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Effect 22 Four Thought
You produce a stack of forty or so business cards and show them to the audience. The cards have various instructions to think of a random number, a color, a popular meal, famous actor, politician, country etc.
Gathering up the cards, you invite another spectator to take part in the experiment. You ask him to verify that each card has instructions to think of different things. He agrees that they do.
Placing the cards face down on the table, you ask him to cut the cards anywhere he likes and take the card he cuts to. The other cards are discarded.
The spectator then concentrates on the four things listed on the card. You then correctly name the four things he was thinking of.
This effect, based on a principle known as dual reality, takes guts to perform but the results are worth it. Dual reality means that the spectator's experience of the effect differs from that of the audience but both experiences are still valid.
As well as taking guts, this effect also requires a fair amount of preparation - in particular making up the special pack of cards with which to perform the trick.
These can be hand written or printed with a computer. Choose whatever method best suits your performance style.
The stack is made up of three different types of cards: two "spectator" cards (shown to the spectators), twenty one "volunteer" cards (shown to your volunteer) and twenty "force" cards (ultimately, the card chosen by your volunteer).
The two spectator cards look like this:
For the "volunteer" cards, make 10 of one and 11 of the other of the following two cards:
Think of the actor Al Pacino Think of the colour yellow Think of the number 192 Think of lasagna and chips
Think of the colour green Think of the composer Beethoven Think of the country France Think of the number 752
And all twenty "force" cards look like this:
Using a sharp craft knife or guillotine, trim about a 1mm wide strip off the bottom (longest) edge of each "force" card.
Then stack them face up in the order below (from bottom to top):
Force card, volunteer card, force card, volunteer card, force card, volunteer card...
and so on, with the top two cards ending up as "volunteer" cards of two different types. The two "spectator" cards then go on top, completing the stack.
To perform the effect, select your volunteer and have him take a seat on stage. Pick up the stack and approach the audience, explaining that you have a special deck of cards that instruct someone to think of particular things like numbers, celebrities, colors and so on.
Show the stack face up to the spectators so they can see the top card. Make sure plenty of spectators get to see the card and as you do so, pick off the top card so they can also see the second card. Turn the deck towards you and remove the second card also, letting everyone see it.
As you walk back to your volunteer, place the two "spectator" cards on the BOTTOM of the deck.
Walk up to your volunteer and show him (but do not hand to him) the rest of the deck. Pick off the top card (now one of the "volunteer" cards) and say
"You didn't get a chance to see. The cards have instructions for you to think of a famous actor, a color, a number ... things like that."
Keeping the second, different "volunteer" card on top of the deck, show it to him saying
" ... and on this card, a famous composer, a country, you get the idea. Forty cards, each asking you to think of different things."
Here you have set up the dual reality: your description of the cards satisfies both the volunteer and the spectators. The spectators think all the other cards are like the ones they saw, the volunteer thinks the spectators saw cards like the ones he can see. There is no reason to think otherwise.
Square up the cards and place them face down in front of your volunteer. Ask him to cut the deck anywhere he likes and remove the card he cuts to and place it face down on the table. Gather up the other cards and place them in your pocket.
Because you have trimmed all of the "force" cards, your volunteer will automatically cut to one of them.
Now turn your back and ask the volunteer to look at his card and think of the things it tells him to.
Continuing the dual reality, the spectators think the volunteer is going to think of a random number, country, composer etc himself. The volunteer understands that he is to think of the things written on his card and because all the cards are different, you are to read his mind and discover which card he chose.
To conclude the effect, say "OK I'm getting it - you're thinking of the color red, the country is Sweden, the composer is Mozart and ... a very popular meal indeed - steak and chips! Correct?"
And of course, you are correct!
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# 7.7: Finding the Equation of a Line
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## The Slope-Intercept and Point-Slope Forms
In the previous sections we have been given an equation and have constructed the line to which it corresponds. Now, however, suppose we're given some geometric information about the line and we wish to construct the corresponding equation. We wish to find the equation of a line.
We know that the formula for the slope of a line is $$m - \dfrac{y_2 - y_1}{x_2-x_1}$$. We can find the equation of a line using the slope formula in either of two ways:
##### 1
If we’re given the slope, $$m$$, and any point $$(x_1, y_1)$$ on the line, we can substitute this information into the formula for slope.
Let $$(x_1, y_1)$$ be the known point on the line and let $$(x,y)$$ be any other point on the line. Then
\begin{aligned} m&=\dfrac{y-y_1}{x-x_1}&\text{ Multiply both sides by } x - x_1\\ m(x-x_1)&=(x-x_1) \cdot \dfrac{y-y_1}{x-x_1}\\ m(x-x_1)&=y-y_1&\text{ For convenience, we'll rewrite the equation. }\\ y-y_1&=m(x-x_1) \end{aligned}
Since this equation was derived using a point and the slope of a line, it is called the point-slope form of a line.
##### 2
If we are given the slope, $$m$$, $$y$$-intercept, $$(0,b)$$, we can substitute this information into the formula for slope.
Let $$(0,b)$$ be the $$y$$-intercept and (x,y) be any other point on the line. Then,
\begin{aligned} m&=\dfrac{y-b}{x-0}\\ m&=\dfrac{y-b}{x}&\text{ Multiply both sides by } x\\ m \cdot x&=\not{x} \cdot \dfrac{y-b}{\not{x}}\\ mx&=y-b&\text{ Solve for } y\\ mx+b&=y&\text{ For convenience, we'll rewrite this equation}\\ y&=mx+b \end{aligned}
Since this equation was derived using the slope and the intercept, it was called the slope-intercept form of a line.
We summarize these two derivations as follows.
##### Forms of the Equation of a Line
We can find the equation of a line if we’re given either of the following sets of information:
The slope, $$m$$, and the $$y$$-intercept, $$(0, b)$$, by substituting these values into:
$$y = mx + b$$
This is the slope-intercept form.
The slope, $$m$$, and any point $$(x_1, y_1)$$, by substituting these values into
$$y-y_1 = m(x-x_1)$$
This is the point-slope form.
Notice that both forms rely on knowing the slope. If we are given two points on the line we may still find the equation of the line passing through them by first finding the slope of the line, then using the point-slope form.
It is customary to use either the slope-intercept form or the general form for the final form of the line. We will use the slope-intercept form as the final form.
## Sample Set A
Find the equation of the line using the given information.
##### Example $$\PageIndex{1}$$
$$m = 6$$, $$y$$-intercept $$(0, 4)$$
Since we're given the slope and the $$y$$-intercept, we'll use the slope-intercept form. $$m = 6, b = 4$$.
$$y = mx + b$$
$$y = 6x + 4$$
##### Example $$\PageIndex{2}$$
$$m = -\dfrac{3}{4}$$, $$y$$-intercept $$(0, \dfrac{1}{8})$$
Since we're given the slope and the $$y$$-intercept, we'll use the slope-intercept form. $$m = \dfrac{-3}{4}$$,
$$b = \dfrac{1}{8}$$
$$y=mx + b$$
$$y = -\dfrac{3}{4}x + \dfrac{1}{8}$$
##### Example $$\PageIndex{3}$$
$$m = 2$$, the point $$(4, 3)$$.
Since we're given the slope and some point, we'll use the point-slope form.
$$y - y_1 = m(x - x_1)$$. Let $$(x_1, y_1)$$ be $$(4,3)$$.
$$y-3=2(x-4)$$ Put this equation in slope-intercept form by solving for $$y$$.
$$y-3=2x-8$$
$$y = 2x-5$$
##### Example $$\PageIndex{4}$$
$$m = -5$$, the point $$(-3, 0)$$
Write the equation in slope-intercept form.
Since we're given the slope and some point, we'll use the point-slope form.
\begin{aligned} y-y_1&=m(x-x_1)&\text{ Let } (x_1,y_1) \text{ be } (-3,0)\\ y-0&=-5[x-(-3)]\\ y&=-5(x+3)&\text{ Solve for } y\\ y&=-5x-15 \end{aligned}
##### Example $$\PageIndex{5}$$
$$m = -1$$, the point $$(0, 7)$$.
Write the equation in slope-intercept form.
We're given the slope and a point, but careful observation reveals that this point is actually the $$y$$-intercept. Thus, we'll use the slope-intercept form. If we had not seen this point was the $$y$$-intercept we would have proceeded with the point-slope form. This would created slightly more work, but still give the same result.
Slope-Intercept Form:
\begin{aligned} y&=mx + b\\ y&= -1x + 7\\ y&= -x + 7 \end{aligned}
Point-Slope Form:
\begin{aligned} y-y_1&=m(x-x_1)\\ y-7&=-1(x-0)\\ y-7&=-x\\ y&=-x+7 \end{aligned}
##### Example $$\PageIndex{6}$$
The two points $$(4, 1)$$ and $$(3, 5)$$.
Write the equation in slope-intercept form.
Since we're given two points, we'll find the slope first.
$$m = \dfrac{y_2-y_1}{x_2-x_1} = \dfrac{5-1}{3-4} = \dfrac{4}{-1} = -4$$
Now, we have the slope and two points, we can use either point and the point-slope form.
Using $$(4, 1)$$
\begin{aligned} y-y_1&=m(x-x_1)\\ y-1&=-4(x-4)\\ y&=-4x+17 \end{aligned}
Using $$(3, 5)$$
\begin{aligned} y-y_1&=m(x-x_1)\\ y-5&=-4(x-3)\\ y-5&=-4x + 12\\ y&=-4x+17 \end{aligned}
We can see that the use of either gives the same result.
## Practice Set A
Find the equation of each line given the following information. Use the slope-intercept form as the final form of the equation.
##### Practice Problem $$\PageIndex{1}$$
$$m=5$$, $$y$$-intercept $$(0,8)$$.
$$y=5x+8$$
##### Practice Problem $$\PageIndex{2}$$
$$m=-8$$, $$y$$-intercept $$(0,3)$$.
$$y=−8x+3$$
##### Practice Problem $$\PageIndex{3}$$
$$m=2$$, $$y$$-intercept $$(0,-7)$$.
$$y=2x−7$$
##### Practice Problem $$\PageIndex{4}$$
$$m=1$$, $$y$$-intercept $$(0,-1)$$.
$$y=x−1$$
##### Practice Problem $$\PageIndex{5}$$
$$m=-1$$, $$y$$-intercept $$(0,-10)$$.
$$y=−x−10$$
##### Practice Problem $$\PageIndex{6}$$
$$m=4$$,the point $$(5,2)$$.
$$y=4x−18$$
##### Practice Problem $$\PageIndex{7}$$
$$m=-6$$,the point $$(-1,0)$$.
$$y=−6x−6$$
##### Practice Problem $$\PageIndex{8}$$
$$m=-1$$,the point $$(-5,-5)$$.
$$y=−x−10$$
##### Practice Problem $$\PageIndex{9}$$
The two points $$(4, 1)$$ and $$(6, 5)$$
$$y=2x−7$$
##### Practice Problem $$\PageIndex{10}$$
The two points $$(−7,−1)$$ and $$(−4,8)$$.
$$y=3x+20$$
## Sample Set B
##### Example $$\PageIndex{7}$$
Find the equation of the line passing through the point $$(4, -7)$$ having slope $$0$$.
We're given the slope and some point, so we'll use the point-slope form. With $$m = 0$$ and $$x_1, y_1)$$ as $$(4, -7)$$, we have:
\begin{aligned} y-y_1 &= m(x-x_1)\\ y-(-7)&=0(x-4)\\ y+7&=0\\ y&=-7 \end{aligned}
This is a horizontal line
##### Example $$\PageIndex{8}$$
Find the equation of the line passing through the point $$(1,3)$$ given that the line is vertical.
Since the line is vertical, the slope does not exist. Thus, we cannot use either the slope-intercept form or the point-slope form. We must recall what we know about vertical lines. The equation of this line is simply $$x=1$$.
## Practice Set B
##### Practice Problem $$\PageIndex{11}$$
Find the equation of the line passing throuhg the point $$(2, 9)$$ having slope $$0$$.
$$y=9$$
##### Practice Problem $$\PageIndex{12}$$
Find the equation of the line passing through the point $$(−1,6)$$ given that the line is vertical.
$$x=−1$$
## Sample Set C
##### Example $$\PageIndex{1}$$
Reading only from the graph, determine the equation of the line.
The slope of the line is $$\dfrac{2}{3}$$, and the line crosses the $$y$$-axis at the point $$(0, -3)$$. Using the slope-intercept form we get:
$$y = \dfrac{2}{3}x - 3$$
## Practice Set C
##### Practice Problem $$\PageIndex{13}$$
Reading only from the graph, determine the equation of the line.
$$y = \dfrac{-2}{3}x + 4$$
## Exercises
For the following problems, write the equation of the line using the given information in slope-intercept form.
##### Exercise $$\PageIndex{1}$$
$$m=3$$, $$y$$-intercept $$(0,4)$$
$$y=3x+4$$
##### Exercise $$\PageIndex{2}$$
$$m=2$$, $$y$$-intercept $$(0,5)$$
##### Exercise $$\PageIndex{3}$$
$$m=8$$, $$y$$-intercept $$(0,1)$$
$$y=8x+1$$
##### Exercise $$\PageIndex{4}$$
$$m=5$$, $$y$$-intercept $$(0,-3)$$
##### Exercise $$\PageIndex{5}$$
$$m=-6$$, $$y$$-intercept $$(0,-1)$$
$$y=−6x−1$$
##### Exercise $$\PageIndex{6}$$
$$m=-4$$, $$y$$-intercept $$(0,0)$$
##### Exercise $$\PageIndex{7}$$
$$m=-\dfrac{3}{2}$$, $$y$$-intercept $$(0,0)$$
$$y = -\dfrac{3}{2}x$$
##### Exercise $$\PageIndex{8}$$
$$m=3, (1,4)$$
##### Exercise $$\PageIndex{9}$$
$$m=1, (3,8)$$
$$y=x+5$$
##### Exercise $$\PageIndex{10}$$
$$m=2, (1,4)$$
##### Exercise $$\PageIndex{11}$$
$$m=8, (4,0)$$
$$y=8x−32$$
##### Exercise $$\PageIndex{12}$$
$$m=−3, (3,0)$$
##### Exercise $$\PageIndex{13}$$
$$m=−1, (6,0)$$
$$y=−x+6$$
##### Exercise $$\PageIndex{14}$$
$$m=−6, (0,0)$$
##### Exercise $$\PageIndex{15}$$
$$m=−2, (0,1)$$
$$y=−2x+1$$
##### Exercise $$\PageIndex{16}$$
$$(0,0), (3,2)$$
##### Exercise $$\PageIndex{17}$$
$$(0,0), (5,8)$$
$$y = \dfrac{8}{5}x$$
##### Exercise $$\PageIndex{18}$$
$$(4,1), (6,3)$$
##### Exercise $$\PageIndex{19}$$
$$(2,5), (1,4)$$
$$y=x+3$$
##### Exercise $$\PageIndex{20}$$
$$(5,−3), (6,2)$$
##### Exercise $$\PageIndex{21}$$
$$(2,3), (5,3)$$
$$y=3$$ (horizontal line)
##### Exercise $$\PageIndex{22}$$
$$(−1,5), (4,5)$$
##### Exercise $$\PageIndex{23}$$
$$(4,1), (4,2)$$
$$x=4$$ (vertical line)
##### Exercise $$\PageIndex{24}$$
$$(2,7), (2,8)$$
##### Exercise $$\PageIndex{25}$$
$$(3,3), (5,5)$$
$$y=x$$
##### Exercise $$\PageIndex{26}$$
$$(0,0), (1,1)$$
##### Exercise $$\PageIndex{27}$$
$$(−2,4), (3,−5)$$
$$y = -\dfrac{9}{5}x + \dfrac{2}{5}$$
##### Exercise $$\PageIndex{28}$$
$$(1,6), (−1,−6)$$
##### Exercise $$\PageIndex{29}$$
$$(14,12), (−9,−11)$$
$$y=x−2$$
##### Exercise $$\PageIndex{30}$$
$$(0,−4), (5,0)$$
For the following problems, read only from the graph and determine the equation of the lines.
##### Exercise $$\PageIndex{31}$$
$$y = \dfrac{2}{5}x + 1$$
##### Exercise $$\PageIndex{33}$$
$$y = \dfrac{1}{4}x + 1$$
##### Exercise $$\PageIndex{35}$$
$$x=−4$$
##### Exercise $$\PageIndex{37}$$
$$y=−3x−1$$
## Exercises for Review
##### Exercise $$\PageIndex{38}$$
Graph the equation $$x-3 = 0$$
##### Exercise $$\PageIndex{39}$$
Supply the missing word. The point at which a line crosses the $$y$$-axis is called the ____
$$y$$-intercept
##### Exercise $$\PageIndex{40}$$
Supply the missing word. The ____ of a line is a measure of the steepness of the line.
##### Exercise $$\PageIndex{41}$$
Find the slope of the line that passes through the points $$(4,0)$$ and $$(−2,−6)$$.
$$m=1$$
##### Exercise $$\PageIndex{42}$$
Graph the equation $$3y = 2x + 3$$
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0
# How many moles are in 3.3 moles of K2S?
Updated: 4/28/2022
Wiki User
11y ago
3.3 moles of K2S
3.3 moles of S-2
6.6 moles of K+1
Wiki User
11y ago
Anonymous
Lvl 1
3y ago
G
Still have questions?
Related questions
### What is the concentration of in 0.15 of M of K2S?
If you have 2 moles of K+ for every mole of K2S and Molarity (M) is Moles per Liter. Then you know that you have .30 M of K2S. The way that you do that is setting up a series of conversion factors like so:(.15moles k2s/liter) x (2 moles of K/ 1 mole of K2S) = .30 moles k/ liter.The moles of K2S cancel out and you are left with moles of K per liter.
### What is the concentration of K in 0.15M of K2S?
If you have 2 moles of K+ for every mole of K2S and Molarity (M) is Moles per Liter. Then you know that you have .30 M of K2S. The way that you do that is setting up a series of conversion factors like so:(.15moles k2s/liter) x (2 moles of K/ 1 mole of K2S) = .30 moles k/ liter.The moles of K2S cancel out and you are left with moles of K per liter.
### How many moles of aluminum are produced from 33 grams of aluminum?
33 grams aluminum (1 mole Al/26.98 grams) = 1.2 moles of aluminum ================
### Calculate the mass in 3.3 moles of potassium sulphide?
The formula mass of potassium sulfide, K2S is 2(39.1)+32.1=110.3 1 mole of potassium sulfide has a mass of 110.3g Mass of 3.3mol of K2S = 3.3 x 110.3 = 364g
### How many moles are contained in 4.67 L sample of gas 33 degrees Celsius and 199 kpa?
how many moles are contained in 4.67 L sample of gas at 33 degrees celcius and 199 kpa
### What is the formula for potassium sulfide?
Formula for potassium sulfide is K2S.
### What is the solution for K2S plus HCl?
K2S + 2HCl --> 2KCl + H2S
### Is k2s an ionic or covalent bond?
K2S or potassium sulfide is an ionic compound.
### What is the correct formula for the ionic compound formed between potassium and sulfur?
Potassium has many sulfide: K2S, K2S4, K2S2, K2S6, K2S3, K2S5. The most known is K2S.
3
### Which one of these have a high boiling point CH3Cl or K2S?
K2S will have high boiling point than CH3Cl.
yes
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# 113_1_midterm_A_sol - EE113 Digital Signal Processing...
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EE113: Digital Signal Processing Prof. Mihaela van der Schaar Midterm Exam Solution(Version A) Spring 2009 1. (a) The modes are the roots of the equatoin λ 2 - 5 λ + 6 = 0, which are λ 1 = 2 and λ 2 = 3. (b) The general homogeneous solution is C 1 2 n + C 2 3 n . The initial conditions are h (0) = 0 and h (1) = 2. We have C 1 + C 2 = 0 2 C 1 + 3 C 2 = 2 and the solutions are C 1 = - 2 and C 2 = 2. The impulse response is h ( n ) = [ - 2(2) n + 2(3) n ] u ( n ). (c) The step response can be computed by u ( n ) * h ( n ) = X k = -∞ u ( n - k )[ - 2(2) k + 2(3) k ] u ( k ) = n X k =0 [ - 2(2) k + 2(3) k ] , n 0 0 , n < 0 = 3 n +1 - 2(2) n +1 + 1 , n 0 0 , n < 0 We can also write it as [3 n +1 - 2(2) n +1 + 1] u ( n ). (d) We can get the step response without using convolution. Let the step response be w ( n ), then we first solve the particular solution, which is w p ( n ) = Ku ( n ). Substituting w p ( n ) = Ku ( n ) into the difference equation, we get Ku ( n ) - 5 Ku ( n - 1) + 6 Ku ( n - 2) = 2 u ( n - 1) For n 2, we have K = 1 and hence w p ( n ) = u ( n ). The homogeneous solution is w h ( n ) = A 1 (2) n + A 2 (3) n , hence the complete step response is w ( n ) = w p ( n )+ w h ( n )) = A 1 (2) n + A 2 (3) n
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113_1_midterm_A_sol - EE113 Digital Signal Processing...
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# Again?
Geometry Level 3
I an isosceles triangle $$ABC$$ with base $$BC$$ the angle at $$A$$ equals $$80^\circ$$. A point $$M$$ is chosen inside the triangle so that angle $$MBC=30 ^\circ$$ and $$MCB=10 ^\circ$$. Find angle $$AMC$$.
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# Structure functions for mesons
I was researching about structure functions and understand how it's a probability density function that describes the distribution of quarks inside hadrons. However, since mesons also have quark gluon interactions, I struggle to see why there can't be a structure function for mesons. Yet when I try to search and find it, only structure functions for hadronic currents come up. Therefore I just want to ask:
Can mesons have structure functions and if not, how could it be possible to describe the distribution and behaviour of quarks inside a meson?
Experimental knowledge of the partonic structure of mesons (the pion and kaon) is very limited due to the lack of a stable pion target! Our current knowledge of the pion structure function in the valence region is obtained primarily from pionic Drell-Yan scattering, and in the pion sea region at low Bjorken-x, from hard diffractive processes measured on e − p collisions at HERA. These data seem to indicate that the pion sea has approximately one-third of the magnitude of the proton sea, while from the parton model one expects the pion sea to be two-thirds of the proton sea. But... at large Bjorken-x virtually nothing is known about the contribution of sea quarks and gluons.
The Electron-Ion Collider (EIC) with an acceptance optimized for forward physics has the potential for accessing pion and kaon structure functions over a large kinematic region through the Sullivan process, where one measures the contribution to the electron Deep Inelastic Scattering (DIS) of the meson cloud of a proton target: the nucleon parton distributions contain a component which can be attributed to the meson cloud.
• Sullivan processes. In these examples, a nucleon’s pion cloud is used to provide access to the pion’s (a) elastic form factor and (b) parton distribution functions. $$t = (k − k′)^2$$ is a Mandelstam variable and the intermediate pion, $$π^∗(P = k − k′)$$, $$P^2 = t$$, is off-shell.
Three informative refs are:
1. Aitkenhead, et al. "Determination of the pion and kaon structure functions." Physical Review Letters 45 (1980) 157 .
2. Collins, & Martin, "Hadron reaction mechanisms." Reports on Progress in Physics 45 (1982) 335. Section 6.1.
3. Aguilar et al, "Pion and kaon structure at the electron-ion collider" European Physical Journal A55 (2019) 1-15.
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# isomorphic proofs
• Nov 23rd 2006, 10:45 AM
PvtBillPilgrim
isomorphic proofs
I need some help with these,
Prove that Z2 x Z3 is isomorphic with Z6.
Prove that Z6 is not isomorphic with S3, although both groups have 6 elements.
Thank you for any assistance.
• Nov 23rd 2006, 10:57 AM
ThePerfectHacker
Quote:
Originally Posted by PvtBillPilgrim
I need some help with these,
Prove that Z2 x Z3 is isomorphic with Z6.
Are you familar with the rule that,
$\mathbb{Z}_n\times \mathbb{Z}_m\simeq \mathbb{Z}_{nm}$ ( $n,m\geq 1$)
If and only if,
$\gcd (n,m)=1$.
$\gcd(2,3)=1$
Now, $\mathbb{Z}_6$ is a cyclic group. The group, $\mathbb{Z}_2\times \mathbb{Z}_3$ is also cyclic because the element $(1,1)$ is a generator, that is, $<(1,1)>=\mathbb{Z}_2\times \mathbb{Z}_3$. Now by the property of cyclic groups of equal cardinality states they are unique up to isomorphism. Q.E.D.
Note that $\mathbb{Z}_6$ is an abelian group. Yet $S_3$ is not (look below and at my other post).
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#### Read Exam Window text version
`Chapter 10MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) A gas at a pressure of 10.0 Pa exerts a force of __________ N on an area of 5.5 m 2. A) 1.8 B) 0.55 C) 5.5 D) 55 E) 18 2) E) 1.8 × 103 3) E) 760.0 4) 1)2) A gas at a pressure of 325 torr exerts a force of __________ N on an area of 5.5 m 2. A) 0.018 B) 2.4 C) 2.4 × 105 D) 593) A pressure of 1.00 atm is the same as a pressure of __________ of mmHg. A) 193 B) 33.0 C) 29.92 D) 1014) The National Weather Service routinely supplies atmospheric pressure data to help pilots set their altimeters. The units the NWS uses for atmospheric pressure are inches of mercury. A barometric pressure of 30.51 inches of mercury corresponds to __________ kPa. A) 16.01 B) 775 C) 103.3 D) 77.50 E) 1.0205) A closed-end manometer was attached to a vessel containing argon. The difference in the mercury levels in the two arms of the manometer was 12.2 cm. Atmospheric pressure was 783 mmHg. The pressure of the argon in the container was __________ mmHg. A) 795 B) 122 C) 882 D) 661 E) 7715)6) A gas vessel is attached to an open-end manometer containing a nonvolatile liquid of density 0.791 g/mL as shown below.6)The difference in heights of the liquid in the two sides of the manometer is 43.4 cm when the atmospheric pressure is 755 mmHg. Given that the density of mercury is 13.6 g/mL, the pressure of the enclosed gas is __________ atm. A) 0.993 B) 1.03 C) 0.987 D) 0.960 E) 0.9901 7) A gas vessel is attached to an open-end manometer filled with a nonvolatile liquid of density 0.993 g/mL as shown below.7)The difference in heights of the liquid in the two sides of the manometer is 32.3 cm when the atmospheric pressure is 765 mmHg. Given that the density of mercury is 13.6 g/mL, the pressure of the enclosed gas is __________ atm. A) 0.993 B) 1.04 C) 1.08 D) 1.01 E) 0.976 8)8) In a Torricelli barometer, a pressure of one atmosphere supports a 760 mm column of mercury. If the original tube containing the mercury is replaced with a tube having twice the diameter of the original, the height of the mercury column at one atmosphere pressure is __________ mm. A) 4.78 × 103 B) 760 C) 121 D) 380 E) 1.52 × 1039) A sample of gas (24.2 g) initially at 4.00 atm was compressed from 8.00 L to 2.00 L at constant temperature. After the compression, the gas pressure was __________ atm. A) 1.00 B) 16.0 C) 2.00 D) 4.00 E) 8.009)10) A sample of a gas (5.0 mol) at 1.0 atm is expanded at constant temperature from 10 L to 15 L. The final pressure is __________ atm. A) 7.5 B) 3.3 C) 15 D) 1.5 E) 0.6710)11) A balloon originally had a volume of 4.39 L at 44 °C and a pressure of 729 torr. The balloon must be cooled to __________°C to reduce its volume to 3.78 L (at constant pressure). A) 38 B) 0 C) 273 D) 546 E) 72.911)12) If 3.21 mol of a gas occupies 56.2 L at 44 °C and 793 torr, 5.29 mol of this gas occupies __________ L under these conditions. A) 30.9 B) 478 C) 61.7 D) 92.6 E) 14.712)13) A gas originally at 27 °C and 1.00 atm pressure in a 3.9 L flask is cooled at constant pressure until the temperature is 11 °C. The new volume of the gas is __________ L. A) 3.7 B) 0.27 C) 3.9 D) 4.1 E) 0.2413)14) If 50.75 g of a gas occupies 10.0 L at STP, 129.3 g of the gas will occupy __________ L at STP. A) 50.8 B) 3.92 C) 5.08 D) 25.5 E) 12.914)2 15) A sample of He gas (2.35 mol) occupies 57.9 L at 300.0 K and 1.00 atm. The volume of this sample is __________ L at 423 K and 1.00 atm. A) 0.709 B) 81.6 C) 41.1 D) 57.9 E) 1.4115)16) A sample of H2 gas (12.28 g) occupies 100.0 L at 400.0 K and 2.00 atm. A sample weighing 9.49 g occupies __________ L at 353 K and 2.00 atm. A) 109 B) 147 C) 68.2 D) 77.3 E) 54.716)17) A sample of an ideal gas (3.00 L) in a closed container at 25.0 °C and 76.0 torr is heated to 300 °C. The pressure of the gas at this temperature is __________ torr. A) 76.5 B) 39.5 C) 2.53 × 10-2 D) 912 E) 146 18) A sample of a gas (1.50 mol) is contained in a 15.0 L cylinder. The temperature is increased from P2 100 °C to 150 °C. The ratio of final pressure to initial pressure [ ] is __________. P1 A) 1.00 B) 1.50 C) 0.882 D) 1.13 E) 0.66717)18)19) A sample of a gas originally at 25 °C and 1.00 atm pressure in a 2.5 L container is allowed to expand until the pressure is 0.85 atm and the temperature is 15 °C. The final volume of the gas is __________ L. A) 2.8 B) 3.0 C) 2.6 D) 2.1 E) 0.3819)20) The reaction of 50 mL of Cl2 gas with 50 mL of CH4 gas via the equation: Cl2 (g) + CH4 (g) HCl (g) + CH3 Cl (g) will produce a total of __________ mL of products if pressure and temperature are kept constant. A) 250 B) 50 C) 200 D) 150 E) 10020)21) The reaction of 50 mL of N2 gas with 150 mL of H2 gas to form ammonia via the equation: N2 (g) + 3H2 (g) 2NH3 (g) will produce __________ mL of ammonia if pressure and temperature are kept constant. A) 100 B) 250 C) 150 D) 200 E) 5021)22) The reaction of 50 mL of Cl2 gas with 50 mL of CH4 gas via the equation: Cl2 (g) + C2 H4 (g) C2 H4 Cl2 (g) will produce a total of __________ mL of products if pressure and temperature are kept constant. A) 25 B) 50 C) 100 D) 150 E) 12522)3 23) The amount of gas that occupies 60.82 L at 31 °C and 367 mmHg is __________ mol. A) 11.6 B) 1.18 C) 0.850 D) 894 E) 0.12023)24) The pressure of a sample of CH4 gas (6.022 g) in a 30.0 L vessel at 402 K is __________ atm. A) 12.4 B) 6.62 C) 2.42 D) 0.414 E) 22.424)25) At a temperature of __________ °C, 0.444 mol of CO gas occupies 11.8 L at 889 torr. A) 106 B) 379 C) 32 D) 73 E) 1425)26) The volume of 0.25 mol of a gas at 72.7 kPa and 15 °C is __________ m 3 . A) 2.2 × 10-1 B) 8.1 × 10-5 C) 4.3 × 10-4 D) 1.2 × 10-4 E) 8.2 × 10-326)27) The pressure exerted by 1.3 mol of gas in a 13 L flask at 22 °C is __________ kPa. A) 18 B) 560 C) 2.4 D) 1.0 E) 25027)28) A 0.325 L flask filled with gas at 0.914 atm and 19 °C contains __________ mol of gas. A) 80.7 B) 12.4 C) 1.48 × 10-2 D) 9.42 E) 1.24 × 10-2 29) A gas in a 325 mL container has a pressure of 695 torr at 19 °C. There are __________ mol of gas in the flask. A) 12.4 B) 1.24 × 10-2 C) 1.48 × 10-2 D) 9.42 E) 80.6 30) A sample of gas (1.9 mol) is in a flask at 21 °C and 697 mmHg. The flask is opened and more gas is added to the flask. The new pressure is 795 mmHg and the temperature is now 26 °C. There are now __________ mol of gas in the flask. A) 0.28 B) 2.1 C) 1.6 D) 3.5 E) 2.928)29)30)31) A sample of gas (1.3 mol) occupies __________ L at 22 °C and 2.5 atm. A) 0.94 B) 3.2 × 10-2 C) 13 D) 31 E) 0.07931)4 32) The volume of 0.65 mol of an ideal gas at 365 torr and 97 °C is __________ L. A) 0.054 B) 9.5 C) 2.4 × 10-2 D) 11 E) 4132)33) The volume occupied by 1.5 mol of gas at 35 °C and 2.0 atm pressure is __________ L. A) 2.2 B) 38 C) 0.026 D) 0.053 E) 1933)34) The mass of nitrogen dioxide contained in a 4.32 L vessel at 48 °C and 141600 Pa is __________ g. A) 5.35 × 104 B) 9.46 × 10-2 C) 10.5 D) 70.5 E) 53.5 35) The density of ammonia gas in a 4.32 L container at 837 torr and 45.0 °C is __________ g/L. A) 3.86 B) 4.22 × 10-2 C) 0.717 D) 0.432 E) 0.194 36) The density of N2 O at 1.53 atm and 45.2 °C is __________ g/L. A) 0.388 B) 2.58 C) 9.99 D) 1.76 E) 18.234)35)36)37) The molecular weight of a gas is __________ g/mol if 3.5 g of the gas occupies 2.1 L at STP. A) 4.6 × 102 B) 5.5 × 103 C) 41 D) 2.7 × 10-2 E) 3737)38) The molecular weight of a gas that has a density of 6.70 g/L at STP is __________ g/mol. A) 150 B) 3.35 C) 73.0 D) 496 E) 0.29838)39) The molecular weight of a gas that has a density of 7.10 g/L at 25.0 °C and 1.00 atm pressure is __________ g/mol. A) 28.0 B) 5.75 × 10-3 C) 14.6 D) 6.85 × 10-2 E) 17439)5 40) The molecular weight of a gas that has a density of 5.75 g/L at STP is __________ g/mol. A) 578 B) 3.90 C) 141 D) 1.73 × 10-3 E) 129 41) The density of chlorine (Cl2 ) gas at 25 °C and 60. kPa is __________ g/L. A) 0.86 B) 1.7 C) 4.9 D) 20 E) 0.5840)41)42) The volume of hydrogen gas at 38.0 °C and 763 torr that can be produced by the reaction of 4.33 g of zinc with excess sulfuric acid is __________ L. A) 2.71 × 10-4 B) 0.592 C) 2.84 D) 1.69 E) 3.69 × 104 43) The volume of HCl gas required to react with excess magnesium metal to produce 6.82 L of hydrogen gas at 2.19 atm and 35.0 °C is __________ L. A) 3.41 B) 6.82 C) 2.19 D) 4.38 E) 13.642)43)44) The volume of fluorine gas required to react with 2.67 g of calcium bromide to form calcium fluoride and bromine at 41.0 °C and 4.31 atm is __________ mL. A) 79.9 B) 210 C) 104 D) 10.4 E) 42044)45) What volume (mL) of sulfur dioxide can be produced by the complete reaction of 3.82 g of calcium sulfite with excess HCl (aq), when the final SO2 pressure is 827 torr at 44.0 °C? A) 1.39 × 10-4 B) 578 C) 1.00 × 10-3 D) 0.106 E) 76145)6 46) Automobile air bags use the decomposition of sodium azide as their source of gas for rapid inflation: 2NaN3 (s) 2Na (s) + 3N2 (g). What mass (g) of NaN3 is required to provide 40.0 L of N2 at 25.0 °C and 763 torr? A) 1.09 B) 160 C) 1.64 D) 107 E) 71.146)47) The Mond process produces pure nickel metal via the thermal decomposition of nickel tetracarbonyl: Ni(CO)4 (l) Ni (s) + 4CO (g). What volume (L) of CO is formed from the complete decomposition of 444 g of Ni(CO)4 at 752 torr and 22.0 °C? A) 63.7 B) 11.0 C) 255 D) 0.356 E) 20.247)48) What volume (L) of NH3 gas at STP is produced by the complete reaction of 7.5 g of H2O according to the following reaction? Mg 3 N2 (s) + 6H2 O (l) 3Mg(OH)2 (aq) + 2NH3 (g) A) 3.1 B) 9.3 C) 0.32 D) 28 E) 1948)49) Ammonium nitrite undergoes thermal decomposition to produce only gases: NH4 NO2 (s) N2 (g) + 2H2 O (g) What volume (L) of gas is produced by the decomposition of 35.0 g of NH4 NO2 (s) at 525 ° C and 1.5 atm? A) 15 B) 72 C) 160 D) 24 E) 4749)50) The thermal decomposition of potassium chlorate can be used to produce oxygen in the laboratory. 2KClO3 (s) 2KCl (s) + 3O2 (g) What volume (L) of O2 gas at 25 °C and 1.00 atm pressure is produced by the decomposition of 7.5 g of KClO3 (s)? A) 11 B) 7.5 C) 4.5 D) 2.2 E) 3.750)7 51) Since air is a mixture, it does not have a "molar mass." However, for calculation purposes, it is possible to speak of its "effective molar mass." (An effective molar mass is a weighted average of the molar masses of a mixture's components.) If air at STP has a density of 1.285 g/L, its effective molar mass is __________ g/mol. A) 31.4 B) 26.9 C) 30.0 D) 28.8 E) 34.451)52) A vessel contained N2 , Ar, He, and Ne. The total pressure in the vessel was 987 torr. The partial pressures of nitrogen, argon, and helium were 44.0, 486, and 218 torr, respectively. The partial pressure of neon in the vessel was __________ torr. A) 521 B) 42.4 C) 19.4 D) 760 E) 23952)53) The pressure in a 12.2 L vessel that contains 2.34 g of carbon dioxide, 1.73 g of sulfur dioxide, and 3.33 g of argon, all at 42 °C is __________ mmHg. A) 395 B) 263 C) 0.347 D) 134 E) 11653)54) A sample of He gas (3.0 L) at 5.6 atm and 25 °C was combined with 4.5 L of Ne gas at 3.6 atm and 25 °C at constant temperature in a 9.0 L flask. The total pressure in the flask was __________ atm. Assume the initial pressure in the flask was 0.00 atm. A) 9.2 B) 1.0 C) 24 D) 2.6 E) 3.754)55) A sample of H2 gas (2.0 L) at 3.5 atm was combined with 1.5 L of N2 gas at 2.6 atm pressure at a constant temperature of 25 °C into a 7.0 L flask. The total pressure in the flask is __________ atm. Assume the initial pressure in the flask was 0.00 atm. A) 2.8 B) 1.0 C) 1.6 D) 0.56 E) 2455)56) In a gas mixture of He, Ne, and Ar with a total pressure of 8.40 atm, the mole fraction of Ar is __________ if the partial pressures of He and Ne are 1.50 and 2.00 atm, respectively. A) 0.357 B) 0.417 C) 0.583 D) 0.238 E) 0.17956)57) A gas mixture of Ne and Ar has a total pressure of 4.00 atm and contains 16.0 mol of gas. If the partial pressure of Ne is 2.75 atm, how many moles of Ar are in the mixture? A) 5.00 B) 6.75 C) 12.0 D) 9.25 E) 11.057)58) A mixture of He and Ne at a total pressure of 0.95 atm is found to contain 0.32 mol of He and 0.56 mol of Ne. The partial pressure of Ne is __________ atm. A) 1.0 B) 1.7 C) 0.60 D) 0.35 E) 1.558)59) A flask contains a mixture of He and Ne at a total pressure of 2.6 atm. There are 2.0 mol of He and 5.0 mol of Ne in the flask. The partial pressure of He is __________ atm. A) 1.86 B) 1.04 C) 0.74 D) 6.5 E) 9.159)8 60) Sodium hydride reacts with excess water to produce aqueous sodium hydroxide and hydrogen gas: NaH (s) + H2 O (l) NaOH (aq) + H2 (g) A sample of NaH weighing __________ g will produce 982 mL of gas at 28.0 °C and 765 torr, when the hydrogen is collected over water. The vapor pressure of water at this temperature is 28 torr. A) 2.93 B) 0.0388 C) 0.960 D) 925 E) 0.92560)61) SO2 (5.00 g) and CO2 (5.00 g) were placed in a 750.0 mL container at 50.0 °C. The total pressure in the container was __________ atm. A) 1.60 B) 0.192 C) 4.02 D) 2.76 E) 6.7861)62) SO2 (5.00 g) and CO2 (5.00 g) are placed in a 750.0 mL container at 50.0 °C. The partial pressure of SO2 in the container was __________ atm. A) 0.192 B) 6.78 C) 4.02 D) 1.60 E) 2.7662)63) SO2 (5.00 g) and CO2 (5.00 g) were placed in a 750.0 mL container at 50.0 °C. The partial pressure of CO2 in the container was __________ atm. A) 1.60 B) 2.76 C) 4.02 D) 0.192 E) 6.7863)64) CO (5.00 g) and CO2 (5.00 g) were placed in a 750.0 mL container at 50.0 °C. The total pressure in the container was __________ atm. A) 1.60 B) 4.02 C) 0.292 D) 10.3 E) 6.3164)65) CO (5.00 g) and CO2 (5.00 g) were placed in a 750.0 mL container at 50.0 °C. The partial pressure of CO in the container was __________ atm. A) 4.02 B) 10.3 C) 0.292 D) 6.31 E) 1.6065)66) CO (5.00 g) and CO2 (5.00 g) were placed in a 750.0 mL container at 50.0 °C. The partial pressure of CO2 in the container was __________ atm. A) 4.01 B) 1.60 C) 10.3 D) 6.31 E) 0.29266)67) The root-mean-square speed of CO at 113 °C is __________ m/s. A) 31.5 B) 317 C) 586 D) 993 E) 58.367)68) A sample of N2 gas (2.0 mmol) effused through a pinhole in 5.5 s. It will take __________ s for the same amount of CH4 to effuse under the same conditions. A) 7.3 B) 5.5 C) 3.1 D) 9.6 E) 4.268)9 69) A sample of O2 gas (2.0 mmol) effused through a pinhole in 5.0 s. It will take __________ s for the same amount of CO2 to effuse under the same conditions. A) 0.23 B) 3.6 C) 5.9 D) 6.9 E) 4.369)70) A sample of He gas (2.0 mmol) effused through a pinhole in 53 s. The same amount of an unknown gas, under the same conditions, effused through the pinhole in 248 s. The molecular mass of the unknown gas is __________ g/mol. A) 19 B) 350 C) 5.5 D) 88 E) 0.1970)71) Using the van der Waals equation, the pressure in a 22.4 L vessel containing 1.00 mol of neon gas at 100 °C is __________ atm. (a = 0.211 L2-atm/mol2 , b = 0.0171 L/mol) A) 0.367 B) 1.00 C) 1.37 D) 1.21 E) 0.73071)72) Using the van der Waals equation, the pressure in a 22.4 L vessel containing 1.50 mol of chlorine gas at 0.00 °C is __________ atm. (a = 6.49 L2-atm/mol2 , b = 0.0562 L/mol) A) 1.50 B) 1.48 C) 0.993 D) 1.91 E) 0.67672)10 Answer Key Testname: CHAPTER 10 PRACTIS QUESTIONS1) 2) 3) 4) 5) 6) 7) 8) 9) 10) 11) 12) 13) 14) 15) 16) 17) 18) 19) 20) 21) 22) 23) 24) 25) 26) 27) 28) 29) 30) 31) 32) 33) 34) 35) 36) 37) 38) 39) 40) 41) 42) 43) 44) 45) 46) 47) 48) 49) 50)D C E C B D B B B E B D A D B C E D A E A B B D A E E E B B C E E C C B E A E E B D E A E E C A B D 11 Answer Key Testname: CHAPTER 10 PRACTIS QUESTIONS51) 52) 53) 54) 55) 56) 57) 58) 59) 60) 61) 62) 63) 64) 65) 66) 67) 68) 69) 70) 71) 72)D E B E C C A C C E E E C D D A C E C D C B12 `
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# Checking Correctness of Table Entries
You’re welcome! I would use parseEquation for that one. Add this to the input (don’t forget to mark the sketch as readOnly):
``````correct = parseEquation(this.latex).differenceFunction("w").evaluateAt(52)=0
correct: correct
``````
The parseEquation reads the latex as an equation and the difference function subtracts the left side and the right side of the equation. Since there should be a “w” variable, if we substitute 52, the difference should be 0. This will allow for different iterations of the equation too i.e. w+w+76=180.
You could also toss in an errorMessage to make sure the correct variable is being used. This goes in the input too.
``````ev1 = simpleFunction(parseEquation(this.latex).lhs, "w").evaluateAt(52)
ev2 = simpleFunction(parseEquation(this.latex).rhs, "w").evaluateAt(52)
errorMessage: when isUndefined(ev1) or isUndefined(ev2) "Only use the w variable" otherwise ""
``````
Could you alternatively have a variable for the differenceFunction, then use that for both the correctness check and for the undefined check?
``````d=parseEquation(this.latex).differenceFunction("w").evaluateAt(52)
correct: d=0
errorMessage: when isUndefined(d) "Only use the w variable" otherwise""
``````
I was trying to do something like that, but couldn’t get it to work. That looks like it should do the trick.
You’re AMAZING!!!
Do you ever put on any clinics or is there a book that has all these commands and what they do? How did you learn all this? I really enjoy creating these activities and want to learn more.
I have one last question for now… how about code to check an expression?
https://teacher.desmos.com/activitybuilder/custom/5ebbff9ec3459f0c74b3fea6
I learned from @Jay! I started by doing the CL scavenger hunt and then just kept learning by making activities. There are webinars that @jay has hosted that are archived somewhere in this forum.
For your question, don’t forget to mark the sketch as `readOnly: true`. For the input, try this:
``````correct = simpleFunction(this.latex, "a", "b").evaluateAt(5,5)=5 and countNumberUsage(this.latex,2)=1 and countNumberUsage(this.latex,3)=1 and countNumberUsage(this.latex)=2
``````
For the evaluateAt, I just chose to sub 5 for a and b, which happens to evaluate to 5. You can pick other numbers too. I also look for one 3, one 2 (countNumberUsage only looks for positive values), and two numbers total.
The @Jay’s webinars are in the Announcements section, but here’s all 5:
Yep! We’re also running clinics by appointment:
And hosting webinars daily:
https://learn.desmos.com/webinars
(for a limited time)
Cwinske,
I have quick follow up for you about this code for checking an equations most of my students wrote down the correct equations which was 2x+76=180, but I had a kid write w=52, which is technical correct but I was hoping to have it in a form “something” = 180 (or 180 = “something”) Do you have any suggestions on how to get it in that format?
Include something like this
``````correct = countNumberUsage(parseEquation(inputName.latex).lhs,180)=1 or countNumberUsage(parseEquation(inputName.latex).rhs,180)=1
``````
or if you just want to check for the 180
`correct=countNumberUsage(inputName,180)=1`
That will just count the numbers on either side. You want to make sure that the side evaluates to 180 always. SimpleFunction on both sides checking that it’s evaluated amount always comes out to 180 should do it.
I just meant in addition to his current check, and was just answering how to check for the “=180”.
Perfect. The complete line of code shared made it seem like a complete check.
Here is an activity where students can either click to drag labels to the correct spots in an image (slide 2) or, where they can enter the correct text label name into a table (in slide 1).
In the table in slide 1, how does the code need to change so that the 3rd column shows a green checkmark when the correct text is entered in the proper cell of the table? Also, is there any way to get the “cell content” command to disregard whether the student puts a word in as uppercase or lowercase? (I will just include in the directions to use all lowercase if not).
By the way, when I get rid of the word “when,” an error occurs.
Code used:
cellContent(1,3): when table1.cellContent(1,2)= “label 1 name here” “” OTHERWISE “N”
Thanks for any help.
For slide 1, you need to change “label 1 name here”, “label 2”, etc. to whatever correct answer you’re looking for. You can add an `or` in to include capitalized.
Here, my intended label is “frog”:
``````cellContent(1,3): when table1.cellContent(1,2)= "frog" or
table1.cellContent(1,2)= "Frog" "✅" otherwise "N"
``````
Thank you. Is it true you cannot have “spaces” in your cell content for the check mark to register- such as in label 2? If I change it to “frog height,” the check mark does not appear, but if I change it to “frog,” it does appear. If so, is there a way to trigger the checkmark for responses that have spaces in them?
Oh I see- I would need to put a second column in the table if there is a second word. Thank you.
As a note, when I did enter a second column in the table (it is now on slide 7, for the 4th row in the table), it is not picking up that “ground” in the first column and “water” in the second column is correct. I am not sure why.
Can you share your activity link? I think I’d actually need to look at it to help with the last question.
No, you don’t need another column. It should be okay if you change it to “frog height”. The answer just has to exactly match the text intended. So, if you want “frog” and a student accidentally enters a space at the beginning or end of “frog” it will be marked incorrect. There shouldn’t be a problem with “frog height”.
Thank you. Slide 7 row 4 is where I have 2 columns to accommodate 2 spaces and “ground” in the 1st and “water” in the second column didn’t work.
Slide 6 row 4 is where ground water with spaces all in 1 column didn’t work.
Change each column in the table to “Format as Text” by clicking the arrow in the header cells. Right now they’re formatted as math, which will remove spaces and reformat as latex when you’re checking content. This will fix your problem of having spaces, and the special case of “ground”, which probably ends up being reformatted and read as “g \round” because “round” is a function in the calculator.
Also, do you intend the labels in the picture to change? Or would you like them to? If so, under what conditions?
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5 Q
2nd
Time Relationships
10 Q
2nd - 3rd
Relationships
9 Q
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Relationships of Time
12 Q
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Basic Facts and Relationships
20 Q
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Family Relationship
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Intro to QAR (Question Answer Relationship)
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13 Q
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Equivalent Mathematical Relationships (EMR)1
10 Q
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22 Q
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Inverse Relationship (Measurement)
7 Q
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10 Q
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Word Relationships
5 Q
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Plant and Animal Relationships Chapter 3
20 Q
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Plant and Animal Relationships Unit Review
13 Q
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## Explore printable Proportional Relationships worksheets for 2nd Grade
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Computer Physics
Time 2.3 hrs
Difficulty Intermediate
Prerequisites Game Engines
Departments Science
Authors Sandra Kuipers
Groupings Individual
Minimum Year Group None
### Blurb
Physics in computer games allow us to simulate 2D and 3D environments. The first steps in computer physics include gravity, friction and collisions.
This work is shared under the following license: Creative Commons BY-SA-NC
### Outline
Learner OutcomesStudents will: ... Competency Focus ... Interdisciplinary Connections ... ReflectionWhat was successful? What needs changing? Alternative Assessments and Lesson Ideas? What other Differentiation Ideas/Plans could be used? ... CreditsAny CC attribution, thanks, credit, etc. Thumbnail Image by Gerd Altmann from Pixabay
5 mins
Making things "Real"
Introduction
• Physics in computer games allow us to make really awesome effects!
• Some of them are realistic, and some of them are "real enough" but also fun.
• Games can have 2D physics, 3D physics, or a combination of both.
• Luckily, using a game engine means we don't have to calculate all the physics by hand.
• But, we do need to understand the basics of how objects interact with each other.
5 mins
Game Physics
Like reality, but not quite
• Video games have a great history of pushing the envelop for what's possible in computing.
• But, even with insanely powerful computers, we can't simulate the entire physical world.
• The following video looks at how game physics relate to real world physics.
• It gives a great overview of the types of things that are and are not possible in games:
(this guy talks slow! try increasing the playback speed to 1.25 or 1.5)
10 mins
Dynamics
Theory
• Computer physics come in many different forms.
• Each type of physics simulates a type of real-world interaction.
• Rigid Body Dynamics is a type of physics for "solid" objects, which means the shape of the object doesn't change when it collides with something. Most video game physics is rigid body, because it's faster for the GPU to calculate.
• Soft Body Dynamics is a type of physics where the object squishes or changes shape when it collides with other objects. Soft body physics can apply to whole objects, like the blob below, or also flat objects like a piece of cloth. Calculating soft body physics is really intensive for the GPU, so it's often not used real-time in video games, but can be used in 3D modelling software like Blender.
• Fluid Dynamics is when we use a computer to simulate the moment of water and other particles. These types of simulations are very intensive, because each particle of "water" is being calculated based on the other particles around them.
5 mins
Collisions
Theory
• Collisions occur anytime two objects in a game "collide" or pass through each other.
• Collisions can be visible, such as a car crashing into another car.
• They can also be invisible, such as triggering an event when you walk into an area.
• Collision detection is the math a game engine uses to determine if a collision has happened.
• Detecting collisions is the heart of a lot of gameplay mechanics.
• What kinds of games can you think of that use collisions? 2D games? 3D games?
• There's some pretty complex math that goes into collision detection.
• Click the image above to see an example. Just scroll through and play with the clickable parts.
• To simplify the math, many objects have an invisible collider around them called a bounding box (or bounding volume).
5 mins
More Physics
Theory
• There are lots of other types of physics in games, which we'll explore in this course:
85 mins
Physics Game
Unity Tutorial
• Learning about game physics is best when it's hands-on.
• Visit the Roll-a-ball Tutorial and work through the steps to create a basic physics game.
25 mins
Evidence
• Your goal in this unit is to create a roll-a-ball game, then modify it in some unique way.
• You could change it in a number of ways: different shapes, larger maps, obstacles to move around.
• Take some time to tinker with the possible options after completing the Unity tutorial.
• Once you're done, upload your entire project folder to Google drive and share the folder as evidence of your learning in this unit.
## Embeds
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# LED circuit
(Redirected from LED as light sensor)
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Simple LED (Light Emitting Diode) circuit diagram
In electronics, an LED circuit is an electrical circuit used to power a light-emitting diode (LED). The circuit must provide sufficient current to light the LED at the required brightness, but must limit the current to prevent damaging the LED. The voltage drop across an LED is approximately constant over a wide range of operating current; therefore, a small increase in applied voltage greatly increases the current. Very simple circuits are used for low-power indicator LEDs. More complex, current source circuits are required when driving high-power LEDs for illumination to achieve correct current regulation.
