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http://what-when-how.com/Tutorial/topic-213m4d/Information-Fusion-in-Signal-and-Image-Processing-23.html
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Image Processing Reference
In-Depth Information
among n possible decisions d 1 ,d 2 ,...,d n . The main steps we have to achieve in order
to build the fusion process are as follows:
1) modeling: this step includes choosing a formalism and expressions for the ele-
ments of information we wish to fuse within this formalism. This modeling can be
guided by additional information (regarding the information and the context or the
field). Let us assume, to give the reader a better idea, that each source S j provides
an element of information represented by M i regarding the decision d i . The form
of M i depends of course on which formalism was chosen. It can, for example, be a
distribution in a numerical formalism, or a formula in a logical formalism;
2) estimation: most models require an estimation phase (for example, all of the
methods that use distributions). Again, the additional information can come into play;
3) combination: this step involves the choice of an operator, compatible with the
modeling formalism that was chosen, and guided by the additional information;
4) decision: this is the final step of fusion, which allows us to go from information
provided by the sources to the choice of a decision d i .
We will not go into further detail about these steps here because it would require
discussing formalisms and technical aspects. This will be the subject of the following
chapters.
The way these steps are organized defines the fusion system and its architecture.
In the ideal case, the decision is made based on all of the M i , for all of the sources
and all of the decisions. This is referred to as global fusion. In the global model, no
information is overlooked. The complexity of this model and of its implementation
leads to the development of simplified systems, but with more limited performances
[BLO 94].
A second model thus consists of first making local decisions for each source sepa-
rately. In this case, a decision d ( j ) is made based on all of the information originating
from the source S j only. This is known as a decentralized decision. Then, in a second
step, these local decisions are fused into a global decision. This model is the obvi-
ous choice when the sources are not available simultaneously. It provides answers
rapidly because procedures are specific to each source, and can easily be adapted to
the addition of new sources. This type of model benefits from the use of techniques
from adaptive control and often uses distributed architectures. It is also referred to
as decision fusion [DAS 96, THO 90]. Its main drawback comes from the fact that
it poorly describes relations between sensors, as well as the possible correlations or
dependences between sources. Furthermore, this model very easily leads to contra-
dictory local decisions ( d ( j )
= k ) and solving these conflicts implies
arbitration on a higher level, which is difficult to optimize, since the original informa-
tion is no longer available. Models of this type are often implemented for real-time
applications, for example in the military.
= d ( k ) for j
Search WWH ::
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# CAHSEE Math 2.4 Statistics, Data, and Probability I
Statistics, Data, and Probability I: Drill Set 2, Problem 4. Which of the following statements is true?
CAHSEE Math Statistics, Data, and Probability I Language English Language Statistics Charts, Graphs, and Tables Statistics, Data Analysis, and Probability 6 Data sets
### Transcript
00:27
So... when a question asks us anything like "which of the following are true",
00:31
it means we just have to roll up our sleeves and go one by one testing each little morsel
00:36
of hot dog they give us to evaluate. Let's just start with A.
00:39
It says, "more than half the students took between 15 and 45 minutes"
00:45
So then we just go back to the pie chart thing and see what it has to say.
00:49
Everything here is in percents, so to make things easier on ourselves, we can just think
00:53
about it as a number of students rather than some fancy fraction.
00:57
And in this case we can throw out answer A immediately because
01:00
57% took 45 minutes to an hour...
01:04
...that is, since over half DIDN'T fit in the 15 to 45 minute time window,
01:09
there's no way that more than half COULD.
01:12
We don't get brownie points for calculating that 10% took 15-30 minutes
01:16
and 23% took 30-45 minutes for a total of 33%.
01:22
So let's throw out A and move on as fast as we can.
01:25
B is sorta the counter-side of A.
01:28
It says LESS than half the students took 45 minutes to an hour to fnish...
01:33
...but we just noted that this number was 57% ... already OVER half.
01:38
So B is a quick easy toss out.
01:40
Move on to C.
01:41
C says: More than a quarter of the class took
01:43
EITHER 15 to 30 minutes to finish, OR did not finish the test at all.
01:49
So here, we've got a SUM. Meaning... we ADD both cases.
01:53
That's what happens when we have a conditional phrase like "OR".
01:57
As in... Hand me a condiment --
01:59
either that half gallon jug of ketchup OR that gallon jug of mustard.
02:04
So in the 15-30 minute group, we see that 10% of the students were captured here --
02:09
and then in the Did Not Finish group, we have 10% as well.
02:12
If we add them, we have a total of 20% of the students in those two groups.
02:17
So C is in fact only 20%, which is LESS than a quarter or 25%.
02:22
So C ain't cuttin' it for us either. Odds are very good D is the answer.
02:27
D says that more than a quarter of the class took between 15 and 45 minutes to finish.
02:33
Well, the 15-30 group was 10% - check.
02:36
And the 30-45 group was 23% - check.
02:40
ADD the two groups to get 33% and yeah -- woo hoo --
02:44
33% is in fact greater than a quarter or 25%
02:49
So the answer is, yes... we can read a pie chart. D.
02:53
Time for a slice.
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# 3 meters to nanometers
3 meters to nanometers calculator converts 3 m (meters) into nm (nanometres) and vice versa easily. Moreover, you can find the answer by simply multiplying 3 m by 1000,000,000 and get it converted into nanometres.
## How many nanometers are there in 3 meters?
Since, 1 m = 1000,000,000 nanometres, therefore, simply multiply 3 meters by 1000000000 to convert it into nanometres.
3 m = 3 * 1000000000 = 3000000000 nm
So, there are 3000000000 nanometres in 3 meters.
## What is the answer of 3 meters converted into nanometres?
3 m is equal to 3000000000 nm, or there are 3000000000 nanometres in 3 meters.
## 3 METERS Conversion to Other Lengths
Nanometers 3e+09 Centimeters 300 Feet 9.843 Inches 118.11 Millimeters 3000 Yards 3.282 Kilometers 0.003 Miles 0.00186451
## Relevant Calculators
3 m to nm is a length conversion calculator that converts 3 meters (m) into nanometres (nm) and vice versa accurately and instantly. Additionally, it converts 3 meters (m) into other units such as kilometers, inches, yards, and more.
m nm
3.00 m 3000000000 nm
3.01 m 3010000000 nm
3.02 m 3020000000 nm
3.03 m 3030000000 nm
3.04 m 3040000000 nm
3.05 m 3050000000 nm
3.06 m 3060000000 nm
3.07 m 3070000000 nm
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3.10 m 3100000000 nm
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3.12 m 3120000000 nm
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3.15 m 3150000000 nm
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3.19 m 3190000000 nm
3.20 m 3200000000 nm
3.21 m 3210000000 nm
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3.27 m 3270000000 nm
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3.30 m 3300000000 nm
3.31 m 3310000000 nm
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3.35 m 3350000000 nm
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3.40 m 3400000000 nm
3.41 m 3410000000 nm
3.42 m 3420000000 nm
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3.44 m 3440000000 nm
3.45 m 3450000000 nm
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3.50 m 3500000000 nm
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3.60 m 3600000000 nm
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3.70 m 3700000000 nm
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3.80 m 3800000000 nm
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3.90 m 3900000000 nm
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3.94 m 3940000000 nm
3.95 m 3950000000 nm
3.96 m 3960000000 nm
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3.98 m 3980000000 nm
3.99 m 3990000000 nm
## More Calculations
4 meters to nanometers 3.1 meters to nanometers 3.2 meters to nanometers 3.3 meters to nanometers 3.4 meters to nanometers 3.5 meters to nanometers 3.6 meters to nanometers 3.7 meters to nanometers 3.8 meters to nanometers 3.9 meters to nanometers
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# Pointing to a 2-dimensional array
I'm trying to write code that has a pointer point to a 2-dimensional array.
My main purpose is for not just one asd array, like I would like to point 5 array each of which is 2 dimensional.
``````int asd1[2][2];
int asd2[2][2];
int *se;
se[0] = asd1;
se[1] = asd2;
``````
-
Have you googled "pass 2d array as parameter"? – Luchian Grigore Mar 31 '12 at 13:13
Use `se = asd[0];`
The reason is that the symbol `asd` yields not a pointer to an int but rather a pointer to a one-dimensional array of ints.
@Mig's solution may be good, too. It depends on what you want. In my experience, it tends to work better when you treat a two-dimensional array of a basic type like int as though it were a one-dimensional of length n*n. (This is expecially true in numerical work, where you are likely to call BLAS and LAPACK, but may be true elsewhere, as well. You probably aren't doing numerical work here, but, well, try @Mig's and mine both, and see which you don't prefer. Good luck.)
-
You can do this:
``````#include<stdio.h>
int main()
{
int asd[2][2] = {{0,1},{2,3}};
int (*se)[2]; // a pointer (*se) to an array (2-element array, but only you know it, not the compiler) of array-of-two-integers [2]
se = asd;
printf("%d %d\n%d %d\n", se[0][0], se[0][1], se[1][0], se[1][1]);
return 0;
}
``````
or:
``````#include<stdio.h>
int main()
{
int asd[2][2] = {{0,1},{2,3}};
int (*se)[2][2]; // a pointer (*se) to a 2-element array (first [2]) of two element array (second [2]) of ints
se = &asd;
printf("%d %d\n%d %d\n", (*se)[0][0], (*se)[0][1], (*se)[1][0], (*se)[1][1]);
return 0;
}
``````
-
You want something like this:
``````int asd[2][2];
int (*se)[2] = asd;
``````
This is equivalent to
``````int (*se)[2] = &asd[0];
``````
because `asd` decays to a pointer to its first element in this context.
The key thing to bear in mind is that the type of `asd[0]` is `int[2]`, not `int*`, so you need a pointer to an `int[2]` (i.e. `int (*)[2]`) and not a pointer to an `int*` (i.e. `int**`).
Incidentally, you can make an `int*` point to the first element of the `asd[0]` if you like:
``````int *p = &asd[0][0]; // or just = asd[0];, because it decays to &asd[0][0];
``````
but accessing the other elements of the 2D array as if it were a 1D array, e.g. `p[2]`, would be undefined behaviour.
As a more general point, it's often better to eschew using raw C-style arrays altogether if you can help it. You might want to investigate `std::array` or `std::vector`, depending on your needs.
-
If you were allocating that array dynamically, you could do something like this:
``````#include <stdio.h>
#include <stdlib.h>
#define SIZE 10
int main() {
int i;
int **asd;
asd = (int **)malloc(sizeof(int *) * SIZE);
for (i = 0; i < SIZE; i++) {
asd[i] = (int*)malloc(sizeof(int) * SIZE);
}
int **se;
se = asd;
se[0][1] = 10;
printf("%d %d\n", se[0][1], asd[0][1]);
for (i = 0; i < SIZE; i++) {
free(asd[i]);
}
free(asd);
return 0;
}
``````
You need a pointer to a pointer, since your array is 2-dimensional:
``````int asd[2][2];
int **se;
se = asd;
``````
Now you should be able to:
``````se[0][1] = 10;
``````
-
No. That'll not work. `asd` is `[[int, int], [int, int]]` no pointer just elements, and `se[n][m]` expects `[int*, int*, …]` pointers to 'sub'-arrays – x539 Mar 31 '12 at 13:20
you can't equate se = asd, that was my problem – berkay Mar 31 '12 at 13:25
@x539 You are right, thanks! I'll edit my answer. – Mig Mar 31 '12 at 13:29
@Mig: The question's a C++ one - whilst you could use `malloc` and `free`, it's very C-style and retro. A more idiomatic C++ way would involve using `std::vector` (i.e. not even `new` and `delete`). – Stuart Golodetz Mar 31 '12 at 13:41
@StuartGolodetz: Ooops, forgot about the C++ tag. – Mig Mar 31 '12 at 19:43
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# Use of statistics in experiment
• Astudious
In summary, the best estimate for the random error σ(X) in a single measurement is given by σ(X)2 ≈ 1/(n-1) * ∑((xi-μ)2) where the sum is over all i. However, this equation pertains to multiple measurements rather than a single measurement. Additionally, it corresponds to the sample variance and may not accurately estimate the value of X's random error for the population as a whole. To predict the value for a new or unknown single data point, the equation should use the sample average of the existing data divided by n.
#### Astudious
I have seen that "the best estimate for the random error σ(X) in a single measurement is given by
σ(X)2 ≈ 1/(n-1) * ∑((xi-μ)2) where the sum is over all i"
I have two questions about this: firstly, how can this pertain to a "single measurement" if it requires the data from multiple measurements (x1, x2, x3, ... xi)? Secondly, this seems to correspond to the sample variance - wouldn't it be a more accurate estimate of the value of X's random error to convert to the variance of the population of X as a whole?
Astudious said:
I have seen that "the best estimate for the random error σ(X) in a single measurement is given by
Where did you see this?
If you look at your equation and plug in n = 1, is the variance defined?
Where did you see this?
If you look at your equation and plug in n = 1, is the variance defined?
http://en.wikipedia.org/wiki/Unbiased_estimation_of_standard_deviation
No, but the variance need not (and perhaps should not) be defined for n=1 - a single measurement by definition cannot have a "spread".
I don't see where it says anything about the variance determined from a single measurement in that article. Where did you see that?
If you want to use several data points that you already have to predict what will happen for a new or unknown single data point, that is the equation you should use.
PS, The correct equation uses the sample average of the existing data in place of μ. If some how you know μ, you can use it, but divide by n rather than n-1.
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# Monty Hall problem
Bertrand’s box paradox is a classic paradox of elementary probability theory. It was first posed by Joseph Bertrand in his Calcul des probabilités, published in 1889.
There are three boxes:
1. a box containing two gold coins,
2. a box containing two silver coins,
3. a box containing one gold coin and one silver coin.
After choosing a box at random and withdrawing one coin at random, if that happens to be a gold coin, it may seem that the probability that the remaining coin is gold is 12; in fact, the probability is actually 23. Two problems that are very similar are the Monty Hall problem and the Three Prisoners problem.
These simple but slightly counterintuitive puzzles are used as a standard example in teaching probability theory. Their solution illustrates some basic principles, including theKolmogorov axioms.
The Monty Hall problem is a brain teaser, in the form of a probability puzzle (Gruber, Krauss and others), loosely based on the American television game show Let’s Make a Deal and named after its original host, Monty Hall. The problem was originally posed in a letter by Steve Selvin to the American Statistician in 1975 (Selvin 1975a), (Selvin 1975b). It became famous as a question from a reader’s letter quoted inMarilyn vos Savant‘s “Ask Marilyn” column in Parade magazine in 1990 (vos Savant 1990a):
Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice?
Vos Savant’s response was that the contestant should switch to the other door. (vos Savant 1990a)
The argument relies on assumptions, explicit in extended solution descriptions given by Selvin (1975b) and by vos Savant (1991a), that the host always opens a different door from the door chosen by the player and always reveals a goat by this action—because he knows where the car is hidden. Leonard Mlodinow stated: “The Monty Hall problem is hard to grasp, because unless you think about it carefully, the role of the host goes unappreciated.” (Mlodinow 2008) It is also assumed that the contestant prefers to win a car, rather than a goat.
Contestants who switch have a 2/3 chance of winning the car, while contestants who stick to their choice have only a 1/3 chance. One explanation notices that 2/3 of the time, the initial choice of the player is a door hiding a goat. The host is then forced to open the other goat door, and the remaining one must, therefore, hide the car. “Switching” only fails to give the car when the player picks the “right” door to begin with, which only has a 1/3 chance.
Many readers of vos Savant’s column refused to believe switching is beneficial despite her explanation. After the problem appeared in Parade, approximately 10,000 readers, including nearly 1,000 with PhDs, wrote to the magazine, most of them claiming vos Savant was wrong (Tierney 1991). Even when given explanations, simulations, and formal mathematical proofs, many people still do not accept that switching is the best strategy (vos Savant 1991a). Paul Erdős, one of the most prolific mathematicians in history, remained unconvinced until he was shown a computer simulation confirming the predicted result (Vazsonyi 1999).
The Monty Hall problem has attracted academic interest from the surprising result and simple formulation. Variations of the Monty Hall problem are made by changing the implied assumptions and can create drastically different consequences. For one variation, if Monty only offers the contestant a chance to switch when the contestant initially chose the door hiding the car, then the contestant should never switch. For another variation, if Monty opens another door randomly and happens to reveal a goat, then it makes no difference (Rosenthal, 2005a), (Rosenthal, 2005b).
The problem is a paradox of the veridical type, because the correct result (you should switch doors) is so counterintuitive it can seem absurd, but is nevertheless demonstrably true. The Monty Hall problem is mathematically closely related to the earlier Three Prisoners problem and to the much older Bertrand’s box paradox.
The problem continues to attract the attention of cognitive psychologists. The typical behaviour of the majority, i.e., not switching, may be explained by phenomena known in the psychological literature as: 1) the endowment effect (Kahneman et al., 1991); people tend to overvalue the winning probability of the already chosen – already “owned” – door; 2) the status quo bias (Samuelson and Zeckhauser, 1988); people prefer to stick with the choice of door they have already made; 3) the errors of omission vs. errors of commission effect (Gilovich et al., 1995); all else considered equal, people prefer that any errors that they are responsible for to have occurred through ‘omission’ of taking action rather than through having taken an explicit action that later becomes known to have been erroneous. Experimental evidence confirms that these are plausible explanations which do not depend on probability intuition (Kaivanto et al., 2014Morone and Fiore, 2007).
### Criticism of the simple solutions
As already remarked, most sources in the field of probability, including many introductory probability textbooks, solve the problem by showing the conditional probabilities the car is behind door 1 and door 2 are 1/3 and 2/3 (not 1/2 and 1/2) given the contestant initially picks door 1 and the host opens door 3; various ways to derive and understand this result were given in the previous subsections. Among these sources are several that explicitly criticize the popularly presented “simple” solutions, saying these solutions are “correct but … shaky” (Rosenthal 2005a), or do not “address the problem posed” (Gillman 1992), or are “incomplete” (Lucas et al. 2009), or are “unconvincing and misleading” (Eisenhauer 2001) or are (most bluntly) “false” (Morgan et al. 1991). Some say that these solutions answer a slightly different question – one phrasing is “you have to announce before a door has been opened whether you plan to switch” (Gillman 1992, emphasis in the original).
The simple solutions show in various ways that a contestant who is determined to switch will win the car with probability 2/3, and hence that switching is the winning strategy, if the player has to choose in advance between “always switching”, and “always staying”. However, the probability of winning by always switching is a logically distinct concept from the probability of winning by switching given the player has picked door 1 and the host has opened door 3. As one source says, “the distinction between [these questions] seems to confound many” (Morgan et al. 1991). This fact that these are different can be shown by varying the problem so that these two probabilities have different numeric values. For example, assume the contestant knows that Monty does not pick the second door randomly among all legal alternatives but instead, when given an opportunity to pick between two losing doors, Monty will open the one on the right. In this situation the following two questions have different answers:
1. What is the probability of winning the car by always switching?
2. What is the probability of winning the car given the player has picked door 1 and the host has opened door 3?
The answer to the first question is 2/3, as is correctly shown by the “simple” solutions. But the answer to the second question is now different: the conditional probability the car is behind door 1 or door 2 given the host has opened door 3 (the door on the right) is 1/2. This is because Monty’s preference for rightmost doors means he opens door 3 if the car is behind door 1 (which it is originally with probability 1/3) or if the car is behind door 2 (also originally with probability 1/3). For this variation, the two questions yield different answers. However as long as the initial probability the car is behind each door is 1/3, it is never to the contestant’s disadvantage to switch, as the conditional probability of winning by switching is always at least 1/2. (Morgan et al. 1991)
Players who STAY have won 49040 cars out of 145987 games yielding a winning percentage of 34%
players who SWITCH have won 68356 cars out of 103063 games yielding a winning percentage of 66%
## Video Transcript:
You’re on a game show and there are three doors in front of you. The host, Monty Hall, says, “Behind one door is a brand new car. Behind the other two doors are goats. Pick a door!” You think, “Well, it doesn’t matter which door I choose, every door has a 1/3 chance of having the car behind it.” So, you choose door number 1. Now it gets interesting. Monty, the host, who knows where the car is, opens door number 2 and reveals a goat. The host always opens a door to reveal a goat. The host says, “If you want, you can switch to door number 3.” What should you do? Stay with your original choice or switch to the other door? All right, so what are you going to do? Stay or switch? Well, it’s a fifty-fifty chance of winning the car in either door. Right? [Wrong!] You actually double your chances of winning the car by switching doors. And that is why the Monty Hall Problem is so evasive!
Choose an explanation to the Monty Hall Problem:
1/3 vs 2/3 – Solution #1 to the Monty Hall Problem
There is a 1/3 chance of the car being behind door number 1 and a 2/3 chance that the car isn’t behind door number 1. After Monty Hall opens door number 2 to reveal a goat, there’s still a 1/3 chance that the car is behind door number 1 and a 2/3 chance that the car isn’t behind door number 1. A 2/3 chance that the car isn’t behind door number 1 is a 2/3 chance that the car is behind door number 3.
100 Doors! – Solution #2 to the Monty Hall Problem
Imagine that instead of 3 doors, there are 100. All of them have goats except one, which has the car. You choose a door, say, door number 23. At this point, Monty Hall opens all of the other doors except one and gives you the offer to switch to the other door. Would you switch? Now you may arrogantly think, “Well, maybe I actually picked the correct door on my first guess.” But what’s the probability that that happened? 1/100. There’s a 99% chance that the car isn’t behind the door that you picked. And if it’s not behind the door that you picked, it must be behind the last door that Monty left for you. In other words, Monty has helped you by leaving one door for you to switch to, that has a 99% chance of having the car behind it. So in this case, if you were to switch, you would have a 99% chance of winning the car.
Pick a Goat – Solution #3 to the Monty Hall Problem
To win using the stay strategy, you need to choose the car on your first pick because you’re planning to stay with your initial choice. The chance of picking the car on your first pick is clearly one out of three. But, in order to win using the switch strategy, you only need to pick a goat on your first pick because the host will reveal the other goat and you’ll end up switching to the car. So you want to use the strategy that lets you win if you choose a goat initially because you’re twice as likely to start by picking a goat.
Scenarios – Solution #4 to the Monty Hall Problem
To understand why it’s better to switch doors, let’s play out a few scenarios. Let’s see what will happen if you were to always stay with your original choice. We’ll play out three scenarios, one for each door that the car could be behind (door number 1, door number 2, or door number 3). And it doesn’t matter which door you start out with, so, to keep it simple, we’ll always start by choosing door number 1.
Stay strategy, scenario 1: the car is behind door number 1. You choose door number 1, then the host reveals a goat behind door number 2 and because you always stay, you stay with door number 1. You win the car! Stay strategy, scenario 2: the car is behind door number 2. You start by picking door number 1, the host reveals a goat behind door number 3, and you’re using the stay strategy so you stay with door number 1. You get a goat and don’t win the car. Stay strategy, scenario 3: the car is behind door number 3. You pick door number 1, the host opens door number 2 to reveal a goat, you stay with door number 1, and you get a goat. So, using the stay strategy, you won the car one out of three times. That means that in any one instance of playing the game, your chance of winning the car if you choose to stay is 1/3 or about 33%.
Now let’s try switching doors. Again, we’ll always start by picking door number 1. Switch strategy, scenario 1: the car is behind door number 1. You choose door number 1, the host opens door number 2 to reveal a goat, you are using the switch strategy so you switch to door number 3. You get a goat. Switch strategy, scenario 2: the car is behind door number 2. You start by picking door number 1, the host opens door number 3 to reveal a goat, you switch to door number 2 and win the car! Switch strategy, scenario 3: the car is behind door number 3. You pick door number 1, the host opens door number 2 to reveal a goat, you switch to door number 3 and win the car again! So, with the switch strategy you won the car 2 out of 3 times. That means, that in any one instance of the game, your chance of winning the car if you choose to switch doors is 2/3 or about 67%.
Therefore, if you play the game three times and stay, on average you’ll win the car once. But if you play the game three times and switch each time, on average you’ll win the car twice. That’s twice as many cars!
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Question: How to calculate moderated z-score (MODZ)
2
2.3 years ago by
Gema Sanz70
Karolinska Institutet
Gema Sanz70 wrote:
Hi,
I'm trying to reproduce the calculation of moderated z-score (MODZ) described in PMID: 29195078 for some microarray data. Calculation is described as follows:
Replicate-consensus signatures (MODZ) L1000 experiments are typically done in 3 or more biological replicates. We derive a consensus replicate signature by applying the moderated z-score (MODZ) procedure as follows. First, a pairwise Spearman correlation matrix is computed between the replicate signatures in the space of landmark genes with trivial self-correlations being ignored (set to 0). Then, weights for each replicate are computed as the sum of its correlations to the other replicates, normalized such that all weights sum to 1. Finally, the consensus signature is given by the linear combination of the replicate signatures with the coefficients set to the weights. This procedure serves to mitigate the effects of uncorrelated or outlier replicates, and can be thought of as a ‘de-noised’ representation of the given experiment’s transcriptional consequences.
I already calculated robust z-score for each gene (n=900) at every sample (n=6, 3 controls, 3 treatments) and I used the transposed matrix (genes as columns, samples as rows) to calculate pairwise spearman correlation using:
``````cor<- cor(tm, use="pairwise.complete.obs","spearman")
``````
but I'm sure I'm missing something because I get error
``````Error in cor(tm, use = "pairwise.complete.obs", "spearman") :
'y' must be numeric
``````
I don't know how to define x or y (I'm quite newby with R), but as I understand, I have to correlate control 1 to control 2, control 2 to control 3 and control 1 to control 3 individually and same for treatments and then compute the weights.
Any ideas about the code or how to proceed will be very much welcome!!!
Gema
modz z-score • 1.9k views
modified 13 months ago by Marouen Ben Guebila0 • written 2.3 years ago by Gema Sanz70
2
You could try:
``````cor <- cor(data.matrix(tm), use="pairwise.complete.obs", "spearman")
``````
Your tm object should only contain numerical values. Most likely it is currently a data-frame object, which is not specifically numerical.
1
Wait, cross-post: C: correlation between genes
1
If you want pairwise sample correlations, then your samples should be columns, with genes as rows
yes it was me, but I wanted to post the whole thing to make things clear about what I need to do
I just get the same error with that code. The only way I found to skip that error and make the code work is to set y as tm, but I don't know the meaning of that
y<- tm cor<- cor(tm, y, use="pairwise.complete.obs","spearman")
Can you display a small version of tm (like first 10 genes)?
2
There is no issue setting y as tm. It just means that it will correlate tm to tm.
If you want gene-to-gene correlations, then:
``````cor(tm, tm, use="pairwise.complete.obs", "spearman")
``````
If you want sample-to-sample correlations, then:
``````cor(t(tm), t(tm), use="pairwise.complete.obs", "spearman")
``````
Sorry, I reached the limit of coments per 6h... this is an example matrix of my data, I have 6 different samples and 928 genes, it is already trasposed
https://imgur.com/a/iZtsm
what I need to get according to the description of the calculation, is a correlation value per gene in each sample, and the output from cor() is only one correlation value per gene pair, I'm not sure how to get that.
Actually, this is a bit weird because the weighted average of zscores isn´t a zscore.
1
Why did you post this as an answer?
1
yes sorry it was just a thought I did not answer the question.
No problem - no need to worry at this point.
3
2.3 years ago by
Devon Ryan95k
Freiburg, Germany
Devon Ryan95k wrote:
As just mentioned on your other post, you need `method="spearman"`, since otherwise R is seeing the unnamed argument as the thing you should be correlating `tm` with. Also, you want `t(tm)`, as Kevin mentioned.
1
Good spot on "spearman" Devon.
``````cor(tm, use="pairwise.complete.obs", method="spearman")
``````
or
``````cor(t(tm), use="pairwise.complete.obs", method="spearman")
``````
Thanks! that solved the "y" problem. As I wrote above what I need to get according to the description of the calculation, is a correlation value per gene in each sample, and the output from cor() is only one correlation value per gene pair, I'm not sure how to get that, I guess I should do a correlation using x for one sample and y for another and do it with all possible comparisons... ?
pairwise Spearman correlation matrix is computed between the replicate signatures in the space of landmark genes with trivial self-correlations being ignored (set to 0). Then, weights for each replicate are computed as the sum of its correlations to the other replicates
From the text I understand that I need one correlation value per gene and per sample and then combine those from replicates.
I have no idea how to do that, they have a script in MATLAB to do it, but I don't know how to extrapolate into R. This is the MATLAB code:
``````function modz_ds = level4_to_level5(zsrep_file, col_meta_file, landmark_file, group_var)
% LEVEL4_TO_LEVEL5 Compute Moderated Z-scores (ModZ) from replicate signatures
% zsrep_file = '/Users/narayan/workspace/cmapM/data/TEST_A375_24H_ZSPCINF_n67x22268.gctx';
% col_meta_file = '/Users/narayan/workspace/cmapM/data/TEST_A375_24H_ZSPCINF.map';
% landmark_file = '/Users/narayan/workspace/cmapM/data_processing/resources/L1000_EPSILON.R2.chip';
% z-scores for all replicate signatures
zsrep = parse_gctx(zsrep_file);
% sample annotations
col_meta = parse_tbl(col_meta_file, 'outfmt', 'record');
% landmark annotations
chip = parse_tbl(landmark_file, 'outfmt', 'record');
%% Generate moderated z-score (ModZ) signatures
% Exclude large outlier zscores
zsrep.mat = clip(zsrep.mat, -10, 10);
% Landmark features
pr_id_lm = {chip(strcmp('LM', {chip.pr_type})).pr_id}';
[~, lm_ridx] = intersect(zsrep.rid, pr_id_lm);
% column ids
cid = {col_meta.cid}';
% Group samples
[rep_gp, rep_idx] = getcls({col_meta.(group_var)}');
num_gp = length(rep_gp);
[num_row, num_col] = size(zsrep.mat);
modz_mat = nan(num_row, num_gp);
for ii=1:num_gp
this_gp = rep_idx == ii;
this_zs = zsrep.mat(:, this_gp);
fprintf(1, '%d/%d %s Computing ModZS %d replicates\n', ii, num_gp, rep_gp{ii}, nnz(this_gp));
% determine weights based on replicate correlations in landmark space
[modz_mat(:, ii), wt, cc] = modzs(this_zs, lm_ridx);
fprintf(1, 'Replicate correlations\n');
disp(cc);
fprintf(1, 'Replicate weights: ');
disp(wt');
end
% Annotate ModZ matrix
modz_ds = mkgctstruct(modz_mat, 'rid', zsrep.rid, 'cid', rep_gp);
[~, uidx] = unique(rep_idx, 'stable');
modz_meta = keepfield(col_meta(uidx), {'rna_well', 'pert_id',...
'pert_iname', 'pert_type',...
'cell_id','pert_idose',...
'pert_itime'});
modz_ds = annotate_ds(modz_ds, modz_meta);
%modz_ds = annotate_ds(modz_ds, row_meta, 'dim', 'row', 'keyfield', 'pr_id');
end
``````
1
I like how obtuse their description is. The actual matlab code is here, which in R is:
``````modzs <- function(m, clipLowWt=TRUE, lowThreshWT=0.1, clipLowCC=TRUE, lowThreshCC=0, metric="avg") {
cc = cor(m, method="spearman") # pair-wise correlations
if(clipLowCC) { # Trim low values
cc[cc<lowThreshCC] = lowThreshCC
}
wt = 0.5 * colSums(cc) # Per-sample values
if(clipLowWt) { # Trim low values
wt[wt<lowThreshWT] = lowThreshWT
}
# Normalize the weights
if(metric=="avg") {
sumWT = sum(abs(wt))
} else {
sumWT = sqrt(sum(wt^2))
}
normWT = wt / sumWT # Normalized weights
# Return the scaled input values
return(m * normWT)
}
``````
I've set the defaults of `modzs()` to match the ones used in matlab. I've not tried this on real data, so I don't have a good sense how reasonable it is.
Note that the input matrix, `m`, has genes as rows and samples as columns.
1
I think it is working! The function doesn't give errors and I got something that I think is the expected output
https://ibb.co/eLtkSc
wow, thanks A LOT!!! this is a start point, I will play around with it! :)
Do you code in MatLab Devon? - I don't!
1
Not since college, which was quite a while back :P
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# Search by Topic
#### Resources tagged with Recording mathematics similar to Round the Four Dice:
Filter by: Content type:
Stage:
Challenge level:
### There are 18 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Recording mathematics
### Round the Four Dice
##### Stage: 2 Challenge Level:
This activity involves rounding four-digit numbers to the nearest thousand.
### Round the Three Dice
##### Stage: 2 Challenge Level:
What happens when you round these three-digit numbers to the nearest 100?
### Round the Dice Decimals 1
##### Stage: 2 Challenge Level:
Use two dice to generate two numbers with one decimal place. What happens when you round these numbers to the nearest whole number?
### How Do You See It?
##### Stage: 1 and 2 Challenge Level:
Here are some short problems for you to try. Talk to your friends about how you work them out.
### Children's Mathematical Writing
##### Stage: 1 and 2
Bernard Bagnall discusses the importance of valuing young children's mathematical representations in this article for teachers.
### Round the Dice Decimals 2
##### Stage: 2 Challenge Level:
What happens when you round these numbers to the nearest whole number?
##### Stage: 1 and 2 Challenge Level:
How could you put these three beads into bags? How many different ways can you do it? How could you record what you've done?
### Multiply Multiples 1
##### Stage: 2 Challenge Level:
Can you complete this calculation by filling in the missing numbers? In how many different ways can you do it?
### Multiply Multiples 2
##### Stage: 2 Challenge Level:
Can you work out some different ways to balance this equation?
### Multiply Multiples 3
##### Stage: 2 Challenge Level:
Have a go at balancing this equation. Can you find different ways of doing it?
### Dice in a Corner
##### Stage: 2 Challenge Level:
How could you arrange at least two dice in a stack so that the total of the visible spots is 18?
### Junior Frogs
##### Stage: 1 and 2 Challenge Level:
Have a go at this well-known challenge. Can you swap the frogs and toads in as few slides and jumps as possible?
### Pupils' Recording or Pupils Recording
##### Stage: 1, 2 and 3
This article, written for teachers, looks at the different kinds of recordings encountered in Primary Mathematics lessons and the importance of not jumping to conclusions!
### Plants
##### Stage: 1 and 2 Challenge Level:
Three children are going to buy some plants for their birthdays. They will plant them within circular paths. How could they do this?
### More Children and Plants
##### Stage: 2 and 3 Challenge Level:
This challenge extends the Plants investigation so now four or more children are involved.
### Knight's Swap
##### Stage: 2 Challenge Level:
Swap the stars with the moons, using only knights' moves (as on a chess board). What is the smallest number of moves possible?
### More Plant Spaces
##### Stage: 2 and 3 Challenge Level:
This challenging activity involves finding different ways to distribute fifteen items among four sets, when the sets must include three, four, five and six items.
### Exploring Number Patterns You Make
##### Stage: 2 Challenge Level:
Explore Alex's number plumber. What questions would you like to ask? What do you think is happening to the numbers?
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# Leggett–Garg inequality
The Leggett–Garg inequality,[1] named for Anthony James Leggett and Anupam Garg, is a mathematical inequality fulfilled by all macrorealistic physical theories. Here, macrorealism (macroscopic realism) is a classical worldview defined by the conjunction of two postulates:[1]
1. Macrorealism per se: "A macroscopic object, which has available to it two or more macroscopically distinct states, is at any given time in a definite one of those states."
2. Noninvasive measurability: "It is possible in principle to determine which of these states the system is in without any effect on the state itself, or on the subsequent system dynamics."
## In quantum mechanics
In quantum mechanics, the Leggett–Garg inequality is violated, meaning that the time evolution of a system cannot be understood classically. The situation is similar to the violation of Bell's inequalities in Bell test experiments which plays an important role in understanding the nature of the Einstein–Podolsky–Rosen paradox. Here quantum entanglement plays the central role.
## Two-state example
The simplest form of the Leggett–Garg inequality derives from examining a system that has only two possible states. These states have corresponding measurement values ${\displaystyle Q=\pm 1}$ . The key here is that we have measurements at two different times, and one or more times between the first and last measurement. The simplest example is where the system is measured at three successive times ${\displaystyle t_{1} . Now suppose, for instance, that there is a perfect correlation ${\displaystyle C_{13}}$ of 1 between times ${\displaystyle t_{1}}$ and ${\displaystyle t_{3}}$ . That is to say, that for N realisations of the experiment, the temporal correlation reads
${\displaystyle C_{13}={\frac {1}{N}}\sum _{r=1}^{N}Q_{r}(t_{1})Q_{r}(t_{3})=1.}$
We look at this case in some detail. What can be said about what happens at time ${\displaystyle t_{2}}$ ? Well, it is possible that ${\displaystyle C_{12}=C_{23}=1}$ , so that if the value at ${\displaystyle t_{1}=\pm 1}$ , then it is also ${\displaystyle \pm 1}$ for both times ${\displaystyle t_{2}}$ and ${\displaystyle t_{3}}$ . It is also quite possible that ${\displaystyle C_{12}=C_{23}=-1}$ , so that the value at ${\displaystyle t_{1}}$ is flipped twice, and so has the same value at ${\displaystyle t_{3}}$ as it did at ${\displaystyle t_{1}}$ . So, we can have both ${\displaystyle Q(t_{1})}$ and ${\displaystyle Q(t_{2})}$ anti-correlated as long as we have ${\displaystyle Q(t_{2})}$ and ${\displaystyle Q(t_{3})}$ anti-correlated. Yet another possibility is that there is no correlation between ${\displaystyle Q(t_{1})}$ and ${\displaystyle Q(t_{2})}$ . That is we could have ${\displaystyle C_{12}=C_{23}=0}$ . So, although it is known that if ${\displaystyle Q=\pm 1}$ at ${\displaystyle t_{1}}$ it must also be ${\displaystyle \pm 1}$ at ${\displaystyle t_{3}}$ , the value at ${\displaystyle t_{2}}$ may as well be determined by the toss of a coin. We define ${\displaystyle K}$ as ${\displaystyle K=C_{12}+C_{23}-C_{13}}$ . In these three cases, we have ${\displaystyle K=1,-3,}$ and ${\displaystyle -1}$ , respectively.
All that was for 100% correlation between times ${\displaystyle t_{1}}$ and ${\displaystyle t_{3}}$ . In fact, for any correlation between these times ${\displaystyle K=C_{12}+C_{23}-C_{13}\leq 1}$ . To see this, we note that
${\displaystyle K={\frac {1}{N}}\sum _{r=1}^{N}\left(Q(t_{1})Q(t_{2})+Q(t_{2})Q(t_{3})-Q(t_{1})Q(t_{3})\right)_{r}.}$
It is easily seen that for every realisation ${\displaystyle r}$ , the term in the parentheses must be less than or equal to unity, so that the result for the sum is also less than (or equal to) unity. If we have four distinct times rather than three, we have ${\displaystyle K=C_{12}+C_{23}+C_{34}-C_{14}\leq 2}$ and so on. These are the Leggett–Garg inequalities. They say something definite about the relation between the temporal correlations of ${\displaystyle \langle Q({\text{start}})Q({\text{end}})\rangle }$ and the correlations between successive times in going from the start to the end.
In the derivations above, it has been assumed that the quantity Q, representing the state of the system, always has a definite value (macrorealism per se) and that its measurement at a certain time does not change this value nor its subsequent evolution (noninvasive measurability). A violation of the Leggett–Garg inequality implies that at least one of these two assumptions fails.
## Experimental violations
One of the first proposed experiments for demonstrating a violation of macroscopic realism employs superconducting quantum interference devices. There, using Josephson junctions, one should be able to prepare macroscopic superpositions of left and right rotating macroscopically large electronic currents in a superconducting ring. Under sufficient suppression of decoherence one should be able to demonstrate a violation of the Leggett–Garg inequality.[2] However, some criticism has been raised concerning the nature of indistinguishable electrons in a Fermi sea.[3][4]
A criticism of some other proposed experiments on the Leggett–Garg inequality is that they do not really show a violation of macrorealism because they are essentially about measuring spins of individual particles.[5] In 2015 Robens et al.[6] demonstrated an experimental violation of the Leggett–Garg inequality using superpositions of positions instead of spin with a massive particle. At that time, and so far up until today, the Cesium atoms employed in their experiment represent the largest quantum objects which have been used to experimentally test the Leggett–Garg inequality.[7]
The experiments of Robens et al.[6] as well as Knee et al.,[8] using ideal negative measurements, also avoid a second criticism (referred to as “clumsiness loophole”[9]) that has been directed to previous experiments using measurement protocols that could be interpreted as invasive, thereby conflicting with postulate 2.
Several other experimental violations have been reported, including in 2016 with neutrino particles using the MINOS dataset.[10]
Brukner and Kofler have also demonstrated that quantum violations can be found for arbitrarily large macroscopic systems. As an alternative to quantum decoherence, Brukner and Kofler are proposing a solution of the quantum-to-classical transition in terms of coarse-grained quantum measurements under which usually no violation of the Leggett–Garg inequality can be seen anymore.[11][12]
Experiments proposed by Mermin[13] and Braunstein and Mann[14] would be better for testing macroscopic realism, but warns that the experiments may be complex enough to admit unforeseen loopholes in the analysis. A detailed discussion of the subject can be found in the review by Emary et al.[15]
## Related inequalities
The four-term Leggett–Garg inequality can be seen to be similar to the CHSH inequality. Moreover, equalities were proposed by Jaeger et al.[16]
## References
1. ^ a b Quantum Mechanics versus macroscopic realism: is the flux there when nobody looks? A. J. Leggett and Anupam Garg. Phys. Rev. Lett. 54, 857 (1985)
2. ^ Testing the limits of quantum mechanics: motivation, state of play, prospects. A. J. Leggett. J. Phys.: Condens. Matter 14, R414-R451 (2002)
3. ^ Electronic structure of superposition states in flux qubits. J. I. Korsbakken, F. K. Wilhelm, and K. B. Whaley, Physica Scripta 137, 4022 (2009). https://link.springer.com/article/10.1007%2Fs10701-011-9598-4
4. ^ Superconducting qubit in a resonator: test of the Leggett-Garg inequality and single-shot readout, A. Palacios-Laloy, PhD thesis (2010). http://iramis.cea.fr/spec/Pres/Quantro/static/wp-content/uploads/2010/10/Palacios-Laloy-Thesis1.pdf
5. ^ Foundations and Interpretation of Quantum Mechanics. Gennaro Auletta and Giorgio Parisi, World Scientific, 2001 ISBN 981-02-4614-5, ISBN 978-981-02-4614-3
6. ^ a b "Ideal Negative Measurements in Quantum Walks Disprove Theories Based on Classical Trajectories". Carsten Robens, Wolfgang Alt, Dieter Meschede, Clive Emary, and Andrea Alberti, Physical Review X 5, 011003 (2015).
7. ^ "Do Quantum Superpositions Have a Size Limit?" , George C. Knee, Physics 8, 6 (2015).
8. ^ "Violation of a Leggett–Garg inequality with ideal non-invasive measurements" , George C. Knee, Stephanie Simmons, Erik M. Gauger, John J.L. Morton, Helge Riemann, Nikolai V. Abrosimov, Peter Becker, Hans-Joachim Pohl, Kohei M. Itoh, Mike L.W. Thewalt, G. Andrew D. Briggs & Simon C. Benjamin, Nature Communications 3 606 (2012).
9. ^ "Addressing the Clumsiness Loophole in a Leggett-Garg Test of Macrorealism". Mark M. Wilde and Ari Mizel, Foundations of Physics 42, 256 (2012).
10. ^
11. ^ Classical world arising out of quantum physics under the restriction of coarse-grained measurements. Johannes Kofler and Caslav Brukner. Phys. Rev. Lett. 99, 180403 (2007), ArXiv 0609079 [quant-ph] Sept. 2006 https://arxiv.org/abs/quant-ph/0609079
12. ^ The conditions for quantum violation of macroscopic realism. Johannes Kofler and Caslav Brukner. Phys. Rev. Lett. 101, 090403 (2008), ArXiv 0706.0668 [quant-ph] June 2007 https://arxiv.org/abs/0706.0668
13. ^ Extreme quantum entanglement in a superpostion of macroscopically distinct states. David Mermin, Phys. Rev. Lett. 65 1838-1840 (1990)
14. ^ Noise in Mermin's n-particle Bell inequality. Braunstein, S.L. and Mann, A., Phys. Rev. A 47, R2427-R2430 (1993)
15. ^ Leggett–Garg inequalities. C. Emary, N. Lambert, and F. Nori, Rep. Prog. Phys. 77, 016001 (2014). https://arxiv.org/abs/1304.5133
16. ^ Bell type equalities for SQUIDs on the assumptions of macroscopic realism and non-invasive measurability. Gregg Jaeger, Chris Viger and Sahotra Sarkar. Phys. Lett. A 210, 5-10 (1996)
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# predict.genlasso: Make predictions given a genlasso object In genlasso: Path algorithm for generalized lasso problems
## Description
This predict method for the genlasso class makes a prediction for the fitted values at new predictor measurements. Hence it is really only useful when the generalized lasso model has been fit with a nonidentity predictor matrix. In the case that the predictor matrix is the identity, it does the same thing as `coef.genlasso`.
## Usage
```1 2``` ```## S3 method for class 'genlasso' predict(object, lambda, nlam, df, Xnew, ...) ```
## Arguments
`object` object of class "genlasso", or an object which inherits this class (i.e., "fusedlasso", "trendfilter"). `lambda` a numeric vector of tuning parameter values at which coefficients should be calculated. The user can choose to specify one of `lambda`, `nlam`, or `df`; if none are specified, then coefficients are returned at every knot in the solution path. `nlam` an integer indicating a number of tuning parameters values at which coefficients should be calculated. The tuning parameter values are then chosen to be equally spaced on the log scale over the first half of the solution path (this is if the full solution path has been computed; if only a partial path has been computed, the tuning parameter values are spaced over the entirety of the computed path). `df` an integer vector of degrees of freedom values at which coefficients should be calculated. In the case that a single degrees of freedom value appears multiple times throughout the solution path, the least regularized solution (corresponding to the smallest value of lambda) is chosen. If a degrees of freedom value does not appear at all in the solution path, the least regularized solution at which this degrees of freedom value is not exceeded is chosen. `Xnew` a numeric matrix X, containing new predictor measurements at which predictions should be made. If missing, it is assumed to be the same as the existing predictor measurements in `object`. `...` additional arguments passed to predict.
## Value
Returns a list with the following components:
`fit` a numeric matrix of predictor values, one column for each value of lambda. `lambda` a numeric vector containing the sequence of tuning parameter values, corresponding to the columns of `fit`. `df` if `df` was specified, an integer vector containing the sequence of degrees of freedom values corresponding to the columns of `fit`.
`coef.genlasso`
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Tropical Varieties And Gfan
Compute the following. You may use the software Gfan.
1. Find a universal Groebner basis for the ideal $I = \langle ac-b^2, bd-c^2\rangle \subset \mathbb{Q}[a,b,c,d]$. How many cones are there of each dimension in the Groebner fan of $I$? Find $\mathcal{T}(I)$ by hand; then check your answer with Gfan.
2. How many edges are there in the $3$-dimensional permutahedron? How would you use Gfan to answer this?
3. Let $f_1 = x+y+z$ and $f_2=x^2+y^2+xz+yz$. Is $\mathcal{T}(f_1) \cap \mathcal{T}(f_2) = \mathcal{T}(\langle f_1, f_2\rangle)$? First try to answer without a computer, then verify your answer with Gfan.
Hint: Gfan is installed on the servers (ima-aga-…). Read the Gfan manual for instructions and examples. Look up the apps: gfan _bases, _topolyhedralfan, _tropicalhypersurface, _tropicalintersection, _tropicalstartingcone, _tropicaltraverse, _tropicalbruteforce, …
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Sciencemadness Discussion Board » Fundamentals » Miscellaneous » Calculus! For beginners, with a ‘no theorems’ approach! Select A Forum Fundamentals » Chemistry in General » Organic Chemistry » Reagents and Apparatus Acquisition » Beginnings » Responsible Practices » Miscellaneous » The Wiki Special topics » Technochemistry » Energetic Materials » Biochemistry » Radiochemistry » Computational Models and Techniques » Prepublication Non-chemistry » Forum Matters » Legal and Societal Issues » Test Forum
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Author: Subject: Calculus! For beginners, with a ‘no theorems’ approach!
blogfast25
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Flow of a fluid between parallel plates:
This is an interesting problem because of how the differential equation is set up.
