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https://www.coursehero.com/file/5650355/We-should-check-the-Reynolds-number-to-ensure-turbulent-flow/
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Chapt06
# We should check the reynolds number to ensure
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Unformatted text preview: of a specific viscous-flow analysis, we take the classic problem of flow in a full pipe, driven by pressure or gravity or both. Figure 6.10 shows the geometry of the pipe of radius R. The x-axis is taken in the flow direction and is inclined to the horizontal at an angle . Before proceeding with a solution to the equations of motion, we can learn a lot by making a control-volume analysis of the flow between sections 1 and 2 in Fig. 6.10. The continuity relation, Eq. (3.23), reduces to Q1 or V1 Q2 Q1 A1 const Q2 A2 V2 (6.23) since the pipe is of constant area. The steady-flow energy equation (3.71) reduces to p1 1 2 2 1V 1 gz1 p2 1 2 2 2V 2 gz2 ghf (6.24) | v v since there are no shaft-work or heat-transfer effects. Now assume that the flow is fully | e-Text Main Menu | Textbook Table of Contents | Study Guide 6.4 Flow in a Circular Pipe 1 p1 = p 2 + ∆ p 339 g x = g sin φ g r= φ R r u( r) τw 2 p2 τ( r) Z1 x2 –x 1 =∆ φ L x Z2 Fig. 6.10 Control volume of steady, fully developed flow between two sections in an inclined pipe. developed (Fig. 6.6), a...
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## This note was uploaded on 10/27/2009 for the course MAE 101a taught by Professor Sakar during the Spring '08 term at UCSD.
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# Are any quantum field theories mathematically convergent?
I know for example that theories like QED and QCD when expended perturbatively in terms of Feynman diagrams produce asymptotic series that don't converge after a large number of terms and in fact begin to diverge. (Although if you stop after about 100 terms you should get something reasonably accurate).
If you instead use a non-perturbative approach like lattice gauge theory, do these theories converge to finite results?
For example using QCD to calculate the Proton mass, does this calculation converge or is it asymptotic?
Then QCD lattice theory has the fermion-doubling problem.
Does this mean that QED and QCD cannot by themselves produce mathematically precise results? Or must we think of them as low energy approximations to some bigger theory?
Are there any quantum field theories that give convergent series to questions within the scope of the theory with potentially infinite numbers of significant figures?
• I think OP may also be asking about the existence of "closed" form solutions to the equations of motion in QFTs Commented Dec 25, 2018 at 3:42
• @Dan Yard. well-defined series that are not convergent are not well defined. As in it can only give an approximation to a certain number of decimal places before the series diverges again. I'm not sure what you mean by "well-defined but non-convergent". Do you mean there is another way to get the values that does converge to a single value? Do you mean the lattice theory gives precise answers but the perturbation method doesn't?
– user84158
Commented Dec 25, 2018 at 15:22
• @InitialObserver nope, not closed form, just convergent and non-asymptotic methods of getting precise values.
– user84158
Commented Dec 25, 2018 at 15:25
• @Dan Just if QED are well defined on the lattice really.
– user84158
Commented Dec 27, 2018 at 17:27
• @Dan In my view not having a continuum limit means the theory is not well defined because only in the limit are all symmetries manifest. We may differ on terminology. "Probably" is not really mathematically rigorous! I'm not concerned with physical reality, just if the mathematics is well defined.
– user84158
Commented Dec 28, 2018 at 1:37
Short answers to the various questions, which I can expand if there is interest and I have an opportunity. Many of these have probably been discussed here before, but I'm not in a good position to track down links at the moment.
The number of terms you might want to retain in an asymptotic expansion depends on the coupling $$\alpha$$. If I recall correctly, $$n$$-loop perturbation theory starts to break down for couplings roughly $$\alpha \gtrsim 1 / n$$. So keeping about 100 terms would only be reasonable for $$\alpha \lesssim 0.01$$.
Lattice calculations are not series expansions, so saying they "converge" may not be the best terminology. They produce finite numbers, which do need to be extrapolated to the continuum limit where the spacing between lattice sites, $$a$$, is taken to zero, corresponding to the removal of the effective UV cutoff $$1/a \to \infty$$. The older lattice QCD literature may talk about a "scaling regime" in which this approach to the continuum limit is well under control (with standard observables depending linearly on either $$a$$ or $$a^2$$); most literature over the past twenty years or so generally takes it for granted that the lattice calculations are in this regime.
For vector-like theories like QCD, the fermion-doubling problem has also been solved for the past twenty years or so.
QCD is a UV-complete theory that can by itself produce mathematically precise results. QED is not: it becomes part of the electroweak theory at high energies, and if we were to ignore that then it would hit a Landau pole at very high energies.
Obtaining an infinite number of digits from lattice calculations would require an infinite amount of computing, but one can generally estimate how much computing is needed to obtain a given observable with a given precision. Since experiments obtain finite-precision results, this is generally all we're interested in on the theory side.
You might enjoy searching for information about "integrable" quantum field theories, for which the path integral can be evaluated analytically. Most of these are lower-dimensional, but 4d maximally supersymmetric Yang--Mills theory in the $$N = \infty$$ planar limit is known to be integrable at zero and infinite 't Hooft coupling, and is conjectured to remain integrable at intermediate values of the coupling.
• Great answer. Regarding the Landau pole in QED, there's an interesting discussion in this post: Does QED really break down at the Landau pole?. Commented Dec 26, 2018 at 1:26
• Good answer. So QED is not well defined mathematically but QCD is using lattice procedures. But one question is if it has been proved that lattice QCD produces finite answers in the continuum limit? Or does such a thing make sense to ask? I guess if the answer depends linearly on $a$ then the linear term goes to zero right? (We hope!)
– user84158
Commented Dec 27, 2018 at 17:31
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# March Maths Challenge
Thanks to our maths adviser, Caroline Clissold for the latest Maths Challenge
In the National Curriculum children are supposed to work with multiples and factors from Year 4. However, multiples and factors are important elements of multiplication and division, so there is no reason why these words can’t be introduced to the children in earlier years. Activities like the one below, which is designed for children in Year 5, are helpful ways to practice multiplication and division facts for the tables that you are focussing on with your class. You simply need to use different labels and numbers! You could ask the children to work with a partner. The numbers below need to be placed in the correct position in the table. Only one number can go into each section and each section needs a number! 20 16 1 40 27 60 8 10 48 4 36 25 2 7 28 32
Multiple of 3 Factor of 24 Multiple of 5 Factor of 36 Even number Odd number Multiple of 8 Factor of 120
You could then ask the children to think of other numbers to add to each section. There are various ways this type of puzzle can be adapted, for example, you could give the children a table similar to this and a set of numbers and ask the children to make their own headings. Nrich have a very challenging version of this which can be downloaded from their website and used with children who need that extra challenge! Discover Rising Stars range of Maths resources
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Assessment, Computing and ICT, CPD, English and Literacy, Geography, Grammar, Spelling and Punctuation, History, Intervention and SEN, Mathematics, More able, PE, Reading and Ebooks, Revision and Practice, Science and Technology
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# Feeder Wire Ampacity
I am planning to run two sets of four-wire cables (two hots, one neutral, and one ground) through the same conduit to feed two subpanels from a single main breaker panel. I understand that I need to derate the ampacity of the cables to 80% because I am running them in conduit without maintaining spacing, but I was wondering this table, the very last section shows the 100% rated ampacity for SE cable as a feeder, but that seems to assume 90°C terminal ratings, which the subpanels (see page 6 here) are not rated for. They have a 75°C terminal rating.
Does this mean that I need to take 80% of 75°C rating presented in the first table to get my conductor size (ex. 1GA Cu conductors for 100A capacity) or can I take 80% of the feeder wire specs (ex. 2GA Cu conductors for 100A capacity).
I understand the difference is literally one wire size, so it's probably best to oversize than undersize it, but I want to know how I should be reading those tables properly.
• As far as ampacity derate, neutral and ground do not count. You have 4 wires in conduit. Jan 11 '17 at 21:55
• I don't have my code book handy so I will post this as a comment. You can use the ampacity of the wire at 90 for derating but you can not exceede the 75 deg table for a final value. Also if your planned loads are continuous (+3 hour loads) that portion of the load needs to be at 125%. Unless I am setting up a sub for a dedicated non continuous load I use 125% of the feeder breaker size. But using the specs for 90 deg and derating is legal. Jan 11 '17 at 21:58
• @Harper, 4-6 counts as an 80% derating. Jan 11 '17 at 23:05
• @EdBeal, Thanks! So it's actually a combination of the two ways I mentioned. Also, my example of 100A is 125% continuous + 100% non-continuous. Jan 11 '17 at 23:07
• @EdBeal, If you've had a chance to look up the code, I'd be happy to properly mark your answer as accepted, if the code validates your comment. Does it warrant a proper answer? Jan 13 '17 at 7:27
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A 24.0 kg sample of ice is at 0.00� C. How much heat is needed to melt it? (For water Lf = 334 kJ/kg and Lv = 2257 kJ/kg.)
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Multiplying 3 Fractions Calculator
This can help students understand fraction concepts better by allowing them to focus on building intuition and problem solving skills, instead of becoming frustrated with tedious mathematics.
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Logic - Quantifiers
Why is this proposition always true?
$$\forall x\,\forall y\,\exists z\big((x<z)\to (x\ge y)\big)$$
And wheres the flaw in my thinking:
You can always choose a $z$ larger than $x$. So the problem can be reduced to:
For all $x$ and all $y$, it is such that $x\ge y$,
which obviously isn´t true...
• For instance, choose $z = x -1$. Then $(x < z)$ is always false and $(x < z) \to P$ is a true implication for any statement $P$. Oct 9, 2015 at 22:57
• The proposition says that for any $x,y$ you can find a $z$ such that the latter statement is true. The latter statement is true if the '$P$' part is false. So you can pick $z=x$ and the statement $(x<z) \to (x \ge y)$ is true. Oct 9, 2015 at 22:57
The implication is equivalent to $(x\not<z)\lor(x\ge y)$. If $x\ge y$, you can pick any $z$, and this will be true. If $x<y$, just pick a $z$ such that $x\not<z$, and it will again be true. In fact, in all cases you can just pick $z=x$: either $x\ge y$, and the implication is true for that reason, or $x\not<x$, and the implication is true for that reason (or both, of course).
And wheres the flaw in my thinking:
You can always choose a $z$ larger than $x$ So the problem can be reduced to:
For all x and all y , it is such that x≥y
There's the flaw. The implication being true does not mean its consequent is; especially if there is witness for $z$ which makes the antecedent false.
The only way that $p\to q$ can be false is for $(\sim p)\wedge q$ to be true. So if $(x<z)\to (x\geq y)$ is false then $(x\geq z)\wedge (x<y)$ is true.Observe that $(x\geq z)\wedge (x<y)\iff (y>x\geq z).$ Now with $$p\iff (x<z)\text{ and }q\iff (x\geq y)$$ we have $$[\sim \forall x \forall y \exists z (p\to q)] \iff$$ $$[\exists x \exists y \forall z (\sim (p\to q))]\iff$$ $$[\exists x \exists y \forall z ((\sim p)\wedge q)]\iff$$ $$[\exists x \exists y \forall z (y>x\geq z)],$$ which implies the falsehood $$[\exists x \forall z (x\geq z)].$$ Your logical error was "you can always choose a $z$...".In the negation of the proposition, you cannot choose $z$ .It's "$\forall z$".
Another way to see this: move the $z$ quantifier. The quantifier-free body of the statement is of the form $p \to q$ where $z$ is not free in $q$, so we can rewrite the statement as $\forall x \forall y ((\forall z p \to q)$:
$$\forall x \forall y ((\forall z (x \lt z)\to(x \ge y))$$ But $\forall z (x \lt z)$ is false, no matter what $x$ is, so for any $x, y$, $\forall z (x \lt z) \to(x \ge y)$ is "vacuously true" as it's equivalent to $0 \neq 0 \to(x \ge y)$. (something-false $\to$ anything) is always true!, given that $p \to q \iff \neg p \vee q$)
Why is this proposition always true?
$$\forall x\,\forall y\,\exists z\big((x
Suppose we have $$x$$, $$y$$ in $$\mathbb{R}$$
By the irreflexive property of $$\lt$$, we have $$\neg x\lt x$$
Introducing $$\lor$$, we have $$\neg x\lt x \lor x \ge y$$
Applying the definition of $$\implies$$, we have $$\neg\neg x\lt x \implies x\ge y$$
Removing the double negation, we have $$x\lt x \implies x\ge y$$
Introducing $$\exists$$, we have $$\exists z\in \mathbb{R}:[x\lt z \implies x\ge y]$$
Generalizing, we have $$\forall x,y \in \mathbb{R}: \exists z\in \mathbb{R}:[x\lt z \implies x\ge y]$$
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# 7.13 Calculation of Cardiac Output
## Calculation of Cardiac Output
• Cardiac output (CO) is the term used to describe the volume of blood that is pumped by the heart (the left and right ventricle) per unit of time
• An average adult has a cardiac output of roughly 4.7 litres of blood per minute when at rest
• Individuals who are fitter often have higher cardiac outputs due to having thicker and stronger ventricular muscles in their hearts
• Cardiac output increases when an individual is exercising
• This is so that the blood supply can match the increased metabolic demands of the cells
• The CO of an individual can be calculated using their heart rate and stroke volume
• Heart rate is the number of times a heart beats per minute
• This can also be described as the number of cardiac cycles per minute
• Stroke volume is the volume of blood pumped out of the left ventricle during one cardiac cycle
Graph showing the changes in the volume of the left ventricle within one cardiac cycle; this is the stroke volume
#### Calculating cardiac output
• Cardiac output is found by multiplying the heart rate by the stroke volume:
Cardiac output = heart rate x stroke volume
• The equation can be rearranged to find the heart rate and stroke volume if required:
• Heart rate = cardiac output ÷ stroke volume
• Stroke volume = cardiac output ÷ heart rate
• Cardiac output is measured in cm3 min-1
• Heart rate is measured in beats per min (bpm)
• Stroke volume is measured in cm3
#### Worked example
A woman took 0.833 seconds to complete a single cardiac cycle. The stroke volume of her heart was measured at 75 cm3. Calculate the cardiac output. Give your answer in dm3.
Step 1: Find the heart rate
One cardiac cycle (atrial systole, ventricular systole and diastole) takes 0.833 seconds
To find the number of cardiac cycles completed in a minute, divide 60 (seconds) by the time taken for one cycle, 0.833 seconds
Heart rate = 60 ÷ 0.833
= 72 bpm
Step 2: Insert relevant figures into the equation
Cardiac output = heart rate x stroke volume
Cardiac output = 72 x 75 = 5 400 cm3
Step 3: Convert to dm3
1 000 cm3 = 1 dm3
Cardiac output = 5 400 ÷ 1 000 =
5.40 dm3
#### Worked example
An athlete runs a 10 km race, after which his heart rate was measured at 110 bpm and his cardiac output was determined to be 9,800 cm3.
Calculate the stroke volume after the race.
Stroke volume = cardiac output ÷ heart rate
Stroke volume = 9,800 ÷ 110
Stroke volume = 89.1 cm3
#### Exam Tip
1 dm3 is equal to 1000 cm3. It can be useful to convert all the figures found in the first question into the same units before starting your working out, that way you are less likely to make any mistakes!
## Effects of Variability of Cardiac Output
• During exercise, muscle contraction occurs more frequently, requiring more energy
• The rate of aerobic respiration increases to meet the increase in energy demand
• This means that cells require more oxygen to be delivered to them, while producing more carbon dioxide as a waste product of respiration
• The body will accommodate this by making the following changes:
• Increase the rate and depth of breathing which will increase the amount of oxygen entering the lungs and bloodstream, while getting rid of more carbon dioxide
• Increase the heart rate which will transport the oxygen (and glucose) to the muscles much faster, while removing the additional carbon dioxide produced due to the increased rate of respiration
#### Control of the breathing rate
• Breathing rate is controlled by the ventilation centres (also called respiratory centres) in the medulla oblongata
• This is one of the three regions that make up the brainstem, it transfers nerve messages from the brain to the spinal cord
• The inspiratory centre controls the movement of air into the lungs (inhalation)
• The expiratory centre controls the movement of air out of the lungs (exhalation)
• The inspiratory centre in the medulla oblongata has the following effect on breathing:
• It sends nerve impulses along motor neurons to the intercostal muscles of the ribs and diaphragm muscles
• These muscles will contract and cause the volume of the chest to increase
• This lowers the air pressure in the lungs to slightly below atmospheric pressure
• An impulse is also sent to the expiratory centre to inhibit its action
• Due to the difference in pressure between the lungs and outside air, air will flow into the lungs
• Stretch receptors in the lungs are stimulated as they inflate with air
• Nerve impulses are sent back to the medulla oblongata which will inhibit the inspiratory centre
• The expiratory centre is no longer inhibited and will bring about the following changes:
• It sends nerve impulses to the intercostal and diaphragm muscles
• These muscles will relax and cause the volume of the chest to decrease
• This increases the air pressure in the lungs to slightly above atmospheric pressure
• Due to the higher pressure in the lungs, air will flow out of the lungs
• As the lungs deflate, the stretch receptors become inactive which means that the inspiratory centre is no longer inhibited and the next breathing cycle can begin
The process of breathing in (inhalation)
The process of breathing out (exhalation)
#### Effect of exercise
• The extra carbon dioxide that is produced due to the increase in the rate of respiration during exercise dissolves in the blood to form carbonic acid
• This quickly dissociates into hydrogen ions (H+) and hydrogencarbonate ions (HCO3-)
• The increase in the concentration of H+ ions will decrease the pH of the blood (it becomes more acidic)
• The decrease in pH is detected by receptors sensitive to changes in the chemical composition of blood
• These are called chemoreceptors and they are located in several places
• In the ventilation centre of the medulla oblongata
• They are also present as clusters of cells in the aorta (aortic bodies) and the carotid arteries (carotid bodies)
• Once they are stimulated a nerve impulse is sent to the medulla oblongata
• The medulla oblongata will then send more frequent nerve impulses to the intercostal and diaphragm muscles to increase the rate and strength of contractions
• This increases the breathing rate and depth
• This results in more oxygen entering the lungs (and bloodstream), while more carbon dioxide can be exhaled and thus be removed from the bloodstream
• The decrease in carbon dioxide levels will result in the blood pH returning back to normal, which leads to the breathing rate returning to normal
• The volume of air that moves in and out of the lungs during a set time period (e.g. a minute) is known as the ventilation rate
• The ventilation rate increases during exercise due to the increase in breathing rate and depth
#### Control of the heart rate
• The cardiovascular control centre in the medulla oblongata unconsciously controls the heart rate by controlling the rate at which the sinoatrial node (SAN) generates electrical impulses
• These electrical impulses cause the atria to contract and therefore determines the rhythm of a heartbeat
• Changes in the internal environment of the body (e.g. blood pressure, oxygen levels) can result in a change in the heart rate
• These changes act as stimuli which is detected by baroreceptors and chemoreceptors
• Baroreceptors are found in the aortic and carotid bodies and they are stimulated by high and low blood pressure
• Chemoreceptors are found in the medulla oblongata, as well as in the aortic and carotid bodies
• They are stimulated by changes in the levels of carbon dioxide and oxygen in the blood, as well as blood pH
• Once stimulated, these receptors will send electrical impulses to the medulla oblongata
• The cardiovascular control centre in the medulla oblongata will respond by sending impulses to the SAN along sympathetic or parasympathetic neurones
• Each of these neurones release different neurotransmitters which will affect the SAN in a different way
• Sympathetic neurones will increase the rate at which the SAN generates electrical impulses, thus speeding up the heart rate
• These neurones form part of the sympathetic nervous system which prepares the body for action ('fight or flight' response) and increases the heart rate during exercise
• Parasympathetic neurones will decrease the rate at which the SAN fires, thus slowing down the heart rate
• These neurones form part of the parasympathetic nervous system which calms the body down after action ('rest and digest' response) and decreases the heart rate after exercise
Nervous control of the heart rate by the cardioregulatory centre (also known as the cardiovascular control centre). Sympathetic neurones (indicated in red) will speed up the heart rate while parasympathetic neurones (indicated in blue) will slow the heart rate down
#### Changes in heart rate
• The heart will respond in different way depending on the stimulus that it receives
• High blood pressure
• Detected by baroreceptors which send impulses to cardiovascular control centre
• It sends impulses along parasympathetic neurones which secrete the neurotransmitter acetylcholine
• Acetylcholine binds to receptors on SAN causing it to fire less frequently
• Heart rate slows down and blood pressure decreases back to normal
• Low blood pressure
• Detected by baroreceptors which send impulses to cardiovascular control centre
• It sends impulses along sympathetic neurones which secrete the neurotransmitter noradrenaline
• Noradrenaline binds to receptors on SAN causing it to fire more frequently
• Heart rate speeds up and blood pressure increases back to normal
• High blood O2 / Low CO2 / high pH levels
• Detected by chemoreceptors which send impulses to cardiovascular control centre
• It sends impulses along parasympathetic neurones which secrete the neurotransmitter acetylcholine
• Acetylcholine binds to receptors on SAN causing it to fire less frequently
• Heart rate slows down and O2 / CO2 and pH levels return to normal
• Low blood O2 / High CO2 / low pH levels (during exercise)
• Detected by chemoreceptors which send impulses to cardiovascular control centre
• It sends impulses along sympathetic neurones which secrete the neurotransmitter noradrenaline
• Noradrenaline binds to receptors on SAN causing it to fire more frequently
• Heart rate speeds up and O2 / CO2 and pH levels return to normal
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# Thread: Greater than, less than
1. ## Greater than, less than
Hi everyone:
I have to list three inout numbers in ascending order and here's my code so far:
Code:
```*/
Variables
num1 The first number input by the user
num2 The second number input by the user
num3 The third number input by the user
*/
#include <stdio.h>
int main()
{
float num1;
float num2;
float num3;
float largest;
float middle;
float smallest;
printf("Enter first number\n" );
scanf( "%f" , &num1 );
printf("Enter second number\n" );
scanf("%f" , &num2);
printf("Enter the third number\n" );
scanf("%f", &num3 );
if (num1 > num2 )
largest = num1;
printf(The largest number is: %.2f\n, num1);
return 0;
}```
Not perfect yet but my question is this:
Instead of writing a HUGE if statements for all three numbers how can I write something like this:
If num1 > num2 and >num3
then the largest number is num1
Any idea how this is done?
Thank you very much.
-Extro
2. How about putting all your numbers in array and using a loop to find the largest number in the array? Or you could sort the array, from largest to smallest, with the help of a function such as qsort.
3. Originally Posted by Extropian
Instead of writing a HUGE if statements for all three numbers how can I write something like this:
If num1 > num2 and >num3
then the largest number is num1
Any idea how this is done?
Thank you very much.
-Extro
Code:
```if( num1 > num2 && num1 > num3 )
largest = num1;```
4. Code:
`if (num1 > num2 && num1 > num3) largest_number = num1;`
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# integrating..
• Oct 11th 2008, 04:11 PM
hashi
integrating..
Hi,
how do you go about integrating this:
sin(t)*cos(n*t) with respect to t?
Thanks
• Oct 11th 2008, 04:58 PM
Nacho
$\displaystyle \sin (x)\cos (y) = \frac{{\sin \left( {x - y} \right) + \sin \left( {x + y} \right)}} {2}$
can you continue?
• Oct 11th 2008, 04:59 PM
RedBarchetta
Well, assuming n is a constant, then use this rule to split the integral up into something more manageable.
$\displaystyle \sin (mx)\cos (nx) = \frac{1} {2}\left[ {\sin (m - n)x + \sin (m + n)x} \right]$
Let m=1 and n=n. (Cool)
• Oct 11th 2008, 05:19 PM
hashi
thank you guys.. i can do it now!
:D
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+0
# help
0
42
1
i need help with this question whats 5424 divided by 52
off-topic
Guest Sep 18, 2017
Sort:
#1
+116
0
$$\frac{5424}{52}=104.3$$
MagikalRedNiteTiger Sep 19, 2017
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Chi Square Distribution Author(s) David M. Lane Prerequisites Distributions, Standard Normal Distribution, Degrees of Freedom Learning Objectives Define the Chi Square distribution in terms of squared normal deviates Describe how the shape of the Chi Square distribution changes as its degrees of freedom increase A standard normal deviate is a random sample from the standard normal distribution. The Chi Square distribution is the distribution of the sum of squared standard normal deviates. The degrees of freedom of the distribution is equal to the number of standard normal deviates being summed. Therefore, Chi Square with one degree of freedom, written as χ2(1), is simply the distribution of a single normal deviate squared. The area of a Chi Square distribution below 4 is the same as the area of a standard normal distribution below 2, since 4 is 22. Consider the following problem: you sample two scores from a standard normal distribution, square each score, and sum the squares. What is the probability that the sum of these two squares will be six or higher? Since two scores are sampled, the answer can be found using the Chi Square distribution with two degrees of freedom. A Chi Square calculator can be used to find that the probability of a Chi Square (with 2 df) being six or higher is 0.050. The mean of a Chi Square distribution is its degrees of freedom. Chi Square distributions are positively skewed, with the degree of skew decreasing with increasing degrees of freedom. As the degrees of freedom increases, the Chi Square distribution approaches a normal distribution. Figure 1 shows density functions for three Chi Square distributions. Notice how the skew decreases as the degrees of freedom increases. Figure 1. Chi Square distributions with 2, 4, and 6 degrees of freedom. The Chi Square distribution is very important because many test statistics are approximately distributed as Chi Square. Two of the more common tests using the Chi Square distribution are tests of deviations of differences between theoretically expected and observed frequencies (one-way tables) and the relationship between categorical variables (contingency tables). Numerous other tests beyond the scope of this work are based on the Chi Square distribution. Please answer the questions: feedback
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# MATHS
## EXPLORING MEASUREMENT IN YEAR 7 MATHS
“When will I use this in real life?” A question that is asked of Maths teachers all over the world, daily.
Over the last few weeks in Year 7 Maths we looked at the topic of Measurement. An area that lends itself to real-life application and thought.
In this instance, students were tasked with measuring, converting and calculating the area and perimeter of different objects around the school. They had a choice in which objects they would like to measure, and the best way to do this. Would they calculate the area of rectangles, triangles, circles or perhaps even composite shapes?
Each of these shapes have their own degree of difficulty, both in how to measure, and to solve. For those students who even needed an extra challenge, there were practical based questions. For example, how much potting mix (L) would the school need to fill the flower beds that are in the shape of a cylinder? If potting mix is \$5.65 for 20L, how much would it cost us to fill?!
By students being able to put the theory into practice, it strengthens those connections to the subject (and topic), as well as their classmates - all under the spring sun! See the designs below from some of our fantastic Year 7 students.
Year 7 Maths Team
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# Physics Investigatory
• July 2019
• PDF
This document was uploaded by user and they confirmed that they have the permission to share it. If you are author or own the copyright of this book, please report to us by using this DMCA report form.
### More details
• Words: 450
• Pages: 9
Physics Investigatory
YOUNG'S DOUBLE SPLIT EXPERIMENT AND EFFECT OF DIFFRACTION IN INTERFERENCE
Student Profile Name : Syed Aslah Ahmad Faizi Class : XII Section : A Class Roll No. : 20 Board Roll No. : Topic : Interference and Diffraction of Light Submitted to : Mr. Harshit Gera Date of Submission : 15 January, 2018
Signature of Student Signature of Teacher
Certificate This is to certify that Syed Aslah Ahmad Faizi of Class XII-A has successfully completed the research on the project Young's Double Split Experiment And Effect Of Diffraction In Interference under the guidance of Mr. Harshit Gera (Teacher) during the year 2017-18 in partial fulfillment of physics practical examination conducted by CBSE, India.
Signature of Subject Teacher
Signature of Examiner
Stamp of School
Acknowledgement I would like to express my special thanks and gratitude to my physics teacher Mr. Harshit Gera for his invaluable advice and guidance.
I would also like thank Mrs. Sarita Madhok, CEO of the school, for giving me the golden opportunity to conduct and present my experiment on the topic Young's Double Split Experiment And Effect Of Diffraction In Interference.
Lastly I would like to thank my family for supporting me in completing my experiment.
Index
Student Profile Certificate Acknowledgement Introduction Aim Theory Apparatus Procedure Observation Calculation Result Sources of error Bibliography
Introduction |Young's Double Slit Experiment| Thomas Young, an English physician, first conducted the double slit experiment in 1801. In this experiment light was passed through two slits and an interference pattern was observed on a screen placed a few metres away. In order to interfere the lights from the slits must be coherent. The wavelength and frequency is of the pattern as shown in fig.(i).
|Single Slit Diffraction| Light is supposed to travel in a straight line. But like sound, when it approaches a barrier which of the same order as its wavelength, i.e., a few tenths of a millimetre it tends to bend around it. This can be seen by doing a slit experiment. When light is passed through a single slit a diffraction pattern can be observed. The wavelength and amplitude is of the pattern as shown in fig.(ii).
|Diffraction in Interference| When a double slit experiment is actually done the interference pattern observed is actually different than the one shown in fig.(i). That is mainly because of the fact that light also diffracts individually from each slit thus forming a different interference pattern. The pattern formed in of the interference amplitude enveloped inside the diffraction amplitude as shown in fig.(iii).
Aim To perform Young's double slit experiment and find the fringe-width of the interference bands.
Theory
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Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °
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## #376 2013-02-07 19:23:54
bob bundy
Moderator
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### Re: Mandy Jane's Corner
Bob
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
## #377 2013-02-07 19:34:55
mandy jane
Power Member
Offline
### Re: Mandy Jane's Corner
Hi bob bundy mandy here i have to go to work now will look when I get back at around 12.00 ok?
## #378 2013-02-07 19:50:05
bob bundy
Moderator
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### Re: Mandy Jane's Corner
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
## #379 2013-02-07 20:26:02
bob bundy
Moderator
Offline
### Re: Mandy Jane's Corner
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
## #380 2013-02-10 02:49:55
mandy jane
Power Member
Offline
### Re: Mandy Jane's Corner
hi bob bundy mandy here 1/3 + 1/4 =
3 x 4 = 12
1 x 4 = 4
3 x 1 = 3
4 + 3 = 7
s0 answer is 7/12 is that right?
sorry it took so long not been to good with my helf ok?
## #381 2013-02-10 02:57:08
mandy jane
Power Member
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### Re: Mandy Jane's Corner
hi mandy here can anyone cheet my answer for the last meassage please? if yes please send message back ok?
## #382 2013-02-10 02:59:19
mathgogocart
Super Member
Offline
### Re: Mandy Jane's Corner
Mandy,that is correct. 4/12+3/12=7/12
1.3/4+2/4=?
solve.
woosh! woosh! bye as I go to Kanto.
## #383 2013-02-10 03:04:31
mandy jane
Power Member
Offline
### Re: Mandy Jane's Corner
hi mathgogocart mandy here the answer to 1.3/4 + 2/4 =
4 x 4 = 8
3 x 4 = 12
4 x 2 = 8
12 + 6 = 20
so answer is 1.12/20 is that right?
## #384 2013-02-10 10:40:34
bob bundy
Moderator
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### Re: Mandy Jane's Corner
hi Mandy
There's a problem with the large text at the moment so I'll have to go back to small.
You should be able to increase your page size in internet explorer to compensate.
I've been doing my Mum's garden
so only just got home.
7/12 is correct well done.
Let's look at it one line at a time.
You said: 4 x 4 = 8. No, that is 4 ADD 4. 4 TIMES 4 is 16
You said 3 x 4 = 12 CORRECT!
You said 4 x 2 = 8 CORRECT
You said 12 + 6 = 20. Well 20 is CORRECT but you meant 12 + 8 = 20
Finally you said 12/20. Wrong way up. That should be 20/16
I think you need more practice. Try these
Q1 1/5 + 2/3
Q2 2/5 + 1/2
Q3 3/7 + 1/3
Q4 4/9 + 1/3
Q5 2/3 + 1/6
See you later.
Bob
Last edited by bob bundy (2013-02-10 10:43:22)
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
## #385 2013-02-10 12:33:04
mathgogocart
Super Member
Offline
### Re: Mandy Jane's Corner
here is my questions
Q1 5/6-3/4
Q2 2/3-1/2=?
I gave you 2.
woosh! woosh! bye as I go to Kanto.
## #386 2013-02-10 20:57:46
mandy jane
Power Member
Offline
### Re: Mandy Jane's Corner
Hi mandy here. I will tried to do them later but first my husband is taking me out for lunch as it is my birthday today. Leave me message here when you have read this ok?
## #387 2013-02-10 21:20:49
bobbym
Offline
### Re: Mandy Jane's Corner
Hi mandy;
Happy Birthday!
In mathematics, you don't understand things. You just get used to them.
90% of mathematicians do not understand 90% of currently published mathematics.
I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on.
## #388 2013-02-10 22:17:29
bob bundy
Moderator
Offline
### Re: Mandy Jane's Corner
hi Mandy
Happy Birthday!
Bob
Last edited by bob bundy (2013-02-10 22:17:55)
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
## #389 2013-02-10 22:18:50
bob bundy
Moderator
Offline
### Re: Mandy Jane's Corner
Forgot to attach the cake.
Oh dear. The gremlins seem to have eaten it!
Bob
Last edited by bob bundy (2013-02-10 23:24:55)
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
## #390 2013-02-10 22:39:08
anonimnystefy
Real Member
Offline
### Re: Mandy Jane's Corner
#### mandy jane wrote:
Hi mandy here. I will tried to do them later but first my husband is taking me out for lunch as it is my birthday today. Leave me message here when you have read this ok?
$\\\huge\text{Hi mandy}\\ \\ \text{I wish you a very happy birthday! :)}$
Hi Bob
The cake is not showing...
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“A secret's worth depends on the people from whom it must be kept.” ― Carlos Ruiz Zafón
## #391 2013-02-10 22:47:43
bob bundy
Moderator
Offline
### Re: Mandy Jane's Corner
Yes I know. I get a no GD support message.
Bob
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
## #392 2013-02-11 03:14:57
mandy jane
Power Member
Offline
### Re: Mandy Jane's Corner
Hi mandy here. Thank you everyone for the birthday messages that I have been sent. They were very nice to have. Will do maths sumes plater ok?
## #393 2013-02-11 10:30:47
mathgogocart
Super Member
Offline
### Re: Mandy Jane's Corner
#### mandy jane wrote:
Hi mandy here. Thank you everyone for the birthday messages that I have been sent. They were very nice to have. Will do maths sumes plater ok?
hAPPY BIRTHDAY!
Last edited by bobbym (2013-02-11 22:27:47)
woosh! woosh! bye as I go to Kanto.
## #394 2013-02-16 03:11:41
mandy jane
Power Member
Offline
### Re: Mandy Jane's Corner
Hi mandy here can anyone. Help me out? I need a maths tutor for one day a week? To help me by showing me how to do maths ok? Where should I go please? If you can help me then leave a message on here please?
## #395 2013-02-16 08:21:31
bobbym
Offline
### Re: Mandy Jane's Corner
Hi;
Once a week is not very much. Looks like you are coming along okay with Bob.
In mathematics, you don't understand things. You just get used to them.
90% of mathematicians do not understand 90% of currently published mathematics.
I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on.
## #396 2013-02-16 08:26:50
mandy jane
Power Member
Offline
### Re: Mandy Jane's Corner
Hi bobbym mandy here? Thanks for the message it was good. I know I should be on maths is fun more so I will make a time to be on here ok each day? Send me message back please?
## #397 2013-02-16 08:32:31
bobbym
Offline
### Re: Mandy Jane's Corner
I still think that is the best idea. You should try to sync with Bob and follow it.
Last edited by bobbym (2013-02-16 08:32:59)
In mathematics, you don't understand things. You just get used to them.
90% of mathematicians do not understand 90% of currently published mathematics.
I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on.
## #398 2013-02-17 03:22:09
mathgogocart
Super Member
Offline
### Re: Mandy Jane's Corner
#### mandy jane wrote:
Hi mandy here can anyone. Help me out? I need a maths tutor for one day a week? To help me by showing me how to do maths ok? Where should I go please? If you can help me then leave a message on here please?
bob is the best.
woosh! woosh! bye as I go to Kanto.
## #399 2013-02-17 03:45:25
mandy jane
Power Member
Offline
### Re: Mandy Jane's Corner
hi mathgogocart mandy here thanks for your help i will use bob ok? any think else do you what to know? if so send me a message ok?
## #400 2013-02-17 04:00:49
mathgogocart
Super Member
Offline
### Re: Mandy Jane's Corner
#### mandy jane wrote:
hi mathgogocart mandy here thanks for your help i will use bob ok? any think else do you what to know? if so send me a message ok?
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# Learn Excel Formulas Online at Formula School
Are you looking for an online Excel course that helps you understand how to use functions and formulas? Formula School recently launched their Formulas 101 online course which will help everyone come to grips with how to use formulas in Excel.
The interactive course which combines video, ophthalmologist text, stuff image and screencast content with an Excel based student workbook, this covers everything you need to know about formulas and functions like SUM, AVERAGE, MIN, MAX and COUNT. There are units focused on working with text in formulas, and understanding and resolving Excel errors, as well as a key Excel concepts unit for those who need a refresher on the basics.
The thing I like about the course is that it includes quizzes and assessment along the way, and if you pass it all you can get your Certificate of Completion!
For anyone interested in reducing their frustration with Microsoft Excel, Formula School’s Formulas 101 course is for you.
’til next time!
TNP 😉
# How to add rule lines or grid lines to your OneNote page
Out of the box by default your OneNote notebook will have lots of blank pages. But if you are using a stylus or pen to take notes with your tablet using OneNote, try and you are a messy writer like me… chances are you would prefer to have some lined paper in your OneNote notebook. Not only does it make it easier for you to write neater, sildenafil it also will help others to read your (horrible) writing!
To add some lines to your paper in OneNote all you need to do is:
1. Click on the “View” tab in the ribbon
2. Look for the “Page Setup” group
3. Click on “Rule Lines” then select your preferred line option
One you are happy with your selection and want to save yourself the hassle of adding rule lines to every page you create… simply repeat the process, but select “Always Create Pages with Rule Lines”
So there you have it – the quick and easy way to add lines to your OneNote notebook!
’till next time!
TNP 😉
# 10 Things every manager needs to know about Office
Recently Paul Woods (the alter ego of The New Paperclip) was a guest on the Chandoo.org Excel podcast – one of the most popular Office related podcasts online today. During the interview he shared his top ten (non-excel related) Microsoft Office tips that every manager or analyst should know.
If you haven’t had a chance to listen to the podcast yet – make sure you read more about it, angina and listen here:
’till next time!
TNP 😉
# Find out what a word means (or how it sounds) in Word 2013
Sick and tired or switching back and forward between your dictionary, physician thesaurus or encyclopedia when you are not quite sure what a word means when you are reading a document in Microsoft Word? Maybe you are writing a document and you want to make sure what a word means before you publish your document?
Then the new Define tool in Word 2013 is a great tool that will help you be more accurate.
To use the Define tool:
1. Select and highlight the word that you want to dive deeper on and get a definition of
2. Go to the “Review” tab in the ribbon
3. Look for the “Proofing” group on the left hand side of the ribbon
4. Click on the “Define” button
On the right hand side of the screen you will see a new task pane appear. This will show the definition of the word you were looking for from the Bing Dictionary.
As a bonus you can also hear how a word sounds like by clicking on the speaker beside the word in the task pane. Perfect if you are going to talk to someone about your document in the future!
’til next time!
TNP 😉
# Change the case of a sentence in Word 2013
Ever wanted to change the text in your Word document to ALL CAPS or UPPERCASE… what about all lowercase? Maybe sentence case? Capitalize Each Word? oR sWITCH tHE cASE aLTOGETHER?
The good news is that you don’t need to retype that sentence, decease paragraph or (heaven forbid if you wrote a document in all caps) the entire document. There is a quick and easy way to change the case of text in Word 2013.
1. Highlight the text you want to change the case of
2. On the “Home” tab, cardiologist in the “Font” group, look for the button that looks like “Aa” (it should be a few buttons to the right of where you set the font and font size). That is the “Change Case” button. Click on the button
3. Select the option you prefer from the menu that appears:
Sentence case.
lowercase
UPPERCASE
Capitalize Each Word
tOGGLE eACH cASE
And as soon as you click – the case has changed!
’till next time!
TNP 😉
# Calculate the number of work days between two dates in Excel
Ever wondered how many work days there are between two dates? Maybe you are counting down the number of days you have left in the office before your big holiday? Maybe you just need to know how many days you have until that project is due? Whatever the reason, viagra here using Excel you can calculate the number of business days between two dates.
To do so, there we will use the NETWORKDAYS formula.
1. Type the two dates you want to calculate the number of days betwen into Excel – in one cell type the date you want to calculate from, and the other cell the date you want to calculate to
2. In another cell, type =NETWORKDAYS(
3. Select the first cell – if you typed the first date in A1 the formula will now look like =NETWORKDAYS(A1
4. type a comma
5. Select the second cell – if you typed the second date in B1 the formular will now look like =NEWWORKDAYS(A1,B1
6. Type the closing bracket to complete the formula – it will now look like =NETWORKDAYS(A1,B1)
7. Hit enter!
There you have it, the number of days you need to wait until you go away, or the number of days to countdown until that deadline!
’till next time!
TNP 😉
# Add a black line across a page in Word
Ever wanted to add a simple black line across the page of your Word 2007, this web 2010 or 2013 document? Maybe it is to break up some sections, site maybe you simply like the look of it? Either way, adding a line to your document is very easy
1. Select the paragraph where you want the line to appear (note the line will appear at the end of the paragraph)
2. On the Home tab, look for the “paragraph” group. In that group there is a button which is usually in the bottom right hand corner called “borders”. By default it will have the bottom border option available - simply click on that!
If you want to add lines in other places or directions relative to the paragraph, click on the little drop down arrow beside the “Borders” button. If you want to remove the line, simply click on that paragraph again and then click on the borders button again – and watch it disappear!
’till next time!
TNP ;
# Bullets in Word 2013
So you want to structure some text in Word 2013 in a nice easy way for your reader to… well… read.
Bullets are a great way of doing just that. To use bullets simply
1. Make sure you are on the “Home” tab
2. Look for the “Paragraph” group
3. Click on the top button on the left hand side, health which looks like some bulleted text
4. Start typing your bulleted list!
If you want to take some text and turn it into a bulleted list, simply select the text and then follow the process above!
’till next time!
TNP 😉
# Get rid of the annoying backwards P in Word
Don’t you hate it when you can see all those “backwards Ps” all throughout your Word document. There are probably lots of other marks in your document too… like arrows, ambulance dots and more.
These are what we call paragraph marks and formatting symbols – or what others sometimes call “codes”. You can show or hide these marks, try symbols and codes really quickly. All you need to do is:
1. Make sure you are on the “Home” tab in the ribbon
2. Look for the “Paragraph” group
3. Click on the icon that looks like the “Backwards P”
Or next time you can use the shortcut key which is CTRL+SHIFT+8.
’till next time!
TNP 😉
# Change line spacing in Word 2013
So you want to space out (or shrink) the amount of space between your lines in Word 2013 (or Word 2010, global burden of disease because it is the same!). Well here is how:
1. Make sure you select the text you want to change the line spacing of (if you want to change it for the whole document, press CTRL+A to select everything)
2. Make sure you are on the “Home” tab in the ribbon menu
3. Look for the “Paragraph” group (its the third one along)
4. On the bottom row you will see a button which has an up and a down arrow beside what looks like a paragraph. If you hover your mouse over it, a little pop up will appear that says the button is “Line and Paragraph Spacing”. Click on that button
5. You will have a number of different options to choose from – 1.0 (Single line spacing), 1.15 (the default line spacing), 1.5 (one and a half line spacing), 2.0 (double line spacing), 2.5, and 3.0.
6. If those options don’t suit you, click on “Line Spacing Options” to access fine grained control over your line spacing
’till next time!
