all my indicators:
//
study("No Volume SVAPO [LazyBear]", shorttitle="SVAPO_NOVOL_LB")
length=input(8, title="SVAPO Period", minval=2, maxval=20)
cutoff=input(1, title="Minimum %o price change", maxval=10, minval=0)
devH=input(1.5, title="Stdev High", maxval=5, minval=0.1)
devL=input(1.3, title="Stdev Low", maxval=5, minval=0.1)
stdevper=input(100, title="Stdev Period", maxval=200, minval=1)
calc_tema(s, length) =>
ema1 = ema(s, length)
ema2 = ema(ema1, length)
ema3 = ema(ema2, length)
3 * (ema1 - ema2) + ema3
calc_linregslope(C, tp) =>
((tp*(sum(cum(1)*C,tp)))-(sum(cum(1),tp)*(sum(C,tp))))/((tp*sum(pow(cum(1),2),tp))-pow(sum(cum(1),tp),2))
calc_OR2(x) =>
y=x // To force expr evaluation
(y == true) or (y[1] == true)
src=close
haOpen=(ohlc4[1] + nz(haOpen[1]))/2
haCl=(ohlc4+haOpen+max(high, haOpen)+min(low, haOpen))/4
haC=calc_tema(haCl, round(length/1.6))
vma=sma(src, length*5)
vave=vma[1]
vmax=2*vave
vc=iff(src<vmax, src, vmax)
vtr=calc_tema(calc_linregslope(src, length), length)
svapo=calc_tema(sum(iff(haC>(nz(haC[1])*(1+cutoff/1000)) and calc_OR2((vtr>=nz(vtr[1]))), vc, iff(haC<(nz(haC[1])*(1-cutoff/1000)) and calc_OR2((vtr>nz(vtr[1]))),-vc,0)), length)/(vave+1),length)
plot(devH*stdev(svapo,stdevper), color=red, style=3)
plot(-devL*stdev(svapo,stdevper), color=green, style=3)
plot(0, color=gray, style=3)
plot(svapo, color=maroon, linewidth=2)
```
Wow !You have so man scripts. Thank u for sharing them
Congrats on achieving #150, LazyBear.
grahvity
Thx
cum(1) takes the sum of 1 = 1 ?
kakola
Try this small script to understand cum().
study("Test Cum() function")
plot(cum(1))
plot(n)
I also included "n", to show you how it is not the same as cum() :)
Hope this makes it clear. | 798 | 2,178 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-51 | latest | en | 0.491737 |
https://iwant2study.org/taskmeister/?option=com_content&view=article&id=814 | 1,642,327,312,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320299852.23/warc/CC-MAIN-20220116093137-20220116123137-00079.warc.gz | 405,539,725 | 15,281 | ### Translations
Code Language Translator Run
### Software Requirements
SoftwareRequirements
Android iOS Windows MacOS with best with Chrome Chrome Chrome Chrome support full-screen? Yes. Chrome/Opera No. Firefox/ Samsung Internet Not yet Yes Yes cannot work on some mobile browser that don't understand JavaScript such as..... cannot work on Internet Explorer 9 and below
### Credits
This email address is being protected from spambots. You need JavaScript enabled to view it.; Francisco Esquembre; Felix J. Garcia Clemente
### Sample Learning Goals
Determinant of a matrix in JavaScript using Laplace expansion
## Determinant of 2x2 matrix
mrdaniel will be showing you how to write a function to calculate the determinant of a square matrix of any size. Let M represent the following 2x2 matrix.
var M = [ [a,b],
[c,d] ];
The following is the procedure to calculate the determinant of a 2x2 matrix.
det(M) = (a*d)-(b*c);
Now, we'll substitute the following values for the variables and write the basic det() function to calculate the determinant of a 2x2 matrix.
var M = [ [1,2],
[3,4] ];
function det(M) {
return (M[0][0]*M[1][1])-(M[0][1]*M[1][0]);
}
// output = -2
## Determinant of 3x3 matrix
The determinant of a 3x3 matrix becomes a bit trickier now. Let M be the following.
var M = [ [a,b,c],
[d,e,f],
[g,h,i] ];
The following is the procedure to calculate the determinant of a 3x3 matrix.
