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# IDW (3D Analyst)
## Summary
Interpolates a raster surface from points using an inverse distance weighted (IDW) technique.
## Usage
• The output value for a cell using inverse distance weighting (IDW) is limited to the range of the values used to interpolate. Because IDW is a weighted distance average, the average cannot be greater than the highest or less than the lowest input. Therefore, it cannot create ridges or valleys if these extremes have not already been sampled (Watson and Philip 1985).
• The best results from IDW are obtained when sampling is sufficiently dense with regard to the local variation you are attempting to simulate. If the sampling of input points is sparse or uneven, the results may not sufficiently represent the desired surface (Watson and Philip 1985).
• The influence of an input point on an interpolated value is isotropic. Since the influence of an input point on an interpolated value is distance related, IDW is not "ridge preserving" (Philip and Watson 1982).
• Some input datasets may have several points with the same x,y coordinates. If the values of the points at the common location are the same, they are considered duplicates and have no affect on the output. If the values are different, they are considered coincident points.
The various interpolation tools may handle this data condition differently. For example, in some cases the first coincident point encountered is used for the calculation; in other cases the last point encountered is used. This may cause some locations in the output raster to have different values than what you might expect. The solution is to prepare your data by removing these coincident points. The Collect Events tool in the Spatial Statistics toolbox is useful for identifying any coincident points in your data.
• The barriers option is used to specify the location of linear features known to interrupt the surface continuity. These features do not have z-values. Cliffs, faults, and embankments are typical examples of barriers. Barriers limit the selected set of the input sample points used to interpolate output z-values to those samples on the same side of the barrier as the current processing cell. Separation by a barrier is determined by line-of-sight analysis between each pair of points. This means that topological separation is not required for two points to be excluded from each other's region of influence. Input sample points that lie exactly on the barrier line will be included in the selected sample set for both sides of the barrier.
• Barrier features are input as polyline features. IDW only uses the x,y coordinates for the linear feature; therefore, it is not necessary to provide z-values for the left and right sides of the barrier. Any z-values provided will be ignored.
• Using barriers will significantly extend the processing time.
• This tool has a limit of approximately 45 million input points. If your input feature class contains more than 45 million points, the tool may fail to create a result. You can avoid this limit by interpolating your study area in several pieces, making sure there is some overlap in the edges, then mosaicking the results to create a single large raster dataset. Alternatively, you can use a terrain dataset to store and visualize points and surfaces comprised of billions of measurement points.
If you have the Geostatistical Analyst extension, you may be able to process larger datasets.
• The input feature data must contain at least one valid field.
## Syntax
Idw_3d (in_point_features, z_field, out_raster, {cell_size}, {power}, {search_radius}, {in_barrier_polyline_features})
Parameter Explanation Data Type in_point_features The input point features containing the z-values to be interpolated into a surface raster. Feature Layer z_field The field that holds a height or magnitude value for each point.This can be a numeric field or the Shape field if the input point features contain z-values. Field out_raster The output interpolated surface raster. Raster Layer cell_size(Optional) The cell size at which the output raster will be created.This will be the value in the environment if it is explicitly set; otherwise, it is the shorter of the width or the height of the extent of the input point features, in the input spatial reference, divided by 250. Analysis Cell Size power(Optional) The exponent of distance. Controls the significance of surrounding points on the interpolated value. A higher power results in less influence from distant points. It can be any real number greater than 0, but the most reasonable results will be obtained using values from 0.5 to 3. The default is 2. Double search_radius(Optional) Defines which of the input points will be used to interpolate the value for each cell in the output raster. There are two ways to specify the specify the searching neighborhood: Variable and Fixed.Variable uses a variable search radius in order to find a specified number of input sample points for the interpolation. Fixed uses a specified fixed distance within which all input points will be used. Variable is the default.The syntax for these parameters are: Variable, number_of_points, maximum_distance, where:number_of_points—An integer value specifying the number of nearest input sample points to be used to perform interpolation. The default is 12 points.maximum_distance—Specifies the distance, in map units, by which to limit the search for the nearest input sample points. The default value is the length of the extent's diagonal. Fixed, distance, minimum_number_of_points, where:distance—Specifies the distance as a radius within which input sample points will be used to perform the interpolation. The value of the radius is expressed in map units. The default radius is five times the cell size of the output raster.minimum_number_of_points—An integer defining the minimum number of points to be used for interpolation. The default value is 0.If the required number of points is not found within the specified distance, the search distance will be increased until the specified minimum number of points is found. When the search radius needs to be increased it is done so until the minimum_number_of_points fall within that radius, or the extent of the radius crosses the lower (southern) and/or upper (northern) extent of the output raster. NoData is assigned to all locations that do not satisfy the above condition. Radius in_barrier_polyline_features(Optional) Polyline features to be used as a break or limit in searching for the input sample points. Feature Layer
## Code Sample
IDW example 1 (Python window)
This example inputs a point shapefile and interpolates the output surface as a TIFF raster.
```import arcpy
from arcpy import env
env.workspace = "C:/data"
arcpy.Idw_3d("ozone_pts.shp", "ozone", "C:/output/idwout.tif", 2000, 2, 10)
```
IDW example 2 (stand-alone script)
This example inputs a point shapefile and interpolates the output surface as a Grid raster.
```# Name: IDW_3d_Ex_02.py
# Description: Interpolate a series of point features onto a
# rectangular raster using Inverse Distance Weighting (IDW).
# Requirements: 3D Analyst Extension
# Import system modules
import arcpy
from arcpy import env
# Set environment settings
env.workspace = "C:/data"
# Set local variables
inPointFeatures = "ca_ozone_pts.shp"
zField = "ozone"
outRaster = "C:/output/idwout01"
cellSize = 2000.0
power = 2
# Check out the ArcGIS 3D Analyst extension license
arcpy.CheckOutExtension("3D")
# Execute IDW
arcpy.Idw_3d(inPointFeatures, zField, outRaster, cellSize,
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# How do I find the distance between two points? [closed]
Let's say I have x1, y1 and also x2, y2.
How can I find the distance between them? It's a simple math function, but is there a snippet of this online?
-
## closed as not constructive by Tuxdude, NatureFriend, Perception, Sankar Ganesh, Michael WildMar 18 '13 at 7:05
As it currently stands, this question is not a good fit for our Q&A format. We expect answers to be supported by facts, references, or expertise, but this question will likely solicit debate, arguments, polling, or extended discussion. If you feel that this question can be improved and possibly reopened, visit the help center for guidance. If this question can be reworded to fit the rules in the help center, please edit the question.
This is ridiculous. Did you even try to search? – Greg Hewgill Mar 8 '11 at 4:37
It didn't work. So I asked here. I figured out why. It's coz I did ^ instead of ** – TIMEX Mar 8 '11 at 4:42
@Greg: His track record says no. @TIMEX: Searching didn't work? Seriously: google.com/search?q=python+distance+points – Glenn Maynard Mar 8 '11 at 4:48
-1 for "is there a snippet of this online?" Seriously, @TIMEX, if searching the web for a code snippet is too hard, now is the time for a change of career. – Johnsyweb Mar 8 '11 at 8:03
I'm surprised this question is closed. It was in my search results for 'python pythagoras' and was how I discovered the existence of math.hypot. – Rob Fisher Mar 18 '13 at 7:51
## 3 Answers
``````dist = sqrt( (x2 - x1)**2 + (y2 - y1)**2 )
``````
As others have pointed out, you can also use the equivalent built-in `math.hypot()`:
``````dist = math.hypot(x2 - x1, y2 - y1)
``````
-
This is, by the way, the distance formula – Andrew Marshall Mar 8 '11 at 4:37
did you mean en.wikipedia.org/wiki/Euclidean_distance ? – Mitch Wheat Mar 8 '11 at 4:38
This isn't how to do the "power" in python? Isn't it **? – TIMEX Mar 8 '11 at 4:40
@TIMEX: Yes it is. The change is now reflected on @MitchWheat's post – inspectorG4dget Mar 8 '11 at 4:43
@RobFisher - explicitly writing this expression may actually be faster than calling `math.hypot` since it replaces a function call with inline bytecodes. – Paul McGuire Jul 16 '13 at 19:58
Let's not forget math.hypot:
``````dist = math.hypot(x2-x1, y2-y1)
``````
Here's hypot as part of a snippet to compute the length of a path defined by a list of x,y tuples:
``````from math import hypot
pts = [
(10,10),
(10,11),
(20,11),
(20,10),
(10,10),
]
ptdiff = lambda (p1,p2): (p1[0]-p2[0], p1[1]-p2[1])
diffs = map(ptdiff, zip(pts,pts[1:]))
path = sum(hypot(*d) for d in diffs)
print path
``````
-
It is an implementation of Pythagorean theorem. Link: http://en.wikipedia.org/wiki/Pythagorean_theorem
-
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# Math-Statistics
posted by Jenny
Suppose you want to test the claim that mean is not equal 3.5. Given a sample size of n = 45 and a level of
significance of a = 0.10, when should you reject H0 ?
A) Reject H0 if the standardized test statistic is greater than 1.96 or less than -1.96
B) Reject H0 if the standardized test statistic is greater than 1.645 or less than -1.645.
C) Reject H0 if the standardized test statistic is greater than 2.575 or less than -2.575
D) Reject H0 if the standardized test statistic is greater than 2.33 or less than -2.33.
1. PsyDAG
Look up Z scores under a table in the back of your statistics text labeled something like "areas under the normal distribution." There you will find that for a two-tailed test, if Z = 1.96, P = .05. Find the values for the other alternatives to answer your question.
I hope this helps.
2. Anonymous
Suppose you want to test the claim that Given a sample size of n = 48 and a level of significance of when should you reject ?
## Similar Questions
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4. ### Psychology/Statistics
For each of the following samples that were given an experimental treatment test whether these samples represent populations that are different from the general population (a) a sample of 10 with a mean of 44, (b) a sample of 1 with …
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10. ### Statistics
Recall that the power (1 – β) of a hypothesis test is the probability that we will reject the null hypothesis given that it is false; that is, the probability that we will correctly reject the null hypothesis. All other things …
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# Dividing Indices
In this worksheet, students divide numbers expressed using indices.
Key stage: KS 3
Curriculum topic: Number
Curriculum subtopic: Understand Integer Powers/Real Roots
Difficulty level:
### QUESTION 1 of 10
An index is sometimes called a power or an exponent.
In 54, the 4 is the index and the 5 is the base.
54 is read as "5 to the power 4" or "5 to the 4" and means 5 x 5 x 5 x 5.
Dividing numbers
When we divide numbers with the same base number, we can just subtract the indices.
57 ÷ 53 = 54
This is because, when we divide we can cancel a 5 on the top with a 5 on the bottom, leaving four 5s on the top.
Simplify:
57 ÷ 55
52
252
12
Simplify:
48 ÷ 44
42
12
44
Simplify:
35 ÷ 34
39
31.25
3
Simplify:
34 ÷ 32
32
36
38
Simplify:
55 ÷ 53
52
258
58
Simplify:
35 ÷ 31
96
34
35
Simplify:
55 ÷ 55
51
0
1
Simplify:
34 ÷ 34
344
31
30
Simplify:
408 ÷ 404
2
402
404
Simplify:
64 ÷ 60
64
640
60
• Question 1
Simplify:
57 ÷ 55
52
• Question 2
Simplify:
48 ÷ 44
44
• Question 3
Simplify:
35 ÷ 34
3
• Question 4
Simplify:
34 ÷ 32
32
• Question 5
Simplify:
55 ÷ 53
52
• Question 6
Simplify:
35 ÷ 31
34
• Question 7
Simplify:
55 ÷ 55
1
• Question 8
Simplify:
34 ÷ 34
30
EDDIE SAYS
This shows that 30 = 1
• Question 9
Simplify:
408 ÷ 404
404
• Question 10
Simplify:
64 ÷ 60
64
EDDIE SAYS
This suggests that 60 = 1
---- OR ----
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Integration Review for Semester Exam
Objective
SWBAT compute indefinite and definite integrals with u-substitution and other inverses of derivative rules. SWBAT use integration to compute area under functions and average value on an interval.
Big Idea
Heading into the semester exam, we take a day to recap the various integration techniques and applications we have learned so far.
15 minutes
Setting the Stage
5 minutes
If you have not done so already, students are always eager to gain any insight or advantage for the exam that we are willing to provide. You might spend a small amount of time attending to this interest, without revealing specifics or otherwise compromising the validity of the exam.
In prior classes my students have been given study guides that in my opinion reveal way too much about the exam, to the point where students have learned to depend exclusively on the study guide in preparing for the exam. Part of the shift in focus to a college level course is in developing and following one’s own plan for preparing for exams. Yet, this is a study skill that many students need to be taught, because merely abandoning students in this endeavor and hoping they will somehow figure it out themselves is not the best instructional approach either.
I share some broad suggestions for how students might prepare:
• Using prior homework assignments and assessments to identify questions that students missed
• Looking up the content of questions in the textbook and our class notes to review and relearn that content
• Re-solve the questions correctly that were initially missed; then, look up additional practice problems on that content in their textbook or online to reinforce their learning
When students ask content-specific questions like “Which types of questions will be emphasized on the exam?” my answer is “I don’t know, ‘ya better prepare for everything we’ve learned this semester.”
I do verbalize a brief summary of the whole semester to help students sequence and put the course content in perspective (below). I encourage students to try to make connections and to look for coherence among the topics. I also ask students to recall and identify topics that they feel they need to study more, without being too helpful or compromising the validity of the exam.
Topics Covered on the Semester Exam
1) A graphical approach to limits
2) Modeling slope graphically with the difference quotient
3) Multiple derivative relationships graphically
4) The difference quotient as the derivative algebraically
5) Computing derivatives algebraically
6) Applying derivatives, especially optimization and related rates
7) Integration as differentiation in reverse
8) Properties and basic applications of integration
Investigation
38 minutes
Pages 2-4 of the In The Classroom file contains 10 integration review questions ranging from computing indefinite and definite integrals, RAM, areas under functions, solving for the constant of integration given a point, and average value.
DIFFERENTIATING THE LESSON: You might customize these questions to address known student weaknesses, or vary the level of difficulty to challenge a wider range of students in your class. There is no way we will finish all of these questions in class today, so you might post the unanswered questions online for students to complete on their own if they choose, and suggest that students discuss and post their answers on the class’s Facebook page.
2 minutes
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1. ## Explanation pls
I came across a snippet
Code:
```#define OPTION1 "option1"
#define OPTION1_BITPOS (0x00000001)
#define OPTION2 "option2"
#define OPTION2_BITPOS (0x00000002)
#define OPTION3 "option3"
#define OPTION3_BITPOS (0x00000004)
//Required to do the bit operations.
#define ALL_BITPOS (0x0001ffff)```
what does 0x000000.... do in general
2. I'm not sure I understand your question. It's just hexadecimal notation. They could have been defined with the decimal constants 1, 2, and 4 - the result would be the same. Or are you asking something else entirely?
3. i came across this to set the appropriate option bits bits dpeending on the input....
but some one explain me about hexamdecimal values dealt here ... and why they are dealt here.
Code:
```A simple C command line utility takes a series of command line options. The options are given to the utility like this : <utility_name> options=[no]option1,[no]options2,[no]option3?... Write C code using bitwise operators to use these flags in the code.
//Each option will have a bit reserved in the global_options_bits integer. The global_options_bits
// integer will have a bit set or not set depending on how the option was specified by the user.
// For example, if the user said nooption1, the bit for OPTION1 in global_options_bits
// will be 0. Likewise, if the user specified option3, the bit for OPTION3 in global_options_bits
// will be set to 1.
#define OPTION1 "option1" // For strcmp() with the passed arguments.
#define OPTION1_BITPOS (0x00000001) // Bit reserved for this option.
#define OPTION2 "option2"
#define OPTION2_BITPOS (0x00000002)
#define OPTION3 "option3"
#define OPTION3_BITPOS (0x00000004)
//Required to do the bit operations.
#define ALL_BITPOS (0x0001ffff)
// Assume you have already parsed the command line option and that
// parsed_argument_without_no has option1 or option2 or option3 (depending on what has
// been provided at the command line) and the variable negate_argument says if the
// option was negated or not (i.e, if it was option1 or nooption1)
if (strcmp((char *) parsed_argument_without_no, (char *) OPTION1) == 0)
{
// Copy the global_options_bits to a temporary variable.
tmp_action= global_options_bits;
if (negate_argument)
{
// Setting the bit for this particular option to 0 as the option has
// been negated.
}
else
{
//Setting the bit for this particular option to 1.
}
// Copy back the final bits to the global_options_bits variable
global_options_bits= tmp_action;
}
else if (strcmp((char *) parsed_argument_without_no, (char *) OPTION2) == 0)
{
//Similar code for OPTION2
}
else if (strcmp((char *) parsed_argument_without_no, (char *) OPTION3) == 0)
{
//Similar code for OPTION3
}
//Now someone who wishes to check if a particular option was set or not can use the
// following type of code anywhere else in the code.
if(((global_options_bits & OPTION1_BITPOS) == OPTION1_BITPOS)
{
//Do processing for the case where OPTION1 was active.
}
else
{
//Do processing for the case where OPTION1 was NOT active.
}```
4. Those 0x000000.... are machine instructions.
5. Originally Posted by nathanpc
Those 0x000000.... are machine instructions.
it just integer numbers in hex notation... of course we could write the machine instructions in this notation as well as any over data stored in the computer...
6. It's just easier to tell what bits are set in 0x1ffff (i.e., the last 17 of them) then it would be with the decimal number 262143.
7. Sometimes those leading zeros are necessary. For example:
0xDCBA might otherwise get interpreted as 0xFFFFDCBA when you meant 0x0000DCBA. Of course you can also write 0xDCBAUL
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# Conditioning-which interpretation is correct?
I am very confused about the following:
Let us define $VaR_u(Y)$ as $\inf \{l:F_Y(l)\geq\ u \}$ where $F_Y$ is the CDF of the random variable Y.
Suppose $A$ and $X$ are independent random variables and suppose we have $VaR_u(AX)=5$. Furthermore, suppose we are interested in the following probability $P(AX>VaR_u(AX)|A=a)$. It seems as if there are two ways to deal with this probability, but surely there must be only one.
First we can take the stance that $VaR_u(AX)$ is a fixed quantity, namely 5. Thus we have $$P(aX>5)=P(aX>VaR_u(AX))$$
On the other hand one could argue that $VaR_u(AX)$ is dependent on the value of $A=a$ such that we have $$P(aX>VaR_u(aX))=1-u$$
I am inclined to believe that the first interpretation is the right one, but how does one 'choose'?
$VaR_u(Y)$ is the quantile function. It is a function of $u$ and of the true underlying distribution of $Y$. It does not depend on any particular value of $Y$ or any samples drawn from the true underlying distribution.
For example, if $u = \frac{1}{2}$, $VaR_u(Y)$ is the median of the underlying distribution of $Y$. If you draw samples from some distribution, it does not change the median of that distribution.
To avoid this confusion, you might consider using a different notation, such as $VaR(u, F) = \inf\{l:F(l)\geq u\}$.
• Thank you so much for clarifying the dependency $VaR_u(Y)$ has on $u$ and the distribution function of Y. This has been bothering me for two days. May 21, 2017 at 18:46
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English | Español
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# Compute $\int _0^\infty\frac{x \sin x}{1+x^2}dx$ with the residue theorem
Compute $\int _0^\infty\frac{x \sin x}{1+x^2}dx$ with the residue theorem
Ok so I have done a couple of these but I'm stuck on this one. I want to use
$$\int_0^\infty \frac{ze^{iz}}{1+z^2}dz$$
and then take the imaginary part of the answer at the one, that is how I have done it so far. But I have to use Jordans Lemma here, right? And I'm not really sure what to do when the lower limit is zero, I have only done it with $-\infty$. Any suggestions?
Consider the contour integral
$$\oint_C dz \frac{z \, e^{i z}}{1+z^2}$$
where $C$ is a semicircle of radius $R$ in the upper half plane. Then the contour integral is equal to
$$\int_{-R}^R dx \frac{x \, e^{i x}}{1+x^2} + i R \int_0^{\pi} d\theta \, e^{i \theta} \, \frac{R e^{i \theta} e^{i R e^{i \theta}}}{1+R^2 e^{i 2 \theta}}$$
Now we want to take the limit as $R \to \infty$ and show that the second integral, which represents the integral about the arc, goes to zero in this limit. We do this by showing that the magnitude of the integral goes to zero. Note that, by the Cauchy-Schwartz inequality,
$$R\left |\int_0^{\pi} d\theta \, e^{i \theta} \, \frac{R e^{i \theta} e^{i R e^{i \theta}}}{1+R^2 e^{i 2 \theta}} \right | \le R\int_0^{\pi} d\theta \, \left | e^{i \theta} \, \frac{R e^{i \theta} e^{i R e^{i \theta}}}{1+R^2 e^{i 2 \theta}}\right | \le \frac{R^2}{R^2-1}\int_0^{\pi} d\theta \, e^{-R \sin{\theta}}$$
Now we make use of the inequality
$$\sin{\theta} \ge \frac{2}{\pi} \theta \quad \forall\theta \in [0,\pi/2]$$
We use the symmetry of the integrand and conclude that the integral about the arc is bounded by
$$\frac{2 R^2}{R^2-1}\int_0^{\pi/2} d\theta \, e^{-2 R \theta/\pi} = \frac{\pi R}{R^2-1}$$
which indeed vanishes as $R \to \infty$.
By the residue theorem, the contour integral is also equal to $i 2 \pi$ times the residue at the pole $z=i$. The pole $z=-i$ is excluded because it is not contained within $C$. Thus,
$$\int_{-\infty}^{\infty} dx \frac{x \, e^{i x}}{1+x^2} = i 2 \pi \frac{i \,e^{-1}}{2 i} = i \frac{\pi}{e}$$
Take the imaginary part, take the positive half of the interval (even integrand), and you get the result from @Jack's solution.
$x\sin x$ is an even function, hence: $$\int_{0}^{+\infty}\frac{x\sin x}{1+x^2}\,dx = \frac{1}{2}\int_{-\infty}^{+\infty}\frac{x\sin x}{1+x^2}\,dx = \frac{1}{2}\cdot\text{Im}\int_{-\infty}^{+\infty}\frac{z e^{iz}}{1+z^2}\,dz$$ and the last integrand function has two simple poles at $z=\pm i$. By integrating along a rectangle countour enclosing $i$ in the upper half-plane and using the residue theorem we have: $$\int_{-\infty}^{+\infty}\frac{z e^{iz}}{1+z^2}\,dz = 2\pi i\cdot\text{Res}\left(\frac{z e^{iz}}{1+z^2},z=i\right) = 2\pi i\cdot \frac{i e^{i^2}}{2i}$$ hence: $$\int_{0}^{+\infty}\frac{x\sin x}{1+x^2}\,dx = \color{red}{\frac{\pi}{2e}}.$$ It is interesting to point out that since $\mathcal{L}(\sin x)=\frac{1}{1+s^2}$ and $\mathcal{L}^{-1}\left(\frac{x}{1+x^2}\right)=\cos(s)$, we have: $$\int_{0}^{+\infty}\frac{x\sin x}{1+x^2}\,dx = \int_{0}^{+\infty}\frac{\cos(s)}{1+s^2}\,ds.$$
• Oh i missed that with the even integrand! The rest of this you wrote I can do but my professor demands that we show that the upper half circle tends to zero as we go to infinity, that for me is always the tricky part. – user269620 Oct 22 '15 at 19:05
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I crave the unexplained, the unusual, and the downright counterintuitive sorts of problems that make you really think about what's going on in a circuit. This issue, I'll tell you about a driver whose output gets bigger when loaded.
I just got a new differential probe.
Whether the probe accurately reports the voltages to which it is exposed, I do not doubt. I'm willing to assume the probe measures things the way its manufacturer says, unless it's broken. What I must always check, though, is the degree to which the probe loads down or distorts the signals in my system when applying the probe.
A trivial setup will suffice for this measurement (Figure 1). Take any driver and run its output directly into your scope with coaxial cables (no probes). Then, while observing the coaxial output, touch a probe onto the output pins of the driver. See what happens.
My setup incorporates an LVDS-style differential driver. The driver feeds a short (1-in.) pair of microstrip traces. The traces go out through SMA connectors, then through 24 inches of RG316 coaxial cables to a scope. At the scope, I connect DC blocking capacitors ahead of the 50-ohm scope inputs. Then I display the resulting differential signal.
When doing this test, the expected result depends on the relation between the input impedance of the probe and the impedance of the circuit under test. Think of the circuit under test as a voltage generator with output voltage v[source] and output impedance Z1. When loaded with a probe having impedance, say, Z2, the expected value of the measured signal should be, according to the resistor-divider theorem:
VMEASURED = VSOURCE × (Z2 / (Z2+Z1))
I'm using a LeCroy D600A-AT 7.5 GHz differential probe with a differential DC input impedance of Z2=4000 ohms.
My 100-ohm differential circuit, terminated at both ends, has an effective differential driving-point impedance of 50 ohms. That happens because, from the perspective of the probe, it "sees" a differential impedance of 100 ohms to the left, at the driver, in parallel with another differential impedance of 100 ohms to the right, at the scope. Two 100-ohm loads in parallel together make a 50-ohm differential driving point impedance.
Given those numbers, I expect this attenuation factor when connecting the probe:
AEXPECTED = (4000 / (4000+50)) = 0.988
If my calculations are correct, the expected loading effect of the probe should shrink the measured signal by only 1.2 percent, a tiny change. Let's try it.
Figure 2 plots the results when measuring a National Semiconductor DS25BR100 driver. This high-powered driver is designed for long-distance transmission applications. It incorporates transmit pre-emphasis, a feature that is turned off for this test. The figure shows the results of experiments conducted with zero, one and two LeCroy probes connected to the output terminals of the device. All measurements are made using the coaxial cable setup in Figure 1. The illustration superimposes both positive and negative signal edges.
Obviously, the probes affect the size of the signal. That's normal. But what totally, completely surprised me in this example is the direction of the change. Look closely. As you add probes the signal grows. Compared to the unloaded amplitude, one probe enlarges the signal by 6.5%. With two, it gets 13% bigger.
What?? I can't believe it. I re-tried the experiment several times under different conditions. I double-checked the stored waveform file names to see if I had them reversed. My assistant looked over the setup. Everything appears right. As far as I can tell, this signal actually GROWS when loaded.
How can that be? The explanation involves a subtle interaction between the common-mode loading of the probe and the differential-mode gain of the driver. There is not a problem with either circuit; it's just how they happen, in this circumstance, to work together.
The DS25BR100 employs a feedback control loop to stabilize its common-mode output voltage (a good idea). The feedback loop reacts to changes in common-mode loading.
What changes might you expect? Well, let's first consider a simple 100-ohm differential load. If you tie a 100-ohm resistor across the output terminals of the driver it provides a 100-ohm termination for differential signals, but draws zero common-mode current. That is, if you exercise such a load with a common-mode signal (same on both sides) no current flows through the resistor -- it's as if the resistor weren't there. The simple 100-ohm load draws no common-mode current.
Similarly, the scope in Figure 1, configured with DC-blocking capacitors, draws zero common-mode current at DC.
The LeCroy differential probe is different. It presents a load of 2K ohms to ground on each side. From a common-mode perspective, that's a 1K load to ground. This load draws a small amount (1.2 mA) of common-mode current from the driver. As differential probes go, that is pretty good. Some high-speed probes draw much more. I think the common-mode current drawn by this differential probe is causing the waveform amplitude artifacts in figure 2.
To validate my thinking, I measured the common-mode output voltage from the driver when the probe was removed, using a high-impedance digital voltmeter, and again with the probe present. The DC droop under that condition amounted to only about 1 mV, indicating an effective common-mode output impedance from the driver of approximately 1 ohm. A practical IC driver would never obtain such a low output impedance without feedback regulation. So, I conclude that the DS25BR100 incorporates an internal feedback loop designed to regulate the common-mode output voltage. (Lee Sledjeski at National Semiconductor confirms my suspicions).
In practice, when you apply a differential probe to the driver, the feedback loop inside the driver raises the common-mode gain to make up for the DC droop caused by the common-mode loading of the probe. As a side effect, the act of raising the common-mode gain also raises the differential-mode gain, causing the signal growth reported here.
To check my assumptions, I loaded the DS25BR100 with a 2.2K-ohm passive metal-film resistor on each side to ground, drawing 1.2 mA of DC common-mode current. Under that condition, the outputs GREW. Then I re-connected the same resistors, this time going from the digital outputs to VCC instead of ground. That arrangement sources common-mode current INTO the driver, instead of taking it out. Guess what—under that condition the outputs SHRANK.
The DS25BR100 is the first case I can recall of a transceiver whose outputs get bigger when loaded. Not all LVDS outputs do this. Figure 2 shows the outputs of a TI DL100-44T. I captured these output waveforms with zero and one probes present (two wouldn't fit on the circuit). This output, when loaded, shrinks as expected by 1.2 percent.
I do not mean to imply that one transceiver is better than the other; only that they are different, and that the difference affects your voltage margin budget. If you are trying to measure output levels with any accuracy it's worth knowing that the DS25BR100 outputs can GROW when probed. That affects your voltage margin budget calculations, and that's worth knowing about.
Best Regards,
Dr. Howard Johnson
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## Cycloidal Gear Form and Clock Wheels
Hello all, this is my first post. My name is Chris, I teach Chemistry 11-18 for a profession but I am an hobby engineer and have a huge interest in clocks and watches - horology.
The gears in clocks are called wheels to a horologist and the profile of the tooth is of a cycloidal form. My understanding of this is that the gear ratio is based on two circles linking together, this is the pitch circle diameter, PCD. To allow the wheels to turn, (because friction against two smooth sides would not be enough) you add the teeth, above the PCD to form the addendum of the tooth and to allow them to mesh with enough depth, you also take something away from the PCD to form the dedendum. The addendum and dedendum form the tooth but it is actually the PCD of both gears which give you the correct gear ratio.
Using a table like the one below, or using an online calculator such as the ones in the links, I can then calculate all the tooth profile details.
From this info, I can then use CAD or solidworks and draw my wheel:
The example below is of a: 0.8Module, 60 teeth, 48mm pitch circle diameter, 50.15484 mm tip diameter, 2.51968mm tooth pitch, 1.259mm tooth width and gap between teeth, 1.0769mm addendum, 1.2598mm dedendum.
My question is that I want to know more about the cycloidal form and actually be able to draw it from points plotted rather than the innacurate form from approximated calculations.
If you looks at the diagram below, the tooth profile is being formed from the drawing on many circles based on the PCD, I believe these are hypocycloidal and epicycloidal curves which generate intersects to plot the tooth:
If the link doesnt work - google "Drawing Cycloidal Curves Chest of Books" : Cycloidal Gears
Basically I want to really understand how this cycloidal curve is formed and using pen and pencil, how I would go about drawing a cyloidal gear like the one I have done in solidworks (using the same dimensions) but drawing the epicycloidal and hypocycloidal curves to "form" the teeth.
There is info here and discussion of the maths, but it loses me very quite quickly and I just cant seem to then transfer this info to actually go and use it to draw a wheel....
Designing Cycloidal Gears
I hope someone could help me out and offer any advice. If this is in the wrong forum, please feel free to move it!
Chris
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+0
# question
0
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Jacob bought new basketball shoes. If the shoes cost \$70 and sales tax is 4%, how much was the tax on the shoes?
Guest Mar 1, 2017
#1
0
Jacob bought new basketball shoes. If the shoes cost \$70 and sales tax is 4%, how much was the tax on the shoes?
\$70 x 4/100 =\$2.80 sales tax.
Guest Mar 1, 2017
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# Present Value and Net Present Value of a Financial Security Homework Help
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# Present Value and Net Present Value of a Financial Security Assignment Help
The Difference between Present Value and Net Present Value of a Financial Security
There are two aspects to financial investment, and one is present value and the other is net present value.
If one person is getting \$ 1000 now, it means that he or she can spend it now as the value of that money will be higher now. Assume that the person decides to spend the money after five years and the value will fall. Present value calculation is an important thing, and this will help in finding out the net present value that a bond should have and this will help a consumer to understand whether they should seek a 0% on financing. The term future value has more importance here. There are high chances that Present value and net present value of a financial security homework help is needed for students as they may find it difficult to solve complex interest calculations
Net present value
This is a complex mathematical formulation that is done in capital budgeting, and this would require a lot of mathematical compounding.
Assume that a person says that he needs \$500 and he would return \$575 next year to you and you should think whether this is a worthwhile option. The first step will be to identify the present value that stands at \$500 and then divide the \$570 with 1.10 that is nearest mathematical value, and the result will be \$518.80, or in other words, you will get 18.80 extra. Assume that the federal rate in the banks that you get if you invest \$500 for one year, i.e. 10%, then the investment that you give to your friend is worthwhile. A student should understand this complex calculations else should seek Present value and net present value of a financial security homework help.
The problems in assignment from a student perspective
A student will need to understand the basics well if they have to successfully do the assignment. They need to understand the complex mathematical formulas also.
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# Question:animation/ plot of surface levels in 3D
## Question:animation/ plot of surface levels in 3D
Maple
Dear Maple helpers. Here a toy-version of my huge problems.
I want to do following:
1). Assuming (0 <a=1/2, b=1/2, c=1/2)
I want to plot (the surface of)
f(x,y,z) = x^a * y^b * z^c in the space (x,y,z) for
f(x,y,z)=50, 100 and 150, and with x=0..40 , y=0..40, z=0..40
I tried with plot3d but it is not possible. How could I do with Maple 14?
2. Assuming (0 <a<1, 0<b<1, 0<c<1), it would be possible to represent a plot in 3D for different values of (a,b,c) using the option of ANIMATION (or something similar)?
Kind regards,
Jean-Jacques
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Babylon Numbers
Can you make a hypothesis to explain these ancient numbers?
Ishango Bone
Can you decode the mysterious markings on this ancient bone tool?
Pattern Recognition
When does a pattern start to exhibit structure? Can you crack the code used by the computer?
Black Box
Why do this problem?
This problem is best suited to tenacious problem solvers who will relish the intellectual challenge in making sense of the numbers. Although determining the function rules will be very challenging, on the technical side the mathematics required to verify the rules is elementary (KS3). Interestingly, the ideas raised will actually have relevance in university number theory and, as such, could provide important insights for those going to university to study mathematics.
Possible approach
This challenge will require experimentation to gain sense of the numbers involved as, after a few trials, it will likely seem that there is little obvious pattern in the numbers produced.
Part of the fun of this problem is to be faced with a straight 'black box' and being left to decide how it works, so you may wish to hold back from offering hints to students.
Once a feel for the numbers is found, students might start to make conjectures concerning the function rules which can then easily be tested: new examples will either add weight to the conjectures or cause them to be rejected. Students might need a spreadsheet or calculator to help in this regard. Alternatively, they could use the Wolfram Alpha computational knowledge engine http://www.wolframalpha.com/.
Once students have solved the problem they should write up their rules clearly to allow others to independently verify them. Of course, it is not possible to prove that the Black Box follows certain rules but, once discovered, evidence will soon build up in their favour.
Finally, we don't suggest using this task with students who are unlikely to enjoy this style of working: reserve it for very keen students who like individual challenges.
Key questions
What sorts of numbers does the problem involve?
After a few trials, are you getting a feel for the sorts of numbers produced?
Do you have any rough ideas for how the numbers are related?
Possible extension
Students could read about the ideas raised. The links will be posted following the publication month for this problem
Possible support
Support questions will be posted following the publication month for this problem.
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# Vocabulary space Positive space Value Negative space Light source
## Presentation on theme: "Vocabulary space Positive space Value Negative space Light source"— Presentation transcript:
Vocabulary space Positive space Value Negative space Light source
Overlap In front Behind Foreshortening perspective Value Light source Highlights Shadows Foreground Middle ground Background
The Space around the world
What is it? Space is all around you! Space is Area. The Space around the world is called Outer Space!
space space - An element of art that refers to the distance or area between, around, above, below, or within things. It can be described as two-dimensional or three-dimensional; as flat, as positive or negative espacio
Positive space refers to the object
Negative space is the area around the object espacio positivo y negativo
Example: The blue boxes are the positive space The black space is the negative space
The area around and between the boxes is negative space The boxes are the object – they are positive space
Example: The cat is the positive space The black area is the ___________
Example: The red boxes are the _________ The blue space is the ___________
Use 2 pieces of colored paper to create an example of positive and negative space. Draw a shape on a piece of paper. Be creative! Try something interesting! Cut out a shape and save both pieces of paper. Glue the shape to a new piece of paper Glue the remaining paper to a new piece of paper And label positive and negative space
space foreground - is the space closest to the viewer, usually the bottom of the artwork middle ground - is the space between the background and foreground, usually in the middle of the artwork Background is the space farthest from the viewer, usually the top of the artwork
space Background middle ground foreground
Background is the space farthest away, at the top middle ground - is the space between the background and foreground foreground - is the space closest, at the bottom of the artwork
space Background is the space farthest away
What if the object was in the sky? Background middle ground foreground What if the object was in the sky? Background is the space farthest away middle ground - is the space between the back and the front foreground - is the space closest
How do artists show space?
Overlap / In front / Behind Size of objects Value / shading Light source Highlights/detail Shadows/detail Foreground Middle ground Background
How do artists show space?
Overlap / In front / Behind Size of objects Value / shading Light source Highlights/detail Shadows/detail Foreground Middle ground Background
How do artists show space?
Overlap / In front / Behind Size of objects Value / shading Light source Highlights/detail Shadows/detail Foreground Middle ground Background
How do artists show space?
Overlap / In front / Behind Size of objects Value / shading Light source Highlights/detail Shadows/detail Foreground Middle ground Background
Draw a picture using all of the ways artists show space!
Create a masterpiece! Draw a picture using all of the ways artists show space! Overlap In front / Behind Size of objects Horizon line Value / shading Light source Highlights/detail Shadows/detail Foreground Middle ground Background Size of objects
Design Assignment Make a sketch of your favorite activity in your journal. This will be a study for your drawing. Examples are: playing soccer, cooking, going to the park, receiving gifts, visiting a family member, etc. The drawing must show “space”, value, foreground, background, middle ground, overlapping, light source, texture, etc.
6th grade assignment Using these exercises: Contour hand sketches
Value scale Invented and frottage texture Positive and Negative example You will create a masterpiece called “Give Me A Hand” Chose the best composition of hands and recreate it on a new piece of paper. Add texture to the shapes on the hands and values to the shapes in the background.
Example of “Give me a hand”
By Kay Dixon
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# LeetCode – Maximum Subarray (Java)
Find the contiguous subarray within an array (containing at least one number) which has the largest sum.
For example, given the array `[−2,1,−3,4,−1,2,1,−5,4]`, the contiguous subarray `[4,−1,2,1]` has the largest sum = 6.
1. Dynamic Programming Solution
The changing condition for dynamic programming is "We should ignore the sum of the previous n-1 elements if nth element is greater than the sum."
```public class Solution { public int maxSubArray(int[] A) { int max = A[0]; int[] sum = new int[A.length]; sum[0] = A[0]; for (int i = 1; i < A.length; i++) { sum[i] = Math.max(A[i], sum[i - 1] + A[i]); max = Math.max(max, sum[i]); } return max; } }```
1. Simple Solution
Mehdi provided the following solution in his comment. The time space is optimized to be O(1).
``` public int maxSubArray(int[] A) { int newsum=A[0]; int max=A[0]; for(int i=1;i<A.length;i++){ newsum=Math.max(newsum+A[i],A[i]); max= Math.max(max, newsum); } return max; }```
This problem is asked by Palantir.
Category >> Algorithms
If you want someone to read your code, please put the code inside <pre><code> and </code></pre> tags. For example:
```<pre><code>
String foo = "bar";
</code></pre>
```
• Norman Amos
This seems simpler:
``` public static int maxSubArray(int[] a) { int r = 0; int l = 0; int rsum = 0; int lsum = 0; int rmax = Integer.MIN_VALUE; int lmax = Integer.MIN_VALUE; for (int i = 0; i rmax) { rmax = rsum; r = i; } if (lsum > lmax) { lmax = lsum; l = a.length - i - 1; } } int rtn = rmax + lmax - rsum; System.out.println("a[" + l + "] + ... + a[" + r + "] = " + rtn); return rtn; } ```
• Avinash Ujjwal
/* Here is the solution using kadane’s algorithm but it only works when there is at least one positive element in the given array. */
public static int maxSubArray(int[] array) {
int maxSum = 0, newSum = 0;
for(int i = 0; i 0) {
newSum = newSum + array[i];
} else {
newSum = 0;
}
maxSum = Math.max(maxSum, newSum);
}
return maxSum;
}
Guess it fails for lot of cases, For Ex input Array:
5, -4, 9, -3500, 25, -3, 12, 5900, 15, -23, 18, 14, 11, 93, 12, 34, 56, -125, 66, -255, 95, 68, 4558, -2, 203
• Eric
only fails with all negative numbers
Extending Mehdi’s solution, you can track the subarray (indices) using 2 more variables startIndex and endIndex (initialize before the for loop and add below code inside the for loop at the end):
if (max == A[i])
{
startIndex = i;
}
else if (max == newsum)
{
endIndex = i;
}
• Allan Yin
I will still work. plz test yourself.
• Dexter
Solution with array indexes:
public static void maxArray(int[] arr){
int prevStart = -1;
int currentStart = prevStart;
int maxHere = 0;
int maxSoFar = 0;
int end = -1;
for(int i = 0; i< arr.length; i++){
if(arr[i] < 0 && maxSoFar == 0){
continue;
}
maxHere += arr[i];
if(currentStart == -1){
currentStart = i;
}
if(maxHere maxSoFar){
maxSoFar = maxHere;
prevStart = currentStart;
end = i;
}
}
System.out.println(“Max Sum:” + maxSoFar);
System.out.println(“Start: ” + prevStart);
System.out.println(“End: ” + end);
}
• Satish
Copy pasted from eclipse ide.I m not sure why it appears this way. Sorry for that
• Satish
Check this out O(1) with position:
public static int maxSubArray(int[] A) {
int newsum=A[0];
int max=A[0];
int j = 0;
int i_max = 0;
for(int i=1; i<A.length; i++){
newsum += A[i];
if (newsum max) {
max = newsum;
i_max = i;
}
}
System.out.println(j + “,” + i_max + ” = ” + max);
return max;
}
• Sam
The naive solution should work on {-1, -2, 5, -2}.
could you double check plz?
• Stephen Boesch
The DP solution is not correct. The condition should be “we ignore the previous n-1 sum if it were less than or equal to 0”.
• Dhwanit Shah
Above solution is not correct. Pls find the rectified complete solution below:
#include
using namespace std;
int max(int a, int b)
{
if(a>b)
return a;
return b;
}
int maxSubArray(int* A, int len) {
int sum = 0;
int maxSum = -1000000; //some very large negative number
for (int i = 0; i < len; i++) {
if (sum < 0){
sum = A[i]; // change
}
else
{
sum += A[i];
}
maxSum = max(maxSum, sum); // change
}
return maxSum;
}
int main() {
int a[] = {-1,-2,3,4,-5,6};
cout<<maxSubArray(a, 6);
return 0;
}
• swm8023
The first solution is correct. It’s same to the last solution
• ryanlr
Put code between <pre> and </pre>
• garukun
Thanks! I was wondering how you guys created code like posting.
• ryanlr
This solution is wrong, I tried it using { -1, -2, 5, -2 }.
• ryanlr
I think your solution is also correct.
• ryanlr
I have added this solution, thanks!
• garukun
Will something like this work? We observe that the if the element about to be added to create the next summation is already greater than the previous summation, then we will ignore the previous summation and start a new maxSum from the current index. The code would look something like:
[code]
public int maxSum(int[] arr) {
int maxSum = Integer.MIN_VALUE;
for (int i = 0, sumSoFar = -1, sum; i sum) {
sumSoFar = arr[i];
} else {
sumSoFar = sum;
}
maxSum = Math.max(maxSum, sumSoFar);
}
return maxSum;
}
// Test input and iterations
[−2,1,−3,4,−1,2,1,−5,4]
i = 0; sum = -3, sumSoFar = -2, maxSum = -2 ;
i = 1; sum = -1, sumSoFar = -1, maxSum = -1 ;
i = 2; sum = -4, sumSoFar = -3, maxSum = -3 ;
i = 3; sum = 1, sumSoFar = 4, maxSum = 4 ;
i = 4; sum = 3, sumSoFar = 3, maxSum = 3 ;
i = 5; sum = 5, sumSoFar = 5, maxSum = 5 ;
i = 6; sum = 6, sumSoFar = 6, maxSum = 6 ;
i = 7; sum = 1, sumSoFar = 1, maxSum = 6 ;
i = 8; sum = 5, sumSoFar = 5, maxSum = 6 ;
[/code]
• Kevin
An array with all negative number will break naive solution, since “(containing at least one number) “
• Jiaming
The “Naive Solution” works, nothing’s wrong with that line, I think it’s not working anymore if you change that row.
• Shaowei Ling
Based on the second method(DP style), I write this for printing the sub array in details. I tested the example in the question and it works.
public int maxSubArray(int[] A) {
int max = A[0];
int[] sum = new int[A.length];
sum[0] = A[0];
int begin = 0;
int end = 0;
if(sum[i] > max) {
max = sum[i];
begin = newBegin;
end = newEnd;
}
}
// print the subArray
for(int i = begin; i <= end; i++) {
System.out.print(A[i]+ " ");
}
System.out.println();
return max;
}
• play_boy
Native Solution is wrong, but you can change some row, and it will be ok!
like this:
public static int maxSubArray(int[] A) {
int sum = 0;
int maxSum = Integer.MIN_VALUE;
for (int i = 0; i < A.length; i++) {
sum += A[i];
if (sum < 0)
sum = A[i]; // change
maxSum = Math.max(maxSum, sum); // change
}
return maxSum;
}
• Mehdi
you can also do it with O(1) space complexity instead of O(n):
``` public int maxSubArray(int[] A) {
int newsum=A[0];
int max=A[0];
for(int i=1;i<A.length;i++){
newsum=Math.max(newsum+A[i],A[i]);
max= Math.max(max, newsum);
}
return max;
}
```
• Yat
Hi, what about the actual subarray that should be returned? Only the max sum is returned in the two solutions above.
• 曹超
because it’s a wrong method
• 曹超
原理就是错的
• Darren
Why does the so-called “Naive Solution” not work? I think the algorithm is correct, and it passes the OJ as well.
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• Calculation resulting from checkbox selection in p...
# Calculation resulting from checkbox selection in pdf
Community Beginner ,
Sep 23, 2017 Sep 23, 2017
Copied
I'm probably way out of my league here, but here's what I need to happen:
I have a fillable pdf with several checkboxes that would result in an amount being added to a total field (Ex: Full page ad / Exhibitor fee is \$500 / Non exhibitor fee is \$750), so the user selects two items (or none). If they check "Full page ad" checkbox, then they select whether or not they are an exhibitor, then the amount (\$500) would go to the total field = \$500. And several more optional add choices, etc.
So, I guess my question(s) are:
If a checkbox is selected, can a number populate in a separate field? And if so, how?
I am a javascript beginner, have only completed the calculations that were simple.
TOPICS
PDF forms
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Adobe Community Professional , Sep 23, 2017 Sep 23, 2017
Then use this:
event.value = 0;
if (this.getField("fullpageexhibitor").value!="Off") event.value=500;
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25 Replies 25
Sep 23, 2017 Sep 23, 2017
Copied
Yes, that's easily done. For example, let's say you have a check-box called "Checkbox1" and you want to populate a field called "Text1" with 500 if it's checked, and zero otherwise.
Use this code as the custom calculation script of Text1:
event.value = (this.getField("Checkbox1").value=="Off") ? 0 : 500;
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Community Beginner ,
Sep 23, 2017 Sep 23, 2017
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WOW! Thank you @try67ā may I ask how to complete it if there are two choices? So if one box is checked, it's 500, and if the other box is checked it's 700?
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Community Beginner ,
Sep 23, 2017 Sep 23, 2017
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I tried using this:
event.value = (this.getField("fullpageexhibitor").value=="Off") ? 0 : 500;
event.value = (this.getField("fullpageadregular").value=="Off") ? 0 : 700;
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Sep 23, 2017 Sep 23, 2017
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No, that won't work... What if both boxes are ticked?
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Community Beginner ,
Sep 23, 2017 Sep 23, 2017
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They can only choose one box (either 500 or 700)
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Sep 23, 2017 Sep 23, 2017
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Then use this:
event.value = 0;
if (this.getField("fullpageexhibitor").value!="Off") event.value=500;
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Community Beginner ,
Sep 23, 2017 Sep 23, 2017
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YES!!!! You're a genius and a scholar and my hero!!!!!!! Thank you try67ā !!!!!
May I ask for another scenario? They also have an input field where they can enter certain booth numbers that are premium prices, so if they enter 301, the price is \$1,000, if they enter 302 the price is \$1,200 -- how would you do that or can that be done? In this case, the user enters a specific number
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Sep 23, 2017 Sep 23, 2017
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You can use something like this:
if (this.getField("boothnumber").valueAsString=="301") event.value=1000;
if (this.getField("boothnumber").valueAsString=="302") event.value=1200;
etc.
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Community Beginner ,
Sep 23, 2017 Sep 23, 2017
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You're my BEST FRIEND! Thank you try67ā!!! That worked, BUT the user can actually enter more than one booth number (maximum 3) in that one field (separated by a comma) -- is it possible to add the amounts?
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Sep 23, 2017 Sep 23, 2017
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Yes, but this is getting more complicated...
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Sep 23, 2017 Sep 23, 2017
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If you wish I can create this script for you, for a small fee. You can send me the full specs to try6767 at gmail.com and we could discuss the details.