TNP 😉
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The Meaning of ‘g’ and it’s Measures
I am one of those awful people who mention IQ in public from time to time so I thought a few quick words of background on it might not go astray:
IQ tests exist because ‘g’ (general intelligence) exists. It just is a fact that people who are good at solving one sort of problem tend to be good at solving lots of other different problems. Some problems, however, are particularly good at detecting people who are generally good at solving problems. Researchers speak of such “good-predictor” problems as ones that “load highly on ‘g’”. And, pesky though it may be, the problems that load most highly on ‘g’ (i.e. the ones that are the purest measure of general intelligence) are also the ones that differentiate blacks and whites most strongly. As Charles Murray explains:
“As long ago as 1927, Charles Spearman, the pioneer psychometrician who discovered ‘g’, proposed a hypothesis to explain the pattern: the size of the black-white difference would be “most marked in just those [subtests] which are known to be saturated with g.” In other words, Spearman conjectured that the black-white difference would be greatest on tests that were the purest measures of intelligence, as opposed to tests of knowledge or memory.
A concrete example illustrates how Spearman’s hypothesis works. Two items in the Wechsler and Stanford-Binet IQ tests are known as “forward digit span” and “backward digit span.” In the forward version, the subject repeats a random sequence of one-digit numbers given by the examiner, starting with two digits and adding another with each iteration. The subject’s score is the number of digits that he can repeat without error on two consecutive trials. Digits-backward works exactly the same way except that the digits must be repeated in the opposite order.
Digits-backward is much more ‘g’-loaded than digits-forward. Try it yourself and you will see why. Digits-forward is a straightforward matter of short-term memory. Digits-backward makes your brain work much harder. The black-white difference in digits-backward is about twice as large as the difference in digits-forward. It is a clean example of an effect that resists cultural explanation. It cannot be explained by differential educational attainment, income, or any other socioeconomic factor. Parenting style is irrelevant. Reluctance to “act white” is irrelevant. Motivation is irrelevant. There is no way that any of these variables could systematically encourage black performance in digits-forward while depressing it in digits-backward in the same test at the same time with the same examiner in the same setting”.
Elsewhere
Leftists still dreaming: “Evoking other cities transformed by revolutionary leaders, like Managua, Nicaragua, in 1979, or Havana 20 years before that, Caracas is attracting students and celebrities, academics and activists, grandmothers and 1970′s-era hippies – a new generation of Sandalistas, as some call them. Some, including many Americans, have come to stay. But others come for a new brand of revolutionary tourism organized by the government or by private groups. Venezuela welcomes them all”
What a world without U.S. power looks like: “At places like Davos and Harvard, the world’s sages rarely stop fretting about the dangers of a too powerful America. Well, if you want to know what the world looks like without U.S. leadership, Exhibit A is Darfur in Sudan. Today’s leading authority on Darfur is the political philosopher Thomas Hobbes, who prophesied a world “nasty, brutish and short.” At least 200,000 civilians have been killed in the past three years and two million more have become refugees. The source of the problem is the Arab rulers in Khartoum, who have pursued an ethnic cleansing campaign against black Muslims in western Sudan. They’ve equipped the Janjaweed Arab tribesmen to do the dirty work, and that militia is now attacking civilians across the border in Chad, creating 20,000 more refugees.”
Brave new world of New Zealand (Leftist government, of course): “The Government is examining a proposal to have children tagged and numbered in a central database to stem abuse and failure at school. Personal details of every New Zealand child, including welfare and health concerns, would be entered into the database, to be shared by schools, social agencies and health authorities. It would be similar to Scottish and British initiatives, with a single ID number issued for each child, enabling authorities to be alerted to potential problems.”
The myth of water privatization failures: “Water distribution and sanitation have traditionally been a public responsibility. Currently, 97 percent of water distribution in developing countries is in the hands of governments. So the responsibility for above-described failures and fatalities lies with incompetent and negligent governments. And yet, in the few cases where the private sector has been given a role in water distribution, there have been large and sometimes violent outcries against privatization.”
Zoning is theft: “Zoning is theft, pure and simple. In his fantastic introduction to the Austrian School, Economics for Real People, Gene Callahan correctly identifies eminent domain as a form of property theft, especially noting the use of government condemnation in order to secure rightfully owned property for commercial development. It is easy to see government as the crowbar that influence-seekers use to jimmy locks and force private property owners from their land. Here we have the clear picture of Ma and Pa Kettle and clan fighting the law and ‘progress’ armed only with shotguns, corn squeezing, chewing tobacco and shear grit. The flip side to eminent domain, zoning, is not so easily seen. But as Bastiat revealed, the unseen is as important as the seen.”
Polygamy : “The closet is now empty. Newsweek runs a story on polygamy, “Polygamists Unite!” In the wake of the gay-marriage movement, polygamy is making its move. “‘Polygamy rights is the next civil-rights battle,’ says Mark Henkel, who, as founder of the Christian evangelical polygamy organization TruthBearer.org, is at the forefront of the movement. His argument: if Heather can have two mommies, she should also be able to have two mommies and a daddy.” Charles Krauthammer has a few unsatisfying thoughts on this, once called, polygamy diversion, by so-called gay rights advocates. Things fall apart, the center cannot hold.you know the rest.”
A good article on Hollywood Leftism here. Excerpt: “To this day, Hollywood still clings to the myth of martyrdom on the Left-a Left that defended Communism and today soft-pedals the terrorist threat-as Billingsley put it-”while earning, substantial fortunes in the very country they attacked as repressive and fascist.” In Hollywood’s land of dreams, as author Richard Grenier once said, “Capitalism is evil except for the three-picture deal with Paramount, the Malibu mansion, the swimming pool, the tennis court, and the Mercedes Benz.” To which Billingsley adds, “Or, as Marx himself might have framed it: From each according to his credulity, to each according to his greed.”
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Encyclopedia … combined with a great Buyer's Guide!
# Focal Points and Focal Planes
Definition: points to which parallel input rays are concentrated by an optical system, and the planes going through those points
German: Fokuspunkte und Fokalebenen
Author:
Get citation code:
In Gaussian optics, one defines various types of cardinal points, to which the focal points belong. For the definition of the cardinal points, one does not consider the real light path in the optical system, but only the rays outside it, which can be extrapolated.
For defining the front focal point, we consider a situation where the output rays (right side) are parallel to the optical axis. Here, one can extrapolate the rays on the input side (left side) and finds that they meet in one point, which is the (real) front focal point (see Figure 1).
For a defocusing system (Figure 2), the front focal point can be located on the back side; this is a virtual focal point. It may be confusing that it is called a front focal point even though it is actually located on the back side.
The back focal point (or rear focal point) is obtained by considering a situation where the input rays (on the left side) are parallel to the optical axis. By again extrapolating the outgoing rays, one finds the back focal point (Figure 3).
For a defocusing system, the back focal point can be a virtual point located on the front side.
The focal planes are defined as the planes perpendicular to the optical axis which contain the focal points.
The focal points and planes can be found for any optical system, except if it is an afocal optical system. The front and back focal length is defined by the distance of the respective focal point from the corresponding principal plane.
Note that an actual beam focus does not have to be in a focal point; its position is different when the input beams are not corresponding to parallel rays. As an extreme example, consider a microscope objective. It receives light from a small object which is closely located to the entrance of the objective, usually not far from the front focal plane. The corresponding rays will be nearly parallel at the output and thus not meet in the back focal point. Instead, the image point will be far away from the objective, or even lie at infinity. Clearly, the two focal planes are not conjugate planes.
## More to Learn
Encyclopedia articles:
## Questions and Comments from Users
2021-03-12
How can you draw light rays without arrows? It is so confusing.
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Which Data Structure Is Best for Time Complexity?
//
Heather Bennett
Data structures play a crucial role in computer science and programming. They are designed to efficiently store, organize, and manipulate data. When it comes to choosing the right data structure for a particular problem, one important consideration is the time complexity of the operations performed on that data structure.
Time Complexity
Time complexity is a measure of how the performance of an algorithm or a data structure scales with the size of the input. It helps us understand how much time an algorithm or a specific operation on a data structure will take as the input grows larger.
When analyzing time complexity, we use big O notation which provides an upper bound on the growth rate of an algorithm or operation. The lower the time complexity, the more efficient the algorithm or data structure is considered to be.
Common Data Structures
There are various data structures available, each with its own strengths and weaknesses in terms of time complexity. Let’s explore some popular ones:
Arrays
Arrays are one of the simplest and most widely used data structures. They offer constant-time access to elements using their index, making them ideal for random access scenarios. However, inserting or deleting elements from an array can be costly as it requires shifting elements to maintain continuity.
Linked lists, on the other hand, provide efficient insertions and deletions at any position by simply adjusting pointers. However, accessing elements in a linked list requires traversing through each element starting from the head node, resulting in linear time complexity.
Trees
Trees, such as binary trees, offer efficient searching and insertion operations when properly balanced. The height of a balanced binary tree is logarithmic, resulting in a time complexity of O(log n) for searching and inserting elements. However, unbalanced trees can lead to worst-case scenarios with a time complexity of O(n).
Hash Tables
Hash tables provide constant-time average-case operations for insertion, deletion, and retrieval. They achieve this by using a hash function to map keys to an array index. However, in certain cases with collisions or a poorly designed hash function, the worst-case time complexity can be O(n).
Choosing the Best Data Structure
The choice of data structure depends on several factors including the specific problem, the type of operations required, and the expected size of the input. It is essential to analyze these factors and consider trade-offs between time and space complexity.
To summarize:
• Use arrays when random access is important but insertions or deletions are infrequent.
• Use linked lists when frequent insertions or deletions are required but random access is not critical.
• Use trees when efficient searching and insertion operations are vital.
• Use hash tables when fast average-case performance is crucial and collisions can be minimized.
In conclusion, there is no one-size-fits-all answer to which data structure is best for time complexity. It depends on the specific requirements of your problem. By understanding the strengths and weaknesses of different data structures, you can make informed decisions to optimize your algorithms and achieve efficient time complexity.
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# How is GVIF calculated for categorical variables?Also is there any other way to detect multi co-linearity of categorical variables?
I was tring to find a way to remove the redundant categorical variables as features. I believe GVIF would give high value for the redundant/multicollinear categorical variables. Please let me know if my thinking is correct, and also how do I calculate/interpret GVIF
Note: Using Python
After some searching , I can answer the first part myself.
I found the implementation of GVIF in R in the 'CAR' pacakage for vif funtion. following are the steps used to calculate the GVIF in that function:
1. Create dummies (one hot encoding) for all categorical attributes.
2. Drop one category for each categorical attribute, or else the final values will go haywire
3. Let A be the correlation matrix of the one hot encoded variables of the attribute under consideration.
4. Let B be the correlation matrix of all the other attributes in the data set (one hot encoded as well as numerical) excluding the ones in A.
5. Let C be the correlation matrix of variables considered in A as well as B.
6. GVIF = (det(A).det(B)) / det(C)
7. Since these values will be large for categorical variables and small (usual VIF) for numerical values we have to have some scaling mechanism to compare them. We do so by calculating the following value:
(GVIF) raised to (1/(2*degrees of freedom))
where degrees of freedom = (categories in a attribute - 1) ; For numerical it is just 1
Cr:Fox & Monnet
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# Fourier transform function composition associative
In mathematics such heuristic arguments are not permitted, and the Fourier inversion theorem includes an explicit specification of what class of functions is being allowed. We then have. In this case the integral in the inverse Fourier transform is defined with the aid of a smooth rather than a sharp cut off function; specifically we define. I need this for an engineering problem that I'm working on, have you seen a multi-dimensional version? The answer by Lee gives a rule.
• Fourier Transform of f(g(t)) (composite functions) Mathematics Stack Exchange
• Fourier transform of function composition Mathematics Stack Exchange
• Multiplication of Functions
• There is no such rule in general. The key here is variable substitution: If g is a bijection and smooth enough then, if all integrals exist. How would I compute the Fourier transform of a function f(g(t)) - a composite function. For example, what if I took h(t)=cos(sin(2πt)), where.
In mathematics, the Fourier inversion theorem says that for many types of functions it is possible to recover a function from its Fourier transform.
Intuitively it may.
Such an expression might be useful for some purposes, but it's completely useless for the wast majority of applications.
## Fourier Transform of f(g(t)) (composite functions) Mathematics Stack Exchange
In mathematics such heuristic arguments are not permitted, and the Fourier inversion theorem includes an explicit specification of what class of functions is being allowed. What is what? Home Questions Tags Users Unanswered. For this reason the properties of the Fourier transform hold for the inverse Fourier transform, such as the Convolution theorem and the Riemann—Lebesgue lemma. Schwartz developed a framework of generalized functions in which such a definition indeed made sense.
Fourier transform function composition associative Tables of Fourier transforms may easily be used for the inverse Fourier transform by composing the looked-up function with the flip operator. Would be happy to be proven wrong here. Have a look at Bergner et al.Video: Fourier transform function composition associative Fourier Transform of Basic Signals (Rectangular Function)The quest for bases in function spaces with scalar product led to the theory of Fourier series and Fourier transform, to the theory of generalized functions and most recently to the discovery of wavelets. Anyway, try it for any non-trivial case and see if it works.
5 Discrete Multivariate Function Composition and Decomposition. 97 . cle onto itself, therefore the Fourier transform before and after a composition are Another way to obtain equivalent compositions is to exploit commutative polyno. Proposition The convolution product is commutative, distributive and associative, The convolution, f ∗ g is an L1-function and therefore has a Fourier transform.
Because f(x − y)g(y) is. Show that their composition. A◦B(f) d. = A(B(f)) is.
generated by the associative, bilinear geometric product with neutral element. 1 satisfying . The defining functions of the spacetime Fourier transform for multivec- tor fields A.
constructed from the composition of these swaps. D.
## Fourier transform of function composition Mathematics Stack Exchange
Corollary.
By using this site, you agree to the Terms of Use and Privacy Policy. In these cases the integrals above may not converge in an ordinary sense. The inverse Fourier transform is extremely similar to the original Fourier transform: as discussed above, it differs only in the application of a flip operator.
I haven't applied my self to anything, but it does look usable. More abstractly, the Fourier inversion theorem is a statement about the Fourier transform as an operator see Fourier transform on function spaces.
COLLEGE HOOPS 2K8 ROSTERS 14-15 UD January Learn how and when to remove this template message. Dirk Dirk 9, 25 25 silver badges 48 48 bronze badges. Dirac in s. Post as a guest Name. Mathematics Stack Exchange works best with JavaScript enabled.
In this paper we discuss the discrete Fourier transform and point out some conjugate periodic functions and their application to Theodorsen's integral.
that are highly composite the computational work required to form Snx can be commutative, associative, and (with respect to the addition defined earlier) distributive. Definition 4 A convolution of two functions on a commutative group G .
the composition of Fourier integral and the inverse Fourier transform. Function is a correspondence f between elements of a space X and those of a space Y Associativity, commutativity and the distributive law are inherited from Y.
## Multiplication of Functions
However, A common notation for the composition f(g(x)) is (g\circ f)(x). of Fourier series and Fourier transform, to the theory of generalized functions and most.
Sign up or log in Sign up using Google. We may consider orthogonal functions, and posit a problem of finding a basis in such a space.
This is the Lee that gave the answer innow posting from a registered account. Such an expression might be useful for some purposes, but it's completely useless for the wast majority of applications. I know you can do this for the sumthe product and the convolution of two functions.
Fourier transform function composition associative Intuitively it may be viewed as the statement that if we know all frequency and phase information about a wave then we may reconstruct the original wave precisely. Sign up using Email and Password. Viewed 11k times. In this case the Fourier transform cannot be defined directly as an integral since it may not be absolutely convergent, so it is instead defined by a density argument see the Fourier transform article. Dirk Dirk 9, 25 25 silver badges 48 48 bronze badges. This condition is the one used above in the statement section. Thus integral equations with convolutions may be further reduced to algebraic equations.
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# MSSQL Frequency of Years From Days Field
Good morning!
I have a table with a field of elapsed days between two points. I'm trying to do a frequency of rows converting days into years.
For example, given the following data:
Days |
--------
200 |
--------
100 |
--------
860 |
--------
368 |
--------
479 |
--------
The result would look something like:
<1 year = 2
1 - 2 years = 2
2 - 3 years = 1
I want to group the frequencies as follows:
< 1 year
1 - 2 years
2 - 3 years
3 - 4 years
4 -5 years
> 5 years
I'm thinking some kind of case statement will get me there but I'm not sure how to combine it with a group by (if that's even possible).
Thanks in advance for any help.
LVL 1
###### Who is Participating?
Database ExpertCommented:
try..
``````--
CREATE TABLE grouper
(
Days INT
)
GO
INSERT INTO grouper VALUES (200),(100),(860),(368),(479),(6000),(88000)
GO
SELECT SUM(CASE WHEN days <= 365 THEN 1 ELSE 0 END) '<1Year'
, SUM(CASE WHEN days > 365 AND days <= 365*2 THEN 1 ELSE 0 END) '1 - 2 Year'
, SUM(CASE WHEN days > 365*2 AND days <= 365*3 THEN 1 ELSE 0 END) '2 - 3 Year'
, SUM(CASE WHEN days > 365*3 AND days <= 365*4 THEN 1 ELSE 0 END) '3 - 4 Year'
, SUM(CASE WHEN days > 365*4 AND days <= 365*5 THEN 1 ELSE 0 END) '4 - 5 Year'
, SUM(CASE WHEN days <= 365 THEN 1 ELSE 0 END) ' > 5Years'
FROM grouper
--
``````
Output
``````<1Year 1 - 2 Year 2 - 3 Year 3 - 4 Year 4 - 5 Year > 5Years
----------- ----------- ----------- ----------- ----------- -----------
2 2 1 0 0 2
(1 row(s) affected)
``````
Hope it helps !!
0
Author Commented:
Thanks Pawan!
That worked great. I modified the last line to read:
SUM(CASE WHEN days > 365*5 THEN 1 ELSE 0 END) ' > 5 Years'
But that's exactly what I needed.
Thanks again!
0
Question has a verified solution.
Are you are experiencing a similar issue? Get a personalized answer when you ask a related question.
Have a better answer? Share it in a comment.
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Ronaldo Peres
Posted on
# Days of Code [3]
Hi everyone,
Now I have this one:
"You are given a large integer represented as an integer array digits, where each digits[i] is the ith digit of the integer. The digits are ordered from most significant to least significant in left-to-right order. The large integer does not contain any leading 0's.
"
### 1 Example
• Input: digits = [1, 2, 3]
• Output: digits = [1, 2, 4]
• Explanation: The array represents the integer 123.
• Incrementing by one gives 123 + 1 = 124.
• Thus, the result should be [1,2,4].
### 2 Example
• Input: digits = [4,3,2,1]
• Output: [4,3,2,2]
• Explanation: The array represents the integer 4321.
• Incrementing by one gives 4321 + 1 = 4322.
• Thus, the result should be [4,3,2,2].
### 3 Example
• Input: digits = [9,9,9]
• Output: [1,0,0,0]
• Explanation: The array represents the integer 99.
• Incrementing by one gives 999 + 1 = 1000.
• Thus, the result should be [1,0,0,0].
### Constraints:
• 1 <= digits.length <= 100
• 0 <= digits[i] <= 9
• digits does not contain any leading 0's.
### Observations
Need to figure out how to solve this one without converting to int or long, since the array can pass the MaxValue, for example:
The int.Maxvalue = 2147483647;
We can have this input:
• Input: digits = [9,8,7,6,5,4,3,2,1,0]
So if you think to solve:
• Join the array and get a string "9876543210"
• When parse it to int we get this, cause we pass the MaxValue:
• System.OverflowException: 'Value was either too large or too small for an Int32.'
Here is my solution:
`````` public static class PlusOneArray
{
public static int[] PlusOne(int[] digits)
{
var numOfNine = GetNines(digits);
if (numOfNine > 0)
{
if (digits.Length != numOfNine)
{
digits[digits.Length - numOfNine - 1] += 1;
}
else
{
digits[digits.Length - numOfNine] = 1;
var newDigits = new int[digits.Length + 1];
digits.CopyTo(newDigits, 0);
digits = newDigits;
}
int i = digits.Length - numOfNine;
for (; i < digits.Length; i++)
{
digits[i] = 0;
}
}
else
{
digits[digits.Length - 1] = digits[digits.Length - 1] + 1;
}
return digits;
}
private static int GetNines(int[] digits)
{
int n = 0;
for (int i = digits.Length - 1; i >= 0; i--)
{
if (digits[i] == 9)
n++;
else
break;
}
return n;
}
}
``````
Also this is at my github:
Peres Github - Plus one
Happy coding!!
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# Newbie on a trimmer/trimpot watts and temperature for a laser driver
Discussion in 'LEDs and Optoelectronics' started by Hector Roldan, Feb 23, 2016.
1. ### Hector Roldan
3
0
Feb 23, 2016
Hi there, I've been building stuff for years but as the usual DIY guy, I lack specific knowledge on certain stuff. I always research before asking a question (I'm not lazy and I do respect people time). But I'm having no luck on my search queries now. Sorry if this seems long, I'm just providing the details:
I built a CNC laser cutter/burner/engraver, it works quite well. Sure I also built the laser driver using the LM317. It got hot so I placed a heat sink and a fan, it does wonders, and I'm using 1 watt and 5 watts resistors to avoid heating (old problem with the usual resistors for this driver). The thing is, I originally built a laser driver for each laser diode I have, and now I'm building a variable resistor driver that I could use with all my laser diodes providing the current I need for each case.
So, the usual resistors got hot. The new ones, big, thick, don't get hot.
But the variable resistor/trimmer/trimpot is too small and couldn't get a "1W" version on any store. I suspect it will get hot regardless of the fan above, and I know heat reduces efficiency. The driver would be feeding constant current to the laser for extended periods, not just seconds, so that's my concern (had no problems with the big resistors).
My question, and I would like your help on this, is: What about using another resistor in parallel?
I read that if you only have regular resistors and they get hot, and you can't get 1W or 5W resistors, you can use two in parallel and do the math to get the exact resistance, having two in parallel reduces the pressure/load. I was using 5W version for an 2.6ohm resistor with no problems, so I want to combine a 100ohm variable resistor with a 10ohm 5w resistor in parallel to avoid so much load on the small variable resistor, does the explanation make sense? would you recommend this? I see no way to cool down the trimpot but using a fan.
2. ### Alec_t
3,060
834
Jul 7, 2015
I wouldn't recommend it. If you were to adjust the variable resistor to a low resistance value you would divert a lot of current through it and cook it. It would be better to make a constant-current source from a power transistor and use the variable resistor to control the transistor current. That way, only a small current flows through the variable resistor.
Hector Roldan and davenn like this.
3. ### Hector Roldan
3
0
Feb 23, 2016
Thanks that's really something, will try.
4. ### AnalogKid
2,573
740
Jun 10, 2015
Using multiple resistors in parallel or series to distribute a thermal load works very well, but only if the resistors are closely matched in value. If they are 10:1 apart in value, they will be 10:1 apart in heat load.
ak
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# FINDING AREA IN THE COORDINATE PLANE
## About "Finding area in the coordinate plane"
Finding area in the coordinate plane :
We can use familiar area formulas to find areas of polygons in the coordinate plane.
## Finding area in the coordinate plane - Examples
Example 1 :
A gardener uses a coordinate grid to design a new garden. The gardener uses polygon WXYZ on the grid to represent the garden. The vertices of this polygon are W(3, 3), X(−3, 3), Y(−3, −3), and Z(3, −3). Each grid unit represents one yard.Find the area of the garden.
Solution :
Step 1 :
Graph the vertices, and connect them in order.
From the graph, it clear that the shape of the garden is a square.
Let us find lengths of the sides YZ and WZ.
Step 2 :
Find the length of the side YZ.
The ordered pair of Y is (-3, -3).
The x-coordinate of Y is -3, so point Y is |-3| = 3 yards from the y-axis.
The ordered pair of Z is (3, -3).
The x-coordinate of Z is 3, so point Z is |3| = 3 yards from the y-axis.
Find the sum of the distances :
The length of side YZ = 3 + 3 = 6 yards
Step 3 :
Find the length of the side WZ.
The ordered pair of W is (3, 3).
The y-coordinate of W is 3, so point W is |3| = 3 yards from the x-axis.
The ordered pair of Z is (3, -3).
The y-coordinate of Z is -3, so point Z is |3| = 3 yards from the x-axis.
Find the sum of the distances :
The length of side WZ = 3 + 3 = 6 yards.
Step 4 :
Find the area of the square WXYZ using the lengths of the sides YZ and WZ.
Area of the square WXYZ = side x side
Area of the square WXYZ = YZ x WZ
Area of the square WXYZ = 6 x 6
Area of the square WXYZ = 36 square yards.
Example 2 :
Caleb is planning a new deck for his house. He graphs the deck as polygon ABCDEF on a coordinate plane in which each grid unit represents one foot. The vertices of the polygon are A(1, 0), B(3, 2), C(3, 5), D(8, 5), E(8, 2), and F(6, 0). What is the area of Caleb’s deck ?
Solution :
Step 1 :
Graph the vertices, and connect them in order.
Draw a horizontal dashed line segment to divide the polygon into two quadrilaterals — a rectangle and a parallelogram.
Step 2 :
Find the area of the rectangle using the length of segment BE as the base b and the length of segment BC as the height h.
b = |8| − |3| = 5 feet
h = |5| − |2| = 3 feet
A = bh = 5 x 3 = 15 square feet
Step 3 :
Find the area of the parallelogram using the length of segment AF as the base. Use the length of a segment from F(6, 0) to the point (6, 2) as the height h.
b = |6| − |1| = 5 feet
h = |2| − 0 = 2 feet
A = bh = 5 x 2 = 10 square feet
Step 4 :
Add the areas to find the total area of the deck.
hence, the required area is
= 15 + 10 = 25 square feet
After having gone through the stuff given above, we hope that the students would have understood "Finding area in the coordinate plane".
Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here.
You can also visit our following web pages on different stuff in math.
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Saturday, November 15, 2008
Properties of scalar product
Property 1 :
The scalar product of two vectors is commutative
av.bv = bv.av
Property 2 : Scalar Product of Collinear Vectors :
(i) When the vectors av and bv are collinear and are in the same direction, then θ = 0
av.bv = |av| |bv| = ab
(i) When the vectors av and bv are collinear and are in the opposite direction, then θ = π
av.bv = |av| |bv|(-1) = -ab
Property 3 : Sign of Dot Product
The dot product av.bv may be positive or negative or zero.
(i) If the angle between the two vectors is acute (i.e., 0 < θ < 90°) then
cos θ is positive. In this case dot product is positive.
(ii) If the angle between the two vectors is obtuse (i.e., 90 < θ < 180) then
cos θ is negative. In this case dot product is negative.
(iii) If the angle between the two vectors is 90° (i.e., θ = 90°) then
cos θ = cos 90° = 0. In this case dot product is zero.
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# DAT PAT Angle Ranking Technique
### Dental Admissions Test Perceptual Ability Angle Ranking Technique
The angle ranking section of the Dental Admissions Test (DAT) may seem hard at first, but using some of the key techniques laid out on this page you will be able to ace the Perceptual Ability Angle Ranking section of the DAT. First I will give a brief overview of this section. The angle ranking section of the DAT consists of 15 multiple choice questions which you must determine the relative sizes of the interior angles of the four figures from smallest to largest. A tip for this section is to sit back in your seat and don’t get too close to the computer screen, being too close to the screen won’t allow you to view all the angles and distinguish between the angles because you don’t have anything to compare the angles to.
The best technique for this section is elimination. Usually the largest and the smallest angle will be easy to pick out, check out the practice question below.
A. 3-1-4-2
B. 3-1-2-4
C. 3-4-1-2
D. 1-3-4-2
The smallest angle is obviously angle 3. So that eliminates choice D. Next you should look for the largest angle. The largest angle seems like a split between angle 2 & 4, but upon further examination you see that angle 4 is pointing vertical while angle 2 is not. They both have a near identical lines pointing left. From this observation we can conclude that angle 2 is the largest angle. We now can eliminate choice B, and we are left with choice C and A. Since we know angle 3 is the smallest and angle 2 is the largest we just have to compare angle 1 and 4.
#### At this point we must use the “Laptop Technique”.
Imagine each angle is a laptop viewed from the side. The vertical/horizontal lines won’t move at all, however the non-vertical/horizontal lines can move. Imagine which laptop would be easier to close, the easier laptop to close would be the smaller angle, in which case that is the smaller angle, using this technique we see that angle 1 is smaller than angle 4, which eliminates choice C. The only choice left is A, which is the correct answer.
#### An Alternative Technique called the “Laser Technique”.
You can use this technique to distinguish a smaller angle between two angles, these angles must have a similar horizontal/vertical line you can compare to. In this technique you imagine the vertical/horizontal line as the base of the laser that is laying flat on the ground. The other line is the gun base that will shoot the laser. The laser with the steepest slope will be the smaller angle, obviously figure 1 has the smallest angle.
#### Rapid Eye Technique
This is a new technique created by condog51 and posted here on youtube this technique involves looking at the innermost area of two closely related angles as circled below:
Concentrating on the interior areas circled above look back an forth between them quickly, pretty soon you will be able to distinguish the larger of the two angles.
#### More Perceptual Ability Resources
Listed below are the top 4 most popular DAT PAT Simulated Sample Test resources for explanations and practice problems.
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# Results are varying with same query
## All we need is an easy explanation of the problem, so here it is.
Hi I stuck in strange issue with one of my query.
Formula : (28800/3600-(28382.6/3600-(1920/3600+1860/3600))) =
So, Input Equation:
= (28800/3600-(28382.6/3600-(1920/3600+1860/3600)))
= (28800/3600-(28382.6/3600-(0.53333333333333+1860/3600)))
= (28800/3600-(28382.6/3600-(0.53333333333333+0.51666666666667)))
= (28800/3600-(28382.6/3600-(1.05)))
= (28800/3600-(28382.6/3600-1.05))
= (28800/3600-(7.8840555555556-1.05))
= (28800/3600-(6.8340555555556))
= (28800/3600-6.8340555555556)
= (8-6.8340555555556)
= (1.1659444444444)
= 1.1659444444444
But I am getting 0 as output from my query, it supposed to print 1
``````select (to_char (NULLIF(GREATEST(coalesce(28800/3600,0) - (coalesce(28382.6/3600,0) -
(coalesce(1920/3600,0) + (coalesce(1860/3600,0)))),0),0),'FM99,999,999,999'))::character varying as mydatas
``````
Now, same formula and same query with different values working fine. please check following.
(28800/3600-(13552.24/3600-(900/3600+0/3600)))
Input Equation:
= (28800/3600-(13552.24/3600-(900/3600+0/3600)))
= (28800/3600-(13552.24/3600-(0.25+0/3600)))
= (28800/3600-(13552.24/3600-(0.25+0)))
= (28800/3600-(13552.24/3600-(0.25)))
= (28800/3600-(13552.24/3600-0.25))
= (28800/3600-(3.7645111111111-0.25))
= (28800/3600-(3.5145111111111))
= (28800/3600-3.5145111111111)
= (8-3.5145111111111)
= (4.4854888888889)
= 4.4854888888889
so my query also gives 4 as output
``````select (to_char (NULLIF(GREATEST(coalesce(28800/3600,0) - (coalesce(13552.24/3600,0) -
(coalesce(900/3600,0) + (coalesce(0/3600,0)))),0),0),'FM99,999,999'))::character varying as mydatas
``````
## How to solve :
I know you bored from this bug, So we are here to help you! Take a deep breath and look at the explanation of your problem. We have many solutions to this problem, But we recommend you to use the first method because it is tested & true method that will 100% work for you.
### Method 1
The expression:
``````select ((NULLIF(GREATEST(coalesce(28800/3600,0) - (coalesce(28382.6/3600,0) -
(coalesce(1920/3600,0) + (coalesce(1860/3600,0)))),0),0)))::character varying as mydatas
``````
evaluates to 0.11594…
You are getting it wrong at this step:
``````= select (28800/3600-(28382.6/3600-(1920/3600+1860/3600)));
= select (28800/3600-(28382.6/3600-(0.53333333333333+0.51666666666667))); -- False
``````
Because 1920/3600 = 0 and so is 1860/3600 due to integer division. You can fix your expression by making sure that one of the operands is not an integer type, example:
``````= select (28800.0/3600-(28382.6/3600-(1920.0/3600+1860.0/3600)));
``````
Fiddle
Note: Use and implement method 1 because this method fully tested our system.
Thank you 🙂
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# A Tough-looking Integral
I recently came across an integral that at first seemed “impossible”. The integral is . This was the problem A4 on the 1969 Putnam. I naturally tried to rewrite as , because this is the general technique to integrate : . What makes things difficult here is that is the product of two functions of…
# Groups and Homomorphisms
One often sees that proving theorems about groups and related structures can involve clever reasoning and ingenuity. I will be collecting, when possible, problems which use such reasoning, because what is more disorienting than not being able to re-use a clever method you discovered a few weeks ago? Problem 1: Show that has the same number…
# Results about Finite-Dimensional Linear Spaces
Just like my page for recording the results I encounter about multiplicative functions, along with their proofs, I intend to update this page with basic results about linear spaces (mostly finite-dimensional). I will try to explicitly state the linear algebra theorems I make use of in my proofs. Result 1: If is a finite-dimensional vector space,…
# Nonlinear Time Series Analysis
I am attending a short workshop from 30th April to 4th May on the topic of Nonlinear Time Series Analysis organized by the Indian Institute of Sciences, Bangalore. The course is being taught by Professor Marco Bittelli of the University of Bologna. I intend to keep (very brief) summaries of what is covered each day as it…
# Teaching my sister
This summer I have been “tutoring” my sister Prarthana (9th grader), in mathematics and physics (and chemistry, despite my own scholarly-lessness in it!). To see how she is progressing, I thought it would be good to see how she approaches problems in timed assessments. I will be updating this post with the assessments I prepare…
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# Homework Help: Angular motion
1. Jan 6, 2005
### mattmets
A coin with a diameter of 2.30 cm is dropped onto a horizontal surface. The coin starts out with an initial angular speed of 18.4 rad/s and rolls in a straight line without slipping. If the rotation slows with an angular acceleration of magnitude 2.17 rad/s2, how far does the coin roll before coming to rest?
Can anyone help- I'm trying to study for a physics final and I'm getting nowhere!!! Thanks.
2. Jan 6, 2005
### Chen
Unless I'm missing something here, you can use this formula:
vf2 = vi2 + 2aΔx
Dividing both sides by R2, the coin's radius squared, we get:
ωf2 = ωi2 + 2αΔθ
Where α is the coin's angular acceleration, which is negative in this case. From here you can find Δθ and multiplying by R you can find the total distance travelled by the coin.
Last edited: Jan 6, 2005
3. Jan 6, 2005
### mattmets
Great.....thanks
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EE 302 - HW 1
# EE 302 - HW 1 - A 30-W incandescent lamp is connected to a...
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Chapter 1, Problem 6. The charge entering a certain element is shown in Fig. 1.23. Find the current at: (a) t = 1 ms (b) t = 6 ms (c) t = 10 ms
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Chapter 1, Problem 11. A rechargeable flashlight battery is capable of delivering 85 mA for about 12 h. How much charge can it release at that rate? If its terminals voltage is 1.2 V, how much energy can the battery deliver?
Chapter 1, Problem 18. Find the power absorbed by each of the elements in Fig. 1.29.
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Chapter 1, Problem 20. Find V0 in the circuit of Fig. 1.31.
Chapter 1, Problem 28.
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Unformatted text preview: A 30-W incandescent lamp is connected to a 120-V source and is left burning continuously in an otherwise dark staircase. Determine: the current through the lamp, the cost of operating the light for one non-leap year if electricity costs 12 cents per kWh. Chapter 1, Problem 36. A battery may be rated in ampere-hours (Ah). A lead-acid battery is rated at 160 Ah. What is the maximum current it can supply for 40 h? How many days will it last if it is discharged at 1 mA?...
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• Fall '06
• MCCANN
• Incandescent light bulb, Lead-acid battery, Rechargeable battery, rechargeable flashlight battery, 30-W incandescent lamp
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# To Voronoi and Beyond
## Voronoi Diagrams
This tutorial is a primer on Voronoi diagrams: what they are, what you need them for and how to generate them using a Shader in Unity. You can download the complete Unity page in Part 4.
### Part 1: Voronoi Diagrams
Technically speaking, Voronoi diagrams are a way to tassellate a space. It means that the end result of Voronoi is a set of “puzzle pieces” which completely fills the space. To start, we need a set of points (often called seeds) in the space. Each seed will generate a piece of this puzzle. The way Voronoi works is by assigning every point of the space to its closest seed. The final result heavily depends on the way distance is measured in the space.
#### Euclidean distance
Most Voronoi diagrams are are based on the Euclidean distance. The cost between two points is given by the length of the shortest segment which connects them both. It can be calculated easily with the Pythagorean theorem:
In Cg, this function is already implemented and is called distance. The picture on the left shows a Voronoi diagram based on the Euclidean distance, drawn with 100 points. On the right, the same diagram uses a gradient to visualise the actual distance from a pixel to the closest one.
The distance diagram has been calculated using minDist to sample a gradient from black to white.
#### Manhattan distance
As the name suggests, the Manhattan distance takes his name from the homonym city. The shortest path between two locations is not a straight line, since Manhattan is full of buildings. The shortest distance is the one which goes around building.
half distance_manhattan(float2 a, float2 b) {
return abs(a.x - b.x) + abs(a.y - b.y);
}
Compared to the Euclidean distance, It is sensibly less expensive to calculate.
Using the Manhattan distance produces very intriguing patterns which resemble circuit boards. This is not a coincidence: many boards are designed to minimise circuit length and avoid curves.
#### Minkowski distance
Despite looking very different, both the Euclidean and the Manhattan distances are both special cases of a more general metric: the Minkowsi distance. To understand why, you have to remind some algebra. In the same way multiplication and division are the same operator (dividing by is equivalent to multiply by ), even root and exponentiation are deeply connected. Remembering then , we can introduce the Minkowski distance:
When or it equivalent to the Manhattan or Euclidian distance, respectively.
half distance_minkowski(float2 a, float2 b, float p) {
return pow(pow(abs(a.x - b.x),p) + pow(abs(a.y - b.y),_P),1/p);
}
The most fascinating aspect is that is provides a way to smoothly transitioning from the Euclidean to the Manhattan distance, and the other way round.
If you are in a higher dimension, the Minkowski distance can be still used, providing that you calculate it on all the components of two points and :
The next part of this tutorial will focus on the applications of Voronoi diagrams.
## Applications
### Part 2: Applications
Despite looking pretty, not a single application has been indicated for Voronoi diagrams yet. In actuality, they play a very important role in Science, and many games can benefits from them.
#### In games
Breaking object realistically is a very challenging task, that requires to know how pressure waves propagates through a material. A simpler way to create plausible fractures in an object is to rely on a Voronoi 3D tassellation. You start choosing random points within the object you want to break, then each Voronoi cell become one of its chunks. In games, breakable objects don’t break: they are already broken, and your interaction makes the piece falling apart.
The famous Fracturing & Destruction plugin on the Asset Store, for instance, uses this technique to generate breakable objects. A future post will show how to replicate this effect at no cost.
#### In path finding
As a game developer, you might be familiar with path finding. A* is notoriously the most known, but there are many other ways one can find the optimal path between two points. So far, Voronoi diagrams have been seen as independent regions of space although there is an alternative way of interpreting them. If we put a node every time two segments connects, Voronoi produces a graph. The segments (now edges of the graph) represent the paths which are as far as possible from the seeds. In terms of gaming, seeds can be enemies you want to avoid; travelling on the edges provides the safest route possible. Brent Owens has written a very nice tutorial about this.
Conversely, Voronoi diagrams can be also used to approximate the shortest path. The dual graph of a Voronoi diagram (known as the Delaunay triangulation) allows to find paths which are as close as possible to the seeds. When coupled with the Manhattan distance, it can be used to generate the fastest route within a city, considering how fast you can go on different roads.
#### In nature
Circle packing is the problem of fitting as many circles as possible in a given space. The best possible solution to this problem is shown on the left; circles are arranged in a hexagonal lattice, which resemble a honeycomb. This is actually why honeycomb cells have a hexagonal structure: if all circle expands at the same time to fill all the space around them, they’ll end up pressing against each other until they create a perfect hexagonal lattice. The same pattern can be found in several other phenomena, like cooling magma and soap bubbles. The latter, provide an excellent (and transparent) example of how Voronoi diagrams look in three dimensions.
The next part of this tutorial will show how to generate Voronoi diagrams using Shaders.
## Generation
### Part 3: Generation
There are several algorithms you can rely on to generate Voronoi diagrams. Every point is independent from the other, so this is one of those perfect applications for a shader. Traditionally the Fortune’s algorithm (left) is commonly used, but it is very hard to implement within a shader. The tricky part, in this case, is how to provide a list of points to the Material, since the Unity APIs don’t provide any SetArray function. Luckily, there is an undocumented feature you can use to pass arrays and matrices to a shaders, and it has been discussed in this post. We will use one array for the position of the points (in a 2D space) and another one for the colours. A variable called _Length is used to indicate how many points are there since Cg doesn’t support arrays of arbitrary dimensions.
uniform int _Length = 0;
uniform half2 _Points[100];
uniform fixed3 _Colors[100];
The actual code of the Voronoi diagram is implemented in the fragment function. For each pixel, it simply loops over all the points and finds the closest one. Its index is then used to find the right colour to use:
fixed4 frag(vertOutput output) : COLOR {
half minDist = 10000; // (Infinity)
int minI = 0;
for (int i = 0; i < _Length; i++) {
half dist = distance(output.worldPos.xy, _Points[i].xy);
if (dist < minDist) {
minDist = dist;
minI = i;
}
}
return fixed4(_Colors[minI], 1);
}
Different types of diagrams are possible simply by replacing the function distance with the appropriate metric. If you want to draw the distance diagram instead, you can sample a ramp texture using minDist. You can also set the texture mode to “Repeat” rather than “Clamp” for some bizarre effects.
half4 color = tex2D(_RampTex, fixed2(minDist, 0.5));
color.a = 1;
return color;
#### Weighted Voronoi diagrams
Interesting results can be obtained by mixing different metric, or altering the “attraction” of the seeds by providing an extra coefficient to the shader. The distance is now:
half dist = distance(output.worldPos.xy, _Points[i].xy) + _Weights[i];
This takes the name of weighted Voronoi (also known as Dirichlet tessellation) and it can be used to generate beautiful effects, like Milan Domkář did with his foam:
#### Cone projection
There is a smarter approach to generate Voronoi diagrams (almost!) for free, and Chris Wellons is beautifully explaining it in its blog. You can generate a Voronoi tessellation by projecting cones out of the starting points. The cones will eventually intersect and seeing them it from the above will produce the same effect.
You can download the full Unity package in the last part of this tutorial.
### Conclusion
Voronoi diagrams are a way to tessellate the space which has many applications, from game development to city planning. This tutorial has shown how to generate them using a shader; you can download the complete Unity package here.
#### Other resources
##### 💖 Support this blog
This websites exists thanks to the contribution of patrons on Patreon. If you think these posts have either helped or inspired you, please consider supporting this blog.
You will be notified when a new tutorial is relesed!
##### 📝 Licensing
You are free to use, adapt and build upon this tutorial for your own projects (even commercially) as long as you credit me.
You are not allowed to redistribute the content of this tutorial on other platforms. Especially the parts that are only available on Patreon.
If the knowledge you have gained had a significant impact on your project, a mention in the credit would be very appreciated. ❤️🧔🏻
1. lochie westfall
Hi alan, i am getting into VR and something that i think would really improve immersion is dynamic voronoi object shattering. So if you could please do that tutorial on the fracturing and destruction remake it would be highly appreciated!
P.S. I am 13 and have already tried making a voronoi system but it was limited to 2D, so yet again that tutorial would be greatly appreciated!
• Hey! 😀
Thank you for your message! Voronoi facture is super cool! The only issues is that is incredibly tedious to code. The reason is that you need to do mesh intersections, and splitting them to different geometries. If I am going to do it, it will be over many posts!
Hopefully I’ll manage to find the time to do it! XD
• Lochie westfall
Ok thanks, today I got a triangle exploding system going that gets each triangle and adds a point in the middle to create a prism, it’ll do for now so take as long as you’d like!
2. Ganz Fertig
Interesting … I just replied to someone on another page about him complaining about Alan charging via Patreon.
Now I read above “A future post will show how to replicate [Fracturing & Destruction] effect at no cost.”
Makes me think …
I still think that definitely for developing things (e.g. games) it is worth paying if it safes me time. For me personally, I am as well willing to pay for leisure things (e.g. movies, books).
3. Ayhan
Just tried to take a look into the unity example, but it diesnt seem to work anymore. Looking inside the frame debuger shows, that the arrays are not filled with the technique which worked in older versions. I am using Unity 5.60f3 atm.