det(M) = a(ei-fh)-b(di-fg)+c(dh-eg); // notice the alternating signs
The algorithm for calculating the determinant of a 3x3 matrix is essentially the following: (1) M[0][0] * determinant of the 2x2 matrix that is excluded from M[0][0]'s row and column. (2) M[0][1] * determinant of the 2x2 matrix that is excluded from M[0][1]'s row and column. (3) M[0][2] * determinant of the 2x2 matrix that is excluded from M[0][2]'s row and column. The algorithm I'm explaining is called Laplace expansion. It technically doesn't need to take the values from the first row of the matrix and multiply them by the remainder matrices-it can actually work if you use any row. This method is also very inefficient for matrices of large size. There are much more efficient ways of calculating the determinant of large square matrices. All of this is explained in the link above. So now to update our det() function to work with 3x3 matrices. We'll rewrite the function to work recursively and our base case will be if M is a 2x2 matrix, which in that case is very simple to calculate the determinant. We'll also need to write a function that deletes a specific row and column from a matrix in order to continue with the aforementioned algorithm, and for this we'll use the JavaScript splice function.
var M = [ [1,2,3],
[4,5,6],
[7,8,9] ];
function det(M) {
if (M.length==2) { return (M[0][0]*M[1][1])-(M[0][1]*M[1][0]); }
return M[0][0]*det(deleteRowAndColumn(M,0)) -
M[0][1]*det(deleteRowAndColumn(M,1)) +
M[0][2]*det(deleteRowAndColumn(M,2));
}
function deleteRowAndColumn(M,index) {
var temp = [];
// copy the array first
for (var i=0; i<M.length; i++) { temp.push(M[i].slice(0)); }
// delete the first row
temp.splice(0,1);
// delete the column at the index specified
for (var i=0; i<temp.length; i++) { temp[i].splice(index,1); }
return temp;
}
// output = 0
Now let's just abstract the code in det(M) into a loop rather than writing out the three separate parts. The Math.pow() function is required in order to determine the sign of the respective part since we aren't explicitly writing out minus or plus.
function det(M) {
if (M.length==2) { return (M[0][0]*M[1][1])-(M[0][1]*M[1][0]); }
for (var i=0; i<M.length; i++) { answer += Math.pow(-1,i)*M[0][i]*det(deleteRowAndColumn(M,i)); }
}
## Determinant of 4x4 and larger matrices
The pattern simply continues now for larger matrices. Let M be the following matrix.
var M = [ [a,b,c,d],
[e,f,g,h],
[i,j,k,l],
[m,n,o,p] ];
The determinant of M will therefore be:
det(M) = a*det([[f,g,h],[j,k,l],[n,o,p]])-b*det([[e,g,h],[i,k,l],[m,o,p]])+...
So because our code from above runs recursively, it should actually work for a matrix of any size 4x4 or larger. Below is the final code to determine the determinant of any size square matrix.
var M = [ [1,2,3,4],
[5,6,7,8],
[9,1,2,3],
[4,5,9,7] ];
function det(M) {
if (M.length==1) { return (M[0][0]);} // to cater to M[0][0] case
if (M.length==2) { return (M[0][0]*M[1][1])-(M[0][1]*M[1][0]); } var answer = 0; for (var i=0; i< M.length; i++) { answer += Math.pow(-1,i)*M[0][i]*det(deleteRowAndColumn(M,i)); } return answer; } function deleteRowAndColumn(M,index) { var temp = []; for (var i=0; i<M.length; i++) { temp.push(M[i].slice(0)); } temp.splice(0,1); for (var i=0; i<temp.length; i++) { temp[i].splice(index,1); } return temp; }
### For Teachers
The determinant of a matrix can be defined in terms of determinant of sub-matrices within the matrix using the formula
$$|A| = \sum_{j=1}^{m}(-1)^{1+j}a_{1j}|A_{1j}|$$
Where $$a_{1j}$$ is the element in first row and column j and $$A_{1j }$$ is the determinant of the sub-matrix obtained from A by removing first row and column j.
Input: Matrix A
Algorithm:
Define the determinant recursively in terms of smaller matrices with the determinant of a 1x1 matrix being itself
|A| = |A| if |A| is 1x1 matrix
$$|A| = \sum_{j=1}^{m}(-1)^{1+j}a_{1j}|A_{1j}|$$ if |A| is 2x2 matrix
$$|A| = \sum_{j=1}^{m}(-1)^{1+j}a_{1j}|A_{1j}|$$ if |A| is 3x3 and higher matrix
Research
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# Testimonials (0)
There are no testimonials available for viewing. Login to deploy the article and be the first to submit your review! | 1,675 | 5,611 | {"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} | 3.75 | 4 | CC-MAIN-2022-05 | latest | en | 0.810014 |
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