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New Here ,
Feb 16, 2022 Feb 16, 2022
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I have several boxes and they can check more than 1, how do I sum up the value of their choices? Thanks.
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Feb 17, 2022 Feb 17, 2022
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There are many different ways to add-up checkbox selections. The exact solution depends on the exact details.
Also, you might start by learning a bit more about scripting PDF checkboxes:
Thom Parker - Software Developer at PDFScripting
Use the Acrobat JavaScript Reference early and often
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New Here ,
Feb 17, 2022 Feb 17, 2022
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Thanks, I created radio checkboxes (for options where they can only choose 1) and regular checkboxes and assigned different values for each checkbox. I was then able to sum up the values based on their choices.
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New Here ,
Jul 18, 2022 Jul 18, 2022
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LATEST
Thanks for this, I just have one question, what do i put in when it says "Line Number"? I tried Textbox29 which is where the total will go, nothing happens
Also I need to total a number of text boxes, is this possible?
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Community Beginner ,
Jul 07, 2019 Jul 07, 2019
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I'm trying to do a similar thing but I'm having trouble adapting the following code to my situation...
event.value = 0;
if (this.getField("CheckBox1#0").value!="Off") event.value=120;
else if (this.getField("CheckBox1#1").value!="Off") event.value=450;
I have a suspicion it's because I'm using multiple checkboxes that have the same name but with different export values (to simulate radio button operation) and the checkbox name has a # in it.
Am I able to replace the checkbox name "CheckBox1#0" with the export value instead?
Also, to make matters worse I have 5 checkboxes that can all be different values (although 2 are the same) and I need the textbox to display the value of the selected checkbox.
I assume it's something like this (although I'm sure the syntax is incorrect)...
event.value = 0;
if (this.getField("CheckBox1#0").value!="Off") event.value=120;
else if (this.getField("CheckBox1#1").value!="Off") event.value=450;
else if (this.getField("CheckBox1#2").value!="Off") event.value=200;
else if (this.getField("CheckBox1#3").value!="Off") event.value=300;
else if (this.getField("CheckBox1#4").value!="Off") event.value=450;
Any help would be appreciated.
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Jul 07, 2019 Jul 07, 2019
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What are the actual export values of these check-box fields?
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Community Beginner ,
Jul 07, 2019 Jul 07, 2019
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Assessment, Capstone, Low, Medium, High
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Jul 07, 2019 Jul 07, 2019
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Then you should use those values in your code, and drop the "#1", "#2", etc. notation. Something like this:
`event.value = 0;if (this.getField("CheckBox1").valueAsString=="Assessment") event.value=120;else if (this.getField("CheckBox1").valueAsString=="Capstone") event.value=450;// etc.else event.value = "";`
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Community Beginner ,
Jul 07, 2019 Jul 07, 2019
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Like magic! Thanks very much.
I had tried that previously but I didn't use the "==". Just one "=".
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Jul 07, 2019 Jul 07, 2019
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That's a common mistake. That assigns one value to another, not compare them.
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New Here ,
Sep 21, 2019 Sep 21, 2019
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Hi there
I've been searching the community and this thread is the closest I found regarding what I'm trying to achieve. I am very rusty with my script knowledge as I haven't practiced it for years; I'm practically beginner level at this point. I have a similar situation where I have a list of services (with checkboxes) that have the same pricing; however, I do want multiple selections added up to a total.
The form allows the client to choose from 3 Service Package deals (Packages 1 to 3). Currently I have these as radio buttons where they can only select one. For the sake of math I'll keep the pricing simple: Packages 1 to 3 are \$5000, \$3000, and \$1000 respectively. Each service package offers discounts on additional services. Package1 offers \$1000 discount, Package2 \$500, and Package3 no discount. The group of radio buttons is named "Group1" and their button choices are 1, 2, & 3 respectively.
There are 5 additional services, Services A to E. The client filling up the form can choose to select as many services, including none. Each service costs \$500. However, depending on the Package deal they selected earlier in the form (Packages 1 to 3), they get 2 (\$1000) or 1 (\$500) services discounted. I would like to autofill the text field named "TotalExtras" so the client would instantly know how much extra on top of the base Package they have chosen.
First Example:
Client selects Package2 for \$3000 (which will discount \$500 from total extras). The client continues to next section and selects three additional services, say A, B & D. So the three choices should total \$1500 (\$500 each). But because they have chosen Package2, the autofill total in the TotalExtras text field should read "\$1000" (\$1500 additional services minus \$500 Package2 discount).
Second Example:
Client selects Package1 for \$5000 (which will discount \$1000 from total extras). The client continues to next section and selects only one additional service, they choose C. The single choice should total \$500. But because they have chosen Package1, the autofill total in the TotalExtras text field should read "\$0" (\$500 additional services minus \$1000 Package1 discount). The total for this TotalExtras field should never be negative and there is no option to go lower than the base price of the chosen Package.
Is this posssible? Or am I asking for too much? Also, am I supposed to have all the checkboxes under one name? They all currently have separate names.
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New Here ,
Jun 07, 2020 Jun 07, 2020
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I have a similar issue... this is the code im trying to use, Im trying to get a check box in adobe to (When checked) to subtract "TTLCalculated" And use that value in the next part of my code, wanted to know if what im doing wrong, im a newbie to javascript, would really appreciate the help!
var DA = Number(this.getField("Downpayment*").valueAsString);
var DA2 = Number(this.getField("TTLCalculated").valueAsString);
event.value = (DA-DA2);
if (this.getField("TTLCalculated").value!="Off") event.value=(DA-DA2);
else if (this.getField("TTLCalculated").value!="Off") event.value=0;
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Jun 08, 2020 Jun 08, 2020
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YOu have exactly the same condition in both parts of the if block. Remove the second if. Just use an else.
Thom Parker - Software Developer at PDFScripting
Use the Acrobat JavaScript Reference early and often
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https://assignmentresearchwriter.com/enmgt-571-managerial-statistics-mathematics-homework-help/
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# Enmgt 571 managerial statistics | Mathematics homework help
ENMGT 571
Managerial Statistics I
Get a plagiarism free copy of this essay from our experts
Enmgt 571 managerial statistics | Mathematics homework help
For as low as \$13/Page
1.
Exam I
Solve the following problems:
I. A computer rating service is commissioned to rank 10 different brands of flat
screen LCD monitors. The rating service is to rank the top three brands in
order, 1 through 3 (12 points)
a) In how many different ways can the computer rating service arrive
at the final ranking?
b) If the rating service can distinguish no difference among the
brands, and therefore arrives at the final ranking by chance, what
is the probability that one particular brand (call it Company Z’s
brand) is ranked first?
c) Suppose that the computer rating service is to choose the top
three from the group of 10, but is not to rank the three? In how
many different ways can the rating service choose the three to be
designated “top of the line” flat screen LCD monitors
d) Consider the same information as part c and assume the rating
service makes its choice by chance. Say there is a company
(company X) that has two brands in the group of 10. What is the
probability that one of company’s X brands is selected in the top
three?
II. An experimenter is studying the effects of temperature, pressure and type of
catalyst on yield from a certain chemical reaction. Three different
temperature, four different pressures and five different catalysts are under
consideration: (12 points)
a) If any particular experimental run involves the use of a single
temperature, pressure and catalyst, how many experimental runs
are possible?
b) How many experimental runs are there that involve use of the
lowest temperature and two lowest pressures along with the five
catalyst?
c) Suppose that five different experimental runs are to be made on
the first day of experimentation. If the five are randomly selected
from among all the possibilities, so that any group of five has the
same probability of selection, what is the probability that a
different catalyst is used on each run?
ENMGT 571
Managerial Statistics I
III.
Exam I
A mathematics professor wishes to schedule an appointment with each of
her eight teaching assistants, four men and four women, to discuss her
calculus course. Suppose all possible orderings of appointments are equally
likely to be selected. (15 points)
a) What is the probability that at least one female assistant is among
the first three of whom the professor meets?
b) What is the probability that after the first five appointments, she has
met with all female assistants?
IV.
A real estate agent has 8 master keys to open several new homes. Only 1
master key will open any given house. If 40% of these homes are usually
left unlocked, (10 Points)
a) what is the probability that the real estate agent can get into a
specific home if the agent selects 3 master keys at random before
leaving the office?
V.
The US Army Corps of Engineers’ study on the DDT contamination of fish
in the Tennessee River (Alabama) revealed the following: 52 percent of the
fish are found between 275 and 300 miles, 39% are found between 305 and
325 miles and 9% are found between 330 and 350 miles (10 points)
a) Given that a contaminated fish is found in a certain distance
upstream, the probability that it is a channel catfish (CC) is
determined from the data as P(CC/275-300)=.775,
P(CC/305-325) = .77 and P(CC/330-350) =.86. If a contaminated
catfish is captured from the Tennessee River, what is the
probability that it was captured 275-300 miles upstream?
b) What is the probability of being a channel catfish (CC)?
2.
A box in a certain supply room contains eleven 40-W light bulbs and nine 75-W
bulbs. Suppose that three bulbs are randomly selected(15 pts)
a. What is the probability that exactly two of the of the selected bulbs are rated 75W?
b. What is the probability that at least one 40W bulb is selected?
c. Given a 40-W bulb is selected on the first draw, what is the probability of
selecting exactly one more 40-W?
ENMGT 571
Managerial Statistics I
3.
Exam I
A market study taken at a local sporting goods store showed that of 30 people
questioned, 18 owned tents, 15 owned sleeping bags, 14 owned camping stoves,
6 owned both tents and camping stoves, and 10 owned both sleeping bags and
camping stoves: (15 pts)
a. What is the probability of owning a tent, owning a camping stove, owning
a sleeping bag, owning both a tent and a camping stove, and owning both
a sleeping bag and a camping stove?
b. What is the probability of owning a tent or a camping stove?
c. What is the probability of owning neither a camping stove or a tent
d. Given a person questioned owns a tent, what is the probability he also
owns a camping stove?
e. Are the events of owning a tent and owning a camping stove mutually
exclusive? Explain briefly?
f. Are the events of owning a sleeping bag and owning a camping stove
independent? Explain Briefly?
g. If four people questioned owned a tent, a sleeping bag, and a camping
stove, then up to how many can own only a camping stove?
4.
A regional telephone company operates three identical relay stations at
different locations. During a one-year period, the number of malfunctions
reported by each station and the causes are shown below. (11 pts)
______Stations:
Problems with electricity supplied
Computer malfunction
Malfunctioning electrical equipment
Caused by other human errors
A
6
4
5
7
B
1
4
4
9
C___
1
2
1
5
a. What is the probability of a Computer Malfunction?
b. What is the probability that station A will experience malfunction
with electrical equipment?
c. Suppose (given) that a malfunction was reported and it was
found to be caused by other human errors. What is the
probability that it came from station C?
d. Given we have a problem at station B, What is the probability
that it is a problem other than a computer malfunction?
Level of Detail: Show all work
Other Requirements: Please answer questions and show all work for 1 (I, II, III, IV, V) and 2 (a, b, c). I already have completed 3 and 4.
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| 1,767
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| 3.5625
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https://stonespounds.com/2867-pounds-in-stones-and-pounds
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| 603,546,447
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2867 pounds in stones and pounds
Result
2867 pounds equals 204 stones and 11 pounds
You can also convert 2867 pounds to stones.
Converter
Two thousand eight hundred sixty-seven pounds is equal to two hundred four stones and eleven pounds (2867lbs = 204st 11lb).
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https://crypto.stackexchange.com/questions/33503/certificateless-cryptography-implementation
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# Certificateless cryptography Implementation
Hello guys I am trying to implement Certificateless Cryptography algorithm with reference to this base paper. Right now I am trying to implement the code in java. There are some points specified in PDF which I didn't understood. I need some help to understand how should I implement them.
The points are as follows (kindly correct me if I misunderstood any point):
1. Pick a generator of $g$ of $Z_p^*$ with order $q$. -> for this step I just followed this link as reference and calculated $g$ as $g^{(p−1)/2}\bmod p=1$ where $g$ is prime number.
2. Choose cryptographic hash in the paper they have specified that they had used SHA-1 to generate hash. I can generate the hash values but I am not able to reproduce the hash functions given in base paper, does $(0,1)^*$ is used to represent binary.
3. Hash usually generates hexadecimal values do I need to convert them into decimal for Raise operation i.e. power
4. Does the hash needs to be of fixed size or not.
5. Sorry to say but in summary I didn't understood the encryption part of this paper. I know stackexchange is not made to explain how it works but I just want a proper guide line so that I can implement the application.
As far I understood the paper (very briefly) it looks like it's talking about the default asymmetric encryption without the CA (certificate authorities). However - it seems there's the SEM part of the system which seems to know the private keys (???) and acts as the revocation authority.
I believe you can use the default Java crypto API without needing to implement it yourself. In fact I'd discourage you to do that until you REALLY know what are you doing, how/what to properly encrypt with the public key, etc. For the start - I'd advice to learn something about the cryptography elements and implementations, what you see in the papers are mathematical models of common crypto elements.
As it goes for your questions:
Pick a generator of g of Z∗p with order q. -> for this step I just followed this link as reference and calculated g as g(p−1)/2modp=1 where g is prime number.
It means to choose you keysize (so - e.g. 2048 bit keys) and generate a private key (p, q) and public key (p*q). Using a KeyPairGenerator should be effective.
Choose cryptographic hash in the paper they have specified that they had used SHA-1 to generate hash. I can generate the hash values but I am not able to reproduce the hash functions given in base paper, does (0,1)∗ is used to represent binary.
Any reasonable hash function will do - so lets assume SHA-1. All numbers are binary for the computer, its only their 'human' representation how you can display them (decimal, hex, ..)
Hash usually generates hexadecimal values do I need to convert them into decimal for Raise operation i.e. power
Hash returns a bit array (byte[]), you can use the BigInteger class to raise it to certain power. Don't forget all operations are under the discrete group (so you do ($g^a$ mod p). About the steps described in the paper - you need to 'sign' the hash. And after the decryption, the signature is compared with a newly computed signature so data integrity is ensured.
As for the topic - I'd advice to spend some time understanding the default cryptography and asymmetric encryption (for this case) and usage practices and it will be much easier for you to follow the paper. The problem with the cryptography is that something is misunderstood or neglected, it will completely break the security.
In all cases - have fun :)
• Thanks for taking your precious time, and sorry for my late reply. As far your point of using java's default cryptography package to simulate the algorithm in paper, I can implement but this is my final year project is dependent on this base paper, my instructor won't accept with java's concept they need the same algorithm to implement. Is there any book you recommend for learning mathematical concept so that I can understand the math behind it and try to implement. Mar 15, 2016 at 8:07
• Then at least see how it is implemented. All the operations are done on the discrete group p ( effectively you do modulo p). However - I see pointless to implement yourself a hash function (sha, md5, whatever) when it needs to keep all cryptographic properties (collision resistance, ...) Mar 15, 2016 at 9:00
• Correct me if I am wrong, you want to tell that I should not create hash function just use the hash wherever it is ask to use in algorithm, or something else. Mar 15, 2016 at 11:07
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# How to calculate steady state with fsolve with white noise in equations ?
4 vues (au cours des 30 derniers jours)
DEVENDER garg le 19 Juil 2020
Modifié(e) : John D'Errico le 19 Juil 2020
I have following equations type
dot(x)=-ax +by^2+n
dot(y)=-cy+dx+w
a,b,c,d are known constants
n and w are gaussian white noise which be defined as n or w =sigma*randn(1,L)
If I dont have n or w .I can easly solve for Steady state using fsolve function or any other function by take dot(x) and dot(y) equal to zero.But how can I handle white noise in steady state solutions?
##### 1 commentaireAfficher -1 commentaires plus anciensMasquer -1 commentaires plus anciens
John D'Errico le 19 Juil 2020
Modifié(e) : John D'Errico le 19 Juil 2020
If a,b,c,d are all known constants, then the steady state solution , thus a limit for x(t) and y(t) as t --> inf. In the absense of noise, you would find this as the solution of the problem
-ax + by^2 == 0
-cy + dx == 0
Now I have no idea if this is your real problem, or just an example one. Usually a real problem is more complicated.
However, as this is a nonlinear problem, then any such nonkinear problem may often have multiple solutions. The one we see here is a simple one. For example, I'll just pick some very simple numbers, and solve the homogeneous problem. Because the problem is so trivial as posed, I can use solve to get all solutions.
xy = solve(-x + y^2 == 0, -y + 2*x == 0)
xy =
struct with fields:
x: [2×1 sym]
y: [2×1 sym]
xy.x
ans =
0
1/4
xy.y
ans =
0
1/2
The nice thing here is solve tells me there are indeed two distinct solutions, whereas fsolve would find one, but not the other, and the solution you find will depend on the starting values.
So there are two solutions to the problem I posed, although without some thought that I don't really want to invest at the moment, I might want to learn if the (0,0) solution is a stable one. My gut tells me it is not a stable one. If so, then any temporary perturbation away from that point, and the solution will drift to the second solution.
Now, you describe a stochastic system, where at any point in time, the differential equation is being hit with a random impulse in some random direction. Is that your question? Are you asking how to analyse such a problem?
If so, then I would suggest that you start with determining if the various potential steady state solutions are stable or not. If one solution is not a stable point, then any perturbation in the system will drive the solution towards the other solution. If you are at a stable solution, then any random impulse will cause the problem to wander around the homogeneously stable solution, in a fuzzy random walk.
For example, suppose you are near/at the (0.25,0.5) homogenous solution for this system? Then imagine your system receives a sudden impulsive perturbation.
xy = solve(-x + y^2 - 0.02 == 0, -y + 2*x + 0.01 == 0)
vpa(xy.x)
ans =
-0.019194109070750548052986782472964
0.25919410907075054805298678247296
vpa(xy.y)
ans =
-0.028388218141501096105973564945927
0.52838821814150109610597356494593
The steady state points are now subtly different. The system would want to wander in that direction, except that the next time step will again see a random impulse in some random direction.
You might want to characterise the solutions in terms of their stability. You might want to characterize the system in terms of the shape of that fuzzy random cloud you would see around a steady state point.
I would also note that if your problem is less specific, and a,b,c,d are some other values, then the stady state points are:
syms a b c d
xy = solve(-a*x + b*y^2 == 0, -c*y + d*x == 0)
xy =
struct with fields:
x: [2×1 sym]
y: [2×1 sym]
xy.x
ans =
0
(a*c^2)/(b*d^2)
>> xy.y
ans =
0
(a*c)/(b*d)
But still you need to do more analysis here, beyond this point.
I think you need to define your problem more clearly, since I'm just guessing as to your real question here, postulating what I might want to do, if that were your real question. And since I have no real expertise in differential equations, and certainly no expertise in stochastic problems of this sort, I'll stop here. You probably need to do some reading before you proceed too far, since I'm fairly sure these problems have been studied to a great deal of depth.
Connectez-vous pour commenter.
### Réponses (1)
That depends on how you define Steady state. If your derivatives are never actually zero, you will never have a fully perfect steady state, so your problem is impossible if the goal is a perfect steady state. Considering that x and y are spatial coordinates, one way of thinking of this is to find which position will give you the smallest possible variance, which, since your noise is zero mean gaussian, will actually be in the place where:
dot(x)= n
dot(y)= w
or, equivalenty:
-ax +by^2 = 0
-cy+dx+w = 0
That is the same solution you would become had you solved the problem completely ignoring the noise. This is related to the Maximum likelihood estimation and it basically says "Considering that my data has zero mean Gaussian noise, which is the most likely solution between all solutions?", and this is the one where your mean "error" is zero and all that's left are the noise fluctuations.
##### 1 commentaireAfficher -1 commentaires plus anciensMasquer -1 commentaires plus anciens
DEVENDER garg le 19 Juil 2020
Thanks for your reply!.I think it will give correct solution if we look at value of x and y after simulation.But I think if we look at value of x^2 or y^2 after simulation it can heavly depend on w*w or n*n. one way,I am thinking of using for loop each one for particular value of vector randn(1,L).Then average the the output result.But in principal it does not look correct to me.
Another one is using differntial equation direclty with time and noise.And manually looking for steady state which has error in the range of gaussian noise as you suggested.
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## College Physics (4th Edition)
The correct answer is: (a) $h_3 = h_1$
The continuity equation is: $A_1~v_1 = A_2~v_2$ By the continuity equation, we know that the flow velocity is the same at $A_1$ and $A_3$ since $A_1 = A_3$. Therefore, the pressure is the same at $A_1$ and $A_3$. Since the pressure is the same, then $h_3 = h_1$. The correct answer is: (a) $h_3 = h_1$
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# solve using the Substitution method y=3x x+y=12
solve using the Substitution method y=3x x+y=12
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# Subject: Analysis - Loan Payments and Amortization
Last-Revised: 16 Feb 2003
Contributed-By: Hugh Chou
This article presents the formula for computing monthly payments on loans. A listing of the full series of payments (principal and interest) that show how a loan is paid off is known as a loan amortization table. This article will explain how these tables are generated for the U.S. system in which interest is compounded monthly.
First you must define some variables to make it easier to set up:
• P = principal, the initial amount of the loan
• I = the annual interest rate (from 1 to 100 percent)
• L = length, the length (in years) of the loan, or at least the length over which the loan is amortized.
The following assumes a typical conventional loan where the interest is compounded monthly. But first, just two more variables to make the calculations easier:
• J = monthly interest in decimal form. This is I / (12 x 100).
• N = number of months over which loan is amortized. This is L x 12.
Finally here is for the big monthly payment (M) formula:
``` J
M = P x ------------------------
1 - ( 1 + J ) ^ -N
```
Please note "1" is just the number one (it does not appear too clearly on some browsers) and "^" is the exponentiation operator (2 ^ 3 = 8).
So to calculate M, you would first calculate 1 + J, then take that to the -N (negative N) power, subtract that from the number 1. Now take the inverse of that (if you have a 1/X button on your calculator push that). Then multiply the result times J and then times P. Sorry for the long way of explaining it, but I just wanted to be clear for everybody.
``` M = P * ( J / (1 - (1 + J) ** -N))
```
So now you should be able to calculate the monthly payment, M. To calculate the amortization table you need to do some iteration (i.e. a simple loop). I will tell you the simple steps:
1. Calculate H = P x J, this is your current monthly interest
2. Calculate C = M - H, this is your monthly payment minus your monthly interest, so it is the amount of principal you pay for that month
3. Calculate Q = P - C, this is the new balance of your principal of your loan.
4. Set P equal to Q and go back to Step 1: You thusly loop around until the value Q (and hence P) goes to zero.
Programmers will see how this makes a trivial little loop to code, but I have found that many people now surfing on the Internet are NOT programmers and still want to calculate their mortgages!
Note that just about every PC or Mac has a spreadsheet of some sort on it, and they are very good tools for doing mortgage analysis. Most of them have a built-in PMT type function that will calculate your monthly payment given a loan balance, interest rate, and the number of terms. Check the help text for your spreadsheet.
Please visit Hugh Chou's web site for a calculator that will generate amortization tables according to the formulas discussed here. He also offers many other calculators:
http://www.hughchou.org/calc/
Previous article is Analysis: Internal Rate of Return (IRR) Next article is Analysis: Paying Debts Early versus Making Investm.. Category is Analysis Index of all articles
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# Octree
Left: Recursive subdivision of a cube into octants. Right: The corresponding octree.
An octree is a tree data structure in which each internal node has exactly eight children. Octrees are most often used to partition a three dimensional space by recursively subdividing it into eight octants. Octrees are the three-dimensional analog of quadtrees. The name is formed from oct + tree, but note that it is normally written "octree" with only one "t". Octrees are often used in 3D graphics and 3D game engines.
## Octrees for spatial representation
Each node in an octree subdivides the space it represents into eight octants. In a point region (PR) octree, the node stores an explicit 3-dimensional point, which is the "center" of the subdivision for that node; the point defines one of the corners for each of the eight children. In a matrix based (MX) octree, the subdivision point is implicitly the center of the space the node represents. The root node of a PR octree can represent infinite space; the root node of an MX octree must represent a finite bounded space so that the implicit centers are well-defined. Note that Octrees are not the same as k-d trees: k-d trees split along a dimension and octrees split around a point. Also k-d trees are always binary, which is not the case for octrees. By using a depth-first search the nodes are to be traversed and only required surfaces are to be viewed.
## History
The use of octrees for 3D computer graphics was pioneered by Donald Meagher at Rensselaer Polytechnic Institute, described in a 1980 report "Octree Encoding: A New Technique for the Representation, Manipulation and Display of Arbitrary 3-D Objects by Computer",[1] for which he holds a 1995 patent (with a 1984 priority date) "High-speed image generation of complex solid objects using octree encoding" [2]
## Application to color quantization
The octree color quantization algorithm, invented by Gervautz and Purgathofer in 1988, encodes image color data as an octree up to nine levels deep. Octrees are used because ${\displaystyle 2^{3}=8}$ and there are three color components in the RGB system. The node index to branch out from at the top level is determined by a formula that uses the most significant bits of the red, green, and blue color components, e.g. 4r + 2g + b. The next lower level uses the next bit significance, and so on. Less significant bits are sometimes ignored to reduce the tree size.
The algorithm is highly memory efficient because the tree's size can be limited. The bottom level of the octree consists of leaf nodes that accrue color data not represented in the tree; these nodes initially contain single bits. If much more than the desired number of palette colors are entered into the octree, its size can be continually reduced by seeking out a bottom-level node and averaging its bit data up into a leaf node, pruning part of the tree. Once sampling is complete, exploring all routes in the tree down to the leaf nodes, taking note of the bits along the way, will yield approximately the required number of colors.
## Octree implementation for point decomposition
The example recursive algorithm outline below (MATLAB syntax) decomposes an array of 3-dimensional points into octree style bins. The implementation begins with a single bin surrounding all given points, which then recursively subdivides into its 8 octree regions. Recursion is stopped when a given exit condition is met. Examples of such exit conditions (shown in code below) are:
• When a bin contains fewer than a given number of points
• When a bin reaches a minimum size or volume based on the length of its edges
• When recursion has reached a maximum number of subdivisions
```function [binDepths,binParents,binCorners,pointBins] = OcTree(points)
binDepths = [0] % Initialize an array of bin depths with this single base-level bin
binParents = [0] % This base level bin is not a child of other bins
binCorners = [min(points) max(points)] % It surrounds all points in XYZ space
pointBins(:) = 1 % Initially, all points are assigned to this first bin
divide(1) % Begin dividing this first bin
function divide(binNo)
% If this bin meets any exit conditions, do not divide it any further.
binPointCount = nnz(pointBins==binNo)
binEdgeLengths = binCorners(binNo,1:3) - binCorners(binNo,4:6)
binDepth = binDepths(binNo)
exitConditionsMet = binPointCount<value || min(binEdgeLengths)<value || binDepth>value
if exitConditionsMet
return; % Exit recursive function
end
% Otherwise, split this bin into 8 new sub-bins with a new division point
newDiv = (binCorners(binNo,1:3) + binCorners(binNo,4:6)) / 2
for i = 1:8
newBinNo = length(binDepths) + 1
binDepths(newBinNo) = binDepths(binNo) + 1
binParents(newBinNo) = binNo
binCorners(newBinNo) = [one of the 8 pairs of the newDiv with minCorner or maxCorner]
... Calculate which points in pointBins==binNo now belong in newBinNo
% Recursively divide this newly created bin
divide(newBinNo)
end
```
## Example color quantization
Taking the full list of colors of a 24-bit RGB image as point input to the Octree point decomposition implementation outlined above, the following example show the results of octree color quantization. The first image is the original (532818 distinct colors), while the second is the quantized image (184 distinct colors) using octree decomposition, with each pixel assigned the color at the center of the octree bin in which it falls. Alternatively, final colors could be chosen at the centroid of all colors in each octree bin, however this added computation has very little effect on the visual result.[5]
```% Read the original RGB image
% Extract pixels as RGB point triplets
pts = reshape(Img,[],3);
% Create OcTree decomposition object using a target bin capacity
OT = OcTree(pts,'BinCapacity',ceil((size(pts,1) / 256) *7));
% Find which bins are "leaf nodes" on the octree object
leafs = find(~ismember(1:OT.BinCount, OT.BinParents) & ...
ismember(1:OT.BinCount,OT.PointBins));
% Find the central RGB location of each leaf bin
binCents = mean(reshape(OT.BinBoundaries(leafs,:),[],3,2),3);
% Make a new "indexed" image with a color map
ImgIdx = zeros(size(Img,1), size(Img,2));
for i = 1:length(leafs)
pxNos = find(OT.PointBins==leafs(i));
ImgIdx(pxNos) = i;
end
ImgMap = binCents / 255; % Convert 8-bit color to MATLAB rgb values
% Display the original 532818-color image and resulting 184-color image
figure
subplot(1,2,1), imshow(Img)
title(sprintf('Original %d color image', size(unique(pts,'rows'),1)))
subplot(1,2,2), imshow(ImgIdx, ImgMap)
title(sprintf('Octree-quantized %d color image', size(ImgMap,1)))
```
File:Black-throated Green Warbler at 246442 colours.jpg
Original image (246442 unique colors)
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USING OUR SERVICES YOU AGREE TO OUR USE OF COOKIES
# Is 98 An Even Number?
• Yes, number 98 is an even number.
• Ninety-eight is an even number, because it can be divided by 2 without leaving a comma spot.
## Is 98 Odd Or Even?
• Number 98 is an even number.
## How To Calculate Even Numbers
• To find the set of even natural numbers, we use 2N where N is any natural number. Any number multiplied with an even number will result in an even number. For example: 0N, 2N, 4N, 6N, ... (where N is any natural number) the result is always an even number.
• The oddness of a number is called its parity, so an odd number has parity 1, while an even number has parity 0.
## Mathematical Information About Numbers 9 8
• About Number 9. Nine is the smallest odd composite number and the minimum composite odd number that is no Fermat pseudoprime. It is the smallest natural number n, for each non-negative integer can be represented as a sum of at most n positive cubes (see Waring's problem), and the smallest positive integer n for which n squares in pairs of different positive edge length exist, the can be put together to form a rectangle. Number Nine is the number which (other than 0) as a single digit checksum generally occurs (in decimal number system) after multiplication by an arbitrary integer always even, and the number which is added to any other (except 0 and -9), as a single digit checksum the same result as the starting number itself - ie it behaves quasi-neutral.
• About Number 8. The octahedron is one of the five platonic bodies. A polygon with eight sides is an octagon. In computer technology we use a number system on the basis of eight, the octal system. Eight is the first real cubic number, if one disregards 1 cube. It is also the smallest composed of three prime number. Every odd number greater than one, raised to the square, resulting in a multiple of eight with a remainder of one. The Eight is the smallest Leyland number.
## What Is An Even Number?
An even number is an integer which is evenly divisible by two. This means that if the integer is divided by 2, it yields no remainder. Zero is an even number because zero divided by two equals zero. Even numbers can be either positive or negative. You can tell if any decimal number is an even number if its final digit is an even number. An integer that is not an even number is an odd number.
Parity is a mathematical term that describes the property of an integer's inclusion in one of two categories: even or odd. An integer is even if it is evenly divisible by two and odd if it is not even. For example, 6 is even because there is no remainder when dividing it by 2. By contrast, 3, 5, 7, 21 leave a remainder of 1 when divided by 2. A formal definition of an even number is that it is an integer of the form n = 2k, where k is an integer. It can then be shown that an odd number is an integer of the form n = 2k + 1.
© Mathspage.com | Privacy | Contact | info [at] Mathspage [dot] com
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Paul Hewitt’s Figuring Physics in the Feb. 2016 issue asked whether it would take a larger velocity change to stop a satellite in a circular orbit or to cause it to escape. An extension of this problem asks: What minimum velocity change is required to crash a satellite into the planet, and how does that compare with the velocity change required for escape? The solution presented here, using conservation principles taught in a mechanics course, serves as an introduction to orbital maneuvers, and can be applied to questions regarding the removal of objects orbiting Earth, other planets, and the Sun.
1.
P.
Hewitt
, “
Figuring Physics: Slow down or speed up?
Phys. Teach.
54
,
69
(
Feb.
2016
).
2.
To find the exact values, set the slope of Eq. (5) to zero and solve the resulting cubic equation. The existence of a maximum can be explained thus: For close orbits, |Δvin| increases from zero with rS as the planet’s angular size decreases, requiring a larger distortion of the satellite’s original circular orbit. For rS >>R, where the planet appears almost “point-like,” |Δvin| tends to the speed required to stop the satellite, vC, which is a decreasing function of rS. Therefore, a maximum must exist in between.
3.
E. D.
Noll
, “
Kepler’s third law for elliptical orbits
,”
Phys. Teach.
34
,
42
43
(
Jan.
1996
.)
4.
P.
Blanco
and
C. E.
Mungan
, “
Satellite splat: An inelastic collision with a surface-launched projectile
,”
Eur. J. Phys.
36
,
045004
(
May
2015
).
5.
A treatment that includes the Earth-to-heliocentric orbit transfer finds a total specific impulse of 16.7 km/s. See
C. E.
Mungan
, “
Relative speeds of interacting astronomical bodies
,”
92
,
7
14
(Summer
2006
).
6.
J.
Amato
, “
Motivating introductory physics students using astronomy and space science
,”
Phys. Teach.
54
,
56
57
(
Jan.
2016
).
7.
E.
Butikov
, “
Orbital maneuvers and space rendezvous
,”
56
,
2582
2594
(
Dec.
2015
).
8.
If you really want to crash a high-orbit satellite into the planet using minimal fuel, execute a two-impulse biparabolic transfer as follows: First send the spacecraft very far from the planet at the escape speed, then (much later) apply a tiny retrograde impulse, causing it to fall back towards the planet on a nearly radial path. See
A.
, “
The elliptic-bi-parabolic planar transfer for artificial satellites
,”
J. Braz. Soc. Mech. Sci. Eng.
25
,
122
128
(
April
2003
); .
9.
For information on these and other missions, see http://solarsystem.nasa.gov/missions/type/orbiter.
10.
To obtain free educational licenses for Systems Tool Kit (STK), visit https://www.agi.com/resources/educational-alliance-program.
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#2483. Pku2279 Mr. Young's Picture Permutations
题目描述
Mr. Young wishes to take a picture of his class. The students will stand in rows with each row no longer than the row behind it and the left ends of the rows aligned. For instance, 12 students could be arranged in rows (from back to front) of 5, 3, 3 and 1 students.
```X X X X X
X X X
X X X
X```
In addition, Mr. Young wants the students in each row arranged so that heights decrease from left to right. Also, student heights should decrease from the back to the front. Thinking about it, Mr. Young sees that for the 12-student example, there are at least two ways to arrange the students (with 1 as the tallest etc.):
``` 1 2 3 4 5 1 5 8 11 12
6 7 8 2 6 9
9 10 11 3 7 10
12 4```
Mr. Young wonders how many different arrangements of the students there might be for a given arrangement of rows. He tries counting by hand starting with rows of 3, 2 and 1 and counts 16 arrangements:
```123 123 124 124 125 125 126 126 134 134 135 135 136 136 145 146
45 46 35 36 34 36 34 35 25 26 24 26 24 25 26 25
6 5 6 5 6 4 5 4 6 5 6 4 5 4 3 3```
Mr. Young sees that counting by hand is not going to be very effective for any reasonable number of students so he asks you to help out by writing a computer program to determine the number of different arrangements of students for a given set of rows.
输入格式
The input for each problem instance will consist of two lines. The first line gives the number of rows, k, as a decimal integer. The second line contains the lengths of the rows from back to front (n1, n2,..., nk) as decimal integers separated by a single space. The problem set ends with a line with a row count of 0. There will never be more than 5 rows and the total number of students, N, (sum of the row lengths) will be at most 30.
输出格式
The output for each problem instance shall be the number of arrangements of the N students into the given rows so that the heights decrease along each row from left to right and along each column from back to front as a decimal integer. (Assume all heights are distinct.) The result of each problem instance should be on a separate line. The input data will be chosen so that the result will always fit in an unsigned 32 bit integer.
样例输入
``````
1
30
5
1 1 1 1 1
3
3 2 1
4
5 3 3 1
5
6 5 4 3 2
2
15 15
0
``````
样例输出
``````
1
1
16
4158
141892608
9694845
``````
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https://mocktestpro.in/mcq/mcqs-on-angle-gauges/
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Engineering Questions with Answers - Multiple Choice Questions
# MCQs on Angle Gauges
1 - Question
Which of the following option is incorrect with respect to angle gauges?
a) Sine bar is better than angle gauges
b) Angle gauges are made of high carbon high chromium steel
c) Angle gauges can measure the angle from 0 to 360 degrees
d) They are available in two sets of 13 and 16 gauges
Explanation: Angle gauges are more accurate than sine bar as it involves trigonometric functions.
2 - Question
How 34’ can be built by using angle gauges?
a) 27’+9’-3’+1’
b) 26’+10’-2’
c) 27’+10’-3’
d) 27’+8’
Explanation: Combinations can be made by only three series of angle gauges. Minutes series is with 1’, 3’, 9’, 27’ angle gauges.
3 - Question
In how many series the gauges can be divided?
a) 1
b) 2
c) 3
d) 4
Explanation: 16 or 13 Gauges can be divided into 3 series i.e. degree, minutes and seconds (fraction of minute).
4 - Question
What is the approximate size of angle gauges?
a) 76mm long and 16 wide
b) 85mm long and 26 wide
c) 16mm long and 75 wide
d) 70mm long and 18 wide
Explanation: Angle gauges are about 3 inch (76.2 mm) long and 5/8 inch (15.87 mm) wide. And their faces are lapped within 0.0002 mm.
5 - Question
What is the accuracy of master angle gauges?
a) 0.1 sec
b) 1 sec
c) 0.25 sec
d) 3 sec
Explanation: Master Angle gauges are laboratory standard and most expensive of all. They measure the angle with an accuracy of 0.25 sec.
6 - Question
Which gauges are present in the first series (degree) of angle gauges?
a) 5°, 10°, 15°, 25° and 40°
b) 1°, 3°, 9°, 27° and 41°
c) 1°, 5°, 9°, 25° and 45°
d) 5°, 10°, 15°, 30° and 45°
Explanation: The standard angles gauges have 5 gauges present in the first series. They consist of 1°, 3°, 9°, 27° and 41° gauges.
7 - Question
How many sets of angle gauges are available?
a) 1
b) 2
c) 3
d) 4
Explanation: Two sets of angle gauges are available. One set with 16 gauges with an accuracy of 1’’ and one set with 13 gauges with 3’’.
8 - Question
Which of the following option is true for the given statements?
Statement 1: Any angle up to 120o4’2’’ be made by a direct combination of angle gauges.
Statement 2: Interferometry can be used to calibrate angle gauges.
a) T, F
b) F, F
c) F, T
d) T, T
Explanation: Direct combination of angle gauges can be made up to angle 81o40.9’. For the larger angle, square block is used with angle gauges.
9 - Question
What are the two grades of angle gauges?
a) Master and tool room
b) Precise and Normal
c) Standard and Industrial
d) High and low
Explanation: Master grade is laboratory standard and made from steel carbide or tungsten carbide. Tool grade is normal industrial purpose angle gauges made from steel.
10 - Question
How angle greater than 90° is measured?
a) By repeating gauges
b) Using square plate
c) Using sine bar
d) Using auto collimator
Explanation: Versatility of angle gauges increases when used with a square plate. Square plates with angle gauge can measure angle greater than 90o.
11 - Question
Which of the following is not a gauge from standard B angle gauges?
a) 0.05’
b) 1’
c) 27’
d) 30
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# Convert [Ropes] to [Palms], (rope to palm)
## LENGTH
1 Ropes
= 80 Palms
*Select units, input value, then convert.
Embed to your site/blog Convert to scientific notation.
Category: length
Conversion: Ropes to Palms
The base unit for length is meters (SI Unit)
[Ropes] symbol/abbrevation: (rope)
[Palms] symbol/abbrevation: (palm)
How to convert Ropes to Palms (rope to palm)?
1 rope = 80 palm.
1 x 80 palm = 80 Palms.
Always check the results; rounding errors may occur.
Definition:
The palm may be either one of two obsolete non-SI units of measurement of length.
In English usage the palm, or small palm, also called handbreadth or handsbreadth, was ..more definition+
In relation to the base unit of [length] => (meters), 1 Ropes (rope) is equal to 6.096 meters, while 1 Palms (palm) = 0.0762 meters.
1 Ropes to common length units
1 rope =6.096 meters (m)
1 rope =0.006096 kilometers (km)
1 rope =609.6 centimeters (cm)
1 rope =20 feet (ft)
1 rope =240 inches (in)
1 rope =6.66666666667 yards (yd)
1 rope =0.00378787878788 miles (mi)
1 rope =6.44329352077E-16 light years (ly)
1 rope =23040.0029027 pixels (PX)
1 rope =3.81E+35 planck length (pl)
Ropes to Palms (table conversion)
1 rope =80 palm
2 rope =160 palm
3 rope =240 palm
4 rope =320 palm
5 rope =400 palm
6 rope =480 palm
7 rope =560 palm
8 rope =640 palm
9 rope =720 palm
10 rope =800 palm
20 rope =1600 palm
30 rope =2400 palm
40 rope =3200 palm
50 rope =4000 palm
60 rope =4800 palm
70 rope =5600 palm
80 rope =6400 palm
90 rope =7200 palm
100 rope =8000 palm
200 rope =16000 palm
300 rope =24000 palm
400 rope =32000 palm
500 rope =40000 palm
600 rope =48000 palm
700 rope =56000 palm
800 rope =64000 palm
900 rope =72000 palm
1000 rope =80000 palm
2000 rope =160000 palm
4000 rope =320000 palm
5000 rope =400000 palm
7500 rope =600000 palm
10000 rope =800000 palm
25000 rope =2000000 palm
50000 rope =4000000 palm
100000 rope =8000000 palm
1000000 rope =80000000 palm
1000000000 rope =80000000000 palm
(Ropes) to (Palms) conversions
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Less Wrong is a community blog devoted to refining the art of human rationality. Please visit our About page for more information.
# inemnitable comments on Typicality and Asymmetrical Similarity - Less Wrong
25 06 February 2008 09:20PM
You are viewing a comment permalink. View the original post to see all comments and the full post content.
Sort By: Old
Comment author: 15 June 2012 09:20:49PM 1 point [-]
I was thinking of it more like: if there's a certain place I can get to in (roughly) 102 hours going 98 mph, and I want to get there in 100 hours, I need to speed up to 100 mph. Similarly, if there's a another place that I can get to in roughly 102 hours going 980 mph, and I want to get to that place in 100 hours, I need to speed up to 1000 mph.
I kind of wanted to clarify that in the original post but I hadn't really thought of a good way to express it at the time.
Furthermore, I think that your interpretation of the example even makes it more clear that it makes sense to think of it in terms of a ratio. In the first case, you've sped up by 2 mph and gotten a gain of about 2 hours, straightforward enough. But in the second case, you've sped up by 20 mph, and only gotten a gain of about 0.2 hours. Here's where I think most people's intuition is probably screaming "whaaaaaaat!?"
But if we think of it in terms of the ratios, then everything fits together nicely again and the screaming intuition voice shuts up. Plus the math we have to do to get to the right answer is a lot easier.
Comment author: 15 June 2012 10:40:13PM 0 points [-]
Oh, I see what you mean now.
(Incidentally, Eliezer's original objection can be resolved by taking logs. Suddenly although the ratios 102/100 and 100/102 are not symmetrical, log(102/100) and log(100/102) are.)
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Topic: High-precision Algorithm
Replies: 2 Last Post: May 12, 2013 5:39 PM
Messages: [ Previous | Next ]
James Waldby Posts: 545 Registered: 1/27/11
Re: High-precision Algorithm
Posted: May 12, 2013 10:48 AM
On Sat, 11 May 2013 15:00:17 -0700, KBH wrote:
>> Does anyone know of a function or software I can use (preferably in Visual Basic, but will use any algorithm) to perform high-precision math calculations, such as ...
>
> 'Thousand Digit Op' will multiply with two integer numbers that are 1000 digits or less and will divide with two integer numbers that are 2000 digits or less. Then it's possible to work with decimal point numbers by dropping the decimal points and keeping track of where the decimal point should be in the result.
>
> http://www.kbhscape.com/integer.htm
>
> The application uses a small amount of instruction in the loops and then a large number of fast loops.
>
> Otherwise, I'm not familar with any open source libraries but there is a shareware UBasic available
> .
Perhaps you (KBH) could provide some timings and cross-checks of
'Thousand Digit Op' relative to Python's built-in large-number
arithmetic. Here is an example of timing some operations, within
following, all of u, v/u, and v%u are about 500 digits long, while
for v/u + v%u.
In [27]: u=3**1048
In [28]: v=5**1431
In [29]: v/u
Out[29]: 159588377228959077124033529846706173377670651019402817413903057980687239665812136895146535461691029393717561194039581042112941167384541312962207141090513753598883354972311476786828567327334557067321195729425873950404649780805993015698785979878971124960712319102981518907826172440288055238618916259743493881382859749952565937542160268965170299634066918779414688423865579464673280918269958244076237834716962819697213949665224639312089807278692420046672759354745098154816407376723302684184662808947654510L
In [30]: v%u
Out[30]: 103326703166979322767585007603325329851383449999195202897493052907961790225869481331767933862961846823885955634647416122394634047149232900536158317485041583325170788355276467455783992346226027047434201755968044448049463697686029887794902906939322808951006983461531977363098808187922066499128461713493537423899150272154451941901015773931881852167807937748056339937566567332680579077984997740268640971147835374932842977963626399422285827509371192145408304567650688258964404464654299922273397388585470015L
In [31]: w = v/u + v%u; print w
262915080395938399891618537450031503229054101018598020311396110888649029891681618226914469324652876217603516828686997164507575214533774213498365458575555336924054143327587944242612559673560584114755397485393918398454113478492022903493688886818293933911719302564513496270924980628210121737747377973237031305282010022107017879443176042897052151801874856527471028361432146797353859996254955984344878805864798194630056927628851038734375634788063612192081063922395786413780811841377602606458060197533124525
In [32]: %timeit w = v/u + v%u
10000 loops, best of 3: 19.3 us per loop
--
jiw
Date Subject Author
5/12/13 James Waldby
5/12/13 KBH
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# Thread: How to solve this pq-formula
1. ## How to solve this pq-formula
t^2+ t+ 1+ sqrt(3)
I dont know how to deal with the sqrt(3)..
Help would be very appreciated!
2. Originally Posted by tinyone
t^2+ t+ 1+ sqrt(3)
I dont know how to deal with the sqrt(3)..
Help would be very appreciated!
Hi tinyone,
It would appear that $1+\sqrt{3}$ is the constant term of the quadratic.
I also assume you are to set the quadratic expression equal to 0 and solve for t.
3. Originally Posted by tinyone
t^2+ t+ 1+ sqrt(3)
I dont know how to deal with the sqrt(3)..
Help would be very appreciated!
If this is equal to 0 then use the quadratic formula. $\sqrt{3}$ is a constant just like 1 is.
$t^2+t+(1+\sqrt{3})$
a = 1
b = 1
c = 1+sqrt(3)
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http://www.gurufocus.com/term/Dividends+Per+Share/NCR/Dividends%2BPer%2BShare/NCR%2BCorporation
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Switch to:
NCR Corp (NYSE:NCR)
Dividends Per Share
\$0.00 (TTM As of Sep. 2016)
NCR Corp's dividends per share for the three months ended in Sep. 2016 was \$0.00. Its dividends per share for the trailing twelve months (TTM) ended in Sep. 2016 was \$0.00. Its Dividend Payout Ratio for the three months ended in Sep. 2016 was 0.00. As of today, NCR Corp's Dividend Yield is 0.00%.
Please click Growth Rate Calculation Example (GuruFocus) to see how GuruFocus calculates Wal-Mart Stores Inc (WMT)'s revenue growth rate. You can apply the same method to get the average dividends per share growth rate.
For more information regarding to dividend, please check our Dividend Page.
Definition
Dividends paid to per common share.
Explanation
1. Dividend payout ratio measures the percentage of the companys earnings paid out as dividends.
NCR Corp's Dividend Payout Ratio for the quarter that ended in Sep. 2016 is calculated as
Dividend Payout Ratio = Dividends Per Share (Q: Sep. 2016 ) / EPS without NRI (Q: Sep. 2016 ) = 0 / 0.69 = 0.00
2. Dividend Yield measures how much a company pays out in dividends each year relative to its share price.
NCR Corp Recent Full-Year Dividend History
Amount Ex-date Record Date Pay Date Type Frequency
NCR Corp's Dividend Yield (%) for Today is calculated as
Dividend Yield = Most Recent Full Year Dividend / Current Share Price = 0 / 38.90 = 0.00 %
Current Share Price is \$38.90.
NCR Corp's Dividends Per Share for the trailing twelve months (TTM) ended in Today is \$0.
* 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.
NCR Corp Annual Data
Dec06 Dec07 Dec08 Dec09 Dec10 Dec11 Dec12 Dec13 Dec14 Dec15 Dividends Per Share 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
NCR Corp Quarterly Data
Jun14 Sep14 Dec14 Mar15 Jun15 Sep15 Dec15 Mar16 Jun16 Sep16 Dividends Per Share 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
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Community Profile
# S AY
### UConn Co-op
13 2013 年以降の合計貢献数
#### S AY's バッジ
Determine whether a vector is monotonically increasing
Return true if the elements of the input vector increase monotonically (i.e. each element is larger than the previous). Return f...
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Bullseye Matrix
Given n (always odd), return output a that has concentric rings of the numbers 1 through (n+1)/2 around the center point. Exampl...
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Select every other element of a vector
Write a function which returns every other element of the vector passed in. That is, it returns the all odd-numbered elements, s...
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Given a and b, return the sum a+b in c.