We have two very large parallel plates, both stationary. Between them a Newtonian fluid flows in one direction. From prior art we can expect the velocity profile u(y) to look like this:
Now we consider a slice of fluid with thickness Δy:
To derive u(y) we first note that there is no acceleration, only uniform motion:
$$\frac{\partial u}{\partial t}=0$$
This means acc. Newton's second law that the balance of forces acting on a slice of moving fluid of thickness Δy must be zero::
$$\Sigma F_x=0$$
The forces acting on the slice are:
1. P<sub>1</sub>ΔyW and P<sub>2</sub>ΔyW. This pressure difference is of course the driving force for the flow.
2. Two shear forces (top and bottom): LWσ(y) and LWσ(y+Δy).
The balance of forces is:
$$P_1\Delta y W-P_2\Delta y W+LW[\sigma(y+\Delta y)-\sigma(y)]=0$$
Rearranged slightly:
$$\frac{\Delta P}{L}=-\frac{\sigma(y+\Delta y)-\sigma(y)}{\Delta y}$$
Take the limit to get a derivative:
$$\frac{\Delta P}{L}=-\lim_{\Delta y \to 0}\frac{\sigma(y+\Delta y)-\sigma(y)}{\Delta y}$$
$$\frac{\Delta P}{L}=-\frac{d\sigma}{dy}$$
The shear stress for a Newtonian fluid is:
$$\sigma=\mu\frac{du}{dy}$$
Plugged in we get:
$$\frac{\Delta P}{L}=-\mu\frac{d^2u}{dy^2}$$
This is a simple second order differential equation which on integration gives:
$$u=-\frac{\Delta P}{2\mu L}y^2+c_1y+c_2$$
c<sub>1</sub> and c<sub>2</sub> are integration constants, the values of which are obtained by applying the following boundary conditions:
$$y=-\frac{h}{2} \implies u=0, y=+\frac{h}{2} \implies u=0$$
Plugging in the values for c<sub>1</sub> and c<sub>2</sub> we get:
$$u=\frac{\Delta Ph^2}{2L\mu}\Big[\frac14-\Big(\frac{y}{h}\Big)^2\Big]$$
For y = 0, we get:
$$u_{max}=\frac{\Delta Ph^2}{8L\mu}$$
For a infinitely long plates we can optionally write:
$$\frac{\Delta P}{L}=\frac{\partial P}{\partial x}$$
Now we need to find the average velocity
$$\bar{u}$$
from:
$$\bar{u}=\frac{1}{h}\int_{-h/2}^{+h/2}u(y)dy$$
This yields:
$$\bar{u}=\frac{\Delta Ph^2}{12L\mu}$$
And so with a substitution:
$$\large{u(y)=6\bar{u}\Big[\frac14-\Big(\frac{y}{h}\Big)^2\Big]}$$
[Edited on 15-5-2016 by blogfast25]
blogfast25
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Derivation of polynomials
Derivatives, differentials and differentiation
More rules for derivation (and differentiation)
Combining the rules and carrying out substitutions
When a derivative becomes zero (optima)
Optimisation problems
Integration (“anti-derivation”)
Basic rules of integration (indefinite integrals)
Integration by parts
The definite integral
Integration between two boundaries
Surface are under a function’s curve
Integration with Wolfram Alpha’s DSolve function
Simple differential equations
Train stopping problem
Hot Coffee: no need to blow on it!
Syphon: tank emptying time
Hydrogen generator (student problem)
Projectile velocity problem
Partial derivatives
Multi-variate optima
Famous second order DEs
Simple Harmonic Oscillator
The Schrödinger Equation
The Heat Equation
Experimental determination of heat conductivity k
Flow between two parallel plates
blogfast25
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WiFi woes?
The field strength of a WiFi signal in a house is governed by the Helmholtz Equation (here in two dimensions):
$$\nabla^2E+\frac{k^2}{n^2}E=f(x,y)$$
Where E is the signal strength, k the wave number of the used radiation and n the refractive index of the (x,y) space to that radiation. f(x,y) is the source function of the radiation.
$$\nabla^2E=\frac{\partial^2E}{\partial x^2}+\frac{\partial^2E}{\partial y^2}$$
If the (x,y) space is sub-divided into:
$$N\times M$$
... grid points, then the signal strengths can be calculated from a giant system of equations of the form:
$$\frac{E(i+1,j)+E(i-1,j)-2E(i,j)}{\Delta x^2}+\frac{E(i,j+1)+E(i,j-1)-2E(i,j)}{\Delta y^2}+\frac{k^2}{n(i,j)^2}E(i,j)=f(i,j)$$
i and j are the index numbers of the grid points.
Here's a fellow who did just that (and more!)
Look at the complicated room plan, the position of the (red point) WiFi source and the tendrils of 'internet goodness' as they snake through the available space!
The link also points to a related Android app by the same author.
aga
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The plot looks nice.
A overlaid plot of the signal in the reverse direction would be interesting.
(internet, as with all meaningful communications, is a two-way thing).
blogfast25
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I wouldn't know about that: I'm IT averse.
blogfast25
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Second order DEs (continued):
We've seen a few examples of 2nd order DEs, now let's have a slightly more systematic glimpse into simple equations of the form:
$$ay''+by'+cy=0 \:\text{where}\:y=f(x)$$
These are known as 2nd order linear homogeneous DEs (with constant coefficients).
We're obviously looking for a solution function that doesn't really change form even after two derivations.
We propose something like (where the suffix p stands for 'particular'):
$$y_p=e^{\lambda x}$$
Derive:
$$y_p'=\lambda e^{\lambda x}=\lambda y_p$$
and again:
$$y_p''=\lambda^2 e^{\lambda x}=\lambda^2y_p$$
Plug these into the original DV:
$$a\lambda^2y_p+b\lambda y_p+cy_p=0$$
The yp drop out and we're left with:
$$a\lambda^2+b\lambda +c=0$$
This is known as the Characteristic Equation. As a quadratic equation it suggests there are two λ meeting that condition:
Example:
$$y''-y'-6y=0\:\text{where}\:y=f(x)$$
With boundary conditions:
$$y(0)=0,y'(0)=5$$
Characteristic equation:
$$\lambda^2-\lambda-6=0$$
which has two Real roots:
$$\lambda=+2, \lambda=-3$$
A theorem (which will remain untreated here) now tells that the Good solution of the DE is a linear combination, as follows:
$$y=c_1e^{2x}+c_2e^{-3x}$$
To prove this works, let's find the integration constants c1 and c2 from the boundary conditions:
$$y(0)=c_1+c_2=0$$
and:
$$y'=2c_1e^{2x}-3c_2e^{-3x}$$
$$y'(0)=2c_1-3c_2=5$$
Together these equations now yield:
$$c_1=+1,c_2=-1$$
So the full solution of the original DE is:
$$y=e^{2x}-e^{-3x}$$
You can now prove the validity of this solution by plugging it into the original DV.
<hr>
That was all kind of easy because the roots of the Characteristic Equation were both Real.
Next time we'll see what happens when that's not the case.
[Edited on 19-5-2016 by blogfast25]
aga
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Quote: Originally posted by blogfast25 That was all kind of easy
Speak for yourself !
blogfast25
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Quote: Originally posted by aga
Quote: Originally posted by blogfast25 That was all kind of easy
Speak for yourself !
Which part did you find hard?
aga
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$$y_p=e^{\lambda x}$$
$$y_p'=\lambda e^{\lambda x}=\lambda y_p$$
The e^x bit basically.
Natural logs in general.
blogfast25
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Quote: Originally posted by aga $$y_p=e^{\lambda x}$$ $$y_p'=\lambda e^{\lambda x}=\lambda y_p$$ The e^x bit basically. Natural logs in general.
$$e=2.71828...$$
Example:
$$e^3=(2.71828)^3=20.0855$$
In general:
$$e^x, e^{ax}, e^{\lambda x}$$
... are all simple exponential functions (a and λ are Real numbers).
Derivation rule (chain rule applies here):
$$(e^{\lambda x})'=e^{\lambda x}\times (\lambda x)'=\lambda e^{\lambda x}$$
So, if:
$$y_p=e^{\lambda x}$$
$$y_p'=\lambda e^{\lambda x}=\lambda y_p$$
$$y_p''=\lambda^2 e^{\lambda x}=\lambda^2y_p$$
Natural logarithm: the anti-function of ex:
Say if:
$$\ln a=b$$
Then:
$$e^b=a$$
Example:
$$\ln (20.0855)=3$$
Because:
$$e^3=20.0855$$
[Edited on 19-5-2016 by blogfast25]
aga
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Bollocks! That sodding Chain Rule !
So, if i can ever get even close to 'getting it'. the function of x is
$$\lambda x$$
and the function of the function of x is
$$e^{\lambda x}$$
hence, following derivative chain rule :
$$(e^{\lambda x})'=e^{\lambda x}\times (\lambda x)'$$
Why is it not a Single un-chained function of x, where
$$f(x) = (e^{\lambda x})$$
?
Edit:
Is it as simple as the fact that x gets times by lambda = one function, then raising 'e' by the result of that function constitues another function ?
[Edited on 19-5-2016 by aga]
blogfast25
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Quote: Originally posted by aga Bollocks! That sodding Chain Rule ! So, if i can ever get even close to 'getting it'. the function of x is $$\lambda x$$ and the function of the function of x is $$e^{\lambda x}$$ hence, following derivative chain rule : $$(e^{\lambda x})'=e^{\lambda x}\times (\lambda x)'$$ Why is it not a Single un-chained function of x, where $$f(x) = (e^{\lambda x})$$ ? Edit: Is it as simple as the fact that x gets times by lambda = one function, then raising 'e' by the result of that function constitues another function ? [Edited on 19-5-2016 by aga]
You're basically correct.
The Chain Rule formally is as follows:
If f is a function of g(x), so:
$$f[g(x)]$$
Then:
$$\big[f[g(x)]\big]'=\frac{df(x)}{dx}=\frac{df(g(x))}{dg(x)} \times \frac{dg(x)}{dx}$$
Much shorter:
$$f(g)\:\text{where}\:g=g(x)$$
Then:
$$\frac{df}{dx}=\frac{df(g)}{dg} \times \frac{dg(x)}{dx}$$
In the example at hand:
$$g(x)=\lambda x$$
$$f=e^{g(x)}=e^{\lambda x}$$
It also happens to be that for exponential functions:
$$\frac{d(e^g)}{dg}=e^g$$
For example:
$$\frac{d(e^x)}{dx}=e^x$$
The e-functions are unique in that respect.
So:
$$f'=e^{\lambda x} \times (\lambda x)'= e^{\lambda x} \times \lambda=\lambda e^{\lambda x}$$
<hr>
Tomorrow: another approach to the chain rule. So hold fire, Hombre!
[Edited on 20-5-2016 by blogfast25]
blogfast25
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Let’s try and smash those chain gremlins that seem to be wound tightly around your bollocks for once and for all (w/o smashing the bollocks themselves)
Where you wrote:
Quote: Originally posted by aga So, if i can ever get even close to 'getting it'. the function of x is $$\lambda x$$ and the function of the function of x is $$e^{\lambda x}$$
... you’ve got the right idea. Knowing what to do with the chain rule boils down to identifying the ‘master function’ (f!) and the ‘subordinate’ one (g!). This is of course a metaphor, not to be taken literally.
Then we’ll derive the ‘master’ first, then the ‘subordinate’ one.
But forget about the derivation for a minute. For now we’ll concentrate on what, in an actual function, plays the part of f and what plays the part of g.
Here are some examples:
1.
$$f=\sqrt{x^2+5}$$
$$g=x^2+5$$
$$f(g)=\sqrt{g}$$
2.
$$f=(2\ln x+8)^{1/3}$$
$$g=2\ln x+8$$
$$f(g)=g^{1/3}$$
3.
$$f=e^{3x^2}$$
$$g=3x^2$$
$$f(g)=e^g$$
Exercises: using the same method and format determine f(g) in the following cases.
1.
$$f=\cos(3\sqrt{x}+7)$$
2.
$$f=\ln \Big(\frac{1}{1+x}\Big)$$
3.
$$f=[\cos (2x)+3]^4$$
4.
$$f=\sqrt [5]{1-2x}$$
5.
$$f=\sin\Big(\ln x+\frac{1}{x^2}\Big)$$
6.
$$f=\sqrt [3] {2x^4-x^2-3}$$
[Edited on 20-5-2016 by blogfast25]
blogfast25
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aga, are you interested in solving your chain rule problem or did you not see the previous post?
aga
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sorry. didn't see it. will start now.
(so much random shizzle piles up so fast these days).
[Edited on 22-5-2016 by aga]
aga
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1.
$$f=\cos(3\sqrt{x}+7)$$
$$g=3\sqrt{x}+7$$
$$f=\cos(g)$$
2.
$$f=\ln \Big(\frac{1}{1+x}\Big)$$
$$g=\frac{1}{1+x}$$
$$f=\ln (g)$$
3.
$$f=[\cos (2x)+3]^4$$
$$g=\cos (2x)+3$$
$$f=g^4$$
4.
$$f=\sqrt [5]{1-2x}$$
$$g=1-2x$$
$$f=\sqrt [5]g$$
5.
$$f=\sin\Big(\ln x+\frac{1}{x^2}\Big)$$
$$g=\ln x+\frac{1}{x^2}$$
$$f=\sin(g)$$
6.
$$f=\sqrt [3] {2x^4-x^2-3}$$
$$g=2x^4-x^2-3$$
$$f=\sqrt [3] g$$
blogfast25
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Well, these were all 100 % correct, showing you have no problem seeing which is the 'master' and which is the 'subordinate' function.
To derive we now apply the Chain Rule. For my examples so far we had:
1.
$$f=\sqrt{x^2+5}$$
$$g=x^2+5$$
$$f=\sqrt{g}$$
2.
$$f=(2\ln x+8)^{1/3}$$
$$g=2\ln x+8$$
$$f=g^{1/3}$$
3.
$$f=e^{3x^2}$$
$$g=3x^2$$
$$f=e^g$$
<hr>
The Chain Rule says:
$$\frac{df}{dx}=\frac{df}{dg}\times \frac{dg}{dx}$$
So let's apply it to these examples:
1.
$$\frac{df}{dg}=(\sqrt{g})'=\frac12g^{-1/2}=\frac{1}{2\sqrt{x^2+5}}$$
(Edit: changed 'x' to 'g')
$$\frac{dg}{dx}=(x^2+5)'=2x+0=2x$$
$$\implies \frac{df}{dx}=\frac{df}{dg}\times \frac{dg}{dx}=f'(x)=\frac{2x}{2\sqrt{x^2+5}}=\frac{x}{\sqrt{x^2+5}}$$
2.
$$\frac{df}{dg}=(g^{1/3})'=\frac13 g^{-2/3}=\frac{1}{3g^{2/3}}=\frac{1}{3(2\ln x+8)^{2/3}}$$
$$\frac{dg}{dx}=(2\ln x+8)'=2\frac{1}{x}+0=\frac{2}{x}$$
$$\implies f'(x)=\frac{df}{dg}\times \frac{dg}{dx}=\frac{2}{3x(2\ln x+8)^{\frac23}}$$
3.
$$\frac{df}{dg}=\big(e^g\big)'=e^g=e^{3x^2}$$
$$\frac{dg}{dx}=(3x^2)'=3(x^2)'=3(2x)=6x$$
$$\implies f'(x)=6xe^{3x^2}$$
<hr>
Note that for very simple composite functions of the form (where a is a constant):
$$f(ax)$$
... the chain rule also becomes very simply:
$$f=(ax)^n \implies f'=an(an)^{n-1}$$
$$f=\cos(ax) \implies f'=-a\sin(ax)$$
$$f=\sin(ax) \implies f'=a\cos(ax)$$
$$f=e^{ax} \implies f'=ae^{ax}$$
<hr>
Now smash the chain gremlins and apply the Chain Rule to the previous exercises!
'Cheatsheet':
http://www.ambrsoft.com/Equations/Derivation/Derivation.htm
[Edited on 23-5-2016 by blogfast25]
aga
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Almost boiled the noodles with this one :
$$\frac{df}{dg}=(\sqrt{x})'=\frac12g^{-1/2}=\frac{1}{2\sqrt{x^2+5}}$$
I am HOPING it should read :
$$\frac{df}{dg}=(\sqrt{g})'=\frac12g^{-1/2}=\frac{1}{2\sqrt{x^2+5}}$$
If not, this rocket mission might not miss Mars, and actually hit it, which would be an environmental disaster.
Assuming a typo, i'll now Baldly go where no Baldy has gone before ...
blogfast25
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Quote: Originally posted by aga Assuming a typo, i'll now Baldly go where no Baldy has gone before ...
AAAAaaaarrghh!!! TERRIBLE typo alert. Harakiri being contemplated now.
[Edited on 23-5-2016 by blogfast25]
aga
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1.
$$f=\cos(3\sqrt{x}+7)$$
$$f=cos(g)$$
$$g=3\sqrt{x}+7$$
1st derivative should be :
$$(\cos(3\sqrt{x}+7))'.(3\sqrt{x}+7)'$$
$$= -sin(3\sqrt{x}+7).(3x^{\frac 12}+7)'$$
$$= -sin(3\sqrt{x}+7).\frac 32 x^{-\frac 12}$$
Fiddle with just to get sqrt instead of fractional powers
$$= -sin(3\sqrt{x}+7).\frac 3{2 \sqrt x}$$
Then go crazy just to confuse bejeesus out of everyone, including myself :
$$= \frac {-3sin(3\sqrt{x}+7)}{2 \sqrt x}$$
Better pause there for a second to see if it's going OK or has all ballsed up already.
aga
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Quote: Originally posted by blogfast25 AAAAaaaarrghh!!! TERRIBLE typo alert. Harakiri being contemplated now.
The Emperor banned seppuku quite some time ago, so you may not.
blogfast25
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1. is entirely correct. Well done.
5 to go.
aga
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Phew!
Beer load a bit high to attempt more right now.
Will certainly mangle the rest tomorrow a.m.
Out of interest, which form did you end up with as the 'correct' answer ?
blogfast25
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Quote: Originally posted by aga . Out of interest, which form did you end up with as the 'correct' answer ?
All last three expressions were 100 % and merely 'synonymous'.
The last one would be my 'aesthetically' preferred one.
Often the preferred version is decided on by what you'll do with it afterwards.. A first derivative is rarely the end of solving a real problem.
aga
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2.
$$f=\ln \Big(\frac{1}{1+x}\Big)$$
$$g=\frac{1}{1+x}$$
$$f=\ln (g)$$
$$\ln \Big(\frac{1}{1+x}\Big)'.\Big(\frac{1}{1+x}\Big)'$$
$$=(1+x).\Big(\frac{1}{1+x}\Big)'$$
$$\Big(\frac{1}{1+x}\Big)'$$
first by power rule :
$$\Big(\frac{1}{1+x}\Big)' = \Big((1+x)^{-1}\Big)' = -1(x+1)^{-2} = \frac {2}{-1(x+1)} = -\frac{2}{x+1}$$
But it is also kinda chainy, so :
$$g=(x+1), f=\frac 1g$$
$$\Big(\frac{1}{1+x}\Big)'.(x+1)'$$
(x+1)' = 1 so that will come out the same. Back to the Question :
$$-\frac{2}{x+1}.(x+1) = -\frac {2(x+1)}{x+1} = -2$$
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Submit a question to our community and get an answer from real people.
Whats the distance between (-4,-9),(-11,-9)
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7
*Notice that the two points are lying on same vertical line (-9), the difference only is on the x-axis --> l-11 - -4l = l-7l = 7
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whats the area of a square shaped parking lot having a side length of 120 feet
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might wanna know about that i am following you and wanna know how to change your profile photo?
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The distance between (-4,-9),(-11,-9) is 7 units. To calculate, find the distance between the x-coordinates and the y-coordinates. Then, square both your answers, add them and take the square root.
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d= [(x2-x1)^2 + (y2-y1)^2] square root.[{-11- (-4)}^2 + {-9- (-9)}^2] square root =. (-7)^2+(0)^2 = 49 square root = 7 units.
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Home > Conversions (Currency) > Conversion tables from/to Argentine Peso > ARS to NOK Conversion Cheat Sheet (Interactive)
From: Step: Decimals: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 You could also enter the values to convert and print directly on the table
[Formula: NOK = ARS x 0.0896460481333] [Printer friendly] [Norwegian Kroner to Argentine Pesos]
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Argentine Pesos to Norwegian KronerExchange Rate and Conversion Table
ARS = 0.0896460481333 NOK rate as of 2021-12-01 02:00:00 GMT (Source)
How to convert from Argentine Pesos to Norwegian Kroner
Since 1 Argentine Peso is equal to 0.0896460481333 Norwegian Kroner, we could say that n Argentine Pesos are equal to 0.0896460481333 times n Norwegian Kroner. In other words, we could use the following formula:
Norwegian Kroner = Argentine Pesos x 0.0896460481333
For example, let's say that we want to convert 2 Argentine Pesos to Norwegian Kroner. Then, we just replace Argentine Pesos in the abovementioned formula with 2:
Norwegian Kroner = 2 x 0.0896460481333
That is, 2 Argentine Pesos are equal to 0.179292096267 Norwegian Kroner.
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# Multiply Rational Expressions Worksheet
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## Multiplying And Dividing Rational Expressions: Apply Problems
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### Multiplying Rational Expressions Worksheet
By color-coding the widespread factors, it’s clear which ones to remove. Now, I can multiply across the numerators and across the denominators by putting them aspect by side. The second step is canceling, or what I like to name slashing! Once we have ‘slashed’ all the like phrases from the top and bottom, we multiply straight across, however do not multiply anything we slashed! So we’re going to slash (r – 3) over (r – 3) and (r – 3) over (r – 3). Here dividing by \(2/3\) is the same factor as multiplying by \(3/2\) (since \(2/3\) and \(3/2\) are inverses of each other).
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# Vertical Circle motion
1. Jun 4, 2012
### jsmith613
This question is about circular motion in a vertical circle
Question 1:
would I be correct in assuming that the magnitude for tension is dependant on
(a) the weight of the object
(b) the position of the object with respect to the horizontal diameter of the circle
So above the 'horizontal diameter' tension is lower than below the 'horizontal diameter' because a component of weight acts towards the circle centre.
When it lies ON the circumference AT the 'horizontal diameter' ONLY tension provides the centripetal force
BELOW the 'horizontal diameter' tension increases because weight opposes the tension force
Is this all correct?
Question 2:
If you look at the free body diagram attached, mg*sin(θ) is present. What does this compoenent of weight do??
Thanks
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2. Jun 4, 2012
### Staff: Mentor
Yes, but also the speed of the object.
Sounds good, as long as you include the effect of speed.
It creates a tangential acceleration.
3. Jun 4, 2012
### jsmith613
in some cases mgsinθ will be in the same direction as velocity and in others it will be in the exact opposite...I presume this is the force that will cause the speed of the object to vary?
4. Jun 4, 2012
### Staff: Mentor
That is correct.
5. Jun 4, 2012
thanks
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# Physics help.
posted by on .
The 78.0 kg climber is supported in the "chimney" by the friction forces exerted on his shoes and back. The static coefficients of friction between his shoes and the wall, and between his back and the wall, are 0.840 and 0.630, respectively. What is the minimum normal force he must exert? Assume the walls are vertical and that friction forces are both at a maximum. Ignore his grip on the rope.
78*g*.63=F1
78*g*.84=F2
F1+F2=78*g
F2-F1=78*g*(.84-.63)
F2=78*g-F1
78*g-2*F2=78*g*(.84-.63)
F2=78*g*(1-.84+.63)/2
F2= 302 N
F1= 463 N
I thought the answer was 302N but the homework website is telling me that I am wrong. Where did I mess up?
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# Chapter 1 - Whole Numbers - Test: 9
171,000
#### Work Step by Step
1. Multiply without the zeros: 57 x 3 = 171 2. Count the zeros (3) and add to the end: 171,000
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http://www.numbersaplenty.com/144201
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Search a number
144201 = 371677
BaseRepresentation
bin100011001101001001
321022210210
4203031021
514103301
63031333
71140261
oct431511
9238723
10144201
1199382
126b549
1350835
143a7a1
hex23349
144201 has 8 divisors (see below), whose sum is σ = 195264. Its totient is φ = 94640.
The previous prime is 144173. The next prime is 144203. The reversal of 144201 is 102441.
Adding to 144201 its reverse (102441), we get a palindrome (246642).
It is a sphenic number, since it is the product of 3 distinct primes.
It is a cyclic number.
It is not a de Polignac number, because 144201 - 25 = 144169 is a prime.
It is a Curzon number.
It is a plaindrome in base 16.
It is an inconsummate number, since it does not exist a number n which divided by its sum of digits gives 144201.
It is not an unprimeable number, because it can be changed into a prime (144203) by changing a digit.
It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 126 + ... + 551.
It is an arithmetic number, because the mean of its divisors is an integer number (24408).
2144201 is an apocalyptic number.
It is an amenable number.
144201 is a deficient number, since it is larger than the sum of its proper divisors (51063).
144201 is an equidigital number, since it uses as much as digits as its factorization.
144201 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 751.
The product of its (nonzero) digits is 32, while the sum is 12.
The square root of 144201 is about 379.7380676203. The cubic root of 144201 is about 52.4392040007.
The spelling of 144201 in words is "one hundred forty-four thousand, two hundred one", and thus it is an iban number.
Divisors: 1 3 71 213 677 2031 48067 144201
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https://www.stat.math.ethz.ch/pipermail/r-help/2001-October/015906.html
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# [R] Re: Helps on "optimize"
Han Lai hlai at whsun1.wh.whoi.edu
Wed Oct 17 21:05:16 CEST 2001
```Sorry about my missing statement. My r-code is to find F-value that maximizes fish yield.
N0=15.4185,M=0.3,t=t,days=days,wt=wt,st=st,st1=st1 in optimize statement is constants or
constant arrays.
Han Lai wrote:
> Hi,
>
> I am trying to write a r-code that maximizes fish yield, which is
> summing over a total of 11 months (t=c(1:11)) for one cohort. My code
> looks like this:
>
> t <- c(1:11)
> days <- 30*t-15
> yt <- 6.6*exp(log(12.3/6.6)*(days-126.3)/(189.5-126.3))
> wt <- exp(log(0.250206)+2.14418*log(yt))
> v0 <- 4; v1 <- 14
> st <- (yt-v0)/(v1-v0); st[yt<=v0] <- 0; st[yt>=v1] <-1;
> st1 <- cumsum(st)
> yield <- function(F,N0,M,t,days,wt,st,st1) {
> sum(wt*N0*exp(-(t-1)*M-F*st1)*(st*F/(M+st*F))*(1-exp(-M-st*F))) }
>
> optimize(yield,interval=c(0,3),max=T,N0=15.4185,M=0.3,t=t,days=days,wt=wt,st=st,st1=st1)
>
> Everything is o.k. untile the last one line, which gives the following
> error message.
>
> Error in sum(wt * N0 * exp(-(t - 1) * M - F * st1) * (st * F/(M + st *
> :
> Argument "t" is missing, with no default
>
>
> Cheers!
> Han
> Han-Lin.Lai at noaa.gov
-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-
r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html
Send "info", "help", or "[un]subscribe"
(in the "body", not the subject !) To: r-help-request at stat.math.ethz.ch
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```
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# What does an eaves gutter look like when it just starts to overflow?
When a downpipe gets blocked we have to design for overflow conditions.
This normally allows for the water to overflow either the front or the back of the gutter.
We assume the water overflows equally along the full length of the gutter.
We then use the weir formula to calculate the depth of water overflowing. shown as “h” in the above diagrams.
Knowing this depth allows us to adjust the flashing, or the distance below the top of the fascia, to prevent any water from entering the building.
However eaves gutters normally have a slope, so the overflowing edge is not level.
This raises the question, if the downstream end is the lowest, wouldn’t all the water be overflowing there?
Meaning that the code, and all our calculations to date have been wrong?
So, I decided to check it out by using computational fluid Dynamics (CFD).
This is a computer program that plots the motion of every single particle in the water, and gives us a reasonably good idea of what will happen.
Its not the best quality but I had to reduce the file size substantially for it to upload.
So, check it out here…..
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# Cinquefoil knot
(Redirected from 5 1 knot)
Cinquefoil
Common name Double overhand knot
Arf invariant 1
Braid length 5
Braid no. 2
Bridge no. 2
Crosscap no. 1
Crossing no. 5
Genus 2
Hyperbolic volume 0
Stick no. 8
Unknotting no. 2
Conway notation [5]
A-B notation 51
Dowker notation 6, 8, 10, 2, 4
Last /Next 4152
Other
alternating, torus, fibered, prime, reversible
In knot theory, the cinquefoil knot, also known as Solomon's seal knot or the pentafoil knot, is one of two knots with crossing number five, the other being the three-twist knot. It is listed as the 51 knot in the Alexander-Briggs notation, and can also be described as the (5,2)-torus knot. The cinquefoil is the closed version of the double overhand knot.
The cinquefoil is a prime knot. Its writhe is 5, and it is invertible but not amphichiral.[1] Its Alexander polynomial is
${\displaystyle \Delta (t)=t^{2}-t+1-t^{-1}+t^{-2}}$,
its Conway polynomial is
${\displaystyle \nabla (z)=z^{4}+3z^{2}+1}$,
and its Jones polynomial is
${\displaystyle V(q)=q^{-2}+q^{-4}-q^{-5}+q^{-6}-q^{-7}.}$
These are the same as the Alexander, Conway, and Jones polynomials of the knot 10132. However, the Kauffman polynomial can be used to distinguish between these two knots.
The name “cinquefoil” comes from the five-petaled flowers of plants in the genus Potentilla.
Edible cinquefoil knot.
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1
Q:
# Josh borrowed \$250 from his mother to buy an electric scooter. Josh will pay her back in 1 year with 3% simple annual interest. How much interest will Josh pay?
A) 7.50 B) 8.50 C) 9.50 D) 10.50
Explanation:
I=prt/100
Subject: Compound Interest
Exam Prep: Bank Exams
Job Role: Bank PO
Q:
If Rs. 2,000 is invested at the rate of 20% per annum, compounded half-yearly, then the amount after 18 months will be:
A) Rs.2628 B) Rs.2662 C) Rs.2600 D) Rs.3200
Explanation:
Filed Under: Compound Interest
Exam Prep: Bank Exams
1 587
Q:
Calculate the principal if an amount of Rs. 441 is received on compound interest at the rate of 5% per annum after 2 years.
A) Rs.400 B) Rs.390 C) Rs.380 D) Rs.350
Explanation:
Filed Under: Compound Interest
Exam Prep: Bank Exams
2 433
Q:
Gitesh took a loan for 4 years at 5% Compound Interest. If the total interest paid was Rs. 431.01, Calculate the principal.
A) Rs. 2000 B) Rs. 2050 C) Rs. 2100 D) Rs. 2150
Explanation:
Filed Under: Compound Interest
Exam Prep: Bank Exams
0 407
Q:
A sum of Rs 20000 becomes Rs 32000 in 12 years, when invested in a scheme of simple interest. If the same sum is invested in a scheme of compound interest with same yearly interest rate (compounding of interest is done yearly), then what will be the amount (in Rs) after 2 years?
A) 21750 B) 22050 C) 23250 D) 24650
Explanation:
Filed Under: Compound Interest
Exam Prep: Bank Exams
0 333
Q:
A sum of Rs 4000 becomes Rs 5800 in 3 years, when invested in a scheme of simple interest. If the same sum is invested in a scheme of compound interest with same yearly interest rate (compounding of interest is done yearly), then what will be the amount (in Rs) after 2 years?
A) 4430 B) 5450 C) 5290 D) 4970
Explanation:
Filed Under: Compound Interest
Exam Prep: Bank Exams
1 366
Q:
The maturity value of a certain sum after two years at 20% p.a. interest compounded annually is Rs. 14,400/-. Find the Principal amount.
A) Rs.9000 B) Rs.9500 C) Rs.10,000 D) Rs.10,500
Explanation:
Filed Under: Compound Interest
Exam Prep: Bank Exams
1 372
Q:
Shams invested Rs. 5000 at 10% per annum compound interest. After n years, Shams received Rs. 1655 more. Find the value of n.
A) 3 years B) 4 years C) 2 years D) 1 year
Explanation:
Filed Under: Compound Interest
Exam Prep: Bank Exams
4 769
Q:
A sum of Rs 6000 becomes Rs 7200 in 2 years, when invested in a scheme of simple interest. If the same sum is invested in a scheme of compound interest with same yearly interest rate (compounding of interest is done yearly), then what will be the amount (in Rs) after 3 years?
A) 7434 B) 8244 C) 7864 D) 7986
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### Introduction
• Algorithm is a basic tool and is the first step of problem-solving that helps in getting the effective solution of a simple problem.
### Definition
• An algorithm is a set of sequential step by step procedure, each of which has a clear-cut meaning and usually written in ordinary/natural English Language to solve a given problem in finite number of steps.
• An algorithm is defined as a complete, unambiguous, finite number of logical steps for solving a specific problem.
### Characteristics/Properties
Every algorithm must satisfy the following criteria i.e., a typical algorithm has following characteristics –
(i) Input: An algorithm must take at least one or more well defined input values.
(ii) Output: An algorithm must provide at least one or more output values.
(iii) Definiteness : No instructions should be repeated i.e., each step of the algorithm must be precisely and unambiguously stated.
(iv) Finiteness : An algorithm must terminate in a finite number of steps.
(v) Effectiveness : An algorithm should be simple and focused on its main objectives. Each step-in algorithm must be effective.
(vi) Generality : The algorithm must be complete in itself so that it can be used to solve problems of a specific type for any input data.
### Features
• No matter what the input values are, an algorithm terminates after executing a finite number of instructions.
• It is a step-by-step instruction for solving a particular task in finite amount of time.
• Algorithm is an effective method for solving a problem expressed as finite sequence of instructions.
• It may be possible to solve to problem in more than one way, resulting in more than one algorithm. The best choice of various algorithms depends on the factors like reliability, accuracy and easy to modify. The most important factor in the choice of algorithm is the time requirement to execute it, after writing code in High-level language with the help of a computer.
• The algorithm which will need the least time when executed is considered the best.
• The algorithms are very easy to understand because it is step wise process.
• Algorithm is programming language independent.
• Algorithm makes the problem simple, clear, correct and effective.
• Every step in an algorithm has its own logical sequence so it is easy to debug.
• An algorithm acts as a blueprint of a program and helps during program development.
• It is easy and good practice to first develop an algorithm of a problem and then convert it into a respective flowchart and then into a computer program.
• Algorithm is mostly effective for simple problems.
• Algorithms use text-based tools.
• Algorithms are time-consuming.
• Complex tasks are difficult to handle using algorithms.
• Difficult to show complex branching and looping in algorithms.
• Understanding complex logic through algorithms feels very difficult.
### Rate of Growth/Classification of an Algorithm
• Growth rate is used to analyse the algorithm.
• The greater the amount of data, larger the number of resources required by an algorithm. Therefore, there is a resource growth rate for a piece of code in the form of function f(n).
• Commonly used rates of growth to analyse the algorithm are –
(i) Constant function algorithm:
• A constant resource need is one where the resource need does not grow as the size of input data grow.
• It is the simplest function with some fixed constant c such as c = 1 or c = 5 etc.
• The constant function is calculated as: –
f(n) = c i.e., for any number n, the constant function f(n) assigns the value c.
• The constant function is used to add two numbers, initialising value to a variable, comparing two numbers.
• Here, the next instructions of most programs are executed once or at most only a few times. If all the instructions of a program have this property, we say that its running time is a constant.
• The graph of such growth rate is represented by a horizontal line.
(ii) Logarithmic function algorithm:
• Logarithmic growth rate is a growth rate where resource need increases by one unit each time the data is doubled.
• The logarithmic function is calculated as: –
f(n) = logb n for constant b > 1
• This function is defined as x= logb n if bx = n, where value b is called base of the logarithm (2 or 10).
• When the running time of a program is logarithmic, the program gets slightly slower as n grows.
• The running time commonly occurs in programs that solve a big problem by transforming it into a smaller problem, cutting the size by some constant fraction.
• When the value of n is a million, log n is a doubled whenever n doubles, log n increases by a constant, but log n does not double until n increases to n2.
(iii) Linear function algorithm:
• The linear function calculated as:
f(n) = n
where an input value n, the linear function f assigns the value of n itself.
• A linear growth rate of an algorithm is a growth rate where the resource needs and amount of data directly proportional to each other.
• The growth rate of linear function is represented by a horizontal line.
• When the running time of a program is linear, it is generally the case that a small amount of processing is done on each input element.
• This is the optimal situation for an algorithm that must process n inputs.
• The example of linear function includes comparing a number to each and every element of size n require n comparisons.
(iv) n. Log n function algorithm:
• This type of function is calculated as:
f(n) = n. log n
• The function that assigns to an input n, the value of n times the logarithm base-two of n.
• A log linear growth rate is slightly curved, i.e., for lower values than higher ones.
• The running time of this algorithm arises for algorithms but solve a problem by breaking it up into smaller sub-problems, solving them independently, and then combining the solutions.
• Here, when n doubles, the running time more than doubles.
• The example of log linear include possible algorithm for sorting n numbers.
• The quadratic function is calculated as:
f(n) = n2
• The function assigns an input value n, the function f assigns the product of n with itself.
• When the running time of an algorithm is quadratic, it is practical for use only on relatively small problems.
• Quadratic running times typically arise in algorithms that process all pairs of data items (mostly n a double nested loop) whenever n doubles, the running time increases four-fold.
• The example of this algorithm includes nested loops, where inner loop implements linear number of instructions and outer loop implemented linear number of times.
(vi) Cubic function algorithm:
• The cubic function is calculated as:
f(n) = n3
The function assigns an input value n the product of n with itself three times.
• This algorithm process triples of data items (mostly in a triple – nested loop) has a cubic running time and is practical for use only on small problems.
• Here, whenever n doubles, the running time increases eight-fold.
• The example includes Matrix Multiplication with three loops.
(vii) Exponential function algorithm:
• The exponential growth rate is one where extra unit of data requires double number of resources.
• It is calculated as:
f(n) = bn
where b is a constant called base and n is called exponent.
• This algorithm with exponential running time is likely to be appropriate for practical use.
• Such algorithms arise naturally as “brute–force” solutions to problems.
• In this algorithm, whenever n doubles, the running time squares.
NB:
The use of n above may be the number of data items to be processed or degree of polynomial or the size of the file to be sorted or searched or the number of nodes in a graph etc.
The most common computing order of complexity are –
O(1) > O(log2 n) > O(n) > O(n. log2 n) > O(n2) > O(n3) > O(2n) > n! > nn
Categories: Algorithm & Design
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New Online Book! Handbook of Mathematical Functions (AMS55)
Conversion & Calculation Home >> Measurement Conversion
Measurement Converter
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: ```skein = 109.728 meter (length) ``` Related Measurements: Try converting from "skein" to arpentcan, arpentlin, astronomical unit, barleycorn, bolt (of cloth), chain (surveyors chain), digitus (Roman digitus), earth to moon (mean distance earth to moon), ell, engineers chain, foot, gradus (Roman gradus), ken (Japanese ken), nail (cloth nail), nautical mile, palm, pica (typography pica), sazhen (Russian sazhen), stadia (Greek stadia), survey foot, or any combination of units which equate to "length" and represent depth, fl head, height, length, wavelength, or width. Sample Conversions: skein = 60,480 agate (typography agate), .00247754 arpentcan, 7.33E-10 astronomical unit, 3 bolt (of cloth), 172,800 bottom measure, 5,917.81 digitus (Roman digitus), 1.10E+17 fermi, 4,937.14 finger, 148.15 gradus (Roman gradus), 59.28 Greek fathom, 1,422.61 Greek palm, 1.16E-14 light yr (light year), 545.45 link (surveyors link), 4,320,000 mil, 312,206.4 point (typography point), .02794337 ri (Japanese ri), 21.82 rod (surveyors rod), .07415036 Roman mile, 362.11 shaku (Japanese shaku), .59275521 stadia (Greek stadia).
Feedback, suggestions, or additional measurement definitions?
Please read our Help Page and FAQ Page then post a message or send e-mail. Thanks!
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Hack 12 Find Relevant Channels and Servers
If you're new to IRC, one of the biggest problems is working out which network to use. Find a perfect match with one of the search engines for IRC channels.
If you haven't got a clue which IRC network to use, picking a random one is not always the best idea. You may find that the one you choose doesn't cater to your interests. A surefire method of picking the right network is to use an IRC search engine.
One popular IRC search engine can be found at http://irc.netsplit.de/channels. This is maintained by Andreas Gelhausen and uses a collection of IRC bots to connect to hundreds of IRC networks and collect data.
3.2.1 Finding Help on IRC
Let's assume that you are new to IRC and need some help, so you enter "irc for beginners" into the search box and click on the Search button. This will search through all of the IRC networks it knows about, looking for channels with topics that match your search terms. The results are shown in Figure 3-1.
Figure 3-1. Using the IRC search engine
The search results are presented as a list of all channels that have topics containing the words you searched for. Each row shows the name of the channel, the name of the network, the time the data was collected, the number of users in the channel, and the topic of the channel. From these results, you can decide which channel you want to join.
If you have an IRC client installed, you can click on the "irc://" link beneath the channel name to automatically connect to that server and join that channel.
3.2.2 Other Interesting Statistics
You can also use the IRC search engine at http://irc.netsplit.de/channels to view statistics about networks and servers. Figure 3-2 shows the growth of the freenode IRC network over the past five years.
Figure 3-2. Plotting the growth of the freenode IRC network
This graph shows the number of users and channels over time. Why not see how your favorite IRC network has grown?
Alex North
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# According to some economists, the July decrease in
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Director
Joined: 26 Mar 2006
Posts: 631
According to some economists, the July decrease in [#permalink]
### Show Tags
13 Jun 2006, 01:33
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According to some economists, the July decrease in umemployment so that it was the lowest in two years suggests that the gradual improvement in job market is continuing.
(A) as underlined
(b) so that it was the lowest two-year rate
(c) to what would be the lowest in two years
(d) to a two-year low level
(e) to the lowest level in two years
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Intern
Joined: 12 Jun 2006
Posts: 48
### Show Tags
13 Jun 2006, 01:57
This would have been a tough one for me on the actual GMAT.....took me a while to figure out.....
I would go with Choice (D) because this seems like the best fit...
(A) and (B) start with 'so that' does'nt fit the sentence right.
(C) Is wordy and in the future tense.
(E) the word 'lowest' is a lil extreme.
SVP
Joined: 30 Mar 2006
Posts: 1728
### Show Tags
13 Jun 2006, 02:20
Will go with D.
A/B/C are awkward.
E - lowest is incorrect as only two years are being compared not all the years .
VP
Joined: 07 Nov 2005
Posts: 1118
Location: India
### Show Tags
13 Jun 2006, 02:31
Going to pick E.
A and B are out b'coz of ' so that it '.Also 'it' seems ambigous.
In C 'would be' makes the construction awkward.
In D , ' to a two year low ' would have been right.
Director
Joined: 27 Feb 2006
Posts: 622
### Show Tags
13 Jun 2006, 02:37
^ E ^
Beyond700 wrote:
According to some economists, the July decrease in umemployment so that it was the lowest in two years suggests that the gradual improvement in job market is continuing.
(A) as underlined
(b) so that it was the lowest two-year rate
(c) to what would be the lowest in two years
(d) to a two-year low level
(e) to the lowest level in two years
A) Out because "so that" and ambigious "it"
B) Same with A
C) "would" is not appropriate here - future ???
D) a two-year low level is a general term. we are looking for a specific expression here !!
Director
Joined: 16 Aug 2005
Posts: 938
Location: France
### Show Tags
13 Jun 2006, 06:22
Beyond700 wrote:
According to some economists, the July decrease in umemployment so that it was the lowest in two years suggests that the gradual improvement in job market is continuing.
(A) as underlined
(b) so that it was the lowest two-year rate
(c) to what would be the lowest in two years
(d) to a two-year low level
(e) to the lowest level in two years
A - use of was incorrect
B - same as A
C - 'what would be' -awkward
D - two-year low level ??? what else can I say
E - should be it
So, my bet is on E
_________________
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Director
Joined: 26 Mar 2006
Posts: 631
### Show Tags
13 Jun 2006, 18:52
OA indeed is 'E'
But OE is bit different than the ones shared in this thread..
VP
Joined: 07 Nov 2005
Posts: 1118
Location: India
### Show Tags
13 Jun 2006, 21:19
Beyond700 wrote:
OA indeed is 'E'
But OE is bit different than the ones shared in this thread..
Can you please post the OE
.
VP
Joined: 02 Jun 2006
Posts: 1260
### Show Tags
14 Jun 2006, 10:51
E... continuing improvement would imply current. to the lowest level implies current.
Senior Manager
Joined: 16 Apr 2006
Posts: 275
### Show Tags
15 Jun 2006, 07:01
To me D also looked correct as it's more concise. But OA is E.
Director
Joined: 26 Mar 2006
Posts: 631
### Show Tags
15 Jun 2006, 23:53
buzzgaurav wrote:
Beyond700 wrote:
OA indeed is 'E'
But OE is bit different than the ones shared in this thread..
Can you please post the OE
.
There you go with OE..
E, the best choice,employs idiomatic construction and uses the precise decrease...
to the lowest level.
Choices A and B are faulty construction. The adverbial so that can modify verbs(eg:-
decreased) but not nouns (eg: the decrease). The meaning of lowest two-year rate in B
is unclear;in any event the phrase distorts the intended meaning of lowest in two years.
In A and B, the referent of it is unclear, as the pronoun could refer to either
unemployment or decrease.
Choice 'C' improperly uses would be to describe a situation that is presented as a current
and known fact. Also, there is no noun for lowest to modify;clearly "the lowest decrease"
is not intended.
In D, the phrase two-year low level is unidiomatic, as well as unclear in its intended meaning
VP
Joined: 07 Nov 2005
Posts: 1118
Location: India
### Show Tags
16 Jun 2006, 01:30
Beyond700 wrote:
buzzgaurav wrote:
Beyond700 wrote:
OA indeed is 'E'
But OE is bit different than the ones shared in this thread..
Can you please post the OE
.
There you go with OE..
E, the best choice,employs idiomatic construction and uses the precise decrease...
to the lowest level.
Choices A and B are faulty construction. The adverbial so that can modify verbs(eg:-
decreased) but not nouns (eg: the decrease). The meaning of lowest two-year rate in B
is unclear;in any event the phrase distorts the intended meaning of lowest in two years.
In A and B, the referent of it is unclear, as the pronoun could refer to either
unemployment or decrease.
Choice 'C' improperly uses would be to describe a situation that is presented as a current
and known fact. Also, there is no noun for lowest to modify;clearly "the lowest decrease"
is not intended.
In D, the phrase two-year low level is unidiomatic, as well as unclear in its intended meaning
Just as I thought ...Thanks for the OE.
GMAT Club Legend
Joined: 01 Oct 2013
Posts: 10158
Re: According to some economists, the July decrease in [#permalink]
### Show Tags
10 Jan 2016, 05:35
Hello from the GMAT Club VerbalBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
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Re: According to some economists, the July decrease in [#permalink] 10 Jan 2016, 05:35
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# NEET Questions Solved
Five resistances are connected as shown in the figure. The effective resistance between the points A and B is
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(3) 15 Ω
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## 12 Calculation Examples
ID Question Answer 1. What is 7% of 89? 6.23 2. Which % 7 out of 76? 9.21% 3. 15 is 25% of what? 60 4. 6 is what % of 23? 26.09% 5. What is 19 increased by 28%? 24.32 6. What is 26 decreased by 53%? 12.22 7. What % of 29 is 15? 51.72% 8. 65% of what is 117? 180 9. 72% of 55 is what? 39.60 10. 51 of what is 98% 52.04 11. What of 67 is 38%? 25.46 12. 31 of 89 is what %? 34.83%
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Solutions Company Statit Training Home
Control Charts for Skewed Time-Based Data Q: I would like to monitor the duration of time that it takes to process a customer service call. However, a histogram of my data reveals that there are often some calls which take much longer than most of the other calls. Will this hurt my control charts? If so, what else can I do? A: Very often, time-related events are somewhat problematical since they have a minimum at zero and are probably clustered around a particular time, but have some valid data points with very long service times. Since this type of data is skewed and non-normal, conventional control charts probably should not be used on the raw data. In particular, an individuals (X) chart should not be used when there are serious departures from normality. Histogram of Original (Skewed) Data Probability Plot of Original (Skewed) Data However, skewed data can be transformed to make it more nearly normal. You can check for normality using the normal probability plot and looking for a nearly-straight line. A common transformation for highly right-skewed data is the fourth root of the original data: In Statit, you can easily create a transformed variable using the Compute/Transform command to create a new variable which is equal to the fourth root of the original variable. This is equal to the number raised to the one-fourth power, i.e., x**0.25. After transforming, you should check for normality again. If the data appears normal, it would then be proper to make an individuals chart or -chart on the transformed data. This should give you some information as to whether service times are "in control." You can "back transform" the values of the center line and control limits into real times again, if desired, to aid in the practical interpretation of these values. Histogram of Transformed Data Probability Plot of Transformed Data If the 4th root transformation doesn’t work well to handle the non-normality, you may need to try other transformations, such as cube root, square root, fifth root, or log, depending on the particular data. Finding the "best" transformation for a data set often takes a bit of experimentation.
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# QlikView Scripting
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Not applicable
I need to do a cumulative sum in the resident table as follows
tab1
a, b, c, d
a, 201101, 1, 0
a, 201101, 1, 1
a, 201102, 1, 1
b, 201101, 1, 0
b, 201102, 1, 1
c, 201101, 1, 0
c, 201102, 1, 0
result
a, b, c, d, e, f
a, 201101, 1, 0, 1, 0
a, 201101, 1, 1, 2, 1
a, 201102, 1, 1, 3, 2
b, 201101, 1, 0, 1, 0
b, 201102, 1, 1, 2, 1
c, 201101, 1, 0, 1, 0
c, 201102, 1, 0, 2, 0
1 Solution
Accepted Solutions
Not applicable
## Re: Cumulative Sum in Load
Hi Robson,
You can see the attached example.
I hope is what you want.
Best regards.
3 Replies
MVP
## Re: Cumulative Sum in Load
Ri Robson,
I think it could look like this
if(peek(a) =a, peek(e)+c,c) as e,
if(peek(a) =a, peek(f)+d,d) as f
resident tab1 order by a ASC, b ASC;
i.e. order your Input table by a, b ascending (like in your example, but force it).