## Basic circuit
The simplest circuit to drive an LED consists of a voltage source and two components connected in series: a current limiting resistor, sometimes called the ballast resistor, and an LED. Optionally, a switch may be introduced to open and close the circuit. Although simple, this circuit is not the most energy efficient circuit to drive an LED, since energy is lost in the resistor. More complicated circuits improve the energy efficiency.
An LED has a voltage drop specified at the intended operating current. Ohm's law and Kirchhoff's circuit laws are used to calculate the appropriate resistor value to obtain the desired current. The value is computed by subtracting the LED voltage drop from the supply voltage, and dividing by the desired operating current. If the supply voltage is equal to the LED's voltage drop, no resistor is needed.
This basic circuit is used in a wide range of applications, including many consumer appliances such as mobile phone chargers.
## Power source considerations
The voltage versus current characteristics of an LED are similar to any diode. Current is approximately an exponential function of voltage according to the Shockley diode equation, and a small voltage change may result in a large change in current. If the voltage is below the threshold or on-voltage no current flows and the result is an unlit LED. If the voltage is too high the current exceeds the maximum rating, overheating and potentially destroying the LED.
It is therefore important that the power source provides an appropriate current. LEDs should only be connected to constant-current sources. Series resistors are a simple way to passively stabilize the LED current. An active constant current regulator is commonly used for high power LEDs, stabilizing light output over a wide range of input voltages which might increase the useful life of batteries. Low drop-out (LDO) constant current regulators also allow the total LED voltage to be a higher fraction of the power supply voltage. Switched-mode power supplies are used in some LED flashlights and household LED lamps.
## Series resistor
Series resistors are a simple way to stabilize the LED current, but energy is wasted in the resistor.
Miniature indicator LEDs are normally driven from low voltage DC via a current-limiting resistor. Currents of 2 mA, 10 mA and 20 mA are common. Sub-mA indicators may be made by driving ultrabright LEDs at very low current. Efficiency tends to reduce at low currents[citation needed], but indicators running on 100 μA are still practical.
In coin cell powered keyring-type LED lights, the resistance of the cell itself is usually the only current limiting device. The cell should not therefore be replaced with a lower resistance type.
LEDs can be purchased with built-in series resistors. These can save printed circuit board space and are especially useful when building prototypes or populating a PCB in a way other than its designers intended. However, the resistor value is set at the time of manufacture, removing one of the key methods of setting the LED's intensity.
### Series resistor calculation
The formula to calculate the correct resistance to use is
$\mbox{resistance} (R) = \frac {\mbox{power supply voltage} (V_s) - \mbox{LED voltage drop}(V_f) } {\mbox{LED current}(I)},$
where power supply voltage (Vs) is the voltage of the power supply, e.g. a 9 volt battery, LED voltage drop (Vf) is the forward voltage drop across the LED, and LED current (I) is the desired current of the LED. The above formula requires the current in amperes, although this value is usually given by the manufacturer in milliamperes, such as 20 mA.
Typically, the forward voltage of an LED is about 1.8–3.3 volts; it varies by the color of the LED. A red LED typically drops 1.8 volts, but voltage drop normally rises as the light frequency increases, so a blue LED may drop around 3.3 volts.
The formula can be explained considering the LED as a ${V_f \over I} \;\Omega$ resistance, and applying Kirchhoff's voltage law (KVL) (R is the unknown quantity):
$V_s=V_r+V_f=R I+V_f$
$R I=V_s-V_f \;$
$R={V_s-V_f \over I}$
## LED arrays
Strings of multiple LEDs are normally connected in series. In one configuration, the source voltage may be greater than or equal to the sum of the individual LED voltages; typically the LED voltages add up to around two-thirds of the supply voltage. A single current-limiting resistor may be used for each string.
The other configuration is to run the sum of the supply voltage at approximately 75 – 85% of the combined LED voltages.[citation needed] This uses the LEDs' combined inherent resistance as a serial resistor. While small voltage drops to each LED generally make no discernible loss of intensity or brightness[citation needed], with sufficient LEDs in series a noticeable drop in brightness begins to show with enough LEDs in the circuit. In assuming that the supply voltage is 12 V, and each LED is 3 V, by using a string of either five or six LEDs whose combined voltage is 15 V or 18 V, they effectively may be under-driven, in favour of absence of power losses in terms of waste heat from resistors as well as simple circuitry.[citation needed]
Parallel operation is also possible but can be more problematic. Parallel LEDs must have closely matched forward voltages (Vf) in order to have similar branch currents and, therefore, similar light output. Variations in the manufacturing process can make it difficult to obtain satisfactory operation when connecting some types of LEDs in parallel.[1]
LED orientation[2]
## LED display
Main article: LED display
LEDs are often arranged in ways such that each LED (or each string of LEDs) can be individually turned on and off.
Direct drive is the simplest-to-understand approach -- it uses many independent single-LED (or single-string) circuits. For example, a person could design a digital clock such that when the clock displays "12:34" on a seven-segment display, the clock would turn on the appropriate segments directly and leave them on until something else needs to be displayed.
However, multiplexed display techniques are more often used than direct drive, because they have lower net hardware costs. For example, most people who design digital clocks design them such that when the clock displays "12:34" on a seven-segment display, at any one instant the clock turns on the appropriate segments of one of the digits -- all the other digits are dark. The clock scans through the digits rapidly enough that it gives the illusion that it is "constantly" displaying "12:34" for an entire minute. However, each "on" segment is actually being rapidly pulsed on and off many times a second. Such multiplexed displays have net lower hardware costs, but the resulting pulsed operation makes the display inevitably dimmer than directly driving the same LEDs independently.
An extension of this technique is Charlieplexing where the ability of some microcontrollers to tri-state their output pins means larger numbers of LEDs can be driven, without using latches. For N pins, it is possible to drive n2-n LEDs
## Polarity
Unlike incandescent light bulbs, which illuminate regardless of the electrical polarity, LEDs will only light with correct electrical polarity. When the voltage across the p-n junction is in the correct direction, a significant current flows and the device is said to be forward-biased. If the voltage is of the wrong polarity, the device is said to be reverse biased, very little current flows, and no light is emitted. LEDs can be operated on an alternating current voltage, but they will only light with positive voltage, causing the LED to turn on and off at the frequency of the AC supply.
Most LEDs have low reverse breakdown voltage ratings, so they will also be damaged by an applied reverse voltage above this threshold. The cause of damage is overcurrent resulting from the diode breakdown, not the voltage itself. LEDs driven directly from an AC supply of more than the reverse breakdown voltage may be protected by placing a diode (or another LED) in inverse parallel.
The manufacturer will normally advise how to determine the polarity of the LED in the product datasheet.[2]
## Pulsed LED operation
Many systems pulse LEDs on and off, by applying power periodically or intermittently. So long as the flicker rate is greater than the human flicker fusion threshold, and the LED is stationary relative to the eye, the LED will appear to be continuously lit. Varying the on/off ratio of the pulses is known as pulse-width modulation. In some cases PWM-based drivers are more efficient than constant current or constant voltage drivers.[3][4] It is also done to allow digital intensity control without a more complex digital-to-analog converter. The Arduino microprocessor boards use this technique to control the on-board LED.
Most LED data sheets specify a maximum DC current that is safe for continuous operation. Often they specify some higher maximum pulsed current that is safe for brief pulses, as long as the LED controller keeps the pulse short enough and then turns off the power to the LED long enough for the LED to cool off.
## LED as light sensor
Mobile phone IrDA
In addition to emission, an LED can be used as a photodiode in light detection. This capability may be used in a variety of applications including ambient light detection and bidirectional communications.[5][6]
As a photodiode, an LED is sensitive to wavelengths equal to or shorter than the predominant wavelength it emits. For example, a green LED is sensitive to blue light and to some green light, but not to yellow or red light.
This implementation of LEDs may be added to designs with only minor modifications in circuitry.[5] An LED can be multiplexed in such a circuit, such that it can be used for both light emission and sensing at different times.[5]
## References
1. ^ "Electrical properties of GaN LEDs & Parallel connections" (PDF). Application Note. Nichia. Archived from the original (PDF) on 2007-08-09. Retrieved 2007-08-13.
2. ^ a b "Plastic infrared light emitting diode" (PDF). Fairchild Semiconductor. 2001-10-31. Retrieved 2009-05-15.
3. ^ Jim Lepkowski, Mike Hoogstra, and Christopher Young. Application note AND8067/D: "NL27WZ04 Dual Gate Inverter Oscillator Increases the Brightness of LEDs While Reducing Power Consumption"
4. ^ [1]
5. ^ a b c Dietz, Paul, William Yerazunis, Darren Leigh (2003). "Very Low-Cost Sensing and Communication Using Bidirectional LEDs" (PDF). Mitsubishi electric research laboratories.
6. ^ Bent, Sarah, Aoife Moloney and Gerald Farrell (2006). "LEDs as both Optical Sources and Detectors in Bi-directional Plastic Optical Fibre Links". Irish Signals and Systems Conference, 2006. IET: 345.
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# SAT Practice Test 5, Section 3: Questions 11 - 15
This is for SAT in Jan 2016 or before.
The following are worked solutions for the questions in the math sections of the SAT Practice Tests found in the The Official SAT Study Guide Second Edition.
It would be best that you go through the SAT practice test questions in the Study Guide first and then look at the worked solutions for the questions that you might need assistance in. Due to copyright issues, we are not able to reproduce the questions, but we hope that the worked solutions will be helpful.
Given:
Twice a number decreased by 3 is 253
To find:
The number
Solution:
Topic(s): Translating words to equations
Let n be the number
Given:
A company manufactures white and black sneakers, both of which are available either high-tops or low-tops
To find:
The number of black sneakers manufactured by the company
Solution:
High-Tops Low-tops Total White 3,600 Black 900 1,500 2,400 Total 4,500 5,500 10,000
We need to get is BlackTotal.
TotalHightops = Total – TotalLowtops = 10,000 – 5,500 = 4,500
BlackHighTops = TotalHighTops – WhiteHighTops = 4,500 – 3,600 = 900
BlackTotal = 900 + 1,500 = 2,400
Given:
The figure
PQRS is a rectangle
Points Q and R lie on the graph y = ax2, where a is a constant
Perimeter of PQRS = 10
To find:
The value of a
Solution:
From the diagram, length of PS = QR = 2
Let the length of PQ = RS = r
Perimeter PQRS = 2 + 2 + r + r = 10
⇒ 4 + 2r =10 ⇒2r =6 ⇒ r = 3
The coordinates of R would be (1,3).
Given that R is on the graph y = ax2, we can substitute x = 1 and y = 3 into the equation.
3 = a(1) 2 ⇒ a = 3
14. Correct answer: 8/3, 2.66, 2.67
Given:
ab + b = a + 2c
a = 2 and c = 3
To find:
The value of b
Solution:
Topic(s): Isolate variable
15. Correct answer: 45/2 < x < 55/2
Given:
The figure
l bisects
45 < y < 55
To find:
One possible value for x
Solution:
Topic(s): Corresponding angles, vertical angles
Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations.
You can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
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## Brown Heaps
A heap is a tree-based data structure that holds comparable values and focuses on the rapid access of smallest or largest values. For simplicity you will only use heaps of numbers in this assignment, but it should be easy to generalize for other types (for which comparison is defined).
### 1The Assignment
Unlike search trees, heaps impose no ordering constraint on the elements in the children: e.g., they do not require that all values in the left child ben be no bigger than all values in the right child of a node. Instead, heaps maintain the invariant that the value at each node is less than (or equal to) the values of its children. Thus, the minimum element in the heap is always its root.These are known as min heaps. Naturally, it is also possible to have max heaps, where the largest element in a heap is at its root.
One way to optimize the worst-case time complexity of the heap operations is to keep the heap balanced. A tree is balanced if the depths of any two leaves differ by at most one. In order to maintain balance in our heap implementation, we will impose a stronger condition: at each node, the number of nodes in the left subheap of a node must be the same or one greater than the number of nodes in the right subheap. We will call heaps with this invariant Brown heaps, defined below.
In what follows, assume there will be no duplicate elements.
data BHeap: | mt | node(value :: Number, left :: BHeap, right :: BHeap) end
### 2Implementing Operations
When implementing operations for this assignment, you will use the above BHeap data definition, subject to the following additional conditions:
• The number of nodes in left is the same or one greater than the number of nodes in right
• value is a number smaller than any value in left or right.
You will need to implement three operations on Brown heaps
• insert :: (Number, BHeap -> BHeap)
insert consumes a Number and a Brown heap and produces a Brown heap consisting of all the elements of the given BHeap in addition to the given Number.
• get-min :: (BHeap%(is-node) -> Number)
get-min consumes a non-empty Brown heap and produces the minimum value in that heap.
• remove-min :: (BHeap%(is-node) -> BHeap)
remove-min consumes a non-empty Brown heap and returns a new Brown heap consisting of all the elements of the given Brown Heap except the element returned by get-min.
All operations in this assignment can be performed in $$O([s -> \log s])$$ time or better, where $$s$$ is the number of elements in the heap. We will be grading on efficiency. Once again, you can assume that there will be no duplicate values in a Brown heap; you will therefore not be asked to insert an element already in a heap. You will need to turn in an analysis stating what efficiency you obtained, and justifying it.
### 3Template Files
We will, as usual, be using Captain Teach:
https://www.captain-teach.org/brown-cs019/
The templates are here:
Initial Tests
Implementation Code
Final Tests
Note: Implementation-dependent testing should be in the implementation file. The final tests file should contain your tests for the three procedures we had you implement.
Keep in mind that the output of insert and remove-min may be implementation-dependent! Think carefully about how you can test all three procedures without directly checking the output of insert or remove-min.
You will need to also determine the run-time complexity of your implementation of insert, get-min, and remove-min.
### 4Handing In
#### 4.1Initial Test Sweep
As with previous assignments, you will submit an initial test sweep. This is due 11:59 PM, Tuesday, November 4th.
https://www.captain-teach.org/brown-cs019/assignments/
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## Envision Math 6th Grade Textbook Answer Key Topic 2.7 Using Expressions to Describe Patterns
Using Expressions to Describe Patterns
How can you write expressions to describe patterns?
Delvin saves a part of everything he earns. The table at the right shows Delvin’s savings pattern.
The INPUT column shows the money he has earned.
The OUTPUT column shows the money he has saved.
Question.
Write an expression to describe the pattern.
Guided Practice
Do you know HOW?
Use the input/output table for 1 and 2.
Question 1.
If the input number is 8, what is the output number?
11
Explanation:
Question 2.
Write an algebraic expression that describes the output pattern.
x + 3
Explanation:
Do you UNDERSTAND?
Question 3.
Suppose that Delvin earned $36 mowing lawns. What input and output entries would you add to his table? Answer: IN:$36; OUT: $18 Explanation: Question 4. Reasonableness Is it reasonable for an output to be greater than the input in the table above? Explain. Answer: See margin. Explanation: Question 5. What is the algebraic expression that describes the output pattern for the table above if the input is x? Answer: $$\frac{1}{2}$$x. Explanation: Independent Practice Use this table for 6 and 7. Question 6. What is the cost of 4 lb, 5 lb, and 10 lb of apples? Answer: Question 7. Write an algebraic expression that describes the output pattern if the input is a variable a. Answer: 2a Explanation: Use this table for 8 and 9. Question 8. Copy and complete the table. Answer: Question 9. Write an algebraic expression that describes the relationship between the input and output values. Answer: x ÷ 3 Explanation: An input/output table is a table of related values. Identify the pattern. What is the relationship between the values? $$\frac{1}{2}$$ (84) = 42 → 42 is half of 84 $$\frac{1}{2}$$ (66) = 33 → 33 is half of 66 $$\frac{1}{2}$$ (50) = 25 → 25 is half of 50. The pattern is: $$\frac{1}{2}$$ (INPUT) = OUTPUT Let x= INPUT. So, the pattern is $$\frac{1}{2}$$ x. Use the pattern to find the missing values. $$\frac{1}{2}$$ (22) = 11 $$\frac{1}{2}$$ (30) = 15 Problem Solving Use the input/output table at right for 10 and 11. Question 10. Hazem keeps $$\frac{1}{3}$$ of the tips he earns. Also, he gets$1 each night to reimburse his parking fee. This information is shown in the input/output table. Write an algebraic expression that describes the output pattern if the input is the variable k.
(k ÷ 3) + 1 or $$\frac{1}{3}$$ k + 1
Explanation:
Question 11.
How much money would Hazem keep in a night if he takes in $36 in tips? Answer:$13
Explanation:
Use the input/output table at right for 12 and 13.
Question 12.
Ms. Windsor’s classroom has a tile floor. The students are making stars to put in the center of 4-tile groups. This input/ output chart shows the pattern. Write an algebraic expression that describes the output pattern if the input is the variable t.
t ÷ 4
Explanation:
Question 13.
Writing to Explain There are 30 rows with 24 tiles in each row on a floor. Explain how to find the number of stars needed to complete the pattern for the floor.
A. b + $2.50 B.$2.50b
C. $b –$2.50
D. b + $2.50 Answer: B.$2.50b
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# Step-by-step Solution
Go!
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## Step-by-step explanation
Problem to solve:
$\frac{d}{dx}\left(x\cdot x\cdot y\right)$
Learn how to solve product rule of differentiation problems step by step online.
$\frac{d}{dx}\left(x^2y\right)$
Learn how to solve product rule of differentiation problems step by step online. Find the derivative using the product rule (d/dx)(xx*y). When multiplying two powers that have the same base (x), you can add the exponents. The derivative of a function multiplied by a constant (y) is equal to the constant times the derivative of the function. The power rule for differentiation states that if n is a real number and f(x) = x^n, then f'(x) = nx^{n-1}.
$2yx$
$\frac{d}{dx}\left(x\cdot x\cdot y\right)$
### Main topic:
Product rule of differentiation
### Time to solve it:
~ 0.02 s (SnapXam)
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Search
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# Articles
SuperUser Account
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# Pulsed Power Measurements
Open as PDF
In many applications, like wireless communication and radar, pulsed (or bursted) RF signals are utilized. Not only is the level of RF power important, but the shape (RF power envelope) of the
waveform can be critical as well (see Figure 1).
Figure 1: An example of a pulsed signal and a few related measurements.
To help have a common language with which to describe the various elements of the waveforms, organizations like the IEEE have provided essential terms related to pulsed signals, along with explanations for each. A review of some of the terms and definitions is provided below.
1. Rise time – The time it takes a signal to change from a specified low value to a specified high value. The low value can be characterized by the proximal line, which is a magnitude reference line located near the bottom of the pulse that is a specified percentage (normally 10%) of the pulse magnitude. Similarly, the high value can be characterized by the distal line, which is a magnitude reference line located near the top of the pulse that is a specified percentage (normally 90%) of the pulse magnitude. Therefore, rise time can also be defined as the time interval from the first crossing of the proximal line to the first crossing of the distal line.
2. Fall time – The time it takes a signal to change from a specified high value to a specified low value. In other words, fall time is the time interval from the last crossing of the distal line to the last crossing of the proximal line.
3. Period – The interval between two successive pulses measured at the same point on both pulses.
Pulse Width – The interval between the first and second signal crossings of the mesial line, which is a magnitude reference line located near the middle of the pulse that is a specified percentage (normally 50%) of the pulse magnitude.
1. – The time a repeating pulse is in its “off” state, which is equal to the pulse period minus the pulse width.
2. Pulse Repetition Frequency (PRF) – PRF is the number of pulses from a repeating signal that are transmitted per second, and therefore is typically measured in pulses per second and captured in Hz.
3. Duty Cycle – Duty cycle is the ratio of the pulse on-time to off-time, and is calculated by taking the ratio of the pulse width over the waveform’s period.
4. Peak Power – The maximum power level of the captured waveform (also defined as the pulse’s peak amplitude).
5. Pulse Power – The average power level across the pulse width.
6. Average Power – The average power level of a pulse period.
7. – Overshoot is magnitude by which the signal exceeds the pulse top amplitude.
8. Droop – The magnitude of the deviation of the signal below the pulse top amplitude within the pulse top.
9. Pulse Top Amplitude – The amplitude of the top line, which represents the second nominal state of a pulse, as defined by IEEE.
10. Pulse Bottom Amplitude – The amplitude of the base line. As defined by IEEE, the base line is defined as two portions of a pulse waveform, which represent the first nominal state from which a pulse departs and to which it ultimately returns.
11. Edge Delay – The time between the left edge of the display and the first mesial transition level of either slope on the waveform. The mesial line is the a magnitude reference line located in the middle of a pulse at a specified percentage (normally 50%) of the pulse magnitude, while the first transition level is the major transition of a pulse waveform between the base line and the top line (commonly called the rising edge).
12. – The time difference between the mesial level of two different pulses on two channels using two sensors.
#### Pulsed Power Test Equipment – Peak Power Meters
Although there are various instruments used to measure pulsed signals, peak power meters and USB peak power sensors are typically the most accurate choices. Boonton power meters like the PMX40 provide up to 16 pulse measurements automatically, covering every one of the key parameters for pulsed signals described above. You can select the right RF test instrument to capture essential pulsed signal metrics for your specific needs at www.boonton.com.
Previous Article Burst Measurements with the RTP Series Measurement Buffer Mode Application
Next Article What Do You Want to Measure – Peak or Average?
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# Dirac Delta Function as Initial condition for 1D Diffusion PDE: ONE or TWO equations(conditions)?
I have 1D diffusion (u(t,x)) PDE with Dirac Delta initial condition.
Question is regarding it's implementation:
Dirac delta func is formally defined as an encapsulation of 2 conditions:
1st condn: function takes value 1 at x=0,
2nd condition:function takes value 0, if x not equal to 0).
Thus, to solve the above PDE for initial condition u(t=0,x)=Dirac_delta(x), is it necessary to treat this dirac delta function initial condition (for that PDE) as ONE or TWO separate conditions (viz. after substituting for t=0 and x=0 giving u(t=0,x=0)=1 solution giving 1st equation for initial condition & substituting t=0 and x=x(i.e non-zero x) giving u(t=0,x)=0 giving 2nd equation for initial condition)??
This means, 1D diffusion PDE with Dirac Delta function (delta(x)) has 1 OR 2 initial conditions, according to for t=0, (x=0 & x not equal to 0)?
Thanks
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Your question is quite a mess. First, please tell us typically the diffusion PDE. Second, please use TEX when typing the equations. – doraemonpaul Jan 15 '13 at 11:29
Your description of the Dirac delta ($\delta_0$) is incorrect. One of possible correct definitions is: $\delta_0$ is the derivative of the Heaviside function $h_0$, which is defined as $h_0(x)=1$ when $x\ge 0$; $h_0(x)=0$ when $x< 0$. The meaning of the derivative is not the classical one; instead, it is a stipulation that the certain integral identities hold: Fundamental theorem of Calculus and integration by parts. In particular, $\int_a^b \delta_0 =h_0(b)-h_0(a)=1$ whenever $a<0<b$.
Now back to counting initial conditions. If we have $u(0,x)=x^2$ for all $x\in\mathbb R$, is this one condition? Or maybe infinitely many, because there's a condition for every $x\in\mathbb R$? I guess we take the position that this is one condition: the distribution of matter represented by $u$ is given at time $t=0$, and that's all we have.
Another example: $u(0,x)=h_0(x)$ for all $x\in\mathbb R$. This is also one initial condition, we just have another function here instead of $x^2$. Since the function $h_0$ is piecewise defined, the condition $u(0,x)=h_0(x)$ amounts to $$u(0,x)=0 \text{ when } x<0 \ \text{ and } \ u(0,x)=1 \text{ when } x\ge 0$$ But this split is just our way of writing $h_0$. It is of no consequence for the mathematical problem.
Similarly, the initial condition $u(0,\cdot) =\delta_0$ is also one condition, which may appear as two (or three) depending on how $\delta_0$ is presented in an application-oriented text.
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@5PM..Thanks for the answer. Could you then tell me, how one implements such a Dirac Delta initial condition on 1D Diffusion PDE (u_sub{t}=D*u_sub{xx}, D:diffusion constant, _sub{x} refers to partial derivative taken wrt x)? Is it implemented only for x=0 (assuming it is δ(0,x)) or rather implemented as an integral(δ(0,x)) fashion? I would like to hear if it can be implementated in x=0 fashion, rather than in integral fashion. Hope its clear. – ems Jan 15 '13 at 22:53
@ems Are you familiar with the concept of a fundamental solution? – user53153 Jan 15 '13 at 22:55
@5PM...Yes. Fundamental solution helps finding the solution for any arbitrary initial condition, provided I have the solution calculated using Dirac Delta initial condition, right? I am trying to find the solution to 1D PDE not by analytical means (which I can find in any textbook), but rather using Differential Transform method (numerical approach, basis function based) – ems Jan 15 '13 at 22:58
@ems For numerical computations you need a nascent delta function, such as the piecewise linear $\eta_\epsilon(x)=\epsilon^{-1} \max(0,1-|x|/\epsilon)$, for small $\epsilon>0$. How small you can take it to be will depend on the particulars of your methods. – user53153 Jan 15 '13 at 23:11
This answer is incomplete reference to the Brownian motion "stocastic Calculus" and the same initial conditions we get the normal distribution
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Welcome to MSE! Was this supposed to be a comment. I realize you don't yet have enough reputation, but this area is for answers. Regards – Amzoti May 3 '13 at 5:29
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# Eureka Math Lesson 15 Homework Answers.
Lesson 16 Homework 4 5 - iblog.dearbornschools.org.
4.5 out of 5. Views: 823.
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Grade 4 Mathematics Module 5, Topic D, Lesson 16. Objective: Use visual models to add and subtract two fractions with the same units. Like (216) Downloadable Resources. Resources may contain links to sites external to the EngageNY.org website. These sites may not be within the jurisdiction of NYSED and in such cases NYSED is not responsible for its content. Grade 4 Mathematics Module 5, Topic.
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GRADE 5 MODULE 5. 4 halves or 2 24. 15 thirds or 5 35. 24 fourths or 6 3.. Lesson 7 Answer Key 5 Homework 1. 216 in3;. Great Minds is a non-profit organization founded in 2007 by teachers and scholars who want to ensure that all students receive a content-rich. Eureka Math, only. Click here to see Eureka Math Tips for Parents Answer Key All of the answer keys on this page are created.
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## allie_bear22 2 years ago check my answer! Which ratio represents the area of the smaller rectangle compared to the area of the larger rectangle? (Figure not drawn to scale)
1. allie_bear22
2. allie_bear22
i got 1/4(x+2)
3. whpalmer4
That's correct.
4. allie_bear22
thnk u
5. Luigi0210
You sure it's not 4/x+2?
6. allie_bear22
hmm thats not an option but 4x(x+2) is
7. Luigi0210
$\frac{ x }{ x+5 }=\frac{ 4*(x) }{ (x+2)*(x+5) }$
8. allie_bear22
is it x/x+2?
9. Luigi0210
Hm, actually nevermind sorry.
10. allie_bear22
lol okay so the other guy was right?
11. Luigi0210
I'm not 100% sure but I'll have to take his word for it
12. allie_bear22
ok thnk u
13. whpalmer4
Area of the smaller rectangle = $$x(x+5)$$ Area of the larger rectangle = $$4x(x^2+7x+10) = 4x(x+2)(x+5)$$ Ratio of areas (smaller to larger) = $\frac{x(x+5)}{4x(x+2)(x+5)} = \frac{x}{4x(x+2)} = \frac{1}{4(x+2)}$
14. whpalmer4
@Luigi0210 was taking the ratio of the sides of the small rectangle and comparing it to the ratio of the sides of the large rectangle, which is not what problem asks....that would be the right setup for determining if the rectangles were proportional in shape, however.
15. whpalmer4
Gotta watch out, multiple-choice questions often have wrong answers that represent results you could get if you set up the problem incorrectly. Don't assume the answer is right just because it is one of the choices :-)
16. Luigi0210
Yup, I realized that right when I lef
17. Luigi0210
*left
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#### single phase transformer
A static device used to transfer ac voltage from one circuit to other circuit through magnetic link without change in frequency.
#### working principle
When an alternating voltage (V1) is applied to the primary windings, an alternating flux (Φ) is set up in the core. This alternating flux links with both the windings and induces EMFs (E1 & E2) in them according to Faraday’s law of electromagnetic induction. E1 is named as primary EMF and E2 is named as secondary EMF.
E1 = – N1 d Φ /dt & E2 = – N2 d Φ / dt
Transformer does not change frequency of AC voltage / current because shape and magnitude of flux remains almost same on either side. Power of transformer (V1I1 = V2I2) remains same on either side. If voltage increases, then the current decreases maintaining the power constant.
#### transformer circuit
1)Electrical circuit which includes primary windings and secondary windings .The winding connected to AC source is called primary winding .The other connected to load is called secondary winding .The windings are made of copper to carry current.
2)Magnetic circuit includes magnetic core .Winding are wound on the core and are not electrically connected each other .The core is made of iron or silicon steel to carry flux .It is laminated to reduce the eddy current losses.
#### step-up transformer
Used to step up the voltage from low voltage level to high voltage level. It has more number of turns of coil on secondary side.
N2 > N1, V2 > V1, I2 < I1
#### step-down transformer
It is used to step down the voltage from high voltage level to low voltage level. It has less number of turns of coil on secondary side.
N2 < N1, V2 < V1 , I2 > I1
#### applications of transformer
1)Power transformers are used for transmission of electric power and usually used at power stations and grid stations.
2)Distribution Transformer are distributing electric power.
3)Auto Transformer are used to get variable voltage and also used as boosters in transformers.
4)Potential Transformers are used to step down voltages which are not measurable by voltmeter.
5)Current Transformers are used to step down line current which are not measurable by ammeter.
#### parts of transformer
There are following essential parts of transformer.
1)Windings
2)Magnetic / Iron Core
3)Transformer Tank
5)Bushings
6)Tappings
7)Vent pipe
Limb :Vertical portion of the core on which wire is wounded
Yoke :Top and bottom horizontal portions of core that provides path for flux.
#### types of transformer
1)Core Type Transformer : Windings are wound on two legs of transformer core .Single magnetic circuit & windings encircle the core. Rectangle shape core with 2-limbs & 2-yokes
2)Shell Type Transformer : Both LV & HV windings placed at central limb. HV coil is between LV coils & LV coils are nearest to top & bottom of yoke. Double magnetic circuit with 3-limbs & 3-yokes. Shell shape core & sandwich windings are used. Used for low voltage purposes. Provides less cooling surface
#### turn ratio of a transformer
Ratio of primary number of turns N1 to secondary number of turns N2 , denoted by while we know that
As frequency and flux remain constant so replace them by constant “a” called turn ratio.
E1 / E2 = N1 / N2 = a
Where N1 / N2 is called turn ratio.
#### voltage transformation equation (k)
k is inverse of turn ratio so it is written as
k = 1 / a
k = N2 / N1
#### ideal transformer at no load
Ideal transformer is practically nothing but its study provides useful tool in the analysis of a practical transformer.
Characteristics
1. Winding resistances (R1 & R2) = Zero
2. Winding reactance (X1 & X2) = Zero
3. Copper and core losses = Zero
4. Leakage flux = Zero
5. Zero voltage regulation = Zero
6. Input = output
7. Efficiency = 100 %
#### phasor diagram of ideal transformer
When an alternating voltage V1 is applied to the primary, it draws a small magnetizing current Im which lags behind the applied voltage by 90°. This alternating current Im produces an alternating flux Ф which is proportional to and in phase with it. The alternating flux links with both the windings and induces e.m.f. E1 in the primary and e.m.f. E2 in the secondary.
The primary e.m.f. E1 is, at every instant, equal to and in opposition to V1 (Lenz’s law). Both e.m.f E1 and E2 lag behind flux by 90°. Since flux f is common to both the windings, it has been taken as the reference phasor. Primary e.m.f. E1 and secondary e.m.f. E2 lag behind the flux f by 90°. Note that E1 and E2 are in phase. But E1 is equal to V1 and 180° out of phase with it.
#### practical transformer
Characteristics are given below
1. Winding resistances (R1 & R2) ≠ Zero
2. Winding reactance (X1 & X2) ≠ Zero
3. Copper and core losses ≠ Zero
4. Leakage flux ≠ Zero
5. Zero voltage regulation ≠ Zero
6. Input ≠ output
7. Efficiency ≠ 100 %
#### phasor diagram of practical transformer at no load
No-load primary current I0 can be resolved into two rectangular components viz.
(i) The component IW in phase with the applied voltage V1. This is known as active or working or iron loss component and supplies the iron loss and a very small primary copper loss.
IW = I0 cos Ф0
(ii) The component Im lagging behind V1 by 90° and is known as magnetizing component. It is this component which produces the mutual flux Фo in the core.
Im = I0 sin Ф0
#### transformer losses
Transformer losses are appear on the form of heat in core and windings
#### hysteresis losses
Associated with magnetization & demagnetization of the core during each half cycle of the flux.
#### eddy current losses
Induced emf in core sets up eddy current in core and hence eddy current losses occur.
#### leakage flux losses
Flux leaking from core to air produces self-inductance in coils which is loss.
#### copper losses
1. Resistive heating losses in the primary and secondary windings of the transformer.
2. Proportional to square of the current flowing in the windings.
3. About 90 % of the total losses.
4. Known as variable losses because they are different at different loads.
Why transformer power in KVA instead of KW?
1)Copper losses depend on current and iron losses depend on voltage and no other losses occur in transformer. Therefore total losses depend on Volt –Ampere (VA).
2)Losses do not depend on angle between volt and current. It means it is independent of power factor.
3)Electrical power on either side of transformer is constant (V1I1 = V2I2), so there is no need of power factor which is only required to compensate losses.
Next page:
-DC Generator
https://www.electricalengineering4u.com/electric-machines/dc-generator/
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Combustible Dust Equivalence Ratio: How to Calculate from Stoichiometry?
Have you ever wondered how to compute the dust equivalence ratio of a solid combustible fuel?
This post demonstrates the math and stoichiometry behind computing combustible dust equivalence ratio.
It begins by developing a mathematical relationship for equivalence ratio based on fuel concentration, air density, and the Fuel-To-Air ratio at stoichiometric conditions.
The second part of this post focuses on determining the stoichiometric Fuel-To-Air ratio for different dusts. Five cases are given for different combustible materials:
Before jumping into the stoichiometric calculations, background on computing the dust equivalence ratio is required.
Combustible Dust Equivalence Ratio
The math required to compute reactive dust equivalence ratio is relatively straight forward.
The equivalence ratio is defined as the Fuel-To-Air-Ratio at the specified concentration, $\left( F/A\right)$, divided by the stoichiometric Fuel-To-Air-Ratio $\left( F/A\right)_{\text{St}}$.
$\mathit{\Phi} = \frac{\left( F/A\right)}{\left( F/A\right)_{\text{St}}}$
The Fuel-To-Air-Ratio can be redefined as the ratio of the fuel concentration, $\mathit{\sigma}_{\text{F}}$, divided by the density of air, $\rho_{\text{a}}$.
$\left(F/A\right) = \left( \mathit{\sigma}_{\text{F}}/\rho_{\text{a}}\right)$
This allows the following relation for combustible dust equivalence ratio to be developed:
$\mathit{\Phi} = \frac{\left( F/A\right)}{\left( F/A\right)_{\text{St}}} = \frac{\left( \mathit{\sigma}_{\text{F}}/\rho_{\text{a}}\right)}{\left( F/A\right)_{\text{St}}} = \frac{\mathit{\sigma}_{\text{F}}}{\rho_{\text{a}}\left( F/A\right)_{\text{St}}}$
This relation has three parameters: the fuel concentration in the dust ($\mathit{\sigma}_{\text{F}}$, kg/m3), the density of air ($\rho_{\text{a}}$, kg/m3), and the stoichiometric Fuel-To-Air-Ratio ($\left( F/A\right)_{\text{St}}$).
The dust concentration is generally taken as the mass of dust divided by the system volume and must be specified in kg/m3.
The density of air can be calculated from the ideal gas law:
$\rho_{\text{a}} = \frac{P}{R_{\text{spec}}T}$
where $P$ is pressure (Pa), $T$ is temperature (K), and $R_{\text{spec}}$ is the specific gas constant for air (universal gas constant divided by molecular weight of air, 287.058 J/kg-K).
At ambient pressure and temperature the density of air can be computed as
$\rho_{\text{a}} = \frac{\text{101325} Pa}{287.058 \text{J/kg-K} \cdot 293 \text{K}} = 1.20 \text{kg/m}^{3}$
The most difficult part of computing the combustible dust equivalence ratio is determining the stoichiometric Fuel-To-Air-Ratio, ($\left( F/A\right)_{\text{St}}$). The follow sections show this calculation for five different test cases.
Case #1: Coal Dust (40% CH4, 60% Ash)
In this case, the dust is assumed to contain only reactive volatiles and inert ash. The concentration of fuel is then the fraction of volatiles multiplied by the overall dust concentration.
$\mathit{\sigma}_{\text{F}} =\psi_{\text{v}}\mathit{\sigma}_{\text{p}}$
where $\mathit{\sigma}_{\text{p}}$ is the overall dust concentration (kg/m3) and $\psi_{\text{v}}$ is the fraction of volatiles.
Now that the fuel concentration is defined, the next step is to determine Fuel-To-Air ratio under stoichiometric conditions.
Table 1 shows the balanced chemical equation and resulting mass of each specie. It is also useful to perform some checks that mass and mols/mass fractions balance out and these are done in the attached worksheet.
CH4 O2 N2 $\rightarrow$ CO2 H2O N2
M (mol) 1.0 2.0 7.523 1.0 2.0 7.523
Mw (g/mol) 16.043 31.999 28.013 44.010 18.015 28.031
Mass (g) 16.043 63.998 210.762 44.010 36.031 210.762
From the balanced chemical equation, the stoichiometric Fuel-To-Air ratio is calculated as the mass of methane divided by the combined mass of oxygen and nitrogen:
$\left( F/A\right)_{\text{St}} = \frac{16.043}{63.998 + 210.762} = 0.0583$
The final relation for combustible dust equivalence ratio can now be determined. Using the specified volatile fraction of 0.4 the equation can be written as
$\mathit{\Phi} = \frac{0.4\cdot\mathit{\sigma}_{\text{p}}}{0.0583\cdot\left(1.2\text{kg/m}^{3}\right)} = 5.717\mathit{\sigma}_{\text{p}}$
again, where $\mathit{\sigma}_{\text{p}}$ is specified in kg/m3. This relation is shown in the plot at the top of this post along with the other materials explained in the following sections.
The stoichiometric dust concentration can be determined by setting the equivalence ratio to 1.0, and gives a value of 175 g/m3.
Case #2: Coal Dust (40% CH4, 60% Carbon)
Often both gas phase volatile reaction and solid phase surface reaction are considered for coal particles.
Assuming a coal particle that is 40% volatile and 60% carbon, the balanced chemical equation given in Table 2 is developed.
Table 2: Balanced chemical equation assuming methane volatiles and surface reaction [Download Worksheet]
CH4 C O2 N2 $\rightarrow$ CO2 H2O N2
M (mol) 1.000 2.004 4.004 15.061 3.004 2.000 15.061
Mw (g/mol) 16.043 12.011 31.999 28.013 44.010 18.015 28.031
Mass (g) 16.043 24.064 128.108 421.896 132.186 36.031 421.896
Now the numerator in the stoichiometric fuel-to-air ratio includes both fuel components
$\left( F/A\right)_{\text{St}} = \frac{16.043 + 24.064}{128.108 + 421.896} = 0.0729$
and the combustible dust equivalence ratio is computed as
$\mathit{\Phi} = \frac{\mathit{\sigma}_{\text{p}}}{0.0729\cdot\left(1.2\text{kg/m}^{3}\right)} = 11.431\mathit{\sigma}_{\text{p}}$
again where $\mathit{\sigma}_{\text{p}}$ is specified in kg/m3. Note that the entire particle now contains fuel instead of just a fraction as shown in Case #1.
Since the entire particle can contribute to combustion the equivalence ratio is substantially larger. The stoichiometric dust concentration under these conditions is 87 g/m3.
Case #3: Organic Dust (CwHxOy)
The same approach can be used for any combustible dust as long as the volatile products and combustion reaction are known.
A general equation for an organic combustible dust is given on Page 60 in the text “Dust Explosion Dynamics” by Dr. Russell Ogle [1], along with a detailed description of how to compute dust equivalence ratio.
$\text{C}_{w}\text{H}_{x}\text{O}_{y} + \left(w + \frac{x}{4} - \frac{y}{2}\right) \left(\text{O}_{2} + 3.76\text{N}_{2}\right) \rightarrow w\text{CO}_{2} + \frac{x}{2}\text{H}_{2}\text{O} + 3.76\left(w + \frac{x}{4} - \frac{y}{2}\right)\text{N}_{2}$
The worksheet for this example allows $w$, $x$, and $y$ to be modified. Example 3.3 of Ogle, 2017 [1] uses the chemical species C1H2.01O0.8, proposed by Medina et al., 2013 [2]. Table 3 gives the balanced chemical equation for this material.
Table 3: Balanced chemical equation for organic dust with the chemical makeup C1H2.01O0.8 [Download Worksheet].
C1H2.01O0.8 O2 N2 $\rightarrow$ CO2 H2O N2
M (mol) 1.000 1.102 4.147 1.000 1.005 4.147
Mw (g/mol) 26.836 31.999 28.013 44.010 18.015 28.031
Mass (g) 26.836 35.279 116.182 44.010 18.105 116.182
Similarly to coal dust the stoichiometric Fuel-To-Air ratio is determined from the mass of fuel and air in the table
$\left( F/A\right)_{\text{St}} = \frac{26.836}{35.279 + 116.182} = 0.177$
resulting in the following relation for equivalence ratio
$\mathit{\Phi} = \frac{\mathit{\sigma}_{\text{p}}}{0.177\cdot\left(1.2\text{kg/m}^{3}\right)} = 4.703\mathit{\sigma}_{\text{p}}$
From this the stoichiometric concentration of C1H2.01O0.8 can be calculated as 212 g/m3. This is within a few percent of the value calculated in the textbook of Ogle, 2017 [1].
Case #4: Aluminum Dust (Al)
The equivalence ratio for metal dusts is calculated using the same approach as organic dusts. Assuming the aluminum reacts with oxygen to create aluminum oxide the balanced equation given in Table 4 is developed.
Al O2 N2 $\rightarrow$ Al2O3 N2
M (mol) 4.0 3.0 11.285 2.0 11.285
Mw (g/mol) 26.981 31.999 28.013 101.96 28.031
Mass (g) 107.924 95.996 316.142 203.92 316.142
The stoichiometric Fuel-To-Air ratio is the mass of aluminum divided by the mass of air
$\left( F/A\right)_{\text{St}} = \frac{107.924}{95.996 + 316.142} = 0.261$
which results in the following relation for dust equivalence ratio.
$\mathit{\Phi} = \frac{\mathit{\sigma}_{\text{p}}}{0.261\cdot\left(1.2\text{kg/m}^{3}\right)} = 3.182\mathit{\sigma}_{\text{p}}$
Using this relation the stoichiometric concentration of aluminum is calculated as 314 g/m3.
Case #5: Iron Dust (Fe)
The last case given in this post is for iron dust and shows that the equivalence ratio decreases with a decrease in metalic dust molecular weight.
Assuming that iron oxide is the combustion product, the balanced chemical equation in Table 5 is developed.
Fe O2 N2 $\rightarrow$ FE\e2O3 N2
M (mol) 4.0 3.0 11.285 2.0 11.285
Mw (g/mol) 55.845 31.999 28.013 159.690 28.031
Mass (g) 223.380 95.996 316.142 319.380 316.142
The stoichiometric Fuel-To-Air ratio is calculated from the table
$\left( F/A\right)_{\text{St}} = \frac{223.380}{95.996 + 316.142} = 0.542$
giving the following relation for dust equivalence ratio.
$\mathit{\Phi} = \frac{\mathit{\sigma}_{\text{p}}}{0.542\cdot\left(1.2\text{kg/m}^{3}\right)} = 1.537\mathit{\sigma}_{\text{p}}$
Due to the lower molecular weight, the stoichiometric concentration of iron is quite a bit higher than aluminum at 650 g/m3.
Conclusion
This post demonstrated how to compute combustible dust equivalence ratio using a simple mathematical relationship and stoichiometry.
The results demonstrate that the equivalence ratio is linearly related to dust concentration and the slope depends on the density of air (pressure and temperature) and stoichiometric Fuel-To-Air ratio.
If you have any questions or would like to help others understand combustible dust fundamentals please share and comment using the buttons/fields at the bottom of the page.
Thanks again and I look forward to sharing more soon!
Below is a list of related textbooks mentioned in this post. Note that the images and links are affiliate links and I will receive a small commission if you make a purchase after clicking.
Dust Explosion DynamicsRussell A. Ogle(US, CAN, GER, UK)
References
[1] R. Ogle, Dust explosion dynamics, Elsevier Inc, 2017.
@BOOK{Ogle2017,
title={Dust Explosion Dynamics},
author={Ogle, Russell},
publisher={Elsevier Inc},
year={2017},
}
@ARTICLE{Medina2013,
}
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# Question on irreducible complex characters
We let $$G$$ be a finite group.
If $$\chi$$ is a complex character of $$G$$, we define $$\overline{\chi}:G \to \mathbb{C}$$ by $$\overline{\chi}(g)=\overline{\chi(g)}$$ for all $$g \in G$$. We write
$$\nu(\chi):= \frac{1}{|G|}\displaystyle\sum_{g \in G}\chi(g^2)$$
for the Frobenius Schur Indicator.
We let Irr($$G$$) denote the set of irreducible complex characters of $$G$$. We want to show that:
$$\displaystyle\sum_{\chi\in Irr(G)}\nu(\chi)\chi(1)=|\{h \in G:h^2=1\}|$$
There is a hint: Define $$\alpha:G \to \mathbb{C}$$ by $$\alpha(g)=|\{h \in G: h^2 = g\}|$$. Prove that $$\alpha$$ is a class function and use that Irr$$(G)$$ is an orthonormal basis of the vector space $$R(G)$$ of class function of $$G$$.
So we first try to show that $$\alpha$$ is a class function, i.e. we want to show that $$|\{h \in G:h^2=g\}|=|\{h \in G:h^2=xgx^{-1}\}|$$, for all $$x,g \in G$$, but I really cannot see how this is true.
As for the second part, assuming that $$\alpha$$ is indeed a class function, we can write $$\alpha$$ (second part of the hint) as $$\alpha=\displaystyle \sum_{\chi \in Irr(g)}\langle\alpha,\chi\rangle\chi = \displaystyle \sum_{\chi \in Irr(g)}\frac{1}{|G|}\displaystyle \sum_{g \in G}\langle\alpha(g),\overline{\chi(g)}\rangle \chi$$
but I am not sure at all how to proceed from here.
This is all in relation with this question
Any help is much appreciated.
• Conjugation by $x$ is an automorphism of $G$. – Lord Shark the Unknown Apr 10 at 14:02
• Thank you. I'm not sure how the result ($\alpha$ being a class function) follows from this. – amator2357 Apr 10 at 14:16
Let us first define the set $$A(g)=\{h \in G: h^2=g\}$$ and $$\alpha(g)=|A(g)|$$, its cardinality. First observe that $$\alpha$$ is a class function, that is, it is constant on conjugacy classes: fix for the moment an $$x \in G$$ and define a map from $$A(g) \rightarrow A(x^{-1}gx)$$ by $$h \mapsto x^{-1}hx$$. This map is well-defined: $$(x^{-1}hx)^2=x^{-1}h^2x=x^{-1}gx$$, so $$x^{-1}hx \in A(x^{-1}gx)$$. The map is also injective: if $$x^{-1}hx=x^{-1}kx$$, then obviously $$h=k$$. And it is surjective: if $$k \in A(x^{-1}gx)$$ then $$xkx^{-1} \in A(g)$$ and $$xkx^{-1}$$ maps to $$k$$. Hence $$\alpha(g)=\alpha(x^{-1}gx)$$ for every $$x \in G$$.
Now $$\alpha$$ is a class function, and it takes non-negative integer values. This does not make it into a character, but since the irreducible characters of $$G$$ form an orthonormal basis for the class functions we can write $$\alpha=\sum_{\chi \in Irr(G)}\nu(\chi)\chi$$, with $$\nu(\chi) \in \mathbb{Z}_{\geq 0}$$. Now we need to show that in fact $$\nu(\chi)=\frac{1}{|G|}\sum_{g \in G}\chi(g^2)$$ From the formula for $$\alpha$$ it follows that $$\nu(\chi)=[\chi,\alpha]=\frac{1}{|G|}\sum_{g \in G}\chi(g)\overline{\alpha(g)}=\frac{1}{|G|}\sum_{g \in G}\chi(g)\alpha(g)$$. Note that $$\chi(g)\alpha(g)=\sum_{\{h \in G: h^2=g\}}\chi(h^2)$$, we get the formula for $$\nu(\chi)$$. Finally, observe that $$\alpha(1)=|\{h \in G: h^2=1\}|$$. So $$\alpha(1)=\sum_{\chi \in Irr(G)}\nu(\chi)\chi(1)= |\{h \in G: h^2=1\}|$$ as wanted.
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Home Teachers' Resources GEOGRAPHY UPDATED SCHEMES OF WORK FORM 2 PDF
# GEOGRAPHY UPDATED SCHEMES OF WORK FORM 2 PDF
SCHEME OF WORK FORM THREE GEOGRAPHY TERM ONE 2011
# WK
NO.
NO.
## TOPIC/
SUBTOPIC
LESSON OBJECTIVES TEACHING / LEARNING ACTIVITIES RESOURCES REFERENCES REMARKS
# 1
## STATISTICAL METHODS.
Compound / Cumulative Bar Graphs.
By the end of the lesson, the learner should be able to:
Highlight steps followed when constructing compound / cumulative bar graphs
Q/A to review basic statistical concepts, simple bar graphs;
Exposition of steps followed when constructing compound e bar graphs.
Certificate Geography
Book III
Pg 1-2
### 2
Compound / Cumulative Bar Graphs. Construct compound / cumulative bar graphs. Copy and complete tables of variables.
Worked example;
Supervised practice.
Graph papers,
Calculators.
Certificate Geography
Book III
Pg 2-3
### 3
Compound / Cumulative Bar Graphs. Interpret compound/ cumulative bar graphs.
Probing questions & brief discussion. Certificate Geography
Book III
Pg 3-4
## Simple Pie charts (Divided Circle).
– construction.
Construct simple pie charts.
Students determine value of angles for the variables.
Guided activity &
Brief discussion.
Calculators. Certificate Geography
Book III
Pg 14
# 2
## Simple Pie charts (Divided Circle).
– interpretation.
Interpret simple pie charts
Guided discovery;
Written exercise.
Calculators. Certificate Geography
Book III
Pg 8-11
# 2
## Proportional circles.
Interpret proportional circles. Teacher explains the steps followed to construct proportional circles.
Students carry out some calculations.
Discussion & Interpretations.
Calculators. Certificate Geography
Book III
Pg 12-14
## Proportional circles.
Construct proportional circles.
Guided discovery;
Written exercise.
Certificate Geography
Book III
Pg 10-11
## MAP WORK.
Terms used &
precautions observed when describing physical features & human activities.
Identify terms used to describe physical features & human activities represented in a map.
Outline precautions observed when describing physical features & human activities.
Oral questions & detailed discussion.
Topographical maps.
Certificate Geography
Book III
Pg 14-16
# 1
Physical features.
Identify physical features. Brain storming on
features on topographical maps.
Drawing illustrative diagrams;
Discussion.
Topographical maps.
Certificate Book III Geography
Pg 16-19
# 2
Relief features. Identify relief features on a map. Locating relief features;
Drawing illustrative diagrams;
Discussion.
Certificate Book III Geography
Pg 19-22
# 3
## Land forms.
Identify landforms on a topographical map using contours.
Identify types of vegetation on a topographical map.
Q/A & brief discussion;
Supervised practical activity.
Topographical maps. Certificate Book III Geography
Pg 18-21
# 4
Watersheds. Identify watersheds on a map. Locating watersheds;
Drawing illustrative diagrams;
Discussion.
Certificate Book III Geography
Pg 23-25
## Vegetation.
Identify types of vegetation.
Q/A & brief discussion: rolling, dissected, hilly landforms & valleys. Topographical maps. Certificate Book III Geography
Pg 25
## Vegetation cover.
Interpret vegetation cover in terms of amount of rainfall received in the area, and the types of soils in the area.
Locate forests, thickets, and riverine trees on a map.
Deduce amount of rainfall received and the likely types of soils in specific parts of an area.
Topographical maps.
Certificate Book III Geography
Pg 25
## Drainage.
Identify natural /artificial hydrographic features in a map.
Make deductions from the hydrographic features evident on the drainage of a given section in a map.
Detailed discussion on drainage of sections in the map.
Deduce amount of rainfall received and nature of underground rocks, etc.
Topographical maps.
Certificate Book III Geography
Pg 27-30
## Drainage patterns.
Describe drainage patterns of an area represented by a map.
Oral questions & brief discussion.
Topographical maps.
Certificate Book III Geography
Pg 30
# 5
## Drainage features.
Identify drainage features on a map.
Describe formation of drainage features.
Outline economic importance of drainage features.
Brief discussion & Q/A.
Topographical maps.
Certificate Book III Geography
Pg 30-31
## Human and economic activities:
Crop farming, livestock rearing &
Mining & fishing.
Identify clues on a map that indicate presence of crop farming and livestock rearing.
Identify clues on a map that indicate presence of mining and fishing.
Q/A & practical activities.
Group activities.
Topographical maps.
Certificate Book III Geography
Pg 31-32
## Tourism.
Identify clues on a map that indicate manufacturing, processing and tourism in an area.
Q/A & practical activities.
Topographical maps.
Certificate Book III Geography
Pg 32-33
### 4
Outline factors influencing settlement.
Identify types of settlement patterns.
Exposition of new concepts.
Probing questions and discussion.
Topographical maps.
Certificate Book III Geography
Pg 34
# 6
## Transport & Communication.
Identify modes of transport & communication of an area as depicted in a topographical map.