7年弱 前
Sum all integers from 1 to 2^n
Given the number x, y must be the summation of all integers from 1 to 2^x. For instance if x=2 then y must be 1+2+3+4=10.
7年弱 前
Swap the first and last columns
Flip the outermost columns of matrix A, so that the first column becomes the last and the last column becomes the first. All oth...
7年弱 前
Determine if input is odd
Given the input n, return true if n is odd or false if n is even.
7年弱 前
Triangle Numbers
Triangle numbers are the sums of successive integers. So 6 is a triangle number because 6 = 1 + 2 + 3 which can be displa...
7年弱 前
select the primes of a vector
Find the prime numbers in a vector
7年弱 前 | 1
Make a checkerboard matrix
Given an integer n, make an n-by-n matrix made up of alternating ones and zeros as shown below. The a(1,1) should be 1. Example...
7年弱 前
Find the sum of all the numbers of the input vector
Find the sum of all the numbers of the input vector x. Examples: Input x = [1 2 3 5] Output y is 11 Input x ...
7年弱 前
Make the vector [1 2 3 4 5 6 7 8 9 10]
In MATLAB, you create a vector by enclosing the elements in square brackets like so: x = [1 2 3 4] Commas are optional, s...
7年弱 前
Times 2 - START HERE
Try out this test problem first. Given the variable x as your input, multiply it by two and put the result in y. Examples:...
7年弱 前
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| 3.421875
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CC-MAIN-2020-10
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| 0.624714
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https://socratic.org/questions/how-do-you-solve-abs-2x-3-5-1
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# How do you solve abs(2x-3)>5?
$x > 4$ OR $x < 1$
$2 x - 3 > 5$ OR $- 2 x + 3 > 5$
$2 x > 8$ OR $- 2 x > 2$
$x > 4$ OR $x < 1$
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CC-MAIN-2020-16
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longest
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https://www.supportyourtech.com/excel/how-to-convert-text-to-number-in-excel-using-formula-a-step-by-step-guide/
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# How to Convert Text to Number in Excel Using Formula: A Step-by-Step Guide
Converting text to numbers in Excel might seem tricky at first, but it’s actually pretty simple once you get the hang of it. Using a formula, you can swiftly change text that looks like a number into an actual number that you can use in calculations. This guide will walk you through the steps, so you’ll be an Excel pro in no time.
## How to Convert Text to Number in Excel Using Formula
In this section, we’ll break down the process into simple, easy-to-follow steps. You’ll learn the exact formula to use and how to apply it to your data.
### Step 1: Open Your Excel Workbook
Open your Excel workbook where you need to convert text to numbers.
Make sure your data is accessible and easy to locate. It’s a good idea to have your text data in a column, so you can quickly apply the formula.
### Step 2: Select the Cell to Enter the Formula
Click on the cell where you want the converted number to appear.
This is where you’ll type your formula. It’s usually a good idea to choose a cell adjacent to your text data to keep things organized.
### Step 3: Enter the Formula =VALUE(A1)
Type in the formula =VALUE(A1), where A1 is the cell containing the text you want to convert.
The VALUE function in Excel converts a text string that looks like a number into an actual number. Ensure you replace A1 with the correct cell reference for your data.
### Step 4: Press Enter
After typing the formula, press Enter on your keyboard.
This step converts the text in cell A1 to a number. The result will appear in the cell where you entered the formula.
### Step 5: Drag the Formula Down
If you have multiple cells to convert, click on the small square at the bottom-right corner of the cell with the formula. Drag it down to apply the formula to other cells.
This will copy the formula to adjacent cells, converting all the text data to numbers in one go. It makes the process quick and efficient.
Once you’ve completed these steps, all your text data will be converted into numbers. This means you can now use these numbers for calculations, graphs, and other Excel functions.
## Tips for Converting Text to Number in Excel Using Formula
• Double-check your cell references to ensure accuracy.
• Use the TRIM function if your text has leading or trailing spaces.
• Apply the VALUE formula to a separate column to keep your original data intact.
• If VALUE doesn’t work, use the DATEVALUE or TIMEVALUE functions for date and time conversions.
• Use the Paste Special feature to convert data in place without a formula.
### What if the VALUE formula doesn’t work?
If the VALUE formula doesn’t work, double-check that your text data is purely numeric and doesn’t include any unwanted characters.
### Can I automate this process?
Yes, you can use VBA (Visual Basic for Applications) to automate this process if you frequently convert text to numbers.
### What does the VALUE function do?
The VALUE function converts text that appears as numbers into actual numbers that Excel can recognize and use in calculations.
### Can I convert dates and times?
Yes, use the DATEVALUE and TIMEVALUE functions to convert text-formatted dates and times into Excel date and time formats.
### Will this formula change my original data?
No, the formula doesn’t change your original data. It creates a new set of data in the cells where you enter the formula.
## Summary
2. Select the Cell to Enter the Formula
3. Enter the Formula =VALUE(A1)
4. Press Enter
5. Drag the Formula Down
## Conclusion
Converting text to numbers in Excel using a formula is a handy skill that can save you a lot of time and headaches. By using the VALUE function, you can ensure that your data is in the correct format for further analysis and calculations.
This guide has provided a step-by-step approach to make the conversion process simple and efficient. So next time you encounter text-formatted numbers, don’t stress! Just follow these steps, and you’ll have your data ready in no time.
If you found this guide helpful, you might want to explore more advanced Excel functions and formulas. There is a wealth of resources available online, from tutorials to forums where you can ask questions and share your experiences. Remember, practice makes perfect, and the more you work with Excel, the more proficient you’ll become. Happy Excel-ing!
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https://busfoundation.org/answers-on-questions/quick-answer-you-have-a-network-that-uses-a-logical-bus-topology-how-do-messages-travel-through-the-network.html
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## What is a logical bus topology?
The bus topology is the type of logical topology in which all the nodes and switches are connected to only one single cable which can be also known as backbone or bus. The nodes are connected like half-duplex mode. In the bus topology, there is a host which is known as a station.
## Which topology connects each network device to a central hub?
In star topology each device in the network is connected to a central device called hub. Unlike Mesh topology, star topology doesn’t allow direct communication between devices, a device must have to communicate through hub.
## In which of the following topologies does each device on the network act as a repeater sending the signal to the next device?
Network Topology 1.2
You might be interested: How Much Does A Round Trip Greyhound Bus Ticket Cost?
Which topology connects each network device to a central hub? Star
In which topology does each device on the network act as a repeater, sending the signal to the next device? Ring
## Which of the following topology connects each device to a neighboring device?
Ring Topology In a ring network, every device has exactly two neighboring devices for communication purpose. It is called a ring topology as its formation is like a ring. In this topology, every computer is connected to another computer.
## What is an example of logical topology?
Logical topologies are bound to the network protocols that direct how the data moves across a network. For example, twisted pair Ethernet is a logical bus topology in a physical star topology layout. While IBM’s Token Ring is a logical ring topology, it is physically set up in a star topology.
## How many types of logical topology are there?
The two logical topologies are broadcast (also known as bus) and sequential (also known as ring).
## What are 2 advantages of a bus topology?
• Easy to connect a computer or peripheral to a linear bus.
• Requires less cable length than a star topology.
• Entire network shuts down if there is a break in the main cable.
• Difficult to identify the problem if the entire network shuts down.
• Not meant to be used as a stand-alone solution.
## What are the two types of ring topology?
Basically, ring topology is divided into the two types which are Bidirectional and Unidirectional. Most Ring Topologies allow packets to only move in one direction, known as a one-way unidirectional ring network. Others allow data to go, either way, called bidirectional.
You might be interested: How To Get A Free Bus Pass For School?
## Which topology requires a multipoint connection?
Explanation: Bus topology requires a multipoint connection.
## Which of the following is the best definition for a LAN?
A local area network ( LAN ) is a collection of devices connected together in one physical location, such as a building, office, or home.
## Which topologies can a can use?
CAN standard supports several topologies. Commonly used topologies are: Line / Bus Topology. Star Topology.
## Do you have a small network that uses a switch to connect multiple devices which physical topology are you using?
You have a small network that uses a switch to connect multiple devices. Which physical topology are you using? Your manager has asked you to implement a wired network infrastructure that will accommodate failed connections. You have a network that uses a logical ring topology.
## How do messages travel in a ring topology?
You have a network that uses a logical ring topology. How do messages travel through the network? a. Messages are sent to all devices connected to the network.
## At which OSI layer does a router operate to forward network messages?
Routers operate on the third layer of the OSI Model, the Network -Control Layer.
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https://www.teacherspayteachers.com/Browse/Search:measuring+angles+with+a+protractor+worksheet?ref=related_search
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showing 1-24 of 387 results
Measuring Angles with a Protractor practice and mastery is easy with this Math Mastery Packet! This set of printable worksheets is perfect for practicing key skills, independent practice, small group work, intervention, homework, and yes..... test prep! Click here to save on the BUNDLE of ALL FOURT
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Also included in: 4th Grade Math Test Prep for the Year Bundle - 43 Packets!
\$2.25
66 Ratings
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PDF (17.29 MB)
Students must identify the angle type (acute, obtuse, or right) and measure 8 angles using a protractor.
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\$1.50
6 Ratings
4.0
PDF (316.71 KB)
These printable math measurement worksheets focus on measuring and sketching angles with a protractor for teaching 4th grade Common Core lesson plans. These activities work well for test prep, as a classroom center or station review, as student homework, for morning work or as a quiz. There are
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\$3.50
1 Rating
4.0
PDF (1.09 MB)
Looking for a fun interactive teaching idea for angle measurement? Well look no further as Measuring Angles with a Protractor Puzzles, for CCSS 4.MD.6, will serve as an exciting lesson plan for 4th grade elementary school classrooms. This is a great resource to incorporate into your unit as a guided
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\$4.90
351 Ratings
4.0
PDF (1.53 MB)
Students have difficulty using protractors to measure angles. This activity is a great starting point for teaching students to use protractors because it takes away the lining up piece (where they struggle) and only requires them to correctly read the angle. Students will also practice drawing an
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\$1.50
127 Ratings
4.0
DOC (355.5 KB)
Here are four pictures to practice measuring angles with a protractor! My students LOVE them! Each picture comes with a chart to name if the angle is obtuse, acute or right. Then they are to estimate how many degree’ they think the angle is, and then finally measure the angle. Each picture and works
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\$3.00
89 Ratings
4.0
PDF (1.64 MB)
This is a Smartboard Interactive Whiteboard activity on Classifying and Measuring Angles. There are 6 pages in this file. Pg 1 Introduction to Acute, Obtuse and Reflex angles. Pg 2 Interactive (flash) sorting activity Pg 3 Measuring acute angles with interactive protractor Pg 4 Measuring obtu
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\$2.00
13 Ratings
4.0
NOTEBOOK (350.41 KB)
An introduction to angles and how to use a protractor to measure angles. Great for grade 2,3,4, students learning about angles and using a protractor. Take thirty seconds to download and check out the free preview to see if it meets your needs, then you will know exactly what you are getting for a
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\$3.00
21 Ratings
4.0
PDF (3.43 MB)
Master that tough skill of measuring angles with protractors! Two five-question quizzes designed to assess student mastery of Texas math TEKS 4.7C: determine the approximate measures of angles in degrees to the nearest whole number using a protractor. Aligned to NEW Texas math standards. Also align
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\$2.50
43 Ratings
4.0
PDF (11.65 MB)
Here is a worksheet for students to use in order to practice measuring angles with a protractor. There are 15 angles for students to practice on. It starts of with "easier" ones and then there are ones that are rotated that challenge students to rotate their protractors as well. There are acute
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\$0.50
40 Ratings
4.0
PDF (73.65 KB)
Can be used as homework, classwork, centers...just about anything! Directions for revealing the secret message: Step #1: Measure each angle on pg 1 and estimate each angle on pg 2. Step #2: Find each of the angle measures in the grid and color them in. ***Each one will appear in the grid mo
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CCSS:
\$2.25
20 Ratings
4.0
PDF (640.38 KB)
4.MD.C.5, 4.MD.C.6, 4.GA.1 21 worksheets with an answer key on the following: Drawing angles when given the degree of the angle. Measuring angles with a protractor. Identifying if an angle is obtuse, acute, or a right angle. PLUS 2 diagrams to help identify angles. Draw and identify lines and
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\$4.00
\$3.75
14 Ratings
3.9
PDF (5.1 MB)
This is a set of 28 task cards that involve measuring angles using a protractor. Task cards are a fun way to engage students and can be used as a card stack, game cards or a scoot. There are a variety of cards that range from simple to complex. Starting with measuring to an even ten, to measuring
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\$2.25
13 Ratings
4.0
ZIP (1.35 MB)
Are you overwhelmed as you try to differentiate your daily math instruction based on “one size fits all” math assessments or exit slips? Don't be! These differentiated 4th Grade assessments or practice sheets for Measuring Angles with a Protractor {4.MD.6} can help you differentiate with more ease!
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Also included in: 4th Grade Measurement ALL STANDARDS BUNDLE
\$3.00
10 Ratings
4.0
PDF (7 MB)
This is a simple worksheet that is perfect for reviewing how to measure angles using a protractor (as well as identifying the type of angle). It's great to use for homework or extra credit. THERE ARE TWO VERSIONS OF THE ANGLE MEASURING ASSIGNMENT PROVIDED WITH THIS PURCHASE. On the front side of
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\$1.00
10 Ratings
4.0
PDF (260.55 KB)
This math station contains 21 total problems in a math station game for students to practice measuring angles with a protractor. Students will need to read a protractor to find the measurement of the angles and then match them up with the correct answer. Angle answers are in increments of 5 degree
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\$3.00
8 Ratings
4.0
PDF (456.98 KB)
This math unit for grade 6+ provides six activity pages in which students use a protractor to measure angles, record the degrees of angle, and solve riddles and other problems. The unit also includes an assessment page in test-prep format. (Find other units by searching "Measurement 6")
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\$1.99
6 Ratings
4.0
PDF (1.44 MB)
A quick, easy way to assess your students’ knowledge of the 4th Grade Common Core math standards. These exit slips can be used as independent formative assessments for the standards. However, they are also the perfect match for the powerpoint units in my 17-unit 4th Grade Common Core math curric
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\$3.25
5 Ratings
4.0
PDF (695.22 KB)
**Search for my Measuring Angles with a Protractor worksheet (free) to utilize when teaching this lesson! It directly correlates with the flipchart and will enhance this lesson even further! This 25-page interactive ActiveInspire Flipchart will provide you with everything you need to teach your st
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\$1.25
3 Ratings
4.0
FLIPCHART (652.31 KB)
To assist students in practicing measuring acute, obtuse and reflex angles with a protractor. Students work with a partner to solve questions and enjoy giving and receiving immediate feedback. This effective learning experience enables students to quickly increase their skill in using a protractor
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\$1.00
2 Ratings
4.0
PDF (1.22 MB)
Students will practice measuring angles with a protractor with this fun activity! Students will use a protractor to measure 10 angles. They will then then find their answer from the answer choices given and glue it in the appropriate place. The angles are measured in 5-degree increments, and angles
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\$1.50
1 Rating
4.0
PDF (14.06 MB)
OVERVIEW: The math video and worksheet included in this resource were created to help students learn, practice, and apply angles concepts for fourth grade. Students will be asked to measure angles using a protractor. *********************************************************************************
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Also included in: Geometry Math Video and Worksheet Bundle
\$2.00
1 Rating
4.0
ZIP (3.33 MB)
This 8-page lesson packet includes everything you need to teach your students to use a protractor to measure angles and to draw them. This packet is designed for 4th grade as it aligns perfectly with CCSS 4.MD.6, but could be used as a challenge lesson for 3rd grade or as a review lesson for 5th gra
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A great resource for students to practice measuring angles with a protractor. Assists students in practicing measuring acute, obtuse and reflex angles with a protractor in a fun and engaging way. Students are up and interacting with each other, giving and receiving immediate feedback. This fast pa
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https://en.khanacademy.org/math/algebra/x2f8bb11595b61c86:linear-equations-graphs/x2f8bb11595b61c86:x-intercepts-and-y-intercepts/e/intercepts-from-table
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### Course: Algebra 1>Unit 4
Lesson 4: x-intercepts and y-intercepts
# Intercepts from a table
## Problem
This table gives a few $\left(x,y\right)$ pairs of a line in the coordinate plane.
$x$$y$
$33$$-22$
$52$$-33$
$71$$-44$
What is the $x$-intercept of the line?
$\left($
$,$
$\right)$
Stuck?
Stuck?
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# loopy getting closer
## loopy getting closer
(OP)
I have discovered that part of my problem was to do with strcmpy so I have scrapped that for now but i still have probs getting the correct message (not as many probs though)
//take 20 throws output message for each throw
#include <stdio.h>
#include <math.h>
#include <string.h>
void main()
{
int
sum=0,
average=0,
throws=20,
score[20],
i, j, k,
total=0;
double standarddev; ???
//char message[11][10];
for( i= 0; i<throws; i++ )
{
printf("Enter score > ");
scanf("%d",&score );
total+=score;
}
average = total/throws;
for(j=0;j<throws;j++ )
sum+= (score[j] - average) * (score[j] - average);
// would this be better:- sum+=pow(score[j] - average,2)
standarddev = sqrt(sum / (throws -1));
for(j=0; j < throws; j++)
{
if(score[j] >= average + 1.5 * standarddev)
printf("\n%d \t %d \texcellent",j+1,score[j]);
if(score[j] >= average + 0.5 * standarddev)
printf("\n %d \t %d \tGood",j+1,score[j]);
if(score[j] >= average - 0.5 * standarddev)
printf("\n%d \t%d \tFair",j+1,score[j]);
if(score[j] >= average - 1.5 * standarddev)
printf("\n%d \t %d \tPoor",j+1,score[j]);
if(score[j] < average - 1.5 * standarddev)
printf("\n%d \t %d \tVery Poor",j+1,score[j]);
}
}
///////// check variables //////////
//printf("\n\n");
//for(k=0; k< throws; k ++)
// {
// printf("\n%d\t%d\t%s",k+1,score[k], message[k]);
// }
// printf("\nsum = %d",sum);
// printf("\tstdv = %d",(int)standarddev);
//printf("\taverage = %d",average);
}
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# 2.6 Geometría, tiempo y dinero
## Unit Goals
• Students reason with shapes and their attributes and partition shapes into equal shares, building a foundation for fractions. They relate halves, fourths, and skip-counting by 5 to tell time, and solve story problems involving the values of coins and dollars.
### Section A Goals
• Identify triangles, quadrilaterals, pentagons, hexagons, and cubes.
• Recognize and draw shapes having specified attributes, such as a given number of angles or a given number of equal faces.
### Section B Goals
• Partition rectangles and circles into halves, thirds, and fourths and name the pieces.
• Recognize 2 halves, 3 thirds, and 4 fourths as one whole.
• Understand that equal pieces do not need to be the same shape.
### Section C Goals
• Tell and write time from analog and digital clocks to the nearest five minutes, using a.m. and p.m.
### Section D Goals
• Find the value of a group of bills and coins.
• Use addition and subtraction within 100 to solve one- and two-step word problems.
### Format
#### Blackline Masters
For activities with important visual aids
pdfdocx (n/a)
#### Teacher Presentation Materials
Teacher presentation materials (with Spanish)
pdfdocx
#### Cumulative Practice Problems
All practice problems from the unit (with Spanish)
pdfdocx
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http://www.mathworks.com/matlabcentral/fileexchange/33448-ant-system-tsp-solver/content/Ant_System_TSP_Solver/nn_shortest_path_tsp.m
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Code covered by the BSD License
# Ant System TSP Solver
by
### Johannes (view profile)
26 Oct 2011 (Updated )
A demo of an Ant System algorithm solving classical Traveling Salesman Problems.
nn_shortest_path_tsp(M)
```function [nn_path, path_length] = nn_shortest_path_tsp(M)
% [nn_path, path_length] = nn_shortest_path_tsp(M)
%
% Applies the nearest neighbor method to compute an efficient
% path for the TSP problem with the city coordinates M.
%
% Input:
% - M - A city map, which is a matrix of 2D city coordinates
%
% Output:
% - nn_path - A vector containing the optimal path
% - path_length - The length of the optimal path
%
% Author: Johannes W. Kruisselbrink
% The number of cities
l = length(M(:,1));
% Compute distance matrix
dmat = -1 * ones(l,l);
for k = 1:l
dmat(k,:) = sqrt(sum((M(k * ones(l,1),:) - M).^2, 2));
end
% Initialize tabu list and path
tabu_list = ones(1,l);
nn_path = zeros(1,l);
% Perform nearest neighbor path finding loop
current_node = 1;
nn_path(1) = current_node;
tabu_list(current_node) = inf;
for j = 2:l
[nearest, nearest_index] = min(tabu_list .* dmat(current_node, :));
next_node = nearest_index;
current_node = next_node;
nn_path(j) = current_node;
tabu_list(current_node) = inf;
end
% Compute path length
path_length = tsp_evaluate_tour(M, nn_path);
end```
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https://www.amle.org/BrowsebyTopic/OrganizationalStructures/OrgDet/TabId/197/ArtMID/823/ArticleID/938/Diagonal-Alignment.aspx
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/
# Diagonal Alignment
### A tool to integrate basic skills across the curriculum
In working with teachers on integration and differentiation, there are a few obstacles that are prevalent in almost every group. Logistics in implementing this approach in a system that is currently used is a common hurdle. Another hurdle is trying to integrate curriculums to determine what should be taught. Diagonal alignment is one way to help teach and reinforce basic skills and concepts throughout an entire year. It's based on the idea that there is vertical alignment by subject matter or by grade level each month. Horizontal alignment refers to the sequence of subject areas within a year.
In a typical school, a simple table can be developed that shows basic concepts taught each month in each subject. When the chart is filled out completely, the teachers can see if there are any natural ties between their subject and another subject to encourage integration. Teachers could also look at the chart and choose to rearrange parts of the curriculum to overlap and integrate, or they can create a spiral effect for skills that cross subjects and can be reinforced every few months. An example might be the idea of measurement being taught in math in August, then science reinforces the concept in October. Or social studies is teaching population growth so math develops an exponential growth unit to tie into the social studies unit.
To keep it simple, start with a structure that is simple to use. First have each subject area determine the basic skills that are vital to the subject area or that students traditionally struggle with. The list could include fractions, percentages, and division for math; map skills, chart development, and geography for social studies; the scientific process and measurement for science; reading and non-fiction writing for language arts. Classes such as art, choir, band, and physical education can and should be included when possible.
Once the list of basic skills is decided upon by each subject area, those teachers create a "cheat sheet" for other teachers. The cheat sheet gives key points for the basic skills and the focus in each of the concepts chosen. The subject area teachers also teach a mini lesson of those vital skills to the staff just prior to the school year. The idea is to share expectations and key points for all non-subject area teachers to make sure there is a consistent message.
After the opening teaching session and the cheat sheets have been delivered, the subject areas begin to teach class. The curriculum of the class does not change in the scope and sequence previously used, but there will need to be a place to teach the basic skills and concepts that were shared with the teaching staff. The skills and concepts need to be taught by the subject expert at the beginning of the year so the base is solid and can be reinforced throughout the rotation of the other classes. A sample rotation is given in figure 1, and the term diagonal alignment is evident when you draw arrows from the content area to the class that will reinforce the basic skills the next month. The order is at the discretion of the staff and may be influenced by other content being taught that may lend itself to a natural integration of subject matter.
Figure 1
Content Area Basic Skills August September October November Language Arts LA basic skills Sci basic skills Math basic skills SS basic skills Social Studies SS basic skills LA basic skills Sci basic skills Math basic skills Math Math basic skills SS basic skills LA basic skills Sci basic skills Science Sci basic skills Math basic skills SS basic skills LA basic skills
Using the previous model of scheduling the rotation of basic skills, in the month of September each teacher will begin to work in the basic skills of other subject areas. The basic skills cheat sheet and communication with other teachers will assist in this process. As an example, language arts determined the basic skills they wanted reinforced were non-fiction writing and reading comprehension. In the month of September, the social studies teachers would teach the required social studies curriculum, but they would integrate either one or both of the concepts of non-fiction writing and reading comprehension. Bringing map skills to math, graphing to science, or measurement to language arts could all be done in the first month of diagonal alignment. The impact has begun. After three months of reinforcing these same areas in different aspects of the curriculum the results will be noticeable.
The rotation could repeat in the second half of the year, or if the subject area teachers feel that the skills are now solid, they can move into another need of the base curriculum. This process is a beginning to a more integrated curriculum. The ultimate goal is to be fully integrated when there are natural ties. Better yet, develop a curriculum that starts with the idea of the bigger picture of the whole curriculum and create an innovative approach to deliver to students.
In the current school year, students return to school mid-August and first semester is done at winter break. School concludes at the end of May. That means the months of December and May are wrapping up semesters and a good time to take advantage of an integrated project-based learning model that combines subject areas to create a larger project. Again, this should lead to a deeper level of curriculum development. Targeted differentiation is a model that is used to develop that higher level of understanding for teachers and is another approach to innovate a school curriculum.
The big question that is often posed is "How do I grade this?" The ideal situation is that we use standards-based assessments with the goal of attainment of the standards. However, many schools still use a grading system that is not standards based. The issue then becomes a matter of communication. Common rubrics or allowing the work done in one subject to be used in another subject for a grade would be possible. Each school or staff should determine what works best.
The logistics of any new program could make or break the attempt. It is vital to keep in mind that learning is the most important aspect of integrating content. Whether you grade, don't grade, share, or don't share doesn't really matter as long as authentic learning is happening. Make the logistics work. Find common ground. Help each other reach the basic standards of your subject area. Remember you do not teach math or science, ultimately you teach kids!
Norm Potter is curriculum supervisor at Twinsburg City Schools, Twinsburg, Ohio.
npotter@twinsburgcsd.org
Published in AMLE Magazine, August 2018.
More on these topics
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# Get a value from the most recent previous record where some text appears in either of two columns
I'm trying to more fully automate the calculation of values and summary data in a spreadsheet I keep about the results of matches in a pool league.
I have a table with lots of information about each match, with the relevant fields being: Date of Match, Winner, Winner Starting Handicap, Winner Ending Handicap, Loser, Loser Starting Handicap, Loser Ending Handicap, Match Start Time.
Handicaps are adjusted at the end of every match, and before the next one. It's a pain to find the most recent past record for a player (could have been a Winner or Loser) and copy his ending handicap from that record to the Starting Handicap (winner or loser) for the one I'm now entering.
I'd like a formula that would find the most recent record (highest date & start time in case he played twice in one day) where he was a winner or a loser, and then get the ending handicap (from the respective winner or loser).
Per teylyn's suggestion, here's a Dropbox link to the file. The relevant tab is Match Results: https://www.dropbox.com/s/1j9c6zsxjd3q4dt/Sample%20for%20Excel%20Question%20on%20Superuser.xlsx?dl=0
I added a blank column L to test things, comparing the results to what's in K to see if they were working, that's why it's there. Forgot to remove it when I put it in Dropbox.
• It would help to see the data structure. Can you post a sample file? Think file sharing service. – teylyn May 4 '15 at 1:10
### Problem Statement
A worksheet has names in Columns E and X. For every row n, EnXn. There are numbers in Column M corresponding to the names in Column E, and numbers in Column AG corresponding to the names in Column X. For any row after the first (let’s say Row 42), we want to get values for K42 and AF42 from previous rows, if possible.
• If E42 is “John”, find the most recent row that contains “John” (in Column E or X). Call that row n. If En = “John”, set K42 equal toMn. If Xn = “John”, set K42 equal toAGn.
• If X42 is “Scott”, find the most recent row that contains “Scott” (in Column E or X). Call that row n. If En = “Scott”, set AF42 equal toMn. If Xn = “Scott”, set AF42 equal toAGn.
### Solution
In the hopes of preserving some sanity, let’s use helper columns; let’s say AR and AS. Assume that (as in the example file), data start in Row 2. Enter
=MAX((\$E\$2:\$E2=\$E3)*(100*ROW(\$E\$2:\$E2)+COLUMN(\$M:\$M)), (\$X\$2:\$X2=\$E3)*(100*ROW(\$X\$2:\$X2)+COLUMN(\$AG:\$AG)))
into AR3 (skipping AR2). End with Ctrl+Shift+Enter, to make it an array formula. Likewise, set AS3 to
=MAX((\$E\$2:\$E2=\$X3)*(100*ROW(\$E\$2:\$E2)+COLUMN(\$M:\$M)), (\$X\$2:\$X2=\$X3)*(100*ROW(\$X\$2:\$X2)+COLUMN(\$AG:\$AG)))
as an array formula. (This is the same as AR3 except the two occurrences of \$E3 have been replaced with \$X3.)
Set K3 to
=IF(\$AR3=0, "?", INDEX(\$A\$1:\$BG\$999, INT(\$AR3/100), MOD(\$AR3,100)))
and AF3 to
=IF(\$AS3=0, "?", INDEX(\$A\$1:\$BG\$999, INT(\$AS3/100), MOD(\$AS3,100)))
(not as array formulas). These are the same except the three occurrences of \$AR3 have been replaced with \$AS3.
And, of course, drag/fill down.
The helper columns find the most recent previous occurrences of the names — ARn finds the most recent previous occurrence of En, and ASn finds the most recent previous occurrence of Xn — basically by finding the maximum, throughout the previous rows, of
(previous_value=this_value) * ROW())
i.e., the highest row number where the name is a match. It then encodes the location where the name was found as
100*ROW() + COLUMN(data_we_want_to_copy)
Both formulas look in both Columns E and X, and return the encoded coordinates of the corresponding Columns M or AG cell. Then the K and AF formulas simply decode the cell address and retrieve the value.
So AR6 is 213 because “John” (E6) was most recently seen on Row 2, and, since he was seen in E2 (rather than X2), we want to copy the value from Column 13 (Column M).
• This a great answer! It solved the problem in a straightforward way (at least as much as possible), it was very clear and explained how you did it, not just "here's the answer", and taught me something I didn't know about excel. I'd say I couldn't ask for more but I'm going to. What did you use to create the pictures of the spreadsheet? I may need to use that technique for my next question. – John Biddle May 4 '15 at 14:38
• That's simple (although a little tedious): press PrtScr (Print Screen) to get a snapshot of the screen as an image in the copy/paste buffer. (There are other ways of doing this, such as the Snipping Tool.) Paste into Microsoft Paint (or any other graphics editor; people seem to like Paint.NET and GIMP, and I've heard that Adobe might have something useful, too.) Then it's straightforward to crop, draw shapes (e.g., circles, ovals, boxes, and lines), copy and paste (e.g., see here and here), and save to a file. – Scott May 4 '15 at 19:46
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## Machines B/C
JoeyC
Member
Posts: 269
Joined: November 7th, 2017, 1:43 pm
Division: C
State: TX
### Re: Machines B/C
Given that a #7-32-5 (UNF) screw is being turned with a 5 inch long wrench, what is the IMA of the action?
Ohayo!
`Dynamic Planet, Protein Modeling, Fast Facts, Thermodynamics`
`Dynamic Planet, Machines, Ornith`
1 Corinthians 13:4-7
Scientia Potentia Est
viditpok
Member
Posts: 12
Joined: August 31st, 2018, 2:29 pm
Division: C
State: FL
### Re: Machines B/C
JoeyC wrote:
September 29th, 2019, 5:42 am
Given that a #7-32-5 (UNF) screw is being turned with a 5 inch long wrench, what is the IMA of the action?
I don't understand the #7-32-5 thing, as when I was looking it up, most results showed 7/32-5/8" or something close to it. Could you please describe what it means?
Orlando Science High School '22
2017-18 Events: Thermodynamics, Crime Busters, Hovercraft, Optics
2018-19 Events: Circuit Lab, Fermi Questions, Forensics, Mission Possible
2019-20 Events: Circuit Lab, Machines, Codebusters
JoeyC
Member
Posts: 269
Joined: November 7th, 2017, 1:43 pm
Division: C
State: TX
### Re: Machines B/C
This can help
The #7-32-5 is a unified screw read out, and works like this-
diameter - number of threads per inch - length of screw
Because it's in unified, all units are in inches, and importantly the diameter has an octothorpe (#) in front of it.
This means that the diameter is not 7 inches, but dictated by this formula,
.06+.013*n, where n is the number. So the diameter is .151 inches, there are 32 threads per inch, meaning pitch (important for screw IMA) is .03125, and the screw is 5 inches long.
Using this information, I can use the Screw IMA equation ; IMA = 2pir/p, or IMA = circumference of total effort arm over pitch, the effort arm being 5 inches because my wrench, effort arm, is 5 inches long, and pitch being 1/32, or .03125.
IMA = 1005.30964915
Ohayo!
`Dynamic Planet, Protein Modeling, Fast Facts, Thermodynamics`
`Dynamic Planet, Machines, Ornith`
1 Corinthians 13:4-7
Scientia Potentia Est
viditpok
Member
Posts: 12
Joined: August 31st, 2018, 2:29 pm
Division: C
State: FL
### Re: Machines B/C
Oh, that makes so much more sense
Thank you so much for the help!
Orlando Science High School '22
2017-18 Events: Thermodynamics, Crime Busters, Hovercraft, Optics
2018-19 Events: Circuit Lab, Fermi Questions, Forensics, Mission Possible
2019-20 Events: Circuit Lab, Machines, Codebusters
AlfWeg
Member
Posts: 96
Joined: July 22nd, 2019, 7:52 am
Division: C
State: MI
### Re: Machines B/C
Who's next? @JoeyC
Last&SeventhYearOlympian
```NorView/CenterVille/UMICH/r/s:
Machines: 2/4/1
DP: 1/1/7
Code:X/16/9
GeoMap:X/9/8
Protein: 2/X/X
```
It's kinda Dull -->AlfWeg's Userpage
JoeyC
Member
Posts: 269
Joined: November 7th, 2017, 1:43 pm
Division: C
State: TX
### Re: Machines B/C
Anyone else I guess?
Alf you can do it if you want.
Ohayo!
`Dynamic Planet, Protein Modeling, Fast Facts, Thermodynamics`
`Dynamic Planet, Machines, Ornith`
1 Corinthians 13:4-7
Scientia Potentia Est
AlfWeg
Member
Posts: 96
Joined: July 22nd, 2019, 7:52 am
Division: C
State: MI
### Re: Machines B/C
JoeyC wrote:
October 13th, 2019, 6:22 pm
Anyone else I guess?
Alf you can do it if you want.
Naw
Last&SeventhYearOlympian
```NorView/CenterVille/UMICH/r/s:
Machines: 2/4/1
DP: 1/1/7
Code:X/16/9
GeoMap:X/9/8
Protein: 2/X/X
```
It's kinda Dull -->AlfWeg's Userpage
Creationist127
Member
Posts: 68
Joined: August 14th, 2018, 3:21 pm
Division: C
State: IN
Location: The Wild Area
### Re: Machines B/C
I'll just go.
There is an inclined plane, with a 26.0* angle of incline. On top of it is a 25.0 kg box, on the cusp of moving. What is the coefficient of static friction between the ramp and the box, and what is the force of friction?
2018: Hovercraft, Thermo, Coaster, Solar System
2019: Thermo, Circuit Lab, Sounds, Wright Stuff
2020: Circuit Lab, WIDI, Wright Stuff, Machines
It's not science if you don't glue your hand to it at least three times.
Umaroth
Member
Posts: 187
Joined: February 10th, 2018, 8:51 pm
Division: C
State: CA
Location: Kraemer Room 504, Troy Room 909, or studying at my dining table
### Re: Machines B/C
Creationist127 wrote:
October 14th, 2019, 7:05 am
I'll just go.
There is an inclined plane, with a 26.0* angle of incline. On top of it is a 25.0 kg box, on the cusp of moving. What is the coefficient of static friction between the ramp and the box, and what is the force of friction?
mgsinθ = μmgcosθ
μ = sin26/cos26 = tan26
μ = 0.488
F = μmgcosθ
F = 107N
Troy SciOly 2019-present
Dank Memes Area Homeschool Juggernaut 2018-present
Kraemer SciOly Pretty-Much Head Coach 2019-present
2020 Tryouts: Circuit, Code, Detector, DP, GeoMaps, Machines
Regionals and GGSO Events: Code, Detector, DP
Umaroth's Userpage
Creationist127
Member
Posts: 68
Joined: August 14th, 2018, 3:21 pm
Division: C
State: IN
Location: The Wild Area
### Re: Machines B/C
Umaroth wrote:
October 14th, 2019, 7:37 am
Creationist127 wrote:
October 14th, 2019, 7:05 am
I'll just go.
There is an inclined plane, with a 26.0* angle of incline. On top of it is a 25.0 kg box, on the cusp of moving. What is the coefficient of static friction between the ramp and the box, and what is the force of friction?
mgsinθ = μmgcosθ
μ = sin26/cos26 = tan26
μ = 0.488
F = μmgcosθ
F = 107N
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## Convert Parts Per Billion (ppb, Percentages And Parts) to Septet (Number Of Performers)
parts per billion (ppb)
Percentages and Parts
septet
Number of Performers
This page features online conversion from parts per billion to septet. These units belong to different measurement systems. The first one is from Percentages And Parts. The second one is from Number Of Performers.
If you need to convert parts per billion to another compatible unit, please pick the one you need on the page below. You can also switch to the converter for septet to parts per billion.
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## Quantity Units
Units: unit, point (1) / pair, brace, yoke / nest, hat trick / half-dozen / decade, dicker / dozen / baker's dozen / score / flock / shock / hundred / great hundred / gross / thousand / great gross
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## Percentages and Parts
Units: percent (%) / permille (‰) / parts per million (ppm) / parts per billion (ppb)
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## Fractions
This section answers a question like "How many one sevenths are there in 1 half?". To get an answer enter 1 under half and see the result under 1/7. See if you can use this section to find out how many one-sixths are there in 15 one-nineths.
Units: half or .5 (1/2) / one third or .(3) (1/3) / quart, one forth or .25 (1/4) / tithe, one fifth or .2 (1/5) / one sixth or .1(6) (1/6) / one seventh or .142857 (1/7) / one eights or .125 (1/8) / one ninth or .(1) (1/9) / one tenth or .1 (1/10) / one sixteenth or .0625 (1/16) / one thirty-second or .03125 (1/32)
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## Metric Prefixes
These prefixes are widely used in Metric System. They apply to any unit, so if you ever see, e.g. kiloapple, you know it's 1000 apples.
Units: yocto (y) / zepto (z) / atto (a) / femto (f) / pico (p) / nano (n) / micro (µ, mc) / milli (m) / centi (c) / deci (d) / deka (da) / hecto (h) / kilo (k) / mega (M) / giga (G) / tera (T) / peta (P) / exa (E) / zetta (Z) / yotta (Y)
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## Number of Performers
Units: solo / duet / trio / quartet / quintet / sextet / septet / octet
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Q:
# What are some common math vocabulary words?
A:
Some common math vocabulary words are line, digit, equation, angle and estimate. These terms are applicable to basic math and more advanced studies, such as geometry or algebra.
## Keep Learning
A line is a straight collection of points that continues indefinitely in both directions. If multiple lines are present, the lines are either intersecting, parallel or skew. Intersecting lines share one common point and continue away from one another. Parallel lines are lines that are in the same plane, they never intersect and always remain the exact same distance apart. Skew lines are lines that are not in the same plane but never intersect.
The 10 numeral symbols in standard math are called digits. The number 4091 contains four digits: 4, 0, 9 and 1.
An equation is any math statement that includes an equals sign. The equal sign signifies that both parts of the equation have the same numerical value. Many equations contain a variable, which is a letter inserted into the equation in the place of a number value. The solution to an equation is made by determining the value that the variable represents.
The v shape that is formed by two lines intersecting is an angle. The point that the two sides intersect at is the vertex of the angle. An angle is measured using degrees. A straight angle is an angle that measures 180 degrees, making it resemble a line.
An estimate is a rough guess or calculation that is formed by looking at the given information and using prior knowledge or simplification to make assumptions.
Sources:
## Related Questions
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# 1 - 3 of 3 results for: STATS300A
## STATS 270:Bayesian Statistics I (STATS 370)
This is the first of a two course sequence on modern Bayesian statistics. Topics covered include: real world examples of large scale Bayesian analysis; basic tools (models, conjugate priors and their mixtures); Bayesian estimates, tests and credible intervals; foundations (axioms, exchangeability, likelihood principle); Bayesian computations (Gibbs sampler, data augmentation, etc.); prior specification. Prerequisites: statistics and probability at the level of Stats300A, Stats305, and Stats310.
Terms: Win | Units: 3 | Grading: Letter or Credit/No Credit
Instructors: Wong, W. (PI)
## STATS 300A:Theory of Statistics I
Finite sample optimality of statistical procedures; Decision theory: loss, risk, admissibility; Principles of data reduction: sufficiency, ancillarity, completeness; Statistical models: exponential families, group families, nonparametric families; Point estimation: optimal unbiased and equivariant estimation, Bayes estimation, minimax estimation; Hypothesis testing and confidence intervals: uniformly most powerful tests, uniformly most accurate confidence intervals, optimal unbiased and invariant tests. Prerequisites: Real analysis, introductory probability (at the level of STATS 116), and introductory statistics.
Terms: Aut | Units: 2-3 | Grading: Letter or Credit/No Credit
## STATS 370:Bayesian Statistics I (STATS 270)
This is the first of a two course sequence on modern Bayesian statistics. Topics covered include: real world examples of large scale Bayesian analysis; basic tools (models, conjugate priors and their mixtures); Bayesian estimates, tests and credible intervals; foundations (axioms, exchangeability, likelihood principle); Bayesian computations (Gibbs sampler, data augmentation, etc.); prior specification. Prerequisites: statistics and probability at the level of Stats300A, Stats305, and Stats310.
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## The Arc Addendum to the Burning Mirror Solution
DRAFT
The Arc Addendum to the Burning Mirror Solution
”The Math Behind Burning Mirrors” contradicts the widely held view that ”circles make poor approximations for parabolas”. For small angles, the geometry of a circle and a parabola converge to within a few parts per thousand. This is a very good approximation for long focal length mirrors. It also shows that the ancients had the facility to make these mirrors with the pendulum and potter’s wheel construction method.
The huge numbers that emerge from the calculations arise because the curves are getting closer to the perfect parabola. The ideal curve has an infinite concentration factor under the approximation used, where the sun is considered a point source. The theoretical numbers are much too high because the sun is not an exact point, but is actually spread over a few degrees of the sky. The sun’s arc parabola calculation brings the concentration factors down to realistic numbers. The formula for the sun image is independent of dish size and simply proportional to the focal length. It relies on the height and distance to the sun combined with the focal length, the longer the f.p. the larger the sun disk image. This approach actually makes the power levels much easier to calculate.
For a perfect parabola with a focal length of 1m, the real sun image will be 9.2mm, regardless of whether the dish is 5cm or 5m wide. A 2m wide dish with a one-meter focal length delivers a concentration factor of approximately 47MegaWatts per square meter, which is considerable. When the focal length is shortened, the intensity increases. For the same dish with a 0.5m focal length the real sun image is about 4mm wide, which produces an intensity of over 200MegaWatts per square meter. Both of these devices are incredibly powerful even when placed alongside the majority of modern lasers. The weakest one is nearly three times more potent than the solar device used in tests to melt stones and vaporize metals.
Mechanical methods can produce these curves because they are relatively deep and do not entail the precision cutting of the shallower curves. Ancient shield making techniques suffice to make the rough shape, followed by a laborious guided grinding and polishing procedure. This manual aspect can be sped up by using a potter’s wheel in a similar fashion to the long focal length devices. Instead of using a moving grinder on a pendulum, a fixed parabolic shaped grinder would be used. These methods are touched on in the Secrets of the Sun Sects. There is not much debate over ancient abilities to make these shapes since they are widely found in the artifacts from shields to bowls.
Archimedes and Syracuse
The persistent problem has always been how the ancients made long focal length mirrors. This is tightly bound to the famous Archimedes story of burning mirrors at Syracuse. The rationale runs that the ancients could not have burned the roman ships because they could not make or did not have parabolic mirrors with long enough focal lengths. Under the arc approximation, it seems that even if they did have near perfect parabolic mirrors with very long focal lengths it would still be very difficult.
Even if the high quality curves of pendulum method are considered perfect, then using the width of the sun means there will be a large disk of light on the target regardless. If the focal length of the dish is 30m the real sun image will be over 27cm across. If the ship is at 50m the sun disk image will be over 46cm wide. These are huge circles of light and would require similarly huge reflectors to provide enough energy to start a fire.
In ”Secrets of the Sun Sects”, the account of Archimedes Burning Mirrors is concluded with this paragraph.
”This short account neatly summarizes the use of burning mirrors as exceptional weaponry of ancient Greece. It also details the usual counter arguments that make it all seem a ‘bit far-fetched’. The ancient method of manufacture makes not only small short-range mirrors possible, but the technique is scalable for larger, longer focal length reflectors exactly as described. In fact, it appears that the longer range mirrors are easier to produce. A two-meter mirror with a sharp hundred-meter focal point is easier to construct than a fifty-centimeter device with a two-meter focal length. As shown, the power also increases radically as the focal point gets longer, which is counter to many modern methods of build. Whilst this possibility does arise, it is probably a red herring in the search for burning mirrors, misdirection is a useful tool for the concealing historian. The commonly held alternative views are more than likely correct. The Carthagians did have catapults and pitch, which is a much easier combination to fire the approaching boats.”
Under the Fusniak approximation, the statements remain true with the exception of the power increasing as the focal length increases, this only holds true under the point of light approximation. There is however, a small window in which the account could have some validity even under this more accurate calculation. It is linked to the limits of the size and types of devices found in the archaeological record and references to problems the Greeks had extending the range of the burning mirrors.
There is a recently discovered manuscript, which appears to be an Arabic translation of a supposedly lost Greek tract on the theory of conic sections. This is thought to have been written by Archimedes during the 2nd century BCE. This provides some interesting clues.
”The manuscript, written around CE 902, is a translation of a Greek manuscript on the code of research of burning mirrors. It outlined an important application of geometry that developed into new concepts on optics by the 10th century. This is probably the oldest copy of the optics manuscript known, though an identical copy made during the 14th Century exists in India. The manuscript references the burning mirrors of the Greeks, who are said to have discovered how to set light to objects thirty cubits away. They wanted to extend this achievement and meet a challenge to set light to objects at a distance of one hundred cubits. In Alexandria, during the 3rd and 2nd centuries BCE, burning mirrors were an important subject of research. Conon of Alexandria, Archimedes, Dosithcus, and Apollonitis are all described as dabbling in the area.”
If one follows the dimensions mentioned in this paper, a few conclusions can be made with the physics. The ”30 cubits” is about 14m, a range that would probably have put Archimedes life in jeopardy as the Romans arrived. However, this would indicate that the sun image would be about 13cm across. In order for a dish to produce fire with that image, it would need over 120 times the area. This results in a very flat dish with a diameter of 150cm, which is just about in line with the shield sizes of the time.
If the mirror range was the slightly safer distance of 30m the dish would need to be about 3m wide. The ”100 cubit” (45m) goal mentioned would require a dish of over 4.5m, which is probably why it remained just an objective. Whilst sun dishes ”twice the height of a man” have been noted in South America, I am not aware of any that large in ancient Greece. These factors leave the solution to the Burning Mirror problem in tact and the Syracuse story a ‘red herring’ in the search for uses of sun dishes in antiquity. The mirrors can retain their Trojan combat use as blinding devices, but are far less likely to be used as solar cannons for Archimedes.
Implications for the Solar Devices
It is worth summarizing the effects the arc approximation has on the extensive range of devices that are described in Secrets of the Sun Sects. The vast majority of the tools have been tested at least on a scale that is practical. There are a few applications that require adjustment for scaling or mechanical reasons.
SOLAR CHAMBERS: About half of the ancient devices are based on the simple premise that dark stones warm up when exposed to sunlight. The ancient solar chambers that utilize this property only require flat reflectors to work and are completely unaffected by the parabolic amendment. This means the simple and inexpensive domestic and industrial cookers remain perfectly viable. Likewise, the water heating, distillation, sterilizing and pumping equipment is still feasible in the ancient world. There is little doubt given all the evidence that the crop drying techniques were applied on a grand scale. All of these items still perform beautifully at both the small and large scale.
PARABOLIC DEVICES: For the parabolic dish devices, the changes only occur on the implementation side, the abilities remain the same. The power is still there at the heart of the burning mirrors, the intense beam is still the most potent entity in antiquity. The relatively shorter focal lengths mean that the use becomes slightly more restricted. In one or two cases, this means the devices have to employ a secondary reflector, which complicates matters slightly. Whilst these improvements have been added to the modern examples in The Sun Devices, when ancient uses were considered simpler was always chosen over complexity. This was primarily because the tool artifacts found are incomplete so less parts, means more likely.
COOKING: The parabolic cooking methods still hold in their entirety, the potency of the devices at this range remain unchanged, fried foods, solar grilled and even boiled remain unchanged. The ancient soldier could still cook his meal with the upturned shield. Any real volume cooking remains within the domain of the solar chambers mentioned above.
OPTICAL: The optical devices are unchanged by the arc addendum, these mirrors will still make excellent components in any reflector telescope ancient or modern. The implications remain that the ancients were scanning the skies with instruments rather than just the naked eye.
METALWORK: On the materials side, most metals could all still be melted, vaporized or worked whilst hot. The smith still had a solar forge in which he could liquefy metals then cast objects along with the ability to bend and fuse others with the intense heat of the beam. He could fuse some metals with a small dish, but not all. There remains the marginally more complicated method, which involves using an iron heated in the larger dish to carry out the same task. Cutting with light is no longer an option without a secondary dish concentrating the first image.
RECYCLING: Recycling methods remain unchanged, though it would not be possible to wander around a dump vaporizing rubbish with ease. This application was aimed primarily at the modern user, but clearly, in the ancient world metal objects would be recycled when damaged.