Check if first column value is identical to previous value, then sum up, else (first appearance of value) start with initial value.
I hope I understood your requirement,
Stefan
Not applicable
## Re: Cumulative Sum in Load
Hi Robson,
You can see the attached example.
I hope is what you want.
Best regards.
Not applicable
## Re: Cumulative Sum in Load
Thanks guys!
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Javascript program to sum the bits in same positions
Hello Everyone,
We can sum the bits in same positions for all the numbers and take modulo with 3. The bits for which sum is not multiple of 3, are the bits of number with single occurrence.
Let us consider the example array {5, 5, 5, 8}. The 101, 101, 101, 1000
Sum of first bits%3 = (1 + 1 + 1 + 0)%3 = 0;
Sum of second bits%3 = (0 + 0 + 0 + 0)%3 = 0;
Sum of third bits%3 = (1 + 1 + 1 + 0)%3 = 0;
Sum of fourth bits%3 = (1)%3 = 1;
Hence number which appears once is 1000
Note: this approach wont work for negative numbers
Add each number once and multiply the sum by 3, we will get thrice the sum of each element of the array. Store it as thrice_sum. Subtract the sum of the whole array from the thrice_sum and divide the result by 2. The number we get is the required number (which appears once in the array).
Array [] : [a, a, a, b, b, b, c, c, c, d]
Mathematical Equation = ( 3*(a+b+c+d) – (a + a + a + b + b + b + c + c + c + d) ) / 2
In more simple words: ( 3*(sum_of_array_without_duplicates) – (sum_of_array) ) / 2
let arr[] = {12, 1, 12, 3, 12, 1, 1, 2, 3, 3}
Required no = ( 3*(sum_of_array_without_duplicates) - (sum_of_array) ) / 2
= ( 3*(12 + 1 + 3 + 2) - (12 + 1 + 12 + 3 + 12 + 1 + 1 + 2 + 3 + 3))/2
= ( 3* 18 - 50) / 2
= (54 - 50) / 2
As we know that set does not contain any duplicate element,
But, std::set is commonly implemented as a red-black binary search tree. Insertion on this data structure has a worst-case of O(log(n)) complexity, as the tree is kept balanced.
Given an array where every element occurs three times, except one element which occurs only once. Find the element that occurs once. Expected time complexity is O(n) and O(1) extra space.
Examples:
Input: arr[] = {12, 1, 12, 3, 12, 1, 1, 2, 3, 3}
Output: 2
In the given array all element appear three times except 2 which appears once.
Input: arr[] = {10, 20, 10, 30, 10, 30, 30}
Output: 20
In the given array all element appear three times except 20 which appears once.
We can use sorting to do it in O(nLogn) time. We can also use hashing, it has the worst case time complexity of O(n), but requires extra space.
The idea is to use bitwise operators for a solution that is O(n) time and uses O(1) extra space. The solution is not easy like other XOR based solutions, because all elements appear odd number of times here. Run a loop for all elements in array. At the end of every iteration, maintain following two values.
ones: The bits that have appeared 1st time or 4th time or 7th time … etc.
twos: The bits that have appeared 2nd time or 5th time or 8th time … etc.
Finally, we return the value of ‘ones’
How to maintain the values of ‘ones’ and ‘twos’?
‘ones’ and ‘twos’ are initialized as 0. For every new element in array, find out the common set bits in the new element and previous value of ‘ones’. These common set bits are actually the bits that should be added to ‘twos’. So do bitwise OR of the common set bits with ‘twos’. ‘twos’ also gets some extra bits that appear third time. These extra bits are removed later.
Update ‘ones’ by doing XOR of new element with previous value of ‘ones’. There may be some bits which appear 3rd time. These extra bits are also removed later.
Both ‘ones’ and ‘twos’ contain those extra bits which appear 3rd time. Remove these extra bits by finding out common set bits in ‘ones’ and ‘twos’.
We can sum the bits in same positions for all the numbers and take modulo with 3. The bits for which sum is not multiple of 3, are the bits of number with single occurrence.
Let us consider the example array {5, 5, 5, 8}. The 101, 101, 101, 1000
Sum of first bits%3 = (1 + 1 + 1 + 0)%3 = 0;
Sum of second bits%3 = (0 + 0 + 0 + 0)%3 = 0;
Sum of third bits%3 = (1 + 1 + 1 + 0)%3 = 0;
Sum of fourth bits%3 = (1)%3 = 1;
Hence number which appears once is 1000
Note: this approach wont work for negative numbers
Below is the implementation of above approach:
`<script>`
` ` `// Javascript program to find the element`
` ` `// that occur only once`
` ` `let INT_SIZE = 32;`
` `
` ` `function` `getSingle(arr, n)`
` ` `{`
` `
` ` `// Initialize result`
` ` `let result = 0;`
` ` `let x, sum;`
` ` `// Iterate through every bit`
` ` `for` `(let i = 0; i < INT_SIZE; i++)`
` ` `{`
` ` `// Find sum of set bits at ith position in all`
` ` `// array elements`
` ` `sum = 0;`
` ` `x = (1 << i);`
` ` `for` `(let j = 0; j < n; j++)`
` ` `{`
` ` `if` `(arr[j] & x)`
` ` `sum++;`
` ` `}`
` ` `// The bits with sum not multiple of 3, are the`
` ` `// bits of element with single occurrence.`
` ` `if` `((sum % 3) != 0)`
` ` `result |= x;`
` ` `}`
` ` `return` `result;`
` ` `}`
`// Driver code`
` ` `let arr = [ 12, 1, 12, 3, 12, 1, 1, 2, 3, 2, 2, 3, 7 ];`
` ` `let n = arr.length;`
` ` `document.write(` `"The element with single occurrence is "` `+ getSingle(arr, n));`
`</script>`
Output
The element with single occurrence is 7
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CC-MAIN-2023-40
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latest
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http://cueflash.com/decks/Chemistry_Review_2
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# cueFlash
## Glossary of Chemistry Review 2
### Deck Info
#### Other Decks By This User
Boyle's Law
p1v1=p2v2 temp is constant; P is inversly proportional to v
Charle's Law
v1/t1=v2/t2 pressure is held constant; V is directly proportional to T
gay-lussac's Law
P1/t1=p2/t2 volume is constant; P is directly proportional to T
Combined Gas Law
p1v1/T1=P2V2/T2
ideal gas law
PV=nRT R=.0821
Dalton's law
Pt=Pa+Pb+Pc
If a flexible container with 2.00 l of gas at 20 degrees celsius, what will happen to the container? What law are you using in this problem?
2.14 l Charle's law;
i the pressure is doubled, what will happen to the volume of a as, assuming the temperature remains constant? Which gas law is this?
the volume will decrease by 1/2; boyle's law
If the volume of a as is tripled, what must have happened to the kelvin temperature, if the pressure remained constant? Which gas law is this?
Temp tripledCharle's law,
If the pressure is tripled and the volume remains constant, what happens to the Kelvin temperature?
it triples as well, gay lussac law
If the pressure of one mole o gas is 1.01 atm and the volume is 22.5 L, what must the temperature of the gas be? Which law do you use to solve this?
227K, Ideal gas law
if the initial pressure is 760 Tor, the initial volume is 23ml and te initial temp is 24 degrees celsius, what will the new temperature be i the pressure is changed to 778 torr, the volume is changed to 56 ml? Which gas law are you using
combined gas, 740 k
list the relative particle size, effect of light, effect of gravity, effect of filtration, and if it's homogeneous or hetero
solution
invisible, no effect of light, gravity, or filtration, homogeneous
list the relative particle size, effect of light, effect of gravity, effect of filtration, and if it's homogeneous or hetero
Colloid
small size, tindall effect, no gravity effect and maybe a filtration effect, hetero
list the relative particle size, effect of light, effect of gravity, effect of filtration, and if it's homogeneous or hetero
Suspension
large size,no effect with light, effected by gravity and filtration, heterogeneous
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http://www.convertit.com/Go/Beverageonline/Measurement/Converter.ASP?From=sazhen&To=depth
| 1,632,177,108,000,000,000
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New Online Book! Handbook of Mathematical Functions (AMS55)
Conversion & Calculation Home >> Measurement Conversion
Measurement Converter
(Help)
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: ```Russian sazhen = 2.10312 length (length) ``` Related Measurements: Try converting from "sazhen" to agate (typography agate), arpentcan, Biblical cubit, cable length, cloth finger, earth to moon (mean distance earth to moon), ell, football field, Greek cubit, Greek fathom, Greek palm, league, light yr (light year), mile, nail (cloth nail), nautical league, nautical mile, point (typography point), stadia (Greek stadia), UK mile (British mile), or any combination of units which equate to "length" and represent depth, fl head, height, length, wavelength, or width. Sample Conversions: sazhen = .0575 bolt (of cloth), 18.4 cloth finger, 9.2 cloth quarter, 1.15 fathom, 6.9 foot, 2.84 gradus (Roman gradus), 1.14 Greek fathom, 82.8 inch, .00326241 li (Chinese li), .00004984 marathon, .00130682 mile, .00113559 nautical mile, .00037338 parasang, 6.82E-17 parsec, 496.8 pica (typography pica), 7.11 Roman foot, 6.94 shaku (Japanese shaku), .01916667 skein, 6.9 survey foot, 2.3 yard.
Feedback, suggestions, or additional measurement definitions?
Please read our Help Page and FAQ Page then post a message or send e-mail. Thanks!
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| 2.84375
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CC-MAIN-2021-39
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http://math.stackexchange.com/questions/tagged/integer-lattices+modules
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| 228,202,121
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# Tagged Questions
2answers
441 views
### Elementary proof that if $A$ is a matrix map from $\mathbb{Z}^m$ to $\mathbb Z^n$, then the map is surjective iff the gcd of maximal minors is $1$
I am trying to find an elementary proof that if $\phi$ is a linear map from $\mathbb{Z}^n\rightarrow \mathbb{Z}^m$ represented by an $m \times n$ matrix $A$, then the map is surjective iff the gcd ...
2answers
227 views
### Show that the matrix $A$ with integer entries is injective on the reals to $\mathbb{R}^m$ iff it is injective on the integer lattice.
Show that the $m \times n$ matrix $A$ with integer entries is an injective linear map from $\mathbb{R}^n$ to $\mathbb{R}^m$ iff it is injective as a linear map from $\mathbb{Z}^n$ to $\mathbb{Z}^m$. ...
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| 2.875
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https://chem.libretexts.org/Core/Analytical_Chemistry/Analytical_Sciences_Digital_Library/JASDL/Courseware/Chemical_Equilibrium/04_Instructor's_Manual/02_Acid-Base_Chemistry/07_Solutions_of_Polyprotic_Acid%2F%2FBase_Systems%2C_Problem_A
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| 616,131,476
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# Solutions of Polyprotic Acid/Base Systems, Problem A
### In-class Problem Set #3
1. Calculate the pH of a 0.127 M solution of ascorbic acid.
Groups quickly see that ascorbic acid has two pKa values.
What does it mean if something has two pKa values? What two chemical expressions represent the two equilibria that are occurring?
Students may be tempted at first to ignore the presence of a second pKa value, but encourage them to investigate what it implies about the acid. Once they have realized that there are two dissociable protons, ask them to write the correct chemical expressions making sure to include the proper charges. This should take about ten minutes. Some groups then use Ka1 to solve for a concentration of H3O+ and Ka2 to solve for a second value of H3O+. If they do so, they realize that the second value is much smaller than the first. But they also apparently have a solution with two different concentrations of H3O+.
What do we know about [H3O+] in the two expressions?
Remind the students that since only one pH value can be measured, the concentration of H3O+ in the two equations must be equal.
Write an expression for [H3O+] in terms of the ascorbic acid species.
Students should have no trouble recognizing that for every one mole of ascorbic acid that is deprotonated, one mole of H3O+ is produced, but they may have some trouble understanding that for every one mole of the fully deprotonated ascorbate formed, two moles of H3O+ are formed.
What is x for the Ka1 expression?
What is x for the Ka2 expression? What does this tell us about the second reaction?
Students should realize very quickly that the second x value is equal to the Ka2. With an x value this small, they should recognize that the second reaction is insignificant and that solving for the pH using just the pKa1 is sufficient. Spend several minutes talking about the general differences between pKa values and when the second one can be ignored. Make sure students are clear that this was an example that began with the fully protonated species of a polyprotic acid.
What is the pH of the solution?
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| 2.828125
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https://www.mathworks.com/matlabcentral/cody/problems/120-radius-of-a-spherical-planet/solutions/231086
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Cody
# Problem 120. radius of a spherical planet
Solution 231086
Submitted on 16 Apr 2013 by Dan
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 = 4*pi; y_correct = 1; assert(isequal(your_fcn_name(x),y_correct))
2 Pass
%% x = 400*pi; y_correct = 10; assert(isequal(your_fcn_name(x),y_correct))
3 Pass
%% x = 40000*pi; y_correct = 100; assert(isequal(your_fcn_name(x),y_correct))
4 Pass
%% x = -4*pi; y_correct = 1i; assert(isequal(your_fcn_name(x),y_correct))
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
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https://www.studypool.com/questions/162682/algebra-homework-please-help
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Sigchi4life
Category:
Mathematics
Price: \$5 USD
Question description
A Broadway musical was attended by a total of 2,838 people, including men, women, and children. The number of women who attended the event was 5 more than 3 times the number of children, while the number of men was 3 less than 4 times the number of women. Write an equation to model this situation. Then, use the equation to find the number of men, women, and children who attended the musical.
(Top Tutor) Daniel C.
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School: UCLA
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https://studyrocket.co.uk/revision/gcse-chemistry-combined-science-aqa/combined-science-rate-of-chemical-change/rates-of-reaction
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Rates of Reaction
Rates of Reaction
What you need to know:
How different environmental factors can affect the rate of reaction.
The rate of a reaction is just how fast a reaction happens!
1. __Temperature __- the higher the temperature, higher the energy, the faster the particles collide, the more likely they are to collide and react, the faster the rate of reaction
2. __Concentration __- The more concentrated a reactant is, the closer the particles are. If the particles are closer together they are more likely to collide and react with one another. Higher concentration increases the rate of reaction.
3. __Catalysts __- Catalysts provide an alternative route for the reactants. This alternative route requires less energy, therefore catalysts increase the rate of reaction.
4. Surface area - The larger the surface area, the more particles are exposed and are more likely to react. The larger the surface area the faster the rate of reaction.
If these don’t quite make sense, look at the next section and come back!
Collision Theory
What you need to know:
What collision theory is.
What are the limitations of collision theory (how is it different to what actually happens in real life).
1. Collision theory is just a way in which we explain how particles react. It’s not ACTUALLY what happens (since what actually happens is very very complicated), but it’s good enough.
2. In collision theory, we describe particles as spheresthat are all whizzing around. How fast they move depends on how much energy they have (so the hotter they are, the faster they move). When two particles collide at high speeds, the two react!
3. Two big things to know about collision theory are:
The more the particles collide, the more likely they are to react. This increases the rate of reaction.
Particles have to collide with enough energy in order to react with one another.
Catalysts
What you need to know:
What a catalyst is.
How a catalyst works, in terms of activation energy.
A catalyst is a substance that is added to a reaction to speed up the reaction. The catalyst is not changed or used up in the reaction.
The type of catalyst needed depends on the reaction, but they all work in very similar ways.
1. They provide an alternative reaction route which requires less energy - the minimum energy the particles need in order to react is called the activation energy. Catalysts decrease this reaction energy so that reactions happen easier and quicker!
Biological catalysts (catalysts used in our bodies) are called enzymes. These help chemical reactions happen in our bodies. Without them we would not be able to survive since the chemical reactions wouldn’t occur as much as we need them to.
Measuring Rates of Reaction
What you need to know:
How can we measure the rate of reaction of different types of reactions.
Whether you use the amount of reactant used or the amount of product formed depends on what the reaction actually is.
For example:
1. If a precipitate (solid) is formed, you can measure how much has been formed
2. This can be done by putting the flask on a piece of paper with an x drawn on. You can time how long it takes for you to not be able to see through to the x anymore (because of the precipitate build up)
3. For some experiments, you can filter off the precipitate and weight it after a certain amount of time.
1. If a precipitate (solid) is formed, you can measure how much has been formed
2. This can be done by putting the flask on a piece of paper with an x drawn on. You can time how long it takes for you to not be able to see through to the x anymore (because of the precipitate build up)
3. For some experiments, you can filter off the precipitate and weight it after a certain amount of time.
Calculating Rates of Reaction
What you need to know:
How to calculate the rate of reaction by using a reaction graph.
There are two main ways to find the rate of a reaction, using the equation above or finding the gradient of a reaction rate graph.
The gradient ‘slope’ of the graph tells you how fast the reaction is. The steeper the gradient, the faster the reaction! For this to work you need to make sure that time is on the x-axis.
Since the reaction rate is changing all the time, we need to pick a time to work out the reaction rate.
As you can see on the graph above, the two lines stop at the same place, this is because they both create the same amount of product. However, the blue line got there first! This is because the blue line has a steeper gradient (a faster reaction rate).
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Questions tagged [hypersonic]
Questions about flight at hypersonic speeds, typically defined as above Mach 5.
40 questions
Filter by
Sorted by
Tagged with
1 vote
45 views
Are shockwaves responsible for heating in very sparse atmospheres?
I recall learning about hypersonic flow in high atmosphere. I think one of the modelling assumptions used is that the relative air flow is so fast, and the average space between molecules is so great, ...
1 vote
91 views
What limits the speed of a hypersonic aircraft flying in the lower atmosphere? Are there estimates for its max possible speed?
I'm guessing it is the heat resistance of the missiles leading edge material. Hypersonic cruise missiles all seem to attain Mach 6 and below. The boost-glide weapons fly faster but up much higher. I ...
• 51
236 views
Hypersonic Boost Glide Range
Looking at an article earlier about hypersonic boost glide vehicles and I noticed how small their L/D ratios were ~2.5/1. If they begin their hypersonic glide at 50km in altitude would they not ...
41 views
Formula for comparing drag on planes flying at different speeds at different altitudes. Ex: Mach .80 at 40,000 ft, vs Mach 5 at 100,000 ft
I am interested in getting a sense of how much extra energy is needed to overcome drag at different speeds and altitudes. For example, for me a baseline would the drag on an airplane flying something ...
• 655
1 vote
161 views
Does lift-to-drag ratio asymptote to 4 (or 6) at high Mach numbers?
The Kuchemann equations for L/D at high mach numbers, approximately verified by wind tunnel tests, were: $$\left(\frac{L}{D}\right)_{max} = \frac{4\cdot(M+3)}{M}$$ and, for wave riders \left(\frac{L}...
• 3,084
1 vote
126 views
How do the shockwaves (supersonic and hypersonic) interact with the structure?
My main question is why spaceplanes are conceptualized as pointy and lengthy, dart-like machines. What would happen if they were more like flying wing? I'm hypothesizing a bit and I am not sure what ...
• 116
103 views
Below what altitudes do shock waves form?
For a shock wave to form, the air must have a certain density. Surely in Thermosphere or even upper parts of Mesospheres the air is very thin. I was wondering if there is a formula related to the Mach ...
72 views
What should the inlet to throat area ratio be for a CD rocket nozzle on an aircraft flying at Mach 10?
I am analysing a rocket CD (convergent-divergent) nozzle at a altitude of 15,000m. I am stuck on how to calculate the areas so that at the throat of the nozzle Mach number equals to one. The ...
129 views
What missile defense systems would be used against hypersonic glide vehicles? [closed]
What missile defense systems would be used against hypersonic glide vehicles and cruise missiles (like Avangard or Sky Star)? What challenges would nations face developing them?
• 1,131
59 views
What would an inlet look like on a hypersonic aircraft? Would it be fixed or movable?
What would an inlet look like on a hypersonic aircraft? Would it be rectangular or like a cone (assymetric). Would it move and if so how?
• 1,131
95 views
What happens when a "ground effect" plane approaches the sound barrier?
Does the shock wave inhibit or contribute to lift? What is the ideal speed to fly near or on the speed of sound while using the ground effect?
340 views
Are there any hypersonic planes in service today?
Are there any hypersonic (Mach 5+) planes in service or officially being built? Or all the talk I here about the SR-72 rumor. Thanks!
• 1,131
1 vote
168 views
What is the speed above which painting an aircraft black (or dark) results in lower skin temperature?
The Concorde is painted white for thermal reasons. If I understand correctly, it is done to reflect as much radiation as possible in order to lower tempereture. But when the skin is really hot (SR-71, ...
• 16.1k
1 vote
112 views
What are the effects of separation at hypersonic rarefied flows?
Shock wave causes an adverse pressure gradient which causes an airfoil to stall. Also because gas density is much lower boundary layer becomes laminar and unable to prevent separation. Correct me if ...
123 views
what is "Shock on shock heating"?
Can anybody enlighten me on the nature and physics behind what is referred to as Shock on Shock heating. Both , "Hypersonic: The Story of the North American X-15 , Jenkins and Landis" and "Lessons ...
• 61
1 vote
585 views
Does an engine that combines an air-breathing rocket and non-air-breathing rocket exist?
Does a rocket engine that combines air-breathing and non-air-breathing modes exist? The advantage would be it would have to carry a smaller tank and less oxygen, leaving more room for payload on a ...
• 8,379
10k views
Why Didn't the USSR Build An X-15?
In the history of aerospace engineering in the United States, the X-15 is lauded as a critical and necessary step on the path to manned spaceflight. Throttle-able chemical rocket propulsion systems, ...
• 1,110
513 views
Why is a blunt trailing edge a better stabilizer at hypersonic speeds?
Here is what I read about the X-15 spaceplane: The X-15 had a thick wedge tail to enable it to fly in a steady manner at hypersonic speeds.[14] This produced a significant amount of drag at lower ...
• 17.1k
1 vote
173 views
What rotors would be best for atmospheric reentry?
What rotor blade set up would be the best for entering Earth atmosphere or to shed speed before landing in place of a parachute? I understand that the longer blades are more efficient, but are not as ...
• 1
1 vote
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# Finite Difference Methods in Financial Engineering
## A Partial Differential Equation Approach
US\$ 132.00 US\$ 118.80
(If any tax is payable it will be calculated and shown at checkout.)
The world of quantitative finance (QF) is one of the fastest growing areas of research and its practical applications to derivatives pricing problem. Since the discovery of the famous Black-Scholes equation in the 1970's we have seen a surge in the number of models for a wide range of products such as plain and exotic options, interest rate derivatives, real options and many others. Gone are the days when it was possible to price these derivatives analytically. For most problems we must resort to some kind of approximate method.
In this book we employ partial differential equations (PDE) to describe a range of one-factor and multi-factor derivatives products such as plain European and American options, multi-asset options, Asian options, interest rate options and real options. PDE techniques allow us to create a framework for modeling complex and interesting derivatives products. Having defined the PDE problem we then approximate it using the Finite Difference Method (FDM). This method has been used for many application areas such as fluid dynamics, heat transfer, semiconductor simulation and astrophysics, to name just a few. In this book we apply the same techniques to pricing real-life derivative products. We use both traditional (or well-known) methods as well as a number of advanced schemes that are making their way into the QF literature:
• Crank-Nicolson, exponentially fitted and higher-order schemes for one-factor and multi-factor options
• Early exercise features and approximation using front-fixing, penalty and variational methods
• Modelling stochastic volatility models using Splitting methods
• Critique of ADI and Crank-Nicolson schemes; when they work and when they don't work
• Modelling jumps using Partial Integro Differential Equations (PIDE)
• Free and moving boundary value problems in QF
Included with the book is a CD containing information on how to set up FDM algorithms, how to map these algorithms to C++ as well as several working programs for one-factor and two-factor models. We also provide source code so that you can customize the applications to suit your own needs.
Wiley; March 2006
464 pages; ISBN 9781118856482
Title: Finite Difference Methods in Financial Engineering
Author: Daniel J. Duffy
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• Contents
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### 83 HFOPDE, chapter 3.6.2
83.1 Problem 1
83.2 Problem 2
83.3 Problem 3
83.4 Problem 4
83.5 Problem 5
_______________________________________________________________________________________
#### 83.1 Problem 1
problem number 734
Problem Chapter 3.6.2.1 from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.
Solve for $$w(x,y)$$
$a w_x + b y^n w_y = c \cos (\lambda x) + k \cos (\mu y)$
Mathematica
$\left \{\left \{w(x,y)\to \frac{a b \lambda \mu c_1\left (\frac{a y-b x}{a}\right )+a k \lambda \sin \left (\frac{b \mu x}{a}\right ) \cos \left (\frac{\mu (a y-b x)}{a}\right )+a k \lambda \cos \left (\frac{b \mu x}{a}\right ) \sin \left (\frac{\mu (a y-b x)}{a}\right )+b c \mu \sin (\lambda x)}{a b \lambda \mu }\right \}\right \}$
Maple
$w \left ( x,y \right ) ={\frac{1}{\mu \,b\lambda \,a} \left ({\it \_F1} \left ({\frac{ay-bx}{a}} \right ) \mu \,b\lambda \,a+\sin \left ( \lambda \,x \right ) c\mu \,b+k\sin \left ({\frac{\mu \, \left ( ay-bx \right ) }{a}}+{\frac{\mu \,bx}{a}} \right ) a\lambda \right ) }$
_______________________________________________________________________________________
#### 83.2 Problem 2
problem number 735
Problem Chapter 3.6.2.2 from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.
Solve for $$w(x,y)$$
$a w_x + b y^n w_y = c \cos (\lambda x+\mu y)$
Mathematica
$\left \{\left \{w(x,y)\to \frac{c \sin \left (\mu \left (\frac{a y-b x}{a}+\frac{b x}{a}\right )+\lambda x\right )+a \lambda c_1\left (\frac{a y-b x}{a}\right )+b \mu c_1\left (\frac{a y-b x}{a}\right )}{a \lambda +b \mu }\right \}\right \}$
Maple
$w \left ( x,y \right ) ={\frac{c}{\lambda \,a+\mu \,b}\sin \left ({\frac{ \left ( \lambda \,a+\mu \,b \right ) x}{a}}+{\frac{\mu \, \left ( ay-bx \right ) }{a}} \right ) }+{\it \_F1} \left ({\frac{ay-bx}{a}} \right )$
_______________________________________________________________________________________
#### 83.3 Problem 3
problem number 736
Problem Chapter 3.6.2.3 from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.
Solve for $$w(x,y)$$
$x w_x + y w_y = a x \cos (\lambda x+\mu y)$
Mathematica
$\left \{\left \{w(x,y)\to \frac{a x \sin \left (x \left (\lambda +\frac{\mu y}{x}\right )\right )+\lambda x c_1\left (\frac{y}{x}\right )+\mu y c_1\left (\frac{y}{x}\right )}{\lambda x+\mu y}\right \}\right \}$
Maple
$w \left ( x,y \right ) ={a\sin \left ( x \left ({\frac{\mu \,y}{x}}+\lambda \right ) \right ) \left ({\frac{\mu \,y}{x}}+\lambda \right ) ^{-1}}+{\it \_F1} \left ({\frac{y}{x}} \right )$
_______________________________________________________________________________________
#### 83.4 Problem 4
problem number 737
Problem Chapter 3.6.2.4 from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.
Solve for $$w(x,y)$$
$a w_x + b \cos ^n(\lambda x) w_y = c\cos ^m(\mu x)+s \cos ^k(\beta y)$
Mathematica
$\text{Timed out}$ Timed out
Maple
$w \left ( x,y \right ) =\int ^{x}\!{\frac{1}{a} \left ( c \left ( \cos \left ( \mu \,{\it \_b} \right ) \right ) ^{m}+s \left ( \cos \left ({\frac{\beta }{a} \left ( \int \! \left ( \cos \left ( \lambda \,{\it \_b} \right ) \right ) ^{n}\,{\rm d}{\it \_b}b+ \left ( -\int \!{\frac{b \left ( \cos \left ( \lambda \,x \right ) \right ) ^{n}}{a}}\,{\rm d}x+y \right ) a \right ) } \right ) \right ) ^{k} \right ) }{d{\it \_b}}+{\it \_F1} \left ( -\int \!{\frac{b \left ( \cos \left ( \lambda \,x \right ) \right ) ^{n}}{a}}\,{\rm d}x+y \right )$
_______________________________________________________________________________________
#### 83.5 Problem 5
problem number 738
Problem Chapter 3.6.2.5 from Handbook of first order partial differential equations by Polyanin, Zaitsev, Moussiaux.
Solve for $$w(x,y)$$
$a w_x + b \cos ^n(\lambda y) w_y = c\cos ^m(\mu x)+s \cos ^k(\beta y)$
Mathematica
$\text{Timed out}$ Timed out
Maple
$w \left ( x,y \right ) =\int ^{y}\!{\frac{ \left ( \cos \left ( \lambda \,{\it \_b} \right ) \right ) ^{-n}}{b} \left ( s \left ( \cos \left ( \beta \,{\it \_b} \right ) \right ) ^{k}+ \left ( \cos \left ({\frac{\mu \, \left ( a\int \! \left ( \cos \left ( \lambda \,{\it \_b} \right ) \right ) ^{-n}\,{\rm d}{\it \_b}-a\int \! \left ( \cos \left ( \lambda \,y \right ) \right ) ^{-n}\,{\rm d}y+bx \right ) }{b}} \right ) \right ) ^{m}c \right ) }{d{\it \_b}}+{\it \_F1} \left ({\frac{-a\int \! \left ( \cos \left ( \lambda \,y \right ) \right ) ^{-n}\,{\rm d}y+bx}{b}} \right )$
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# What is Half of 7 8? Here’s How to Find the Answer
What is half of 7 8?” you might wonder. The answer is two things: one is 1/2 of 7/8, and the other is 7/16 of an inch. This makes seventeen-eighths of an inch, the same as seven and a sixteenth of an eighth. This formula is also helpful for converting fractions. This way, you’ll know how much of each size is left over, and you’ll know how much of either one is leftover.
## Decimalization a number gives a common fraction.
What is half of seven-eight? You can find the answer by using the decimal part of the number. Using the decimal part of the number gives you a common fraction. The last digit is the thousandth digit. If you want a calculator to produce the same result, you must use the third digit. If the number is less than half of seven-eight, you’ll get a mixed number.
## You can halve, multiply by two, or divide by five.
What is half of 7 8? It’s 1/3 of a cup. Then, you need to multiply this by two. You can also divide by five and eight by two. This way, you get the same answer. In the second case, you’ll need to divide by four, and so on. You’ll be able to locate the answer quickly if you follow the steps above. Try this method if you can’t discover the solution.
You may wonder what half of seven-eight is. There’s no hard and fast rule, but the following steps will help you find the answer. First, you’ll need to divide the whole number by the reciprocal. In the third case, you’ll need to divide it by two, and this is exactly what you need to do. Then, you’ll need to multiply the answer by three to make it equivalent to a fourth.
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Línea de servicio de Sensores Industriales
Atención al cliente para protección contra explosiones
Did You Figure it Out? Get the Answer to the Riddle from E-News Issue No. 1, February 2019
2019-02-12
Boat Trip
How long would Lena be on the water?
On a hot and sunny summer day, Lena decides to take a trip with her boat. It takes 5 hours for her to row her boat down the river. If she continues to row at the same pace, she will need 6 hours to row back up the river. Now imagine that Lena is traveling the same distance with her boat on a lake (without a current).
How long would she be on the water if her boat traveled at a constant speed?
Solution
It is a uniform motion, so the following physical relationship is valid:
Speed x time = distance or s ∙ t = d
The time t” is calculated accordingly: t = d/s
For our example this means:
5 = d/(S + s) → 5(S + s) = d, and
6 = d/(S - s) → 6(S - s) = d
Where “S” the speed of the boat and “s” is the speed of the current.
By setting the two equal to one another, you get:
5(S + s) = 6(S - s)
5S + 5s = 6S - 6s
11s = S
So, the speed of the boat is eleven times greater than the speed of the current.
t = d/s and the distance d” can be calculated using
d 5(S + s) = 6(S - s),
which yields d = 5 [ S + (1/11)S ] = 6 [ S - (1/11)S ] = (60/11)S
and therefore
t = d/s = 60/11
Because you have to consider the distance there and back, the result is:
t = 120/11 = 10 (10/11)h = 10h 54,5 seconds
It is logical that a trip without a current is shorter, because traveling with the current is shorter than traveling against the current.
e-news
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¡Descubra nuestra revista online! En ella le esperan interesantes historias de éxitos, informes de aplicaciones, entrevistas y noticias regionales.
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# LA MATERIA CLASSIFYING MATTER
## Presentation on theme: "LA MATERIA CLASSIFYING MATTER"— Presentation transcript:
LA MATERIA CLASSIFYING MATTER
CCNN 2º ESO FQ 3º ESO María Ginés
VOCABULARY STATE CHANGES CLASSIFYING MATTER SEPARATING METHODS
Vaporization Evaporation Boiling (also called ebullition) Condensation Boiling point Freezing Melting Melting point Sublimation Deposition Pure substance Mixture Homogeneous mixture Heterogeneous mixture Compound Simple substance Sieving Magnetization Filtering Decanting Crystallization Distillation STATE CHANGES CLASSIFYING MATTER SEPARATING METHODS
MATERIA Aquello que tiene masa y ocupa un espacio FUNDAMENTALES
presenta PROPIEDADES CUANTITATIVAS o MAGNITUDES FÍSICAS PROPIEDADES CUALITATIVAS pueden ser No dependen de otras: Longitud (m) Masa (kg) Tiempo (s) Temperatura (K) se pueden No permiten identificarlas No se pueden medir Comparar con un patrón DERIVADAS es MEDIR GENERALES ESPECÍFICAS de forma Dependen de tras: velocidad (m/s) Densidad (kg/m3) DIRECTA INDIRECTA No identifican la sustancia: Temperatura/ masa/volumen/ Superficie (etc.) Sí la identifican: densidad/ PF / PE/ Con un instrumento: longitud/tiempo/ Velocidad/etc. A partir de una fórmula: volumen/superficie
STATES OF MATTER GAS LIQUID SOLID Matter is made up of tiny particles
are explained by STATES OF MATTER Matter is made up of tiny particles Each particle is in constant motion KINETIC THEORY deposition condensation freezing GAS LIQUID SOLID melting vaporization sublimation Estados Físicos Characteristics of states physics
STATE CHANGE GRAPH
VOCABULARY: STATE CHANGES
Vaporization: is the state change from liquid to gas. Evaporation: is the state change from liquid to gas. It happens at room temperature and only on the surface of the liquid. The rate of evaporation increases when the temperature increases. Boiling (also called ebullition): is the state change from liquid to gas at a constant temperature. All the liquid changes state and while it happens, the temperature doesn’t change. Condensation: is the state change from gas to liquid. Boiling point: is the temperature at which a liquid boils. It is different for each substance. Boiling point = Condensing point
VOCABULARY: STATE CHANGES
Freezing: Is the state change from liquid to solid. Melting: Is the state change from solid to liquid. Melting point: Is the temperature at which a solid turns into a liquid. It is different for each substance. Melting point = Freezing point Sublimation: Is the transition from solid to gas. Deposition: Is the state change from gas to solid.
HETEROGENEOUS MIXTURES
Are made up by two or more substances at not fixed proportions They can be separated by physical methods MATTER A sample of matter that consists of only one component. PURES SUBSTANCES MIXTURES HOMOGENEOUS MIXTURES HETEROGENEOUS MIXTURES COMPOUNDS SIMPLE SUBSTANCES They are made up by two or more elements combined in a fixed proportion The properties of a compound are completely different that their elements. We can break down by chemical means They have uniform composition and properties throughout We cannot see their components with a naked eye They do not have uniform composition and properties throughout We can see their components They are made up by atoms of the same element We cannot break down in any way
SEPARATING MIXTURES: SIEVING
cay clay gravel sand
SEPARATING MIXTURES: MAGNETIZATION
iron salt
SEPARATING MIXTURES: FILTERING
sand filter paper funnel water
SEPARATING MIXTURES: DECANTING
decanting funnel oil r water water
SEPARATING MIXTURES: CRYSTALLIZATION
concentration filtering cristallization
SEPARATING MIXTURES: DISTILLATION
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# The period length of the decimal expansion of a fraction
### by Helmut Richter
It is explained how, for a given natural number, the period length of the decimal fraction of the reciprocal of that number can be determined without actually performing the division. The article begins with examples in the decimal number system, but the theory is then displayed for arbitrary bases of number systems. The reader should be able to apply the statements in the beginning to other bases as well.
First of all, we observe that factors 2 and 5 in the denominator change neither the period length nor the sequence of digits in the period, their influence can always be separated into an extra summand, e.g.: 1/12 = 1/3 - 1/4 or 1/70 = 5/7 - 7/10, and the decimal expansions of 1/4 and 7/10 terminate. The effect is that this extra summand garbles the first few digits so that the period does not start at the decimal point. Another way to handle this case is to multiply the fraction with the least power of 10 that makes the factors 2 and 5 in the denominator disappear: 100/12 = 25/3 = 8 + 1/3, then to look at the decimal fraction, and to shift the decimal point to the left again to compensate for the power of 10 introduced in the first step.
Now, we are only interested in the cases where the denominator has no factor 2 or 5, i.e. ends in a digit 1, 3, 7, or 9. It is easy to see that in this case the period starts right at the decimal point:
```0.294347 294347 ... = 294347 / 999999
0.142857 142857 ... = 142857 / 999999 = 1 / 7
```
Multiplying with a constant leaves the period intact. If this constant happens to be a factor of the denominator, the period may be shortened, but even then the decimal fraction is still periodic with the previous period length:
```0.002457 002457 ... = 2457 / 999999 = 1 / 407
0.149877 149877 ... = 149877 / 999999 = 61 / 407
0.135135 135135 ... = 135135 / 999999 = 55 / 407 = 5 / 37
0.135 135 135 135 ... = 135 / 999 = 5 / 37
```
This way to look at decimal fractions gives us the clue to period lengths:
For denominator q, the period length is the smallest k such that q divides 10k-1. E.g. for 41, the period length is 5 because 41 divides 99999.
From this we get some useful consequences:
• For any denominator q = p1a1 * p2a2 * . . . (prime factor decomposition),
the period length is a factor of
phi(q) = (p1-1)p1a1-1 * (p2-1)p2a2-1 * . . . (number of those integers smaller than q that have no common factor with q). As a special case:
• For a prime denominator q, the period length is a factor of q-1.
• The period length of the least common multiple of q1 and q2 is the least common multiple of the two period lengths.
Examples:
9997 = 13 * 769, hence the period length of 9997 is a factor of 12 * 768 = 9216. Indeed, it is 192, which is 9216/48.
We could have found that also the other way round:
The period length of 13 is 6, that of 769 is 192, so the period length of 9997 is the LCM of 6 and 192, namely 192.
The proofs of the three statements are omitted here. The first (and hence also the second) is an immediate consequence of Fermat's Little Theorem, and the third is easy to see if one computes the greatest common divisor of two different 10k-1 by Euclid's algorithm.
Now we are nearly done:
1. For a given q, divide it into powers of primes. For each prime power, determine the period length, then take the LCM of all these.
2. For a prime power pk, determine first two things:
• The period length L of p.
• The highest power pj such that the period length of pj is also L.
Then the period length of pk is pk-j * L if k>j, otherwise L. If the denominator is a factor of of b-1 where b ist the base of the number system, this rule may not be true for the first step: in this case let L be the period length of p2 instead; the reader may try for himself the case p=2 in base b=3.
This follows relatively easy from the above observations if j>1. In case j=1 (the normal case), it is also true, but more tedious to prove.
Some examples: The period length of 3 (in base 10) is 1, of 9 as well, of 27 not. Hence 27 has period length 3, 81 has 9 and so on. Here the case j>1 is somewhat "artificial" as the square of 3 is a factor of b-1=9. A more "natural" case would be one where p and b-1 have no common divisor, e.g. p=7 in base 18: not only 7 but 73 divides 183-1, so that the period length of 73 in that base is the same as that of 7, namely 3. For base 10, i.e. for decimal numbers, the smallest such "natural" examples are p=487 and p=56598313 where p2 has the same period length (p-1 in both cases) as p itself. Interesting are of course only those cases where the smallest bk-1 that is a multiple of p is also a multiple of a higher power of p. More examples for other bases are: 10932 is a factor of 2364-1, 112 is a factor of 35-1, 52 is a factor of 74-1, 134 is a factor of 2394-1, 54 is a factor of 4434-1, and 56 is a factor of 10684-1.
3. Remains the case of a prime p. The only candidates are the factors of p-1. In other words, we have to find K such that the period length is (p-1)/K. In which cases K is even can be determined by quadratic residues: K is even if and only if the base b of the number system is a quadratic residue modulo p. For base 10, this is the case if and only if the distance of p to the nearest multiple of 40 is one of 1, 3, 9, or 13. For other K, one would have to look for the solvability of higher-order roots modulo 10. Here are some examples of primes, ordered by K:
K Examples of primes p such that the decimal period length of 1/p is (p-1)/K
3 103, 127, 139, 331, 349, 421, 457, 463, 607, 661, 673, 691, 739, 829, 967, 1657, 1669, 1699, 1753, 1993
4 53, 173, 277, 317, 397, 769, 773, 797, 809, 853, 1009, 1013, 1093, 1493, 1613, 1637, 1693, 1721
5 11, 251, 1061, 1451, 1901, 1931, 2381, 3181, 3491, 3851, 4621, 4861, 5261, 6101, 6491, 6581, 6781
6 79, 547, 643, 751, 907, 997, 1201, 1213, 1237, 1249, 1483, 1489, 1627, 1723, 1747, 1831, 1879, 1987
7 211, 617, 1499, 2087, 2857, 6007, 6469, 7127, 7211, 7589, 9661, 10193, 13259, 13553, 14771, 18047
8 41, 241, 1601, 1609, 2441, 2969, 3041, 3449, 3929, 4001, 4409, 5009, 6089, 6521, 6841, 8161, 8329
9 73, 1423, 1459, 2377, 2503, 3457, 7741, 9433, 10891, 10909, 16057, 17299, 17623
10 281, 521, 1031, 1951, 2281, 2311, 2591, 3671, 5471, 5711, 6791, 7481, 8111, 8681, 8761, 9281
11 353, 3499, 10429, 13619, 15269, 20219, 20593
12 37, 613, 733, 1597, 2677, 3037, 4957, 5197, 5641, 7129, 7333, 7573, 8521, 8677, 11317, 14281
13 2393, 15497, 18149, 18617, 20021, 25819, 26183
14 449, 1289, 3557, 4397, 4999, 5209, 6203, 6637, 7043, 8387, 10613, 11369, 13147, 13399, 14323
15 3061, 4021, 7621, 11491, 14851, 19381, 22651
16 1889, 5521, 8849, 9521, 9649, 12689, 13649, 17681, 17729, 18049
17 137, 9419, 25127, 27541, 51341, 80207, 85103
18 2467, 3187, 4357, 4483, 5689, 7039, 7237, 8317, 9973, 10243, 10711, 10729, 11071, 12763, 13267
19 16189, 23447, 29527, 30971, 37811, 79687, 88807
20 641, 3121, 13921, 23041, 23561, 26681, 29921
24 1321, 6481, 11689, 16729, 19081, 24001, 31849
25 101, 15101, 17851, 19501, 39251, 46901, 59651
28 757, 9689, 12853, 14533, 15877, 20693, 55889
30 1231, 11311, 14551, 21991, 23911, 27751, 28111
33 859, 40063, 42703, 63097, 167113, 173713, 181303
34 239, 11969, 12071, 13907, 26759, 34511, 58889
44 1409, 3169, 20681, 42197, 43517, 68597, 84437
54 271, 21871, 23599, 32401, 36559, 40879, 43201
92 1933, 51613, 150053, 470489, 639493, 1108693, 1651493
This list is exhaustive for primes under 2000: if a prime p is under 2000 and does not appear in this list, its period length is either p-1 (if K odd, see above for the criterion) or (p-1)/2. For each of the K, there are no other possible p between the listed ones; the point where each list ends is, however, quite unsystematic (but it will not end before 2000 is reached).
The largest values of K belong to the large prime factors of numbers that consist only of ones ("repunits"), e.g. 265371653, which is a factor of 1111111111111, has a period length of 13 which leads to K=20413204. The largest K known belongs to the largest repunit which is completely factorised; this is currently the prime number (101031-1)/9 with a period length of 1031 and hence K near 1.0777*101027.
How period lengths for prime numbers of moderate size can be determined with modest means (for p<100000 even with a non-programmable pocket calculator), is described in a separate article.
© Helmut Richter published here 1997-04-10; last update 1998-05-07
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https://s-nako.work/2020/04/bucket-sort-counting-sort-and-radix-sort/
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# Bucket sort, counting sort and radix sort.
## GOAL
To understand the difference and pros and cons of “bucket sort”, “counting sort” and “radix sort” and implement them.
## Bucket sort
Bucket sort is one of the sorting algorithm that has “bucket” to put numbers in.
```import random
import itertools
length = 50 #length of the list
input_list = [random.randint(0, 99) for x in range(length)]
#input: list that has 0~99 int elements
#output: sorted list
def bucket_sort(input_list):
bucket = [[] for x in range(100)]
for i in input_list:
bucket[i].append(i)
return list(itertools.chain.from_iterable(bucket))
print(input_list)
print(bucket_sort(input_list))```
```#output
[6, 27, 26, 38, 90, 80, 69, 14, 65, 53]
[6, 14, 26, 27, 38, 53, 65, 69, 80, 90]```
Complexity: time O(n), space O(r) as n is length of the list, r is range. In the worst case, space complexity is O(r+n)
Pros: Fast algorithm, stable sort
Cons: big memory, the range of input number should be limited.
## Counting sort
Counting sort is one of the derivation of bucket sort. Create Bucket and put the number of “occurrences” in it. Iterate the input list counting the number of occurrences.
```import random
length = 10
input_list = [random.randint(0, 99) for x in range(length)]
def counting_sort(input_list):
bucket = [0 for x in range(100)]
for i in input_list:
bucket[i] += 1
output = []
for idx, num in enumerate(bucket):
for i in range(num):
output.append(idx)
return output
print(input_list)
print(counting_sort(input_list))```
```#output
[84, 33, 72, 10, 31, 4, 4, 46, 89, 52]
[4, 4, 10, 31, 33, 46, 52, 72, 84, 89]```
Complexity: time O(n), space O(r) as n is length of the list, r is range.
Pros: Fast algorithm, fixed size memory
Cons: unstable, big memory, the range of input number should be limited.
## Radix sort
Radix sort is one of the derivation of bucket sort. Convert each elements into n-adic number and put each digit into the bucket.
```import random
import itertools
length = 10
input_list = [random.randint(0, 99) for x in range(length)]
def radix_sort(input_list): # n = 10
for i in (range(2)): #digit is 2
bucket= [[] for i in range(10)]
for num in input_list:
index = (num//(10**i)) % 10
bucket[index].append(num)
input_list = list(itertools.chain.from_iterable(bucket)).copy()
return input_list
print(input_list)
print(radix_sort(input_list))```
```#output
[26, 4, 7, 48, 71, 31, 95, 20, 94, 55]
[4, 7, 20, 26, 31, 48, 55, 71, 94, 95]```
Complexity: time O(n), space O(d) as n is length of the list, d is number of digits. In the worst case, space complexity is O(d+n)
Pros: Fast algorithm, small memory, stable sort
Cons: the range of input number should be limited
Because it takes some time to calculate division and modulo, using shift operation of binary numbers is efficient.
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http://betterlesson.com/lesson/640171/where-s-the-math-analyzing-our-kaleidoscope-images
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# Where's The Math? Analyzing our Kaleidoscope Images
Unit 5: Congruence and Similarity
Lesson 16 of 16
## Big Idea: The angle between mirrors is related to the number of reflections. The basis of our Kaleidoscope images.
Print Lesson
Standards:
Subject(s):
Math, Geometry, Similarity and Congruence, Reflections, Kaleidoscope, Graphing, rotations,
60 minutes
### Mauricio Beltre
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http://clay6.com/qa/23948/a-combination-of-lens-is-formed-by-keeping-two-identical-equiconvex-lens-of
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# A combination of lens is formed by keeping two identical equiconvex lens of focal length f kept in contact and the space between them filled with liquid of refractive index $\mu_2 =\large\frac{4}{3}$. Focal length of the combination is
$(a)\;f \\ (b)\;\frac{f}{2} \\ (c)\;\frac{4}{3} f \\ (d)\;\frac{3f}{4}$
In effect we have a combination of 3 lens , the liquid between the two double convex lens forming a double concave lens.
We need to find the focal length of this liquid lens, through it has the same radius of curvature R.