Practical activities.
Brief discussion.
Topographical maps.
Certificate Book III Geography
Pg 35
# 6
## Enlargement & Reduction of maps.
Enlarge / reduce a map or a section of a map using the square method.
Explanations & practical activities.
Topographical maps.
Certificate Book III Geography
Pg 39-40
## Sections and profiles.
Define the term section as used in mapwork.
Explain importance of sections in interpreting maps.
Draw various types of sections and profiles.
Exposition & brief discussion.
Guided practical activities.
Topographical maps.
Certificate Book III Geography
Pg 41-44
# 7
2
## Vertical Exaggeration (VE),
&
Intervisibility.
Calculate VE of a map.
Determine the steepness of a slope between two given points.
Determine whether two points in a map are intervisible.
Guided calculations.
Guided practical activities.
Practical activity to determine intervisibility.
Topographical maps.
Certificate Book III Geography
Pg 45-47
### 3,4
TEST & MID-TERM BREAK
# 8
## WEATHERING
Definition
&
Agents of weathering.
Define weathering.
Identify agents of weathering.
Explain factors that influence weathering.
Brain storming;
Q/A & detailed discussion.
Certificate Book III Geography
Pg 54-55
## Physical weathering.
Define physical weathering.
Explain major forms of physical weathering.
Oral questions, brief discussion. Certificate Book III Geography
Pg 55-59
## Chemical weathering.
Explain ways in which chemical weathering takes place. Probing questions & discussion. Certificate Book III Geography
Pg 60-61
# 9
## Biological weathering.
Importance of weathering.
Identify agents of biological weathering.
Explain importance of weathering.
Probing questions & discussion. Certificate Book III Geography
Pg 61-64
## MASS WASTING.
Factors influencing mass wasting.
Define the term mass wasting & mass movement.
Explain factors that influence mass wasting.
Q/A: review weathering.
Discussion: factors affecting nature or speed of mass wasting & movement.
Certificate Book III Geography
Pg 64-66
## Slow mass wasting.
Describe slow processes of mass wasting. Discussion: soil creep & talus creep.
Q/A: factors that may trigger slow mass wasting.
Certificate Book III Geography
Pg 66-68
## Rapid mass wasting.
Describe rapid processes of mass wasting. Exposition, Q/A & discussion. Certificate Book III Geography
Pg 68-71
# 10
## Effects of mass wasting.
Explain effects of mass wasting on physical and human environment. Brief discussion.
Highlight positive and negative effects.
Certificate Book III Geography
Pg 71-72
# 10
## THE HYDROLOGI-CAL CYCLE.
Input & output processes of the hydrological cycle.
Identify input and output processes of the hydrological cycle.
Q/A: forms of precipitation.
Q/A & brief discussion on the major output processes.
Chart – the hydrological cycle.
Certificate Book III Geography
Pg 73
### 3
Internal transfer processes of the hydrological cycle.
Describe movement of water in the atmosphere, land and oceans. Exposition of new terms & brief discussion.
Certificate Book III Geography
Pg 73
### 4
Storage processes of the hydrological cycle.
Significance of the hydrological cycle.
Identify surface storages, ground water storage and the cyrosphere.
Explain the significance of the hydrological cycle.
Exposition of new terms & brief discussion. CCertificate Book III Geography
Pg 73-74
# 11
## ACTION OF RIVERS.
Terms related to action of rivers.
Define terms related to action of rivers.
Exposition of new terms & Q/A with reference to diagrams and contour maps.
Contour maps, diagrams in textbooks.
Certificate Book III Geography
Pg 74-76
## River erosion.
Describe the processes of river erosion. Exposition of new terms,
Explanations & detailed
Discussion.
Certificate Book III Geography
Pg 76-77
# 11
## Factors affecting river erosion.
Explain factors affecting the rate of river erosion. Probing questions,
Brief discussion & illustrative diagrams.
Illustrative diagrams.
Certificate Book III Geography
Pg 77-78
## River transportation.
Describe processes of river transportation of load.
Brief discussion with exposition of new concepts.
Certificate Book III Geography
Pg 76-80
# 12
## River deposition.
Describe processes of river deposition of load.
Exposition of new concepts;
Brief discussion.
Certificate Book III Geography
Pg 80-85
# 12-
13
END OF TERM ONE ASSESSMENT TEST
SCHEME OF WORK FORM THREE GEOGRAPHY TERM TWO 2011
# WK
NO.
NO.
## TOPIC/
SUBTOPIC
LESSON OBJECTIVES TEACHING / LEARNING ACTIVITIES RESOURCES REFERENCES REMARKS
# 1
## The youthful stage of a river.
Identify features associated with the youthful stage of a river.
Detailed discussion and drawing of illustrative diagrams. Diagrams in textbooks.
Certificate Book III Geography
Pg 92, 77-83
## Mature stage of a river.
Identify features associated with the mature stage of a river.
Give examples of such features in Kenya.
Detailed discussion, drawing of illustrative diagrams. Diagrams in textbooks.
Certificate Book III Geography
Pg 83-84, 92
# 2
## The old stage of a river.
Identify features associated with the old stage of a river.
Give examples of such features in Kenya.
Exposition, detailed discussion and drawing of illustrative diagrams & fieldwork. Illustrative diagrams.
Certificate Book III Geography
Pg 85-92
### River capture.
Define the term river capture.
Identify conditions favouring river capture.
Give examples of river captures in Kenya.
Exposition, detailed discussion and drawing of illustrative diagrams. Certificate Book III Geography
Pg 93-94
## River rejuvenation.
Cite reasons for river rejuvenation.
Describe landforms that result from river rejuvenation.
Discussion with exposition and explanation of new terms. Illustrative diagrams. Certificate Book III Geography
Pg 94-95
# 3
### Drainage patterns.
Define the term drainage pattern.
Describe various drainage patterns.
Cite examples of specific drainage patterns in Kenya.
Review types of drainage patterns;
Detailed discussion on types of drainage patterns;
Illustrative diagrams &
Give examples of specific drainage pattern.
Illustrative diagrams. Certificate Book III Geography
Pg 95-99
### Drainage systems.
Give types of drainage systems. Exposition;
Brief discussion.
Certificate Book III Geography
Pg 99-101
## Significance of rivers and associated features.
Explain the significance of rivers and associated features to humankind and a country.
Q/A, brief discussion, giving examples pf important specific features. Certificate Book III Geography
Pg 101-2
# 4
## LAKES
Lakes formed by tectonic movements and downwarping.
Explain formation of rift lakes.
Explain formation of lakes by downwarping.
Q/A to review formation of the Rift Valley.
Discuss & cite examples of such lakes.
Maps showing distribution of lakes.
Certificate Book III Geography
Pg 177-8
## Lakes formed by volcanicity.
Explain formation of crater lakes and lava – dammed lakes.
Q/A to review volcanic action;
Brief discussion.
Maps showing distribution of lakes. Certificate Book III Geography
Pg 178-180
# 4
## Lakes resulting from erosion.
Explain formation of as a result of erosion.
Q/A and
Brief discussion.
Maps showing distribution of lakes. Certificate Book III Geography
Pg 180
## Lakes resulting from glaciation.
Explain formation of lakes resulting from glaciation.
Cite examples of lakes resulting from glaciation.
Brief discussion.
Students refer to maps.
Maps showing distribution of lakes. Certificate Book III Geography
Pg 180
# 5
## Lakes formed by deposition.
&
other modes.
Identify types of lakes formed by deposition.
Cite examples of lakes formed by other modes.
Brief discussion, citing examples & locating them in the map. Maps showing distribution of lakes. Certificate Book III Geography
Pg 181-2
## Significance of lakes.
Explain the significance of lakes to humankind and the country. Oral questions and brief discussion.
Topic assessment. / Assignment.
Bk III Cert. Geography
Pg 182-4
## OCEANS, SEAS & COASTS
Distinction between oceans and seas.
Nature of sea water.
Distinguish between oceans and seas.
Compare and contrast seas and oceans.
Describe nature of sea water.
Exposition. Q/A & discussion.
World map.
Bk III Cert. Geography
Pg 104-5
## Water movements in oceans.
Give reasons for vertical and horizontal movements of water. Probing questions & explanations. Cert. Bk III Geography
Pg 106-
# 6
## Major ocean currents.
State characteristics of major ocean currents.
Give examples of some ocean currents.
Discussion and oral questions.
Cert. Bk III Geography
Pg 182-4
## Tides.
Describe causes of tides.
Identify types of tides.
Probing questions & explanations. Cert. Geography
Pg 108-110
# 6
## Waves.
Describe formation of waves.
Identify types of waves.
Explanations and oral questions.
Bk III Cert. Geography
Pg 111-3
## Wave action & resultant features.
Describe processes of wave erosion.
Identify features resulting from wave action.
Explanations, probing questions & discussion.
Refer to diagrams.
Illustrative
Diagrams.
Bk III Cert. Geography
Pg 113-6
# 7
### 1-2
TEST & MID TERM BREAK
## Wave transportation.
Describe features resulting from wave transportation.
Review river transportation;
Brief discussion.
Bk III Cert. Geography
Pg 116-7
## Wave deposition.
Describe deposition by waves. Review river deposition.
Brief discussion.
Bk III Cert. Geography
Pg 117-120
# 8
## Features resulting from wave deposition.
Describe features resulting from wave deposition.
Review river deposition.
Brief discussion.
Bk III Cert. Geography
Pg 120-122
## Types of coasts:
Submerged coasts.
Identify types of submerged coasts.
Explanations & illustrative diagrams.
Bk III Cert. Geography
Pg 123-7
## Emerged coasts.
Identify types of emerged coasts. Explanations & illustrative diagrams. Bk III Cert. Geography
Pg 127-9
# 9
## Coral coasts.
Identify conditions necessary for growth of polyps.
Identify types of coral reefs.
Discussion & illustrative diagrams. Bk III Cert. Geography
Pg 130-134
## Significance of oceans, seas and coastal features.
Explain the significance of oceans, seas and coastal features to humankind and to a country / region.
Q/A, brief discussion. Cert. Geography
Bk III
Pg 134-6
## ACTION OF WIND AND WATER IN ARID AREAS.
Processes of wind erosion.
Describe processes of wind erosion.
Discussion: abrasion, attrition & deflation processes of erosion.
Exposition of new concepts.
Map of Africa showing distribution of arid zones.
Cert. Geography
Bk III
Pg 137-139
## Features resulting from wind erosion.
Identify features resulting from wind erosion. Explanations & drawing illustrative diagrams.
Cert. Geography
Bk III
Pg 139-142
# 10
## Wind transportation.
Explain the ways in which wind transports its load.
State factors affecting wind transportation.
Exposition, explanations & illustrative diagrams. Cert. Geography
Bk III
Pg 143-4
### 3,4
Features resulting from wind deposition. Identify features resulting from wind deposition.
Explain formation of wind deposition.
Exposition, explanations & drawing illustrative diagrams. Cert. Geography
Bk III
Pg 144-7
# 11
## Action of water in arid areas.
describe action of water in arid areas. Brain storming;
Brief discussion.
Cert. Geography
Bk III
Pg 147
## Resultant features of water action in arid areas.
Identify features resulting from action of water in arid areas.
Exposition, explanations and illustrative diagrams.
Illustrative diagrams. Cert. Geography
Bk III
Pg 147-152
## Significance of features in arid areas.
Explain the significance of features formed by water and wind action in arid zones.
Brain storming;
Brief discussion.
Cert. Geography
Bk III
Pg 152–3
# 12-13
END OF SECOND TERM ASSESSMENT TEST
SCHEME OF WORK FORM THREE GEOGRAPHY TERM THREE 2011
# WK
NO.
NO.
## TOPIC/
SUBTOPIC
LESSON OBJECTIVES TEACHING / LEARNING ACTIVITIES RESOURCES REFERENCES REMARKS
# 1
## ACTION OF WATER IN LIMESTONE AREAS.
Surface and underground water.
Describe processes leading to surface and underground water.
Probing questions on sources of water, infiltration of water, etc.
Brief discussion.
Cert. Geography
Bk III
Pg 154
## Occurrence of underground water.
Explain factors, which affect the occurrence of underground water.
Identify features resulting from underground water.
Explain the importance of underground water.
Exposition & explanations.
Probing questions.
Cert. Geography
Bk III
Pg 155-8
## Significance of underground water.
Outline significance of underground water. Brief discussion and probing questions. Cert. Geography
Bk III
Pg 158
## Karst landscape.
Describe development of a Karst landscape.
Explanations and illustrative diagrams. Cert. Geography
Bk III
Pg 159
# 2
## Karst features.
Identify Karst features on the surface and underground.
Explain the significance of Karst features.
Explanations and illustrative diagrams;
Q/A & brief discussion.
Cert. Geography
Bk III
Pg 160-163
# 2
## GLACIATION
Types of glaciers.
Define the terms glaciation and glacier.
Identify types of glaciers.
Brief discussion.
Cert. Geography
Bk III
Pg 1164-5
## Processes of glaciation and resultant features.
Describe the processes of glacial erosion, transportation and deposition.
Probing questions &
Drawing illustrative diagrams.
Illustrative diagrams. Cert. Geography
Bk III
Pg 166
# 3
## Glaciation in highland areas.
Identify features formed by glaciation in highland areas.
Describe formation of features by glaciation in highlands.
Exposition & explanations.
Drawing diagrams.
Illustrative diagrams. Cert. Geography
Bk III
Pg 166-8
## Glaciation in lowland areas.
Identify features formed by glaciation in lowland areas.
Describe formation of features by glaciation in lowland areas.
Exposition & explanations.
Drawing diagrams.
Illustrative diagrams. Cert. Geography
Bk III
Pg 172-3
# 4
## Significance of glaciation.
Highlight significance of glaciation. Q/A and brief discussion. Cert. Geography
Bk III
Pg 175-6
## SOIL.
Soil constituents.
Define the term soil.
Describe composition of soil.
Q/A: review soil profile, soil structure, etc.
Brief discussion.
Cert. Geography
Bk III
Pg 193-195
## Soil formation.
Explain processes through which soil is formed. Q/A: review weathering.
Detailed discussion & illustrative diagrams.
Cert. Geography
Bk III
Pg 195-6
# 5
## Factors of soil formation.
Explain factors influencing soil formation. Q/A & discussion. Cert. Geography
Bk III
Pg 196-8
## Properties of soil.
Identify characteristics of soil. Q/A & discussion &
practical activities e.g. determining soil pH; water retention capacity, porosity, etc.
Cert. Geography
Bk III
Pg 198-202
## Soil degeneration within a locality.
Identify types & causes of soil degeneration. Q/A: loss of soil fertility.
Detailed discussion.
Fieldwork.
# 6
## Soil degeneration.
Identify types & causes of soil degeneration in a specific area.
Fieldwork.
Cert. Geography
Bk III
Pg 208-210
## Agents of soil degeneration.
Identify agents of soil degeneration. Brain storming;
Brief discussion.
Cert. Geography
Bk III
Pg 210-1
## Effects of soil erosion.
Outline effects of soil erosion. Brain storming;
Brief discussion.
Cert. Geography
Bk III
Pg 211-2
## Zonal order.
Identify types of soils of the zonal order.
State characteristics of various types of soils of the zonal order.
Exposition & explanations.
Cert. Geography
Bk III
Pg 212
# 7
## Azonal order.
Identify azonal order soils.
State characteristics of various types of soils.
Exposition & explanations. Cert. Geography
Bk III
Pg 213
### 3-4
TEST & MID – TERM BREAK
# 8
## Intrazonal order.
Identify types of soils of the intrazonal order.
State characteristics of various intrazonal order soils.
Exposition & explanations. Cert. Geography
Bk III
Pg 214
## Soil conservation & management.
Identify soil conservation & management measures. Brief discussion & assignment. Cert. Geography
Bk III
Pg 215-8
## AGRICULTURE
Factors influencing Agriculture.
Explain the factors influencing Agriculture.
Q/A: review various aspects of Agriculture, its importance, etc.
Probing questions on factors affecting Agriculture; biotic, edaphic, economic, climate, etc.
Cert. Geography
Bk III
Pg 223-8
## Types of Agriculture.
Identify various types of farming systems and methods.
State characteristics of each type of farming system and method.
Q/A, detailed discussion.
Cite examples where each type of farming is successful.
Cert. Geography
Bk III
Pg 229-238
# 9
## Tea farming in Kenya.
Identify major tea-growing areas in Kenya.
State conditions necessary for tea growing.
Describe cultivation, processing and marketing of tea in Kenya.
Outline some achievements of KTDA.
Highlight some problems facing tea farming in Kenya.
Brain storming;
Probing questions;
Brief discussion.
Cert. Geography
Bk III
Pg 241-8
## Sugar cane growing in Kenya.
Identify major sugar cane-growing areas in Kenya.
State conditions necessary for sugar cane growing.
Describe cultivation, processing and marketing of sugar cane in Kenya.
Highlight some problems facing sugar cane farming in Kenya.
Brain storming;
Probing questions;
Brief discussion;
Assignments.
Cert. Geography
Bk III
Pg 248-254
## Maize growing in Kenya.
Identify major maize-growing areas in Kenya.
State conditions necessary for maize growing.
Describe cultivation, processing and marketing of maize in Kenya.
Highlight some problems facing maize farming in Kenya.
Brain storming;
Probing questions;
Brief discussion.
Cert. Geography
Bk III
Pg 254-260
## Cocoa in Ghana.
Identify major cocoa growing areas in Ghana.
State conditions necessary for cocoa growing.
Describe cultivation, processing and marketing of cocoa in Ghana.
Highlight some problems facing cocoa growing in Ghana.
Oral questions, brief discussion & explanations. Cert. Geography
Bk III
Pg 260-265
# 10
## Oil palm in Nigeria.
Identify major oil palm-growing areas in Nigeria.
State conditions necessary for oil palm growing in Nigeria.
Describe cultivation, processing and marketing of oil palm in Nigeria.
Highlight some problems facing oil palm growing in Nigeria.
Brain storming;
Probing questions;
Brief discussion.
Cert. Geography
Bk III
Pg 265-269
## Coffee in Kenya.
Identify major coffee growing areas in Kenya.
State conditions necessary for coffee growing.
Describe cultivation, processing and marketing of coffee in Kenya.
Highlight some problems facing coffee farming in Kenya.
Q/A, brief discussion
& Assignment.
Cert. Geography
Bk III
Pg 271-4
# 10
## Coffee in Brazil.
Identify conditions favourable for coffee farming in Brazil.
Outline problems facing coffee farming in Brazil.
Q/A & explanations & Assignment. Cert. Geography
Bk III
Pg 274-5
## Wheat growing in Kenya & Canada.
Identify conditions favouring wheat growing in Kenya & Canada.
Q/A & explanations. Cert. Geography
Bk III
277-282
# 11
## Horticulture in Kenya.
Identify conditions favouring growing horticulture in Kenya.
Give examples of horticultural crops and their importance.
Describe marketing of horticultural crops in Kenya.
Highlight problems facing horticulture in Kenya.
Brain storming;
Probing questions;
Brief discussion.
Cert. Geography
Bk III
283
## Horticulture in the Netherlands.
Identify conditions favouring horticulture in the Netherlands.
Describe marketing of horticultural crops in the Netherlands.
Highlight problems facing horticulture in the Netherlands.
Probing questions & brief discussion. Cert. Geography
Bk III
Pg 288
## Commercial dairy & beef farming in Kenya & in Denmark.
Outline conditions favouring commercial dairy & beef farming in Kenya.
Identify dairy / beef cattle breeds reared in Kenya.
Identify problems facing commercial dairy & beef farming in Kenya and the responses of the government to these problems.
Probing questions, brief discussion & assignment. Cert. Geography
Bk III
Pg 290
## Beef farming in Argentina.
Identify conditions favouring beef farming in Argentina.
Describe the organization of beef farms in Argentina.
Explanations, probing questions & discussion. Cert. Geography
Bk III
Pg 306
# 12-13
END OF YEAR TEST
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Cody
# Problem 2377. Area of a disk
Solution 1071838
Submitted on 4 Dec 2016 by MIN HYEOK LEE
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 = 2; y_correct = 12.57; assert(isequal(crcl_area(x),y_correct))
d = 2 12 e = 1×2 logical array 1 0 f = 12.5700 452.3900 y = 12.5700
2 Pass
x = 12; y_correct = 452.39; assert(abs(y_correct-crcl_area(x)) <= 0.01)
d = 2 12 e = 1×2 logical array 0 1 f = 12.5700 452.3900 y = 452.3900
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Episode #125 of the Stack Overflow podcast is here. We talk Tilde Club and mechanical keyboards. Listen now
# All Questions
5 questions
Filter by
Sorted by
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31 views
### Correspondence between time series models in continuous vs. discrete time
I am interested in an overview over the connection and correspondence between time series models in continuous vs. discrete time in finance. E.g. take ARMA(p,q) or GARCH(s,r) or ARMA(p,q)-GARCH(s,r) ...
145 views
### Continuous time Fourier representation
I have learned that the Fourier transform of a continuous-time unit-periodic stochastic process is: x(t) = \sum\limits_{k=-\infty}^{\infty} a_k e^{i2\pi kt} \quad \quad \text{ where } \quad \quad ...
66 views
### Simulating a (discretized) Cox process via binomial sampling
Let X be a Cox process (doubly-stochastic Poisson process) driven by a Poisson process with fixed intensity(rate) $\lambda=50$ , and choose some small time interval $dt=0.01$ . Is the proper way to ...
I have a continuous variable, $P_t$ whose evolution is unknown. However, I obtain a history of it i.e. $P_0, P_{dt}, P_{2dt}, ...... , P_T$. For a continuous process variable, I know that the rate of ...
The negative binomial distribution with parameters $p\in(0,1)$ and $t>0$ is sometimes defined as the distribution of the number of failures before the $t$th success. This is supported on the set \$\{...
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https://www.teachoo.com/1320/468/Ex-13.2--5---A-cylindrical-pillar-is-50-cm-in-diameter/category/Ex-13.2/
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Ex 13.2
Chapter 13 Class 9 Surface Areas and Volumes
Serial order wise
This video is only available for Teachoo black users
Get live Maths 1-on-1 Classs - Class 6 to 12
### Transcript
Ex 13.2, 5 A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of Rs.12.50 per m2. [ = 22/7] Pillar is in the form a cylinder Curved surface area of pillar = 2 rh Height (h) cylindrical pillar = 3.5 m Radius (r) of the circular end of pillar = 50/2 = 25 cm r = 25 1/(100 ) m Curved surface area of pillar = 2 rh = (2 22/7 0.25 3.5)m2 = (44 x 0.125) m2 = 5.5 m2 Cost of painting 1 m2 area = Rs 12.50 Cost of painting 5.5 m2 area = Rs (5.5 12.50) = Rs 68.75 Therefore, the cost of painting the CSA of the pillar is Rs 68.75
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https://www.ks.uiuc.edu/Research/namd/mailing_list/namd-l.2013-2014/3574.html
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# Re: Re: Re: about the vdw Interpolation calculation in NAMD source code
From: 卢禹锟 (luyukun_at_itp.ac.cn)
Date: Thu Nov 20 2014 - 20:35:48 CST
Hi,
The distance in my code is rij=sqrt(p_ij_x*p_ij_x+...), exactly the pair distance without any shift.
Could you list more directly how to return a switch vdw energy and force by using:
BigReal vdw_d = A * table_four_i[0] - B * table_four_i[4];
BigReal vdw_c = A * table_four_i[1] - B * table_four_i[5];
BigReal vdw_b = A * table_four_i[2] - B * table_four_i[6];
BigReal vdw_a = A * table_four_i[3] - B * table_four_i[7];
and also the qi*qj*erfc(rij)/rij with interpolation expression using:
BigReal slow_d = table_four_i[8 SHORT(+ 4)];
BigReal slow_c = table_four_i[9 SHORT(+ 4)];
BigReal slow_b = table_four_i[10 SHORT(+ 4)];
BigReal slow_a = table_four_i[11 SHORT(+ 4)];
Best
> -----原始邮件-----
> 发件人: "Jim Phillips" <jim_at_ks.uiuc.edu>
> 发送时间: 2014年11月21日 星期五
> 收件人: namd-l_at_ks.uiuc.edu, "卢禹锟" <luyukun_at_itp.ac.cn>
> 抄送:
> 主题: Re: Re: namd-l: about the vdw Interpolation calculation in NAMD source code
>
>
> I guess that you are missing that r2list[k] = r^2 + r2_delta, not r^2.
>
> The FULL and SLOW macros only affect splitting for multiple timestepping,
> which doesn't affect VDW forces at all.
>
> Jim
>
>
> On Thu, 20 Nov 2014, 卢禹锟 wrote:
>
> > Hi,
> >
> > Thank you at first! Actually, I want to calculate the all atom force and the coarse grained force acting on virtual site at the same time. The input of the parameter is surprisingly easy for me to understand and i didn't stuck long in this problem. But I feel confusion in how the interpolation method return the theoretical formula.
> >
> > The coarse grained interaction function i want to add is just the 12-6 LJ potential form plus with switch term.
> >
> > The electrostatic part is qi*qj*erfc(rij)/rij, but I have to use the interpolation method for efficiency.
> >
> > I guess the following code is the key interpolation form in calculating vdw ( in ComputeNonbondedBase2.h):
> > BigReal vdw_d = A * table_four_i[0] - B * table_four_i[4];
> > BigReal vdw_c = A * table_four_i[1] - B * table_four_i[5];
> > BigReal vdw_b = A * table_four_i[2] - B * table_four_i[6];
> > BigReal vdw_a = A * table_four_i[3] - B * table_four_i[7];
> >
> > the vdw energy term is calculated in this way:
> > vdwenergy of ij_pair = - ( ( ( diffa * vdw_d * (1/6.)+ vdw_c * (1/4.)) * diffa + vdw_b *(1/2.)) * diffa + vdw_a );
> >
> > but I find this energy does not equal to A/rij^12-B/rij^6 or (A/rij^12-B/rij^6)*switch_function(rij):
> >
> > from left to right: patch_atomindex_i, patch_atomindex_j, interpolation value, analytic value
> > 558 149: -7.48422e-11 -4.6242e-10
> > 558 150: -1.50365e-12 -2.78378e-10
> > 558 520: -5.98947e-06 -1.20509e-05
> > 558 521: -8.25325e-10 -1.26156e-09
> > 558 522: -5.30187e-10 -9.66423e-10
> > 558 40: -1.38634e-05 -2.38436e-05
> > 558 41: -2.01247e-07 -6.87222e-07
> >
> > (the relative difference far exceed the interpolation error).
> >
> > For the electrostatic part, I completely have no idea in the relation between FULL() or SLOW() macros and the calculation of qi*qj*erfc(rij)/rij.
> >
> > I know the theory of the fast and slow force splitting method in namd, but how the interpolation return the desired formula is thus a headache for me.
> >
> > Best Regards
> >
> >> -----原始邮件-----
> >> 发件人: "Jim Phillips" <jim_at_ks.uiuc.edu>
> >> 发送时间: 2014年11月20日 星期四
> >> 收件人: namd-l_at_ks.uiuc.edu, "卢禹锟" <luyukun_at_itp.ac.cn>
> >> 抄送:
> >> 主题: Re: namd-l: about the vdw Interpolation calculation in NAMD source code
> >>
> >>
> >> What is the functional form (including parameters) of the new LJ potential
> >> you want? If it is A f(r) + B g(r) and you want to replace the existing
> >> LJ potential for all atom pairs then you just change f(r) and g(r) from
> >> r^-12 and r^-6 to your new form, and figure out how to read in your
> >> parameters, which would be somewhat more complex. We might be able to
> >> support this more easily in the future, but it will help to know what
> >> kinds of things people actually want to do.
> >>
> >> Jim
> >>
> >>
> >> On Thu, 20 Nov 2014, 卢禹锟 wrote:
> >>
> >>> Dear all,
> >>
> >>
> >> I try to modified the namd code to implement a new type force field. I want some help in understanding the nonbonded force calculation like VDW in namd. I have found some local experts, but none of them show interesting in such detail.
> >>
> >>
> >> The source code involving nonbonded calculation (ComputeNonbondedBase.h ComputeNonbondedBase2.h) show a very abstract interpolation method and a large number of macro definitions, completely beyond my ability to fully understand it.
> >>
> >>
> >> Could some one give me some hint in how this is realized in namd. For example, using the interpolation to calculate a new LJ potential energy.
> >>
> >>
> >> Best Regards
> >>
> >>
> >>
> >>
> >>
> >>
> >>
> >>
> >> --
> >> 卢禹锟(YuKun Lu)
> >>
> >>
> >> Address: Institute of Theoretical Physics, CAS,
> >> Zhong Guan Cun East Street 55 #,
> >> P. O. Box 2735, Beijing 100190, P. R. China
> >>
> >>
> >> Email：lcyqky_at_Gmail.com
> >> luyukun_at_itp.ac.cn
> >
> >
> > --
> > 卢禹锟(YuKun Lu)
> >
> >
> > Address: Institute of Theoretical Physics, CAS,
> > Zhong Guan Cun East Street 55 #,
> > P. O. Box 2735, Beijing 100190, P. R. China
> >
> >
> > Email：lcyqky_at_Gmail.com
> > luyukun_at_itp.ac.cn
> >
> >
> >
> >
```--
Address: Institute of Theoretical Physics, CAS,
Zhong Guan Cun East Street 55 #,
P. O. Box 2735, Beijing 100190, P. R. China
Email：lcyqky_at_Gmail.com
luyukun_at_itp.ac.cn
```
This archive was generated by hypermail 2.1.6 : Wed Dec 31 2014 - 23:23:02 CST
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# Prove that $\sum_{1\le i<j\le n}\frac{x_ix_j}{(1-x_i)(1-x_j)} \le \frac{n(n-1)}{2(2n-1)^2}$
Let $$n \ge 2$$ be a an integer and $$x_1,...,x_n$$ are positive reals such that $$\sum_{i=1}^nx_i=\frac{1}{2}$$ Prove that $$\sum_{1\le i
Here is the source of the problem (in french) here
Edit:
I'll present my best bound yet on $$\sum_{1\le i This formula was derived in @GCab's Answer.
First let $$a_k=x_k/(1-x_k)$$ so we want to prove $$\left(\sum_{k=1}^na_k\right)^2-\sum_{k=1}^na_k^2\le \frac{n(n-1)}{(2n-1)^2}$$ But since $$\frac{x_k}{1-x_k}<2x_k\implies \sum_{k=1}^na_k<1 \quad (1)$$ Hence $$\left(\sum_{k=1}^na_k\right)^2\le \sum_{k=1}^na_k$$ Meaning $$\left(\sum_{k=1}^na_k\right)^2-\sum_{k=1}^na_k^2\le\sum_{k=1}^na_k(1-a_k)$$
Now consider the following function $$f(x)=\frac{x}{1-x}\left(1-\frac{x}{1-x}\right)$$
$$f$$ is concave on $$(0,1)$$ and by the tangent line trick we have $$f(x)\le f'(a)(x-a)+f(a)$$
set $$a=1/2n$$ to get $$a_k(1-a_k)\le\frac{4n^2\left(2n-3\right)}{\left(2n-1\right)^3}\left(x_k-\frac{1}{2n}\right)+ \frac{2(n-1)}{(2n-1)^2}$$ Now we sum to finish $$\sum_{k=1}^na_k(1-a_k)\le \frac{2n(n-1)}{(2n-1)^2}$$
Maybe by tweaking $$(1)$$ a little bit we can get rid of this factor of $$2$$
• From a math olympiad, I'll post the original problem just wait a minute.
– PNT
May 7 at 20:37
• Is it from an active (on-going) contest? In that case see math.meta.stackexchange.com/q/16774/42969 May 7 at 20:52
• No, you can read the date in the paper. It says 6 May so it's yesterday
– PNT
May 7 at 20:53
• @PNT Could you give some more information on the contest? I.e. what knowledge is assumed / What is the level of the contest? Who is it aimed at? This could greatly restrict the possible techniques May 8 at 9:24
• It’s aimed for high school students, for knowledge I think it requires just basic inequalities like CS or AM-GM. @Dr.Mathva
– PNT
May 8 at 12:26
Edit. Eventhough I had in mind the version of the ACM I needed, it all messed up when I tried to find a reference. As for the valid version, I will refer to Vasile Cirtoaje's Algebraic Inequalities. Old and new Methods, p. $$267$$.
We will employ a powerful technique developed by Vasile Cirtoaje back in $$2006$$ called the Arithmetic Compensation Method (see, for instance, this document).
Let to this end $$F(x_1, \ldots, x_n):=\sum_{1\leqslant i which is clearly symmetric and continuous on $$S:=\left\{(x_1, \ldots,x_n)\mid \sum_{i=1}^n x_i=\frac12, \forall i: x_i\geqslant 0\right\}$$. We will now refer to the Remark 1.1 from the document linked above, which basically states that
If $$F(x_1,x_2,x_3,\ldots,x_n)>F\left(\frac{x_1+x_2}2, \frac{x_1+x_2}{2}, x_3,\ldots,x_n\right)\label{(i)}\tag{i}$$ implies $$F(x_1,x_2,x_3,\ldots,x_n)\leqslant F(0, x_1+x_2, x_3, \ldots, x_n)$$, then $$F(x_1,x_2,x_3,\ldots,x_n)\leqslant \underset{1\leqslant k\leqslant n}\max F\left(\frac1{2k}, \ldots, \frac1{2k},0,\ldots ,0\right)$$
Notice that (\ref{(i)}) is equivalent to (where we will denote, for brevity, $$y:=\frac{1}2(x_1+x_2)$$) \begin{align*} \frac{x_1x_2}{(1-x_1)(1-x_2)}-\frac{y^2}{(1-y)^2}+\left(\frac{x_1}{1-x_1}+\frac{x_2}{1-x_2}-\frac{2y}{1-y}\right)\sum_{i=3}^n \frac{x_i}{1-x_i}>0\\ \iff \frac{y^2-x_1x_2}{(1-x_1)(1-x_2)(1-y)^2}\left[2y-1+2(1-y)\sum_{i=3}^n \frac{x_i}{1-x_i}\right]>0\\ \iff 2y-1+2(1-y)\sum_{i=3}^n \frac{x_i}{1-x_i}>0 \end{align*}
Where the last equivalence follows from $$y^2-x_1x_2\geqslant 0$$. We will now turn to the second inequality: \begin{align*} F(x_1,x_2,x_3,\ldots,x_n)- F(0, x_1+x_2, x_3, \ldots, x_n)\leqslant0\\ \iff \frac{x_1x_2}{(1-x_1)(1-x_2)}+\left(\frac{x_1}{1-x_1}+\frac{x_2}{1-x_2}-\frac{2y}{1-y}\right)\sum_{i=3}^n \frac{x_i}{1-x_i}\leqslant0\\ \iff \frac{x_1x_2}{(1-x_1)(1-x_2)(1-2y)}\left[1-2y+2(y-1)\sum_{i=3}^n \frac{x_i}{1-x_i}\right]\leqslant 0\\ \end{align*}
Which is easily seen to follow from (\ref{(i)}). Thus, we conclude that \begin{align*} F(x_1,x_2,x_3,\ldots,x_n)&\leqslant \underset{1\leqslant k\leqslant n}\max F\left(\frac1{2k}, \ldots, \frac1{2k},0,\ldots ,0\right)\\ &= \underset{1\leqslant k\leqslant n}\max \binom{k}{2}\frac{1}{(2k-1)^2}\\ &= \underset{1\leqslant k\leqslant n}\max \frac{k(k-1)}{2(2k-1)^2} \end{align*}
But $$f: x\mapsto \frac{x(x-1)}{2(2x-1)^2}$$ is increasing on $$[1,\infty)$$, and, hence, the result follows.
Remark. Based on the difficulty of the other contest problems, this solution seems a bit overkill to me, but I have failed in the attempt to find a simpler method, since most well-known inequalities work in the other direction...
• Nitpicking: As I understand it, the used “AC method” requires the domain $S$ to be compact. May 8 at 14:45
• I'm confused... Don't we want to prove $\leq \frac{k(k-1)}{2(2k-1)^2}$? May 13 at 12:21
• +1, wonderful proof and method. Would you mind I ask you a question regarding the paper ijpam.eu/contents/2012-80-3/6/6.pdf you quoted before? Did the author make a small mistake in Equation (3.6) in that the summand thereof should be $\displaystyle\frac1{(mx_i-x_1x_2y)(mx_i-t^2y)}$? It seems the author misses the factor $mx_i-t^2y$ on the denominator.
– Hans
May 16 at 1:24
• Just wondering: For the proof of the ACM, the iteration process can in pinciple be infinitely long, or? Starting with $x_1>...>x_n$, we have $y_1=y_2=\frac{x_1+x_2}{2}$ and $x_1>y_1=y_2>y_3>...y_n=x_n$. Repeating the step successively with $y_2$ and $y_3$ will give a nested sequence $z_{1,k},...,z_{n,k}$, where $z_{m,0}=x_m$ and $z_{m,1}=y_m$, with the properties $$x_1>z_{1,k}>...>z_{n,k}>x_n \, .$$ $z_{1,k}$ decreases monotonically, and $z_{n,k}$ increases monotonically in $k$, while both being bounded below/above. Thus we have convergence, but why does it converge to the same number? May 16 at 13:36
• @Diger: Did you see Lemma 2.1-2.3 in ijpam.eu/contents/2012-80-3/6/6.pdf?
– Hans
May 16 at 20:09
You should be able to look for the maximum, by the method of Lagrange. We set $$F(x_1,...,x_n) \equiv \sum_{i\neq j} \frac{x_i x_j}{(1-x_i)(1-x_j)} \stackrel{\text{wts.}}{\leq} \frac{n(n-1)}{(2n-1)^2} \, ,$$ and define the Lagrange-function as $$L(x_1,...,x_n) = F(x_1,...,x_n) - \lambda \left(x_1+...+x_n-1/2\right) \,.$$ At an extremum we have $$\nabla L = 0 \qquad , \qquad \nabla=\begin{pmatrix} \partial_{x_1} \\ \vdots \\ \partial_{x_n} \\ \partial_\lambda\end{pmatrix} \, ,$$ in coordinates \begin{align}\frac{2}{(1-x_k)^2} \sum_{\substack{i=1 \\ i\neq k}}^n \frac{x_i}{1-x_i} &= \lambda \quad\text{for}\quad k=1,...,n \tag{1} \\ \sum_{i=1}^n x_i &= 1/2 \tag{2} \, .\end{align} Multiplying Equation (1) by $$\frac{(1-x_k)^2}{2}$$ and adding $$\frac{x_k}{1-x_k}$$ to both sides gives $$c=\sum_{i=1}^n \frac{x_i}{1-x_i} = \frac{\lambda}{2} (1-x_k)^2 + \frac{x_k}{1-x_k} \tag{3}$$ where the sum on the LHS is now independent on $$k$$ and we can consider it as a constant $$c$$. We now sum (3) from $$k=1$$ to $$n$$ \begin{align} nc&=\frac{\lambda}{2} \left( n - 1 + \sum_{k=1}^n x_k^2\right) + c \\ \Rightarrow \quad c&= \frac{\lambda}{2} + \frac{\lambda}{2(n-1)} \sum_{k=1}^n x_k^2 > \frac{\lambda}{2} \tag{4} \end{align} which we will need to use in a second. Rearranging Equation (3) gives a monic polynomial in $$x_k$$ $$x_k^3 - 3x_k^2 + \left( 3 - \frac{2+2c}{\lambda} \right)x_k + \frac{2c}{\lambda} - 1 = 0 \, . \tag{5}$$ A cubic can have either only 1 real solution, in which case there is nothing to show, since then $$x_1=...=x_n$$ follows, or it can have 3 real solutions. In this case there appears to be a great variety of combinations of the three roots for the various $$x_k$$. However, since the product of the three roots is equal to $$1-\frac{2c}{\lambda}<0$$ by Equation (4) and (5), either all three roots are negative (discard), or only one root is negative and we are left with two choices for the various $$x_k$$. But then, since the sum of all three roots is equal to $$3$$ by Equation (5) and one root is negative, the sum of the two positive roots must be strictly greater than $$3$$. This implies, that at least one of them is strictly greater than $$1/2$$, which stands in contradiction to Equation (2). Thus, we are left with only one valid real solution satisfying Equation (2) and $$x_1=...=x_n=x$$ follows. It is then easy to see that $$x=1/2n$$.
It remains to show, that this extremal point is actually a maximum. The hessian of $$L$$ is given in coordinates by $$\frac{1}{2} \, {\rm Hess}_{k,l} (L) = \frac{2\delta_{k,l}}{(1-x_k)^3} \sum_{i\neq k} \frac{x_i}{1-x_i} + \frac{1-\delta_{k,l}}{(1-x_k)^2(1-x_l)^2} \qquad \text{for} \qquad k,l=1,...,n \\ \frac{1}{2} \, {\rm Hess}_{n+1,k} (L) = \frac{1}{2} \, {\rm Hess}_{k,n+1} (L) = -1 \qquad \text{for} \qquad k=1,...,n \\ \frac{1}{2} \, {\rm Hess}_{n+1,n+1} (L) = 0 \, .$$ With $$x_k=x=1/2n$$ the first line becomes $$\frac{1}{2} \, {\rm Hess}_{k,l} (L) = \frac{1-\frac{\delta_{k,l}}{n}}{\left(1-\frac{1}{2n}\right)^4} \qquad \text{for} \qquad k,l=1,...,n$$ and in matrix-form this looks like $$\frac{1}{2} \, {\rm Hess} (L) = \frac{1}{\left(1-\frac{1}{2n}\right)^4} \begin{pmatrix} 1-1/n & 1 & \dots & 1 & -1 \\ 1 & 1-1/n & \dots & 1 & -1 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 1 & 1 & \dots & 1-1/n & -1 \\ -1 & -1 & \dots & -1 & 0 \end{pmatrix} \, .$$ Then, defining the variational vector $$\epsilon=(\epsilon_1,...,\epsilon_n,\epsilon_\lambda)^t$$, we find for the second variation $$\frac{1}{2} \sum_{k,l=1}^{n+1} \epsilon_k \, {\rm Hess}_{k,l} (L) \, \epsilon_l = - \frac{\sum_{k=1}^n \epsilon_k^2/n}{\left(1-\frac{1}{2n}\right)^4}<0$$ $$\forall \epsilon \in {\mathbb R}^{n+1}$$ satisfying $$\sum_{k=1}^n \epsilon_k = 0 \, ,$$ which is a consequence of Equation (2).
• I'm not familiar with lagrange multipliers. Are you sure there is no other method using Jensen or tangent line?
– PNT
May 15 at 13:50
• Probably there must be some trivial solution, if I interpret @G Cab's "hint" correctly, but unfortunately he doesn't want to share ;). While his hint is pretty obvious, it did not help me in any way to find such simple solution. Lagrange multipliers is the first thing I think of, when trying to maximize/minimize something, given a constraint. May 15 at 14:14
• +1. Nice Lagrange multiplier proof.
– Hans
May 23 at 1:07
Hint:
Indicating as $$S_n , \, T_n$$ $$\begin{array}{l} T_n = \sum\limits_{1 \le i < j \le n} {a_i a_j } \\ S_n = \sum\limits_{1 \le i,j \le n} {a_i a_j } = \sum\limits_{1 \le i \le n} {\sum\limits_{1 \le j \le n} {a_i a_j } } = \sum\limits_{1 \le i \le n} {a_i } \sum\limits_{1 \le j \le n} {a_j } = \left( {\sum\limits_{1 \le i \le n} {a_i } } \right)^2 \\ \end{array}$$ then you also have that $$\begin{array}{l} S_n = \sum\limits_{1 \le i,j \le n} {a_i a_j } = \sum\limits_{1 \le i < j \le n} {a_i a_j } + \sum\limits_{1 \le i = j \le n} {a_i a_j } + \sum\limits_{1 \le j < i \le n} {a_i a_j } = \\ = 2T_n + \sum\limits_{1 \le i \le n} {a_i ^2 } \\ \end{array}$$
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Clean Pennies With Vinegar
Directions
First , you mix the vinegar an the salt together,
Then, you put 5 pennies in the bowl and count to 10 slowly,
Finally you wash the pennies of with water.
Question/problem
Will the pennies get clean if you put them in the salt and vinegar mixture
Hypothisis
If you put the pennies in vinegar then they will get clean.
Materials
*A few old, not so shiny pennies
* 1/4 cup of white vinegar
* 1 teaspoon of salt
* Non metal bowl
* Paper towls
Control Group ( something that is constant or unchanged )
Something that is not changed is the * Amount of vinegar
* Amount of salt
Independent Variable ( something you change )
Something that you can change is put a little bit more vinegar and see if the pennies get cleaner.
Dependent Variable ( something you measure )
You measure how clean the pennies get.
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# Math
posted by on .
A lighthouse is east of a Coast Guard patrol boat. The Coast Guard station is 20 km north of the lighthouse. The radar officer aboard the boat measures the angle between the lighthouse and the station to be 23°. How far is the boat from the station?
• Math - ,
draw a diagram to see that
20/d = sin23°
• Math - ,
47.11
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https://smashingessays.com/a-study-of-10-different-weight-loss-programs-involved-500-subjects-each-of-the-10-programs-had-50-subjects-in-it-the-subjects-were-followed-for-12-months-weight-change-for-each-subject-was-recorded/
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Get help from the best in academic writing.
# A study of 10 different weight loss programs involved 500 subjects. Each of the 10 programs had 50 subjects in it. The subjects were followed for 12 months. Weight change for each subject was recorded. The researcher wants to test the claim that all ten programs are equally effective in weight loss.
A study of 10 different weight loss programs involved 500 subjects. Each of the 10 programs had 50 subjects in it. The subjects were followed for 12 months. Weight change for each subject was recorded. The researcher wants to test the claim that all ten programs are equally effective in weight loss..
STAT 200
Answer all 20 questions. Make sure your answers are as complete as possible, particularly when it asks for you to show your work. Answers that come straight from calculators, programs or software packages without any explanation will not be accepted. If you need to use technology (for example, Excel, online or hand-held calculators, statistical packages) to aid in your calculation, you must cite the sources and explain how you get the results. For example, state the Excel function along with the required parameters when using Excel; describe the detailed steps when using a hand-held calculator; or provide the URL and detailed steps when using an online calculator, and so on.
1. A study of 10 different weight loss programs involved 500 subjects. Each of the 10 programs had 50 subjects in it. The subjects were followed for 12 months. Weight change for each subject was recorded. The researcher wants to test the claim that all ten programs are equally effective in weight loss.
(a) Which statistical approach should be used?
(i) confidence interval
(ii) t-test
(iii) ANOVA
(iv) Chi square
(b) Explain the rationale for your selection in (a). Specifically, why would this be the appropriate statistical approach?
2. A study was conducted to determine whether the mean braking distance of four-cylinder cars is greater than the mean braking distance of six-cylinder cars. A random sample of 20 four- cylinder cars and a random sample of 20 six-cylinder cars were obtained, and the braking distances were measured.
(a) What would be the appropriate hypothesis test for this analysis?
(i) t-test for two independent samples
(ii) t-test for dependent samples
(iii) z-test for population mean
(iv) correlation
(b) Explain the rationale for your selection in (a). Specifically, why would this be the appropriate statistical approach?
3. A history professor took a sample of 10 final exam scores from a class of 30 students. The 10 scores are shown in the table below:
(a) What is the sample mean?
(b) What is the sample standard deviation? (Round your answer to two decimal places)
(c) If you leveraged technology to get the answers for part (a) and/or part (b), what technology did you use? If an online applet was used, please list the URL, and describe the steps. If a calculator or Excel was used, please write out the function.
4. There are 15 members on the board of directors for a Fortune 500 company. If they must select a chairperson, a first vice chairperson, a second vice chairperson, and a secretary.
(a) How many different ways the officers can be selected?
(b) Please describe the method used and the reason why it is appropriate for answering the question. Just the answer, without the description and reason, will receive no credit.
5. Sara has eight new summer outfits. She plans to pack three of the new summer outfits in her trip to Tokyo.
(a) How many different ways can the three summer outfits be selected?
(b) Please describe the method used and the reason why it is appropriate for answering the question. Just the answer, without the description and reason, will receive no credit.
6. Consider selecting one ball at a time from a box which contains 10 red, 6 yellow and 4 blue balls. What is the probability that the first ball is yellow and the second ball is also yellow? Express the probability in fraction format. (Show all work. Just the answer, without supporting work, will receive no credit.)
(a) Assuming the ball selection is without replacement.
(b) Assuming the ball selection is with replacement.
7. Let random variable x represent the number of heads when a fair coin is tossed two times.
(a) Construct a table describing the probability distribution.
(b) Determine the mean and standard deviation of x. Show all work. Just the answer, without supporting work, will receive no credit.
8. Mimi plans make a random guess at 10 true-or-false questions. Answer the following questions:
(a) Let X be the number of correct answers Mimi gets. As we know, the distribution of X is a binomial probability distribution. What is the number of trials (n), probability of successes (p) and probability of failures (q), respectively?
(b) Find the probability that she gets at most 5 correct answers. (Round the answer to 3 decimal places.
(c) To get the answers for part (b), what technology did you use? If an online applet was used, list the URL and describe the steps. If a calculator or Excel was used, write out the function.
Refer to the following information for Questions 9 and 10.
The heights of pecan trees are normally distributed with a mean of 10 feet and a standard deviation of 2 feet.
9. Show all work. Just the answer, without supporting work, will receive no credit.
(a) What is the probability that a randomly selected pecan tree is between 9 and 12 feet tall? (Round the answer to 4 decimal places)
(b) Find the 75th percentile of the pecan tree height distribution. (Round the answer to 2 decimal places)
10. Show all work. Just the answer, without supporting work, will receive no credit.
(a) For a sample of 64 pecan trees, state the standard deviation of the sample mean (the “standard error of the mean”). (Round your answer to three decimal places)
(b) Suppose a sample of 64 pecan trees is taken. Find the probability that the sample mean heights is between 9.5 and 10 feet. (Round your answer to four decimal places)
11. A survey showed that 720 of the 1000 adult respondents believe in global warming.
(a) Construct a 90% confidence interval estimate of the proportion of adults believing in global warming. (Round the lower bound and upper bound of the confidence interval to three decimal places) Include description of how confidence interval was constructed.
(b) Describe the results of the survey in everyday language.
12. A city built a new parking garage in a business district. For a random sample of 64 days, daily fees collected averaged \$2,000, with a standard deviation of \$400.
(a) Construct a 90% confidence interval estimate of the mean daily parking fees collected. (Round the lower bound and upper bound of the confidence interval to two decimal places) Include description of how confidence interval was constructed.
(b) Describe the confidence interval in everyday language.
13. An AP Statistics teacher claims that the AP Statistics grade distribution is as follows:
Suppose that a sample of 100 students taking AP Statistics class yields the observed counts shown below:
Use a 0.10 significance level to test the claimed AP Statistics grade distribution is correct.
(a) Identify the appropriate hypothesis test and explain the reasons why it is appropriate for analyzing this data.
(b) Identify the null hypothesis and the alternative hypothesis.
(c) Determine the test statistic. (Round your answer to two decimal places)
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http://mathoverflow.net/questions/118169/is-there-a-relationship-between-entropy-of-a-fininte-distrete-probability-distrib
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# Is there a relationship between Entropy of a fininte distrete probability distribution and the squre sum of the values of probability mass function of that distribution?
Sorry for the long title. What I mean is that for two vectors (a_1,...,a_n) and (b_1,...,b_n) with the property $a_i,b_i \geq 0$ and $\sum a_i =\sum b_i =1$.
If $-\sum a_ilog(a_i) > -\sum b_ilog(b_i)$ implies $\sum a_i^2 < \sum b_i^2$ or something similar?
-
Certainly not as you ask it. If h(a)>h(b) implied |a|_2<|b|_2, then h(b)>h(a) would imply |b|_2<|a|_2 and you would get h(a)>h(b) if and only if |a|_2<|b|_2. This would mean that the entropy and the L2 norm are measuring the same thing (they're not). – Anthony Quas Jan 6 '13 at 4:07
@Anthony I am not sure, if it can be ture under this specific condiction. – gstar2002 Jan 6 '13 at 10:42
and I think for the case n = 2, it is true. – gstar2002 Jan 6 '13 at 21:51
Of course, in general it is not true that inequality between entropies implies the same inequality between $\ell^2$ norms. Although this is true for $n=2$, it fails already for $n=3$ (as the entropy $H(p_1,p_2,p_3)$ is obviously not constant on the level curve determined by conditions $\sum p_i=1$ and $\sum p_i^2=Const$).
Nonetheless, there is a deep link between these two quantities. In order to explain it, it is better to somewhat change the viewpoint. Namely, given two probability distributions $P=(p_1,\dots,p_n)$ and $Q=(q_1,\dots,q_n)$ (for simplicity I assume that all $p_i,q_i$ are strictly positive), the corresponding Kullback-Leibler deviation is defined as $$I(Q|P) = \sum p_i \log \frac{p_i}{q_i} \;,$$ and the so-called information energy as $$\chi^2(Q,P) = \sum \biggl(\frac{p_i}{q_i}-1\biggr)^2 q_i \;.$$ Both these quantities measure "closedness" of $P$ and $Q$ (although they are not distances) and are monotone invariants of the pair $(Q,P)$ (they do not increase under quotient maps). If $Q=U_n$ is the uniform distribution, then these quantities are, up to linear rescaling, precisely the entropy and the $\ell^2$-norm of $P$, respectively, as $$I(U_n|P) = \log n - H(P)$$ and $$\chi^2(U_n,P) = n \sum p_i^2 - 1 \;.$$
Now, the link between $I$ and $\chi^2$ is provided by the fact that in naturally defined infinitesimal limits the Kullback-Leibler deviation $I$ and the information energy $\chi^2$ coincide and produce the Fisher information metric. One can read more about this in the corresponding wiki articles, in the old book Statistical decision rules and optimal inference by Čencov (AMS, 1982), or in more recent publications on information geometry.