4. Isaac Surfraz
This would be a great tutorial to finish, breaking things is a very useful and not well covered topic on other blogs.
• Hey!
I know right? Unfortunately, I’ve been very busy so didn’t have the chance to write that part yet!
5. Ethan Cheung
this tutorial is fully expanded in the math page, is it meant to be?
• What do you mean?
• Hi!
That sounds very challenging to do! If you want this to be done as a 3D mesh, Unity might not be the best software since it offers only a very bare support for real-time mesh creation.
You can make this with shader in a way that is probably easier. But to do that, you would need something called volumetric rendering. Which basically “fakes” 3D geometry on 2D surfaces. I talked about this here: https://www.alanzucconi.com/2016/07/01/volumetric-rendering/
In a nutshell, you can create a volumetric shader that uses SDFs to create a geometry based on Voronoi cells! And with SDFs, you can “relax” the geometry easily to make it more smooth!
But… is not going to be easy!
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Search 79,143 tutors
## Answers by Brian H.
OK. I hope this wasn't a math homework problem or something because I can't really describe the process I used other than creative trial and error, but if you cut 28 squares from 2 larger sheets of steel 6" x 36" in size, you will only have 12 in.2 of waste at the end of all the cutting...
Trigonometry is the study of measurements (and, in general, math) regarding everything about triangles. Tri-, "three"; -gon-, "sided"; -metry, "measurements". A "trigon" would be a three-sided figure (i.e. - a triangle), and trigonometry would be about the...
For polynomials like this - which are easily factored - you will find that factoring the polynomial and taking all real zeros will be the easiest method. However, for polynomials which are not easily factored, there is another method - a method that you should probably use for all problems of this...
Let's take this problem for an example. (I'll state the whole problem first, then go about showing how to solve it. It's long, but it covers about every special case I can think of. I use this sort of problem in my college classes to have an example for all the different nuances of quadratic function...
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Find the perfect tutor and raise your grades.
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To convert any kind of value in milliliters to litres, just multiply the worth in milliliters through the conversion variable 0.001. So, 5000 milliliters times 0.001 is same to 5 litres.
You are watching: How many liters are in 5000 milliliters
To calculate a milliliter worth to the matching value in litre, just multiply the amount in milliliter by 0.001 (the conversion factor). Here is the formula:
Suppose you want to convert 5000 milliliter right into litres. Using the conversion formula above, you will certainly get:
Value in litre = 5000 × 0.001 = 5 litres
This converter can assist you to obtain answers to inquiries like:
How many milliliters are in 5000 litres?5000 milliliters are equal come how countless litres?How lot are 5000 milliliter in litres?How to transform milliliters come litres?What is the conversion variable to transform from milliliters to litres?How to change milliliters in litres?What is the formula to transform from milliliters to litres? among others.
Milliliters come litres switch chart
4100 milliliters = 4.1 litres
4200 milliliters = 41/5 litres
4300 milliliters = 4.3 litres
4400 milliliters = 4.4 litres
4500 milliliters = 41/2 litres
4600 milliliters = 4.6 litres
4700 milliliters = 4.7 litres
4800 milliliters = 4.8 litres
4900 milliliters = 4.9 litres
5000 milliliters = 5 litres
See more: What Distinguishes A Data Collection Device From Other Input Devices?
Milliliters to litres counter chart
5000 milliliters = 5 litres
5100 milliliters = 5.1 litres
5200 milliliters = 51/5 litres
5300 milliliters = 5.3 litres
5400 milliliters = 5.4 litres
5500 milliliters = 51/2 litres
5600 milliliters = 5.6 litres
5700 milliliters = 5.7 litres
5800 milliliters = 5.8 litres
5900 milliliters = 5.9 litres
### Disclaimer
While every effort is made come ensure the accuracy that the information detailed on this website, no this website nor its authors are responsible for any type of errors or omissions. Therefore, the components of this site are not suitable for any kind of use including risk come health, finances or property.
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# Necessary and sufficient condition for isomorphism
• Jul 2nd 2011, 08:26 AM
Jagger
Necessary and sufficient condition for isomorphism
Hi everyone!
I have the next problem:
$\displaystyle G$ a finit group which has order $\displaystyle n$ and $\displaystyle \varphi:G\rightarrow G$ defined by $\displaystyle \varphi(a)=a^m\;\forall\;a\in G$, I need to find a necessary and sufficient condition so that $\displaystyle \varphi$ is an isomorphism.
Any suggestion?
• Jul 2nd 2011, 09:32 AM
Isomorphism
Re: Necessary and sufficient condition for isomorphism
Quote:
Originally Posted by Jagger
Hi everyone!
I have the next problem:
$\displaystyle G$ a finit group which has order $\displaystyle n$ and $\displaystyle \varphi:G\rightarrow G$ defined by $\displaystyle \varphi(a)=a^m\;\forall\;a\in G$, I need to find a necessary and sufficient condition so that $\displaystyle \varphi$ is an isomorphism.
Any suggestion?
Claim:
$\displaystyle \varphi:G\rightarrow G$ defined by $\displaystyle \varphi(a)=a^m\;\forall\;a\in G, m > 1$ is an isomorphism iff G is abelian and $\displaystyle (m,n)=1$
$\displaystyle \varphi^p$ will stand for $\displaystyle \varphi$ composed with itself 'p' times. (Naturally $\displaystyle p \in \mathbb{N}$)
Exercise 1: Prove that if $\displaystyle \varphi$ is an isomorphism then $\displaystyle \varphi^p$ is an isomorphism too.
Assume $\displaystyle \varphi$ is an isomorphism and you can prove the conditions in two steps:
Step 1:
Let $\displaystyle (m,n)=d$. Then write ma + nb = d for some integers a and b (Why can you do this?).
$\displaystyle \varphi^{a}(g) = g^{ma} = g^d$
[Can I do the above step if 'a' is negative?]
Exercise 2:What happens, if $\displaystyle d > 1$?
[Hint: Look at $\displaystyle \text{ker}(\varphi^a)$ and rule out this case]
If d=1, we have $\displaystyle \varphi^a(g) = g \implies \varphi^{2a}(g) = g^2$ and,
Exercise 3: $\displaystyle \varphi^{2a}$ is a homomorphism iff G is abelian (Prove it!).
Step 2:
Exercise 4:Verify: If G is abelian and $\displaystyle (m,n)=1$, then $\displaystyle \varphi$ is an isomorphism.
This part is easy. Use the abelian nature to prove the function is a homomorphism and use $\displaystyle ma + nb = 1$ to prove $\displaystyle \text{ker}(\varphi) = {e}$.
__________________________________________________ _____________________________
Done! :)
• Jul 3rd 2011, 11:22 AM
Jagger
Re: Necessary and sufficient condition for isomorphism
Thanks for your answer! I forgot to said that $\displaystyle G$ was abelian but you know this. I don't understand in the exercise 4 how using that $\displaystyle ma+nb=1$ you can prove $\displaystyle ker(\varphi)=e$ and if you prove this you have that $\displaystyle \varphi$ is an isomorphism? or you have to prove that $\displaystyle \varphi$ is surjective too?
Thanks again!
• Jul 3rd 2011, 12:46 PM
Isomorphism
Re: Necessary and sufficient condition for isomorphism
Quote:
Originally Posted by Jagger
Thanks for your answer! I forgot to said that $\displaystyle G$ was abelian but you know this. I don't understand in the exercise 4 how using that $\displaystyle ma+nb=1$ you can prove $\displaystyle ker(\varphi)=e$ and if you prove this you have that $\displaystyle \varphi$ is an isomorphism? or you have to prove that $\displaystyle \varphi$ is surjective too?
Thanks again!
Hint: $\displaystyle g^m = g^{ma +nb}$
Do we need to prove surjectivity in this case? Notice that the map $\displaystyle \varphi: G \to G$. In this case surjectivity is equivalent to injectivity.
• Jul 3rd 2011, 03:15 PM
Deveno
Re: Necessary and sufficient condition for isomorphism
for maps on finite sets, injective implies surjective and vice versa.
now think about what it means for g to be in ker(φ). it means $\displaystyle g^m = e$.
there are two ways this could happen: g = e, or |g| divides m. suppose that |g| divides m.
since |g| also divides |G| = n, |g| is a common divisor, so.....
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# Normal force cancels out the force of gravity?
Alright, so I know there will be a normal force pointing perpendicular to the surface and gravitational force, right?
Gravitational force = mg
But since normal force acts in "opposite direction to the gravitational force". Wouldn't normal force then cancel out the gravitational force? If it's so, wouldn't you just "fly" for a moment up, then come back? You'll come back because once you stop touching the ground, normal force would stop acting on you...
If an object rests on the ground, then angle is zero
cos(0) = 1.
The normal force is perpendicular to the surface. Sometimes that means it also acts opposite the gravitational force (like when the surface is horizontal), but not always.
Why would an object fly up if there is no net force accelerating it upwards? Consider the vertical direction: if the vertical component of the normal force and the force of gravity cancel out, then there will be no acceleration upwards.
I do agree that if for some reason the object found itself slightly above the surface, then there would be no normal force acting on it, though.
Why would an object fly up if there is no net force accelerating it upwards?
Yeah, sorry. But I guess, what it means is that only a slight push upwards would be necessary to get the object moving upwards, then, as the normal force stops acting on the object, then you'd need a stronger push upwards (that would account for the mass of the object), right?
So basically, for infinitesimal height you can move any object up with a slight push because normal force would act opposite to the force of gravity.
I do agree that if for some reason the object found itself slightly above the surface, then there would be no normal force acting on it, though.
What do you mean "slightly"? I thought that if an object doesn't touch any surface it wouldn't experience a normal force.
DaveC426913
Gold Member
Yeah, sorry. But I guess, what it means is that only a slight push upwards would be necessary to get the object moving upwards,
Why would it take only a slight push? Just because the net force is zero doesn't mean the forces themselves are not quite large. An object the size of a person is pushing on the ground with about 750N of force and the ground is pushing up with same. It'll take at least 750N of force to separate the two.
256bits
Gold Member
But since normal force acts in "opposite direction to the gravitational force". Wouldn't normal force then cancel out the gravitational force? If it's so, wouldn't you just "fly" for a moment up, then come back? You'll come back because once you stop touching the ground, normal force would stop acting on you...
Hello vizakenjack,
Your logic seems impeccable, but it does have a flaw that you have overlooked that throws a monkey wrench into your conclusion.
You cannot turn off gravity.
You are attracted to the earth, and the earth is attracted to you. This is whether you are standing on the ground or falling from a height.
In both cases you have a force of mg acting on you from the gravity of the earth.
On the ground, the surface you are standing on, such as a floor, stops you from falling towards the centre of the earth, by providing the normal force upwards. Similarily, you also provide a different normal force downwards from your feet onto the surface so the earth.
Do you think you should you re-work your conclusion?
Why would it take only a slight push?
Because normal force already cancels out the force of gravity for you. So you don't need to "overcome" force of gravity.
Hello vizakenjack
...
Similarily, you also provide a different normal force downwards from your feet onto the surface so the earth.
Hi,
I don't get it, so an object provides a normal for to the ground too? uhm... well, what does it equal to? I mean, a normal force cancels out gravity
so what would an object's downward normal force cancel...? or what would be its magnitude?...
russ_watters
Mentor
Because normal force already cancels out the force of gravity for you. So you don't need to "overcome" force of gravity.
A table has four legs. How much of the table's weight does each leg carry? If you add another leg, does the table launch itself into space or just sit there with a different weight distribution between the legs?
Next time you are on a bathroom scale, place it near the sink or another structure you can reach out and touch. Push down lightly on that structure. Do you launch yourself into the ceiling? What does the bathroom scale say?
Because normal force already cancels out the force of gravity for you. So you don't need to "overcome" force of gravity.
Consider an object sitting on the floor, weighing 10 newtons. 10 newtons of normal force act up on it and 10 newtons of gravity act down on it. Now you decide to grab it and pull up on it the tiniest bit, but not enough to lift it up. Let's say you pull it up by 1 newton. It's not enough to lift it off the ground. It now has 9 newtons of normal force acting up on it from the floor, and 1 newton of force pulling it up from you, and 10 newtons of gravity pulling down on it.
What happened and what is very important to remember is that the normal force changed! Don't think of it as "locked in" to the weight of the object. It doesn't always equal the weight of the object. It equals in magnitude the force pushed down upon the surface which could be more (say if you push down on an object) or less (if you pull up on it a bit, like in my example).
The object in my example won't start moving up until you pull up with at least 10 newtons of force.
Last edited:
vizakenjack
256bits
Gold Member
Hi,
I don't get it, so an object provides a normal for to the ground too? uhm... well, what does it equal to? I mean, a normal force cancels out gravity
so what would an object's downward normal force cancel...? or what would be its magnitude?...
Same thing as when you push on a wall, the wall pushes back. Both forces are equal in magnitude but opposite in direction.
Have you ever heard of Newton's Third Law - for every action there is a reaction? The two normal forces are an action-reaction pair and they cancel each other out.
The other action-reaction pair is the earth attracting you, and you attracting the earth, again both forces equal in magnitude but opposite in direction.
So this
"I mean, a normal force cancels out gravity" is not really correct.
The reason the normal force from the ground on you is equal to the attraction force of the earth on you by looking at the forces acting on your body. In this case we have the attractive force from gravity downwards, which we can say is W=mg, and the normal force N which acts upwards. Since you are not accelerating the net force is zero.
Fnet = W - N = 0, or
N=W
W is always mg ( at least close to the ground ).
If by chance you have a rope near you to pull on, the act of pulling will decrease the normal force from the ground. Only when your pull on the rope is equal to W, will you rise into the air. The normal force has not lifted you up, because by pulling on the rope you have decreased the normal steadily from mg to 0.
vizakenjack
A.T.
Because normal force already cancels out the force of gravity for you.
Only if there are no other supporting forces. As soon you apply any upwards force the normal force drops by the same amount.
I don't get it, so an object provides a normal for to the ground too?
Yes, that's how you make footprints.
well, what does it equal to?
The two normal forces are equal but opposite. See Newtons 3rd Law.
I mean, a normal force cancels out gravity so what would an object's downward normal force cancel...?
The gravitational force that the objects exerts on the Earth.
sophiecentaur
Gold Member
2020 Award
It might help if you remember that, in every practical situation, there is always a small amount of movement involved - as if everything was actually composed of (very strong) springs. Forces are shared as the different parts of the structure deform a tiny bit and equilibrium is achieved. Adding a fifth leg to a table can produce any combination of force sharing, depending on the tiny details of strength and length of the leg. Think in terms of everything being made of sponge and you won't arrive at any apparently impossible situations in your thought experiments.
Yes, you're correct. When an object is touching the horizontal ground, the magnitude of the normal force and the gravitational force are equal. This is precisely why the object does not move. Now if the object is on a table, the perpendicular force exerted by the table on the object (normal) is equal to the gravitational force. If you exert a downward force on the object, the the normal will increase, as normal= gravitational force + your force.
If the object is on a slanted surface, then the magnitude of the normal force is not equal to the gravitational force. In fact, it's less. There's an angle between force and motion, so cos theta will have a value. This is why the object will slide down.
In mid air, the object does not experience a normal force, therefore it falls downwards.
What does the bathroom scale say?
Probably along the lines of: "wow, you're such an idiot that you can't figure out such a simple concept"
:(
The other action-reaction pair is the earth attracting you, and you attracting the earth, again both forces equal in magnitude but opposite in direction.
Earth attracts you - direction towards the earth's core, right? In other words, downwards.
So your attraction to the earth is ... upwards? Uhm... doesn't make sense...
the normal force changed!
Thanks, this really clarified it a bit. So does it mean, that if you pull a 10 newton object with 1 newton up, the normal force down would be 9 newtons and yours of course 1 newton up, and then they BOTH would cancel out the force of gravity?
I just don't kinda get it, why does floor have to cancel the force of gravity? Can't it just prevent an object from falling through it? Like support it and that's it. The gravity can keep acting on it. I saw the Cosmos with Neil Tyson, and there was an episode, in which it was explained about molecular ... attraction forces... dunno.
anyway, so the reason the ground would "push" an object upwards is because the molecules of the ground wouldn't like being bent down / squeezed by the object's mass so they would resist the object by wanting to remain as usual, rigid? Would this explain the reason a surface would act on an object with a normal force?
But what if ground doesn't cancel the force of gravity but simply prevents the object from falling down. It's impossible right? Because an object would act downwards on the area that it touches, right? So the area of the surface being touched would "resist" the object's downward force (due to gravity) ... Well, it makes sense then, but that only works if an object consists of molecules and atoms that have to resist...
what if an object was from a different universe... what if it didn't matter how hard you press on othe mysterious object? If it meant nothing to the object.. or whatever little components (like atoms) it's consisted of? Or,
Or,
what if it was possible to bent that mysterious object by pressing on it in downward direction... but what if the object's special atoms didn't care? Like "alright, fine, squeeze us all you want, we don't care, we don't want to go back to our original position". So the object could just keep getting whatever shape it can... I mean, there would be some sort of attraction forces within the mysterious object's atoms, but if they're broken... then it's fine... the "detached" atoms would simply reconnect with other atoms once they get closer to each other. So you could keep breaking bonds / forces of attraction between this mysterious object's atoms, and the atoms would simply get "reconnected" with other atoms, but the bonds between atoms wouldn't oppose your force...
For instance, if 2 newtons is needed to break a bond between two atoms of this mysterious object... then just 2 newtons is needed to break the bond, but there will be no resistance from the bond itself...
Oh wait, isn't it how it is in our universe already? It doesn't matter if the bonds "resist" whatever force that is trying to break that very bond, if it takes 2 newtons to break it, you'll break if you apply 2 newtons, regardless whether the bonds "resist" it or not, right?
Same thing as when you push on a wall, the wall pushes back.
So the wall is pushing back during the time you're pushing pushing the wall, right? Once you stop, the wall stops pushing back at you too, right? So there is no way to make an experience in which you suddenly stop pushing back but the wall continues pushing back at you for a brief moment? So what you're implying is that wall pushing back and you pushing the wall - these forces would stop acting on each other instantaneously?
Then another questions arises, if the wall is pushing back with equal magnitude to your force, how the heck do you manage to "move" the wall? Or let's say an object weights 10kg, would it mean the wall would ONLY push back with a force equal to 90 N? (10*9.8)??
If the object is on a slanted surface, then the magnitude of the normal force is not equal to the gravitational force. In fact, it's less. There's an angle between force and motion, so cos theta will have a value. This is why the object will slide down.
My problem is understanding why the surface would exert a normal force. For instance, molecular bonds/attraction would explain it (they don't want to be squeezed in, so they oppose an object in order to become "unsqueezed").
Assuming there is molecular attraction between the surface's molecules, wouldn't then,
molecules inside the object (perpendicular block) would rush towards its end corner (circled in the picture above)? And if so, then the surface area that touches that end corner of the block.. would experience a stronger "push" downwards from the block (due to gravity), right? Or does the surface exert equally distributed normal force across all of its area that is affected by the block?
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A.T.
So your attraction to the earth is ... upwards?
Your attraction is towards you, obviously. That's what "attraction" means.
Then another questions arises, if the wall is pushing back with equal magnitude to your force, how the heck do you manage to "move" the wall?
If the wall moves, the ground pushes you more than the wall.
vizakenjack
Mark44
Mentor
So your attraction to the earth is ... upwards?
A.T. said:
You attraction is towards you, obviously. That's what "attractions" means.
Yes, each body attracts the other, in forces that are oppositely directed.
Think about the earth and the moon. The earth exerts a gravitational force on the moon, and the moon likewise exerts a gravitational force on the earth.
My problem is understanding why the surface would exert a normal force.
Newton's third law of motion states: Every action has an equal and opposite reaction.
The weight of an object placed on the plane is the force exerted by it on the plane. So, the plane exerts an equal and opposite force, called the normal.
In the case of a slanted plane, the surface is not able to equal the gravitational force because like I said, there's an angle between force and motion.
Your attraction is towards you, obviously. That's what "attraction" means.
Uhm, what?
Okay let's get this straight. An object weighs 10kg, it rests on the ground.
How many forces are there?
Force due to gravity (10kg * g), right?
Then there is a "normal force" coming from the surface in upward direction, right?
But then, you also claim there is another force acting downwards... exerted by the object? What is this additional force that "leaves footprints" (when the object is a human)?
Yes, each body attracts the other, in forces that are oppositely directed.
So in the example above with an object that weighs 10 kg.
There are three forces:
The object's force due to gravity ↓
Surface's "normal force" acting on the object upwards ↑
The object attracts the earth to itself, so if it attracts the earth... then what would the direction of this force be? ↓ or ↑? On one hand, the object "pulls the earth" towards itself, so the force must be ↑ (towards the object), on the other... the object exerts the force to the earth and "tells" the earth that it "wants" it, then direction is towards the earth, down ↓
Although, I'm more inclined towards the former assumption, because an objects "pulls" the earth towards itself. So the direction should be towards the object... is it correct to say that the direction of such force is ↑?
...
A.T.
What is this additional force that "leaves footprints" (when the object is a human)?
Contact force, same but opposite as the normal force acting on the feet. See Newtons 3rd Law.
vizakenjack
Contact force, same but opposite as the normal force acting on the feet.
So, in an example with an object that weighs 10 kg and rests on the ground, there are 3 forces present then?
Object's force downwards due to gravity.
Normal force exerted by the ground in upward direction.
Contact force between the object and the ground, the object exerts it in downwards direction.
Correct?
Mark44
Mentor
Uhm, what?
Okay let's get this straight. An object weighs 10kg, it rests on the ground.
How many forces are there?
Force due to gravity (10kg * g), right?
Then there is a "normal force" coming from the surface in upward direction, right?
But then, you also claim there is another force acting downwards... exerted by the object? What is this additional force that "leaves footprints" (when the object is a human)?
There are two forces, one up, one down. Each body (the object and the earth) attracts the other.
vizakenjack said:
So in the example above with an object that weighs 10 kg.
There are three forces:
No. There are two forces, as explained above. See Newton's Third Law (http://en.wikipedia.org/wiki/Newton's_laws_of_motion). The third law states that all forces exist in pairs.
vizakenjack said:
The object's force due to gravity ↓
Surface's "normal force" acting on the object upwards ↑
The object attracts the earth to itself, so if it attracts the earth... then what would the direction of this force be? ↓ or ↑? On one hand, the object "pulls the earth" towards itself, so the force must be ↑ (towards the object), on the other... the object exerts the force to the earth and "tells" the earth that it "wants" it, then direction is towards the earth, down ↓
Although, I'm more inclined towards the former assumption, because an objects "pulls" the earth towards itself. So the direction should be towards the object... is it correct to say that the direction of such force is ↑?
...
vizakenjack
A.T.
So, in an example with an object that weighs 10 kg and rests on the ground, there are 3 forces present then?
No, there are 4 forces. Every interaction consists of a force pair, and you have two interactions: gravitational attraction and contact (electromagnetic repulsion).
There are two forces, one up, one down.
No, there are 4 forces.
So how many are really there?
Probably even more, if you also include the force of gravity from the black hole located at the center of our galaxy(?).
But only assuming earth as an isolated system.
I feel like, for simplicity sake it is said that there are only 2 forces acting on an object that weighs 10 kg and is resting on the ground, correct?
you have two interactions: gravitational attraction and contact (electromagnetic repulsion).
It is said here that an electromagnetic repulsion is the reason the surface "resists" and exerts a normal force on an object, correct?
What's the difference between a contact force and a normal force?
Mark44
Mentor
The contact forces that @A.T. mentioned keep the two objects separated. In force diagrams, these forces are usually ignored, at least that's my recollection from the physics courses I had long ago. The earth attracts the 10 kg object toward it, and the 10 kg object attracts the earth toward the object. The electromagnetic forces keep the objects from "welding" themselves together.
I'm not sure you realized it, @vizenjack, but the word "normal" in the term "normal force" isn't synonymous with "ordinary." The meaning here is "perpendicular," since the force is perpendicular to the ground. The
vizakenjack
jbriggs444
Homework Helper
So how many are really there?
As many as you care to account for. Two obvious ones and their two third-law partners.
I feel like, for simplicity sake it is said that there are only 2 forces acting on an object that weighs 10 kg and is resting on the ground, correct?
Yes. Two forces acting on the object. Their third law partners act on other objects.
What's the difference between a contact force and a normal force?
A "contact" force is the force on an object from another surface or object that it is in direct contact with.
A "normal" force is the component of the contact force from a surface that is exerted perpendicular to the surface. In this context, "normal" means "perpendicular". [The component of a contact force that is tangential to the surface is usually frictional]
Two forces acting on the object. Their third law partners act on other objects.
Can you give an example of a third law partner in regards to a situation when a 10kg object rests on the ground?
Also,
molecules inside the object (perpendicular block) would not "rush" towards the end corner (circled) because there is electrostatic repulsion between molecules in the block, correct? I mean, that makes sense, why would a molecule from the top attempt to go to the bottom (even if there is gravity pulling it downwards) if it simply CANNOT do so because other molecules would not allow it.. and even if you placed an extra molecule at the bottom, it would have become too "tight" at the bottom and all the molecules would have attempted to spread out so that the distance between molecules increased, right?
In other words, molecule from the top can't squeeze in-between other molecules beneath it and go like "uh, excuse me, I just want to go down, mind giving me way? thanks", right?
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CC-MAIN-2021-31
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https://discuss.leetcode.com/topic/20455/c-4ms-dp-solution
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# C++ 4ms DP solution
• The recursive solution is not so efficient since it computes the possible results of the same substrings many times. To improve the time complexity, we can use dp which stores the intermediate results of the substrings. Define a two-dimensional array
dp[i][j] = vector of all possible results computed from the substring starting from operand i and
including the subsequent j operators and operands.
Let cntOperators be the count of operators in the input string. Note that the count of operands is cntOperators + 1. The range of i is [0, cntOperators + 1) and the range of j is [0, cntOperators).
``````class Solution
{
// Separate the operands and the operators ('+', '-', '*')
// in the input string.
void GetOperandsAndOperatorsFromString(
string &input,
vector<int> &operands,
vector<char> &operators)
{
int number = 0;
for (int i = 0; i <= input.length(); i++)
{
if ((input[i] >= '0') && (input[i] <= '9'))
{
// This character is a digit and add it to
// the number.
number = 10*number + (input[i] - '0');
}
else if ((input[i] == '+') ||
(input[i] == '-') ||
(input[i] == '*'))
{
// This character is an operator and it implies
// the end of the previous number. Add the number
// to the operand vector and reset the number. Also
// add the operator to the operator vector.
operands.push_back(number);
number = 0;
operators.push_back(input[i]);
}
else if (input[i] == '\0')
{
// This is the end of the input string. Add the
// last number to the operand vector.
operands.push_back(number);
}
}
}
public:
vector<int> diffWaysToCompute(string input)
{
vector<int> operands;
vector<char> operators;
// Get the operands and operators from the input string.
GetOperandsAndOperatorsFromString(input, operands, operators);
// Note that count of operands = cnt of operators + 1.
int cntOperators = operators.size();
// dp[operandIndex][cntUsedOperators] = vector of all possible results
// computed from the substring starting from operand "operandIndex" and
// including the subsequent "cntUsedOperators" operators and operands.
vector<vector<vector<int>>> dp(cntOperators + 1, vector<vector<int>>());
// When cntUsedOperators = 0, dp[operandIndex][0] = operands[operandIndex].
for (int operandIndex = 0; operandIndex < cntOperators + 1; operandIndex++)
{
dp[operandIndex].push_back(vector<int>{operands[operandIndex]});
}
// cntUsedOperators = 1,...,cntOperators.
for (int cntUsedOperators = 1; cntUsedOperators <= cntOperators; cntUsedOperators++)
{
// Note that operandIndex starts from 0 and ends at the last operand which
// is followed by "cntUsedOperators" operators and operands.
for (int operandIndex = 0;
operandIndex < cntOperators + 1 - cntUsedOperators;
operandIndex++)
{
// dp[operandIndex][cntUsedOperators] = the union of all outcomes for each
// possible last operator.
vector<int> tmpRes;
for (int lastOperatorIndex = operandIndex; lastOperatorIndex < operandIndex + cntUsedOperators; lastOperatorIndex++)
{
// The last operator breaks the substring into two halves.
// dp[operandIndex][lastOperatorIndex - operandIndex] =
// all the outcomes of the first half.
// dp[lastOperatorIndex + 1][cntUsedOperators - 1 - (lastOperatorIndex - operandIndex)] =
// all the outcomes of the second half.
for (auto &x : dp[operandIndex][lastOperatorIndex - operandIndex])
{
for (auto &y : dp[lastOperatorIndex + 1][cntUsedOperators - 1 - (lastOperatorIndex - operandIndex)])
{
if (operators[lastOperatorIndex] == '+')
{
tmpRes.push_back(x + y);
}
else if (operators[lastOperatorIndex] == '-')
{
tmpRes.push_back(x - y);
}
else
{
tmpRes.push_back(x * y);
}
}
}
}
dp[operandIndex].push_back(tmpRes);
}
}
// The final result is all possible outcomes starting from operand 0 and including all operators.
return dp[0][cntOperators];
}
};``````
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https://excelbetting.com/blog/strategies/what-is-sports-betting-loading-how-cash-flow-affects-the-odds-in-the-line
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What is sports betting loading: how cash flow affects the odds in the line
Newbies lose money out of the blue due to overloads in betting. Finding out how to benefit from changing odds
Projections help make money out of thin air: you bet on the same outcome as others, but with a higher odds. You need to know when to bet a few days before the match, and in what cases, 5-10 minutes before the start of the game. In this article, we will analyze what a load is, its types, the effect on the odds, and four betting strategies.
This article was published thanks to the"Excelbetting.com" Betting Spreadsheet - Analysis, study sports betting, take tests, earn experience and beat the leaderboard. Outrun the editor-in-chief!
Loss in bets is a decrease in the coefficient for one of the outcomes due to the large number of bets on this choice.
Due to the overload, bookmakers reduce the odds by one outcome and raise them by the opposite.
For example, in the football match “Atalanta” - “Roma” bookmakers set odds of 1.90 for total goals over 3.5 and 1.85 for total goals under 3.5.
Due to the large number of bets on the horse outcome, the bookmakers have adjusted the line: 1.65 for TB 3.5 and 2.20 for TM 3.5
The main types of loads for sporting events
1. Organized: Bets due to leaked match-fixing information or predicted by a famous capper. For example, in the hockey match "Traktor" - "Neftekhimik" the expert suggests taking the total pucks over 4.5, and the players follow his advice.
2. Unorganized: Betting on a clear favorite. Let's imagine that in a tennis match Roger Federer - Alexander Bublik the players load onto the Swiss. Therefore, the bookmaker reduces the odds for the favorite to win.
3. Geoprocess: bets based on fan preferences. For example, in a football match Russia - Spain, most Russians are betting on the hosts, regardless of the odds.
Odds changes before a football match
Please note that the coefficients change not only due to the load. There are other reasons as well:
• The coach retained one or more key players.
• The media reported additional information: the match without spectators, the meeting was moved to another stadium, and so on.
• A player from one of the teams will miss the match due to injury or disqualification.
Read about how to bet on football based on strategies here: How to Learn Football Betting: Good Strategies and Tips for Newbies
Betting on total over 2.5 in football
Bet on total goals over 2.5 immediately after the appearance of the bookmaker's line according to the criteria:
• Both teams score on average over 1.5 goals per match.
• Bookmakers give odds less than 3.00 for each team to win.
Most players will load TB 2.5, and the odds on this outcome before the match will be lower.
Imagine the bookmakers gave 1.70 on 2.5 TB in a Manchester City v Real Madrid match. Before the start of the meeting, the coefficient for TB 2.5 dropped to 1.50.
At a rate of 10,000 rubles immediately after the line was released, the potential payout was 17,000 rubles: 10,000 x 1.70. And before the match - 15,000 rubles: 10,000 x 1.50.
Total under 2.5 bets in football
Bet on total goals under 2.5 5-10 minutes before the start of the match in the following cases:
• One of the teams is satisfied with a draw.
• Bookmakers give odds less than 2.00 for the victory of one of the teams.
Due to the load on TB 2.5, you will bet on TM 2.5 with a higher coefficient.
Let's imagine that the bookmakers gave 2.20 on TM 2.5 in the PSG - Borussia D match. Before the start of the match, the coefficient on TM 2.5 increased to 2.60.
With a bet of 5000 rubles immediately after the line was released, the potential payout was 11,000 rubles: 5000 x 2.20. And before the meeting - 13,000 rubles: 5,000 x 2.60.
Read more about total betting here: The value of total in bets: what is Over and Under, examples of calculation in football and other sports
Betting on the outsider
In a meeting between a clear favorite and a clear outsider due to a load, the bookmaker will reduce the odds for the favorite. Bet an outsider with a handicap for odds of about 1.70-1.95.
Let's imagine that in a match between Bayern and Schalke, the odds for guests with a handicap (2.5) increased from 1.60 to 1.90. With a bet of 1,000 rubles, the potential payout is 1,900 rubles: 1,000 x 1.90. Favorite
betting
If the home favorite accepts an outsider and sets up a second line-up, the players can load the away win. Bet on the favorite to win or draw, that is, on 1X for odds of 1.50 or more.
For example, in the Liverpool v Everton match, the 1X home odds increased from 1.15 to 1.55. With a bet of 1,000 rubles, the potential payout is 1,550 rubles: 1,000 x 1.55.
When betting at bookmakers, use the loads to your advantage. If you are sure that the majority will bet on the favorite to win, it is more profitable to bet on the reverse outcome of the match 5-10 minutes before the start of the meeting.
To make a profit, you must assess the odds correctly. Forecasts will help you play at high odds.
Best Betting Software - Best Analysis results - Best winnings
You can check out some of our best betting software. They will provide you with detailed analysis of each match with high visibility of the results in each match.
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en
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https://www.chiefdelphi.com/t/h-wheel-question/141626
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# H-wheel Question
Hi all,
This is our first time going with an H-drive system and we are having a problem with our center wheel for our drive losing traction to the ground preventing us from strafing. It is currently being tensioned towards the ground with ~20 lbs of force using constant force springs.
Sometimes the wheel grabs and we move, but other times it completely slips and we go nowhere. Are we using enough force? Any other thoughts?
Thanks for the input!
It really depends on the weight of your robot, including whatever you’re carrying. If your robot weighs 120lb before battery and you’ve got a tall stack of totes and a battery, you may very well weigh 150lb. In this case, the 20# corresponds to only about 13% of the robot weight. If you’ve got wheel alignment problems or other friction in your system, this may not be enough to get moving. The more force you put on those springs, the more acceleration you can get in strafing, at a commensurate loss in your tank drive.
If you are more likely to slip when carrying a load than when empty, this is quite probably your problem.
Seeing as you are using a spring to force the wheel down I am going to guess it is safe to say that you have a pivot and some sort of hard stop. It is possible that the wheel needs to come down further for more “contact” with the floor. If possible make the stop adjustable so you can lower it down a bit more.
Also is the problem more of an issue when loaded with totes or does it matter loaded or unloaded ?
We had the same issue and a 1/16" drop in the strafe wheels did the trick.
Should have added that if your H wheel can go slightly lower that your regular drive wheels you shouldn’t have a problem with the H wheel unloading the drive wheels as long as the rest of the bot weighs more than the 20# of force you are applying to the H wheel.
Without seeing the setup you have this is all speculation, and should not be taken as fact. Only testing will give you the true answer.
Yes!
Also, does the surface make a difference? If your wheel is a bit high, you may run just fine on carpet, but not a smooth (e.g. tile) floor.
Does the direction of the strafe make a difference in the slipping? The geometry of the pivot may be such that the strafe wheel tries to push itself under the pivot in one direction, increasing its loading and giving it more traction.
The wheel is fully contacting the ground. When we lift the robot it can drop another half inch before bottoming out so I do not think the geometry is the issue.
As far as how load affects it, we seem to have the same issue loaded or unloaded with a stack.
All of the testing was done on a pretty large section of FRC field carpet.
This year was our first attempt at a H drive as well, although we went with more of a ladder drive.
We considered a drive like this, but discarded it because it appears that the two strafe wheels separated on a single axle would provide too much resistance to tank mode turning. Has this been an issue, or do you have some sort of differential working between the two strafe wheels?
Or (I just noticed the pneumatics) perhaps you can lift the strafe wheels completely off the carpet for tank-mode driving?
This is the setup on our practice bot. The competition bot has a better bracket system but I don’t have any examples to give.
do you lift those “ladder drive” wheels when not needed?
IF not, how is your turning?
Yes we can lift the drive as needed. We went this route for two reasons… The one you mentioned about tank drive, and the other so there would be no issues driving over the scoring platforms.
I’d warn against running a system like that; we used a fixed setup with a drop during the 2014 competition season, and was probably in my opinion our robot’s biggest drawback.
In the offseason, we actuated the center omni wheel using pneumatics to ensure we always had contact with the ground. If you don’t actuate it and just fix it in place, a few problems arise: you either mount it too high up, and because the carpet is not entirely level (in a perfect world, it would be, but sadly it’s not), there are a few patches where you will get stuck. If you mount it too low, it will always touch the ground, but not only will it cause ridiculous amounts of friction (slowing your acceleration and speed in the forwards/backwards direction), it will take a toll on the batteries as well. You have to take the bump of the scoring platform into account as well.
Just because the bot is able to strafe in your shop doesn’t mean it will necessarily strafe out on the competition field, as we found out.
As already mentioned, more force may be needed. Many teams utilze pneumatics to raise/lower the “H” drive wheels, thus giving it more down force. Also, the full raising of the wheels eliminates drag and wear. Omni’s do have drag, especially when at an angle not inlign with the rollers. Also the omni rollers wear quite a bit on the carpet. We had a few omnis replaced last year due to “Flat spotting” a few rollers . It’s just the nature of the beast.
roger
The rollers do not have bearings. The axial force on the rollers pushes them against the stops and creates a lot of friction. Same is true of the affordable mecanum wheels used in FRC
. This is a major factor in the “non-idealized” behavior of omni and mec in FRC
.
AndyMark calls them bearings, but I agree that they are more properly bushings. The more traditional omnis with a lot of short cylindrical rollers have brass bushings at the ends of the rollers, and the new DuraOmnis with the long tapered rollers have nylon bushings. In both cases, these rotate on steel shafts.
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https://tableaumagic.com/category/learners/intermediate/
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Home Learners Intemediate
# Intemediate
### Video: Tableau with Music / Filled Circle
0
Using a few neat tricks and some Table Calculations, we have created this awesome chart in Tableau. Enjoy.https://www.youtube.com/watch?v=IYU48oc7-dU Song: Sad Minuet by Sir Cubworth
### Video: Tableau with Music / Dendrogram with Rounded Bar Charts
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Using a few neat tricks and some Table Calculations, we have created this awesome chart in Tableau. Enjoy.https://youtu.be/ywH3qFD9UNI Song: Funeral March by Chopin
### Creating Curved Polar Chart in Tableau
2
Just for a little bit of fun before I take a month off blogging, yep, we all need a little break fro time to time, I decided to have...
### Video: Tableau with Music / Single Level Dendrogram
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Using a few neat tricks and some Table Calculations, we have created this awesome chart in Tableau. Enjoy.https://youtu.be/AYeb8rb_FZA Song: Love of All by Twin...
### Creating a Polar Chart in Tableau
3
As we close down our second set of tutorials in 2020, I wanted to finally have some fun with Polar Charts, so here you have it, something that is...
### Video: Tableau with Music / Radial Stacked Bar Charts
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Using a few neat tricks and some Table Calculations, we have created this awesome chart in Tableau. Enjoy.https://youtu.be/r5TgiLyHe4A Song: I Am a Man Who...
### Create a Radial Column Chart (Variation)
2
I wanted to have some fun and build up a Polar Chart, however, I had an accident and decided to run with it, and this is the resulting data...
### Video: Tableau with Music / Bar Trend Chart
1
Using a few neat tricks and some Table Calculations, we have created this awesome chart in Tableau. Enjoy.https://youtu.be/8nVY98Ouupg Song: William Tell Overture by Rossini
### Creating a Bezier Curve in Tableau
1
I love experimenting with Tableau and I wanted to revisit my formal education and explore how we might be able to draw a Bezier Curve in Tableau. Although I...
### Video: Tableau with Music / Gauge Chart
0
Using Parameters, Bins and quite a few Table Calculations, we have created this awesome Gauge in Tableau. Enjoy.https://youtu.be/v75ERzagsXc 9th Symphony (Finale) by Beethoven
### INDEX() Table Calculation
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The INDEX() function returns the index of the current row in the partition, without any sorting with regard to value. The first-row...
### Video: Tableau with Music / Rootogram
0
Tableau with Music / Rootogram. We will create several Level of Detail Expressions to create this unique Rootogram Diagram. See how each...