REFINING: It is noted that the smith and the ancient alchemist were probably the same person in deep antiquity. He still had the power and ability to readily experiment with refining techniques. He was more restricted in where he could practice this art unless furnished with a large flat dish to wander around. Instead of just pointing the dish at a variety of stones, each would have to be placed within the deep dish to find out if there were any useful effects or products to be had. This new material synthesis concept holds for ancient as well as modern. The largest problem here is that the limits on volumes are more restricted since fresnel arrangements of mirrors cannot be used easily.
STONEWORK: The closely linked areas of stone working and ceramics are the most affected techniques. All of the methods tested still work in exactly the same way provided the object is small. Ceramic pots or stones can have a glaze applied quickly and neatly within the confines of the dish. It is only the larger objects that become more awkward to work. Huge rocks can still be shattered by heat, simply by pointing the beam from a shield-sized dish toward the desired fracture point and pouring water on afterward.
When a huge object is to be glazed where it is standing, there are issues with respect to the dimensions of the dish relative to the focal length. These can be overcome in two ways. The first involves using a larger dish than originally envisioned. The second involves the primitive or double dish Cassegrain set ups described. The advantage of this more complicated arrangement is that it can do the fine work at high powers and apply the finishes at slightly lower power without restrictions. The primitive version involves using a flat panel on the object to reflect light onto a short f.p. dish. The result is an off center beam with lower power but flexibility in use. This is the same method as Lindroth proved with a small 30cm dish and a 2mm beam. It cuts via vaporization. Back from its maximum power, it can also glaze stones with ceramic paints or the natural mica within the stone.
The last is a true Cassegrain device akin to the designs of the modern patented solar cutters/polishers. The first dish points directly at the sun and directs the light to an inline dish. This second mirror reflects the beam back down through a hole in the center of the first. Both dishes are effectively concentrating the light to very high powers, which will cut through just about anything. Obviously, the power can be reduced for other tasks by using the device at a short distance from the true f.p. The issue with this device is that whilst it is relatively simple looking, the geometry is not. Dishes have been found with the necessary hole, dimensions and curves though the tripod for the second dish has not. There are some objects from antiquity that could do the task, but it is much preferable to find them all in one piece or at least in the vicinity. This more complicated set up and geometry may help explain why the stone cutting technique was so easily lost.
GEMS: Gem processing remains an easy and lucrative sideline for the solar artisan. The approximation does not alter the speed, ease or new/old techniques in this arena. The natural gems are simply placed in the beam for partial or complete transforms.
Experimental Confirmation and Failings
The details of the calculation have been checked by much better physicists and engineers than myself and none of them spotted the arc amendment. It seems to have been forgotten behind the headline that spherical reflectors actually make excellent long focal length mirrors and the astounding effects produced in test. To be fair, the guys were more intrigued by the effects and the possibilities raised by cheap long focal length mirrors. The tests were carried out with mirrors with focal lengths of a few meters at most. Invariably those built with the ancient method were small with high fp to dish ratios. This was because aim was to test the theory that the angle of pendulum dictates the accuracy of the curve to a parabola. Deep dishes can be made with standard mechanical methods now and in antiquity, there was no need to test these. All results seemed to fall into line with the predictions of the calculation, given the quality of the devices.
The test for the burning mirrors of Syracuse simply followed on from making small mirrors that could start fires easily. Under the point of light approximation, there was no reason to think that at greater distances the beam would be less intense. Under the arc approximation, the dish has to increase in a proportion equal to the increase in size of the focal point to maintain the potency. As the calculations above show, this does still permit a mirror just at the limit of the technology to burn a ship, but it was right at the limit of ancient mirror construction techniques as well.
The results were materially quite amazing, vaporized rather than melted metals, glass rather than fractured stones, all aspects at achieved at very high temperatures. The problem in experiment seemed to be keeping the power down rather than not enough. The prospect of even higher powers at lower angles seemed to be confirmed by scaled down versions. What we were in fact doing was wandering between the improvements in power caused by more accurate parabola construction and extremely high energy inputs from the larger cruder devices.
Conclusion
The sun’s arc approximation adds limits to the upper power of these devices; it does not change what has been done or what is possible. The ancient method of constructing relatively flat curves still allows for a greater degree of flexibility. The accuracy of the curves to the ideal parabola is still more than adequate for the purposes of burning mirrors. The wide range of applications was still available to the ancient artisans and scientists. The devices are still incredibly useful today.
The new calculation method removes the tendency of the point source calculation to increase to infinity in a neat and elegant way. It puts more realistic power figures on the devices that can be confirmed across all sizes and focal lengths. There is still a matrix of mirror sizes and focal length that need constructing and testing, hopefully, some will try. The new approximation suggested has greatly simplified the method of calculating the potency of these devices, for which the author is grateful.
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CVE 341-Assignement 7_CHAPTER 12
# CVE 341-Assignement 7_CHAPTER 12 - (Answers y 2 =0 a bottom...
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CV Q.1) Water flowing in a wide ch channel. If the flow depth is 1.2 flow is chocked over the bump, a Q.2) Water flow in a wide rectan depth of 1m. Estimate (a) The water depth y (b) The bump height Q.3) A trapezoidal channel with 0.0005 and carries a discharge of a) What is the specific en b) Is the flow subcritical c) What is the alternate d d) What is the critical dep Q.4) Water flows with a velocity What is the change in water surfa elevation produced by a gradual would be the water surface if the Q.5) Referring to Problem 4, wh upstream flow condition? What i without changing the upstream c American University of Sharjah Civil Engineering Department VE 341 – WATER RESOURCES ASSIGNMENT 7 hannel encounters a 22cm high bump at the botto m and the velocity is 2.5 m/s before the bump, d and discuss. (Answers y c =0.97m ngular channel approaches a 10 cm-high bump a 2 over the bump t that will cause the crest flow to be critical.
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Unformatted text preview: (Answers y 2 =0. a bottom width of 5 m and side slopes t = 2 runs f 50 m 3 /s at normal depth. Take Manning’s n as nergy of the flow in the channel? or supercritical? depth to the normal depth? pth in the channel? (Answers E=3.07m, Fr=0.359, y=1 y of 3.0 m/s and a depth of 3.0 m in a rectangula face elevation produced by a gradual upward cha l upward change in bed elevation (hump) of 30 cm ere were a gradual drop of 30 cm? (Answers 0.205m hat is the maximum height of the setup that woul is the minimum width that could exist with a 0.3 condition, knowing that the upstream width is 6.0 (Answers 0.428m om of the determine if the m & E c =1.46m 3 /s) at 1.5 m/s and a .859m & 0.197m) s on a slope of 0.021 (metric). 1.12m, y c =1.72m) ar channel. (a) ange in bed m? (b) What drop, 0.1m rise) ld not affect the 3-m step-up .0 m? , 2.105m, 5.64m)...
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Students can download 4th Maths Term 1 Chapter 4 Measurements Ex 4.1 Questions and Answers, Notes, Samacheer Kalvi 4th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.
## Tamilnadu Samacheer Kalvi 4th Maths Solutions Term 1 Chapter 4 Measurements Ex 4.1
Convert into centimeter
Question 1.
3 m = _____ cm
3 m = 3 × 100 = 300 cm
Question 2.
37 m = _____ cm
37 m = 37 × 100 = 3700 cm
Question
3.5 m 9 cm = _____ cm
5 m 9 cm = 5 × 100 + 9 = 500 + 9 = 509 cm
Question 4.
7 m 35 cm = _____ cm
7 m 35 cm = 7 × 100 + 35 = 700 + 35 = 735 cm
Convert into metre
Question 1.
600 cm = ____ m
600 cm = 600 ÷ 100 = 6m
Question 2.
3600 cm = ____ m
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# Trigonometry
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31/03/2022 0 0
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# Contextualization
## Introduction to Rational and Irrational Numbers
Mathematics is a fascinating subject, with each concept building upon one another to create a comprehensive understanding of the world around us. One such concept is the distinction between rational and irrational numbers.
The number system that we use in our everyday life is a composition of many different types of numbers. Rational numbers are those that can be expressed as a fraction or a ratio of two integers. This includes whole numbers, integers, and fractions. For example, 5, -3, and 1/2 are all rational numbers.
On the other hand, irrational numbers are those that cannot be expressed as a simple fraction. Their decimal representation goes on forever without repeating. Examples include π (pi) and √2 (the square root of 2).
This distinction between rational and irrational numbers is one of the fundamental ideas in mathematics and has a profound impact on many areas, including algebra, calculus, and even our understanding of the physical world.
## Significance of Rational and Irrational Numbers
Rational and irrational numbers are not just abstract concepts that mathematicians like to think about. They have real-world applications and are used in many practical situations.
For instance, irrational numbers are used in engineering and physics to calculate measurements and quantities. They are also used in architecture and design to create aesthetically pleasing structures.
Rational numbers, on the other hand, are found in everyday life. From cooking recipes to calculating distances, we use rational numbers constantly, even if we don't realize it.
Understanding the distinction between rational and irrational numbers is not only important for success in math class but also for understanding the world around us.
## Resources
To help you dive deeper into the topics of rational and irrational numbers, here are some trusted resources that you can refer to:
1. Khan Academy: Provides detailed video lessons and quizzes on rational and irrational numbers.
2. Math is Fun: A comprehensive guide to rational and irrational numbers with interactive examples and challenges.
3. BBC Bitesize: Offers a clear and concise overview of rational and irrational numbers with a focus on their real-world applications.
4. Wolfram MathWorld: An extensive encyclopedia of mathematics that provides in-depth articles on different concepts, including rational and irrational numbers.
Remember, these resources are just a starting point. Feel free to explore other sources to enrich your understanding of rational and irrational numbers.
# Practical Activity
## Objective of the Project
The main objective of this project is to encourage students to deepen their understanding of rational and irrational numbers while developing their research, collaboration, and creative problem-solving skills. By the end of the project, students should be able to:
1. Differentiate between rational and irrational numbers.
2. Identify and classify real-world examples of rational and irrational numbers.
3. Understand the historical and theoretical background of rational and irrational numbers.
4. Present their findings in a written report and a creative visual representation.
## Detailed Description of the Project
In groups of 3 to 5, students will research and discuss the concepts of rational and irrational numbers. They will delve into the theoretical underpinnings, explore real-world applications, and discover significant contributors to the field. Moreover, they will create a visual representation of their understanding and findings, which could be in the form of a poster, a video, a slideshow, or a website.
The students will be given two weeks to complete the project, with an estimated workload of 12 to 15 hours per student. The first week will be dedicated to research and discussion, while the second week will focus on creating the visual representation and writing the report.
## Necessary Materials
• Internet access for research.
• Library access for additional resources.
• Notebooks and writing materials for note-taking and brainstorming.
• Art supplies for creating the visual representation.
## Detailed Step-by-Step for Carrying out the Activity
1. Form Groups and Assign Roles (1 hour): Students should form groups of 3 to 5 and assign roles among themselves. Roles could include researcher, writer, presenter, creative director, etc.
2. Research (8-10 hours): Each group should spend the first week researching rational and irrational numbers. They should look for real-world examples, historical context, and theoretical explanations. They can use the resources provided in the introduction, as well as any other reliable sources they find.
3. Discussion (2-3 hours): After conducting their research, the group should spend time discussing their findings. They should ensure that everyone in the group understands the concepts and is ready to explain them.
4. Creation of Visual Representation (2-3 hours): Using their understanding and findings, the group should create a visual representation of rational and irrational numbers. This could be a poster, a video, a slideshow, or a website.
5. Report Writing (4-5 hours): The final step is to write a report detailing their research, discussion, visual representation, and conclusions. The report should be divided into four main sections: Introduction, Development, Conclusions, and Used Bibliography.
• Introduction: The students should contextualize the theme, explain its relevance, and state the objective of the project.
• Development: Here, the students should detail the theory behind rational and irrational numbers, explain the activity in detail, and present and discuss their findings.
• Conclusion: The students should revisit the main points of the project, explicitly state the learnings obtained and the conclusions drawn about rational and irrational numbers.
• Bibliography: The students should list all the resources they used during the project.
6. Presentation (30 minutes to 1 hour): Each group will present their visual representation and report to the class. The presentation should be clear, engaging, and informative.
## Project Deliverables
At the end of the project, each group will submit:
1. A visual representation of rational and irrational numbers.
2. A written report detailing their research, discussion, and conclusions about rational and irrational numbers.
Through this multifaceted project, students will not only deepen their understanding of rational and irrational numbers but also develop important skills such as research, collaboration, problem-solving, and effective communication.
Math
# Contextualization
Scatter plots, also known as scatter diagrams or scatter graphs, are mathematical tools used to investigate the relationship between two sets of data. These plots are a visual representation of data points that show how much one variable is affected by another. They are particularly useful when there is a large amount of data and you want to identify any patterns or correlations.
In a scatter plot, each dot represents a single data point, with the position of the dot indicating the values for the two variables. The closer the dots are to a straight line, the stronger the relationship between the two variables. If the line slopes upwards from left to right, it indicates a positive correlation, while a downward slope signifies a negative correlation. A flat line indicates no correlation.
Scatter plots are not only useful for visualizing data, but they also have a practical application in the real world. They are widely used in science, engineering, finance, and many other fields to understand the relationship between two variables and make predictions based on this relationship. For example, they can be used to predict how the price of a product will change based on its demand, or how the temperature will affect the growth of a plant.
# Importance of Scatter Plots
Scatter plots are a fundamental tool in data analysis and are one of the first steps in understanding the relationship between two variables. They allow us to see patterns and trends in the data that may not be apparent from just looking at the raw numbers. This makes them an important tool for scientists, researchers, and anyone who deals with large amounts of data.
In addition, scatter plots can also be used to model data. This means that once we have identified a pattern or trend in the data, we can use this to make predictions about future data points. This is particularly valuable in fields such as finance, where being able to predict future trends can help make better investment decisions.
Understanding scatter plots and how to interpret them is therefore not only a useful mathematical skill but also an important skill in many real-world applications. By the end of this project, you will be able to confidently create and interpret scatter plots, and use them to make predictions and model data.
# Resources
2. Interactive Scatter Plot Tutorial
3. BBC Bitesize: Scatter Graphs
4. Math is Fun: Scatter Plots
5. Book: "Statistics and Data Analysis for the Behavioral Sciences", by Dana S. Dunn, Suzanne Mannes, and Stephen G. West.
You will find these resources helpful in understanding the theory and practical application of scatter plots.
# Practical Activity
## Objective of the Project:
The main objective of this project is to enable students to create and interpret scatter plots. The students will work in groups to collect data, construct a scatter plot, interpret the plot to identify relationships, and use the plot to make predictions.
## Detailed Description of the Project:
In this project, students will work in groups of 3 to 5 to collect data on two variables of their choice. They will then plot this data on a scatter plot, interpret the plot, and use it to make predictions. The data can be collected from any reliable source or can be gathered by students themselves (for example, by conducting a survey). The project will be conducted over a period of one week, with each group expected to spend approximately 4 to 6 hours on the project.
## Necessary Materials:
• A computer or laptop with internet access for research and data analysis
• A notebook for recording data and observations
• Graphing paper or a computer program for creating scatter plots
• A ruler or a computer program for plotting the data accurately
• Calculator (for calculating statistical parameters, if necessary)
## Detailed Step-by-Step for Carrying out the Activity:
1. Choose a Topic: Start by choosing a topic for the project. This can be anything that has two measurable variables that you can collect data on. For example, you could choose the number of hours of study and the test score, the temperature and the number of ice cream cones sold, or the amount of rainfall and the number of plants in a garden.
2. Collect Data: Once you have chosen your topic, start collecting data on your two variables. This can be done by conducting a survey, researching online, or using data from a reliable source.
3. Organize and Analyze Data: Once you have collected your data, organize it in a table or spreadsheet. Then, calculate any necessary statistical parameters, such as the mean or standard deviation, that you may need later.
4. Create the Scatter Plot: Using your organized data, create a scatter plot. This can be done on paper or using a computer program. Make sure to label your axes and include a title.
5. Interpret the Scatter Plot: Look at your scatter plot and try to identify any patterns or relationships. Is the relationship between the two variables positive, negative, or none? How strong is the relationship? Are there any outliers?
6. Make Predictions: Based on your scatter plot, make some predictions. For example, if your scatter plot shows a positive relationship between hours of study and test score, you could predict that someone who studies for 10 hours will get a higher test score than someone who studies for 5 hours.
7. Write the Report: Finally, write a detailed report of your project. This report should include an introduction (where you explain the project and its relevance), a development section (where you detail the theory behind scatter plots, explain the steps you took to create your plot, and discuss your findings), a conclusion (where you summarize what you learned from the project), and a bibliography (where you list the sources you used for the project). Remember, this report should be written in a clear, concise, and engaging way.
## Project Deliverables:
At the end of this project, each group is expected to submit a written report and a scatter plot. The scatter plot should be neat, accurate, and clearly labeled. The report should be written in a clear, concise, and engaging way, and should include an introduction, a development section, a conclusion, and a bibliography.
The introduction should provide context for the project, explain the chosen topic, and state the objective of the project. The development section should detail the theory behind scatter plots, explain the steps taken to create the scatter plot, and discuss the findings. The conclusion should summarize the main points of the project and state what the group learned from the project. Finally, the bibliography should list all the sources used in the project.
The report should be a reflection of the group's understanding of scatter plots, their ability to collect and analyze data, and their problem-solving and teamwork skills. The scatter plot should be a clear and accurate representation of the data, and should show the group's ability to interpret and use the plot to make predictions.
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Math
# Contextualization
## Introduction to Spatial Geometry and the Volume of the Prism
Geometry is the mathematical study of shapes and their properties. In our journey of understanding this branch of mathematics, we've explored the concepts of lines, angles, and polygons. Now, we're going to delve into the fascinating world of spatial geometry, where we deal with three-dimensional shapes.
One crucial concept in spatial geometry is the concept of volume. Volume is the amount of space that a three-dimensional shape, like a prism, occupies. It is measured in cubic units, such as cubic meters (m^3), cubic centimeters (cm^3), or cubic inches (in^3).
A prism is a three-dimensional solid with two identical, parallel bases that are connected by rectangular faces. The bases are always the same shape and the same size. The height of the prism is the perpendicular distance between the two bases. The volume of a prism is the product of the area of one of its bases and its height.
To calculate the volume of a prism, we use a simple formula: Volume = Base Area x Height. By understanding this formula, we can quickly determine the volume of any prism, regardless of its size or shape.
## Importance of Volume Calculation in Real Life
The concept of volume, especially that of a prism, is not just an abstract mathematical concept. It has several practical applications in our everyday lives and various fields of work.
For instance, architects and engineers use the concept of volume to determine the amount of space a building will occupy. This helps them plan and design structures more efficiently. Similarly, in construction, workers need to calculate the volume of materials like concrete or gravel to know how much they need for a project.
Moreover, understanding volume can help in tasks as simple as cooking. When you're following a recipe and need to figure out how much space a particular ingredient will occupy, you're essentially calculating its volume.
## Reliable Resources for Further Understanding
For a deeper understanding of the concept of volume of a prism and its applications, you can refer to the following resources:
Using these resources, you can not only gain a better understanding of the concept but also explore its real-world applications.
# Practical Activity
## Objective of the Project
The objective of this project is to not only apply the formula for calculating the volume of a prism but also to deepen your understanding of this concept by constructing various prisms using everyday materials and comparing their volumes.
## Detailed Description of the Project
In groups of 3 to 5, students will construct different prisms using materials like cardboard, paper, or plastic, and calculate their volumes. The prisms can be of any shape (triangular, rectangular, hexagonal, etc.) as long as they fit the definition of a prism. You will then compare the volumes of these prisms, discuss your findings, and present them in a comprehensive report.
## Necessary Materials
1. Cardboard or any other material that can be used to create prisms.
2. Ruler or measuring tape.
3. Scissors.
4. Glue or tape.
5. Protractor (if you're making prisms with non-rectangular bases).
6. Calculator.
## Detailed Step-by-Step for Carrying Out the Activity
1. Formation of Groups: Form groups of 3 to 5 students. Each group will be assigned different types of prisms to construct and calculate their volumes.
2. Research and Planning: Begin by researching the properties of the assigned type of prism. Understand its shape, the formula for calculating its volume, and its real-world applications. Plan how you are going to construct the prism.
3. Prism Construction: Using the materials provided, construct the assigned prism. Ensure that the dimensions of your prism are accurate.
4. Volume Calculation: Calculate the volume of your prism using the formula: Volume = Base Area x Height.
5. Documentation: Document the steps you took to construct the prism and calculate its volume. Also, note down any observations or difficulties you faced during the process.
6. Repeat Steps 2-5: Repeat steps 2 to 5 for each type of prism assigned to your group.
7. Comparison and Discussion: Compare the volumes of the different prisms you constructed. Can you find any patterns or relationships? Discuss your findings with the rest of the group.
8. Report Writing: Based on your findings and discussions, write a comprehensive report on your project. The report should be structured as follows:
• Introduction: Contextualize the theme, its relevance, and real-world application. State the objective of this project.
• Development: Detail the theory behind the volume of a prism, explain the steps of your project, and discuss your findings. Include any images or diagrams that can help illustrate your work.
• Conclusion: Summarize the main points of the project, state the learnings obtained, and draw conclusions about the project.
• Used Bibliography: Indicate the sources you relied on to work on the project.
## Project Deliveries and Duration
This project should be completed within a month. Each group will deliver a constructed prism, documented process, and a comprehensive report. The report should not only detail the steps you took and the results you obtained but also reflect on the learnings you gained from the project. It should be properly structured, well-written, and well-presented, with clear and concise language. It should also include visual aids, such as diagrams or photographs, to enhance understanding.
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Math
# Contextualization
## Introduction to Logarithms
Logarithms are an important concept in mathematics that play a significant role in various fields, including science, engineering, and finance. They are a way of expressing numbers that are too large or too small to be conveniently written or manipulated in their usual form. The concept of logarithms was first introduced by John Napier in the early 17th century and later developed by mathematicians such as Johannes Kepler and Henry Briggs.
A logarithm is the inverse operation of exponentiation. In simple terms, a logarithm is the power to which a number (called the base) must be raised to give another number. For example, in the equation 10^2 = 100, the '2' is the logarithm of 100. This is because 10 raised to the power of 2 equals 100. In this case, the logarithm is said to have a base of 10.
The logarithm with base 10 (written as log10) is called the common logarithm. Another commonly used base is the natural logarithm, which has a base of the mathematical constant 'e' (approximately 2.718). Logarithms can also have different bases, such as 2 or any other positive number.
## Importance and Applications of Logarithms
Logarithms are used to simplify complex calculations, especially those involving large numbers or numbers with many decimal places. They can also transform multiplicative operations into additive ones, making calculations easier. Logarithms have numerous applications in real-world scenarios, some of which include:
1. Exponential growth and decay: Logarithms can be used to model exponential growth and decay processes, such as population growth and radioactive decay.
2. Sound and light intensity: Logarithmic scales, such as the Richter scale for measuring earthquake magnitudes or the decibel scale for sound intensity, are used to compare values that span a wide range.
3. pH scale: The pH scale, which measures the acidity or alkalinity of a solution, is logarithmic.
4. Computer science: Logarithms are used in computer science and information theory to calculate the complexity of algorithms and to measure data compression.
In this project, we will delve into the world of logarithms, understanding their fundamental properties, learning to solve logarithmic equations, and exploring their real-world applications.
## Suggested Resources
2. Math is Fun: Logarithms
3. Brilliant: Logarithms
5. Book: "Precalculus Mathematics in a Nutshell: Geometry, Algebra, Trigonometry" by George F. Simmons
These resources provide a solid introduction to logarithms, offer numerous examples and practice exercises, and delve into their applications in the real world. Don't hesitate to use them as a starting point for your research and exploration of this fascinating mathematical concept.
# Practical Activity
## Objective of the Project:
This activity aims to provide students with a hands-on experience in understanding and working with logarithms. The students will explore the properties of logarithms, learn to solve logarithmic equations, and apply logarithms to real-world problems.
## Detailed Description of the Project:
This group project will involve students in a series of engaging and interactive tasks. The tasks will include:
1. Exploration of Logarithmic Properties: Students will explore the properties of logarithms, including the Product Rule, Quotient Rule, and Power Rule. This will involve simple calculations and problem-solving exercises.
2. Solving Logarithmic Equations: Students will learn how to solve logarithmic equations by using the properties of logarithms. They will be provided with a variety of equations to solve.
3. Application of Logarithms: Students will apply their knowledge of logarithms to solve real-world problems. They will be given scenarios where logarithms can be used, and they will have to formulate and solve the corresponding logarithmic equations.
## Necessary Materials:
• Paper and Pencils
• Calculators (optional)
## Detailed Step by Step for Carrying out the Activity:
1. Logarithmic Properties Exploration: Each group will be given a set of logarithmic properties to explore. The group members will work together to understand and apply these properties in solving simple logarithmic problems.
2. Solving Logarithmic Equations: The groups will be provided with a set of logarithmic equations to solve. They will use their understanding of logarithmic properties to solve these equations step by step.
3. Application of Logarithms: The groups will be given a set of real-world problems where logarithms can be applied. They will have to identify the logarithmic equation that represents the problem and solve it to find the solution.
4. Group Discussion and Conclusion: After completing the tasks, each group will discuss their findings and understanding of logarithms. They will then prepare a report summarizing their work and findings.
## Project Deliverables:
1. Written Report: The report should be structured as follows:
• Introduction: Describe the concept of logarithms, their relevance and real-world applications, and the objective of this project.
• Development: Detail the theory behind logarithms, the activities performed, the methodology used, and the obtained results. Include explanations of the logarithmic properties, solving logarithmic equations, and the application of logarithms in the real world. Discuss the process of group work, the challenges faced, and how they were overcome.
• Conclusions: Conclude the report by summarizing the main points, the learnings obtained, and the conclusions drawn about the project.
• Bibliography: Indicate the sources used to gather information or to aid in understanding the logarithmic concepts and solving the problems.
2. Presentation: Each group will present their findings to the class. The presentation should include a brief overview of logarithms, a discussion of the activities and methodology used, and a summary of the results and learnings.
This project is expected to take one week, with each group spending approximately three to five hours on it. It will not only test your understanding of logarithms but also your ability to work collaboratively, think critically, and solve problems creatively. Enjoy your journey into the world of logarithms!
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New Online Book! Handbook of Mathematical Functions (AMS55)
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Convert From: (required) Click here to Convert To: (optional) Examples: 5 kilometers, 12 feet/sec^2, 1/5 gallon, 9.5 Joules, or 0 dF. Help, Frequently Asked Questions, Use Currencies in Conversions, Measurements & Currencies Recognized Examples: miles, meters/s^2, liters, kilowatt*hours, or dC.
Conversion Result: ```Indian rupee on 4-28-2017 = 1.55579948774241E-02 United States dollar on 4-28-2017 (currency)``` Related Measurements: Try converting from "INR" to AUD (Australian dollar on 4-28-2017), BND (Bruneian dollar on 4-28-2017), BTN (Bhutanese ngultrum on 4-28-2017), CAD (Canadian dollar on 4-28-2017), DEM (German mark on 4-28-2017), DKK (Danish krone on 4-28-2017), ESP (Spanish peseta on 4-28-2017), FJD (Fijian dollar on 4-28-2017), GIP (Gibraltar pound on 4-28-2017), ISK (Icelandic krona on 4-28-2017), JMD (Jamaican dollar on 4-28-2017), MXN (Mexican peso on 4-28-2017), NZD (New Zealand dollar on 4-28-2017), OMR (Omani rial on 4-28-2017), PEN (Peruvian nuevo sol on 4-28-2017), PLN (Polish zloty on 4-28-2017), PTE (Portuguese escudo on 4-28-2017), SEK (Swedish krona on 4-28-2017), SGD (Singapore dollar on 4-28-2017), TWD (Taiwanese new dollar on 4-28-2017), or any combination of units which equate to "currency" and represent currency. Sample Conversions: INR = .02769311 AWG (Aruban guilder on 4-28-2017), .03111599 BZD (Belizean dollar on 4-28-2017), .02126 CAD (Canadian dollar on 4-28-2017), .01547195 CHF (Swiss franc on 4-28-2017), 2.76 DJF (Djiboutian franc on 4-28-2017), .10624688 DKK (Danish krone on 4-28-2017), .08491047 FIM (Finnish markka on 4-28-2017), .01202625 FKP (Falkland pound on 4-28-2017), .01202625 GBP (British pound sterling on 4-28-2017), .10656642 HRK (Croatian kuna on 4-28-2017), .01124714 IEP (Irish pound on 4-28-2017), .05637762 ILS (Israeli new shekel on 4-28-2017), .15439361 MAD (Moroccan dirham on 4-28-2017), .02267975 NZD (New Zealand dollar on 4-28-2017), 2.86 PTE (Portuguese escudo on 4-28-2017), .49785584 tick, .03708355 TND (Tunisian dinar on 4-28-2017), .46993811 TWD (Taiwanese new dollar on 4-28-2017), .01555799 USD (United States dollar on 4-28-2017), .04185039 XCD (East Caribbean dollar on 4-28-2017).
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https://byjus.com/question-answer/diet-problem-a-dietician-wishes-to-mix-two-types-of-foods-in-such-a-way/
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Question
# (Diet problem): A dietician wishes to mix two types of foods in such a way that vitamin contents of the mixture contain atleast $$8$$ units of vitamin A and $$10$$ units of vitamin C. Food I contains $$2$$ units/kg of vitamin A and $$1$$ unit/kg of vitamin C. Food II contains $$1$$ unit/kg of vitamin A and $$2$$ units/kg of vitamin C. It costs $$Rs 50$$ per kg to purchase Food I and $$Rs. 70$$ per kg to purchase Food II. Formulate this problem as a linear programming problem to minimise the cost of such a mixture.
Solution
## Let the mixture contain $$x$$ kg of Food I and $$y$$ kg of Food II. Clearly, $$x \geq 0,y \geq 0$$. We make the following table from the given data:ResourcesFoodI$$(x)$$FoodII$$(y)$$RequirementVitamins A (units/ kg)$$2$$$$1$$$$8$$Vitaminc C (units/ kg)$$1$$$$2$$$$10$$Cost (Rs/ kg)$$50$$$$70$$Since the mixture must contain at least $$8$$ units of vitamin A and $$10$$ units of vitamin C, we have the constraints:$$2x + y \geq 8$$$$x + 2y \geq 10$$Total cost $$Z$$ of purchasing $$x$$ kg of food I and $$y$$ kg of Food II is$$Z = 50x 70y$$Hence, the mathematical formulation of the problem is:Minimise $$Z = 50x + 70y ... (1)$$subject to the constraints :$$2x + y \geq 8 ... (2)$$$$x + 2y \geq 10 ... (3)$$$$x, y \geq 0 ... (4)$$Let us the graph the inequalities $$(2)$$ to $$(4)$$. The feasible region determined by the system is shown in the Fig. Here again, observe that the feasible region is unbounded.Let us evaluate $$Z$$ at the corner points $$A(0,8), B(2,4)$$ and $$C(10,0)$$.Corner Point$$Z = 50x + 70y$$$$(0, 8)$$$$560$$$$(2, 4)$$$$380\leftarrow Minimum$$$$(10, 0)$$$$500$$In the table, we find that smallest value of $$Z$$ is $$380$$ at the point $$(2,4)$$. Can we say that the minimum value of $$Z$$ is $$380$$? Remember that the feasible region is unbounded. Therefore, we have to draw the graph of the inequality$$50x + 70y < 380$$ i.e., $$5x + 7y < 38$$to check whether the resulting open half plane has any point common with the feasible region. From the Fig, we see that it has no points in common.Thus, the minimum value of $$Z$$ is $$380$$ attained at the point $$(2, 4)$$. Hence, the optimal mixing strategy for the dietician would be to mix $$2$$ kg of Food I and $$4\ kg$$ of Food II,and with this strategy, the minimum cost of the mixture will be $$Rs. 380$$.Mathematics
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https://terrytao.wordpress.com/2015/08/06/equidistribution-for-multidimensional-polynomial-phases/
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The equidistribution theorem asserts that if ${\alpha \in {\bf R}/{\bf Z}}$ is an irrational phase, then the sequence ${(n\alpha)_{n=1}^\infty}$ is equidistributed on the unit circle, or equivalently that
$\displaystyle \frac{1}{N} \sum_{n=1}^N F(n\alpha) \rightarrow \int_{{\bf R}/{\bf Z}} F(x)\ dx$
for any continuous (or equivalently, for any smooth) function ${F: {\bf R}/{\bf Z} \rightarrow {\bf C}}$. By approximating ${F}$ uniformly by a Fourier series, this claim is equivalent to that of showing that
$\displaystyle \frac{1}{N} \sum_{n=1}^N e(hn\alpha) \rightarrow 0$
for any non-zero integer ${h}$ (where ${e(x) := e^{2\pi i x}}$), which is easily verified from the irrationality of ${\alpha}$ and the geometric series formula. Conversely, if ${\alpha}$ is rational, then clearly ${\frac{1}{N} \sum_{n=1}^N e(hn\alpha)}$ fails to go to zero when ${h}$ is a multiple of the denominator of ${\alpha}$.
One can then ask for more quantitative information about the decay of exponential sums of ${\frac{1}{N} \sum_{n=1}^N e(n \alpha)}$, or more generally on exponential sums of the form ${\frac{1}{|Q|} \sum_{n \in Q} e(P(n))}$ for an arithmetic progression ${Q}$ (in this post all progressions are understood to be finite) and a polynomial ${P: Q \rightarrow \/{\bf Z}}$. It will be convenient to phrase such information in the form of an inverse theorem, describing those phases for which the exponential sum is large. Indeed, we have
Lemma 1 (Geometric series formula, inverse form) Let ${Q \subset {\bf Z}}$ be an arithmetic progression of length at most ${N}$ for some ${N \geq 1}$, and let ${P(n) = n \alpha + \beta}$ be a linear polynomial for some ${\alpha,\beta \in {\bf R}/{\bf Z}}$. If
$\displaystyle \frac{1}{N} |\sum_{n \in Q} e(P(n))| \geq \delta$
for some ${\delta > 0}$, then there exists a subprogression ${Q'}$ of ${Q}$ of size ${|Q'| \gg \delta^2 N}$ such that ${P(n)}$ varies by at most ${\delta}$ on ${Q'}$ (that is to say, ${P(n)}$ lies in a subinterval of ${{\bf R}/{\bf Z}}$ of length at most ${\delta}$).
Proof: By a linear change of variable we may assume that ${Q}$ is of the form ${\{0,\dots,N'-1\}}$ for some ${N' \geq 1}$. We may of course assume that ${\alpha}$ is non-zero in ${{\bf R}/{\bf Z}}$, so that ${\|\alpha\|_{{\bf R}/{\bf Z}} > 0}$ (${\|x\|_{{\bf R}/{\bf Z}}}$ denotes the distance from ${x}$ to the nearest integer). From the geometric series formula we see that
$\displaystyle |\sum_{n \in Q} e(P(n))| \leq \frac{2}{|e(\alpha) - 1|} \ll \frac{1}{\|\alpha\|_{{\bf R}/{\bf Z}}},$
and so ${\|\alpha\|_{{\bf R}/{\bf Z}} \ll \frac{1}{\delta N}}$. Setting ${Q' := \{ n \in Q: n \leq c \delta^2 N \}}$ for some sufficiently small absolute constant ${c}$, we obtain the claim. $\Box$
Thus, in order for a linear phase ${P(n)}$ to fail to be equidistributed on some long progression ${Q}$, ${P}$ must in fact be almost constant on large piece of ${Q}$.
As is well known, this phenomenon generalises to higher order polynomials. To achieve this, we need two elementary additional lemmas. The first relates the exponential sums of ${P}$ to the exponential sums of its “first derivatives” ${n \mapsto P(n+h)-P(n)}$.
Lemma 2 (Van der Corput lemma, inverse form) Let ${Q \subset {\bf Z}}$ be an arithmetic progression of length at most ${N}$, and let ${P: Q \rightarrow {\bf R}/{\bf Z}}$ be an arbitrary function such that
$\displaystyle \frac{1}{N} |\sum_{n \in Q} e(P(n))| \geq \delta \ \ \ \ \ (1)$
for some ${\delta > 0}$. Then, for ${\gg \delta^2 N}$ integers ${h \in Q-Q}$, there exists a subprogression ${Q_h}$ of ${Q}$, of the same spacing as ${Q}$, such that
$\displaystyle \frac{1}{N} |\sum_{n \in Q_h} e(P(n+h)-P(n))| \gg \delta^2. \ \ \ \ \ (2)$
Proof: Squaring (1), we see that
$\displaystyle \sum_{n,n' \in Q} e(P(n') - P(n)) \geq \delta^2 N^2.$
We write ${n' = n+h}$ and conclude that
$\displaystyle \sum_{h \in Q-Q} \sum_{n \in Q_h} e( P(n+h)-P(n) ) \geq \delta^2 N^2$
where ${Q_h := Q \cap (Q-h)}$ is a subprogression of ${Q}$ of the same spacing. Since ${\sum_{n \in Q_h} e( P(n+h)-P(n) ) = O(N)}$, we conclude that
$\displaystyle |\sum_{n \in Q_h} e( P(n+h)-P(n) )| \gg \delta^2 N$
for ${\gg \delta^2 N}$ values of ${h}$ (this can be seen, much like the pigeonhole principle, by arguing via contradiction for a suitable choice of implied constants). The claim follows. $\Box$
The second lemma (which we recycle from this previous blog post) is a variant of the equidistribution theorem.
Lemma 3 (Vinogradov lemma) Let ${I \subset [-N,N] \cap {\bf Z}}$ be an interval for some ${N \geq 1}$, and let ${\theta \in{\bf R}/{\bf Z}}$ be such that ${\|n\theta\|_{{\bf R}/{\bf Z}} \leq \varepsilon}$ for at least ${\delta N}$ values of ${n \in I}$, for some ${0 < \varepsilon, \delta < 1}$. Then either
$\displaystyle N < \frac{2}{\delta}$
or
$\displaystyle \varepsilon > 10^{-2} \delta$
or else there is a natural number ${q \leq 2/\delta}$ such that
$\displaystyle \| q \theta \|_{{\bf R}/{\bf Z}} \ll \frac{\varepsilon}{\delta N}.$
Proof: We may assume that ${N \geq \frac{2}{\delta}}$ and ${\varepsilon \leq 10^{-2} \delta}$, since we are done otherwise. Then there are at least two ${n \in I}$ with ${\|n \theta \|_{{\bf R}/{\bf Z}} \leq \varepsilon}$, and by the pigeonhole principle we can find ${n_1 < n_2}$ in ${Q}$ with ${\|n_1 \theta \|_{{\bf R}/{\bf Z}}, \|n_2 \theta \|_{{\bf R}/{\bf Z}} \leq \varepsilon}$ and ${n_2-n_1 \leq \frac{2}{\delta}}$. By the triangle inequality, we conclude that there exists at least one natural number ${q \leq \frac{2}{\delta}}$ for which
$\displaystyle \| q \theta \|_{{\bf R}/{\bf Z}} \leq 2\varepsilon.$
We take ${q}$ to be minimal amongst all such natural numbers, then we see that there exists ${a}$ coprime to ${q}$ and ${|\kappa| \leq 2\varepsilon}$ such that
$\displaystyle \theta = \frac{a}{q} + \frac{\kappa}{q}. \ \ \ \ \ (3)$
If ${\kappa=0}$ then we are done, so suppose that ${\kappa \neq 0}$. Suppose that ${n < m}$ are elements of ${I}$ such that ${\|n\theta \|_{{\bf R}/{\bf Z}}, \|m\theta \|_{{\bf R}/{\bf Z}} \leq \varepsilon}$ and ${m-n \leq \frac{1}{10 \kappa}}$. Writing ${m-n = qk + r}$ for some ${0 \leq r < q}$, we have
$\displaystyle \| (m-n) \theta \|_{{\bf R}/{\bf Z}} = \| \frac{ra}{q} + (m-n) \frac{\kappa}{q} \|_{{\bf R}/{\bf Z}} \leq 2\varepsilon.$
By hypothesis, ${(m-n) \frac{\kappa}{q} \leq \frac{1}{10 q}}$; note that as ${q \leq 2/\delta}$ and ${\varepsilon \leq 10^{-2} \delta}$ we also have ${\varepsilon \leq \frac{1}{10q}}$. This implies that ${\| \frac{ra}{q} \|_{{\bf R}/{\bf Z}} < \frac{1}{q}}$ and thus ${r=0}$. We then have
$\displaystyle |k \kappa| \leq 2 \varepsilon.$
We conclude that for fixed ${n \in I}$ with ${\|n\theta \|_{{\bf R}/{\bf Z}} \leq \varepsilon}$, there are at most ${\frac{2\varepsilon}{|\kappa|}}$ elements ${m}$ of ${[n, n + \frac{1}{10 |\kappa|}]}$ such that ${\|m\theta \|_{{\bf R}/{\bf Z}} \leq \varepsilon}$. Iterating this with a greedy algorithm, we see that the number of ${n \in I}$ with ${\|n\theta \|_{{\bf R}/{\bf Z}} \leq \varepsilon}$ is at most ${(\frac{N}{1/10|\kappa|} + 1) 2\varepsilon/|\kappa|}$; since ${\varepsilon < 10^{-2} \delta}$, this implies that
$\displaystyle \delta N \ll 2 \varepsilon / \kappa$
and the claim follows. $\Box$
Now we can quickly obtain a higher degree version of Lemma 1:
Proposition 4 (Weyl exponential sum estimate, inverse form) Let ${Q \subset {\bf Z}}$ be an arithmetic progression of length at most ${N}$ for some ${N \geq 1}$, and let ${P: {\bf Z} \rightarrow {\bf R}/{\bf Z}}$ be a polynomial of some degree at most ${d \geq 0}$. If
$\displaystyle \frac{1}{N} |\sum_{n \in Q} e(P(n))| \geq \delta$
for some ${\delta > 0}$, then there exists a subprogression ${Q'}$ of ${Q}$ with ${|Q'| \gg_d \delta^{O_d(1)} N}$ such that ${P}$ varies by at most ${\delta}$ on ${Q'}$.
Proof: We induct on ${d}$. The cases ${d=0,1}$ are immediate from Lemma 1. Now suppose that ${d \geq 2}$, and that the claim had already been proven for ${d-1}$. To simplify the notation we allow implied constants to depend on ${d}$. Let the hypotheses be as in the proposition. Clearly ${\delta}$ cannot exceed ${1}$. By shrinking ${\delta}$ as necessary we may assume that ${\delta \leq c}$ for some sufficiently small constant ${c}$ depending on ${d}$.
By rescaling we may assume ${Q \subset [0,N] \cap {\bf Z}}$. By Lemma 3, we see that for ${\gg \delta^2 N}$ choices of ${h \in [-N,N] \cap {\bf Z}}$ such that
$\displaystyle \frac{1}{N} |\sum_{n \in I_h} e(P(n+h) - P(n))| \gg \delta^2$
for some interval ${I_h \subset [0,N] \cap {\bf Z}}$. We write ${P(n) = \sum_{i \leq d} \alpha_i n^i}$, then ${P(n+h)-P(n)}$ is a polynomial of degree at most ${d-1}$ with leading coefficient ${h \alpha_d n^{d-1}}$. We conclude from induction hypothesis that for each such ${h}$, there exists a natural number ${q_h \ll \delta^{-O(1)}}$ such that ${\|q_h h \alpha_d \|_{{\bf R}/{\bf Z}} \ll \delta^{-O(1)} / N^{d-1}}$, by double-counting, this implies that there are ${\gg \delta^{O(1)} N}$ integers ${n}$ in the interval ${[-\delta^{-O(1)} N, \delta^{-O(1)} N] \cap {\bf Z}}$ such that ${\|n \alpha_d \|_{{\bf R}/{\bf Z}} \ll \delta^{-O(1)} / N^{d-1}}$. Applying Lemma 3, we conclude that either ${N \ll \delta^{-O(1)}}$, or that
$\displaystyle \| q \alpha_d \|_{{\bf R}/{\bf Z}} \ll \delta^{-O(1)} / N^d. \ \ \ \ \ (4)$
In the former case the claim is trivial (just take ${Q'}$ to be a point), so we may assume that we are in the latter case.
We partition ${Q}$ into arithmetic progressions ${Q'}$ of spacing ${q}$ and length comparable to ${\delta^{-C} N}$ for some large ${C}$ depending on ${d}$ to be chosen later. By hypothesis, we have
$\displaystyle \frac{1}{|Q|} |\sum_{n \in Q} e(P(n))| \geq \delta$
so by the pigeonhole principle, we have
$\displaystyle \frac{1}{|Q'|} |\sum_{n \in Q'} e(P(n))| \geq \delta$
for at least one such progression ${Q'}$. On this progression, we may use the binomial theorem and (4) to write ${\alpha_d n^d}$ as a polynomial in ${n}$ of degree at most ${d-1}$, plus an error of size ${O(\delta^{C - O(1)})}$. We thus can write ${P(n) = P'(n) + O(\delta^{C-O(1)})}$ for ${n \in Q'}$ for some polynomial ${P'}$ of degree at most ${d-1}$. By the triangle inequality, we thus have (for ${C}$ large enough) that
$\displaystyle \frac{1}{|Q'|} |\sum_{n \in Q'} e(P'(n))| \gg \delta$
and hence by induction hypothesis we may find a subprogression ${Q''}$ of ${Q'}$ of size ${|Q''| \gg \delta^{O(1)} N}$ such that ${P'}$ varies by most ${\delta/2}$ on ${Q''}$, and thus (for ${C}$ large enough again) that ${P}$ varies by at most ${\delta}$ on ${Q''}$, and the claim follows. $\Box$
This gives the following corollary (also given as Exercise 16 in this previous blog post):
Corollary 5 (Weyl exponential sum estimate, inverse form II) Let ${I \subset [-N,N] \cap {\bf Z}}$ be a discrete interval for some ${N \geq 1}$, and let ${P(n) = \sum_{i \leq d} \alpha_i n^i}$ polynomial of some degree at most ${d \geq 0}$ for some ${\alpha_0,\dots,\alpha_d \in {\bf R}/{\bf Z}}$. If
$\displaystyle \frac{1}{N} |\sum_{n \in I} e(P(n))| \geq \delta$
for some ${\delta > 0}$, then there is a natural number ${q \ll_d \delta^{-O_d(1)}}$ such that ${\| q\alpha_i \|_{{\bf R}/{\bf Z}} \ll_d \delta^{-O_d(1)} N^{-i}}$ for all ${i=0,\dots,d}$.
One can obtain much better exponents here using Vinogradov’s mean value theorem; see Theorem 1.6 this paper of Wooley. (Thanks to Mariusz Mirek for this reference.) However, this weaker result already suffices for many applications, and does not need any result as deep as the mean value theorem.
Proof: To simplify notation we allow implied constants to depend on ${d}$. As before, we may assume that ${\delta \leq c}$ for some small constant ${c>0}$ depending only on ${d}$. We may also assume that ${N \geq \delta^{-C}}$ for some large ${C}$, as the claim is trivial otherwise (set ${q=1}$).
Applying Proposition 4, we can find a natural number ${q \ll \delta^{-O(1)}}$ and an arithmetic subprogression ${Q}$ of ${I}$ such that ${|Q| \gg \delta^{O(1)}}$ and such that ${P}$ varies by at most ${\delta}$ on ${Q}$. Writing ${Q = \{ qn+r: n \in I'\}}$ for some interval ${I' \subset [0,N] \cap {\bf Z}}$ of length ${\gg \delta^{O(1)}}$ and some ${0 \leq r < q}$, we conclude that the polynomial ${n \mapsto P(qn+r)}$ varies by at most ${\delta}$ on ${I'}$. Taking ${d^{th}}$ order differences, we conclude that the ${d^{th}}$ coefficient of this polynomial is ${O(\delta^{-O(1)} / N^d)}$; by the binomial theorem, this implies that ${n \mapsto P(qn+r)}$ differs by at most ${O(\delta)}$ on ${I'}$ from a polynomial of degree at most ${d-1}$. Iterating this, we conclude that the ${i^{th}}$ coefficient of ${n \mapsto P(qn+r)}$ is ${O(\delta N^{-i})}$ for ${i=0,\dots,d}$, and the claim then follows by inverting the change of variables ${n \mapsto qn+r}$ (and replacing ${q}$ with a larger quantity such as ${q^d}$ as necessary). $\Box$
For future reference we also record a higher degree version of the Vinogradov lemma.
Lemma 6 (Polynomial Vinogradov lemma) Let ${I \subset [-N,N] \cap {\bf Z}}$ be a discrete interval for some ${N \geq 1}$, and let ${P: {\bf Z} \rightarrow {\bf R}/{\bf Z}}$ be a polynomial ${P(n) = \sum_{i \leq d} \alpha_i n^i}$ of degree at most ${d}$ for some ${d \geq 1}$ such that ${\|P(n)\|_{{\bf R}/{\bf Z}} \leq \varepsilon}$ for at least ${\delta N}$ values of ${n \in I}$, for some ${0 < \varepsilon, \delta < 1}$. Then either
$\displaystyle N \ll_d \delta^{-O_d(1)} \ \ \ \ \ (5)$
or
$\displaystyle \varepsilon \gg_d \delta^{O_d(1)} \ \ \ \ \ (6)$
or else there is a natural number ${q \ll_d \delta^{-O_d(1)}}$ such that
$\displaystyle \| q \alpha_i \|_{{\bf R}/{\bf Z}} \ll \frac{\delta^{-O(1)} \varepsilon}{N^i}$
for all ${i=0,\dots,d}$.