No - infinitesimal here means that time goes to 0. The number of points remains fixed, but instead of a single distribution on $n$ points one looks at a family of distributions parametrized, say, by the interval $[0,1]$ (in other words, about a path in the space of distributions). – R W Jan 16 '13 at 0:32
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# Finding the longest palindromic substring
I have the following code that works correctly. I am wondering if there is a simpler way to go about implementing my solution and/or if there is anything that looks non-standard in my code. If anyone wants to offer criticism I'm all ears.
### Question 2 main function and helper functions.
"""Given a string a, find the longest palindromic substring contained in a.
Your function definition should look like question2(a), and return a string."""
# Gives substrings of s in decending order.
def substrings(s):
# Declare local variable for the length of s.
l = len(s)
# Here I chose range over xrange for python version compatibility.
for end in range(l, 0, -1):
for i in range(l-end+1):
yield s[i: i+end]
# Define palindrome.
def palindrome(s):
return s == s[::-1]
# Main function.
def Question2(a):
for l in substrings(a):
if palindrome(l):
return l
# Simple test case.
print Question2("stresseddesserts")
# stresseddesserts
• Just one thing – change comments from #... above functions to docstrings inside functions. – kyrill Apr 8 '17 at 14:33
In terms of efficiency, I like the code.
I'd just fix a few stylistic issues:
• Naming: l, a, s don't say anything about what the variable means. Also, you named your function palindrome, but that's returning a boolean, so I'd name it is_palindrome. The i in the substring function is a bit puzzling to me: you have a variable named end, why not call the other one start?
• As @kyrill mentioned, you should probably change comments above the functions to docstrings.
• The comment Declare local variable for the length of s. is useless.
• I'd instead add another comment explaining why it works. If you have a function simply called substrings I expect it to return the substrings. This code only works because it returns the substrings starting from the longest to the shortest. I'd at least change the comment descending order (which may mean for example lexicographical order or alphabetical order) to from longest to shortest. Better be explicit.
• I know, it's a bit long, but since you're using a generator, I'd rename the function substrings to get_next_substring.
• As @zondo mentioned Question2 is not a PEP8 compliant name, but I'm not sure if the automatic grader will need it to be called like that.
• get_next_substring would be a misleading name. The function does not return the next substring – when you call it, you get a generator. And that generator returns the next substring when you call its __next__ method. So I think substrings is a more apt name. – kyrill Apr 8 '17 at 18:28
• @kyrill What you say is correct of course, but seeing how the function is used, it really looks like it's returning a list and that's not the case, that's why I think substrings is not a good enough name. As you say, also get_next_substring is not perfect, but I couldn't come up with anything better. – ChatterOne Apr 8 '17 at 20:04
• I don't see why returning a generator instead of a list means that substrings is a misleading name. You would still do for substring in substrings(...), and other similar things. A generator is still a sequence in many respects. – zondo Apr 8 '17 at 21:02
I saw your substrings function and expected to tell you to use yield instead of creating a list, but you already do. I also like your comment on why you use range. If you want to be completely compatible with Python 3, though, you'll need to use parentheses for print:
print(Question2("stresseddesserts"))
The only other thing I'll mention is that Question2 does not follow official Python naming guidelines as defined in PEP 8. According to that document, function names should use snake_case. Your assignment also uses lowercase, so I'm a little surprised you changed it especially since your other functions are lowercase.
All in all, the program is concise, efficient, and organized. Well done.
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Mary, a veterinary student, has been assigned an experiment : GMAT Critical Reasoning (CR)
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# Mary, a veterinary student, has been assigned an experiment
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Mary, a veterinary student, has been assigned an experiment in mammalian physiology that would require her to take a healthy, anesthetized dog and subject it to a drastic blood loss in order to observe the physiological consequences of shock. The dog would neither regain consciousness nor survive the experiment. Mary decides not to do this assignment.
Mary’s decision most closely accords with which one of the following principles?
(A) All other things being equal, gratuitously causing any animal to suffer pain is unjustified.
(B) Taking the life of an animal is not justifiable unless doing so would immediately assist in saving several animal lives or in protecting the health of a person.
(C) The only sufficient justification for experimenting on animals is that future animal suffering is thereby prevented.
(D) Practicing veterinarians have a professional obligation to strive to prevent the unnecessary death of an animal except in cases of severely ill or injured animals whose prospects for recovery are dim.
(E) No one is ever justified in acting with the sole intention of causing the death of a living thing, be it animal or human.
If you have any questions
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27 Oct 2005, 12:46
B
Current Student
Joined: 28 Dec 2004
Posts: 3384
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Schools: Wharton'11 HBS'12
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### Show Tags
27 Oct 2005, 13:01
B
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27 Oct 2005, 13:09
i agree ....B is the best choice, IMO for lack of a better option
Not sure how you can infer the second half of the choice.... unless it would "immediately assist in saving several animal lives or in protecting the health of a person"
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27 Oct 2005, 14:55
between A and B I will choose A.
_________________
hey ya......
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28 Oct 2005, 00:47
go D go...
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28 Oct 2005, 15:19
B for me !
28 Oct 2005, 15:19
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## How Do I Only Make One Thrid Of This???
By Tor1985 Updated 15 May 2010 , 1:05am by anxietyattack
Tor1985 Posted 14 May 2010 , 3:24pm
post #1 of 4
I feel so dumb right now...lack of sleep and kids bugging me is making it impossible for me to cut this recipe in third...Can someone possible help me out??
2/3 cup + 3 tablespoons water, divided
1/4 cup Meringue Powder
12 cups sifted confectioners' sugar, (about 3 lbs.) divided
1 1/4 cups solid vegetable shortening
3 tablespoons light corn syrup
3/4 teaspoon salt
3/4 teaspoon no-color almond extract
3/4 teaspoon clear vanilla extract
1/2 teaspoon no-color butter flavor
I halfed it last time but just came short...I don't want to make that much as I only need a little!
Thanks
3 replies
indydebi Posted 14 May 2010 , 4:44pm
post #2 of 4
A simple "do the math" .... everything times 1/3:
2/3 cup + 3 tablespoons water, divided
...... 2/3 x 1/3 = 2/9 (slightly under 1/4) + 1 Tbsp
....... Note: 1 Tbsp= 1/16 of a cup
1/4 cup Meringue Powder
...... 1/4 x 1/3 = 1/12 cup
...... note: 1/6 cup is 2 Tbsp + 2 tsp; so 1/12 of a cup (half of 1/6 of a cup) will be 1 Tbsp + 1 tsp
12 cups sifted confectioners' sugar, (about 3 lbs.) divided
...... 12 x 1/3 = 4 cups
1 1/4 cups solid vegetable shortening
...... 5/4 x 1/3 = 5/12 cup
...... note: 1/6 cup = 2/12 cup = 2 Tbsp + 2 tsp;
ergo 1/12 cup (half of 2/12) = 1 Tbsp + 1 tsp;
ergo 5/12 cup = 5 Tbsp + 5 Tsp = 6 Tbsp + 2 tsp.
6 Tbsp = 3/8 cup so .......
5/12 cup = 3/8 cup plus 2 tsp
3 tablespoons light corn syrup
...... 3 x 1/3 = 1
3/4 teaspoon salt
...... 3/4 x 1/3 = 3/12 = 1/4
3/4 teaspoon no-color almond extract
...... 3/4 x 1/3 = 3/12 = 1/4
3/4 teaspoon clear vanilla extract
...... 3/4 x 1/3 = 3/12 = 1/4
1/2 teaspoon no-color butter flavor
HEre is the link to conversation chart I used in my shop (had it taped to my refrigerator for easy reference!) . Scroll to the bottom: http://www.botanical.com/botanical/cvcookix.html
Tor1985 Posted 14 May 2010 , 5:07pm
post #3 of 4
Thanks so much!
I coverted most of them, it was just really hard to concentrate with my kids bugging me!
Thanks again!
anxietyattack Posted 15 May 2010 , 1:05am
post #4 of 4
I'm pretty ok at math but I'm always second guessing myself So I use this link: http://www.fruitfromwashington.com/Recipes/scale/recipeconversions.php
works perfectly.
p.s.
Sorry I'm late to the party lol
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Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.
“) training_loop_vae(ds_train)
test_batch <- as_iterator(ds_test) %>% iter_next() encoded <- encoder(test_batch[[1]][1:1000]) test_var <- tf$$math$$reduce_variance(encoded, axis = 0L) print(test_var %>% as.numeric() %>% round(5)) } “
## Experimental setup and data
The idea was to add white noise to a deterministic series. This time, the Roessler system was chosen, mainly for the prettiness of its attractor, apparent even in its two-dimensional projections:
Like we did for the Lorenz system in the first part of this series, we use deSolve to generate data from the Roessler equations.
Then, noise is added, to the desired degree, by drawing from a normal distribution, centered at zero, with standard deviations varying between 1 and 2.5.
Here you can compare effects of not adding any noise (left), standard deviation-1 (middle), and standard deviation-2.5 Gaussian noise:
Otherwise, preprocessing proceeds as in the previous posts. In the upcoming results section, we’ll compare forecasts not just to the “real”, after noise addition, test split of the data, but also to the underlying Roessler system – that is, the thing we’re really interested in. (Just that in the real world, we can’t do that check.) This second test set is prepared for forecasting just like the other one; to avoid duplication we don’t reproduce the code.
## Results
The LSTM used for comparison with the VAE described above is identical to the architecture employed in the previous post. While with the VAE, an fnn_multiplier of 1 yielded sufficient regularization for all noise levels, some more experimentation was needed for the LSTM: At noise levels 2 and 2.5, that multiplier was set to 5.
As a result, in all cases, there was one latent variable with high variance and a second one of minor importance. For all others, variance was close to 0.
In all cases here means: In all cases where FNN regularization was used. As already hinted at in the introduction, the main regularizing factor providing robustness to noise here seems to be FNN loss, not KL divergence. So for all noise levels, besides FNN-regularized LSTM and VAE models we also tested their non-constrained counterparts.
#### Low noise
Seeing how all models did superbly on the original deterministic series, a noise level of 1 can almost be treated as a baseline. Here you see sixteen 120-timestep predictions from both regularized models, FNN-VAE (dark blue), and FNN-LSTM (orange). The noisy test data, both input (x, 120 steps) and output (y`, 120 steps) are displayed in (blue-ish) grey. In green, also spanning the whole sequence, we have the original Roessler data, the way they would look had no noise been added.
Despite the noise, forecasts from both models look excellent. Is this due to the FNN regularizer?
Looking at forecasts from their unregularized counterparts, we have to admit these do not look any worse. (For better comparability, the sixteen sequences to forecast were initiallly picked at random, but used to test all models and conditions.)
What happens when we start to add noise?
#### Substantial noise
Between noise levels 1.5 and 2, something changed, or became noticeable from visual inspection. Let’s jump directly to the highest-used level though: 2.5.
Here first are predictions obtained from the unregularized models.
Both LSTM and VAE get “distracted” a bit too much by the noise, the latter to an even higher degree. This leads to cases where predictions strongly “overshoot” the underlying non-noisy rhythm. This is not surprising, of course: They were trained on the noisy version; predict fluctuations is what they learned.
Do we see the same with the FNN models?
Interestingly, we see a much better fit to the underlying Roessler system now! Especially the VAE model, FNN-VAE, surprises with a whole new smoothness of predictions; but FNN-LSTM turns up much smoother forecasts as well.
“Smooth, fitting the system…” – by now you may be wondering, when are we going to come up with more quantitative assertions? If quantitative implies “mean squared error” (MSE), and if MSE is taken to be some divergence between forecasts and the true target from the test set, the answer is that this MSE doesn’t differ much between any of the four architectures. Put differently, it is mostly a function of noise level.
However, we could argue that what we’re really interested in is how well a model forecasts the underlying process. And there, we see differences.
In the following plot, we contrast MSEs obtained for the four model types (grey: VAE; orange: LSTM; dark blue: FNN-VAE; green: FNN-LSTM). The rows reflect noise levels (1, 1.5, 2, 2.5); the columns represent MSE in relation to the noisy(“real”) target (left) on the one hand, and in relation to the underlying system on the other (right). For better visibility of the effect, MSEs have been normalized as fractions of the maximum MSE in a category.
So, if we want to predict signal plus noise (left), it is not extremely critical whether we use FNN or not. But if we want to predict the signal only (right), with increasing noise in the data FNN loss becomes increasingly effective. This effect is far stronger for VAE vs. FNN-VAE than for LSTM vs. FNN-LSTM: The distance between the grey line (VAE) and the dark blue one (FNN-VAE) becomes larger and larger as we add more noise.
## Summing up
Our experiments show that when noise is likely to obscure measurements from an underlying deterministic system, FNN regularization can strongly improve forecasts. This is the case especially for convolutional VAEs, and probably convolutional autoencoders in general. And if an FNN-constrained VAE performs as well, for time series prediction, as an LSTM, there is a strong incentive to use the convolutional model: It trains significantly faster.
With that, we conclude our mini-series on FNN-regularized models. As always, we’d love to hear from you if you were able to make use of this in your own work!
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# General Properties of Quadratic Equation
We will discuss here about some of the general properties of quadratic equation.
We know that the general form of quadratic equation is ax^2 + bx + c = 0, where a is the co-efficient of x^2, b is the coefficient of x, c is the constant term and a ≠ 0, since, if a = 0, then the equation will no longer remain a quadratic
When we are expressing any quadratic equation in the form of ax^2 + bx + c =0, we have in the left side of the equation a quadratic expression.
For example, we can write the quadratic equation x^2 + 3x = 10 as x^2 + 3x – 10 = 0.
Now we will learn how to factorize the above quadratic expression.
x^2 + 3x - 10
= x^2 + 5x - 2x - 10
= x(x + 5) -2 (x + 5)
= (x + 5)(x – 2),
Therefore, x^2 + 3x – 10 = (x + 5)(x – 2) ............ (A)
Note: We know that mn = 0 implies that, either (i) m = 0 or n = 0 or (ii) m = 0 and n = 0. It is not possible that both of m and n are non-zero.
From (A) we get,
(x + 5)(x – 2) = 0, then any one of x + 5 and x - 2 must be zero.
So, factorizing the left side of the equation x^2 + 3x – 10 = 0 we get, (x + 5)(x – 2) = 0
Therefore, any one of (x + 5) and (x – 2) must be zero
i.e., x + 5 = 0 ................ (I)
or, x – 2 = 0 .................. (II)
Both of (I) and (II) represent linear equations, which we can solve to get the value of x.
From equation (I), we get x = -5 and from equation (II), we get x = 2.
Therefore the solutions of the equation are x = -5 and x = 2.
We will solve a quadratic equation in the following way:
(i) First we need to express the given equation in the general form of the quadratic equation ax^2 + bx + c = 0, then
(ii) We need to factorize the left side of the quadratic equation,
(iii) Now express each of the two factor equals to 0 and solve them
(iv)The two solutions are called the roots of the given quadratic equation.
Notes: (i) If b ≠ 0 and c = 0, one root of the quadratic equation is always zero.
For example, in the equation 2x^2 - 7x = 0, there is no constant term. Now factoring the left side of the equation, we get x(2x - 7).
Therefore, x(2x - 7) = 0.
Thus, either x = 0 or, 2x – 7 = 0
either x = 0 or, x = 7/2
Therefore, the two roots of the equation 2x^2 - 7x = 0 are 0, 7/2.
(ii) If b = 0, c = 0, both the roots of the quadratic equation will be zero. For example, if 11x^2 = 0, then dividing both sides by 11, we get x^2 = 0 or x = 0, 0.
`
Introduction to Quadratic Equation
Formation of Quadratic Equation in One Variable
General Properties of Quadratic Equation
Methods of Solving Quadratic Equations
Problems on Quadratic Equations
Word Problems Using Quadratic Formula
Examples on Quadratic Equations
Word Problems on Quadratic Equations by Factoring
Worksheet on Formation of Quadratic Equation in One Variable
Worksheet on Nature of the Roots of a Quadratic Equation
Worksheet on Word Problems on Quadratic Equations by Factoring
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# Talk:1516: Win by Induction
Jump to: navigation, search
Am I missing something or does Randall not quite understand how Pokemon works? (Or is intentionally misrepresenting it for the sake of the joke) Pokemon don't come out with their own pokeball with them-- the pokemon aren't magically created. In theory, if someone were to give a pokemon its own pokemon, a chain could occur, but it would be limited to the number of pokemon previously caught. The pokemon are born in the wild and are captured inside pokeballs-- not created from them.
Is the alt text a reference to double-yolkers (eggs with two yolks)? They're only about 1 in every 1000 but it seems like an obvious reference. --Fenn (talk) 08:32, 24 April 2015 (UTC)
Makes sense to me. I didn't even think of double yolks until you mentioned it here. 173.245.50.89 09:04, 24 April 2015 (UTC)BK201
Seconded. --188.114.110.52 14:34, 24 April 2015 (UTC)
I'd think it's a reference to the rate of twins, which is currently almost exactly 1/30 (and on the rise) [1] 173.245.56.186 17:45, 24 April 2015 (UTC)Merkky173.245.56.186 17:45, 24 April 2015 (UTC)
The explanation currently says that doubling makes it uncountably infinite. I'm pretty sure that doubling at each step (or every few steps) is still a countable infinite set. Proof here: http://practicaltypography.com/the-infinite-pixel-screen.html (see section "The internet demands a recount", because the first attempt is wrong). We can also prove it using the same argument as when proving that N x N is countable infinite (making zig-zag), but in this case making a breadth-first search of the tree of Pikachus: map 1 to the first Pikachu, map 2 and 3 to the two Pikachus at the second level, map 4, 5, 6, 7 to the four Pikachus at the third level, map (2^(n-1))…((2^n) - 1) to the 2^(n-1) Pikachus at level n. 108.162.229.177 (talk) (please sign your comments with ~~~~)
Saw this too late. Yes, I agree, and I have fixed it accordingly. --Stephan Schulz (talk) 09:28, 24 April 2015 (UTC)
The problem being that we don't have an exact number for how many steps include double Pikachus. Granted, this is just a problem of practice, not theory. 173.245.50.88 12:37, 24 April 2015 (UTC)
"infinite, but countable" {Cough.} Someone doesn't understand infinity. Perhaps they meant "enumerable". 108.162.250.155 09:29, 24 April 2015 (UTC)ū
Someone doesn't understand countability. 141.101.89.217 09:46, 24 April 2015 (UTC)
enumeration is counting, in the simplest sense. "To name one by one; specify, as if in a list". That said, the whole of infinite whole numbers CAN be counted, just not by a human and not within a reasonable amount of time. --188.114.110.52 14:34, 24 April 2015 (UTC)
"The front most Pikachu speaks." Hey, look, it has those little lines to show it's speaking, not the blank white space behind it. Duh. 108.162.250.155 09:32, 24 April 2015 (UTC)
Looks like Megan is looking at her watch as well. Mention in transcript/explanation? Fenn (talk) 09:34, 24 April 2015 (UTC)
Are Megan and Cueball supposed to fight each other? It seems like Cueball still has his closed Pokéball in his hands. Is it then Megan's Pokéball that has evolved into all these Pikachu? And is it because she waits for her Pokémon to be ready to fight Cueball, that she checks her watch? I do not know anything about the Pokémon game/world. But it seems to me that some part of this setup is unexplained by the above... --Kynde (talk) 11:23, 24 April 2015 (UTC)
Friendly reminder: Grammatically speaking, Pokémon are like sheep or deer. Singular and plural are both written the same. One Pikachu, many Pikachu, all the Pikachu. You'd be surprised at how much rage forgetting this causes in certain corners of the Internet. 141.101.99.42 (talk) (please sign your comments with ~~~~)
What doesn't make sense to me is how this could continue indefinitely – after all, each of those Pikachu must have caught its own Pikachu beforehand. I don't see any infinite loop here, just a bunch of Pikachu that already had one another caught itselves. 141.101.96.217 10:13, 24 April 2015 (UTC)
The word "induction" could also be intended to have a double meaning, referring also to electromagnetic induction. Pikachu is, after all, and electric pokémon. 141.101.105.194 (talk) (please sign your comments with ~~~~)
Yes, I think this is right. Something about Maxwell's equations and induction. 173.245.54.203 (talk) (please sign your comments with ~~~~)
From an engineering standpoint, in my opinion, Pikachu act more like biological capacitors (stored electric charge at potentially high voltage able to deliver large discharge currents) than inductors ("storing" magnetic energy via constant current, able to deliver high voltage when interrupted, like the ignition coil for an older automotive engine). I'm not too familiar with the Pokémon in-game/in-show universe, but I would imagine the Nurse Jenny corps could use electric Pokémon such as Pikachu (or Raichu) like defibrillators for cardiac events! --BigMal // 173.245.50.177 11:42, 24 April 2015 (UTC)
There are certain moves, including some that Pikachu can learn, that appear to be based on induction (Thunder Wave and Shock Wave). Besides, they build up charge in their bodies from somewhere; I'd suspect induction from the surrounding environment is what charges them up. --188.114.110.52 14:34, 24 April 2015 (UTC)
There's a point floating about how infinity doesn't imply completion. For instance, the number of all even integers is infinite, yet any given integer "only has a 50% chance of being even", so the series is quite obviously incomplete. This article seems to tend towards the idea (in diction) that an infinite number of pikachu would result in a win based on a 'logical' premise, without referring specificially to the terms of it's assumption. Xerxesbeat (talk) 11:38, 24 April 2015 (UTC)
What happens if the Pikachu in the ball is recursing - picking himself? That doesn't fit the 30-40 double yolk thing, but would explain an infinite series. Food for thought. Megan is bored, waiting for the fight to start. I thought the game was supposed to begin when the players choose, though, so I don't understand why the wait is happening at all. 108.162.221.151 (talk) (please sign your comments with ~~~~)
I doubt this is an intentional part of the joke, but the strongest Ground-type moves (Earthquake, Precipice Blades, etc.) are multi-target, hitting all foes in a 1v5 situation such as Horde Battles. In theory, a strong enough super effective move from Cueball's lead would still end the battle in one turn. 173.245.56.176 12:04, 24 April 2015 (UTC)
Not Land's Wrath, Dig, or Earth Power, which are strong ground-type moves.173.245.48.126 13:05, 24 April 2015 (UTC)
Actually, Land's Wrath is multi-target. (The ones you named are also weaker than Earthquake and Precipice Blades, so the original comment stands regardless. Although a lucky Magnitude is more powerful than any of those.) --108.162.221.98
I normally get a hearty chuckle out of Randall's graphical musings, but this one had me scratching my head. Fortunately, ExplainXKCD always comes to the rescue! After reading this page, my first thought was: Pokéception! 13:17, 24 April 2015 (UTC)
Induction
Two other possibilities: one, in a bit of googling, it would appear that there is a type of Pokémon evolution called induced evolution, which involves stones of some kind? Alternately, we can use the term induction in the sense of soneone being inducted into a group. In this case, Megan has trained her Pikachu to be a Pokémaster. (Perhaps by arranging for it to be inducted into a rarified "gym"? I confess, I know nothing about the show.) 173.245.56.196 13:11, 24 April 2015 (UTC)
I'm surprised no one mentioned that Pokémon is a game a long time before becoming a show. Although it was because of the animated series that Pikachu became "special" among the hundreds of other cute critters.
Also, no mention to the russian matryoshka dolls? Come on... Closest other xkcd I recall is https://xkcd.com/878/ 198.41.230.68 (talk) (please sign your comments with ~~~~)
Axiom of choice
Could this be to do with the axiom of choice from set theory? From my understanding, it's a fundamental axiom of set theory that says 'given a set of sets, it's possible to choose one element from each of those sets'. "Choosing" is in this case a specific operation that can be performed on an element.
One specific detail about the axiom is that all sets under consideration must be nonempty; that is, they must contain at least one element. So I think this is analogous to the situation of a Pokemon trainer owning multiple (full) Pokeballs: his Pokeballs are a collection of non-empty sets from which he is now trying to choose a single element ("Pikachu, I choose you!").
Under normal circumstances, he can do this without invoking the axiom of choice because he knows the names of all his Pokemon and so can select one from each set. In this case, he could prove his ability to make the choice simply by releasing all of his Pokemon from their balls one at a time. (The Pokemon's name is actually irrelevant, because simply releasing the Pokemon counts as a choice).
However, the situation becomes more complex if it turns out that his Pokemon also possess Pokeballs, because now his ability to make the choice is uncertain. In this situation, there could be infinitely many Pikachus, and so he can't definitely select a Pikachu from all the Pokeballs under his control. In a situation like this, a mathematician would invoke the axiom of choice.
However, it seems that Cueball is actually having a go at it using an inductive method of choice: first by choosing a Pikachu, then having each Pikachu choose a Pikachu. If the number of Pikachus carrying Pokeballs is finite, then eventually, this will demonstrate that the choice can be made and so the axiom of choice is unnecessary. However, if it's infinite, then this will generate a neverending stream of Pikachus. In the latter case, the game never begins, because you can't begin a Pokemon battle until all participants have chosen Pokemon. Most likely, the other players would simply abandon the game, which Cueball could claim as a victory. Hawthorn (talk) 13:52, 24 April 2015 (UTC)
This sentence is nonsensical: When Trainers do battle, the anime's dub has immersed the phrase "<Pokémon's name>, I choose you!" into popular culture memory, which is accompanied by throwing the ball containing the selected Pokémon to the ground, which releases the Pokémon at full size. 108.162.219.161 17:51, 24 April 2015 (UTC)
Should it be noted that the Pikachu is drawn without its tail? It would normally a have lightning bolt shaped tail that appears to the side or from behind its head. (Trivia or other note?) Azule (talk) 15:22, 24 April 2015 (UTC)
In Pokemon games from Gold and up, pokemon are able to hold items, including pokeballs. While in the game, once a pokeball is filled it is no longer available to select as an item, this comic would seem to imply the possible 'inception' scenario of having a pokemon hold an active pokeball (as the games have already shown that a pokeball can go into a pokeball). --173.245.54.193 14:13, 24 April 2015 (UTC)
ahem... "pokeception" short for "pocket inception" - I can't be the first one to coin this (?) - Brettpeirce (talk) 16:33, 24 April 2015 (UTC)
With Megan looking at her watch and Cueball holding the ball, I think we're meant to understand that Megan IS the Pokémon Cueball intends to use against Pikachu.108.162.221.153 19:12, 24 April 2015 (UTC)
Since Cueball has a closed ball in hand he has yet to choose a Pokemon. Tjus Megan cannot be his. She must have thrown the first Pikachu ball. Should be changed in explanation.Kynde (talk) 20:31, 24 April 2015 (UTC)
It is possible the "win by induction" is from the Pikachu's opponent inferring the series in infinite, and conceding. 19:56, 24 April 2015 (UTC)
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# How to know when a vortex shedding pattern has been onset?
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December 27, 2016, 17:57 How to know when a vortex shedding pattern has been onset? #1 Senior Member Hector Redal Join Date: Aug 2010 Location: Madrid, Spain Posts: 243 Rep Power: 16 Hi, I would like to know how to measure / determine when a vortex karman shedding pattern has been established in an horizontal flow past a circular cylinder. What parameter is measured so as to determine when the vortex shedding has been onset? I was tryiing to simulate a flow past a circular cylinder with a horizontal velocity, and initially I was measuring the frequency of the vertical velocity. I assumed that when the frequency of this vertical velocity is greater than zero, the karman vortex shedding has been onset. But, after having read this reference, http://www.leb.eei.uni-erlangen.de/w...1/pdf/0116.pdf, I have some doubts about what really I was measuring. Maybe, the condition for the onset of the vortex shedding is not having frequency of the vertical velocity different than zero, but observing a repeating pattern of swirling vortices. Then, only by observation of the flow, this can be determine, can't it? Any reference I can take a look at is welcome / appreciated. Thanks. Best regards, Hector.
December 27, 2016, 18:56 #2 Senior Member Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,793 Rep Power: 71 You should determine not only a positive frequancy but a well determinated stationary frequency. That happens after the numerical transient, when the alternance of counterotating vortices is established.
December 28, 2016, 08:59
#3
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Hector Redal
Join Date: Aug 2010
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Quote:
Originally Posted by FMDenaro You should determine not only a positive frequancy but a well determinated stationary frequency. That happens after the numerical transient, when the alternance of counterotating vortices is established.
I can observed a stationary frequency in the flow past a circular cylinder, but my question is that if the pattern observed is a real vortex shedding.
I have uploaded several snapshots of the transient flow I am observing, ordered in increasing time step: from 1 (initial time step) to 5 (final time step).
Is the pattern obserbed what is called a vortex shedding pattern?
Best regards,
Hector.
Attached Images
CylinderT1.jpg (39.3 KB, 25 views) CylinderT2.jpg (39.6 KB, 17 views) CylinderT3.jpg (39.3 KB, 18 views) CylinderT4.jpg (39.3 KB, 15 views) CylinderT5.jpg (39.8 KB, 17 views)
December 28, 2016, 10:34 #4 Super Moderator Alex Join Date: Jun 2012 Location: Germany Posts: 3,406 Rep Power: 48 I am not quite sure what you are trying to achieve. Are you interested in the time it takes until the onset of vortex shedding? Then you should have a much clearer definition for "onset of vortex shedding". Maybe the instant of time when a specific fraction of the final amplitude (vertical velocity, vertical force on cylinder...) is reached? Edit: here are a few images from one of my tutorials. Heated cylinder, laminar, Re=200, Pr=1. The color represents the temperature, the lines are streamlines. 02.jpg symmetric flow field 03.jpg initial instability 05.jpg growing instability 06.jpg growing even further 07.jpg statistical steady-state
December 28, 2016, 10:37 #5 Senior Member Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,793 Rep Power: 71 I suggest to plot the vorticity field and the vector field...the streamlines would help, too.
January 2, 2017, 16:47
#6
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Hector Redal
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Hi,
I have plot the streamlines of the flow that I am simulating.
I have obtained them with Paraview program. I am not very familiar with it, so I am not obtained very much detailed and accurate pictures. Sorry for this.
I can see a single vortex shedding from the rear part of the cylinder, and one single vortex close to the rear part of the cylinder.
I assume that vortex sheeding is onset when two vortex are shed from the cylinder. Is my understanding correct?
So based on this assumption, the vortex shedding has not still been set on. Right?
Best regards,
Hector.
Attached Images
StreamlinesT_1.jpg (195.6 KB, 11 views) StreamlinesT_2.jpg (176.5 KB, 11 views) StreamlinesT_3.jpg (170.5 KB, 8 views) StreamlinesT_4.jpg (173.1 KB, 9 views) StreamlinesT_5.jpg (176.3 KB, 11 views)
January 2, 2017, 17:16 #7 Super Moderator Alex Join Date: Jun 2012 Location: Germany Posts: 3,406 Rep Power: 48 Is the undisturbed flow supposed to be a uniform velocity in x-direction? I think you will need to tell us more about your simulation setup. Start with a detailed description of all boundary conditions. Continue with solver settings (and the type of solver you use), physical parameters/dimensionless numbers and include a few pictures of the mesh you used.
January 2, 2017, 17:32 #8 Senior Member Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,793 Rep Power: 71 I see a strange pattern ... the flow is not entering uniformly, it seems you have an angle... furthermore, the developing of the shedding seems still not complete. 1) how about the inflow velocity? 2) how about the time in your simulation? you should consider that a rough estimation of the non-dimensional time for the developed vortex shedding is of the order of the Re number.
January 2, 2017, 17:33
#9
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Hector Redal
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The undisturbed flow is a uniform velocity in the x-direction.
The force gravity is in the y direction.
The cylinder is at a hotter temperature than the fluid with non-slip condition.
The non-dimensional parameters are Re=10, Ri=1.3, Pr=0.72.
Attached you can find a piture of the mesh I am using.
Attached Images
Mesh.jpg (177.8 KB, 9 views)
January 2, 2017, 17:36
#10
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Filippo Maria Denaro
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Quote:
Originally Posted by HectorRedal The undisturbed flow is a uniform velocity in the x-direction. The force gravity is in the y direction. The cylinder is at a hotter temperature than the fluid with non-slip condition. The non-dimensional parameters are Re=10, Ri=1.3, Pr=0.72. Attached you can find a piture of the mesh I am using.
Something is wrong with the inflow setting... there is not reason that the velocity vector has an angle along x....
I strongly suggest to run a case without force gravity and coupling with temperature to check if your code works fine.
January 2, 2017, 17:47
#11
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Hector Redal
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Quote:
Originally Posted by FMDenaro I see a strange pattern ... the flow is not entering uniformly, it seems you have an angle... furthermore, the developing of the shedding seems still not complete. 1) how about the inflow velocity? 2) how about the time in your simulation? you should consider that a rough estimation of the non-dimensional time for the developed vortex shedding is of the order of the Re number.
I am including a broader picture of the streamlines.
The Reynolds number of the simulation is 10. Richardson number 1.3
Attached Images
January 2, 2017, 17:50
#12
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Hector Redal
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Quote:
Originally Posted by FMDenaro Something is wrong with the inflow setting... there is not reason that the velocity vector has an angle along x.... I strongly suggest to run a case without force gravity and coupling with temperature to check if your code works fine.
When the force gravity is zero, the flow pattern is symetric along the x-direction.
The velocity vector has an angle along the x direction when is near the cylinder.
I think this is correct.
Far away from the cylinder the velocity flow is horizontal.
January 2, 2017, 17:52
#13
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Hector Redal
Join Date: Aug 2010
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Quote:
Originally Posted by FMDenaro I see a strange pattern ... the flow is not entering uniformly, it seems you have an angle... furthermore, the developing of the shedding seems still not complete. 1) how about the inflow velocity? 2) how about the time in your simulation? you should consider that a rough estimation of the non-dimensional time for the developed vortex shedding is of the order of the Re number.
Which is the characteristic in the flow that makes you think the vortex shedding is not complete?
Is it because we can only see one single vortex shedding from the cylinder?
January 2, 2017, 17:57
#14
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Filippo Maria Denaro
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Quote:
Originally Posted by HectorRedal Which is the characteristic in the flow that makes you think the vortex shedding is not complete? Is it because we can only see one single vortex shedding from the cylinder?
without gravity, we know that the von Karman instability appears at Re=40-50. At Re=10 I expect two steady counterotating vortex.
https://en.wikipedia.org/wiki/K%C3%A..._vortex_street
January 2, 2017, 18:09
#15
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Hector Redal
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Quote:
Originally Posted by FMDenaro without gravity, we know that the von Karman instability appears at Re=40-50. At Re=10 I expect two steady counterotating vortex. https://en.wikipedia.org/wiki/K%C3%A..._vortex_street
Exactly, that is the pattern I am obtaining.
Attached you can see the streamlines I am getting when no gravity at all (Re 20). At Re10, the two steady counterotating vortexes are smaller.
The previous pictures were for the case where gravity is taken into account.
Attached Images
Streamlines without gravity.jpg (11.3 KB, 12 views)
January 2, 2017, 18:17
#16
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Filippo Maria Denaro
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Quote:
Originally Posted by HectorRedal Exactly, that is the pattern I am obtaining. Attached you can see the streamlines I am getting when no gravity at all (Re 20). At Re10, the two steady counterotating vortexes are smaller. The previous pictures were for the case where gravity is taken into account.
ok, the coupling with the temperature field is quite strong for your Ri number that you could also completely destroy the shedding...you should check some test-case in literature for such a case.
January 5, 2017, 05:06
#17
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Hector Redal
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Quote:
Originally Posted by flotus1 I am not quite sure what you are trying to achieve. Are you interested in the time it takes until the onset of vortex shedding? Then you should have a much clearer definition for "onset of vortex shedding". Maybe the instant of time when a specific fraction of the final amplitude (vertical velocity, vertical force on cylinder...) is reached? Edit: here are a few images from one of my tutorials. Heated cylinder, laminar, Re=200, Pr=1. The color represents the temperature, the lines are streamlines. Attachment 52885 symmetric flow field Attachment 52886 initial instability Attachment 52888 growing instability Attachment 52889 growing even further Attachment 52890 statistical steady-state
I would like to determine which is the value of the Richardson number that makes the vortex shedding pattern appear in the flow.
When the Richardson number is zero, there are two counterrotating vortices in the rear part of the cylinder. As the richardson number increases, these two vortices should break and the vortex patter should appear. I would like to determine this Richardson number.
I don't have such clear picture on when this happens.
Does this happen when a periodic vertical velicity is observed in the right-most part of the domain (independently of the amplitud)?
I mean, at the initial time, the velocity is horizontal at all points of the domain. As the simulation runs, a vertical velocity is observed at the point (x=L, y =H/2). This verticial velocity oscilates as the simulation runs, reaching a maximum value when the transient ends.
Does a periodic value of the vertical velocity mean there are vortices in the flow?
January 5, 2017, 06:02
#18
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Hector Redal
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Quote:
Originally Posted by FMDenaro ok, the coupling with the temperature field is quite strong for your Ri number that you could also completely destroy the shedding...you should check some test-case in literature for such a case.
I have found the following reference where the vortex shedding formation and mechanism is described: Near-wake effects of a heat input on the vortex-shedding mechanism. R. Kieft *, C.C.M. Rindt, A.A. van Steenhoven, International Journal of Heat and Fluid Flow 28 (2007) 938–947.
It appears a good reference to take a look at. I will go through it and see if I can come to a conclusion.
Thanks for your kind support, as always.
January 5, 2017, 06:42 #19 Senior Member Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,793 Rep Power: 71 I think you should consider checking your code step-by-step. First, be sure the temperature equation is well solve by running a case where it is just a passive scalar. You can check by menas of the iso-temperature curves. Then, active the gravity forcing (I think is directed along the y axis) for small Ri values. I would expect that the inflow velocity is not perturbed like I see in your previous figures. Of course, the literature you found is a further step to assess the code.
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# User:IssaRice/Computability and logic/Diagonalization lemma
The diagonalization lemma, also called the Godel-Carnap fixed point theorem, is a fixed point theorem in logic.
## Rogers's fixed point theorem
Let $f$ be a total computable function. Then there exists an index $e$ such that $\varphi_e \simeq \varphi_{f(e)}$.
(simplified)
Define $d(x) = \varphi_x(x)$ (this is actually slightly wrong, but it brings out the analogy better).
Consider the function $f\circ d$. This is partial recursive, so $f\circ d \simeq \varphi_i$ for some index $i$.
Now $\varphi_{f(d(i))} \simeq \varphi_{\varphi_i(i)}$ since $f\circ d \simeq \varphi_i$. This is equivalent to $\varphi_{d(i)}$ by definition of $d$. Thus, we may take $e = d(i)$ to complete the proof.
## Diagonalization lemma
(semantic version)
Let $A$ be a formula with one free variable. Then there exists a sentence $G$ such that $G$ iff $A(\ulcorner G\urcorner)$.
Define $\mathrm{diag}(x)$ to be $\ulcorner C(\ulcorner C\urcorner)\urcorner$ where $x = \ulcorner C\urcorner$. In other words, given a number $x$, the function $\mathrm{diag}$ finds the formula with that Godel number, then diagonalizes it (i.e. substitutes the Godel number of the formula into the formula itself), then returns the Godel number of the resulting sentence.
Let $B$ be $A(\mathrm{diag}(x))$, and let $G$ be $B(\ulcorner B\urcorner)$.
Then $G$ is $A(\mathrm{diag}(\ulcorner B\urcorner))$, by substituting $x = \ulcorner B\urcorner$ in the definition of $B$.
We also have $\mathrm{diag}(\ulcorner B\urcorner) = \ulcorner B(\ulcorner B\urcorner)\urcorner$ by definition of $\mathrm{diag}$.
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# deterministic automaton
Also found in: Wikipedia.
## deterministic automaton
(theory)
A finite-state automaton in which the overall course of the computation is completely determined by the program, the starting state, and the initial inputs. The class of problems solvable by such automata is the class P (see polynomial-time algorithm).
The following article is from The Great Soviet Encyclopedia (1979). It might be outdated or ideologically biased.
## Deterministic Automaton
a mathematical model of a system whose states change discretely with time in such a way that every state of the system is completely defined by the previous state and the input signal. A deterministic automaton is formally described in the form of the function f(Si, aj) = ak, where si is the input signal and aj is the previous state. A typical example of a deterministic automation is a digital computer, in which the state of all registers and cells is determined by their previous state and by the input signals. Deterministic automatons are a natural form for describing the logical structure of discrete computing devices. Conversion to nondeterministic automatons is possible both by the introduction of the probabilities of a change of states and by the free selection of the next state.
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Some fact: For each automaton an equivalent deterministic automaton can be created.
For an assertion graph G, we construct a corresponding finite-state automaton [M.sub.G] such that [L.sup.*](G) = [[summation].sup.*] - [L.sup.*]([M.sub.G]) and [L.sup.[omega]](G) = [[summation].sup.[omega]] - [L.sup.[omega]]([M.sub.G]), the number of states in [M.sub.G] is only twice as many as that in G, and a deterministic assertion graph is transformed to a deterministic automaton. On the other hand, for an arbitrary finite-state automaton M, we construct a corresponding assertion graph [G.sub.M] such that [L.sup.*]([G.sub.M]) = [[summation].sup.*] - [L.sup.*](M) and [L.sup.[omega]]([G.sub.M]) = [[summation].sup.[omega]] - [L.sup.[omega]](M), the number of states in [G.sub.M] is the same as that in M, and a deterministic automaton is transformed to a deterministic assertion graph.
Hopcroft (1971) has given an algorithm that computes the minimal automaton of a given deterministic automaton. The running time of the algorithm is O([absolute value of A] x n log n) where [absolute value of A] is the cardinality of the alphabet and n is the number of states of the given automaton.
Given a deterministic automaton A, Hopcroft's algorithm computes the coarsest congruence which saturates the set F of final states.
Cal designed an essentially deterministic automaton (Figure 9).
It is shown that an elementary soliton graph defines a deterministic automaton iff it reduces to a graph not containing even-length cycles.
1985] built on a string S is a deterministic automaton able to recognize all the substrings of S.
In contrast, a deterministic automaton has a single run on w.
Proof: For each n [greater than or equal to] 0 we construct a deterministic automaton that accepts all words of length at most n in the complement of [Q.sub.k].
Let [T.sub.U] be the ordinary trie representing the set U, seen as a finite deterministic automaton (Q, [delta], [epsilon], T) where the set of states is Q = Pref (u) (prefixes of words in u), the initial state is [epsilon], the set of final states is T = [A.sup.*] [intersection] Pref (u) and the transition function [delta] is defined on Pref (U) x A by
Let D = (Q, [q.sub.0], [Delta], F) be a deterministic automaton. Let [f.sub.w] be the function induced by a word w on Q.
Because they are deterministic automatons, computers struggle to generate numbers that are truly random.
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# filtfilt
Zero-phase digital filtering
## Syntax
``y = filtfilt(b,a,x)``
``y = filtfilt(sos,g,x)``
``y = filtfilt(d,x)``
## Description
example
````y = filtfilt(b,a,x)` performs zero-phase digital filtering by processing the input data, `x`, in both the forward and reverse directions. After filtering the data in the forward direction, `filtfilt` reverses the filtered sequence and runs it back through the filter. The result has the following characteristics: Zero phase distortion.A filter transfer function equal to the squared magnitude of the original filter transfer function.A filter order that is double the order of the filter specified by `b` and `a`. `filtfilt` minimizes start-up and ending transients by matching initial conditions. Do not use `filtfilt` with differentiator and Hilbert FIR filters, because the operation of these filters depends heavily on their phase response.```
````y = filtfilt(sos,g,x)` zero-phase filters the input data, `x`, using the second-order section (biquad) filter represented by the matrix `sos` and the scale values `g`.```
````y = filtfilt(d,x)` zero-phase filters the input data, `x`, using a digital filter, `d`. Use `designfilt` to generate `d` based on frequency-response specifications.```
## Examples
collapse all
Zero-phase filtering helps preserve features in a filtered time waveform exactly where they occur in the unfiltered signal.
Use `filtfilt` to zero-phase filter a synthetic electrocardiogram (ECG) waveform. The function that generates the waveform is at the end of the example. The QRS complex is an important feature in the ECG. Here it begins around time point 160.
```wform = ecg(500); plot(wform) axis([0 500 -1.25 1.25]) text(155,-0.4,'Q') text(180,1.1,'R') text(205,-1,'S')```
Corrupt the ECG with additive noise. Reset the random number generator for reproducible results. Construct a lowpass FIR equiripple filter and filter the noisy waveform using both zero-phase and conventional filtering.
```rng default x = wform' + 0.25*randn(500,1); d = designfilt('lowpassfir', ... 'PassbandFrequency',0.15,'StopbandFrequency',0.2, ... 'PassbandRipple',1,'StopbandAttenuation',60, ... 'DesignMethod','equiripple'); y = filtfilt(d,x); y1 = filter(d,x); subplot(2,1,1) plot([y y1]) title('Filtered Waveforms') legend('Zero-phase Filtering','Conventional Filtering') subplot(2,1,2) plot(wform) title('Original Waveform')```
Zero-phase filtering reduces noise in the signal and preserves the QRS complex at the same time it occurs in the original. Conventional filtering reduces noise in the signal, but delays the QRS complex.
Repeat the above using a Butterworth second-order section filter.
```d1 = designfilt('lowpassiir','FilterOrder',12, ... 'HalfPowerFrequency',0.15,'DesignMethod','butter'); y = filtfilt(d1,x); subplot(1,1,1) plot(x) hold on plot(y,'LineWidth',3) legend('Noisy ECG','Zero-Phase Filtering')```
This is the function that generates the ECG waveform.
```function x = ecg(L) %ECG Electrocardiogram (ECG) signal generator. % ECG(L) generates a piecewise linear ECG signal of length L. % % EXAMPLE: % x = ecg(500).'; % y = sgolayfilt(x,0,3); % Typical values are: d=0 and F=3,5,9, etc. % y5 = sgolayfilt(x,0,5); % y15 = sgolayfilt(x,0,15); % plot(1:length(x),[x y y5 y15]); % Copyright 1988-2002 The MathWorks, Inc. a0 = [0,1,40,1,0,-34,118,-99,0,2,21,2,0,0,0]; % Template d0 = [0,27,59,91,131,141,163,185,195,275,307,339,357,390,440]; a = a0 / max(a0); d = round(d0 * L / d0(15)); % Scale them to fit in length L d(15)=L; for i=1:14, m = d(i) : d(i+1) - 1; slope = (a(i+1) - a(i)) / (d(i+1) - d(i)); x(m+1) = a(i) + slope * (m - d(i)); end end```
## Input Arguments
collapse all
Transfer function coefficients, specified as vectors. If you use an all-pole filter, enter `1` for `b`. If you use an all-zero (FIR) filter, enter `1` for `a`.
Example: `b = [1 3 3 1]/6` and ```a = [3 0 1 0]/3``` specify a third-order Butterworth filter with a normalized 3-dB frequency of 0.5π rad/sample.
Data Types: `double`
Input signal, specified as a real-valued or complex-valued vector, matrix, or N-D array. `x` must be finite-valued. `filtfilt` operates along the first array dimension of `x` with size greater than 1.
Example: `cos(pi/4*(0:159))+randn(1,160)` is a single-channel row-vector signal.
Example: `cos(pi./[4;2]*(0:159))'+randn(160,2)` is a two-channel signal.
Data Types: `double`
Complex Number Support: Yes
Second-order section coefficients, specified as a matrix. `sos` is a K-by-6 matrix, where the number of sections, K, must be greater than or equal to 2. If the number of sections is less than 2, then `filtfilt` treats the input as a numerator vector. Each row of `sos` corresponds to the coefficients of a second-order (biquad) filter. The ith row of `sos` corresponds to ```[bi(1) bi(2) bi(3) ai(1) ai(2) ai(3)]```.
Example: `s = [2 4 2 6 0 2;3 3 0 6 0 0]` specifies a third-order Butterworth filter with a normalized 3-dB frequency of 0.5π rad/sample.
Data Types: `double`
Scale factors, specified as a vector.
Data Types: `double`
Digital filter, specified as a `digitalFilter` object. Use `designfilt` to generate a digital filter based on frequency-response specifications.
Example: ```d = designfilt('lowpassiir','FilterOrder',3,'HalfPowerFrequency',0.5)``` specifies a third-order Butterworth filter with a normalized 3-dB frequency of 0.5π rad/sample.
Data Types: `double`
## Output Arguments
collapse all
Filtered signal, returned as a vector, matrix, or N-D array.
## References
[1] Gustafsson, F. “Determining the initial states in forward-backward filtering.” IEEE® Transactions on Signal Processing. Vol. 44, April 1996, pp. 988–992.
[2] Mitra, Sanjit K. Digital Signal Processing. 2nd Ed. New York: McGraw-Hill, 2001.
[3] Oppenheim, Alan V., Ronald W. Schafer, and John R. Buck. Discrete-Time Signal Processing. 2nd Ed. Upper Saddle River, NJ: Prentice Hall, 1999.
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New Online Book! Handbook of Mathematical Functions (AMS55)
Conversion & Calculation Home >> Measurement Conversion
Measurement Converter
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Convert From: (required) Click here to Convert To: (optional) Examples: 5 kilometers, 12 feet/sec^2, 1/5 gallon, 9.5 Joules, or 0 dF. Help, Frequently Asked Questions, Use Currencies in Conversions, Measurements & Currencies Recognized Examples: miles, meters/s^2, liters, kilowatt*hours, or dC.
Conversion Result: ```electric potential = 1 volt (electric potential) ``` Related Measurements: Try converting from "electromotive force" to abvolt, statvolt, volt, or any combination of units which equate to "mass length squared / time cubed electric current" and represent electric potential, electromotive force, induced emf, potential, potential difference, voltage, or voltage potential. Sample Conversions: electromotive force = 100,000,000 abvolt, .00333564 statvolt, 1 volt.
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Singular inf matrix probs for gmm w/b...
Message/Author
Bruce A. Cooper posted on Monday, November 07, 2011 - 12:23 pm
Hi Linda -
I'm estimating a GMM with MLR for pain occurrence as a binary variable across 7 months. I'm getting conceptually very interesting results, but in trying to finalize the 3-class and especially the 4-class solutions, I keep getting errors that result from (at least partly), classes that keep shifting in order. One problem is that the L and Q slope estimates are sometimes astoundingly big (and I don't know how they are scaled). For example, in the most recent problematic model, the problem class has S= 539.098 and Q= -269.345 and the program fixed Q (threshold is 1.723). But what is that S?!? But S is large for another class also (8.117). Basically, the primary problem class has no pain at baseline, 100% with pain at month 2, then a steep decline in the probability of pain over months 3-7. Any help with these estimates and how to keep the classes in order would be great! Thanks, Bruce
Linda K. Muthen posted on Monday, November 07, 2011 - 1:51 pm
Bruce A. Cooper posted on Monday, November 14, 2011 - 11:28 am
Thanks, Linda -
So the 2nd class in my 4-C model starts at close to a P=0, then goes to P=1 at T2, then returns to P=0 by T4. They represent a meaningful class. (1) How can I model the group? I can't fix the S and Q to e(logit) = infinity or minus infinity. (2) Can I estimate a mixture model for these data with a piecewise approach? If so, what would the model look like, taking off from
i s q | bpn1m0@0 bpn1m1@1 bpn1m2@2 bpn1m3@3 bpn1m4@4 bpn1m5@5 bpn1m6@6 ;
I specified a model with the 2nd piece beginning at T4 and got interesting est & plots.
i1 s1 q1 | bpn1m0@0 bpn1m1@1 bpn1m2@2 bpn1m3@3 bpn1m4@4 bpn1m5@5 bpn1m6@6 ;
i2 s2 q2 | bpn1m0@0 bpn1m1@0 bpn1m2@0 bpn1m3@1 bpn1m4@2 bpn1m5@3 bpn1m6@4 ;
Before I encourage a migraine, does this look right?
Thanks!
Bruce
Linda K. Muthen posted on Tuesday, November 15, 2011 - 9:32 am
You should not do anything. You should simply describe the class.
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Solved
# sql count if
Posted on 2014-04-09
Medium Priority
304 Views
I am needing to look at a table and see if columns have a match. I want to see if multiple employees have been clocked onto the same work order and operation. I have the columns odd_no, Oper_no and empid. for example I may have the work order number 123456 with operation number 10. I want to COUNT how many different employees worked on that operation..
0
Question by:sharris_glascol
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LVL 66
Expert Comment
ID: 39988988
``````SELECT COUNT(DISTINCT empid)
FROM your_table
WHERE odd_no = 123456 AND oper_no = 10
``````
You'll want to look at the full data set just to make sure the above is correct, and any duplicate values are being handled
``````SELECT odd_no, oper_no, emp_id
FROM your_table
WHERE odd_no = 123456 AND oper_no = 10
ORDER BY odd_no, oper_no, emp_id
``````
0
Author Comment
ID: 39988993
what if I want to look at all work orders and operations to see how many employees where clocked into the same one and not just one exact work order?
0
LVL 66
Expert Comment
ID: 39989017
Not entirely sure what you mean by 'the same one', but give this a whirl..
List...
``````SELECT odd_no, oper_no, emp_id
FROM your_table
ORDER BY odd_no, oper_no, emp_id
``````
Count...
``````SELECT odd_no, oper_no, COUNT(emp_id) as empl_id_count
FROM your_table
GROUP BY odd_no, oper_no
ORDER BY odd_no, oper_no
``````
0
Author Comment
ID: 39989044
Here is what i need to count
work order operation number employee number count
12345 10 2321 2
2468 20 1234 2
12345 10 1234 2
2468 20 2321 2
13579 10 2321 1
This is kinda what I am looking for....
0
LVL 66
Expert Comment
ID: 39989172
Nice mockup data. Next time please add that to your original question, so we don't have to spend time flushing out requirements.
Based on the set above, give this a whirl, and add an ORDER BY clause to sort it by whatever is most readable..
``````SELECT
odd_no as work_order,
oper_no as operation_number,
emp_id as employee_number,
COUNT(emp_id) as empl_id_count
FROM your_table
GROUP BY odd_no, oper_no, emp_id
``````
0
LVL 49
Accepted Solution
PortletPaul earned 2000 total points
ID: 39990666
Both sample data and expected results are best practice when asking this type of question. At this point I think we have the expected result but can deduce the sample data.