For double convex lens,
$\large\frac{1}{f} =(\mu -1) \bigg[ \large\frac{1}{R} +\frac{1}{R}\bigg]$
$\therefore \large\frac{1}{f} =(\frac{3}{2}-1) \bigg[ \large\frac{1}{R} +\frac{1}{R}\bigg]$
$\therefore f=R$
For the liquid lens :
$\large\frac{1}{f'} =(\mu _2 -1) \bigg[ -\large\frac{1}{R} -\frac{1}{R}\bigg]$
$\qquad= \large\frac{1}{3} \bigg(\large\frac{-2}{R}\bigg)$
$\large\frac{1}{f'}=\large\frac{-2}{3f}$
Now the focal length of combination
$\large\frac{1}{F} =\frac{1}{f} +\frac{1}{f} +\frac{1}{f}$
$\qquad= \large\frac{2}{f} -\frac{2}{3f}$
$\qquad= \large\frac{4}{3f}$
$\therefore F= \large\frac{3f}{4}$
Hence d is the correct answer.
answered Jan 16, 2014 by 1 flag
in case of liquid lens the considered refractive index should be refrective index of liquid with respect to surrounding, and surrounding in case of combination with lens will be lens , as light enters from lens to liquid. so it would be 1.333/1.5
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https://au.mathworks.com/matlabcentral/cody/problems/1702-maximum-value-in-a-matrix/solutions/1903947
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Cody
# Problem 1702. Maximum value in a matrix
Solution 1903947
Submitted on 18 Aug 2019 by Hasna Thajudeen
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = [1 2 3; 4 5 6; 7 8 9]; y_correct = 9; assert(isequal(your_fcn_name(x),y_correct))
2 Pass
x = -10:0; y_correct = 0; assert(isequal(your_fcn_name(x),y_correct))
3 Pass
x = 17; y_correct = 17; assert(isequal(your_fcn_name(x),y_correct))
4 Pass
x = magic(6); y_correct = 36; assert(isequal(your_fcn_name(x),y_correct))
5 Pass
x = [5 23 6 2 9 0 -1]'; y_correct = 23; assert(isequal(your_fcn_name(x),y_correct))
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https://www.mycoursehelp.com/QA/let-x1xn-be-iid-gammaa%C3%9F-with-a-known-fin/15391/1
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Chat Now
# .Let X1,…Xn be iid gamma(a,ß) with a known. Find the best unbiased estimator of 1/ß 2.Let
### .Let X1,…Xn be iid gamma(a,ß) with a known. Find the best unbiased estimator of 1/ß 2.Let
.Let X1,…Xn be iid gamma(a,ß) with a known. Find the best unbiased estimator of 1/ß 2.Let W1,…,Wk be unbiased estimators of a parameter ? with Var(Wi) =si^2 and Cov(Wi ,Wj)=0 if i?j. Show that, of all estimators of the form ?aiWi, where the ai are constant and E(?aiWi)= ?, The estimator W* =(?Wi/ si^2)/ (?1/ si^2) has the minimum variance and show also VarW*=1/(?1/ si^2)
Abhinav 02-Dec-2019
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http://www.sparknotes.com/math/calcbc2/computingintegrals/section4/
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# Computing Integrals
### Contents
page 1 of 2
Page 1
Page 2
#### Partial Fraction Decomposition
We have not yet discussed how to integrate rational functions (recall that a rational function is a function of the form f (x)/g(x) , where f , g are polynomials). The method that allows us to do so, in certain cases, is called partial fraction decomposition.
Here we demonstrate this procedure in the case where the denominator g(x) is a product of two distinct linear factors. This method can easily be generalized to the case where g is a product of arbitrarily many distinct linear factors. The cases where g has repeated linear factors or factors of degree 2 are slightly more complicated and will not be considered.
The first step is to divide the polynomial f by the polynomial g to obtain
= h(x) +
where h(x) and r(x) are polynomials, with the degree of r strictly less than the degree of g . There is a result called the division algorithm that guarantees that we can do this. Since we know how to integrate polynomials, we are left with figuring out how to integrate r(x)/g(x) . Multiplying the numerator and denominator by a constant, we may assume that g(x) is of the form g(x) = (x - a)(x - b) . Since the degree of r is less that 2 , we may write it as r(x) = cx + d .
We want to write r(x)/g(x) in the form
+
since we know how to integrate functions of this form (by change of variables, for example). Multiplying the equation
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# Investigating how height affects time taken for a falling object to reach ground level.
Extracts from this document...
Introduction
Physics Coursework 2003:
Investigating How Height Affects Time Taken for A Falling Object to Reach Ground Level.
Introduction: In this investigation, how height will affect the time taken for a steel ball bearing to reach the ground will be investigated.
It is was Isaac Newton that first discovered gravity and wrote laws defining it. His Second Law of Motion states that the Resultant Force on an object (F) is equal to the Mass of the body (m) times its acceleration (a), or .
The weight (W) of a body is the force of gravity acting on it, which gives it acceleration (g) if it is falling freely close to the earth’s surface. If the body was to have a mass (m) Newton’s 2nd Law of Motion could calculate its weight. Given that and Newton’s Law becomes .
In April of 2003, in a method similar to that, which will be conducted in this investigation, the acceleration of gravity was concluded to be 9.81.
Using the knowledge mentioned above, several equation of motion have been created.
Middle
0.082
0.3
0.247
0.061
0.2
0.202
0.041
0.1
0.143
0.020
0.0
0.000
0.000
This is what the expected graph of Height vs. time should look like:
Height (m)
Time (secs)
The following page shows what the expected graph for Height vs. time2 should look like. It is a straight line passing through the origin, thus proving the prediction .
The expected gradient, m, should be equal to ½ g, or, 4.905ms-2. It is actually 4.926ms-2, which is only 0.021 ms-2 out or 0.428%. This is probably due to the rounding of decimal places when drawing the graph and human error in plotting the points (i.e. not exactly accurate to 3 decimal places.)
Apparatus & Diagram:
Safety: As there is a very minimal risk in this investigation, no safety measures need to be taken.
It is planned to drop the ball from a height of 1m and decrease in intervals of 0.1m. At each height 5 readings will be recorded and then the mean result will be calculated. This makes the results more reliable (and better for use in calculation like working out g or the mass of the steel ball.)
The Results will be recorded in a table like this:
Height (cm) Time taken for ball to reach ground (seconds) Mean Result Mean Result2 1st 2nd 3rd 4th 5th 100 90 80 70 60 50 40 30 20 10 00
Conclusion
.
Note: The factor that affected the acceleration was g, (which, on earth, is ) is the mass of the planet, for Earth this is constant.
The results of the investigation are consistent with the prediction. The relationship of was proved in the similarity of the graphs on page 4 & 8, they had almost the exact same gradient, only 0.072ms-2 in difference (or 1.462%) it was also very similar to the mathematical prediction of the gradient (½g) again only 0.051 ms-2 out.
Evaluation:In this investigation, all results are held to be very reliable. When the data was being collected, sophisticated technology was used which measured time accurately and reliably to the nearest thousandth of a second. All recorded results were in very close proximity of each other, so that 0.006seconds was the maximum difference observed.
There were no anomalies observed. All points on the graph on page 8 are not only close to the line of best fit, they are actually on it.
The results in this investigation are believed to be very reliable; as a result no changes need to be made to the procedure.
Physics Coursework October 2003 Andrew Lavery S2F
This student written piece of work is one of many that can be found in our AS and A Level Mechanics & Radioactivity section.
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# 841 dry quarts in liters
## Conversion
841 dry quarts is equivalent to 926.126812823315 liters.[1]
## Conversion formula How to convert 841 dry quarts to liters?
We know (by definition) that: $1\mathrm{dryquart}\approx 1.101220942715\mathrm{liter}$
We can set up a proportion to solve for the number of liters.
$1 dryquart 841 dryquart ≈ 1.101220942715 liter x liter$
Now, we cross multiply to solve for our unknown $x$:
$x\mathrm{liter}\approx \frac{841\mathrm{dryquart}}{1\mathrm{dryquart}}*1.101220942715\mathrm{liter}\to x\mathrm{liter}\approx 926.1268128233149\mathrm{liter}$
Conclusion: $841 dryquart ≈ 926.1268128233149 liter$
## Conversion in the opposite direction
The inverse of the conversion factor is that 1 liter is equal to 0.00107976573634822 times 841 dry quarts.
It can also be expressed as: 841 dry quarts is equal to $\frac{1}{\mathrm{0.00107976573634822}}$ liters.
## Approximation
An approximate numerical result would be: eight hundred and forty-one dry quarts is about nine hundred and twenty-six point one two liters, or alternatively, a liter is about zero times eight hundred and forty-one dry quarts.
## Footnotes
[1] The precision is 15 significant digits (fourteen digits to the right of the decimal point).
Results may contain small errors due to the use of floating point arithmetic.
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# Find all solutions of the equation 2^x+3^y=z^2
1. Feb 23, 2012
### .d9n.
1. The problem statement, all variables and given/known data
Find all solutions in non-negative integers x, y, z of the equation
2^x + 3^y= z^2
2. Relevant equations
3. The attempt at a solution
not entirely sure how to go about it, im assuming maybe logarithms?
2=log(base Z) (2^x+3^y)
or maybe substitution
let
u=2^x so x=log(base 2)(u)
v=3^y so y=log(base 3)(v)
w=z^2 so 2=log(base z)(w)
u+v=w → 2^(log(base 2)(u))+3^(log(base 3)(v))=2^(log(base z)(w))
2. Feb 23, 2012
### morphism
Where is this from? The first thing I tried worked, but it produced a fairly long-winded solution so I'm reluctant to share it in case there's a simpler method... But here goes: Begin by considering your equation modulo 8, keeping in mind that z can't be divisible by 2.
3. Mar 1, 2012
### .d9n.
wasn't too sure what you mean with the modulo 8,
but this is what i have so far
so z cant be even as its equal to an odd plus an even number = odd
so when z=0, x=3 then z=3 and when y=2, x=4 then z=5
so if z^2-3^y has to be a perfect square then y has to be even and a power of 2 i.e.
2^x=z^2-3^y → 2^x = (z-3^2^w)^2
4. Mar 2, 2012
### .d9n.
so far i have
2^x + 3^y = z^2 [1]
where x, y, and z are non-negative integers.
We can see that [1] is a congruence modulo 8:
3^y = z^2 (mod 8) [2]
When z=1 and and y=2 we have
3^2 = 1 (mod 8) [3]
and by Fermat's theorem. So this means that y must be even.
As y is even, we can write y = 2k for some integer k. So we have:
3^2k + 2^x = z^2
z^2 - 3^2k = 2^x
(z + 3^k)(z - 3^k) = 2^x [4]
This means that the left hand side must be a power of 2, and therefore each factor i.e. (z+3^k) must also be a power of 2. But its not possible to have both factors on the left hand side divisible by 8, as this would imply that their difference
(z + 3^k) - (z - 3^k) = 2*3^k
would be divisible by 8, which is clearly impossible. The only power of 8 that is not a multiple of 8 is 8^0 = 1. So therefore one of the factors must be 1, and this is obviously the smaller factor; we have:
z- 3^k = 1
z = 3^k + 1 [5]
If we substitute that into [4], we obtain:
(3^k+1+3^k )(3^k+1-3^k )=2^x
2∙3^k + 1 = 2^x
2∙3^k = 2^x - 1 [6]
and this means that 2^x - 1 must be twice a power of 3.
So if k = 0, we have 2^x - 1 = 2, which doesn’t work. If k = 1, we have:
6 = 2^x - 1
where the LHS comes out even and the RHS comes out odd, so think i may have gone wrong somewhere, any ideas?
5. Mar 9, 2012
### morphism
Sorry, I'm too tired to read all that, so let me indicate what I'd meant with my hint.
Looking at the equation mod 8 and mod 3, and using the fact that $2\nmid z$ and $3\nmid z$, we can conclude that x and y must be even. Thus the equation $$(2^{x/2})^2 + (3^{y/2})^2 = z^2$$ is primitive Pythagorean, and the solutions to such equations are fully known (google it).
If you write down a parametrization for these Pythagorean triples, and use the fact that the only integer solutions to $$2^a = 3^b + 1$$ are $(a,b) \in \{(1,0), (2,1)\}$ (easy to prove), you should be able to obtain all positive integral solutions (x,y,z) to the original equation. (I got (3,0,3) and (4,2,5) as the only solutions.)
6. Mar 12, 2012
### .d9n.
this is what i have so far, but think i may have gone wrong somewhere, any ideas?
We are looking for solutions of:
2^x + 3^y = z^2 [1]
where x, y, and z are non-negative integers.
We can see that [1] is a congruence modulo 8:
3^y = z^2 (mod 8) [2]
When z=1 and and y=2 we have
3^2 = 1 (mod 8) [3]
and by Fermat's theorem. So this means that y must be even.
As y is even, we can write y = 2k for some integer k. So we have:
3^2k + 2^x = z^2
z^2 - 3^2k = 2^x
(z + 3^k)(z - 3^k) = 2^x
We can see that [1] is a congruence modulo 3:
2^x = z^2 (mod 3) [4]
When z=1 and and y=2 we have
2^x = 1 (mod 3) [5]
and by Fermat's theorem. So this means that x must be even.
As x is even, we can write x = 2k for some integer l. So we have:
3^y + 2^2l = z^2
z^2 - 2^2l = 3^y
(z + 2^l)(z - 2^l) = 3^y
So we are left with
2^2l + 3^2k = z^2
(2^l )^2+ (3^k )^2=z^2 [6]
Which is a Pythagorean triple, where a^2+b^2=c^2. Where a=2^(x/2),b=3^(y/2) and c=z, we can assume that a,b,c are coprime due to Fermat’s Lemma.
We know that c has to be odd because if we assume c is even, then there exists another value C such that c=2∙C. Also that c^2 is divisible by 4 since
c^2=(2C)^2 =4C^2
We know a and b must be odd because a,b,c are coprime. As a and b are odd there must exist values A and B such that
a=2A+1,b=2B+1
but a^2+b^2 cannot be divisible by 4 since
a^2+b^2=(2A+1)^2+(2B+1)^2=4A^2+4A+1+4B^2+4B+1=4(A^2+A+B^2+B)+2
we have a contradiction so c is odd.
So we have a^2=(c+b)(c-b)and c+b and c-b must be even since c and b are odd. So therefore there must exist u,v,w such that
a=2u,c+b=2v,c-b=2w
which means that (2u)^2=(2v)(2w) dividing both sides by 4 gives us u^2=vw.
If we assume v and w are not coprime, then there exists d such that d>1 and d divides both v,w, then d divides both v+w and v-w, but
c+b +c-b=2v+2w
so 2c=2v+2w which means that c=v+w so d divides c.
Also c+b-(c+b)=2v-2w. So 2b=2v-2w which means that b=v-w, so d divides b. Which is a contradiction as c and b are coprime from [6].
So we reject our assumption and v and w are coprime. By properties of coprimes we know that v and w are squares of themselves. So there exists p,q such that
v=p^2,w=q^2
and we have our solutions since
c=v+w=p^2+q^2
b=v-w=p^2-q^2
a=2u=2pq (As u^2=vw which means u=pq)
We know p,q are relatively prime and opposite parity as z is odd, so we are left with.
(p^2+q^2)=(2pq)^2+(p^2-q^2 )^2
so z=c=v+w=p^2+q^2,3^(y/2)=b=v-w=p^2-q^2 and 2^((x/2) )=a=2u=2pq
7. Mar 28, 2012
### morphism
Okay, so we know that 2^(x/2), 3^(y/2) and z are a primitive Pythagorean triple. This means that we can write 2^(x/2)=2nm, 3^(y/2)=n^2-m^2 and z=n^2+m^2 for some integers n and m, with only n even (say). From the first of these equations we find that m=1 and n=2^(x/2 - 1). But then the second equation becomes 3^(y/2) = 2^(x-2) - 1.
This is an equation of the form 2^a = 3^b + 1, and we are looking for integral solutions to such an equation. This forces $b\geq0$ (for otherwise 2^a won't be an integer), which in turn implies that $a\geq1$. I claim that $a \leq 2$ as well. Indeed, if a>2, then 3^b+1=0 mod 8, and this leads to a contradiction. Thus a is either 1 or 2. From here it's easy to see that the only possible solutions (a,b) are (1,0) and (2,1).
Returning to our problem, we see that $(x-2,y/2) \in \{(1,0), (2,1)\}$. You can take it from here.
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## 116c- Lecture 6
April 18, 2008
We revisited the proof of the Schröder-Bernstein theorem and showed how arguments using recursion can provide explicit fixed points for the required map. Recall that if $f:A\to B$ and $g:B\to A$ are injective, we consider the monotone map $\pi:{\mathcal P}(A)\to{\mathcal P}(A)$ given by $\pi(X)=A\setminus g[B\setminus f[X]]$, since if $X$ is a fixed point of $\pi$, then $A\setminus X=g[B\setminus f[X]]$, and we obtain a bijection $h:A\to B$ by setting $h(x)=f(x)$ if $x\in X$ and $h(x)=g^{-1}(x)$ if $x\in A\setminus X$.
We also presented a combinatorial proof considering “paths” along the graphs of $f$ and $g$ (surely folklore, but apparently first recorded by Paul Cohen) and Cantor’s original argument (using choice).
We then started the proof of the equivalence (in ${\sf ZF}$) of several versions of choice:
1. The well-ordering principle (our official version of ${\sf AC}$).
2. The existence of choice functions $f:{\mathcal P}(X)\setminus\{\emptyset\}\to X$ for any set $X$.
3. Zorn’s lemma.
4. Trichotomy: Given any sets $A$ and $B$, one of them injects into the other. (Called trichotomy as it gives that either $|A|=|B|$, $|A|<|B|$ or $|B|<|A|$.)
5. $k$-trichotomy (for a fixed $2\le k\in\omega$): Given any $k$ sets, at least one of them injects into another.
(The proof that (5) implies (1) will be given in Tuesday.)
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Switch to:
GuruFocus has detected 3 Warning Signs with Coca-Cola Femsa SAB de CV \$KOF.
More than 500,000 people have already joined GuruFocus to track the stocks they follow and exchange investment ideas.
Coca-Cola Femsa SAB de CV (NYSE:KOF)
Total Liabilities
\$6,559 Mil (As of Sep. 2016)
Coca-Cola Femsa SAB de CV's total liabilities for the quarter that ended in Sep. 2016 was \$6,559 Mil.
Coca-Cola Femsa SAB de CV's quarterly total liabilities declined from Mar. 2016 (\$6,566.37 Mil) to Jun. 2016 (\$6,452.73 Mil) but then increased from Jun. 2016 (\$6,452.73 Mil) to Sep. 2016 (\$6,559.01 Mil).
Coca-Cola Femsa SAB de CV's annual total liabilities declined from Dec. 2013 (\$7,959.63 Mil) to Dec. 2014 (\$7,344.72 Mil) and declined from Dec. 2014 (\$7,344.72 Mil) to Dec. 2015 (\$6,180.58 Mil).
Definition
Total liabilities are the liabilities that the company has to pay others. It is a part of the balance sheet of a company that shareholders do not own, and would be obligated to pay back if the company liquidated.
Coca-Cola Femsa SAB de CV's Total Liabilities for the fiscal year that ended in Dec. 2015 is calculated as
Total Liabilities = Total Current Liabilities + Total Long Term Liabilities = Total Current Liabilities + (Long-Term Debt + Other Long-Term Liab. = 1785.63059474 + (3706.00365562 + 257.182359282 + NonCurrentDeferredLiabilities + PensionAndRetirementBenefit + Minority Interest) + 65.7894736842 + 132.457702582 + 233.514552186) = 6,181
Total Liabilities = Total Assets (A: Dec. 2015 ) - Total Shareholders Equity (A: Dec. 2015 ) = 12317.1603318 - 6136.58199372 = 6,181
Coca-Cola Femsa SAB de CV's Total Liabilities for the quarter that ended in Sep. 2016 is calculated as
Total Liabilities = Total Current Liabilities + (Total Long Term Liabilities) = Total Current Liabilities + (Long-Term Debt + Other Long-Term Liab. = 1936.22814858 + (3627.59566817 + 732.607204473 + NonCurrentDeferredLiabilities + PensionAndRetirementBenefit + Minority Interest) + 0 + 0 + 262.580806086) = 6,559
Total Liabilities = Total Assets (Q: Sep. 2016 ) - Total Shareholders Equity (Q: Sep. 2016 ) = 12480.0972791 - 5921.08545179 = 6,559
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Related Terms
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Coca-Cola Femsa SAB de CV Annual Data
Dec07 Dec08 Dec09 Dec10 Dec11 Dec12 Dec13 Dec14 Dec15 Dec16 Total Liabilities 3,507 3,134 3,459 3,346 3,754 5,010 7,960 7,345 6,181 7,665
Coca-Cola Femsa SAB de CV Quarterly Data
Sep14 Dec14 Mar15 Jun15 Sep15 Dec15 Mar16 Jun16 Sep16 Dec16 Total Liabilities 8,214 7,345 7,023 6,916 6,696 6,181 6,566 6,453 6,559 7,665
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# Marco Gavanelli – Università di Ferrara, Italy Marco Alberti – Universidade nova de Lisboa, Portugal Evelina Lamma – Università di Ferrara, Italy.
## Presentation on theme: "Marco Gavanelli – Università di Ferrara, Italy Marco Alberti – Universidade nova de Lisboa, Portugal Evelina Lamma – Università di Ferrara, Italy."— Presentation transcript:
Marco Gavanelli – Università di Ferrara, Italy Marco Alberti – Universidade nova de Lisboa, Portugal Evelina Lamma – Università di Ferrara, Italy
Abductive Logic Programming ALP = KB = logic program (set of clauses) A = atoms without definitions, can be assumed IC = Integrity constraints (usually, implications) KB GKB IC Reasoning from effects to causes
Diagnosis headache:- flu. headache:- period. headache:- hangover. flu, vaccine -> false. hangover -> drank. period, sex(male) -> false. ?- headache. Yes, fluMore? ; Yes, period
Event Calculus holdsat(Fluent,Time):- initially(Fluent), not(clipped(0,Fluent,Time)). holdsat(Fluent,Time):- initiates(Action,Fluent), happens(Action,T1), not(clipped(T1,Fluent,Time)). clipped(T1,Fluent,T2):- terminates(Action), happens(Action,T), T1 { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/4/1465552/slides/slide_4.jpg", "name": "Event Calculus holdsat(Fluent,Time):- initially(Fluent), not(clipped(0,Fluent,Time)).", "description": "holdsat(Fluent,Time):- initiates(Action,Fluent), happens(Action,T1), not(clipped(T1,Fluent,Time)). clipped(T1,Fluent,T2):- terminates(Action), happens(Action,T), T1
Abductive Event Calculus holdsat(Fluent,Time):- initially(Fluent), not(clipped(0,Fluent,Time)). holdsat(Fluent,Time):- initiates(Action,Fluent), happens(Action,T1), not(clipped(T1,Fluent,Time)). clipped(T1,Fluent,T2):- terminates(Action), happens(Action,T), T1 { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/4/1465552/slides/slide_5.jpg", "name": "Abductive Event Calculus holdsat(Fluent,Time):- initially(Fluent), not(clipped(0,Fluent,Time)).", "description": "holdsat(Fluent,Time):- initiates(Action,Fluent), happens(Action,T1), not(clipped(T1,Fluent,Time)). clipped(T1,Fluent,T2):- terminates(Action), happens(Action,T), T1
Sound negation p(1). p(f(X)):- q(X). q(2). ?- not(p(Y)). yes,Y\=1, Y\=f(2)
Abduction = constraint solving [Kowalski, Toni, Wetzel 98] headache :- flu. headache :- hangover. flu, vaccine -> false Constraint Store ?-, headache.vaccine flu Constraint Solver fail
Abduction = constraint solving [Kowalski, Toni, Wetzel 98] headache :- flu. headache :- hangover. headache :- period. flu, vaccine -> false. hangover -> drank. period, sex(male) -> false. Constraint Store ?-, headache.vaccine flu Constraint Solver fail
Constraint Handling Rules (CHR) Declarative language for defining constraint solvers Simplification rules: c1, c2,..., cn guard | body activated if some constraints in the store match with c1, c2,..., cn and guard is true removes c1, c2,..., cn from the store and executes body Propagation rules: c1, c2,..., cn ==> guard | body activated if some constraints in the store match with c1, c2,..., cn and guard is true executes body
Example: leq (less or equal) reflexivity@ leq(X,X) true. antisymmetry@ leq(X,Y), leq(Y,X) X=Y. transitivity@ leq(X,Y), leq(Y,Z) ==> leq(X,Z). leq(A,B), leq(B,C), leq(C,A) leq(A,B), leq(B,C), leq(C,A), leq(A,C) leq(A,B), leq(B,A), A=C A=B, A=C
Abduction in CHR [Abdennadher, Christiansen, Dahl] Abducibles mapped to CHR constraints headache :- flu. headache :- hangover. flu, vaccine -> false
Abduction in CHR headache :- flu. headache :-... flu, vaccine ==> false Constraint Store ?-, headache.vaccine flu fail
Abduction in CHR headache :- hangover. headache :-... drank. hangover ==> drank Constraint Store ?- headache. hangover success drank
Abduction in CHR headache :- period. headache :-... sex(male). Constraint Store period, sex(male) ==> false ! CHR: invalid syntax "sex(male)" ! Undeclared constraint sex/1 in head of rule
Problem: Implementation (CHR) Syntax Declarative Semantics Operational Semantics Implementation
Operational semantics Propagation a(X)a(Y), b -> c(X=Y, b) -> c Case analysis (X=Y, b) -> cX=Y, (b -> c) \/ X\=Y Equality rewriting p(A,B,C)=p(D,E,F) A=D, B=E, C=F Unfolding p(X) -> Goalp(X):- a.p(X):-b. a -> Goal.b -> Goal... Constraint solving
Abduction in CHR (SCIFF) headache :- flu. headache :-... Constraint Store ?-, headache.vaccine flu fail ic( )vaccine, flu -> false
Abduction in CHR (SCIFF) headache :- flu. headache :-... Constraint Store ?-, headache.vaccine flu fail ic( )vaccine, flu -> false ic( flu -> false)
Transitions Propagation transition (+ case analysis): abd(X), ic([abd(Y)|Rest]-> Head) ==> copy(ic([abd(Y) |Rest ]-> Head), ic([abd(Y)|Rest]-> Head)), reif_unify(X,Y,Boolean), ( Boolean=1, ic(Rest->Head) ; Boolean=0). ic([abd(vaccine),abd(flu)]-> false) ic([abd(period),sex(male)]-> false) ic([abd(hangover)]-> drank) abd(hangover)
No hashing Does not use CHRs hashing Solution: abducibles are represented with redundant information: abd(Functor, Arity, Atom) So to abduce atom X : abd(X):- functor(F, A, X), abd(F, A, X). E.g., if I abduce atom mother(X,john), the constraint store contains abd(mother, 2, mother(X,john))
Hashing Propagation transition (+ case analysis): abd(F,A,X), ic([abd(F,A,Y)|Rest]-> Head) ==> copy(ic([abd(Y) |Rest ]-> Head), ic([abd(Y)|Rest]-> Head)), reif_unify(X,Y,Boolean), ( Boolean=1, ic(Rest->Head) ; Boolean=0). Requires the first two arguments identical
Hashing Propagation transition (+ case analysis): abd(F,A,X), ic([abd(F,A,Y)|Rest]-> Head) ==> copy(ic([abd(Y) |Rest ]-> Head), ic([abd(Y)|Rest]-> Head)), reif_unify(X,Y,Boolean), ( Boolean=1, ic(Rest->Head) ; Boolean=0). ic([abd(vaccine,1,vaccine),abd(flu,1,flu)]-> false) ic([abd(period,1,period),sex(male)]-> false) ic([abd(hangover,1,hangover)]-> drank) abd(hangover,1,hangover)
Postpone choices new CHR constraint nondeterministic(Goal) says that Goal can open a choice point, so should be called as late as possible. Two phases, declared by a CHR constraint phase/1 : phase(deterministic) : only deterministic goals are executed phase(nondeterministic) : exactly ONE nondeterministic goal can be executed switch2det @ phase(nondeterministic), nondeterministic(G) call(G), phase(deterministic). switch2nondet @ phase(deterministic) phase(nondeterministic)
Postponing nondet. Propagation transition (+ case analysis): abd(F,A,X), ic([abd(F,A,Y)|Rest]-> Head) ==> copy(ic([abd(Y) |Rest ]-> Head), ic([abd(Y)|Rest]-> Head)), reif_unify(X,Y,B), ( B=1, ic(Rest->Head) ; B=0).
Postponing nondet. Propagation transition (+ case analysis): abd(F,A,X), ic([abd(F,A,Y)|Rest]-> Head) ==> copy(ic([abd(Y) |Rest ]-> Head), ic([abd(Y)|Rest]-> Head)), reif_unify(X,Y,B), (B == 1 -> ic(Rest, Head) ; B == 0 -> true ; nondeterministic((B#=1,ic(Rest,Head)) ; B#=0)) ).
Results ExperimentSCIFF 2005 SCIFF 2011 Auction protocol2.27 s0.37 s Block world45.0 s15.7 s A l LoWS Feeble Conformance 84.4 s36.8 s A l LoWS non- conformant 3.7 s3.3 s
Conclusions CHR implementation of an abductive proof-procedure Sound, complete, sound treatment of negation Well integrated with constraint solving, CLP(FD), CLP(R), universally quantified variables, quantifier restrictions, etc. Easy to extend for other features (see other talk after coffee break)
Thank you for your attention! Questions?
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Baseball for the Thinking Fan
You are here > Home > Baseball Newsstand > Baseball Primer Newsblog > Discussion
Baseball Primer Newsblog
— The Best News Links from the Baseball Newsstand
## Friday, August 15, 2014
#### How Bullpens Took Over Modern Baseball
The broader trend that goes back half a century is clear. In 1964 (four years after the save rule first came to baseball), teams used an average of 2.58 pitchers per game, including the starter; today, they’re using 3.92 pitchers per game. In ’64, relievers tossed an average of 2.64 innings per game; today, they’re throwing an eyelash more than three innings per game.
...
As a result, managers choose a few relievers from a phalanx of fireballers, then go get a few more if some of them break down.
In other words, the pitchers might be on the mound for fewer and fewer pitches, but the trend of harder throwers looks like it’s here to stay.
Doing the math, the average reliever in 1964 threw 1.67 IP/G as opposed to 1.03 IP/G today.
Joyful Calculus Instructor Posted: August 15, 2014 at 01:03 PM | 64 comment(s) Login to Bookmark
Tags: jonah keri, relief pitching
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1. shoewizard Posted: August 15, 2014 at 11:45 PM (#4772257)
Simple solution to a lot of todays woes in the game. No more than two relievers may appear in one inning, and no more than 3 relievers in back to back innings. ( so if you use two relievers in the the 7th you can only use 1 in the 8th)
Wont be a dramatic shift, but just enough to move the needle and also add some interesting strategy.
That and of course keep hitters in the box
2. madvillain Posted: August 16, 2014 at 12:11 AM (#4772262)
Nah, those rules will never fly imo. Less invasive (and thus simpler in a more important way) solutions to length of play are available. What if a reliever gets injured? What if he's wildly ineffective? Too many variables in play imo. Just put hitters and pitchers on the clock and go to booth replay, boom you just cut down the average game 20 minutes.
3. Posted: August 16, 2014 at 12:14 AM (#4772264)
Trying to read the article...and the first four paragraphs put me to sleep... what does it ultimately say?
4. Posted: August 16, 2014 at 12:27 AM (#4772265)
1. First chart talks about increase of top speed of pitchers...basically pitchers know they are going to go no more than an inning and are putting more effort at 100%. It's nice seeing confirmation of what most people suspect that there are more pitchers throwing high heat.
2. There is a huge problem using ops+ to analyze changes from year to year, but still it's a good chart to show the difference between performance of relievers and starters. I would argue that the best conclusion to draw from that chart is that there is a good possibility that pitchers who would have been starters are now becoming relievers. Where in the past you could rank teams best pitchers as 5 of a teams 6 best pitchers were starters, it's now probable that on average 3 out of 6 are starters.
I think discussing relief usage is a good topic.
5. Pasta-diving Jeter (jmac66) Posted: August 16, 2014 at 12:28 AM (#4772267)
Trying to read the article...and the first four paragraphs put me to sleep... what does it ultimately say?
if you look at Figure 3 (they're not numbered, but it's the 3rd fig). the OPS+ allowed by relievers has plummeted downwards in the last 50 years--in the 60's and early 70's the OPS+ allowed by relievers was worse than that of starters, but now, it's just the opposite. OPS+ by relievers is around 94, while that for starters is around 103. Which suggests (I hate to say it) the modern use of relievers is effective (yech)
6. Posted: August 16, 2014 at 09:59 AM (#4772330)
Which suggests (I hate to say it) the modern use of relievers is effective (yech)
which anyone who has looked at the issue for more than five minutes has known for a long time. I've been arguing it at least since 2007 when I showed similar charts to the ones in this article at a conference in San Francisco.
Part of the reason relievers have become so effective is that the expansion of their use without a change in roster size means that managers have fewer bench options from which to choose. Because you need more relievers in today's usage model (and the effectiveness of the model means that there is no internal incentive to move away from them), you wind up carrying seven relievers - and either four or five bench players, one of whom (the backup catcher) you probably aren't going to use unless you must.
Here's what I would do:
1. Increase roster size to 30, of which only 25 can be designated as active for a day's game;
2. The five-man taxi squad can have no more than two pitchers;
3. No more than 10 pitchers can be active for any one game.
Would a couple more hitters on the bench make a difference? I don't know, but I'd like to see it tried. I don't particularly like rules that say you can't use more than "x" pitchers in an inning because that puts a manager in a position where he "can't" act. I'd rather see if there's a way to give a manager more options to counter a reliever.
-- MWE
7. McCoy Posted: August 16, 2014 at 10:07 AM (#4772334)
My solution has always been no substitutions during the inning unless it is for an injury. If you want to bring in a reliever, hitter, or a defensive sub you have to do so after 3 outs and before the start of the next half inning. If a player gets pulled because of an injury he must sit out at least 3 games if he is a positional player, 6 games if he is a reliever, or go on the DL. Problem solved.
The playoffs would require different rules. If you pull a player because of injury he cannot play the rest of the series. If you pull a player during the series because of injury and MLB thinks you pulled them even though they weren't injured you forfeit your first round draft pick, player possibly gets suspended, manager/coaches get suspended, GM gets suspended, owner gets suspended, and the team gets hit with a huge fine.
8. Posted: August 16, 2014 at 10:20 AM (#4772335)
Would a couple more hitters on the bench make a difference?
I doubt it. For every Manny Mota there were a dozen Ed Armbristers, guys we only remember for reasons that had nothing to do with their hitting prowess.
9. Posted: August 16, 2014 at 11:29 AM (#4772358)
In ’64, relievers tossed an average of 2.64 innings per game; today, they’re throwing an eyelash more than three innings per game
Doing the math, the average reliever in 1964 threw 1.67 IP/G as opposed to 1.03 IP/G today
So bullpens face just one more batter per game now than 50 years ago; but individual relievers face two batters fewer.
One thing that has clearly disappeared is the quick hook, which accounts for some of this dynamic. IOW, despite a huge drop in CGs, bullpens aren't on balance pitching a lot more of games, because starters so rarely leave early. This doctrine has become pretty inflexible (and perhaps for excellent reasons of staff management overall).
Take two seasons almost at random, by Cub pitchers having undistinguished years: Dick Ellsworth 1964 vs. Jeff Samardzija 2013. Both were regular rotation starters all year with below-average ERAs and W-L records. Ellsworth made 36 starts, Samardzija 33.
Ellsworth's starts ranged between 11.1 and 1.2 innings, 50 and 12 batters faced. He had four starts with more than 40 BF, and five with fewer than 20 BF (parts of the range Samardzija didn't even approach).
Samardzija, by contrast, threw in a range between 9 and 3.1, 33 and 22 BF. He threw 24 starts with BF between 25 and 30 inclusive. By contrast, Ellsworth in 1964 had only eight such 25-30 starts.
The standardization in starter usage nowadays is to me even more striking than the bullpen usage. A starter goes out nowadays, and unless something extremely freaky happens, he will get to 100 pitches and be gone soon thereafter, almost without regard to the score of the game (which includes pinch-hitting strategy, of course; in 1964 I'd say it was much more common to hit for a starter if you were behind at all in the middle innings, no matter what his pitch count at that point).
10. valuearbitrageur Posted: August 16, 2014 at 11:58 AM (#4772383)
What if a reliever gets injured? What if he's wildly ineffective? Too many variables in play imo
The manager can still yank the reliever even if team is over his limit, but it's a balk.
Just forcing relievers to face two batters would be a significant change, it would eliminate running relievers out back to back to maximize matchups, and speed slowest part of game.
Your suggestions are good, they speed game but don't address offensive imbalance or high K rates.
11. PASTE does not get put on waivers in August Posted: August 16, 2014 at 12:50 PM (#4772410)
3. No more than 10 pitchers can be active for any one game.
Only one starter will be active, so you're still allowing for a nine man bullpen here. Unless your unspoken suggestion is that the team can only change its designated pitchers once every week or two.
12. Pat Rapper's Delight (as quoted on MLB Network) Posted: August 16, 2014 at 01:39 PM (#4772427)
One thing that has clearly disappeared is the quick hook, which accounts for some of this dynamic.
I was talking about that recently on another forum. Drives me nuts that managers will leave an ineffective starter (perhaps at the beginning stages of nursing an injury...) to absorb a line like 4-2/3 12 8 8 4 2 just so he can "get his innings in" and "save the 7-man bullpen" because bullpen roles are now so regimented that they don't know what to do if the game dictates they should bring in a reliever before the 5th inning. So they do nothing.
13. PASTE does not get put on waivers in August Posted: August 16, 2014 at 02:00 PM (#4772433)
As a believer in keeping workloads as stable and consistent as possible, I would be inclined to keep a pitcher who has nothing in the game until he throws 90 pitches, in the interests of acclimating his mind and body to going 90-110 pitches every fifth day like clockwork (so long as he isn't hurt, of course). You could pull him from the game and have him throw pitches in the bullpen a while before he takes a shower, but usually in the case where he's getting lit up the game is effectively over by the third inning anyway.
I don't know whether any of the managers who prefer to let a struggling starter struggle think along these lines, or if they all just do it to "save the bullpen".
14. valuearbitrageur Posted: August 16, 2014 at 02:23 PM (#4772437)
I don't know whether any of the managers who prefer to let a struggling starter struggle think along these lines, or if they all just do it to "save the bullpen".
Kirk Gibson does it because his starters "owe him 100 pitches". Despite throwing 44 starts and 300 innings at a 3.00 ERA (130 ERA+ ish) over his first 2 seasons, the next year Daniel Hudson got in Gibby's dog house early because in his first 9 starts only went 6 innings or more 4 times. The other starts he went 5 IP (83 pitches), 4 IP (90 pitches), 3.2 IP (100 pitches), and 1.2 IP twice (40 & 50 pitches). He blew his arm out on the 50 pitch outing.
By my count Hudson still is in deficit 90 pitches to Gibby, which I'm sure he'll demand if and when Daniel returns from back to back Tommy John surgeries.
15. Posted: August 16, 2014 at 02:24 PM (#4772438)
1. Increase roster size to 30, of which only 25 can be designated as active for a day's game;
2. The five-man taxi squad can have no more than two pitchers;
3. No more than 10 pitchers can be active for any one game.
I've been pushing something similar 28 instead of 30, with 25 being designated as active for a series. It's not as drastic of a suggestion that you are putting out there, and I also didn't have a roster requirement. My thought process would be that each team would put their two starting pitchers that wouldn't pitch in the series on the inactive roster, along with any position player that is having a day to day injury or just not good enough, teams would probably then replace one of the starting pitchers with a reliever giving most teams 3 starting pitchers and 8 relievers for a series, but at least one of the spots from the dropped starting pitcher will hopefully be filled with a bat.
I actively oppose any rule that would argue for forcing a pitcher to pitch to a number of batters, that is too much loss of managerial control in my opinion.
16. Lance Reddick! Lance him! Posted: August 16, 2014 at 02:25 PM (#4772439)
Part of the reason relievers have become so effective is that the expansion of their use without a change in roster size means that managers have fewer bench options from which to choose. Because you need more relievers in today's usage model (and the effectiveness of the model means that there is no internal incentive to move away from them), you wind up carrying seven relievers - and either four or five bench players, one of whom (the backup catcher) you probably aren't going to use unless you must.
A return to the 11-man staff while maintaining a 7-man bullpen can be achieved organically by the first team to think of the obvious:
If starters are only being asked to throw 100 pitches, why do they still need the fourth day of rest we decided to give them when they were throwing 130? Shorten the rotation, with the fifth starter becoming a long reliever (or acting in a platoon with the fourth starter), and keep six dedicated short relievers. Voila, bench bat.
3. No more than 10 pitchers can be active for any one game.
Only one starter will be active, so you're still allowing for a nine man bullpen here. Unless your unspoken suggestion is that the team can only change its designated pitchers once every week or two.
His spoken suggestion, directly above that one: 2. The five-man taxi squad can have no more than two pitchers.
17. Posted: August 16, 2014 at 02:32 PM (#4772441)
If starters are only being asked to throw 100 pitches, why do they still need the fourth day of rest we decided to give them when they were throwing 130?
Because your body still needs the time to heal.
18. Lance Reddick! Lance him! Posted: August 16, 2014 at 02:40 PM (#4772443)
The body needs time to recover, but why does it need *as much* when it's been taxed less in the first place? The answer is that it doesn't, or else one-inning relievers would be getting shelved for days at a time.
19. McCoy Posted: August 16, 2014 at 02:51 PM (#4772444)
Thank you, doctor.
20. Posted: August 16, 2014 at 02:52 PM (#4772445)
The body needs time to recover, but why does it need *as much* when it's been taxed less in the first place? The answer is that it doesn't, or else one-inning relievers would be getting shelved for days at a time
The difference between 130 and 110 isn't that big of a difference, basically it means that a modern pitcher probably threw 180 pitches in the game while a pitcher from the 70's threw 210 pitches. It's still a workout. And of course pitchers in the day were getting career ending injuries at a higher rate than modern pitchers.
21. Ziggy: The Platonic Form of Russell Branyan Posted: August 16, 2014 at 05:39 PM (#4772498)
Since the back end of the bullpen is basically AAA guys anyway, why don't teams (more often) artificially increase their roster size by shuffling tired relievers back and forth through AAA? Just tell the minor league manager "don't use these guys, we're recalling them in three days, they need their rest". Hell, it doesn't have to be AAA, since they're never going to pitch there, let it be rookie ball or something. In fact, you don't even need to have them physically go to the minor league team, officially assign them there but keep them physically with the big league club so they can be "called up" whenever necessary.
This has the added advantage that your relievers don't accrue as much service time. Maybe not a big deal for lots of them (bad relievers don't usually make much anyway), but it'll make a difference in some cases.
22. tfbg9 Posted: August 16, 2014 at 05:40 PM (#4772500)
This thread represents to me why this site is still worth coming to. Good thread. Just baseball.
23. Posted: August 16, 2014 at 05:44 PM (#4772504)
ince the back end of the bullpen is basically AAA guys anyway, why don't teams (more often) artificially increase their roster size by shuffling tired relievers back and forth through AAA? Just tell the minor league manager "don't use these guys, we're recalling them in three days, they need their rest".
When a player gets sent down, barring sending someone to the disabled list, they have to stay down for 10 days.
24. Pasta-diving Jeter (jmac66) Posted: August 16, 2014 at 05:47 PM (#4772507)
When a player gets sent down, barring sending someone to the disabled list, they have to stay down for 10 days.
rather specifically to prevent teams from doing what was suggested by Ziggy
25. Ziggy: The Platonic Form of Russell Branyan Posted: August 16, 2014 at 05:49 PM (#4772508)
Well rats. It seemed like such a good idea.
26. McCoy Posted: August 16, 2014 at 06:37 PM (#4772520)
You could still do it but even if the rule didn't exist it would have limited value. You don't really want AAA relievers throwing innings at the major league level. If you have 7 or 8 middle relievers with options of comparable ability and skill you could shuttle them all between MLB and the minors but the options would be the key. How many teams are going to have a bunch of relievers with options on their 25 man roster or on their 40 man roster.
27. Posted: August 16, 2014 at 07:20 PM (#4772535)
You don't really want AAA relievers throwing innings at the major league level
NOW you tell me. Hang on while I copy that sentence and send it to Ron Washington.
28. puck Posted: August 16, 2014 at 07:26 PM (#4772538)
And of course pitchers in the day were getting career ending injuries at a higher rate than modern pitchers.
Wait, is this known? It seems far from an "of course" piece of knowledge.
29. Posted: August 16, 2014 at 07:27 PM (#4772540)
Wait, is this known? It seems far from an "of course" piece of knowledge.
Considering there was no Tommy John surgery, or other ways of making it back to the league, it seems pretty obvious.
30. snapper (history's 42nd greatest monster) Posted: August 16, 2014 at 07:36 PM (#4772542)
I've been pushing something similar 28 instead of 30, with 25 being designated as active for a series. It's not as drastic of a suggestion that you are putting out there, and I also didn't have a roster requirement. My thought process would be that each team would put their two starting pitchers that wouldn't pitch in the series on the inactive roster, along with any position player that is having a day to day injury or just not good enough, teams would probably then replace one of the starting pitchers with a reliever giving most teams 3 starting pitchers and 8 relievers for a series, but at least one of the spots from the dropped starting pitcher will hopefully be filled with a bat.
I actively oppose any rule that would argue for forcing a pitcher to pitch to a number of batters, that is too much loss of managerial control in my opinion.
Hate this idea. You're just giving them MORE RPs to use. Every team will choose to replace the two SPs with 2 more RPs.
1) No more than 11 pitchers on the 25-man roster. Ever.
2) Each pitcher used must face 3 batters. If a pitcher is injured and must leave before the 3 batters, he may not be used for 7 days.
31. Posted: August 16, 2014 at 07:43 PM (#4772547)
1) No more than 11 pitchers on the 25-man roster. Ever.
2) Each pitcher used must face 3 batters. If a pitcher is injured and must leave before the 3 batters, he may not be used for 7 days.
I can support the first thing, but can't support the second. Roster construction doesn't really affect the game in front of the fans for the most part, but the second thing absolutely does, and I hate it. One batter, one pitcher isn't really that big of a deal. Only reason to propose it is for time of game, and realistically speaking, we are talking less than two minutes, per game on average...you want to save time, you concentrate on the batting box antics.
32. JJ1986 Posted: August 16, 2014 at 07:53 PM (#4772552)
The easiest way to adjust pitching changes to speed up the game is to remove the warmup pitches. The pitcher gets one to test the mound and then is ready to go.
33. valuearbitrageur Posted: August 16, 2014 at 08:03 PM (#4772557)
Each pitcher used must face 3 batters. If a pitcher is injured and must leave before the 3 batters, he may not be used for 7 days.
Why not just make it similar to a balk, batter takes first, all baserunners advance? A manager can't angle shoot that very easily, if ever, even if he planned to intentionally walk the batter, he's still advancing other baserunners.
34. valuearbitrageur Posted: August 16, 2014 at 08:22 PM (#4772566)
The easiest way to adjust pitching changes to speed up the game is to remove the warmup pitches. The pitcher gets one to test the mound and then is ready to go.
Another good idea. Obviously they would fully warm up in the bullpen first so should not a safety issue and doesn't just speed play. Since manager has to warm up relievers more often to prepare for use and their first pitches are more tentative, also reduces reliever effectiveness.
35. Posted: August 16, 2014 at 08:28 PM (#4772571)
The easiest way to adjust pitching changes to speed up the game is to remove the warmup pitches. The pitcher gets one to test the mound and then is ready to go.
You can tell when someone posts something that has probably never played the game. They might not need eight pitches, but just one? That is insane.
Let's be perfectly honest. No rule that is going to reduce commercial time is ever going to happen. Sometimes when I see these suggestions, I wonder if anyone making them are actually baseball fans.
36. JJ1986 Posted: August 16, 2014 at 09:33 PM (#4772600)
You can tell when someone posts something that has probably never played the game. They might not need eight pitches, but just one? That is insane.
Pitchers get mostly ready in the bullpen. If he needs 7 more pitches out there, then he's ready to come in a minute or two later.
37. Howie Menckel Posted: August 16, 2014 at 09:39 PM (#4772604)
I have mentioned before my 1970 Orioles example, but it's worth it again I think (the shoulder seasons are pretty similar) because to anyone under 40 this might seem unfathomable - or at least unfathomable that us current posters actually remember such a time. team went 108-54 and won World Series.
Your SPs are 20-game winners Cuellar, McNally and HOF Palmer starting 40-40-39 games for 119.
Then Phoebus starts 21 of 27 G to get to 140, and Hardin starts 19 of 36 to get to 159 starts (Phoebus had the 4th and final slot til late June, had a bad outing, and Hardin pretty much took over the 4th SP role til Sept when Phoebus - who had only tossed 18 total IP in July/Aug (might also be injury) - got more chances).
The other 3 starts were by Marcelino Lopez - 1 in Aug and 2 in Sept for a team that won the pennant by 15 G.
In the bullpen, no one threw as many as 62 IP and the entire pen only needed to throw fewer than 400 IP thanks to 60 CG - and more than that were nearly completed.
Richert, DHall (4 IP in August), Watt, Leonhard (who rarely pitched in the 2nd half), Lopez, Drabowsky (traded to KC in June) and Hardin threw almost all the R IP. Only 12 pitchers hurled for these Orioles all season - the 12th was Fred Beene, who relieved 4 times (2 in July, 2 in Sept).
The Orioles alternated between a 9- and 10-man staff that season.
Then they dumped Phoebus, prospect Enzo Hernandez, Beene, and Severinsen on Padres in the offseason for Pat Dobson - who joined the 3 above as Orioles 20-game winners in 1971.
38. Walt Davis Posted: August 16, 2014 at 09:44 PM (#4772607)
Since the back end of the bullpen is basically AAA guys anyway, why don't teams (more often) artificially increase their roster size by shuffling tired relievers back and forth through AAA?