Proof: We induct on ${d}$. For ${d=1}$ this follows from Lemma 3 (noting that if ${\|P(n)\|_{{\bf R}/{\bf Z}}, \|P(n_0)\|_{{\bf R}/Z} \leq \varepsilon}$ then ${\|P(n)-P(n_0)\|_{{\bf R}/{\bf Z}} \leq 2\varepsilon}$), so suppose that ${d \geq 2}$ and that the claim is already proven for ${d-1}$. We now allow all implied constants to depend on ${d}$.
For each ${h \in [-2N,2N] \cap {\bf Z}}$, let ${N_h}$ denote the number of ${n \in [-N,N] \cap {\bf Z}}$ such that ${\| P(n+h)\|_{{\bf R}/{\bf Z}}, \|P(n)\|_{{\bf R}/{\bf Z}} \leq \varepsilon}$. By hypothesis, ${\sum_{h \in [-2N,2N] \cap {\bf Z}} N_h \gg \delta^2 N^2}$, and clearly ${N_h = O(N)}$, so we must have ${N_h \gg \delta^2 N}$ for ${\gg \delta^2 N}$ choices of ${h}$. For each such ${h}$, we then have ${\|P(n+h)-P(n)\|_{{\bf R}/{\bf Z}} \leq 2\varepsilon}$ for ${\gg \delta^2 N}$ choices of ${n \in [-N,N] \cap {\bf Z}}$, so by induction hypothesis, either (5) or (6) holds, or else for ${\gg \delta^{O(1)} N}$ choices of ${h \in [-2N,2N] \cap {\bf Z}}$, there is a natural number ${q_h \ll \delta^{-O(1)}}$ such that
$\displaystyle \| q_h \alpha_{i,h} \|_{{\bf R}/{\bf Z}} \ll \frac{\delta^{-O(1)} \varepsilon}{N^i}$
for ${i=1,\dots,d-1}$, where ${\alpha_{i,h}}$ are the coefficients of the degree ${d-1}$ polynomial ${n \mapsto P(n+h)-P(n)}$. We may of course assume it is the latter which holds. By the pigeonhole principle we may take ${q_h= q}$ to be independent of ${h}$.
Since ${\alpha_{d-1,h} = dh \alpha_d}$, we have
$\displaystyle \| qd h \alpha_d \|_{{\bf R}/{\bf Z}} \ll \frac{\delta^{-O(1)} \varepsilon}{N^{d-1}}$
for ${\gg \delta^{O(1)} N}$ choices of ${h}$, so by Lemma 3, either (5) or (6) holds, or else (after increasing ${q}$ as necessary) we have
$\displaystyle \| q \alpha_d \|_{{\bf R}/{\bf Z}} \ll \frac{\delta^{-O(1)} \varepsilon}{N^d}.$
We can again assume it is the latter that holds. This implies that ${q \alpha_{d-2,h} = (d-1) h \alpha_{d-1} + O( \delta^{-O(1)} \varepsilon / N^{d-2} )}$ modulo ${1}$, so that
$\displaystyle \| q(d-1) h \alpha_{d-1} \|_{{\bf R}/{\bf Z}} \ll \frac{\delta^{-O(1)} \varepsilon}{N^{d-2}}$
for ${\gg \delta^{O(1)} N}$ choices of ${h}$. Arguing as before and iterating, we obtain the claim. $\Box$
The above results also extend to higher dimensions. Here is the higher dimensional version of Proposition 4:
Proposition 7 (Multidimensional Weyl exponential sum estimate, inverse form) Let ${k \geq 1}$ and ${N_1,\dots,N_k \geq 1}$, and let ${Q_i \subset {\bf Z}}$ be arithmetic progressions of length at most ${N_i}$ for each ${i=1,\dots,k}$. Let ${P: {\bf Z}^k \rightarrow {\bf R}/{\bf Z}}$ be a polynomial of degrees at most ${d_1,\dots,d_k}$ in each of the ${k}$ variables ${n_1,\dots,n_k}$ separately. If
$\displaystyle \frac{1}{N_1 \dots N_k} |\sum_{n \in Q_1 \times \dots \times Q_k} e(P(n))| \geq \delta$
for some ${\delta > 0}$, then there exists a subprogression ${Q'_i}$ of ${Q_i}$ with ${|Q'_i| \gg_{k,d_1,\dots,d_k} \delta^{O_{k,d_1,\dots,d_k}(1)} N_i}$ for each ${i=1,\dots,k}$ such that ${P}$ varies by at most ${\delta}$ on ${Q'_1 \times \dots \times Q'_k}$.
A much more general statement, in which the polynomial phase ${n \mapsto e(P(n))}$ is replaced by a nilsequence, and in which one does not necessarily assume the exponential sum is small, is given in Theorem 8.6 of this paper of Ben Green and myself, but it involves far more notation to even state properly.
Proof: We induct on ${k}$. The case ${k=1}$ was established in Proposition 5, so we assume that ${k \geq 2}$ and that the claim has already been proven for ${k-1}$. To simplify notation we allow all implied constants to depend on ${k,d_1,\dots,d_k}$. We may assume that ${\delta \leq c}$ for some small ${c>0}$ depending only on ${k,d_1,\dots,d_k}$.
By a linear change of variables, we may assume that ${Q_i \subset [0,N_i] \cap {\bf Z}}$ for all ${i=1,\dots,k}$.
We write ${n' := (n_1,\dots,n_{k-1})}$. First suppose that ${N_k = O(\delta^{-O(1)})}$. Then by the pigeonhole principle we can find ${n_k \in I_k}$ such that
$\displaystyle \frac{1}{N_1 \dots N_{k-1}} |\sum_{n' \in Q_1 \times \dots \times Q_{k-1}} e(P(n',n_k))| \geq \delta$
and the claim then follows from the induction hypothesis. Thus we may assume that ${N_k \geq \delta^{-C}}$ for some large ${C}$ depending only on ${k,d_1,\dots,d_k}$. Similarly we may assume that ${N_i \geq \delta^{-C}}$ for all ${i=1,\dots,k}$.
By the triangle inequality, we have
$\displaystyle \frac{1}{N_1 \dots N_k} \sum_{n_k \in Q_k} |\sum_{n' \in Q_1 \times \dots \times Q_{k-1}} e(P(n',n_k))| \geq \delta.$
The inner sum is ${O(N_k)}$, and the outer sum has ${O(N_1 \dots N_{k-1})}$ terms. Thus, for ${\gg \delta N_1 \dots N_{k-1}}$ choices of ${n' \in Q_1 \times \dots \times Q_{k-1}}$, one has
$\displaystyle \frac{1}{N_k} |\sum_{n_k \in Q_k} e(P(n',n_k))| \gg \delta. \ \ \ \ \ (7)$
We write
$\displaystyle P(n',n_k) = \sum_{i_k \leq d_k} P_{i_k}(n') n_k^i$
for some polynomials ${P_{i_k}: {\bf Z}^{k-1} \rightarrow {\bf R}/{\bf Z}}$ of degrees at most ${d_1,\dots,d_{k-1}}$ in the variables ${n_1,\dots,n_{k-1}}$. For each ${n'}$ obeying (7), we apply Corollary 5 to conclude that there exists a natural number ${q_{n'} \ll \delta^{-O(1)}}$ such that
$\displaystyle \| q_{n'} P_{i_k}(n') \|_{{\bf R}/{\bf Z}} \ll \delta^{-O(1)} / N_k^{i_k}$
for ${i_k=1,\dots,d_k}$ (the claim also holds for ${i_k=0}$ but we discard it as being trivial). By the pigeonhole principle, there thus exists a natural number ${q \ll \delta^{-O(1)}}$ such that
$\displaystyle \| q P_{i_k}(n') \|_{{\bf R}/{\bf Z}} \ll \delta^{-O(1)} / N_k^{i_k}$
for all ${i_k=1,\dots,d_k}$ and for ${\gg \delta^{O(1)} N_1 \dots N_{k-1}}$ choices of ${n' \in Q_1 \times \dots \times Q_{k-1}}$. If we write
$\displaystyle P_{i_k}(n') = \sum_{i_{k-1} \leq d_{k-1}} P_{i_{k-1},i_k}(n_1,\dots,n_{k-2}) n_{k-1}^{i_{k-1}},$
where ${P_{i_{k-1},i_k}: {\bf Z}^{k-2} \rightarrow {\bf R}/{\bf Z}}$ is a polynomial of degrees at most ${d_1,\dots,d_{k-2}}$, then for ${\gg \delta^{O(1)} N_1 \dots N_{k-2}}$ choices of ${(n_1,\dots,n_{k-2}) \in Q_1 \times \dots \times Q_{k-2}}$ we then have
$\displaystyle \| \sum_{i_{k-1} \leq d_{k-1}} q P_{i_{k-1},i_k}(n_1,\dots,n_{k-2}) n_{k-1}^{i_{k-1}} \|_{{\bf R}/{\bf Z}} \ll \delta^{-O(1)} / N_k^{i_k}.$
Applying Lemma 6 in the ${n_{k-1}}$ and the largeness hypotheses on the ${N_i}$ (and also the assumption that ${i_k \geq 1}$) we conclude (after enlarging ${q}$ as necessary, and pigeonholing to keep ${q}$ independent of ${n_1,\dots,n_{k-2}}$) that
$\displaystyle \| q P_{i_{k-1},i_k}(n_1,\dots,n_{k-2}) \|_{{\bf R}/{\bf Z}} \ll \frac{\delta^{-O(1)}}{N_{k-1}^{i_{k-1}} N_k^{i_k}}$
for all ${i_{k-1}=0,\dots,d_{k-1}}$ (note that we now include that ${i_{k-1}=0}$ case, which is no longer trivial) and for ${\gg \delta^{O(1)} N_1 \dots N_{k-2}}$ choices of ${(n_1,\dots,n_{k-2}) \in Q_1 \times \dots \times Q_{k-2}}$. Iterating this, we eventually conclude (after enlarging ${q}$ as necessary) that
$\displaystyle \| q \alpha_{i_1,\dots,i_k} \|_{{\bf R}/{\bf Z}} \ll \frac{\delta^{-O(1)}}{N_1^{i_1} \dots N_k^{i_k}} \ \ \ \ \ (8)$
whenever ${i_j \in \{0,\dots,d_j\}}$ for ${j=1,\dots,k}$, with ${i_k}$ nonzero. Permuting the indices, and observing that the claim is trivial for ${(i_1,\dots,i_k) = (0,\dots,0)}$, we in fact obtain (8) for all ${(i_1,\dots,i_k) \in \{0,\dots,d_1\} \times \dots \times \{0,\dots,d_k\}}$, at which point the claim easily follows by taking ${Q'_j := \{ qn_j: n_j \leq \delta^C N_j\}}$ for each ${j=1,\dots,k}$. $\Box$
An inspection of the proof of the above result (or alternatively, by combining the above result again with many applications of Lemma 6) reveals the following general form of Proposition 4, which was posed as Exercise 17 in this previous blog post, but had a slight misprint in it (it did not properly treat the possibility that some of the ${N_j}$ could be small) and was a bit trickier to prove than anticipated (in fact, the reason for this post was that I was asked to supply a more detailed solution for this exercise):
Proposition 8 (Multidimensional Weyl exponential sum estimate, inverse form, II) Let ${k \geq 1}$ be an natural number, and for each ${j=1,\dots,k}$, let ${I_j \subset [0,N_j]_{\bf Z}}$ be a discrete interval for some ${N_j \geq 1}$. Let
$\displaystyle P(n_1,\dots,n_k) = \sum_{i_1 \leq d_1, \dots, i_k \leq d_k} \alpha_{i_1,\dots,i_k} n_1^{i_1} \dots n_k^{i_k}$
be a polynomial in ${k}$ variables of multidegrees ${d_1,\dots,d_k \geq 0}$ for some ${\alpha_{i_1,\dots,i_k} \in {\bf R}/{\bf Z}}$. If
$\displaystyle \frac{1}{N_1 \dots N_k} |\sum_{n \in I_1 \times \dots \times I_k} e(P(n))| \geq \delta \ \ \ \ \ (9)$
for some ${\delta > 0}$, then either
$\displaystyle N_j \ll_{k,d_1,\dots,d_k} \delta^{-O_{k,d_1,\dots,d_k}(1)} \ \ \ \ \ (10)$
for some ${1 \leq j \leq d_k}$, or else there is a natural number ${q \ll_{k,d_1,\dots,d_k} \delta^{-O_{k,d_1,\dots,d_k}(1)}}$ such that
$\displaystyle \| q\alpha_{i_1,\dots,i_k} \|_{{\bf R}/{\bf Z}} \ll_{k,d_1,\dots,d_k} \delta^{-O_d(1)} N_1^{-i_1} \dots N_k^{-i_k} \ \ \ \ \ (11)$
whenever ${i_j \leq d_j}$ for ${j=1,\dots,k}$.
Again, the factor of ${N_1^{-i_1} \dots N_k^{-i_k}}$ is natural in this bound. In the ${k=1}$ case, the option (10) may be deleted since (11) trivially holds in this case, but this simplification is no longer available for ${k>1}$ since one needs (10) to hold for all ${j}$ (not just one ${j}$) to make (11) completely trivial. Indeed, the above proposition fails for ${k \geq 2}$ if one removes (10) completely, as can be seen for instance by inspecting the exponential sum ${\sum_{n_1 \in \{0,1\}} \sum_{n_2 \in [1,N] \cap {\bf Z}} e( \alpha n_1 n_2)}$, which has size comparable to ${N}$ regardless of how irrational ${\alpha}$ is.
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CC-MAIN-2023-23
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https://www.codeproject.com/Articles/652209/The-Snake?fid=1842309&df=90&mpp=10&noise=1&prof=True&sort=Position&view=Quick&spc=Relaxed
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13,148,214 members (50,872 online)
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#### Stats
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Posted 11 Sep 2013
# The Snake
, 15 Sep 2013
Rate this:
Moving the snake by a timer and some labels
## Introduction
This article is about make a motion by a timer that changes the location of some labels that simulate a snake. The snake should get some targets (that placed random) and has a special time to do this. it couldn't go out of the field and become bigger and faster by spending the time.
the best part that i ask too see them:
• The idea that i used too make a smooth and flexible snake
• How did i solve the delay problem of key down event
• * How to code for multiple key down (for example Alt + f4) ->i explained at last part of Using the code
## Background
The labels are square and we can not use them to draw a smooth moving snake so i put some of them together And made a flexible rectangle. It is what I've done to design the snake. the labels follow each other with 1 unit distance, if we use more distance in the turns the snake doesn't look good enough so i tried to make it as i said by these codes:
here:
At first i set the firstly location of head label of snake body and use as first amount of x and y variable.
```int x = 50;
int y = 100;
```
a is the variable that shows the direction of snake moving that gets in a special key down event, and x,y are the variable that get or set the locations of the snake body. the codes work this way:
1. For example label6 is the head and labels 7,8,9... are the next body labels that follow the head(label6). At first i set first location of label6:
` <span style="font-size: small;">label6.Location = new Point(x, y);</span> `
2. and we suppose the snake is going to right direction (so the event should set a = "r"), the other thing that this event check is that if the user pressed a direction opposite to the moving direction? and if it was true it doesn't change the direction because a snake could not do this.
```protected override void OnKeyDown(KeyEventArgs keyEvent)
{
<span class="Apple-tab-span" style="white-space: pre;"> </span>string key;
<span style="font-size: 9pt;"><span class="Apple-tab-span" style="white-space: pre;"> </span>key = keyEvent.KeyCode.ToString();</span>```
```//checking the direction and set the new direction<span style="font-size: 9pt;">
</span><span class="Apple-tab-span" style="white-space: pre;"> </span>if (key == "Right")
{
if (a != "l")
a = "r";
}
if (key == "Left")
{
if (a != "r")
a = "l";
}
if (key == "Up")
{
if (a != "d")
a = "u";
}
if (key == "Down")
{
if (a != "u")
a = "d";
<span class="Apple-tab-span" style="white-space: pre;"> </span> }
<span class="Apple-tab-span" style="white-space: pre;"> </span>}
```
3. get back to the labels: so the first label location set, so now we should set the next label location with 1px distance so we do it as the direction of moving like this:
```<span style="font-size: 14px;"> </span><span style="font-size: 14px;"> <span class="Apple-tab-span" style="white-space: pre;"> </span> </span>
if (a == "r")
{
x = x + 1;
}
if (a == "l")
{
x = x - 1;
}
if (a == "u")
{
y = y - 1;
}
if (a == "d")
{
y = y + 1;
}
label5.Location = new Point(x, y);```
4. and the other labels 1px distance to previous one...
```if (a == "r")
{
x = x + 1;
}
if (a == "l")
{
x = x - 1;
}
if (a == "u")
{
y = y - 1;
}
if (a == "d")
{
y = y + 1;
}
label4.Location = new Point(x, y);
if (a == "r")
{
x = x + 1;
}
if (a == "l")
{
x = x - 1;
}
if (a == "u")
{
y = y - 1;
}
if (a == "d")
{
y = y + 1;
}
```
So the snake moving this way.
For database I just used three text files that keep the record, but I delete the type that user can't access them that much easy but i preferred to don't use sql data base for this project.
The random part is the targets location that have random places that should be inside of the snake field:
```Random fisrtX = new Random(DateTime.Now.Second);
Random firstY = new Random(DateTime.Now.Millisecond);```
When the snake catch the target we need new location so I used this.
```int secondX = fisrtX.Next(10, 390);
int secondY = firstY.Next(10, 370); ```
## Using the code
First of all install the Fonts that are beside the project to open it completely.
At the first form (form1) you can see a simple timer that change the back colors of some labels.
At the 2nd, 3rd, 4th forms you can see 3 timers that:
timer1 do the motion parts by getting the side button that user pressed and setting the location of the leader label. The leader label is the first label that moves and other labels will follow each other, this kind of programming make the visual part of snake flexible and real.
timer2 is for game time that when the snake catch the targets it will be added.
timer3 is for making the snake faster and make the game harder, by making the snake longer and adding more labels to follow that make it faster.
At last i wanna explain more about the On key down override, why i use it? and why i didn't use Key down event from the form? and explain the problems that i face when i writing this code and solutions:
1. If you see the game forms, you can see there is no text box, button and any thing that have tab index and i just used Labels instead of button. the reason was the key down event that works when your are selected the form page, but at first one of buttons had been selected and during the game, the user nay select one button that caused to disable key down code so for solution i used some labels.
2. When i my program completed i understand a big problem and that was the delay between key down and the snake moving that make the Hard part boring. so i tried this code and understood that it has very quick feed back in key down and moving
3.* The most big problem that i face in this project was the using of Alt+f4 by user to exit the game! what was wrong? The Timers and the game !!! if the timers didn't stop the game won't be stopped so used a code on the exit button to stop the timers, but what should i do with Alt+f4. you know what will happened if the user exits by these button during the game. yes when he was at main page a message box shows that you lose!!! so i tried to code for the Alt+f4 by the code that i introduced... i set more codes in the source to check them and use (on key down override)
```string keyCode = "";
if (Alt == "Alt: True" && keyCode != "18")//check the alt is pressed or no
{
keyCode = keyCode + "18";//this is used to not to get the old alt pressed againg
}
if (KeyValue == "KeyValue: 115")//this is the key value of f4 button
{
keyCode = keyCode + "115";
}
if (keyCode == "18115")//chech if alt and f4 are pressed together
{
timer1.Stop();
timer2.Stop();
timer3.Stop();
}
```
*** i used more explanations and comments in the Form2 (easy part) so to check the codes use this form
## Points of Interest
The best part of this project that make me proud is the design of the snake that make if flexible in turns and it's moving and the limitation of getting targets that how much it should be close to catch the targets, I tried over and over to get the proper result.
And as i said other good parts of this project is key down void that i used instead of form key down it make the program faster to feedback the user key pressed. so use that instead of key down event in different objects.
## Share
Student Iran (Islamic Republic of)
I'm programming since 2009. i started it from school and c# language and then i learnt working sql and now i'm working on asp.net projects
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Last Visit: 31-Dec-99 18:00 Last Update: 24-Sep-17 15:25 Refresh 1
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| 3.015625
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CC-MAIN-2017-39
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| 0.916548
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https://socratic.org/questions/how-do-you-find-the-center-and-the-radius-of-the-circle-whose-equation-is-x-2-y-
| 1,585,546,023,000,000,000
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How do you find the center and the radius of the circle whose equation is x^2+y^2-4x-2y-4=0?
Jan 11, 2016
centre = (2 , 1 ) and radius = 3
Explanation:
comparing the general form for the equation of a circle :
${x}^{2} + {y}^{2} + 2 g x + 2 f y + c = 0$
with the question given
# x^2 + y^2- 4x -2y - 4 = 0
comparing terms we can see that 2g = - 4 hence g = - 2
similarly 2f = - 2 hence f = - 1 and c = - 4
the centre of the circle = (- g , - f ) = (2 , 1 ) and
$r = \sqrt{{g}^{2} + {f}^{2} - c}$
$\Rightarrow r = \sqrt{{\left(- 2\right)}^{2} + {\left(- 1\right)}^{2} - \left(- 4\right)} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$
=
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| 4.34375
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CC-MAIN-2020-16
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longest
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en
| 0.676516
|
https://es.mathworks.com/help/pde/ug/pde.transientstructuralresults.interpolatevelocity.html
| 1,686,348,161,000,000,000
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# interpolateVelocity
Interpolate velocity at arbitrary spatial locations for all time or frequency steps for dynamic structural model
## Syntax
``intrpVel = interpolateVelocity(structuralresults,xq,yq)``
``intrpVel = interpolateVelocity(structuralresults,xq,yq,zq)``
``intrpVel = interpolateVelocity(structuralresults,querypoints)``
## Description
````intrpVel = interpolateVelocity(structuralresults,xq,yq)` returns the interpolated velocity values at the 2-D points specified in `xq` and `yq` for all time or frequency steps.```
example
````intrpVel = interpolateVelocity(structuralresults,xq,yq,zq)` uses the 3-D points specified in `xq`, `yq`, and `zq`.```
````intrpVel = interpolateVelocity(structuralresults,querypoints)` uses the points specified in `querypoints`.```
## Examples
collapse all
Interpolate velocity at the geometric center of a beam under a harmonic excitation.
Create a transient dynamic model for a 3-D problem.
`structuralmodel = createpde("structural","transient-solid");`
Create the geometry and include it in the model. Plot the geometry.
```gm = multicuboid(0.06,0.005,0.01); structuralmodel.Geometry = gm; pdegplot(structuralmodel,"FaceLabels","on","FaceAlpha",0.5) view(50,20)```
Specify Young's modulus, Poisson's ratio, and the mass density of the material.
```structuralProperties(structuralmodel,"YoungsModulus",210E9, ... "PoissonsRatio",0.3, ... "MassDensity",7800);```
Fix one end of the beam.
`structuralBC(structuralmodel,"Face",5,"Constraint","fixed");`
Apply a sinusoidal displacement along the `y`-direction on the end opposite the fixed end of the beam.
```structuralBC(structuralmodel,"Face",3, ... "YDisplacement",1E-4, ... "Frequency",50);```
Generate a mesh.
`generateMesh(structuralmodel,"Hmax",0.01);`
Specify the zero initial displacement and velocity.
```structuralIC(structuralmodel,"Displacement",[0;0;0], ... "Velocity",[0;0;0]);```
Solve the model.
```tlist = 0:0.002:0.2; structuralresults = solve(structuralmodel,tlist);```
Interpolate velocity at the geometric center of the beam.
```coordsMidSpan = [0;0;0.005]; intrpVel = interpolateVelocity(structuralresults,coordsMidSpan);```
Plot the `y`-component of velocity of the geometric center of the beam.
```figure plot(structuralresults.SolutionTimes,intrpVel.vy) title("Y-Velocity of the Geometric Center of the Beam")```
## Input Arguments
collapse all
Solution of the dynamic structural analysis problem, specified as a `TransientStructuralResults` or `FrequencyStructuralResults` object. Create `structuralresults` by using the `solve` function.
Example: ```structuralresults = solve(structuralmodel,tlist)```
x-coordinate query points, specified as a real array. `interpolateVelocity` evaluates velocities at the 2-D coordinate points `[xq(i),yq(i)]` or at the 3-D coordinate points `[xq(i),yq(i),zq(i)]`. Therefore, `xq`, `yq`, and (if present) `zq` must have the same number of entries.
`interpolateVelocity` converts query points to column vectors `xq(:)`, `yq(:)`, and (if present) `zq(:)`. It returns velocities as an `FEStruct` object with the properties containing vectors of the same size as these column vectors. To ensure that the dimensions of the returned solution are consistent with the dimensions of the original query points, use the `reshape` function. For example, use ```intrpVel = reshape(intrpVel.ux,size(xq))```.
Data Types: `double`
y-coordinate query points, specified as a real array. `interpolateVelocity` evaluates velocities at the 2-D coordinate points `[xq(i),yq(i)]` or at the 3-D coordinate points `[xq(i),yq(i),zq(i)]`. Therefore, `xq`, `yq`, and (if present) `zq` must have the same number of entries. Internally, `interpolateVelocity` converts query points to the column vector `yq(:)`.
Data Types: `double`
z-coordinate query points, specified as a real array. `interpolateVelocity` evaluates velocities at the 3-D coordinate points `[xq(i),yq(i),zq(i)]`. Therefore, `xq`, `yq`, and `zq` must have the same number of entries. Internally, `interpolateVelocity` converts query points to the column vector `zq(:)`.
Data Types: `double`
Query points, specified as a real matrix with either two rows for 2-D geometry or three rows for 3-D geometry. `interpolateVelocity` evaluates velocities at the coordinate points `querypoints(:,i)`, so each column of `querypoints` contains exactly one 2-D or 3-D query point.
Example: For 2-D geometry, ```querypoints = [0.5,0.5,0.75,0.75; 1,2,0,0.5]```
Data Types: `double`
## Output Arguments
collapse all
Velocities at the query points, returned as an `FEStruct` object with the properties representing spatial components of velocity at the query points. For query points that are outside the geometry, `intrpVel` returns `NaN`. Properties of an `FEStruct` object are read-only.
## Version History
Introduced in R2018a
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Do everything better
Do everything better
# How the Razzle-Dazzle Game Takes All Your Money
Evil WeekEvil WeekWelcome to Evil Week, our annual chance to delve into all the slightly sketchy hacks we'd usually refrain from recommending. Want to weasel your way into free drinks, play elaborate mind games, or, er, launder some money? We've got all the info you need to successfully be unsavory.
This game is illegal. Razzle, aka Razzle-Dazzle, is a scam posing as a carnival or street game, where you roll marbles in a box collecting points. Popping up in Havana, New York, New Orleans, and curbs and parking lots across the U.S., it’s like Three-Card Monte, but more complicated, and better disguised as a real game.
That’s part of the appeal, and that’s why a good scammer can keep suckers playing it, paying a couple of bucks a turn until they’ve lost thousands of dollars. Here’s a beautiful demonstration by mathematician James Grime of Numberphile, and magician Brian Brushwood of Scam Nation.
### How it supposedly works
The scammer calls you over to his stand, and shows you his wooden tray full of little holes, each marked with a number from 1 to 6. You get to roll marbles into the holes, add up those numbers, then consult a chart to see how many points you get. And if you hit ten points, you’ll win a big prize: a Switch, or an iPhone, or a TV, or all your money back. Let’s look at that chart!
Say you roll the marbles, and the numbers add up to 45. Well then: you’ve won 5 points, and you’re halfway to winning! (The “H.P.” on this chart is less clear, but one magic trick seller believes the scammer promises to give your money back if you win on the next roll.)
It’s a bit dizzying, but so are most games when you start out. You don’t have to learn any strategy, just play the odds. So you pay a buck to play the first round. Just to see how quickly the points add up. And wow, yeah, you got some points right away! You’ll win that TV with like five bucks!
You keep playing, spending more to make more. Soon you’re very close to winning.
At one point you make a roll that doesn’t add points, but it does add a new potential prize to the pool. It also doubles the cost per turn. Sure, fine, as soon as you win you’ll make all your money back. So you keep playing.
Until you’re out of money. No prize. How did your luck turn so fast?
To feel the energy of the game, watch a demonstration that’s a little more razzle-dazzle than above, from the scam-busting TV show The Real Hustle.
### How it actually works
How did you earn points so fast, then not at all? Because you never actually earned those points. The scammer lied and pretended you’d rolled the right scores. But your odds of actually rolling those scores are very low.
Razzle boards basically work like rolling a bunch of six-sided dice. (Some versions do use dice.) If you’ve played enough craps or Monopoly, you know that rolling really high or low totals is less likely than rolling middle totals. Your odds of rolling a 7 are 1/6, but your odds of rolling a 2 are 1/36.
Razzle works on the same principle, but you’re rolling eight “dice.” Your odds of rolling, say, 28, are about 1/12. Your odds of rolling an 8 are literally less than one in a million. Seriously, you can test it on this calculator. All the lowest and highest totals are very unlikely, and the middle totals are very likely.
Guess which kind of total gives you points.
That crazy yellow chart hides the odds, because it puts the totals out of order. Here’s how it would look laid out in order:
Now it’s easy to see that all the points come from the numbers at the end. And as this graph illustrates, the numbers at the end are very rare. The most likely point-getting rolls, 17 and 39, are a 1 in 160 chance each. The least likely, 8 and 48, are a 1 in 1.6 million chance. The odds of winning any points on a roll are 1 in 36, and of course most of those rolls get you just half a point.
You’d have to roll hundreds or thousands of times to rack up enough points. The scammer doesn’t have to weight the marbles, or even sneakily arrange the numbers on the holes. As long as things on the board are pretty random, you’ll never hit your total before you run out of money.
Play for a very long time, and you would eventually win. But there’s also a 1 in 12 chance of hitting that special 29, which doubles your prize, but also doubles the cost to play. So about every 12 rolls, you start losing money twice as fast. By the time you caught up, you’d spend way more than the cost of a TV.
### No no no wait wait but I scored those points early on!
The scammer told you that you scored points. He’s been running this game for a while, and he’s learned how to add up numbers really fast. He’s also learned how to add them up wrong, quicker than you notice. And if you do notice, you notice that he “accidentally” gave you points you didn’t earn. So you don’t call it out. You hope he makes the same mistake again. And you make sure he never makes a mistake in his favor.
But the scammer doesn’t have to make a mistake in his favor. He can honestly tell you when you win points, because you barely ever win points. You can’t catch him in the scam, because the only lie he ever has to tell is that you had any chance of winning.
If you avoid rolling 29 for a long time, and you seem to be easily fooled by his fake totals, the scammer could fake a 29 to start you doubling your money faster. But he could also just wait, and let the odds take control.
All he has to do is keep you interested by handing you fake points now and then.
### How cool it is
There’s something cool about this, right? Scams and gambling are sexy! Ocean’s Eleven! American Hustle! Casino Royale!
Razzle games have been around for generations, run by carnies, streetside scammers, casinos, nightclubs, and the mafia. Cops went on the take for it. Mob families fought for territory rights. Magic shops show off Razzle sets, though they’re cagey about whether they actually sell this tool designed specifically for fraud. The game has a lot of...razzle dazzle.
Like most gambling, it’s less glamorous in person. Here’s a news report of some poor schmuck who wouldn’t let reporters show his face, so they just pointed a camera at his crappy sneakers:
Not so much razzle there. Just stealing a guy’s sneaker money in a parking lot.
A 1981 Department of Justice report on Razzle-Dazzle says the game can fool “affluent, intelligent marks.” It offers several options for shutting operators down:
While the basic criminal charge against a Razzle Dazzle operator would be gambling, it may be possible to press others including fraud, false pretense, larceny, or theft by deception. Law enforcement officers making related arrests should look for records indicating substantial losses by players as well as other items used by the operator which could be analyzed by the Gambling Subunit of the FBI’s crime laboratory.
So if you were eyeing the Razzle-Dazzle model in that magic store, maybe keep your money.
Staff Writer, Lifehacker | Nick has written for Gawker, Valleywag, the Daily Dot, and Urlesque. He currently runs the scripted comedy podcast "Roommate From Hell."
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# Stats & Ethics Practice
```Stats Practice
Using the following sample data, find the following measures.
Measures of Central Tendency
Calculate the mean, median and mode:
Mean ________
Median _______
Mode _______
Measures of Variance
Calculate the range of the data _______
Now find the standard deviation of the data…
______________________
Sum of squared deviations
Number of scores
1.
2.
3.
4.
5.
6.
Find the mean of the data
Find the difference between each number and the mean
Square each individual difference
Add up all the squared numbers
Divide by # of data points (This value is referred to as the variance…ave difference btwn data points & mean)
Find the square root of the quotient from #5
Data
Deviation from the mean
Squared deviation
7
15
20
4
8
5
4
Mean of Data:
Sum of Squares:
Standard Deviation = _______________
What is the difference between descriptive and inferential statistics?
What is the standard for deciding if a result is statistically significant? What does that MEAN???
Want some more practice??? Ok! Here we go…
Data:
Identify The Following…
Mean ________
Median ________
Range ________
Standard Deviation ________
Data
Deviation from the mean
Mean of Data:
Sum of Squares:
Mode ________
Squared deviation
Applying the Normal Curve Percentages!!!
Assume a distribution of aptitude test scores forms a normal curve with a mean of 100 and a standard deviation of 15.
Within which standard deviation will most of the scores fall?
If a student scores a 120, within which standard deviation will that score fall?
If a student scores within the second deviation, what is the possible range of the student’s score?
Assume your class took a final exam in psychology in which the scores produced a normal curve with a mean score of
80 and a standard deviation of 5.
68% of the scores on the final exam would fall between _______ and _______.
If the student scores within two standard deviations from the mean, what is the possible range of the student’s
scores?
What percentage of students may have scored either higher than 90 or lower than 70?
Ethics!!!
Read each situation below and decide which ethical guideline is not being followed. More than one answer may apply!!!
Informed Consent (IC)
Protection from Physical or Emotional Harm (H)
Confidentiality (C)
Debriefing (D)
1.
A teacher in your school gives you a mandatory anonymous drug use survey to complete in class and tells you
she cannot let you know why you are completing the survey because it would throw off her results.
__________________________________________________________________________________
2. You agree to participate in an experiment that is designed to measure your ability to lie in various
circumstances. Under the direction of the researcher, you make false statements to your mother, your best
friend, and your favorite teacher. The guilt you feel after lying to these influential and important people has
you questioning your morals and values.
__________________________________________________________________________________
3. A psychologist in your town is invited to speak at career day at your school. You have been seeing the
psychologist for more than a year for depression and attempted suicide. At career day the psychologist speaks
of working with teen patients who are depressed and have attempted suicide and cites a few examples of his
cases. Although he uses no names, you feel he is talking about you and run from the room embarrassed.
__________________________________________________________________________________
4. At the conclusions of a study testing memory and mood, you are released by the researcher, paid a small fee,
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# 20 Vowel Team Worksheets Free Printable
The “Ee” Vowel Team Worksheets vowel team ea worksheet, vowel team posters for kids, vowel team worksheets pdf, vowel team poster fundations, vowel team ea videos, via: 99worksheets.com
Numbering Worksheets for Kids. Kids are usually introduced to this topic matter during their math education. The main reason behind this is that learning math can be done with the worksheets. With an organized worksheet, kids will be able to describe and explain the correct answer to any mathematical problem. But before we talk about how to create a math worksheet for kids, let’s have a look at how children learn math.
In elementary school, children are exposed to a number of different ways of teaching them how to do a number of different subjects. Learning these subjects is important because it would help them develop logical reasoning skills. It is also an advantage for them to understand the concept behind all mathematical concepts.
To make the learning process easy for children, the educational methods used in their learning should be easy. For example, if the method is to simply count, it is not advisable to use only numbers for the students. Instead, the learning process should also be based on counting and dividing numbers in a meaningful way.
The main purpose of using a worksheet for kids is to provide a systematic way of teaching them how to count and multiply. Children would love to learn in a systematic manner. In addition, there are a few benefits associated with creating a worksheet. Here are some of them:
Children have a clear idea about the number of objects that they are going to add up. A good worksheet is one which shows the addition of different objects. This helps to give children a clear picture about the actual process. This helps children to easily identify the objects and the quantities that are associated with it.
This worksheet helps the child’s learning. It also provides children a platform to learn about the subject matter. They can easily compare and contrast the values of various objects. They can easily identify the objects and compare it with each other. By comparing and contrasting, children will be able to come out with a clearer idea.
He or she will also be able to solve a number of problems by simply using a few cells. He or she will learn to organize a worksheet and manipulate the cells. to arrive at the right answer to any question.
This worksheet is a vital part of a child’s development. When he or she comes across an incorrect answer, he or she can easily find the right solution by using the help of the worksheets. He or she will also be able to work on a problem without having to refer to the teacher. And most importantly, he or she will be taught the proper way of doing the mathematical problem.
Math skills are the most important part of learning and developing. Using the worksheet for kids will improve his or her math skills.
Many teachers are not very impressed when they see the number of worksheets that are being used by their children. This is actually very much true in the case of elementary schools. In this age group, the teachers often feel that the child’s performance is not good enough and they cannot just give out worksheets.
However, what most parents and educators do not realize is that there are several ways through which you can improve the child’s performance. You just need to make use of a worksheet for kids. elementary schools.
As a matter of fact, there is a very good option for your children to improve their performance in math. You just need to look into it.
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## ››Convert pound/square inch [gauge] to bar
psig bar
The above form works if you are measuring differential pressure, such as the difference in psi between two points.
If you are measuring relative to vacuum and want to resolve the pressure relative to the atmosphere, then you should use the form below.
Note that psig can measure differential pressure in some applications and absolute pressure in others, so you need to know which one fits your calculation.
psig (relative to atmosphere)
bar (relative to vacuum)
How many psig in 1 bar? The answer is 14.503773800722.
We assume you are converting between pound/square inch [gauge] and bar.
You can view more details on each measurement unit:
psig or bar
The SI derived unit for pressure is the pascal.
1 pascal is equal to 0.00014503773800722 psig, or 1.0E-5 bar.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between pounds/square inch and bars.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of psig to bar
1 psig to bar = 0.06895 bar
10 psig to bar = 0.68948 bar
20 psig to bar = 1.37895 bar
30 psig to bar = 2.06843 bar
40 psig to bar = 2.7579 bar
50 psig to bar = 3.44738 bar
100 psig to bar = 6.89476 bar
200 psig to bar = 13.78951 bar
## ››Want other units?
You can do the reverse unit conversion from bar to psig, or enter any two units below:
## Enter two units to convert
From: To:
## ››Definition: Pound/square inch
Psig (pound-force per square inch gauge) is a unit of pressure relative to the surrounding atmosphere. By contrast, psia measures pressure relative to a vacuum (such as that in space). At sea level, Earth's atmosphere actually exerts a pressure of 14.7 psi. Humans do not feel this pressure because internal pressure of liquid in their bodies matches the external pressure. If a pressure gauge is calibrated to read zero in space, then at sea level on Earth it would read 14.7 psi. Thus a reading of 30 psig on a tire gauge represents an absolute pressure of 44.7 psi.
## ››Definition: Bar
The bar is a measurement unit of pressure, equal to 1,000,000 dynes per square centimetre (baryes), or 100,000 newtons per square metre (pascals). The word bar is of Greek origin, báros meaning weight. Its official symbol is "bar"; the earlier "b" is now deprecated, but still often seen especially as "mb" rather than the proper "mbar" for millibars.
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more!
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# System identification/ Filter estimation to mimic frequency equalizer of audio with Scipy
At the current problem I'm working on, I have two signals: One "original" signal that contains audio (voice). The second signal is the same audio file but edited with a frequency equalizer, for example to make all frequencies between $649-677\textrm{ Hz}$ louder.
What I'm trying to achieve now is to "teach" a SCIPY filter to mimic this behavior. Therefore I created a function with two butterworth-filters and try to curve_fit the $x$-data to the $y$-data. The intuition is, that like this the system learns the optimal frequency intervals to mimic the equalizer. Obviously it doesn't work as supposed (no curve fit achieved most of the time. If I achieve a curve fit and apply the filter to the original data, it didn't change at all). Any suggestions?
import numpy as np
from scipy.optimize import curve_fit
from scipy import signal
xdata = voice_1
ydata = voice_2
# Function to optimize, supposed that two bandpass filters are enough
def func(x, p1,p2,p3,p4):
b, a = signal.butter(2,[p1,p2],btype="band")
x = signal.lfilter(b,a,x)
b, a = signal.butter(2,[p3,p4],btype="band")
x = signal.lfilter(b,a,x)
return x
# Optimize function to achieve a curve-fit and therefore get the right frequency response in p1,p2,p3,p4
popt, pcov = curve_fit(func, xdata, ydata,p0=(0.5,0.5,0.5,0.5))
What I also tried was creating a xsignal containing of white-noise, creating a ysignal where I changed some frequency levels and calculating a transfer function H as FFT(ysignal)/FFT(xsignal). As soon as I apply this function to other data than white noise the results seem to be wrong as well.
• "seem to be wrong": how do you figure? could you plot e.g. the in-, reference and output PSDs? Jun 13, 2016 at 10:53
• Jomona, you could try a time domain equailizer using your two signals and the Wiener-Hopf equations - see dsp.stackexchange.com/questions/31318/… and mathworks.com/matlabcentral/fileexchange/… . I think this would work quite well for you and be a "least squared error" solution. Jun 13, 2016 at 12:33
• "As soon as I apply this function to other data than white noise the results seem to be wrong as well." Are you looking at magnitude and phase of the output separately? Jun 13, 2016 at 15:41
• Thanks for all your comments. Regarding the last one: I make an IFFT and get the signal back into the time-domain. There I only look at the Amplitude over time. @ Dan Boschen: Thanks for the link. This got me to think about the filtering solely in the time-domain. Maybe I can come up with a solution here. Jun 13, 2016 at 16:38
• I know this is not an answer, but i can't comment yet as i need a 50 reputation to do so... I also know that this is an old post, so i'm hoping the OP can post an answer here. I researched this quite a lot and tried the FFT method which simply did not work because of the noise present in each files, but i have the same exact question as you and it seems to be exactly the solution to my problem, but i'm having a hard time applying the resulting eq... Here is my original question: dsp.stackexchange.com/questions/90194/… I have tried a fe Dec 9, 2023 at 17:57
Thanks for all your comments. I finally found the answer this article and implemented it in python:https://www.dsprelated.com/freebooks/filters/Time_Domain_Filter_Estimation.html It works pretty well, if you first find the transfer function h(t) with raw and equalized white noise and then apply the function as a filter to any audio data:
import numpy as np
import scipy
def learn_equalizer(signal_x, signal_y, detail_level):
# Create Toeplitz Matrix
X = scipy.linalg.toeplitz(first_col, first_row)
# Implement the formula for finding the optimal solution numerically
X = X[0:len(X)-detail_level,]
h = np.dot(X.T,X)
h = scipy.linalg.inv(h)
h = np.dot(h,X.T)
equalizer = np.dot(h,signal_y)
return equalizer
Finally looking at the curves let's consider two plots: One with the raw and equalized signal, to show their differences. The other plot shows the equalized signal and the "mimicked" equalizer. Here you see how good the learned filter is, when applied to real sounds after being trained on white noise.
Signal comparison:
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http://www.piclist.com/techref/postbot.asp?by=thread&id=%5BEE%5D%3A+bridge+circuit+or+equiv+for+PIC+A%2FD+input&w=body&tgt=post&at=20010228015653a
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Searching \ for '[EE]: bridge circuit or equiv for PIC A/D input' in subject line. ()
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Search entire site for: 'bridge circuit or equiv for PIC A/D input'.
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'[EE]: bridge circuit or equiv for PIC A/D input'
2001\02\28@015653 by
You guys were so good on the last question, I just have to ask another.
As a senior design project, I am making a temperature sensor for up to
500C range with SiC. The SiC chunk will vary in resistance from 18k to
120k over the 0-500C range. Now I need someway to turn the resistance
into voltage to input to my PIC A/D. My first try was like this...
Bridge Configuration
------------------------------------------------
| | |
+ Rmin Rmax Rthree
Vsource |--Vref+ |--Vref-
|-------A/D in
- Rmax Rmax Rfour
| | |
-----------------------------------------------
Rmin=18k
Rmax=120k
Rthree = variable; %ohms - This is our SiC chunk
Rfour=120k
Vsource =5; %volts
BUT I tested today with a variable in place of R3 and it didn't work
good at all. I guess I need to use the entire 5V range to get all the
resolution I can get. My question is, what is my best bet to do it with
minimum electronics? I had hoped this would be good enough because I
could get away with only a 5V source. I was thinking about ---
------------------------------------------------
| |
+ SiC chunk 120k
5vdc |---to +opamp |---to-opamp
- 120k 18k
| |
-----------------------------------------------
and having the differential opamp for maybe 2x gain (should mean signal
out from 0 to about 4.6V) But I was wondering if someone had an idea
for a better solution. Thanks for any help guys..
MJ
--
http://www.piclist.com hint: The PICList is archived three different
ways. See http://www.piclist.com/#archives for details.
'[EE]: bridge circuit or equiv for PIC A/D input'
2001\03\01@174803 by
A satandard ohmmeter is built using a constant current source and the
resistor. You may have to add a buffer amplifier and/or heavy decoupling
across Rx (i.e. a good quality ~1uF capacitor) to satisfy the A/D input
impedance requirements.
18k to 120k wants a current source that gives 5V across 120k i.e. Ic =
5/120E3 ~= 41 uA. This is easily implemented with an opamp. It will not
work from 5V, you will need at least 6V for it using a rail to rail opamp
whose input range includes Vcc probably.
Peter
--
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More... (looser matching)
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## 550533
550,533 (five hundred fifty thousand five hundred thirty-three) is an odd six-digits composite number following 550532 and preceding 550534. In scientific notation, it is written as 5.50533 × 105. The sum of its digits is 21. It has a total of 2 prime factors and 4 positive divisors. There are 367,020 positive integers (up to 550533) that are relatively prime to 550533.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 6
• Sum of Digits 21
• Digital Root 3
## Name
Short name 550 thousand 533 five hundred fifty thousand five hundred thirty-three
## Notation
Scientific notation 5.50533 × 105 550.533 × 103
## Prime Factorization of 550533
Prime Factorization 3 × 183511
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 2 Total number of prime factors rad(n) 550533 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 550,533 is 3 × 183511. Since it has a total of 2 prime factors, 550,533 is a composite number.
## Divisors of 550533
4 divisors
Even divisors 0 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 734048 Sum of all the positive divisors of n s(n) 183515 Sum of the proper positive divisors of n A(n) 183512 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 741.979 Returns the nth root of the product of n divisors H(n) 2.99998 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 550,533 can be divided by 4 positive divisors (out of which 0 are even, and 4 are odd). The sum of these divisors (counting 550,533) is 734,048, the average is 183,512.
## Other Arithmetic Functions (n = 550533)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 367020 Total number of positive integers not greater than n that are coprime to n λ(n) 183510 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 45253 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 367,020 positive integers (less than 550,533) that are coprime with 550,533. And there are approximately 45,253 prime numbers less than or equal to 550,533.
## Divisibility of 550533
m n mod m 2 3 4 5 6 7 8 9 1 0 1 3 3 4 5 3
The number 550,533 is divisible by 3.
## Classification of 550533
• Arithmetic
• Semiprime
• Deficient
### Expressible via specific sums
• Polite
• Non-hypotenuse
• Square Free
### Other numbers
• LucasCarmichael
## Base conversion (550533)
Base System Value
2 Binary 10000110011010000101
3 Ternary 1000222012010
4 Quaternary 2012122011
5 Quinary 120104113
6 Senary 15444433
8 Octal 2063205
10 Decimal 550533
12 Duodecimal 226719
16 Hexadecimal 86685
20 Vigesimal 38g6d
36 Base36 bssl
## Basic calculations (n = 550533)
### Multiplication
n×i
n×2 1101066 1651599 2202132 2752665
### Division
ni
n⁄2 275266 183511 137633 110107
### Exponentiation
ni
n2 303086584089 166859166398269437 91861477454738467959921 50572774767589532981379187893
### Nth Root
i√n
2√n 741.979 81.9586 27.2393 14.0655
## 550533 as geometric shapes
### Circle
Radius = n
Diameter 1.10107e+06 3.4591e+06 9.52175e+11
### Sphere
Radius = n
Volume 6.98938e+17 3.8087e+12 3.4591e+06
### Square
Length = n
Perimeter 2.20213e+06 3.03087e+11 778571
### Cube
Length = n
Surface area 1.81852e+12 1.66859e+17 953551
### Equilateral Triangle
Length = n
Perimeter 1.6516e+06 1.3124e+11 476776
### Triangular Pyramid
Length = n
Surface area 5.24961e+11 1.96645e+16 449508
## Cryptographic Hash Functions
md5 b36a3b5cd3b958155ec8d6afb8c10552 16497a41c417b57183605bbf314d9865c38c9c66 0acb5c582a68ccf91c408f17c60608ff25e8a2b1a34d0cc4b416e8ee9112cd3d 0fcc0d1cbc0c6d01c0f610ddf1bf671e0fe5bef393f9657d66dc82e9e2b16c67810b6c9e75f93fa6dd7213ccf947061a3d28a4f8725ac678ac0698e9b918f933 802ce4ff1aa73efc301bdc3f027abbd92f156e25
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There are 366 different Starters of The Day, many to choose from. You will find in the left column below some starters on the topic of Tables. In the right column below are links to related online activities, videos and teacher resources.
A lesson starter does not have to be on the same topic as the main part of the lesson or the topic of the previous lesson. It is often very useful to revise or explore other concepts by using a starter based on a totally different area of Mathematics.
Main Page
### Tables Starters:
5.5 Times Table: Write out the 5.5 times table as far as possible.
Aunt Sophie's Post Office: Work out the number of stamps needed to post a parcel.
Hot Summer Test: Write out a large times table. Get as far as possible in 5 minutes.