Jim is ever so close but I believe you need
COUNT() OVER()
to match that expected result. Like this:
`````` CREATE TABLE Your_Table
([odd_no] int, [Oper_no] int, [empid] int)
;
INSERT INTO Your_Table
([odd_no], [Oper_no], [empid])
VALUES
(12345, 10, 1234),
(12345, 10, 2321),
(13579, 10, 2321),
(2468, 20, 2321),
(2468, 20, 1234)
;
**Query 1**:
SELECT
odd_no AS work_order
, oper_no AS operation_number
, empid AS employee_number
, COUNT(empid) OVER (PARTITION BY odd_no, oper_no) AS empl_id_count
FROM your_table
ORDER BY
odd_no,
oper_no,
empid
**[Results][2]**:
| WORK_ORDER | OPERATION_NUMBER | EMPLOYEE_NUMBER | EMPL_ID_COUNT |
|------------|------------------|-----------------|---------------|
| 2468 | 20 | 1234 | 2 |
| 2468 | 20 | 2321 | 2 |
| 12345 | 10 | 1234 | 2 |
| 12345 | 10 | 2321 | 2 |
| 13579 | 10 | 2321 | 1 |
http://sqlfiddle.com/#!3/4002c/1
``````
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• 1. Mathematical Prayer Dear Lord. We thank you a hundredfold for the love and care that you have given us. May we in return to you your good works by multiplying it with love and respect, adding more faith, subtracting the unworldly behavior and evil works and dividing your given talents to others so we can sum it all and be united as one in your family. In this we pray, Amen
• 2. Highest Pointer, Lowest Grade,
• 3. How to play: The class will be divided into three groups. Each group will choose a representative. The representative will be the one to hold the illustration board and chalk to write and raise the group’s answer.
• 4. Goal The group must raise and answer the questions correctly. The group who got the got the correct answer and raise it first will be rank as first. After the game. Their rankings will be summed up and the group who got the lowest score wins the game and the group who got the highest score loses the game.
• 5. FINDING THE LCD
• 6. Find the LCD OF… 12 and 24 ANSWER: 24
• 7. XYZ and X3Y4Z2 Answer : X3Y4Z2
• 8. M andM+2 Answer: m(m+2)
• 9. X+5 and x-7 ANSWER: (X+5)(X-7)
• 10. X2 -16 and X+4 ANSWER: (X+4)(X-4)
• 11. I Know How to Follow Instructions
• 12. Objectives: Use the given instructions to simplify the complex fraction by following what it is said in the activity sheet. Material: activity sheet, manila paper and pentel pen Procedure: The class will be divided into three groups. Each group will be given an activity sheet with a questionnaire. They must follow the instruction to be able to solve the problem. The activity will only be given 15 minutes to finish the task. Display your work on the manila paper and one member must share their insights about their work.
• 13. Complex Fraction A fraction that has a rational expression either in the numerator, denominator or both.
• 14. Method 1: Write the numerator as one fraction and the denominator as one. Then invert to reciprocal the denominator. Multiply then simplify.
• 15. Method 2: Multiply the numerator and the denominator by the LCD of the fractions within the numerator and the denominator then simplify.
• 16. Simplify the following complex fractions (Show the two methods in simplifying the complex fractions): Group 1: Group 2: Group 3: + +
• 17. Seatwork • Answer p. 124, Test I (Practice and Application) nos. 1-3: only
• 18. Assignment Answer p. 124, Test II (Practice and Application) nos. 6-10 only. Write it in your notebook.
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# hour
Physical unit
Unit name hour
Unit symbol ${\ displaystyle \ mathrm {h}}$
Physical quantity (s) Time , span of time
Formula symbol ${\ displaystyle t; \, T; \; \, \ tau, \, \ mathrm {T}}$
dimension ${\ displaystyle {\ mathsf {T}}}$
system Approved for use with the SI
In SI units ${\ displaystyle \ mathrm {1 \, h = 3600 \; s}}$
Named after Old high German stunta
Derived from Day
The hour (from Old High German stunta 'standing', 'stay', 'fixed point in time', 'short period', 'hour') denotes the twenty-fourth part of a day . In addition to a division into 24 equal parts, there are also other hour terms.
The Latin word is hora , hence the symbol h or h.
For the exact differentiation of an hour (with 60 minutes) e.g. B. of a lesson (often 45 minutes) is also referred to as a hour .
## The unit of measurement is hour
The hour is a unit of time . The unit symbol is h (from Latin hora ). The hour does not belong to the International System of Units (SI), but is approved for use with the SI. This makes it a legal unit of measurement . Due to the non- decimal subdivision, a conversion into seconds is first necessary for scientific calculations.
1 hour = 60 minutes = 3600 ( SI ) seconds
Since today's atomic clocks can measure time very precisely and the rotation speed of the earth varies, the hour was redefined as a second, which connects atomic time with universal astronomical time .
## Counting and dividing hours
The 24-hour counting of a whole day is first attested in ancient Egypt and was later used in ancient Greece around the third century BC, where the 24-hour system was derived from the angle . From there it spread over the whole (old) world until the turn of the times .
Various methods of counting hours, i.e. numbering , have been used throughout history :
The beginning of minute one is called the full hour , for example 08:00:00; it is also called "stroke eight". The term comes from the fact that (in the example above) “the eighth hour is full”. The widespread hour divisions of half an hour , quarter of an hour and the times (chronologically) "quarter eight" (07:15), "half eight" (07:30), "quarter to eight" or "quarter to eight" (07: 45) and "quarter past eight" (08:15).
## Other definitions of the length of the hour
Foldable pocket calendar (approx. 1400; SBB-PK, Lib. Pic. A 72) - complete monthly cycle with signs of the zodiac and list of hours with daylight per month.
The term 'hour' - in addition to today's physical-chronometric term - is also used for the historical chronological and astronomical time systems :
• Equal hours : An hour length is subject to its own definitions and is not tied to a 24-hour division.
• Temporal hours (Roman hours) : The day arc of the bright day ( sunrise to sunset) was divided into twelve parts and transferred accordingly to the night. Due to the different lengths of day and night in the course of the year, the length of the hours (between 30 and 90 minutes) also change continuously.
• In the horae canonicae , the first hour of the day ( prime ) began between 3 a.m. and 9:30 a.m. and the first hour of the night between 3 p.m. and 9:30 p.m. The English term afternoon refers to the 9th hour ( ninth ), which has been striking at 12 noon since around 1300.
• Seasonal hours : Similar to the temporal hours, the day and night are divided into periods of fluctuating duration, but deviate from a continuous 24-hour division.
• Equinox hour : An equinox hour is the twenty-fourth part of the sunny day and is independent of the day and night arcs that change with the seasons.
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# Principles of Macroeconomics
Topics: Gross domestic product, Measures of national income and output, Purchasing power parity Pages: 2 (934 words) Published: November 2, 2014
Data Exercise One
ECON 201: Principles of Macroeconomics
September 5, 2014
To live in an economy that is not negatively impacted by recession, downsizing, or business capsizing would be ideal. The unfortunate reality is that we are faced with economic situations that will be either helpful or hurtful to us all. Over the last few quarters between 2013 and 2014 the U.S. Bureau of Economic Analysis (BEA), conducted an analysis that reflects the changes in GDP. During this time the Nominal GDP was much greater than the Real GDP. Expenditures Approach to Calculating GDP
-638175208978500From 2013 through 2014 the Nominal GDP was greater because the values during that time were not adjusted. It is understood that Nominal GDP is the value of GDP in current dollar and Real GDP is the value of those dollars after adjustments or changes in prices. For the most recent quarter, the percentage change was -\$20.1. The difference is within the Net Exports of Goods and Services. Below is a table that reflects such. Income Approach to Calculating GDP
2nd QTR 2013 3rd QTR 2013 4th QTR 2013 1st QTR 2014
GDP 16,619.2 16,872.3 17,078.3 17,044.0
GNP 16,711.2 17,103.1 17,321.2 17,255.0
Net National Product 14,221.3 14,462.9 14,650.6 14,556.3
National Income 15,511.5 14,650.5 14,770.2 14,733.7
Personal Income 14,131.3 14,247.4 14,311.7 14,484.7
Gross Domestic Product or GDP is the total value of final goods and services produced in a given year. GDP is comprised of four basic categories. Those categories are Consumption Expenditures, Private Investment Expenditures, Government Purchases and Net Exports. Gross National Product or GNP is the total of final goods and services produced in a given year by another country. The difference between the two can be easily identified by understanding that GNP include foreign net income opposed to considering net exports and imports. Based on the table provided above, to determine GNP from GDP you have to include the value of...
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# What is Point in Geometry ? Point in Geometry Definition, Types, and Examples
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## Definition
In geometry, a point is a fundamental unit without any quantitative attributes such as size: width, length, height and depth etc. The point only represents a position or location of a variable in plane marked by a dot (.) and is preferably denoted with a capital letter (P, Q, A etc.).
## Line or Line Segment
Only when point is extended infinitely or indefinitely in two directions, then point is termed as line or more precisely line segment. A line has infinite length without any width and height i.e., a line is one-dimensional. In 2-dimensional Euclidean geometry, a point is represented by a pair of numbers which is always ordered e.g., (x,y). Here x represents the horizontal, whereas y indicates the vertical component of a plane. In 3-dimensional space, a point is represented by an order of triplet e.g., (x,y,z) in which the third component is added to the space in which a point is located.
#### Real-life Examples
For better understanding, the real-life examples of a point can be a star in the sky, the pointed end of a needle, or tip of a ball point etc.
## Dimensions of a Point
As discussed above, a point is free of any dimension as it doesn’t own any measurable property. A point itself is zero dimensional and can possess quantitative property only when it is extended in at least one direction; thereby producing a line. One value is required in order to identify the point on line which will provide one dimension or making a line one dimensional.
Similarly, if point is extended to two directions, two values are required to find the point on plane; which creates two dimensions (2D) as seen in circle, square etc. Likewise, if a line is stretched to entirely different direction other than horizontal or vertical components, a new third value is identifiable to find the location of point, giving it three dimensions (3D) as found in sphere, cube etc.
#### Real-life Examples
For better understanding, the real-life examples of 2D figures are chessboard, a piece of paper or a pizza. The 3D figures could be basketball, cylinder or an egg etc.
### Type of Points
A point can be classified into a few types which are as explained below:
• Collinear Points
• Non-collinear Points
• Concurrent Points
• Coplanar Points
• Non-coplanar Points
• Equidistant Points
Collinear Points
The word ‘collinear point’ derived from the word ‘col’ (together) and ‘linear’ (straight line) which means the points which are occurring together in a linear way.
For Example. There are four points D, E, F, G, occurring on a same straight line.
#### Real-life Examples.
For better understanding, the real-life examples of collinear points are numbers on a ruler, birds sitting in a row on a wire etc.
### Non-collinear Points
In contrast to the above type, non-collinear points are those which are not occurring together on a straight line. It can also be understood with the concept; as if a straight line cannot be drawn through a few points, then points are non-collinear in nature.
For Example. There are four points D, E, F, G, not occurring on a same straight line.
#### Real-life Examples.
For better understanding, the real-life examples of non-collinear points are sprinklers on a cake, coffee beans in a jar i.e., scattered positions of points.
### Concurrent Points
A concurrent point is that point on which two or more lines are intersected.
For Example. Consider the points, D, E, F, and G, making line DE and FG, such that these intersect at point H. Then the point H is known as point of intersection or concurrent point.
#### Real-life Examples.
For better understanding, the real-life examples of concurrent points are intersecting points of a kite, spider web, spokes of bicycle etc.
### Coplanar Points
The word ‘coplanar’ is derived from the word ‘co’ and ‘plane’ which means, the points which lie on the single plane. In geometry, a plane is known as a 2-D flat surface which extends indefinitely, on which a point, line and three dimensions can lie. In real life, a wall or a piece of paper are the examples of plane.
• There should be at least three points on a plane to be called as coplanar.
For Example. There are four points, D, E, F and G lying on a common plane.
#### Real-life Examples.
For better understanding, the real-life examples of coplanar points are the hands-on of clock, objects on a table, grids of a graph paper all lie on a single plane.
### Non-coplanar Points
Unlike coplanar points, the non-coplanar points are those which do not lie on the same plane. There should be at least four or more points which could be called as non-planar only if these are not lying on a single plane. Usually non-coplanar points exist in case of three-dimensional figures.
For Example. There are four points, D, E, and F lying on a common plane, However, the point is said to be non-coplanar as it does not lie on a single plane with others.
#### Real-life Examples.
For better understanding, the real-life example is the point G on a wall with plane A would be non-coplanar to the points E and F on a wall with plane B.
### Equidistant Points
A point is said to be equidistant from others if it is at similar distance from other points.
For Example. The points E in the given triangle is known as equidistant point which is at similar distance from F and G respectively.
#### Real-life Example.
For better understanding, the real-life example of equidistant point is the central point of a sky wheel is equidistant from all other points.
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How many positive integers less than 30 are either a multiple of 2, an odd prime number, or the sum of a positive multiple of 2 and an odd prime?
A. 29
B. 28
C. 27
D. 25
E. 23
[Reveal] Spoiler: OA
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enigma123 wrote:
How many positive integers less than 30 are either a multiple of 2, an odd prime number, or the sum of a positive multiple of 2 and an odd prime?
(A) 29
(B) 28
(C) 27
(D) 25
(E) 23
Any idea how to solve this guys?
30 sec approach:
Any odd non-prime, greater than 1, can be obtained by the sum of an odd prime and a positive even number. So this set plus the set of odd primes basically makes the set of all odd numbers greater than 1 in the range. Now, the set of all odd numbers greater than 1 together with the set of all even numbers makes the set of all numbers from 1 to 30, not inclusive, so total of 28 numbers.
Answer: B.
To illustrate:
# of even numbers in the range is (28-2)/2+1=14: 2, 4, 6, ..., 28;
# of odd primes in the range is 9: 3, 5, 7, 11, 13, 17, 19, 23, and 29;
# of integers which are the sum of a positive multiple of 2 and an odd prime is 5: 9=7+2, 15=13+2, 21=19+2, 25=23+2 and 27=23+4;
Total: 14+9+5=28. You can see that we have all numbers from 1 to 30, not inclusive: 2, 3, 4, 5, 6, ...., 29.
Hope it's clear.
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Many thanks Bunuel - you mean to say answer is B. I take it's a typo at your end
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enigma123 wrote:
Many thanks Bunuel - you mean to say answer is B. I take it's a typo at your end
Yes, 28 is answer B.
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How many positive integers less than 30 are either a multiple
of 2, an odd prime number, of the sum of a positive multiple
of 2 and an odd prime?
(A) 29
(B) 28
(C) 27
(D) 25
(E) 23
is there any shortcut method to solve this type of problem??
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Any odd number can be expressed as 2k+1 or 2k+(3-2) or 2(K-1)+3. Thus, with the prime number 3, we can express all the odd numbers.
Since, 1 i is the only number that cannot be expressed, answer is numbers <30 =29-1.
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asax wrote:
Any odd number can be expressed as 2k+1 or 2k+(3-2) or 2(K-1)+3. Thus, with the prime number 3, we can express all the odd numbers.
Since, 1 i is the only number that cannot be expressed, answer is numbers <30 =29-1.
Definitely very clever. I spent 2 minutes going the long way until I realized that.
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Re: How many positive integers less than 30 are either a [#permalink]
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Bunuel wrote:
enigma123 wrote:
How many positive integers less than 30 are either a multiple of 2, an odd prime number, or the sum of a positive multiple of 2 and an odd prime?
(A) 29
(B) 28
(C) 27
(D) 25
(E) 23
Any idea how to solve this guys?
30 sec approach:
Any odd non-prime, greater than 1, can be obtained by the sum of an odd prime and a positive even number. So this set plus the set of odd primes basically makes the set of all odd numbers greater than 1 in the range. Now, the set of all odd numbers greater than 1 together with the set of all even numbers makes the set of all numbers from 1 to 30, not inclusive, so total of 28 numbers.
Answer: B.
To illustrate:
# of even numbers in the range is (28-2)/2+1=14: 2, 4, 6, ..., 28;
# of odd primes in the range is 9: 3, 5, 7, 11, 13, 17, 19, 23, and 29;
# of integers which are the sum of a positive multiple of 2 and an odd prime is 5: 9=7+2, 15=13+2, 21=19+2, 25=23+2 and 27=23+4;
Total: 14+9+5=28. You can see that we have all numbers from 1 to 30, not inclusive: 2, 3, 4, 5, 6, ...., 29.
Hope it's clear.
Hey Bunuel,
How can this be the entire list?
# of integers which are the sum of a positive multiple of 2 and an odd prime is 5: 9=7+2, 15=13+2, 21=19+2, 25=23+2 and 27=23+4;
Shouldnt be:
2(1)+3<30
2(1)+5<30
2(1)+7<30
2(1)+11<30
....
2(1)+23<30
Now
2(2)+3<30
2(2)+5<30
2(2)+7<30
2(2)+11<30
....
2(2)+23<30
etc
Your list didn't include all those? What am I missing?
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Re: How many positive integers less than 30 are either a [#permalink]
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alphabeta1234 wrote:
Bunuel wrote:
enigma123 wrote:
How many positive integers less than 30 are either a multiple of 2, an odd prime number, or the sum of a positive multiple of 2 and an odd prime?
(A) 29
(B) 28
(C) 27
(D) 25
(E) 23
Any idea how to solve this guys?
30 sec approach:
Any odd non-prime, greater than 1, can be obtained by the sum of an odd prime and a positive even number. So this set plus the set of odd primes basically makes the set of all odd numbers greater than 1 in the range. Now, the set of all odd numbers greater than 1 together with the set of all even numbers makes the set of all numbers from 1 to 30, not inclusive, so total of 28 numbers.
Answer: B.
To illustrate:
# of even numbers in the range is (28-2)/2+1=14: 2, 4, 6, ..., 28;
# of odd primes in the range is 9: 3, 5, 7, 11, 13, 17, 19, 23, and 29;
# of integers which are the sum of a positive multiple of 2 and an odd prime is 5: 9=7+2, 15=13+2, 21=19+2, 25=23+2 and 27=23+4;
Total: 14+9+5=28. You can see that we have all numbers from 1 to 30, not inclusive: 2, 3, 4, 5, 6, ...., 29.
Hope it's clear.
Hey Bunuel,
How can this be the entire list?
# of integers which are the sum of a positive multiple of 2 and an odd prime is 5: 9=7+2, 15=13+2, 21=19+2, 25=23+2 and 27=23+4;
Shouldnt be:
2(1)+3<30
2(1)+5<30
2(1)+7<30
2(1)+11<30
....
2(1)+23<30
Now
2(2)+3<30
2(2)+5<30
2(2)+7<30
2(2)+11<30
....
2(2)+23<30
etc
Your list didn't include all those? What am I missing?
First of all we are asked about the number of positive integers less than 30, which are a multiple of 2 OR an odd prime number OR the sum of a positive multiple of 2 and an odd prime.
Next, EACH numbers from 1 to 30, not inclusive is a multiple of 2 OR an odd prime number OR the sum of a positive multiple of 2 and an odd prime. So, the list is 2, 3, 4, 5, ..., 29 (total of 28 numbers).
So, which number is not included in the list?
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Re: How many positive integers less than 30 are either a [#permalink]
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24 Aug 2012, 12:27
Bunuel wrote:
alphabeta1234 wrote:
30 sec approach:
Any odd non-prime, greater than 1, can be obtained by the sum of an odd prime and a positive even number. So this set plus the set of odd primes basically makes the set of all odd numbers greater than 1 in the range. Now, the set of all odd numbers greater than 1 together with the set of all even numbers makes the set of all numbers from 1 to 30, not inclusive, so total of 28 numbers.
Answer: B.
To illustrate:
# of even numbers in the range is (28-2)/2+1=14: 2, 4, 6, ..., 28;
# of odd primes in the range is 9: 3, 5, 7, 11, 13, 17, 19, 23, and 29;
# of integers which are the sum of a positive multiple of 2 and an odd prime is 5: 9=7+2, 15=13+2, 21=19+2, 25=23+2 and 27=23+4;
Total: 14+9+5=28. You can see that we have all numbers from 1 to 30, not inclusive: 2, 3, 4, 5, 6, ...., 29.
Hope it's clear.
Hey Bunuel,
How can this be the entire list?
# of integers which are the sum of a positive multiple of 2 and an odd prime is 5: 9=7+2, 15=13+2, 21=19+2, 25=23+2 and 27=23+4;
Shouldnt be:
2(1)+3<30
2(1)+5<30
2(1)+7<30
2(1)+11<30
....
2(1)+23<30
Now
2(2)+3<30
2(2)+5<30
2(2)+7<30
2(2)+11<30
....
2(2)+23<30
etc
Your list didn't include all those? What am I missing?
Bunuel's Response:
First of all we are asked about the number of positive integers less than 30, which are a multiple of 2 OR an odd prime number OR the sum of a positive multiple of 2 and an odd prime.
Next, EACH numbers from 1 to 30, not inclusive is a multiple of 2 OR an odd prime number OR the sum of a positive multiple of 2 and an odd prime. So, the list is 2, 3, 4, 5, ..., 29 (total of 28 numbers).
So, which number is not included in the list?[/quote]
Hey Bunuel,
Thanks for pointing out my mistake the same numbers that are generated by 2K+odd prime are also included in the same list as the odd primes.
In other words
A=# of even numbers between 1 and 29, inclusive
B=# of odd primes between 1 and 29, inclusive
C=# of 2K+odd_prime, between 1 and 29, inclusive
AUBUC=A+B+C-AB-AC-BC-ABC+N
AB=0, since there are no numbers both even and odd primes between 1 and 29, inclusive
AC=0, since there are no numbers both even and 2K+odd_prime(=odd) between 1 and 29, inclusive
ABC=0 since no numbers are even, and odd prime and a 2K+odd_prime
and N=1, since only 1 fits the criteria of being niether an even number, neither an odd prime, and neither a 2K+odd_prime
My question I guess is for BC, numbers both an odd prime and 2K+odd_prime. Is there a way to tell, without actually listing out all the numbers that meet this condition and checking ?
Thank you!
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Re: How many positive integers less than 30 are either a [#permalink]
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22 Sep 2012, 13:56
I can't believe that what made this problem difficult was a "typo error" in the question statement!!!!
Instead of "... number, of the sum of a positive multiple..." is "... number, OR the sum of a positive...
Thank you for clarifying!!! =)
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26 Jun 2013, 01:25
Bumping for review and further discussion*. Get a kudos point for an alternative solution!
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Re: How many positive integers less than 30 are either a [#permalink]
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26 Jun 2013, 08:26
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enigma123 wrote:
How many positive integers less than 30 are either a multiple of 2, an odd prime number, of the sum of a positive multiple of 2 and an odd prime?
A. 29
B. 28
C. 27
D. 25
E. 23
Qquestion: 0<x<30
so, 1<=x<=29
leave x=1 alone for a while, and consider everything else i.e. 2<=x<=29
integer either multiple of 2 that will be almost half the no's (14)
odd prime no, and sum of a positive multiple of 2 and an odd prime => Rest everything else has to be either a prime no or the sum of some multiple of 2(Those 14 no we got earlier)and a odd no
only for x=1, it is neither even, nor prime and definitely not the sum.
Thus ans = total no's - 1 = 29 - 1 = 28
Ans: B
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Re: How many positive integers less than 30 are either a [#permalink]
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04 Oct 2013, 22:09
Bunuel wrote:
enigma123 wrote:
How many positive integers less than 30 are either a multiple of 2, an odd prime number, or the sum of a positive multiple of 2 and an odd prime?
(A) 29
(B) 28
(C) 27
(D) 25
(E) 23
Any idea how to solve this guys?
30 sec approach:
Any odd non-prime, greater than 1, can be obtained by the sum of an odd prime and a positive even number. So this set plus the set of odd primes basically makes the set of all odd numbers greater than 1 in the range. Now, the set of all odd numbers greater than 1 together with the set of all even numbers makes the set of all numbers from 1 to 30, not inclusive, so total of 28 numbers.
Answer: B.
To illustrate:
# of even numbers in the range is (28-2)/2+1=14: 2, 4, 6, ..., 28;
# of odd primes in the range is 9: 3, 5, 7, 11, 13, 17, 19, 23, and 29;
# of integers which are the sum of a positive multiple of 2 and an odd prime is 5: 9=7+2, 15=13+2, 21=19+2, 25=23+2 and 27=23+4;
Total: 14+9+5=28. You can see that we have all numbers from 1 to 30, not inclusive: 2, 3, 4, 5, 6, ...., 29.
Hope it's clear.
I did not understand the last condition ?
sum of a positive multiple of 2 and an odd prime ?
it can be possible: 7=5+2 ???
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Re: How many positive integers less than 30 are either a [#permalink]
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05 Oct 2013, 04:12
sunny3011 wrote:
Bunuel wrote:
enigma123 wrote:
How many positive integers less than 30 are either a multiple of 2, an odd prime number, or the sum of a positive multiple of 2 and an odd prime?
(A) 29
(B) 28
(C) 27
(D) 25
(E) 23
Any idea how to solve this guys?
30 sec approach:
Any odd non-prime, greater than 1, can be obtained by the sum of an odd prime and a positive even number. So this set plus the set of odd primes basically makes the set of all odd numbers greater than 1 in the range. Now, the set of all odd numbers greater than 1 together with the set of all even numbers makes the set of all numbers from 1 to 30, not inclusive, so total of 28 numbers.
Answer: B.
To illustrate:
# of even numbers in the range is (28-2)/2+1=14: 2, 4, 6, ..., 28;
# of odd primes in the range is 9: 3, 5, 7, 11, 13, 17, 19, 23, and 29;
# of integers which are the sum of a positive multiple of 2 and an odd prime is 5: 9=7+2, 15=13+2, 21=19+2, 25=23+2 and 27=23+4;
Total: 14+9+5=28. You can see that we have all numbers from 1 to 30, not inclusive: 2, 3, 4, 5, 6, ...., 29.
Hope it's clear.
I did not understand the last condition ?
sum of a positive multiple of 2 and an odd prime ?
it can be possible: 7=5+2 ???
2 is not an odd prime.
But 7 CAN be written as the sum of a positive multiple of 2 and an odd prime: 7 = 4 + 3.
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Re: How many positive integers less than 30 are either a [#permalink]
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16 Oct 2013, 23:26
Bunuel wrote:
enigma123 wrote:
How many positive integers less than 30 are either a multiple of 2, an odd prime number, or the sum of a positive multiple of 2 and an odd prime?
(A) 29
(B) 28
(C) 27
(D) 25
(E) 23
Any idea how to solve this guys?
30 sec approach:
Any odd non-prime, greater than 1, can be obtained by the sum of an odd prime and a positive even number. So this set plus the set of odd primes basically makes the set of all odd numbers greater than 1 in the range. Now, the set of all odd numbers greater than 1 together with the set of all even numbers makes the set of all numbers from 1 to 30, not inclusive, so total of 28 numbers.
Answer: B.
To illustrate:
# of even numbers in the range is (28-2)/2+1=14: 2, 4, 6, ..., 28;
# of odd primes in the range is 9: 3, 5, 7, 11, 13, 17, 19, 23, and 29;
# of integers which are the sum of a positive multiple of 2 and an odd prime is 5: 9=7+2, 15=13+2, 21=19+2, 25=23+2 and 27=23+4;
Total: 14+9+5=28. You can see that we have all numbers from 1 to 30, not inclusive: 2, 3, 4, 5, 6, ...., 29.
Hope it's clear.
In this # of integers which are the sum of a positive multiple of 2 and an odd prime ,.. why didnt we count 7=5+2 and 13=11+2,19=13+4 .. ??? these all are Sum of multiple of 2 and odd primes. ????
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Re: How many positive integers less than 30 are either a [#permalink]
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17 Oct 2013, 02:09
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Expert's post
ishdeep18 wrote:
Bunuel wrote:
enigma123 wrote:
How many positive integers less than 30 are either a multiple of 2, an odd prime number, or the sum of a positive multiple of 2 and an odd prime?
(A) 29
(B) 28
(C) 27
(D) 25
(E) 23
Any idea how to solve this guys?
30 sec approach:
Any odd non-prime, greater than 1, can be obtained by the sum of an odd prime and a positive even number. So this set plus the set of odd primes basically makes the set of all odd numbers greater than 1 in the range. Now, the set of all odd numbers greater than 1 together with the set of all even numbers makes the set of all numbers from 1 to 30, not inclusive, so total of 28 numbers.
Answer: B.
To illustrate:
# of even numbers in the range is (28-2)/2+1=14: 2, 4, 6, ..., 28;
# of odd primes in the range is 9: 3, 5, 7, 11, 13, 17, 19, 23, and 29;
# of integers which are the sum of a positive multiple of 2 and an odd prime is 5: 9=7+2, 15=13+2, 21=19+2, 25=23+2 and 27=23+4;
Total: 14+9+5=28. You can see that we have all numbers from 1 to 30, not inclusive: 2, 3, 4, 5, 6, ...., 29.
Hope it's clear.
In this # of integers which are the sum of a positive multiple of 2 and an odd prime ,.. why didnt we count 7=5+2 and 13=11+2,19=13+4 .. ??? these all are Sum of multiple of 2 and odd primes. ????
Because 7, 13, and 19 (all primes) are included in the second set (dd primes).
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Re: How many positive integers less than 30 are either a [#permalink]
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23 Dec 2013, 23:03
What are the actual 2 numbers that answer this question? I know 1 is one of them, but I can't think of the other one...I used to think it was 0 but technically 0 is neither positive nor negative...
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Re: How many positive integers less than 30 are either a [#permalink]
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24 Dec 2013, 00:33
catalysis wrote:
What are the actual 2 numbers that answer this question? I know 1 is one of them, but I can't think of the other one...I used to think it was 0 but technically 0 is neither positive nor negative...
I think you misinterpreted the question. It asks: "how many positive integers less than 30 are ..."
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Re: How many positive integers less than 30 are either a [#permalink]
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24 Dec 2013, 10:48
Bunuel wrote:
catalysis wrote:
What are the actual 2 numbers that answer this question? I know 1 is one of them, but I can't think of the other one...I used to think it was 0 but technically 0 is neither positive nor negative...
I think you misinterpreted the question. It asks: "how many positive integers less than 30 are ..."
Hi Bunuel - Sorry, I think I misworded my original question. I know the answer is 28, which means 28 numbers less than 30 meet the constraints given. However, I was just curious which values are the numbers that do NOT meet the constraints.
However, I have kind of answered my own question because now I realize that there are only 29 integers to choose from (1-29 inclusive), not 30 like I had originally thought, because 0 is not a positive integer and 30 cannot be included because the question asks for numbers less than 30. Therefore, it makes sense that 1 is the only integer that does not meet the constraints and I should not be looking for a second number. (29 possible integers - 1 integer that does not meet the constraints = 28 integers that meet the constraints, just like the answer says)
Hope this makes sense...
Re: How many positive integers less than 30 are either a [#permalink] 24 Dec 2013, 10:48
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BoundaryMeshRegion Error: Boundary surface is not closed
I am trying to create a BoundaryMeshRegion comprised of many cubes with random locations, in a relatively small region.
I have had success implementing my technique for a smaller number of cubes (say, 30), but have run into an error which sometimes pops up after repeated runs of the same number. With an increase in the number of cubes, this error frequently pops up, of the form (ignore the blah blah blah...that is just to eliminate a ton of numbers):
BoundaryMeshRegion::bsuncl: The boundary surface is not closed because the edges Line[{{{blah},{blah}, {blah}}}] only come from a single face.
I have tried to fix this using the suggestion in this thread with very limited success: Understanding BoundaryMeshRegion error
As for the exact method in which these cubes are created (inside of a Do loop and appending the results out one by one, etc., etc.), I need it to be done this way, as this is the technique which must be used as a part in a separate, much more complicated simulation.
Is there anything that can be done to eliminate this problem, such that any number of cubes can be used (whilst maintaining the spirit of this block-building technique)? Best of luck!
P.S., I highly recommend running the same number of cubes repeatedly. I thought I had fixed it, but after another few simulations, the error popped up again.
Here is the code.
Clear["Global*"];
cubeNumber = 40;
cubeTable = {};
pos0 = Table[{RandomReal[{0.5, 0.7}], RandomReal[{0.5, 0.7}],
RandomReal[{0, 0.1}]}, cubeNumber]
Do[
cubes =
Graphics3D[
Cube[{pos0[[k]][[1]], pos0[[k]][[2]], pos0[[k]][[3]]}, 0.05]];
AppendTo[cubeTable, cubes];
, {k, 1, cubeNumber}];
cubicMesh =
BoundaryDiscretizeRegion[
RegionUnion[
Flatten[Table[
DiscretizeGraphics[cubeTable[[k]]], {k, Length[cubeTable]}]]]]
At first we use the OP's codes but remove Graphics3D and Part.
Clear["Global*"];
cubeNumber = 40;
cubeTable = {};
pos0 = Table[{RandomReal[{0.5, 0.7}], RandomReal[{0.5, 0.7}],
RandomReal[{0, 0.1}]}, cubeNumber];
Do[cubes = Cube[pos0[[k]], 0.05];
AppendTo[cubeTable, cubes];, {k, 1, cubeNumber}];
After that we again recommend to use NDSolveFEM Or OpenCascadeLink to make the BoundaryMeshRegion effective.(Compare with BoundaryDiscretizeRegion[regs])
Needs["NDSolveFEM"];
regs = cubeTable // RegionUnion;
bm = ToBoundaryMesh[regs];
bmr = BoundaryMeshRegion[bm]
bmr // Volume
RegionBoundary[bmr] // Area
bm["Wireframe"]
regs = cubeTable // RegionUnion;
Needs["OpenCascadeLink"];
(* Needs["NDSolveFEM"]; *)
bmr = BoundaryMeshRegion[bm]
bmr // Volume
bm["Wireframe"]
0.00348886.
I know that you said that your approach is dictated by other limitations, but there are still quite a few things you could improve in your code. Here is refactored example:
SeedRandom[1];
cubeNumber = 40;
pos0 =
RandomVariate[
UniformDistribution[{{0.5, 0.7}, {0.5, 0.7}, {0, 0.1}}],
cubeNumber
];
cubeTable = Cube[#, 0.05]& /@ pos0;
Region[RegionUnion @@ cubeTable]
At least you could consider:
1. Generating the positions in one call to RandomVariate rather than a table.
2. Mapping Cube over the list of positions rather than using a table. Note also that your Cube[{pos0[[k]][[1]], pos0[[k]][[2]], pos0[[k]][[3]]}, 0.05] is completely equivalent to Cube[pos0[[k]], 0.05]`.
3. Consider avoiding the discretization of graphics and perhaps use regions instead.
• thank you! The positions are necessarily generated one at a time in my other code (a monte carlo sim), and their positions dictate boundary conditions in a physical process in the next iteration of the monte carlo sim. So basically I need to create and then update a mesh of cubes after each step in my Do loop so that the next particle "sees" the new cube that is there. Any idea on how one could achieve this?
– Zach
Commented Aug 10, 2022 at 4:07
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# 470
Are you smarter than a mathematician? - December 3, 2011
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Or you could simply copy his answer without verifying it :P
Side Note: Patrick Vennebush is running an online version of his favorite game.
You can play the game using the following form:
The game is limited to the first 100 players.
The rules are simple:
• Pick a positive integer (i.e., 1, 2, 3, 4, ...).
• The winner is the person who picks the smallest number not chosen by anyone else.
Also, Hello Sam R. :-)
I think that it's "more correct" to consider 0 as a natural number (Set theory wise)
it definitely makes notation shorter
NU{0} and N, verus N and N+
I'm not sure about this game--but I'd like to imagine that mathematicians would get the right answer on the Monty Hall paradox.
Found this on the Internet:
1. there is an obscure number
2. then the obscure number becomes not obscure because of there is the smallest obscure number
I think that's roughly the same as the Berry Paradox: http://en.wikipedia.org/wiki/Berry_paradox
Aha, the interesting number paradox. The first number to have no interesting properties has the interesting property of being the first number to have no interesting properties. :D
what about the set of all numbers that aren't the smallest number of any definable collection?
Should we call this a Gödel number?
No. This would be an anti-Zorn number.
True. The Gödel number would be the smallest anti-Zorn number.
And it might, or might not, exist.
This is why Fuzzy Sets were created.
Interesting. I had never come across Fuzzy sets--but they seem obvious to me.
I do like fuzzy logic--and fuzzy dice.
but there are uncoutable many uninteresting numbers in R.
I'm not so sure about that...
You can only name a countable infinite number of numbers (our language is countable), we can state that those others numbers have the interesting property that you can't name them xD
I came across a video of Ken Jennings on Are You Smarter Than a Fifth Grader on YouTube the other day. You can't actually peek or copy on the \$1,000,000 question.
You were still funny but technically inaccurate - isn't that a cardinal sin to a mathematician?? ;)
More likely an ordinal sin.
It's only ordinal if you are counting!
An SMS game show in my country used the "smallest unique number" game for selecting winners. The problem was that players were sent back a message stating if the number they picked was lower or higher than the current winning number. They probably did that to encourage people to send multiple messages (thus paying more for them), but someone abused those reports by connecting a bunch of phones to a computer and calculating the right number. That way, that person (actually a bunch of undercover people instead of him) repeatedly won the game day after day. Then he got sued.
True story.
The game is redundant. The answer is "No".
Perhaps "trivial" would be a better word for it.
Update!!
Also here: claim for update! :)
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If you make a mistake or the comment doesn't show up properly, email me and I'll gladly fix it :-).
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# Savings vs Investment: Understanding the Difference
(*)(*)(*)(*)(*) ( 4.2, 96 students)
• Price per Classroom\$2.00
• Grade Levels 7, 8, 9, 10
• Topics Investment
• Duration 20 Minutes
• Auto Scored? Yes
• Teacher Evaluation Needed? No
### Activity Description
In this hands-on worksheet activity, students will gain practical insights into the dynamics of savings and investments over varying timeframes. The primary objective is to understand how the same initial amount allocated to both savings and investments can perform differently based on the investment duration and to understand the risk involved in investments. There are three schemes that students will compare -
1. Savings scheme
2. Investment scheme that gives positive returns
3. Investment scheme that gives negative returns
Activity Steps:
1. The student will decide the amount of money and the decided money will be allocated for three schemes equally (one savings scheme and two investment schemes)
2. Students will select different investment durations using a slider (0 - 20 years)
3. For different durations given, students will analyze which scheme gives the best returns.
Key Takeaways:
Students will understand that savings schemes always give predictable returns and are of low risk, while investment schemes involve a risk that sometimes may give negative returns, but they are capable of giving multifold returns.
### Learning Objective
• Recognize the difference in performance between savings and investments.
• Understand how the duration of an investment can impact its returns.
### Teacher Tips
Included with the activity, you can view the tips to clarify student's doubts or to evaluate answers (for a teacher scored worksheet).
## More Activities You Might Like
\$2.00/Classroom
### Savings: Taking Inflation Into Account
Students will understand that if they don't invest their money, the value of money will reduce over time due to inflation.
• Saving, Investment
• 7, 8, 9, 10 Auto-Graded
• ### Simple Interest: Saving For Your Dream House
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### Simple Interest: Saving For Your Dream House
Students will use the simple interest formula and its variations to find the interest, principal to be deposited, and the interest rate they should be looking for to get \$100,000 after 10 years for their dream house.
• Numbers & Operations, Fractions, Ratios & Proportions, Saving, Investment
• 6, 7, 8, 9, 10 Auto-Graded
• ### Start Saving Early
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### Start Saving Early
Students will explore the total interest that can be earned by launching saving endeavors at different ages with a goal of retiring at the age of 65.
• Saving, Investment
• 7, 8, 9, 10 Auto-Graded
• ### Invest For The Long Term - Risk Management
\$2.00/Classroom
### Invest For The Long Term - Risk Management
Students will analyze the share prices of a company called Chipotle Mexican Grill over a period of eight years, and through prompt questions, they will understand the benefit of investing for the long term.
• Investment
• 7, 8, 9, 10 Auto-Graded
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# ENGI 8673 Subsea Pipeline Engineering
Lecture 09: Mechanical Design – Pressure Containment: Part 2
Shawn Kenny, Ph.D., P.Eng. Assistant Professor Faculty of Engineering and Applied Science Memorial University of Newfoundland spkenny@engr.mun.ca
D..Update – Available DNV OS & RP Educational Purposes Only DNV OS-F101 Submarine Pipeline Systems DNV-RP-C205 Environmental Conditions and Environmental Loads DNV-RP-F105 Free Spanning Pipelines DNV-RP-F109 On-bottom Stability Design of Submarine Pipelines DNV-RP-F111 Interference between Trawl Gear and Pipelines 2 © 2008 S. P. Kenny.Eng. Ph. ENGI 8673 Subsea Pipeline Engineering – Lecture 09 .
. Ph. P.D. Kenny. ENGI 8673 Subsea Pipeline Engineering – Lecture 09 .Eng.Lecture 09 Objective To examine mechanical behaviour for pressure containment of thick-walled pipelines 3 © 2008 S.
Thick-Walled Cylinders Characteristics Through thickness (radial) variation of stress Valid for D/t ≤ 20 t ≥ 0. P.10 inner pipe radius Pe Theory of Elasticity Lamé Formula • Force equilibrium • ε-δ relationships • Compatibility equation • σ-ε relationships 4 © 2008 S. Ph.D..Eng. Pi ENGI 8673 Subsea Pipeline Engineering – Lecture 09 . Kenny.
Eng.) Symmetry No shear stress τ rθ = 0 Equilibrium equation dσ r σ r − σθ + + Fr = 0 dr r ε-δ relationship εr = du 1 dv u dv 1 du v + . γ rθ = + − . Kenny. P. Ph. ENGI 8673 Subsea Pipeline Engineering – Lecture 09 .D. εθ = dr r dθ r dr r dθ r 5 © 2008 S..Thick-Walled Cylinders (cont.
D.) Open End & Unconstrained σz = 0 Hooke’s law εr = εθ = du 1 = (σ r − νσ θ ) dr E u 1 = (σ θ − νσ r ) r E γ rθ = 0 6 © 2008 S.Thick-Walled Cylinders (cont.Eng.. P. Ph. ENGI 8673 Subsea Pipeline Engineering – Lecture 09 . Kenny.
Kenny.) Rearranging and Equating σr = σθ = E E ⎛ du u⎞ +ν ⎟ (ε r + νεθ ) = 1−ν 2 1 − ν 2 ⎜ dr r⎠ ⎝ E E ⎛u du ⎞ +ν (εθ + νε r ) = 1−ν 2 1−ν 2 ⎜ r dr ⎟ ⎝ ⎠ Substitute into Force Equilibrium dσ r σ r − σθ + + Fr = 0 dr r 7 © 2008 S.D.Eng. P. Ph..Thick-Walled Cylinders (cont. ENGI 8673 Subsea Pipeline Engineering – Lecture 09 .
D.Thick-Walled Cylinders (cont. P.Eng. Ph..) Equilibrium Equation of Motion d 2u 1 du u + − 2 =0 2 dr r dr r Solution ⇒ u = c1r + c2 r Radial and Tangential Stress Equations σr = σθ = E 1−ν 2 E 1−ν 2 ⎡ ⎛ 1−ν ⎞⎤ c1 (1 + ν ) − c2 ⎜ 2 ⎟ ⎥ ⎢ ⎝ r ⎠⎦ ⎣ ⎡ ⎛ 1−ν ⎞⎤ c1 (1 + ν ) + c2 ⎜ 2 ⎟ ⎥ ⎢ ⎝ r ⎠⎦ ⎣ 8 © 2008 S. Kenny. ENGI 8673 Subsea Pipeline Engineering – Lecture 09 .
Ph.Thick-Walled Cylinders (cont..D. Kenny. ENGI 8673 Subsea Pipeline Engineering – Lecture 09 .) Constant Longitudinal Strain σr = σθ = E 1−ν 2 E 1−ν 2 ⎡ ⎛ 1−ν ⎞⎤ c1 (1 + ν ) − c2 ⎜ 2 ⎟ ⎥ ⎢ ⎝ r ⎠⎦ ⎣ ⎡ ⎛ 1−ν c1 (1 + ν ) + c2 ⎜ 2 ⎢ ⎝ r ⎣ ⎞⎤ ⎟⎥ ⎠⎦ σ r + σθ = εz = − ν E 2Ec1 1−ν (σ r + σ θ ) = constant ∴ plane sections remain plane 9 © 2008 S. P.Eng.
Ph.Eng..Thick-Walled Cylinders (cont. P. Kenny. ENGI 8673 Subsea Pipeline Engineering – Lecture 09 .D.) Recall Open End and P Unconstrained Cylinder ∫ σ ( 2π r ) dr = c π ( b − a ) = 0 b 2 2 a z 3 e a Pi b ∴c3 = σz = 0 Substitute Pressure Boundary Conditions 1 − ν a 2 pi − b 2 pe c1 = E b2 − a2 2 2 1 + ν a b ( pi − pe ) c2 = E b2 − a2 10 © 2008 S.
Eng. Ph. Kenny.. ENGI 8673 Subsea Pipeline Engineering – Lecture 09 .D.Thick-Walled Cylinders (cont. P.) Lamé Equations a 2 pi − b 2 pe ( pi − pe ) a b − σr = b2 − a2 b2 − a2 r 2 2 2 Pe a ( ) Pi b a 2 pi − b 2 pe ( pi − pe ) a b + σθ = b2 − a2 b2 − a2 r 2 2 2 ( ) 2 2 2 2 1 − ν a pi − b pe r 1 + ν ( pi − pe ) a b + u= 2 2 E b −a E b2 − a2 r ( ) ( ) τ max = σθ − σ r 2 ( pi − pe ) a2b2 = (b 2 − a2 r 2 ) Conditions for largest τmax? 11 © 2008 S.
σθ τmax on 45° Pe a Pi b Pressure to Initiate Yielding (Tension) τ max = py 12 σy 2 2 (b = − a2 σ y 2b 2 ENGI 8673 Subsea Pipeline Engineering – Lecture 09 ) © 2008 S.Thick-Walled Cylinders (cont.D.Eng..) Principal Stresses σr. . Ph. P. Kenny.
Eng. Ph. Kenny..Hoop Stress Comparison D/t Ratio 13 © 2008 S. ENGI 8673 Subsea Pipeline Engineering – Lecture 09 .D. P.
P. Ph.D.. Kenny.Hoop Stress Comparison Through Thickness 14 © 2008 S.Eng. ENGI 8673 Subsea Pipeline Engineering – Lecture 09 .
ν = 0. P.Eng.762.D. Pe = 2MPa. ENGI 8673 Subsea Pipeline Engineering – Lecture 09 . Pi = 20MPa.3 Thin wall theory tmin = 18. Kenny. φ = 0.2mm tmin ≥ Lamé tmin = 19.1mm 1 2 t 1 t − + =0 2 φ σ y + pe ⎞ Do Do ⎛ 2 ⎜1 + ⎟ pi − pe ⎠ ⎝ 15 © 2008 S.Example 9-01 Calculate the minimum wall thickness using thin wall theory.80 E = 205GPa. σy = 448MPa.0mm tmin ( pi − pe ) Do ≥ 2φσ y + pi pi Do 2φσ y Barlow tmin = 21.. Ph. Barlow’s equation and Lamé equation using the outside diameter Do = 0.
(2003)..Eng.D. 544p. S.K. and Fenster. P. ENGI 8673 Subsea Pipeline Engineering – Lecture 09 . 4th Edition.References Ugural. Advanced Strength and Applied Elasticity. Prentice Hall.C. Kenny. 16 © 2008 S. A. Ph.
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# [1]胡启洲,邓卫,田新现.基于四维消耗的公交线网优化模型及蚁群算法[J].东南大学学报(自然科学版),2008,38(2):304-308.[doi:10.3969/j.issn.1001-0505.2008.02.024] Hu Qizhou,Deng Wei,Tiam Xinxian.Optimization model of public traffic network and ant algorithm with four dimensions consumption[J].Journal of Southeast University (Natural Science Edition),2008,38(2):304-308.[doi:10.3969/j.issn.1001-0505.2008.02.024] 点击复制 基于四维消耗的公交线网优化模型及蚁群算法() 分享到： var jiathis_config = { data_track_clickback: true };
38
2008年第2期
304-308
2008-03-20
## 文章信息/Info
Title:
Optimization model of public traffic network and ant algorithm with four dimensions consumption
1 东南大学交通学院, 南京 210096; 2 平顶山教育学院, 平顶山 467000
Author(s):
1 School of Transportation, Southeast University, Nanjing 210096, China
2 Pingdingshan Education College, Pingdingshan 467000, China
Keywords:
U491.13
DOI:
10.3969/j.issn.1001-0505.2008.02.024
Abstract:
Aimed at the optimization problem for public traffic line network, a new method and algorithm of public traffic line network optimization is presented using the concept of four dimensions consumption. Based on the definition of four dimensions consumption concept, including time, space, environment and energy, constraint conditions and objective function are analyzed on three facets: points, lines and range. Taking the benefits maximization and costs minimization and sustainable development as the objectives a network optimization model is developed. A multi-objective linear programming model of public transportation optimization is established. Ant algorithm is used to plan the public transport network. The best layout structure of public traffic line network and the best operational efficiency of public traffic system can then be achieved. The application results show that the multi-objective linear programming model not only meets the transport demands, but also implements the environment protection objective and improves the utilization ratio of urban public traffic line network. The optimization results accord with the real situation. The method is feasible.
## 参考文献/References:
[1] 王炜,杨新苗,陈学武.城市公共交通系统规划方法与管理技术[M].北京:科学出版社,2002.
[2] Wang Wei,Wang Fumin.Scheme design technique for urban traffic management planning[J].Journal of Southeast University:English Edition,2005,21(3):353-358.
[3] Jin Wenlong.A dynamical system model of the traffic assignment problem[J].Transportation Research Part B,2007,41(1):32-48.
[4] 吴世江,史其信,陆化普.基于交通效率的城市公共交通路网布局模型[J].土木工程学报,2005,38(1):117-120.
Wu Shijiang,Shi Qixin,Lu Huapu.Traffic efficiency based urban public traffic network distribution model[J].China Civil Engineering Journal,2005,38(1):117-120.(in Chinese)
[5] Brons M,Nijkamp P,Pels E,et al.Efficiency of urban public transit:a meta analysis[J]. Transportation,2005,32(1):1-21.
[6] Szeto W Y,Lo H K.Transportation network improvement and tolling strategies:the issue of intergeneration equity[J]. Transportation Research Part A,2006,40(3):227-243.
[7] Poorzahedy H,Abulghaseml F.Application of ant system to network design problem[J].Transportation,2005,32(3):251-273.
[8] 陆化普,王建伟,李江平,等.城市交通管理评价体系[M].北京:人民交通出版社,2003.
[9] 王秋平.大城市快速路系统线网规划研究[R].国际运输与物流学术论文集.成都:西南交通大学出版社,2005.
[10] 陈森发.复杂系统建模理论与方法[M].南京:东南大学出版社,2005.
[11] Berechman J,Paaswell R E.Evaluation,prioritization and selection of transportation investment projects in New York City[J].Transportation,2005,32(3):223-249.
[12] 东南大学交通学院.银川市城区公共交通规划报告[R].南京:东南大学交通学院,2005.
## 相似文献/References:
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https://doubtnut.com/question-answer/if-a-b-c-are-rational-and-no-two-of-them-are-equal-then-the-equations-b-cx2-c-ax-a-b0-and-ab-cx2-bc--53797210
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or
# If a, b, c are rational and no two of them are equal, then the equations <br> (b-c)x^(2)+(c-a)x+(a-b)=0 <br> and, a(b-c)x^(2)+ b(c-a)x+c(a-n)=0
Question from Class 11 Chapter Quadratic Expressions And Equations
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have rational roots and exactly one them is commonwill be such that at least one has rational rootshave at least one root common. no common root
A
Solution :
Clearly, the sum of the coefficients in both the equations is zero. Therefore, x = 1 is a common root of the two equations. Let `alpha and beta` be the other roots of the two equations. <br> Then, <br> `alpha xx 1 = (a-b)/(b-c) and beta xx 1 = (c(a-b))/(a(b-c))` <br> `rArr" "alpha = (a-b)/(b-c) and beta = (c(a-b))/(a(b-c))` <br> `rArr" "alpha, beta in Q and a ne beta" "[because "a, b, c are distinct rationals"]` <br> Hence, options (a) is true. `
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http://www.developerit.com/2013/07/02/automatically-triggering-standard-spaceship-controls-to-stop-its-motion
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# Automatically triggering standard spaceship controls to stop its motion
Filed under:
|
##### love2d
I have been working on a 2D top-down space strategy/shooting game. Right now it is only in the prototyping stage (I have gotten basic movement) but now I am trying to write a function that will stop the ship based on it's velocity. This is being written in Lua, using the Love2D engine.
My code is as follows (note- object.dx is the x-velocity, object.dy is the y-velocity, object.acc is the acceleration, and object.r is the rotation in radians):
``````function stopMoving(object, dt)
local targetr = math.atan2(object.dy, object.dx)
if targetr == object.r + math.pi then
local currentspeed = math.sqrt(object.dx*object.dx+object.dy*object.dy)
if currentspeed ~= 0 then
object.dx = object.dx + object.acc*dt*math.cos(object.r)
object.dy = object.dy + object.acc*dt*math.sin(object.r)
end
else
if (targetr - object.r) >= math.pi then
object.r = object.r - object.turnspeed*dt
else
object.r = object.r + object.turnspeed*dt
end
end
end
``````
It is implemented in the update function as:
``````if love.keyboard.isDown("backspace") then
stopMoving(player, dt)
end
``````
The problem is that when I am holding down backspace, it spins the player clockwise (though I am trying to have it go the direction that would be the most efficient at getting to the angle it would have to be) and then it never starts to accelerate the player in the direction opposite to it's velocity.
What should I change in this code to get that to work?
EDIT :
I'm not trying to just stop the player in place, I'm trying to get it to use it's normal commands to neutralize it's existing velocity.
I also changed math.atan to math.atan2, apparently it's better. I noticed no difference when running it, though.