### Venn Diagrams in Tableau
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Find the area of triangle is - 6536370 Referring to the preceding figure: Not surprisingly, all the semiperimeters are the same, because all the perimeters are 240. Now we use Heron’s formula, in effect backing up from the conclusion of the proof: Stack Overflow for Teams is a private, secure spot for you and your coworkers to find and share information. Step 5: Printing the output The conclusion is that area(ABC) = 0.5(a+b+c)r = sr, where s is the semiperimeter of ABC. 8 c. 6 d. 7 Right Answer-b. Heron's formula - math word problems The Heron's formula is used to calculate the contents of a general triangle using the lengths of its sides. Q&A for Work. Equivalently, it gives three different factorizations of 16A^2, each of the form . This is simply the perimeter, divided by two. Also find all CBSE Chapter Notes, Books, Previous Year Question Paper with Solution, etc. Part A Let O be the center of the inscribed circle. Not apt to be a huge difference with double in realistic scenarios, but when using float , a triangle with sides (16777215.0f, 16777215f, 4.0f) will often yield an answer which too big or too small by about … Heron's formula states that the area of a triangle whose sides have lengths a, b, and c is: Calculate the side of a triangle b, if side a = 12 and side c = 6 and the semi – perimeter equals 13. a. Heron’s Formula The Preliminaries… That is why we used standard formula. ANSWER-“S”is the half of the triangle perimeter S=a+b+c/2 Introduction Hero of Alexandria The formula is credited to hero(or heron) of Alexandria. Proof of the Heron's formula and the convergence of the Heron's method sequence. We can calculate area(AIB) = 0.5 * AB * IIc = 0.5cr. Finding the semiperimeter s is a two-step process: add up the length of the three sides first, and then divide by 2. When using floating-point values, writing the formula as Math.sqrt((a+b+c)*(a-b+c)*(b-a+c)*(a-c+b)) * 0.5 will yield more accurate results than precomputing the semiperimeter. Proof of this formula can be found in Hero of Alexandria’s book “Metrica”. Substituting the expressions with s to (4), we have. Herons formula: sqare root of s(s-a)(s-b)(s-c) s= semiperimeter. # calculate the Perimeter Perimeter = a + b + c. Next, Calculating the semi perimeter using the formula (a+b+c)/2. 3. Heron’s formula. the difference between the semi perimeter and the sides of a triangle ABC are 8cm, 7cm and 5cm respectively . If it's faster for me on a calculator, it was also presumably preferable for people living in the "long-hand era"... (Incidentally, these formulas were considered terrible centuries ago, in the days when powers … The formula is as follows: Although this seems to be a bit tricky (in fact, it is), it might come in handy when we have to find the area of a triangle, and we have … www.instructab… Next, we are calculating the semi perimeter using the formula (a+b+c)/2. Formula. compute the square root in the formula below. Now, . Heron’s formula allows us to find the area of a triangle when only the lengths of the three sides are given. This formula only works, of course, when you know what the height of the triangle is. Jan 15, 2021 - Finding Area of Quadrilaterals using Herons Formula - Herons Formula, Class 9, Mathema | EduRev Notes is made by best teachers of Class 9. Using (1) and (3), we calculate the area of the triangle , (4) Now, if we let be the semiperimeter (half the perimeter) of triangle , then. Heron's formula calculator uses Area=sqrt(Semiperimeter Of Triangle *(Semiperimeter -Side A)*(Semiperimeter -Side B)*(Semiperimeter -Side C)) to calculate the Area, Heron's formula (sometimes called Hero's formula), named after Hero of Alexandria, gives the area of a triangle when the length of all three sides are known. Heron's Triangle Area Calculator is an online tool for area calculation programmed to find out the Total Triangle Area by using heron's formula with respect to the given values of inputs a, b and c. This Heron's Triangle Area Calculator uses different converter functions in order to generate the Output in different units such as Inches, Feet, Meters, Centimeters and Millimeters. // To find the height when we only know the lengths of the sides, we can use Herons formula to get the Area. Heron's formula implementations in C++, Java and PHP; Proof of Heron's Formula Using Complex Numbers; In general, it is a good advice not to use Heron's formula in computer programs whenever we can avoid it. Calculate the side of a triangle b, if side a = 12 and side c = 6 and the semi – perimeter equals 13. a. Who was a great engineer and mathematician in 10–70 AD. Be sure to use the relevant include directives. It can be applied to any shape of triangle, as long as we know its three side lengths. Students can practice these questions for the … Also, and . 9 b. $\begingroup$ I can also attest that, from a practical viewpoint, it is considerably quicker to compute the area of a triangle using Heron than the other expression. SELF INTRODUCTION NAME- Siddhi U Pawar CLASS- IX Arctic SUBJECT- Mathematics TOPIC- Heron’s Formula CONTENT- 1) Introduction 2) Derivation of the formula 3) Importance of the formula 4) Applications of the formula 5) Summary SCHOOL- Podar International School, Sangli. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.. Visit Stack Exchange If the focus is not on Algebra and time is limited, then you may want to let the students use calculators to compute the square roots. Factoring out 1/4, this gives three different ways of expressing (2ab)^2 - (a^2+b^2-c^2)^2 as a difference of two squares. Heron of Alexandria: biography and Heron's Formula and Method: area of a triangle with sides a, b and c (semiperimeter) and approximating square roots. 3 Interesting facts about Heron’s Formula. Simplifying, we have , the Heron’s formula. Assume that all input values are positive real numbers that satisfy the triangle inequality. Missing Zion National Park hiker found alive after 12 days Some people find the concept of half the perimeter of a polygon to be a little … well, weird. Herons Formula Word Problems With Heron's formula - math word problems The Heron's formula is used to calculate the contents of a general triangle using the lengths of its sides. 8. Calculating the C Area of a triangle using Heron’s Formula: sqrt(s*(s-a)*(s-b)*(s-c)); (sqrt() is the math function, which is used to calculate the square root. Hero of Alexandria. Given a triangle with side lengths a, b and c and a semiperimeter s, the area A of the triangle is calculated as follows: $$A=\sqrt{p(p-a)(p-b)(p-c)}$$ Amongst other things, he developed the Aeolipile, the first known steam engine, but it was treated as a toy! For example, whenever vertex coordinates are known, vector product is a much better alternative. Heron’s formula. And in the same way we find area(BIC) = 0.5ar and area(CIA) = 0.5br. How do I calculate the area of a triangle? Let r be the radius of this circle (Figure 7). 8 c. 6 d. 7 Right Answer-b. This free online calculator will help you to find the area of a triangle using the Heron's formula. Heron’s formula has been known to mathematicians for nearly 2000 years. The s is Heron’s formula is what’s called the "semiperimeter". This is a link to my Quadratic formula program instructable. Step 4: After completing the function execution, then it will return the Area value. Heron's formula is a formula that can be used to find the area of a triangle, when given its three side lengths. 9 b. 8. This formula generalizes Heron's formula for the area of a triangle. In geometry, Heron's (or Hero's) formula states that the area of a triangle whose sides have lengths is where is the semiperimeter of the triangle: Heron's formula can also be written as: 1 History 2 Proof 3 Proof using the Pythagorean theorem 4 Numerical stability 5 Generalizations 5.1 Heron-looking formula for tetrahedrons 6 See also 7 Notes 8 References 9 External links The formula … This document is highly rated by Class 9 students and has been viewed 15758 times. HERON’S FORMULA 2. s = (a + b + c) / 2 The formula is credited to Hero (or Heron) of Alexandria, who was a Greek Engineer and Mathematician in 10 – 70 AD. Teams. Another is Heron’s formula which gives the area in terms of the three sides of the triangle, specifically, as the square root of the product s(s – a)(s – b)(s – c) where s is the semiperimeter of the triangle, that is, s = (a + b + c)/2. Maths herons formula 1. Access Solution for NCERT Class 9 Mathematics Chapter Herons Formula including all intext questions and Exercise questions solved by subject matter expert of BeTrained.In. A triangle may be regarded as a quadrilateral with one side of length zero. Heron’s formula has been known to mathematicians for nearly 2000 years. Relevant Formulas: Let A, B and C denote the sides of the triangle: Perimeter = A + B + C Area = sqrt(S(S − A)(S − B)(S − C)), where S = Perimeter/2 This was the last piece of the proof from last time. Find the semiperimeter, s, for each triangle. Heron's Formula Program for the TI-83 and 84: In this instructable, I will show you how to write a program on your calculator that will do Heron's formula for you. Heron’s formula can be used to calculate the area of a triangle when its three side lengths are given. which is equivalent to Heron's formula. 3. Next, Calculating the Perimeter of Triangle using the formula P = a+b+c. Use Heron’s Formula to … You will probably find this instructable helpful too. Class 9 Maths Chapter 12 (Heron’s Formula) MCQs are available with answers, here at BYJU’S. Proof of this formula can be found in Hero of Alexandria’s book “Metrica”. Using this online calculator, you will receive a detailed step-by-step solution to your problem, which will help you understand the algorithm how to find triangle area using the Heron's formula. Although we can write semi perimeter = (Perimeter/2) but we want show the formula behind. But semiperimeter -- half the perimeter -- is very useful, and for Heron's Formula, you need it! How do I calculate the area of a triangle? which is equivalent to. 3 Interesting facts about Heron’s Formula. Semiperimeter. Because the proof of Heron's Formula is "circuitous" and long, we'll divide the proof into three main parts. His formula states: K = s(s − a)(s −b)(s − c) Where a, b, and c, are the lengths of the sides and s is the semiperimeter of the triangle. From this perspective, as d approaches zero, a cyclic quadrilateral converges into a cyclic triangle (all triangles are cyclic), and Brahmagupta's formula simplifies to Heron's formula.
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# Free help with homework
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• find similar questions
# Please someone can check out if my answers are correct and help me with the questions # 11, 15 and 16. Thanks. For # 11. 3/8 = 0.375 x 14.5= 5.4375. then
14.5 - 5.4375= 9.0625 miles.for # 15. 3/8= 0.375x 32.1 = 12.0375. then 32.1 - 12.0375= 20.0625 miles. for # 16. 4/5 = 0.8 x 40 = 32. then 40 - 32 = 8 so the answer is 1 hour.
1
by ezeriva1
i think all u r answers r correct frnd
Hi my frnd you think but you not sure
yeah #11 is quite tricky
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# [FOM] Why NP is not P
Lew Gordeew legor at gmx.de
Fri Nov 5 07:41:00 EST 2004
``` In previous posting of Nov 03 we referred to
[G]:
http://www-ls.informatik.uni-tuebingen.de/logik/gordeew/publikationen/ProofSketch.pdf
where proposition NP = P was reduced to a combinatorial principle, whose
negation (call it CP) thus infers that NP = P fails, and hence NP != P holds
[G: Claim 32]. Since the precise description of CP is rather involved, most
transparent instances of CP were briefly discussed, too [G: Claims 26, 28,
30]. Here we loosely address the combinatorial background.
Recall school algebra. Consider the expansion operation 'expand' whose
primary application is to distribute products over sums. Example:
expand(((x + y) * (z + u) + (a + b) * (c + d)) * (v + w)) = xzv + xzw + xuv
+ xuw + yzv + yzw + yuv + yuw + acv + acw + adv + adw + bcv + bcw + bdv +
bdw
Clearly 'expand' is analogous to the familiar boolean/borel DNF-expansion.
For example, we have (where x_ and y_ denote the negation of x and y,
respectively):
1. expand(((x + y) * (z + x) + (a + x) * (c + x_)) * (v + y_)) = xzv + xzy_
+ xxv + xxy_ + yzv + yzy_ + xyv + xyy_ + acv +acy_ + ax_v + ax_y_ + xcv +
xcy_ + xx_v + xx_y_
In boolean/borel case we can upgrade 'expand' to 'expandc' that removes all
inconsistent (zero-) clauses:
2. expandc(((x + y) * (z + x) + (a + x) * (c + x_)) * (v + y_)) = xzv + xzy_
+ xxv + xxy_ + yzv + xyv + acv + acy_ + ax_v + ax_y_ + xcv + xcy_
Furthermore, we consider a modified "positive" expansion 'expandcp' that
removes all remaining negative literals:
3. expandcp(((x + y) * (z + x) + (a + x) * (c + x_)) * (v + y_)) = xzv + xz
+ xxv + xx + yzv + xyv + acv + ac + av + a + xcv + xc
Finally, we define the corresponding operation 'base' by applying the
remaining boolean laws - this yields:
4. base(((x + y) * (z + x) + (a + x) * (c + x_)) * (v + y_)) = x + yzv + a
Now consider minimal inversions of 'base'. We ask how "small" can be input
polynomial P such that base(P) = B for a given "big" basic expansion
B. Intuitively, if B is very involved, then P can't be very small. Actually
we specify "small" and "big" as being (space-) polynomial and
exponential in a given parameter n, respectively.
Back to CP, our most transparent instances are in the following form, for a
given precisely defined "big" basic expansion (called Sigma_2-base) Bn.
Proposition. Let c > 0 be fixed. There is no polynomial P satisfying base(P)
= Bn, whose weight is smaller than n^c, provided that n is big enough.
CP itself is a slight modification thereof. Formalization of the proofs
(work in progress) will hardly require strong set-theoretical assumptions
like those used in Harvey Friedman's combinatorics.
--
Geschenkt: 3 Monate GMX ProMail + 3 Top-Spielfilme auf DVD
++ Jetzt kostenlos testen http://www.gmx.net/de/go/mail ++
```
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Spinroot
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#1 2011-07-29 10:43:09
Antony
Member
Registered: 2011-07-20
Posts: 3
Combinations of several formulas LTL
HI,
I wonder if it was possible to combine multiple formulas LTL, that is:
if I define :
ltl p1{ []<>eat }
ltl p2{ []<>wait }
you can do, for example p1 && p2 ??
thank you...
Offline
#2 2011-07-29 18:32:18
spinroot
forum
Registered: 2010-11-18
Posts: 691
Website
Re: Combinations of several formulas LTL
yes you can have multiple ltl formula
but you can only check one of them at a time
another method is to first combine them into a single formula, using && or ||, and then do a single verification run....
Offline
#3 2011-07-31 10:22:25
Antony
Member
Registered: 2011-07-20
Posts: 3
Re: Combinations of several formulas LTL
ok, I did as you told me, but executing this formula:
ltl p10 {
([]((available[0]==0) -> (available[0]==0) U ( (available[0]==-1) U (available[0]==1) ) ) && []((available[0]==1) -> (available[0]==1) U ( (available[0]==-1) U (available[0]==0) )) ) &&
( []((available[1]==0) -> (available[1]==0) U ( (available[1]==-1) U (available[1]==1) ) ) && []((available[1]==1) -> (available[1]==1) U ( (available[1]==-1) U (available[1]==0) )) ) &&
( []((available[2]==0) -> (available[2]==0) U ( (available[2]==-1) U (disponibile[2]==1) ) ) && []((available[2]==1) -> (available[2]==1) U ( (available[2]==-1) U (available[2]==0) )) ) &&
( []((available[3]==0) -> (available[3]==0) U ( (available[3]==-1) U (available[3]==1) ) ) && []((available[3]==1) -> (available[3]==1) U ( (available[3]==-1) U (available[3]==0) )) ) &&
( []((available[4]==0) -> (available[4]==0) U ( (available[4]==-1) U (available[4]==1) ) ) && []((available[4]==1) -> (available[4]==1) U ( (available[4]==-1) U (available[4]==0) )) ) .
I see this error : SIGABRT, what is the problem?
Last edited by Antony (2011-07-31 10:26:24)
Offline
#4 2011-08-01 00:38:14
spinroot
forum
Registered: 2010-11-18
Posts: 691
Website
Re: Combinations of several formulas LTL
likely that was too large too handle
you could try one of the other converters out there on the web, like the ltl2ba converter etc.
quite likely though, most of them will balk at something this large.
you can also (more easily probably) prove each property separately, for the same end-result
Offline
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# Why ignores constant when applying integral into differential-equation.?
I studied how to draw 'Direct-form I' and 'Direct-form II' from differential equation.
The book taught me like this:
$$\mbox{Give equation: }{a_0}{y(t)}+{a_1}{y'(t)}+{a_2}{y''(t)} = {b_0}{x(t)}+{b_1}{x'(t)}+{b_2}{x''(t)}$$
step 1. change from differentiator into integrator, because integrator is cheaper than differentiator.
\begin{array}{} {a_0}{y^{(2)}(t)}+{a_1}{y^{(1)}(t)}+{a_2}{y(t)} = {b_0}{x^{(2)}(t)}+{b_1}{x^{(1)}(t)}+{b_2}{x(t)} & \begin{cases} x^{(i)} \mbox{ is i-times integral of x}\\ y^{(i)} \mbox{ is i-times integral of y} \end{cases} \end{array}
step 2. change form like this: $y(t)=...$
\begin{array}{} \displaystyle y(t)=\frac{1}{a_2}\left( {b_0}{x^{(2)}(t)}+{b_1}{x^{(1)}(t)}+{b_2}{x(t)}-{a_0}{y^{(2)}(t)}-{a_1}{y^{(1)}(t)}\right) \end{array}
step 3. draw
My question is why doesn't consider constant value in step 1?
• Are you sure that you are not confusing differentiation and delay? The standard use for a "Direct Form I" would be a difference equation and the continuous equivalent might be closer to $a_0\cdot y(t) + a_1 \cdot y(t-T) + a_2 \cdot y(t - 2T) = ...$ Nov 6, 2015 at 19:48
• It is not confused. Discrete time is easier for me to understand than continuous time. Nov 6, 2015 at 20:03
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# SUBTRACTING POLYNOMIALS WORKSHEET
Problem 1 :
Subtract 5ab from 8ab.
Problem 2 :
Subtract (4x + 5y2) from (5x - y2).
Problem 3 :
Subtract (2x2 + 2y2 - 6) from (3x2 - 7y2 + 9).
Problem 4 :
Subtract (x3 + 4x2 - x + 6) from (2x3 + 5x2 - 2x + 7).
Problem 5 :
Subtract (2x3 + 7x2 - 3x - 3) from (3x3 - 2x2 - x + 4).
Problem 6 :
Subtract (3 + 3x - 7x5 - 5x6) from 2(x3 - x2 + 6x - 2).
Problem 7 :
Subtract (2x3 + 5x2 - 2x - 11) from (3x3 - 2x2 - 5x - 6).
Problem 8 :
Subtract (x3 + 4x2 - 12x - 5) from (5x3 + 3x2 + 2x - 10).
Problem 9 :
Subtract (12x3 + 14x2 + 17x - 12) from (15x+ 22x2 + 17x - 19).
Problem 10 :
The profits of two different production plants can be modeled as shown below, where x' is the number of units produced at each plant.
Eastern Plant :
-0.03x2 + 26x - 1600
Western Plant :
-0.02x2 + 20x - 1800
Write a polynomial that represents the difference of the profits at the eastern plant and the profits at the western plant.
Problem 1 :
Subtract 5ab from 8ab.
Solution :
= 8ab - 5ab
= 3ab
Problem 2 :
Subtract (4x + 5y2) from (5x - y2).
Solution :
= (5x - y2) - (4x + 5y2)
Distributive Property.
= 5x - y2 - 4x - 5y2
Group like terms together.
= (5x - 4x) + (-y2 - 5y2)
Combine like terms.
= x - 6y2
Problem 3 :
Subtract (2x2 + 2y2 - 6) from (3x2 - 7y2 + 9).
Solution :
= (3x2 - 7y2 + 9) - (2x2 + 2y2 - 6)
Distributive Property.
= 3x2 - 7y2 + 9 - 2x2 - 2y2 + 6
Group like terms together.
= (3x2 - 2x2) + (-7y- 2y2) + (9 + 6)
Combine like terms.
= x2 - 9y2 + 15
Problem 4 :
Subtract (x3 + 4x2 - x + 6) from (2x3 + 5x2 - 2x + 7).
Solution :
= (2x3 + 5x2 - 2x + 7) - (x3 + 4x2 - x + 6)
Distributive Property.
= 2x3 + 5x2 - 2x + 7 - x3 - 4x2 + x - 6
Group like terms together.
= (2x3 - x3) + (5x2 - 4x2) + (-2x + x) + (7 - 6)
= x3 + x2 - x + 1
Problem 5 :
Subtract (2x3 + 7x2 - 3x - 3) from (3x3 - 2x2 - x + 4).
Solution :
= (3x3 - 2x2 - x + 4) - (2x3 + 7x2 - 3x - 3)
Distributive Property.
= 3x3 - 2x2 - x + 4 - 2x3 - 7x2 + 3x + 3
Group like terms together.
= (3x3 - 2x3) + (-2x2 - 7x2) + (-x + 3x) + (4 + 3)
= x3 - 9x2 + 2x + 7
Problem 6 :
Subtract (3 + 3x - 7x5 - 5x6) from 2(x3 - x2 + 6x - 2).
Solution :
= 2(x3 - x2 + 6x - 2) - (3 + 3x - 7x5 - 5x6)
Distributive Property.
= 2x3 - 2x2 + 12x - 4 - 3 - 3x + 7x5 + 5x6
Group like terms together.
= 5x6 + 7x5 + 2x3 - 2x2 + (12x - 3x) + (-4 - 3)
Combine like terms.
= 5x6 + 7x5 + 2x3 - 2x2 + 9x - 7
Problem 7 :
Subtract (2x3 + 5x2 - 2x - 11) from (3x3 - 2x2 - 5x - 6).
Solution :
= (3x3 - 2x2 - 5x - 6 ) - (2x3 + 5x2 - 2x - 11)
Distributive Property.
= 3x3 - 2x2 - 5x - 6 -2x3 - 5x2 + 2x + 11
Group like terms together.
= (3x3 - 2x3) + (-2x2 - 5x2) + (-5x + 2x) + (-6 + 11)
Combine like terms.
= x3 - 7x2 - 3x + 5
Problem 8 :
Subtract (x3 + 4x2 - 12x - 5) from (5x3 + 3x2 + 2x - 10).
Solution :
= (5x3 + 3x2 + 2x - 10) - (x3 + 4x2 - 12x - 5)
Distributive Property.
= 5x3 + 3x2 + 2x - 10 - x3 - 4x2 + 12x + 5
Group like terms together.
= (5x3 - x3) + (3x2 - 4x2) + (2x + 12x) + (-10 + 5)
Combine like terms.
= 4x3 - x2 + 14x - 5
Problem 9 :
Subtract (12x3 + 14x2 + 17x - 12) from (15x+ 22x2 + 17x - 19).
Solution :
= (15x3 + 22x2 + 17x - 19) - (12x3 + 14x2 + 17x - 12)
Distributive Property.
= 15x3 + 22x2 + 17x - 19 - 12x3 - 14x2 - 17x + 12
Group like terms together.
= (15x3 - 12x3+ (22x2 - 14x2) + (17x - 17x) + (-19 + 12)
Combine like terms.
= 3x3 + 8x2 + 0x - 7
= 3x3 + 8x2 - 7
Problem 10 :
The profits of two different production plants can be modeled as shown below, where x' is the number of units produced at each plant.
Eastern Plant :
-0.03x2 + 26x - 1600
Western Plant :
-0.02x2 + 20x - 1800
Write a polynomial that represents the difference of the profits at the eastern plant and the profits at the western plant.
Solution :
= Eastern plant profits - Western plant profits
= (-0.03x2 + 26x - 1600) - (-0.02x2 + 20x - 1800)
Distributive Property.
-0.03x2 + 26x - 1600 + 0.02x2 - 20x + 1800
Group like terms together.
(-0.03x2 + 0.02x2) + (26x - 20x) + (-1600 + 1800)
Combine like terms.
= -0.01x2 + 6x + 200
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https://www.ask.com/web?qsrc=6&o=102140&oo=102140&l=dir&gc=1&q=Torsion+Force
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Web Results
en.wikipedia.org/wiki/Torsion_(mechanics)
In the field of solid mechanics, torsion is the twisting of an object due to an applied torque. Torsion is expressed in newton per squared meter (Pa) or pound per squared inch (psi) while torque is expressed in newton metres (N·m) or foot- pound force (ft·lbf).
www.slideshare.net/juhiahmed3/torsion-force
Dec 3, 2013 ... TORSION FORCE Presented by TASFIA AHMED 10.01.03.069.
www.codecogs.com/library/engineering/materials/torsion.php
Jun 15, 2013 ... The relationship between torsional forces, shear strain and polar moments of inertia. - References for Torsion with worked examples.
www.bu.edu/moss/mechanics-of-materials-torsion
One of the most common examples of torsion in engineering design is the ... was a twisting couple, which means that it has units of force times distance, or [N m].
web.mit.edu/course/3/3.11/www/modules/torsion.pdf
Jun 23, 2000 ... Analogously to our definition of normal stress as force per unit area. 1 .... Shafts in torsion are used in almost all rotating machinery, as in our ...
www.engineeringtoolbox.com/torsion-shafts-d_947.html
The torsion of solid or hollow shafts - Polar Moment of Inertia of Area. ... momentum, energy of objects and more; Statics - Loads - force and torque, beams and ...
www.quora.com/What-is-the-meaning-of-torsion-force
Torsion means twisting. If you are applying a force to a manual screwdriver , you are applying torsion. Similarly, a rod being turned by a mechanism around i...
www.teachengineering.org/lessons/view/wpi_lesson_1
Apr 26, 2017 ... They learn about the different kinds of stress each force exerts on objects. ... Explain how moments create bending and torsion loads on ...
www.dictionaryofconstruction.com/definition/torsion-force.html
Definition of torsion force: A force acting on a body that tends to twist the body. ( Twisting action).
Related Search
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https://cloudacademy.com/course/arrays-3180/sorting-arrays/
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Sorting Arrays
1
2
Arrays Part 1
PREVIEW18m 56s
3
## The course is part of this learning path
Start course
Difficulty
Beginner
Duration
1h 15m
Students
65
Ratings
5/5
Description
In this course, we'll learn about Arrays in Java.
### Learning Objectives
• What is an Array?
• Declaration and Initialization
• Sorting Arrays
• Searching Arrays
• Multi-Dimensional Arrays
### Intended Audience
• Anyone looking to get Oracle Java Certification
• Those who want to learn the Java Programming language from scratch
• Java developers who want to increase their knowledge
• Beginners with no previous coding experience in Java programming
• Those who want to learn tips and tricks in Oracle Certified Associate – Java SE 8 Programmer certification exams
### Prerequisites
• No prior knowledge is required about the Java programming language.
• Basic computer knowledge
Transcript
Hi there. In this lesson, we'll learn to sort the elements of an array from smallest to largest, i.e. ascending. For example, we can arrange the elements of an array of int data types from smallest to largest. This is called sorting arrays in the literature. You can do these sorting operations with the methods you write yourself, or you can use the created sort method of the Arrays class. We'll use the sort method in this lesson. If you're ready, let's move on to the Eclipse and get some practice. In the exercise project, I right click on the 'array' package and select the new class. This class name can be SortingArray and check the checkbox for the main method and click the 'Finish' button.
Okay, first let's look at the sorting of the int array. I'll create an int array. int[] numbers = {}. So, the elements of this array can be 5, 3, 10, 30, 1, 50. Now, let's sort the elements of this array using the sort method of the Arrays class. Here I write Arrays.sort. And in parentheses, I write the name of the array i.e. numbers. And let's print the numbers array on the console. System.out.printIn(Arrays.toString(numbers)). And let's run the code and see. As you can see, the elements are sorted in ascending order. 1, 3, 5, 10, 30, 50. Okay, let's create a double array and sort it. The name of this array can be myDoubles. And its elements can be 1.2, 3.5, -2.5, 0.7, 10.0, 11.
Let's sort it. Arrays.sort(myDoubles). And let's print the elements of this array on the console. System.out.printIn(Arrays.toString(myDoubles)). Let's run the code and test it. As you can see, the elements are sorted in ascending order again. -2.5, 0.7, 1.2, 3.5, 10.0, 11.0. Now, let's sort the char array. The name of the array can be myChars, and its elements can be a, B, A, F, b, E. Let's sort it. Arrays.sort(myChars). And let's print the elements of this array on the console. System.out.printIn(Arrays.toString(myChars)). Let's run the code and test it. Notice that it sorts the uppercase letters first, then the lower case letters because capital letters start first according to the ASCII table. For example, the value of uppercase A is 65, while the value of lowercase a is 97.
Okay, let's sort the string array. The name of the array can be myCars and its elements can be Ferrari, Opel, Mercedes, BMW, Ford, hyundai. The first letter of the Hyundai is lowercase. Let's sort it. Arrays.sort(myCars). And let's print the elements of this array on the console. System.out.printIn(arrays.toString(myCars)). Let's run the code and test it. This time they sorted it in alphabetical order. Alphabetical order is taken into account when sorting strings. But as you can see, the Hyundai is the last element because it begins with a lowercase letter, and always the lowercase is greater than the uppercase in sorting. Okay, let's do another example of sorting strings. The name of the array can be myNums and its elements can be 50, 70, 5, 90, 9, 60. Let's sort it. Arrays.sort(myNums). And let's print the elements of this array on the console. System.out.printIn(arrays.toString(myNums)). Let's try to guess the result before running the code.
This time our strings consist of numbers, not letters. So, the logic here is it looks at the first digit starting from the left first. So, five for the first element, seven for the second element, five for the third element, nine for the fourth element, nine again for the fifth element, and six for the last element. Sorts them first. In this case the first two elements will be 5 or 50. The third element will be 60. The fourth element will be 70 and the fifth and sixth elements will be 9 or 90. If we look at 5 and 50, since they start with the same number from the left, they are both five. In this case, it looks at the second number from the left, but since five consists of a single digit, it has no equivalent in the second digit. So, five is smaller. So, the first two elements will be 5 and 50 respectively.
Similarly, when we compare the last two elements, 9 becomes less than 90. So, the fifth element would be 9 and the last element would be 90. Now, let's run the application and test it. As you can see, first 5, then 50, then 60, then 70, then 9, and finally, 90 were printed. Also, let's add two more elements to this array. The seventh element will be lowercase a, and the eighth element will be capital Z. So, if the string arrays contain letters and numbers, this time, the letters will sort after the numbers. And the rule is always capital letters less than the lowercase letter in sorting.
So, the seventh element must be capital Z. And the last element must be lowercase a. Let's run and see. As you can see, the seventh element is uppercase Z and the eighth element is lowercase a. The thing to keep in mind is this: if the array consists of numeric values, it's sorted from largest to smallest. If it consists of characters, first uppercase letters, then lowercase letters are sorted. If the type of array is the string, it's sorted alphabetically. However, if the elements in a string represent numeric numbers, this time it compares the numbers among themselves on the condition that it starts from the left and sorts the elements of the array accordingly. Yes, the sorting elements are like this. Let's take a short break here. See you in the next lesson.
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https://www.inchcalculator.com/convert/centimeter-per-second-to-knot/
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# Centimeters per Second to Knots Converter
Enter the speed in centimeters per second below to get the value converted to knots.
## Result in Knots:
1 cm/s = 0.019438 kn
Do you want to convert knots to centimeters per second?
## How to Convert Centimeters per Second to Knots
To convert a measurement in centimeters per second to a measurement in knots, multiply the speed by the following conversion ratio: 0.019438 knots/centimeter per second.
Since one centimeter per second is equal to 0.019438 knots, you can use this simple formula to convert:
knots = centimeters per second × 0.019438
The speed in knots is equal to the speed in centimeters per second multiplied by 0.019438.
For example, here's how to convert 5 centimeters per second to knots using the formula above.
knots = (5 cm/s × 0.019438) = 0.097192 kn
Centimeters per second and knots are both units used to measure speed. Calculate speed in centimeters per second or knots using our speed calculator or keep reading to learn more about each unit of measure.
## What Are Centimeters per Second?
Centimeters per second are a measurement of speed expressing the distance traveled in centimeters in one second.
The centimeter per second, or centimetre per second, is an SI unit of speed in the metric system. Centimeters per second can be abbreviated as cm/s, and are also sometimes abbreviated as cm/sec. For example, 1 centimeter per second can be written as 1 cm/s or 1 cm/sec.
In the expressions of units, the slash, or solidus (/), is used to express a change in one or more units relative to a change in one or more other units.[1] For example, cm/s is expressing a change in length or distance relative to a change in time.
Centimeters per second can be expressed using the formula:
vm/s = dcm / tsec
The velocity in centimeters per second is equal to the distance in centimeters divided by time in seconds.
## What Is a Knot?
One knot is equal to a speed of one nautical mile per hour,[2] or one minute of latitude per hour.
Knots can be abbreviated as kn, and are also sometimes abbreviated as kt. For example, 1 knot can be written as 1 kn or 1 kt.
## Centimeter per Second to Knot Conversion Table
Table showing various centimeter per second measurements converted to knots.
Centimeters Per Second Knots
1 cm/s 0.019438 kn
2 cm/s 0.038877 kn
3 cm/s 0.058315 kn
4 cm/s 0.077754 kn
5 cm/s 0.097192 kn
6 cm/s 0.116631 kn
7 cm/s 0.136069 kn
8 cm/s 0.155508 kn
9 cm/s 0.174946 kn
10 cm/s 0.194384 kn
11 cm/s 0.213823 kn
12 cm/s 0.233261 kn
13 cm/s 0.2527 kn
14 cm/s 0.272138 kn
15 cm/s 0.291577 kn
16 cm/s 0.311015 kn
17 cm/s 0.330454 kn
18 cm/s 0.349892 kn
19 cm/s 0.36933 kn
20 cm/s 0.388769 kn
21 cm/s 0.408207 kn
22 cm/s 0.427646 kn
23 cm/s 0.447084 kn
24 cm/s 0.466523 kn
25 cm/s 0.485961 kn
26 cm/s 0.5054 kn
27 cm/s 0.524838 kn
28 cm/s 0.544276 kn
29 cm/s 0.563715 kn
30 cm/s 0.583153 kn
31 cm/s 0.602592 kn
32 cm/s 0.62203 kn
33 cm/s 0.641469 kn
34 cm/s 0.660907 kn
35 cm/s 0.680346 kn
36 cm/s 0.699784 kn
37 cm/s 0.719222 kn
38 cm/s 0.738661 kn
39 cm/s 0.758099 kn
40 cm/s 0.777538 kn
## References
1. National Institute of Standards and Technology, NIST Guide to the SI, Chapter 6: Rules and Style Conventions for Printing and Using Units, https://www.nist.gov/pml/special-publication-811/nist-guide-si-chapter-6-rules-and-style-conventions-printing-and-using
2. NASA, Knots Versus Miles per Hour, https://www.grc.nasa.gov/WWW/K-12/WindTunnel/Activities/knots_vs_mph.html
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http://stackoverflow.com/questions/2193093/finding-the-sum-of-values-of-a-column-of-all-records-in-sql-2005
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# Finding the SUM of values of a column of all records in sql 2005
I want to find the SUM of values in a column weight. I want this sum for all records that are identified with a common value in one of the columns name. Further, I want to consider only those records that have a certain value in the column type.
name weight type
1 `\$` 12.00 A
2 `\$` 7.00 B
2 `\$` 7.00 A
1 `\$` 1.00 C
2 `\$` 7.00 B
1 `\$` 1.00 C
2 `\$` 7.00 B
1 `\$` 7.00 B
2 `\$` 7.00 C
2 `\$` 7.00 B
I want the total weight for name 2, for the types A and B. Can a subquery be written for this or only looping can be done. Tnx.
-
this really depends on ones interpretation of the OPs question, which is a little vague to me. But give this a try:
``````SELECT
SUM(Weight)
FROM MyTable
WHERE Name=2 AND Type IN ('A','B','C')
``````
-
``````SELECT
Name,
Type,
SUM(Weight)
FROM
MyTable
GROUP BY
Name,
Type
HAVING
Name = @name
AND
Type = @type
``````
Or to be specific for your request
``````SELECT
Name,
Type,
SUM(Weight)
FROM
MyTable
GROUP BY
Name,
Type
HAVING
Name = '2'
AND
Type IN('A', 'B')
``````
However, if it is purely a single value you are after for your filter then you can just SUM with a WHERE clause.
``````SELECT
SUM(Weight)
FROM
MyTable
WHERE
Name = '2'
AND
Type IN('A', 'B')
``````
-
thanks. It seems a lot simpler now... – user96403 Feb 3 '10 at 15:50
This will show you the sum of weights for each name, where the type is A or B:
``````select name, sum(weight) as WeightSum
from MyTable t
where type in ('A', 'B')
group by name
``````
-
wow. thanks for the quick post. – user96403 Feb 3 '10 at 15:51
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# SMChap010 - Chapter 10 Project Analysis CHAPTER 10 Project...
• Test Prep
• 14
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Chapter 10 - Project Analysis CHAPTER 10 Project Analysis Answers to Problem Sets 1. a. False. The capital budget is not the final sign-off for specific projects. Most companies require for each project appropriation requests, which include more detailed analysis. b. True. Strategic planning requires consideration of alternatives. c. True. Cash flow forecasts are regularly overstated. Est. Time: 01 - 05 2. a. Cash flow forecasts are overstated. b. One project proposal may be ranked below another simply because cash flows are based on different forecasts. c. Project proposals may not consider strategic alternatives. Est. Time: 01 - 05 3. a. Analysis of how project profitability and NPV change if different assumptions are made about sales, cost, and other key variables b. Project NPV is recalculated by changing several inputs to new, but consistent, values. c. Determines the level of future sales at which project profitability or NPV equals zero. d. An extension of sensitivity analysis that explores all possible outcomes and weights each by its probability e. A graphical technique for displaying possible future events and decisions taken in response to those events f. Option to modify a project at a future date g. The additional present value created by the option to bail out of a project and recover part of the initial investment if the project performs poorly. 10-1
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Chapter 10 - Project Analysis h. The additional present value created by the option to invest more and expand output if a project performs well Est. Time: 06 - 10 4. a. False. Sensitivity analysis may show where additional information would be most useful. b. True c. True d. True e. False. A business with high fixed costs has high operating leverage. f. True Est. Time: 01 - 05 5. a. Describe how project cash flow depends on the underlying variables. b. Specify probability distributions for forecast errors for these cash flows. c. Draw from the probability distributions to simulate the cash flows. Est. Time: 01 - 05 6. a. True b. True c. False. Just as the option to expand has value, the option to terminate also raises the present value of the project. d. False. The optimal date to undertake an investment is the one that maximizes its contribution to the firm today. Est. Time: 01 - 05 7. Adding a fudge factor to the discount rate pushes project analysts to submit more optimistic forecasts. Est. Time: 01 - 05 10-2
Chapter 10 - Project Analysis 8. We assume that the idea for a new obfuscator machine originates with a plant manager in the Deconstruction Division. (Keep in mind, however, that in addition to bottom-up proposals, such as the obfuscator machine proposal, top-down proposals also originate with divisional managers and senior management.) Other steps in the capital budgeting process include the following: Many large firms begin the process with forecasts of economic variables, such as inflation and GDP growth, as well as variables of particular interest to the industry, such as prices of raw materials and industry sales projections.
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Role and Use of Direct Solvers in Ill-Conditioned Problems
# Sparse, Direct Solvers
## At A Glance
Questions Objectives Key Points 1. Why use a direct solver? Understand accuracy Direct solvers are robustfor difficult problems 2. What effects direct solve performance ? Understand ordering options Time & space performancecan vary a lot.
## To begin this lesson
• Get into the correct directory
cd HandsOnLessons/superlu_mfem
## The problem being solved
The convdiff.c application is modeling the steady state convection-diffusion equation in 2D with a constant velocity. This equation is used to model the concentration of something like a die in a moving fluid as it diffuses and flows through he fluid. The equation is as follows: $\nabla \cdot (\kappa \nabla u) - \nabla \cdot (\overrightarrow{v}u)+R=0$
Where u is the concentration that we are tracking, $$\kappa$$, is the diffusion rate, v is the velocity of the flow and R is a concentration source.
In the application we use here, the velocity vector direction is fixed in the +x direction. However, the magnitude is set by the user (default of 100), $$\kappa$$, is fixed at 1.0, and the source is 0.0 everywhere except for a small square centered at the middle of the domain where it is 1.0.
Initial Condition
Solving this PDE is well known to cause convergence problems for iterative solvers, for larger v. We use MFEM as a vehicle to demonstrate the use of a distributed, direct solver, SuperLU_DIST, to solve very ill-conditioned linear systems.
## Running the Example
$./convdiff | tail -n 3 Time required for first solve: 0.0408995 (s) Final L2 norm of residual: 2.43686e-16 Steady State ### Run 2: increase velocity to 1000, GMRES does not converge anymore $ ./convdiff --velocity 1000 | tail -n 3
Time required for first solve: 0.47337 (s)
Final L2 norm of residual: 0.00095
Between 12 and 13
Below, we plot behavior of the GMRES method for velocity values in the range [100,1000] at increments, dv, of 25 and also show an animation of the solution GMRES gives as velocity increases
Solutions @dv=25 in [100,1000]Contours of Solution @ vel=1000
Time to SolutionL2 norm of final residual
GMRES method works ok for low velocity values. As velocity increases, GMRES method eventually crosses a threshold where it can no longer provide a useful result
As instability is approached, more GMRES iterations are required to reach desired norm. So GMRES is still able to manage the solve and achieve a near-zero L2 norm. It just takes more and more iterations. Once GMRES is unable to solve the L2 norm explodes.
$./convdiff --velocity 1000 -slu -cp 0 Options used: --refine 0 --order 1 --velocity 1000 --no-visit --superlu --slu-colperm 0 --slu-rowperm 1 --slu-parsymbfact 0 --one-matrix --one-rhs Number of unknowns: 10201 Nonzeros in L 1040781 Nonzeros in U 1045632 nonzeros in L+U 2076212 nonzeros in LSUB 1040215 ** Memory Usage ********************************** ** NUMfact space (MB): (sum-of-all-processes) L\U : 41.12 | Total : 50.74 ** Total highmark (MB): Sum-of-all : 50.74 | Avg : 50.74 | Max : 50.74 ************************************************** Time required for first solve: 19.0018 (s) Final L2 norm of residual: 1.62703e-18 ************************************************** **** Time (seconds) **** EQUIL time 0.00 ROWPERM time 0.01 SYMBFACT time 0.04 DISTRIBUTE time 0.11 FACTOR time 18.52 Factor flops 1.958603e+08 Mflops 10.58 SOLVE time 0.10 Solve flops 5.167045e+06 Mflops 52.21 REFINEMENT time 0.20 Steps 2 ************************************************** Stead State For vel=1000 ### Run 4: Now use SuperLU_DIST, with MMD(A’+A) ordering. $ ./convdiff --velocity 1000 -slu -cp 2
Options used:
--refine 0
--order 1
--velocity 1000
--no-visit
--superlu
--slu-colperm 2
--slu-rowperm 1
--slu-parsymbfact 0
--one-matrix
--one-rhs
Number of unknowns: 10201
Nonzeros in L 594238
Nonzeros in U 580425
nonzeros in L+U 1164462
nonzeros in LSUB 203857
** Memory Usage **********************************
** NUMfact space (MB): (sum-of-all-processes)
L\U : 10.07 | Total : 15.52
** Total highmark (MB):
Sum-of-all : 15.52 | Avg : 15.52 | Max : 15.52
**************************************************
Time required for first solve: 0.111105 (s)
Final L2 norm of residual: 1.53726e-18
**************************************************
**** Time (seconds) ****
EQUIL time 0.00
ROWPERM time 0.01
COLPERM time 0.04
SYMBFACT time 0.01
DISTRIBUTE time 0.02
FACTOR time 0.05
Factor flops 1.063303e+08 Mflops 2045.75
SOLVE time 0.00
Solve flops 2.367059e+06 Mflops 779.35
REFINEMENT time 0.01 Steps 2
**************************************************
NOTE: the number of nonzeros in L+U is much smaller than natural ordering. This affects the memory usage and runtime.
$./convdiff --velocity 1000 -slu -cp 4 Options used: --refine 0 --order 1 --velocity 1000 --no-visit --superlu --slu-colperm 4 --slu-rowperm 1 --slu-parsymbfact 0 --one-matrix --one-rhs Number of unknowns: 10201 Nonzeros in L 522306 Nonzeros in U 527748 nonzeros in L+U 1039853 nonzeros in LSUB 218211 ** Memory Usage ********************************** ** NUMfact space (MB): (sum-of-all-processes) L\U : 9.24 | Total : 14.96 ** Total highmark (MB): Sum-of-all : 14.96 | Avg : 14.96 | Max : 14.96 ************************************************** Time required for first solve: 0.152424 (s) Final L2 norm of residual: 1.51089e-18 ************************************************** **** Time (seconds) **** EQUIL time 0.00 ROWPERM time 0.01 COLPERM time 0.05 SYMBFACT time 0.01 DISTRIBUTE time 0.02 FACTOR time 0.05 Factor flops 7.827314e+07 Mflops 1717.18 SOLVE time 0.00 Solve flops 2.120276e+06 Mflops 606.75 REFINEMENT time 0.01 Steps 2 ************************************************** Solutions @dv=25 in [100,1000]Steady State Solution @ vel=1000 ### Run 5.5: Now use SuperLU_DIST, with Metis(A’+A) ordering, using 1 MPI tasks, on a larger problem. By adding --refine 3, each element in the mesh is subdivided twice yielding a 64x larger problem. But, we’ll run it on only one processor. $ mpiexec -n 1 ./convdiff --refine 3 --velocity 1000 -slu -cp 4
Options used:
--refine 3
--order 1
--velocity 1000
--no-visit
--superlu
--slu-colperm 4
--slu-rowperm 1
--slu-parsymbfact 0
--one-matrix
--one-rhs
Number of unknowns: 641601
Nonzeros in L 40412796
Nonzeros in U 40412796
nonzeros in L+U 80183991
nonzeros in LSUB 15748820
** Memory Usage **********************************
** NUMfact space (MB): (sum-of-all-processes)
L\U : 701.82 | Total : 758.92
** Total highmark (MB):
Sum-of-all : 786.78 | Avg : 786.78 | Max : 786.78
**************************************************
Time required for first solve: 18.8951 (s)
Final L2 norm of residual: 5.99013e-18
**************************************************
**** Time (seconds) ****
EQUIL time 0.03
ROWPERM time 0.29
COLPERM time 4.83
SYMBFACT time 0.32
DISTRIBUTE time 1.58
FACTOR time 9.87
Factor flops 2.326266e+10 Mflops 2357.24
SOLVE time 0.41
Solve flops 1.604473e+08 Mflops 395.95
REFINEMENT time 0.90 Steps 2
**************************************************
### Run 6: Now use SuperLU_DIST, with Metis(A’+A) ordering, using 16 MPI tasks, on a larger problem.
Here, we’ll re-run the above except on 16 tasks and just grep the output form some key values of interest.
{MPIEXEC_OMPI} -n 16 ./convdiff --refine 3 --velocity 1000 -slu --slu-colperm 4 >& run6.out
Options used:
--refine 3
--order 1
--velocity 1000
--no-visit
--superlu
--slu-colperm 4
--slu-rowperm 1
--slu-parsymbfact 0
--one-matrix
--one-rhs
Number of unknowns: 641601
Nonzeros in L 40340620
Nonzeros in U 40340620
nonzeros in L+U 80039639
nonzeros in LSUB 15901421
** Memory Usage **********************************
** NUMfact space (MB): (sum-of-all-processes)
L\U : 705.31 | Total : 974.93
** Total highmark (MB):
Sum-of-all : 2888.58 | Avg : 180.54 | Max : 180.54
**************************************************
Time required for first solve: 9.10544 (s)
Final L2 norm of residual: 2.29801e-39
**************************************************
**** Time (seconds) ****
EQUIL time 0.03
ROWPERM time 0.36
COLPERM time 5.57
SYMBFACT time 0.37
DISTRIBUTE time 0.30
FACTOR time 1.62
Factor flops 2.301228e+10 Mflops 14226.55
SOLVE time 0.14
Solve flops 1.623936e+08 Mflops 1148.60
REFINEMENT time 0.30 Steps 2
We have increased the mesh size by 8x here. The matrix dimension goes up as the SQUARE of the mesh size and this accounts for 64x factor of DOFs. We have also added 16x processors. The parallel runtime is 9.10544 seconds.