North Pole Test: This starter requires you to write out a difficult times table.
Not Multiples: Write down the numbers from a list which are not multiples of a given number.
Refreshing Revision: It is called Refreshing Revision because every time you refresh the page you get different revision questions.
Strange Tables: A challenge to learn an unfamiliar times table involving decimals.
Subtract Quickulations: Calculations appear on the screen every few seconds.
Table Legs: Learn an unusual times table from the strategic finger moving up and down the 'Table Leg'!
Table Spiders: Multiply the number on the spider's back by the numbers next to its legs.
Timed Tables: How fast can you answer 24 mixed times tables questions?
#### Number Grids
Investigate the properties of number with these interactive number grids.
Transum.org/go/?to=grids
### Curriculum for Tables:
#### Year 5
Pupils should be taught to multiply and divide numbers mentally drawing upon known facts more...
### Exam-Style Questions:
There are almost a thousand exam-style questions unique to the Transum website.
### Feedback:
Comment recorded on the 28 September 'Starter of the Day' page by Malcolm P, Dorset:
"A set of real life savers!!
Keep it up and thank you!"
Comment recorded on the 25 June 'Starter of the Day' page by Inger.kisby@herts and essex.herts.sch.uk, :
"We all love your starters. It is so good to have such a collection. We use them for all age groups and abilities. Have particularly enjoyed KIM's game, as we have not used that for Mathematics before. Keep up the good work and thank you very much
Best wishes from Inger Kisby"
Comment recorded on the s /Coordinate 'Starter of the Day' page by Greg, Wales:
"Excellent resource, I use it all of the time! The only problem is that there is too much good stuff here!!"
Comment recorded on the 19 November 'Starter of the Day' page by Lesley Sewell, Ysgol Aberconwy, Wales:
"A Maths colleague introduced me to your web site and I love to use it. The questions are so varied I can use them with all of my classes, I even let year 13 have a go at some of them. I like being able to access Starters for the whole month so I can use favourites with classes I see at different times of the week. Thanks."
Comment recorded on the 1 August 'Starter of the Day' page by Peter Wright, St Joseph's College:
"Love using the Starter of the Day activities to get the students into Maths mode at the beginning of a lesson. Lots of interesting discussions and questions have arisen out of the activities.
Thanks for such a great resource!"
Comment recorded on the 18 September 'Starter of the Day' page by Mrs. Peacock, Downe House School and Kennet School:
"My year 8's absolutely loved the "Separated Twins" starter. I set it as an optional piece of work for my year 11's over a weekend and one girl came up with 3 independant solutions."
Comment recorded on the 8 May 'Starter of the Day' page by Mr Smith, West Sussex, UK:
"I am an NQT and have only just discovered this website. I nearly wet my pants with joy.
To the creator of this website and all of those teachers who have contributed to it, I would like to say a big THANK YOU!!! :)."
Comment recorded on the 14 September 'Starter of the Day' page by Trish Bailey, Kingstone School:
"This is a great memory aid which could be used for formulae or key facts etc - in any subject area. The PICTURE is such an aid to remembering where each number or group of numbers is - my pupils love it!
Thanks"
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"Thank you very much for providing these resources for free for teachers and students. It has been engaging for the students - all trying to reach their highest level and competing with their peers while also learning. Thank you very much!"
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"I think this is a brilliant website as all the students enjoy doing the puzzles and it is a brilliant way to start a lesson."
Comment recorded on the 19 June 'Starter of the Day' page by Nikki Jordan, Braunton School, Devon:
"Excellent. Thank you very much for a fabulous set of starters. I use the 'weekenders' if the daily ones are not quite what I want. Brilliant and much appreciated."
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"Thank you for sharing such a great resource. I was about to try and get together a bank of starters but time is always required elsewhere, so thank you."
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"I use Starter of the Day as a registration and warm-up activity for my Year 6 class. The range of questioning provided is excellent as are some of the images.
I rate this site as a 5!"
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Could we have some on angles too please?"
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"Dear Transum,
I love you website I use it every maths lesson I have with every year group! I don't know were I would turn to with out you!"
Comment recorded on the 26 March 'Starter of the Day' page by Julie Reakes, The English College, Dubai:
"It's great to have a starter that's timed and focuses the attention of everyone fully. I told them in advance I would do 10 then record their percentages."
Comment recorded on the 17 November 'Starter of the Day' page by Amy Thay, Coventry:
"Thank you so much for your wonderful site. I have so much material to use in class and inspire me to try something a little different more often. I am going to show my maths department your website and encourage them to use it too. How lovely that you have compiled such a great resource to help teachers and pupils.
Thanks again"
Comment recorded on the 1 February 'Starter of the Day' page by Terry Shaw, Beaulieu Convent School:
"Really good site. Lots of good ideas for starters. Use it most of the time in KS3."
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"This is an excellent website. We all often use the starters as the pupils come in the door and get settled as we take the register."
Comment recorded on the 11 January 'Starter of the Day' page by S Johnson, The King John School:
"We recently had an afternoon on accelerated learning.This linked really well and prompted a discussion about learning styles and short term memory."
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"Find the starters wonderful; students enjoy them and often want to use the idea generated by the starter in other parts of the lesson. Keep up the good work"
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"This resource has made a great deal of difference to the standard of starters for all of our lessons. Thank you for being so creative and imaginative."
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Comment recorded on the 1 May 'Starter of the Day' page by Phil Anthony, Head of Maths, Stourport High School:
"What a brilliant website. We have just started to use the 'starter-of-the-day' in our yr9 lessons to try them out before we change from a high school to a secondary school in September. This is one of the best resources on-line we have found. The kids and staff love it. Well done an thank you very much for making my maths lessons more interesting and fun."
Comment recorded on the 23 September 'Starter of the Day' page by Judy, Chatsmore CHS:
"This triangle starter is excellent. I have used it with all of my ks3 and ks4 classes and they are all totally focused when counting the triangles."
Comment recorded on the 3 October 'Starter of the Day' page by S Mirza, Park High School, Colne:
"Very good starters, help pupils settle very well in maths classroom."
### Notes:
Times Tables is the common term referring to the multiples of numbers 2 to 12 (or 2 to 10). Having a quick recall of these tables is an important pre-requisite for studying other aspects of mathematics and for coping with personal finance and other area of everyday live involving numbers.
People of any age can improve their skills in recalling table facts. They should learn then as they would learn a song or a dance. You need to know your times tables forwards, backwards and all mixed up. Spend time learning them well and you'll reap the benefits in future.
Here on this website we have developed many activities that help pupils learn their times tables and as then revise them in different ways so that the recall becomes easier and easier. Some of the activities are games and quizzes while others help pupils spot the patterns in the times tables in many different ways.
Here's a plan for learning a new times table in only five days!
### Tables Teacher Resources:
Number Grids: Investigate the properties of number with these interactive number grids.
Table Legs: Transum believes that this is the best way of developing a quick recall of the times tables.
Flash Tables: A never ending sequence of times tables questions to be projected on to a whiteboard or screen.
Quickulations: A mental arithmetic visual aid that displays random calculations then after a few seconds displays the answers.
Sieve of Eratosthenes: A self checking, interactive version of the Sieve of Eratosthenes method of finding prime numbers.
### Tables Activities:
Number Grids: Investigate the properties of number with these interactive number grids.
Times Tables: A collection of activities to help you learn your times tables in only 5 days.
Fast Factors: Match the numbers with the answer to the times table. A timed activity to improve instant recall of key table facts.
Quotientmaster: A fast-paced activity to help you practise a Times Table by developing an ability to quickly recall quotients.
Tables Grab: A one or two player game. The objective is to grab all the multiples of the chosen times table faster than the other player.
Tables Dash: Revise multiplication facts by racing across the screen to match the times tables question with the correct answer without getting hit by lightning.
Tables Conga: Use the arrow keys to collect all the multiples in order while avoiding the Conga Virus!
Beat The Clock: It is a race against the clock to answer 30 mental arithmetic questions. There are nine levels to choose from.
Fizz Buzzer: The digital version of the popular fizz buzz game. Press the buzzers if they are factors of the counter.
Convoluted: Find the runs of four multiples in order as quickly as you can.
Grid Arithmetic: Fill in a multiplication grid with the answers to simple multiplication and division questions.
Times Square: Practise your times tables with this self-checking multiplication grid.
Stamp Sticking: Drag stamps onto the envelopes to make the exact postage as shown at the top left of each envelope.
Expedite: Drag the numbered cards to produce a multiplication fact. Complete twenty mixed times tables questions to earn a trophy.
TablesMaster: How fast can you answer times table questions? This activity provides feedback to help you improve.
Hard Times: The hardest multiplication facts (according to Transum research) are presented in the form of pairs games.
Sieve of Eratosthenes: A self checking, interactive version of the Sieve of Eratosthenes method of finding prime numbers.
### Tables Videos:
Times Tables in 10 minutes: Jill Mansergh uses a number stick to teach the 17 times table in less than ten minutes.
Nine Times Fingers: Learn your 9 times table fast using your fingers!
Maths Fact Fluency: A four-step process for taking pupils on the path from acquisition to automaticity
Memorise the Multiplication Table: A quick overview of methods for learning all one hundred multiplication facts from 1x1 to 10x10.
### Tables Worksheets/Printables:
Fast Factors Board: A printable page to be used with the Fast Factors Activity.
Table Spiders: A number of pages each containing table spiders with gaps to be filled.
Multiplication Tables Worksheets: Printable times tables worksheet for all of the tables from two to twelve.
Links to other websites containing resources for Tables are provided for those logged into 'Transum Mathematics'. Subscribing also opens up the opportunity for you to add your own links to this panel. You can sign up using one of the buttons below:
### Search
The activity you are looking for may have been classified in a different way from the way you were expecting. You can search the whole of Transum Maths by using the box below.
### Other
Is there anything you would have a regular use for that we don't feature here? Please let us know.
#### Sieve of Eratosthenes
A self checking, interactive version of the Sieve of Eratosthenes method of finding prime numbers.
Transum.org/go/?to=sieve
### Teaching Notes:
Many Transum activities have notes for teachers suggesting teaching methods and highlighting common misconceptions. There are also solutions to puzzles, exercises and activities available on the web pages when you are signed in to your Transum subscription account. If you do not yet have an account and you are a teacher, tutor or parent you can apply for one by completing the form on the Sign Up page.
Have today's Starter of the Day as your default homepage. Copy the URL below then select
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Saturday, August 12, 2017
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## 共享
2x-3y=48
3x+5y=15
2x-3y=48,3x+5y=15
2x-3y=48
2x=3y+48
x=\frac{1}{2}\left(3y+48\right)
x=\frac{3}{2}y+24
3\left(\frac{3}{2}y+24\right)+5y=15
\frac{9}{2}y+72+5y=15
\frac{19}{2}y+72=15
\frac{19}{2}y=-57
y=-6
x=\frac{3}{2}\left(-6\right)+24
x=-9+24
x=15
x=15,y=-6
2x-3y=48
3x+5y=15
2x-3y=48,3x+5y=15
\left(\begin{matrix}2&-3\\3&5\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}48\\15\end{matrix}\right)
inverse(\left(\begin{matrix}2&-3\\3&5\end{matrix}\right))\left(\begin{matrix}2&-3\\3&5\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-3\\3&5\end{matrix}\right))\left(\begin{matrix}48\\15\end{matrix}\right)
\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-3\\3&5\end{matrix}\right))\left(\begin{matrix}48\\15\end{matrix}\right)
\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-3\\3&5\end{matrix}\right))\left(\begin{matrix}48\\15\end{matrix}\right)
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{5}{2\times 5-\left(-3\times 3\right)}&-\frac{-3}{2\times 5-\left(-3\times 3\right)}\\-\frac{3}{2\times 5-\left(-3\times 3\right)}&\frac{2}{2\times 5-\left(-3\times 3\right)}\end{matrix}\right)\left(\begin{matrix}48\\15\end{matrix}\right)
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{5}{19}&\frac{3}{19}\\-\frac{3}{19}&\frac{2}{19}\end{matrix}\right)\left(\begin{matrix}48\\15\end{matrix}\right)
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{5}{19}\times 48+\frac{3}{19}\times 15\\-\frac{3}{19}\times 48+\frac{2}{19}\times 15\end{matrix}\right)
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}15\\-6\end{matrix}\right)
x=15,y=-6
2x-3y=48
3x+5y=15
2x-3y=48,3x+5y=15
3\times 2x+3\left(-3\right)y=3\times 48,2\times 3x+2\times 5y=2\times 15
6x-9y=144,6x+10y=30
6x-6x-9y-10y=144-30
-9y-10y=144-30
-19y=144-30
-19y=114
y=-6
3x+5\left(-6\right)=15
3x-30=15
3x=45
x=15
x=15,y=-6
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# CS502 Assignment No.2
Fundamentals of Algorithms (CS502)
Assignment#02
Total marks = 20
Lectures Covered:Thisassignment covers Lecture # 23 to 26.
Objectives:
Objectives of this assignment are:
• To implement Activity Selection Algorithm in real life problems
• Generation of Variable-length codes using Huffman Coding Algorithm
Instructions:
Please read the following instructions carefully before solving & submitting the assignment:
1. The assignment will not be accepted after due date.
2. Zero marks will be awarded to the assignment that does not open or the file is corrupt.
3. The assignment file must be an MS Word (.doc/.docx) file format; Assignment will not be accepted in any other format.
4. Zero marks will be awarded to the assignment if copied (from other student or copied from handouts or internet).
5. Zero marks will be awarded to the assignment if the Student ID is not mentioned in the assignment file.
For any query about the assignment, contact only at CS502@vu.edu.pk
Please do not post queries related to assignment on MDB.
Question # 1: 10Marks (5+5)
In the context of Activity Selection Problem, consider the following set of activities:
Activity A B C D E F G H I J K L M N O Start Time 2 4 6 1 6 3 1 8 5 6 7 4 2 5 7 Finish Time 9 5 7 3 8 5 2 9 8 9 9 8 3 6 8
You are required to find the optimal solution (usingGreedy Algorithm) for the following two greediness approaches:
1. Select the activities that start first and schedule them(5 marks)
2. Select the activities that finish first and schedule them(5 marks)
Note:
No need to provide any pseudo/working code. Just provide the results for each greediness approach of the following two steps in the below mentioned tabular form:
1. Sorted Activities
2. Final Selected Activities
Activity StartTime Finish Time
Question # 2: 10 Marks
Consider the following scenario in which a set of Alphabets along with their frequencies is given. You are required to generate the output binary tree and find the Variable-length codes for the given Alphabets using the provided Huffman Coding Algorithm.
Total File Length: 210
Frequency table:
A B C D E F 10 20 30 40 50 60
Huffman Coding Algorithm:
1. Consider all pairs: <frequency, symbol>.
2. Choose the two lowest frequencies, and make them brothers, with the root having the
combined frequency.
1. Iterate.
Note:
No need to provide all the steps of the binary tree generation. Only mention the Final Binary Tree and the Variable-length codes for the given Alphabets in tabular form.
Your solution should be strictly according to the below-mentioned template.
Solution Template:
Alphabets: A, B, C, D, E, F
Total File Length: 100
Frequency table:
A B C D E F 45 13 12 16 9 5
Output Tree:
Letter to be encoded A B C D E F Frequency 45 13 12 16 9 5 Variable-length code 0 101 100 111 1101 1100
Good Luck
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### Replies to This Discussion
sir kindly share the idea of q 2 also
plz koi help kryn... Q.1 part 1 ka idea bh dy dyn??
starting time sort kr k us mese activity select kesy hogi?
There will be four tables;
1. Sorted activities on the basis of Start Time.
2. Final selected activities.
3. Sorted activities on the basis of Finish Time.
4. Final selected activities....in question :1
sir ne btaya
bilkul. start time k inc order main sort karain phir normal procedure chalaen.
Check Respected Frnds Cs502 Q 2 Solution
Attachments:
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likhn mi n lecture suna hy woh to left side py ketha hy youtube se
thanksx
ap k kn sa semester hy
3rd mcs ka
1st question ho gya?????
yes kr rha hn us k solution ata hhy
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Explore BrainMass
# Bivariate Regression in Excel
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!
(a) Make an Excel scatter plot. What does it suggest about the population correlation between X and Y?
(b) Make an Excel worksheet to calculate SSxx , SSyy , and SSxy. Use these sums to calculate the sample correlation coefficient. Check your work by using Excel's function =CORREL(array1,array2). Find t.05 for a two-tailed test for zero correlation. (d) Calculate the t test statistic. Can you reject rho = 0? (e) Use Excel's function =TDIST(t,deg_freedom,tails) to calculate the two-tail p-value.
Part-Time Weekly Earnings (\$) by College Students
Hours Worked (X) Weekly Pay (Y)
10 93
15 171
20 204
20 156
35 261
© BrainMass Inc. brainmass.com June 3, 2020, 9:16 pm ad1c9bdddf
https://brainmass.com/statistics/regression-analysis/bivariate-regression-174322
#### Solution Summary
This solution answers questions regarding regression analysis in Excel.
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Mathbox for BJ < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > bj-abeq2 Structured version Visualization version GIF version
Theorem bj-abeq2 32898
Description: Remove dependency on ax-13 2282 from abeq2 2761. (Contributed by BJ, 23-Jun-2019.) (Proof modification is discouraged.)
Assertion
Ref Expression
bj-abeq2 (𝐴 = {𝑥𝜑} ↔ ∀𝑥(𝑥𝐴𝜑))
Distinct variable group: 𝑥,𝐴
Allowed substitution hint: 𝜑(𝑥)
Proof of Theorem bj-abeq2
Dummy variable 𝑦 is distinct from all other variables.
StepHypRef Expression
1 ax-5 1879 . . 3 (𝑦𝐴 → ∀𝑥 𝑦𝐴)
2 bj-hbab1 32896 . . 3 (𝑦 ∈ {𝑥𝜑} → ∀𝑥 𝑦 ∈ {𝑥𝜑})
31, 2cleqh 2753 . 2 (𝐴 = {𝑥𝜑} ↔ ∀𝑥(𝑥𝐴𝑥 ∈ {𝑥𝜑}))
4 abid 2639 . . . 4 (𝑥 ∈ {𝑥𝜑} ↔ 𝜑)
54bibi2i 326 . . 3 ((𝑥𝐴𝑥 ∈ {𝑥𝜑}) ↔ (𝑥𝐴𝜑))
65albii 1787 . 2 (∀𝑥(𝑥𝐴𝑥 ∈ {𝑥𝜑}) ↔ ∀𝑥(𝑥𝐴𝜑))
73, 6bitri 264 1 (𝐴 = {𝑥𝜑} ↔ ∀𝑥(𝑥𝐴𝜑))
Colors of variables: wff setvar class Syntax hints: ↔ wb 196 ∀wal 1521 = wceq 1523 ∈ wcel 2030 {cab 2637 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1762 ax-4 1777 ax-5 1879 ax-6 1945 ax-7 1981 ax-9 2039 ax-10 2059 ax-11 2074 ax-12 2087 ax-ext 2631 This theorem depends on definitions: df-bi 197 df-or 384 df-an 385 df-tru 1526 df-ex 1745 df-nf 1750 df-sb 1938 df-clab 2638 df-cleq 2644 df-clel 2647 This theorem is referenced by: bj-abeq1 32899 bj-abbi2i 32901 bj-abbi2dv 32905 bj-clabel 32908 bj-ru1 33058
Copyright terms: Public domain W3C validator
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EASEL BY TPT
# Easter Egg Fillers Freebie: 2 Digit Addition
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### Description
Want the kiddos to have some Easter Egg fun and practice some important skills?
This freebie gives the children practice with 2 digit addition, and gives them some fun as well!
This resource is made to go with those fun plastic Easter Eggs that you can pick up at your local dollar store.
It includes two pages of adding 2 digit numbers. There's also a blank page for plugging in your own skills. This set is a sampling of a larger set: Easter Egg Fillers.
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### Standards
to see state-specific standards (only available in the US).
Understand that the two digits of a two-digit number represent amounts of tens and ones. Understand the following as special cases:
Add within 100, including adding a two-digit number and a one-digit number, and adding a two-digit number and a multiple of 10, using concrete models or drawings and strategies based on place value, properties of operations, and/or the relationship between addition and subtraction; relate the strategy to a written method and explain the reasoning used. Understand that in adding two-digit numbers, one adds tens and tens, ones and ones; and sometimes it is necessary to compose a ten.
Fluently add and subtract within 100 using strategies based on place value, properties of operations, and/or the relationship between addition and subtraction.
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# A Tutorial on Identification of Nonlinear ARX and Hammerstein-Wiener Models
This example shows identification of single-input-single-output (SISO) nonlinear black box models using measured input-output data. The example uses measured data from a two-tank system to explore various model structures and identification choices.
### The Input-Output Data Set
In this example, the data variables stored in the `twotankdata.mat` file will be used. It contains 3000 input-output data samples of a two-tank system generated using a sample time of 0.2 seconds. The input u(t) is the voltage [V] applied to a pump, which generates an inflow to the upper tank. A rather small hole at the bottom of this upper tank yields an outflow that goes into the lower tank, and the output y(t) of the two tank system is then the liquid level [m] of the lower tank. We create an IDDATA object `z` to encapsulate the loaded data:
This data set is also used in the nonlinear grey box example "Two Tank System: C MEX-File Modeling of Time-Continuous SISO System" where physical equations describing the system are assumed. This example is about black box models, thus no knowledge of the actual physical laws is assumed.
```load twotankdata z = iddata(y, u, 0.2, 'Name', 'Two tank system'); ```
### Splitting the Data and Plotting
Split this data set into 3 subsets of equal size, each containing 1000 samples.
We shall estimate models with `z1`, and evaluate them on `z2` and `z3`. `z1` is similar to `z2`, but not `z3`. Therefore in some sense, `z2` would help us evaluate the interpolation ability of the models, while `z3` would help us evaluate their extrapolation ability.
```z1 = z(1:1000); z2 = z(1001:2000); z3 = z(2001:3000); plot(z1,z2,z3) legend('Estimation','Validation 1', 'Validation 2') ```
### Regressor Selection
The first step in estimating nonlinear black box models is to pick the regressor set. The regressors are simple formulas based on time-delayed input/output variables, the simplest case being the variables lagged by a small set of consecutive values. For example, if "u" is the name of an input variable, and "y" the name of an output variable, then an example regressor set can be { y(t-1), y(t-2), u(t), u(t-1), u(t-2) }, where "t" denotes the time variable. Another example involving polynomial terms could be {abs(y(t-2)), y(t-2)*u(t-1), u(t-4)^2}. More complex, arbitrary formulas in the delayed variables are also possible.
In the System Identification Toolbox™, the regressors sets don't have to be created explicitly; a large collection can be generated implicitly by just specifying their form (such as polynomial order), the contributing variables and their lags. Specification objects `linearRegressor`, `polynomialRegressor` and `customRegressor` are used for this purpose. For example R = linearRegressor({'y','u'},{[1 2 4], [2 10]}) implies the regressor set {y(t-1), y(t-2), y(t-4), u(t-2), u(t-10)}. Similarly, R = polynomialRegressor({'y'},{[1 2]},3) implies 3rd order polynomial regressors in variable 'y' with lags 1 and 2, that is, the set {y(t-1)^3, y(t-2)^3}. More configuration choices are available such as using only the absolute values, and/or allowing mix of lags in the regressor formulas.
### Linear Regressor Specification Using an Order Matrix
When the model uses only linear regressors that are based on contiguous lags, it is often easier to specify them using an order matrix, rather than using a `linearRegressor` object. The order matrix [na nb nk], defines the numbers of past outputs, past inputs and the input delay used in the regressor formulas. This is the same idea that is used when estimating the linear ARX models (see ARX, IDPOLY). For example, NN = [2 3 4] implies that the output variable uses lags (1,2) and the input variable uses the lags (4,5,6) leading to the regressor set {y(t-1), y(t-2), u(t-4), u(t-5), u(t-6)}.
Usually model orders are chosen by trial and error. Sometimes it is useful to try linear ARX models first in order to get hints about the best orders. So let us first try to determine the optimal orders for a linear ARX model, using the functions `arxstruc` and `selstruc`.
```V = arxstruc(z1,z2,struc(1:5, 1:5,1:5)); % select best order by Akaike's information criterion (AIC) nn = selstruc(V,'aic') ```
```nn = 5 1 3 ```
The AIC criterion has selected nn = [na nb nk] = [5 1 3], meaning that, in the selected ARX model structure, y(t) is predicted by the 6 regressors y(t-1),y(t-1),...,y(t-5) and u(t-3). We will try to use these values when estimating nonlinear models.
### Nonlinear ARX (IDNLARX) Models
In an IDNLARX model, the model output is a nonlinear function of regressors, like y(t) = Fcn(y(t-1),y(t-1),...,y(t-5), u(t-3)).
IDNLARX models are typically estimated with the syntax:
```M = NLARX(Data, Orders, Fcn). ```
It is similar to the linear ARX command, with the additional argument "Fcn" specifying the type of nonlinear function. A Nonlinear ARX model produces its output by using a 2-stage transformation: 1. Map the training data (input-output signals) to a set of regressors. Numerically, the regressor set is a double matrix R, with one column of data for each regressor. 2. Map the regressors to a single output, y = Fcn®, where Fcn() is a scalar nonlinear function.
To estimate an IDNLARX model, after the choice of model orders, we need to pick the type of nonlinear mapping function Fcn() to use. Let us first try a Wavelet Network (wavenet) function.
```mw1 = nlarx(z1,[5 1 3], wavenet); ```
The Wavelet Network uses a combination of wavelet units to map the regressors to the output of the model. The model `mw1` is an @idnalrx object. It stores the nonlinear mapping function (wavenet here) in its `OutputFcn` property. The number of these units to use can be specified in advance, or we can leave it to the estimation algorithm to determine an optimal value automatically. The property NonlinearFcn.NumberOfUnits stores the chosen number of units in the model.
```NLFcn = mw1.OutputFcn; NLFcn.NonlinearFcn.NumberOfUnits ```
```ans = 3 ```
Examine the result by comparing the response of the model `mw1` to the measured output in the dataset `z1`.
```compare(z1,mw1); ```
### Using Model Properties
The first model was not quite satisfactory, so let us try to modify some properties of the model. Instead of letting the estimation algorithm automatically choose the number of units in the WAVENET function, this number can be explicitly specified.
```mw2 = nlarx(z1,[5 1 3], wavenet(8)); ```
Default InputName and OutputName have been used.
```mw2.InputName mw2.OutputName ```
```ans = 1×1 cell array {'u1'} ans = 1×1 cell array {'y1'} ```
The regressors of the IDNLARX model can be viewed using the GETREG command. The regressors are made from `mw2.InputName`, `mw2.OutputName` and time delays.
```getreg(mw2) ```
```ans = 6×1 cell array {'y1(t-1)'} {'y1(t-2)'} {'y1(t-3)'} {'y1(t-4)'} {'y1(t-5)'} {'u1(t-3)'} ```
Note that the order matrix is equivalent to the linear regressor set created using the specification object, as in:
```R = linearRegressor([z1.OutputName; z1.InputName],{1:5, 3}); mw2_a = nlarx(z1, R, wavenet(8)); getreg(mw2_a) ```
```ans = 6×1 cell array {'y1(t-1)'} {'y1(t-2)'} {'y1(t-3)'} {'y1(t-4)'} {'y1(t-5)'} {'u1(t-3)'} ```
`mw2_a` is essentially the same model as `mw2`. The use of a specification object allows more flexibility; use it when you do not want to use continguous lags, or have a minimum lag in output variables that is greater than one.
### Assigning Regressors to Mapping Functions
The output of an IDNLARX model is a static function of its regressors. Usually this mapping function has a linear block, a nonlinear block, plus a bias term (output offset). The model output is the sum of the outputs of these blocks and the bias. In order to reduce model complexity, only a subset of regressors can be chosen to enter the nonlinear block and the linear block. The assignment of the regressors to the linear and nonlinear components of the mapping function is performed using the RegressorUsage property of the model.
```RegUseTable = mw2.RegressorUsage ```
```RegUseTable = 6×2 table y1:LinearFcn y1:NonlinearFcn ____________ _______________ y1(t-1) true true y1(t-2) true true y1(t-3) true true y1(t-4) true true y1(t-5) true true u1(t-3) true true ```
RegUseTable is a table that shows which regressors are used as inputs to the linear and the nonlinear components of the Wavelet Network. Each row corresponds to one regressor. Suppose we want to use only those regressors that are functions of input variable 'u1' in the nonlinear component. This configuration can be achieved as follows.
```RegNames = RegUseTable.Properties.RowNames; I = contains(RegNames,'u1'); RegUseTable.("y1:NonlinearFcn")(~I) = false; ```
As an example, consider a model that uses polynomial regressors of order 2 in addition to the linear ones used by `mw2`. First, generate a polynomial regressor specification set.
```P = polynomialRegressor({'y1','u1'},{1:2, 0:3},2) ```
```P = <strong>Order 2 regressors in variables y1, u1</strong> Order: 2 Variables: {'y1' 'u1'} Lags: {[1 2] [0 1 2 3]} UseAbsolute: [0 0] AllowVariableMix: 0 AllowLagMix: 0 TimeVariable: 't' ```
Now add the specification `P` to the model's Regressors property.
```mw3 = mw2; % start with the mw2 structure for new estimation mw3.Regressors = [mw3.Regressors; P]; % add the polynomial set to the model structure RegNames = mw3.RegressorUsage.Properties.RowNames; Ind = contains(RegNames,'u1'); mw3.RegressorUsage.("y1:NonlinearFcn")(~Ind) = false; ```
Finally, update the parameters of the model to fit the data by minimizing 1-step ahead prediction errors.
```mw3 = nlarx(z1, mw3) ```
```mw3 = <strong>Nonlinear ARX model with 1 output and 1 input</strong> Inputs: u1 Outputs: y1 Regressors: 1. Linear regressors in variables y1, u1 2. Order 2 regressors in variables y1, u1 Output function: Wavelet Network with 5 units Sample time: 0.2 seconds Status: Estimated using NLARX on time domain data "Two tank system". Fit to estimation data: 96.72% (prediction focus) FPE: 3.583e-05, MSE: 3.438e-05 ```
### Evaluating Estimated Models
Different models, both linear and nonlinear, can be compared together in the same COMPARE command.
```mlin = arx(z1,[5 1 3]); % linear ARX model % compare(z1,mlin,mw2,mw3) ```
Model validation can be carried out by running the COMPARE command on the validation datasets. First, validate the models by checking how well their simulated responses match the measured output signal in `z2`.
```compare(z2,mlin,mw2,mw3) ```
Similarly, validate the estimated models against the dataset z3.
```compare(z3,mlin,mw2,mw3) ```
An analysis of the model residues may also be performed to validate the model quality. We expect the residues to be white (uncorrelated with itself except at lag 0) and uncorrelated with the input signal. The resideus are the 1-step ahead prediction errors. The residues can be visualized with bounds on insignificant values by using the RESID command. For example on the valiation dataset `z2`, the residues for the three identified models are shown below.
```resid(z2,mlin,mw2,mw3) legend show ```
The residues are all uncorrelated with the input signal of `z2` (z2.InputData). However, they exhibit some auto-correlation for the models `mlin` and `mw3`. The model `mw2` performs better on both (auto- and cross-correlation) tests.
Function PLOT may be used to view the nonlinearity responses of various IDNLARX models. The plot essentially shows the characteristics of the nonlinear mapping function Fcn().
```plot(mw2,mw3) % plot nonlinearity response as a function of selected regressors ```
### Nonlinear ARX Model with SIGMOIDNET Function
In place of WAVENET, other nonlinear functions can be used in the model. Try the SIGMOIDNET function, which maps the regressors to the output using a sum of sigmoid units (plus a linear and an offset term). Configure the sigmoidnet to use 8 units of sigmoid (dilated and translated by different amounts).
```ms1 = nlarx(z1,[5 1 3], sigmoidnet(8)); compare(z1,ms1) ```
Now evaluate the new model on the data set z2.
```compare(z2,ms1); ```
and on the data set z3.
```compare(z3,ms1); ```
### Other Nonlinear Mapping Functions in IDNLARX Model
CUSTOMNET is similar to SIGMOIDNET, but instead of the sigmoid unit function, users can define their own unit function in MATLAB files. This example uses the gaussunit function.
```type gaussunit.m ```
```function [f, g, a] = gaussunit(x) %GAUSSUNIT customnet unit function example % %[f, g, a] = GAUSSUNIT(x) % % x: unit function variable % f: unit function value % g: df/dx % a: unit active range (g(x) is significantly non zero in the interval [-a a]) % % The unit function must be "vectorized": for a vector or matrix x, the output % arguments f and g must have the same size as x, computed element-by-element. % Copyright 2005-2006 The MathWorks, Inc. % Author(s): Qinghua Zhang f = exp(-x.*x); if nargout>1 g = - 2*x .* f; a = 0.2; end % FILE END ```
For the customnet based model, use only the third, fifth and the sixth regressors as input to the nonlinear component of the customnet function.
```mc1_ini = idnlarx('y1','u1', [5 1 3], customnet(@gaussunit)); Use = false(6,1); Use([3 5 6]) = true; mc1_ini.RegressorUsage{:,2} = Use; mc1 = nlarx(z1,mc1_ini); ```
Tree partition function is another mapping function that can be used in the Nonlinear ARX model. Estimate a model using treepartition function:
```mt1 = nlarx(z1,[5 1 3], treepartition); ```
Compare responses of models `mc1` and `mt1` to data `z1`.
```compare(z1, mc1, mt1) ```
### Using the Network Object from Deep Learning Toolbox
If you have the Deep Learning Toolbox™ available, you can also use a neural network to define the mapping function Fcn() of your IDNLARX model. This network must be a feed-forward network (no recurrent component).
Here, we will create a single-output network with an unknown number of inputs (denote by input size of zero in the network object). The number of inputs is set to the number of regressors during the estimation process automatically.
```if exist('feedforwardnet','file')==2 % run only if Deep Learning Toolbox is available ff = feedforwardnet([5 20]); % use a feed-forward network to map regressors to output ff.layers{2}.transferFcn = 'radbas'; ff.trainParam.epochs = 50; % Create a neuralnet mapping function that wraps the feed-forward network OutputFcn = neuralnet(ff); mn1 = nlarx(z1,[5 1 3], OutputFcn); % estimate network parameters compare(z1, mn1) % compare fit to estimation data end ```
### Hammerstein-Wiener (IDNLHW) Models
A Hammerstein-Wiener model is composed of up to 3 blocks: a linear transfer function block is preceded by a nonlinear static block and/or followed by another nonlinear static block. It is called a Wiener model if the first nonlinear static block is absent, and a Hammerstein model if the second nonlinear static block is absent.
IDNLHW models are typically estimated with the syntax:
```M = NLHW(Data, Orders, InputNonlinearity, OutputNonlinearity). ```
where Orders = [nb bf nk] specifies the orders and delay of the linear transfer function. InputNonlinearity and OutputNonlinearity specify the nonlinear functions for the two nonlinear blocks, at the input and output end of the linear block. M is an @idnlhw model. The linear Output-Error (OE) model corresponds to the case of trivial (unit gain) nonlinearities, that is, if the input and the output nonlinear functions of an IDNLHW model are both unit gains, the model structure is equivalent to a linear OE model.
### Hammerstein-Wiener Model with the Piecewise Linear Nonlinear Function
The PWLINEAR (piecewise linear) nonlinear function is useful in general IDNLHW models.
Notice that, in Orders = [nb nf nk], nf specifies the number of poles of the linear transfer function, somewhat related to the "na" order of the IDNLARX model. "nb" is related to the number if zeros.
```mhw1 = nlhw(z1, [1 5 3], pwlinear, pwlinear); ```
The result can be evaluated with COMPARE as before.
```compare(z1,mhw1) ```
Model properties can be visualized by the PLOT command.
Click on the block-diagram to choose the view of the input nonlinearity (UNL), the linear block or the output nonlinearity (YNL).
For the linear block, it is possible to choose the type of plots within Step response, Bode plot, Impulse response and Pole-zero map.
```plot(mhw1) ```
The break points of the two piecewise linear functions can be examined. This is a two row matrix, the first row corresponds to the input and the second row to the output of the PWLINEAR function.
```mhw1.InputNonlinearity.BreakPoints ```
```ans = Columns 1 through 7 2.6179 3.3895 4.1611 4.9327 5.6297 6.3573 7.0850 -2.2681 -1.6345 0.1033 2.2066 3.7696 4.9984 5.8644 Columns 8 through 10 7.8127 8.2741 9.2257 6.4075 6.5725 9.1867 ```
### Hammerstein-Wiener Model with Saturation and Dead Zone Nonlinearities
The SATURATION and DEADZONE functions are physics inspired nonlinear functions. They can be used to describe actuator/sensor saturation scenario and dry friction effects.
```mhw2 = nlhw(z1, [1 5 3], deadzone, saturation); compare(z1,mhw2) ```
The absence of a nonlinearity is indicated by the UNITGAIN object, or by the empty "[]" value. A unit gain is a just a feed-through of an input signal to the output without any modification.
```mhw3 = nlhw(z1, [1 5 3], 'deadzone', unitgain); % no output nonlinearity mhw4 = nlhw(z1, [1 5 3], [],'saturation'); % no input nonlinearity ```
The limit values of DEADZONE and SATURATION can be examined after estimation.
```mhw2.InputNonlinearity.ZeroInterval mhw2.OutputNonlinearity.LinearInterval ```
```ans = -0.9974 4.7573 ans = 0.2603 0.5846 ```
The initial guess of ZeroInterval for DEADZONE or LinearInterval for SATURATION can be specified while estimating the model.
```mhw5 = nlhw(z1, [1 5 3], deadzone([0.1 0.2]), saturation([-1 1])); ```
### Hammerstein-Wiener Model - Specifying Properties
The estimation algorithm options can be specified using the NLHWOPTIONS command.
```opt = nlhwOptions(); opt.SearchMethod = 'gna'; opt.SearchOptions.MaxIterations = 7; mhw6 = nlhw(z1, [1 5 3], deadzone, saturation, opt); ```
An already-estimated model can be refined by more estimation iterations.
Evaluated on the estimation data `z1`, `mhw7` should have a better fit than `mhw6`.
```opt.SearchOptions.MaxIterations = 30; mhw7 = nlhw(z1, mhw6, opt); compare(z1, mhw6, mhw7) ```
### Hammerstein-Wiener Model - Use Other Nonlinear Functions
The nonlinear functions PWLINEAR, SATURATION and DEADZONE have been mainly designed for use in IDNLHW models. They can only estimate single variable nonlinearities. The other more general nonlinear functions, SIGMOIDNET, CUSTOMNET and WAVENET, can also be used in IDNLHW models. However, the non-differentiable functions TREEPARTITION and NEURALNET cannot be used because the estimation of IDNLHW models require differentiable nonlinear functions.
Get trial now
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https://quick-advices.com/which-gives-amount-or-extent-of-a-surface/
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# Which gives amount or extent of a surface?
## Which gives amount or extent of a surface?
Area (symbolized A ) is a two-dimensional quantity representing amount or extent of surface.
What means of surface area?
: the amount of area covered by the surface of something The lake has roughly the same surface area as 10 football fields.
What are the units for surface area?
In the International System of Units (SI), the standard unit of area is the square metre (written as m2), which is the area of a square whose sides are one metre long. A shape with an area of three square metres would have the same area as three such squares.
### Is surface area a TSA or CSA?
Curved Surface Area (CSA) – It includes the area of all the curved surfaces. Lateral Surface Area (LSA) – It includes the area of all the surface excluding the top and bottom areas. Total Surface Area (TSA) – It includes the area of all the surfaces of the object including the bases.
How do you calculate surface area?
Multiply the length and width, or c and b to find their area. Multiply this measurement by two to account for both sides. Add the three separate measurements together. Because surface area is the total area of all of the faces of an object, the final step is to add all of the individually calculated areas together.
What unit do you use for surface area?
square units
Surface area is measured in square units.
#### How do you find the area of a surface?
How do I find the surface area of a square?
To find the area of a rectangle, multiply its height by its width. For a square you only need to find the length of one of the sides (as each side is the same length) and then multiply this by itself to find the area.
How do I know whether I have TSA or CSA?
when it’s asked to find the area of a sheet required to make a cubical or cuboidal box, then u will use TSA. formula for cuboid is 2(lb+bh+lh) and for cube is 6a² and a is the length of the side. there is one more formula which is CSA+1 base. it include the area of CSA and 1 base.
## How do I calculate an area?
To work out the area of a square or rectangle, multiply its height by its width. If the height and width are in cm, the area is shown in cm². If the height and width are in m, the area is shown in m².
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# Tutorial 8 - Three-way branch migration first passage times¶
Multistrand's "First Passage Time" mode can be used to get statistics on how long it takes to reach a certain state, or one of a set of defined states. Here we compare two sequences for toehold-mediated three-way strand displacement. Starting in with the incoming strand binding by the toehold only, we run simulations until either the incoming strand falls off again (a "failure") or the incumbent top strand is displaced and dissociates (a "success"). This is specified via Dissoc_Macrostate StopConditions that look for the incoming strand being in a complex by itself, or the incumbent strand being in a complex by itself, respectively. We perform only 100 trials for each sequence, because that already takes around 10 minutes (most of which is design A). Once that's done, for each sequence we have 100 trial results, each tagged with either "failure" or "success". We can look at histograms of how long such trajectories took, or alternatively tabulate what fraction of trajectories completed by a certain time.
In [2]:
from multistrand.objects import *
from multistrand.options import Options
from multistrand.system import SimSystem
import numpy as np
import matplotlib
import matplotlib.pylab as plt
In [3]:
# Multistrand uses numerical arguments to specific the type of macrostate being defined,
# for StopCondition and Macrostate definitions:
Exact_Macrostate = 0 # match a secondary structure exactly (i.e. any system state that has a complex with this exact structure)
Bound_Macrostate = 1 # match any system state in which the given strand is bound to another strand
Dissoc_Macrostate = 2 # match any system state in which there exists a complex with exactly the given strands, in that order
Loose_Macrostate = 3 # match a secondary structure with "don't care"s, allowing a certain number of disagreements
Count_Macrostate = 4 # match a secondary structure, allowing a certain number of disagreements
We will do Multistrand simulations on two sequences, Design A and Design B. These are taken from Schaeffer's PhD thesis, figure 7.4.
Design A has a hairpin and is slow / fails often, whereas Design B is fast. (In Schaeffer's figure 7.4, Design A has the same sequence, but Design B is "ACCGCACCACGTGGGTGTCG".)
In [5]:
def setup_threewaybm_comparison(trials=100, toehold_seq = "GTGGGT", bm_design_A = "ACCGCACGTCCACGGTGTCG", bm_design_B = "ACCGCACGTCACTCACCTCG"):
toehold = Domain(name="toehold",sequence=toehold_seq,length=len(toehold_seq))
branch_migration_A = Domain(name="bm_A", sequence=bm_design_A, seq_length=len(bm_design_A))
branch_migration_B = Domain(name="bm_B", sequence=bm_design_B, seq_length=len(bm_design_B))
substrate_A = toehold + branch_migration_A
substrate_B = toehold + branch_migration_B
incumbent_A = Strand(name="incumbent",domains=[branch_migration_A.C])
incumbent_B = Strand(name="incumbent",domains=[branch_migration_B.C])
incoming_A = substrate_A.C
incoming_B = substrate_B.C
start_complex_A = Complex(strands=[incoming_A, substrate_A, incumbent_A], structure=".(+)(+)")
start_complex_B = Complex(strands=[incoming_B, substrate_B, incumbent_B], structure=".(+)(+)")
complete_complex_A = Complex(strands=[incumbent_A],structure=".")
complete_complex_B = Complex(strands=[incumbent_B],structure=".")
failed_complex_A = Complex(strands=[incoming_A],structure="..")
failed_complex_B = Complex(strands=[incoming_B],structure="..")
# the "success" state
complete_sc_A = StopCondition("complete",[(complete_complex_A,Dissoc_Macrostate,0)])
complete_sc_B = StopCondition("complete",[(complete_complex_B,Dissoc_Macrostate,0)])
# the "failure" state
failed_sc_A = StopCondition("failed",[(failed_complex_A,Dissoc_Macrostate,0)])
failed_sc_B = StopCondition("failed",[(failed_complex_B,Dissoc_Macrostate,0)])
o1 = Options(simulation_mode="First Passage Time", parameter_type="Nupack",
substrate_type="DNA", num_simulations = trials, simulation_time=1.0,
dangles = "Some", temperature = 310.15, join_concentration=1e-6, # 1 uM concentration
start_state=[start_complex_A], rate_scaling='Calibrated')
o1.stop_conditions = [complete_sc_A,failed_sc_A]
o2 = Options(simulation_mode="First Passage Time", parameter_type="Nupack",
substrate_type="DNA", num_simulations = trials, simulation_time=1.0,
dangles = "Some", temperature = 310.15, join_concentration=1e-6, # 1 uM concentration
start_state=[start_complex_B], rate_scaling='Calibrated')
o2.stop_conditions = [complete_sc_B,failed_sc_B]
return o1,o2
Once the simulations has run, we will want to show histograms of how long they took.
In [6]:
def plot_histograms_complete_vs_failed( result_list, colors=['b','m'], figure=1 ):
# separate based on which stop condition ended the simulation ((( what if simulation ran out of time? )))
times_complete = np.array([i.time for i in result_list if i.tag == 'complete'])
times_failed = np.array([i.time for i in result_list if i.tag == 'failed'])
neither = [i for i in result_list if not i.tag == 'complete' and not i.tag == 'failed']
if len(neither)>0 :
print "some trajectories did not finish, one way nor the other..."
for i in neither :
assert (i.type_name=='Time')
assert (i.tag == None )
if len(times_complete)>0 and len(times_failed)>0 :
min_time = np.min( [np.min(times_complete),np.min(times_failed)] )
max_time = np.max( [np.max(times_complete),np.max(times_failed)] )
else:
if len(times_complete)>0 :
min_time = np.min(times_complete)
max_time = np.max(times_complete)
if len(times_failed)>0 :
min_time = np.min(times_failed)
max_time = np.max(times_failed)
plt.figure( figure )
plt.hold(False)
if len(times_complete)>0 :
plt.hist( times_complete, 50, range=(min_time,max_time), color=colors[0], label="complete" )
plt.hold(True)
if len(times_failed)>0 :
plt.hist( times_failed, 50, range=(min_time,max_time), color=colors[1],rwidth=.5, label="failed")
plt.title("Completion times for successful and failed trajectories, Design A")
plt.xlabel("First Passage Time (s)",fontsize='larger')
plt.ylabel("# of Trajectories",fontsize='larger')
plt.yticks(fontsize='larger',va='bottom')
plt.xticks(fontsize='larger')
plt.legend(loc=0)
plt.show()
In [7]:
def plot_histograms_two_designs( result_lists, colors=['magenta','b'], figure=1 ):
times = []
times.append(np.array([i.time for i in result_lists[0] if i.tag == 'complete']))
times.append(np.array([i.time for i in result_lists[1] if i.tag == 'complete']))
min_time = np.min( [np.min(times[0]),np.min(times[1])] )
max_time = np.max( [np.max(times[0]),np.max(times[1])] )
# the above might fail if any list is empty; min, max of empty is undefined
plt.figure( figure )
plt.hold(False)
plt.hist( times[1], 50, range=(min_time,max_time), color=colors[1], label="Design B")
plt.hold(True)
plt.hist( times[0], 50, range=(min_time,max_time), color=colors[0], label="Design A", rwidth=.5 )
plt.title("Successful trajectories in two designs")
plt.xlabel("First Passage Time (s)",fontsize='larger')
plt.ylabel("# of Trajectories",fontsize='larger')
plt.yticks(fontsize='larger',va='bottom')
plt.xticks(fontsize='larger')
plt.legend(loc=0)
plt.show()
In [8]:
def plot_completion_graph( result_lists, colors=['b','r'], figure=1, labels=['Design A', 'Design B'] ):
times = []
percents = []
for rl in result_lists:
n = len(rl)
t = [i.time for i in rl if i.tag == 'complete']
t.sort()
times.append(np.array(t))
p = np.array(range(1,len(t)+1))
p = 100 * p / n # percentage of all trials, including ones that don't complete
percents.append( p )
plt.figure( figure )
plt.hold(False)
for t,p,c,label in zip(times,percents,colors,labels):
plt.plot( t, p, color = c, linewidth=2.0, label=label )
plt.hold(True)
plt.xlabel("Simulation Time (s)",fontsize='larger')
plt.ylabel("% of Trajectories Complete",fontsize='larger')
plt.yticks([0,20,40,60,80,100],("0%","20%","40%","60%","80%","100%"),fontsize='larger',va='bottom')
plt.xticks(fontsize='larger')
plt.title( "Percentage of Total Trajectories Complete by a Given Time" )
plt.legend(loc=0)
plt.show()
In [9]:
def plot_completion_graph_complete_vs_failed( result_list, colors=['b','m'], figure=1, labels=['complete','failed'] ):
n = len(result_list) # number of trials
times = []
percents = []
for rl in ['complete','failed']:
t = [i.time for i in result_list if i.tag == rl]
t.sort()
times.append(np.array(t))
p = np.array(range(1,len(t)+1))
p = 100 * p / n
percents.append( p )
# the last data points of each type should sum to 1.0 if all trajectories either fail or complete (rather than time out)
plt.figure( figure )
plt.hold(False)
for t,p,c,label in zip(times,percents,colors,labels):
plt.plot( t, p, color = c, linewidth=2.0, label=label )
plt.hold(True)
plt.xlabel("Simulation Time (s)",fontsize='larger')
plt.ylabel("% of Trajectories Complete",fontsize='larger')
plt.yticks([0,20,40,60,80,100],("0%","20%","40%","60%","80%","100%"),fontsize='larger',va='bottom')
plt.xticks(fontsize='larger')
plt.title( "Percentage of Total Trajectories Complete by a Given Time, Design A" )
plt.legend(loc=0)
plt.show()
Now we are ready to run the simulations. First we set up the molecules and options. You might want to try different numbers of trials, or run it with different sequences.