This is exactly what teams do.
Relievers throw about 490-500 innings a year and there are 7 bullpen slots. So that's 70 innings per slot.
In 2013, only 35 individual pitchers even hit the 70 IP mark. Another 57 made it to 60. We're at an average of 3 relievers per team at this point and we're already a bit short of the 210 innings we need out of those 3 slots.
Needless to say, everything after that is a mess as there are another 36 guys who hit the 50 mark. We're already 4 slots in and we're sitting at probably around 250 of the 280 innings we need out of them.
So we've got to fill 250 relief innings out of 3 bullpen slots and no single pitcher is even going to hit 50 innings. All told last year, there were 327 pitchers, at least 95% of their appearances in relief, who pitched at least 5 innings. You can add a few guys who get bounced out of the rotation that are missed by the 95% thing.
There's infinite churn out there for replacing tired and injured guys. It seems rather likely that a large percentage of reliever DL trips are "tired veteran that I want to keep but don't have an option on".
Here's Robert Coello who threw 17 innings for the Angels in 2013 (and some in earlier years too). His first appearance for them was on May 12. Within that first WEEK, he made 5 appearances and threw 7 innings. He then had 4 full days off then made 3 appearances, 3.1 IP in a week; then 4 appearances 3.2 IP in a week, then one more before going down (or DL or somewhere). The guy was around for roughly 27 games, made 13 appearances and 15 IP -- conveniently exactly 1/6 of a season. So if you extrapolated that out, that's 78 games and 90 innings, a massive load in today's game.
Then they brought him back in Sept.
It's hard to find a good reliever who can give you 90 innings a year. It is not so hard to find 6 Robert Coellos. For example, Micheal Roth on the 2013 Angels. From game 11-25, he made 6 relief appearances and one (looks emergency) start -- 9.1 innings total, 6 in relief including 2 2-IP stints. He popped back on 13 May, threw 2.1 innings of relief then 2 innings on 15 May then back down. Pops back up in July, makes 4 sppearances in 8 games, then a few days off, then 2 appearances in 2 days. I can't say for sure but he was on the roster for possibly as few as 32 games and made 14 relief appearances covering 17 innings -- or a 70 game, 85 inning pace.
JC Gutierrez, 30, seems to have started the season as the Royals' mop-up reliever and had just 25 appearances and 29 IP through 90+ games. He was traded to the Angels at the deadline. Over the next 29 games for the Angels, he made 16 appearances and 15 IP -- again an 80-90 IP pace. He also made 11 appearances over their last 22 games. Cory Rasmus was picked up in late Aug and made 6 appearances in 8 games, got a few days off, then 10 appearances in 20 - total 16 appearances, 15 IP in 32 games.
Baseball hasn't completely gotten rid of the relievers who throw 80+ innings a year exactly, they just cobble them together from the detritus of the waiver wire and AAA.
39. Posted: August 16, 2014 at 09:45 PM (#4772608)
Pitchers get mostly ready in the bullpen. If he needs 7 more pitches out there, then he's ready to come in a minute or two later.
They get their body ready, every mound, every night is different. Depending on humidity, the specific groundscrew, and in some cases, the mounds are tailored based upon the hometeam starting pitcher...so yes, it takes more than one throw to get a proper feel for the mound. This isn't pickup softball, these are elite professionals who can tell the difference between 1/4 ounce weight of the bat etc.
40. Posted: August 16, 2014 at 09:49 PM (#4772609)
I have mentioned before my 1970 Orioles example, but it's worth it again I think (the shoulder seasons are pretty similar) because to anyone under 40 this might seem unfathomable - or at least unfathomable that us current posters actually remember such a time. team went 108-54 and won World Series.
The thing about mentioning the Orioles, is that even in their day and age, it was an outlier that EVERYONE at the time talked about. Grab the 1968 Tigers, and look at how many of their pitchers are pitching effectively in 1972. In fact grab any team from the late 60's and you will see the same thing. Pitchers just didn't have careers in the 70's. For every Nolan Ryan you can find, you could find a half dozen John Fulghams.
41. bobm Posted: August 16, 2014 at 10:00 PM (#4772618)
The standardization in starter usage nowadays is to me even more striking than the bullpen usage. A starter goes out nowadays, and unless something extremely freaky happens, he will get to 100 pitches and be gone soon thereafter, almost without regard to the score of the game (which includes pinch-hitting strategy, of course
From B-R:
Starts by Pitch Count
```Year P<=89 90<=P<=99 100<=P<=109 P>=110 UNK Total
1991 1537 765 820 1085 1 4208
2013 1248 1494 1482 637 1 4862
```
% of Starts (rounded) by Pitch Count
```Year P<=89 90<=P<=99 100<=P<=109 P>=110 UNK Total 90<=P<=109
1991 37% 18% 19% 26% 0% 100% 38%
2013 26% 31% 30% 13% 0% 100% 61%
```
42. Walt Davis Posted: August 16, 2014 at 10:03 PM (#4772622)
The OPS+ chart is reminiscent of one I did a long, long time ago for an article here on reliever usage -- for which you can find some of the unformatted HTML code here (http://www.baseballthinkfactory.org/primate_studies/discussion/wdavis_2003-01-15_0) :-) ... don't ask me.
It's no longer relevant I suppose but in the early days of this shift it seemed important to look at 1-inning relievers vs. multi-inning. Some of that drop he shows from mid-90s to early 00s is, if I recall, much more dramatic if you try to remove the mop-up guys. That is, the guys you'd actually use in a close game were substantially more dominant relative to starters than that chart makes it appear.
It's somewhat moot now that pretty much everybody is a 1-inning guy but in the transition from fireman to closer there were still a fair number of set-up guys, etc. pitching multiple innings (I think I used 1.1 IP/appearance as my cutoff). I used ERA+ at the time (don't think we had OPS+ by pitcher in those days) but my memory is that for short relievers, it was on the order of a 115-120 ERA+ average (vs. 100ish for starters) and the overall reliever vs. starter comp was pulled back to about 108-110 by horrible mop-up guys.
teams used an average of 2.58 pitchers per game, including the starter; today, they’re using 3.92 pitchers per game. In ’64, relievers tossed an average of 2.64 innings per game; today, they’re throwing an eyelash more than three innings per game.
I hope this sinks in because this is the major change we've seen in bullpen usage. Starters aren't, on average, leaving games that much earlier -- about 1/3 of an inning although I suspect that's dropping a bit again. It's the number of relievers used to cover those last 3 innings that has skyrocketed but, as MWE, this guy and even me back in the day pointed out, that seems to be because it's really effective.
The other change is the one discussed above that starters are essentially required to go at least 5 innings in every start. So while the mean IP/start hasn't changed a lot, the variation around that mean has been substantially reduced. And of course the shift from 4-day rotations to 5-man rotations reducing the number of starts per season.
But do keep that in mind next time somebody points out that all this "babying" of pitchers doesn't seemed to have reduced injuries. On a per-start basis, they aren't being babied -- they're pitching nearly the same IP/start. Why the drop in the number of starts hasn't had more obvious benefit is a good question but in the old days, unless it was Fergie or similar, those CG they threw were balanced by getting yanked in the 3rd when it wasn't their day.
43. The District Attorney Posted: August 16, 2014 at 10:15 PM (#4772632)
Good Lord, I've only ever heard of one of the four guys mentioned in #38. (Which itself is not ideal from the fan point of view, IMHO. I don't want it to be like football where an individual player might have such a small and specialized role that I don't even know his name.)
I think that:
• In life, you're less likely to get hurt when you work out frequently and less strenuously, as opposed to infrequently and more strenuously. If you combined a four-man rotation with a slightly quicker hook than currently (i.e., the "danger zone" is more like 80 pitches, rather than 100), you'd almost surely win more games, and I think you'd see injuries decline if anything -- certainly not increase. Of course, I can't prove this. But it makes sense to me, and I don't think anyone can disprove it either. Baseball is unfortunately not a science where you get to run controlled experiments. I do at least firmly believe science supports my first statement.
• As far as the bullpen goes, there's no doubt that when a guy knows he's only going to pitch one inning, he can go all-out, and will pitch better in that one inning. However, there is a tradeoff involved. In order to make it possible to do that with every reliever, you have to carry fewer hitters. Hitters also benefit from rest (albeit not as much as pitchers), and there is more to be gained from platooning hitters than from platooning relievers. I can justify carrying "extra" relievers if the purpose is to spread around the relief workload and make each guy better. I don't think that the LOOGY who's going to pitch 30 innings is worth the tradeoff.
• I don't care about the rules nearly as much, but I guess I'll offer an opinion. Limiting relief use via the rules does feel artificial to me. I'd like to see both a smaller roster and no warmup throws from the mound, but it is probably pointless to even bring them up, because one costs the union jobs, and the other costs everyone money. So I end up thinking that practically speaking, the way to speed up the game is to eliminate in-game dawdling, which has no beneficiaries.
44. 6 - 4 - 3 Posted: August 16, 2014 at 10:44 PM (#4772640)
Call the strikezone as in the rulebook
=> increase the likelihood of a taken pitch being called strike
=> decrease the likelihood of a player reaching via walk
=> batters more aggressive early in the count
=> starters throw fewer pitches, last longer
=> fewer walks and strikeouts
=> lions and zebras live together in perfect harmony
45. Posted: August 16, 2014 at 10:55 PM (#4772641)
Call the strikezone as in the rulebook
=> increase the likelihood of a taken pitch being called strike
=> decrease the likelihood of a player reaching via walk
=> batters more aggressive early in the count
=> starters throw fewer pitches, last longer
=> fewer walks and strikeouts
=> lions and zebras live together in perfect harmony
Not sure what this helps. As it stands strike outs are increasing, walks are decreasing already. Not sure we want anything that would accelerate those actions.
46. Hysterical & Useless Posted: August 16, 2014 at 10:58 PM (#4772643)
Pitching changes provide commercial revenue, as well as a chance to get a snack/take a leak/update your scorecard. Rule changes to limit pitching changes are not needed and would be detrimental to the game. Virtually the entire problem with game time/pace is that every hitter nowadays thinks he's Mike ####### Hargrove. Why don't we petition MLB to have the umpires enforce the rules? In baseball, players can only ASK for time to be called; the umpire is not obligated to grant the request, and they should be instructed NOT to grant it after every ####### pitch. If the hitter steps out and time isn't called, pitcher can lob one to the catcher and it should be called a strike. Simple. No more 15 pitch innings lasting 10 minutes.
47. Betts, Bogaerts, and D Price(GGC) Posted: August 17, 2014 at 10:23 AM (#4772744)
Good Lord, I've only ever heard of one of the four guys mentioned in #38. (Which itself is not ideal from the fan point of view, IMHO. I don't want it to be like football where an individual player might have such a small and specialized role that I don't even know his name.)
This. The Red Sox used 48 players last year and 26 of them were pitchers. Contrast this with 1963, were they used 31 players, 16 of them pitchers. They were a bad team in 1963 and world champs last year. I think part of this was it was harder back then to call up players from the minors, but that's still an increase in players used of roughly 50%
48. McCoy Posted: August 17, 2014 at 10:56 AM (#4772755)
Last year the Cubs used 56 players with 31 of them being pitchers. In 1879 they used two pitchers for the entire season and in only one game did the starter not complete the game!
49. Posted: August 17, 2014 at 11:17 AM (#4772766)
1)No more than 11 pitchers on the 25-man roster. Ever.
Wait, wouldn't such a rule dictate that mean Adam Dunn would never again pitch to a batter? Why do you hate America?
50. Posted: August 17, 2014 at 11:22 AM (#4772770)
Last year the Cubs used 56 players with 31 of them being pitchers. In 1879 they used two pitchers for the entire season and in only one game did the starter not complete the game!
In that same year, the pennant winning Providence Grays also used only two pitchers for the entire season. A 19 year old pitcher, Monte Ward, won more games than the Cubs all by himself, going 47-19 in 587 innings.
51. snapper (history's 42nd greatest monster) Posted: August 17, 2014 at 12:30 PM (#4772812)
I can support the first thing, but can't support the second. Roster construction doesn't really affect the game in front of the fans for the most part, but the second thing absolutely does, and I hate it.
Why do you hate it? I can't stand seeing a RP used for one batter.
If you can't get both righties and lefties out, you don't belong in MLB. To me, anything that reduces specialization is good for the game.
And, yes, I want the DH abolished. If you're too old, slow, fat or crippled to play the field, you don't belong in MLB either.
52. Posted: August 17, 2014 at 12:42 PM (#4772819)
Why do you hate it? I can't stand seeing a RP used for one batter.
Because you are taking away managerial options. I don't like limiting things just because something is un-pleasing. It's fundamentally changing the concept of the game. Baseball already has a roster limit that managers have to abide by, no return to the game once they left, this provides enough strategy to make teams be wary of too much changing out of players. If you limit the number of 'pitchers' on the roster, you have already done enough to eliminate some one batter appearances, but you haven't completely eliminated the option, I prefer the game to have options.
If you can't get both righties and lefties out, you don't belong in MLB. To me, anything that reduces specialization is good for the game.
I'm the opposite, I can't stand that every team has basically 3 utility players and no speed burners or power bench bats from the right side etc. The lack of specialization leads to lower quality play.
53. snapper (history's 42nd greatest monster) Posted: August 17, 2014 at 01:15 PM (#4772841)
Because you are taking away managerial options. I don't like limiting things just because something is un-pleasing. It's fundamentally changing the concept of the game. Baseball already has a roster limit that managers have to abide by, no return to the game once they left, this provides enough strategy to make teams be wary of too much changing out of players. If you limit the number of 'pitchers' on the roster, you have already done enough to eliminate some one batter appearances, but you haven't completely eliminated the option, I prefer the game to have options.
Yeah, I don't watch baseball for the managing. The less I know about what the managers are thinking, the happier I am.
I'm the opposite, I can't stand that every team has basically 3 utility players and no speed burners or power bench bats from the right side etc. The lack of specialization leads to lower quality play.
But, that's true because of the specialization of relief pitching. I don't care for pinch-running specialists, but the lack of bench bats, and platoon bats is directly the result of 7-8 man bullpens.
Force GMs/managers to have a 6 man bench, and you'll get a lot more variety on that bench. And, none of those players, even a platoon 1B or LF, will be as specialized as a LOOGY or ROOGY.
54. Posted: August 17, 2014 at 01:32 PM (#4772852)
But, that's true because of the specialization of relief pitching. I don't care for pinch-running specialists, but the lack of bench bats, and platoon bats is directly the result of 7-8 man bullpens.
I know that, but it's still a loss of specialization. The first part that I agreed with from the post you quoted was a suggestion of limiting the roster number of pitchers. This expands the benches a little bit, and also brings down the number of available relievers which limits the likelihood of the one batter appearances, but doesn't eliminate it as a possibility.
55. snapper (history's 42nd greatest monster) Posted: August 17, 2014 at 01:38 PM (#4772856)
I know that, but it's still a loss of specialization. The first part that I agreed with from the post you quoted was a suggestion of limiting the roster number of pitchers. This expands the benches a little bit, and also brings down the number of available relievers which limits the likelihood of the one batter appearances, but doesn't eliminate it as a possibility.
Right, but I think it's a good loss of specialization. Just like eliminating the DH is a good loss of specialization.
To me, the goal should be that any attempt at specialization should come with you having to pay the natural price for the advantage you gain.
If you want to use a LOOGY to get out a very dangerous lefty, fine, but he's going to be exposed to the RH batting next. If you want to get a big bat/no glove guy in the lineup, fine, but you're going to have to live with his glove for at least 6 or 7 innings. If you want to put an all glove/no bat guy at CF or SS, you have to let him bat 2 or 3 times. If you want to play a platoon at a position, that guy has to face RPs of the opposite hand, at least some of the time.
It's all nicely symmetric, except the LOOGY/ROOGY and DH situations, where you get specialization w/o cost.
56. nick swisher hygiene Posted: August 17, 2014 at 01:52 PM (#4772864)
I'm with snapper on this one.......basically, "in-game strategy" is pretty dull. it consists of a few simple sets of choices which baseball fans obsess over.
but that said, I'd rather see the timing rules that are already part of the game enforced before making rule changes.
to step up the pace of the game, stop the stepping out and stepping off; if that doesn't work, THEN take bigger steps.
57. Posted: August 17, 2014 at 02:25 PM (#4772882)
Just for the record, in 2013 there were 1163 relief appearances that was one batter only. This averages out to just under 39 per team(with the most being 73 for the Giants and fewest being 16 for the Nationals.
Obviously this is happening more than it was in the past, but even as recently as 1980, teams averaged over 17 one batter faced performances per year. I support trying to reduce that number of times one batter relievers are used, but I just can't support eliminating that option.
58. snapper (history's 42nd greatest monster) Posted: August 17, 2014 at 02:28 PM (#4772884)
This averages out to just under 39 per team
So, every other game we're wasting 4-6 minutes on an OOGY. Seems like a good place to lop 3 minutes off the average game.
59. Posted: August 17, 2014 at 03:01 PM (#4772908)
I just think that there are other ways to fix that problem than eliminating it as an option.
60. snapper (history's 42nd greatest monster) Posted: August 17, 2014 at 03:02 PM (#4772910)
I just think that there are other ways to fix that problem than eliminating it as an option.
OK, but why is that option important to baseball? Managers, for generations, basically operated as if hey didn't have the option, and baseball didn't seem to suffer.
61. Posted: August 17, 2014 at 03:14 PM (#4772917)
I'd be fine with a rule against mid-inning pitching changes. There are already constraints: you can't pull back a pitcher once introduced. Simply extend that stipulation. It's not an unfair tactical constraint, it's just a different one.
62. snapper (history's 42nd greatest monster) Posted: August 17, 2014 at 03:15 PM (#4772920)
I'd be fine with a rule against mid-inning pitching changes. There are already constraints: you can't pull back a pitcher once introduced. Simply extend that stipulation. It's not an unfair tactical constraint, it's just a different one.
Agree about the fairness, but I think an outright ban goes too far. If a guy starts and inning and has nothing, he shouldn't be left out there to get pummeled. That's why I like the 3 batter rule. I could be sold on 4 batters.
63. Posted: August 17, 2014 at 03:21 PM (#4772925)
OK, but why is that option important to baseball? Managers, for generations, basically operated as if hey didn't have the option, and baseball didn't seem to suffer.
Because I just don't see what the purpose is of removing options that exist within the fundamental rules. There is a cost to using a one batter faced option, and that is the loss of availability of a relief pitcher later in the game. That is sufficient penalty. If you combine it with a rule limiting teams roster size based upon makeup(no more than 11 pitchers) and it's a real detriment to using a specialist reliever. It's basically one bullet that you have to choose the best time to use.
Just like if you have a great pinch hitter, you are limited on when the best time to utilize him. I just cannot support rules that limit the between the lines management. Before the game limitations sure, equipment limitations most definitely, but not saying 'we have a special rule, specifically for this situation, simply because we don't like seeing this situation' it seems like a petty rule in my opinion.
64. PASTE does not get put on waivers in August Posted: August 18, 2014 at 07:44 AM (#4773292)
Agree about the fairness, but I think an outright ban goes too far. If a guy starts and inning and has nothing, he shouldn't be left out there to get pummeled. That's why I like the 3 batter rule. I could be sold on 4 batters.
Ah, that's my cue to point out yet again how well a nice, simple rule that the pitcher can be removed midinning only once he's allowed a run of his own would be.
There's never any problem, in my mind, with changing pitchers between innings--so if you want to put a LOOGY in there with two outs in the inning to strike Pedro Alvarez out and then change to a real pitcher for the next inning, that's no problem. You're not interrupting the action.
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# Thread: 3 children, probability of boy/girl = 0.5 EASY!
1. ## 3 children, probability of boy/girl = 0.5 EASY!
Here is the problem (high school) Probability of getting a boy/girl is 0.5. And a family has 3 children. What is the probability that the oldest is a boy?
Also what is the probability that the oldest is a boy and the youngest is a girl?
2. ## Re: 3 children, probability of boy/girl = 0.5 EASY!
Originally Posted by karldiesen
Here is the problem (high school) Probability of getting a boy/girl is 0.5. And a family has 3 children. What is the probability that the oldest is a boy?
Also what is the probability that the oldest is a boy and the youngest is a girl?
There are eight strings of B's and G's of length three.
Half of them begin with a B. So what is the first answer?
Two of them begin with a B and end with a G. So?
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https://wiki.hpc.tulane.edu/trac/wiki/cypress/Programming/BiConjugateGradient?version=1
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Version 1 (modified by cmaggio, 9 years ago) ( diff )
migrated content from ccs wiki
# Bcg.c
```/*
Large Scale Computing
solve Ax=b
General Sparse Matrix A
solve_cg solve Ax=b with BCG method
this returns the number of iterations, or negative if failed.
*/
#include<stdio.h>
int solve_bcg(int dim, /* dimension */
int nnz, /* # of non-zeros in the matrix */
double *nzval, /* array of nonzero values */
int *colidx, /* array of column indices of the nonzeros */
int *rowptr, /* the first column of nonzero in each row */
/* rowptr[j] stores the location of nzval[] and colidx[], */
/* which starts row j. This has dim+1 entry that is nnz */
double *x, /* solition */
double *b, /* right hand side */
double tol); /* tolerance */
```
```/*
Large Scale Computing
solve Ax=b
General Sparse Matrix A
*/
#include<stdio.h>
#include<stdlib.h>
#include<math.h>
int solve_bcg(int dim, /* dimension */
int nnz, /* # of non-zeros in the matrix */
double *nzval, /* array of nonzero values */
int *colidx, /* array of column indices of the nonzeros */
int *rowptr, /* the first column of nonzero in each row */
double *x, /* solition */
double *b, /* right hand side */
double tol){ /* tolerance */
double *r;
double *p;
double *Ap;
double *rt;
double *pt;
double *ATpt;
double *nzvalT; /* AT array of nonzero values */
int *colidxT; /* AT array of column indices of the nonzeros */
int *rowptrT; /* AT the first column of nonzero in each row */
double rr,ptAp,rr1,alpha,beta;
int i,j,count;
int col, relpos;
int *marker;
/* Allocate Working Spaces */
r = (double *)calloc(dim, sizeof(double));
rt = (double *)calloc(dim, sizeof(double));
p = (double *)calloc(dim, sizeof(double));
pt = (double *)calloc(dim, sizeof(double));
Ap = (double *)calloc(dim, sizeof(double));
ATpt = (double *)calloc(dim, sizeof(double));
nzvalT = (double *)calloc(nnz, sizeof(double));
colidxT = (int *)calloc(nnz, sizeof(int));
rowptrT = (int *)calloc(dim+1, sizeof(int));
if ((r == NULL) || (rt == NULL) ||
(p == NULL) || (pt == NULL) ||
(Ap == NULL) || (ATpt == NULL) ||
(nzvalT == NULL) || (colidxT == NULL) ||
(rowptrT == NULL)){
printf("memory allocation failed\n");
return -1;
}
/*
Transpose A
*/
/* Allocate marker */
marker = (int *)calloc(dim, sizeof(int));
for (i = 0 ; i < dim ; i++) marker[i] = 0;
/* Get counts of each column of A, and set up column pointers */
for (i = 0 ; i < dim ; i++){
for (j = rowptr[i] ; j < rowptr[i+1]; j++) ++marker[colidx[j]];
}
rowptrT[0] = 0;
for (j = 0 ; j < dim ; j++) {
rowptrT[j+1] = rowptrT[j] + marker[j];
marker[j] = rowptrT[j];
}
/* Transfer the matrix into the compressed column storage. */
for (i = 0 ; i < dim ; i++) {
for (j = rowptr[i] ; j < rowptr[i+1] ; j++) {
col = colidx[j];
relpos = marker[col];
colidxT[relpos] = i;
nzvalT[relpos] = nzval[j];
++marker[col];
}
}
free(marker);
/*
*/
/* compute r0 */
for (i = 0 ; i < dim ; i++){
r[i] = b[i];
for (j = rowptr[i] ; j < rowptr[i+1] ; j++){
r[i] -= nzval[j] * x[colidx[j]];
}
}
rr = 0.0;
for (i = 0 ; i < dim ; i++){
p[i] = r[i]; /* p0 = r0 */
rt[i] = r[i]; /* rt0 = r0 */
pt[i] = r[i]; /* pt0 = r0 */
rr += r[i] * r[i]; /* rr = r.r */
}
/* bcg iteration */
count = 0;
while(rr > tol){
for (i = 0 ; i < dim ; i++){
/* Ap = A*p */
Ap[i] = 0.0;
for (j = rowptr[i] ; j < rowptr[i+1] ; j++){
Ap[i] += nzval[j] * p[colidx[j]];
}
/* ATpt = A^T*pt */
ATpt[i] = 0.0;
for (j = rowptrT[i] ; j < rowptrT[i+1] ; j++){
ATpt[i] += nzvalT[j] * pt[colidxT[j]];
}
}
// alpha = r.r / pt.Ap
ptAp = 0.0;
for (i = 0 ; i < dim ; i++){
ptAp += pt[i] * Ap[i];
}
alpha = rr / ptAp;
//Beta
rr1 = 0.0;
for (i = 0 ; i < dim ; i++){
x[i] += alpha * p[i];
r[i] -= alpha * Ap[i];
rt[i] -= alpha * ATpt[i];
rr1 += r[i] * rt[i];
}
beta = rr1 / rr;
for (i = 0 ; i < dim ; i++){
p[i] = r[i] + beta * p[i];
pt[i] = rt[i] + beta * pt[i];
}
rr = rr1;
count++;
}
/* Deallocate Working Spaces */
free(r);
free(p);
free(Ap);
free(rt);
free(pt);
free(ATpt);
free(nzvalT);
free(colidxT);
free(rowptrT);
return count;
}
```
Note: See TracWiki for help on using the wiki.
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# Maths2Measure's Shop
Average Rating4.26
(based on 113 reviews)
I trained as a teacher of secondary Mathematics and evolved into a Computing teacher as well. I offer mainly Mathematics resources written for the current Australian Curriculum (NSW). Luckily Mathematics is recognisable all over the world, so I hope that teachers in different countries can use some of my work.
117k+Views
I trained as a teacher of secondary Mathematics and evolved into a Computing teacher as well. I offer mainly Mathematics resources written for the current Australian Curriculum (NSW). Luckily Mathematics is recognisable all over the world, so I hope that teachers in different countries can use some of my work.
#### Area of right-angled triangles
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This is a set of 2 activities helping students understand how to explore the relationship between right-angled triangles and rectangles, leading to finding the area of a right-angled triangle. The first activity has 2 pages that need to be printed on separate pages because students will need to cut out and assemble rectangles. The second activity needs a page of cm graph paper. Full solutions are also provided. (Australian Curriculum) NSW MA3-10MG selects and uses the appropriate unit to calculate area, including areas of squares, rectangles and triangles
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(0)
Two doubled sided worksheets with solutions. Students work out the answers to the questions by applying their knowledge of the main four compass directions. Suitable for a lesson or as an introduction to compass directions. (Australian Curriculum) NSW MA3-17MG locates and describes positions on maps using grid-reference system
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#### Missing shapes
(3)
A Maths worksheet with answers, where students need to work out the pattern and draw the missing shape. A good preliminary activity before the introduction of number patterns. (Australian Curriculum) NSW MA3-8NA analyses and creates geometric and number patterns, constructs and completes number sentences, and locates points on the Cartesian plane
#### Financial Maths Word Search
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A literacy activity with answers, for students to familiarise themselves with the vocabulary associated with the topic of Financial Mathematics. Can be used at the end of a lesson as a review.
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A double sided worksheet with solutions. Students need to gather data on pet ownership from the class. This data is then organised into a table. The students then use the data to answer questions and are able to distinguish between the categorical and numerical data gathered. Dot and column graphs are used to show data. Worksheet could be used as a lesson to show students the difference between these two types of data. (Australian Curriculum) NSW MA3-18SP uses appropriate methods to collect data and constructs, interprets and evaluates data displays, including dot plots, line graphs and two-way tables
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#### Algebra trails
(0)
Students need to know their algebra facts when completing these worksheets. There are 2 double-sided worksheets with solutions. The four operations are used with the positive and negative algebraic terms. These worksheets could be used with Gifted and Talented students. (Australian Curriculum) NSW MA4-8NA generalises number properties to operate with algebraic expressions
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(0)
A set of 5 graded single sided worksheets with solutions. A suitable resource for students who need more practice. The first column could be worked on in class and the second column left for the students to finish for homework. Australian Curriculum (NSW) MA4-8NA generalises number properties to operate with algebraic expressions. If you like my work, please write a review to help others, and check out my shop for more useful activities.
#### Matchstick shapes
(0)
A double sided worksheet with solutions. Students can use matchsticks to help them form different 2-dimensional shapes. They need to think more deeply to minimise the number of matchsticks they use. A good activity for students to discuss in pairs. (Australian Curriculum) NSW MA3-15MG manipulates, classifies and draws two-dimensional shapes, including equilateral, isosceles and scalene triangles, and describes their properties.
#### Left and right directions
(0)
A double sided worksheet and a single sided worksheet, both with solutions. Students need to use their knowledge about the directions of left and right in order to solve some problems. Also includes some turning 90 degrees to the left or right. Suitable as revision of these concepts at the end of a lesson or as a homework task. (Australian Curriculum) NSW MA3-17MG locates and describes positions on maps using grid-reference system
#### Matchstick patterns
(0)
A double sided worksheet with answers. Students can use matchsticks to help them construct patterns, make predictions and describe patterns. A good activity for students working in pairs to discuss their solutions. (Australian Curriculum) NSW MA3-8NA analyses and creates geometric and number patterns, constructs and completes number sentences, and locates points on the Cartesian plane
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https://community.powerbi.com/t5/Desktop/KPI-goal-set-to-previous-quarter/td-p/1399502
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Showing results for
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Helper II
## KPI goal set to previous quarter
Hi,
I have some KPIs and I would like to set the goal to be the previous quarter. Our quarters run from April to March, so unless I'm mistaken, I cant use the previousquarter formula.
My data is in a table with multiple columns as below.
so for example, for HospAdSH i want it to show 141 for 2020 Q1 (it does this), and 234 as the goal (the Q4 2019 sum)
3 REPLIES 3
Solution Specialist
Hi,
You can create a calculated column to have your Quartere starting from April.
Use the below link for the same.
Regards,
manikumar
Super User IV
@Rickfear , You can use datesqtd. It just looks for qtr. You can use that Ir your Qtr start fron Jan, Apr, Jul and Oct
examples
// With TI
QTD Sales = CALCULATE(SUM(Sales[Sales Amount]),DATESQTD(('Date'[Date])))
Last QTD Sales = CALCULATE(SUM(Sales[Sales Amount]),DATESQTD(dateadd('Date'[Date],-1,QUARTER)))
Last complete QTD Sales = CALCULATE(SUM(Sales[Sales Amount]),DATESQTD( ENDOFQUARTER(dateadd('Date'[Date],-1,QUARTER))))
Last to last QTD Sales = CALCULATE(SUM(Sales[Sales Amount]),DATESQTD(dateadd('Date'[Date],-2,QUARTER)))
Next QTD Sales = CALCULATE(SUM(Sales[Sales Amount]),DATESQTD(dateadd('Date'[Date],1,QUARTER)))
//Without TI
Qtr Start Date = DATEADD(STARTOFYEAR('Date'[Date],"3/31"),QUOTIENT(DATEDIFF('Date'[Start Of Year], 'Date'[Date],MONTH),3)*3,MONTH)
Qtr Month No = DATEDIFF('Date'[Qtr Start Date],'Date'[Date],MONTH)+1
Qtr Rank = RANKX(all('Date'),'Date'[Qtr Start date],,ASC,Dense)
This Qtr = CALCULATE(sum('order'[Qty]), FILTER(ALL('Date'),'Date'[Qtr Rank]=max('Date'[Qtr Rank])))
Last Qtr = CALCULATE(sum('order'[Qty]), FILTER(ALL('Date'),'Date'[Qtr Rank]=max('Date'[Qtr Rank])-1))
This QTD = CALCULATE(sum('order'[Qty]), FILTER(ALL('Date'),'Date'[Qtr Rank]=max('Date'[Qtr Rank]) && [Qtr Month No ] <=max([Qtr Month No ])))
Last QTD = CALCULATE(sum('order'[Qty]), FILTER(ALL('Date'),'Date'[Qtr Rank]=max('Date'[Qtr Rank])-1 && [Qtr Month No ] <=max([Qtr Month No ])))
Power BI — Qtr on Qtr with or Without Time Intelligence
https://medium.com/@amitchandak.1978/power-bi-qtd-questions-time-intelligence-2-5-d842063da839
To get the best of the time intelligence function. Make sure you have a date calendar and it has been marked as the date in model view. Also, join it with the date column of your fact/s. Refer :
Proud to be a Super User!
Helper II
Thanks both, I ended up going in a different direction. I added a column to the end called Qnum which goes 1,2,2,2,3,3,3,4,4,4 etc. I then added additonal goal columns with the formula =SUMIF(\$Q:\$Q,[@Qnum]-1,C:C). This now gives me my required goals to use.
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http://www.listserv.uga.edu/cgi-bin/wa?A2=ind0509e&L=sas-l&D=0&P=24514
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```Date: Fri, 30 Sep 2005 22:41:45 -0400 Reply-To: "Richard A. DeVenezia" Sender: "SAS(r) Discussion" From: "Richard A. DeVenezia" Subject: Re: SAS graph Comments: To: sas-l@uga.edu df ss wrote: > Hi all, > I have a data like this > ID m1_L m1_M m1_H m2_L m2_M m2_H > 1 1 2 3 1 2 3 > 2 2 3 4 2 3 4 > > I would like using the data to draw a picture like > this > > 1 |--|--| |--|--| > 2 |--|--| |--|--| > m1 m2 > > I'm not quite if SAS has the function to do so. If > anyone knows that, pleas let me know. It would be a > great help. > Thanks a lot A little data transformation gets you close in one plot. In multiple plots, search the archives for scattermatrix.. -------------------------- data foo; input ID m1_L m1_M m1_H m2_L m2_M m2_H; cards; 1 1 2 3 1 2 3 2 2 3 4 2 3 4 run; data fooie/view=fooie; set foo; array bulb m1_L--m2_H; do _n_ = 1 to dim(bulb); vname = vname (bulb[_n_]); length cat1 cat2 \$5; cat1 = scan (vname,1,'_'); cat2 = scan (vname,2,'_'); value = bulb[_n_]; if cat1 ne lag(cat1) then group+1; if mod (group,2)=0 then value+10; OUTPUT; end; keep id cat1 cat2 value group; run; goptions reset=all ftext='Arial' htext=18pt; symbol1 v="|" i=join h=32pt w=3; axis1 order=2 to 1 by -1 offset=(40pct); axis2 order=1 to 4, 11 to 14 offset=(10pct) label=none value=none; proc gplot data=fooie; plot id*value=group / vaxis=axis1 haxis=axis2 nolegend; title "Fooie"; run; -------------------------- Richard A. DeVenezia http://www.devenezia.com/ ```
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# What Is an Enterprise Value (EV)?
## Introduction
An Enterprise Value (EV) is a measure of a company's absolute value. It is a calculation used to determine the total worth of a business. EV takes into account all aspects of a business, such as its market capitalization, liabilities, minority ownership, and cash and cash equivalents. EV is also sometimes referred to as a company's market capitalization.
Understanding EV is important because it is used to help investors determine the value of a company. By looking at the EV calculation, investors can compare the value of different companies to get a better understanding of the company they are potentially investing in. This information can then be used to help them decide if the company is worth investing in or if it is overvalued.
## Overview of Enterprise Value
Enterprise value (EV) is a metric used to measure a business’ total value. It is essentially the amount that a company would be worth if acquired in its entirety and takes into account the capital structure of a business (debt, equity, assets and liabilities). It is a more accurate representation of a company’s true worth than market capitalization and is the metric used in most mergers and acquisitions.
### Components of EV
EV is calculated by taking a business’ market capitalization and adding the value of the debt it has taken on, subtracting its cash holdings, and adding in any other non-operational assets. The formula for calculating enterprise value is:
• EV = (Market Cap + Debt + Minority Interest - Cash Holdings + Non-Operational Assets)
### EV vs Equity Value
Equity value (EV) is often confused with equity value, which is the total value of all the equity shares of a company. Equity value is related to enterprise value, as it is calculated by taking the total enterprise value and subtracting any debt and minority interest the business holds. The formula for calculating equity value is:
• Equity Value = Enterprise Value - Debt
### EV vs Market Capitalization
EV is a more accurate representation of a business’ worth than market capitalization, which is simply the total value of all the company’s shares outstanding. The formula for calculating Market Capitalization is:
• Market Capitalization = Number of Shares Outstanding x Price Per Share
Market capitalization does not take into account debt, cash, and other non-operational assets and liabilities, whereas the enterprise value does. Therefore, the enterprise value is a more accurate representation of a company’s true value than the market capitalization.
## Calculating Enterprise Value
The enterprise value (EV) of a company is a measure of the overall value of the company, used to determine how much a buyer might be willing to pay for the company. The enterprise value formula is the sum of a company’s market capitalization, debt, preferred stock, and minority interest minus cash and other liquid assets.
### Enterprise Value Formula
EV = Market Capitalization + Debt + Preferred Stock + Minority Interest - Cash & Other Liquid Assets
### What is Included in the Calculation
• Market Capitalization - The total market value of a company's issued common stock.
• Debt - The outstanding debt of a company.
• Preferred Stock - The value of a company's preferred stock.
• Minority Interest - The portion of the equity of a company that is owned by outside shareholders.
• Cash & Other Liquid Assets - Amounts owed to a company such as accounts receivable, inventory, prepayments, and other liquid assets.
### What is Excluded in the Calculation
• Intangible Assets - Assets such as patents, trademarks, and intellectual property.
• Non-controlling Interest - Part ownership in a subsidiary that is not controlled by the parent company.
• Non-operating Assets - Assets such as investment securities and real estate.
## Different Ways to Measure EV
Enterprise value (EV) is a measure of a company's market capitalization, or the cost of all a company's outstanding securities and liabilities, including both equity and debt. Companies often use EV, rather than market capitalization, as a key performance indicator and can measure it in a variety of ways. Three of the most popular methods include the Financial Statement Method, the Comparable Method, and the Discounted Cash Flow (DCF) Method.
### Financial Statement Method
The Financial Statement Method is the simplest way of calculating EV as it uses balance sheet amounts as the primary calculations for EV. This method looks at a snapshot of the company's books and adds its market capitalization and its debt and then subtracts any cash and investments on hand. This method derives an accurate baseline measure of enterprise value.
### Comparable Method
The Comparable Method of calculating EV is based on the values of other companies in the same industry. As part of this method, analyst use multiples such as EV to EBITDA or EV to EBIT to compare the EV of potential investments. This method is often used in takeover decisions.
### DCF Method
TheDCF Method utilizes a discounted cash flow analysis to estimate the value of the firm. This method takes projected cash flow, discounts it to its present value, and produces an estimate of the company’s value.
## Enterprise Value Multiples
Enterprise value (EV) is a measure of a company's total worth that includes equity market capitalization as well as debt, cash, and other minority interests. EV multiples provide a useful tool for comparing companies of different sizes and in different industries. EV multiples are calculated by dividing the enterprise value of a company by a measure of earning power or operating income.
### EV / EBITDA
The Enterprise Value to Earnings Before Interest, Tax, Depreciation and Amortization (EV/EBITDA) multiple is commonly used to compare companies of unequal sizes in the same industry. EV/EBITDA is a useful measure as EBITDA is typically unaffected by its capital structure, which allows for a true comparison of company value irrespective of capital structure.
### EV / EBIT
The Enterprise Value to Earnings Before Interest and Tax (EV/EBIT) multiple is used to compare company values across industries. Unlike EBITDA, which excludes the effects of financing decisions, EBIT includes both the financial costs and benefits associated with a company's capital structure. This multiple is useful for comparison of diversified businesses with different capital structures.
### EV / Sales
The Enterprise Value to Sales (EV/Sales) multiple is used to compare companies of different sizes as sales are generally unaffected by capital structure decisions. EV/Sales multiples are typically amongst the lowest of the EV multiples and are commonly used to compare companies with large differences in their financial structure or operating performance.
## Factors Affecting Enterprise Value (EV)
Enterprise Value (EV) is typically used as a measure of a company's value and can be used to analyze potential investments or acquisitions. Although EV is relatively straightforward to calculate, understanding the factors which affect it is key to properly valuing a company. The most common elements that influence EV include a company’s performance, its capital structure, and the context of the industry.
### Performance of Company
The performance of a company is likely the biggest indicator of EV. Quality of the management, along with overall financial condition, will have a major impact on EV despite any fluctuations in the stock market or external economic conditions. As a result, companies with a strong performance and positive fundamentals will typically have higher EV's, while struggling companies will have lower EV's. Additionally, the growth rate and profitability of a company can also positively affect its EV by signalling to investors that it has potential for further returns.
### Capital Structure
Another major factor that affects EV is a company's capital structure. Equity holders will typically want a higher EV because the higher EV can make their share of the EV larger. This might encourage a company to increase its debt in order to increase its EV, which can be both beneficial and detrimental to a company depending on the type of debt and its management. Equity holders will also want to make sure that the company's debt is relatively low in order to justify the higher EV.
### Context of Industry
Finally, the context of the industry is also highly relevant to a company's EV. For example, industries with higher growth rates or those subject to frequent regulatory change will have higher EV's, whereas mature markets or sectors that are slow to change will have lower EV's. Additionally, companies operating in industries with high barriers to entry, such as healthcare or technology, will likely have higher EV's than those in more open markets since these companies can effectively control their own pricing and profits.
## Conclusion
Enterprise Value (EV) is an important measure of the value of a company that considers both equity and debt. It is calculated by subtracting cash and cash equivalents from a company’s market cap, then adding its total debt, preferred equity and minority interests. EV can be a valuable metric for understanding a company’s financial situation and is used to compare companies with different capital structure to give a more true picture of their financial stability.
Understanding EV and using it to compare companies can provide insight into a company’s financial health and likelihood of success. It is also an important metric to use in mergers and acquisitions to get a comprehensive understanding of a company and its potential value. By taking into account equity, debt and other factors that can add or detract from a company’s overall value, EV provides a more accurate assessment of a company’s true worth.
### Benefits of understanding EV
• EV can provide a better understanding of the financial health of a company.
• It allows for a more accurate comparison between companies with different capital structures.
• It is important for mergers and acquisition decisions.
• It provides a more complete assessment of a company's true worth.
Overall, EV is a convenient metric for understanding the comprehensive value of a company and can provide an accurate comparison between different companies with different capital structures. Understanding EV can help give insight into a company's financial health, which can be beneficial for making more informed decisions.
Expert-built startup financial model templates
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Solving Best Time to Buy and Sell Stock with Transaction Fee in go. Please try yourself first to solve the problem and submit your implementation to LeetCode before looking into solution.
Problem Description
You may complete as many transactions as you like, but you need to pay the transaction fee for each transaction. You may not buy more than 1 share of a stock at a time (ie. you must sell the stock share before you buy again.)
Return the maximum profit you can make.
Example 1:
``````Input: prices = [1, 3, 2, 8, 4, 9], fee = 2
Output: 8
Explanation: The maximum profit can be achieved by:
Selling at prices[3] = 8
• `0 < prices.length <= 50000`.
• `0 < prices[i] < 50000`.
• `0 <= fee < 50000`.
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# Search by Topic
#### Resources tagged with Multiplication & division similar to The Patent Solution:
Filter by: Content type:
Stage:
Challenge level:
### There are 47 results
Broad Topics > Calculations and Numerical Methods > Multiplication & division
### Like Powers
##### Stage: 3 Challenge Level:
Investigate $1^n + 19^n + 20^n + 51^n + 57^n + 80^n + 82^n$ and $2^n + 12^n + 31^n + 40^n + 69^n + 71^n + 85^n$ for different values of n.
### As Easy as 1,2,3
##### Stage: 3 Challenge Level:
When I type a sequence of letters my calculator gives the product of all the numbers in the corresponding memories. What numbers should I store so that when I type 'ONE' it returns 1, and when I type. . . .
### Thirty Six Exactly
##### Stage: 3 Challenge Level:
The number 12 = 2^2 × 3 has 6 factors. What is the smallest natural number with exactly 36 factors?
### Eminit
##### Stage: 3 Challenge Level:
The number 8888...88M9999...99 is divisible by 7 and it starts with the digit 8 repeated 50 times and ends with the digit 9 repeated 50 times. What is the value of the digit M?
### Largest Number
##### Stage: 3 Challenge Level:
What is the largest number you can make using the three digits 2, 3 and 4 in any way you like, using any operations you like? You can only use each digit once.
### X Marks the Spot
##### Stage: 3 Challenge Level:
When the number x 1 x x x is multiplied by 417 this gives the answer 9 x x x 0 5 7. Find the missing digits, each of which is represented by an "x" .
### Oh! Hidden Inside?
##### Stage: 3 Challenge Level:
Find the number which has 8 divisors, such that the product of the divisors is 331776.
##### Stage: 3 Challenge Level:
Visitors to Earth from the distant planet of Zub-Zorna were amazed when they found out that when the digits in this multiplication were reversed, the answer was the same! Find a way to explain. . . .
### Slippy Numbers
##### Stage: 3 Challenge Level:
The number 10112359550561797752808988764044943820224719 is called a 'slippy number' because, when the last digit 9 is moved to the front, the new number produced is the slippy number multiplied by 9.
### Remainders
##### Stage: 3 Challenge Level:
I'm thinking of a number. When my number is divided by 5 the remainder is 4. When my number is divided by 3 the remainder is 2. Can you find my number?
### Diggits
##### Stage: 3 Challenge Level:
Can you find what the last two digits of the number $4^{1999}$ are?
### I'm Eight
##### Stage: 1, 2, 3 and 4 Challenge Level:
Find a great variety of ways of asking questions which make 8.
### Long Multiplication
##### Stage: 3 Challenge Level:
A 3 digit number is multiplied by a 2 digit number and the calculation is written out as shown with a digit in place of each of the *'s. Complete the whole multiplication sum.
##### Stage: 3 Challenge Level:
If you take a three by three square on a 1-10 addition square and multiply the diagonally opposite numbers together, what is the difference between these products. Why?
### Skeleton
##### Stage: 3 Challenge Level:
Amazing as it may seem the three fives remaining in the following `skeleton' are sufficient to reconstruct the entire long division sum.
### A One in Seven Chance
##### Stage: 3 Challenge Level:
What is the remainder when 2^{164}is divided by 7?
### Countdown
##### Stage: 2 and 3 Challenge Level:
Here is a chance to play a version of the classic Countdown Game.
##### Stage: 2 and 3 Challenge Level:
Watch our videos of multiplication methods that you may not have met before. Can you make sense of them?
### Powerful Factorial
##### Stage: 3 Challenge Level:
6! = 6 x 5 x 4 x 3 x 2 x 1. The highest power of 2 that divides exactly into 6! is 4 since (6!) / (2^4 ) = 45. What is the highest power of two that divides exactly into 100!?
### Two Many
##### Stage: 3 Challenge Level:
What is the least square number which commences with six two's?
### Times Right
##### Stage: 3 and 4 Challenge Level:
Using the digits 1, 2, 3, 4, 5, 6, 7 and 8, mulitply a two two digit numbers are multiplied to give a four digit number, so that the expression is correct. How many different solutions can you find?
### Factorial
##### Stage: 4 Challenge Level:
How many zeros are there at the end of the number which is the product of first hundred positive integers?
### 3388
##### Stage: 3 Challenge Level:
Using some or all of the operations of addition, subtraction, multiplication and division and using the digits 3, 3, 8 and 8 each once and only once make an expression equal to 24.
### Factoring Factorials
##### Stage: 3 Challenge Level:
Find the highest power of 11 that will divide into 1000! exactly.
### More Mods
##### Stage: 4 Challenge Level:
What is the units digit for the number 123^(456) ?
### Magic Potting Sheds
##### Stage: 3 Challenge Level:
Mr McGregor has a magic potting shed. Overnight, the number of plants in it doubles. He'd like to put the same number of plants in each of three gardens, planting one garden each day. Can he do it?
### Number Rules - OK
##### Stage: 4 Challenge Level:
Can you convince me of each of the following: If a square number is multiplied by a square number the product is ALWAYS a square number...
### Repeaters
##### Stage: 3 Challenge Level:
Choose any 3 digits and make a 6 digit number by repeating the 3 digits in the same order (e.g. 594594). Explain why whatever digits you choose the number will always be divisible by 7, 11 and 13.
### Multiples Sudoku
##### Stage: 3 Challenge Level:
Each clue in this Sudoku is the product of the two numbers in adjacent cells.
### Round and Round and Round
##### Stage: 3 Challenge Level:
Where will the point stop after it has turned through 30 000 degrees? I took out my calculator and typed 30 000 ÷ 360. How did this help?
##### Stage: 3 Challenge Level:
Powers of numbers behave in surprising ways. Take a look at some of these and try to explain why they are true.
### A First Product Sudoku
##### Stage: 3 Challenge Level:
Given the products of adjacent cells, can you complete this Sudoku?
### Ones Only
##### Stage: 3 Challenge Level:
Find the smallest whole number which, when mutiplied by 7, gives a product consisting entirely of ones.
### 2010: A Year of Investigations
##### Stage: 1, 2 and 3
This article for teachers suggests ideas for activities built around 10 and 2010.