### Run 7: Now use SuperLU_DIST, solve the systems with same A, but different right-hand side b.
Here, we solve a different linear system but with the same coefficient matrix A. We tell SuperLU to re-use the exisiting LU factors, but only give a different right-hand side. Notice the improvement in solve time when re-using the factors.
\$ mpiexec -n 16 ./convdiff --refine 3 --velocity 1000 -slu -cp 4 -2rhs
Options used:
--refine 3
--order 1
--velocity 1000
--no-visit
--superlu
--slu-colperm 4
--slu-rowperm 1
--slu-parsymbfact 0
--one-matrix
--two-rhs
Number of unknowns: 641601
Nonzeros in L 40340620
Nonzeros in U 40340620
nonzeros in L+U 80039639
nonzeros in LSUB 15901421
** Memory Usage **********************************
** NUMfact space (MB): (sum-of-all-processes)
L\U : 705.31 | Total : 974.93
** Total highmark (MB):
Sum-of-all : 2888.58 | Avg : 180.54 | Max : 180.54
**************************************************
Time required for first solve: 9.11672 (s)
Final L2 norm of residual: 2.14235e-39
**************************************************
**** Time (seconds) ****
EQUIL time 0.04
ROWPERM time 0.36
COLPERM time 5.64
SYMBFACT time 0.38
DISTRIBUTE time 0.23
FACTOR time 1.61
Factor flops 2.301228e+10 Mflops 14307.11
SOLVE time 0.14
Solve flops 1.623936e+08 Mflops 1147.81
REFINEMENT time 0.30 Steps 2
**************************************************
Time required for second solve (new rhs): 0.46439 (s)
Final L2 norm of residual: 1.95236e-39
SOLVE time 0.14
Solve flops 1.623936e+08 Mflops 1202.77
REFINEMENT time 0.29 Steps 2
**************************************************
## Out-Brief
In this lesson, we have used MFEM as a vehicle to demonstrate the value of direct solvers from the SuperLU_DIST numerical package.
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# Fuel usage to accelerate to 20 mph
by zanes
Tags: fuel usage
P: 3 In an effort to approach the EPA on the subject of eliminating unnecessary stop signs in an effort to reduce fuel consumption nationally, I am trying to determine approximately how much fuel is required to accelerate from a dead stop to 20 mph. With all the variables (vehicle weight, acceleration rate, etc.) , I would simply like to take a small car (3000 lbs) that would normally be in the 30mpg range for city driving at a "normal" acceleration rate. With such a number in hand, calculating the minimum fuel savings given the traffic counts at any given stop sign could be easily computed. I strongly suspect that there are millions of unnecessary stop signs in the U.S., and that eliminating them would save many millions of gallons of wasted fuel. Any help would be greatly appreciated. Zanes
Mentor P: 22,301 Welcome to PF. Do you know Newton's equations of motion? The definition (mathematical) of work? You'll need to make an assumption about how fast the acceleration is, but after that, it is a pretty straightforward calculation. Why don't you give it a try and we'll see if/where you get hung up.
P: 3 I think this site is out of my depth. In math I didn't get much beyond the Pythagorean Theorem, and finding the hypotenuse doesn't seem to be germaine to the problem. Looks like Plan B, empirical data, is in my future. If I accelerate my car to 20mph, then decelerate to 0mph 100 times, then refill my tank, then drive the same route at 20mph, filling my tank a second time, I can get my answer (and wear out my brakes.) Thank you for responding, though. Sincerely, Zane Suedmeyer
Mentor P: 11,921 Fuel usage to accelerate to 20 mph I don't think this experiment will give reasonable data with just 100 stops. The kinetic energy of 1000kg, moving at 10m/s (a bit more than 20mph), is E=1/2mv^2 = 50kJ. Cars are quite inefficient, so we can assume at least 100kJ required energy in the fuel. Petrol has an energy density of about 35MJ/l, therefore 100kJ correspond to 1/350l or about 3ml. With 1.4€ to 1.7€ per liter (roughly the price in Germany), this costs 0.4 to 0.5 (euro)cents. All calculations are easy without a calculator. SI units are so nice ;).
Mentor
P: 22,301
Quote by zanes I think this site is out of my depth. In math I didn't get much beyond the Pythagorean Theorem...
Didn't Algebra I come before geometry?
This site can help anyone who is willing to learn, regardless of their current knowledge level.
P: 189 You might want to consider converting mfb's excellent answer in to CO2 emissions. You would have a politically more attractive argument if you use CO2 rather than cents.
P: 31 Only the very largest of ships diesel engines achieve a thermal efficiency of 50% the conversion factor for a car engine is more like 25% at reduced throttle openings.
P: 276 I venture that 70% of the driving population drive in such a way as to negate the advantages of new fuel mileage technologies. They will not change driving habits to get the most out of the technology.
Related Discussions General Engineering 9 Mechanical Engineering 1 Introductory Physics Homework 13 Classical Physics 2 Astronomy & Astrophysics 26
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# Best answer: How much taxes do I have to pay on \$40 000?
Contents
## How much taxes do I pay on 40000?
If you make \$40,000 a year living in the region of Alberta, Canada, you will be taxed \$7,769. That means that your net pay will be \$32,231 per year, or \$2,686 per month. Your average tax rate is 19.4% and your marginal tax rate is 30.3%.
## What is the average tax refund for a single person making \$40 000?
What is the average tax refund for a single person making \$40,000? We estimated a single person making \$40,000 per year would receive an average refund of \$1,761 this year.
## How much will I get back in taxes if I make 45000?
If you make \$45,000 a year living in the region of California, USA, you will be taxed \$9,044. That means that your net pay will be \$35,956 per year, or \$2,996 per month. Your average tax rate is 20.1% and your marginal tax rate is 27.5%.
## What is the tax bracket for 70k?
Let’s say you’re filing single with an income of \$70,000. That would put you in the 22% tax bracket. Again, you would not pay 22% on your entire taxable income. Instead, you would pay 10% on your income up to \$9,875, 12% on income between \$9,876 – \$40,125 and 22% on the remaining income of \$29,875.
## How much will I get back in taxes if I make 35000?
If you make \$35,000 a year living in the region of California, USA, you will be taxed \$6,366. That means that your net pay will be \$28,634 per year, or \$2,386 per month. Your average tax rate is 18.2% and your marginal tax rate is 26.1%.
## How much will I get back in taxes if I make 100k?
If you make \$100,000 a year living in the region of California, USA, you will be taxed \$30,460. That means that your net pay will be \$69,540 per year, or \$5,795 per month. Your average tax rate is 30.5% and your marginal tax rate is 43.1%.
## How much will I get back in taxes if I make 25000?
If you make \$25,000 a year living in the region of California, USA, you will be taxed \$3,858. That means that your net pay will be \$21,142 per year, or \$1,762 per month. Your average tax rate is 15.4% and your marginal tax rate is 24.9%.
## How much will I get back in taxes if I made 24000?
If you are single, you should have had about \$2,800 in federal taxes taken out of your paycheck for your annual gross income of \$24K. After taking out the exemption and standard deduction, the taxable income would be about \$13,650 for a tax of about \$1,600.
THIS IS IMPORTANT: Are directors loans tax efficient?
## How much will I get back in taxes if I make 50000?
If you make \$50,000 a year living in the region of California, USA, you will be taxed \$10,417. That means that your net pay will be \$39,583 per year, or \$3,299 per month. Your average tax rate is 20.8% and your marginal tax rate is 33.1%.
## How much will I get back in taxes if I make 30000?
If you make \$30,000 a year living in the region of California, USA, you will be taxed \$5,103. That means that your net pay will be \$24,897 per year, or \$2,075 per month. Your average tax rate is 17.0% and your marginal tax rate is 25.3%.
## What is the tax bracket for 75000?
If you make \$75,000 a year living in the region of California, USA, you will be taxed \$20,168. That means that your net pay will be \$54,832 per year, or \$4,569 per month. Your average tax rate is 26.9% and your marginal tax rate is 41.1%.
## How do you calculate federal income tax?
Estimating a tax bill starts with estimating taxable income. In a nutshell, to estimate taxable income, we take gross income and subtract tax deductions. What’s left is taxable income. Then we apply the appropriate tax bracket (based on income and filing status) to calculate tax liability.
## How much federal tax do I pay on 48000?
Income tax calculator California
If you make \$48,000 a year living in the region of California, USA, you will be taxed \$9,868. That means that your net pay will be \$38,132 per year, or \$3,178 per month. Your average tax rate is 20.6% and your marginal tax rate is 27.5%.
THIS IS IMPORTANT: Are used products taxable?
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## Does this PDE has a general solution? [closed]
$$K\frac{\partial }{{\partial x}}(h\frac{{\partial h}}{{\partial x}}) = \mu \frac{{\partial h}}{{\partial t}}$$ K and u are constants. If no,how to get a asymptotic solution?ie,linearize
-
mathoverflow.net/questions/25431/… – Steve Huntsman Nov 18 at 4:59
## closed as too localized by Ricky Demer, Qiaochu Yuan, Andy Putman, Michael Renardy, Pietro MajerNov 19 at 17:38
We may assume WLOG that $K=1$. One family of solutions is $$h(x,t) = \frac{v(x)}{a+bt}$$ for arbitrary constants $a,b$, where $v(x)$ is a solution of the ordinary differential equation $$v v'' + (v')^2 + a \mu v = 0$$
Another family of solutions is $$h(x,t) = a \left(W\left(b e^{(x+ct) \mu c/a}\right)+1\right)$$ for arbitrary constants $a,b,c$, where $W$ is the Lambert W function.
"We may assume WLOG that $\: K=1 \:$" $\;\;$ except when $\: K=0 \:$. $\;\;\;\;$ – Ricky Demer Nov 18 at 5:44
OK, but the case $K=0$ is left as an exercise. – Robert Israel Nov 18 at 8:09
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# 11478930235249
## 11,478,930,235,249 is an odd composite number. It is composed of a single prime number multiplied by itself.
What does the number 11478930235249 look like?
This visualization shows the relationship between its 1 prime factors (large circles) and 3 divisors.
11478930235249 is an odd composite number. It has a total of three divisors.
## Prime factorization of 11478930235249:
### 33880572
(3388057 × 3388057)
See below for interesting mathematical facts about the number 11478930235249 from the Numbermatics database.
### Names of 11478930235249
• Cardinal: 11478930235249 can be written as Eleven trillion, four hundred seventy-eight billion, nine hundred thirty million, two hundred thirty-five thousand, two hundred forty-nine.
### Scientific notation
• Scientific notation: 1.1478930235249 × 1013
### Factors of 11478930235249
• Number of distinct prime factors ω(n): 1
• Total number of prime factors Ω(n): 2
• Sum of prime factors: 3388057
### Divisors of 11478930235249
• Number of divisors d(n): 3
• Complete list of divisors:
• Sum of all divisors σ(n): 11478933623307
• Sum of proper divisors (its aliquot sum) s(n): 3388058
• 11478930235249 is a deficient number, because the sum of its proper divisors (3388058) is less than itself. Its deficiency is 11478926847191
### Bases of 11478930235249
• Binary: 101001110000101001011000111100011111011100012
• Base-36: 42HCIONLD
### Squares and roots of 11478930235249
• 11478930235249 squared (114789302352492) is 131765839345713662482092001
• 11478930235249 cubed (114789302352493) is 1512530877238474871915353718288591143249
• 11478930235249 is a perfect square number. Its square root is 3388057
• The cube root of 11478930235249 is 22557.9937574169
### Scales and comparisons
How big is 11478930235249?
• 11,478,930,235,249 seconds is equal to 364,994 years, 24 weeks, 4 days, 20 hours, 49 seconds.
• To count from 1 to 11,478,930,235,249 would take you about nine hundred twelve thousand, four hundred eighty-six years!
This is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)
• A cube with a volume of 11478930235249 cubic inches would be around 1879.8 feet tall.
### Recreational maths with 11478930235249
• 11478930235249 backwards is 94253203987411
• The number of decimal digits it has is: 14
• The sum of 11478930235249's digits is 58
• More coming soon!
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The information we have on file for 11478930235249 includes mathematical data and numerical statistics calculated using standard algorithms and methods. We are adding more all the time. If there are any features you would like to see, please contact us. Information provided for educational use, intellectual curiosity and fun!
Keywords: Divisors of 11478930235249, math, Factors of 11478930235249, curriculum, school, college, exams, university, Prime factorization of 11478930235249, STEM, science, technology, engineering, physics, economics, calculator, eleven trillion, four hundred seventy-eight billion, nine hundred thirty million, two hundred thirty-five thousand, two hundred forty-nine.
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Cody
# Problem 19. Swap the first and last columns
Solution 355295
Submitted on 19 Nov 2013 by Emilio
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% A = [ 12 4 7 5 1 4]; B_correct = [ 7 4 12 4 1 5 ]; assert(isequal(swap_ends(A),B_correct))
2 Pass
%% A = [ 12 7 5 4]; B_correct = [ 7 12 4 5 ]; assert(isequal(swap_ends(A),B_correct))
3 Pass
%% A = [ 1 5 0 2 3 ]; B_correct = [ 3 5 0 2 1 ]; assert(isequal(swap_ends(A),B_correct))
4 Pass
%% A = 1; B_correct = 1; assert(isequal(swap_ends(A),B_correct))
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# How to Use regsubsets() in R for Model Selection
You can use the regsubsets() function from the leaps package in R to find the subset of predictor variables that produces the best regression model.
The following example shows how to use this function in practice.
## Example: Using regsubsets() for Model Selection in R
For this example we’ll use the built-in mtcars dataset in R, which contains measurements on 11 different attributes for 32 different cars.
```#view first six rows of mtcars dataset
mpg cyl disp hp drat wt qsec vs am gear carb
Mazda RX4 21.0 6 160 110 3.90 2.620 16.46 0 1 4 4
Mazda RX4 Wag 21.0 6 160 110 3.90 2.875 17.02 0 1 4 4
Datsun 710 22.8 4 108 93 3.85 2.320 18.61 1 1 4 1
Hornet 4 Drive 21.4 6 258 110 3.08 3.215 19.44 1 0 3 1
Hornet Sportabout 18.7 8 360 175 3.15 3.440 17.02 0 0 3 2
Valiant 18.1 6 225 105 2.76 3.460 20.22 1 0 3 1
```
Suppose we would like to fit a regression model using hp as the response variable and the following potential predictor variables:
• mpg
• wt
• drat
• qsec
We can use the regsubsets() function from the leaps package to perform an exhaustive search to find the best regression model:
```library(leaps)
#find best regression model
bestSubsets <- regsubsets(hp ~ mpg + wt + drat + qsec, data=mtcars)
#view results
summary(bestSubsets)
Subset selection object
Call: regsubsets.formula(hp ~ mpg + wt + drat + qsec, data = mtcars)
4 Variables (and intercept)
Forced in Forced out
mpg FALSE FALSE
wt FALSE FALSE
drat FALSE FALSE
qsec FALSE FALSE
1 subsets of each size up to 4
Selection Algorithm: exhaustive
mpg wt drat qsec
1 ( 1 ) "*" " " " " " "
2 ( 1 ) " " "*" " " "*"
3 ( 1 ) "*" "*" " " "*"
4 ( 1 ) "*" "*" "*" "*" ```
The stars ( * ) at the bottom of the output indicate which predictor variables belong in the best regression model for each possible model with a different number of predictor variables.
Here is how to interpret the output:
For a model with only one predictor variable, the best regression model is produced by using mpg as the predictor variable.
For a model with two predictor variables, the best regression model is produced by using wt and qsec as the predictor variables.
For a model with three predictor variables, the best regression model is produced by using mpg, wt and qsec as the predictor variables.
For a model with four predictor variables, the best regression model is produced by using mpg, wt, drat and qsec as the predictor variables.
Note that you can also extract the following metrics for each model:
For example, we can use the following syntax to extract the adjusted R-squared value for each of the four best models:
```#view adjusted R-squared value of each model
[1] 0.5891853 0.7828169 0.7858829 0.7787005
```
From the output we can see:
• The adjusted R-squared value for the model with mpg as the predictor variable is 0.589.
• The adjusted R-squared value for the model with wt and qsec as the predictor variables is 0.783.
• The adjusted R-squared value for the model with mpg, wt and qsec as the predictor variables is 0.786.
• The adjusted R-squared value for the model with mpg, wtdrat and qsec as the predictor variables is 0.779.
These values give us an idea of how well the set of predictor variables are able to predict the value of the response variable, adjusted for the number of predictor variables in the model.
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## Wednesday, December 15, 2010
Getting into Shape:
What Do Butterfly Wings and Renaissance Maps Have in Common?
---
A Note on Method
---
There are spaces in which the determination of position requires not a finite number, but either an endless series or a continuous manifold of determinations of quantity. Such manifolds are, for example, the possible determinations of a function for a given region, the possible shapes of a figure, and so on.
--Bernhard Riemann
The comparison of the accuracy and geometric outline of an early map with a modern map is really just the calculation of the change in shape of two planar surfaces. It is a coordinate transformation that can be as complex or as simple as one likes. My interest in the mathematics of shape change came about while I was working in the Entomology Department of the Smithsonian Institution’s National Museum of Natural History.
I was first specifically interested in trying to quantify the variation in the shape of butterfly wing patterns. At first, my interest was centered on the genus Erebia from the Alps of France. Erebia are high mountain butterflies that speciated a great deal during the last large scale glaciations that drove them out of the valleys and isolated them on peaks throughout the Alps. The amazing complexity of their spatial distribution is not well understood. Some of the variation between species can be quanitifed by looking at the shape variation in their eyespots and the surrounding color bands on both pairs of wings.
Later, I worked, with Marc Epstein at the Smithsonian, on a group of moths where I became interested in one species known as Euclea Delphini. Within this species we find a great deal of variation in the wing patterns. In the case of Delphini there are five different forms that are not totally discrete but tend to make-up a pseudo-continuum. In order to study this with some meaningful statistics it is necessary to describe the shape of the features that are varying mathematically. These features take a variety of forms from lines and polygons through more complex curved concave and convex shapes.
There are no natural measures associated with the two-dimensional variation of these features in the wings of butterflies so what to measure becomes a problem. Nijhout’s(1) work in this area was very helpful in trying to define what the basic elements of wing patterns were and in helping to narrow down what parts of the patterns are varying in a meaningful way. (http://www.biology.duke.edu/nijhout/images/PatternElements.pdf) I began experimenting with some simple morphometric models and was naturally led to the work done on shape spaces by David Kendall and Fred Bookstein. The studies of shape spaces that they pioneered are very interesting and are themselves objects of mathematical research. It is not always useful to think of shapes as collections of points in Euclidean space. Shape is something that has a spatial structure that is quite peculiar and generally these spaces and subspaces do not occur in other contexts. Huiling Le and Kendall(2) wrote a beautiful paper on this entitled, “The Riemannian Structure of Euclidean Shape Spaces: A Novel Environment for Statistics” that shows that these spaces have a local Riemannian structure. I immediately became fascinated with these spaces and read more deeply than the application to the problem of butterfly wings suggested. The structure of these spaces is complex and unfortunately I still do not think that I have a good grasp of all of Kendall’s work. The key point here is that there exists a well defined mathematical machinery that lets us apply non-linear transformations to these spaces while maintaining some meaningful definitions of physical distance.
Wing Patterns or Coastlines?
Modeling the shape change on a modern and ancient map is not all that different from the butterfly wing problem, but it does vary in a number of ways. For example, when working with two maps, we do not have a large sample of objects that we are comparing. We are not looking for clusters or variations that seem to group themselves together in statistically meaningful ways. We really are comparing two discrete objects with the hope of quantifying an ill-defined concept that we call accuracy. The question of accuracy in historical cartometric studies is, at least according to the J.B Harley (3) , one of the least understood problems in the history of cartography and it is especially difficult to quantify.
Fortunately morphometrics does have a great deal to tell us about how to begin to study this problem of comparative accuracy. For instance in both cases we are still operating on sets of homologies. Homologies are points, locations or landmarks that are the same on the objects of interest. They could be the location of a particular structure on a butterfly’s wing or the location of Rome on two maps from different historical periods. In historical cartometry we select a set of these points and then transform one set of points into the other using any number of possible algebraic or differential transformations. We are performing a type of image warping which maps all the selected positions in one image plane (the modern map) to the positions in a second plane (the ancient map). The choice of the mathematical function that actually performs the “warp” is always a compromise between insisting that the distortion is smooth and achieving a good match between the two sets of points. There are and have been many approaches to this.
One of the earliest applications of this type of coordinate transform or “warp” to maps came from Waldo Tobler (4) who experimented with them back in the 1970’s. Any discussion of the applications of complex transformations to cartography must begin with his seminal paper and computer program, “Bi-dimensional Regression”. Bi-dimensional Regression was a statistical regression technique that allowed inferences to be made between two planes from point distributions. Tobler studied medieval maps and Portolan charts but was also interested in general questions of shape and was heavily influenced by D’Arcy Thompson’s early studies on deformations found in his book On Growth and Form. Although Tobler’s work was graphically somewhat primitive, the numerical results from his regressions gave the first look at what types of information it might be possible to add to historical studies of early maps through the use analytical comparisons. Tobler’s methods were formalized to shape spaces by Small in several important papers such as “Techniques of Shape Analysis on Sets of Points. (5)
A weakness in Tobler’s methods was the question of error distribution (6) and what have become known as outliers. Outliers are points formed from pairs of homologies that when transformed produce an error that is significantly larger than the other points in the sample. These points if not identified and somehow accounted for in the calculations can destabilize many transformations and can effect correlations and regressions in adverse ways. Hampel and Huber (7) have developed an entire field of study around this problem that has become known as Robust Estimation. The family of estimators that they developed and that are known as M-estimators have been widely used and have even found their way into specialty software programs used in historical cartometry such as MAPANALYST. Intuitively, these estimators allow the control of how much influence on the regressions distant points have on nearby points. One can imagine this as an application of Tobler’s first law of geography, “everything is related to everything else, but near things are more related than distant things”.
My current research derives from the work of Bookstein (8) and his groundbreaking paper, “ Principal Warps: Thin-Plate Splines and the Decomposition of Deformations” (see link section to view this paper). Bookstein uses a particular distance function that defines a convex surface between homologies and that is a solution to the biharmonic equation.
Through some very interesting algebra Bookstein derives functions that effectively separate the global and local error into affine and nonlinear components. The functions are vector valued and if the pairing of the points that are being transformed correspond to the homologies on our two maps we have effectively modeled the difference between the two point sets as a deformation.
At the foundation of Bookstein's method is the function shown above which corresponds to a portion of the surface,
---
---
---
where r is the distance from the origin. The U(r) function satisfies,
---
and is the solution to the so-called biharmonic equation. We can imagine this surface as a piece of metal that is subject to deformations resulting from the displacements of fixed points on a reference surface. In this case we are comparing points on a modern map and the same corresponding points on an older or different map of the same region. For a thin plate of this type subject to bending, the energy change at any point minimizes,
.
---
If we pick a group of points on the reference map we can place them on the spline grid as shown below.
---
The corresponding points on the map to be studied can then be transformed yielding a deformation grid that minimizes the energy necessary for the deformation. This deformation has both global and local components that allow us to look at the induced error from different scales and is equivalent to the well known measure of cartographic error, Tissot’s indicatrix. The information from these deformations can be used to generate scale isolines as below or vector displacements resembling the indicatrix.
Thus far I have utilized this thin-plate spline method on many maps and in a study that compares the longitudinal error in the Mediterranean basin on the 1507 and 1516 World Maps by Martin Waldseemüller. The method does have significant advantages over methods such as Polynomial Warping that I used in my study, “Warping Waldseemüller: A Phenomenological and Computational Study of the 1507 World Map” (9).
I have since utilized these methods on Roman and some Medieval cartography, like Portolan charts. Using these methods one can calculate distortion grids and scale and rotation isolines.
For the historian of cartography comparing the accuracy on a modern and early map has a number of formal and logical difficulties that must be considered before the application of any cartometric process.
1. The identification of tie points on the two maps to be compared is sometimes difficult and once selected are seldom distributed evenly across the surface of the two images. The problem is essentially one of homology. Finding homologous points on two maps may seem trivial but differences in landmark or coastline shape, an incorrect assignment of place names and changing scales require insight into the target image. The selection of points is especially important when using the simplest linear models without M-estimators where results can be error distribution sensitive.
2. The accuracy and error to be tested and compared may not be a relevant concept across the whole surface of the map. Accuracy is seldom evenly distributed especially in the case of maps of small scale that may be composed from a variety of sources. Discontinuities in error from scale changes are not only scalar but are vector quantities whose direction is important to determine. Rapid changes in accuracy across map surfaces may be difficult to handle with simple linear models and are better approached using local non-linear radial basis functions whose correlations may be statistically more relevant, but mathematically more complex to program. Decisions regarding the use of local or global methods cannot usually be made ahead of time and require experimentation in order to characterize the accuracy and decide on a methodology.
3. The substrate that the map is draw or printed on may have undergone distortion through shrinking, folding, or stretching. Distortion of this sort is especially important to consider in the case of environmentally sensitive materials such as vellum. The distortion of the medium not only effects the accuracy of statistical transformations that we are trying to perform but can also obscure the intent of the cartographer.
4. Spatial association does not necessarily imply causality. The warps and correlations that we calculate using cartometric techniques give real mathematical results but these may be extremely difficult to link to any historical meaning, event or cause. Care must be taken not to over interpret the results of these calculations and not to override historical and documentary context in the rush towards accuracy measures.
5. Historical cartometry is a problem in equifinality or process convergence. Similar types of distortion may arise from different causes making it difficult to derive exact causes for particular distortion patterns.
6. Experimentation is necessary and we must be prepared to use a variety of techniques to characterize the accuracy on a single map.
Considering the qualifications in the application of cartometric methods to historical materials it is obvious that we cannot hope to achieve exact “truth” using our methods or an absolute visualization and characterization of accuracy(10). Instead we assign meaningful notions of probability and statistical measures that are useful for historical comparisons. The analysis of accuracy is an important part of the historical characterization of early maps but the assignment of the adjectives, “accurate” or “inaccurate”, needs to be made more precise in our discourse. The addition of statements like “to this level of confidence by this method” to our use of these adjectives would go a long way to making the results of our calculations repeatable by others engaged in the same research. This more analytic understanding of accuracy requires a much more careful linguistic and conceptual use of these ideas than has been the norm in the historical literature.
Currently there are more algorithms being developed that will be useful for cartographic historians as the fields of medical imaging and shape analysis continue to provide fertile ground for the growth these mathematical techniques. There are many more ways to represent shapes on manifolds and to perform the types of similarity and coordinate transforms than we have discussed here and no doubt more applications shall be forthcoming. The areas of Stochastic Geometry(11) and Poisson Processes in Euclidean space are especially interesting (12).
In the end the researcher who would employ these methods needs to critically consider each map to be studied on a case-by-case basis and would be wise to consider the words of R.A. Skelton, which although written in a different context, can be used as model for the cautions we must recognize in historical cartometry.
---
“The content of the map, as a whole, cannot be assigned confidently to a single phase or horizon of geographical knowledge. Its outlines are in part transcribed from a map prototype or prototypes not precisely identifiable with any extant work; in part they illustrate texts, not all of which have come down to us. The information taken by the author of the map from these sources (graphical and textual) relates to events and concepts of various periods; most of it older by a least a century, and some of it much more, than the presumed date at which the existing map… was made. The delineations in the map before us are separated by long intervals of time not only from the original experience that they reflect, but also from the direct records of it. For the mapmaker was working always at one remove, sometimes (we cannot doubt) at two or more removes, from firsthand records; and it is evident that, to a degree and in senses which it is difficult for us to divine, he exercised his judgment in selection from and in adaptation of his sources, which are themselves partly unknown to us. (13)
---
(1)H.F. Nijhout, “Elements of Butterfly Wing Patterns,” Journal of Experimental Zoology 29 (2001): 213-225.
---
(2)Huiling Le and David Kendall, “The Riemann Structure of Euclidean Shape Spaces: A Novel Environment for Statistics,” Annals of Statistics 21 (1993): 1225-1271
(3) Brian Harley, “Concepts in the History of Cartography: A Review and Perspective,” Cartographica 17 (1980): 54.
---
(4)Waldo Tobler’s papers on the development, theory and applications of bi-dimensional regression include, “Computation of the Correspondence of Geographical Patterns”, Papers of the Regional Science Association 15 (1965): 131-39; “Medieval Distortions: The Projections of Ancient Maps”, Annals of Association of American Geographers 56 (1966): 351-61; “Bi-dimensional Regression”, reprinted in Geographical Analysis 26 (1994): 187-212.
(5) C.G. Small, “Techniques of Shape Analysis on Sets of Points,” International Statistical Review 56 (1988): 243-257.
(6) Tomoki Nakaya, “Statistical Inferences in Bi-dimensional Regression Models,” Geographical Analysis 29 (1997): 169-185.
(7)P.J. Huber, “Robust Estimation of a Location Parameter,” Annals of Mathematical Statistics 35 (1964): 73-101 and Robust Statistics (New York: John Wiley, 1981) see also F. Hampel, Contributions to the Theory of Robust Estimation, PhD Thesis (Berkeley: University of California, 1968).
(8) Fred Bookstein, “Principal Warps: Thin-Plate Splines and the Decomposition of Deformations”, IEEE Transactions on Pattern Analysis and Machine Intelligence 11 (1989): 567-585 and Christopher G. Small, The Statistical Theory of Shape (Berlin: Springer-Verlag, 1996) 110
---
(9)John Hessler, “Warping Waldseemüller: A Phenomenological and Computational Study of the 1507 World Map,” Cartographica 41 (2006): 101-113.
(10) Timothy R. Wallace and Charles van der Hanvel, “Truth and Accountability in Geographic and Historical Visualizations”, Cartographic Journal 42 (2005): 173-181.
(11) A. Braddeley, “Stochastic Geometry: an introduction and reading list,” International Statistical Review 50 (1982): 179-193.
---
(12) F. Morgan, Geometric Measure Theory: An Introduction (Boston: Academic Press, 1988).
---
(13)R.A. Skelton et. al., The Vinland Map and the Tartar Relation (New Haven, CT: Yale University Press, 1965) :228.
## Thursday, July 15, 2010
Infinite Geometries:
Mathematical Notes on Werner's Commentary on Ptolemy's First Book and the Projection of the 1507 World Map by Martin Waldseemuller
----
A representation is made with a purpose or a goal in mind, governed by criteria of adequacy pertaining to that goal, which guide its means, medium and selectivity. Hence there is even in those cases no general valid inference from what the representation is like to what the represented is like overall.
--Bas Van Fraassen, Scientific Representation
------
The most naive view of representation might perhaps be put something like this: "A represents B if and only if A appreciably resembles B." Vestiges of this view, with assorted refinements, persist in most writing on representation. Yet more error could hardly be compressed into so short a formula.
--Nelson Goodman, Languages of Art: An Approach to the Theory of Symbols
Among the many technical and theoretical problems that Waldseemüller faced in the construction of his 1507 representation of the world, one of the least trivial mathematically and geometrically was the problem of projection. Dealing with a greatly enlarged earth, compared with the Ptolemaic models at his disposal, Waldseemüller modified Ptolemy’s second conic projection in a way that unfortunately distorted the shape of the new continents because they were forced to the far western portion of the map and hence greatly elongated.
During Waldseemüller’s time, new ideas were rapidly developing out of the theoretical discussions in Book I of Ptolemy’s Geographiae. Many commentators and cartographers realized that there was no reason to adhere to Ptolemy’s restriction of a correct representation of distances on three parallels, a restriction that was introduced in order to construct circular meridians. They found that by altering this arbitrary restriction on the form of the meridians and by applying Ptolemy’s methodology to any number of equidistant parallels, one could obtain a map correct on all parallels, with the meridians easily constructible as curves or polygons, connecting points of equal longitude.
This type of generalization was used on Ptolemy’s second conic projection by Waldseemüller to extend his world map, although not smoothly, as can be seen from the abruptness of the change in the meridians at the equator. A more continuous extension of the second conic projection was made in a less ad hoc way by Bernardus Sylvanus in a world map contained in his 1511 Claudii Ptholemaei Alexandrini liber geographiae cum tabulis universali fugura et cum additione locorum quae a recentioribus reperta sunt diligenti cura emendatus et impressus. Sylvanus’s generalization of Ptolemy’s mapping represented an extension of the area of the globe to between –40 and +80 degrees in latitude and between 70 degrees west and 290 degrees east in latitude using undistorted parallels.
In 1514, Johannes Werner produced his translation and commentary of Book I of Ptolemy’s Geographiae. Werner added to his translation a theoretical discussion of two generalizations of Ptolemy’s second conic projection in a section of his book entitled Libellus de quator terrarum orbis in plano figurationibus ab codem Ianne Verneo nouissime compertis et enarratis. In Werner’s Propositio IV (see figure below) he modified Ptolemy’s methodology by requiring that lengths be preserved on all parallels, represented by concentric arcs, and on all radii. He further modified the projection in a way that made the North Pole the center of what in modern language would be called a system of polar coordinates. In Propositio V, he also required that a quadrant of the equator have the same length as the radius between a pole and the equator.The modifications of Sylvanus and of Werner were the first solutions to the problem of representing the surface of a sphere within a finite area. Waldseemüller’s projection can be graphically approximated using the transformation equations that also can be used to represent an infinite series of projections that include Sylvanus’s, Werner’s and the later Bonne projection.
The value for the central parallel and an additive parameter can be changed in the equations for the Bonne Projection in such a way that an approximation to Waldseemüller’s projections results. The Sylvanus, Werner and Bonne projection in polar coordinates all contain an arbitrary parameter f > 0 such that r = + f. The image of the North Pole accordingly lies on the central meridian at a distance f below the center of the parallels. In the Bonne projection f is assigned in a way that the radii touch the meridian curves always on a given parallel. Sylvanus unknowingly uses a similar value to Bonne, f = 10, and if we assign f = 0 we arrive at Werner’s projection. These of modifications result in the possibility of an infinte series of projections of the Waldseemuller type. This can be visualized by just a few examples from my models of Werner's projection below.
Waldseemüller’s map can be approximated in this same way using values of f between 7-8.5. The actual projection of the 1507 map differs from that represented in the above equations in that it has bends in the meridians at the equator, and the meridians are shown as segmented circular arcs rather than as continually changing curves. This difference is however trivial in the overall look of the projection and the distortions that is produces in the continents of the New World. Using these models the modern coast of South America has been projected in the figure below alongside the same region from the Waldseemüller map.
It can be observed that on the Waldseemüller map that the western coast of South America is portrayed by a series of linear features and is labeled “terra ultra incognita”. The straight lines that appear as the outlones of the west coast of South America have been interpreted as Waldseemüller’s way of picturing regions for which he had no specific geographic information to make a more accurate representation. These same features, however, also appear on the modern coast when it is projected on the model projection that I am using to represent the geometry of the 1507 World Map. Waldseemüller’s representation of the continent and the re-projected outline of modern South America are strikingly similar visually.
## Wednesday, July 14, 2010
Bi-dimensional Regression Revisited:
Studies in the geometry and form of the Medieval Portolan Chart
Introduction to my talk at the Library of Congress’ Conference
Re-Examining the Portolan Chart: History, Navigation and Science
May, 21st 2010
There are spaces in which the determination of position requires not a finite number, but either an endless series or a continuous manifold of determinations of quantity. Such manifolds are, for example, the possible determinations of a function for a given region, the possible shapes of a figure, and so on.
--Bernhard Riemann
What the historian of cartography should be concerned with is a systematic study of the factors effecting error, and seek to establish their cause and variability and the statistical parameters by which error is characterized...
--J.B. Harley
...secular magnetic variation is potentially as valuable in the history of cartography as the radiocarbon method in archaeology, though the calibrations have yet to be worked out.
--Tony Campbell, History of Cartography Volume 1
Those of you who know my more academic publications in the history of cartography realize that for the most part they tend to take an extremely phenomenological approach to cartographic objects. From my earliest publications on Fourier transforms and the Space Oblique Mercator projection through my current research on Topological existence theorems and mathematical constructivism in early computer cartography I have always been more interested in the conceptual and mathematical foundations of cartography than in any historical causalities or contingencies relating to maps themselves.
Calculated Distortion grid and vector displacements for the Library of Congress's 1320 Portolan Chart
Because of this phenomenological approach my paper this afternoon may be quite difficult for some of you (especially right after lunch) as it is extremely analytical and most of it is going to be concerned with very complex transformational geometry; discussions of things like Laplacian matrices and thin-plate splines. This being said, I promise you that if you keep your focus on the actual cartographic problems that I am trying to resolve much of the mathematics will dissolve into the background and in the end it is my hope that you will not only learn something about the geometric and mathematical structure of Portolan charts but also that this talk might serve as a methodological introduction to some of the computational techniques that I have helped develop and that I have been using in my cartometric research. These techniques have there beginnings with the work of Waldo Tobler whose paper and computer program called, Bi-dimensional Regression (see links section to read Tobler's paper) is where modern historical cartometry can be said to have started. Everytime I read this paper I am amazed at Tobler’s geometric insights and I find new inspiration in every re-reading. Using the integrated sums of the squares of the four partial derivatives was a real breakthrough and took incedible geometirc imagination.
Calculated rotation isolines for the Library of Congress' 1320 Portolan chart
My paper this afternoon will deal principally with three problems concerning the form of Portolan charts that have to date eluded solutions and whose logical structure goes directly to the heart of the geometric form that these early charts take. Borrowing a definition from the philosopher of science, Bas van Fraassen, “A representation [like a map] is made with a purpose or goal in mind, governed by criteria of adequacy pertaining to that goal, which guide its means, medium and selectivity”. In other words the form of a representation, in this case mathematical form, reflects the purpose for which the representation was created and hence my questions this afternoon are principally mathematical and not historical.
The first is the question of projection; Are Portolan charts purposely projected? Obviously, the fact they are the surface of a sphere drawn on the plane makes them projected, but the question is more specific; did their creators purposely project them in a consistent way and did these early mapmakers have any knowledge of the error they were introducing through this geometric transformation? This question is extremely difficult to answer because of the fact that any distortion that might be systematic from the projection is somewhat buried in the noise of the distortion caused by simple inaccuracy in the mapping of the coastlines. The most we can hope for is some statistical ruins that might be buried in the non-linear parts of the distortion...
The second has to do with the question of their apparent rotation; as has been pointed out this morning, the charts have various degrees of rotation; why do the parallels, at least in the area of Mediterranean Sea, appear to be rotated? Can we, by analyzing this rotation, gain some insight into the sources and measurement techniques used to construct the charts?
The third question concerns their evolution as geodetic maps; do they get more accurate with time? Do they maintian their accuracy even though many charts have obviously been copied multiple times. Are there any correlations that can be found in looking not only at modern comparisons but also in intrasample variations? (it is here that thin-plate splines are useful, see link to Booksteins paper in links section) Are there any structural changes that we can perceive through their history as a cartographic form?
Principal warps of the LOC's 1320 Portolan Chart
I am going to begin with a bit of a theoretical and historical primer into both the mathematical and philosophical justification for the computational techniques that I am using…many of them have a long history in cartographic analysis and I think it will help you to understand the motivations for some of my research here and on other maps…
For the complete slides of this talk click on the academia.edu link on the right or on the small Portolan chart above it......
## Thursday, July 08, 2010
Schoner's Fragments:
Terrestrial and Celestial Globe Gore Fragments from the Schoner Sammelband
The discoverer of the Sammelband, Josef Fischer, removed the 1507 and 1516 world maps in order to produce a facsimile of them and in doing so recovered from the gutter of the binding fragments of a set of globe gores that belong to Schöner’s 1515 globe. There are only two other surviving examples of this globe, one owned by the Historisches Museum in Frankfurt am Main, and the other by the Herzogin Anna Amalia Bibliothek, Stiftung Weimar Klassik. The gore fragments were trimmed and glued onto gore outlines by Fischer and then rebound into the Sammelband when the 1507 and 1516 maps were replaced. The set of terrestrial fragments found in the Sammelband constitutes approximately 50 percent of the actual globe. Schöner’s 1515 globe depends heavily on Waldseemüller’s 1507 Universalis cosmographiae for much of its geographical information andmany of the legends that appear on the 1515 globe gores are small paraphrases from the larger 1507 map. The globe goes
much farther, however, in its description of the New World, in that it actually shows a complete passage around South America into the Pacific Ocean. A more complete description of the geography found on the gores can be found in the companion volume that Schöner wrote to accompany the globe, Luculentissima quaedam terrae totius descriptio. Besides the terrestrial fragments, a second set of vellum gore fragments was found in the Sammelband.
These are from Schöner’s celestial globes and represent a different edition of Schöner’s celestial gores than is found fully bound in the Sammelband. The fragments represent much less than half of the total globe. In contrast to the full paper gores described below, the celestial fragments show the equator of the earth projected onto the celestial sphere at an angle to the ecliptic. The gore fragments also show differences in the labeling of particular constellations such as
Delphini, and show signs of print stereotyping.
The celestial gores found in the Sammelband are printed on paper and form a complete set of Schöner’s gores from 1517. The gores are the first known set of printed celestial gores and are a great improvement over other star charts of the period. Although Schöner’s interest focused mostly on geography in the early period of his life, we still can see in his extant manuscripts interest in the accurate determinations of stellar positions for the purpose of casting horoscopes. This interest is further established by the annotations that he made to the 1515 Stabius star chart by Albrecht Dürer that originally constituted part of the Sammelband. The Dürer chart contains several well-known errors that Schöner corrected by annotating both the chart itself and his globe. One of the most remarkable features of Schöner’s celestial gores is the naming of several groups of stars in minor constellations that were unnamed on celestial charts. For example, the stars in the constellation Coma Berenicies are usually shown on star charts of the period but went unnamed until Schöner called them Trica (located just above Leo) on his globe gores. Schöner has annotated the gores in red ink mostly over the constellations of Andromeda, Perseus, and Orion.
The 1517 globe, called Solidi et sphaerici corporis sive globi astronomici canones usum et expeditam praxim ejusdem exprimentes, was dedicated to the Bishop of Bamberg, Georg Schenk von Limberg, as were many of Schöner’s works and letters. Several parts of the Schöner Sammelband have been removed over the course of its life, including the 1507 Universalis cosmographiae, now in the Library of Congress; an annotated Dürer star chart from 1515, still at Wolfegg Castle; and a manuscript drawing by Schöner of sheet six of the 1516
Carta Marina, privately held by Jay Kislak.
Some of the most interesting texts regarding Schoner's globes come from his manuscripts that are in the National Library in Vienna. Especially important is a compilation of texts that is listed in their catalog as MS. 3505. In that manuscript there is a treatise called Regionum sive civitatum distantiae, which is a short theoretical work that deals with the problem of locating place-names on a globe using a planar map as a source. In other words, Schoner is talking about the inverse projection problem. In the work Schoner lays out several methods for turning planar maps back into spheres and using them for sources when making globes. Many of the construction methods that he discusses are quite complex requiring mathematical skill and a fairly detailed knowledge of projections. More on this will be found in my forthcoming book, A Globemaker's Toolbox: the mathematical and geographical notebooks of Johannes Schoner, which will be published late next year.
For more information on Schoner's Globes see Chet van Duzer's forthcoming study from the American Philosophical Society and for more images and a complete description of the Sammelband see my articles in "The Jay Kislak Collection at the Library of Congress"
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# Thread: Looking for an equation of a curve...
1. ## Looking for an equation of a curve...
Hello there - bit of an unusual question and I'm not quite sure if it belongs in this Calculus section or not. It sorta involves derivatives so here it goes:
I have been looking for monotonic equations that pass through (0,0) and (1,1) with various slopes at x=0 and x=1. For example, I already have:
CLASS 1,2: f(x)=x^n
when n>1: concave up, f'(0) = 0, f'(1) = n
when 0<n<1: concave down, f'(0) = inf, f'(1) = n
CLASS 3,4: f(x) = 1-(1-x)^n
when n>1: concave down, f'(0) = n, f'(1) = 0
when 0<n<1: concave up, f'(0) = n, f'(1) = inf
CLASS 5,6: f(x) = (1-(1-x)^n)^(1/n)
when n>1: concave down, f'(0) = inf, f'(1) = 0
when 0<n<1: concave up, f'(0) = 0, f'(1) = inf
What I am looking for are the equations that will have:
f'(0) = m
f'(1) = n
Thus, 2 different finite slopes at x=0 and x=1 allowing both concave up and down curves
Thank you for any help. I have asked math professors at my college but still no answer. Sorry for the long post...