In [34]:
# The following two options are equivalent, since they are defaults
# o1,o2 = setup_threewaybm_comparison()
o1,o2 = setup_threewaybm_comparison(trials=100, toehold_seq = "GTGGGT", bm_design_A = "ACCGCACGTCCACGGTGTCG", bm_design_B = "ACCGCACGTCACTCACCTCG")
# The same, but a longer toehold:
# o1,o2 = setup_threewaybm_comparison(trials=100, toehold_seq = "GTGGGTAGGT", bm_design_A = "ACCGCACGTCCACGGTGTCG", bm_design_B = "ACCGCACGTCACTCACCTCG")
Actually do the simulation. On my laptop, they take about 10 minutes for 100 trials. You might want to just do 10, for speed -- but the histograms below will look much better if you run 100 or more!
Note that there is no need to update the energy model in between runs, because conditions are the same. Each simulation trial prints the complexes present when it stops including its energy, each preceeded by that simulation's random number seed.
In [35]:
print
print "Running Design A"
s=SimSystem(o1)
s.start()
print
print "Running Design B"
s=SimSystem(o2)
s.start()
Running Design A
6604224628510739414: [1] '12:Automatic_12,14:incumbent': -22.215213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
......((((((((((((((((((((+))))))))))))))))))))
6604224628510739414: [0] '6:Automatic_12*': -3.37
CGACACCGTGGACGTGCGGTACCCAC
....((((((....))))))......
1703676841: [1] '12:Automatic_12,14:incumbent': -22.215213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
......((((((((((((((((((((+))))))))))))))))))))
1703676841: [0] '6:Automatic_12*': -3.39
CGACACCGTGGACGTGCGGTACCCAC
....(((((......)))))......
100207477: [1] '12:Automatic_12,14:incumbent': -22.215213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
......((((((((((((((((((((+))))))))))))))))))))
100207477: [0] '6:Automatic_12*': -1.16
CGACACCGTGGACGTGCGGTACCCAC
.......((((((.....))..))))
1300460798: [1] '12:Automatic_12,14:incumbent': -22.215213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
......((((((((((((((((((((+))))))))))))))))))))
1300460798: [0] '6:Automatic_12*': -2.39
CGACACCGTGGACGTGCGGTACCCAC
.....(((((....))))).......
1921764337: [1] '12:Automatic_12,14:incumbent': -22.215213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
......((((((((((((((((((((+))))))))))))))))))))
1921764337: [0] '6:Automatic_12*': -3.37
CGACACCGTGGACGTGCGGTACCCAC
....((((((....))))))......
646007698: [1] '12:Automatic_12,14:incumbent': -21.425213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
.......(((((((((((((((((((+))))))))))))))))))).
646007698: [0] '6:Automatic_12*': -3.39
CGACACCGTGGACGTGCGGTACCCAC
....(((((......)))))......
317465885: [1] '14:incumbent': -1.92
CGACACCGTGGACGTGCGGT
.....(((........))).
317465885: [0] '6:Automatic_12*,12:Automatic_12': -27.435213819
CGACACCGTGGACGTGCGGTACCCAC+GTGGGTACCGCACGTCCACGGTGTCG
..((((((((((((((((((((((((+))))))))))))))))))))))))..
579133852: [1] '14:incumbent': -2.39
CGACACCGTGGACGTGCGGT
.....(((((....))))).
579133852: [0] '6:Automatic_12*,12:Automatic_12': -27.435213819
CGACACCGTGGACGTGCGGTACCCAC+GTGGGTACCGCACGTCCACGGTGTCG
..((((((((((((((((((((((((+))))))))))))))))))))))))..
2011192973: [1] '14:incumbent': -0.28
CGACACCGTGGACGTGCGGT
...(((.......)))....
2011192973: [0] '6:Automatic_12*,12:Automatic_12': -28.325213819
CGACACCGTGGACGTGCGGTACCCAC+GTGGGTACCGCACGTCCACGGTGTCG
.(((((((((((((((((((((((((+))))))))))))))))))))))))).
1849152600: [1] '12:Automatic_12,14:incumbent': -22.215213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
......((((((((((((((((((((+))))))))))))))))))))
1849152600: [0] '6:Automatic_12*': -3.39
CGACACCGTGGACGTGCGGTACCCAC
....(((((......)))))......
1250869512: [1] '12:Automatic_12,14:incumbent': -21.425213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
.......(((((((((((((((((((+))))))))))))))))))).
1250869512: [0] '6:Automatic_12*': -3.39
CGACACCGTGGACGTGCGGTACCCAC
....(((((......)))))......
1939215445: [1] '12:Automatic_12,14:incumbent': -22.215213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
......((((((((((((((((((((+))))))))))))))))))))
1939215445: [0] '6:Automatic_12*': 0.8
CGACACCGTGGACGTGCGGTACCCAC
..(..((((......)))))......
589042999: [1] '12:Automatic_12,14:incumbent': -21.425213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
.......(((((((((((((((((((+))))))))))))))))))).
589042999: [0] '6:Automatic_12*': -3.37
CGACACCGTGGACGTGCGGTACCCAC
....((((((....))))))......
1517655462: [1] '12:Automatic_12,14:incumbent': -22.215213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
......((((((((((((((((((((+))))))))))))))))))))
1517655462: [0] '6:Automatic_12*': -3.39
CGACACCGTGGACGTGCGGTACCCAC
....(((((......)))))......
566557742: [1] '12:Automatic_12,14:incumbent': -20.395213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
......(((((((((((((((((((.+.)))))))))))))))))))
566557742: [0] '6:Automatic_12*': -3.37
CGACACCGTGGACGTGCGGTACCCAC
....((((((....))))))......
1816938696: [1] '12:Automatic_12,14:incumbent': -21.425213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
.......(((((((((((((((((((+))))))))))))))))))).
1816938696: [0] '6:Automatic_12*': -3.37
CGACACCGTGGACGTGCGGTACCCAC
....((((((....))))))......
397395160: [1] '12:Automatic_12,14:incumbent': -22.215213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
......((((((((((((((((((((+))))))))))))))))))))
397395160: [0] '6:Automatic_12*': -2.9
CGACACCGTGGACGTGCGGTACCCAC
....((((........))))......
1945134143: [1] '12:Automatic_12,14:incumbent': -21.425213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
.......(((((((((((((((((((+))))))))))))))))))).
1945134143: [0] '6:Automatic_12*': -1.92
CGACACCGTGGACGTGCGGTACCCAC
.....(((........))).......
452052276: [1] '14:incumbent': 1.38
CGACACCGTGGACGTGCGGT
..((...))..((.....))
452052276: [0] '6:Automatic_12*,12:Automatic_12': -28.325213819
CGACACCGTGGACGTGCGGTACCCAC+GTGGGTACCGCACGTCCACGGTGTCG
.(((((((((((((((((((((((((+))))))))))))))))))))))))).
1537260995: [1] '12:Automatic_12,14:incumbent': -21.425213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
.......(((((((((((((((((((+))))))))))))))))))).
1537260995: [0] '6:Automatic_12*': -2.41
CGACACCGTGGACGTGCGGTACCCAC
.....((((......)))).......
1694886059: [1] '12:Automatic_12,14:incumbent': -22.215213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
......((((((((((((((((((((+))))))))))))))))))))
1694886059: [0] '6:Automatic_12*': -3.37
CGACACCGTGGACGTGCGGTACCCAC
....((((((....))))))......
1078272123: [1] '12:Automatic_12,14:incumbent': -22.215213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
......((((((((((((((((((((+))))))))))))))))))))
1078272123: [0] '6:Automatic_12*': 2.17
CGACACCGTGGACGTGCGGTACCCAC
((....)).....(((...)))....
1086126865: [1] '12:Automatic_12,14:incumbent': -22.215213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
......((((((((((((((((((((+))))))))))))))))))))
1086126865: [0] '6:Automatic_12*': -3.37
CGACACCGTGGACGTGCGGTACCCAC
....((((((....))))))......
243081636: [1] '12:Automatic_12,14:incumbent': -22.215213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
......((((((((((((((((((((+))))))))))))))))))))
243081636: [0] '6:Automatic_12*': -3.39
CGACACCGTGGACGTGCGGTACCCAC
....(((((......)))))......
49972049: [1] '12:Automatic_12,14:incumbent': -20.395213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
......(((((((((((((((((((.+.)))))))))))))))))))
49972049: [0] '6:Automatic_12*': -3.37
CGACACCGTGGACGTGCGGTACCCAC
....((((((....))))))......
517598746: [1] '12:Automatic_12,14:incumbent': -22.215213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
......((((((((((((((((((((+))))))))))))))))))))
517598746: [0] '6:Automatic_12*': -3.39
CGACACCGTGGACGTGCGGTACCCAC
....(((((......)))))......
1650440717: [1] '12:Automatic_12,14:incumbent': -22.215213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
......((((((((((((((((((((+))))))))))))))))))))
1650440717: [0] '6:Automatic_12*': -3.37
CGACACCGTGGACGTGCGGTACCCAC
....((((((....))))))......
1932474339: [1] '12:Automatic_12,14:incumbent': -21.425213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
.......(((((((((((((((((((+))))))))))))))))))).
1932474339: [0] '6:Automatic_12*': -3.39
CGACACCGTGGACGTGCGGTACCCAC
....(((((......)))))......
585069599: [1] '12:Automatic_12,14:incumbent': -22.215213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
......((((((((((((((((((((+))))))))))))))))))))
585069599: [0] '6:Automatic_12*': -2.9
CGACACCGTGGACGTGCGGTACCCAC
....((((........))))......
1503746406: [1] '12:Automatic_12,14:incumbent': -21.425213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
.......(((((((((((((((((((+))))))))))))))))))).
1503746406: [0] '6:Automatic_12*': -0.5
CGACACCGTGGACGTGCGGTACCCAC
.(..(((((......)))))...)..
1330835290: [1] '12:Automatic_12,14:incumbent': -17.825213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
.......((((((((((((((((...+...)))))))))))))))).
1330835290: [0] '6:Automatic_12*': -3.39
CGACACCGTGGACGTGCGGTACCCAC
....(((((......)))))......
792985827: [1] '12:Automatic_12,14:incumbent': -22.215213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
......((((((((((((((((((((+))))))))))))))))))))
792985827: [0] '6:Automatic_12*': -3.37
CGACACCGTGGACGTGCGGTACCCAC
....((((((....))))))......
1755277292: [1] '12:Automatic_12,14:incumbent': -21.425213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
.......(((((((((((((((((((+))))))))))))))))))).
1755277292: [0] '6:Automatic_12*': -3.39
CGACACCGTGGACGTGCGGTACCCAC
....(((((......)))))......
1401272251: [1] '12:Automatic_12,14:incumbent': -21.425213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
.......(((((((((((((((((((+))))))))))))))))))).
1401272251: [0] '6:Automatic_12*': -2.39
CGACACCGTGGACGTGCGGTACCCAC
.....(((((....))))).......
1225201953: [1] '12:Automatic_12,14:incumbent': -22.215213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
......((((((((((((((((((((+))))))))))))))))))))
1225201953: [0] '6:Automatic_12*': -1.67
CGACACCGTGGACGTGCGGTACCCAC
.(..((((((....)))))).)....
432674135: [1] '14:incumbent': -2.91
CGACACCGTGGACGTGCGGT
....(((((......)))))
432674135: [0] '6:Automatic_12*,12:Automatic_12': -26.545213819
CGACACCGTGGACGTGCGGTACCCAC+GTGGGTACCGCACGTCCACGGTGTCG
...(((((((((((((((((((((((+)))))))))))))))))))))))...
34651511: [1] '14:incumbent': -2.91
CGACACCGTGGACGTGCGGT
....(((((......)))))
34651511: [0] '6:Automatic_12*,12:Automatic_12': -28.325213819
CGACACCGTGGACGTGCGGTACCCAC+GTGGGTACCGCACGTCCACGGTGTCG
.(((((((((((((((((((((((((+))))))))))))))))))))))))).
127525177: [1] '14:incumbent': -2.91
CGACACCGTGGACGTGCGGT
....(((((......)))))
127525177: [0] '6:Automatic_12*,12:Automatic_12': -26.545213819
CGACACCGTGGACGTGCGGTACCCAC+GTGGGTACCGCACGTCCACGGTGTCG
...(((((((((((((((((((((((+)))))))))))))))))))))))...
1743918023: [1] '14:incumbent': -2.41
CGACACCGTGGACGTGCGGT
.....((((......)))).
1743918023: [0] '6:Automatic_12*,12:Automatic_12': -28.325213819
CGACACCGTGGACGTGCGGTACCCAC+GTGGGTACCGCACGTCCACGGTGTCG
.(((((((((((((((((((((((((+))))))))))))))))))))))))).
1461588800: [1] '12:Automatic_12,14:incumbent': -22.215213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
......((((((((((((((((((((+))))))))))))))))))))
1461588800: [0] '6:Automatic_12*': 0.8
CGACACCGTGGACGTGCGGTACCCAC
...(((.......))).(...)....
951600502: [1] '12:Automatic_12,14:incumbent': -22.215213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
......((((((((((((((((((((+))))))))))))))))))))
951600502: [0] '6:Automatic_12*': -2.39
CGACACCGTGGACGTGCGGTACCCAC
.....(((((....))))).......
1561590077: [1] '14:incumbent': 1.99
CGACACCGTGGACGTGCGGT
...(.....)..(.....).
1561590077: [0] '6:Automatic_12*,12:Automatic_12': -25.975213819
CGACACCGTGGACGTGCGGTACCCAC+GTGGGTACCGCACGTCCACGGTGTCG
..((((((((((((((((((((((..+..))))))))))))))))))))))..
1088734281: [1] '14:incumbent': -2.89
CGACACCGTGGACGTGCGGT
....((((((....))))))
1088734281: [0] '6:Automatic_12*,12:Automatic_12': -27.435213819
CGACACCGTGGACGTGCGGTACCCAC+GTGGGTACCGCACGTCCACGGTGTCG
..((((((((((((((((((((((((+))))))))))))))))))))))))..
1284210019: [1] '14:incumbent': -2.89
CGACACCGTGGACGTGCGGT
....((((((....))))))
1284210019: [0] '6:Automatic_12*,12:Automatic_12': -27.435213819
CGACACCGTGGACGTGCGGTACCCAC+GTGGGTACCGCACGTCCACGGTGTCG
..((((((((((((((((((((((((+))))))))))))))))))))))))..
751360865: [1] '12:Automatic_12,14:incumbent': -22.215213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
......((((((((((((((((((((+))))))))))))))))))))
751360865: [0] '6:Automatic_12*': -3.37
CGACACCGTGGACGTGCGGTACCCAC
....((((((....))))))......
428065467: [1] '12:Automatic_12,14:incumbent': -22.215213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
......((((((((((((((((((((+))))))))))))))))))))
428065467: [0] '6:Automatic_12*': -3.37
CGACACCGTGGACGTGCGGTACCCAC
....((((((....))))))......
908868015: [1] '12:Automatic_12,14:incumbent': -22.215213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
......((((((((((((((((((((+))))))))))))))))))))
908868015: [0] '6:Automatic_12*': 0.54
CGACACCGTGGACGTGCGGTACCCAC
.........((..........))...
234392080: [1] '12:Automatic_12,14:incumbent': -22.215213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
......((((((((((((((((((((+))))))))))))))))))))
234392080: [0] '6:Automatic_12*': -3.39
CGACACCGTGGACGTGCGGTACCCAC
....(((((......)))))......
1078337795: [1] '14:incumbent': -2.41
CGACACCGTGGACGTGCGGT
.....((((......)))).
1078337795: [0] '6:Automatic_12*,12:Automatic_12': -24.655213819
CGACACCGTGGACGTGCGGTACCCAC+GTGGGTACCGCACGTCCACGGTGTCG
....((((((((((((((((((((((+))))))))))))))))))))))....
1317715975: [1] '12:Automatic_12,14:incumbent': -22.215213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
......((((((((((((((((((((+))))))))))))))))))))
1317715975: [0] '6:Automatic_12*': -3.39
CGACACCGTGGACGTGCGGTACCCAC
....(((((......)))))......
1076957156: [1] '14:incumbent': -2.91
CGACACCGTGGACGTGCGGT
....(((((......)))))
1076957156: [0] '6:Automatic_12*,12:Automatic_12': -26.075213819
CGACACCGTGGACGTGCGGTACCCAC+GTGGGTACCGCACGTCCACGGTGTCG
.((((((((((((((((((((((((.+.)))))))))))))))))))))))).
322084270: [1] '12:Automatic_12,14:incumbent': -21.425213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
.......(((((((((((((((((((+))))))))))))))))))).
322084270: [0] '6:Automatic_12*': -3.37
CGACACCGTGGACGTGCGGTACCCAC
....((((((....))))))......
1695574596: [1] '12:Automatic_12,14:incumbent': -22.215213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
......((((((((((((((((((((+))))))))))))))))))))
1695574596: [0] '6:Automatic_12*': -3.39
CGACACCGTGGACGTGCGGTACCCAC
....(((((......)))))......
324074100: [1] '12:Automatic_12,14:incumbent': -22.215213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
......((((((((((((((((((((+))))))))))))))))))))
324074100: [0] '6:Automatic_12*': -2.9
CGACACCGTGGACGTGCGGTACCCAC
....((((........))))......
551939943: [1] '12:Automatic_12,14:incumbent': -22.215213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
......((((((((((((((((((((+))))))))))))))))))))
551939943: [0] '6:Automatic_12*': -2.9
CGACACCGTGGACGTGCGGTACCCAC
....((((........))))......
1426420060: [1] '14:incumbent': -2.39
CGACACCGTGGACGTGCGGT
.....(((((....))))).
1426420060: [0] '6:Automatic_12*,12:Automatic_12': -26.545213819
CGACACCGTGGACGTGCGGTACCCAC+GTGGGTACCGCACGTCCACGGTGTCG
...(((((((((((((((((((((((+)))))))))))))))))))))))...
509003315: [1] '14:incumbent': -2.91
CGACACCGTGGACGTGCGGT
....(((((......)))))
509003315: [0] '6:Automatic_12*,12:Automatic_12': -28.325213819
CGACACCGTGGACGTGCGGTACCCAC+GTGGGTACCGCACGTCCACGGTGTCG
.(((((((((((((((((((((((((+))))))))))))))))))))))))).
2051029996: [1] '12:Automatic_12,14:incumbent': -20.395213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
......(((((((((((((((((((.+.)))))))))))))))))))
2051029996: [0] '6:Automatic_12*': -3.39
CGACACCGTGGACGTGCGGTACCCAC
....(((((......)))))......
290195647: [1] '14:incumbent': -2.39
CGACACCGTGGACGTGCGGT
.....(((((....))))).
290195647: [0] '6:Automatic_12*,12:Automatic_12': -27.435213819
CGACACCGTGGACGTGCGGTACCCAC+GTGGGTACCGCACGTCCACGGTGTCG
..((((((((((((((((((((((((+))))))))))))))))))))))))..
1245806206: [1] '14:incumbent': -2.91
CGACACCGTGGACGTGCGGT
....(((((......)))))
1245806206: [0] '6:Automatic_12*,12:Automatic_12': -27.435213819
CGACACCGTGGACGTGCGGTACCCAC+GTGGGTACCGCACGTCCACGGTGTCG
..((((((((((((((((((((((((+))))))))))))))))))))))))..
1521638287: [1] '14:incumbent': -2.41
CGACACCGTGGACGTGCGGT
.....((((......)))).
1521638287: [0] '6:Automatic_12*,12:Automatic_12': -27.435213819
CGACACCGTGGACGTGCGGTACCCAC+GTGGGTACCGCACGTCCACGGTGTCG
..((((((((((((((((((((((((+))))))))))))))))))))))))..
176578821: [1] '12:Automatic_12,14:incumbent': -22.215213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
......((((((((((((((((((((+))))))))))))))))))))
176578821: [0] '6:Automatic_12*': -3.37
CGACACCGTGGACGTGCGGTACCCAC
....((((((....))))))......
1629703447: [1] '12:Automatic_12,14:incumbent': -22.215213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
......((((((((((((((((((((+))))))))))))))))))))
1629703447: [0] '6:Automatic_12*': -3.39
CGACACCGTGGACGTGCGGTACCCAC
....(((((......)))))......
1936991502: [1] '12:Automatic_12,14:incumbent': -22.215213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
......((((((((((((((((((((+))))))))))))))))))))
1936991502: [0] '6:Automatic_12*': -2.39
CGACACCGTGGACGTGCGGTACCCAC
.....(((((....))))).......
1851141978: [1] '12:Automatic_12,14:incumbent': -22.215213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
......((((((((((((((((((((+))))))))))))))))))))
1851141978: [0] '6:Automatic_12*': -3.37
CGACACCGTGGACGTGCGGTACCCAC
....((((((....))))))......
715855049: [1] '12:Automatic_12,14:incumbent': -22.215213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
......((((((((((((((((((((+))))))))))))))))))))
715855049: [0] '6:Automatic_12*': -3.39
CGACACCGTGGACGTGCGGTACCCAC
....(((((......)))))......
339746679: [1] '14:incumbent': -2.89
CGACACCGTGGACGTGCGGT
....((((((....))))))
339746679: [0] '6:Automatic_12*,12:Automatic_12': -28.325213819
CGACACCGTGGACGTGCGGTACCCAC+GTGGGTACCGCACGTCCACGGTGTCG
.(((((((((((((((((((((((((+))))))))))))))))))))))))).
1447254580: [1] '12:Automatic_12,14:incumbent': -22.215213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
......((((((((((((((((((((+))))))))))))))))))))
1447254580: [0] '6:Automatic_12*': -1.92
CGACACCGTGGACGTGCGGTACCCAC
.....(((........))).......
218874352: [1] '12:Automatic_12,14:incumbent': -22.215213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
......((((((((((((((((((((+))))))))))))))))))))
218874352: [0] '6:Automatic_12*': -3.37
CGACACCGTGGACGTGCGGTACCCAC
....((((((....))))))......
1437549229: [1] '12:Automatic_12,14:incumbent': -22.215213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
......((((((((((((((((((((+))))))))))))))))))))
1437549229: [0] '6:Automatic_12*': -3.37
CGACACCGTGGACGTGCGGTACCCAC
....((((((....))))))......
538700323: [1] '12:Automatic_12,14:incumbent': -22.215213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
......((((((((((((((((((((+))))))))))))))))))))
538700323: [0] '6:Automatic_12*': -3.39
CGACACCGTGGACGTGCGGTACCCAC
....(((((......)))))......
980747773: [1] '12:Automatic_12,14:incumbent': -22.215213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
......((((((((((((((((((((+))))))))))))))))))))
980747773: [0] '6:Automatic_12*': -2.39
CGACACCGTGGACGTGCGGTACCCAC
.....(((((....))))).......
1220919013: [1] '12:Automatic_12,14:incumbent': -19.505213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
......((((((((((((((((((..+..))))))))))))))))))
1220919013: [0] '6:Automatic_12*': -3.39
CGACACCGTGGACGTGCGGTACCCAC
....(((((......)))))......
491074223: [1] '14:incumbent': -2.89
CGACACCGTGGACGTGCGGT
....((((((....))))))
491074223: [0] '6:Automatic_12*,12:Automatic_12': -27.435213819
CGACACCGTGGACGTGCGGTACCCAC+GTGGGTACCGCACGTCCACGGTGTCG
..((((((((((((((((((((((((+))))))))))))))))))))))))..
256168903: [1] '12:Automatic_12,14:incumbent': -22.215213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
......((((((((((((((((((((+))))))))))))))))))))
256168903: [0] '6:Automatic_12*': 3.02
CGACACCGTGGACGTGCGGTACCCAC
.....(.(....))............
1351504537: [1] '12:Automatic_12,14:incumbent': -22.215213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
......((((((((((((((((((((+))))))))))))))))))))
1351504537: [0] '6:Automatic_12*': -2.9
CGACACCGTGGACGTGCGGTACCCAC
....((((........))))......
578962909: [1] '12:Automatic_12,14:incumbent': -22.215213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
......((((((((((((((((((((+))))))))))))))))))))
578962909: [0] '6:Automatic_12*': -2.39
CGACACCGTGGACGTGCGGTACCCAC
.....(((((....))))).......
499999699: [1] '14:incumbent': -2.42
CGACACCGTGGACGTGCGGT
....((((........))))
499999699: [0] '6:Automatic_12*,12:Automatic_12': -28.325213819
CGACACCGTGGACGTGCGGTACCCAC+GTGGGTACCGCACGTCCACGGTGTCG
.(((((((((((((((((((((((((+))))))))))))))))))))))))).
1839462860: [1] '12:Automatic_12,14:incumbent': -21.425213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
.......(((((((((((((((((((+))))))))))))))))))).
1839462860: [0] '6:Automatic_12*': -3.39
CGACACCGTGGACGTGCGGTACCCAC
....(((((......)))))......
693936086: [1] '12:Automatic_12,14:incumbent': -22.215213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
......((((((((((((((((((((+))))))))))))))))))))
693936086: [0] '6:Automatic_12*': -1.69
CGACACCGTGGACGTGCGGTACCCAC
.(..(((((......))))).)....
946618205: [1] '12:Automatic_12,14:incumbent': -21.425213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
.......(((((((((((((((((((+))))))))))))))))))).
946618205: [0] '6:Automatic_12*': -3.39
CGACACCGTGGACGTGCGGTACCCAC
....(((((......)))))......
17312865: [1] '14:incumbent': -2.91
CGACACCGTGGACGTGCGGT
....(((((......)))))
17312865: [0] '6:Automatic_12*,12:Automatic_12': -27.435213819
CGACACCGTGGACGTGCGGTACCCAC+GTGGGTACCGCACGTCCACGGTGTCG
..((((((((((((((((((((((((+))))))))))))))))))))))))..
543609672: [1] '14:incumbent': -2.41
CGACACCGTGGACGTGCGGT
.....((((......)))).
543609672: [0] '6:Automatic_12*,12:Automatic_12': -28.325213819
CGACACCGTGGACGTGCGGTACCCAC+GTGGGTACCGCACGTCCACGGTGTCG
.(((((((((((((((((((((((((+))))))))))))))))))))))))).
602524189: [1] '12:Automatic_12,14:incumbent': -22.215213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
......((((((((((((((((((((+))))))))))))))))))))
602524189: [0] '6:Automatic_12*': -3.39
CGACACCGTGGACGTGCGGTACCCAC
....(((((......)))))......
1845717590: [1] '14:incumbent': -2.91
CGACACCGTGGACGTGCGGT
....(((((......)))))
1845717590: [0] '6:Automatic_12*,12:Automatic_12': -27.435213819
CGACACCGTGGACGTGCGGTACCCAC+GTGGGTACCGCACGTCCACGGTGTCG
..((((((((((((((((((((((((+))))))))))))))))))))))))..
1226319464: [1] '12:Automatic_12,14:incumbent': -22.215213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
......((((((((((((((((((((+))))))))))))))))))))
1226319464: [0] '6:Automatic_12*': -2.41
CGACACCGTGGACGTGCGGTACCCAC
.....((((......)))).......
795853589: [1] '14:incumbent': -2.89
CGACACCGTGGACGTGCGGT
....((((((....))))))
795853589: [0] '6:Automatic_12*,12:Automatic_12': -27.435213819
CGACACCGTGGACGTGCGGTACCCAC+GTGGGTACCGCACGTCCACGGTGTCG
..((((((((((((((((((((((((+))))))))))))))))))))))))..
595977454: [1] '12:Automatic_12,14:incumbent': -21.425213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
.......(((((((((((((((((((+))))))))))))))))))).
595977454: [0] '6:Automatic_12*': -3.37
CGACACCGTGGACGTGCGGTACCCAC
....((((((....))))))......
1029342980: [1] '14:incumbent': -2.42
CGACACCGTGGACGTGCGGT
....((((........))))
1029342980: [0] '6:Automatic_12*,12:Automatic_12': -28.325213819
CGACACCGTGGACGTGCGGTACCCAC+GTGGGTACCGCACGTCCACGGTGTCG
.(((((((((((((((((((((((((+))))))))))))))))))))))))).
1050871073: [1] '14:incumbent': -2.89
CGACACCGTGGACGTGCGGT
....((((((....))))))
1050871073: [0] '6:Automatic_12*,12:Automatic_12': -27.435213819
CGACACCGTGGACGTGCGGTACCCAC+GTGGGTACCGCACGTCCACGGTGTCG
..((((((((((((((((((((((((+))))))))))))))))))))))))..
530568369: [1] '14:incumbent': -2.91
CGACACCGTGGACGTGCGGT
....(((((......)))))
530568369: [0] '6:Automatic_12*,12:Automatic_12': -28.325213819
CGACACCGTGGACGTGCGGTACCCAC+GTGGGTACCGCACGTCCACGGTGTCG
.(((((((((((((((((((((((((+))))))))))))))))))))))))).
894194274: [1] '12:Automatic_12,14:incumbent': -22.215213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
......((((((((((((((((((((+))))))))))))))))))))
894194274: [0] '6:Automatic_12*': -3.37
CGACACCGTGGACGTGCGGTACCCAC
....((((((....))))))......
243414330: [1] '12:Automatic_12,14:incumbent': -21.425213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
.......(((((((((((((((((((+))))))))))))))))))).
243414330: [0] '6:Automatic_12*': -3.37
CGACACCGTGGACGTGCGGTACCCAC
....((((((....))))))......
961724006: [1] '12:Automatic_12,14:incumbent': -22.215213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
......((((((((((((((((((((+))))))))))))))))))))
961724006: [0] '6:Automatic_12*': -3.39
CGACACCGTGGACGTGCGGTACCCAC
....(((((......)))))......
1174277070: [1] '12:Automatic_12,14:incumbent': -22.215213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
......((((((((((((((((((((+))))))))))))))))))))
1174277070: [0] '6:Automatic_12*': -2.39
CGACACCGTGGACGTGCGGTACCCAC
.....(((((....))))).......
735503385: [1] '12:Automatic_12,14:incumbent': -22.215213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
......((((((((((((((((((((+))))))))))))))))))))
735503385: [0] '6:Automatic_12*': -3.37
CGACACCGTGGACGTGCGGTACCCAC
....((((((....))))))......
1207741803: [1] '12:Automatic_12,14:incumbent': -22.215213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
......((((((((((((((((((((+))))))))))))))))))))
1207741803: [0] '6:Automatic_12*': -3.39
CGACACCGTGGACGTGCGGTACCCAC
....(((((......)))))......
1418025509: [1] '12:Automatic_12,14:incumbent': -22.215213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
......((((((((((((((((((((+))))))))))))))))))))
1418025509: [0] '6:Automatic_12*': -3.39
CGACACCGTGGACGTGCGGTACCCAC
....(((((......)))))......
1178591573: [1] '12:Automatic_12,14:incumbent': -22.215213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
......((((((((((((((((((((+))))))))))))))))))))
1178591573: [0] '6:Automatic_12*': -3.39
CGACACCGTGGACGTGCGGTACCCAC
....(((((......)))))......
1077731645: [1] '12:Automatic_12,14:incumbent': -22.215213819
GTGGGTACCGCACGTCCACGGTGTCG+CGACACCGTGGACGTGCGGT
......((((((((((((((((((((+))))))))))))))))))))
1077731645: [0] '6:Automatic_12*': -2.9
CGACACCGTGGACGTGCGGTACCCAC
....((((........))))......
Running Design B
-4621967137565713499: [1] '15:incumbent': 0.15
CGAGGTGAGTGACGTGCGGT
..(.((.....)).).....
-4621967137565713499: [0] '7:Automatic_13*,13:Automatic_13': -19.695213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
...(((((((.(((((((((((((..+..))))))))))))).)))))))...
388830634: [1] '15:incumbent': 0.89
CGAGGTGAGTGACGTGCGGT
....((.....)).......
388830634: [0] '7:Automatic_13*,13:Automatic_13': -25.845213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
..((((((((((((((((((((((((+))))))))))))))))))))))))..
310607269: [1] '15:incumbent': 1.99
CGAGGTGAGTGACGTGCGGT
..(.(.......).).....
310607269: [0] '7:Automatic_13*,13:Automatic_13': -26.735213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
.(((((((((((((((((((((((((+))))))))))))))))))))))))).
1903955352: [1] '15:incumbent': 0.15
CGAGGTGAGTGACGTGCGGT
..(.((.....)).).....
1903955352: [0] '7:Automatic_13*,13:Automatic_13': -22.835213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
....((((((((((((((((((((((+))))))))))))))))))))))....
550162319: [1] '15:incumbent': 0.0
CGAGGTGAGTGACGTGCGGT
....................
550162319: [0] '7:Automatic_13*,13:Automatic_13': -24.905213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
...(((((((((((((((((((((((+)))))))))))))))))))))))...
656524252: [1] '15:incumbent': 0.0
CGAGGTGAGTGACGTGCGGT
....................
656524252: [0] '7:Automatic_13*,13:Automatic_13': -25.845213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
..((((((((((((((((((((((((+))))))))))))))))))))))))..
831440664: [1] '15:incumbent': 1.1
CGAGGTGAGTGACGTGCGGT
............(....)..
831440664: [0] '7:Automatic_13*,13:Automatic_13': -26.735213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
.(((((((((((((((((((((((((+))))))))))))))))))))))))).
1364734972: [1] '15:incumbent': 0.0
CGAGGTGAGTGACGTGCGGT
....................
1364734972: [0] '7:Automatic_13*,13:Automatic_13': -26.735213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
.(((((((((((((((((((((((((+))))))))))))))))))))))))).
1079349712: [1] '15:incumbent': 0.15
CGAGGTGAGTGACGTGCGGT
..(.((.....)).).....
1079349712: [0] '7:Automatic_13*,13:Automatic_13': -25.845213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
..((((((((((((((((((((((((+))))))))))))))))))))))))..
812960041: [1] '15:incumbent': 0.0
CGAGGTGAGTGACGTGCGGT
....................
812960041: [0] '7:Automatic_13*,13:Automatic_13': -24.905213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
...(((((((((((((((((((((((+)))))))))))))))))))))))...
1929238377: [1] '15:incumbent': 0.89
CGAGGTGAGTGACGTGCGGT
....((.....)).......
1929238377: [0] '7:Automatic_13*,13:Automatic_13': -25.275213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
.(((((((((((((((((((((((..+..))))))))))))))))))))))).
1174170184: [1] '15:incumbent': 0.15
CGAGGTGAGTGACGTGCGGT
..(.((.....)).).....
1174170184: [0] '7:Automatic_13*,13:Automatic_13': -26.735213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
.(((((((((((((((((((((((((+))))))))))))))))))))))))).
1232192394: [1] '15:incumbent': 1.22
CGAGGTGAGTGACGTGCGGT
.......(.((.....)).)
1232192394: [0] '7:Automatic_13*,13:Automatic_13': -25.845213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
..((((((((((((((((((((((((+))))))))))))))))))))))))..
2036680884: [1] '15:incumbent': 1.99
CGAGGTGAGTGACGTGCGGT
..(.(.......).).....
2036680884: [0] '7:Automatic_13*,13:Automatic_13': -26.735213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
.(((((((((((((((((((((((((+))))))))))))))))))))))))).
1312301610: [1] '15:incumbent': 0.0
CGAGGTGAGTGACGTGCGGT
....................
1312301610: [0] '7:Automatic_13*,13:Automatic_13': -24.385213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
..((((((((((((((((((((((..+..))))))))))))))))))))))..
433859118: [1] '15:incumbent': 2.73
CGAGGTGAGTGACGTGCGGT
....(.......).......
433859118: [0] '7:Automatic_13*,13:Automatic_13': -26.735213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
.(((((((((((((((((((((((((+))))))))))))))))))))))))).
502191336: [1] '15:incumbent': 1.68
CGAGGTGAGTGACGTGCGGT
......(.....).......
502191336: [0] '7:Automatic_13*,13:Automatic_13': -25.845213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
..((((((((((((((((((((((((+))))))))))))))))))))))))..
8051621: [1] '15:incumbent': 0.39
CGAGGTGAGTGACGTGCGGT
...........((.....))
8051621: [0] '7:Automatic_13*,13:Automatic_13': -26.735213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
.(((((((((((((((((((((((((+))))))))))))))))))))))))).
1274141817: [1] '15:incumbent': 0.0
CGAGGTGAGTGACGTGCGGT
....................
1274141817: [0] '7:Automatic_13*,13:Automatic_13': -25.845213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
..((((((((((((((((((((((((+))))))))))))))))))))))))..
797148701: [1] '15:incumbent': 0.0
CGAGGTGAGTGACGTGCGGT
....................
797148701: [0] '7:Automatic_13*,13:Automatic_13': -26.735213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
.(((((((((((((((((((((((((+))))))))))))))))))))))))).
1351792591: [1] '15:incumbent': 0.0
CGAGGTGAGTGACGTGCGGT
....................
1351792591: [0] '7:Automatic_13*,13:Automatic_13': -26.735213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
.(((((((((((((((((((((((((+))))))))))))))))))))))))).
616745900: [1] '15:incumbent': 0.39
CGAGGTGAGTGACGTGCGGT
...........((.....))
616745900: [0] '7:Automatic_13*,13:Automatic_13': -26.735213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
.(((((((((((((((((((((((((+))))))))))))))))))))))))).
2009355622: [1] '15:incumbent': 1.1
CGAGGTGAGTGACGTGCGGT
............(....)..
2009355622: [0] '7:Automatic_13*,13:Automatic_13': -26.735213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
.(((((((((((((((((((((((((+))))))))))))))))))))))))).
1384412553: [1] '15:incumbent': 0.39
CGAGGTGAGTGACGTGCGGT
...........((.....))
1384412553: [0] '7:Automatic_13*,13:Automatic_13': -24.905213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
...(((((((((((((((((((((((+)))))))))))))))))))))))...
161822546: [1] '15:incumbent': 0.0
CGAGGTGAGTGACGTGCGGT
....................
161822546: [0] '7:Automatic_13*,13:Automatic_13': -26.735213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
.(((((((((((((((((((((((((+))))))))))))))))))))))))).
1996505209: [1] '15:incumbent': 1.09
CGAGGTGAGTGACGTGCGGT
............(.....).
1996505209: [0] '7:Automatic_13*,13:Automatic_13': -26.735213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
.(((((((((((((((((((((((((+))))))))))))))))))))))))).
1444056789: [1] '15:incumbent': 1.1
CGAGGTGAGTGACGTGCGGT
............(....)..
1444056789: [0] '7:Automatic_13*,13:Automatic_13': -25.845213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
..((((((((((((((((((((((((+))))))))))))))))))))))))..
44035297: [1] '15:incumbent': 0.0
CGAGGTGAGTGACGTGCGGT
....................
44035297: [0] '7:Automatic_13*,13:Automatic_13': -26.735213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
.(((((((((((((((((((((((((+))))))))))))))))))))))))).
1107154586: [1] '15:incumbent': 0.0
CGAGGTGAGTGACGTGCGGT
....................
1107154586: [0] '7:Automatic_13*,13:Automatic_13': -26.735213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
.(((((((((((((((((((((((((+))))))))))))))))))))))))).
1620170972: [1] '15:incumbent': 0.39
CGAGGTGAGTGACGTGCGGT
...........((.....))
1620170972: [0] '7:Automatic_13*,13:Automatic_13': -25.275213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
.(((((((((((((((((((((((..+..))))))))))))))))))))))).
1503780799: [1] '15:incumbent': 0.0
CGAGGTGAGTGACGTGCGGT
....................
1503780799: [0] '7:Automatic_13*,13:Automatic_13': -25.845213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
..((((((((((((((((((((((((+))))))))))))))))))))))))..
1522727870: [1] '15:incumbent': 3.63
CGAGGTGAGTGACGTGCGGT
..(......)..(....)..
1522727870: [0] '7:Automatic_13*,13:Automatic_13': -23.445213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
...(((((((((((((((((((((..+..)))))))))))))))))))))...
742727299: [1] '13:Automatic_13,15:incumbent': -20.625213819
GTGGGTACCGCACGTCACTCACCTCG+CGAGGTGAGTGACGTGCGGT
......((((((((((((((((((((+))))))))))))))))))))
742727299: [0] '7:Automatic_13*': 1.45
CGAGGTGAGTGACGTGCGGTACCCAC
((..((.....))...))........
1041740653: [1] '15:incumbent': 0.15
CGAGGTGAGTGACGTGCGGT
..(.((.....)).).....
1041740653: [0] '7:Automatic_13*,13:Automatic_13': -25.845213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
..((((((((((((((((((((((((+))))))))))))))))))))))))..
404904914: [1] '15:incumbent': 0.15
CGAGGTGAGTGACGTGCGGT
..(.((.....)).).....
404904914: [0] '7:Automatic_13*,13:Automatic_13': -26.735213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
.(((((((((((((((((((((((((+))))))))))))))))))))))))).
1924011278: [1] '15:incumbent': 1.09
CGAGGTGAGTGACGTGCGGT
............(.....).
1924011278: [0] '7:Automatic_13*,13:Automatic_13': -24.905213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
...(((((((((((((((((((((((+)))))))))))))))))))))))...
1181974547: [1] '15:incumbent': 1.09
CGAGGTGAGTGACGTGCGGT
............(.....).
1181974547: [0] '7:Automatic_13*,13:Automatic_13': -26.735213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
.(((((((((((((((((((((((((+))))))))))))))))))))))))).
2145366691: [1] '15:incumbent': 0.0
CGAGGTGAGTGACGTGCGGT
....................
2145366691: [0] '7:Automatic_13*,13:Automatic_13': -25.845213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
..((((((((((((((((((((((((+))))))))))))))))))))))))..
766270485: [1] '15:incumbent': 0.0
CGAGGTGAGTGACGTGCGGT
....................
766270485: [0] '7:Automatic_13*,13:Automatic_13': -25.845213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
..((((((((((((((((((((((((+))))))))))))))))))))))))..
921410960: [1] '15:incumbent': 1.99
CGAGGTGAGTGACGTGCGGT
..(.(.......).).....
921410960: [0] '7:Automatic_13*,13:Automatic_13': -25.275213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
.(((((((((((((((((((((((..+..))))))))))))))))))))))).
712711488: [1] '15:incumbent': 0.39
CGAGGTGAGTGACGTGCGGT
...........((.....))
712711488: [0] '7:Automatic_13*,13:Automatic_13': -23.445213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
...(((((((((((((((((((((..+..)))))))))))))))))))))...
2107116500: [1] '15:incumbent': 0.39
CGAGGTGAGTGACGTGCGGT
...........((.....))
2107116500: [0] '7:Automatic_13*,13:Automatic_13': -26.735213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
.(((((((((((((((((((((((((+))))))))))))))))))))))))).
2031133742: [1] '15:incumbent': 0.0
CGAGGTGAGTGACGTGCGGT
....................
2031133742: [0] '7:Automatic_13*,13:Automatic_13': -24.905213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
...(((((((((((((((((((((((+)))))))))))))))))))))))...
366325459: [1] '15:incumbent': 1.22
CGAGGTGAGTGACGTGCGGT
.......(.((.....)).)
366325459: [0] '7:Automatic_13*,13:Automatic_13': -26.735213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
.(((((((((((((((((((((((((+))))))))))))))))))))))))).
469499357: [1] '15:incumbent': 2.73
CGAGGTGAGTGACGTGCGGT
...........(.......)
469499357: [0] '7:Automatic_13*,13:Automatic_13': -25.845213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
..((((((((((((((((((((((((+))))))))))))))))))))))))..
1161848385: [1] '15:incumbent': 0.39
CGAGGTGAGTGACGTGCGGT
...........((.....))
1161848385: [0] '7:Automatic_13*,13:Automatic_13': -25.845213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
..((((((((((((((((((((((((+))))))))))))))))))))))))..
789786259: [1] '15:incumbent': 0.0
CGAGGTGAGTGACGTGCGGT
....................
789786259: [0] '7:Automatic_13*,13:Automatic_13': -26.735213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
.(((((((((((((((((((((((((+))))))))))))))))))))))))).
794179123: [1] '15:incumbent': 0.0
CGAGGTGAGTGACGTGCGGT
....................
794179123: [0] '7:Automatic_13*,13:Automatic_13': -26.735213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
.(((((((((((((((((((((((((+))))))))))))))))))))))))).
1130250897: [1] '15:incumbent': 0.15
CGAGGTGAGTGACGTGCGGT
..(.((.....)).).....
1130250897: [0] '7:Automatic_13*,13:Automatic_13': -26.735213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
.(((((((((((((((((((((((((+))))))))))))))))))))))))).
1090866970: [1] '15:incumbent': 1.09
CGAGGTGAGTGACGTGCGGT
............(.....).
1090866970: [0] '7:Automatic_13*,13:Automatic_13': -26.735213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
.(((((((((((((((((((((((((+))))))))))))))))))))))))).
1543842306: [1] '15:incumbent': 0.39
CGAGGTGAGTGACGTGCGGT
...........((.....))
1543842306: [0] '7:Automatic_13*,13:Automatic_13': -26.735213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
.(((((((((((((((((((((((((+))))))))))))))))))))))))).
135878279: [1] '13:Automatic_13,15:incumbent': -20.625213819
GTGGGTACCGCACGTCACTCACCTCG+CGAGGTGAGTGACGTGCGGT
......((((((((((((((((((((+))))))))))))))))))))
135878279: [0] '7:Automatic_13*': 0.35
CGAGGTGAGTGACGTGCGGTACCCAC
...((..(.((.....)).)..))..
183153459: [1] '15:incumbent': 0.0
CGAGGTGAGTGACGTGCGGT
....................
183153459: [0] '7:Automatic_13*,13:Automatic_13': -25.275213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
.(((((((((((((((((((((((..+..))))))))))))))))))))))).
357172939: [1] '15:incumbent': 1.09
CGAGGTGAGTGACGTGCGGT
............(.....).
357172939: [0] '7:Automatic_13*,13:Automatic_13': -26.735213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
.(((((((((((((((((((((((((+))))))))))))))))))))))))).
1123300: [1] '15:incumbent': 0.15
CGAGGTGAGTGACGTGCGGT
..(.((.....)).).....
1123300: [0] '7:Automatic_13*,13:Automatic_13': -24.905213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
...(((((((((((((((((((((((+)))))))))))))))))))))))...
1625367343: [1] '15:incumbent': 2.48
CGAGGTGAGTGACGTGCGGT
(.....).....(....)..
1625367343: [0] '7:Automatic_13*,13:Automatic_13': -25.845213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
..((((((((((((((((((((((((+))))))))))))))))))))))))..
1953642256: [1] '15:incumbent': 0.0
CGAGGTGAGTGACGTGCGGT
....................
1953642256: [0] '7:Automatic_13*,13:Automatic_13': -26.735213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
.(((((((((((((((((((((((((+))))))))))))))))))))))))).
1205454187: [1] '15:incumbent': 0.39
CGAGGTGAGTGACGTGCGGT
...........((.....))
1205454187: [0] '7:Automatic_13*,13:Automatic_13': -26.735213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
.(((((((((((((((((((((((((+))))))))))))))))))))))))).
442541274: [1] '15:incumbent': 0.15
CGAGGTGAGTGACGTGCGGT
..(.((.....)).).....
442541274: [0] '7:Automatic_13*,13:Automatic_13': -24.905213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
...(((((((((((((((((((((((+)))))))))))))))))))))))...
1579483543: [1] '15:incumbent': 0.39
CGAGGTGAGTGACGTGCGGT
...........((.....))
1579483543: [0] '7:Automatic_13*,13:Automatic_13': -24.485213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
.((((((((((((((((((((((((.+.)))))))))))))))))))))))).
205951714: [1] '15:incumbent': 0.39
CGAGGTGAGTGACGTGCGGT
...........((.....))
205951714: [0] '7:Automatic_13*,13:Automatic_13': -25.845213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
..((((((((((((((((((((((((+))))))))))))))))))))))))..
24510184: [1] '15:incumbent': 1.62
CGAGGTGAGTGACGTGCGGT
........((.....))...
24510184: [0] '7:Automatic_13*,13:Automatic_13': -26.735213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
.(((((((((((((((((((((((((+))))))))))))))))))))))))).
810168037: [1] '15:incumbent': 0.0
CGAGGTGAGTGACGTGCGGT
....................
810168037: [0] '7:Automatic_13*,13:Automatic_13': -23.155213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
.((((((.((((((((((((((((((+)))))))))))))))))).)))))).
1104871124: [1] '15:incumbent': 0.39
CGAGGTGAGTGACGTGCGGT
...........((.....))
1104871124: [0] '7:Automatic_13*,13:Automatic_13': -26.735213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
.(((((((((((((((((((((((((+))))))))))))))))))))))))).
269954864: [1] '15:incumbent': 1.09
CGAGGTGAGTGACGTGCGGT
............(.....).
269954864: [0] '7:Automatic_13*,13:Automatic_13': -26.735213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
.(((((((((((((((((((((((((+))))))))))))))))))))))))).
2047683604: [1] '15:incumbent': 0.0
CGAGGTGAGTGACGTGCGGT
....................
2047683604: [0] '7:Automatic_13*,13:Automatic_13': -25.845213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
..((((((((((((((((((((((((+))))))))))))))))))))))))..
1641170672: [1] '15:incumbent': 0.39
CGAGGTGAGTGACGTGCGGT
...........((.....))
1641170672: [0] '7:Automatic_13*,13:Automatic_13': -24.905213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
...(((((((((((((((((((((((+)))))))))))))))))))))))...
311887117: [1] '15:incumbent': 2.49
CGAGGTGAGTGACGTGCGGT
.......(.(.......).)