### So It's Times!
##### Stage: 2 and 3 Challenge Level:
How will you decide which way of flipping over and/or turning the grid will give you the highest total?
### Missing Multipliers
##### Stage: 3 Challenge Level:
What is the smallest number of answers you need to reveal in order to work out the missing headers?
### The Remainders Game
##### Stage: 2 and 3 Challenge Level:
A game that tests your understanding of remainders.
### More Magic Potting Sheds
##### Stage: 3 Challenge Level:
The number of plants in Mr McGregor's magic potting shed increases overnight. He'd like to put the same number of plants in each of his gardens, planting one garden each day. How can he do it?
### One O Five
##### Stage: 3 Challenge Level:
You can work out the number someone else is thinking of as follows. Ask a friend to think of any natural number less than 100. Then ask them to tell you the remainders when this number is divided by. . . .
### Integrated Product Sudoku
##### Stage: 3 and 4 Challenge Level:
This Sudoku puzzle can be solved with the help of small clue-numbers on the border lines between pairs of neighbouring squares of the grid.
### Expenses
##### Stage: 4 Challenge Level:
What is the largest number which, when divided into 1905, 2587, 3951, 7020 and 8725 in turn, leaves the same remainder each time?
### The Unmultiply Game
##### Stage: 2, 3 and 4 Challenge Level:
Unmultiply is a game of quick estimation. You need to find two numbers that multiply together to something close to the given target - fast! 10 levels with a high scores table.
### Vedic Sutra - All from 9 and Last from 10
##### Stage: 4 Challenge Level:
Vedic Sutra is one of many ancient Indian sutras which involves a cross subtraction method. Can you give a good explanation of WHY it works?
### Going Round in Circles
##### Stage: 3 Challenge Level:
Mathematicians are always looking for efficient methods for solving problems. How efficient can you be?
### A Chance to Win?
##### Stage: 3 Challenge Level:
Imagine you were given the chance to win some money... and imagine you had nothing to lose...
### What Is the Question?
##### Stage: 3 and 4 Challenge Level:
These pictures and answers leave the viewer with the problem "What is the Question". Can you give the question and how the answer follows?
### Galley Division
##### Stage: 4 Challenge Level:
Can you explain how Galley Division works?
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15
Apr 14
## When Multiplication is Repeated Subtraction
Mathematics professors have told me I am wrong when I have said multiplication is repeated subtraction.
The reason they tell me I am wrong is simple. Mathematics professors have been sucked in by the MIRA meme that multiplication IS repeated addition.
If multiplication is repeated addition, then division is repeated subtraction.
End of story. Or is it?
Integral multiplication of a multiplicand by a POSITIVE multiplier, can be calculated by the repeated ADDITION of the multiplicand from zero, as many times as the multiplier says.
Integral multiplication of a multiplicand by a NEGATIVE multiplier, can be calculated by the repeated SUBTRACTION of the multiplicand from zero, as many times as the multiplier says.
It seems people have either forgotten, or never knew, multiplication distributes over subtraction as well as addition.
So just as two multiplied by positive three 2 × (+3) may be calculated as 0 + 2 + 2 + 2, which equals positive six, two multiplied by negative three 2 × (3) may be calculated as 0 - 2 - 2 - 2, which equals negative six.
Thus multiplication is repeated subtraction, as well as repeated addition. It just depends on whether your integral multiplier is positive or negative.
===================
EXTENDED ARTICLE
The laws of mathematics require multiplication be undertaken via repeated subtraction just as they require multiplication be undertaken via repeated addition.
This is because the Distributive Law does not discriminate between subtraction and addition. Integral multiplication distributes over both addition and subtraction.
Brahmagupta wrote Brāhma Sphuta-siddhānta (BSS) in 628 CE. In this book, the first to document the rules of zero, Brahmagupta gave the rules of ‘saṅkalana’, or addition. Here, he defined zero to be the sum of a positive number and negative number of equal magnitude, सम-ऐक्यम् खम् (BSS, Chapter 18:30a).
Thus in modern notation, 0 = +n + n. So given zero is described as, for example, +3 + 3, if we take away 3 then +3 remains and if from +3 + 3 (zero), we take away +3 then 3 remains. Thus based on what may be implied from the mnemonic Sanskrit verses (shlokas) of Brahmagupta, a negative number may be defined as 0 – +n = n.
An example may be 0 – +3 = 3. Therefore, we can write the statements, +3 = 0 + 1 + 1 + 1 and 3 = 0 – 1 – 1 – 1. Given we know multiplication distributes over both addition and subtraction, 2 × +3 = 2 × (0 + 1 + 1 + 1) and this in turn becomes the equation 0 + 2 + 2 + 2 = +6.
By the Distributive Law, 2 × 3 = 2 × (0 – 1 – 1 – 1) and so as ‘multiplication is repeated addition’, we must say ‘multiplication is repeated subtraction’ as 2 × 3 = 0 – 2 – 2 – 2 = 6.
Thus, based on the Sanskrit of Brahmagupta, our definition of multiplication on the integers ab, is better described as: (the total of) a either added to or subtracted from ZERO, b times, as per sign of b.
As at 16 June, 2014, there are only 3 Google search results for the phrase “multiplication is repeated subtraction”. One is an error and should read “division is repeated subtraction”, while the other two are the author’s.
This contrasts with more than 74,000 Google search results for the phrase “multiplication is repeated addition”. www.google.com/search?q=%22multiplication+is+repeated+addition%22
There has been a lot of fuss in recent years about whether Multiplication IS Repeated Addition (MIRA) and should be defined as repeated addition. Despite the hundreds of thousands of words written on this topic, it appears nobody has proven MIRA wrong with this "Proof by Contradiction" until now.
Multiplication cannot be defined as repeated addition because the same laws of mathematics mean just as multiplication may INVOLVE repeated addition, multiplication may INVOLVE repeated subtraction.
Multiplication on the integers may involve either repeated addition or repeated subtraction, depending on the sign of the multiplier, yet cannot be defined solely as one or the other.
Please Enter Your Feedback and/or Question Here
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http://reference.wolfram.com/legacy/v5/Built-inFunctions/ListsAndMatrices/ListConstruction/CoefficientArrays.html
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This is documentation for Mathematica 5, which was
based on an earlier version of the Wolfram Language.
CoefficientArrays CoefficientArrays[polys, vars] gives the arrays of coefficients of the variables vars in the polynomials polys. CoefficientArrays gives a list containing SparseArray objects, which can be converted to ordinary arrays using Normal. If CoefficientArrays[polys, vars] gives , , , ... , then polys can be reconstructed as + . vars + . vars . vars + ... . Any element of polys of the form lhs == rhs is taken to correspond to the polynomial lhs - rhs. CoefficientArrays[polys, , , ... ] takes all expressions in polys that match any of the to be variables. CoefficientArrays[polys] is equivalent to CoefficientArrays[polys, Variables[polys]]. The length of the list CoefficientArrays[polys, vars] is one more than the total degree of polys. The are sparse arrays with ranks i + 1. The first element has the same length as the list polys. If polys is a single polynomial rather than a list, is also not a list. For linear equations, the solution to Thread[polys==0] is given by LinearSolve[, -]. For nonlinear equations, the are not unique. CoefficientArrays by default assigns non-zero coefficients only to monomials where the variables appear in the same order as vars. CoefficientArrays[polys, vars, Symmetric->True] makes all the symmetric in all their indices. The resulting arrays will generally be less sparse. See Section 3.7.12. See also: CoefficientList, SparseArray, Solve. New in Version 5.0.
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https://www.physicsforums.com/threads/problem-with-a-sum-of-series.121297/
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# Problem with a sum of series
1. May 18, 2006
### rahl__
Hi, I need to find the sum of such series:
$$\sum_{n=1}^{\infty}\sin{\frac{\pi}{2^n}}$$
i know that it's sum is less than $$\pi$$ but i dont know how to find the exact value.
thanks in advance for any help or clues
Last edited: May 18, 2006
2. May 18, 2006
### NateTG
What have you tried doing?
3. May 19, 2006
### rahl__
first of all i'd like to correct myself as i don't really need to find that sum. i was just wondering whether my mathematical knowledge is big[?] enough to solve this problem, so what should I have really asked about is: what method would you choose to find that sum.
I have used the comparative criterion (precisely this inequality: $$sin {x}\leq x$$)to find out that this series is convergent and that it's sum is equal or less than $$\pi$$, but i dont know what to do next. could you tell me what is the level of difficulty of this problem? is the solution rather complicated or can it be presented in a few lines? or which mathematical terms should i know in order to solve it on my own?
Last edited: May 19, 2006
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PracticeTest
# PracticeTest - ∈ R n × n is the equation A B 2 = A 2 2...
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Linear Algebra Midterm Summer 2011 1. Consider the system of equations x 1 + 2 x 2 + 3 x 3 + 2 x 4 = 5 4 x 1 + 8 x 2 + x 3 + x 4 + 6 x 5 = 10 3 x 1 + 6 x 2 + x 3 + 2 x 4 + 5 x 5 = 15 2 x 1 + 4 x 2 + x 3 + 9 x 4 + 10 x 5 = 30 This can be succintly written as A x = b and represented in matrix form as [ A | b ] R 4 × 6 . Given that rref([ A | b ]) = 1 2 0 0 0 27 0 0 1 0 0 - 18 0 0 0 1 0 16 0 0 0 0 1 - 16 (a) Find ker A as a span of vectors. (You’ll need to consider rref([ A | 0 ]) here.) (b) Find im A as a span of vectors. (c) Find a particular solution s of the system A x = b . 1
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(d) Use parts (a) and (c) to find the set of all possible solutions K = s + ker A , and write this in the form K = { s + t 1 v 1 + ··· + t k v k | t 1 ,...,t k R } where the vectors v i are the basis vectors of ker A . (e) Use part (d) with t 1 = ··· = t k = 3 to find a second solution s 2 of A x = b and verify that this is indeed a solution. 2. Let A R n × n satisfy A 2 = A . Show that if all entries of A are nonzero, then A is not invertible. [Hint: It’s easier to prove the contrapositive of this statement.] 2
3. If a and b are real numbers, we know that ( a + b ) 2 = a 2 + 2 ab + b 2 . If A,B
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Unformatted text preview: ∈ R n × n , is the equation ( A + B ) 2 = A 2 +2 AB + B 2 still true? If so, prove it, if not, find examples of matrices A and B for which this fails. 4. Let A = 2 4 8 4 5 1 7 9 3 . (a) Find ker A and im A as spans of vectors. 3 (b) Find all solutions of the system A x = b , where b = 6 9 16 . 5. Consider the basis β = ±² 1 1 ³ , ² 1 2 ³´ for R 2 . Find the representation [ x ] β of the vector x = ²-2 5 ³ in this basis. 6. Let ρ = { e 1 , e 2 } be the standard basis for R 2 and let β = ±² 1 1 ³ , ² 1 ³´ be another basis for R 2 . If T A ∈ L ( R 2 , R 2 ) is the counterclockwise rotation through π/ 4, with associated matrix A = " 1 √ 2-1 √ 2 1 √ 2 1 √ 2 # find the matrix representation [ T A ] ρ,β of T A in these bases. 4...
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# Posts Tagged Physics
### Showing My Math
So here’s the deal. I’m working on a new project and I want to make sure at least some of the details have actual technical grounding. I’m okay with a little handwavium, it’s probably unavoidable, but I want to at least have some grounding in reality. Problem is, I’m not all that great at actual technical grounding, as the last physics class I took was non-AP physics in high school where I barely got a B. So I might occasionally make these posts, I might make just this one, in an attempt to crowdsource some of my equations. The questions I have are whether I’ve got the right equations, and whether I’m using them correctly, and to also play around with LaTeX a little. But mostly the first two things.
So here’s today’s problem: Given a cylindrical space ship with an internal radius of 6km, how fast must it be rotating to create a centripetal acceleration equivalent to earth gravity for someone standing on the inside surface? I didn’t know any of these equations, but found them at this rather helpful forum post. First, we must find the speed at that 6km point that would produce an acceleration of 9.8m/s²:
$a=\frac{v^2}{r}$
$a=9.8 \frac{m}{s^2}$
$r=6000 m$
$9.8=\frac{v^2}{6000}$
$9.8*6000=v^2$
$58800=v^2$
$v=242.49 \frac{m}{s}$
That number sounds awfully damn fast, but consider the speed of rotation of the earth at sea level on the equator is roughly 465 m/s. Next step, at least what I’m assured is the next step, is converting this into radians/second:
$W=\frac{v}{r}$
$v=242.49$
$r=6000$
$W=\frac{242.49}{6000}=0.0404$
Finally this can be converted to revolutions per minute. The conversion formula I found is:
$1 \frac{rad}{s} = \frac{60}{2pi} = \frac{30}{pi} rpm$
$0.0404 \frac{rad}{s} = \frac{30*0.0404}{pi}=\frac{1.212}{pi}=0.386 rpm$
Therefore the ship is rotating at a rate slightly faster than once every three minutes. What I didn’t expect is that, since the rate of rotation is a constant, centripetal acceleration increases linearly from the axis of rotation. I’m so accustomed to formulas for gravity having squares all over the place, but this isn’t, strictly speaking, gravity. It’s an acceleration equal to gravity. So at half the distance from the axis of rotation, we can work backwards with W as a constant…
$W=\frac{v}{r}$
$0.0404=\frac{v}{3000}$
$v=0.0404*3000=121.2\frac{m}{s}$
$a=\frac{v^2}{r}$
$a=\frac{121.2^2}{3000} = \frac{14689.44}{3000} = 4.9 \frac{m}{s^2}$
Which is equivalent to half gravity.
My next trick will be to find a formula that describes the rate of descent for a body falling through linearly increasing gravity. That’s less likely to come up in-story, but more for my own curiosity.
Edit: Some further poking around (which, I’m ashamed to say, has mostly been at Wikipedia so far) suggests that 2rpm is about the maximum rotation that most humans can adjust to with no ill effects, so my rotation of nearly 1/6 that rate is shockingly safe in and of itself. So that’s good to know. Now if only it didn’t have a “citation needed” tag.
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data - DATA PAGE Emission Spectrum of Hydrogen Color of...
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DATA PAGE Emission Spectrum of Hydrogen Color of Line Noted Wavelength (nanometers) Red 650 Green 480 Blue 430 Violet 410 Emission Spectrum of Mercury Color of Line Noted Wavelength (nanometers) Red 690 Orange 600 Green 540 Blue 430 Violet 400 Emission Spectrum of Neon Color of Line Noted Wavelength (nanometers) Red 652, 658, 685, 695, 705 Orange 600, 620, 635, 636, 646 Green 528, 534, 560 Blue 465 Emission Spectrum of Incandescent Light Bulb Emission Spectrum of Fluorescent Light Bulb Color of Line Noted
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Red Orange Yellow Green Blue Violet Color of Line Noted Red Orange Yellow Green Blue Violet Emission Spectrum of Sunlight Color of Line Noted Wavelength (nanometers) Red 580-620 Orange 570-590 Green 490-570 Blue 410-460 Treatment of Data Calculation of Hydrogen Energies: Energy = (hc)/ λ where h is Plank’s constant (6.626 x 10 -34 Js), c is the velocity of light (2.998 x 10 8 m/s), and λ is the wavelength in meters. Therefore, the calculations of the hydrogen energies from the data are as follows:
Energy violet = [(6.626 x 10 -34 Js) (2.998 x 10 8 m/s)] / [(410 nm) (1 m/ 10 9 nm)] = 4.845 x 10 -19 J Energy blue = [(6.626 x 10 -34 Js) (2.998 x 10 8 m/s)] / [(430 nm) (1 m/ 10 9 nm)] = 4.620 x 10 -19 J Energy green = [(6.626 x 10 -34 Js) (2.998 x 10 8 m/s)] / [(480 nm) (1 m/ 10 9 nm)] = 4.138 x 10 -19 J Energy red = [(6.626 x 10 -34 Js) (2.998 x 10 8 m/s)] / [(650 nm) (1 m/ 10 9 nm)] = 3.056 x 10 -19 J The calculations of the hydrogen energies from the visible spectrum are as follows: Energy violet = [(6.626 x 10 -34 Js) (2.998 x 10 8 m/s)] / [(410.2 nm) (1 m/ 10 9 nm)] = 4.843 x 10 -19 J Energy blue = [(6.626 x 10 -34 Js) (2.998 x 10 8 m/s)] / [(434.1 nm) (1 m/ 10 9 nm)] = 4.576 x 10 -19 J Energy green = [(6.626 x 10 -34 Js) (2.998 x 10 8 m/s)] / [(486.1 nm) (1 m/ 10 9 nm)] = 4.087 x 10 -19 J Energy red = [(6.626 x 10 -34 Js) (2.998 x 10 8 m/s)] / [(656.3 nm) (1 m/ 10 9 nm)] = 3.027 x 10 -19 J From the calculations, it is shown that the collected data is very similar to the actual data of the visible spectrum. Therefore, the energies obtained from the collected data are fairly accurate. Calculation of Initial Orbit Number: To calculate the initial orbit number, n initial, for the Bohr model of the hydrogen spectrum, the follow equation must be used: ΔE = 2.179 x 10 -18 [(1/n 2 final ) – (1/n 2 initial )] joules All of the lines of the hydrogen spectrum that lie in the visible region have a final orbit number of two. Therefore, n final can be interpreted as n final = 2. ΔE is the hydrogen energies in relation to the various colors. n inital of violet :
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Therefore, the initial orbit number of violet is six because according to the Bohr model the value of the principal quantum number can only be integers between one and infinity. n
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This note was uploaded on 02/05/2012 for the course BIO 102 taught by Professor Avery during the Spring '11 term at FGCU.
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## Contents
Gunning Fog, Flesch Reading Ease, and Flesch-Kincaid are reading level algorithms that can be helpful in determining how readable your content is. Reading level algorithms only provide a rough guide, as they tend to reward short sentences made up of short words. Whilst they're rough guides, they can give a useful indication as to whether you've pitched your content at the right level for your intended audience.
## Interpreting the Results
This service analyses the readability of all rendered content. Unfortunately, this will include navigation items, and other short items of content that do not make up the part of the page that is intended to be the subject of the readability test. These items are likely to skew the results. The difference will be minimal in situations where the copy content is much larger than the navigation items, but documents with little content but lots of navigation items will return results that aren't correct.
Philip Chalmers of Benefit from IT provided the following typical Fog Index scores, to help ascertain the readability of documents.
Typical Fog Index Scores
Fog Index Resources
6 TV guides, The Bible, Mark Twain
8 - 10 Most popular novels
10 Time, Newsweek
11 Wall Street Journal
14 The Times, The Guardian
15 - 20 Academic papers
Over 20 Only government sites can get away with this, because you can't ignore them.
Over 30 The government is covering something up
The following table contains the readability results for http://boosterpackforlife.com.
Summary Value
Total sentences 214
Total words 3386
Average words per Sentence 15.82
Words with 1 Syllable 1994
Words with 2 Syllables 866
Words with 3 Syllables 335
Words with 4 or more Syllables 191
Percentage of word with three or more syllables 15.53%
Average Syllables per Word 1.62
Gunning Fog Index 12.54
Flesch Reading Ease 53.48
## Gunning-Fog Index
The following is the algorithm to determine the Gunning-Fog index.
• Calculate the average number of words you use per sentence.
• Calculate the percentage of difficult words in the sample (words with three or more syllables).
• Add the totals together, and multiply the sum by 0.4.
• Algorithm: (average_words_sentence + number_words_three_syllables_plus) * 0.4
The result is your Gunning-Fog index, which is a rough measure of how many years of schooling it would take someone to understand the content. The lower the number, the more understandable the content will be to your visitors. Results over seventeen are reported as seventeen, where seventeen is considered post-graduate level.
## Flesch Reading Ease
The following is the algorithm to determine the Flesch Reading Ease.
• Calculate the average number of words you use per sentence.
• Calculate the average number of syllables per word.
• Multiply the average number of syllables per word multiplied by 84.6 and subtract it from the average number of words multiplied by 1.015.
• Subtract the result from 206.835.
• Algorithm: 206.835 - (1.015 * average_words_sentence) - (84.6 * average_syllables_word)
The result is an index number that rates the text on a 100-point scale. The higher the score, the easier it is to understand the document. Authors are encouraged to aim for a score of approximately 60 to 70.
## Flesch-Kincaid grade level
The following is the algorithm to determine the Flesch-Kincaid grade level.
• Calculate the average number of words you use per sentence.
• Calculate the average number of syllables per word.
• Multiply the average number of words by 0.39 and add it to the average number of syllables per word multiplied by 11.8.
• Subtract 15.50 from the result.
• Algorithm: (0.39 * average_words_sentence) + (11.8 * average_syllables_word) - 15.9
The result is the Flesch-Kincaid grade level. Like the Gunning-Fog index, it is a rough measure of how many years of schooling it would take someone to understand the content. Negative results are reported as zero, and numbers over twelve are reported as twelve.
## Reading Level Algorithms
Readability is the measure of how easy it is to read and comprehend a document. Readability tests were first developed in the 1920s in the United States. They are mathematical formulas, designed to determine the suitability of books for American students at a certain age, or grade level. Automating the process was intended to make it easier for tutors, librarians, and publishers to determine whether a book would be suitable for its intended audience. The formulas are based around the average words to a sentence, and the average syllables used per word. As such, they tend to reward short sentences made up of short words.
Being mathematically based, readability tests are unable to determine the likelihood that the document is comprehensible, interesting, or enjoyable. It's possible to obtain good readability scores with gobbledygook, providing the content contains short sentences made up of monosyllabic words. We'll leave the question as to why the word "monosyllabic" has five syllables for another day. Layout and design are also important factors to the readability of a document that cannot be determined using readability tests. Documents aimed at a higher level may require background knowledge, which cannot be determined by the tests.
For a document to be easily understood, the writing style should be clear and simple. This involves a writing style that is direct, and familiar to the intended reader. The structure of the document should be logical, unambiguous, and avoid redundant words.
Many of these factors cannot be measured using readability tests. Instead, readability tests provide a prediction of the reading ease for a document. Sentence length and polysyllabic words do have a direct impact on the readability of documents, albeit a surface measure of the characteristics of the text. They provide an indication that the content may be too dense with a quantifiable measure. The results should be used in conjunction with good writing style guidelines.
Guideline 14 of the Web Content Accessibility Guidelines requires that documents are clear and simple. Readability tests can provide a rough guide to the likelihood of a document being clearly understood. This service is to provide content authors with a guide to the readability of their website.
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The OEIS mourns the passing of Jim Simons and is grateful to the Simons Foundation for its support of research in many branches of science, including the OEIS.
The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A179460 Numbers m for which 2*A179382(m)=A002326(m-1). 3
2, 3, 5, 6, 7, 9, 10, 13, 14, 15, 17, 19, 21, 22, 27, 29, 30, 31, 33, 34, 35, 39, 41, 42, 49, 50, 51, 54, 55, 57, 61, 63, 65, 66, 69, 70, 71, 73, 75, 79, 82, 85, 86, 87, 89, 90, 91, 93, 97, 99, 101, 102, 103, 104, 105, 106, 107, 114, 115, 121, 122, 125, 126, 129, 133, 135 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS m is in the sequence iff the set {1,2,...,2^(2*m-2)} considered in reduced residue system modulo 2*m-1 contains the same number of odd and even integers. LINKS Table of n, a(n) for n=1..66. EXAMPLE 5 in the sequence since modulo 2*5-1=9 we have {1,2,4,8,16,32}={1,2,4,8,7,5} and the last set contains 3 odd and 3 even elements. MATHEMATICA fQ[n_] := Block[{r = Union@ PowerMod[2, Range[0, 2 n - 2], 2 n - 1]}, Length@ r == 2 Count[ OddQ@ r, True]]; Select[ Range@ 138, fQ] (* Robert G. Wilson v, Aug 26 2010 *) CROSSREFS Cf. A002326, A179382. Sequence in context: A038161 A158746 A062470 * A344281 A171886 A343238 Adjacent sequences: A179457 A179458 A179459 * A179461 A179462 A179463 KEYWORD nonn AUTHOR Vladimir Shevelev, Jul 14 2010 EXTENSIONS More terms from Robert G. Wilson v, Aug 26 2010 STATUS approved
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Last modified June 13 17:01 EDT 2024. Contains 373391 sequences. (Running on oeis4.)
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## BFC237011155 Datasheet(PDF) 12 Page - Vishay Siliconix
Part No. BFC237011155 Description DC Film Capacitors MKT Radial Potted Type Download 18 Pages Scroll/Zoom 100% Maker VISHAY [Vishay Siliconix] Homepage http://www.vishay.com Logo
## BFC237011155 Datasheet(HTML) 12 Page - Vishay Siliconix
12 / 18 page MKT 370Vishay BCcomponentsDC Film CapacitorsMKT Radial Potted Typewww.vishay.comFor technical questions, contact: dc-film@vishay.comDocument Number: 2810888Revision: 04-Aug-09Maximum RMS current (sinewave) as a function of frequencyThe maximum RMS current is defined by Iac = ω x C x Uac.Uac is the maximum AC voltage depending on the ambient temperature in the curves “Max. RMS voltage and AC current as afunction of frequency”.Max. AC voltage as a function of frequencyMax. AC voltage as a function of frequency102101100f (Hz)103102101104105Tamb ≤ 85 °C, 630 VdcCompact size102101100f (Hz)10310210110410585 °C < Tamb ≤ 105 °C, 630 VdcCompact sizeTangent of loss angle as a function of frequency(typical curve)Insulation resistance as a function of the ambient temperature(typical curve)Max. DC and AC voltage as a function of frequencyfor voltage 63 VMax. DC and AC voltage as a function of frequencyfor voltages > 63 Vf (Hz)102103104105103102101Curve 1: C = 0.33 µFCurve 2: 0.33 µF, C = 1.2 µFCurve 3: 1.2 µF, C = 3.9 µFCurve 4: 3.9 µF, C = 6.8 µFCurve 5: C = 6.8 µF54321Tamb (°C)- 500501001021031041051.21.00.80.60.40.20.0- 60- 202060100Tamb (°C)1.21.00.80.60.40.20.0- 60- 202060100Tamb (°C)
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# What Fraction Is Equivalent to Four Fifths?
Some of the fractions that are equivalent to four fifths include: 8/10, 12/15, 16/20, 20/25, 24/30, 28/35, 32/40, 36/45 and 40/50. All of these fractions are divisible by 4/5, which is their simplest form.
Q&A Related to "What Fraction Is Equivalent to Four Fifths"
Eight tenths, twelve fifteenths, and sixteen twentieths are some. http://wiki.answers.com/Q/What_fractions_are_equiv...
The fraction four fifths converted to a number with decimals is 0.8... http://www.chacha.com/question/what-is-equivalent-...
If you mean four twelfths, divide by 2: two sixths. then divide by 2 again. one third. http://answers.yahoo.com/question/index?qid=201201...
Four and one-eight = 4 1/8 Five and two-fifths = 5 2/5 Converting each mixed fraction to improper fraction, we get: 4 1/8 = (4*8+1) / 8 = 33/8 5 2/5 = (5*5+2) / 5 = 27/5 Therefore http://www.webanswers.com/education/homework-quest...
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1. Identify the numerator (top number) and denominator (bottom number) of any fraction in the equation. For example, in the fraction "four-fifths, or 4/5, ...
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Purchase Solution
# 3-Regular Polyhedron: How many square faces can the polyhedron have?
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5. The faces of a 3-regular polyhedron are all squares or hexagons. How many square faces can the polyhedron have? Does the number of square faces uniquely determine the polyhedron? (Please see attachment)
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A 3-regular polyhedron is investigate. The solution is detailed and well presented. The response received a rating of "5" from the student who posted the question.
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First of all, we have to state precisely that we are talking about convex polyhedrons.
For every convex polyhedron (either regular or not) it is valid the Euler relation:
v + f - e = 2 ( 1)
where
v = no. of vertices
f = no. of faces
e = no. of edges
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##### Geometry - Real Life Application Problems
Understanding of how geometry applies to in real-world contexts
This quiz test you on how well you are familiar with solving quadratic inequalities.
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# 2 ONE- DIMENSIONAL MOTION
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1 2 ONE- DIMENSIONAL MOTION Chapter 2 One-Dimensional Motion Objectives After studying this chapter you should be able to derive and use formulae involving constant acceleration; be able to understand the concept of force; be able to use Newton's Laws of Motion in various contexts; know how to formulate and solve equations of motion; be able to use the principles of conservation of momentum. 2.0 Introduction The physical world is full of moving objects. Kinematics is the study of motion; dynamics is the study of forces that produce motion. In this chapter the mathematics for describing motion is developed and then the links between the forces acting and the change in motion are described. To describe the motion of real objects you usually need to make simplifying assumptions. Perhaps the most important simplification in applied mathematics is ignoring the size and shape of an object. In Chapter 1 the notion of replacing a real object by a point was introduced. For example, in defining Newton's Law of Gravitation for the force acting on an object near to the earth's surface, the object and the earth were considered as points. Now the normal terminology is to consider objects as particles, and this is then called the particle model. This simplification provides a starting point for many problems, but it does mean that some features of the motion of objects have to be ignored. For example, consider the description of the motion of a tennis ball. At first sight the ball may appear to follow the typical parabolic path of any object thrown near the earth's surface (such motions are studied in detail in Chapter 5). However, a closer study of the motion shows that the ball will be spinning, causing it to swing and dip. The particle model will be good enough to describe the overall parabolic motion but the effects of spin will have to be neglected. A less simple model which includes the features of size and shape would be required to describe the effects of spin. 15
2 Consider the motion of a snooker ball. It is possible to make the ball slide or roll on the table or to move with a combination of both types of motion. For which type of motion will the particle model be most appropriate? What features of the motion of a rolling snooker ball will be neglected with the particle model? This chapter will concentrate on the description of objects which move along a straight line. The objects will be modelled as particles, and represented on diagrams by 'thick' points. 2.1 How to represent motion Displacement Typical of the questions to be answered are: if a ball is thrown vertically upwards, how long does it take to fall to the floor? What is its velocity as it hits the floor? To answer questions like these you need to find the position and velocity of the ball as functions of time. The first step is to represent the ball as a particle. The position of a particle moving in a straight line at any given instant of time is represented on a straight line by a single point. In order to describe the exact position you choose a directed axis with a fixed origin 0 and a scale as shown opposite. The position of the particle relative to the origin is called the displacement and is often denoted by the letter s measured in metres (m). The choice of the origin in a problem will depend on what motion is being modelled. For example, in the problem of throwing a ball in the air an origin at the point of release would be a sensible choice. s s O Displacement - time graphs As a particle moves the displacement changes so that s is a function of time, t. A graphical method of showing motion is the displacement-time graph which is a plot of s against t. As an example, the figure shows the displacement-time graph for the motion of a ball thrown in the air and falling to the floor. s (m) From the graph a qualitative description of how the position of the particle varies in time can be given. The ball starts at the origin and begins to move in the positive s direction (upwards) to a maximum height of 5 m above point of release. It then falls to the floor which is 2 m below the point of release. It takes just over 2 seconds to hit the floor but this part of the motion has not been shown t (s) 16
3 Activity 1 graphs Interpreting displacement - time Discuss the motion represented by each of the displacement - time graphs shown here. s s t t s s t t Velocity Once the position of a particle has been specified its motion can be described. But other quantities, such as its speed and acceleration, are often of interest. For example, when travelling along a road in a car it is not the position that is of interest to the police but the speed of the car! The statement that the speed of a car on the M1 is 60 mph means that if the speed remains unchanged then the car travels for 60 miles in one hour. However the statement gives no information about the direction of motion. The statement that the velocity of a car on the M1 is 60 mph due north tells us two things about the car. First its speed is 60 mph (the magnitude) and the car is heading due north (the direction). Quantities which have magnitude and direction are called vectors and these are discussed more fully in Chapters 3 and 4. The average velocity of a particle over a given time period, T say, is probably familiar to you and is defined by average velocity = displacement time taken. 17
4 Such a definition does not describe the many changes in velocity that may occur during the motion of the particle. In the time interval T from t = t 0 to t = t 0 + T the distance travelled is s( t 0 +T) st ( 0 ) so the average velocity is st ( 0 +T) st 0 T ( ) Now the quantity that is more interesting is not the average velocity of the particle but the instantaneous velocity. You will have seen from the definition of differentiation in the Foundation Core that the link between average changes and instantaneous changes is the derivative. As the time interval T tends to zero the ratio st ( 0 +T) st 0 T ( ) tends towards the derivative of s(t). Thus velocity is defined in the following way. If s(t) is the displacement of a particle then its velocity is defined by v = ds dt In the SI system of units velocity is measured in metres per second written as ms 1. Activity 2 Limits of average velocity The displacement of a particle is given by s = t 2 + 2t. Calculate the average velocity of the particle during each of the time intervals from t = 1 to t =2 from t = 1 to t =1.1 from t = 1 to t =1.01 from t = 1 to t =
5 Estimate the value that the average velocity is tending towards as t 1. Does this value agree with ds dt when t = 1? s The definition of the velocity as a derivative can be interpreted geometrically as the slope of the tangent to the displacement - time graph. For example, consider the displacement - time graph opposite for the ball thrown into the air: The slopes of the tangents to the graph at each of the points t = 0.5, t = 1 and t = 1. 5 are equal to the velocities at these points. When t = 0.5 (point A) the slope of the tangent is given by = 5 ms 1. At t = 1 (point B) the slope is zero and at t = 1. 5 (point C) the slope is 5 ms 1. You can now say more about the motion of the ball. At point A, the ball is going upwards with speed 5 ms 1. At point B the ball is instantaneously at rest and it is at its highest point. At point C the ball is falling to the floor with speed 5 ms 1. For one-dimensional motion the sign of the velocity indicates the direction of motion of the particle. If v > 0 then s is increasing with time since ds dt > 0. If v < 0 then s is decreasing with time since ds dt < 0. The magnitude of the velocity of a particle is called its speed. For example, if v = 3 ms 1 then you can say that the particle moves with a speed 3 ms 1 in the direction of s decreasing. B 0.5 A 2.5 C s increasing v > 0 direction of motion s increasing v < 0 direction of motion v (ms -1 ) t 10 Velocity - time graphs If you find the velocity, v, at several times, t, then a graph of the velocity against time is called a velocity - time graph. The figure shows the velocity - time graph for the motion of a ball thrown into the air and falling to the floor t (s) The ball begins its motion with a speed of 10 ms 1 and this speed falls to zero during the first second of the motion. The speed then increases to approximately 12 ms 1 when the ball hits the floor. 19
6 Activity 3 Interpreting velocity - time graphs The diagram shows the velocity -time graph for Graph 1 in Activity 1. v (ms -1 ) Describe the motion of the particle from this graph. Sketch the velocity - time graphs for the other displacement - time graphs in Activity 1. 0 t (s) Describe the motion of the particle in each case. Acceleration Chapter 1 identified the change in motion as an important quantity in the link between force and motion. Hence in many situations it is not the velocity that is important but the change in velocity. This is described by the acceleration. You have seen that the velocity is defined as the rate of change of position; the acceleration is defined in a similar way to the velocity. If v(t) is the velocity of a particle at time t then the acceleration of the particle is defined by a = dv dt. In the SI system of units, acceleration is measured in metres per second per second, written as ms 2. The acceleration can be obtained from the slope of the velocity - time graph. For example, for the motion of the ball thrown into the air, the diagram shows the acceleration - time graph. The acceleration is constant with magnitude 10 ms 2. The negative sign implies that the velocity decreases continuously with time. 10 a (ms -2 ) 0 s = 5 ms -1 t (s) On a diagram velocities are shown with single arrows and accelerations with double arrows. a = 10 ms -2 O s 20
7 Activity 4 Interpreting acceleration - time graphs Sketch the acceleration - time graphs from your velocity - time graphs in Activity 3. Describe the motion of the particle in each case. Relationships between displacement, velocity and acceleration In this section you have seen that the motion of a particle along a straight line can be described by a displacement s(t). The velocity and acceleration of the particle are then given by v = ds dt and a = dv dt respectively. Graphical descriptions of motion are given by displacement - time, velocity - time and acceleration - time graphs. Now in mechanics problems the acceleration is often known and the velocity and displacement have to be found. This is achieved by integration. You will know from your knowledge of pure mathematics that integration is equivalent to finding the area under a graph. Activity 5 Finding a velocity and displacement a (ms -2 ) The graph opposite shows the acceleration of a particle over 4 seconds of its motion. The particle starts from rest, at t = 0. Estimate the velocity of the particle at the end of 0.5, 1, 1.5, 2, 2.5, 3, 3.5, 4 seconds t (s) Use your results to sketch a velocity - time graph for the particle. From your velocity - time graph estimate the distance travelled during the 4 seconds of motion between t = 0 and t = 4. The equation for the graph above is a = 2 + t. By integrating a to find v and then integrating v to find s, check your answer. 21
8 Exercise 2A 1. Sketch displacement - time and velocity - time graphs for the following: (a) A car starts from rest and increases its velocity steadily to 10 ms 1 in 5 seconds. The car holds this velocity for another 10 seconds and then slows steadily to rest in a further 10 seconds. (b) A ball is dropped on a horizontal floor from a height of 3 m. The ball bounces several times before coming to rest. 5. A road has a sharp bend around which approaching traffic cannot be seen. The width of the road is 6 m and the speed limit is 30 mph. The bend is on the near side of the road. How far from the bend should a pedestrian cross the road to avoid an accident? Assume the man walks at 3 mph. car pedestrian 2. (c) A person jumps out of an aircraft and falls until the parachute opens. The person glides steadily to the ground. v (ms -1 ) t (s) The diagram represents the motion of an object for 30 seconds. Calculate the acceleration for each of the following intervals: 6. A particle is set in motion at time t = 0 and its position is subsequently given by s = 4 + 5t 2t 2. (a) Calculate the velocity of the particle after 1 second and after 2 seconds. What is the speed and direction of motion at each of these times? (b) Find the time at which the particle is instantaneously at rest. (c) Calculate the acceleration of the particle at t = 1 s and t = 2 s. (d) Describe the motion of the particle. 7. Repeat Problem 6 for (a) s = t 2 4t +1 (a) 0 < t < 10 (b) 10 < t < 15 (c) 15 < t < 30 Calculate the displacement of the object over the 30 seconds. 3. t(s) v( ms 1 ) (a) Plot these figures on a velocity - time graph. (b) Verify that for 0 t 4, the figures are consistent with constant acceleration. (b) s = 3 +18t 7.5t 2 + t The acceleration of a particle is given by a = 10 ms 2. At the instant t = 1 s, the particle is at position s = 2 m and has velocity 3 ms -1. (a) Find the velocity and displacement of the particle as functions of time. (b) Calculate the position and velocity of the particle when t = 2 s. (c) Calculate an estimate for the acceleration at t = 6. (d) Calculate an estimate for the displacement after 8 seconds. 4. During the launch of a rocket the velocity was noted every second for 10 seconds and the following table of values obtained. t(s) v( kph) Estimate the distance travelled by the rocket during the first 10 seconds of its motion. 22
9 2.2 Modelling motion under constant acceleration There are several simple formulae which can be used when dealing with motion under constant (also known as uniform) acceleration. velocity Discuss common types of motion where you think that constant acceleration is likely to occur. The diagram represents the motion of an object with initial velocity u and final velocity v after t seconds has elapsed. The gradient of the line is calculated from the expression v u t v - u v u. t time Since the gradient is equal to the value of the acceleration, a, then a = v u. t This can be rewritten as v = u + at (1) The area under the velocity - time graph is equal to the displacement of the object. Using the rule for the area of a trapezium gives s = ( u + v)t 2 (2) ( u + v) Note that is the average velocity of the object, so (2) is 2 the algebraic form of the result that the displacement is equal to the average velocity multiplied by the time. Example A motorbike accelerates at a constant rate of 3 ms 2. Calculate (a) (b) the time taken to accelerate from 20 mph to 40 mph. the distance in metres covered during this time. Solution You can use equation (1) to find the time and then equation (2) to find the distance travelled. But first you must convert mph to ms 1. 23
10 Using the conversion rule 5 miles 8 kilometres, gives mph = 16 kph = ms 1 = ms 1. Hence 20 mph = 8.89 ms 1 and 40 mph = ms 1. (a) (b) Using equation (1) the time taken to accelerate from 8.89 ms 1 to ms 1 at 3 ms 2 is given by t = = 2.96 s (to 3 sig. fig.). 3 Using equation (2) the distance travelled in this time is s = 2.96 = 39.5 m. 2 There are two further useful formulae which can be obtained from (1) and (2). Writing (1) in the form t = v u a and substituting into (2) gives s = u + v v u, 2 a so s = v2 u 2 2a. v 2 u 2 = ( v u) ( v+u) Remember This expression can be rearranged to give 24 v 2 = u 2 + 2as (3) Similarly, substituting for v from (1) into (2) gives s = u + u + at t 2 or s = ut at 2 (4) Formula (3) is useful when the time t has not been given or is not required, while formula (4) is useful when the final velocity v has not been given nor is required.
11 Example A car accelerates from a velocity of 16 ms 1 to a velocity of 40 ms 1 in a distance of 500 m. Calculate the acceleration of the car. Solution Using (3) 40 2 = a a = = ms Example A car decelerates from a velocity of 36 ms 1. The magnitude of the deceleration is 3 ms 2. Calculate the time required to cover a distance of 162 m. Solution When an object is slowing down it is said to be decelerating. You then use equations (1) - (4) but with a negative value for a. In this problem set a = 3. Let t seconds be the time required to cover 162 m. Using (4), 162 = 36t ( 3) t2. Rearranging this gives Dividing by 1.5 gives Factorizing gives 1. 5t 2 36t +162 = 0. t 2 24t +108 = 0. ( t 6) ( t 18)=0, so that t = 6 or18. Discuss the significance of the two solutions to the quadratic equation. Which is the required time? 25
12 Activity 6 Stopping distance on the road The Highway Code gives the following data for overall stopping distances of vehicles at various speeds. Speed (mph) Thinking Braking Stopping Distance Distance Distance (metres) (metres) (metres) What is meant by the thinking distance? Show that the deceleration during braking is roughly the same at each of the three speeds. Use a graph plotter to fit a suitable curve through the data for speed and stopping distance. Use your results to estimate the speed corresponding to a stopping distance of 150 metres. The amber warning light on traffic signals is intended to give drivers time to slow before a red stop light. Time the duration of amber lights in your locality. Do they give sufficient warning at the speed limit in operation on the road? Exercise 2B 1. A car accelerates uniformly from a speed of 50 kph to a speed of 80 kph in 20 seconds. Calculate the acceleration in ms For the car in Question 1, calculate the distance travelled during the 20 seconds. 3. A train signal is placed so that a train can decelerate uniformly from a speed of 96 kph to come to rest at the end of a platform. For passenger comfort the deceleration must be no greater than 0.4 ms 2. Calculate (a) the shortest distance the signal can be from the platform; (b) the shortest time for the train to decelerate. 4. A rocket is travelling with a velocity of 80 ms 1. The engines are switched on for 6 seconds and the rocket accelerates uniformly at 40 ms 2. Calculate the distance travelled over the 6 seconds. 5. In 1987 the world record for the men's 60 m race was 6.41 seconds. (a) Assuming that the race was carried out under constant acceleration, calculate the acceleration of the runner and his speed at the end of the race. (b) Now assume that in a 100 m race the runner accelerates for the first 60 m and completes the race by running the next 40 m at the speed you calculated in (a). Calculate the time for the athlete to complete the race. 6. The world record for the men s 100 m was 9.83 s in Assume that the last 40 m was run at constant speed and that the acceleration during the first 60 m was constant. (a) Calculate this speed. (b) Calculate the acceleration of the athlete. 26
13 7. Telegraph poles, 40 m apart stand alongside a railway line. The times taken for a locomotive to pass the two gaps between three consecutive poles are 2.5 seconds and 2.3 seconds respectively. Calculate the acceleration of the train and the speed past the first post. 8. The world record for the women s 60 m and 100 m are respectively 7.00 seconds and seconds. Analyse this information using the method given in Question A set of traffic lights covers road repairs on one side of a road in a 30 mph speed limit area. The traffic lights are 80 m apart so time must be allowed to delay the light changing from green to red. Assuming that a car accelerates at 2 ms 2 what is the least this time delay should be? 10. A van travelling at 40 mph skids to a halt in a distance of 15 m. Find the acceleration of the van and the time taken to stop, assuming that the deceleration is uniform. 2.3 How do bodies move under gravity? For many centuries it was believed that: (a) (b) heavier bodies fell faster than light ones and the speed of a falling body was constant. Discuss why such views were held and suggest ways in which they could be refuted. Galileo Galilei ( ) was the first person to state clearly that all objects fall with the same constant acceleration. Since there were no accurate timers available in the seventeenth century, he demonstrated this principle by timing balls which were allowed to roll down inclined planes. Activity 7 Galileo s rolling ball experiment You will need balls of different masses, a table, some blocks, four metre rulers and a stop watch. (a) (b) (c) Use the blocks to set up the table as shown in the diagram so that it takes about 4 seconds for the ball to roll down. Fix the rulers to make a channel down which the ball can roll. Measure the time it takes the ball to roll distances of 0.5, 0.75, 1, 1.25, 1.5, 1.75 and 2.0 m. Repeat the experiment and find the average time in each case. Find a relationship between the distance travelled, s metres, and time taken, t seconds. Do your results support Galileo s statement? Modern determination of the acceleration of falling bodies gives values in the region of 9.81 ms 2 although the value varies slightly over the surface of the Earth. The magnitude of this acceleration is denoted by g. A common approximation is to take g =10 ms 2 and this value will be used in this text for solving problems. 27
14 Since the acceleration acts towards the Earth s surface, its sign must be opposite to that of any velocities which are upwards. When dealing with problems involving motion under gravity, you can use the formulae for constant acceleration developed in the previous section. Example A ball is thrown vertically upwards with an initial speed of 30 ms 1. Calculate the height reached. Solution at the top of the flight the velocity of the ball is zero h Since the ball slows down, take a = g= 10 ms 2. Using (3) 0 = h where h is the maximum height reached. Thus h = = 45 m (to 2 significant figures). Example A ball is thrown vertically upwards with a speed of 40 ms 1. Calculate the time interval between the instants that the ball is 20 m above the point of release. Solution Using (4) with a = g= 10 ms 2 20 = 40t t 2 where t seconds is the time passed since the ball was thrown up. 5t 2 40t + 20 = 0 t 2 8t + 4 = 0. Using the quadratic formula t = 8 ± 48 2 = 7.62 or The ball is 20 m above the point of release twice, at t = 0.54 s (on way up) and t = 7.62 s (on way down). The required time interval is = 7.08 seconds. 28
15 Two useful formulae which can be used on a body falling from rest through a height h metres can be found by putting u = 0, a = 10 and s = h in equation (4) to give t = 2h g = h 5 (5) and by putting u = 0, a = 10 and s = h in equation (3) to give v = 2gh = 20h (6) where t is the time of fall in seconds, v ms 1 is the final velocity and g = 10 ms 2. Activity 8 Estimating the value of g You will need a bouncy ball and a metre rule for this activity. Drop the ball onto a hard floor from a height of 2 m. Measure the height of the rebound. Repeat this several times and average your results. Now drop the ball from the rebound height and measure the new height of rebound. Repeat this procedure several times and average your results. Keep measuring new rebound heights for two further cases. h 1 h 2 h3 h4 Now drop the ball from 2 m and measure the time elapsed until the fourth bounce with the floor. Repeat several times and average your results. Show that the total time, t, up to the fourth bounce is t = 2h 1 g + 2 2h 2 g + 2 2h 3 g + 2 2h 4 g. Use this, together with your measurements, to calculate a value for g. Exercise 2C (Take g = 10 ms 2 ) 1. A ball is dropped on to level ground from a height of 20 m. (a) Calculate the time taken to reach the ground. The ball rebounds with half the speed it strikes the ground. (b) Calculate the time taken to reach the ground a second time. 29
16 2. A stone is thrown down from a high building with an initial velocity of 4 ms 1. Calculate the time required for the stone to drop 30 m and its velocity at this time. 3. A ball is thrown vertically upwards from the top of a cliff which is 50 m high. The initial velocity of the ball is 25 ms 1. Calculate the time taken to reach the bottom of the cliff and the velocity of the ball at that instant. 4. The diagram show three positions of a ball which has been thrown upwards with a velocity of u ms 1 C B A Position A is the initial position. Position B is halfway up. Position C is at the top of the motion. Copy the diagram and for each position put on arrows where appropriate to show the direction of the velocity. On the same diagram put on arrows to show the direction of the acceleration. 5. An aircraft is flying at a height of 4 km when it suddenly loses power and begins a vertical dive. The pilot can withstand a deceleration of 5 g before becoming unconscious. What is the lowest height that the pilot can pull out of the dive? 6. If the Earth is assumed to be a perfect sphere then the acceleration due to gravity at a height h m above the surface of the Earth is given by k R + h ( ) 2 where R is the radius of the Earth in metres and k is a constant. (a) Given that R = 6400 km and at the Earth's surface g = 9.8 ms 2, estimate the value of k. (b) Use your calculator to find the height at which the acceleration differs by 1% from its value at the Earth s surface. (c) Use a graphic calculator to draw the variation of acceleration with height. 7. When a ball hits the ground it rebounds with half of the speed that it had when it hit the ground. If the ball is dropped from a height h, investigate the total distance travelled by the ball. 8. One stone is thrown upwards with a speed of 2 ms 1 and another is thrown downwards with a speed of 2 ms 1. Both are thrown at the same time from a window 5 m above ground level. (a) Which hits the ground first? (b) Which is travelling fastest when it hits the ground? (c) What is the total distance travelled by each stone? 2.4 What causes changes in motion? Think about the following types of motion: an athlete running around a bend in a 200 m race; a ball being thrown over a fence; a car braking to a halt; a rocket accelerating in space; a snowball picking up snow as it rolls on level ground. In each case there is a change in motion. Discuss what these changes are in each case. A change in motion is caused by a force. In medieval times it 30
17 was thought that a force was required to keep a body in motion and that the only state which corresponded to an absence of forces was a state of rest. The true relationship between forces and motion was stated by Newton using ideas of Galileo. In the absence of any forces there must be no change in the motion of the body, that is, the body must be at rest or moving with uniform velocity. Although first stated by Galileo, this is now generally known as Newton s First Law of Motion. Newton s First Law A body remains in a state of rest or moves with uniform motion, unless acted on by a force. Activity 9 The chute experiment. You will need a level table, 2 metre rules, a billiard ball, a chute and a stopwatch. You can make a chute out of a piece of folded cardboard. Use the metre rules to make a channel on the table. Allow the ball to roll down the chute so that it takes about 3 seconds to travel 1 m. Mark the point on the chute from which you release the ball. Releasing the ball from this mark for each run, time the ball to roll 0.25, 0.5, 0.75 and 1 m. Draw a displacement - time graph. What forces act on the marble as it rolls on the table? Are your results consistent with Newton s First Law? If the total force acting on a body is zero (for example, if equal and opposite forces act on the same body), then the body remains at rest or moves in a state of uniform motion. Newton s First Law defines what happens to the motion of a body if no force is present. When a force acts on a body, there is a change in motion. Firstly, you need a clear definition of how to measure the motion of a body. Newton did a great deal of experimental work on this and came to the conclusion that the motion of a body was measured by the product of its mass and its velocity. This quantity is known as the momentum of the body. Momentum = m v 31
18 A force produces a change in the momentum of a body, through a combination of changes in mass and/ or velocity. Discuss the change in momentum in each of the cases given at the beginning of the section. The physical law relating change in motion and the force acting on a body is given by Newton s Second Law. Newton s Second Law The rate of change of momentum of a body is proportional to the applied force. In the case where the mass of the body is constant, this leads to the result that the force is proportional to the product of the mass and the acceleration of the body. By choosing appropriate units to measure mass, acceleration and force, the constant of proportionality can be made equal to one so that F = ma where m is in kilograms ( kg) a is in metres per second per second ( ms 2 ) F is in newtons. One newton is the force sufficient to produce an acceleration of one ms 2 in a body of mass one kg. The abbreviation for a newton is N. The approximate magnitude of some typical forces are force exerted by an adult arm 250 N force exerted by Earth s gravity on an adult 700 N force exerted by a car engine 2000 N Example A car of mass 1100 kg can accelerate from rest to a speed of 30 mph in 12 seconds. Calculate the force required. Solution Using the conversion 10 mph = ms 1 gives mph = 13.3 ms 1 Assuming that the car accelerates at a constant rate, then using v = u + at with v = 13.3, u = 0 and t = 12 gives a = = 1.11 ms 2.