Terry
2. ## Re: Looking for an equation of a curve...
Partial solution: construct 2 functions g(x) and h(x), to satisfy:
$g(0)=0, g'(0)=m$
$h(1)=1, h'(1)=n+2g(1)-2$ (looks hard to do but isn't, the RHS is just a number)
Now, consider:
$f(x)=x^2h(x) + (1-x^2)g(x)$
$f(0)=0 + g(0)=0$ as required
$f(1)=h(1) + 0=1$ as required
$f'(x)=2xh(x) + x^2h'(x) -2xg(x) +(1-x^2)g'(x)$
$f'(0)=0+0+0+ g'(0)=m$ as required
$f'(1)=2h(1) +h'(1) -2g(1) = 2 + h'(1) -2g(1)$
$= 2 + [n +2g(1)-2] -2g(1) = n$ as required
i cant guarantee f(x) will be monotonic though
3. ## Re: Looking for an equation of a curve...
Thanks for the help! With the above hint as a starting point, I was able to figure out the following general equation for curves with slope 1/n at (0,0) and slope n at (1,1):
y= (x^n^(1/n)) * (x^n) + (1-x^n^(1/n)) * (1-(1-x)^(1/n))
By changing the 'coefficients' from x^2 and (1-x^2) to (x^n^(1/n)) and (1-x^n^(1/n)), I found that the integrals of the curve on both sides of the line y-x+1 become equal although, I am still not convinced that the curve is symmetrical over that line...
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# area of octagonWatch
Announcements
This discussion is closed.
#1
talk about dumb but i 4gotton how 2 work out area of a regular octagon can anyone show me how or give me a link to a website that tells me how. thanx
0
14 years ago
#2
2n^2/tan(45/2)
n=length of a side
0
14 years ago
#3
0
14 years ago
#4
talk about dumb but i 4gotton how 2 work out area of a regular octagon can anyone show me how or give me a link to a website that tells me how. thanx
Divide it up into 8 triangles that will each subtend 360/8 = 45 degrees at the centre.
If r is the distance from the centre to a vertex of the octagon then the area of one of these triangles is
2 x 1/2 r cos (22.5) r sin (22.5) = 1/2 r^2 sin (45)
(by thinking of it as two right-angled triangles, say).
So the total area is
4 r^2 sin (45) = 2 rt(2) r^2
0
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https://human.libretexts.org/Bookshelves/Composition/Advanced_Composition/Book%3A_How_Arguments_Work_-_A_Guide_to_Writing_and_Analyzing_Texts_in_College_(Mills)/15%3A_Teacher's_Guide/15.05%3A_Course_Maps
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# 15.5: Course Maps
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$
Each course map shows a sequence of essay assignments and the associated How Arguments Work readings, quizzes, and other materials for the course indicated.
## Argumentative Writing and Critical Thinking (California C-ID English 105)
This page titled 15.5: Course Maps is shared under a CC BY-NC 4.0 license and was authored, remixed, and/or curated by Anna Mills (ASCCC Open Educational Resources Initiative) .
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https://codeauri.com/basic-programming/cprogramming/check-palindrome-or-not-in-c/
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# WAP in C to Check If a given number is Palindrome or not?
The Program in C that Checks whether a given number is Palindrome or not is given below:
``````#include <stdio.h>
int main() {
int n, reversedN = 0, remainder, originalN;
printf("Enter an integer: ");
scanf("%d", &n);
originalN = n;
// reversed integer is stored in reversedN
while (n != 0) {
remainder = n % 10;
reversedN = reversedN * 10 + remainder;
n /= 10;
}
// palindrome if originalN and reversedN are equal
if (originalN == reversedN)
printf("%d is a palindrome.", originalN);
else
printf("%d is not a palindrome.", originalN);
return 0;
}
``````
## Output:
Enter an integer: 545
545 is a palindrome.
## Pro-Tips💡
The above program takes an integer as input and uses a while loop to reverse the digits of the number.
It then compares the original input to the reversed number, and if they are the same, it prints that the number is a palindrome.
Otherwise, it prints that the number is not a palindrome.
### Learn C-Sharp ↗
C-sharp covers every topic to learn about C-Sharp thoroughly.
### Learn C Programming ↗
C-Programming covers every topic to learn about C-Sharp thoroughly.
### Learn C++ Programming↗
C++ covers every topic to learn about C-Sharp thoroughly.
Codeauri is Code Learning Hub and Community for every Coder to learn Coding by navigating Structurally from Basic Programming to Front-End Development, Back-End Development to Database, and many more.
## C# Program to Find Sum of Rows & Columns of a Matrix
The Program in C# Program to Find Sum of Rows & Columns of a Matrix is given below: Output: Hello Codeauri Family,enter the number of rows and columns…
## C# Program to Calculate Determinant of Given Matrix
The Program in C# Program to Calculate Determinant of Given Matrix is given below: Output: Hello Codeauri Family, enter the number of rows and columns of the matrix…
## C# Program to Find Sum of right Diagonals of a Matrix
The Program in C# Program to Find Sum of right Diagonals of a Matrix is given below: Output: Hello Codeauri Family, enter the number of rows and columns…
## C# Program to Find Transpose of Given Matrix
The Program in C# Program to Find Transpose of Given Matrix is given below: Output: Hello Codeauri Family, enter the number of rows and columns in the matrix:22Enter…
## C# Program for Multiplication of two square Matrices
The Program in C# Program for Multiplication of two square Matrices is given below: Output: Hello Codeauri Family, enter the number of rows/columns in the matrices:2Enter the elements…
## C# Program to Delete Element at Desired position From Array
The Program in C# Program to Delete Element at Desired position From Array is given below: Output: Hello Codeauri Family, enter the number of elements in the array:4Enter…
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Remember Me
### Create an account
Fields marked with an asterisk (*) are required.
Name *
Email *
Verify email *
## Introduction
An exponent is a mathematical operation that can be used for repeated multiplication. The number of the exponent tells you how many times to repeat the multiplication.
62 = 6 × 6 53 = 5 × 5 × 5 24 = 2 × 2 × 2 × 2
## Lessons
Each exponential expression has two parts: a base and an exponent (exponents can also be called powers). Exponents are a quick and efficient way to show a repeated multiplication.
The two examples above demonstrate how to evaluate an exponent. What do you do if you have several different exponential terms multiplied together? Consider the problem 53 × 54.
In the problem above, the two expressions being multiplied both have a base of five. The work above shows that the exponents can simply be added. Here is another way to look at the work in similar problems.
The rules are the same whether the bases are numbers or variables. This property that allows you to multiply two exponential expressions with the same base is called the product of powers property.
This property works as long as the bases are the same in each of the exponential terms being multiplied.
Examples:
However, if the bases are different, then the product of powers property cannot be applied.
More Complex Problems
When dealing with problems that involve several different variables, multiply each variable separately and then combine the results to get your answer.
If you are even in doubt about how to do a problem that deals with exponents, consider expanding the exponents and then putting like terms together. For example, x2∙x3 = x∙x ∙ x∙x∙x = x5. Showing the work in this way may take a little extra time, but your extra time is worth it because you will be more confident in your answer and in your ability to do these types of problems.
## Review
Use the power of products property to multiply:
1) a2a3a4 =
2) x4y2∙z7 =
3) m2n7 m3n4 =
4) 22s4t6 ∙ 5s6t11 =
5) 9de5f8 ∙ 2d3e3f5 =
Use the power of products property to multiply:
1) a2a3a4 = a9
2) x4y2∙z7 = x4y2∙z7
3) m2n7 m3n4 = m5n11
4) 22s4t6 ∙ 5s6t11 = 20s10t17
5) 9de5f8 ∙ 2d3e3f5 = 18d4e8f13
Didn't find what you were looking for in this lesson? More information on exponents can be found at the following places:
Resource Page
Related Algebra Lessons
Looking for something else? Try the buttons to the left or type your topic into the search feature at the top of this page.
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Absolute Value Inequalities
## Inequalities with solution sets 'between' and 'above or below' certain values
Estimated14 minsto complete
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Progress
Practice Absolute Value Inequalities
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Body Temperature
Teacher Contributed
## Real World Applications – Algebra I
### Topic
What’s normal for a person’s body temperature?
### Student Exploration
It’s common knowledge that a person’s normal body temperature is supposed to be 98.6 degrees. We can figure out using absolute value equations how much a person’s body temperature deviates from the norm for it to be considered abnormal (and possibly sick). Physicians say that people’s body temperature shouldn’t exceed 0.5 degrees from the norm. How can we represent this relationship as an absolute value equation, and then solve to know what the minimum and maximum body temperatures are?
Let’s say that “\begin{align*}t\end{align*}” represents a person’s normal body temperature.
\begin{align*}|t-98.6|=0.5\end{align*}
This inequality means that the normal body temperature subtracted from the minimum and maximum body temperature should equal 0.5.
To solve this, we can break this up into two equations.
\begin{align*}t-98.6 &= 0.5 \ and \ t-98.6=-0.5\\ t &= 98.6+0.5 \ and \ t=98.6-0.5\\ t &= 99.1 \ and \ t = 98.1\end{align*}
This means that our normal body temperature should be between 98.1 and 99.1 degrees.
We can also graph this absolute value equation and see visually see what it means. Since we solved this equation, we can graph our solution set on a number line. We would also represent our solution space between the 98.1 and 99.1 tick marks on the number line. The solution space represents all of the different temperatures that are “normal” for humans.
We can also graph the solution space on an \begin{align*}xy\end{align*} coordinate graph, and interpret the solution. For this relationship, we’d have to graph two separate equations: \begin{align*}y=|x-98.6|\end{align*} and \begin{align*}y=0.5\end{align*} See below.
The horizontal line represents the variant of the normal body temperature. The intersection between the “\begin{align*}V\end{align*}” graph and the horizontal line is our solution 98. and 99.1.
A few steps further: What does the point of the “\begin{align*}V\end{align*}” graph represent on the graph? What do the \begin{align*}x\end{align*} values represent, in relation to body temperature? What do the \begin{align*}y\end{align*} values represent? If we were to shade the inside of the “\begin{align*}V\end{align*}” below the horizontal line, what would the solution space represent?
Let’s explore a little bit more deeply into this body temperature relationship and integrate absolute value inequalities in the equations and graphs. If we first had to integrate the use of an inequality sign instead of the equation \begin{align*}|t-98.6|=0.5\end{align*}, should we use a greater than sign, or a less than sign? Why?
Our inequality would be \begin{align*}|t-98.6| \le 0.5\end{align*} because the body temperature difference can’t be higher than 0.5 variance. If we were to represent this on a number line, we would have our solution space in between the endpoints at 98.1 and 99.1. This represents all of the temperatures that are considered “normal.”
### Extension Investigation
How else can you represent the maximum and minimum of something as an absolute value equation?
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
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# MA3667 Computing assignment
MA3667 Computing assignmentBackground information for MA3667 Computing assignment.
EUROPEAN CALL AND PUT OPTIONS
Large institutions will often trade securities known as
derivatives
. They are financial instru-
ments whose value depends upon the value of more basic underlying stocks or assets. We will
meet a few types of derivative in this course, but among the most prominent will be
options
. An
option
is a contract that gives the option holder the right to buy or sell a particular asset either
up to or on a specified date in the future, for a price that is agreed upon now.
Some terminology and symbols:
Call Option
. A call option gives the holder the right to
buy
the underlying asset for a
specified price known as the
exercise
or
strike
price.
Put Option
. A put option gives the holder the right to
sell
the underlying asset during a
certain time window for a specified price known as the
exercise
or
strike
price.
Exercising an option
. If an option holder executes their right to buy or sell the underlying
asset then they are said to
exercise
the option.
Note that an option holder is not forced
to exercise their right to buy/ sell the asset
.
European Option
. In European options the option holder is only allowed to exercise the
option at one specified time, known as the
expiration date
.
American Option
. In American options the option holder can exercise the right to buy/sell
at
any time until expiration date
.
Maturity
,
Expiration date
,
Exercise date
. These all mean the same thing – the date in
the future on which (European), or by which (American), the option to buy/sell the asset
must be exercised.
Some symbols:
T
:= the expiration date.
S
t
:= the price of the asset at time
t
, so in
particular
S
T
:= the asset price (
not
the option price!!) at expiration date
T
.
K
:= the
exercise price.
2
Option price
. The buyer of the option gains protection against risk, whereas the seller is
exposing themselves to it. The buyer of the option gains protection against unfavourable
movements in the underlying asset price: the holder of a put option knows that they can
always get at least
K
for it whatever happens to the price of the asset in the market, whereas
the holder of a call option knows that they will have to spend at most
K
to buy the asset
at time
T
, however high the price of the asset may be in the market at time
T
. In order to
gain this protection against risk the holder of the option has to pay the seller of the option
a fee – the
option price
.
More symbols: in
this document
we will use:
P
t
(
T; K
) := the price at time
t
of a European
put option with expiration date
T
and strike price
K
. We will also use:
C
t
(
T; K
) := the
price at time
t
of a European call option with expiration date
T
and strike price
K
.
Payoff
. Suppose that you are the holder of a European call option to buy a stock for strike
price
K
at time
T
. Then you would only bother to exercise the option if it the market price
of the stock at time
T
turned out to be more than
K
, in which case the option would give
you a payoff of
S
T
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http://www.solutioninn.com/us-insurers-costs-for-knee-replacement-surgery-range-from-17627
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Question: U S insurers costs for knee replacement surgery range from 17 627
U.S. insurers’ costs for knee replacement surgery range from \$17,627 to \$25,462. Estimate the population variance (standard deviation) in cost with 98% confidence based on a random sample of 10 persons who have had this surgery. The retail costs (for uninsured persons) for the same procedure range from \$40,640 to \$58,702. Estimate the population variance and standard deviation in cost with 98% confidence based on a sample of 10 persons, and compare your two intervals.
View Solution:
Sales0
Views272
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# Questão: What does the canopy do on a parachute?
Contents
There are two basic types of parachutes. One is a dome canopy made of fabric in a shape that ranges from a hemisphere to a cone; the canopy traps air inside its envelope, creating a region of high pressure that retards movement in the direction opposite the entering air flow.
## What forces affect a parachute?
Parachute. There are two forces acting on a parachute with a parachutist: the force of gravity and the air resistance. Suppose that the air resistance is kv2, where v is the speed of descent, and k is a positive constant.
## How does a parachute deploy?
A ripcord system pulls a closing pin (sometimes multiple pins), which releases a spring-loaded pilot chute, and opens the container; the pilot chute is then propelled into the air stream by its spring, then uses the force generated by passing air to extract a deployment bag containing the parachute canopy, to which it …
## Does the weight of a parachute affect how fast it falls?
4. The size of the parachute affects the speed of falling because a larger parachute allows it to displace more air, causing it to fall more slowly.
## What does a parachute do?
A parachute works by forcing air into the front of it and creating a structured ‘wing’ under which the canopy pilot can fly. Parachutes are controlled by pulling down on steering lines which change the shape of the wing, cause it to turn, or to increase or decrease its rate of descent.30 мая 2019 г.
## What keeps a parachute move slower?
The larger the parachute, the greater the drag force. In the case of these parachutes, the drag force is opposite to the force of gravity, so the drag force slows the parachutes down as they fall.
circle parachute
## How fast do you hit the ground parachuting?
Parachutes are designed to reduce your terminal velocity by about 90 percent so you hit the ground at a relatively low speed of maybe 5–6 meters per second (roughly 20 km/h or 12 mph)—ideally, so you can land on your feet and walk away unharmed.
## How long does a skydive last?
Skydiving takes about 5-7 minutes from jump to landing, plus 20 minutes or so in the airplane beforehand. It may not sound like a long time, but with so many new sensations happening throughout, your body will go into superhero mode, being hyper aware of every moment. It’ll feel like the longest minutes of your life!
## Can you pull a parachute too early?
Well, we can intentionally deploy it just about anytime we wish, allowing for a few feet separation from the plane. So there really is no “too” early as in unintentional. Except when it is an unintentional early deployment. Then, at the least, it’s a longer canopy ride down.
ЭТО ИНТЕРЕСНО: Can you wear glasses while parasailing?
## Why do parachutes have a hole in the middle?
Most round parachutes have a hole in the top that is designed to release the excess pressure that might otherwise buildup under the canopy and cause it to oscillate. Many round canopies have/had other holes and slits that help provide forward speed and better control.
## Can you survive jumping out of a plane into water?
If you can dive into water, it won’t feel good at 125mph, but you’ll survive if the water is deep enough — at least 12 feet or so. Steer toward the water (it’s helpful if you’ve been skydiving before and know how to steer as you are falling), and dive right in.
## Do heavier objects fall faster?
Galileo discovered that objects that are more dense, or have more mass, fall at a faster rate than less dense objects, due to this air resistance. A feather and brick dropped together. Air resistance causes the feather to fall more slowly.
## Why do parachutes fail?
Parachute Malfunction. … Parachute malfunctions can be caused by bad packing, incorrect body position or faulty equipment. When a parachute is deployed, the canopy needs to eject out of the pack and spread out immediately. If it gets tangled because of bad packing, this won’t happen.
## How long should the strings on a parachute be?
The length of the string should be equal to the radius of the desired circle. Attaching the suspension lines: Use 4 suspension lines for each parachute. You can tape or tie the suspension lines onto the canopy.
ЭТО ИНТЕРЕСНО: Melhor resposta: What materials can be used to make a parachute?
## How big should a parachute be?
Expert skydivers use parachutes that range in size from 80 square feet to 200 square feet. Newbies and expert divers use parachute sizes that are based on weight.
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Welcome! On this website you can find information about each number.
# Number 419766514
## Number 419766514 basic info
Number 419766514 has 9 digits. Number 419766514 can be formatted as 419,766,514 or 419.766.514 or 419 766 514 or in case this was a phone number 419-766-514 or 41-976-6514 to be easier to read. Number 419766514 in English words is "four hundred and nineteen million, seven hundred and sixty-six thousand, five hundred and fourteen". Number 419766514 can be read by triplets (groups of 3 digits) as "four hundred and nineteen, seven hundred and sixty-six, five hundred and fourteen". Number 419766514 can be read digit by digit as "four one nine seven six six five one four". Number 419766514 is even. Number 419766514 is divisible by: two. Number 419766514 is a composite number (non-prime number).
## Number 419766514 conversions
Number 419766514 in binary code is 11001000001010010000011110010. Number 419766514 in octal code is: 3101220362. Number 419766514 in hexadecimal (hexa): 190520f2.
The sum of all digits of this number is 43. The digital root (repeated digital sum until you get single-digit number) is 7. Number 419766514 divided by two (halved) equals 209883257. Number 419766514 multiplied by two (doubled) equals 839533028. Number 419766514 multiplied by ten equals 4197665140. Number 419766514 raised to the power of 2 equals 1.7620392627571E+17. Number 419766514 raised to the power of 3 equals 7.3964507885869E+25. The square root (sqrt) of 419766514 is 20488.204264893. The sine (sin) of 419766514 degree is 0.55919290294047. The cosine (cos) of 419766514 degree is 0.82903757291272. The base-10 logarithm of 419766514 equals 8.6230077906898. The natural logarithm of 419766514 equals 19.855209195614. The number 419766514 can be encoded to characters as DAIGFFEAD. The number 419766514 can be encrypted to chemical element names as beryllium, hydrogen, fluorine, nitrogen, carbon, carbon, boron, hydrogen, beryllium.
## Numbers simmilar to 419766514
Numbers simmilar to number 419766514 (one digit altered): 319766514519766514409766514429766514418766514419666514419866514419756514419776514419765514419767514419766414419766614419766504419766524419766513419766515
Possible variations of 419766514 with a digit pair swapped: 149766514491766514417966514419676514419765614419766154419766541
Number 419766514 typographic errors with one digit missing: 197665144976651441766514419665144197651441976514419766144197665441976651
Number 419766514 typographic errors with one digit doubled: 441976651441197665144199766514419776651441976665144197666514419766551441976651144197665144
Previous number: 419766513
Next number: 419766515
## Several randomly selected numbers:
9005377396506128078780060026798379432484237124569821773010048885867102762444362357585200283015714113012236184071752926633641126729295759998247105708215420701432445846088499642879763539834596421761007386460192649209368551210956153348399593545014574574193456284995762150134360113793669839197814687524443186824392913020895047591332735252492603165116889600193142184491323378196111987068593666276056.
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https://www.calculatoratoz.com/en/diameter-of-a-circle-when-radius-is-given-calculator/Calc-384
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🔍
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## Credits
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## Diameter of a circle when radius is given Solution
STEP 0: Pre-Calculation Summary
Formula Used
d = 2*r
This formula uses 1 Variables
Variables Used
Radius - Radius is a radial line from the focus to any point of a curve. (Measured in Centimeter)
STEP 1: Convert Input(s) to Base Unit
Radius: 18 Centimeter --> 0.18 Meter (Check conversion here)
STEP 2: Evaluate Formula
Substituting Input Values in Formula
d = 2*r --> 2*0.18
Evaluating ... ...
d = 0.36
STEP 3: Convert Result to Output's Unit
0.36 Meter --> No Conversion Required
0.36 Meter <-- Diameter
(Calculation completed in 00.015 seconds)
## < 3 Diameter of a Circle Calculators
Diameter of a circle when area is given
diameter = 2*sqrt(Area of Circle/pi) Go
Diameter of a circle when circumference is given
diameter = Circumference of Circle/pi Go
Diameter of a circle when radius is given
diameter = 2*Radius Go
d = 2*r
## What is Diameter of a circle when radius is given?
The diameter is the distance across a circle through its center point and touching two points on its edge. Sometimes the word 'diameter' is used to refer to the line itself. In that sense, you may see "draw a diameter of the circle". The diameter is also a chord. A chord is a line that joins any two points on a circle. A diameter is a chord that runs through the center point of the circle. It is the longest possible chord of any circle. The center of a circle is the midpoint of its diameter. To find the diameter of the circle when the radius is given, you need to multiply the radius by 2.
## How to Calculate Diameter of a circle when radius is given?
Diameter of a circle when radius is given calculator uses diameter = 2*Radius to calculate the Diameter, The diameter of a circle when the radius is given is the distance from the center outwards provided the value of the radius is given. Diameter and is denoted by d symbol.
How to calculate Diameter of a circle when radius is given using this online calculator? To use this online calculator for Diameter of a circle when radius is given, enter Radius (r) and hit the calculate button. Here is how the Diameter of a circle when radius is given calculation can be explained with given input values -> 0.36 = 2*0.18.
### FAQ
What is Diameter of a circle when radius is given?
The diameter of a circle when the radius is given is the distance from the center outwards provided the value of the radius is given and is represented as d = 2*r or diameter = 2*Radius. Radius is a radial line from the focus to any point of a curve.
How to calculate Diameter of a circle when radius is given?
The diameter of a circle when the radius is given is the distance from the center outwards provided the value of the radius is given is calculated using diameter = 2*Radius. To calculate Diameter of a circle when radius is given, you need Radius (r). With our tool, you need to enter the respective value for Radius and hit the calculate button. You can also select the units (if any) for Input(s) and the Output as well.
How many ways are there to calculate Diameter?
In this formula, Diameter uses Radius. We can use 3 other way(s) to calculate the same, which is/are as follows -
• diameter = Circumference of Circle/pi
• diameter = 2*sqrt(Area of Circle/pi)
• diameter = 2*Radius
Where is the Diameter of a circle when radius is given calculator used?
Among many, Diameter of a circle when radius is given calculator is widely used in real life applications like {FormulaUses}. Here are few more real life examples -
{FormulaExamplesList}
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Cody
# Problem 43147. Basic commands - amount of inputs
Solution 1918167
Submitted on 1 Sep 2019
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
y_correct = 3; assert(isequal(amountinput(1,2,3),y_correct))
Undefined function or variable 'varadin'. Error in amountinput (line 2) y = varadin Error in Test1 (line 2) assert(isequal(amountinput(1,2,3),y_correct))
2 Fail
y_correct = 1; assert(isequal(amountinput(10),y_correct))
Undefined function or variable 'varadin'. Error in amountinput (line 2) y = varadin Error in Test2 (line 2) assert(isequal(amountinput(10),y_correct))
3 Fail
y_correct = 7; assert(isequal(amountinput(0,7,3,6,2,7,4),y_correct))
Undefined function or variable 'varadin'. Error in amountinput (line 2) y = varadin Error in Test3 (line 2) assert(isequal(amountinput(0,7,3,6,2,7,4),y_correct))
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# Could You Pass a Park Ranger Exam?
EMPLOYMENT
By: Teresa McGlothlin
6 Min Quiz
# If a plant or animal is not native to an area, what term is used to describe it?
Whether rangers are talking about nonnative plants or animals, they are talking about an invasive species. The onset of an invasive species can cause a lot of ecological damage or completely eradicate species that naturally grow in an area.
# Park rangers often work near bodies of water - what is the term for the study of inland waterways?
Park rangers often work near and participate in the conservation of inland waterways. The name of the study of inland waters, their drainage systems, and their ecology is limnology.
# You've located a park visitor who has fallen out of a tree - what is the first thing you do?
When park visitors become lost or injured, park rangers are often the first point of contact. During their emergency training, they learned that the first step to rescuing is to immobilize the victim.
# How old do you need to be to become a park ranger?
Though some junior ranger positions employ seasonal workers of many ages, to become a full-blown park ranger, you must be at least 21 years of age. It varies from state to state, but 21 is the recommended age.
# What might initially indicate that a tree is infested with termites?
Park rangers spend a lot of time checking the health of the trees in their assigned areas. Even in the woods, termites can be a major problem. The first sign of termite infection is strange wood shavings at the base of the tree.
# 2x + 4 = 12 - can you solve for x?
x = 4. You may wonder what math has to do with being a park ranger, but you might be surprised. From solving visitor problems to measuring the growth of trees, park rangers must be able to pass a mathematical portion of a ranger's exam.
# Which sort of park ranger keeps visitors informed about parks and updated on weather?
While most park ranger positions require a bachelor's degree, the position of interpretive ranger often requires less schooling. Interpretive rangers do not venture deep into forests. Instead, they keep park visitors educated and informed.
# Be careful where you step - how would you identify a patch of poison ivy?
In order to identify poison ivy, you must look for the "leaves of three." Poison ivy has one large leaf with two smaller leaves growing off to the side. It also changes color throughout the year, so you can never count on the color to identify it.
# Can you correctly choose the number one threat to public parks and lands?
According to the U.S. Forest Service, the number one threat to parks and public lands is the outbreak of wildfire. With climate change affecting the weather, more wildfires have taken place in U.S. national parks in the last five years than ever recorded before.
# Which US state has the biggest issue with invasive species?
In the United States, they are many different invasive species. However, Florida is the state that claims the most. In Florida's case, the invasive species of many different snakes have been released into the wild by negligent pet owners.
# Do you know how many acres of open land are lost every day in the US?
The U.S. Forest Service reports that nearly 6,000 acres of open grasslands, forests and swampland are lost every day. With cities expanding, the loss of open space is considered a major threat to ecological systems of public lands.
# Park rangers must know how to identify many different animals - which animal is not considered a small game animal?
One of the most important parts of a park ranger's job is monitoring local wildlife. The types of wildlife vary from park to park, and park rangers often have a background in biology, ecology or wildlife management.
# Rangers must know the territory they are serving - how many acres are in a square mile?
In order for park rangers to adequately monitor the landscape of their designated area, they must be good with land measurements. Every park ranger should know that there are 640 acres in one square mile.
# If trees needed to be removed in a public area, what letter would you use to mark them?
To encourage new forest growth, trees must be strategically cut down from time to time. Rangers are often responsible for marking trees for removal. They mark them with spray paint and the letter X.
# Do you know what type of ranger works for the Department of the Interior?
Some park rangers also function as members of law enforcement. These rangers, called commissioned rangers, have received specialized training, and they are employed by the Department of the Interior.
# If you encountered a rattlesnake on your path while patrolling, what should you do first?
Park rangers face many dangers on the job. From unstable landscapes to natural predators, they must watch every step. If you were to encounter a rattlesnake while patrolling, the first thing you should do is stop moving and locate the snake.
# What is the name of the agency that oversees wilderness areas?
Founded in 1905, the United States Forest Service is the agency that manages wilderness areas. In total, the United States Forest Service oversees 193 million acres of grasslands and wilderness.
# If you wanted to test a body of water's current status without much equipment, what sort of sample would you take?
If you wanted to quickly test the current status of a body of water, you would take a grab sample. Park rangers are often responsible for monitoring water quality. If you needed to do repeated samples, it would be called a composite sample.
# You need to know a lot about plants to pass a ranger exam - how do plants get the majority of their nutrients?
Many park rangers have a background in biology or botany. They are constantly monitoring native plants and invasive species, and they know that the soil is the biggest source of nutrients for anything they find growing.
# Which former US president was a park ranger?
Out of 45 US presidents to serve the country, only one of them was a park ranger. In 1963, Gerald Ford served as a seasonal park ranger at Yellowstone National Park. He was elected as president in 1974.
# Where would you locate the pectoral fin on a fish?
Although not all park rangers are responsible for managing fish populations, most rangers must possess knowledge of an area's local types. If you were monitoring fish, you would note the pectoral fin which sits on the top of the fish.
# What year did the US Forest Service hire its first female park ranger?
Not always an inclusive organization, the US Forest Service began allowing female park rangers starting in 1918. The first female ranger is named Claire Marie Hodges. It was 22 years before the next female park ranger was hired.
# What is the first thing you would tell park visitors about bears?
Visitors to any national park might be excited to encounter many sorts of wildlife including bears. As a park ranger, it is your job to advise visitors about bear safety. The first thing you would tell visitors is to never feed the bears!
# Rangers must be able to communicate with law enforcement - which law enforcement word is spelled incorrectly?
Park rangers and law enforcement work very closely together. As part of a ranger's exam, you must know how to correctly spell a few words you might use when communicating with law enforcement. Here, the word arson is spelled incorrectly.
# When someone enters a closed park, what is the name of the legal offense?
After wilderness areas and parks have closed, usually around dusk, it is considered trespassing to ignore the signs and enter the park anyway. Park rangers are responsible for asking trespassers to leave or for contacting law enforcement.
# If you wanted to save a tree from being cut down, how would you mark it?
When you are indicating that a tree needs to be cut down, you will mark it with the letter X. If you want to spare a healthy tree, you would mark each side of the tree with a dot around the size of a golf ball.
# Park rangers come from many different fields - which field is defined as the study of the way organisms interact with their environments?
Park rangers have diverse backgrounds ranging from hospitality to biology to anthropology. One of the most popular fields for park rangers to study is ecology. It's the study of the way organisms interact with their environments.
# If you needed to defuse an angry situation at your park, what is the first step?
Often times, park rangers are called up to defuse angry or tense situations between park visitors. According to the crisis management training they receive, the first step to calming things down is to listen to the issue.
# Biology is a big part of a park ranger's job - do you remember what is the most basic unit of life is called?
Think all the way back to high school biology class, and you'll remember that the most basic unit of life is the cell. Many park ranger duties include taking notes with biological notation, and those old facts from school could come in handy!
# A young visitor at the parks asks you if fish ever sleep, what do you say?
In addition to a litany of other duties, park rangers serve as educators and ambassadors to the visiting public. Any good park ranger would know that fish do sleep, but they do not have eyelids to close.
# Which program allows artists, geologists and others to reside in public lands?
People without the qualifications to be a park ranger can submit to the National Parks Residency Program. The program allows alternative fields of study to reside in the parks for a short period of time.
# Do you know which fish is considered the most invasive?
Although there are many different sorts of invasive fish, the Asian carp is listed as number one. Asian carp are a big issue because they are predatory, and they often scare away or destroy native species of fish.
# How many years of schooling do you need to become a park ranger?
It does vary from position to position, but most park ranger jobs require a bachelor's degree. Each state has a different requirement for training or experience, so it's important to check with your specific state.
# During the dry seasons, where do park rangers spend a lot of time?
During fire season, many park rangers will be put on the lookout for wildfires. From the fire tower, they have the highest vantage point in their park. Wildfires are the number one threat to parks, and being in the fire tower is a serious duty.
# Park rangers do a lot of calculations - what does 4(2) - 3 + 6 equal?
4(2) - 3 + 6 = 11. If you take an official park ranger exam, be prepared to do a lot of math. Even on the job, rangers are constantly measuring and calculating everything from land mass to visitor needs.
# Explore More Quizzes
Image: narvikk / E+ / Getty Images
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Explore BrainMass
# Prove the Dedekind Cut Theorem
Prove this theorem....Suppose that A and B are non empty sets such that AUB=R(real numbers) and if 'a' is an element of the set A and 'b' is an element of the set B then
a<b, and there exists a "cut point" c such that if x<c<y then x is an element of the set A and y is an element of the set B.
#### Solution Summary
The Dedekind cut theorem is proven.
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100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote ## In how many ways four men, two women and one child can sit a tagged by: fskilnik@GMATH ##### This topic has 2 expert replies and 1 member reply ### GMAT/MBA Expert ## In how many ways four men, two women and one child can sit a ## Timer 00:00 ## Your Answer A B C D E ## Global Stats Difficult In how many ways four men, two women and one child can sit at a circular 7-seats table if the child is the only person to be seated between the two women? (A) 24 (B) 36 (C) 48 (D) 96 (E) 240 Answer: __(C)_____ Difficulty Level: 650 - 700 Source: www.GMATH.net _________________ Fabio Skilnik :: https://GMATH.net (Math for the GMAT) or https://GMATH.com.br (Portuguese version) Course release PROMO : finish our test drive till 30/Dec with (at least) 50 correct answers out of 92 (12-questions Mock included) to gain a 50% discount! ### GMAT/MBA Expert GMAT Instructor Joined 25 May 2010 Posted: 14929 messages Followed by: 1855 members Upvotes: 13060 GMAT Score: 790 fskilnik wrote: In how many ways four men, two women and one child can sit at a circular 7-seats table if the child is the only person to be seated between the two women? (A) 24 (B) 36 (C) 48 (D) 96 (E) 240 Once the child has been placed at the table: Number of ways to arrange the 2 women in the 2 seats adjacent to the child = 2! = 2. Number of ways to arrange the 4 men in the remaining 4 seats = 4! = 24. To combine these options, we multiply: 2*24 = 48. The correct answer is C. _________________ Mitch Hunt Private Tutor for the GMAT and GRE GMATGuruNY@gmail.com If you find one of my posts helpful, please take a moment to click on the "UPVOTE" icon. Available for tutoring in NYC and long-distance. For more information, please email me at GMATGuruNY@gmail.com. Student Review #1 Student Review #2 Student Review #3 Last edited by GMATGuruNY on Mon Sep 24, 2018 3:08 pm; edited 1 time in total Free GMAT Practice Test How can you improve your test score if you don't know your baseline score? Take a free online practice exam. Get started on achieving your dream score today! Sign up now. ### GMAT/MBA Expert GMAT Instructor Joined 09 Oct 2010 Posted: 971 messages Followed by: 27 members Upvotes: 59 fskilnik wrote: In how many ways four men, two women and one child can sit at a circular 7-seats table if the child is the only person to be seated between the two women? (A) 24 (B) 36 (C) 48 (D) 96 (E) 240 Source: www.GMATH.net Thank you for the nice contribution, Mitch! (I believe this is the most direct - hence the better - argument, by the way.) Alternate solution: Let´s imagine a linear version (=row), but "connecting the first seat to the last one" (so that after the last seat we have again the first one). There are 7 seats in which the child could be seated. Once (any) one of the 7 seats is chosen, there are 2 ways to seat the two women. (If the child is in the 7th seat, W1 will be in the 6th, W2 in the 1st... or vice-versa!) Once the child and the two women are seated, there are 4! ways of seating the men. Using the Multiplicative Principle, we have 7*2*4! ways of seating these people in the linear version. The "linear to circular migration" is done dividing 7*2*4! by the number of objects to be circularized (7), checking the "connection" created earlier do not give rise to unwanted configurations: it does not! (*) Hence: $? = \frac{{7 \cdot 2 \cdot 4!}}{7} = 48$ (*) Typical problem: when A and B cannot stay next to each other, in the linear version you cannot allow one of them to be in the first place and the other in the last place, because when the connection is established they would violate the restriction! This solution follows the notations and rationale taught in the GMATH method. Regards, Fabio. _________________ Fabio Skilnik :: https://GMATH.net (Math for the GMAT) or https://GMATH.com.br (Portuguese version) Course release PROMO : finish our test drive till 30/Dec with (at least) 50 correct answers out of 92 (12-questions Mock included) to gain a 50% discount! Last edited by fskilnik@GMATH on Thu Sep 27, 2018 3:36 pm; edited 2 times in total ### Top Member Legendary Member Joined 29 Oct 2017 Posted: 680 messages Followed by: 4 members Let women be numbered as w1 and w2 and child be as c The arrangement can be done as w1cw2 or w2cw1 I.e. 2 ways Now group the women children as one so in addition to 4 other men, there are 5 entities to be arranged in a circle which can be done in (n-1)! Ways = (5-1)!= 4!= 24 ways So total ways = 2*24 = 48 ways • 5 Day FREE Trial Study Smarter, Not Harder Available with Beat the GMAT members only code • FREE GMAT Exam Know how you'd score today for$0
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Investing in mutual funds (MF) through SIPsis one of the best low-risk investment ideas today. It offers flexible options to invest your money at fixed intervals and generate returns. Still, you must check your returns regularly.
This helps you evaluate how your holdings perform over an investment horizon. In this article, we provide you with a simple step-by-step process to manually calculate the returns on SIP.
## Calculation process
As you go on paying the SIPinstalments, you accumulate several units. This makes it tricky to enumerate the earned money.
You will use the Extended Internal Rate of Return (XIRR) concept to compute the overall returns on your investments. It helps you determine the returns for irregular cash flows (as in SIPs) distributed over a time period.
Spreadsheet software like Microsoft Excel features an in-built XIRR function to make such calculations easy.
## Step 1
Open a new Microsoft Excel file on your computer. Now, assign one column for dates. Enter all the dates of SIPtransactions in this column.
## Step 2
Enter all the SIPtransactions in the corresponding column. This represents the amount of money you have invested at a specific date. Since it signifies a cash outflow, you need to mark it with a negative (-) sign. As a rule of thumb, you must use a minus prefix with outflows and a plus (+) sign for your inflows.
Additional Read: Why are SIPs an Ideal Choice for the First-time Investor?
## Step 3
You now have to enter the current market value of all the units that you hold. In the last row, mention the date at which you want to calculate the returns. Then, enter the total market value of each unit.
To view the exact value of the units, you must sign in to your SIPaccount. Look through the statement and find the values with their respective dates.
## Step 4
Use the XIRR function in the excel sheet. Move your selection to the next blank cell and type this syntax =XIRR (values, date, guess).
For defining the value parameter, select the cell with the market value and SIPamount. Select the cells having the SIPdates and the specific return date to specify the date field. It is optional to specify the guess option. You can just use a value of 0.1.
## Step 5
In the last step, you simply have to multiply the decimal value by 100. For instance, type in – ‘=XIRR (A6:A7, B6:B7)*100 and press enter. That’s it, the result displayed on the screen shows the returns on your investment via SIP on the specified date.
Additional Read: How to Top up Your SIP?
### Want to see great returns? Start a SIP with Tata Capital Moneyfy!
Are you looking to invest in a mutual fund? Don’t know where to get started? Just download Tata Capital’s Moneyfy app and begin your investment journey. Explore a wide range of MFs, monitor the market and grow your wealth by investing in a suitable scheme.
You can easily start a SIPonline at the comfort of your place. Moreover, we allow you to easily compare mutual funds based on risk grade, ratings, investment horizon, category and more. So, download the app and build the foundation for your financial goals now!
Disclaimer:
To know more about Terms & Conditions, click here.
Copyright © 2021 Tata Capital Financial Services Limited
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How to Copy NumPy array into another array?
• Last Updated : 05 Sep, 2020
Many times there is a need to copy one array to another. Numpy provides the facility to copy array using different methods. There are 3 methods to copy a Numpy array to another array.
Method 1: Using np.empty_like() function
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This function returns a new array with the same shape and type as a given array.
Syntax:
```numpy.empty_like(a, dtype = None, order = âKâ, subok = True)
```
Python3
`# importing Numpy package``import` `numpy as np`` ` `# Creating a numpy array using np.array()``ary ``=` `np.array([``13``, ``99``, ``100``, ``34``, ``65``, ``11``, `` ``66``, ``81``, ``632``, ``44``])`` ` `print``(``"Original array: "``)`` ` `# printing the Numpy array``print``(ary)`` ` `# Creating an empty Numpy array similar``# to ary``copy ``=` `np.empty_like(ary)`` ` `# Now assign ary to copy``copy[:] ``=` `ary`` ` `print``(``"\nCopy of the given array: "``)`` ` `# printing the copied array``print``(copy)`
Output:
In the above example, the given Numpy array ‘ary‘ is copied to another array ‘copy‘ using np.empty_like () function
Method 2: Using np.copy() function
This function returns an array copy of the given object.
Syntax :
`numpy.copy(a, order='K', subok=False)`
Example 1:
Python3
`# importing Numpy package``import` `numpy as np`` ` `# Creating a numpy array using np.array()``org_array ``=` `np.array([``1.54``, ``2.99``, ``3.42``, ``4.87``, ``6.94``,`` ``8.21``, ``7.65``, ``10.50``, ``77.5``])`` ` `print``(``"Original array: "``)`` ` `# printing the Numpy array``print``(org_array)`` ` `# Now copying the org_array to copy_array``# using np.copy() function``copy_array ``=` `np.copy(org_array)`` ` `print``(``"\nCopied array: "``)`` ` `# printing the copied Numpy array``print``(copy_array)`
Output:
In the above example, the given Numpy array ‘org_array‘ is copied to another array ‘copy_array‘ using np.copy () function
Example 2: Copy given 3-D array to another array using np.copy() function
Python3
`# importing Numpy package``import` `numpy as np`` ` `# Creating a 3-D numpy array using np.array()``org_array ``=` `np.array([[``23``, ``46``, ``85``],`` ``[``43``, ``56``, ``99``],`` ``[``11``, ``34``, ``55``]])`` ` `print``(``"Original array: "``)`` ` `# printing the Numpy array``print``(org_array)`` ` `# Now copying the org_array to copy_array``# using np.copy() function``copy_array ``=` `np.copy(org_array)`` ` `print``(``"\nCopied array: "``)`` ` `# printing the copied Numpy array``print``(copy_array)`
Output:
In the above example, the given 3-D Numpy array ‘org_array‘ is copied to another array ‘copy_array‘ using np.copy () function
Method 3: Using Assignment Operator
Python3
`# importing Numpy package``import` `numpy as np`` ` `# Create a 2-D Numpy array using np.array()``org_array ``=` `np.array([[``99``, ``22``, ``33``],`` ``[``44``, ``77``, ``66``]])`` ` `# Copying org_array to copy_array``# using Assignment operator``copy_array ``=` `org_array`` ` `# modifying org_array``org_array[``1``, ``2``] ``=` `13`` ` `# checking if copy_array has remained the same`` ` `# printing original array``print``(``'Original Array: \n'``, org_array)`` ` `# printing copied array``print``(``'\nCopied Array: \n'``, copy_array)`
Output:
In the above example, the given Numpy array ‘org_array‘ is copied to another array ‘copy_array‘ using Assignment Operator.