© Game Development or respective owner
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• #### Love2D engine for Lua; What about 3D?
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Lua has been really awesome to learn, it's so simple. I really enjoy scripting languages, and I had an equally enjoyable time learning Python. The Love engine, http://love2d.org/, is really awesome, but I'm looking for something that can handle 3D as well. Is there anything that accommodates 3D in… >>> More
• #### How to use batch rendering with an entity component system?
as seen on Game Development - Search for 'Game Development'
I have an entity component system and a 2D rendering engine. Because I have a lot of repeating sprites (the entities are non-animated, the background is tile based) I would really like to use batch rendering to reduce calls to the drawing routine. What would be the best way to integrate this with… >>> More
• #### Turning off antialiasing in Löve2D
as seen on Stack Overflow - Search for 'Stack Overflow'
I'm using Löve2D for writing a small game. Löve2D is an open source game engine for Lua. The problem I'm encountering is that some antialias filter is automatically applied to your sprites when you draw it at non-integer positions. love.graphics.draw( sprite, x, y ) So when x or y is not round… >>> More
• #### How can i run my .LÖVE game directly from the lua interpreter?
as seen on Game Development - Search for 'Game Development'
I've just started with LOVE and LUA , i'm interested in LOVE because i want to play around with something different from my dayjob(i'm a webdeveloper) and since it uses LUA and is interpreted , i though it would be a great way to try out the API. but i couldn't find how to run my .LÖVE game directly… >>> More
• #### How should I share variables between instances/classes?
as seen on Game Development - Search for 'Game Development'
I'm making a game using LOVE, so everything is programmed in Lua. I've been experimenting with using classes and object orientation recently. I've found out that a nice system to use is having most of the game's code in different classes, and having a table of instances with all of the instances of… >>> More
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https://numberworld.info/1001111111
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# Number 1001111111
### Properties of number 1001111111
Cross Sum:
Factorization:
7 * 7 * 11 * 13 * 142873
Divisors:
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
3babbe47
Base 32:
tqnfi7
sin(1001111111)
-0.58845037016167
cos(1001111111)
0.80853333997838
tan(1001111111)
-0.72779975922503
ln(1001111111)
20.724376331119
lg(1001111111)
9.0004822814915
sqrt(1001111111)
31640.339931802
Square(1001111111)
### Number Look Up
Look Up
1001111111 (one billion one million one hundred eleven thousand one hundred eleven) is a amazing figure. The cross sum of 1001111111 is 8. If you factorisate 1001111111 you will get these result 7 * 7 * 11 * 13 * 142873. 1001111111 has 24 divisors ( 1, 7, 11, 13, 49, 77, 91, 143, 539, 637, 1001, 7007, 142873, 1000111, 1571603, 1857349, 7000777, 11001221, 13001443, 20430839, 77008547, 91010101, 143015873, 1001111111 ) whith a sum of 1368161424. The figure 1001111111 is not a prime number. 1001111111 is not a fibonacci number. The number 1001111111 is not a Bell Number. The figure 1001111111 is not a Catalan Number. The convertion of 1001111111 to base 2 (Binary) is 111011101010111011111001000111. The convertion of 1001111111 to base 3 (Ternary) is 2120202202201021022. The convertion of 1001111111 to base 4 (Quaternary) is 323222323321013. The convertion of 1001111111 to base 5 (Quintal) is 4022241023421. The convertion of 1001111111 to base 8 (Octal) is 7352737107. The convertion of 1001111111 to base 16 (Hexadecimal) is 3babbe47. The convertion of 1001111111 to base 32 is tqnfi7. The sine of the figure 1001111111 is -0.58845037016167. The cosine of the number 1001111111 is 0.80853333997838. The tangent of 1001111111 is -0.72779975922503. The root of 1001111111 is 31640.339931802.
If you square 1001111111 you will get the following result 1002223456567654321. The natural logarithm of 1001111111 is 20.724376331119 and the decimal logarithm is 9.0004822814915. You should now know that 1001111111 is special figure!
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https://brandalyzer.blog/2012/06/30/adstock-grps/
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AdStock is a simple mathematical model of how advertising builds and decays. It is invented by Simon Broadbent as he studied Milward Brown’s ad awareness data.
2. Used in marketing-mix modelling to come up with advertising ROIs, etc.
3. Helps you decide when to be off-air and when to be on-air
4. Helps you understand the advertising decay behaviour
Let us take awareness as a parameter to understand the concept of AdStock. As a consumer watches an advertisement for the first time, let us assume that consumer gains certain awareness of the brand, category, etc. Now, when the same consumer watches the advertisement for the second time, the advertisement builds on the awareness. The advertisement hopefully will strengthen the awareness, recall, preferences, etc. So, advertising builds on itself and that is why we call it as a campaign building.
The normal GRP data doesn’t take into account the build and decay rates. So it doesn’t take into account the residual effect of advertising, though a company doesn’t advertise in a specific period. AdStock is nothing but the GRP data taking into account of the build and decay of advertising, which is more sensible in marketing applications.
As explained, the AdStock GRPs are the GRPs weighted for the advertising build and decay rates.
Let us look at case to optimize the scheduling strategy for an advertisement. For this case, the advertisement is assumed to have a half-life of 6 weeks (hypothetical). This will come out for a decay rate of 12.24% as shown in the table below.
We have four options of scheduling, each using roughly the same (1200-1500 GRPs) amount of GRPs. Once we translate these raw GRPs into AdStock GRPs, it will help us decide which scheduling strategy is the most optimum as explained below.
The AdStock GRPs are adjusted based on the decay rate. For example, the number 469 in Wk 2 is arrived by: (250 of Wk2) plus (250*87.8) (decayed GRPs of Week 1) = 469.
Similarly, 662 in Week 3 is arrived by: (250 of Wk 3) + (250*87.8) (decay of Wk 2) +(250*77.0)(decay of Wk 1)= 469
From the above, it is clear that Option 1 gives the maximum ROI. The other parameter important for selection of an option is the off-air time. Which of the above options gives me the maximum off-air time (when you don’t air the advertisement)?
From the above table, it is clear that Option 1 gives the maximum off-air time for the advertisement by still maintaining more than 500 GRPs. In the above example, 500 GRPs is considered as the threshold and if it goes below, then the advertisement has to come on-air.
To sum it up, AdStock helps marketers understand ‘When to advertise‘? AdStock is commonly used in scheduling, marketing-mix modelling, etc.
Any comments on this regard are most welcome.
Thank you.
## 3 thoughts on “AdStock GRPs”
1. Tom
Is it possible for you to show in Excel, how the decay rate is found..guessing probably from regression..and how to convert weekly adstock to monthly? Thanks
2. Nils Randrup
Since you don’t define ROI then your first example does not make sense. Your option 1 costs more to run (Total GRP 1,500) and it generates a lower Adstock GRP score (4,301) whereas option 4 costs less (Total GRP 1,200) and has a higher Adstock GRP score (4,671). So without more information it seems like your conclusion is actually wrong …. And I wonder why …. So either clarify or change your recommendation.
3. Nils Randrup
Also half-life scores we normally use in practice is 3-4 weeks and not 6.
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https://republicofsouthossetia.org/question/according-to-cavalieri-s-principle-if-a-cone-and-a-pyramid-have-the-same-height-which-of-the-fol-16372380-15/
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## According to Cavalieri’s Principle, if a cone and a pyramid have the same height, which of the following answer choices is true?
Question
According to Cavalieri’s Principle, if a cone and a pyramid have the same height, which of the following answer choices is true?
Question 12 options:
A)They have equal volumes.
B)They have equal surface areas.
C)They have equal volumes if their bases have equal areas.
D)They have equal surface areas if their bases have equal areas.
in progress 0
3 weeks 2021-09-08T04:02:36+00:00 2 Answers 0
## Answers ( )
Option (C).
Step-by-step explanation:
If a cone and a pyramid have same height,
Volume of a cone =
=
Volume of a pyramid =
=
As we can see if area of the bases of cone and pyramid are same, their volumes will be same.
Cavalier’s Principle states, two solids having same heights and all cross sections of equal areas at the equal distances from the base, will have the same volumes.
Therefore, as per Cavalier’s Principle, option (C) will be the answer.
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https://math.stackexchange.com/questions/590997/orders-of-poles-zeros-of-an-even-elliptic-function
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# Orders of poles/zeros of an even elliptic function
I am reading a proof of the fact that every even elliptic function $f$ with periods $1$ and $\tau$ is a rational function of the Weierstrass $\wp$ function.
The proof seems to use this fact often, but I don't understand where it comes from:
If $f$ is an even elliptic function with a zero (or a pole) at $0$, $1/2$, $\tau/2$, or $(1+\tau)/2$, the order of the zero (or pole) is even.
What I do know is that for elliptic functions [not necessarily even], (a) the number of poles (up to congruency) is $\ge 2$, and (b) the number of zeros (up to congruency) is the same as the number of poles. I don't understand why evenness of $f$ forces the order of the lattice points and half periods to be even. Why can't for example, one have a simple pole at $0$ and a simple pole at $1/2$? Apologies if this is very obvious...
So, if I have stated the above fact correctly, does this imply the following?
If $f$ is an even elliptic function, it has even order.
If $\omega$ is a period of $f$ and $f$ is even, then we have $f(\omega/2+z) = f(\omega/2+z-\omega) = f(-\omega/2+z) = f(\omega/2-z)$, which shows that the order of vanishing of $f$ at $\omega/2$ is even.
If $f$ is even then $f'$ is odd, $f''$ is even, $f'''$ is odd, and so on. In particular, $f'(0) = f'''(0) = \ldots = 0$. So if $f$ is nonzero then the vanishing order of $f$ at $0$ cannot be odd, so it must be even.
Similarly, if $z \mapsto f(\omega/2+z)$ is even, its order of vanishing at $0$, which is the order of vanishing of $f$ at $\omega/2$, is even.
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https://creativenumerology.com/tag/week-43/
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# WEEKLY FORECAST – OCTOBER 22, 2021
2021 is a 5 Year (2+0+2+1=5). TO CALCULATE YOUR PERSONAL NUMBER FOR 2021, simply add 5 to your month and day of birth. (Do NOT include your year of birth in this calculation).
Example birthday: September 23:
Month = 9
Day = 2+3 = 5
Year: 2+0+2+1 = 5
9 = 1+9 = 10.
Keep adding until you reach a single number.
1+0=1.
So, in this example, the Year Number for 2021 is 1.
THROUGHOUT 2021, USE YOUR YEAR NUMBER TO READ YOUR MONTHLY AND WEEKLY FORECASTS.
SCROLL DOWN TO YOUR PERSONAL WEEKLY FORECAST
WEEK 43 runs from Friday October 22nd to Thursday the 28th.
NOTE 1: WEEK 43 is a 7 Week (4+3=7) – and the 4th week of a 10-week cycle in which all the week numbers begin with 4.
4 = difficulty (rather than simplicity), obstruction, details, correcting errors, organizing, delays, detours, law, order, method, system, determination, hard work, finding order in chaos, and hopefully, breakthrough. 4 affects our sense of identity, belonging, and competence.
3 = people, population, popularity, words, images, and all forms of COMMUNICATION: truth, lies, clarity, doubt; inspiration, hopelessness; entertainment, and boredom. 3 can help you to see the inner reality behind the outer appearance – and find creative solutions to destructive problems.
7 = the power of the mind. A 7 WEEK is a learning week, and WEEK 43 reveals how easily the mind can be tricked into believing words and images that are set up to deceive or distract. Propaganda is one of the biggest problems on Earth right now and falls under 7‘s negative aspects of secrets, lies, conspiracies (real ones), and baseless conspiracy theories. 7 brings out our fear of being wrong or losing dignity. However, 7 also seeks truth whether it proves us wrong or not.
In this 5 Year of change, and this 7 Week of the mind, significant changes of mind are likely. As George Bernard Shaw put it: progress is impossible without change, and those who cannot change their minds cannot change anything.”
Introspective 7 is a spiritual number because it studies and analyzes the nature – the spirituality – the inner workings – of something or someone. This week, intuition runs high as the masculine mind (Spirit) and feminine emotions (Will) more easily work together instead of repelling each other.
It’s OK to be afraid. These are frightening times. The natural emotion of fear is also something we must gain a deeper understanding of. People still speak of fear as if it has only one level – instead of being the complex multi-layered basis of ALL our emotions. The emotional range starts with our very first instinct – survival. Right now, the collective survival instinct is switched firmly on. When fear is correctly understood and channeled, evolutionary changes occur. The same is true of anger. But it must always be remembered that there is never a need to harm yourself or anyone else when feeling your truth and expressing your emotions.
October 2021 is a 6 MONTH (10+2+0+2+1=1+5=6), which represents BALANCE, FAIRNESS, EQUALITY, and all things JUDICIAL. And just look at how matters of JUSTICE and LAW are now front-and-center in the world. This is most relevant at a time when the threat of fascism is tightening its grip. Remember that Hitler and the Nazis changed the laws of Germany in order to make their genocidal actions legal. And because 2022 will be a 6 Year in the world, this focus on justice and legality, is just beginning. In October every year, we start to feel the theme or focus of the upcoming year.
WEEK 43 teaches us about the power of INTENT and helps us discover what our overall intent actually is. Not just what’s in our minds, but what’s in our hearts, too. Is your intent peaceful? Aggressive? Loving? Hateful? Afraid? Indifferent? In denial? To know how you truly feel is a major step on the evolutionary path. From there you can make decisions with some inkling of what you intend to do, how you intend to live, and what changes you will have to make in the process.
NOTE 2: Wednesday, October 27, is the 300th day of the year), followed by 64 days in which all the “day numbers” begin with 3, making this a productive time in terms of sorting TRUTH from LIES. 3 is the number of COMMUNICATION and has the ability to spread through time. 3 represents generation after generation of people, each with their individual part to play in the story of life. By nature, we are all story tellers. That is how we communicate. That’s how we relate to each other.
NOTE 3: Additional insight can be gained this week by reading the 3, 4, 5, 6, and 7 forecasts as well as your own.
“We are not what happened to us, we are what we wish to become.” ~Carl Jung
### PLEASE SUPPORT THIS WORK by making a donation – large or small. This will help to offset the costs of this ad-free site and keep it available to everyone. Thank you somuch. DONATE HERE
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#### YOUR WEEKLY FORECAST – BEGINNING FRIDAY, OCTOBER 22, 2021
WEEKLY FORECAST / 1 YEAR
Power is in the air – different kinds of power – each claiming to be more powerful than the other. If handled correctly, a power drama in your own life can help you change direction towards a more fulfilling path. You are learning about the games people play – and when power is used without understanding its effects, unnecessary problems arise.
WEEKLY FORECAST / 2 YEAR
True friends are never plentiful, but we also have acquaintances who could turn out to be true friends. Notice the expectation that exists between certain people who think they are friends but are only acquaintances. Friendship is a form of love, but when it is expected rather than felt, the love becomes stifled. When you realize that you are equals and don’t owe each other anything, genuine friendship can form.
WEEKLY FORECAST / 3 YEAR
Do not take anything for granted. Look for truth in opposing points of view. Appearances can be deceiving, so independently check the facts and take a more self-reliant stance in general. If you are confronted with facts you would rather not hear, or a reality you would rather not accept, do not create unnecessary problems for yourself by spinning the truth and denying the obvious. There is a fine line between “positive thinking” and denial.
### PLEASE SUPPORT THIS WORK by making a donation – large or small. This will help to offset the costs of this ad-free site and keep it available to everyone. Thank you somuch. DONATE HERE
WHO ARE YOU? WHY ARE YOU HERE? WHERE ARE YOU GOING? This profile will help you to understand your purpose for being here at this precarious evolutionary time in the human journey, what you have to work on, and what you have to work with.
You can get a whole year of Monthly/Weekly/Daily Forecasts (with all the calculations made for you) in your CREATIVE NUMEROLOGY YEAR BOOK. Written specifically to help you steer your way through these tense, extraordinary, unpredictable, and evolutionary times, each book contains inspiring monthly, weekly, and daily readings for your specific yearly cycle. This is a 9-book collectable set. Buy one book – or more – or buy all 9 books and get one free. (Free Book does not apply to Kindle).
Buy my book, LIFE CYCLES: your emotional journey to freedom and happiness,
Available in PAPERBACK or KINDLE…
LIFE CYCLES – FRENCH VERSION
LES CYCLES DE LA VIE – Broché
Acheter CYCLES DE VIE – ICI
WEEKLY FORECAST / 4 YEAR
Let your heart soften where it has been hurt and hardened. Don’t be pushed into moving at an unnatural pace. In this cycle, Free Will means being able to operate in a way that feels appropriate to the situation, and slowly enough to notice the details that will help you make sense of it all. By focusing on what really matters at this precise time, you will gain insight that can improve many areas of your life.
WEEKLY FORECAST / 5 YEAR
If something seems tedious, boring, or superficial, it is finally time to let go of an old routine, habit, or frame of mind. Your effort to ‘go forward’ and ‘reverse’ at the same time is causing the ‘drag’ you are feeling. You won’t get very far trying to make a splash in the shallow end of life. If you hold on to the past and try to move forward at the same time, all you will get is standstill. You KNOW your real fulfillment is in the deeper waters.
WEEKLY FORECAST / 6 YEAR
Consider the possible reactions of others. There is a time to exert your power and a time when it is best kept under wraps. This is one of those defining moments when you must choose which it will be. You are learning the difference between mere tactics and real strategy, and you can now regain some of your “lost Will” by calculating the results of a certain move before you make it. This is powerful problem-solving energy.
### PLEASE SUPPORT THIS WORK by making a donation – large or small. This will help to offset the costs of this ad-free site and keep it available to everyone. Thank you somuch. DONATE HERE
WHO ARE YOU? WHY ARE YOU HERE? WHERE ARE YOU GOING? This profile will help you to understand your purpose for being here at this precarious evolutionary time in the human journey, what you have to work on, and what you have to work with.
You can get a whole year of Monthly/Weekly/Daily Forecasts (with all the calculations made for you) in your CREATIVE NUMEROLOGY YEAR BOOK. Written specifically to help you steer your way through these tense, extraordinary, unpredictable, and evolutionary times, each book contains inspiring monthly, weekly, and daily readings for your specific yearly cycle. This is a 9-book collectable set. Buy one book – or more – or buy all 9 books and get one free. (Free Book does not apply to Kindle).
Buy my book, LIFE CYCLES: your emotional journey to freedom and happiness,
Available in PAPERBACK or KINDLE…
LIFE CYCLES – FRENCH VERSION
LES CYCLES DE LA VIE – Broché
Acheter CYCLES DE VIE – ICI
WEEKLY FORECAST / 7 YEAR
As you dig deeper for answers, you are seeing things in a new light, and this can increase your power. However, accepting reality could be a test of your self-respect: a test of whether you are willing to receive, with dignity, the gift of wisdom – or, for the sake of pride or convenience, you will choose to remain ‘safely’ in the dark. Be prepared to change your mind about something of which you were once so sure.
WEEKLY FORECAST / 8 YEAR
Be aware of all your feelings, otherwise guilt will use your denied emotions to make you feel unworthy of your own needs. There is a survival vibration in your life right now as your Will continues to free itself, and you should know that you are not running out of time or space. You do not have to lose something in order to gain something. But in order to open the way forward, you really do have to finish something … and let it go.
WEEKLY FORECAST / 9 YEAR
Here you are – evolving – in a world that seems completely crazy! Be honest about how you really feel, even if you yourself feel a little crazy at times! Your feelings are trying to steer you to your right place, but guilt is trying to numb your feelings. Your masculine mind must accept your feminine emotions as its true desire, rather than judge them as weak or ‘negative’. Just imagine the power generated when mind and emotion willingly and lovingly combine their energies and work together.
### PLEASE SUPPORT THIS WORK by making a donation – large or small. This will help to offset the costs of this ad-free site and keep it available to everyone. Thank you somuch. DONATE HERE
ORDER A PERSONAL READING.
WHO ARE YOU? WHY ARE YOU HERE? WHERE ARE YOU GOING? This profile will help you to understand your purpose for being here at this precarious evolutionary time in the human journey, what you have to work on, and what you have to work with.
You can get a whole year of Monthly/Weekly/Daily Forecasts (with all the calculations made for you) in your CREATIVE NUMEROLOGY YEAR BOOK. Written specifically to help you steer your way through these tense, extraordinary, unpredictable, and evolutionary times, each book contains inspiring monthly, weekly, and daily readings for your specific yearly cycle. This is a 9-book collectable set. Buy one book – or more – or buy all 9 books and get one free. (Free Book does not apply to Kindle).
Buy my book, LIFE CYCLES: your emotional journey to freedom and happiness,
Available in PAPERBACK or KINDLE…
LIFE CYCLES – FRENCH VERSION
LES CYCLES DE LA VIE – Broché
Acheter CYCLES DE VIE – ICI
Join me on and
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# Homework Help: Prove an integral representation of the zero-order Bessel function
1. Jul 4, 2014
### Dale12
1. The problem statement, all variables and given/known data
In section 7.15 of this book: Milonni, P. W. and J. H. Eberly (2010). Laser Physics.
there is an equation (7.15.9) which is an integral representation of the zero-order Bessel function:
$J_0(\alpha\rho)=\frac{1}{2\pi}\int^{2\pi}_{0}e^{i[\alpha(xcos{\phi}+ysin{\phi})]}d\phi$
This equation could also be found in this paper:
Durnin, J. (1987). "Exact solutions for nondiffracting beams. I. The scalar theory." Journal of the Optical Society of America A 4(4): 651.
2. Relevant equations
here $x=\rho cos{\phi}, y=\rho sin{\phi}$.
3. The attempt at a solution
Rerwite it as:
$J_0(\alpha\rho)=\frac{1}{2\pi}\int^{2\pi}_{0}e^{i[\alpha(\rho cos^2{\phi}+\rho sin^2{\phi})]}d\phi$
$J_0(\alpha\rho)=\frac{1}{2\pi}\int^{2\pi}_{0}e^{i[\alpha\rho]}d\phi$
and after integral of \phi, this becomes
$J_0(\alpha\rho)=e^{i[\alpha\rho]}?$
Also I tried to look it up in the handbook of mathematics by Abramowitz, M. but failed to find this equation, except one like this:
$J_0(t)=\frac{1}{\pi}\int^{\pi}_0 e^{itcos{\phi}}d\phi$
this integral from 0 to $\pi$ could be rewritten to $2\pi$
$J_0(t)=\frac{1}{2\pi}\int^{2\pi}_0 e^{-itcos{\phi}}d\phi$
as http://math.stackexchange.com/quest...ic-integral-int-02-pi-e-2-pi-i-lambda-cost-dt
describes.
yet, this is not what I want.
http://mathworld.wolfram.com/BesselFunctionoftheFirstKind.html
http://en.wikipedia.org/wiki/Bessel_function
Last edited: Jul 5, 2014
2. Jul 22, 2014
### Dale12
Well, I have found that the error appears in step 2.
It should be phi' instead of phi.
after that, it should be rho*cos(phi'-phi) above exp.
and then the integral would be J0.
Thanks anyway!
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# Locus, I discovered a new method to define a circle!
I discovered a new way to define a circle. Make a cone out of a paper or anything else and observe it's shadow in the sun .First, hold the cone as such it's base is parallel to the ground and the axis of the cone is perpendicular to the ground. The shadow will be a triangle.Then, tilt the top of the cone such that it makes an angle from the vertical and go on increasing it ,you will notice that the apex of your triangle(the previous shadow) will slowly demolish and what you will get is a complete circle. Definition : When a cone is kept in the sun , and it is tilted with an increasing angle from the vertical , it's shadow disintegrates from a triangle into a circle.
I don't know if it is of any use but I just came across it and thought it best to share..Please give your views about this @nihar mahajan,@Calvin lin,@Brian Charlesworth@sharky kesa
Note by Raven Herd
5 years, 5 months ago
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How does this relate with what you know about Conics ?
Staff - 5 years, 5 months ago
I did not mention my definition , I have edited it now. Also , I do not understand your question and I have not studied conics yet.
- 5 years, 5 months ago
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# Can I cast Quarantine Field with Brisela in play for my opponent?
Can I cast Quarantine Field with Brisela, Voice of Nightmares in play for my opponent?
I just tried, Magic Online wouldn't let me.
But with X as 2 Quarantine Field has a CMC of 6.
Is this right?
• As far as I remember you should be able to cast it. But my knowledge of magic rules is somewhat rusty and outdated. – tsuma534 Sep 15 '16 at 12:58
You can cast Quarantine Field while your opponent controls Brisela, Voice of Nightmares, as long as you choose 1 or greater for X.
First, before you even begin to cast a spell, the game checks to see if you are allowed to do so. The game ignores any effects that require information determined during the spell’s proposal (this is an exception described in 601.2). This means that we ignore Brisela's effect for now.
601.2. [...] A player must be legally allowed to cast the spell to begin this process (see rule 601.3), ignoring any effect that would prohibit that spell from being cast based on information determined during that spell’s proposal. [...]
Second, we follow the steps for casting a spell as described by 601.2a through 601.2i.
• 601.2a - Put the spell on the stack.
• 601.2b - Choose a value for X.
• ...
• 601.2e - Check if the spell can legally be cast. (This time, no exception is made for effects that require information determined during the spell’s proposal.)
When we reach 601.2e, the spell is on the stack and has a value chosen for X. According to 202.3c, the converted mana cost of a spell on the stack includes X.
202.3c When calculating the converted mana cost of an object with an {X} in its mana cost, X is treated as 0 while the object is not on the stack, and X is treated as the number chosen for it while the object is on the stack.
If you choose a value of 2 for X while casting Quarantine field, then the converted mana cost will be 6. 6 is not "3 or less", and so the spell is legal.
The Gatherer page for Brisela has this ruling:
For spells with {X} in their mana costs, use the value chosen for X to determine if the spell’s converted mana cost is 3 or less. For example, your opponent could cast Burn from Within (a spell with mana cost {X}{R}) with X equal to 3, but not with X equal to 2.
You should have been able to cast the spell, so you should report this as a bug.
Yes, you can cast Quarantine Field. Rule 202.3c states:
202.3c When calculating the converted mana cost of an object with an {X} in its mana cost, X is treated as 0 while the object is not on the stack, and X is treated as the number chosen for it while the object is on the stack.
And Rule 601.2b specifies that the XX have a value as soon as you state that you want to play that spell:
[...] If the spell has a variable cost that will be paid as it’s being cast (such as an {X} in its mana cost; see rule 107.3), the player announces the value of that variable. [...]
• Can you explain the rule that allows you to even start casting it though? As 202.3c says, x is 0 while the object is not on the stack. So while it's in your hand, it has a CMC of 2, so it seems like it's an illegal option to say "I cast this". It would only become legal once it gets to the stack. – GendoIkari Sep 15 '16 at 13:40
• "Casting" is moving an object from any zone (usually your hand) to the stack. When you do that, you immediately specify what options you choose for that card: any "Choose one:" modes, any X in the casting cost - everything that determines the final shape of the spell and its cost. This is an atomic action: one moment you have a card with CMC 2 in your hand, the next you have a spell with CMC 6 on the stack. Brisela bars neither of those. – steenbergh Sep 15 '16 at 13:45
You should be able to cast it.
From the gatherer's ruling on Brisela:
For spells with {X} in their mana costs, use the value chosen for X to determine if the spell’s converted mana cost is 3 or less. For example, your opponent could cast Burn from Within (a spell with mana cost {X}{R}) with X equal to 3, but not with X equal to 2.
Meaning you can cast Quarantine Field for any cost greater than 0 as it will have a cmc of at least 4
The converted mana cost you refer to applies only while the spell is on the stack. While it's in your hand, the cmc is 2, with X being zero. Thus, Brisela prevents you from even casting it.
• Welcome to the site, and thanks for your answer! For most cases you'd be right - for example if you discard Quarantine Field using Blast of Genius, its CMC is 2. But for OP's case (can you cast the spell?) the rules explicitly have an exception: see the rulings for definitive proof, and the accepted answer for the specifics. – Benjamin Cosman Sep 15 '16 at 22:18
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A241130 T(n,k)=Number of nXk 0..2 arrays with no element equal to exactly two horizontal and vertical neighbors, with new values 0..2 introduced in row major order 6
%I #4 Apr 16 2014 09:53:15
%S 1,2,2,4,9,4,11,54,54,11,30,325,723,325,30,82,1965,9773,9773,1965,82,
%T 224,11876,132369,295584,132369,11876,224,612,71793,1792237,8974020,
%U 8974020,1792237,71793,612,1672,434007,24269723,272418756,611547441,272418756
%N T(n,k)=Number of nXk 0..2 arrays with no element equal to exactly two horizontal and vertical neighbors, with new values 0..2 introduced in row major order
%C Table starts
%C ....1........2...........4..............11................30.................82
%C ....2........9..........54.............325..............1965..............11876
%C ....4.......54.........723............9773............132369............1792237
%C ...11......325........9773..........295584...........8974020..........272418756
%C ...30.....1965......132369.........8974020.........611547441........41661463219
%C ...82....11876.....1792237.......272418756.......41661463219......6369134821236
%C ..224....71793....24269723......8270609664.....2838552905280....973866422990563
%C ..612...434007...328645291....251096788713...193403197619803.148910067386335603
%C .1672..2623694..4450319537...7623351508315.13177423185201303
%C .4568.15861001.60263615333.231446775293031
%H R. H. Hardin, <a href="/A241130/b241130.txt">Table of n, a(n) for n = 1..97</a>
%F Empirical for column k:
%F k=1: a(n) = 2*a(n-1) +2*a(n-2) for n>4
%F k=2: a(n) = 6*a(n-1) +2*a(n-2) -9*a(n-3) -10*a(n-4) +8*a(n-5) for n>6
%F k=3: [order 26]
%e Some solutions for n=3 k=4
%e ..0..1..0..2....0..1..1..2....0..0..1..0....0..1..0..1....0..1..0..1
%e ..1..0..2..2....0..2..2..0....1..1..2..0....0..2..1..2....2..2..1..2
%e ..0..0..0..2....2..1..1..0....0..2..1..2....2..1..0..2....1..1..2..0
%Y Column 1 is A021006(n-3)
%K nonn,tabl
%O 1,2
%A _R. H. Hardin_, Apr 16 2014
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Last modified December 7 12:37 EST 2023. Contains 367656 sequences. (Running on oeis4.)
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chapter ix. example comparing the b-representation of a cubic f with c its values at knot averages. c from * a practical guide to splines * by c. de boor c integer i,id,j,jj,n,nm4 real bcoef(23),d(4),d0(4),dtip1,dtip2,f(23),t(27),tave(23),x c the taylor coefficients at 0 for the polynomial f are data d0 /-162.,99.,-18.,1./ c c set up knot sequence in the array t . n = 13 do 5 i=1,4 t(i) = 0. 5 t(n+i) = 10. nm4 = n-4 do 6 i=1,nm4 6 t(i+4) = float(i) c do 50 i=1,n c use nested multiplication to get taylor coefficients d at c t(i+2) from those at 0 . do 20 j=1,4 20 d(j) = d0(j) do 21 j=1,3 id = 4 do 21 jj=j,3 id = id-1 21 d(id) = d(id) + d(id+1)*t(i+2) c c compute b-spline coefficients by formula (9). dtip1 = t(i+2) - t(i+1) dtip2 = t(i+3) - t(i+2) bcoef(i) = d(1) + (d(2)*(dtip2-dtip1)-d(3)*dtip1*dtip2)/3. c c evaluate f at corresp. knot average. tave(i) = (t(i+1) + t(i+2) + t(i+3))/3. x = tave(i) 50 f(i) = d0(1) + x*(d0(2) + x*(d0(3) + x*d0(4))) c print 650, (i,tave(i), f(i), bcoef(i),i=1,n) 650 format(45h i tave(i) f at tave(i) bcoef(i)// * (i3,f10.5,2f16.5)) stop end
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## Cell potential calculation
$\Delta G^{\circ} = -nFE_{cell}^{\circ}$
Aijun Zhang 1D
Posts: 53
Joined: Tue Oct 10, 2017 7:13 am
### Cell potential calculation
Just a quick question.
When we calculate cell potential for a reaction composed of two half reactions. Do we always subtract the smaller standard reduction potential from the larger one? Therefore the final result is always positive?
Thank you!
Wenjie Dong 2E
Posts: 53
Joined: Fri Jun 23, 2017 11:40 am
### Re: Cell potential calculation
I guess you should always reverse the less reactive one.
Anna Li 2E
Posts: 21
Joined: Sat Jul 22, 2017 3:00 am
### Re: Cell potential calculation
Galvanic cells generate electrical energy from spontaneous redox reactions, meaning that the deltaG (gibb's free energy) should be negative as we know that indicates spontaneous reactions. When reactions are spontaneous that means that E MUST be positive. So when you are adding/subtracting cell potentials, you should reverse the E value that keeps the total E as a positive number.
Nathan Tu 2C
Posts: 46
Joined: Fri Sep 29, 2017 7:07 am
### Re: Cell potential calculation
If the cell is not a galvanic cell, or rather the case in any cell, subtract the E anode from E cathode which means subtracting the E of the oxidation reaction from the E of the reduction reaction. If the answer is not positive, than the cell is not a galvanic cell.
Ya Gao
Posts: 52
Joined: Fri Sep 29, 2017 7:04 am
Been upvoted: 1 time
### Re: Cell potential calculation
There are two ways to calculate the cell potential. 1st: list out the anode and cathode reaction and copy the corresponding cell potential to these two reaction. Cell potential of this reaction equals to cell potential of the cathode reaction minus the cell potential of the anode reaction.
2nd: list out the anode and cathode reaction and reverse the anode reaction and therefore, the cell potential for the anode reaction has to be reversed as well. Cell potential of this reaction equals to cell potential of the cathode adds the cell potential of the anode.
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Home > Error Bars > Bar Graphs With Error Bars Excel
# Bar Graphs With Error Bars Excel
## Contents
But when it isn't, Excel gives us some useful tools to convey margins of error and standard deviations. If you work in a field that needs to reflect an accurate range The point markers have been retained here to illustrate the construction of the step chart, but they would be removed in the final chart. Here is our chart with error bars of one standard deviation. Change Fixed Value to Custom. Check This Out
There are several ways to enter values: fixed values, a percentage of the point's value, a number of standard deviations, the standard error of the plotted points, and custom values. Helpful (0) Reply options Link to this post by Baltser, Baltser Apr 19, 2011 1:39 PM in response to Alexa M Level 1 (0 points) Apr 19, 2011 1:39 PM in This shows our sample chart with positive and negative X and Y error bars, with a fixed value of 0.75. By dividing the standard deviation by the square root of N, the standard error grows smaller as the number of measurements (N) grows larger.
## Excel 2013 Error Bars
In the provided example, you couldn't just drop a standard deviation calculation into cell b4, for example, as it only includes one piece of sample data. 2. Top of Page Share Was this information helpful? To use custom values to determine the error amount, click Custom, and then do the following: Click Specify Value. The Chart Tools contextual tab activates. 2.
This displays the Chart Tools, adding the Design, Layout, and Format tabs. I want to change each of the 6 bars individually for the std deviation of males walking, females walking, etc. Do one of the following: On the Layout tab, in the Analysis group, click Error Bars, and then click None. How To Insert Error Bars In Excel Mac Up next Excel Graphs With Error Bars Tutorial By Nestor Matthews - Duration: 14:12.
This chart shows the standard error of the plotted points as the error bar value for each point. The dialog box will now shrink and allow you to highlight cells representing the standard error values: When you are done, click on the down arrow button and repeat for the On the other hand, at both 0 and 20 degrees, the values range quite a bit. The Error Bar dialogs, shown below, are not at all complicated.
But I thinking I have found another easier solution..in the positive value: first enter the std error or std dev value, follow by comma, space then enter the std error or How To Add Error Bars In Excel 2013 Apple Info Site Map Hot News RSS Feeds Contact Us Copyright © Apple Inc. Let's take, for example, the impact energy absorbed by a metal at various temperatures. The +/- value is the standard error and expresses how confident you are that the mean value (1.4) represents the true value of the impact energy.
## Bar Graph With Error Bars Excel Mac
Apple disclaims any and all liability for the acts, omissions and conduct of any third parties in connection with or related to your use of the site. Terms of Use Updated Privacy Policy Cookie Usage ERROR The requested URL could not be retrieved The following error was encountered while trying to retrieve the URL: http://0.0.0.10/ Connection to 0.0.0.10 Excel 2013 Error Bars The resulting error bars, should be unique to each bar in the chart. Bar Graph With Error Bars Excel 2010 Loading...
In this example, these error bars have been removed. http://greynotebook.com/error-bars/bar-chart-error-bars-excel.php Custom Error Bar Values If you have a more complicated set of values for your error bars, you enter them using the Custom (+) and Custom (-) boxes. Sign in 533 47 Don't like this video? Share: Categories: Advanced Excel Tags: Standard Deviation Excel Graph | Comments Written by Tepring Crocker Tepring Crocker is a freelance copywriter and marketing consultant. Vertical Error Bars In Excel
Result: Note: if you add error bars to a scatter chart, Excel also adds horizontal error bars. Science Class Online 19,843 views 5:01 Multiple Bar Graphs in Excel - Duration: 12:21. Please try again later. this contact form Notice the shortcuts to quickly display error bars using the Standard Error, a 5% value or 1 standard deviation. 3.
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## With the standard error calculated for each temperature, error bars can now be created for each mean.
No, but you can include additional information to indicate how closely the means are likely to reflect the true values. The resulting error bars, then, are also unique to each bar in the chart. More precisely, the part of the error bar above each point represents plus one standard error and the part of the bar below represents minus one standard error. How To Add Error Bars In Excel 2010 In this example, it would be a best guess at what the true energy level was for a given temperature.
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These are chat archives for ramda/ramda
6th
Dec 2018
Daniel Pereira
@dpereira411
Dec 06 2018 01:11
is there something like map, but where i can look into next item?
Daniel Pereira
@dpereira411
Dec 06 2018 01:16
or previous
ill try with reduce :p
Dec 06 2018 01:19
Check out aperture
Daniel Pereira
@dpereira411
Dec 06 2018 01:19
lovely :D
ramda is so awesome
Ben Briggs
@ben-eb
Dec 06 2018 09:12
Following on from last night's thought:
``````const W = flip(chain)(identity);
const square = W(multiply);
square(3) // 9``````
Alexander Lichter
@manniL
Dec 06 2018 09:37
@ben-eb Isn't that the same as `unnest`?
``````const R = require('ramda')
const W1 = R.flip(R.chain)(R.identity)
const W2 = R.unnest
const square1 = W1(R.multiply)
const square2 = W2(R.multiply)
console.log(square1(3), square2(3)) // 9, 9``````
Ben Briggs
@ben-eb
Dec 06 2018 09:39
@manniL Looks like it yeah! Thanks :)
Alexander Lichter
@manniL
Dec 06 2018 09:40
NP :)
Also realized this is documented like that :see_no_evil:
"Shorthand for R.chain(R.identity), which removes one level of nesting from any Chain."
Ben Briggs
@ben-eb
Dec 06 2018 09:42
@manniL So we don't really need to `flip` `chain` at all
`const W = chain(identity)` is also how `unnest` is defined
Alexander Lichter
@manniL
Dec 06 2018 09:43
@ben-eb Right! :)
Alexander Lichter
@manniL
Dec 06 2018 10:51
Imagine having a list of lists with coordinates `[ [1, 2], [8, 4], ...]`. I want to transform each list into an object with `x` and `y` (quite simple) but also adding a consecutive letter from the alphabet.
My approach is:
``````R.pipe(
R.zipWith((letter, coords) => [letter, ...coords], alphabet), // alphabet is just a list ['a','b',...]
R.map(R.zipObj(['letter', 'x', 'y']))
)``````
But I still feel it could look cleaner. Any ideas?
(Important info: I generate the coordinates from text input before, one pair per line and so on)
Full code:
`````` R.pipe(
R.map(
R.pipe(
R.split(', '),
R.map(Number),
)
),
R.zipWith((letter, coords) => [letter, ...coords], alphabet),
R.map(R.zipObj(['letter', 'x', 'y']))
)``````
Ben Briggs
@ben-eb
Dec 06 2018 14:28
@manniL `R.zipWith(R.prepend, alphabet)`
Alexander Lichter
@manniL
Dec 06 2018 14:35
@ben-eb Thank you! :clap: That makes totally sense
Scott Sauyet
@CrossEye
Dec 06 2018 14:37
@mannil: Let me, guess, Advent of Code?
Alexander Lichter
@manniL
Dec 06 2018 14:38
Exactly :D
@CrossEye
Already finished both tasks but wanted to make the code cleaner :)
Scott Sauyet
@CrossEye
Dec 06 2018 14:38
Saw that problem just before I went to bed last night. Haven't tried it yet. But dreamed about how bad one of my earlier solutions was.
:smile:
Alexander Lichter
@manniL
Dec 06 2018 14:39
Oh, I don't even want to take a look on my solutions from last year :joy:
They were really messy :sweat_smile:
@CrossEye Have you solved the other tasks from the last days already? :)
Scott Sauyet
@CrossEye
Dec 06 2018 14:40
Yes, first five days.
Alexander Lichter
@manniL
Dec 06 2018 14:40
Nice! Do you have the code on GitHub?
Scott Sauyet
@CrossEye
Dec 06 2018 14:40
I haven't participated in previous years.
Alexander Lichter
@manniL
Dec 06 2018 14:40
I'd love to take a look and learn a few things :)
Scott Sauyet
@CrossEye
Dec 06 2018 14:41
In a secret gist. Some of it is too embarrassing to share at the moment. I'll try to clean it up, but perhaps these shouldn't be shared until after it's all over.
Just looking at how bad yesterday's solution was. (I really did wake up at 3AM thinking about how bad it was!)
I also started by working in plain JS, until I felt too hampered and introduced Ramda.
Alexander Lichter
@manniL
Dec 06 2018 14:43
I think you can share it already. If ppl want to cheat, there are so many solution on reddit they could use.
Ah, I see! I wanted to solve all things with Ramda and "as close to the style" as possible this year
Scott Sauyet
@CrossEye
Dec 06 2018 14:44
Are there? I haven't really paid attention.
Scott Sauyet
@CrossEye
Dec 06 2018 14:46
I think I'll just stay away. :wink:
Alexander Lichter
@manniL
Dec 06 2018 14:46
I like to look at them after I solved the day, just bc I'm interested how ppl solved it in other languages
Scott Sauyet
@CrossEye
Dec 06 2018 14:47
yes, I might do that too... but I want to think about and clean mine up before that.
Alexander Lichter
@manniL
Dec 06 2018 14:47
Looking forward to it :)
Scott Sauyet
@CrossEye
Dec 06 2018 14:48
I've done problems from various places, CodeWars, Euler, etc. And I like the feature where the forum discussions open up about the solutions once you've solved them.
Alexander Lichter
@manniL
Dec 06 2018 14:49
Yeah, that's actually a nice way. But for a "public contest" it's hard to realize
Scott Sauyet
@CrossEye
Dec 06 2018 14:50
I never try to do these at midnight my time, so I'm not in the contest (at least the leaderboard one; I don't know if there's others.)
Alexander Lichter
@manniL
Dec 06 2018 14:50
Well, I'd have to wake up quite early for them, so I'm out as well :D
and I'm not into solving the puzzles quickly this year
Scott Sauyet
@CrossEye
Dec 06 2018 14:52
I've never been particularly speedy at this in any case. So I'd probably fare badly. But these puzzles are somewhat fun. I found Day 5 the easiest to code so far, but my algorithm was horribly inefficient. I'm rewriting now before I move on to Day 6.
Alexander Lichter
@manniL
Dec 06 2018 14:55
Last year I was fairly ok, but never in the leaderboards. Top 1k often though
Alexander Lichter
@manniL
Dec 06 2018 15:49
I'm looking for a more elegant way to do sth. like this (function is used in a reducer):
``````const willBeUsedAsReduceFunction = (shifts, [actionOne, actionTwo]) => {
if (isShiftAction(actionOne) && !isShiftAction(actionTwo)) {
shifts.push(actionOne)
}
return shifts
}``````
Tymoteusz Czech
@Tymek
Dec 06 2018 16:26
Hi. I'd like to ask what's expected specificity in Ramda, i.e. what belongs to the core and what should be left to other helpers. I'm asking, because I have a case where having `testObject` I would like to check if it contains all keys from `spec` object (recursive), that is: if everything from `spec` is `equal` in the same `path` in `testObj`
Ben Briggs
@ben-eb
Dec 06 2018 17:02
@manniL Check out `R.both`
Oh, wait. Maybe not because you have two different variables to inspect
Alexander Lichter
@manniL
Dec 06 2018 17:03
@ben-eb yes, unfortunately :see_no_evil:
Ben Briggs
@ben-eb
Dec 06 2018 17:04
I'm wondering if there's a pointfree way of expressing this:
``````const x = {
fn: always(42)
};
evolve({
fn: o => compose(inc, o)
})(x).fn() // 43``````
It's not really necessary for it to be pointfree I'm wondering if it's possible :)
Alexander Lichter
@manniL
Dec 06 2018 17:13
it's possible I think
Ben Briggs
@ben-eb
Dec 06 2018 17:13
@Tymek Did you try `whereEq` ?
``````whereEq({
a: {
b: 'c'
}
})({
a: {
b: 'c'
}
}) // true``````
Alexander Lichter
@manniL
Dec 06 2018 17:14
@ben-eb So close :see_no_evil:
``````const res = evolve({
fn: compose(inc, call)
})(x).fn //43``````
But you can't call the fn ^^#
Ben Briggs
@ben-eb
Dec 06 2018 17:14
Yeah, that's the gotcha :) We want to delay evaluation
Alexander Lichter
@manniL
Dec 06 2018 17:15
``````const res = evolve({
fn: compose(always, inc, call)
})(x).fn()``````
Like that? :D
Ben Briggs
@ben-eb
Dec 06 2018 17:15
Nice, thanks! :+1:
Alexander Lichter
@manniL
Dec 06 2018 17:16
You are welcome :)
Tymoteusz Czech
@Tymek
Dec 06 2018 17:16
@ben-eb o. thx! Somehow I missed it.
Ben Briggs
@ben-eb
Dec 06 2018 17:19
@manniL Ahhh it's not quite the same
@manniL If we then call the func with some args then it falls over
Alexander Lichter
@manniL
Dec 06 2018 17:20
@ben-eb Dang! :see_no_evil:
In which case(s)?
Ben Briggs
@ben-eb
Dec 06 2018 17:20
@manniL Yeah the idea is to take some existing object method, apply the original method with its args and then run our function afterwards
@manniL
``````const x = {
fn: multiply(2)
};
evolve({
fn: o => compose(inc, o)
})(x).fn(2)``````
Should be 5
Johnny Hauser
@m59peacemaker
Dec 06 2018 17:22
@CrossEye thanks
Ben Briggs
@ben-eb
Dec 06 2018 17:22
@Tymek You're welcome :)
Johnny Hauser
@m59peacemaker
Dec 06 2018 17:24
Since the hareactive chat has no activity.... does anyone here know much about it - more importantly the theory behind it, or perhaps even better, the Conal Elliot concept of FRP, i.e. reactive-banana?
There is a lot that I like about "Classic FRP", if that's the right way to describe it, but I can't figure out how they are composing "time".
In a lot of JS observable libraries, there is `zip`, which is somewhat of a barrier https://en.wikipedia.org/wiki/Barrier_(computer_science)
It's like `lift`ing the values of the observables, but associating them with a barrier of their times
Zip is like this to me:
``````const time = barrier([ a, b, c ])
const value = lift (sum) ([ a, b, c ])
const zipped = when (time) (value)``````
Johnny Hauser
@m59peacemaker
Dec 06 2018 17:29
But I can't imagine how to implement such a thing, nor understand or find anything about how the classic frp folks are doing that.
Ben Briggs
@ben-eb
Dec 06 2018 17:36
@manniL I worked it out. It's just `map(inc)`
@manniL Alternately you could use `B` combinator; `const B = f => g => x => f(g(x));` - which is the same thing
@manniL I guess the only problem is that it's still not as good as the original implementation because it now has to be a unary function (whereas the other preserved arity)
Ben Briggs
@ben-eb
Dec 06 2018 18:23
For completeness' sake:
``````const x = {
fn: multiply(2)
};
evolve({
fn: map(inc)
})(x).fn(2) // 5``````
Kurt Milam
@kurtmilam
Dec 06 2018 18:31
@ben-eb I think `o(inc)` will also work if you want to save a couple of keystrokes :D
It works for your example - not sure whether it fails on any others.
Dec 06 2018 18:31
Kurt Milam
@kurtmilam
Dec 06 2018 18:32
:thumbsup:
The `oink` combinator.
Dec 06 2018 18:32
hahaha
Ben Briggs
@ben-eb
Dec 06 2018 18:32
:joy:
Why does `o` work and `compose` doesn't?
Dec 06 2018 18:33
:tophat: :sparkles: Magic
Because `compose` is variadic, while `o` is binary and will wait for that second argument
So `compose` with a single argument will return an uncurried version of the function you passed in, while `o` will return a function that's witing for the second half of the composition
Ben Briggs
@ben-eb
Dec 06 2018 18:35
Ahhh, thanks :) Still it means in this case that `x => compose(inc, x)` is better if we assume `x` is not a unary function, but `o(inc)` is better if it is
Kurt Milam
@kurtmilam
Dec 06 2018 18:39
if pigs could fly (sorry, delayed combinator joke)
Ben Briggs
@ben-eb
Dec 06 2018 18:51
@manniL I'm not sure if it's better (it's probably not as easy to grok) but I had a go at your earlier question:
``````const isShiftAction = propEq('shift', true);
const isNotShiftAction = complement(isShiftAction);
const appendTo = flip(append)
const willBeUsedAsReduceFunction = shifts =>
ifElse(
both(
compose(isNotShiftAction, last)
),
always(shifts)
)
willBeUsedAsReduceFunction([])([{ shift: true }, {}])``````
Now we're not mutating `shifts` :)
Scott Sauyet
@CrossEye
Dec 06 2018 19:08
Note that `o` is technically ternary: `(b → c) → (a → b) → a → c`. That is
``o(f, g, x) //=> f(g(x))``
But I don't think I've ever called it that way. It's always either `o(f)(g)(x)` or `o(f, g)(x)`.
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Question
# A health inspector was interested in determining if the noise levels of hospital corridors in...
A health inspector was interested in determining if the noise levels of hospital
corridors in an urban area is greater than 59 decibels. To test this, the noise levels at
84 randomly chosen hospitals from the urban area were measured in decibels. The
sample mean noise level was 61.2 decibels, and the population standard deviation is
7.9 decibels. Use the five step method to test the inspectors claim using α = .025.
#### Homework Answers
Answer #1
To Test :-
H0 :-
H1 :-
Test Statistic :-
Z = 2.5523 2.55
Test Criteria :-
Reject null hypothesis if
Result :- Reject null hypothesis
There is sufficient evidence to support the claim that the noise levels of hospital corridors in an urban area is greater than 59 decibels.
Know the answer?
Your Answer:
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# 0 ms C++ solution
• ``````class Solution {
public:
string complexNumberMultiply(string a, string b) {
// (a1 + a2 i) * (b1 + b2 i) = (a1b1 - a2b2) + (a1b2 + a2b1) i
size_t pos = a.find('+');
int a1 = stoi(a.substr(0, pos)), a2 = stoi(a.substr(pos + 1, a.size() - pos - 2));
pos = b.find('+');
int b1 = stoi(b.substr(0, pos)), b2 = stoi(b.substr(pos + 1, b.size() - pos - 2));
int c1 = a1 * b1 - a2 * b2, c2 = a1 * b2 + a2 * b1;
}
};
``````
• Using 'stoi' instead of 'atoi' is really a good skill~
It saves 3ms runtime :)
thx~
• @Vincenberg said in 0 ms C++ solution:
``````class Solution {
public:
string complexNumberMultiply(string a, string b) {
// (a1 + a2 i) * (b1 + b2 i) = (a1b1 - a2b2) + (a1b2 + a2b1) i
size_t pos = a.find('+');
int a1 = stoi(a.substr(0, pos)), a2 = stoi(a.substr(pos + 1, a.size() - pos - 2));
pos = b.find('+');
int b1 = stoi(b.substr(0, pos)), b2 = stoi(b.substr(pos + 1, b.size() - pos - 2));
int c1 = a1 * b1 - a2 * b2, c2 = a1 * b2 + a2 * b1;
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# Swift Program to Calculate Average Using Arrays
In this article, we will learn how to write a swift program to calculate averages using an array. Average is defined as the ratio of the sum of the elements present in the given sequence to the total number of elements available in the given sequence. The general formula of average is −
Average = (p1+p2+p3+..+pn)/n
Here we use the following methods to calculate the average using array −
• Using pre−define functions
• Without using predefine functions
## Method 1: Using Pre-Define Functions
To find the average of the given array, we use reduce() method to find the sum of all the elements present dnd for a total number of elements present in the given array we use the count property. The reduce(_:_:) method is used to combine the given sequence according to the given closure.
### Syntax
func reduce<result>(_initial result: result,
_nextResult:(result, Self.Element) throws->result)rethrows->result
Here, the initial parameter is used for the initial accumulating value and passed in the nextResult the first time the closure is executed. Whereas nextResult is a closure that adds the accumulating value and the item of the sequence into a new accumulating value, it is further used in the next call of the nextResult closure.
### Algorithm
• Step 1 − Create an array of integer type.
• Step 2 − Find the sum of all the elements of the array using reduce function.
var arraySum = myArray.reduce(0, +)
• Step 3 − Calculate the array’s total number of elements using count property
var length = myArray.count
• Step 4 − Find the average of the given array by dividing the sum and the total number of array.
var average = Double(arraySum)/Double(length)
• Step 5 − Print the output.