19 Applying Newton's Second Law gives F = ma = = 1220 N (to 3 significant places). Example A force of magnitude 20 N is applied to a particle of mass 4 kg for 6 seconds. Given that the initial velocity of the body is 15 ms 1, (a) (b) calculate the acceleration, a, of the body; calculate its velocity, v, after 6 seconds. After 6 seconds a force, F, is applied to bring the body to rest in a further 125 m. (c) Calculate the magnitude of the force. Solution (a) Using Newton s Second Law, 20 = 4a a = 5 ms 2 (b) (c) Using (1), v = = 45 ms 1 Since the particle is slowing down, the acceleration in equation (3) will be negative so put a = A. Then using (3) 0 = A 125. Solving for A, A = 452 = 8.1 ms Using Newton s Second Law, F = = 32.4 N. Activity 10 Forces produced by car engines (a) (b) Collect data on typical accelerations and masses of cars. Calculate the average force produced during the acceleration. Use the data for braking distances shown in Activity 6 to calculate the average force applied during braking. 33
20 Since all objects fall with an acceleration of g (neglecting air resistance) near the Earth s surface, the force acting on an object of mass m is given by F = mg. This force is known as the weight of the object. Exercise 2D 1. A force of 50 N is applied to a particle of mass 4 kg for 5 seconds only. (a) Calculate the acceleration for the first 5 seconds. (b) Write down the acceleration for the next 5 seconds. (c) Calculate the distance travelled during the first 10 seconds given that the particle was at rest initially. 2. A world class sprinter can accelerate from rest to 10 ms 1 in about 2 seconds. Estimate the magnitude of the force required to produce this acceleration. 3. The brakes of a train are required to bring it to rest from a speed of 80 kph in a distance of 500 m. The mass of the train is 200 tonnes. (a) Calculate the average deceleration. (b) Calculate the average force required to be exerted by the brakes. 4. hopper belt bed The diagram shows a conveyor belt which is designed to convey coal dust from a hopper to a bed. Initially, there is no dust on the belt. The force required to drive the belt in this case is 200 N and the velocity of the belt is 2 ms 1. The length of the belt is 40 m. Coal is now allowed to fall on the belt at a rate of 8 kg per second. The force driving the belt is adjusted so that the velocity stays at 2 ms 1. Draw a diagram of force against time for the first 30 seconds of operation. 5. A toy car of mass 0.04 kg is propelled from rest by an engine which provides a pulling force of N lasting for 8 seconds. (a) Calculate the acceleration. (b) Calculate the velocity after 8 seconds. If the speed of the toy car decreases uniformly to zero during the next 32 seconds, find the total distance travelled by the car. 6. A rocket has a mass of 40 tonnes of which 80% is fuel. The rocket motor develops a thrust of 1200 kn at all times. (a) Calculate the acceleration when the rocket is full of fuel. (b) Calculate the acceleration just before the fuel is exhausted. (c) What is the acceleration after the fuel is exhausted? 7. A parachute reduces the speed of a parachutist of mass 70 kg from 40 ms 1 to 10 ms 1 in 3 seconds. Calculate the average force exerted by the parachute. 8. A car travelling at 30 mph is brought to rest in 3 seconds during a collision. Calculate the average force exerted on the car during the collision. Assume mass of driver is 70 kg and that of the car is 1100 kg. 34
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### CHAPTER We find the average speed from average speed = d/t = (230 km)/(3.25 h) =
CHAPTER 1. We find the average speed from average speed = d/t = (30 km)/(3.5 h) = 70.8 km/h.. We find the time from average speed = d/t; 5 km/h = (15 km)/t, which gives t = 0.60 h (36 min). 3. We find
### 5 Day 5: Newton s Laws and Kinematics in 1D
5 Day 5: Newton s Laws and Kinematics in 1D date Friday June 28, 2013 Readings Knight Ch 2.4-7, Ch 4.6, 4.8 Notes on Newton s Laws For next time: Knight 5.3-8 lecture demo car on a track, free-fall in
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A174207 Depleted stream of the natural numbers written in base 2: delete even occurrences of digit 0 and odd occurrences of digit 1. 3
0, 1, 1, 0, 1, 0, 1, 1, 1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 0, 1, 0, 0 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,1 COMMENTS The numbers in base 2: 0, 1, 10, 11, 100, 101, 110, 111, 1000..... The infinite stream of digits: 0110111001011101111000... Delete even (2nd, 4th, 6th...) occurrences of 0 (replaced by ~): 011~1110~10111~11110~0.. Delete odd (1st, 3rd, 5th,...) occurrences of 1 (replaced by ~): 0~1~~1~0~10~1~~1~1~0~0.. Digits remaining define the sequence: 0, 1, 1, 0, 1, 0, 1, 1, 1, 0, 0,.... LINKS MAPLE nb := 100 ; L := [0] ; for n from 1 to nb do nn := ListTools[Reverse](convert(n, base, 2)) ; L := [op(L), op(nn)] ; end do; Lout := [] ; ie :=1 ; io :=1 ; for i from 1 to nops(L) do if op(i, L) =0 then if ie>0 then if type(ie, 'odd') then printf("%d, ", 0) ; end if; end if; ie := ie+1 ; else if io>0 then if type(io, 'even') then printf("%d, ", 1) ; end if; end if; io := io+1 ; end if; end do: # R. J. Mathar, Dec 06 2011 MATHEMATICA nMax = 60; bits = Join @@ Table[IntegerDigits[n, 2], {n, 0, nMax}]; pos0 = Position[bits, 0] // Flatten; even0 = Partition[pos0, 2][[All, 2]]; bits2 = ReplacePart[bits, Alternatives @@ even0 -> "~"]; pos1 = Position[bits2, 1] // Flatten; odd1 = Partition[pos1, 2][[All, 1]]; DeleteCases[ ReplacePart[ bits2, Alternatives @@ odd1 -> "~"], "~"] (* Jean-François Alcover, Nov 18 2016 *) CROSSREFS Cf. A174203, A174204 (base 10), A174205, A174206, A174208-A174210 Sequence in context: A073070 A099104 A066829 * A048820 A144101 A127000 Adjacent sequences: A174204 A174205 A174206 * A174208 A174209 A174210 KEYWORD easy,nonn,base AUTHOR Paolo P. Lava and Giorgio Balzarotti, Mar 15 2010 STATUS approved
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# Get position from Isometric TileMap
I really getting frustated about that, i cant get it to work… but in default TileMap(ex: topdown 2d) is working fine.
(get this code from internet)
get position and store in a dictionary:
``````tiles = new Dictionary<Vector3, WorldTile>();
foreach (Vector3Int pos in Tilemap.cellBounds.allPositionsWithin)
{
var localPlace = new Vector3Int(pos.x, pos.y, pos.z);
if (!Tilemap.HasTile(localPlace)) continue;
var tile = new WorldTile
{
LocalPlace = localPlace,
WorldLocation = Tilemap.CellToWorld(localPlace),
TileBase = Tilemap.GetTile(localPlace),
TilemapMember = Tilemap,
Name = localPlace.x + "," + localPlace.y,
Cost = 1 // TODO: Change this with the proper cost from ruletile
};
}
``````
my test, click on tile and set color to green:
``````private void Update () {
if (Input.GetMouseButtonDown(0))
{
Vector3 point = Camera.main.ScreenToWorldPoint(Input.mousePosition);
var worldPoint = new Vector3Int(Mathf.FloorToInt(point.x), Mathf.FloorToInt(point.y), 0);
var tiles = GameTiles.instance.tiles; // This is our Dictionary of tiles
if (tiles.TryGetValue(worldPoint, out _tile))
{
print("Tile " + _tile.Name + " costs: " + _tile.Cost);
_tile.TilemapMember.SetTileFlags(_tile.LocalPlace, TileFlags.None);
_tile.TilemapMember.SetColor(_tile.LocalPlace, Color.green);
}
}
}
``````
but this get other tile or none… plis help
Hey,
Did you get any solution for this? @antoniomonteiro
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https://www.tutorialspoint.com/python-program-to-find-common-array-elements
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# Python program to find common array elements
While considering the multi-dimensional arrays as an example, there is a method that is capable of finding the common elements present within a multi-dimensional array - intersection_update().
This method is used in order to find the common or intersecting elements present within the same array which is multi-dimensional in nature. Let us consider an input output scenario and then proceed with a program.
### Input Output Scenarios
Consider a 2D array which is multi-dimensional in nature.
arr = [[1, 2, 3, 4], [3, 4, 5, 6], [7, 8, 3, 4], [4, 9, 8, 3], [4, 3, 10, 12]]
• The above array consists of 5 sub arrays.
• We can clearly observe that the elements “ 3 ” and “ 4 ” are present all the sub arrays of “ arr ”.
• So, the elements “ 3 ” and “ 4 ” are considered to be the common elements of the 2D array arr.
### Example
In this example, we are going to find the common elements present within a multi dimensional array using the method intersection_update().
• Consider a two dimensional array from which the common elements can be found.
• Declare a parameterized method which can find the common elements taking the 2D array as a parameter.
• Within the method, initiate the set array with 0 and assign the value to a variable.
• Traverse the elements of the array using a loop.
• With the help of the method intersection_update(), find the common elements one after another while traversing.
def common_elements(arr):
result = set(arr[0])
for x in arr[1:]:
result.intersection_update(x)
return list(result)
# main section
if __name__ == "__main__":
arr = [[1, 2, 3, 4], [3, 4, 5, 6], [7, 8, 3, 4], [4, 9, 8, 3], [4, 3, 10, 12]]
res = common_elements(arr)
if len(res) > 0:
print ("The common elements present within the array are: ",res)
else:
print ("There are no common elements present within the array !!")
### Output
The output of the above program is as follows −
The common elements present within the array are: [3, 4]
## Finding Common Elements in two Different Arrays
The NumPy intersect1d() method can be used to find the common elements of two one-dimensional arrays. This is similar to the intersect_update() method, which deals with multi-dimensional arrays. To better understand this concept, let's look at an example.
### Input Output Scenarios
Consider two arrays which are one dimensional in nature.
arr1 = [1, 2, 3, 4]
arr2 = [3, 4, 5, 6]
• We can clearly see that the elements “ 3 ” and “ 4 ” are present in both arrays arr1 and arr2.
• So, we can conclude that the common elements of the arrays arr1 and arr2 are 3 and 4.
### Example
In the following example, we are going to find the common elements present within multiple one dimensional arrays using the method intersect1d() of the numpy module.
import numpy as n
arr1 = [1, 2, 3, 4]
print("The first array is: ")
print(arr1)
arr2 = [3, 4, 5, 6]
print("The second array is: ")
print(arr2)
narr1 = n.array(arr1)
narr2 = n.array(arr2)
print("The common elements within the given arrays are: ")
print(n.intersect1d(narr1, narr2))
### Output
The output of the above program is as follows −
The first array is:
[1, 2, 3, 4]
The second array is:
[3, 4, 5, 6]
The common elements within the given arrays are:
[3 4]
In this way, depending on the type of arrays, the procedure can be followed. If the array considered is a multi-dimensional array, then the procedure followed in the first program will be followed.
If the arrays considered are one-dimensional arrays, then the procedure followed in the second program will be followed. This is how the common elements of one or more arrays are found.
Updated on: 05-May-2023
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• KDR
• Helping Hand
2014-05-01T21:01:18-04:00
Since -2 appears twice, it is called a double root. This is showed with the expression
(x+2)². Squaring the expression gives the double root part. With a zero at 2, another expression is required:
(x-2).
In order to find a polynomial with these roots, both expressions need to be multiplied:
(x+2)(x+2)(x-2).
Starting with the two (x+2)'s, we need to use the FOIL (First Outer Inner Last) method of distributing, so this is how to distribute this:
(x*x) + (x*2) + (x*2) + (2*2).
Fully simplified, you now have (x² + 4x +4)(x-2). This time, distributing will be a bit more complicated, due to the fact that there are no longer only 4 terms. Now the distribution looks like so:
(x²*x) +(x²*-2) + (4x*x) + (4x*-2) +(x*4) +(4*-2).
When each equation within parentheses is completed, we get:
x^3 - 2x² + 4x² - 6x + 4x - 8.
After combining like terms, your answer is x³ + 2x² - 2x - 8.
thank you so much!
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# How to Calculate Mortgage Payoff
Co-authored by Carla Toebe, Real Estate Broker
Updated: December 31, 2019
The method for precisely determining the rate of amortization, which is the amount needed to pay off a particular mortgage loan, will vary depending on factors like the type of loan, its terms, and what options are exercised by the borrower. However, there is a standard formula used for calculating the loan payoff amount of a mortgage based on the principal, the interest rate, the number of payments made, and the number of payments remaining. This article provides detailed information that will assist you in calculating your mortgage payoff amount based on the terms of your loan. Contact your lender to confirm that your calculation is correct based on the particulars of your mortgage.
### Part 1 of 2: Knowing the Essentials
1. 1
Understand why your mortgage payoff amount does not equal your current balance. Your statement says your most recent balance is \$12,250, and yet your payoff amount (the amount it would take to close out the mortgage) is listed as \$12,500. How can that be?
• You may assume that some complicated financial formulas are at play (ones that cost you money, of course), but in fact the answer is quite simple: mortgages are paid in arrears. That is, you pay June’s interest with July’s payment.[1]
• This process starts at the beginning, at closing. If you close in June, for instance, your first payment won’t be due until August, with July’s interest included.
2. 2
Gather the information needed for your calculations. In order to determine the payoff amount, either using a calculation program or on your own, you need to know a handful of basic figures regarding your mortgage. All of these figures should be available on your statement or other loan documents. These include:
• The total amount borrowed when you took out the loan (for example, \$200,000).
• The annual interest rate (for example, 3%, or 0.03). To do the calculations yourself, you will need to divide this number by twelve (0.03 / 12 = 0.0025), because mortgage interest compounds monthly.
• The total number of payments for the life of the loan, which for monthly payments is the number of years times twelve (for example, 20 years = 240 payments).
• The total number of years / payments remaining, and the number paid so far (for example, 15 years = 180 payments made; 5 years = 60 payments remaining).
3. 3
Consider online calculators if you’d rather not exercise your math muscles. The calculations aren’t all that complicated, but punching a few numbers in and pressing “calculate” is certainly easier. A search for “mortgage payoff calculator” will provide several useful results. Try out a few to ensure accuracy.
• Please note, however, that you may find slight variations in your final results, probably only a few cents, but enough that you should always confirm your payoff amount with your lender before attempting to make a final payment.
• Using two such online calculators for a \$200,000 loan at 3% annual interest over 20 years, with five years remaining, produces payoff results of \$61,729.26 and \$61,729.33, respectively.[2][3]
4. 4
Contact your mortgage lender if you plan to make a final payment. Avoid making what you think is a mortgage payoff only to find out that you still owe a few dollars or cents that keeps your mortgage alive and accruing interest.
• Contact your lender by phone or online; lenders with online account management likely have a page that allows you to make the request. You will likely receive your payoff amount after a week.
• You will be asked to choose a specific date for which to determine the payoff.
• If you want to make the final payment, you will need to complete some version of payment form (online or by mail) with the precise amount due and the time frame in which this amount is valid as a final payoff.
• Such requests can be made solely for informational purposes as well. Some lenders may even include a payoff amount on your monthly statement.
### Part 2 of 2: Making the Calculations
1. 1
Lay out the formula carefully. It looks rather unwieldy at first, but the math is relatively straightforward once you’ve inserted your figures. Just be sure to copy the formula exactly or your results may vary significantly. The formula is:[4]
• B = L [(1 + c)^n - (1 + c)^p] / [(1 + c)^n (- 1)] , in which:
• B = payoff balance due (\$)
• L = total loan amount (\$)
• c = interest rate (annual rate / 12)
• n = total payments (years x 12 for monthly payments)
• p = number of payments made so far
2. 2
Insert your figures. Using the same example as for the online calculators, a 20-year, \$200,000 mortgage at 3% interest with five years to go, appears thusly:
• B = 200,000 [(1 + 0.0025)^240 - (1 + 0.0025)^180] / [(1 + 0.0025)^240 (-1)]
• Remember that the 3% annual interest rate (0.03) is divided by twelve because it compounds monthly (c = 0.03 / 12 = 0.0025).
• Twenty years of monthly payments is 240 (n = 20 x 12 = 240), and fifteen years of payments so far is 180 (p = 15 x 12 = 180).
3. 3
Power up your numbers. After adding 1 to 0.0025 the three times it appears in the formula, you will need to raise the resulting 1.0025 to the 240th power (twice) and to the 180th power (once). A good calculator will come in handy here.
• 1.0025^240 = 1.82075499532
• 1.0025^180 = 1.56743172467
• B = 200,000(1.82075499532 - 1.56743172467) / (1.82075499532 - 1)
4. 4
Subtract from the inside. Once you accomplish this step, the formula will appear much more manageable and will be that much closer to completion.
• 1.82075499532 - 1.56743172467 = 0.253323270652
• 1.82075499532 - 1 = 0.82075499532
• B = 200,000(0.253323270652 / 0.82075499532)
5. 5
Divide, multiply, and conquer. Now you can finish things off and solve for B, your payoff amount.
• 0.253323270652 / 0.82075499532 = 0.308646638883
• 200,000 x 0.308646638883 = 61729.3277766
• Therefore, your payoff amount is \$61,729.33.
## Community Q&A
Search
• Question
How do you calculate a mid-month mortgage payoff?
Carla Toebe
Real Estate Broker
Carla Toebe is a Real Estate Broker in Washington. She has been an active real estate broker since 2005, and founded the real estate agency CT Realty LLC in 2013.
Real Estate Broker
To calculate mid month, multiply the monthly payment for the insurance premiums, interest, taxes, homeowner insurance, and anything else that is lumped into the monthly mortgage payments by the number of days until the close divided by the number of days in the month.
• Question
I owe 31900.00 on mortgage initially. The rate is 6.5 and I have 11yrs & 10 months to pay off, what would my payoff be
Carla Toebe
Real Estate Broker
Carla Toebe is a Real Estate Broker in Washington. She has been an active real estate broker since 2005, and founded the real estate agency CT Realty LLC in 2013.
Real Estate Broker
There are many other factors that are needed to answer this question. You need what the total term of the loan is, the original loan amount, the total number of payments in the term and the total number of payments made so far.
• Question
I have a mortgage with a private individual. I am going to pay off the mortgage. I need an ammortization schedule where I can plug in the amount of my monthly payments (\$800) for the last 2 years at 7% interest.
Carla Toebe
Real Estate Broker
Carla Toebe is a Real Estate Broker in Washington. She has been an active real estate broker since 2005, and founded the real estate agency CT Realty LLC in 2013.
Real Estate Broker
You can find free ammortization schedules on line. Just search for them. These will allow you to plug in your values.
• How do I tell how long my mortgage payments will take when paying off my home?
• How do I calculate a mortgage payoff when I know the loan amount, interest rate, balance, and the amount of time?
• How do I determine my payoff amount?
200 characters left
## Tips
• You must get a quote for the mortgage payoff before you sell a property. The money you make on the sale will first go to covering the payoff, then towards all of the other associated fees and closing costs that the seller is responsible for. After these costs, you will then receive anything leftover.[5]
Thanks!
## Warnings
• While you can make the calculations yourself, it always best to get the payoff number from the lender themselves. If your calculations are off even by a few cents, you may end up keeping the mortgage open, and you will continue to accrue interest.
Thanks!
## Video.By using this service, some information may be shared with YouTube.
Real Estate Broker
This article was co-authored by Carla Toebe. Carla Toebe is a Real Estate Broker in Washington. She has been an active real estate broker since 2005, and founded the real estate agency CT Realty LLC in 2013.
Co-authors: 8
Updated: December 31, 2019
Views: 80,335
Article SummaryX
Mortgage payoff is the remaining amount you need to pay on your mortgage, including interest. To calculate your mortgage payoff, you’ll need to know the total amount you borrowed, your annual interest rate, the total number of payments for the whole duration of the loan, and the total number of payments remaining. There are a number of online calculators that will work this out for you. All you need to do is enter your figures and take note of your mortgage payoff. For more tips from our Financial co-author, including the formula to manually calculate your mortgage payoff, read on!
Thanks to all authors for creating a page that has been read 80,335 times.
• MZ
Maricela Zambrano
May 16, 2017
"All the information was very helpful, very informative and in plain simple English, easy to understand. Thank you."
• SL
Suzanne Lesser
Mar 20, 2018
"Thanks! This explained what kind of charges I'd be looking at if I tried to pay off my mortgage! "
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# How come if $\ i\$ not of the following form, then $12i + 5$ must be prime?
I know if $\ i\$ of the following form $\ 3x^2 + (6y-3)x - y\$ or $\ 3x^2 + (6y-3)x + y - 1, \ \ x,y \in \mathbb{Z}^{+},i \in \mathbb Z_{\ge 0}$,
then $\ 12i + 5\$ must be composite number, because:
$12(3x^2 + (6y-3)x + y - 1) + 5 = 36x^2 + (72y - 36)x + (12y - 7) = (6x + 12y - 7)(6x + 1)$
How come if $\ i\$ not of the following form $\ 3x^2 + (6y-3)x - y\$ and $\ 3x^2 + (6y-3)x + y - 1\$,
then $12i + 5$ must be prime? $\ x,y \in \mathbb{Z}^{+},i \in \mathbb Z_{\ge 0}$.
For example: $\ 5 = \ 3*1^2 + (6*1-3)*1 - 1\$ ,when $\ x = y = 1\$, as proved before,$\ 5*12+5\$ must be composite number;
$\ 0, \ 1. \ 2.\ 3.\ 4\$ can't of the following form $\ 3x^2 + (6y-3)x - y\$ and $\ 3x^2 + (6y-3)x + y - 1\$,
$\ 12*0 + 5 = 5\$, $\ 5$ is prime,$\ 12*1 + 5 = 17\$, $\ 17$ is prime, and so on,
• What makes you think this is true for all such $i$? Apr 30, 2014 at 11:30
• This has been proved,but that prove unreadable,waiting a prove of coolness.
– mike
Apr 30, 2014 at 11:33
• Lemme reply Gerry Myerson here, that proof not in english and not in readable math language, that will mess you up.
– mike
Apr 30, 2014 at 11:39
Suppose that $12i+5$ is composite, say $12i+5 = ab$. Then looking modulo $6$, we get that one of the elements $a,b$ is $1 \pmod{6}$ and the other is $-1 \pmod{6}$; assume w.l.o.g. that $a$ is $1 \pmod{6}$, and write $a = 6r+1$ and $b = 6s-1$.
Observe that $r$ and $s$ must have different parity. Indeed, if they are both even or both odd, then computing modulo $12$, we would get that $ab \equiv -1 \pmod{12}$, which is false.
If $s>r$, then we set $x = r$, and $s = r + (2y-1)$ for some positive number $y$. Then compute $ab$, to get $i = 3x^2 + (6y-3)x + y - 1$.
If $s<r$, then we set $x = s$, and $r = s + (2y-1)$ for some positive number $y$. Then compute $ab$, to get $i = 3x^2 + (6y-3)x - y$.
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# Feet per second to microns per minute conversion
1 ft/s = 3048780.49 µ/min
## How to convert feet per second to microns per minute?
To convert feet per second to microns per minute, multiply the value in feet per second by 3048780.49.
You can use the formula :
microns per minute = feet per second × 3048780.49
1 foot per second is equal to 3048780.49 microns per minute.
## Feet per second to microns per minute conversion table
Feet per second Microns per minute
1 ft/s 3048780.49 µ/min
2 ft/s 6097560.98 µ/min
3 ft/s 9146341.47 µ/min
4 ft/s 12195121.96 µ/min
5 ft/s 15243902.45 µ/min
6 ft/s 18292682.94 µ/min
7 ft/s 21341463.43 µ/min
8 ft/s 24390243.92 µ/min
9 ft/s 27439024.41 µ/min
10 ft/s 30487804.9 µ/min
20 ft/s 60975609.8 µ/min
30 ft/s 91463414.7 µ/min
40 ft/s 121951219.6 µ/min
50 ft/s 152439024.5 µ/min
60 ft/s 182926829.4 µ/min
70 ft/s 213414634.3 µ/min
80 ft/s 243902439.2 µ/min
90 ft/s 274390244.1 µ/min
100 ft/s 304878049 µ/min
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## Credits
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## Diagonal of Decagon across four sides given height Solution
STEP 0: Pre-Calculation Summary
Formula Used
diagonal_across_4_sides = (Height/((sqrt(5+2*sqrt(5)))))*sqrt(5+2*sqrt(5))
d4 = (h/((sqrt(5+2*sqrt(5)))))*sqrt(5+2*sqrt(5))
This formula uses 1 Functions, 1 Variables
Functions Used
sqrt - Squre root function, sqrt(Number)
Variables Used
Height - Height is the distance between the lowest and highest points of a person standing upright. (Measured in Meter)
STEP 1: Convert Input(s) to Base Unit
Height: 12 Meter --> 12 Meter No Conversion Required
STEP 2: Evaluate Formula
Substituting Input Values in Formula
d4 = (h/((sqrt(5+2*sqrt(5)))))*sqrt(5+2*sqrt(5)) --> (12/((sqrt(5+2*sqrt(5)))))*sqrt(5+2*sqrt(5))
Evaluating ... ...
d4 = 12
STEP 3: Convert Result to Output's Unit
12 Meter --> No Conversion Required
12 Meter <-- Diagonal across four sides
(Calculation completed in 00.000 seconds)
## < 9 Diagonal of Decagon across four sides Calculators
Diagonal of Decagon across four sides given area
diagonal_across_4_sides = (sqrt((2*Area)/(5*sqrt(5+2*sqrt(5)))))*sqrt(5+2*sqrt(5)) Go
Diagonal of Decagon across four sides given diagonal across three sides
diagonal_across_4_sides = ((2*Diagonal across three sides)/(sqrt(14+6*sqrt(5))))*sqrt(5+2*sqrt(5)) Go
Diagonal of Decagon across four sides given diagonal across two sides
diagonal_across_4_sides = ((2*Diagonal across two sides)/(sqrt(10+2*sqrt(5))))*sqrt(5+2*sqrt(5)) Go
Diagonal of Decagon across four sides given inradius
Diagonal of Decagon across four sides given height
diagonal_across_4_sides = (Height/((sqrt(5+2*sqrt(5)))))*sqrt(5+2*sqrt(5)) Go
Diagonal of Decagon across four sides given diagonal across five sides
diagonal_across_4_sides = (Diagonal across five sides/((1+sqrt(5))))*sqrt(5+2*sqrt(5)) Go
Diagonal of Decagon across four sides given circumradius
Diagonal of Decagon across four sides given perimeter
diagonal_across_4_sides = (Perimeter/10)*sqrt(5+2*sqrt(5)) Go
Diagonal of Decagon across four sides
diagonal_across_4_sides = Side*sqrt(5+2*sqrt(5)) Go
### Diagonal of Decagon across four sides given height Formula
diagonal_across_4_sides = (Height/((sqrt(5+2*sqrt(5)))))*sqrt(5+2*sqrt(5))
d4 = (h/((sqrt(5+2*sqrt(5)))))*sqrt(5+2*sqrt(5))
## What is a decagon?
Decagon is a polygon with ten sides and ten vertices. A decagon, like any other polygon, can be either convex or concave, as illustrated in the next figure. A convex decagon has none of its interior angles greater than 180°. To the contrary, a concave decagon (or polygon) has one or more of its interior angles greater than 180°. A decagon is called regular when its sides are equal and also its interior angles are equal.
## How to Calculate Diagonal of Decagon across four sides given height?
Diagonal of Decagon across four sides given height calculator uses diagonal_across_4_sides = (Height/((sqrt(5+2*sqrt(5)))))*sqrt(5+2*sqrt(5)) to calculate the Diagonal across four sides, The Diagonal of decagon across four sides given height formula is defined as a straight line joining two opposite corners of the decagon , where diagonal = diagonal across 4 sides of decagon , height = height of decagon. Diagonal across four sides and is denoted by d4 symbol.
How to calculate Diagonal of Decagon across four sides given height using this online calculator? To use this online calculator for Diagonal of Decagon across four sides given height, enter Height (h) and hit the calculate button. Here is how the Diagonal of Decagon across four sides given height calculation can be explained with given input values -> 12 = (12/((sqrt(5+2*sqrt(5)))))*sqrt(5+2*sqrt(5)).
### FAQ
What is Diagonal of Decagon across four sides given height?
The Diagonal of decagon across four sides given height formula is defined as a straight line joining two opposite corners of the decagon , where diagonal = diagonal across 4 sides of decagon , height = height of decagon and is represented as d4 = (h/((sqrt(5+2*sqrt(5)))))*sqrt(5+2*sqrt(5)) or diagonal_across_4_sides = (Height/((sqrt(5+2*sqrt(5)))))*sqrt(5+2*sqrt(5)). Height is the distance between the lowest and highest points of a person standing upright.
How to calculate Diagonal of Decagon across four sides given height?
The Diagonal of decagon across four sides given height formula is defined as a straight line joining two opposite corners of the decagon , where diagonal = diagonal across 4 sides of decagon , height = height of decagon is calculated using diagonal_across_4_sides = (Height/((sqrt(5+2*sqrt(5)))))*sqrt(5+2*sqrt(5)). To calculate Diagonal of Decagon across four sides given height, you need Height (h). With our tool, you need to enter the respective value for Height and hit the calculate button. You can also select the units (if any) for Input(s) and the Output as well.
How many ways are there to calculate Diagonal across four sides?
In this formula, Diagonal across four sides uses Height. We can use 9 other way(s) to calculate the same, which is/are as follows -
• diagonal_across_4_sides = Side*sqrt(5+2*sqrt(5))
• diagonal_across_4_sides = (Diagonal across five sides/((1+sqrt(5))))*sqrt(5+2*sqrt(5))
• diagonal_across_4_sides = ((2*Diagonal across three sides)/(sqrt(14+6*sqrt(5))))*sqrt(5+2*sqrt(5))
• diagonal_across_4_sides = ((2*Diagonal across two sides)/(sqrt(10+2*sqrt(5))))*sqrt(5+2*sqrt(5))
• diagonal_across_4_sides = (Perimeter/10)*sqrt(5+2*sqrt(5))
• diagonal_across_4_sides = (sqrt((2*Area)/(5*sqrt(5+2*sqrt(5)))))*sqrt(5+2*sqrt(5))
• diagonal_across_4_sides = (Height/((sqrt(5+2*sqrt(5)))))*sqrt(5+2*sqrt(5))
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# Biostatistics
The science that applies statistical theory and mathematical principals to research in medicine, biology, environmental science, health and related fields.
## epidemiology
A population has a mean of 60 and a standard deviation of 5. A random sample of 16 measurements is drawn from this population. Describe the sampling distribution of the sample means by computing its mean. Assume that the population is infinite.
• 60
• 75
• 70
• 65
The area from the ROC curve to the base: used as an indicator of the efficacy of a test in terms of sensitivity and specificity – can be used to compare performance of various tests.
• Area under the curve
If you are going to propose a hypothesis, it’s customary to write a ______________.
• statement
It is possible to fit a polynomial of any number of terms to a set of data.
• polynomial fitting
The normality assumption is at the core of a majority of standard statistical procedures, and it is important to be able to test this assumption.
• Lilliefors test
The heights female college students are normally distributed with mean of 68 inches and standard deviation of 3 inches. If 25 students are randomly drawn from the population, what would be the standard error of the resulting sampling distribution of the means?
• 0.5
• 0.6
• 0.1
• 0.3
If I…(do this to an independent variable)….then (this will happen to the _________________).
• dependent variable
As the sample size increases, the mean of the _____ of the mean will approach the population mean of μ.
• sampling distribution
A statistic is unbiased if... ... the mean of the sampling distribution IS _____________ to the value of the population mean.
• greater
• lesser
• equal
A population has a mean of 60 and a standard deviation of 5. A random sample of 16 measurements is drawn from this population. Describe the sampling distribution of the sample means by computing its standard deviation. Assume that the population is infinite.
• 1.96
• 1.55
• 1.35
• 1.25
The mean of sampling distribution of the proportion, P, is a special case of the _____.
• sampling distribution of the mean
In hypothesis testing, the critical value is the threshold for significance.
• critical value
Inverse of the standard error of the estimate or a derivative of this inverse.
• Efficiency of an estimate
The mean of the sampling distribution of the mean formula is is μM = μ, where μM is the _____ of the mean.
• mean of the sampling distribution
The heights female college students are normally distributed with mean of 68 inches and standard deviation of 3 inches. If 25 students are randomly drawn from the population, what would be the expected mean of the resulting sampling distribution of the means?
• 68
• 90
• 45
• 88
One of the most important measures of dispersion, the _____________ is the difference between the maximum and minimum values of a distribution.
• range
• variance
• IQR
• standard deviation
The __________ statistic, named after its creators, tests the hypothesis that there is no autocorrelation of one time lag present in the errors obtained from forecasting
• Durbin-Watson (DW)
If you had to choose a "best statistic" to describe a population, which of the following would be best.
• low bias, low variability
• low bias, high variability
• high bias, high variability
• high bias, low variability
____________ are used to draw inferences about a population from a sample.
• Inferential statistics
What does this symbol represent?
• sample
• variable
• population proportion
• ratio
This is a statement of what a statistical hypothesis test is set up to establish.
• alternative hypothesis
The process of reaching to a decision after considering probabilities of various outcomes and value judgments regarding the utility of those outcomes.
• Decision analysis
For data sets having a normal, bell-shaped distribution, approximately 68 percent of the data values are within 1 standard deviation of the mean; approximately 95 percent are within 2 standard deviations of the mean; and approximately 99.7 percent (nearly all) are within 3 standard deviations of the mean.
• empirical rule
The mean of the sampling distribution of the mean is the same as the _____.
• population mean
In a population of five university students with GPAs of 2.5, 2.3, 1.7, 1.4, and 1.1, a sample of three students are considered. What would be the standard deviation of the resulting sampling distribution?
• 0.88
• 0.56
• 0.22
• 0.32
The process of going into the deep of a phenomenon, data-set, thought, etc., and looking at its various components.
• Analysis
When the sample is small, the sampling distribution of a proportion will have an approximate normal distribution.
• True
• False
When two or more comparisons or other statistical tests of hypothesis are done on the same set of data, the total probability of alpha error can increase much beyond the prefixed level such as 5%. This is known as "blinding".
• True
• False
The average speed of 1500 vehicles traveled on a stretch of highway that day is 67 miles per hour with a standard deviation of 3.5 miles per hour. If 100 vehicles are randomly selected as samples, what would be the standard error of the resulting sampling distribution of sample means?
• 0.39
• 0.34
• 0.56
• 0.78
A standardized measure of the association or mutual dependence between two variables, say, X and Y.
• correlation coefficient
A hypothesis is an ____________ about something in the world around you.
• educated guess
Cause-specific rate is the rate obtained when numerator is restricted to a particular cause (e.g., of morbidity or of mortality).
• True
• False
A person with disease classified as without disease. In place of disease, false negativity can be for any other attribute.
• False negative
A good hypothesis statement should: an “if” and “________________” statement.
• then
The Gallup Poll asked a random sample of 1785 adults whether they attended church during the past week. Let p-hat be the proportion of people in the sample who attended church. A newspaper report claims that 40% of all U.S. adults went to church last week. Suppose this claim is true. Calculate the standard deviation of the sampling distribution.
• 0.2514
• 0.2585
• 0.0116
• 0.5264
A trial with the objective to examine if a new regimen is different from another regimen by more than a prespecified medically unimportant margin.
• Equivalence trial
The 300 customers who called the call center spend an average of 45 minutes on hold, with a standard deviation of 12 minutes. What is the standard error of the sampling distribution for a sample of 150 randomly selected customers
• 0.50
• 0.58
• 0.79
• 0.69
A _____ is a value which is generated from a population.
• parameter
The simple arithmetic average of a distribution of variable values (or scores), the _____________ provides a single, concise numerical summary of a distribution.
• mean
• median
• mode
The heights of children of exceptionally tall (or short) parents “regress” to the mean of the population
• regression
Variety of causes of death: some people meet death slowly such as by cancer, and some sudden such as by myocardial infarction.
• Death spectrum
28% of all Woodrow students believe Monday will be snow day. You take a sample of 50 students and find that 15 of them believe Monday will be a snow day. What does 28 represent?
• E
• N
• S
• P
If a forecast variable Y is regressed against several explanatory variables X1, X2, . . , Xk, then the estimated Y value is designated Y.
• multiple correlation coefficient
Using computer to solve problems without understanding the implications of the underlying procedure. This is known as "black box approach".
• True
• False
This is simply the SEE divided by the average of the dependent variable.
• coefficient of variation
A procedure of combining evidence in different reports on the same aspect. If different trials on the same regimen report varying efficacy, they can be combined to come to a unified conclusion, which may command substantially more confidence than result of any one of the individual trials.
• Meta-analysis
This consists of calculations that provide information about levels of variability within a regression model and form a basis for tests of significance.
• ANOVA
It’s good science to let people know if your study results are solid, or if they could have happened by chance. The usual way of doing this is to test your results with a _____________.
• p-value
Identification data of a document containing the authors‘ name, title, publication name, volume, publication date, page numbers, etc.
• Citation
When there are more scores toward one end of the distribution than the other, this results in _____________.
• positive
• above average
• negative
• skew
A difference in time between an observation and a previous observation.
• lag
The technique of estimating a smooth trend, usually by taking weighted averages of observations.
• smoothing
A _____________ is the probability that the observed result, or a result more extreme, could be obtained if the null hypothesis is true.
• p-value
A summary of the death and survival pattern of a group of people—generally for the entire population of an area, but can be used for patients of a particular disease also.
• Life table
A sample proportion is where a random sample of objects n is taken from a population P.
• True
• False
The ______________, also called root mean squared error (RMSE), is the square root of the mean squared residual term from the ANOVA table of the summary output.
• standard error of the estimate (SEE)
A sampling distribution is a graph of a statistics for your sample data.
• True
• False
A characteristics that is assessed only in two categories such as ascites present or absent (or yes/no), or gender as male or female.
• Binary variable
The technique of multiple regression is an extension of simple regression.
• multiple regression
The average speed of 1500 vehicles traveled on a stretch of highway that day is 67 miles per hour with a standard deviation of 3.5 miles per hour. If 100 vehicles are randomly selected as samples, what would be the mean of the resulting sampling distribution of sample means?
• 67
• 68
• 69
• 70
_____ refers tot he ability to draw conclusions about the characteristics of the population as a whole based on the results of data collected from a sample.
• Generalizability
The set of characteristics such as age, disease and severity, which are necessary in a subject to be considered eligible for inclusion in the study.
• Inclusion criteria
The ______________ is the score of a distribution residing at the 50th percentile, separating the top and bottom 50 percent of scores.
• mode
• median
• mean
A non parametric test for comparing central tendency in three or more groups.
• Kuskal-Wallis test
The first could be called missed diagnosis and the second as misdiagnosis. In place of healthy/diseased this could be any other categorization.
• Misclassification
A form of regression analysis where the observations are measured at the same point in time or over the same time period but differ along another dimension.
• cross-sectional model
The statistical procedure to discover a construct out of data that can possibly explain the variation and relationship among different variables.
• Factor analysis
A _____ is a selected individual or group representing the full set of members of a certain group of interest.
• population
The ___________ is calculated by dividing MSR (mean squared regression) by MSE (mean squared error), or explained variance by unexplained variance.
• F ratio or F statistic
The set of words that describes the essential features of a study.
• Keywords
Bibliography is a list of citations of the related literature.
• True
• False
A person without disease classified as with disease. In place of disease, false positivity can be for any other attribute
• False positive
The _____________ hypothesis is always the accepted fact.
• null
The sampling distribution of a proportion is when you repeat your survey for all possible samples of the population.
• True
• False
The _____ theorem tells us that if we have a large number of independent, identically distributed variables, the distribution will approximately follow a normal distribution.
• central limit
The probability of occurrence of one of two or more mutually exclusive events is the sum of the probabilities of their individual occurrence
The Gallup Poll asked a random sample of 1785 adults whether they attended church during the past week. Let p-hat be the proportion of people in the sample who attended church. A newspaper report claims that 40% of all U.S. adults went to church last week. Suppose this claim is true. Calculate the standard deviation of the sampling distribution of p-hat.
• 0.25652
• 0.0045
• 0.8987
• 0.0116
This condition exists when the errors do not have a constant variance across an entire range of values.
• heteroscedasticity
It equals the change in Y for each unit change in X.
• slope
A good hypothesis statement should: Have design _______________.
• criteria
____________ is the Excel function that calculates the kurtosis of a data set’s distribution.
• KURT
_____ are a collection of statistical tools which are used to quantitatively describe or summarize a collection of data.
• Descriptive statistics
How well the actual observations fit into a specified pattern.
• Goodness of fit
A ______________ is a set of ordered observations of a phenomenon at equally spaced time points.
• time series
The remaining portion of life at any age that would be spent without any morbidity
• Healthy life expectancy
The Gallup Poll asked a random sample of 1785 adults whether they attended church during the past week. Let p-hat be the proportion of people in the sample who attended church. A newspaper report claims that 40% of all U.S. adults went to church last week. Suppose this claim is true. What is the mean of the sampling distribution of p-hat?
• 0.55
• 0.66
• 0.5
• 0.4
An international organisation of producers and consumers of medical research that helps to clarify the research achievements, particularly health care interventions such as drugs, diet alteration and behavior change
• Cochrane Collaboration
The tendency of getting poor output or poor outcome when the inputs or efforts are poor.
• Garbage-in, garbage-out syndrome
In a population of five university students with GPAs of 2.5, 2.3, 1.7, 1.4, and 1.1, a sample of three students are considered. What would be the mean of the resulting sampling distribution?
• 1.8
• 3.6
• 2.5
• 4.9
The statistical procedure to classify units or individuals into groups such that the units are similar within each group but dissimilar across groups: generally used when the number and nature of the groups are not known.
• Cluster analysis
linear combination of individual forecasts to assist in obtaining a more accurate forecast.
• composite regression model
Increasing the sample size of an opinion poll will ___________ the variability of the estimates made from the data collected in the poll.
• decrease
• reduce
• stabilize
When the error terms remaining after application of a forecasting method show autocorrelation, it indicates that the forecasting method has not removed all of the pattern from the data.
• autocorrelated errors
A plot of the residuals versus a z value (or cumulative normal percentile) derived from the normal probability distribution for the ranking location of the residual.
• normal probability plot
A distribution is the arrangement of data by the values of one variable in order, from ______________.
• left to right
• low to high
• right to left
• high to low
An extraneous factor that could be an explanation of the outcome of interest in addition to the factor under study so that its effect can not be differentiated from the other: such as dietary factors when examining relationship between smoking and cervical cancer. Presence of unaccounted confounders decreases the validity of a study.
• Confounder
28% of all Woodrow students believe Monday will be snow day. You take a sample of 50 students and find that 15 of them believe Monday will be a snow day. What does 50 represent?
• I
• P
• N (capital)
• n
If x objects have a certain characteristic then the sample proportion “p” is: p = x*n.
• True
• False
The result stating that the chance of a summative measure such as sample mean following a Gaussian distribution rapidly increases in almost all practical situations as the number of individuals in a sample increases (i.e., sample size becomes large).
• Central Limit Theorem
The probability of occurrence of an event such as disease when some a-priori information such as sign-symptoms are known: denoted by P(A/B) where after slash (/) sign is what is known a-priori.
• Conditional probability
______________ are so called because the assumptions underlying their use are fewer and weaker than those associated with parametric tests
• Nonparametric or distribution-free tests
The frequency of desired outcome per unit of resource inputs such as time, money and manpower.
• Efficiency
Similar course of the disease process in the two regimens under comparison: also evaluated in terms of comparable bioavailability of drug products, say, within 80% to 125% with respect to area under the concentration curve and Cmax.
• Bioequivalence
In the game of Scrabble, each player begins by drawing 7 tiles from a bag containing 100 tiles. There are 42 vowel, 56 consonants, and 2 blank tiles in the bag. Cait chooses an SRS of 7 tiles. Let p-hat be the proportion of vowels in her sample.
• Yes
• No
Census is the survey of the entire population.
• True
• False
In regression analysis, a ______________ is one that takes the values 0 or 1 to indicate the absence or presence of some categorical effect (month, quarter, strike, fire loss) that may be expected to shift the outcome.
• dummy variable
If you are using a significance level of .05, a ______________ allots all 5 percent to testing the statistical significance in the one direction of interest.
• one-tailed test
A ____________ shows the number of observations falling into each of several ranges of values, which are typically portrayed as histograms.
• distribution
Regardless of the shape of the population distribution, this theorem states that the sampling distribution of the mean of n independent sample values will approach the normal distribution as the sample size increases.
• central limit theorem
You can have the Y-axis on a logarithmic scale instead of a linear one.
• log scale
A _____ is a subset drawn from a larger population.
• sample
A _____ quantity is a quantity without a physical unit and is thus a pure number.
• dimensionless
A ______________ is calculated by subtracting the forecast value from the actual value to give an error value for each forecast period. In forecasting, this term is commonly used as a synonym for residual.
• forecast error
Hypothesis should be testable, either by experiment or ________________.
• observation
The 300 customers who called the call center spend an average of 45 minutes on hold, with a standard deviation of 12 minutes. What is the expected average of the sampling distribution for a sample of 150 randomly selected customers?
• 55
• 45
• 50
• 60
Hypothesis testing in statistics is a way for you to _______________ of a survey or experiment to see if you have meaningful results.
• test the results
A good hypothesis statement should: be based on information in ______________ research.
• prior
If we could take many such samples, the collection of possible values of the statistic would follow its _______________.
• sampling distribution
A less scientific but a quick method to arrive at a consensus among experts.
• Delphi method
A prospective study of a cohort for a specified period, generally to observe the occurrence of an outcome of interest, and thereby determine the incidence.
• Cohort study
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CC-MAIN-2024-30
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https://maptools.com/videos/Compass/NorthReferenceExercises
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## Using your Compass with a Map - North Reference Exercises A few north reference conversions for you to try.
This video presents 5 north reference conversion problems for you to solve. Each problem will state the conversion to be preformed and provide a declination digram to use. If you want to solve the problem on your own, pause the video, do the conversion, and then continue playing the video to see the solution. The problems will explore different combinations of Grid and Magnetic North on either side of True North. They will also get a bit trickier as they progress. Once you feel that you have the hang of this, I’d suggest you dig out a map of your local area and try a few north reference conversions using your declination values.