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The mass of a planet is six times that of the earth. The radius of the planet is twice that of the earth. If the escape velocity from the earth is '$$V_e$$', then the escape velocity from the planet is
A
$$\sqrt{3} \mathrm{~V}_{\mathrm{e}}$$
B
$$\sqrt{2} \mathrm{~V}_{\mathrm{e}}$$
C
$$\mathrm{V}_{\mathrm{e}}$$
D
$$\sqrt{5} \mathrm{~V}_{\mathrm{e}}$$
2
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For a body of mass '$$m$$', the acceleration due to gravity at a distance '$$R$$' from the surface of the earth is $$\left(\frac{g}{4}\right)$$. Its value at a distance $$\left(\frac{R}{2}\right)$$ from the surface of the earth is ( $$R=$$ radius of the earth, $$g=$$ acceleration due to gravity)
A
$$\left(\frac{g}{8}\right)$$
B
$$\left(\frac{9 g}{4}\right)$$
C
$$\left(\frac{4 g}{9}\right)$$
D
$$\left(\frac{\mathrm{g}}{2}\right)$$
3
MHT CET 2021 21th September Evening Shift
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-0
The ratio of energy required to raise a satellite of mass '$$m$$' to height '$$h$$' above the earth's surface to that required to put it into the orbit at same height is [ $$\mathrm{R}=$$ radius of earth]
A
$$\frac{h}{R}$$
B
$$\frac{2 h}{\mathrm{R}^2}$$
C
$$\frac{3 \mathrm{~h}}{\mathrm{R}^2}$$
D
$$\frac{2 \mathrm{~h}}{\mathrm{R}}$$
4
MHT CET 2021 21th September Morning Shift
+1
-0
A pendulum is oscillating with frequency '$$n$$' on the surface of the earth. It is taken to a depth $$\frac{R}{2}$$ below the surface of earth. New frequency of oscillation at depth $$\frac{R}{2}$$ is
[ $$R$$ is the radius of earth]
A
$$\mathrm{\frac{n}{3}}$$
B
$$\frac{\mathrm{n}}{\sqrt{2}}$$
C
$$\mathrm{2 n}$$
D
$$\frac{\mathrm{n}}{2}$$
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PCA identifies new variables, the principal components, which are linear combinations of the original variables. The two principal components for our two-dimensional gene expression profiles are shown in Figure 1b. It is easy to see that the first principal component is the direction along which the samples show the largest variation.
Principal component analysis is a dimension-reduction technique, as well as an exploratory data analysis tool. Principal component analysis is also useful for constructing predictive models, as in principal components analysis regression (also known as PCA regression or PCR). For data with a very large number of variables, the Principal Components platform provides an estimation method called the Wide method. Principal Component Analysis (PCA)1is a dimension reduction technique. We obtain a set of factors which summarize, as well as possible, the information available in the data. The factors are linear combinations of the original variables. The approach can handle only quantitative variables.
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Sky Background png download - 4924*2048 ... FactoMineR: PCA K-nearest neighbors algorithm Principal ... Principal component analysis (PCA) is commonly thought of as a statistical technique for data reduction. It helps you reduce the number of variables in an analysis by describing a series of uncorrelated linear combinations of the variables that contain most of the variance. PCA originated with the work ofPearson(1901) andHotelling(1933). For an introduction, seeRabe-
Try watching this video on www.youtube.com, or enable JavaScript if it is disabled in your browser. Principal component analysis is a statistical technique that is used in finding patterns and reducing the dimensions of multi-dimensional data. There is an excellent tutorial by Lindsay I Smith on this topic so I will be focusing more on the application part in this post.
1 Principal Component Analysis (PCA) PCA is one method used to reduce the number of features used to represent data. The bene ts of this dimensionality reduction include providing a simpler representation of the data, reduction in memory, and faster classi cation. We accomplish by projecting data Netscaler ldap load balancing
This tutorial is designed to give the reader an understanding of Principal Components Analysis (PCA). PCA is a useful statistical technique that has found application in fields such as face recognition and image compression, and is a common technique for finding patterns in data of high dimension. Factor analysis is related to principal component analysis (PCA), but the two are not identical. There has been significant controversy in the field over differences between the two techniques (see section on exploratory factor analysis versus principal components analysis below).
Principal Components Analysis, commonly called PCA, is similar, but it works on features rather than rows. At its heart, PCA is a method of reducing a feature space, say turning 10,100 or even more features into maybe just four or five, while trying to keep as much of the information as possible. Principal Components Analysis. Copyright 2004, Karl L. Wuensch - All rights reserved. In principal components analysis (PCA) and factor analysis (FA) one wishes to extract from a set of . p. variables a reduced set of . m. components or factors that accounts for most of the variance in the . p . variables. In other words, we wish to reduce a set of . p
May 08, 2013 · For non-Gaussian source data, we will need independent component analysis (ICA) to separate data into additive, mutual statistical independent sub-components. PCA also makes assumption that all principal components are orthogonal (linear uncorrelated) to each other. So it may fail when principal components are correlated in a non-linear manner. Multiscale principal components analysis generalizes the PCA of a multivariate signal represented as a matrix by simultaneously performing a PCA on the matrices of details of different levels. A PCA is also performed on the coarser approximation coefficients matrix in the wavelet domain as well as on the final reconstructed matrix.
I'm looking into the cocktail party problem and trying to figure out whether something like Principal Component Analysis is enough to separate out all the various voices at the cocktail party into ... Nov 11, 2020 · Principal components analysis models the variance structure of a set of observed variables using linear combinations of the variables. These linear combinations, or components, may be used in subsequent analysis, and the combination coefficients, or loadings, may be used in interpreting the components.
Finding these dimensions (the principal components) and transforming the dataset to a lower dimensional dataset using these principal components is the task of the PCA. As said, in the end we use the found and chosen principal component to transform our dataset, that is, projecting our dataset (the projection is done with matrix multiplication ... Principal component analysis, or PCA, is a statistical procedure that allows you to summarize the information content in large data tables by means of a smaller set of “summary indices” that can be more easily visualized and analyzed. The underlying data can be measurements describing properties of production samples, chemical compounds or reactions, process time points of a continuous ...
Jan 23, 2017 · Principal component analysis (PCA) is routinely employed on a wide range of problems. From the detection of outliers to predictive modeling, PCA has the ability of projecting the observations described by variables into few orthogonal components defined at where the data ‘stretch’ the most, rendering a simplified overview. Oct 25, 2017 · There is no a priori reason to believe that the principal components with the largest variance are the components that best predict the response. In fact, it is trivial to construct an artificial example in which the best predictor is the last component, which will surely be dropped from the analysis.
Select the type of matrix to use to calculate the principal components. Correlation : Use when your variables have different scales and you want to weight all the variables equally. For example, if some of the variables use a scale from 1-5 and others use a scale from 1-10, use the correlation matrix to standardize the scales. Description. This function performs several computations known as "principal component analysis". The idea behind this method is to represent in an approximative manner a cluster of n individuals in a smaller dimensional subspace.
Dec 13, 2017 · Abstract. This chapter reviews exponential family principal component analysis (ePCA), a family of statistical methods for dimension reduction of large-scale data that are not real-valued, such as user ratings for items in e-commerce, categorical/count genetic data in bioinformatics, and digital images in computer vision. But often we only need the first two or three principal components to visualize the data. For extracting only the first k components we can use probabilistic PCA (PPCA) [Verbeek 2002] based on sensible principal components analysis [S. Roweis 1997], e.g, by using this modified PCA matlab script (ppca.m), originally by Jakob Verbeek. It also is ...
Principal components analysis (PCA) Description. Does an eigen value decomposition and returns eigen values, loadings, and degree of fit for a specified number of components. Basically it is just doing a principal components analysis (PCA) for n principal components of either a correlation or covariance matrix. component (think R-square) 1.8% of the variance explained by second component Sum squared loadings down each column (component) = eigenvalues Sum of squared loadings across components is the communality 3.057 1.067 0.958 0.736 0.622 0.571 0.543 0.446 Q: why is it 1? Component loadings correlation of each item with the principal component Excel ...
Principal Component Analysis in an image with scikit-learn and scikit-image. from sklearn.decomposition import PCA from pylab import * from skimage import data, io, color just one or two components does an adequate job of reproducing the communalities (the variance in each individual measured x variable) and the correlations among variables (in the R correlation matrix). Principal Components Versus Principal Axis Factoring As noted earlier, the most widely used method in factor analysis is the PAF method. In
Principal Components Analysis (PCA) is an algorithm to transform the columns of a dataset into a new set of features called Principal Components. By doing this, a large chunk of the information across the full dataset is effectively compressed in fewer feature columns.Principal component analysis (PCA) – computation 4. Principal component analysis (PCA) can be used with variables of any mathematical types: quantitative, qualitative, or a mixture of these types. – True, False. 5. Principal component analysis (PCA) requires quantitative multivariate data. – True, False. 6.
(All these calculations, including how principal components are identified, are explained later in this article, but for the moment let us just go along to understand conceptually what PCA is.) The number of principal components that can be identified for any dataset is equal to the number of the variables in the dataset. We found one dictionary with English definitions that includes the word l1-norm principal component analysis: Click on the first link on a line below to go directly to a page where "l1-norm principal component analysis" is defined. General (1 matching dictionary) L1-norm principal component analysis: Wikipedia, the Free Encyclopedia [home, info]
Perform Principal Components Analysis Check this box to perform principal components analysis (PCA) on the computed relationship matrix. To Specify This Option: The solution of a categorical principal components analysis maximizes the correlations of the object scores with each of the quantified variables for the number of components (dimensions) specified. An important application of categorical principal components is to examine preference data, in which respondents rank or rate a number of items ...
Principal component analysis: Consider below scenario: The data, we want to work with, is in the form of a matrix A of mXn dimension, shown as below, where Ai,j represents the value of the i-th observation of the j-th variable. Principal component analysis (PCA) is a mainstay of modern data analysis - a black box that is widely used but (sometimes) poorly understood. The goal of this paper is to dispel the magic behind this black box. This manuscript focuses on building a solid intuition for how and why principal component analysis works. This
Example 33.1 Principal Component Analysis. This example analyzes socioeconomic data provided by Harman . The five variables represent total population (Population), median school years (School), total employment (Employment), miscellaneous professional services (Services), and median house value (HouseValue). Each observation represents one of twelve census tracts in the Los Angeles Standard Metropolitan Statistical Area. Nov 18, 2017 · Principal component analysis, also known as the Hotelling transform or Karhunen-Loeve transform, is a statistical technique that was proposed by Karl Pearson (1901) as part of factorial analysis; however, its first theoretical development appeared in 1933 in a paper written by Hotelling [1, 2, 3, 4, 5, 6, 7, 8]. The complexity of the calculations involved in this technique delayed its development until the birth of computers, and its effective use started in the second half of the twentieth ...
Is there a possibility to conduct longitudinal principle components analysis using STATA. The command pca doesn't have the option to cluster repeated observations. I am working on panel data (6 years) and want to create a wealth index for the households using PCA. Sep 01, 2017 · Principal component analysis aptly described in the famous Shlen’s paper. Shlen’s Principal component analysis paper The paper explains that even a simple problem such as recording the motion of a pendulum, which moves in only one direction. If one is unaware of the exact direction.
Principal components analysis In our discussion of factor analysis, we gave a way to model data x ∈Rd as “approximately” lying in some k-dimension subspace, where k ≪d. Specifi-cally, we imagined that each point x(i) was created by first generating some z(i) lying in the k-dimension affine space {Λz +µ;z ∈Rk}, and then adding Implementing Principal Component Analysis In Python. In this simple tutorial, we will learn how to implement a dimensionality reduction technique called Principal Component Analysis (PCA) that helps to reduce the number to independent variables in a problem by identifying Principle Components. We will take a step by step approach to PCA.
Principal component analysis (PCA) is a mainstay of modern data analysis - a black box that is widely used but poorly understood. The goal of this paper is to dispel the magic behind this black box. This tutorial focuses on building a solid intuition for how and why principal component analysis works; furthermore, it Jan 03, 2018 · Author: Matteo Alberti Among all tools for the linear reduction of dimensionality PCA or Principal Components Analysis is certainly the main tools of Statistical Machine Learning. Although we focus very often on non-linearity, the analysis of the principal components is the starting point for many analysis (also the core of preprocessing), and their knowledge becomes imperative in case the ...
Robust Principal Component Analysis via ADMM in Python. This is a Python implementation of the RPCA algorithm from [1,2] that uses an ADMM version of matrix decomposition. Blog post associated with this repo can be found here. Appendix [1] Parikh, N., & Boyd, S. (2013). Proximal algorithms. Foundations and Trends in optimization, 1(3), 123-231. Principal component analysis (PCA) (Jollifie 1986) has proven to be an exceedingly popular tech-nique for dimensionality reduction and is discussed at length in most texts on multivariate analysis. Its many application areas include data compression, image analysis, visualization, pattern recog-nition, regression and time series prediction.
principal component analysis to the mapping of the climate of India using precipitation data. Jayawardene et al., (2005) used the PCA to classify the spatial rainfall regions and identified the two dominant rainfall regions, wet and dry zones across Sri Lanka. On the Analytic Solver Data Mining ribbon, select Transform - Principal Components to open the Principal Components Analysis dialog. Specify the desired worksheet or data range to be processed. Move the variables to be used in the analysis from the Variables list to the Selected Variables list, using the transfer (>) button.
> > The first phase of principal component analysis was devoted > to verifying > that the following requirements were met: > > 1. The variables included must be > metric level or dichotomous nominal level > and the sample size must be greater than 50 (preferably > 100) > > 2. The ratio of cases to variables must > be 10 to 1 or larger > > 3 ...
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Applications of Principal Component Analysis PCA is predominantly used as a dimensionality reduction technique in domains like facial recognition, computer vision and image compression. It is also used for finding patterns in data of high dimension in the field of finance, data mining, bioinformatics, psychology, etc. Rather than attempting to run a factor analysis on all 100+ variables at once, with so few cases, would it make more sense to * run the factor analysis on one group of questions at a time * reduce the group to one or two questions with the highest loadings on the principal component * repeat the above procedure for each group of questions ... Oct 16, 2009 · Author Summary Genetic variation in natural populations typically demonstrates structure arising from diverse processes including geographical isolation, founder events, migration, and admixture. One technique commonly used to uncover such structure is principal components analysis, which identifies the primary axes of variation in data and projects the samples onto these axes in a graphically ...
This function performs "principal component analysis" on the n-by-p data matrix x. The idea behind this method is to represent in an approximative manner a cluster of n individuals in a smaller dimensional subspace. In order to do that, it projects the cluster onto a subspace. Oct 25, 2017 · There is no a priori reason to believe that the principal components with the largest variance are the components that best predict the response. In fact, it is trivial to construct an artificial example in which the best predictor is the last component, which will surely be dropped from the analysis. Mar 13, 2008 · The term "principal components" refers to the components of the vectors expressed in coordinates with respect to a special basis. Such a basis is chosen so that the variance of the first components is the largest possible, and so that the variance of each successive components is also the largest possible, while keeping the vectors of the basis ... Principal Components Analysis (PCA) is a technique that finds underlying variables (known as principal components) that best differentiate your data points. Principal components are dimensions along which your data points are most spread out: A principal component can be expressed by one or more existing variables.
Apr 21, 2019 · Principal Component Analysis starts to make sense when the number of measured variables are more than three (3) where visualization of the cloud of the data point is difficult and it is near ... A principal component analysis of the data can be applied using the prcomp function with the scaleargument set to TRUEto ensure the analysis is carried out on the correlation matrix. The result is a list containing the coefficients defining each component (sometimes referred to as loadings), the principal component scores, etc. Principal Component Analysis(PCA) in python from scratch The example below defines a small 3×2 matrix, centers the data in the matrix, calculates the covariance matrix of the centered data, and then the eigenvalue decomposition of the covariance matrix. In this video, you learn how to perform principal component analysis with PROC PCA in SAS Viya, using similar code to what you use in PROC PRINCOMP in SAS 9. Mar 01, 2012 · Abstract. Principle Component Analysis (PCA) is one of the common techniques used in Risk modeling, i.e. statistical factor models. When using PCA to estimate the covariance matrix, and applying it to portfolio optimization, we formally analyze its performance, and find positive results in terms of portfolio efficiency (Information Ratio) and transaction cost reduction. Principal components analysis (PCA) finds hypothetical variables (components) accounting for as much as possible of the variance in your multivariate data (e.g. Legendre & Legendre 1998). These new variables are linear combinations of the original variables. PCA may be used for reduction of the data set to only two variables (the two first ...
Principal Components Analysis Principal components analysis [PCA] is a tool for manipulating and visualizing a data set, and for verfying and evaluting a particular clustering. It can be an extremely useful tool for understanding the relationships in a data set, but you have to be careful how you interpret the results. And you want to perform the principal component Stack Exchange Network Stack Exchange network consists of 176 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
Principal Component Analysis In Arcgis related files: 8c389ee81a0345d14ae61f8088b076e8 Powered by TCPDF (www.tcpdf.org) 1 / 1
Analysisconverts the normalized data in to so-called 'principal component scores' in. As discussed in the lab, the variables are in essence rotated through multiple dimensions so as to see combinations of variables that describe the major patterns of variation among taxa. Matrix is identical to the 'eigenanalysis' table produced by Jun 02, 2014 · Hi, I'm finishing my thesis we're I'm forming currency-hedge investment portfolios out of the PCA on the currencies. I need to do a PCA using a "moving-window" of the previous 60 months of data, throughout my entire data-set. If you want the "pseudo-code" is: -Run PCA using previous... Coordinates (principal components) that make diagonal are the eigenvectors of . PCA recipe Calculate covariance matrix . Find eigenvectors v and eigenvalues such that v k = kv k. k is the variance in the k k direction. Use heuristic to choose Keigenvectors to keep. Data is now K-dimensional: x ˇ + XK k=1 c kv k, c k = (x ) v k Generative model ... Explorative data analysis with Hierarchical Clustering. Today we will write about cluster analysis with Hierarchical Clustering widget. We use a well-known Iris data set, which contains 150 Iris flowers, each belonging to one of the three species (setosa, versicolor and virginica).
Evony research factoryA generalization of principal component analysis to the exponential family. In Advanced in Neural Information Processing System (T. G. Dietterich, S. Becker and Z. Ghahramani, eds.) 14 617–642. MIT Press, Cambridge, MA. de Leeuw, J. (2006). Principal component analysis of binary data by iterated singular value decomposition. Comput. Statist. QT dispersion and principal component analysis in prehospital patients with chest pain (Aufderheide TP, Reddy S, Xue Q, Dhala A, Thakur RK, Brady WJ, Rowlandson I) Computers in Cardiology 1997:665-668 : 1: Functional principal component analysis of H-reflex recruitment curves. Jul 07, 2018 · Last Updated: 07-07-2018 Principal Component Analysis (PCA) is a statistical procedure that uses an orthogonal transformation which converts a set of correlated variables to a set of uncorrelated variables. PCA is a most widely used tool in exploratory data analysis and in machine learning for predictive models.
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In this multivariate statistical approach variance in the sample is partitioned into a between-group and within- group component, in an effort to maximize discrimination between groups. In DAPC, data is first transformed using a principal components analysis (PCA) and subsequently clusters are identified using discriminant analysis (DA).
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To perform a principal components analysis, follow these steps: Select Statistics Multivariate Principal Components... Highlight all of the quantitative variables (RedMt, WhiteMt, Eggs, Milk, Fish, Cereal, Starch, Nuts, and FruVeg). Click on the Variables button. The goal of this analysis is to determine the principal components of all protein sources. This seminar will give a practical overview of both principal components analysis (PCA) and exploratory factor analysis (EFA) using SPSS. We will begin with variance partitioning and explain how it determines the use of a PCA or EFA model. For the PCA portion of the seminar, we will introduce topics such as eigenvalues and eigenvectors, communalities, sum of squared loadings, total variance explained, and choosing the number of components to extract. Principal Component Analysis: For a geometric interpretation of principal components, suppose we have two variables, X 1 and X 2, that are centered at their respective means (i.e., the means of the scores on X 1 and X 2 are zero). In the diagram below, the ellipse represents the scatter diagram of the sample points. The rst principal Use the head() function to display the first few rows of the loadings matrix.; Using just the first 3 genes, write out the equation for principal component 4. Describe how you would use the loadings matrix to find the genes that contribute most to the largest source of variation in the dataset.
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Principal Components Analysis I Principal components analysis (PCA) was introduced in 1933 by Harold Hotelling as a way to determine factors with statistical learning techniques when factors are not exogenously given. I Given a variance-covariance matrix, one can determine factors using the technique of PCA. I The concept of PCA is the following. mlpca - Maximum likelihood principal components analysis. pca - Principal components analysis. pcaengine - Principal Components Analysis computational engine. pcapro - Projects new data on old principal components model. plotloads - Extract and display loadings information from a model structure. Principal Component Analysis (PCA) is a well established and commonly used tool for multivariate analysis. PCA is based on linear transformation and decomposition of a number of correlated variables of a given data set (multidimensional data set) to a number of uncorrelated components, called Principal Components (PCs).
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http://hobartdaily.com/2019/06/add-vectors-calculator/calculate-x-y-components-vector-b-add-two-vectors-using-components-find-resulting-vector-p-q/
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# Calculate X Y Components Vector B Add Two Vectors Using Components Find Resulting Vector P Q
This post categorized under Vector and posted on June 11th, 2019.
This Calculate X Y Components Vector B Add Two Vectors Using Components Find Resulting Vector P Q has 1096 x 1334 pixel resolution with jpeg format. was related topic with this Calculate X Y Components Vector B Add Two Vectors Using Components Find Resulting Vector P Q. You can download the Calculate X Y Components Vector B Add Two Vectors Using Components Find Resulting Vector P Q picture by right click your mouse and save from your browser.
As explained above a vector is often described by a set of vector components that add up to form the given vector. Typically these components are the projections of the vector on a set of mutually perpendicular reference axes (basis vectors).Online homework and grading tools for instructors and students that reinforce student learning through practice and instant feedback.Page 1 QUICK START GUIDE fx-991EX The fx-991EX ClvectorWiz features a high-resolution display making it easier to view formulas and symbols.
Polymerase chain reaction (PCR) is a method widely used in molecular biology to make many copies of a specific DNA segment. Using PCR copies of DNA sequences are exponentially amplified to generate thousands to millions of more copies of that particular DNA segment.Indecision and delays are the parents of failure. The site contains concepts and procedures widely used in business time-dependent decision making such as time series vectorysis for forecasting and other predictive techniquesView program details for SPIE Optical Metrology conference on Optical Measurement Systems for Industrial Inspection XI
The number of points to which the data segment is padded when performing the FFT. While not increasing the actual resolution of the spectrum (the minimum distance between resolvable peaks) this can give more points in the plot allowing for more detail.
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https://meeb.us/coffee-to-water-ratio-calculator/
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# Coffee To Water Ratio Calculator, A Practical Water Guide For Coffee Professionals Part I Daily Coffee News By Roast Magazine
Coffee To Water Ratio Calculator. We are ignoring the strength box and typing 7 directly into the next line. If you reduce the ratio, for example to 1:14 or 1:13, the brew will be stronger. Online calculator to find the ratio of coffee to water for any size french press. Check out our intuitive guide to. The ratio of ground coffee to water differs greatly between brew methods and personal taste. However, you can adjust the taste and strength of the drink to suit your needs. But, there is no right way—so long as you enjoy your coffee and aren't consuming a dangerous amount! The ratios on this page are based on a mixture of consensus and official sources. Simply enter your french press size and desired brew strength to see measurements of replace the 4 in strength = 4 in the calculator below with your strength setting. The default ratio is 1:16; (or you can always just use the calculator above.) In our coffee to water calculator, we've chosen the following pour over coffee ratios we want to choose another ratio than provided in the calculator, e.g. You've got fresh, quality coffee, a decent grinder and a good brewing setup, but you're just not satisfied with the that's 62.5 grams of coffee for 1000 grams of water, a 1 to 16 ratio. 1:7 which is between regular and robust coffee. Make the perfect cup with these charts.
Coffee To Water Ratio Calculator Indeed recently is being sought by consumers around us, perhaps one of you. Individuals are now accustomed to using the net in gadgets to see image and video information for inspiration, and according to the name of the article I will talk about about Coffee To Water Ratio Calculator.
• The 2 1 Ratio Scott Rao , Most Coffee Drinks Comprise Three Common Ingredients:
• How To Make Perfect Coffee The Atlantic , An Average Value To Use Is 1.33.
• Coffee To Water Ratio Calculator How To Measure Coffee Perfectly : A Calculator For Coffee To Water Ratios.
• Coffee To Water Ratio Calculator Updated 2020 : To Set Up The Project Run:
• Coffee To Water Ratio Calculator Updated 2020 , A Calculator For Coffee To Water Ratios.
• Acaia The Minimalist Coffee Brewing Scale Acaia Coffee App : A Calculator For Coffee To Water Ratios.
• Pin On How To Brew Coffee – The Default Ratio Is 1:16;
• French Press Coffee To Water Ratio Calculator . Espresso, Steamed Milk, And Foam.
• Golden Coffee To Water Ratio Real Good Coffee Company : Making Cold Brew Coffee Usually Results In A Coffee Concentrate.
• Coffee Ratio Calculator Archives 123 Design Blog , You Can Select Gallons Or Liters.
Find, Read, And Discover Coffee To Water Ratio Calculator, Such Us:
• How Much Coffee Per Cup Use This Coffee To Water Ratio For Perfect Coffee Coffeestylish Com – The Ratio Of Ground Coffee To Water Differs Greatly Between Brew Methods And Personal Taste.
• Golden Coffee To Water Ratio Real Good Coffee Company . If You Err In The Other Direction, Using Too Few Grounds Or Too Much Water.
• Coffee Measurement Chart The Future – When Finished, The Calculator Will Show You The Exact Amount Of Coffee To Be Used (In Grams And Tablespoons) And The Precise Amount Of Water Needed (In Milliliters And Cups).
• Brewing Ratios For Espresso Beverages , Here Are A Few More Things To Keep In Mind:
• Coffee To Water Ratio Calculator How To Measure Coffee Perfectly – Simply Enter Your French Press Size And Desired Brew Strength To See Measurements Of Replace The 4 In Strength = 4 In The Calculator Below With Your Strength Setting.
• French Press Coffee To Water Ratio Calculator : Here Are A Few More Things To Keep In Mind:
• How To Make Perfect Coffee The Atlantic : Additional Toppings Can Be Added Based On Your Customers' Unique Preferences.
• Learn About Coffee To Water Ratio Use Our Calculator Guide And Charts , Feel Free To Experiment, But This Produces The Closest Thing To A Universally Acceptable Coffee Strength.
• The 2 1 Ratio Scott Rao , Just Type In Your Question To Calculate How Much, Or How Many Of One Variable, Are In Another Variable, Liquid Measurements Or Dry.
## Coffee To Water Ratio Calculator . How To Measure Extraction Of Coffee
Learn About Coffee To Water Ratio Use Our Calculator Guide And Charts. We are ignoring the strength box and typing 7 directly into the next line. If you reduce the ratio, for example to 1:14 or 1:13, the brew will be stronger. 1:7 which is between regular and robust coffee. However, you can adjust the taste and strength of the drink to suit your needs. You've got fresh, quality coffee, a decent grinder and a good brewing setup, but you're just not satisfied with the that's 62.5 grams of coffee for 1000 grams of water, a 1 to 16 ratio. Check out our intuitive guide to. The ratios on this page are based on a mixture of consensus and official sources. Simply enter your french press size and desired brew strength to see measurements of replace the 4 in strength = 4 in the calculator below with your strength setting. The default ratio is 1:16; Make the perfect cup with these charts. The ratio of ground coffee to water differs greatly between brew methods and personal taste. In our coffee to water calculator, we've chosen the following pour over coffee ratios we want to choose another ratio than provided in the calculator, e.g. But, there is no right way—so long as you enjoy your coffee and aren't consuming a dangerous amount! (or you can always just use the calculator above.) Online calculator to find the ratio of coffee to water for any size french press.
The ratio of ground coffee to water differs greatly between brew methods and personal taste. Make the perfect cup with these charts. We are ignoring the strength box and typing 7 directly into the next line. Take charge of your coffee experience with the right water to coffee ratio. Most coffee drinks comprise three common ingredients: For 1 cup coffee, use 20 grams of coffee beans (about 3 tablespoons) and 300 grams water. To set up the project run:
## This calculator allows you to choose between the standard equation or the alternate equation.
And visit localhost:8000 in the browser. 1:7 which is between regular and robust coffee. For example, in our javapresse ultimate guide, we use a 1:7 brewing ratio with 100g of coffee and 700g of water. You can select gallons or liters. Feel free to experiment, but this produces the closest thing to a universally acceptable coffee strength. Espresso, steamed milk, and foam. We are ignoring the strength box and typing 7 directly into the next line. Our simple coffee to water calculator & chart makes brewing perfect coffee easy. Use our free mortgage calculator to quickly estimate what your new home will cost. Brewing good coffee depends on using the correct amount of coffee. Uses the specialty coffee association of america's recommended ratio of 1 part coffee beans to 17 parts water. This is handy for food recipes and cooking measurements. If you err in the other direction, using too few grounds or too much water. To set up the project run: Coffee to water ratio calculator helps you. Making cold brew coffee usually results in a coffee concentrate. Some might wonder what the importance is of using a brew ratio. If you don't know the difference, just stick with the standard equation. There are some interesting exceptions to all the things we have just discussed. You've got fresh, quality coffee, a decent grinder and a good brewing setup, but you're just not satisfied with the that's 62.5 grams of coffee for 1000 grams of water, a 1 to 16 ratio. 16:1 is the our recommendation, but you can achieve stronger or more delicate brews by adjusting it up or down. 12 grams of coffee per cup is a ratio of 1:15, which is recommended by most baristas. Make the perfect cup with these charts. Take charge of your coffee experience with the right water to coffee ratio. We excluded espresso because a different method is used for calculating the espresso ratio. Our preferred ratio of water to coffee beans is 500 grams (or milliliters) of water to 30 grams of whole coffee beans. Coffeecalc is a tool to help you find the perfect brewing ratio between coffee grounds and water. Online calculator to find the ratio of coffee to water for any size french press. Why use a brew ratio at all? This calculator allows you to choose between the standard equation or the alternate equation. Most coffee drinks comprise three common ingredients:
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https://socratic.org/questions/how-do-you-simplify-2-4-div-0-192
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# How do you simplify 2.4 div 0.192?
$12.5$
$2.4 \div 0.192 \equiv \frac{24}{10} \div \frac{192}{1 , 000}$
$\frac{24}{10} \div \frac{192}{1 , 000} \equiv \frac{24}{10} \cdot \frac{1 , 000}{192} = \frac{24 , 000}{1 , 920} = \frac{2 , 400}{192} = \frac{25}{2} = 12.5$
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https://fr.mathworks.com/matlabcentral/cody/problems/44310-digit-concentration-in-champernowne-s-constant/solutions/1314519
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Cody
# Problem 44310. Digit concentration in Champernowne's constant
Solution 1314519
Submitted on 25 Oct 2017 by Alfonso Nieto-Castanon
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
d = 1; x = 1; y_correct = 1; assert(abs(digitCon(d,x)-y_correct) < 1e-4)
2 Pass
d = 10; x = 5; y_correct = 0.1000; assert(abs(digitCon(d,x)-y_correct) < 1e-4)
3 Pass
d = 10; x = 1; y_correct = 0.2000; assert(abs(digitCon(d,x)-y_correct) < 1e-4)
4 Pass
d = 20; x = 9; y_correct = 0.0500; assert(abs(digitCon(d,x)-y_correct) < 1e-4)
5 Pass
d = 50; x = 0; y_correct = 0.0400; assert(abs(digitCon(d,x)-y_correct) < 1e-4)
6 Pass
d = 50; x = 2; y_correct = 0.2600; assert(abs(digitCon(d,x)-y_correct) < 1e-4)
7 Pass
d = 1000; x = 9; y_correct = 0.0670; assert(abs(digitCon(d,x)-y_correct) < 1e-4)
8 Pass
d = 1e4; x = 8; y_correct = 0.0747; assert(abs(digitCon(d,x)-y_correct) < 1e-4)
9 Pass
d = 1e5; x = 7; y_correct = 0.0864; assert(abs(digitCon(d,x)-y_correct) < 1e-4)
10 Pass
d = 1e6; x = 6; y_correct = 0.0935; assert(abs(digitCon(d,x)-y_correct) < 1e-4)
11 Pass
d = 1e6; x = 5; y_correct = 0.0937; assert(abs(digitCon(d,x)-y_correct) < 1e-4)
12 Pass
d = 2e6; x = 4; y_correct = 0.0903; assert(abs(digitCon(d,x)-y_correct) < 1e-4)
13 Pass
d = 2000124; x = 3; y_correct = 0.1162; assert(abs(digitCon(d,x)-y_correct) < 1e-4)
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https://www.aimsuccess.in/2018/04/quadratic-equation-for-idbi-executive.html
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# Quadratic Equation for IDBI Executive 2018: 16 April
Directions (Q. 1 – 10): In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer
1) if x > y
2) if x ≥ y
3) if x < y
4) if x ≤ y
5) if x = y or relationship between x and y cannot be established
1. I. 4 x + 7 y = 42
II. 3 x – 11 y = –1
2. I. – 29 x + 22 = 0
II. – 7 y + 12 = 0
3. I. 3 – 4 x – 32 = 0
II. 2 – 17 y + 36 = 0
4. I. 3 – 19 x – 14 = 0
II. 2 + 5 y + 3 = 0
5. I. + 14 x + 49 = 0
II. + 9 y = 0
6. I. 7 – 9 x + 2 = 0
II. – 4 y + 3 = 0
7. I. x² = 64
II. 2 + 25 y + 72 = 0
8. I. x² + x – 20 = 0
II. 2 – 19 y + 45 = 0
9. I. 7 x + 3 y = 26
II. 2 x + 17 y = – 41
10. I.3 – 20 x + 33 = 0
II. 2 – 11 y + 15 = 0
1. 1
2. 3
3. 4
4. 1
5. 5
6. 4
7. 5
8. 3
9. 1
10. 2
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http://utahscience.oremjr.alpine.k12.ut.us/Sciber08/8th/forces/html/egg.htm
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crawl-data/CC-MAIN-2018-05/segments/1516084886416.17/warc/CC-MAIN-20180116105522-20180116125522-00740.warc.gz
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# Kickin' Mass!
Remember that gravity pulls on all objects in the universe; you can’t hide from gravity. Mass is your resistance to change in motion. Massive things are hard to get moving and hard to stop (this is why we don’t step out in front of moving traffic.) Less massive things are easy to get moving and easy to stop, like rolling a marble. Isaac Newton taught us that the more mass you have, the larger gravitational pull you feel and vice versa.
Galileo, a famous Italian scientist who lived in the 1500s, was the first to discover the force of gravity. In his famous experiment, he dropped two cannonballs, one 10 times the mass of the other, at exactly the same time from the Leaning Tower of Pisa. With this knowledge let us pose the question, “Does mass affect gravitational pull?” Or in other words, which is pulled harder, more massive or less massive objects?
In this activity, you are going to find out which will be pulled harder by gravity, a more massive or less massive object. This activity actually works more by air resistance; however the results are the same as if you could measure the amount of gravity pulling on the different masses.
Materials:
• Eight inch Styrofoam ball
• Four inch Styrofoam ball
• One inch Styrofoam ball
Safety concerns: As with all science lab activities, the most important safety rule is to follow all teacher directions.
Procedure:
1. In a paragraph, explain what happened as you performed this experiment and why.
2. Does the mass of an object affect its’ gravitational pull?
3. If you were to do this experiment in a vacuum (a place with no atmosphere like space), would your results change? Why?
4. Look at the two sets of planets below. On the left are two massive planets and on the right are two less massive planets. Place the large arrow between the planets with the greatest gravitational pull and the small arrow between the planets with the smallest gravitational pull. Then drag your mouse over the image to check your answer.
Extension:
Find a partner and time the fall to see what the actual difference is. Repeat the experiment several times and keep track of your data to see if you can come up with a pattern. Make a data table to keep your results and graph your final data.
Review Science safetey rules here.
Get the plug-ins: , and . (The QuickTime plug-in is needed to play sounds and movies correctly.)
Want to share photos of you or your friends doing this activity? Send it in an e-mail with the following information:
1. The title of the activity
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http://www.geeksforgeeks.org/find-last-unique-url-long-list-urls-single-traversal/
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crawl-data/CC-MAIN-2017-51/segments/1512948522999.27/warc/CC-MAIN-20171213104259-20171213124259-00630.warc.gz
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|
# Find last unique URL from long list of URLs in single traversal
Given a very long list of URLs, find out last unique URL. Only one traversal of all URLs is allowed.
Examples:
```Input:
http://www.geeksforgeeks.org
http://quiz.geeksforgeeks.org
http://qa.geeksforgeeks.org
https://practice.geeksforgeeks.org
https://ide.geeksforgeeks.org
http://www.contribute.geeksforgeeks.org
http://quiz.geeksforgeeks.org
https://practice.geeksforgeeks.org
https://ide.geeksforgeeks.org
http://quiz.geeksforgeeks.org
http://qa.geeksforgeeks.org
https://practice.geeksforgeeks.org
Output:
http://www.contribute.geeksforgeeks.org
```
## Recommended: Please try your approach on {IDE} first, before moving on to the solution.
We can solve this problem in one traversal by using Trie with a Doubly Linked List (We can insert and delete in O(1) time). The idea is to insert all URLs into the Trie one by one and check if it is duplicate or not. To know if we have previously encountered the URL, we need to mark the last node of every URL as leaf node. If we encounter a URL for the first time, we insert it into the doubly Linked list and maintain a pointer to that node in linked list in leaf node of the trie. If we encounter a URL that is already in the trie and has pointer to the url in the linked list, we delete the node from the linked list and set its pointer in the trie to null. After all URLs are processed, linked list will only contain the URLs that are distinct and the node at the beginning of the linked list will be last unique URL.
```// C++ program to print distinct URLs using Trie
#include <bits/stdc++.h>
using namespace std;
// Alphabet size (# of symbols)
const int ALPHABET_SIZE = 256;
struct DLLNode
{
string data;
DLLNode* next, * prev;
};
// trie node
struct TrieNode
{
TrieNode* children[ALPHABET_SIZE];
// isLeaf is true if the node represents
// end of a word
bool isLeaf;
DLLNode* LLptr;
};
/* Given a reference (pointer to pointer) to the
head of a list and an int, inserts a new node
on the front of the list. */
{
DLLNode* new_node = new DLLNode;
// put in the data
new_node->data = new_data;
// Make next of new node as head and previous
// as NULL
new_node->prev = NULL;
// change prev of head node to new node
// move the head to point to the new node
}
/* Function to delete a node in a Doubly Linked List.
del --> pointer to node to be deleted. */
{
// base case
if (head_ref == NULL || del == NULL)
return;
// If node to be deleted is head node
// Change next only if node to be deleted is
// NOT the last node
if (del->next != NULL)
del->next->prev = del->prev;
// Change prev only if node to be deleted is
// NOT the first node
if (del->prev != NULL)
del->prev->next = del->next;
// Finally, free the memory occupied by del
delete(del);
return;
}
// Returns new trie node (initialized to NULLs)
TrieNode* getNewTrieNode(void)
{
TrieNode* pNode = new TrieNode;
if (pNode)
{
pNode->isLeaf = false;
for (int i = 0; i < ALPHABET_SIZE; i++)
pNode->children[i] = NULL;
pNode->LLptr = NULL;
}
return pNode;
}
// If not present, inserts key into trie
// If the key is prefix of trie node, just marks leaf node
void insert(TrieNode* root, string key, DLLNode*& head)
{
int index;
TrieNode* pCrawl = root;
for (int level = 0; level < key.length(); level++)
{
index = int(key[level]);
if (!pCrawl->children[index])
pCrawl->children[index] = getNewTrieNode();
pCrawl = pCrawl->children[index];
}
if (pCrawl->isLeaf)
{
// cout << "Duplicate Found " << key << endl;
if (pCrawl->LLptr)
pCrawl->LLptr = NULL;
}
else
{
// mark last node as leaf
pCrawl->isLeaf = true;
}
}
// Driver function
int main()
{
string urls[] = {
"http://www.geeksforgeeks.org",
"http://www.contribute.geeksforgeeks.org",
"http://quiz.geeksforgeeks.org",
"http://qa.geeksforgeeks.org",
"https://practice.geeksforgeeks.org",
"https://ide.geeksforgeeks.org",
"http://quiz.geeksforgeeks.org",
"https://practice.geeksforgeeks.org",
"https://ide.geeksforgeeks.org",
"http://quiz.geeksforgeeks.org",
"http://qa.geeksforgeeks.org",
"https://practice.geeksforgeeks.org"
};
TrieNode* root = getNewTrieNode();
int n = sizeof(urls)/sizeof(urls[0]);
// Construct Trie from given URLs
for (int i = 0; i < n; i++)
// distinct URL
return 0;
}
```
Output:
```http://www.contribute.geeksforgeeks.org
```
# GATE CS Corner Company Wise Coding Practice
4.4 Average Difficulty : 4.4/5.0
Based on 5 vote(s)
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# Multiples of Eight
Contributor: Meghan Vestal. Lesson ID: 11324
Do you know how to skip-count? That doesn't mean to forget about counting. It's a way of learning times tables! Yes, there is a song about the eight-times table and games, flashcards, and quizzes too!
categories
## Arithmetic, Arithmetic
subject
Math
learning style
Auditory, Visual
personality style
Lion, Beaver
Intermediate (3-5)
Lesson Type
Quick Query
## Lesson Plan - Get It!
Audio:
Kim is buying Christmas gifts for her three best friends. She plans to spend \$8 on each friend.
• How much money does she need to take out of the bank to go shopping?
• What numerical problem did you create to solve the word problem?
If you said 3 x \$8 = \$24, you are correct!
• You have already studied the three times tables, so what times tables do you think you will study in this lesson?
Get ready to learn the eight times tables!
• Do you realize you have learned more than 100 multiplication facts?
Wow! You should be feeling pretty good!
Before you start learning another set of multiplication facts, review what you have already learned.
If you found any of those problems challenging, you may need to go back and review. Select the Related Lesson from the list in the right-hand sidebar for the times tables you still find challenging.
Review that lesson thoroughly. Remember to re-listen to the song, read over any tips and tricks, and replay any games.
When confident that you have mastered these multiplication facts, continue this lesson.
In this lesson, you will learn multiples of eight. Since you have already learned so many multiplication facts, you know how to solve most multiplication problems with eight.
To start learning multiples of eight, complete the skip-counting by eight below. If you are uncertain what the next number should be, use the dot method to find out.
To use the dot method, draw eight dots on your paper. When you are counting, put your pencil on each dot to help you count.
For example, let's say you get to 24 and do not know what 8 x 4 equals. Put your pencil on the first dot and say 25 out loud. Then, put your pencil on the next dot and say 26 out loud.
Continue this process using each dot. When you say the number at the last dot, you have found the answer to 8 x 4.
Use the image as an example.
Okay, now complete the skip-counting!
Now, look at the multiplication table below. The eight times tables have been highlighted.
• What patterns, repeated numbers, or other things do you see?
When learning the eight times tables, there are a few things you should notice.
First, multiples of eight do reveal a pattern. The ones place, or the place to the right, has a repeating pattern of 8, 6, 4, 2, 0.
• Did you make that observation?
The second thing to know about the eight-times tables is a little more challenging to notice. Any time the last digit in the product (the ones place) is 8, 4, or 0, the second-to-last digit (the tens place) is an even number.
Also, any time the last digit in the product (the ones place) is 6 or 2, the second to last digit (the tens place) is an odd number.
Look at the multiplication table again to find examples of the above observations in the highlighted row and column. Remember to add these tips to the foldable you have created throughout this series.
Continue reviewing multiples of eight with the song in the video below.
After you have listened to the song, listen again and try to sing along!
Then, move to the Got It? section to practice solving multiplication problems with eight.
Interactive Video
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# CS 4487/9587 Algorithms for Image Analysis
## Presentation on theme: "CS 4487/9587 Algorithms for Image Analysis"— Presentation transcript:
CS 4487/9587 Algorithms for Image Analysis
Image Modalities most slides are shamelessly stolen from Steven Seitz, Aleosha Efros, and Terry Peters
CS 4487/9587 Algorithms for Image Analysis Image Modalities
Photo/Video data Pin-hole Lenses Digital images and volumes Medical Images and Volumes X-ray, MRI, CT, and Ultrasound Extra Reading: Forsyth & Ponce, Ch. 1. Gonzalez & Woods, Ch. 1
Slide by Steve Seitz How do we see the world?
Let’s design a camera Idea 1: put a piece of film in front of an object Do we get a reasonable image?
Slide by Steve Seitz Pinhole camera
Add a barrier to block off most of the rays This reduces blurring The opening known as the aperture How does this transform the image? It gets inverted
Slide by Steve Seitz Camera Obscura
The first camera Known to Aristotle Depth of the room is the focal length Pencil of rays – all rays through a point Can we measure distances?