### Example
Following Swift program to calculate the average of the array using pre-defined functions.
import Foundation
import Glibc
// Creating an array of integer type
var myArray : [Int] = [3, 5, 6, 74, 32, 7, 31]
// Finding the sum of the given array elements
var arraySum = myArray.reduce(0, +)
// Calculating the total number of
// elements present in the array
var length = myArray.count
// Finding the average
var average = Double(arraySum)/Double(length)
print("Array:", myArray)
print("Average:", average)
### Output
Array: [3, 5, 6, 74, 32, 7, 31]
Average: 22.571428571428573
Here in the above code, we create an array of integer types. First, we find the sum of all the array elements using reduce(0,+) where 0 is the initial value which will add with the first element of the given array, and then the result will add to the next value and this process will continue till the last element. Now we calculate the total number of elements using the count property, then find the average by dividing the sum by the total number of elements and print the output.
## Method 2: Without using Pre-Define Functions
We can also find the average of the given array using loops. Here we add all the elements of the given array using for loop and then find the average.
### Algorithm
• Step 1 − Create an array of integer type.
• Step 2 − Find the sum of all the elements of the array using for loop.
for x in 0..<myArray.count{
arraySum += myArray[x]
}
• Step 3 − Calculate total number of elements of the array using count property.
var length = myArray.count
• Step 4 − Find average of the given array by dividing sum and total number of array.
var average = Double(arraySum)/Double(length)
• Step 5 − Print the output.
### Example
Following Swift program to calculate the average of the array using loops.
import Foundation
import Glibc
// Creating an array of integer type
var myArray : [Int] = [4, 56, 78, 21, 5, 6, 76, 3, 21, 1]
// Finding the sum of the given array elements
var arraySum = 0
for x in 0..<myArray.count{
arraySum += myArray[x]
}
// Calculating the total number of
// elements present in the array
var length = myArray.count
// Finding the average
var average = Double(arraySum)/Double(length)
print("Array:", myArray)
print("Average:", average)
In the following example, we check whether the given array is empty or not using conditional statement.
### Output
Array: [4, 56, 78, 21, 5, 6, 76, 3, 21, 1]
Average: 27.1
Here in the above code, we create an array of integer type. First we find the sum of the given array elements using for loop. Here for loop iterate through each element and add them together and store result into a variable. Now we find the total number of arrays present in the given array using count property, then find average by dividing sum by the total number of elements, and print the output.
## Conclusion
So this is how we can find the average using array. Here we can either use pre-define function or using loops to find the average of the given array.
Updated on: 29-Dec-2022
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# Puzzling Regression Anatomy
01 Jul 2020
Consider a regression model with the following causal structure:
The variable x1 affects y directly and also indirectly via x2. The following R code implements the model and simulates a corresponding data set.
set.seed(1)
n = 10000
beta1 = 1; beta2=1
x1 = rnorm(n,0,1)
x2 = x1+rnorm(n,0,1)
y = beta1*x1 + beta2*x2 + rnorm(n,0,1)
Assume we want to consistently estimate the direct linear effect beta1 from x1 on y. To do so, we can simply estimate a multiple linear regression where we add x2 as a control variable:
coef(lm(y~x1+x2))
## (Intercept) x1 x2
## 0.007553765 0.998331323 1.000934100
But what does it intuitively mean to add x2 as control variable? The Frisch-Waugh-Lovell Theorem implies that we get the same estimator for beta1 as in the multiple regression above by the following procedure:
# y.tilde is residual of regression
# y on x2
y.tilde = resid(lm(y~x2))
# x1.tilde is residual of regression
# x1 on x2
x1.tilde = resid(lm(x1~x2))
# Regression y.tilde on x1.tilde
# we get the same estimate for beta1
# as in the multiple regression with x1 and x2
coef(lm(y.tilde ~ x1.tilde))
## (Intercept) x1.tilde
## -5.104062e-17 9.983313e-01
Hence, controlling for x2 means that we essentially regress the residual variations of y and x1 that cannot be linearly explained by x2 on each other. So far this seems intuitive.
The interesting thing is that one gets the same estimate for beta1 also with one of the following two regressions below (but only the regression above also yields correct standard errors):
# Approach A
lm(y.tilde ~ x1)
# Approach B
lm(y ~ x1.tilde)
Approach A regresses the residual variation of y that cannot be linearly predicted by x2 on x1. Approach B regresses y on the residual variation of x1 that cannot be linearly predicted by x2.
Only one approach yields a consistent estimate of beta1. Make a guess which one…
Let’s check:
# A: Inconsistent
coef(lm(y.tilde ~ x1))[2]
## x1
## 0.4860465
# B: Consistent
coef(lm(y ~ x1.tilde))[2]
## x1.tilde
## 0.9983313
So only approach B works. Angrist and Pischke (2009) refer to it as regression anatomy. For me that result was a bit puzzling for a long time because my intuitive interpretation of what it means to control for x2 was more in line with approach A. I first want to shed light on that intuition and explain why approach A does not work. Afterward I want to give some intuition for the working approach B.
## An intuition for control variables and why approach A fails
I have different intuitions what controlling for x2 means in the linear regression:
y = beta0 + beta1*x1 + beta2*x2 + eps
One of my intuitions is the following:
“By controlling for x2, we essentially subtract the variation that can be linearly explained by x2 from y, i.e. up to an estimation error we subtract beta2*x2.”
This interpretation suggests that approach A should work, but that approach fails to get a consistent estimate for beta1. So is the intuition above wrong? Not completely, but the qualification “up to an estimation error” causes trouble for approach A. Consider the following code.
# Modified approach A
y.tilde2 = y - beta2*x2
coef(lm(y.tilde2 ~x1))[2]
## x1
## 0.9992699
It is a modified version of approach A. It computes the residual variation y.tilde2 by directly subtracting beta2*x2 from y. Now we get a consistent estimator of beta1 when regressing y.tilde2 on x1.
But approach A differs because we subtract beta2.hat*x2 from y where beta2.hat is estimated in the first stage regression:
# Same result as original approach A
beta2.hat = coef(lm(y~x2))[2]
beta2.hat # inconsistent estimate of beta2
## x2
## 1.510801
y.tilde = y-beta2.hat*x2
coef(lm(y.tilde ~x1))[2] # inconsistent estimate of beta1
## x1
## 0.4860465
The problem with approach A is that we don’t estimate beta2.hat consistently in the regression of y on x2. Instead, since x1 and x2 are correlated, beta2.hat also captures some of the direct effect of x1 on y. This means in y.tilde we have already removed some of the effect from x1 on y that we want to estimate. Therefore approach A yields an estimator for beta1 that is biased towards 0.
Remark: In the original computation of approach A, we also subtract the estimated constant from the initial regression when computing y.tilde, but that has no effect on the slope coefficient in the second stage regression.
Interestingly, in some empirical papers an approach similar to approach A is performed, i.e. one first computes residuals of y from a first regression and then regresses those residuals on another set of explanatory variables. But the computation above shows that one should really be careful with this approach, since it only works if the first regression yields consistent estimates.
Let us consider an example where such an approach would work. Consider the following modified model:
We now have an additional variable z that affects x2 but is uncorrelated with x1.
z = rnorm(n,0,1)
x2 = x1+z+rnorm(n,0,1)
y = beta1*x1 + beta2*x2 + rnorm(n,0,1)
We now conduct a variation of approach A where y.tilde3 are the residuals of an instrumental variable regression of y on x2 using z as instrument:
library(AER)
reg1 = ivreg(y~x2|z)
coef(reg1)[2] # consistent beta2.hat
## x2
## 1.006464
y.tilde3 = resid(reg1)
coef(lm(y.tilde3 ~ x1))[2] # consistent beta1.hat
## x1
## 0.9756005
We now see that regressing y.tilde3 on x1 yields a consistent estimator of beta1.
### Why does approach B work
Let us now discuss why approach B works. Given our causal structure I find it more intuitive to first discuss why a similar approach works to consistently estimate beta2.
# x2.tilde is residual from regression
# of x2 on x1
x2.tilde = resid(lm(x2~x1))
# consistent estimate of beta2
coef(lm(y ~ x2.tilde))[2]
## x2.tilde
## 0.996786
Here I have the following intuition why it works. Intuitively, to consistently estimate the causal effect of x2 on y we need to distill variation of x2 that is uncorrelated with x1. If we regress x2 on x1, the residuals x2.tilde of this regression are by construction uncorrelated with x1. They describe the variation of x2 that cannot be linearly predicted by x1. That is exactly the variation of x2 needed to consistently estimate beta2.
The equivalent procedure also works to estimate beta1 consistently:
x1.tilde = resid(lm(x1~x2))
coef(lm(y ~ x1.tilde))[2] # consistent
## x1.tilde
## 0.98519
So even though x2 does not influence x1 we can similarly distill in x1.tilde the relevant variation in x1 that is uncorrelated with x2. For the regression anatomy it is irrelevant which causal direction has generated the correlation between x1 and x2.
### Final remarks
I find it amazing that over many years I still often learn new intuitions for basic econometric concepts like multiple linear regression. Currently, I think introducing multiple regression via the Frisch-Waugh-Lovell theorem and the regression anatomy can be much more helpful to build intuition in an applied empirical course than covering the matrix algebra. (Of course, it is a different story if you want to prove econometric theorems.) For an example of such a course, you can check out the open online material (videos, quizzes, interactive R exercises) of my course Market Analysis with Econometrics and Machine Learning.
## References
Angrist, Joshua D., and Jörn-Steffen Pischke. 2009. Mostly Harmless Econometrics: An Empiricist’s Companion.
Published on 01 Jul 2020
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# Recent content by Pwnage101
1. ### Why doesnt macquarie have a faculty of law?
Replace " 'Department(s)' " with " 'Division(s)' ".
2. ### curve sketching help!
Open circle at 0.
3. ### solve tanx=x
True - http://faculty.washington.edu/rag/CLASSES/m308/HANDOUTS/example.html.
4. ### solve tanx=x
I am of the opinion that "the more solutions, the better" with regards to solving problems in mathematics in different ways. It really helps you get used to various different methods, and in doing so you learn new techniques, so don't worry whether or not someone has already posted, your input...
5. ### solve tanx=x
There is no solution in the given interval. You could sketch both functions tan(x) and x on the same axes and see why. A more algebraic approach could include: Clearly x=0 is a solution, since tan(0)=0. Now, let f(x)=tanx-x ---> f'(x)=[sec(x)]^2-1 Now, for 0<x<pi /2, we...
6. ### Integrate e^2x cos (e^2x)
Not sure if there's an easier way, but you could: 1). Open the provided link http://www.codecogs.com/latex/eqneditor.php (perhaps bookmark this) 2). Use this equation editor 3). When finished, copy the text into the reply box, with "[_tex]" before and "[_/tex]" [with no underscore ("_")]...
7. ### Advantages of Actuarial Studies at UNSW
Someone of that stature would be a Professor by now, and there are no other ones listed here: http://www.asb.unsw.edu.au/schools/actuarialstudies/facultyandstaff/Pages/academicstaff.aspx
8. ### Advantages of Actuarial Studies at UNSW
No - Macquarie's program was established 30 years before UNSW's. The lecturer who developed the program at UNSW was a graduate and former lecturer of Macquarie (not that it matters). Source: http://www.asb.unsw.edu.au/schools/Pages/MichaelSherris.aspx.
9. ### Help me prove this standard form please
1).Use the substituion x=a(sec(u)) 2). To integrate sec(u) with respect to u note that d/du{sec(u)+tan(u)}=sec(u)[sec(u)+tan(u)], hence multiplying sec(u) by {sec(u)+tan(u)}/{sec(u)+tan(u)} will make things nice. 3). To transform the result from the variable u to x, draw up a right angled...
10. ### Advantages of Actuarial Studies at UNSW
True. Reason is that people have been graduating from Macquarie with Actuarial Studies degrees for over 40 years, while UNSW has only offered the course for ~10 years (not sure on exact number). In terms of number of graduates from each University nowadays, i'd say they are quite similar.
11. ### actuarial studies
Not entirely true. Law is no longer offered as a double degree, but the most popular double degrees with Actuarial Studies at Macquarie seem to be Applied Finance or Economics. It is also offered as a double degree with mathematics or statistics.
12. ### actuarial studies
No. Just touch up on probability and loan repayments and you should be fine.
13. ### binomial theorem/ combinations question
Trebla provides a nice algebraic proof of a well known result from Pascal's triangle. A non-algebraic (combinatorics) proof would go something similar to the following. Assume n and k are positive integers, with n>k. Consider Tom, who has (n-1) (distinct, distinguishable) blue marbles and 1...
\textup{Second last line is incorrect. It should be } z^{-1}\left[=\frac{r$$\cos\theta\textcolor{red}{-}\sin\theta$$}{r^{2}}\right]= \frac{\overline{z}}{r^{2}}. Edited, thanks to Drongoski for picking up the typo.
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# Stoichiometry - House Method
## Presentation on theme: "Stoichiometry - House Method"— Presentation transcript:
Stoichiometry - House Method
1. Start with a balanced chemical equation. Na2CO3 + Ca(OH) NaOH + CaCO3
Stochiometry - House Method
2. Place the given information above the proper compounds in the equation; except for moles. Moles will be written below the proper compounds!! 120 g X g Na2CO3 + Ca(OH) NaOH + CaCO3
Stochiometry - House Method
3. Draw a simple house around each compound used in the problem. Add moles to the downstairs of each house. 120 g X g Na2CO3 + Ca(OH) NaOH + CaCO3 mole mole
Stochiometry - House Method
4. Draw arrows to show the path of the conversion from beginning to end. 120 g X g Na2CO3 + Ca(OH) NaOH + CaCO3 mole mole
Stochiometry - House Method
5. Set up your factor label conversions in the direction of the arrows. 120 g X g Na2CO3 + Ca(OH) NaOH + CaCO3 mole mole 120 g Na2CO3 X 1 mol Na2CO3 X mol NaOH X g NaOH = g Na2CO mol Na2CO mol NaOH Notice that there are no numbers in front of the mol to mol conversion…YET !
Stochiometry - House Method
6. The numbers in front of the mol to mol conversion are the addresses (coefficients) of the compounds. 120 g X g 1 Na2CO3 + Ca(OH) NaOH + CaCO3 1 mole mole 2 1 120 g Na2CO3 X 1 mol Na2CO3 X mol NaOH X g NaOH = g Na2CO mol Na2CO mol NaOH
Stochiometry - House Method
7. Solve the math. 120 g X g Na2CO3 + Ca(OH) NaOH + CaCO3 mole mole 2 1 120 g Na2CO3 X 1 mol Na2CO3 X mol NaOH X g NaOH = g Na2CO mol Na2CO mol NaOH 91 g NaOH
Stoichiometry- In Review…..
-calc. of quantities in BALANCED chemical equations.
Mole Ratio/ Mole Bridge
- Must have balanced equation - Use coefficients from balanced equation - Used to get from one substance to another Set- Up: # mole of unknown substance #mole of given substance # of moles = Coeff. from balanced equation
For Stoichiometry Problems (A Step-by-Step Guide!!!):
1. Balance the equation 2. Write given and unknowns with the houses. 3. If given is not in moles, convert to moles. 4. Do MOLE BRIDGE 5. If solving to moles, you're done!! 6. If solving to any other unit, do one more conversion to solve to the proper unit.
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# State diagram
11:47
A state diagram for a door that can only be opened and closed
A state diagram is a type of diagram used in computer science and related fields to describe the behavior of systems. State diagrams require that the system described is composed of a finite number of states; sometimes, this is indeed the case, while at other times this is a reasonable abstraction. Many forms of state diagrams exist, which differ slightly and have different semantics.
## Overview
State diagrams are used to give an abstract description of the behavior of a system. This behavior is analyzed and represented by a series of events that can occur in one or more possible states. Hereby "each diagram usually represents objects of a single class and track the different states of its objects through the system".[1]
State diagrams can be used to graphically represent finite-state machines (also called finite automata). This was introduced by Claude Shannon and Warren Weaver in their 1949 book The Mathematical Theory of Communication. Another source is Taylor Booth in his 1967 book Sequential Machines and Automata Theory. Another possible representation is the state-transition table.
## Directed graph
A directed graph
A classic form of state diagram for a finite automaton (FA) is a directed graph with the following elements (Q, Σ, Z, δ, q0, F):[2][3]
• Vertices Q: a finite set of states, normally represented by circles and labeled with unique designator symbols or words written inside them
• Input symbols Σ: a finite collection of input symbols or designators
• Output symbols Z: a finite collection of output symbols or designators
The output function ω represents the mapping of ordered pairs of input symbols and states onto output symbols, denoted mathematically as ω : Σ × QZ.
• Edges δ: represent transitions from one state to another as caused by the input (identified by their symbols drawn on the edges). An edge is usually drawn as an arrow directed from the present state to the next state. This mapping describes the state transition that is to occur on input of a particular symbol. This is written mathematically as δ : Q × ΣQ, so δ (the transition function) in the definition of the FA is given by both the pair of vertices connected by an edge and the symbol on an edge in a diagram representing this FA. Item δ(q, a) = p in the definition of the FA means that from the state named q under input symbol a, the transition to the state p occurs in this machine. In the diagram representing this FA, this is represented by an edge labeled by a pointing from the vertex labeled by q to the vertex labeled by p.
• Start state q0: (not shown in the examples below). The start state q0 ∈ Q is usually represented by an arrow with no origin pointing to the state. In older texts,[2][4] the start state is not shown and must be inferred from the text.
• Accepting state(s) F: If used, for example for accepting automata, F ∈ Q is the accepting state. It is usually drawn as a double circle. Sometimes the accept state(s) function as "Final" (halt, trapped) states.[3]
For a deterministic finite automaton (DFA), nondeterministic finite automaton (NFA), generalized nondeterministic finite automaton (GNFA), or Moore machine, the input is denoted on each edge. For a Mealy machine, input and output are signified on each edge, separated with a slash "/": "1/0" denotes the state change upon encountering the symbol "1" causing the symbol "0" to be output. For a Moore machine the state's output is usually written inside the state's circle, also separated from the state's designator with a slash "/". There are also variants that combine these two notations.
For example, if a state has a number of outputs (e.g. "a= motor counter-clockwise=1, b= caution light inactive=0") the diagram should reflect this : e.g. "q5/1,0" designates state q5 with outputs a=1, b=0. This designator will be written inside the state's circle.
### Example: DFA, NFA, GNFA, or Moore machine
S1 and S2 are states and S1 is an accepting state or a final state. Each edge is labeled with the input. This example shows an acceptor for binary numbers that contain an even number of zeros.
### Example: Mealy machine
S0, S1, and S2 are states. Each edge is labeled with "j / k" where j is the input and k is the output.
## Harel statechart
Diagram showing how Harel's Statecharts contributed to object-oriented methods and notation
Harel statecharts,[5] invented by computer scientist David Harel, are gaining widespread usage since a variant has become part of the Unified Modeling Language (UML).[non-primary source needed] The diagram type allows the modeling of superstates, orthogonal regions, and activities as part of a state.
Classic state diagrams require the creation of distinct nodes for every valid combination of parameters that define the state. This can lead to a very large number of nodes and transitions between nodes for all but the simplest of systems (state and transition explosion). This complexity reduces the readability of the state diagram. With Harel statecharts it is possible to model multiple cross-functional state diagrams within the statechart. Each of these cross-functional state machines can transition internally without affecting the other state machines in the statechart. The current state of each cross-functional state machine in the statechart defines the state of the system. The Harel statechart is equivalent to a state diagram but it improves the readability of the resulting diagram.
## Alternative semantics
There are other sets of semantics available to represent state diagrams. For example, there are tools for modeling and designing logic for embedded controllers.[6] These diagrams, like Harel's original state machines,[7] support hierarchically nested states, orthogonal regions, state actions, and transition actions.[8]
## State diagrams versus flowcharts
Newcomers to the state machine formalism often confuse state diagrams with flowcharts. The figure below shows a comparison of a state diagram with a flowchart. A state machine (panel (a)) performs actions in response to explicit events. In contrast, the flowchart (panel (b)) does not need explicit events but rather transitions from node to node in its graph automatically upon completion of activities.[9]
Nodes of flowcharts are edges in the induced graph of states. The reason is that each node in a flowchart represents a program command. A program command is an action to be executed. So it is not a state, but when applied to the program's state, it results in a transition to another state.
In more detail, the source code listing represents a program graph. Executing the program graph (parsing and interpreting) results in a state graph. So each program graph induces a state graph. Conversion of the program graph to its associated state graph is called "unfolding" of the program graph.
The program graph is a sequence of commands. If no variables exist, then the state consists only of the program counter, which keeps track of where in the program we are during execution (what is the next command to be applied).
In this case before executing a command the program counter is at some position (state before the command is executed). Executing the command moves the program counter to the next command. Since the program counter is the whole state, it follows that executing the command changed the state. So the command itself corresponds to a transition between the two states.
Now consider the full case, when variables exist and are affected by the program commands being executed. Then between different program counter locations, not only does the program counter change, but variables might also change values, due to the commands executed. Consequently, even if we revisit some program command (e.g. in a loop), this doesn't imply the program is in the same state.
In the previous case, the program would be in the same state, because the whole state is just the program counter, so if the program counterpoints to the same position (next command) it suffices to specify that we are in the same state. However, if the state includes variables, then if those change value, we can be at the same program location with different variable values, meaning in a different state in the program's state space. The term "unfolding" originates from this multiplication of locations when producing the state graph from the program graph.
A self transition is a transition where the initial and the final state are the same.
A representative example is a do loop incrementing some counter until it overflows and becomes 0 again. Although the do loop executes the same increment command iteratively, so the program graph executes a cycle, in its state space is not a cycle, but a line. This results from the state being the program location (here cycling) combined with the counter value, which is strictly increasing (until the overflow), so different states are visited in sequence, until the overflow. After the overflow the counter becomes 0 again, so the initial state is revisited in the state space, closing a cycle in the state space (assuming the counter was initialized to 0).
The figure above attempts to show that reversal of roles by aligning the arcs of the state diagrams with the processing stages of the flowchart.
You can compare a flowchart to an assembly line in manufacturing because the flowchart describes the progression of some task from beginning to end (e.g., transforming source code input into object code output by a compiler). A state machine generally has no notion of such a progression. The door state machine shown at the top of this article, for example, is not in a more advanced stage when it is in the "closed" state, compared to being in the "opened" state; it simply reacts differently to the open/close events. A state in a state machine is an efficient way of specifying a particular behavior, rather than a stage of processing.
## Other extensions
An interesting extension is to allow arcs to flow from any number of states to any number of states. This only makes sense if the system is allowed to be in multiple states at once, which implies that an individual state only describes a condition or other partial aspect of the overall, global state. The resulting formalism is known as a Petri net.
Another extension allows the integration of flowcharts within Harel statecharts. This extension supports the development of software that is both event driven and workflow driven.
• David Harel
• DRAKON
• SCXML an XML language that provides a generic state-machine based execution environment based on Harel statecharts.
• UML state machine
• YAKINDU Statechart Tools is a software for modeling state diagrams (Harel statecharts, Mealy machines, Moore machines), simulation, and source code generation.
## References
1. ^
2. ^ a b Taylor Booth (1967) Sequential Machines and Automata Theory, John Wiley and Sons, New York.
3. ^ a b John Hopcroft and Jeffrey Ullman (1979) Introduction to Automata Theory, Languages, and Computation, Addison-Wesley Publishing Company, Reading Mass, ISBN 0-201-02988-X
4. ^ Edward J. McClusky, Introduction to the Theory of Switching Circuits, McGraw-Hill, 1965
5. ^ David Harel, Statecharts: A visual formalism for complex systems. Science of Computer Programming, 8(3):231–274, June 1987.
6. ^ Tiwari, A. (2002). Formal Semantics and Analysis Methods for Simulink Stateflow.
7. ^ Harel, D. (1987). A Visual Formalism for Complex Systems. Science of Computer Programming , 231–274.
8. ^ Alur, R., Kanade, A., Ramesh, S., & Shashidhar, K. C. (2008). Symbolic analysis for improving simulation coverage of Simulink/Stateflow models. International Conference on Embedded Software (pp. 89–98). Atlanta, GA: ACM.
9. ^ Samek, Miro (2008). Practical UML Statecharts in C/C++, Second Edition: Event-Driven Programming for Embedded Systems. Newnes. p. 728. ISBN 978-0-7506-8706-5.
By: Wikipedia.org
Edited: 2021-06-18 17:51:07
Source: Wikipedia.org
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I wonder how to add regression line equation and R^2 on the ggplot. My code is:
```library(ggplot2)
df <- data.frame(x = c(1:100))
df\$y <- 2 + 3 * df\$x + rnorm(100, sd = 40)
p <- ggplot(data = df, aes(x = x, y = y)) +
geom_smooth(method = "lm", se=FALSE, color="black", formula = y ~ x) +
geom_point()
p
```
Mar 22, 2022 3,081 views
## 1 answer to this question.
Let us look at one of the possible solutions
```# EQUATION AND R-SQUARED Get AS STRING
lm_eqn <- function(df){
m <- lm(y ~ x, df);
equation <- substitute(italic(y) == a + b %.% italic(x)*","~~italic(r)^2~"="~r2,
list(a = format(unname(coef(m)[1]), digits = 2),
b = format(unname(coef(m)[2]), digits = 2),
r2 = format(summary(m)\$r.squared, digits = 3)))
as.character(as.expression(equation));
}
p1 <- p + geom_text(x = 25, y = 300, label = lm_eqn(df), parse = TRUE)
```
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# Newton's Laws of Motion MCQ Quiz - Objective Question with Answer for Newton's Laws of Motion - Download Free PDF
Last updated on Nov 11, 2022
## Latest Newton's Laws of Motion MCQ Objective Questions
#### Newton's Laws of Motion Question 1:
Work done upon a body is _______.
1. only a vector quantity
2. only a scalar quantity
3. both vector and scalar
4. neither vector nor scalar
5. None of the above/More than one of the above
Option 2 : only a scalar quantity
#### Newton's Laws of Motion Question 1 Detailed Solution
The correct answer is only a scalar quantity.
Concept:
• Work is said to be done by a force when the body is displaced actually through some distance in the direction of the applied force.
• Since the body is being displaced in the direction of F, therefore work done by the force in displacing the body through a distance s is given by
$$W = \vec F \cdot \vec s$$
Or, W = Fs cos θ
• Thus work done by a force is equal to the scalar or dot product of the force and the displacement of the body.
EXPLANATION:
• Work done upon a body is only a scalar quantity. Therefore option 2 is correct.
• Scalar quantities are the measurements that strictly refer to the magnitude of the medium.
• There are absolutely no directional components in a scalar quantity only the magnitude of the medium. For example, length, speed, work, mass, density, etc.
Scalar quantities Vector quantities The physical quantities which have only magnitude and no direction are called scalar quantities or scalars. The physical quantities which have both magnitude and direction and obey the laws of vector addition are called vector quantities or vectors. A scalar quantity can be specified by a single number, along with the proper unit. A vector quantity is specified by a number with a unit and its direction. Examples: Mass, volume, density, time, temperature, electric current, etc. Examples Displacement, velocity, force, momentum, etc.
#### Newton's Laws of Motion Question 2:
______ is the external agency applied on a body to change its state of rest or uniform motion.
1. Heat
2. Power
3. Energy
4. Force
5. Heat and Energy
Option 4 : Force
#### Newton's Laws of Motion Question 2 Detailed Solution
• Force is the external agency applied on a body to change its state of rest or uniform motion.
Key Points
• Newton's first law of motion:
• Newton's first law states that if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force.
• This postulate is known as the law of inertia.
• The law of inertia was first formulated by Galileo Galilei for horizontal motion on Earth and was later generalized by René Descartes.
• In thermodynamics, heat is energy in transfer to or from a thermodynamic system, by mechanisms other than thermodynamic work or transfer of matter.
• Power is defined as the rate of doing work.
• It is the ratio of work done (W) and time taken (t).
• The SI power unit is the watt (W), in honor of the physicist James Watt.
• In physics, energy is the quantitative property that must be transferred to a body or physical system to perform work on the body or to heat it.
#### Newton's Laws of Motion Question 3:
A particle of mass 1 kg is projected at an angle of 30° with horizontal velocity v = 40 m/s. The change in linear momentum of the particle after time = 1 s will be (g = 10 m/s2)
1. 7.5 kg-m/s
2. 15 kg-m/s
3. 10 kg-m/s
4. 20 kg-m/s
Option 3 : 10 kg-m/s
#### Newton's Laws of Motion Question 3 Detailed Solution
Concept:
• Newton's second law of motion states that " the rate of change of momentum of a body is directly proportional to the applied force on it.
• $$F = K\frac {Δ P}{Δ T}$$
• where k is proportionality constant and equal to 1, ΔP = change in momentum, Δt = change in time
• P = MV.
• where, M = mass and V = velocity.
Calculation:
Given, m = 1 kg, g = 10 m/s2Δt = 1 s
Then F = mg = 1 × 10 = 10 N
Now from equation, $$F = \frac {Δ P}{Δ T}$$
ΔP = F × Δt
⇒ ΔP = 10 × 1 = 10 kg-m/s
#### Newton's Laws of Motion Question 4:
Two particles are released from the same height at an interval of 1 s. How long after the first particle begins to fall will the two particles be 10 m apart? (g = 10 m/s2)
1. 1.5 s
2. 2 s
3. 1.25 s
4. 2.5 s
Option 1 : 1.5 s
#### Newton's Laws of Motion Question 4 Detailed Solution
Concept:
For a body moving in a straight line with constant acceleration, then
• v = u + at
• s = ut + $$\frac12$$at2
• v2 = u2 + 2as
Where u = initial velocity, v = final velocity, a = acceleration, s = displacement, t = time
Acceleration due to gravity, g = 9.8 m/s2
Calculation:
Let after time t, the distance between the particles is 10 m,
Then the distance traveled by the first particle is,
initial speed, u = 0, a = g = 10 m/s2 ( given)
$$s_1 = 0 × t + \frac 12gt^2 = \frac 12 gt^2$$ --- (1)
Now, after 1 s, second particle is released then the t' = t-1, u = 0,
Then the distance traveled by the second particle is,
$$s_2 = 0 × (t-1) + \frac 12g(t-1)^2 = \frac 12 g(t-1)^2$$ --- (2)
Now, s1 - s2 = 10 m,
Then, from equation 1 and 2
$$⇒ \frac 12 ×10×t^2 - \frac12×10× (t-1)^2 = 10$$
⇒ t2 - (t2 + 1 - 2t) = 2
t = 1.5 s
#### Newton's Laws of Motion Question 5:
When the speed of car is doubled, then what will be the braking force of the car to stop it in the same distance.
1. Two times
2. Half
3. One-fourth
4. Four times
Option 4 : Four times
#### Newton's Laws of Motion Question 5 Detailed Solution
The correct option is Four-times.
Key Points
• If a car's speed doubles over the same distance, four times its previous speed, braking force is needed to bring it to a halt.
• The time it takes to stop increases with speed.
• Thus, speeding lengthens your stopping distance and increases the power of collision.
• Keep in mind that every parameter stays the same.
• Work done = braking force × distance
• Work done by brakes = loss of kinetic energy
Braking Force:
• The force that slows the car when the driver depresses the brake pedal is known as the braking force.
• It is impossible to directly quantify this well-known force since it operates on tires that are in touch with the ground, without which nothing would be possible.
Distance:
• Distance is a measurement of how far apart two things or points are, either numerically or occasionally qualitatively.
• The distance can refer to a physical length in physics or to an estimate based on other factors in common usage.
Speed:
• The speed at which an object's location changes in any direction.
• The distance traveled in relation to the time it took to travel that distance is how speed is defined.
• Since speed simply has a direction and no magnitude, it is a scalar quantity.
## Top Newton's Laws of Motion MCQ Objective Questions
#### Newton's Laws of Motion Question 6
When a bus starts suddenly, the passengers are pushed back. This is an example of which of the following?
1. Newton's first law
2. Newton's second law
3. Newton's third law
4. None of Newton's laws
Option 1 : Newton's first law
#### Newton's Laws of Motion Question 6 Detailed Solution
The correct answer is Newton's first law.
CONCEPT:
• Newton’s first law of motion: It is also called the law of inertia. Inertia is the ability of a body by virtue of which it opposes a change.
• According to Newton’s first law of motion, an object will remain at rest or in uniform motion in a straight line unless acted upon by an external force.
• The inertia of rest: When a body is in rest, it will remain at rest until we apply an external force to move it. This property is called inertia of rest.
• The inertia of motion: When a body is in a uniform motion, it will remain in motion until we apply an external force to stop it. This property is called inertia of motion.
EXPLANATION:
• When a bus suddenly starts moving, the passengers fall backward due to the law of inertia of rest or 1st law of Newton.
• Because the body was in the state of rest and when the bus suddenly starts moving the lower body tends to be in motion, but the upper body still remains in a state of rest due to which it feels a jerk and falls backward. Hence option 1 is correct.
Laws of Motion given by Newton are as follows:
Law of Motion Statement First Law of motion An object continues to remain at rest or in a state of uniform motion along a straight line unless an external unbalanced force acts on it. The second law of motion The rate of change of momentum is proportional to the applied force and the change of momentum occurs in the direction of the force. Third law of motion Every action force has an equal and opposite reaction force which acts simultaneously.
#### Newton's Laws of Motion Question 7
Due to an acceleration of 2 m/s2, the velocity of a body increases from 20 m/s to 30 m/s in a certain period. Find the displacement (in m) of the body in that period.
1. 650
2. 125
3. 250
4. 325
Option 2 : 125
#### Newton's Laws of Motion Question 7 Detailed Solution
CONCEPT:
• Equation of motion: The mathematical equations used to find the final velocity, displacements, time, etc of a moving object without considering force acting on it are called equations of motion.
• These equations are only valid when the acceleration of the body is constant and they move on a straight line.
• There are three equations of motion:
V = u + at
V2 = u2 + 2 a S
$${\text{S}} = {\text{ut}} + \frac{1}{2}{\text{a}}{{\text{t}}^2}$$
Where, V = final velocity, u = initial velocity, s = distance traveled by the body under motion, a = acceleration of body under motion, and t = time taken by the body under motion.
EXPLANATION:
Given - Acceleration (a) = 2 m/s2, Final velocity (v) = 30 m/s and Initial velocity (u) = 20 m/s
Displacement = s
We know,
⇒ v2 - u2 = 2as (Equation of Motion)
⇒ s = (v2- u2)/ (2 x a)
⇒ s = (302-202) / (2 x 2)
⇒ s = 500 /4
⇒ s = 125 m
#### Newton's Laws of Motion Question 8
What is the mass of an object that requires a force of 90 N to accelerate at a rate of 2.6 m/s2?
1. 44.6 kg
2. 34.6 kg
3. 54.6 kg
4. None of these
Option 2 : 34.6 kg
#### Newton's Laws of Motion Question 8 Detailed Solution
CONCEPT:
• Force: The interaction which after applying on a body changes or try to change the state of motion or state of rest is called force.
• Force is denoted by F and the SI unit is Newton (N).
Force (F) = Mass (m) × acceleration (a)
CALCULATION
Given - Force (F) = 90 N and acceleration (a) = 2.6 m/s2
• The force required to accelerate a body of mass “m” with acceleration “a” is given by:
⇒ F = ma
⇒ 90 = m × 2.6
⇒ m = 90/2.6
⇒ m = 34.6 kg
#### Newton's Laws of Motion Question 9
Which one of the following has maximum inertia?
1. An atom
2. A molecule
3. A one-rupee coin
4. A cricket ball
Option 4 : A cricket ball
#### Newton's Laws of Motion Question 9 Detailed Solution
CONCEPT:
• Newton’s first law of motion: It is also called the law of inertia. Inertia is the ability of a body by virtue of which it opposes a change.
• According to Newton’s first law of motion, an object will remain at rest or in uniform motion in a straight line unless acted upon by an external force.
• The inertia of rest: When a body is in rest, it will remain at rest until we apply an external force to move it. This property is called inertia of rest.
• The inertia of motion: When a body is in a uniform motion, it will remain in motion until we apply an external force to stop it. This property is called inertia of motion.
EXPLANATION:
• The heavier the object is harder to move it or the greater the amount of force required to move it hence higher the inertia.
• Since higher mass has higher inertia. In all the above 4 options, a cricket ball has maximum mass so it will have maximum inertia. Hence option 4 is correct.
#### Newton's Laws of Motion Question 10
A car accelerates uniformly from 18 km/h to 36 km/h in 5 s. what is the acceleration in m/s?
1. 0.5 m/s2
2. 3 m/s2
3. 1 m/s2
4. 2 m/s2
Option 3 : 1 m/s2
#### Newton's Laws of Motion Question 10 Detailed Solution
Concept:
Acceleration:
• The rate of change of velocity is called the acceleration of the body. i.e.,$$a = \frac{{{v_2} - {v_1}}}{{{t_2} - {t_1}}}$$
• Where V2 = final velocity of the object, V1 = initial velocity of the object, t2 = final time, and t1 = initial time
• It is a vector quantity.
• Its direction is the same as that of change in velocity (Not of the velocity).
Calculation:
Given - u = 18 km/h = 5 m/s v = 36 km/h = 10 m/s and t = 5 s
• As acceleration is written as
• $$\Rightarrow a = \frac{{{v_2} - {v_1}}}{{{t_2} - {t_1}}}$$
• $$\Rightarrow a = \frac{{{10} - {5}}}{5}=1m/s^2$$
Hence the correct answer is 1m/s2
#### Newton's Laws of Motion Question 11
Newton’s first law is also known as _______.
1. Law of rotation
2. Law of displacement
3. Law of momentum
4. None of the option is correct
Option 4 : None of the option is correct
#### Newton's Laws of Motion Question 11 Detailed Solution
There are three Newton’s laws of motion.
• Newton’s first law is also known as the law of inertia. It states that ‘an object will continue to stay at rest and an object in motion will remain in motion in the same direction and speed unless an unbalanced force acts upon it’.
• Newton’s second law of motion states that ‘for a constant mass, force is equal to the product of mass and acceleration. Force is also equal to the change in momentum per change in time.
• Newton’s third law of motion states that ‘for every action, there is an equal and opposite reaction.
#### Newton's Laws of Motion Question 12
Work done upon a body is _______.
1. only a vector quantity
2. only a scalar quantity
3. both vector and scalar
4. neither vector nor scalar
Option 2 : only a scalar quantity
#### Newton's Laws of Motion Question 12 Detailed Solution
The correct answer is only a scalar quantity.
Concept:
• Work is said to be done by a force when the body is displaced actually through some distance in the direction of the applied force.
• Since the body is being displaced in the direction of F, therefore work done by the force in displacing the body through a distance s is given by
$$W = \vec F \cdot \vec s$$
Or, W = Fs cos θ
• Thus work done by a force is equal to the scalar or dot product of the force and the displacement of the body.
EXPLANATION:
• Work done upon a body is only a scalar quantity. Therefore option 2 is correct.
• Scalar quantities are the measurements that strictly refer to the magnitude of the medium.
• There are absolutely no directional components in a scalar quantity only the magnitude of the medium. For example, length, speed, work, mass, density, etc.
Scalar quantities Vector quantities The physical quantities which have only magnitude and no direction are called scalar quantities or scalars. The physical quantities which have both magnitude and direction and obey the laws of vector addition are called vector quantities or vectors. A scalar quantity can be specified by a single number, along with the proper unit. A vector quantity is specified by a number with a unit and its direction. Examples: Mass, volume, density, time, temperature, electric current, etc. Examples Displacement, velocity, force, momentum, etc.
#### Newton's Laws of Motion Question 13
If the force applied on an object is doubled and mass to half. What will be the ratio of its accelerations?
1. 1 : 2
2. 2 : 1
3. 1 : 4
4. 4 : 1
Option 3 : 1 : 4
#### Newton's Laws of Motion Question 13 Detailed Solution
CONCEPT:
• Force: The interaction which after applying on a body changes or try to change the state of motion or state of rest is called force.
• Force is denoted by F and the SI unit is Newton (N).
Force (F) = Mass (m) × acceleration (a)
EXPLANATION:
Given - Let the initial force applied on an object = F, the final force = 2F, initial mass of the object = M, and the final mass = M/2.
• The force required to accelerate a body of mass “m” with acceleration “a” is given by:
⇒ F = ma
• Initial acceleration is
⇒ a1 = F/M ------------ (1)
• Final acceleration is
⇒ a2 = 2F / (M/2). ------------ (2)
On dividing equation 1 and 2, we get
$$⇒ \frac{{{a_1}}}{{{a_2}}} = \frac{{F/M}}{{\frac{{2F}}{{\left( {M/2} \right)}}}} = \frac{1}{4} = 1:4$$
#### Newton's Laws of Motion Question 14
A car is running with a velocity of 20 m/s and suddenly the brakes is applied to stop the car. Find the total work done to stop the car if the mass of the car is 200 kg.
1. 20 k J
2. 40 K J
3. –40 K J
4. 0
Option 3 : –40 K J
#### Newton's Laws of Motion Question 14 Detailed Solution
CONCEPT:
• Kinetic energy: The energy of a particle due to its velocity is called kinetic energy. It is denoted by KE.
KE = ½ m V2 where m is mass of the particle and V is velocity of the particle.
• Work energy theorem: The work done by all the forces on a system is equal to change in kinetic energy of the system.
Work done by all the forces (W) = final KE – Initial KE
EXPLANATION:
Given that: Mass of the car (m) = 200 kg
Initial velocity of the car (u) = 20 m/s
Final velocity of the car (V) = 0
Change in KE = Final KE – Initial KE = ½ m V2 – ½ m u2 = ½ × 200 × (0 – 202) = - 40000 J = - 40 KJ
According to work energy theorem:
Work done by all the forces (W) = final KE – Initial KE = - 40 KJ
#### Newton's Laws of Motion Question 15
The work to be done to increase the speed of a 0.5 kg ball from 4 m/s to 8 m/s is:
1. 8 J
2. 12 J
3. 16 J
4. 4 J
Option 2 : 12 J
#### Newton's Laws of Motion Question 15 Detailed Solution
CONCEPT:
Work done:
It is the dot product of Force and Displacement.
Work Done (W) = Force (F) × Displacement (S)
Kinetic energy:
The energy of a particle due to its velocity is called kinetic energy.
It is denoted by K.E.
$$K.E=\frac{1}{2}mV^2$$
where m = mass of the particle and V = velocity of the particle.
Work energy theorem:
The work done by all the forces on a system is equal to change in kinetic energy of the system.
Work done by all the forces (W) = Final K.E – Initial K.E
CALCULATION:
Given that:
Initial velocity (u) = 4 m/s, final velocity (v) = 8 m/s and
Mass (m) = 0.5 kg
Applying the work-energy theorem:
Work done = Change in Kinetic Energy:
ΔK.E = (K.E)2 - (K.E)1
$$\Rightarrow \frac{1}{2}m(v^2-u^2)$$
$$\Rightarrow \frac{1}{2}\times\;0.5\times(8^2-4^2)$$
∴ ΔK.E is 12 J
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# Mrs. Spata’s Fourth Grade Blog
## Expanded Form for Multiplication
on November 13, 2008
This is another “tool” for our toolbelt. Everyone MUST try this strategy!!
There are three steps to using expanded form. The first is to multiply the bottom factor to the ones of the top factor. The second step is to multiply the bottom factor to the tens of the top factor. The third step is to add the two products to each other.
For example:
27 ** Go to the math EnVision website to watch a tutorial
_____
28
+ 80
___
108
### 7 Responses to “Expanded Form for Multiplication”
1. sspata says:
Another example 34×2=
step one: what is 4×2=
step two: what does 30×2=
**remember that the 3 is in the tens place so
it really is a 30.
step three: add the two products 8+60=
Hope that helps. :o)
2. beck says:
I think that the expanded form is pretty easy now.
3. sspata says:
Becky – I am glad that you like the strategy now. It just took a little practice… Great work!!!
4. Kim Hardwick says:
I put a comment on the blog late Friday night. Did you happen to see it, Mrs. Spata? It related to the report the students are working on.
Thanks,
Kim Hardwick (Emma’s Mom)
5. sspata says:
Ms. Hardwick-
I wrote you back on Saturday. Check under “Native American Day”. Sorry that you didn’t see it. Let me know if you need any more information about it. 🙂
6. SARA says:
Hi Mrs. Spata=] first I had truoble with the expanded form. know I have been doinig some word problems and finaly got it =] =]!!!!!!!!!!!!!!!! Oh my head said thank you and you rock see you on thrusday :o)
7. sspata says:
Great job!!!
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Remember that the MOST he can watch is 10, so whatever he watches has to be less than or equal to 10.
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Here’s how I’d do it: You know that the sprayer spews out .5 gallons in 2.5 minutes. Figure out how much it spews in 1 minute like this: .5 gal = 2.5 minutes Divide both sides by 2.5 minutes 1 gal = .5 gal/2.5 minutes or .2 gallons/minute Now, you can make t = ...
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Hmm. And you've gone through the chapter and can find no information about any of these things? If not, my guess is the textbook is using different words to mean the same things...like instead of saying "equality" they might use a different word. Try checking ...
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compare and contrast the ways in which solving a linear inequality is similar / dissimilar to solving a linear equation please explain
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Explain the scientific interpretation of the fossils found in the crandle of humankind?
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Question 3. 3. Harm may be understood as: (Points : 1) financial, physical, and emotional harm. quantifiable harm. direct but not third-party harm. all of the above.
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physically, emotionally, or financially
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Using multiple sources, defend your thesis on how and why therapeutic and forensic roles can or cannot be reconciled. Ensure that you argue for a particular side of the issue (not arguing both sides) and that your argument is well-supported by scholarly literature, ...
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The case describes an appellate legal opinion or court decision involving expert witness testimony. When a case is appealed, it goes to an appellate or to a higher court. The appellate court then reviews the findings of the lower court, which in this case was the trial court. ...
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300
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The Supreme Court has ruled that prosecutors cannot be subject to civil suits against them: A even in cases of egregious rulebreaking B unless the case is later overturned C because trial judges are always on the prosecutors side D because it is important that prosectors win ...
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i think it is b
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an early learning center that emphasizes "product" over "process" would. a.threaten most mainstream children. b. be child- directed, reflecting self- selection of activities and materials. c. ignore the principles of developmentally appropriate practices. d...
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which one of the following statements is NOT true when a partnership is created between a parent and a teacher a. parents are cpable of making major contributions to rheir childs education b.teachers share responsibility and pwer c. teachers inform parents,who are peripheral ...
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line L bisects and is perpendicular to line AB and line CD. Line AB = CD line AB = 140 and line LM = 24 what is the lenght of line NQ
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2% testosterone qs ad petroleum jelly 240g. How many ml of testosterone is needed to make this compound?
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The Basketball team came to play a game. First they went onto the court. which sentence completes the sentence. A. Before they went to the game, they ate a small meal. B. then at half time they were loosing the game. c. They all drove to the game with coach. D. Then they threw...
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Review chapter 12, section 2 of Writing for Success. What is the most important concept you learned
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two point charges of 2 c and -3c are separated by 6cm what is the magnitude of the electrostatic force between them? and is the force attractive or repulsive? explain and what happens to the magnitude of the force when the distance is increased by a factor of 2 to 12cm
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algebra 114
an investor invested \$800 in two mutual funds. One fund earned 7% and the other earned 3% profit. If the total profit was \$48, how much was invested in each mutual fund.
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1) Select the abbreviation that tells how the italicized noun clause is used. S-subject, DO-direct object, SC-subject complement, OP-object of preposition Our study efforts should go to (whatever subject is most in need of improvement.) *my answer is DO 2)Select the ...
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1-the graduate selection committee wants to select the top 10% of applicants. On a standardized test with a mean of 500 and a standard diviation of 100, what would be the cutoff score for selecting the top 10% of applicants, assuming that test is normally distributed?
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1. ## integral substitution
evaluate integral for interval between x=0 and x=1
INT: (x^2)(1+x)^0.5 dx
thanks
2. Originally Posted by daaavo
evaluate integral for interval between x=0 and x=1
INT: (x^2)(1+x)^0.5 dx
thanks
Try a u sub
let $u=x+1$ then $du=dx$ and $u-1=x$
using the above and substition
$\int x^2(x+1)^{\frac{1}{2}}dx=\int(u-1)^2 u^{\frac{1}{2}}du$
expanding the binomial we get...
$\int (u^2-2u+1)u^{\frac{1}{2}}du=\int u^{\frac{5}{2}}-2u^{\frac{3}{2}}+u^{\frac{1}{2}}du$
From here it is off to the races...
3. Originally Posted by daaavo
evaluate integral for interval between x=0 and x=1
INT: (x^2)(1+x)^0.5 dx
After substitution $u^2 = 1 + x$ the integral becomes $2\int_1^{\sqrt 2 } {u^2 \left( {u^2 - 1} \right)^2 \,du}.$
Now this is routine. After some straightforward calculations your integral is equal to $\frac{4}
{{105}}\left( {11\sqrt 2 - 4} \right).$
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# Can I learn real analysis on my own?
Can I Learn Real Analysis on My Own?
# Can I Learn Real Analysis on My Own?
If you are a mathematics enthusiast, you might have heard of real analysis. Real analysis is the branch of mathematics that deals with the study of real numbers and their properties. It might sound intimidating at first, but with enough persistence and motivation, you can learn real analysis on your own.
## FAQs
### 1. Is Learning Real Analysis on My Own Possible?
The answer is yes, you can learn real analysis on your own. However, be prepared to put in a lot of effort and hard work since real analysis is not an easy subject to grasp. Besides the fact that it’s just plain harder, the way you learn real analysis is not by memorizing formulas or algorithms and plugging things in. Rather, you need to read and reread definitions and proofs until you understand the larger concepts at work, so you can apply those concepts in your own proofs. It is essential to have a strong foundation in calculus and mathematical proofs since real analysis builds upon these topics.
### 2. What are the Resources Available for Self-Study?
There are several resources available online for self-study in real analysis. Some of them are:
See also Is economics very math heavy?
You can also refer to textbooks such as “Principles of Mathematical Analysis” by Walter Rudin and “Real Analysis” by Royden and Fitzpatrick.
### 3. How Should I Approach Self-Study in Real Analysis?
Here are a few tips to help you approach self-study in real analysis:
• Start with the basics: Begin with the fundamentals of real analysis such as limits, continuity, differentiability, etc. and work your way up.
• Understand the definitions: Definitions are crucial in real analysis, so make sure you understand them inside-out.
• Read the proofs carefully: Real analysis is all about proofs, and reading them carefully will help you understand the concepts better.
• Practice, practice, practice: Practice is essential in mathematics, and real analysis is no different. Try solving as many problems as possible to get better at the subject.
### 4. What are the Challenges of Learning Real Analysis on Your Own?
Self-studying in real analysis can be challenging, and here are some of the common challenges:
• Time-consuming: Real analysis requires a lot of time and effort to understand the concepts and solve problems.
• Lack of guidance: When learning on your own, the absence of a teacher or mentor could pose a challenge.
• Comprehending proofs: It’s not uncommon for students to have a hard time understanding theorems and proofs used in real analysis.
• Abstract concepts: Real analysis deals with abstract concepts, which can be challenging to understand.
## Conclusion
In conclusion, self-studying in real analysis can be tough, but with perseverance and dedication, it is possible to learn the concepts and excel in the subject. Start with the basics, read and reread definitions and proofs, and practice solving as many problems as you can. Make use of the resources available online and seek help from teachers or peers whenever necessary. Remember, learning mathematics is a journey, and the key is to keep moving forward.
See also What is applied math useful for?
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# IPMT function
This article describes the formula syntax and usage of the IPMT function in Microsoft Excel.
## Description
Returns the interest payment for a given period for an investment based on periodic, constant payments and a constant interest rate.
## Syntax
IPMT(rate, per, nper, pv, [fv], [type])
The IPMT function syntax has the following arguments:
• Rate Required. The interest rate per period.
• Per Required. The period for which you want to find the interest and must be in the range 1 to nper.
• Nper Required. The total number of payment periods in an annuity.
• Pv Required. The present value, or the lump-sum amount that a series of future payments is worth right now.
• Fv Optional. The future value, or a cash balance you want to attain after the last payment is made. If fv is omitted, it is assumed to be 0 (the future value of a loan, for example, is 0).
• Type Optional. The number 0 or 1 and indicates when payments are due. If type is omitted, it is assumed to be 0.
Set type equal to If payments are due 0 At the end of the period 1 At the beginning of the period
## Remarks
• Make sure that you are consistent about the units you use for specifying rate and nper. If you make monthly payments on a four-year loan at 12 percent annual interest, use 12%/12 for rate and 4*12 for nper. If you make annual payments on the same loan, use 12% for rate and 4 for nper.
• For all the arguments, cash you pay out, such as deposits to savings, is represented by negative numbers; cash you receive, such as dividend checks, is represented by positive numbers.
## Example
Copy the example data in the following table, and paste it in cell A1 of a new Excel worksheet. For formulas to show results, select them, press F2, and then press Enter. If you need to, you can adjust the column widths to see all the data.
Data Description 10.00% Annual interest 1 Period for which you want to find the interest paid. 3 Years of loan \$8,000 Present value of loan Formula Description Live Result =IPMT(A2/12, A3, A4*12, A5) Interest due in the first month for a loan with the terms in A2:A5. (\$66.67) =IPMT(A2, 3, A4, A5) Interest due in the last year for a loan with the same terms, where payments are made yearly. (\$292.45)
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# How do we measure the probability of outcomes in a quantum computer if qubits collapse when measured?
I understand the idea behind Quantum Computing being that we superpose all possible states of a system of qubits and amplify the probabilities of whatever it is that we want depending on the circuit we are making. However when using something like Qiskit the output comes as the frequency of observations, so for example, the $$\left|00\right>$$ state was measured 150 times, and the $$\left|01\right>$$ state 300 times. At first, I thought that when you measured a system of qubits you would find it for example in the $$\left|00\right>$$ state. Still, then if you measured it again right after you would be able to find it in another state like $$\left|01\right>$$, so you just have to measure lots of times (like 1000) to more or less get an idea of what the probability of each output is. This is also what would make quantum computing so efficient since you run the circuit once and then you only have to measure repeatedly. However, qubits apparently collapse when measured, so if you measure the $$\left|00\right>$$ state and you measure again, it will always be in a $$\left|00\right>$$ state. How do we work out the output of a quantum circuit then? Do we have to repeat the whole process of the circuit 1000 times? Wouldn't that make it as inefficient as classical computing?