Problem 1
Convert a bearing of 285° Grid to a Magnetic bearing.
Remember, when you are working with the declination diagram, bearings increase in a clockwise direction.
Magnetic North is 16 degrees East of True North. Grid North is 1.5 degrees East of True North.
The angle between Grid North and Magnetic North is the 16 degrees from Magnetic to True, less the 1.5 degrees from True to Grid, or 14 and a half degrees. Let’s round that up to 15 degrees.
Now draw the 285° Grid bearing onto the diagram. Start at Grid North and go in a clockwise direction.
The answer we are seeking is the angle from Magnetic North to the target.
In this case the Magnetic bearing we are seeking is smaller than the Grid bearing we were given.
So we subtract 15° from the 285° Grid value to get the answer of 270° Magnetic.
Now lets make our short cut rules for this situation.
Problem 2
Convert a bearing of 67° Magnetic to a True bearing.
Magnetic North is 12 degrees West of True North. Grid North is 1 degree 42 minutes West of True North.
The angle between True North and Magnetic North is 12 degrees.
Draw the 67° Magnetic bearing onto the diagram.
The answer we are seeking is the angle from True North to the target.
In this case the True bearing we are seeking is smaller than the Magnetic bearing we were given.
So we subtract the 12° difference from the 67° Magnetic value to get the answer of 55° True.
Problem 3
Convert a bearing of 108° Grid to a Magnetic bearing.
Magnetic North is 9 degrees East of True North. Grid North is 25 minutes East of True North.
The angle between Grid North and Magnetic North is the 9 degrees from Magnetic to True, less the 25’ minutes from True to Grid. That would be 8 degrees 35 minutes, which we’ll round up to 9°.
Draw the 108° Grid bearing onto the diagram.
The answer we are seeking is the angle from Magnetic North to the target.
In this case the Magnetic bearing we are seeking is smaller than the Grid bearing we were given.
So we subtract the 9° from the 108° Grid value to get the answer of 99° Magnetic.
Problem 4
Convert a bearing of 11° Grid to a Magnetic bearing.
Magnetic North is 18 degrees East of True North. Grid North is 1 degree 15 minutes East of True North.
The angle between Grid North and Magnetic North is the 18 degrees from Magnetic to True, less the 1° 15’ from True to Grid. That would be 16° 45’, which we’ll round up to 17°.
Draw the 11° Grid bearing onto the diagram.
The answer we are seeking is the angle from Magnetic North to the target. Remember the angle need to go clockwise from Magnetic North to the target bering.
In this case the Magnetic bearing we are seeking is smaller than the Grid bearing we were given.
So we subtract the 17° from the 11° Grid value. This gives us a result of minus 6 degrees. Bearings need to be given in terms of a 0 to 360 degree circle. So add 360° to the -6° to get the answer of 354° Magnetic.
Problem 5
Convert a bearing of 350° Magnetic to a Grid bearing.
Magnetic North is 14 degrees East of True North. Grid North is 1 degree 45 minutes West of True North.
The angle between Grid North and Magnetic North is the 14 degrees from Magnetic to True, plus the 1° 45’ from True to Grid. That would be 15° 45’, which we’ll round up to 16°.
Draw the 350° Magnetic bearing onto the diagram.
The answer we are seeking is the angle from Grid North to the target.
In this case the Grid bearing we are seeking is larger than the Magnetic bearing we were given.
So we add the 16° to the 350° Grid value. This gives us a result of 366 degrees. Bearings need to be given in terms of a 0 to 360 degree circle. So subtract 360° from the 366° to get the answer of 6° Magnetic.
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CC-MAIN-2022-27
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https://minuteshours.com/43-37-hours-in-hours-and-minutes
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# 43.37 hours in hours and minutes
## Result
43.37 hours equals 43 hours and 22.2 minutes
You can also convert 43.37 hours to minutes.
## Converter
Forty-three point three seven hours is equal to forty-three hours and twenty-two point two minutes.
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https://api.qgis.org/api/2.12/Node_8h_source.html
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QGIS API Documentation 2.12.0-Lyon
Node.h
Go to the documentation of this file.
1 /***************************************************************************
2 Node.h - description
3 -------------------
4 copyright : (C) 2004 by Marco Hugentobler
5 email : [email protected]
6 ***************************************************************************/
7
8 /***************************************************************************
9 * *
10 * This program is free software; you can redistribute it and/or modify *
12 * the Free Software Foundation; either version 2 of the License, or *
13 * (at your option) any later version. *
14 * *
15 ***************************************************************************/
16
17 #ifndef NODE_H
18 #define NODE_H
19
20 #include "Point3D.h"
21
23 class ANALYSIS_EXPORT Node
24 {
25 protected:
30 public:
31 Node();
32 Node( const Node& n );
33 ~Node();
34 Node& operator=( const Node& n );
36 Node* getNext() const;
38 Point3D* getPoint() const;
40 void setNext( Node* n );
42 void setPoint( Point3D* p );
43 };
44
45 inline Node::Node() : mPoint( 0 ), mNext( 0 )
46 {
47
48 }
49
50 inline Node::~Node()
51 {
52 delete mPoint;
53 }
54
55 inline Node* Node::getNext() const
56 {
57 return mNext;
58 }
59
60 inline Point3D* Node::getPoint() const
61 {
62 return mPoint;
63 }
64
65 inline void Node::setNext( Node* n )
66 {
67 mNext = n;
68 }
69
70 inline void Node::setPoint( Point3D* p )
71 {
72 mPoint = p;
73 }
74
75 #endif
Node is a class used by Line3D.
Definition: Node.h:23
Node()
Definition: Node.h:45
Point3D is a class to represent a three dimensional point.
Definition: Point3D.h:23
Node * getNext() const
Returns a pointer to the next element in the linked list.
Definition: Node.h:55
~Node()
Definition: Node.h:50
void setPoint(Point3D *p)
Sets a new pointer to an associated Point3D object.
Definition: Node.h:70
Point3D * getPoint() const
Returns a pointer to the Point3D object associated with the node.
Definition: Node.h:60
void setNext(Node *n)
Sets the pointer to the next node.
Definition: Node.h:65
Node * mNext
Pointer to the next Node in the linked list.
Definition: Node.h:29
Point3D * mPoint
Pointer to the Point3D object associated with the node.
Definition: Node.h:27
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https://casestudy.sbs/case-study/addition-problem-solving-worksheets-for-grade-4
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## 4th Grade Math Word Problems Worksheets
Math word problem worksheets for grade 4.
These word problem worksheets place 4th grade math concepts into real world problems that students can relate to.
We encourage students to read and think about the problems carefully, by:
• providing mixed word problem worksheets
• including irrelevant data within word problems so students must understand the context before applying a solution
## The four operations
Mixed addition and subtraction word problems
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Mixed 4 operations
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## Time & money word problems for 4th grade
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## Mixed word problems
Mixed word problems for grade 4
Sample Grade 4 Word Problem Worksheet
## More word problem worksheets
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## Grade 4 Word Problem Worksheets
Free printable fourth grade math worksheets to help your students improve their ability to solve word problems! Applying mathematical concepts to solve word problems can be challenging. These free worksheets are great repetition for your students! Click on the images below to download the word problem worksheets. They’re perfect for daily math warm ups! Don’t forget to check out the word problem task cards. Print them off, laminate them and use them for early finishers during maths!
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## Daily Word Problems - Level D (4th Grade)
This page contains a large collection of daily math word problems at a 4th grade level (Level D). Many problems are multi-step, and cover a wide-range of skills, including: multi-digit multiplication, division, operations with fractions and decimals, place value, reading graphs, and measurement.
Logged in members can use the Super Teacher Worksheets filing cabinet to save their favorite worksheets.
Download and print daily math review problems for 4th grade. The Math Buzz set includes all types of problems, including measurement, charts, rounding, place value, operations with whole numbers, decimals, and operations with fractions.
STW has more Daily Word Problem sets for other grade levels. Have a look!
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The addition word problem worksheets presented here involve performing addition operations with regrouping and without regrouping. Our extensive and well-researched word problem worksheets feature real-life scenarios that involve single-digit addition, two-digit addition, three-digit addition, and addition of large numbers. These pdf handouts are designed to provide ample practice for elementary school children. Free worksheets are included.
These printable practice worksheets involve simple addition of single-digit numbers. Read the word problems and perform addition operations to arrive at the answers.
Addition Word Problems: Sum up to 20
Featured in these worksheets are engaging word problems whose sums add up to 20. Addends may have a combination of single-digit and two-digit numbers.
A number of real-life scenarios in the form of word problems featured in the addition worksheets here involve single digit and two-digit addends.
Two-digit Addition Problems - No Regrouping
The word problems in this section do not require regrouping or carrying. Find the answers to the word problems that feature two-digit addends.
Two-digit Addition Problems - With Regrouping
All two-digit addition word problems presented in this set of worksheets here require regrouping (carry over). Follow the place value columns to sum up the two-digit addends.
Theme based Word Problems
Presented here are worksheets with three colorful themes - Fall Season, Aquarium and Theme Park. Read the questions and solve the word problems. Answer keys are included.
A total of 15 addition word problems spread over three PDF worksheets presented here require you to sum up three-digit addends with the two-digit addends.
Enhance your arithmetic skills. Read the word problems and sum up three-digit addends in these printable worksheets. Some problems may require regrouping. Answer key included in each worksheet.
The word problems presented in the worksheets here feature large numbers with addends up to eight digits.
Related Worksheets
» Subtraction Word Problems
» Multiplication Word Problems
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## Worksheet on Word Problems on Addition
In 4th grade worksheet on word problems on addition, all grade students can practice the questions on word problems based on addition. This exercise sheet on addition can be practiced by the students to get more ideas to solve the worksheet on word problems on addition.
1. In an examination, 75,236 students passed and 14,892 students failed. Find how many students appeared for the examination.
2. There are 3,786 men, 3,672 women and 1,508 children in a village. Find the total population of the village.
3. There are 5,873 male and 6,389 female primary teachers in a city. Find the total number of teachers in the city.
4. In a school there are the following numbers of students: 127 in grade one, 120 in grade two, 110 in grade three, 100 in grade four and 93 in grade five. Find the total number of students in the school.
5. A man plucked 435 mangoes from one tree, 450 mangoes from the second tree and 295 mangoes from the third tree. Find the total number of mangoes plucked from the trees.
6. A school library has 3,730 books in French, 2,531 books in English, and 5,368 books in other languages. How many books are there in library?
7. There are 37,536 bags of wheat, 35,380 bags of rice and 25,240 bags of gram in a store. Find the total number of bags in the store.
8. A school management spent \$ 26,756 on teachers pay, \$ 325,378 on building construction, \$ 5,780 on games and \$ 8,235 on other work. Find the total amount spent on different items by the school management.
9. A man plucked 375 apples from one tree, 504 apples from the second tree and 229 apples from the third tree. Find the total number of apples plucked from the trees.
10. A toy factory manufactured 52,253 toys in January, 50,375 toys in February and 608,368 toys in March. How many toys were manufactured in the above mentioned months in total?
11. Mr. Jones goes to the shop to buy furniture for his house. He buys a double bed for \$57930 and a sofa for \$39860. What is the total amount spent by Mr. Jones at the shop.
12. According to a census, there were 45779 females in a town. The number of males is 2336 more. What is the total population of the town?
13. 7883 people visited the zoo on Friday, 6596 on Saturday and 8224 on Sunday. How many people visited the zoo on these 3 days?
14. Ron is studying aboard. During his holidays, he travels to his home town. To reach his home town he covered 4562 km by air and 1057 km by bus. What is the total distance travelled by him to reach the home town?
15. Ron deposited \$13645 in bank on Tuesday and \$6880 on Wednesday. What is the amount of money deposited by him in bank in these 2 days?
16. Shelly went to buy furniture for her room and had \$15000. She selects two different tables at the shop –
(i) a square table costing \$9640
(ii) a round table costing \$7890.
She has already bought a chair for \$5999. Which table would she be able to buy?
17. An overhead tank has 4670 litres of water. If 2770 litres of water is added more to the tank, what is the total amount of water in the tank? Give any two ways in which we can conserve water.
18. During a cricket match, 3670 people watched the match from South End, 2270 from North End and 2850 from East End. How many people watched the match?
1. 90128
2. 8966
3. 12262
4. 550
5. 1180
6. 11629
7. 98156
8. \$ 366149
9. 1108
10. 710996
14. 5619 km
16. Round table
17. 7440 litres
● Four Fundamental Operations - worksheets
Worksheet on Word Problems on Addition.
Worksheet on Subtraction.
Worksheet on Mixed Addition and Subtraction.
Worksheet on Word Problems on Addition an d S ubtraction.
Worksheet on Estimating Sums and Differences.
Worksheet on Multiplication.
Worksheet on Multiplication of a Number by a 2-Digit Number.
Worksheet on Multiplication of a Number by a 3-Digit Number.
Worksheet on Estimating Products.
Worksheet on Word Problems on Multiplication.
Worksheet on Division.
Worksheet on Division Facts.
Worksheet on Estimating the Quotient.
Worksheet on Dividing Numbers.
Worksheet on Division by Two-Digit Numbers.
Worksheet on Word Problems on Division.
Worksheet on Four Fundamental Operations.
Worksheet on Systems of Numeration.
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Give a new lease of life to your math practice with our free pdf addition word problems worksheets. Let children solve these pdfs by addition of whole numbers, both with regrouping (carrying) and without regrouping. These word problems on addition provide abundant practice on addition within 10; sums up to 20; and 2-digit, 3-digit, 4-digit, and large (multi-digit) numbers.
These printable worksheets on addition word problems are suitable for kindergarten through grade 5 children.
## List of Addition Word Problems Worksheets
• Addition within 10 Word Problems
• Addition within 20 Word Problems
Addition within 10 Word Problems Worksheets
Perfect for visual learners, this compilation requires kindergarten and 1st grade kids to read the word problems and count the number of objects to find the sum.
Addition within 20 Word Problems Worksheets
Introduce these printable sums up to 20 word problem pdfs to give kids the pick-me-up they need while doing addition within 20. These beginner-level resources are great for kindergarten, grade 1, and grade 2 kids.
Analytically read real-world scenarios and find the sum of two 2-digit numbers. Our addition word problem worksheets on adding numbers within 100 are a delicious addition treat for 2nd grade children!
Instruct children in 3rd grade and 4th grade to solve these 3-digit by 3-digit addition word problems worksheets and watch them ace addition up to 1000 by regrouping wherever necessary.
Addition is math at its most playful, and with our printable word problems for addition worksheets involving 4-digit numbers around, grade 3 and grade 4 learners won't need to look any further.
Related Printable Worksheets
▶ Subtraction Word Problems
▶ Multiplication Word Problems
▶ Division Word Problems
▶ Word Problems
#### IMAGES
4. Adding large numbers (problem solving)
6. 30 Addition Word Problems Year 4 ~ ESL Worksheets Kids
#### VIDEO
2. Solving an Equation by the Distributive Property
6. addition word problems for class 1| addition problems grade 1| story sums for class 1| story sums
1. How Free Grade School Worksheets Enhance Learning at Home
In today’s digital age, free grade school worksheets have become an invaluable resource for parents and educators alike. These worksheets provide a wealth of educational benefits that enhance learning at home.
2. Free Grade School Worksheets: A Valuable Resource for Parents and Teachers
In today’s digital age, finding valuable resources to aid in a child’s education has become increasingly important. Free grade school worksheets have emerged as a popular tool for parents and teachers alike.
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When it comes to helping your child excel in math, providing them with engaging and interactive learning tools is crucial. Free printable 5th grade math worksheets are an excellent resource that can make learning enjoyable while reinforcing...
5. 4th Grade Math Word Problems Worksheets
Math word problem worksheets for grade 4. These word problem worksheets place 4th grade math concepts into real world problems that students can relate to.
The following collection of free 4th grade maths word problems worksheets cover topics including addition ... Solve Inequalities—Step-by-Step
Don't forget to check out the word problem task cards. Print them off, laminate them and use them for early finishers during maths!
Addition problem solving worksheets · Addition Subtraction Solve Word Problems Math Worksheets Bar Model/Tape Diagram · 3rd Grade Math
9. Daily Word Problems
Multiply 2-digit by 1-digit numbers together. Use addition, subtraction, and multiplication to calculate costs. Solve a word problem involving fractions. 4th
10. Addition and Subtraction Word Problems
What will be the balance in his account once the check is released? Answer = ______. Free Math Worksheets @ http
Addition word problem worksheets contain abundant practice sheets with regrouping and no regrouping addends. Multi-digit word problems are included.
12. Worksheet on Word Problems on Addition
Find the total number of teachers in the city. 4. In a school there are the following numbers of students: 127 in grade one, 120 in grade two, 110
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# Weight of Zinc, solid
## zinc, solid: convert volume to weight
### Weight of 1 cubic centimeter of Zinc, solid
carat 34.15 ounce 0.24 gram 6.83 pound 0.02 kilogram 0.01 tonne 6.83 × 10-6 milligram 6 830
#### How many moles in 1 cubic centimeter of Zinc, solid?
There are 104.47 millimoles in 1 cubic centimeter of Zinc, solid
### The entered volume of Zinc, solid in various units of volume
centimeter³ 1 milliliter 1 foot³ 3.53 × 10-5 oil barrel 6.29 × 10-6 Imperial gallon 0 US cup 0 inch³ 0.06 US fluid ounce 0.03 liter 0 US gallon 0 meter³ 1 × 10-6 US pint 0 metric cup 0 US quart 0 metric tablespoon 0.07 US tablespoon 0.07 metric teaspoon 0.2 US teaspoon 0.2
• For instance, calculate how many ounces, pounds, milligrams, grams, kilograms or tonnes of a selected substance in a liter, gallon, fluid ounce, cubic centimeter or in a cubic inch. This page computes weight of the substance per given volume, and answers the question: How much the substance weighs per volume.
#### Foods, Nutrients and Calories
MINI CINNAMON DANISH, UPC: 701730050159 contain(s) 382 calories per 100 grams (≈3.53 ounces) [ price ]
14 foods that contain Choline, from glycerophosphocholine. List of these foods starting with the highest contents of Choline, from glycerophosphocholine and the lowest contents of Choline, from glycerophosphocholine
#### Gravels, Substances and Oils
CaribSea, Marine, Aragonite, Florida Crushed Coral weighs 1 153.3 kg/m³ (71.99817 lb/ft³) with specific gravity of 1.1533 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Peat Hypnum moss, packaged weighs 593 kg/m³ (37.01978 lb/ft³) [ weight to volume | volume to weight | price | density ]
Volume to weightweight to volume and cost conversions for Refrigerant R-408A, liquid (R408A) with temperature in the range of -51.12°C (-60.016°F) to 60°C (140°F)
#### Weights and Measurements
One millisecond is defined as a period of time equal to one thousandth (1/1000) of a second.
Dynamic viscosity is a measure of how resistive a fluid is to flow. Viscosity describes how thick a fluid is
st/cm³ to gr/metric tbsp conversion table, st/cm³ to gr/metric tbsp unit converter or convert between all units of density measurement.
#### Calculators
Volume of a sphere calculator with surface area to volume ratio
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https://uk.mathworks.com/matlabcentral/cody/problems/43069-find-the-volume-of-cone/solutions/2058638
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Cody
# Problem 43069. Find the volume of cone
Solution 2058638
Submitted on 16 Dec 2019 by Asif Newaz
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
r = 6/sqrt(pi); h = 1; y = coneVolume(r,h) y_correct = 12; assert(abs(y-y_correct)<1e-3)
y = 12 y = 12
2 Pass
r = 4.7; h = 2.3; y = coneVolume(r,h) y_correct = 53.2050; assert(abs(y-y_correct)<1e-3)
y = 53.2050 y = 53.2050
3 Pass
r = 1.1; h = 1.21; y = coneVolume(r,h) y_correct = 1.5332; assert(abs(y-y_correct)<1e-3)
y = 1.5332 y = 1.5332
4 Pass
r = 10.7; h = 100; y = coneVolume(r,h) y_correct = 11989.36476; assert(abs(y-y_correct)<1e-3)
y = 1.1989e+04 y = 1.1989e+04
5 Pass
r = pi^3; h = exp(1); y = coneVolume(r,h) y_correct = 2736.6694; assert(abs(y-y_correct)<1e-3)
y = 2.7367e+03 y = 2.7367e+03
6 Pass
r = sqrt(5); h = sqrt(817); y = coneVolume(r,h) y_correct = 149.66135; assert(abs(y-y_correct)<1e-3)
y = 149.6613 y = 149.6613
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http://mathhelpforum.com/algebra/143900-conversions.html
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1. Conversions?
I'm in the ninth grade and I got this assignment. I'm 99% sure my teacher has the wrong answer. So, just tell me what you would get for this problem? Thanks.
The problem is that you have two hospitals with a certain amount of rooms and two cleaning products with different prices. I won't post the entire problem, just the converting part since that is where I apparently went wrong.
Cleaning Product 1:
spectrokill: 15 000$/m^3 (plus 2000$ for a machine to use it)
covers 50m^2/L
Cleaning Product 2:
javel: 20$/L covers 40m^2/L What I did: I thought you would have to convert the spectrokill into$/L, since 1m^3= 1kL which is = to 1000 L
I did 15 000 / 1000 to get 15$/L and then compared with the other product for each hospital. My teacher told me you have to find the cubic root of 1500 to get$24.66 but I don't understand why you would need to do that, since you're measuring volume and not distance.
2. I did 1500 / 1000 to get 15$/L It should be 1.5$/L
3. ooops
I wasn't looking at the problem. it was priced 15 000\$/m^3, not 1500.
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https://gamedev.stackexchange.com/questions/20255/continuous-physics-engines-collision-detection-techniques
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Continuous Physics Engine's Collision Detection Techniques
I'm working on a purely continuous physics engine, and I need to choose algorithms for broad and narrow phase collision detection. "Purely continuous" means I never do intersection tests, but instead want to find ways to catch every collision before it happens, and put each into "planned collisions" stack that is ordered by TOI.
Broad Phase The only continuous broad-phase method I can think of is encasing each body in a circle and testing if each circle will ever overlap another. This seems horribly inefficient however, and lacks any culling.
I have no idea what continuous analogs might exist for today's discrete collision culling methods such as quad-trees either. How might I go about preventing inappropriate and pointless broad test's such as a discrete engine does? I also would like to be able to see collisions more than 1 frame ahead to.
Narrow Phase
I've managed to adapt the narrow SAT to a continuous check rather than discrete, but I'm sure there's other better algorithms out there in papers or sites you guys might have come across.
What various fast or accurate algorithm's do you suggest I use and what are the advantages / disatvantages of each?
Final Note:
I say techniques and not algorithms because I have not yet decided on how I will store different polygons which might be concave, convex, round, or even have holes. I plan to make a decision on this based on what the algorithm requires (for instance if I choose an algorithm that breaks down a polygon into triangles or convex shapes I will simply store the polygon data in this form).
• I recommend you check these books amazon.com/Real-Time-Collision-Detection-Interactive-Technology/… amazon.com/… Commented Dec 1, 2011 at 8:59
• I'm sorry to add, but why not use Box2D? It's been ported to nearly every language. If you don't plan on using it, why not browse it's source so that you can see how it manages it's collosion? Commented Dec 7, 2011 at 15:03
I am really just throwing ideas around here. Assuming you have (at the very least) the current position and next position; for each frame.
You would need two separate broad phases, followed by your narrow phase:
• One that figures out that a collision will occur.
• One that figures out roughly where the collision actually occurs (e.g. a broad phase/inaccurate SAT)
• Finally your narrow phase would improve the result of the second broad phase.
You could look into spatial hashing (using the next position, not current) for the initial broad phase. This would partition your problem space nicely into groups of collision candidates.
Do a binary multi-sample using the circle intersection method you described. In other words:
left = current
right = next
midpoint = (left + right) / 2
loop a desired amount of times tweaked to the accuracy you want:
is a collision occuring at midpoint?
right = midpoint
else?
left = midpoint
midpoint = (left + right) / 2
pointOfCollision = midpoint
That accuracy tweak could also take distance into consideration - I think using the 'length squared' of next - current would get a pixel-perfect result.
Narrow Phase
Do a binary multi-sample using something like PMask - the logic will be exactly the same as above; just using a different collision routine.
Finally
You will be able to work out the time-of-intersection from pointOfCollision, current and your current speed and acceleration (assuming you have a reasonable integrator).
• So for secondary broad phase detection, are you suggesting I get the midpoint of the circle's travel path, and test if it is inside the circle being tested against? I was thinking I could simply create an equation that gives the two circles distance from each other over time, and seeing if at any time the distance equals 0. Commented Dec 1, 2011 at 19:09
• Also, what does Pmask do exactly? the site doesn't really explain =/. Commented Dec 1, 2011 at 19:25
• @Griffin your first comment might work - see if you can figure out. I am basically doing a binary search over a collision space... PMask is pretty clever. See a unsigned 64-int as a 8x8 grid of pixels (on/off) - a simple AND (binary) determines if a collision is occurring (non-zero); you need to do some clever bitshifting first, but that's the idea. Read the source for more info; it's hard to explain here (or rather, my explanation would suck) - you really need to refer to the source. Commented Dec 1, 2011 at 22:54
Alright, I've seen you've updated your question so as to be more specific. I'll try and help you out some more.
Essentially, you divide your screen into equal-sized grids. Then, if an object lies within a grid, you add it to a "bucket" in a 1D hash table.
That's your first check done. If objects aren't in the same bucket, it would be impossible for them to intersect.
Continuing with that, you now have a list of buckets with objects (potentially) in them. You can do another broad-phase check here by either:
A.) Dividing this bucket into 4 other buckets, and checking the resulting 1D hash table. If they aren't in the same bucket, no collision.
Or:
B.) Doing a simple distance check and keeping the width and/or height of the object in mind to ensure accuracy.
But what about when you potentially have a collision?
Then I would recommend something along the lines of this. It's essentially a kind of mix between polygonal collision (for complex shapes) or rectangle/circle for less complex shapes.
Also, if you really want to "catch collisions before they happen and store them" then you can always do something like this:
If two objects are in the same bucket, then they might collide.
Furthermore, are the objects close enough that they may collide soon? (Taking into account velocity, object size and distance)
If the answer to both is yes, then go ahead and store it to do an intersection test later.
Well, unfortunately I have lost track of my "All Collision Types and What They Are Used For" Handbook. :)
However, even though this is an extremely broad queston, I'll get you started.
There's a good (answered) question pertaining to something like this here.
As well as an article by the people who made N and N+ over here.
Not to mention, you've got the good ol' fallback Per-pixel Collision.
I sincerely doubt that anyone will have a list handy of each and every type of collision, but this should help get you started.
However, I should mention that the type of collision you need (and will end up using) largely depends on the type of game you are creating. That's why you find tutorials - most people assume you have an idea of what you want, so they help you in that specific area. I realize that most of my links are tutorials on a specific subject, but I think a tutorial will honestly help you more. A list is one thing, but if you read about each bulletpoint yourself, you can come to a more educated decision that will likely suit your needs more specifically.
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http://math.stackexchange.com/questions/33286/degree-of-frobenius
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# Degree of Frobenius
Let $k$ be an algebraically closed field of characteristic $p>0$ and $K/k$ be a function field, i.e. $K$ is finite over $k(t)$. Consider the field extension $K \subseteq K^{1/p}$. Why does it have degree $p$?
In the case $K = k(t)$ this is clear, because then $K^{1/p} = k(t^{1/p})$ and $t^{1/p}$ has degree $p$ over $K$, since $x^p - t$ is irreducible over $K$ and has $t^{1/p}$ as a root. But I don't know how to deal with the general case.
Remark: The equality $[K^{1/p} : K] = p$ is used in Hartshorne's book in the context of the Frobenius morphism of curves.
-
Since, as you write, $[k(t)^{1/p}:k(t)]=p$, the result follows from $[K:k(t)]=[K^{1/p}:k(t)^{1/p}]$ by multiplicativity of degrees. $[K:k(t)]=[K^{1/p}:k(t)^{1/p}]$ holds since the map $x\mapsto x^p$ is an isomorphism $K^{1/p}\to K$ which restricts to an isomorphism $k(t)^{1/p}\to k(t)$ - i.e. the extensions $K^{1/p}\supset k(t)^{1/p}$ and $K\supset k(t)$ are isomorphic.
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https://www.performanceboats.com/gn7-dyno/10270-cam-lift-question.html
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cam lift question
# Thread: cam lift question
1. ## cam lift question
If I have a pair of heads where the valve springs/valve combo has a max lift of .700, the solid roller cam has .715 lift, and valve lash is set at .025, is it possible that it will work, or does it not work like that?
.715
- .025
.690
.690 actual lift with a valvetrain combo rated for a max of .700
I realize that it would need to be checked during engine assembly to make sure that it is all working correctly, I'm just wondering if the math of it works like that.
Is there a relative easy fix if the valve train doesn't work right due to slightly too much lift, or are you supposed to replace springs and valves? (valve spring pocket machining? re-grind cam for lower lift? <-- how much lower lift?)
Thanks again!
2.
3. interesting. where does rockerratio play in this? .715 x 1.7?
<----also wants to learn.
4. ## NO
NO ............................Phuck.....Oh yeah, for sure,,...............maybe....
5.
6. yeah, basically its your cam(.715) x rocker ratio(example 1.7) =1.215"
then subract your lash for net lift. hers where i got this
http://www.circletrack.com/techartic.../photo_06.html
losta good info here -Jas
7. Originally Posted by hotrod56cars
If I have a pair of heads where the valve springs/valve combo has a max lift of .700, the solid roller cam has .715 lift, and valve lash is set at .025, is it possible that it will work, or does it not work like that?
.715
- .025
.690
.690 actual lift with a valvetrain combo rated for a max of .700
I realize that it would need to be checked during engine assembly to make sure that it is all working correctly, I'm just wondering if the math of it works like that.
Is there a relative easy fix if the valve train doesn't work right due to slightly too much lift, or are you supposed to replace springs and valves? (valve spring pocket machining? re-grind cam for lower lift? <-- how much lower lift?)
Thanks again!
Hey hot rod, The cam spec you are listing being .715 lift of valve off the seat. On that cam the acctual or net amount of lift at the cam would be .420. The rocker factors in as the lift at valve. Therefore .420 x 1.7 = .715. Here is a double check for you. .715 devided by rocker ratio. A dial indicator at each ind of the rocker tells these numbers. Lastly a cam of that size wants a minimum .040 coil clearance. YOU MY FRIEND ARE IN THE DANGER ZONE!!!!!!!!!!!!!!! I don't know if you have purchased your'e rockers yet but if you do the math with the info above a slower ratio rocker is a easier fix. IE .420 x 1.6=.672 lift at valve. Very good spec cam.IMO M
8. Ol guy is right, and that would be a VERY tight combination, best dealt with by less lift, one way (different cam) or another (lower ratio rockers).
9. ok not to hijack but to understand, my cam is .680 lift how would i know what rocker to use to atchieve that lift would it be in the part number?
10. Originally Posted by bordsmnj
ok not to hijack but to understand, my cam is .680 lift how would i know what rocker to use to atchieve that lift would it be in the part number?
Hey bord, not knowing the motor Ford or chevy it's a hard call But the stock ratio is the factor the cam grinder comes up with for the lift at the valve,Chev uses a 1.70 rocker so the lift at cam is .400. Ford 429-460 uses a 1.73 rocker so the lift at cam would be .393 Then you multiply by rocker ratio and that determines the lift at valve. Again the cam grinders design cams to work with stock ratio. then its up to the motor builder to take some lift out or add some by rocker ratio changes. I was racing a 429 ford and would order my cam choice knowing I was going to machine the head for chevy 1.70 rockers, The chevy valve train is much easier to play with and less costly in the long run. Hope this helps. M
11. ah ha! beautiful. thanks for clearing that little mystery up. i looked at a cam i have here and said to myself there's just no way those lobes are .700 tall,lol. i'll keep that 1.7 in my mind. -jas
12. Originally Posted by hotrod56cars
If I have a pair of heads where the valve springs/valve combo has a max lift of .700, the solid roller cam has .715 lift, and valve lash is set at .025, is it possible that it will work, or does it not work like that?
.715
- .025
.690
.690 actual lift with a valvetrain combo rated for a max of .700
I realize that it would need to be checked during engine assembly to make sure that it is all working correctly, I'm just wondering if the math of it works like that.
Is there a relative easy fix if the valve train doesn't work right due to slightly too much lift, or are you supposed to replace springs and valves? (valve spring pocket machining? re-grind cam for lower lift? <-- how much lower lift?)
Thanks again!
That's a bit tight and would likely be a problem. Depending on the application, you may be able to buy +.100" longer retainers to give you a bit more room. You'll likely need different springs to get proper seat/over the nose spring pressure. Likely cheaper than new valves.
If I can, I like to set my spring height about .100" taller than my valve lift. This gives you room to shim the spring if it starts to get a bit soft and still allow enough room to avoid coil bind. Good luck.
13. Someone might add the insignificant fact that valve spring PRESSURE may not be correct for the application. Required pressure is dependant on lobe profile (how radical the ramps are), application, rpm's it regularly sees, valve weight, etc.
You need .060" minimum before spring coil bind. Just because a spring is spec'ed at .700 has little to do with how much lift it will handle before coil bind. Different manufacturers (of springs) rate their springs different (before coil bind). (I don't even think that sentence made sense)
But, what do I know.
Wags
14. Originally Posted by wagspe208
Someone might add the insignificant fact that valve spring PRESSURE may not be correct for the application. Required pressure is dependant on lobe profile (how radical the ramps are), application, rpm's it regularly sees, valve weight, etc.
You need .060" minimum before spring coil bind. Just because a spring is spec'ed at .700 has little to do with how much lift it will handle before coil bind. Different manufacturers (of springs) rate their springs different (before coil bind). (I don't even think that sentence made sense)
But, what do I know.
Wags
tHIS DOESN'T EVEN WARRANT A RESPONSE!!! I'll just back out and say listen to wags Good luck guys. P.M. me If you want My thoughts. Instant balls just add a keyboard. M
15. If the heads came as a package from say Brodix for example, and the spring package was for .700 lift, this would typically be .075-.100 from coil bind. You say your cam has a gross theoretical of .715, factor in a typical lash of .020, plus valve train deflection (with some decent parts) and you will be lucky if your getting .650 actual at the valve.
I would say you will be more than fine, but need to check the heads to make sure they are indeed set-up for .700 lift (coil to coil bind and seal to retainer clearance) and are of the proper pressure for the cam.
Valvetrain. I'm looking at a set of Merlins, 030620-3, assembled, new, recomended for a solid roller cam, max valve lift .700
Originaly I was going to go with 030620-4 with a max valve lift of .800, but unfortunately the Summit Credit Card doesn't have the balance for that.
My cam is .715 lift with a 1.7 rocker arm ratio.
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# All Solutions: the ph will lie between 0 and 14
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2 2 only 1; 0.01 molar hydrochloric acid will have a p of 2, and an infinitely dilute solution of hydrochloric acid (or any acid or base, strong or weak) will have a p of 7. Thus first aid for spills of acids or bases is dilution, not neutralization Weak Acids; pk a of the acid is between 2 and 12 If you make a 1 molar solution of a weak acid like acetic acid, only a fraction of the acetic acid molecules will donate their most acidic proton to water. The resulting solution will be less acidic than a solution of a strong acid of the same concentration. More precisely, 1 molar acetic acid, whose pk a is about 4.8, will have a p of only 2.4, which is less acidic than 1 molar hydrochloric acid, whose p will be 0. A shortcut to the p of 2.4 for 1 molar acetic acid is to see that 2.4 equals 1/2 of 4.8, the pk a of acetic acid. This is because the essential mathematical operation for the calculation of the p of a 1 molar solution of a weak acid is taking the square root of the acid ionization constant; dividing the log of a number by 2 is equivalent to taking the square root of the number. If you are willing to round the acid ionization constants for weak acids to the nearest power of ten, the math gets even easier. Thus if you take 5 as the pk a for acetic acid you can estimate the p of 1 molar acetic acid to be 2.5, a value that is for most purposes just as useful as 2.4. You can thus estimate that a 1 molar solution of a slightly stronger acid such as formic acid, pk a = 4, will have a p of 2, and that a 1 molar solution of a somewhat weaker acid such as phenol, pk a = 10, will have a p of about 5. One molar solutions of still weaker acids will be still less acidic, and you can be confident that solutions of ethanol or sugar, pk a about 14, will be no more acidic than pure water. In general, an aqueous solution of a substance whose most acidic proton has a pk a greater than 14 will be neutral. Just as you know that a 0.01 molar solution of Cl is less acidic than a 1 molar solution of Cl you would expect a 0.01 molar solution of acetic acid to be less acidic than a 1 molar solution of this acid. When you do the math you find that 0.01 molar acetic acid will have a p of 3.4, which is 1 p unit to the basic side of 2.4, the p of 1 molar acetic acid. The shortcut, apparently, is that for each factor of 100 by which the solution is diluted the p is 1 unit closer to 7. This is consistent with the idea that the more dilute the solution the more nearly its p will be 7. This also reflects the fact that 1/2[log 100] = 1/2[2] =1. You might now anticipate that the p of 0.1 molar acetic acid should be half way between 2.4 and 3.4, or /2[1] = 2.9; you would be right. Do p in Your ead; w1.0 Addison Ault 1/15/2008, 3:41 PM
3 3 Summary of solutions of acids. Strong acids are completely ionized; 1 molar solutions will have a p of 0, and less concentrated solutions will be less acidic. Weak acids are not completely ionized and 1 molar solutions will have a p equal to 1/2[pK a )] Less concentrated solutions of weak acids will be still less acidic, and if the acid is very weak the solution will be no more acidic than pure water. Table 1 presents a summary of several examples and reveals the pattern. Table 1 Acid pk a p of 1.0 molar solution p of 0.01 molar solution hydrogen chloride formic acid acetic acid anilinium ion pyridinium ion ammonium ion phenol ethanol O C O- 3 C O C O- N + N + N O- formic acid acetic acid anilinium pyridinium ammonium phenol ion ion ion pk a Solutions of Bases: the p will lie between 7 and 14 If you add a base, a source of hydroxide ion, to pure water you will have a basic solution. You can also produce basic solutions by adding substances that consume hydronium ion; salts of weak acids are such substances. Strong bases Do p in Your ead; w1.0 Addison Ault 1/15/2008, 3:41 PM
4 4 Strong bases are hydroxide ion donors that are completely ionized in water. The most common example of such a base is sodium hydroxide, and a 1 molar solution of sodium hydroxide will be 1 molar in hydroxide ion. The solution will therefore be 1x10 14 molar in hydronium ion, and will thus have a p of 14. Potassium hydroxide behaves in the same way. You can see that less concentrated solutions of strong bases will be less basic. For example, 0.1 molar sodium hydroxide will have a p of 13, and 0.01 molar NaO will have a p of 12. Finally, just as is true of an infinitely dilute solution of an acid, an infinitely dilute solution of a base will have a p of 7. Salts of weak acids and strong bases You have been told that the solutions of the salt of a weak acid and a strong base will be basic. An example of such a salt is sodium acetate, the salt of acetic acid, a weak acid, and sodium hydroxide, a strong base. The acetate ions will react to a small extent with water to form a small number of acetic acid molecules and an equally small number of hydroxide ions. The sodium ions will not react at all with water, and so the solution of 1 mole of sodium acetate in 1 liter of water will be slightly more basic than pure water due to the production of hydroxide ions; you have probably done a problem that shows that 1 molar sodium acetate should have a p of about 9.4. A shortcut to the value of 9.4 for the p of 1 molar sodium acetate is to see that just as the p of 1 molar acetic acid is 2.4 p units on the basic side of 0, the p of 1 molar sodium acetate is 2.4 p units on the basic side of 7. This result is, again, based on the essential mathematical operation for the calculation of the p of a 1 molar solution of the salt of a weak acid and a strong base. This operation is to take the square root of [(ionization constant of the acid) x (ion product of water)]. Taking the shortcut, then, the p is 1/2[pK a + pk w ]. For sodium acetate this is 1/2[ ] = = 9.4. If we were to round the pka value for acetic acid to 5, we would then estimate the p of 1 molar sodium acetate to be 9.5. If the weak acid is a little stronger, having a pk a of, say, 4, you can estimate that a 1 molar solution of its sodium salt should have a p of 4/2 + 7 = 9. The salt of the stronger weak acid undergoes hydrolysis to a lesser extent, and solutions of its salt are less basic. A 1 molar solution of this acid would have a p of 4/2 = 2. In general, the more acidic the weak acid the less basic will be its salts. Do p in Your ead; w1.0 Addison Ault 1/15/2008, 3:41 PM
5 5 Picking a weak acid more or less at random, you could predict that a 1 molar solution of fluoroacetic acid, pk a = 2.6, should have a p of about 1.3, and a 1 molar solution of sodium fluoroacetate should have a p of about 8.3. You can see that taking this trend to the extreme gives us something you know to be true: a 1 molar solution of hydrogen chloride should have a p of 0, and a 1 molar solution of sodium chloride should have a p of 7. Finally, just as less concentrated solutions of acids are less acidic so less concentrated solutions of bases are less basic, both types of solutions approaching a p of 7 at infinite dilution. Weak Bronsted bases A substance that can act as a proton acceptor toward water will produce an aqueous solution whose p is between 7 and 14. Because the pyridinium ion is about as acidic as acetic acid, pyridine will be as basic as acetate ion. Since the behavior of pyridine will be the same as that of acetate ion the math will be the same, and the p of 1 molar pyridine should be about 5/2 + 7 = 9.5. Summary of solutions of bases. An aqueous solution of a weak base will have a p between 7 and 14. A stronger base will give a more basic solution. Less concentrated solutions will be less basic. Table 2 presents a summary of several examples and reveals the pattern. Table 2 Base pk a of the conjugate acid p of 1.0 molar solution p of 0.01 molar solution sodium acetate pyridine aniline ammonia sodium phenoxide sodium hydroxide sodium ethoxide sodium amide Do p in Your ead; w1.0 Addison Ault 1/15/2008, 3:41 PM
6 6 Ethoxide ion and amide ion are so basic that they react completely with water to produce hydroxide ion. Thus 1 mole of either ethoxide ion or amide ion in 1 liter of water will give 1 mole of hydroxide ion, and a solution whose p will be 14. Solutions of Salts: The p will tend toward 7 Salts are ionic substances in which the ions are neither the proton or the hydroxide ion. Depending upon the relative tendencies of the two ions to react with water in such a way that protons or hydroxide ions are formed, solutions of salts may be either somewhat acidic, somewhat basic, or nearly neutral. There are four possibilities, and when you see which possibility is represented, you can quickly decide whether the solution will be slightly acidic, slightly basic, or just about neutral. If you know the approximate pk a values of the relevant ions you can estimate quite well the p of the solution. Salts of a strong acid and a strong base: the solution will be neutral Salts of a strong acid and a strong base, substances such as sodium chloride, potassium bromide, sodium nitrate, and potassium iodide will all give solutions with a p of practically 7. Neither ion shows any tendency to react with water to produce either hydronium ion or hydroxide ion. Salts of a strong acid and a weak base: the solution will be acidic Salts of strong acids and weak bases, substances such as pyridinium chloride and ammonium bromide, will give acidic solutions, as we have seen. In these salts the anion does not react with water to give hydroxide ion, but the cation does react with water to produce hydronium ion. The solution will therefore be acidic. Table 1 illustrates this possibility. Salts of a weak acid and a strong base: the solution will be basic Salts of weak acids and strong bases, substances such as sodium acetate and sodium phenoxide, will give basic solutions, as we have also seen. In these salts the cation does not react with water to give hydronium ion, but the anion does react with water to produce hydroxide ion. The solution will therefore be basic. Table 2 illustrates this possibility. Salts of weak acids and weak bases: the solution can be either slightly acidic or slightly basic Do p in Your ead; w1.0 Addison Ault 1/15/2008, 3:41 PM
7 7 Salts of weak acids and weak bases can give solutions that are either slightly acidic or slightly basic, depending upon the relative strengths of the acids and bases that have combined to give the salt. If the weak acid is stronger than the weak base, the solution will be slightly acidic; if the weak base is stronger than the weak acid, the solution will be slightly basic. The numerical value of the p is half way between the pk a values of the weak acid and the conjugate acid of the weak base. The case of ammonium phenoxide provides an example. You can see from Table 1 that 1 molar ammonium chloride will give a solution of p = 4.5, and from Table 2 that 1 molar sodium phenoxide will give a solution of p = 12. Thus you see that the phenoxide ion as a base is stronger than the ammonium ion as an acid. When they are put into competition in the same solution the phenoxide ion will win and he solution will be slightly basic; the p of the solution will be about 9.5, a p half way between the pk a of 9 for the ammonium ion and 10 for phenol. Other examples could include anilinium acetate, where the anilinium ion is a stronger acid than acetate ion is a base (solutions of this salt will have a p of 5), and ammonium acetate, where the ammonium ion as an acid and the acetate ion as a base are equal in strength, with the result that solutions of ammonium acetate will have a p of about 7. These examples and several others are presented in Table 3. An example of relevance in biochemistry is provided by the amino acid glycine. Glycine, or aminoacetic acid, exists as an inner salt, and so in analogy with ammonium acetate the p of 1 molar glycine will be half way between the pk a of the carboxyl group of glycine (2.4) and the pk a of the ammonium group of glycine (9.6). The half way p is [ ]/2 = 12/2 = 6. This half way p is also called the isoelectric p, or the isoelectric point. on the p scale. It is the p at which, over a period of time each glycine molecule is, on the average, neutral. ` + N C C O O pk a = 9.6 pk a = N C C O O conjugate acid of glycine glycine p in pure water = 6.2 Do p in Your ead; w1.0 Addison Ault 1/15/2008, 3:41 PM
8 8 It is generally true that solutions of amino acids without acidic or basic side chains will give aqueous solutions with a p of 6, a p half way between the pk a of the carboxyl group (close to 2.5) and the pk a of the ammonium group (close to 9.5). Summary of solutions of salts of weak acids and weak bases Solutions of salts will be weakly acidic, neutral, or weakly basic, depending upon which ion is the stronger acid or the stronger base. The exact p will be the average of the two relevant pk a values. The p of solutions of several salts are presented in Table 3. Table 3 Salt pk a of acid pk a of conj acid of base p of 1 molar solution sodium chloride 7 ammonium phenoxide anilinium acetate ammonium acetate pyridinium acetate glycine alanine Solutions of Conjugate Pairs Sometimes you will prepare an aqueous solution that contains both an acid, such as acetic acid, and its conjugate base, in this case an acetate salt, perhaps sodium acetate. When you do this, the p of the resulting solution will be close to the pk a of acetic acid, the exact value of the p depending upon the ratio of the concentrations of the members of the conjugate pair. The isoconjugate p If you should dissolve equimolar amounts of acetic acid and sodium acetate in water, the ratio of acetic acid molecules to acetate ions, the members of this conjugate pair, will be 1 to 1, and the p of the solution will equal the 4.8, the pk a of acetic acid, the acidic member of this conjugate pair. For the Do p in Your ead; w1.0 Addison Ault 1/15/2008, 3:41 PM
9 9 conjugate pair acetic acid and acetate ion, the p of 4.8 is the isoconjugate p or isoconjugate point on the p scale. Deviations from the isoconjugate p If you prepare a solution in which the acidic member of the conjugate pair predominates, the p of the solution will be on the acidic side of the isoconjugate p. Conversely, if you prepare a solution in which the basic member of the conjugate pair predominates the p of the solution will be on the basic side of the isoconjugate p. The degree to which the p of the solution deviates from the isoconjugate p equals the log of the ratio of the members of the conjugate pair. If the ratio is 10 to 1, the deviation is 1 p unit, and smaller ratios will give smaller deviations. And, of course, as the ratio approaches 1 to 1, the deviation will approach 0; the log of 1 is 0, and the p will equal the pk a. The math here follows the enderson-asselbalch equation, but if you round the pk a values to integers and ratios to powers of 10 you can do the math in your head to get approximate values that are quite adequate. Table 4 illustrates the pattern for deviations from the isoconjugate p. Table 4 ratio of members of conjugate pair log of ratio of members of conjugate pair deviation from isoconjugate p 1000 to to to to to to Buffers You may have recognized the solutions that we have just described as buffered solutions, solutions that resist a change in p either by dilution or by the addition of small amounts of acids and bases. These solutions resist a change in p by dilution because the p is determined by the ratio of the concentrations of the members of the conjugate pair. Although dilution will change the concentrations of both members of the conjugate pair, dilution will not change their ratio. Do p in Your ead; w1.0 Addison Ault 1/15/2008, 3:41 PM
10 10 These solutions will also resist a change in p upon the addition of small amounts of either acid or base. Continuing with the acetic acid/acetate ion solution, so long as added acid or base does not reduce either acetate ion or acetic acid below 1/10 of the total of acetate ion plus acetic acid, the ratio of the members of this conjugate pair will stay within the range of 10:1 to 1:10 and the p of the solution will stay within 1 p unit of the isoconjugate p. We call solutions such as this, which resist these efforts to change the p, buffered solutions. Table 5 presents examples of acetic acid/acetate ion buffers. The isoconjugate p of this pair is about 4.8. Table 5 acetic acid to acetate ion ratio p of the solution 10 to to to to to to to Do p in Your ead; w1.0 Addison Ault 1/15/2008, 3:41 PM
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