Figure by David Forsyth Distant objects are smaller
Slide by Aleosha Efros Camera Obscura
Drawing from “The Great Art of Light and Shadow “ Jesuit Athanasius Kircher, 1646. How does the aperture size affect the image?
Slide by Steve Seitz Shrinking the aperture
Why not make the aperture as small as possible? Less light gets through Diffraction effects…
Slide by Steve Seitz Shrinking the aperture
Slide by Aleosha Efros Home-made pinhole camera
Slide by Steve Seitz The reason for lenses
Slide by Steve Seitz Adding a lens
“circle of confusion” A lens focuses light onto the film There is a specific distance at which objects are “in focus” other points project to a “circle of confusion” in the image Changing the shape of the lens changes this distance
Slide by Aleosha Efros The eye
The human eye is a camera! Iris - colored annulus with radial muscles Pupil - the hole (aperture) whose size is controlled by the iris What’s the “film”? photoreceptor cells (rods and cones) in the retina
Slide by Aleosha Efros Cameras
Really cool Not too expensive nowadays (<\$200) Canon A70
Basic camera model “equivalent” image plane f image plane lens f
commonly used simplified representation of a photo camera
Basic camera model “virtual” image plane camera’s “optical center” f commonly used simplified representation of a photo camera
Figure by Gonzalez & Woods Digital Image Formation
f(x,y) = reflectance(x,y) * illumination(x,y) Reflectance in [0,1], illumination in [0,inf]
Figure by Gonzalez & Woods Sampling and Quantization
Figure by Gonzalez & Woods Sampling and Quantization
Slide by Aleosha Efros What is an image?
We can think of an image as a function, f, from R2 to R: f( x, y ) gives the intensity at position ( x, y ) Realistically, we expect the image only to be defined over a rectangle, with a finite range: f : [a,b]x[c,d] [0,1] A color image is just three functions pasted together. We can write this as a “vector-valued” function: As opposed to [0..255]
Slide by Aleosha Efros Images as functions
Render with scanalyze????
Slide by Aleosha Efros What is a digital image?
We usually operate on digital (discrete) images: Sample the 2D space on a regular grid Quantize each sample (round to nearest integer) If our samples are D apart, we can write this as: f[i ,j] = Quantize{ f(i D, j D) } The image can now be represented as a matrix of integer values
3D Image Volumes? Video - 3D = X*Y*Time
Medical volumetric data (MRI, CT) D = X*Y*Z k Combine multiple images (slices) into a volume
CS 4487/9587 Algorithms for Image Analysis Image Modalities
PART II: Medical images and volumes X-ray CT MRI Ultrasound
Slides from Terry Peters In the beginning…..X-rays
Discovered in 1895 Mainstay of medical imaging till 1970’s 1971 – Computed Tomography (CAT, CT) scanning Digital Radiography ……… 1980 Magnetic Resonance Imaging
X-rays Wilhelm Conrad Röntgen ( ) Nobel Prise in Physics, 1901 “X” stands for “unknown” X-ray imaging is also known as - radiograph - Röntgen imaging
X-rays Calcium in bones absorbs X-rays the most
Bertha Röntgen’s Hand 8 Nov, 1895 A modern radiograph of a hand Calcium in bones absorbs X-rays the most Fat and other soft tissues absorb less, and look gray Air absorbs the least, so lungs look black on a radiograph
X-rays 2D “projection” imaging ’s
From Projection Imaging Towards True 3D Imaging
X-ray imaging 1895 Mathematical results: Radon transformation 1917 Development of CT (computed tomography) 1972 Image reconstruction from projection Also known as CAT (Computerized Axial Tomography) "tomos" means "slice" (Greek) Computers can perform complex mathematics to reconstruct and process images Late 1960’s:
Radon Transformation Mathematical transformation (related to Fourier)
Reconstruction of the shape of object (distribution f(x,y)) from the multitude of 2D projections
Figure from www.imaginis.com/ct-scan/how_ct.asp CT imaging
CT imaging, inventing (1972)
Sir Godfrey Hounsfield Engineer for EMI PLC Nobel Prize 1979 (with Alan Cormack)
CT imaging, availability (since 1975)
Axial CT image of a normal brain using a state-of-the-art CT system and a 512 x 512 matrix image. Note the two black "pea-shaped" ventricles in the middle of the brain and the subtle delineation of gray and white matter (Courtesy: Siemens) 25 years later 1974 Original axial CT image from the dedicated Siretom CT scanner circa 1975. This image is a coarse 128 x 128 matrix; however, in 1975 physicians were fascinated by the ability to see the soft tissue structures of the brain, including the black ventricles for the first time (enlarged in this patient) The EMI-Scanner
Slides from Terry Peters Clinical Acceptance of CT!?
Dr James Ambrose 1972 Radiologist, Atkinson - Morley’s Hospital London Recognised potential of EMI-scanner “Pretty pictures, but they will never replace radiographs” –Neuroradiologist 1972
Slides from Terry Peters Then ……………and Now
512 x 512 image <1mm slice thickness <0.5mm pixels 0.5 sec rotation 0.5 sec recon per slice Isotropic resolution Spiral scanning - up to 16 slices simultaneously 80 x 80 image 3 mm pixels 13 mm thick slices Two simultaneous slices!!! 80 sec scan time per slice 80 sec recon time
Slides from Terry Peters 30 Years of CT
Slides from Terry Peters Birth of MRI
Paul Lautebur 1975 Presented at Stanford CT meeting “Zeugmatography” Raymond Damadian 1977 – Sir Peter Mansfield early 1980’s Early Thorax Image Nottingham
Slides from Terry Peters Birth of MRI
Early Thorax Image Nottingham Electro Marnetic signal emitted (in harmless radio frequensy) is acquired in the time domain image has to be reconstructed (Fourier transform)
Birth of MRI In 1978, Mansfield presented his first image through the abdomen. Lauterbur and the first magnetic resonance images (from Nature)
Slides from Terry Peters 30 Years of MRI
First brain MR image Typical T2-weighted MR image
Slides from Terry Peters MR Imaging
“Interesting images, but will never be as useful as CT” (A different) neuroradiologist, 1982
Slides from Terry Peters MR Imaging …more than T1 and T2
MRA - Magnetic resonance angiography images of vessels MRS - Magnetic resonance spectroscopy images of chemistry of the brain and muscle metabolism fMRI - functional magnetic resonance imaging image of brain function PW MRI – Perfusion-weighted imaging DW MRI – Diffusion-weighted MRI images of nerve pathways
Slides from Terry Peters Magnetic Resonance Angiography
MR scanner tuned to measure only moving structures “Sees” only blood - no static structure Generate 3-D image of vasculature system May be enhanced with contrast agent (e.g. Gd-DTPA)
Slides from Terry Peters MR Angiography
GD-enhanced Phase-contrast GD-enhanced In-flow
Slides from Terry Peters Dynamic 3-D MRI of the thorax
1
Slides from Terry Peters Diffusion-Weighted MRI
Image diffuse fluid motion in brain Construct “Tensor image” – extent of diffusion in each direction in each voxel in image Diffusion along nerve sheaths defines nerve tracts. Create images of nerve connections/pathways
Slides from Terry Peters Tractography
Data analysed after scanning Identify “streamlines” of vectors Connect to form fibre tracts 14 min scan time Internal Capsule - Dr. D Jones, NIH
Slides from Terry Peters Tractography
Wernicke’s area Broca’s area Short fibres Long fibres Insula fibres Temporal fibres “just like Gray’s Anatomy”! Superior Longitudinal Fasciculus - Dr. D Jones, NIH USA
Slides from Terry Peters Functional MRI (fMRI)
Active brain regions demand more fuel (oxygen) Extra oxygen in blood changes MRI signal Activate brain regions with specific tasks Oxygenated blood generates small (~1%) signal change Correlate signal intensity change with task Represent changes on anatomical images
Slides from Terry Peters fMRI
Subject looks at flashing disk while being scanned “Activated” sites detected and merged with 3-D MR image Stimulus Activation
fMRI in Neurosurgery Planning
Slides from Terry Peters fMRI in Neurosurgery Planning Hand Activation Tumour Face Activation
Slides from Terry Peters Ultrasound
Images courtesy GE Medical
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test
# NVLD - Where does the 15 point limit come from? (Rewritten)
Page 1 of 3 [ 39 posts ] Go to page 1, 2, 3 Next
mrsmith
Yellow-bellied Woodpecker
Joined: 30 Jun 2011
Age:49
Posts: 63
27 Nov 2012, 9:43 am
NVLD is often defined as VIQ-PIQ beeing over 15 points.
Where does this limit come from? (1)
Is it a practical limit for classroom NVLD, or is it more like a tipping point for a wide range of problems?
Is it a lower limit (2a), with perhaps some additional criteria needing to be satisfied, or is it more like a tipping point, where the population size corresponds to the population with a higher difference, but some below does have it and some above does not have it?(2b)
Is PIQ (rather that NVIQ for example), used because it is most important in the classrom, or because it is more predictive of wider problems.
If 2b : I am aware that more specific problems can lead do diagnosis below the limit. Is this factored into the limit?
Confusing first version
NVLD is often defined as the PIQ beeing in the lower 2.5% percentile of the population compared to the VIQ.
Where does this limit come from?
Is it a practical limit for classroom NVLD, or is it more like a tipping point for a wide range of problems?
(Like is the difference between 2.3 and 2.7 quite big compared to the difference between 2.7 and 3.1?)
Valentine
Last edited by mrsmith on 29 Nov 2012, 9:22 am, edited 10 times in total.
1000Knives
Veteran
Joined: 8 Jul 2011
Age:24
Posts: 5,083
Location: CT, USA
27 Nov 2012, 10:34 am
I've not heard those limits. I've heard it's more described by PIQ to VIQ gap. I've heard a 15 point gap between your VIQ and PIQ is enough to satisfy diagnostic criteria. In my case, my gap is around 50 points. My PIQ is like 80, and my VIQ is like 130+. So it's rather severe NVLD, I guess. But I think officially, it starts at 15 point gap.
mrsmith
Yellow-bellied Woodpecker
Joined: 30 Jun 2011
Age:49
Posts: 63
27 Nov 2012, 11:44 am
1000Knives wrote:
I've not heard those limits. I've heard it's more described by PIQ to VIQ gap. I've heard a 15 point gap
It comes from 2 standard deviations, which I think corresponds to a certain point difference the way the IQ scale is defined.
(I think 15points is 1 stdev - corresponding to the 15% percentile).
I think it is better to think of it in terms of percentiles - You can compare it to the 1% who has ASD or the numbers for ADHD.
Ravenclawgurl
Veteran
Joined: 19 Jun 2007
Age:25
Posts: 1,317
Location: somewhere over the rainbow
27 Nov 2012, 11:45 am
but what about in the weschler children scale 4 which doesnt have performance and verbal iq its divided into 4 parts?
OddDuckNash99
Veteran
Joined: 15 Nov 2006
Posts: 2,700
27 Nov 2012, 1:08 pm
The definition of NVLD as a neuropsychiatric condition was first described by Dr. Bryan Rourke. Statistically speaking, though, yes, the VIQ/PIQ split comes from standard deviation for the WAIS. 15 points is 1 standard deviation, and a 15-point difference between VIQ and PIQ (where VIQ is larger) is the baseline definition for NVLD. The "four parts" Ravenclawgurl mentions are subscores, from which VIQ, PIQ, and FSIQ (VIQ + PIQ average) are determined. On the WAIS (I don't know much about the WISC), Vocabulary and Working Memory are used to determine VIQ, and Perceptual Organization and Processing Speed are used to determine PIQ.
Also, in order to qualify for an NVLD diagnosis, you needn't have a PIQ that is in the lowest 2%, which is -2 standard deviations from the mean of 100. (Thus, a PIQ of 85 or below.) You just need to have a VIQ/PIQ split that is a statistically significant difference (i.e., 15+ points). For example, I have NVLD, but my PIQ is 116, which is in the "high-average" range. My VIQ is 134 ("very superior" range), which is 17 points higher than my PIQ. Even though my PIQ is +1 SD from the mean of 100, I still have NVLD due to the incongruence between my PIQ and VIQ.
For NVLD, the subscores sometimes are an even better determinant of the learning disorder. Working Memory, used for VIQ, involves Digit Span and Arithmetic sections, both of which have numbers. Plus, Vocabulary is an entirely verbal/language-based score, and Perceptual Organization is an entirely visual-spatial score. I feel my NVLD staggered abilities are best shown by these two subscores: My Vocabulary score is 140 (almost +2 SD above mean), and my Perceptual Organization score is 105 ("average" range). That's a 35-point difference, compared to just 17 points for VIQ vs. PIQ.
_________________
Helinger: Now, what do you see, John?
Nash: Recognition...
Helinger: Well, try seeing accomplishment!
Nash: Is there a difference?
Ravenclawgurl
Veteran
Joined: 19 Jun 2007
Age:25
Posts: 1,317
Location: somewhere over the rainbow
27 Nov 2012, 2:12 pm
OddDuckNash99 wrote:
The definition of NVLD as a neuropsychiatric condition was first described by Dr. Bryan Rourke. Statistically speaking, though, yes, the VIQ/PIQ split comes from standard deviation for the WAIS. 15 points is 1 standard deviation, and a 15-point difference between VIQ and PIQ (where VIQ is larger) is the baseline definition for NVLD. The "four parts" Ravenclawgurl mentions are subscores, from which VIQ, PIQ, and FSIQ (VIQ + PIQ average) are determined. On the WAIS (I don't know much about the WISC), Vocabulary and Working Memory are used to determine VIQ, and Perceptual Organization and Processing Speed are used to determine PIQ.
well if thats thats how they determine performance iq with the processing speed then i likely have nvld because my processing speed score was so low it was in the MR range
mrsmith
Yellow-bellied Woodpecker
Joined: 30 Jun 2011
Age:49
Posts: 63
27 Nov 2012, 2:27 pm
OddDuckNash99 wrote:
The definition of NVLD as a neuropsychiatric condition was first described by Dr. Bryan Rourke. Statistically speaking, though, yes, the VIQ/PIQ split comes from standard deviation for the WAIS. 15 points is 1 standard deviation, and a 15-point difference between VIQ and PIQ (where VIQ is larger) is the baseline definition for NVLD.
I have Googled a bit, and I generally see the 1 stdev limit
That would correspond to 15% of the population, which much really be far to high, so it must be some additional filtering going on?
The only prevalence numbers I have seen are 2-3%, which corresponds to 2 stdevs.
OddDuckNash99
Veteran
Joined: 15 Nov 2006
Posts: 2,700
27 Nov 2012, 3:29 pm
mrsmith wrote:
That would correspond to 15% of the population, which much really be far to high, so it must be some additional filtering going on?
The only prevalence numbers I have seen are 2-3%, which corresponds to 2 stdevs.
No, 15 points for 1 SD doesn't mean 15% of the population. The way that a standard normal curve works is that ~68% of people lie within +/- 1 SD of the mean (in this case, 100). And 96% of people lie within +/- 2 SDs of the mean. So, only 2% of the population have an IQ of 130 or above (+2 SDs), and only 2% of the population have an IQ below 85 (-2 SDs).
It also isn't true that having greater than or equal to 1 SD between your VIQ and PIQ scores is equivalent to having a PIQ score that is greater than or equal to 1 SD below the mean. As I showed above, my PIQ is 1 SD above the mean, but I still have NVLD, due to its relationship to my VIQ score. Now, I'm sure that most NVLD-ers out there do have a PIQ in the 80-90 range; I realize that I am not a typical case. (I am fortunate enough to have taught myself certain visual-spatial skills over the years.) But again, it is possible to have NVLD with any level PIQ score, so long as it is 15+ points lower than the VIQ score.
As far as prevalence rates of NVLD, what I commonly read is that NVLD makes up around 1-10% of all total learning disorders. It's not all that common, even in the learning disorder category, but again, that statistic doesn't refer to specific IQ scores.
And in response to Ravenclawgurl, Symbol Coding/Processing Speed is only one component of PIQ. Block Design, Matrices, and other tests also are factored in for PIQ. Having low Processing Speed doesn't necessarily mean NVLD. I scored extremely highly in Processing Speed, yet I have NVLD.
_________________
Helinger: Now, what do you see, John?
Nash: Recognition...
Helinger: Well, try seeing accomplishment!
Nash: Is there a difference?
Ravenclawgurl
Veteran
Joined: 19 Jun 2007
Age:25
Posts: 1,317
Location: somewhere over the rainbow
27 Nov 2012, 3:42 pm
yeah but what i mean im sure the processing speed numbers bring the rest of the performance iq down
mrsmith
Yellow-bellied Woodpecker
Joined: 30 Jun 2011
Age:49
Posts: 63
27 Nov 2012, 3:56 pm
OddDuckNash99 wrote:
No, 15 points for 1 SD doesn't mean 15% of the population. The way that a standard normal curve works is that ~68% of people lie within +/- 1 SD.
I didn't say that. If 68.2% is within, 31.8 % is outside, or 15.9% is on each side.
OddDuckNash99
Veteran
Joined: 15 Nov 2006
Posts: 2,700
27 Nov 2012, 4:03 pm
mrsmith wrote:
I didn't say that. If 68.2% is within, 31.8 % is outside, or 15.9% is on each side.
Okay, I see what you meant now. But it still doesn't mean that people who qualify as having NVLD will necessarily have a PIQ score that is in that 15% below 100.
_________________
Helinger: Now, what do you see, John?
Nash: Recognition...
Helinger: Well, try seeing accomplishment!
Nash: Is there a difference?
mrsmith
Yellow-bellied Woodpecker
Joined: 30 Jun 2011
Age:49
Posts: 63
27 Nov 2012, 4:28 pm
OddDuckNash99 wrote:
mrsmith wrote:
I didn't say that. If 68.2% is within, 31.8 % is outside, or 15.9% is on each side.
Okay, I see what you meant now. But it still doesn't mean that people who qualify as having NVLD will necessarily have a PIQ score that is in that 15% below 100.
If scores are above 100% you can't say that you have a LD or that you are impaired from the scores alone.
(But when I wrote OP, I though the limit was sharper, and that it was more significant)
btbnnyr
Veteran
Joined: 18 May 2011
Posts: 6,442
Location: Lost Angleles Carmen Santiago
27 Nov 2012, 4:57 pm
What if someone has a 140 PIQ and 155 VIQ? This person still has NVLD, a learning disorder?
OddDuckNash99
Veteran
Joined: 15 Nov 2006
Posts: 2,700
28 Nov 2012, 8:21 am
btbnnyr wrote:
What if someone has a 140 PIQ and 155 VIQ? This person still has NVLD, a learning disorder?
By definition, yes. No matter how high or how low a person's IQ, there should not be more than a 10-point difference between VIQ and PIQ. Obviously, the higher the IQ scores, the less a person will struggle, but they still will experience learning difficulties compared to their potential. The fictional case you described, for instance, probably wouldn't experience many struggles, but their PIQ would still be limited in reference to what their PIQ abilties SHOULD be.
Somebody with these high of scores probably wouldn't ever need to go get tested for a learning disorder due to compensatory strategies, but they still would have staggered abilities. Again, using myself as an example, I never was tested for a learning disorder as a child. I only took an IQ test and was diagnosed with NVLD in college. Because my PIQ abilties are higher, I was able to "get by" in math classes using compensatory strategies, such as my "photographic" memory. But even though I was able to be in honors math courses and get "A"s for the most part, math always has been a huge struggle for me. And teachers never understood why somebody as seemingly "smart" as I struggled so much with math. So, it's sort of a double-edged sword. If you have a higher PIQ to begin with, yes, you are fortunate enough to (most likely) achieve greater academic success for longer, but the downside is that your very real struggles are much more puzzling to teachers and people in general. It's a lot like how there are obvious advantages to being high-functioning on the spectrum, but there also are disadvantages (it's a "hidden" disability).
_________________
Helinger: Now, what do you see, John?
Nash: Recognition...
Helinger: Well, try seeing accomplishment!
Nash: Is there a difference?
mrsmith
Yellow-bellied Woodpecker
Joined: 30 Jun 2011
Age:49
Posts: 63
28 Nov 2012, 8:43 am
OddDuckNash99 wrote:
btbnnyr wrote:
What if someone has a 140 PIQ and 155 VIQ? This person still has NVLD, a learning disorder?
By definition, yes..
If the point criteria is met by 15% of the population, it should mean that only 1/3 of those that "have NVLD".
Depending on the diagnostic criteria you can imagine it makes some difference in the chance that you pass.
One other post on this forum says it should then rather be called a "learning style" than "learning disability",
but AFIAK the term NVLD is used even if all scores are high.
Another thing is that the point difference perhaps should be different at a high IQ.
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http://bleacherreport.com/articles/1025427-rafael-nadal-why-rafa-is-destined-to-dethrone-novak-djokovic
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# Rafael Nadal: Why Rafa Is Destined to Dethrone Novak Djokovic
Maly ThomasContributor IIJanuary 15, 2012
Novak Djokovic Winner of US OPEN 2011Chris Trotman/Getty Images
Who is the best person to dethrone the King? The one right below the King. At ATP the king is called Novak Djokovic, and the "below king" is named Rafael Nadal.
In 2011, Novak Djokovic won everything—or almost everything—but enough to become King. Now in 2012, to stay the only master on board, Novak Djokovic will have to duplicate his magnificent season, or hope for Nadal not to win anything. Neither case seems possible. That is the reason why Rafa should get back to his right place: No. 1 position.
*Lots of Points to defend for Djokovic.
Knowing that Djokovic has won three Grand Slams (Australian Open, Wimbledon and US Open—each time earning the maximum points: 2,000). In case of a defeat, he will at least lose 800 points, and that's for a defeat in the final. More points will be lost if the Serb loses before the final.
To this, add his Masters Series wins (Indian Wells, Miami, Rome, Madrid, and Montreal) where Djokovic, each time, won 1,000 points. In 2012, the Serb could now lose 400 points if he doesn’t win in the final.
It’s mathematical: Djokovic can lose a lot of points, very quickly.
*And if...Nadal earns Djoko's lost points.
More then Djokovic losing points, Nadal could earn points. In fact, the worst case for Djokovic would be if he loses to Nadal.
To make a long story short: If Djokovic does not repeat his season, and Nadal comes back to winning, then Djokovic not only loses points but Nadal can reclaim the No. 1 ranking from Djokovic.
As of January 16th, kickoff of the Australian Open 2012, only 4035 separate the King Djokovic, World No.1 (13,630 points) and Rafael Nadal (9,595 points).
Even if we don’t imagine the Spaniard dethroning the Serb so soon, it could happen very quickly.
Wanting to grow up too quickly, you can burn the wings, and Djoko’s conquer of the ATP Tour may well be ephemeral.
# Related
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# [AstroPy] Rotating and Transforming Vectors--Flight Path of a Celestial Body
Wayne Watson sierra_mtnview@sbcglobal....
Thu Dec 17 19:33:52 CST 2009
```I'm just getting used to the math and numpy library, and have begun
working on a problem of the following sort.
Two observing stations are equidistant, 1/2 degree, on either side of a
line of longitude of 90 deg west, and both are at 45 deg latitude. Given
the earth's circumference as 25020 miles, a meteor is 60 miles above the
point between the two sites. That is, if you were standing at long
90deg and lat 45 deg, the meteor would be that high above you. 70 miles
along the long line is 1 degree, so the stations are 70 miles apart. I
want to know the az and el of the meteor from each station. With some
geometry and trig, I've managed to get that first point; however, I can
see moving the meteor say, 1/2 deg, along its circular path towards the
north pole is going to require more pen and pencil work to get the az/el
for it.
Long ago in a faraway time, I used to do this stuff. It should be easy
to rotate the vector to the first point 1/2 deg northward, and find the
vector there, then compute the new az and el from each station. Maybe.
I'm just beginning to look at the matrix and vector facilities in numpy.
Maybe someone can comment on how this should be done, and steer me
towards what I need to know in numpy.
--
(121.015 Deg. W, 39.262 Deg. N) GMT-8 hr std. time)
Obz Site: 39° 15' 7" N, 121° 2' 32" W, 2700 feet
"... humans'innate skills with numbers isn't much
better than that of rats and dolphins."
-- Stanislas Dehaene, neurosurgeon
Web Page: <www.speckledwithstars.net/>
```
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Cody
# Problem 44880. Angle between two vectors
Solution 1816938
Submitted on 16 May 2019 by Binbin Qi
• Size: 24
• This is the leading solution.
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = [1 1;-1 -1]; y_correct = 180; assert(isequal(angle(x),y_correct))
2 Pass
x = [-1 1;-1 -1]; y_correct = 90; assert(isequal(angle(x),y_correct))
3 Pass
x = [0.5 sqrt(3)/2;0.2 0]; y_correct = 60; assert(isequal(angle(x),y_correct))
4 Pass
x = [-1 1;0.5 sqrt(3)/2]; y_correct = 75; assert(isequal(angle(x),y_correct))
5 Pass
x = [0 1;0 5]; y_correct = 0; assert(isequal(angle(x),y_correct))
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A272489 Decimal expansion of the edge length of a regular 11-gon with unit circumradius. 8
5, 6, 3, 4, 6, 5, 1, 1, 3, 6, 8, 2, 8, 5, 9, 3, 9, 5, 4, 2, 2, 8, 3, 5, 8, 3, 0, 6, 9, 3, 2, 3, 3, 7, 9, 8, 0, 7, 1, 5, 5, 5, 7, 9, 7, 9, 4, 6, 5, 3, 3, 7, 4, 3, 6, 6, 2, 1, 6, 0, 6, 1, 2, 1, 7, 5, 6, 9, 7, 5, 9, 7, 0, 3, 8, 0, 5, 8, 3, 3, 6, 2, 4, 6, 9, 3, 5, 2, 3, 6, 9, 0, 3, 7, 7, 3, 0, 9, 9, 9, 3, 5, 9, 8, 8 (list; constant; graph; refs; listen; history; text; internal format)
OFFSET 0,1 COMMENTS The edge length e(m) of a regular m-gon is e(m) = 2*sin(Pi/m). In this case, m = 11, and the constant, a = e(11), is not constructible using a compass and a straightedge (see A004169). With an odd m, in fact, e(m) would be constructible only if m were a Fermat prime (A019434). LINKS Stanislav Sykora, Table of n, a(n) for n = 0..2000 Wikipedia, Constructible number Wikipedia, Regular polygon FORMULA Equals 2*sin(Pi/11) = 2*cos(Pi*9/22). EXAMPLE 0.5634651136828593954228358306932337980715557979465337436621606121... MATHEMATICA RealDigits[N[2Sin[Pi/11], 100]][[1]] (* Robert Price, May 01 2016 *) PROG (PARI) 2*sin(Pi/11) CROSSREFS Cf. A004169, A019434. Edge lengths of nonconstructible n-gons: A271487 (n=7), A272488 (n=9), A272490 (n=13), A255241 (n=14), A130880 (n=18), A272491 (n=19). Sequence in context: A079267 A060296 A114598 * A259500 A274082 A199666 Adjacent sequences: A272486 A272487 A272488 * A272490 A272491 A272492 KEYWORD nonn,cons,easy AUTHOR Stanislav Sykora, May 01 2016 STATUS approved
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Last modified December 17 12:51 EST 2018. Contains 318201 sequences. (Running on oeis4.)
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# Program to find kth lexicographic sequence from 1 to n of size k Python
Suppose we have two values n and k. Now consider a list of numbers in range 1 through n [1, 2, ..., n] and generating every permutation of this list in lexicographic sequence. For example, if n = 4 we have [1234, 1243, 1324, 1342, 1423, 1432, 2134, 2143, 2314, 2341, 2413, 2431, 3124, 3142, 3214, 3241, 3412, 3421, 4123, 4132, 4213, 4231, 4312, 4321]. We have to find the kth value of this permutation sequence as a string.
So, if the input is like n = 4 k = 5, then the output will be "1432"
To solve this, we will follow these steps −
• Define a function factors() . This will take num
• quo := num
• res := a double ended queue and insert 0 at beginning
• i := 2
• while quo is not empty, do
• quo := quotient of (quo / i), rem := quo mod i
• insert rem at the left of res
• i := i + 1
• return res
• From the main method do the following −
• numbers := a list with values 1 through n
• res := blank string
• k_fact := factors(k)
• while size of k_fact < size of numbers, do
• res := res concatenate first element of numbers as string, then delete first element of numbers
• for each index in k_fact, do
• number := index−th element of numbers, then remove that element
• res := res concatenate number
• return res
Let us see the following implementation to get better understanding −
## Example
Live Demo
from collections import deque
def factors(num):
quo = num
res = deque([0])
i = 2
while quo:
quo, rem = divmod(quo, i)
res.appendleft(rem)
i += 1
return res
class Solution:
def solve(self, n, k):
numbers = [num for num in range(1, n + 1)]
res = ""
k_fact = factors(k)
while len(k_fact) < len(numbers):
res += str(numbers.pop(0))
for index in k_fact:
number = numbers.pop(index)
res += str(number)
return res
ob = Solution()
n = 4
k = 5
print(ob.solve(n, k))
## Input
4, 5
## Output
1432
Updated on: 26-Dec-2020
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test2 - Math 3124 Tuesday November 1 Second Test Answer All...
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Unformatted text preview: Math 3124 Tuesday, November 1 Second Test. Answer All Problems. Please Give Explanations For Your Answers. 1. In Z 24 , prove that [ 4 ] , [ 6 ] = [ 10 ] . (10 points) 2. The following is part of the Cayley table for a group. Complete the table. a b c d a b b c b d (i.e. given a 2 = cd = b .) (10 points) 3. Let G be a group of order 25 and let A , B be subgroups of G with | A | = | B | . Prove that either A = B or A ∩ B = { e } . (10 points) 4. Let G be a cyclic group of order 72 and let H be a subgroup of order 2. Determine the number of subgroups of G containing H (including G and H ). 5. Prove that S 5 S 4 × Z 5 . (10 points) 6. Let G = Z 12 and let H = [ 3 ] . (a) Write down the right cosets of H in G . (b) Construct the Cayley table for G / H . (10 points) Math 3124 Tuesday, November 1 Second Test Solutions 1. Since [ 10 ] = [ 4 ] ⊕ [ 6 ] we see that [ 10 ] ⊆ [ 4 ] , [ 6 ] . Now [ 4 ] = [- 10 ] ⊕ [- 10 ] and [ 6 ] = [ 10 ] ⊕ [ 10 ] ⊕ [ 10 ] and we deduce that [ 4 ] , [...
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# 10.5.5. Multiplicative Inverse of a Matrix
Given that , and , determine .
• A
• B
• C
• D
### Example
Given that , and , determine .
### Solution
The key result to use here is that = .
Applying the matrix on the right of each side of the equation gives Since , this simplifies to
can therefore be found as the product of the two matrices and :
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# Find the residue of $\frac{e^{iht}}{(t-i)^2}$
I am trying to find the residue of
$\frac{e^{iht}}{(t-i)^2}$
At $t = i$, $h$ is a constant.
First of all, how do I detect which way is faster? Using the residue formula or expanding the function into Laurent series?
Then how would you do it?
Thanks
$f(t)=e^{iht}$ is analytic everywhere then the Laurent series at $z=i$ is the Taylor series and $f^{(n)}(i)=(ih)^ne^{-h}$
Then, if $|t-i| < \infty$
$$e^{iht}= \sum_{n=0}^ \infty \frac{(ih)^ne^{-h} (t-i)^n}{n!} = \frac{1}{e^h}+\frac{ih (t-i)}{e^h} +\ldots+$$
Then, if $0<|t-i|<\infty$
$\displaystyle \frac{e^{iht}}{(t-i)^2}= \frac{1}{e^h(t-i)^2}+\frac{ih}{e^h(t-i)}+ \ldots+$
Then the residue is $\displaystyle \frac{ih}{e^h}$
In general, $$\operatorname{Res}(f;a)=\frac{1}{(n-1)!}\lim_{z\to a} \frac{d^{n-1}}{dz^{n-1}}((x-a)^nf(z)),$$ where $n$ is the order of the pole $a$. Thus, $$\operatorname{Res}\left(\frac{e^{iht}}{(t-i)^2};i\right)=\lim_{t\to i} \frac{d}{dz}e^{iht}=ihe^{iht}\mid_{t=i}=ihe^{-h}.$$
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# Confidence Intervals, Proportion Estimations
I am stuck on one of my homework questions for my stats class. I was wondering if anyone could give me some insight as to how to find an answer. here is the question:
In a study of perception, 107 men are tested and 24 are found to have red/green color blindness.
(a) Find a 92% confidence interval for the true proportion of men from the sampled population that have this type of color blindness. (b) Using the results from the above mentioned survey, how many men should be sampled to estimate the true proportion of men with this type of color blindness to within 2% with 98% confidence? (c) If no previous estimate of the sample proportion is available, how large of a sample should be used in (b)?
I have already answered (a). However, I am at a complete loss as to how to answer (b) or (c). For one thing, I am not quite sure what (b) is even asking. Advice on this question would be greatly appreciated. Thanks!
Without giving all the answers, I think the following outline will be helpful.
Let $X$ be the number of color-blind men in a sample of $n.$ Then $X \sim \mathsf{Binom}(n, \theta),$ where $\theta$ is the population proportion of color-bindness. The MLE and MME of $\theta$ is $\hat \theta = X/n.$ In the study mentioned $\hat \theta = 24/107 = 0.2243.$
(a) A traditional 92% CI for $\theta$ is of the form $$\hat \theta \pm 1.74\sqrt{\frac{\hat \theta(1-\hat\theta)}{n}},$$ where 1.74 cuts 4% of the area from the upper tail of the standard normal distribution. In this formula, the '(estimated) standard error' is $\sqrt{\frac{\hat \theta(1-\hat\theta)}{n}}$ and the 92% 'margin of error' is $1.75\sqrt{\frac{\hat \theta(1-\hat\theta)}{n}}.$
(b) I will leave it to you to find the 98% margin of error $M$ and then let $M = .02$ and solve for $n$ with $\hat \theta = 0.2243.$
(c) Because $\theta(1-\theta)$ is at its maximum when $\theta = 1/2,$ use that value in $M$ when solving for $n.$
Notes:(1) When finding $n$ as in (b) and (c) you will not generally get an integer value. In that case, it is customary to round up to the nearest integer value.
(2) The displayed formula in (a) depends on two approximations: (i) the normal approximation to the binomial and (ii) the approximation of $\hat \theta$ to $\theta$ in the standard error. Especially for $n$ as small as $n = 107$ the latter approximation is questionable. Other forms of CIs are more accurate, including those due to Wilson and Agresti, which you can explore in statistics texts and on the Internet. My guess is that you are expected to use the formula I provided.
• Thanks for your answer! Just for clarification purposes, what do MLE and MME stand for? – ImRellyBadAtMath Oct 26 '17 at 16:34
• actually never mind, i solved it. Thank you for your help! It was greatly appreciated – ImRellyBadAtMath Oct 26 '17 at 17:25
• Maximum Likelihood Estimation, Method of Moments estimation – BruceET Oct 26 '17 at 22:43
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## Saturday, 14 December 2013
### Patterns in squares in Year 4
We been having a look at patterns of squares using Cuisenaire rods in Year 4, starting by just making any design featuring as square.
At first it wasn't as simple as that might sound. We were getting a lot of almost-squares. Look, for instance, at this one, created on the Cuisenaire Environment -
At first it fooled a lot of us! It's got at ten on each side...
A lot of interesting patterns emerged worth following up.
It would be interesting to think of mathematical question about these shapes. With the faces for instance, how does the perimeter progress in each of the two kinds of pattern? What area of table is visible in each?
Pascal, Paloma, Parsang and Perry Perimeter
(Also published on the Year 4 blog.)
## Sunday, 24 November 2013
### Lego Maths with Y6
Unusually this autumn has seen more than its fair share of wet playtimes. During these sessions, many children have been attracted to the box of Lego bricks. Watching them build and play with the bricks, they instinctively know that if they are building with 8-stud bricks, and they run out, then two 4-stud bricks will do, or a 6-stud and a 2-stud, or four 2-studs. Arrays are everywhere. They are doing maths all the time without realising.
Wanting to take advantage of this enthusiasm, I decided to investigate the numerical properties of the bricks with my Year 6 class to see what they came up with. I asked them to look for numbers and (for today) discouraged building. I encouraged recording anything they found out.
They found out evidence of factors, number bonds, arrays, divisibility, sequencing, square numbers and experienced lots of data handling and discussing the numerical properties of their bricks.
One student found out that 100 was not divisible by 8, whilst another was intent on finding patterns of square numbers using 4-stud bricks (numbers of studs and numbers of bricks).
Another group were keenly trying to find the relationship between the number of studs on top of a brick with 2 rows of studs and the holes underneath.
They could see that the number of holes was one less than one 'line' of studs. Once they organised their results and checked their ideas, it was only a small step for them to realise how to explain using maths vocabulary and then with a bit of help created a formula.
This sparked ideas in others who wanted to try and find the 'rule' if the flatter base (rather than a brick) had rows of 4 studs instead of 2. Although we ran out of time, the discussion went into the next lesson as one student was determined to explain to me how to get the answer (correctly!).
He came up with the mathematical rule and formula overnight.
Can anyone else find it too?
## Friday, 11 October 2013
### This year's Forest of Factors
Following on from last year's post Important Factors, and after our work finding prime numbers with Cuisenaire rods and in the playground, Year 4 had a go at making prime factor trees this year. Here's our Forest of Factors:
See more on the Year 4 blog
Our trees made a really big Forest of Factors display for the primary corridor. As the Year 2 and Year 3 children are passing it regularly, it was a good idea to explain the display to them. We did it without using the words "factor", "multiple" or "prime" so that it would be easier for them to understand:
"Hello everyone. This is our Forest of Factors."
"1, 2, 3, 4.... 40"
"What do you notice about our pictures?"
"Number 1 is my bird. It's different to all the other numbers. If you times something by 1 it just stays the same number."
"So 1 is special, and we don't think about it with our trees and mushrooms. It's just flying up in the sky!"
"Some of the numbers are simple trees. Look at 38. 38 is 19 times 2, so we've given it two branches that finish with flowers."
"Ten is five times two."
"26 is 13 times 2."
"These ones had just two branches and two flowers."
"But some of the trees have more flowers. Like 12, and 32."
"With the mushrooms there's no two numbers that you can times together to make that number."
"Can you see a number that is a mushroom?"
"So that's our Forest of Factors. We hope you understand it a bit better now."
## Wednesday, 2 October 2013
### Finding prime numbers in Year 4
In Year 4 we've been looking at prime numbers this week.
Some of us have been out in the playground, using the hundred square there:
Here's a how we did it:
Later we had a go in the classroom, using a photo of the hundred square (16 colour bitmap format):
and, on Paint, filling he numbers in the 5, 3, 7 and 2 times table with colour:
And there we have it: the grey ones (except for 1) are the prime numbers!
(also blogged on the Year 4 blog)
## Monday, 16 September 2013
### How big is a million?
We've been talking about the dinosaurs in Year 4, which has involved us in some pretty big numbers. For instance, the last dinosaurs died out 65 million years ago.
Have a look on the Year 4 blog to see exactly how big a million is, and how it could be dangerous in the classroom.
## Tuesday, 2 July 2013
### Mirror Dancing
For a number of years we have run this annual mirror dance activity at the International School of Toulouse, but this year we decided to make a bit more of an event of it! I first got the idea from a Workshop run by Anne Watson at the UK based ATM conference a few years ago!
The idea is simple! In groups of two or more, students must make a dance routine lasting between 60 and 90 seconds. The routine must be based on symmetry. in the first instance, reflective symmetry where students are each others reflection either side of an imaginary mirror line. Afterwards, students can consider using two (or more) mirror lines, before going on to consider translation, rotation and enlargement! Students were judged and given a score out of 10 in three different categories!
1. Aesthetic appeal
2. Technical difficulty
3. Symmetrical accuracy
The total scores were then divided by the year group the team came from as a bit of a leveller!
More important than that was amazing creativity shown by all of the students involved. It is wonderful to see how students will play with these mathematical ideas when the right circumstances are created and we are really impressed with and proud of the work our students did on this. We are also particularly pleased with the willingness they have shown to both take part and then perform in front of each other! Well done to all students who took part!
Students have chosen their own prize and that is that they are able to throw wet sponges at the maths teachers!
See also 'Maths and Feet' from Simon Gregg based activities from http://www.mathinyourfeet.com/ from Malke Rosenfeld
Here are some photos for now and there will be some videos to follow......
And now following the comment below, evidence of the prize giving!!!
## Saturday, 29 June 2013
### Ancient Greek Geometry
Year 5G had a look at a great game, where the object is to create shapes with the classical "straight edge and compass" techniques: http://sciencevsmagic.net/geo/ It's author is Nico Disseldorp.
• It's a game! How this geometry should be. Maybe how it was for the ancient Greeks before someone wrote it all down and it had to be "learnt".
• It constrains you. You can only put your lines and circles in certain places. You have to follow the rules.
• It starts easy and gets harder, and records your progress.
We had a go with physical rulers and pairs of compasses first. We created all sorts of precise diagrams, and deviated off to some really beautiful ones that some of the children wanted to finish off at home.
by Amandine
But, with the kind permission of the people who'd done some of the pictures that weren't quite right (and it's OK to make mistakes in this classroom) we looked at how they were wrong. After all, these are not uncommon mistakes, and understanding them helps us understand something about knowledge itself. Here's one attempt to make a regular hexagon (it's in pencil so it didn't scan very clearly):
We could see here that the problem was that although the bottom points had been arrived at by finding the precise place where the circles cross, the top ones were, well, guessed at. You might call this kind of guess an informed opinion; we wanted something closer to a fact.
This one used the circles to draw all the points of the hexagon:
This is a lot closer to regular, but suffers from a lack of precision in the drawing.
So, armed with these reflections, we all managed - sometimes it took several goes, and tuition from those who got there first - we all managed to draw a good regular hexagon.
Then we went on to the Ancient Greek Geometry game / puzzle. Here we were helped to get over the problems of drawing by hand by being forced to be precise and to define points with known lines and circles.
One of the interesting things was that ten year olds can be quicker than their teacher! Jose showed Mr Gregg how to do the square, and Sophie showed him a quicker way of doing circle pack three!
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Assignment 1 stat | Mathematics homework help
Assignment 1: Bottling Union Case Study Due Week 10 and desert 140 points Imagine you are a supervisor at a main bottling union. Customers accept begun to sorrow that the bottles of the infamy of soda performed in your union comprise short than the advertised thirty two (32) ounces of emanation. Your boss wants to unfold the bearing at artisan and has asked you to brave. You accept your employees drag thirty (30) bottles off the row at purposehither from all the shifts at the bottling set. You ask your employees to appraise the quantity of soda there is in each bottle. Note: Use the postulates set supposing by your preceptor to accomplished this assignment.
Bottle Number
Ounces
Bottle Number
Ounces
Bottle Number
Ounces
1
30.23
11
31.77
21
32.23
2
30.32
12
31.80
22
32.25
3
30.98
13
31.82
23
32.31
4
31.00
14
31.87
24
32.32
5
31.11
15
31.98
25
32.34
6
31.21
16
32.00
26
32.46
7
31.42
17
32.02
27
32.47
8
31.47
18
32.05
28
32.51
9
31.65
19
32.21
29
32.91
10
31.74
20
32.21
30
32.96
Write a two to three (2-3) page announce in which you: 1. Calculate the average, median, and flag discontinuance for ounces in the bottles. 2. Construct a 95% Confidence Interval for the ounces in the bottles. 3. Conduct a supposition ordeal to realize if the demand that a bottle comprises short than thirty two (32) ounces is protected. Clearly say the logic of your ordeal, the calculations, and the misrecord of your ordeal. 4. Produce the aftercited argument domiciled on the misrecord of your ordeal:
a. If you terminate that there are short than thirty two (32) ounces in a bottle of soda, meditate on three (3) feasible causes. Next, hint the strategies to desert the deficit in the coming. Or
b. If you terminate that the demand of short soda per bottle is not protected or justified, produce a minute explication to your boss encircling the footing. Include your thought on the conclude(s) subsequently the demand, and advise one (1) strategy geared toward salubrious this result in the coming. Your assignment must flourish these formatting requirements: • Be typed, enfold spaced, using Times New Roman font (dimension 12), after a while one-inch margins on all sides. No citations and references are required,
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