Cross-posted on physics.SE
• I think that it might be helful for you to read some introductory paper about quantum computing Jul 5, 2023 at 13:58
Do we have to repeat the whole process of the circuit 1000 times? Yes, the idea of (many) quantum circuits is that they produce the desired output with some probability, and thus to ensure the success probability being high enough you run the entire thing multiple times.
Wouldn't that make it as inefficient as classical computing? No, because efficient quantum algorithms take this into account. Roughly speaking, the point is that increasing the success probability of an algorithm (that has a sufficiently high success probability to begin with) requires polynomially many additional runs, and therefore does not affect exponential advantages (if any exists).
I think you are mixing two important concepts here :
• The first one is measurement, indeed when realizing a quantum process, before measurement your system is likely to be in some superposition. More formally :
$$|\psi\rangle=\sum_{i=0}^{2^n}a_i|i\rangle= C|0^{\otimes n}\rangle$$
From a circuit $$C$$, applied on the initial n-quibit basis state $$|0^{\otimes n}\rangle$$.
When measuring, we, in the "classical realm" are going to interact with $$|\psi\rangle$$, wich is still in the "quantum realm", in superposition. A fundamental concept in quantum mechanics, is the notion of wave function collapse https://en.wikipedia.org/wiki/Wave_function_collapse, when interacting with a classical observer, $$|\psi\rangle$$ will collapse to a classical state, wich in this case will be one of the $$|i\rangle$$ (supposing it is your measurement basis) with the probability $$p_i=|a_i|^2$$
Wich is given by the Born rule https://en.wikipedia.org/wiki/Born_rule, another key concept.
After that measurement, if you again measure your state, the state you will get as an output will depend on the new basis in wich you will measure it, to get intuition about this you need to get familiar with observables https://en.wikipedia.org/wiki/Observable. If you measure it in the same basis, you will indeed get the same state. Moreover, in reality (experiments), sometimes it is not even possible to retrieve a state after a measurement, sometimes the state is "destroyed" during the measurement, and what you get is just the bit-string corresponding to the basis state measured.
• Then their is quantum computations, it is important to understand what you want to get from processing quantum information. Sometimes indeed, you'll need to realise a certain number of measurement in order to get the results (bitstring) that you want (hence need to realise a couple of time your circuit), it usually depends on the probability of getting the right bit-string at the end of your process, the efficiency/complexity of you quantum computation will depend on this probability, sometimes, for specific tasks, you'll get what you want with a single measurement, for this, you can check https://en.wikipedia.org/wiki/Deutsch%E2%80%93Jozsa_algorithm.
To give a less theoretical example, quantum generative modelling https://arxiv.org/abs/2111.12738, wich is a classical machine learning inspired quantum process, wich aims at generating ouputs that mimic a specific model, here a fun example of famous classical generative modelling https://this-person-does-not-exist.com/en (where the model is the way human faces are). In the quantum version, measurement and process come into play in differents ways, first you need to train a quantum circuit (as in classical machine learning) for it to tailor the specific model/distribution, wich requires measurements. Once your model is trained (and suppose it's well trained), a single measurement can generate an outcome that is relevant for the purpose of the process, wich is in this case, generate new bit-strings that mimic the model it was trained for (similarly as new faces were generated in the classical examples).
I tried to give you hints of what sounded weird in your question, the bottom line being that theoretical study of quantum computing is key to get the right intuition on what could or could not be more efficient than classical computing. Quantum computing is far from being just "superpose all possible states of a system of qubits and amplify the probabilities of whatever it is that we want depending on the circuit we are making".
Hope this was a bit helpfull, cheers.
• Thanks for your answer! One final question, so then once a system of qubits is measured in the 00 state any further measurements will give the same result? I ask because in other discussions about the wave function I've read that "After it has collapsed into an eigenstate, the particle's wave function will then evolve in accordance with the Schrodinger equation. A subsequent measurement of the same attribute could result in collapse into a different eigenstate and hence a different measured value.". If this is true then after collapse a second measurement could result in for example 11. Jul 6, 2023 at 11:11
• Here the problem is related with observables. When you measure a system you measure an observable. Once you have chosen wich observable to measure (wich basis in my answer), your system will end up in one of the eigenstate of the observable (one of the basis states). If, before redoing any measurment on you state, your system does not undergo any evolution/interaction, basically satying the same, another measurment for the same observable/basis will yield the same result. The only way in this case to get another result would be to change the observable to measure (change the measurment basis) Jul 6, 2023 at 16:40
• In the case where, after being measured, your system undergo another unitary evolution (that can be defined by a circuit or an hamiltonian), then yes your state will evolve according to shrodingers equation and depending on the evolution, can end up in any new superposition of eigenstates of the original observable. Jul 6, 2023 at 16:44
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https://antimeta.wordpress.com/2005/02/01/consistency-and-platonism/
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## Consistency and Platonism
1 02 2005
I noticed an odd thing this evening about set theoretic platonists. They seem to accept a proposed axiom as true just in case we can never prove it consistent with the previously accepted axioms, and they reject the ones (like V=L) that are equiconsistent with the previous system. This sounds odd at first, because of course we’d like to believe only things that actually are consistent with the axioms that we already have, so it seems that axioms of the latter sort would be better candidates than the former. But this doesn’t seem to be the practice of most set theorists.
If I understand John Steel right, this is because the equiconsistent axioms add no new interesting mathematical structures, while the others do. After all, if one is interested in studying the structure of ZF+V=L, these all exist inside structures of (say) ZF+”there exists a measurable”, because each of the latter structures gives a structure for the former by the relative consistency proof. However, structures of ZF+V=L don’t give rise to structures that satisfy some of these large cardinal axioms, so we lose that interesting area of mathematical research.
In addition, there seems to be another problem with consistency for platonists, in that whatever system of axioms is “true” (according to the platonists) may turn out not to be consistent. For instance, given a universe satisfying ZFC, either it contains a model of ZFC (in which case we can take the smallest one) or else it contains no model of ZFC. Now, based on Gödel’s Completeness Theorem, there is a model for a theory iff that theory is consistent, so if the universe contains no model of ZFC (as it mustn’t, if we “choose” the universe to be the smallest model of ZFC inside the “actual” universe), then ZFC is inconsistent. If the universe satisfies some large cardinal axioms, then these can be added to the picture as well.
What exactly this inconsistency means is a bit confusing to me. After all, it doesn’t seem like it could mean that there’s “actually” a proof of “0=1” from ZFC, but rather just that the universe “thinks” that some sequence of formulas actually is such a proof. If the model of ZFC that thinks ZFC is inconsistent is a set model within the actual universe, then it’s clear that this proof must “actually” either be infinitely long or use some steps or axioms that the model mistakenly thinks are valid but “actually” aren’t. But for the “actual” universe, I’m not sure what this could mean. Thus, I can see why a platonist who believes that ZFC is “actually true” might be forced to believe as a result that Con(ZFC) is “actually true” as well, and thus Con(ZFC+Con(ZFC)), and so on. This sequence of claims is I believe weaker even than the claim that there exists even a weakly inaccessible cardinal, but I can see how the analogy might hold. The set theorist is forced to accept a sequence of claims, each of which she knows can never be shown to be consistent merely given the (relatively) uncontroversial claims of ZFC. Thus, higher consistency strength, together with a lack of an obvious disproof, can be seen as a good guide to truth, allowing the set theorist to ascend the hierarchy of large cardinals. The fact that the large cardinals are linearly ordered only makes this seem more convincing.
However, I think if one doesn’t take the platonist view about ZFC, then one isn’t forced into this chain of reasoning. For the fictionalist (or formalist, or intuitionist, or whatever), it doesn’t make sense to say that ZFC is “actually” true, and so Gödel’s Completeness Theorem doesn’t force us into any awkward positions for denying Con(ZFC) – especially since we’re withholding judgement on whether Con(ZFC) itself is “actually true” also. If “ZFC is consistent” is interpreted in its natural language sense, we can believe it to be literally true while remaining non-committal about the set-theoretic claim Con(ZFC). Thus, we are never forced to take even one step up the hierarchy of consistency strengths.
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https://www.mathway.com/examples/basic-math/data-measurement-and-statistics/finding-the-lower-or-first-quartile?id=640
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# Basic Math Examples
, , , , , , ,
Step 1
There are observations, so the median is the mean of the two middle numbers of the arranged set of data. Splitting the observations either side of the median gives two groups of observations. The median of the lower half of data is the lower or first quartile. The median of the upper half of data is the upper or third quartile.
The median of the lower half of data is the lower or first quartile
The median of the upper half of data is the upper or third quartile
Step 2
Arrange the terms in ascending order.
Step 3
Find the median of .
The median is the middle term in the arranged data set. In the case of an even number of terms, the median is the average of the two middle terms.
Remove parentheses.
Cancel the common factor of and .
Factor out of .
Factor out of .
Factor out of .
Cancel the common factors.
Factor out of .
Cancel the common factor.
Rewrite the expression.
Divide by .
Convert the median to decimal.
Step 4
The lower half of data is the set below the median.
Step 5
The median for the lower half of data is the lower or first quartile. In this case, the first quartile is .
The median is the middle term in the arranged data set. In the case of an even number of terms, the median is the average of the two middle terms.
Remove parentheses.
Cancel the common factor of and .
Factor out of .
Factor out of .
Factor out of .
Cancel the common factors.
Factor out of .
Cancel the common factor.
Rewrite the expression.
Divide by .
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http://oeis.org/A259177
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The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A259177 Triangle read by rows T(n,k) which is a bisection of A237593. 11
1, 2, 1, 2, 1, 3, 2, 3, 1, 1, 4, 2, 1, 4, 2, 1, 5, 2, 2, 5, 2, 1, 1, 6, 3, 1, 1, 6, 2, 1, 2, 7, 3, 1, 2, 7, 3, 2, 1, 8, 3, 1, 1, 2, 8, 3, 1, 1, 2, 9, 4, 1, 1, 2, 9, 3, 2, 1, 2, 10, 4, 2, 1, 2, 10, 4, 1, 2, 2, 11, 4, 1, 1, 1, 3, 11, 4, 2, 1, 1, 2, 12, 5, 2, 1, 1, 2, 12, 4, 2, 1, 1, 3, 13, 5, 1, 1, 2, 3, 13, 5, 2, 1, 2, 2, 14 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS Row n has length A003056(n) hence column k starts in row A000217(k). Row n is a permutation of the n-th row of A237591 for some n, hence the sequence is a permutation of A237591. Row sums give A000027. Mirror of A259176. LINKS EXAMPLE Written as an irregular triangle the sequence begins: 1; 2; 1, 2; 1, 3; 2, 3; 1, 1, 4; 2, 1, 4; 2, 1, 5; 2, 2, 5; 2, 1, 1, 6; 3, 1, 1, 6; 2, 1, 2, 7; 3, 1, 2, 7; 3, 2, 1, 8; 3, 1, 1, 2, 8; 3, 1, 1, 2, 9; ... Illustration of initial terms (side view of the pyramid): Row _ 1 _|_| 2 _|_ _| 3 _|_|_ _| 4 _|_|_ _ _| 5 _|_ _|_ _ _| 6 _|_|_|_ _ _ _| 7 _|_ _|_|_ _ _ _| 8 _|_ _|_|_ _ _ _ _| 9 _|_ _|_ _|_ _ _ _ _| 10 _|_ _|_|_|_ _ _ _ _ _| 11 _|_ _ _|_|_|_ _ _ _ _ _| 12 _|_ _|_|_ _|_ _ _ _ _ _ _| 13 _|_ _ _|_|_ _|_ _ _ _ _ _ _| 14 _|_ _ _|_ _|_|_ _ _ _ _ _ _ _| 15 _|_ _ _|_|_|_ _|_ _ _ _ _ _ _ _| 16 |_ _ _|_|_|_ _|_ _ _ _ _ _ _ _ _| ... The above structure represents the first 16 levels (starting from the top) of one of the side views of the infinite stepped pyramid described in A245092. For another side view see A259176. . Illustration of initial terms (partial front view of the pyramid): Row _ 1 |_|_ 2 _|_ _|_ 3 |_| |_ _|_ 4 _|_| |_ _ _|_ 5 |_ _|_ |_ _ _|_ 6 _|_| |_| |_ _ _ _|_ 7 |_ _| |_| |_ _ _ _|_ 8 _|_ _| |_|_ |_ _ _ _ _|_ 9 |_ _| _|_ _| |_ _ _ _ _|_ 10 _|_ _| |_| |_| |_ _ _ _ _ _|_ 11 |_ _ _| |_| |_|_ |_ _ _ _ _ _|_ 12 _|_ _| |_| |_ _| |_ _ _ _ _ _ _|_ 13 |_ _ _| _|_| |_ _| |_ _ _ _ _ _ _|_ 14 _|_ _ _| |_ _|_ |_|_ |_ _ _ _ _ _ _ _|_ 15 |_ _ _| |_| |_| |_ _| |_ _ _ _ _ _ _ _|_ 16 |_ _ _| |_| |_| |_ _| |_ _ _ _ _ _ _ _ _| ... A part of the hidden pattern of the symmetric representation of sigma emerges from the partial front view of the pyramid described in A245092. For another partial front view see A259176. For the total front view see A237593. MATHEMATICA (* function f[n, k] and its support functions are defined in A237593 *) a259177[n_, k_] := f[n, 2*k] TableForm[Table[a259177[n, k], {n, 1, 16}, {k, 1, row[n]}]] (* triangle *) Flatten[Table[a259177[n, k], {n, 1, 26}, {k, 1, [n]}]] (* sequence data *) (* Hartmut F. W. Hoft, Mar 06 2017 *) PROG (PARI) row(n) = (sqrt(8*n + 1) - 1)\2; s(n, k) = ceil((n + 1)/k - (k + 1)/2) - ceil((n + 1)/(k + 1) - (k + 2)/2); T(n, k) = if(k<=row(n), s(n, k), s(n, 2*row(n) + 1 - k)); a259177(n, k) = T(n, 2*k); for(n=1, 26, for(k=1, row(n), print1(a259177(n, k), ", "); ); print(); ) \\ Indranil Ghosh, Apr 21 2017 (Python) from sympy import sqrt import math def row(n): return int(math.floor((sqrt(8*n + 1) - 1)/2)) def s(n, k): return int(math.ceil((n + 1)/k - (k + 1)/2)) - int(math.ceil((n + 1)/(k + 1) - (k + 2)/2)) def T(n, k): return s(n, k) if k<=row(n) else s(n, 2*row(n) + 1 - k) def a259177(n, k): return T(n, 2*k) for n in range(1, 27): print([a259177(n, k) for k in range(1, row(n) + 1)]) # Indranil Ghosh, Apr 21 2017 CROSSREFS Cf. A000203, A000217, A003056, A024916, A175254, A196020, A236104, A237270, A237271, A237591, A237593, A244580, A245092, A249351, A259176, A259179, A261350. Sequence in context: A281574 A191784 A261350 * A304036 A173442 A112309 Adjacent sequences: A259174 A259175 A259176 * A259178 A259179 A259180 KEYWORD nonn,tabf AUTHOR Omar E. Pol, Aug 15 2015 STATUS approved
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https://ximpledu.com/en-us/powers-of-i/
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# Powers of i
How to solve the powers of i problems: formulas, examples, and their solutions.
## Formulas: i, i2, i3, i4
i1 = i
i2 = -1
i3 = -i
i4 = 1
These formulas are used
to find the higher powers of i.
## Example 1: Simplify i23
To use the formula [i4 = 1],
divide the exponent 23 by 4.
The quotient is 5.
The remainder is 3.
So 23 = 4⋅5 + 3.
So i23 = i4⋅5 + 3.
i4⋅5 + 3 = i4⋅5i3
Product of powers
i4⋅5 = (i4)5
Power of a power
i4 = 1
i3 = -i
Then (given) = 15⋅(-i).
15 = 1
So (given) = -i.
## Example 2: Simplify i86
Divide the exponent 86 by 4.
The quotient is 21.
The remainder is 2.
So 86 = 4⋅21 + 2.
So i86 = i4⋅21 + 2.
i4⋅21 + 2 = i4⋅21i2
i4⋅21 = (i4)21
i4 = 1
i2 = -1
Then (given) = 121⋅(-1).
121 = 1
So (given) = -1.
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http://riskheads.org/insurance-premium-modelling-introducing-multiple-insured-parties/
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Insurance Premium Modelling – Introducing Multiple Insured Parties
Last month, we took a look at the way an insurer would have to operate if they only insured a single party. As we discovered, it would be a hard business to run and not a very profitable one if an insurer only dealt with a single incidence of risk. They’d need to keep reserves equivalent to the maximum possible pay out under the policy and the premiums would be unlikely to support this kind of ring fencing of cash.
Fortunately, that’s not how insurance works – insurance is all about combining similar risks from many parties in order to share some of that risk and for insurers to be able to charge reasonable premiums and still make a profit.
The Loss Formula for Multiple Parties
Our simple loss formula needs an update at this point; it’s going to look something like this:
(Where X is the loss and N is the number of losses in the year.)
The trouble with this formula is that it is clear that the loss is going to vary from year to year; some years will be better for our insurer in that they will beat the odds and their insured parties will have fewer accidents than the average and some years will be worse when they also beat the odds but their insured parties have more accidents than the average.
There are two things we need to know for this; the first is the distribution of loss and the second is the frequency of loss for the risk.
This is where the maths gets a bit more complicated.
The Distribution of Loss
This will normally be calculated using a Pareto distribution. Where the α depends on the specific risk. For our car insurer they’d have to calculate this distribution based on the chances of theft, fire, mechanical failure, etc. and then combine the results.
The Frequency of Loss
The frequency of loss can be calculated in many ways but one of the simplest ways to do so is with a Poisson distribution. This only works well in the event that losses are unrelated and relatively rare so it might not be suitable for all types of premium calculation.
These two formulae can then be used to work out the total expected losses in multiple years; to some degree of certainty. You can then treat the expected loss in any given year as an average of the possible losses across multiple years.
However, there’s still more complexity to add to this model and that’s what we’ll be looking at in the next part of this series next month.
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https://fr.slideshare.net/iansalinas3/dll-feb-1014-2020docx
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# dll Feb 10-14, 2020.docx
Olongapo City National High School
9 Nov 2022
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### dll Feb 10-14, 2020.docx
• 1. DAILY LESSON LOG School BARRETTO NATIONAL HIGH SCHOOL Grade Level 8 Teacher AMOR B. SALINAS Learning Area MATHEMATICS Teaching Date and Time February 3-7,2020 Quarter FOURTH February 10, 2020 February 11, 2020 February 12, 2020 February 13, 2020 February 14, 2020 I. OBJECTIVES A. Content Standard The learner demonstrates understanding of key concepts of inequalities in a triangle, and parallel and perpendicular lines. B. Performance Standard The learner is able to communicate mathematical thinking with coherence and clarity in formulating, investigating, analyzing, and solving real- life problems involving triangle inequalities, and parallelism and perpendicularity of lines using appropriate and accurate representations.. C. Learning Competency/Objectives Write the LC code for each 51.The learner proves properties of parallel lines cut by a transversal. (M8GE-IVd-1) 51.The learner proves properties of parallel lines cut by a transversal. (M8GE-IVd-1) 51. The learner proves properties of parallel lines cut by a transversal. (M8GE-IVd-1) AM-Election of SSG Officers PM- Pre Valentine Activity Recollection Grade 10 II. CONTENT Parallel and Perpendicular Lines Parallel and Perpendicular Lines 51.Parallel and Perpendicular Lines 51.The learner proves properties of parallel lines cut by a transversal. (M8GE-IVd-1) 51.The learner proves properties of parallel lines cut by a transversal. (M8GE-IVd-1) The learner proves properties of parallel lines cut by a transversal. (M8GE-IVd-1) II. LEARNING RESOURCES A. References 1. Teachers Guide page K to 12 Basic Curriculum Guide page 144 K to 12 Basic Curriculum Guide page 144 K to 12 Basic Curriculum Guide page 144 2. Learners Material page Geometry III. 2009 p. 73 Geometry III. 2009 p. 73 Geometry III. 2009 p. 73 3. Textbook page 4. Additional Materials from learning resource B. Other Learning Resources II. LEARNING PROCEDURES A. Reviewing previous lesson /senting the new lesson. Gear Up “Gear Up” activity Describe Me! B. Establishing a purpose for the lesson Presentation of Objectives Presentation of Objectives Presentation of Objectives C. Presenting examples / instances of the new lesson. Presentation of illustrative examples. Presentation of illustrative examples. Presentation of illustrative examples. D. Discussing new concepts and practicing new skill #1 Guided Practice Sagot Mo! Show Mo! Solve Me! Solve Me!
• 2. E. Discussing new concepts and practicing new skill #2 Check What You Know Student Activity! Prove Me! F. Developing mastery (leads to formative Assessment 3) Name Me! Find Me! Prove Me! G. Finding practical application of concepts and skills in daily living Draw Me! Challenge Yourself! Guided Practice (See attach LCTG) H. Making generalization and abstractions about the lesson. See attach DLP See attach DLP See attach DLP I. Evaluating Learning Challenge Yourself! Assessment Find Me! J. Additional activities for application or remediation. Journal Writing Journal Writing Journal Writing VI. REMARKS VII. REFLECTION A. No. of learners who earned 80% in the evaluation B. No. of learners who requires additional activities for remediation who scored below 80% in the evaluation C. Did the remedial lessons work? No. of learners who have caught up with the lesson. D. No. of learners who continue to require remediation. E. Which of my teaching strategies worked well? Why did this work? F. What difficulties did I encounter which my principal or supervisor can help me solve? G. What innovation or localized materials did I use / discover which I wish to share with other. Prepared: AMOR SALINAS Math Teacher Checked: EDNA O. ORINE Head Teacher III Noted: SOLEDAD E. POZON, Ed.D. Principal IV
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http://www.mymathsblog.co.uk/cumulative-frequency/
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# Cumulative Frequency
### Overview
In order to understand and use cumulative frequency, you need knowledge covered in Data Collection, Interpreting and Representing Data and Range and Averages. None of the concepts are too difficult. You need to be able to complete cumulative frequency tables, draw box-plots and be able to calculate range, median, lower and upper quartiles. You then need to pull these skills together to compare data sets. I’ve tried to pull this altogether in one question.
### Cumulative Frequency Question
Mr. Wenger wanted to choose his penalty taker scientifically. He decided to record the speed of shots recorded by his squad. It soon became clear that Robert and Theo had the fastest shot. Robert’s and Theo’s last 100 shot speeds were:
a) Complete the cumulative frequency table.
b) Graph the cumulative frequencies for Robert’s and Theo’s shot speeds and estimate the median, lower quartile, upper quartile and inter quartile range for both players’ shot speeds.
c) Draw box-plots for Robert’s and Theo’s shot speeds.
d) Advise Mr. Wenger which player he should select for penalty taking duties (assuming that shot speed is the most important factor). Explain why you would give this advice.
### Cumulative Frequency Question Approach
a) In completing the cumulative frequency table, make sure that the cumulative frequency is equal to the sum of individual frequencies, in this case one hundred.
b) You can plot at least 2 cumulative frequencies on the same graph. This help you to compare sets of data. You can estimate the median value by drawing lines on your graph. In range and averages we saw that median = (n+1)/2 th value. However when we have a large number of data items (which is the usually the case with cumulative frequency questions) we simplify the formula to median = (n/2)th value.
The median value = (n/2) :- where n = number of values, so in this question the meidan = 100/2 = 50th value
The lower quartile value = (n/4) :- where n= number of values, so in this question the lower quartile value = 100/4 = 25th value
The upper quartile = (3n/4), so in this question the upper quartile value = (3×100)÷4 = 75th value.
So to get the median, lower quartile and upper quartiles draw lines from 50,25 and 75 respectively from the Cumulative Frequency axis on the graph and read off the values on the Shot Speed axis. If this is not clear, just look at the graph in the answer below and you will see what I mean.
The inter-quartile range = Upper quartile — Lower quartile.
c) A box-plot is drawn on a scale in this format:-
Box-Plot (Box-and-Whisker Diagram)
The only way to learn these box-plots (sometimes known as box-and-whisker diagrams is to actually do a few.
d) Usually the data in the questions will be prepared so that the conclusions to be drawn are clear. So don’t try and be clever, just use common sense to give your answer but most importantly in your explanation demonstrate your knowledge and understanding of this subject.
a)
b)
Cumulative Frequency Graph
Shot Speeds (kph):-
Robert Theo
Median 87 81
Lower Quartile 81 76
Upper Quartile 95 92
Inter-quartile range 14 16
c) Box-plots for Robert and Theo shot speed (kph):-
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https://betterlesson.com/lesson/639199/planetary-poster
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# Planetary Poster
49 teachers like this lesson
Print Lesson
## Objective
Students will explore the objects in our Solar System by building a poster representing the Sun, eight planets, dwarf planets, asteroid belt, Kuiper Belt, and the Oort Cloud.
#### Big Idea
Explore scale with your students as they construct a poster depicting our Solar System.
## NGSS Background
This lesson is based on California's Middle School Integrated Model of NGSS.
MS-ESS1-1 Earth's Place in the Universe
PE: MS-ESS1-3 Analyze and Interpret data to determine scale properties of objects in the Solar System.
DCI: ESS1.B Earth and the Solar System - The Solar System consists of the Sun and a collection of objects including planets, their moons, and asteroids that are held in orbit around the Sun by its gravitational pull on them.
SP2: Developing and Using Models - The models in this lesson are the drawings of the eight planets. These models serve a the student's understanding of placement and position in our Solar System. The inner planets need to be modeled correctly by making them seem very small in relation to the Sun and the outer gas giants and they need to be relatively close to the Sun. The outer planets need to be correctly modeled by drawing them large in a decreasing order of size, with Jupiter being the largest and Neptune being smallest. Their distances should be exaggerated in comparison to the inner planets.
CCC: Scale, proportion, and quantity - scale is critical in this lesson. The correct scale of our Solar system must depict the inner planets as small and close to the Sun, while the outer planets are large and spread across greater distances. It is a students misconception to draw the planets at the same size and equally spread across their poster paper.
This lesson is built to follow Solar System Sentence Strip, which gets the students thinking about scale distances in a guided ware are you. The students are tasked with creating a poster of the Solar System which provides accurate colors and scale. Before the students are allowed to work on their final draft of the poster they must draw a rough draft the must be approved by the teacher first. An approved rough draft does not need color, but must represent the planets at a correct scale.
This activity provides the students with more freedom to express what they know about planetary distance and scale. Most students can draw the eight planets with details to complete this assignment. The difficulty for students is to correctly model the planets' size and distance to scale. The learning for this lesson occurs when the student's rough drafts are rejected and they must return to consider precision of scale in order to model the inner and the outer planets correctly.
## Set-Up
15 minutes
Materials
1. blank scratch paper
2. poster paper
3. colored pencils
Students will create a poster depicting the Sun and the eight planets in our Solar System. The requirements for this poster are:
1. The poster must include four facts about each planet and the Sun (9 fact boxes total).
2. Use accurate colors (no purple Sun).
3. No crayons or makers. Colored pencils only.
4. Color must cover the entire poster (no blank space).
5. Must address scale (distance vs. size) see TIP below.
6. Must have a title.
7. All objects must be labeled.
8. Rough draft must be approved by the teacher before a student can start on the final draft.
TIP: As a class we discuss scale. Most posters that the students are exposed to have some sort of sacrifice made by the creator. Either the planets are resized to show detail or distances are reduced to fit everything in. My students are expected to address the issue of scale in their poster by either drawing their poster to the correct scale size AND correct scale distance to the best of their ability.
I tell them that I have seen two basic types of posters, one where the students draw the Sun in the center and position the planets orbiting around that center and the other places the Sun off to one side and the planets stretch off to the side. I provide two examples. I typically don't like to give examples, as most students tend to reproduce the example and not do any original work. In this case I make sure to provide an incomplete example to assist them to think about scale.
I encourage my students to do extra research and include more detail as extra credit. Some examples are listed below:
Extra Credit
1. Moons
2. Asteroid belt
3. Kuiper Belt
4. Oort Cloud
5. Dwarf planets
6. Orbital paths
7. Comets
8. Rings around Jupiter, Uranus, and Neptune (Saturn's rings are required)
9. More fact boxes
## Student Activity
120 minutes
In order to complete this assignment the students have to do a fair amount of research on our Solar System. Fortunately our textbooks cover this well. I do allow my students to use their school issued Chromebooks to conduct Internet searches, but I limit it to only one day. I have found that students will spend copious amounts of time researching great astronomy websites, but never commit pencil to paper.
Students doing internet research on their Chromebooks.
The rough drafts must be approved by me first. This causes two things to happen, (1) cut down on the amount of poster paper I have to pass out by limiting silly mistakes, and (2) forces the students to plan the scale within their poster. The rough draft does not need to be in color.
TIP: The hardest part of this assignment for students is accurately depicting scale. Students want to make all the planets the same size and/or want to evenly distribute them across the paper. By not approving the rough draft I am able to control this on the final draft.
Students using Solar System Sentence Strip as a reference.
I typically allow my students about two weeks to complete their posters, and provide two days of class time. I require the poster to be done with colored pencils and I know that many of my students only have access to colored pencils if I provide them.
An example of what I typically reject as a rough draft due to a poor example of scale.
Note that Neptune is smaller that Earth.
Student Work Sample
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• Home
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• Symmetry Activities For Preschoolers To Build Math’s Skills
# Symmetry Activities For Preschoolers To Build Math’s Skills
In the journey of early education, symmetry emerges as a captivating concept that not only sparks the imagination but also lays the foundation for essential math skills. Engaging preschoolers in symmetry activities not only introduces them to the beauty of balance but also cultivates a fundamental understanding of mathematical concepts. In this comprehensive guide, we’ll dive deeper into the significance of symmetry in early childhood education, exploring a collection of interactive and fun symmetry activities designed to be hands-on and tailored for the curious minds of preschoolers.
## The Importance of Symmetry in Early Childhood Education
Symmetry, in its essence, is a fundamental mathematical concept that introduces children to the idea of balance, equality, and geometric patterns. In the early stages of cognitive development, activities centered around symmetry contribute significantly to the acquisition of several key skills:
1. Geographical Awareness:
2. Symmetry activities require preschoolers to visualize and understand how shapes and objects can be mirrored or reflected. This enhances their geographic awareness, a crucial skill for later mathematical understanding.
3. Fine Motor Skills:
4. Hands-on symmetry activities involve tasks like folding paper, arranging shapes, or drawing. These activities contribute to the development of fine motor skills as preschoolers manipulate objects and engage in precise movements.
5. Cognitive Skills:
6. Exploring symmetry encourages cognitive development by promoting logical thinking and problem-solving. As preschoolers engage in activities that require them to predict and complete symmetrical images, they are exercising their cognitive abilities.
7. Creativity:
8. Symmetry activities for kids provide a creative outlet for preschoolers. Whether it’s creating symmetrical art or building structures, these activities foster imagination and self-expression.
9. Mathematical Foundation:
10. Understanding symmetry lays the groundwork for more advanced mathematical concepts in the future. It introduces children to geometric principles, setting the stage for broader mathematical understanding.
## Deeper Exploration of Symmetry Activities
1. Mirror, Mirror on the Paper:
2. Provide each preschooler with a small handheld mirror. Encourage them to explore symmetry by holding the mirror against various objects and observing the reflected images. This hands-on activity not only introduces the concept of symmetry but also enhances visual perception and fine motor skills.
3. Butterfly Symmetry Painting:
4. Fold a piece of paper in half and place a few drops of paint on one side. Press the paper together and unfold it to reveal a symmetrical butterfly pattern. Preschoolers can explore color mixing and symmetry simultaneously, creating vibrant and visually appealing artwork. This activity goes beyond symmetry and introduces concepts like color blending.
5. Symmetry with Nature Walk:
6. Take preschoolers on a nature walk and collect leaves, flowers, or small twigs. Back in the classroom, unfold a large sheet of paper and invite the children to create symmetrical art by arranging their nature treasures on one side and folding the paper to create a mirrored image on the other. This activity not only incorporates nature into the learning process but also reinforces the idea that symmetry exists in the world around us.
7. Interactive Symmetry Drawing:
8. Engage preschoolers in interactive drawing by folding a paper in half and drawing a shape or pattern on one side. Encourage them to predict and complete the symmetrical image on the other side. This activity enhances cognitive skills, creativity, and an understanding of symmetry. As children participate in this activity, they are not merely observing symmetry but actively creating it, enhancing their sense of agency in the learning process.
9. Playful Symmetry with Shapes:
10. Cut out various shapes like circles, squares, and triangles from colored paper. Allow preschoolers to fold the shapes and explore symmetry by matching the sides to create symmetrical designs. This activity not only introduces shapes and symmetry but also encourages tactile learning. The tactile aspect engages multiple senses, providing a holistic learning experience.
11. Mirror Me Dance:
12. Turn learning into a playful movement activity by playing “Mirror Me.” Pair up preschoolers and designate one as the leader. The leader performs simple movements, and the partner mirrors these actions. This interactive symmetry activity not only promotes physical coordination but also reinforces the concept of mirror images. It bridges the gap between abstract mathematical concepts and physical movement, making learning dynamic and engaging.
13. Building Symmetrical Structures:
14. Provide preschoolers with building blocks or construction materials. Encourage them to create symmetrical structures by duplicating the design on both sides. This hands-on activity not only enhances geographical awareness but also introduces the idea of balance in the building. As preschoolers experiment with creating symmetrical structures, they develop an understanding of equilibrium and balance, fundamental concepts in both mathematics and architecture.
15. Symmetry Snack Time:
16. Transform snack time into a symmetry exploration. Cut fruits or sandwiches into symmetrical halves and invite preschoolers to identify the matching sides. This tasty activity combines learning with a practical understanding of symmetry. Connecting symmetry to everyday activities like eating reinforces the idea that mathematical concepts are woven into our daily lives.
17. Symmetrical Collage Creations:
18. Supply preschoolers with a variety of materials like colored paper, feathers, and buttons. Ask them to create a symmetrical collage by placing matching elements on both sides of a folded sheet of paper. This artistic activity encourages creativity and reinforces symmetry concepts. It goes beyond traditional fun symmetry activities, allowing children to express themselves artistically while applying mathematical principles.
19. Symmetry in Our Faces:
20. Explore symmetry by discussing facial features. Provide mirrors and guide preschoolers to identify the symmetrical aspects of their own faces. This self-awareness activity not only introduces symmetry but also fosters a sense of identity. Understanding symmetry in the context of their own faces makes the concept more relatable and personal for preschoolers.
As we immerse preschoolers in the delightful world of interactive symmetry activities, we not only nurture their creativity but also lay the groundwork for a solid understanding of mathematical concepts. These hands-on and interactive activities seamlessly blend fun with learning, fostering a love for exploration and discovery. Through symmetry, preschoolers not only grasp foundational math skills but also develop cognitive abilities, fine motor skills, and a heightened sense of geographical awareness. The activities outlined in this guide serve as gateways to a world where symmetry is not just a mathematical concept but a source of joy, creativity, and early academic success.
So, let the exploration of symmetry begin, as preschoolers embark on a journey where learning is an adventure, and every symmetrical discovery contributes to the beautiful mosaic of their educational foundation. With a deeper understanding of the significance of symmetry in early childhood education, parents and educators can appreciate how these activities contribute to a child’s holistic development and prepare them for future academic success. In expanding upon the importance of symmetry, we highlight its multiple benefits, emphasizing its role in mental, motor, and creative development. This deeper exploration provides a context for parents and educators, offering insights into the broader impact of symmetry activities on a child’s overall growth. As we dive into each activity, we uncover the layers of learning embedded within, making the case for symmetry as a powerful educational tool in the preschool classroom and at home.
EuroKids goes beyond traditional teaching methods, embracing innovative approaches to make learning a delightful journey. Our specially designed interactive symmetry activities not only introduce mathematical concepts but also spark a lifelong love for learning. As we encourage preschoolers to engage in creative and interactive tasks, EuroKids becomes a partner in building a strong foundation for future academic excellence. In every activity, we inspire curiosity, ensuring each child’s early learning experiences are joyful and enriching.
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# 5. The Cost of Capital Part 2
Overview
To calculate the cost of equity for MarkerCo, we're going to use the Capital Asset Pricing Model (CAPM), a common but not uncontested technique among analysts.
To explore more Kubicle data literacy subjects, please refer to our full library.
Summary
1. Lesson Goal (00:04)
The goal of this lesson is to use the Capital Asset Pricing Model to calculate the cost of equity.
2. Understanding Types of Risk (00:33)
The Capital Asset Pricing Model, or CAPM, is a model that can be used to calculate the cost of equity. CAPM assumes that any investment contains two types of risk: systematic risk and unsystematic risk. Systematic risk is risk associated with any investment, such as the risk of an economic recession. Unsystematic risk is risk that is specific to the investment in question, such as the effect of rising oil prices on an energy company.
3. Principles of CAPM (01:11)
CAPM calculates the cost of equity by considering three variables. First is the risk-free rate of return, representing the return required for accepting systematic risk. Second is the additional rate of return for investing in more volatile equity markets, representing the return required for unsystematic risk. Third is the volatility of the stock in question compared to the market, representing the return required for accepting uncertainty.
This produces the following formula for the cost of equity:
$$r_E = r_F + (r_M - r_F)\beta$$
In the formula above:
• $$r_E$$ is the cost of equity we want to calculate
• $$r_F$$ is the risk-free rate of return. This is typically the rate of return on 10-year government bonds
• $$r_M$$ is the expected return of the market as a whole
• $$\beta$$, or beta, represents the volatility of the stock in question compared to the market. If this value is greater than one, the stock is more volatile than the market, if it less than one, the stock is less volatile than the market.
4. Issues with CAPM (02:38)
There are some issues with CAPM. For example, the risk-free rate should theoretically be constant, but the rate on government bonds can change. The expected market return can also fluctuate significantly. Some analysts prefer to consider the Internal Rate of Return instead of using CAPM to calculate the cost of equity.
Transcript
In the previous lesson we examined the formula for the Weighted Average Cost of Capital, or WACC. We then focused on the hardest value in this formula to calculate, which is often the cost of equity. In this lesson, we're going to use the Capital Asset Pricing Model to calculate the cost of equity, and separately examine ways that analysts often get around using WACC by implenting IRR calculations when valuing a business. The Capital Asset Pricing Model is developed under the assumption that investments contain two types of risks. The first type of risk is systematic risk, which is risk that cannot be diversified away. This is risk from events such as recessions, interest rate hikes, etc that effect the whole market. The second type of risk is unsystematic risk, which is risk specific to individual investments, which can be diversified away. An example of unsystematic risk might be a rise in oil prices that affect a company like Exxon Mobil. To account for both types of risk, the CAPM Model adopts the following approach to calculating the cost of capital. It starts with the risk-free rate of return, which is assumed typically to be a 10-year government bond. This is essentially the systematic risk. It then calculates a premium to be paid by investing in the equity markets, and this additional premium is the expected return of the market, rM, minus the risk-free rate of capital, rF.
In step three, we apply a value called beta, to the market risk premium, and beta measures how the stock in question fluctuates compared to the market. If beta is less than one, this means that the stock price is less volatile than the market. And this would typically be the case for energy or water utilities, which are pretty stable businesses. If beta is greater than one, then the stock is more volatile than the market, and this would certainly be the case for a lot of technology or biotechnology stocks. Putting these steps together, we can develop a formula for the cost of equity, which is equal to the risk-free rate plus the market risk premium, which is the expected market return, minus the risk-free rate, multiplied by beta.
It's important to stress at this point that the Capital Asset Pricing Model is quite controversial. It's not used by every analyst, and it does have some obvious weaknesses. For example, the risk-free rate is based on long-term government bonds, which can of course change in price. What's more, the expected market return can also change very frequently. In the following chart, I showed the one year return on the S and P 500 since 2012, and as you can see, the one year return fluctuates dramatically from just under 30% to minus 7.5%.
As a consequence, many analysts have chosen to ignore the CAPM Model and focus instead on the Internal Rate of Return. As you may have seen in previous courses, the internal rate of return is an effective way of understanding the return on any money invested in either an asset or a company, and in the coming lessons I'll show you in our Excel model how to use both the Capital Asset Pricing Model and Internal Rate of Return when valuing a business.
Financial Modeling Essentials
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#### Exercise 4
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# Number 21890
### Properties of number 21890
Cross Sum:
Factorization:
2 * 5 * 11 * 199
Divisors:
1, 2, 5, 10, 11, 22, 55, 110, 199, 398, 995, 1990, 2189, 4378, 10945, 21890
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 2 (Binary):
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
Base 32:
lc2
sin(21890)
-0.57908850761822
cos(21890)
0.81526468115852
tan(21890)
-0.71030736520478
ln(21890)
9.9937851905169
lg(21890)
4.3402457615679
sqrt(21890)
147.95269514274
Square(21890)
### Number Look Up
Look Up
21890 which is pronounced (twenty-one thousand eight hundred ninety) is a very unique figure. The cross sum of 21890 is 20. If you factorisate the figure 21890 you will get these result 2 * 5 * 11 * 199. 21890 has 16 divisors ( 1, 2, 5, 10, 11, 22, 55, 110, 199, 398, 995, 1990, 2189, 4378, 10945, 21890 ) whith a sum of 43200. 21890 is not a prime number. 21890 is not a fibonacci number. The number 21890 is not a Bell Number. The number 21890 is not a Catalan Number. The convertion of 21890 to base 2 (Binary) is 101010110000010. The convertion of 21890 to base 3 (Ternary) is 1010000202. The convertion of 21890 to base 4 (Quaternary) is 11112002. The convertion of 21890 to base 5 (Quintal) is 1200030. The convertion of 21890 to base 8 (Octal) is 52602. The convertion of 21890 to base 16 (Hexadecimal) is 5582. The convertion of 21890 to base 32 is lc2. The sine of the figure 21890 is -0.57908850761822. The cosine of the figure 21890 is 0.81526468115852. The tangent of the number 21890 is -0.71030736520478. The square root of 21890 is 147.95269514274.
If you square 21890 you will get the following result 479172100. The natural logarithm of 21890 is 9.9937851905169 and the decimal logarithm is 4.3402457615679. that 21890 is very great number!
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http://www.genietutorial.com/blog/2017/09/19/neptune-company-produces-toys-and-other-items-for-use-in-beach-and-resort-areas-a-small-inflatable-toy-has-come-answer/
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Exercise 4-12 Target Profit and Break-Even Analysis; Margin of Safety; CM Ratio [LO1, LO3, LO5, LO6, LO7]
Menlo Company distributes a single product. The company’s sales and expenses for last month follow:
Total Per Unit Sales \$1,092,000 \$70 Variable expenses 764,400 49 Contribution margin 327,600 \$21 Fixed expenses 264,600 Net operating income \$63,000
Requirement 1: What is the monthly break-even point in units sold and in sales dollars?
Monthly break-even point units Sales \$
Requirement 2: Without resorting to computations, what is the total contribution margin at the break-even point?
Total contribution margin at the break-even point \$
Requirement 3: How many units would have to be sold each month to earn a target profit of \$96,600? Use the formula method.
Units sold units
Requirement 4: Refer to the original data. Compute the company’s margin of safety in both dollar and percentage terms. (Round your percentage value to 2 decimal places. Omit the “\$” and “%” signs in your response.)
Dollars Percentage Margin of safety \$ %
Requirement 5: What is the company’s CM ratio? If sales increase by \$91,000 and there is no change in fixed expenses, by how much would you expect monthly net operating income to increase?(Omit the “\$” and “%” signs in your response.)
CM ratio % Increased net operating income \$
Problem 4-31 Changes in Fixed and Variable Costs; Target Profit and Break-Even Analysis [LO4, LO5, LO6]
Neptune Company produces toys and other items for use in beach and resort areas. A small, inflatable toy has come onto the market that the company is anxious to produce and sell. The new toy will sell for \$2.9 per unit. Enough capacity exists in the company’s plant to produce 30,900 units of the toy each month. Variable costs to manufacture and sell one unit would be \$1.84, and fixed costs associated with the toy would total \$48,631 per month.
The company’s Marketing Department predicts that demand for the new toy will exceed the 30,900 units that the company is able to produce. Additional manufacturing space can be rented from another company at a fixed cost of \$2,432 per month. Variable costs in the rented facility would total \$2.03 per unit, due to somewhat less efficient operations than in the main plant.
Requirement 1:
Contribution margin \$
(b) Compute the total fixed costs to be covered if more than 30,900 units are produced. (Omit the “\$” sign in your response)
Total fixed costs to be covered by remaining sales \$
(c) Compute the monthly break-even point for the new toy in units and in total sales dollars. (Round your answers to the nearest whole number. Omit the “\$” sign in your response.)
Monthly break-even point in unit sales units Monthly break-even point in dollar sales \$
Requirement 2: How many units must be sold each month to make a monthly profit of \$10,962?
Units to be sold units
Requirement 3: If the sales manager receives a bonus of 20 cents for each unit sold in excess of the break-even point, how many units must be sold each month to earn a return of 21% on the monthly investment in fixed costs? (Round your answer to the nearest whole number.)
Total units to be sold units
Problem 6-15 Comprehensive Problem with Labor Fixed [LO1, LO2, LO3, LO4]
Far North Telecom, Ltd., of Ontario, has organized a new division to manufacture and sell specialty cellular telephones. The division’s monthly costs are shown below:
Manufacturing costs: Variable costs per unit: Direct materials \$88 Variable manufacturing overhead \$3 Fixed manufacturing overhead costs (total) \$220,400 Selling and administrative costs: Variable 12% of sales Fixed (total) \$163,000
Far North Telecom regards all of its workers as full-time employees and the company has a long-standing no layoff policy. Furthermore, production is highly automated. Accordingly, the company includes its labor costs in its fixed manufacturing overhead. The cellular phones sell for \$330 each. During September, the first month of operations, the following activity was recorded:
Units produced 3,800 Units sold 3,000
rev: 02-09-2011
references
Section Break Problem 6-15 Comprehensive Problem with Labor Fixed [LO1, LO2, LO3, LO4]
1.
Problem 6-15 Requirement 1
Requirement 1:
Unit product cost \$
Unit product cost \$
rev: 02-09-2011
references
Worksheet Learning Objective: 06-01 Explain how variable costing differs from absorption costing and compute unit product costs under each method. Learning Objective: 06-04 Understand the advantages and disadvantages of both variable and absorption costing. Problem 6-15 Requirement 1 Learning Objective: 06-02 Prepare income statements using both variable and absorption costing. Difficulty: Medium Learning Objective: 06-03 Reconcile variable costing and absorption costing net operating incomes and explain why the two amounts differ.
2.
Problem 6-15 Requirement 2
Requirement 2: Prepare an absorption costing income statement for September. (Input all amounts as positive values. Omit the “\$” sign in your response.)
\$ \$
rev: 02-09-2011
references
Worksheet Learning Objective: 06-01 Explain how variable costing differs from absorption costing and compute unit product costs under each method. Learning Objective: 06-04 Understand the advantages and disadvantages of both variable and absorption costing. Problem 6-15 Requirement 2 Learning Objective: 06-02 Prepare income statements using both variable and absorption costing. Difficulty: Medium Learning Objective: 06-03 Reconcile variable costing and absorption costing net operating incomes and explain why the two amounts differ.
3.
Problem 6-15 Requirement 3
Requirement 3: Prepare a contribution format income statement for September using variable costing. (Input all amounts as positive values except net operating loss which should be indicated by a minus sign. Omit the “\$” sign in your response.)
\$ Variable expenses: \$ Fixed expenses: \$
rev: 02-09-2011
references
Worksheet Learning Objective: 06-01 Explain how variable costing differs from absorption costing and compute unit product costs under each method. Learning Objective: 06-04 Understand the advantages and disadvantages of both variable and absorption costing. Problem 6-15 Requirement 3 Learning Objective: 06-02 Prepare income statements using both variable and absorption costing. Difficulty: Medium Learning Objective: 06-03 Reconcile variable costing and absorption costing net operating incomes and explain why the two amounts differ.
4.
Problem 6-15 Requirement 4
Requirement 4: Assume that the company must obtain additional financing in order to continue operations. As a member of top management, would you prefer to rely on the statement in (2) above or in (3) above when meeting with a group of prospective investors?
rev: 02-09-2011
Absorption costing statement Variable costing statement
references
Multiple Choice Learning Objective: 06-01 Explain how variable costing differs from absorption costing and compute unit product costs under each method. Learning Objective: 06-04 Understand the advantages and disadvantages of both variable and absorption costing. Problem 6-15 Requirement 4 Learning Objective: 06-02 Prepare income statements using both variable and absorption costing. Difficulty: Medium Learning Objective: 06-03 Reconcile variable costing and absorption costing net operating incomes and explain why the two amounts differ.
5.
Problem 6-15 Requirement 5
Requirement 5: Reconcile the absorption costing and variable costing net operating incomes in requirement 2 and 3 above. (Negative amounts should be indicated by a minus sign. Omit the “\$” sign in your response.)
\$ : Fixed manufacturing overhead cost deferred \$
For instant digital download of the above solution, Please click on the “PURCHASE” link below to get the tutorial for Problem 4-31 Changes in Fixed and Variable Costs; Target Profit and Break-Even Analysis_Answer
Problem 6-15 Comprehensive Problem with Labor Fixed_Answer
Exercise 4-12 Target Profit and Break-Even Analysis; Margin of Safety CM Ratio_Answer
For instant digital download of the above solution or tutorial, please click on the below link and make an instant purchase. You will be guided to the PAYPAL Standard payment page wherein you can pay and you will receive an email immediately with a download link.
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| 3.1875
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CC-MAIN-2018-09
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https://finance.yahoo.com/news/collectors-universe-inc-nasdaq-clct-160032591.html
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U.S. Markets closed
# Is Collectors Universe, Inc.’s (NASDAQ:CLCT) 31% Better Than Average?
One of the best investments we can make is in our own knowledge and skill set. With that in mind, this article will work through how we can use Return On Equity (ROE) to better understand a business. By way of learning-by-doing, we’ll look at ROE to gain a better understanding of Collectors Universe, Inc. (NASDAQ:CLCT).
Over the last twelve months Collectors Universe has recorded a ROE of 31%. One way to conceptualize this, is that for each \$1 of shareholders’ equity it has, the company made \$0.31 in profit.
Want to help shape the future of investing tools and platforms? Take the survey and be part of one of the most advanced studies of stock market investors to date.
### How Do I Calculate Return On Equity?
The formula for return on equity is:
Return on Equity = Net Profit ÷ Shareholders’ Equity
Or for Collectors Universe:
31% = 4.629 ÷ US\$15m (Based on the trailing twelve months to September 2018.)
It’s easy to understand the ‘net profit’ part of that equation, but ‘shareholders’ equity’ requires further explanation. It is the capital paid in by shareholders, plus any retained earnings. You can calculate shareholders’ equity by subtracting the company’s total liabilities from its total assets.
### What Does ROE Signify?
Return on Equity measures a company’s profitability against the profit it has kept for the business (plus any capital injections). The ‘return’ is the yearly profit. A higher profit will lead to a higher ROE. So, all else being equal, a high ROE is better than a low one. That means ROE can be used to compare two businesses.
### Does Collectors Universe Have A Good Return On Equity?
One simple way to determine if a company has a good return on equity is to compare it to the average for its industry. However, this method is only useful as a rough check, because companies do differ quite a bit within the same industry classification. Pleasingly, Collectors Universe has a superior ROE than the average (13%) company in the Consumer Services industry.
That is a good sign. In my book, a high ROE almost always warrants a closer look. For example you might check if insiders are buying shares.
### The Importance Of Debt To Return On Equity
Companies usually need to invest money to grow their profits. The cash for investment can come from prior year profits (retained earnings), issuing new shares, or borrowing. In the first two cases, the ROE will capture this use of capital to grow. In the latter case, the use of debt will improve the returns, but will not change the equity. Thus the use of debt can improve ROE, albeit along with extra risk in the case of stormy weather, metaphorically speaking.
### Collectors Universe’s Debt And Its 31% ROE
While Collectors Universe does have some debt, with debt to equity of just 0.20, we wouldn’t say debt is excessive. When I see a high ROE, fuelled by only modest debt, I suspect the business is high quality. Conservative use of debt to boost returns is usually a good move for shareholders, though it does leave the company more exposed to interest rate rises.
### The Key Takeaway
Return on equity is one way we can compare the business quality of different companies. In my book the highest quality companies have high return on equity, despite low debt. If two companies have around the same level of debt to equity, and one has a higher ROE, I’d generally prefer the one with higher ROE.
But ROE is just one piece of a bigger puzzle, since high quality businesses often trade on high multiples of earnings. It is important to consider other factors, such as future profit growth — and how much investment is required going forward. So I think it may be worth checking this free this detailed graph of past earnings, revenue and cash flow .
If you would prefer check out another company — one with potentially superior financials — then do not miss this free list of interesting companies, that have HIGH return on equity and low debt.
To help readers see past the short term volatility of the financial market, we aim to bring you a long-term focused research analysis purely driven by fundamental data. Note that our analysis does not factor in the latest price-sensitive company announcements.
The author is an independent contributor and at the time of publication had no position in the stocks mentioned. For errors that warrant correction please contact the editor at editorial-team@simplywallst.com.
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https://engineermind.in/2017/11/ohms-law/
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# OHM’S Law
## KIRCHOFF’S CURRENT AND VOLTAGE LAWS ( KCL and KVL)
Dark Light
At constant temperature, the current flowing through a conductor is directly proportional to the potential difference(p.d) in volts across the two ends of the given conductor and inversely proportional to the resistance (R) in ohms (Ω) between the ends of the same conductor.
[Ipropto P.D/Rpropto Volts/Ohms]
In all practical problems of electrical calculations, it is assumed that the temperature rise is within limits, so those electrical properties such as insulation and conduction properties of the Given material are not destroyed.Hence Ohm’s Law Mathematically stated as,
[I=V/R]
[V=IR]
[R=V/I]
[I=(V_{R}-V_{Q})/R=V/R (A)]
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## Six Greatest Electrical Engineering Marvels
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