311887117: [0] '7:Automatic_13*,13:Automatic_13': -21.555213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
..(((((((((((((((((((((.((+)).)))))))))))))))))))))..
2008276168: [1] '15:incumbent': 2.95
CGAGGTGAGTGACGTGCGGT
.(..........).......
2008276168: [0] '7:Automatic_13*,13:Automatic_13': -25.845213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
..((((((((((((((((((((((((+))))))))))))))))))))))))..
1692747787: [1] '15:incumbent': 0.0
CGAGGTGAGTGACGTGCGGT
....................
1692747787: [0] '7:Automatic_13*,13:Automatic_13': -26.735213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
.(((((((((((((((((((((((((+))))))))))))))))))))))))).
1875471352: [1] '15:incumbent': 0.39
CGAGGTGAGTGACGTGCGGT
...........((.....))
1875471352: [0] '7:Automatic_13*,13:Automatic_13': -25.845213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
..((((((((((((((((((((((((+))))))))))))))))))))))))..
590133089: [1] '15:incumbent': 1.62
CGAGGTGAGTGACGTGCGGT
........((.....))...
590133089: [0] '7:Automatic_13*,13:Automatic_13': -24.905213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
...(((((((((((((((((((((((+)))))))))))))))))))))))...
603278334: [1] '15:incumbent': 1.22
CGAGGTGAGTGACGTGCGGT
.......(.((.....)).)
603278334: [0] '7:Automatic_13*,13:Automatic_13': -26.735213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
.(((((((((((((((((((((((((+))))))))))))))))))))))))).
1944751909: [1] '13:Automatic_13,15:incumbent': -20.625213819
GTGGGTACCGCACGTCACTCACCTCG+CGAGGTGAGTGACGTGCGGT
......((((((((((((((((((((+))))))))))))))))))))
1944751909: [0] '7:Automatic_13*': -0.07
CGAGGTGAGTGACGTGCGGTACCCAC
...(((..............)))...
1362699087: [1] '15:incumbent': 1.09
CGAGGTGAGTGACGTGCGGT
............(.....).
1362699087: [0] '7:Automatic_13*,13:Automatic_13': -25.845213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
..((((((((((((((((((((((((+))))))))))))))))))))))))..
486244723: [1] '15:incumbent': 0.15
CGAGGTGAGTGACGTGCGGT
..(.((.....)).).....
486244723: [0] '7:Automatic_13*,13:Automatic_13': -25.845213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
..((((((((((((((((((((((((+))))))))))))))))))))))))..
1567934247: [1] '15:incumbent': 0.39
CGAGGTGAGTGACGTGCGGT
...........((.....))
1567934247: [0] '7:Automatic_13*,13:Automatic_13': -25.845213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
..((((((((((((((((((((((((+))))))))))))))))))))))))..
234026863: [1] '15:incumbent': 3.42
CGAGGTGAGTGACGTGCGGT
..(......).((.....))
234026863: [0] '7:Automatic_13*,13:Automatic_13': -26.735213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
.(((((((((((((((((((((((((+))))))))))))))))))))))))).
495148915: [1] '15:incumbent': 0.0
CGAGGTGAGTGACGTGCGGT
....................
495148915: [0] '7:Automatic_13*,13:Automatic_13': -26.735213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
.(((((((((((((((((((((((((+))))))))))))))))))))))))).
1163455431: [1] '15:incumbent': 0.0
CGAGGTGAGTGACGTGCGGT
....................
1163455431: [0] '7:Automatic_13*,13:Automatic_13': -26.735213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
.(((((((((((((((((((((((((+))))))))))))))))))))))))).
163029701: [1] '15:incumbent': 2.53
CGAGGTGAGTGACGTGCGGT
.......(......).....
163029701: [0] '7:Automatic_13*,13:Automatic_13': -25.845213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
..((((((((((((((((((((((((+))))))))))))))))))))))))..
731810015: [1] '15:incumbent': 0.39
CGAGGTGAGTGACGTGCGGT
...........((.....))
731810015: [0] '7:Automatic_13*,13:Automatic_13': -25.845213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
..((((((((((((((((((((((((+))))))))))))))))))))))))..
494916153: [1] '15:incumbent': 0.15
CGAGGTGAGTGACGTGCGGT
..(.((.....)).).....
494916153: [0] '7:Automatic_13*,13:Automatic_13': -25.845213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
..((((((((((((((((((((((((+))))))))))))))))))))))))..
1437536349: [1] '15:incumbent': 0.89
CGAGGTGAGTGACGTGCGGT
....((.....)).......
1437536349: [0] '7:Automatic_13*,13:Automatic_13': -26.735213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
.(((((((((((((((((((((((((+))))))))))))))))))))))))).
1214650420: [1] '15:incumbent': 1.22
CGAGGTGAGTGACGTGCGGT
.......(.((.....)).)
1214650420: [0] '7:Automatic_13*,13:Automatic_13': -21.885213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
...(((((((((((((((((.(((((+))))).)))))))))))))))))...
1355466057: [1] '15:incumbent': 0.39
CGAGGTGAGTGACGTGCGGT
...........((.....))
1355466057: [0] '7:Automatic_13*,13:Automatic_13': -25.275213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
.(((((((((((((((((((((((..+..))))))))))))))))))))))).
180019418: [1] '15:incumbent': 0.39
CGAGGTGAGTGACGTGCGGT
...........((.....))
180019418: [0] '7:Automatic_13*,13:Automatic_13': -25.845213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
..((((((((((((((((((((((((+))))))))))))))))))))))))..
1712880915: [1] '15:incumbent': 0.78
CGAGGTGAGTGACGTGCGGT
((..........))......
1712880915: [0] '7:Automatic_13*,13:Automatic_13': -25.845213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
..((((((((((((((((((((((((+))))))))))))))))))))))))..
2124199206: [1] '15:incumbent': 0.0
CGAGGTGAGTGACGTGCGGT
....................
2124199206: [0] '7:Automatic_13*,13:Automatic_13': -26.735213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
.(((((((((((((((((((((((((+))))))))))))))))))))))))).
2103406152: [1] '15:incumbent': 1.1
CGAGGTGAGTGACGTGCGGT
............(....)..
2103406152: [0] '7:Automatic_13*,13:Automatic_13': -26.735213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
.(((((((((((((((((((((((((+))))))))))))))))))))))))).
1164703847: [1] '15:incumbent': 0.39
CGAGGTGAGTGACGTGCGGT
...........((.....))
1164703847: [0] '7:Automatic_13*,13:Automatic_13': -26.735213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
.(((((((((((((((((((((((((+))))))))))))))))))))))))).
1879677723: [1] '15:incumbent': 0.15
CGAGGTGAGTGACGTGCGGT
..(.((.....)).).....
1879677723: [0] '7:Automatic_13*,13:Automatic_13': -25.845213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
..((((((((((((((((((((((((+))))))))))))))))))))))))..
1706137248: [1] '15:incumbent': 0.0
CGAGGTGAGTGACGTGCGGT
....................
1706137248: [0] '7:Automatic_13*,13:Automatic_13': -26.735213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
.(((((((((((((((((((((((((+))))))))))))))))))))))))).
520508907: [1] '15:incumbent': 0.15
CGAGGTGAGTGACGTGCGGT
..(.((.....)).).....
520508907: [0] '7:Automatic_13*,13:Automatic_13': -26.735213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
.(((((((((((((((((((((((((+))))))))))))))))))))))))).
159237719: [1] '15:incumbent': 0.15
CGAGGTGAGTGACGTGCGGT
..(.((.....)).).....
159237719: [0] '7:Automatic_13*,13:Automatic_13': -26.735213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
.(((((((((((((((((((((((((+))))))))))))))))))))))))).
493539578: [1] '15:incumbent': 0.0
CGAGGTGAGTGACGTGCGGT
....................
493539578: [0] '7:Automatic_13*,13:Automatic_13': -26.735213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
.(((((((((((((((((((((((((+))))))))))))))))))))))))).
2105409990: [1] '15:incumbent': 2.91
CGAGGTGAGTGACGTGCGGT
...(............)...
2105409990: [0] '7:Automatic_13*,13:Automatic_13': -25.845213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
..((((((((((((((((((((((((+))))))))))))))))))))))))..
220939564: [1] '15:incumbent': 3.63
CGAGGTGAGTGACGTGCGGT
..(......)..(....)..
220939564: [0] '7:Automatic_13*,13:Automatic_13': -25.845213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
..((((((((((((((((((((((((+))))))))))))))))))))))))..
890723705: [1] '15:incumbent': 1.1
CGAGGTGAGTGACGTGCGGT
............(....)..
890723705: [0] '7:Automatic_13*,13:Automatic_13': -25.845213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
..((((((((((((((((((((((((+))))))))))))))))))))))))..
1901457778: [1] '15:incumbent': 0.0
CGAGGTGAGTGACGTGCGGT
....................
1901457778: [0] '7:Automatic_13*,13:Automatic_13': -25.845213819
CGAGGTGAGTGACGTGCGGTACCCAC+GTGGGTACCGCACGTCACTCACCTCG
..((((((((((((((((((((((((+))))))))))))))))))))))))..
All the above information is collected into the options objects o1.interface.results and o2.interface.results. Before plotting histograms, let's look to see what kind of data is there. (This next line ought to be skipped if you have run a lot of trials!)
In [36]:
o1.interface.results
Out[36]:
[(6604224628510739414, 17, 0.000160850907002, 'failed', result_type='status_line' ),
(1703676841, 17, 1.92616617991e-05, 'failed', result_type='status_line' ),
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(1945134143, 17, 7.00135599426e-05, 'failed', result_type='status_line' ),
(452052276, 17, 0.000114969373847, 'complete', result_type='status_line' ),
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(1077731645, 17, 2.11169834848e-05, 'failed', result_type='status_line' )]
That information is printed in a more understandable format, if you ask via print.
In [37]:
result_list1 = o1.interface.results
result_list2 = o2.interface.results
# look at a few things by hand, just to check
print result_list1[0]
Trajectory Seed [6604224628510739414]
Result: Normal
Completion Time: 0.000160850907002
Completion Tag: failed
The easiest thing to do is to separate results based on whether strand displacement occurred, or not.
In [38]:
times_complete1 = np.array([i.time for i in result_list1 if i.tag == 'complete'])
times_failed1 = np.array([i.time for i in result_list1 if i.tag == 'failed'])
print "Design A: %d trajectories total, %d completed, %d failed." % (len(result_list1), len(times_complete1), len(times_failed1))
times_complete2 = np.array([i.time for i in result_list2 if i.tag == 'complete'])
times_failed2 = np.array([i.time for i in result_list2 if i.tag == 'failed'])
print "Design B: %d trajectories total, %d completed, %d failed." % (len(result_list2), len(times_complete2), len(times_failed2))
Design A: 100 trajectories total, 28 completed, 72 failed.
Design B: 100 trajectories total, 97 completed, 3 failed.
Now let's go ahead and see the histograms. OK, they don't look so interesting with just 10 trials. Go back and change that to 100 and try again, if you didn't already...
In [39]:
%matplotlib inline
plot_histograms_complete_vs_failed(o1.interface.results)
In [40]:
plot_completion_graph_complete_vs_failed(o1.interface.results)
In [41]:
plot_histograms_two_designs([o1.interface.results, o2.interface.results])
In [42]:
plot_completion_graph([o1.interface.results, o2.interface.results])
In [ ]:
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CC-MAIN-2020-16
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longest
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en
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https://leetcodesol.com/leetcode-add-two-numbers-problem-solution-c-sharp.html
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text/html
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crawl-data/CC-MAIN-2023-50/segments/1700679100448.65/warc/CC-MAIN-20231202172159-20231202202159-00156.warc.gz
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# Leetcode Add Two Numbers problem solution in C#
Feb 12, 2023
In the Leetcode Add Two Numbers problem solution in C# programming You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order, and each of their nodes contains a single digit. Add the two numbers and return the sum as a linked list.
You may assume the two numbers do not contain any leading zero, except the number 0 itself.
## Leetcode Add Two Numbers problem solution in C# programming
``````public class Solution {
public ListNode AddTwoNumbers(ListNode l1, ListNode l2) {
ListNode temp = null, res = null, prev = null;
int sum = 0, carry = 0;
while(l1!=null || l2!=null){
sum = carry + (l1!=null ? l1.val : 0) + (l2!=null ? l2.val : 0);
carry = (sum >= 10) ? 1 : 0;
sum = sum % 10;
temp = new ListNode(sum);
if(res == null){
res = temp;
}
else
{
prev.next = temp;
}
prev = temp;
if(l1!=null)
l1 = l1.next;
if(l2!=null)
l2 = l2.next;
}
if(carry>0){
temp.next = new ListNode(carry);
}
return res;
}
}``````
#### By Neha Singhal
Hi, my name is Neha singhal a software engineer and coder by profession. I like to solve coding problems that give me the power to write posts for this site.
| 341
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CC-MAIN-2023-50
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http://www.dbforums.com/showthread.php?1636895-Can-you-solve-my-problem
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# Thread: Can you solve my problem?
1. Registered User
Join Date
Oct 2008
Posts
46
## Unanswered: Can you solve my problem?
I have data like this:
ID Col1 Col2 Col3
1 BS-S001 BS-S004 NULL
2 BS-S005 BS-S011 NULL
3 BS-S011 BS-S020 NULL
4 BS-S021 BS-S024 NULL
5 BS-S024 BS-S036 NULL
I need to script to update Col3 like following output:
ID Col1 Col2 Col3
1 BS-S001 BS-S004 NULL
2 BS-S005 BS-S011 Gap
3 BS-S011 BS-S020 Gap
4 BS-S021 BS-S024 Overlap
5 BS-S024 BS-S036 Overlap
anybody have idea how to verify col1 and Col2 and update Col3.
2. Registered User
Join Date
Nov 2003
Location
London
Posts
169
What's the criteria for GAP and Overlap?
*edit* I think you're wearing a pink hat MCrowley...am I right? :P
3. Registered User
Join Date
Jan 2003
Location
Massachusetts
Posts
5,862
I can probably solve your problem, but just to make it interesting, you must answer a question for me: Tell me what color hat I am wearing.
While you are thinking about that one, can you shed a little more light on the conditions for "Gap" and "Overlap"? It is not terribly intuitive from the sample data.
4. Registered User
Join Date
Oct 2008
Posts
46
col1 is starting num and col2 is ending num
If the starting and ending secuance is diff so i have to update col3
Look at the col1 and col2 's value.
in first record its good....start from 001 and ednd to 004
in second record start from 005 and end 011....but in 3rd record also start from 011 so its duplicate so i have to update col3.
And also, if there is unsequance no. eg. after start 001 end 005 and next record start from 008 and end 010......there is missing 009 so its also gap.
5. Registered User
Join Date
Jan 2003
Location
Massachusetts
Posts
5,862
huh?
So, the first record is always NULL by definition, unless it begins at 002?
If the current record (by ID) starts with the same number, or before the end of the last record, it should be marked "Overlap"?
If the current record (by ID) starts with a number that is greater than 1 + (end of last record), then it should be marked "Gap"?
But the description says
Originally Posted by rajan142
next record start from 008 and end 010......there is missing 009 so its also gap.
Isn't 009 included in that record?
I am still missing something pretty basic here.....
Oh, and SQLSlammer, you are close. But Mauve is the correct name for the color. ;-)
6. Registered User
Join Date
Oct 2008
Posts
46
Lets explain this way:
I have 3 columns in a single table:
BEGDOC,ENDDOC, Remark (these are column name)
BS-S001 BS-S004 NULL >> in this record doc start form 001 and end 004
We don't have to do anything in this record
BS-S006 BS-S011 Gap >> in this record doc start from 006 and end 011 in this record doc 005 is gaping so it has to update on Remark column.
And also
BS-S011 BS-S024 Overlap >> on this record doc 011 is overlap
BS-S024 BS-S036 Overlap >> on this record doc 024 is overlap
7. Registered User
Join Date
Jan 2003
Location
Massachusetts
Posts
5,862
Is you original sample data correct, then? It looks nothing like what you have just posted.
8. Registered User
Join Date
Oct 2008
Posts
46
Here are my data:
DocID BEGDoc ENDDoc Remarks
1 BS-S001 BS-S099 NULL >> It will not Update
2 BS-S100 BS-S110 NULL >> It will not Update
3 BS-S120 BS-S150 NULL >> It has to Update with "Gap"
4 BS-S140 BS-S190 NULL >> It has to Update with "Overlap"
9. Registered User
Join Date
Oct 2008
Posts
46
I think we have to use CURSOR.
I have 1175574 row(s) in that table
10. Registered User
Join Date
Jan 2009
Posts
2
-- pls take a backup of the table first..
update X set remark ='overlap' where docid in(
select b.id from X a join X b on b.docid = a.docid+1 and b.begdoc <= a.enddoc)
-- if the above query is taking long, we can try a WHILE loop if you know the maximum value of DocID
11. Registered User
Join Date
Oct 2008
Posts
46
I gonna try this. But i have two criteria, Gap and Overlap
12. Registered User
Join Date
Oct 2008
Posts
46
I got the Overlap
Can you try Gap also?
13. Registered User
Join Date
Jan 2009
Posts
2
use earlier query, but change the logic to
set a record's Remark to 'Gap' if its Begdoc > the previous record's Enddoc + 1
14. Registered User
Join Date
Oct 2008
Posts
46
I try this but could not get exact record.
update cds_csv set Remarks ='Gap' where DocID in(
select b.DocID from cds_csv a join cds_csv b on b.DocID = a.DocID+1 and a.ENDDOC<=b.BEGDOC)
15. Registered User
Join Date
May 2006
Posts
29
| 1,286
| 4,450
|
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| 2.734375
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en
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https://www.gradesaver.com/textbooks/math/algebra/elementary-algebra/chapter-1-some-basic-concepts-of-arithmetic-and-algebra-chapter-1-review-problem-set-page-38/62
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## Elementary Algebra
Published by Cengage Learning
# Chapter 1 - Some Basic Concepts of Arithmetic and Algebra - Chapter 1 Review Problem Set - Page 38: 62
#### Answer
$32$
#### Work Step by Step
Substituting $x= -4$ and $y= 3$, the given expression, $3x+7y-5x+y ,$ evaluates to \begin{array}{l}\require{cancel} 3(-4)+7(3)-5(-4)+3 \\\\= -12+21+20+3 \\\\= 9+20+3 \\\\= 29+3 \\\\= 32 .\end{array}
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
| 179
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| 4.3125
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CC-MAIN-2019-04
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latest
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en
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https://studydaddy.com/question/show-work-suppose-that-the-random-variable-x-follows-a-binomial-distribution-wit
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crawl-data/CC-MAIN-2019-30/segments/1563195525973.56/warc/CC-MAIN-20190719012046-20190719034046-00492.warc.gz
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|
Waiting for answer This question has not been answered yet. You can hire a professional tutor to get the answer.
QUESTION
# Show work! Suppose that the random variable x follows a binomial distribution with p = 0.40 and n = 14. Question 1: What is the probability that x =
Show work!
Suppose that the random variable x follows a binomial distribution with p = 0.40 and n = 14.
Question 1: What is the probability that x = 4?
Question 2: What is the probability that x = 7?
Question 3: What is the probability that x = 9?
Question 4: What is the probability that x = 11?
Question 5: What is the probability that x ≤ 2?
Question 6: What is the probability that x > 3?
Question 7: What is the probability that x ≥ 3?
Question 8: What is the probability that x ≤ 13?
| 220
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| 2.828125
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https://www.teacherspayteachers.com/Product/STAAR-EOC-Algebra-1-Checkpoint-A10B-A10C-2478823
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|
# STAAR EOC Algebra 1 – Checkpoint A.10B & A.10C
Subjects
Resource Types
Product Rating
4.0
File Type
PDF (Acrobat) Document File
0.41 MB | 4 pages
### PRODUCT DESCRIPTION
Checkpoints are an excellent resource to monitor your students’ progress for the STAAR EOC Algebra 1 exam. This checkpoint covers category 1 student expectations A.10B & A.10C
A.10B – Multiply polynomials of degree one and degree two. Supporting Standard
A.10C – Determine the quotient of a polynomial of degree one and polynomial of degree two when divided by a polynomial of degree one and polynomial of degree two when the degree of the divisor does not exceed the degree of the dividend. Supporting Standard
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STAAR-ALGEBRA-1-EOC-CHECKPOINT – A11A-A11B
STAAR-ALGEBRA-1-EOC-CHECKPOINT-A12C-A12D
STAAR ALGEBRA 1 EOC CHECKPOINT – A.12A & A.12B
STAAR ALGEBRA 1 EOC SPIRAL REVIEW – 1ST SEMESTER
STAAR EOC Algebra 1 – Reporting Category 2 Checkpoint Bundle
STAAR EOC Algebra 1 – Checkpoint A.8A
STAAR EOC Algebra 1 – Checkpoint A.6B & A.6C
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https://www.mcs.anl.gov/research/projects/otc/InteriorPoint/abstracts/Luo-Sun.html
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## An Analytic Center Based Column Generation Algorithm For Convex Quadratic Feasibility Problems
### Z.-Q. Luo and J. Sun
ABSTRACT: We consider the analytic center based column generation algorithm for the problem of finding a feasible point in a set defined by a finite number of convex quadratic inequalities. At each iteration, the algorithm computes an approximate analytic center of the set defined by the intersection of quadratic inequalities generated in the previous iterations. If this approximate analytic center is a solution, then the algorithm terminates; otherwise a quadratic inequality violated at the current center is selected and a new quadratic cut (defined by a convex quadratic inequality) is placed near the approximate center. As the number of cuts increases, the set defined by their intersection shrinks and the algorithm eventually finds a solution of the problem. Previously, similar analytic center based column generation algorithms were studied first for the linear feasibility problem and later for the general convex feasibility problem. Our method differs from these early methods in that we use quadratic cuts" in the computation instead of linear cuts. Moreover, our method has a polynomial worst case complexity of $O(n\ln \frac{1}{\varepsilon})$ on the total number of cuts to be used, where $n$ is the number of convex quadratic inequalities in the problem and $\varepsilon$ is the radius of the largest ball contained in the feasible set. In contrast, the early column generation methods using linear cuts can only solve the convex quadratic feasibility problem in pseudo-polynomial time.
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http://www.lighthouse3d.com/very-simple-libs/vsressurfrevlib-very-simple-surface-of-revolution-lib/
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Surface of Revolution is a simple process to create new geometry. Spheres, cylinders, cones, and torus, all can be created using this approach. This library provides the functionality required to build such geometry. Some geometry is already prebuilt as sohwn in the image below.
In order to create a surface of revolution we need a 2D profile, i.e., a curve or set of line segments, and an axis about which the profile will revolve. For instance, considering the chess pawn presented in the figure above, the profile is as presented in the image below. In here the axis is always the Y axis.
the 2D profile is a set of points in the XY plane. For instance consider the following profile (green line on the left) and associated surface of revolution.
Codewise this surface can be defined with the following function:
`void create (float *p, int numP, int sides, int closed, float smoothCos);`
The function receives an array of floats `p` where the points are specified in 2D (XY plane); `numP` provides the number of points; `sides` determines the number of sides in the figure, the greater the number the smoother the surface; `closed` is a flag indicating if the 2D profile is closed or not, if closed then the first point is added to the end of the list; `smoothCos` parameter is used to preserve edges, all edges with a cosine (in radians) larger than the value provided will be smoothed (note the hard edges on the top and bottom of the above figure, and the smooth lateral edges).
The example surface above can be defined as:
```VSSurfRevLib mySurfRev;
...
float p1[] = { 1,2, 1,1, 2,1, 2,2};
mySurfRev.create(p1, 4, 4096, 1, 1.0f);
```
A set of functions is also provided to generate common geometric shapes:
```
void createSphere(float radius, int divisions);
void createTorus(float innerRadius, float outerRadius, int rings, int sides);
void createCylinder(float height, float radius, int sides, int stacks);
void createCone(float height, float baseRadius, int sides);
void createPawn();
```
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https://nl.mathworks.com/matlabcentral/answers/2063922-i-want-to-get-a-scaled-vector-field-vec-f-x-2-x-i-y-2-y-j-the-fig-should-be-like-the-attached?s_tid=prof_contriblnk
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# I want to get a scaled vector field $\vec F=(x^2-x)i+(y^2-y)j$. The Fig should be like the attached Fig:
2 views (last 30 days)
Atom on 25 Dec 2023
I want to get a scaled vector field $\vec F=(x^2-x)i+(y^2-y)j$. The Fig should be like the attached Fig:
[x,y] = meshgrid(-2:.16:2,-2:.16:2);
Fx = (x.*x-x);
Fy=y.*y-y;
figure;
quiver(x,y,Fx,Fy,'k','linewidth',1.2)
hold on
x=-2:0.01:2;
y=1-x;
plot(x,y,'k','linewidth',2);
Note that we have $div \vec F>0$ for $x+y>1$, $div \vec F<0$ for $x+y<1$ and $div \vec F=0$ on the line $x+y=1$ .
##### 3 CommentsShow 1 older commentHide 1 older comment
Atom on 25 Dec 2023
@John D'Errico Sorry, I have edited. Still I am not getting the desire plot.
Dyuman Joshi on 25 Dec 2023
What is the vector field scaled to?
The arrows of the vector field in the reference image appear to be of the same length.
Sulaymon Eshkabilov on 25 Dec 2023
Is this what you are trying to obtain:
[x, y] = meshgrid(-2:0.25:2); % x and y values
F_x = 1*(x.^2 - x); % The vector field of x
F_y = 1*(y.^2 - y); % The vector field of y
F = F_x+F_y;
% Normalize the vectors
magnitude = sqrt(F_x.^2 + F_y.^2);
F_x_Nor = F_x ./ magnitude;
F_y_Nor = F_y ./ magnitude;
% Plot the scaled vector field
quiver(x, y, F_x_Nor, F_y_Nor);
hold on
X=-2:0.1:2;
Y=1-X;
plot(X,Y,'k','linewidth',2);
xlabel('x');
ylabel('y');
title('Scaled Vector Field: F = (x^2 - x) + (y^2 - y)');
axis([-2 2 -2 2])
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https://ncatlab.org/nlab/show/regular+differential+operator
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Idea
# Idea
Grothendieck has developed a deep version of differential calculus, based on a linearization of $O_X$-bimodules. It is also related to the (de Rham) descent data for the stack of $O_X$-modules over the simplicial scheme resolving the diagonal of $X$. As abstract descent data correspond to the flat connections for the corresponding monad, this was historically the first case in which this correspondence was noted; in positive characteristics Grothendieck called the corresponding descent data for the de Rham site “costratifications”, the corresponding connection is nowdays called Grothendieck connection, see Berthelot, Ogus 1978 for the fundamentals and the related notion of crystal (algebraic geometry).
This corresponds to looking at a sequence of infinitesimal neighborhoods of the diagonal. This geometrical principle can be applied to other categories; it is the basis of the study of jet schemes and close in spirit to some constructions in synthetic differential geometry.
## Definitions
Given a commutative unital ring $R$, a filtration $M_n$ ($n\geq -1$) on a $R$-$R$-bimodule $M$ is a differential filtration if the commutator $[r,P]$ for any $P$ in $M_n$ and $r$ in $R$ is in $M_{n-1}$, and $M_{-1} = 0$. A bimodule is differential if it has an exhaustive ($\cup_N M_n = M$) differential filtration. Every $R$-$R$-bimodule has a differential part, i.e. the maximal differential submodule of $M$.
Regular differential operators, as defined by Grothendieck, are the elements of the differential part $Diff(R,R)$ of $Hom(R,R)$ i.e. a maximal differential subbimodule in $Hom(R,R)$. The operators in $Diff(R,R)_n$ are called the differential operators of degree $\leq n$. If $R\to B$ is a ring morphism, then the differential part of $B$ via its natural $R$-$R$-bimodule structure is also an object of $R\backslash \mathrm{Ring}$; in particular $Diff(R,R)$ is a ring and $R\hookrightarrow Diff(R,R)$ is an embedding of rings.
More generally (and in the affine case equivalently), for a $S$-scheme $X$, let $P^n_{X/S}$ denote the sheaf $(O_X\otimes_{f^{-1}(O_S)} O_X)/I^{n+1}$, where $I$ is the ideal of the diagonal (this makes sense since the diagonal morphism is an immersion, cf. EGAI, 5.3.9.), and $f:X\rightarrow S$ the structure morphism. Consider $P^n_{X/S}$ as $O_X$-module via the morphism $O_X\rightarrow O_X\otimes_{f^{-1}(O_S)} O_X$, $a\mapsto a\otimes 1$.
For $O_X$-modules $E,F$, $Diff_S(F,E)_n$ is defined to be $Hom_{O_X}(P_{X/S}^n\otimes_{O_X} F, E)$. Note that $P^n_{X/S}$ has two canonical structures as $O_X$-module given by the projections $p_i: X \times_S X\rightarrow X$. The tensor product $P_{X/S}^n \otimes_{O_X} F$ is understood to be constructed via $p_1$ and considered as an $O_X$-module via $p_0$.
## Properties
In the affine case, and in characteristics zero, the sheaf of regular differential operators is locally isomorphic to the Weyl algebra. For that simple case, a good reference is Coutinho 1995.
## Noncommutative analogues
Regular differential operators have been nontrivially generalized to noncommutative rings (and schemes) by V. Lunts and A. L. Rosenberg, as well as to the setting of braided monoidal categories. Their motivation is an analogue of a Beilinson-Bernstein localization theorem for quantum groups. The category of differential bimodules is categorically characterized in their work as the minimal coreflective topologizing monoidal subcategory of the abelian monoidal category of $R$-$R$-bimodules which is containing $R$. In the case of noncommutative rings, Lunts-Rosenberg definition of differential operators has been recovered from a different perspective in the setup of noncommutative algebraic geometry represented by monoidal categories; the emphasis is on the duality between infinitesimals and differential operators, see Maszczyk 2006.
## References
• P. Berthelot, A. Ogus, Notes on crystalline cohomology, Princeton Univ.P. 1978. vi+243, ISBN0-691-08218-9
• S. C. Coutinho, A primer of algebraic $D$-modules, London Math. Soc. Stud. Texts 33, Cambridge University Press, Cambridge, 1995. xii+207 pp.
• Tomasz Maszczyk, Noncommutative geometry through monoidal categories, arXiv:0611806
For regular differential operators in Lie theory see
• I. N. Bernstein, I. M. Gelfand, S. I. Gelfand, Differential operators on the base affine space and a study of $\mathfrak{g}$-modules, In: I. M. Gelfand, ed., Lie groups and their representations, 21–64. Adam Hilger, 1975
Lunts-Rosenberg work in noncommutative setting
• V. A. Lunts, A. L. Rosenberg, Differential operators on noncommutative rings, Selecta Math. (N.S.) 3:3 (1997) 335–359 (doi)
• V. A. Lunts, A. L. Rosenberg, Localization for quantum groups, Selecta Math. (N.S.) 5:1 (1999) 123–159 (doi).
• V. A. Lunts, A. L. Rosenberg, Differential calculus in noncommutative algebraic geometry I. D-calculus on noncommutative rings, MPI 1996-53 pdf
• V. A. Lunts, A. L. Rosenberg, Differential calculus in noncommutative algebraic geometry II. D-Calculus in the braided case. The localization of quantized enveloping algebras, MPI 1996-76 pdf
MO questions
Last revised on July 24, 2024 at 17:13:13. See the history of this page for a list of all contributions to it.
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https://brainmass.com/math/basic-algebra/graphical-addition-30251
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Share
Explore BrainMass
Find the voltage V where V=V1+V2 using graphical addition and analytical methods, given that V1=3sin(theta+20 degrees) and V2=5sin(theta-40degrees).
Solution Preview
V1 = 3*sin(theta+20 degrees)
V2 = 5*sin(theta-40degrees)
Let us assume
theta+20 = phi
THerefore,
V1 = 3*sin(phi)
V2 = 5*sin(phi-60 degree)
Hence the phase difference between two volatages is ...
Solution Summary
A graphical addition problem is solved. A diagram is included with this fully-worked solution.
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http://psychology.wikia.com/wiki/Random_digit_dialing?oldid=48906
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Random digit dialing
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
34,196pages on
this wiki
Random digit dialing (RDD) is a method for selecting people for involvement in telephone statistical surveys by generating telephone numbers at random. Random digit dialing has the advantage that it includes unlisted numbers that would be missed if the numbers were selected from a phone book. In populations where there is a high telephone-ownership rate, it can be a cost efficient way to get complete coverage of a geographic area.
RDD is widely used for statistical surveys, including election opinion polling [1] and selection of experimental control groups [2].
An important consideration in random digit dialing surveys is bias introduced by non-responders. Non-response bias can be a problem if responders differ from non-responders for the measured variables. For example, nonresponders may not have been contacted because they work multiple minimum-wage jobs. [3] Various techniques are used to reduce the non-response rate, such as multiple call attempts, monetary incentives, advance letters, and leaving messages on answering machines, because reducing the non-response rate may directly reduce non-response bias. [4] In addition, when trying to calculate total error estimates, response rate calculations can be imprecise because it can be difficult to determine whether certain telephone numbers are interviewable. [5] [6] Random digit dialing (RDD) is a method for selecting people for involvement in telephone statistical surveys by generating telephone numbers at random. Random digit dialing has the advantage that it includes unlisted numbers that would be missed if the numbers were selected from a phone book. In populations where there is a high telephone-ownership rate, it can be a cost efficient way to get complete coverage of a geographic area.
RDD is widely used for statistical surveys, including election opinion polling [7] and selection of experimental control groups [8].
An important consideration in random digit dialing surveys is bias introduced by non-responders. Non-response bias can be a problem if responders differ from non-responders for the measured variables. For example, nonresponders may not have been contacted because they work multiple minimum-wage jobs. [9] Various techniques are used to reduce the non-response rate, such as multiple call attempts, monetary incentives, advance letters, and leaving messages on answering machines, because reducing the non-response rate may directly reduce non-response bias. [10] In addition, when trying to calculate total error estimates, response rate calculations can be imprecise because it can be difficult to determine whether certain telephone numbers are interviewable. [11] [12]
When the desired coverage area matches up closely enough with country codes and area codes, random digits can be chosen within the desired area codes. In cases where the desired region doesn't match area codes (for instance, electoral districts), surveys must rely on telephone databases, and must rely on self-reported address information for unlisted numbers. Increasing use of mobile phones, number portability, and VoIP have begun to decrease the ability for RDD to target specific areas within a country and achieve complete coverage.
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https://www.aqua-calc.com/calculate/volume-to-weight/substance/zinc-op-ii-cp--blank-nitrate-blank-hexahydrate
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# Weight of Zinc(II) nitrate hexahydrate
## zinc(ii) nitrate hexahydrate: convert volume to weight
### Weight of 1 cubic centimeter of Zinc(II) nitrate hexahydrate
carat 10.33 ounce 0.07 gram 2.07 pound 0 kilogram 0 tonne 2.07 × 10-6 milligram 2 065
#### How many moles in 1 cubic centimeter of Zinc(II) nitrate hexahydrate?
There are 6.94 millimoles in 1 cubic centimeter of Zinc(II) nitrate hexahydrate
### The entered volume of Zinc(II) nitrate hexahydrate 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
• 1 cubic meter of Zinc(II) nitrate hexahydrate weighs 2 065 kilograms [kg]
• 1 cubic foot of Zinc(II) nitrate hexahydrate weighs 128.91374 pounds [lbs]
• Zinc(II) nitrate hexahydrate weighs 2.065 gram per cubic centimeter or 2 065 kilogram per cubic meter, i.e. density of zinc(II) nitrate hexahydrate is equal to 2 065 kg/m³; at 15°C (59°F or 288.15K) at standard atmospheric pressure. In Imperial or US customary measurement system, the density is equal to 128.9137 pound per cubic foot [lb/ft³], or 1.1936 ounce per cubic inch [oz/inch³] .
• Melting Point (MP), Zinc(II) nitrate hexahydrate changes its state from solid to liquid at 36.4°C (97.52°F or 309.55K)
• Zinc(II) nitrate hexahydrate is a colorless and odorless solid having a tetragonal crystalline structure.
• Also known as: Zinc nitrate hexahydrate.
• Molecular formula: Zn(NO3)2 ⋅ 6H2O
Elements: Hydrogen (H), Nitrogen (N), Oxygen (O), Zinc (Zn)
Molecular weight: 297.49 g/mol
Molar volume: 144.063 cm³/mol
CAS Registry Number (CAS RN): 10196-18-6
• Bookmarks: [ weight to volume | volume to weight | price | mole to volume and weight | mass and molar concentration | density ]
• A material, substance, compound or element with a name containing, like or similar to Zinc(II) nitrate hexahydrate:
• 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
CRACKED PEPPER FRIES, UPC: 041303066607 contain(s) 188 calories per 100 grams (≈3.53 ounces) [ price ]
13383 foods that contain Vitamin E (alpha-tocopherol). List of these foods starting with the highest contents of Vitamin E (alpha-tocopherol) and the lowest contents of Vitamin E (alpha-tocopherol), and Recommended Dietary Allowances (RDAs) for Vitamin E (Alpha-Tocopherol)
#### Gravels, Substances and Oils
CaribSea, Freshwater, Instant Aquarium, Sunset Gold weighs 1 505.74 kg/m³ (94.00028 lb/ft³) with specific gravity of 1.50574 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 ]
Gallium(III) oxide, alpha form [Ga2O3] weighs 6 440 kg/m³ (402.03607 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | mass and molar concentration | density ]
Volume to weightweight to volume and cost conversions for Sunflower oil with temperature in the range of 10°C (50°F) to 140°C (284°F)
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## UE Efficient methods in optimization
Diplômes intégrant cet élément pédagogique :
### Descriptif
The subject of this half-semester course are more advanced methods in convex optimization. It consists of 6 lectures, 2 x 1,5 hours each, and can be seen as continuation of the course “Non-smooth methods in convex optimization”.
This course deals with:
Evaluation : A two-hours written exam (E1) in December. For those who do not pass there will be another two-hours exam (E2) in session 2 in spring.
• Topic 1: convex analysis
• Topic 2: convex programming
• Basic notions: vector space, affine space, metric, topology, symmetry groups, linear and affine hulls, interior and closure, boundary, relative interior
• Convex sets: definition, invariance properties, polyhedral sets and polytopes, simplices, convex hull, inner and outer description, algebraic properties, separation, supporting hyperplanes, extreme and exposed points, recession cone, Carathéodory number, convex cones, conic hull
• Convex functions: level sets, support functions, sub-gradients, quasi-convex functions, self-concordant functions
• Duality: dual vector space, conic duality, polar set, Legendre transform
• Optimization problems: classification, convex programs, constraints, objective, feasibility, optimality, boundedness, duality
• Linear programming: Farkas lemma, alternative, duality, simplex method
• Algorithms: 1-dimensional minimization, Ellipsoid method, gradient descent methods, 2nd order methods
• Conic programming: barriers, Hessian metric, duality, interior-point methods, universal barriers, homogeneous cones, symmetric cones, semi-definite programming
• Relaxations: rank 1 relaxations for quadratically constrained quadratic programs, Nesterovs π/2 theorem, S-lemma, Dines theorem Polynomial optimization: matrix-valued polynomials in one variable, Toeplitz and Hankel matrices, moments, SOS relaxations
### Syllabus
This course deals with
• Topic 1: convex analysis
• Topic 2: convex programming
• Basic notions: vector space, affine space, metric, topology, symmetry groups, linear and affine hulls, interior and closure, boundary, relative interior
• Convex sets: definition, invariance properties, polyhedral sets and polytopes, simplices, convex hull, inner and outer description, algebraic properties, separation, supporting hyperplanes, extreme and exposed points, recession cone, Carathéodory number, convex cones, conic hull
• Convex functions: level sets, support functions, sub-gradients, quasi-convex functions, self-concordant functions
• Duality: dual vector space, conic duality, polar set, Legendre transform
• Optimization problems: classification, convex programs, constraints, objective, feasibility, optimality, boundedness, duality
• Linear programming: Farkas lemma, alternative, duality, simplex method
• Algorithms: 1-dimensional minimization, Ellipsoid method, gradient descent methods, 2nd order methods
• Conic programming: barriers, Hessian metric, duality, interior-point methods, universal barriers, homogeneous cones, symmetric cones, semi-definite programming
• Relaxations: rank 1 relaxations for quadratically constrained quadratic programs, Nesterovs π/2 theorem, S-lemma, Dines theorem Polynomial optimization: matrix-valued polynomials in one variable, Toeplitz and Hankel matrices, moments, SOS relaxations
### Pré-requis recommandés
Linear algebra: matrices, vector spaces, linear functions
### Compétences visées
At the end of the course, the student will be able to convert optimization problems into a standard form amenable to a solution by a solver.
### Bibliographie
S. Boyd, L. Vandenberghe. Convex Optimization. Cambridge University Press. 2004. http://stanford.edu/~boyd/cvxbook/
A. Ben-Tal, A. Nemirovski. Lectures on modern convex optimization. SIAM, Philadelphia, 2001.
R.T. Rockafellar. Convex analysis. Princeton University Press, Princeton, 1970
J.-B. Lasserre. An Introduction to Polynomial Optimization. Cambridge University Press, Cambridge, 2015
### Informations complémentaires
Méthode d'enseignement : En présence
Lieu(x) : Grenoble
Langue(s) : Anglais
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##### Project - Physics
Projectile Motion \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ Lab Starts here: Lab Manual: Here is the Lab manual: Projectile_Motion Download Projectile_Motion Simulation to be used for this Lab: projectile-motion_Phet https://phet.colorado.edu/sims/html/projectile-motion/latest/projectile-motion_en.html ---- Exceptional (20-10) Acceptable (10-5)Marginal (5-0)A Correct explanation of the concepts/answers. Correct explanation. Results can be reduplicated accurately. Every question is answered. Good and supportive teamwork. ---- Acceptable (10-5) Material is lacking in meaningful explanation or wrong explanation/ missing answers. Results cannot be duplicated. Poor demonstration of teamwork. ---- Marginal (5-0) Few or no answers/explanations to questions. No submission. A very low level of results. Misbehavior with the other group members. Lab Instructions V0.0 January 29, 2018 Projectile Motion . 1 Objective To experimentally study projectile motion. 2 Overview Recall that a projectile is an object that is in free-fall, i.e., it moves while being subjected to just the Earth’s gravity. When a projectile remains close to the surface of the earth and it’s trajectory spans a distance much less than the earth’s circumference, its horizontal and ver- tical motion get decoupled. It’s horizontal motion (say, x-direction) proceeds with constant velocity, while it’s vertical motion (say, y-direction) proceeds as motion with constant ac- celeration ~a = (g, pointing vertically down), where g = 9.8m/s2. The projectile’s trajectory, x vs.y curve, can then be readily derived from it’s x vs.t and y vs.t equations, by eliminating t. In this experiment, you will have a shooter that can launch a projectile at one of three possible, but unknown, speeds, and a continuously adjustable angle, which you can read-off from the built-in protractor. The main idea of this experiment is to launch a projectile, first, to determine its unknown speed, by measuring the (x,y) coordinates at which the projectile lands, given it’s angle of launch θ. Next, choose a different target, and predict the angle of launch θ that would land the projectile at the chosen target. Finally, experimentally, verify your prediction by checking if the projectile indeed hits the target. 3 Apparatus Simulation: https://phet.colorado.edu/sims/html/projectile-motion/latest/projectile- motion_en.html For use by the physics department, Ohlone College, Fremont, CA. Lab Instructions V0.0 January 29, 2018 5 Overall experimental set up Figure 1: Experimental Setup For use by the physics department, Ohlone College, Fremont, CA. Figure 2: Coordinate System and Trajectory Lab Instructions V0.0 January 29, 2018 1. Choose your coordinate system: axis and origin. The cross-hair marks the exact position of the launch 2. Carefully measure the initial vertical position y0 and horizontal position x0 of the cross-hair. measure distances relative to x0,. Record (x0, y0) with the experimental data. 6 Part 1: Determine the initial launch speed v0 for a set of launch angles. 6.1 Experimental data and calculations 1. Follow the steps below, to determine the launch speed for N = 3 different launch angles θ = {15◦, 45◦, 75◦} 2. For each launch angle θi, i = 1, ...,N, launch the projectile for M = 3 to 5 times. For each shot, record the (xj,yj),j = 1, ...,M coordinates of the impact point. 3. Compute the average coordinates, (< xi >,< yi >), for each θi. For use by the physics department, Ohlone College, Fremont, CA. 4. Recall that the equation for the trajectory of the projectile is given by: y = y0 + tan θi(x−x0) − 1 2 g (x−x0)2 v0i2 cos2 θi (1) Derive the above equation in a separate section in your lab report. 5. Using x0, y0, θi, x =< xi > and y =< yi >, calculate the initial launch speed, v0i, from the equation of the trajectory above. Thus, you will have a single launch speed for each θi. Note: Assume the you dont know the speed even though you can see that in simulation Lab Instructions V0.0 January 29, 2018 Table 1: Data for Part 1 x0 =?,y0 =?, θi xj yj < xi > < yi > v0i θ1 =? M data points M data points single value single value single value θ2 =? M data points M data points single value single value single value ... ... ... ... ... ... θN =? M data points M data points single value single value single value 6.2 Analysis and discussion 1. Does the launch speed depend significantly upon the launching angle? Is it reasonable? 2. Show that for y = y0, maximum range is achieved at θ = 45 ◦. Approximately, for what angle do you achieve maximum range? Do you expect it to be different from 45◦? If so, why? 7 Part 2: Predict the launch angle required to hit a chosen target 7.1 Experimental data and calculations 1. Choose a target that is different from Part 1. 2. Position the target .The idea is to choose yT , the y− coordinate of the target to be different from the y coordinate in Part 1 (Check Appendix). 3. You might have to redefine your axis and origin. Carefully measure the coordinates (xT , yT ) of the target, and if needed, y0 and x0 again. 4. FIRST CALCULATE (See next step). Shoot only AFTER you calculate, and only at the predicted angle. 5. Predict the correct launch angle: Solve the equation of the trajectory for the unknown θ. Hint: try to re-write the equation in terms of a single trigonometric function, instead of two trigonometric functions. 6. The predicted angle and shoot! For use by the physics department, Ohlone College, Fremont, CA. Table 2: Data for Part 2 x0 y0 xT yT θpred Did θpred work? Lab Instructions V0.0 January 29, 2018 8 Conclusion Briefly summarize your findings. For use by the physics department, Ohlone College, Fremont, CA. Appendix Projectile_Motion -online version Objective Overview Apparatus Precaution Overall experimental set up Part 1: Determine the initial launch speed v0 for a set of launch angles. Experimental data and calculations Analysis and discussion Part 2: Predict the launch angle required to hit a chosen target Experimental data and calculations Conclusion projectile motion part2_5
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# SAT Math : How to find the length of an arc
## Example Questions
### Example Question #8 : Sectors
Figure not drawn to scale.
In the figure above, circle C has a radius of 18, and the measure of angle ACB is equal to 100°. What is the perimeter of the red shaded region?
Possible Answers:
36 + 20π
18 + 10π
18 + 36π
36 + 36π
36 + 10π
Correct answer:
36 + 10π
Explanation:
The perimeter of any region is the total distance around its boundaries. The perimeter of the shaded region consists of the two straight line segments, AC and BC, as well as the arc AB. In order to find the perimeter of the whole region, we must add the lengths of AC, BC, and the arc AB.
The lengths of AC and BC are both going to be equal to the length of the radius, which is 18. Thus, the perimeter of AC and BC together is 36.
Lastly, we must find the length of arc AB and add it to 36 to get the whole perimeter of the region.
Angle ACB is a central angle, and it intercepts arc AB. The length of AB is going to equal a certain portion of the circumference. This portion will be equal to the ratio of the measure of angle ACB to the measure of the total degrees in the circle. There are 360 degrees in any circle. The ratio of the angle ACB to 360 degrees will be 100/360 = 5/18. Thus, the length of the arc AB will be 5/18 of the circumference of the circle, which equals 2πr, according to the formula for circumference.
length of arc AB = (5/18)(2πr) = (5/18)(2π(18)) = 10π.
Thus, the length of arc AB is 10π.
The total length of the perimeter is thus 36 + 10π.
The answer is 36 + 10π.
### Example Question #1 : How To Find The Length Of An Arc
In the circle above, the angle A in radians is
What is the length of arc A?
Possible Answers:
Correct answer:
Explanation:
Circumference of a Circle =
Arc Length
### Example Question #10 : Sectors
In the figure above, and are diameters of the cirlce, which has a radius of . What is the sum of the lengths of arcs and ?
Possible Answers:
Correct answer:
Explanation:
The formula for arclength is .
You know that so and must both equal .
Since
,
the sum the lengths of arcs and must equal .
### Example Question #1 : How To Find The Length Of An Arc
Figure NOT drawn to scale
Refer to the above figure. Evaluate .
Possible Answers:
Correct answer:
Explanation:
and , being secant segments of a circle, intercept two arcs such that the measure of the angle that the segments form is equal to one-half the difference of the measures of the intercepted arcs - that is,
Setting and solving for :
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Browse Learn ET2411
# Advanced Concepts of GD&T 1994 2-day ET2411
Providing you have a basic understanding of geometric dimensioning and tolerancing fundamentals, this course teaches the advanced concepts of GD&T as prescribed in the ASME Y14.5M-1994 Standard.
Utilizing the expertise of world-renowned GD&T expert Alex Krulikowski, this course offers an in-depth explanation of advanced GD&T topics like composite tolerancing, tolerance analysis, datum selection, non-rigid part dimensioning, and many more key dimensioning topics, including the system approach for part dimensioning. Newly acquired learning is reinforced throughout the class with more than 250 practice problems.
Each attendee receives a robust collection of learning resources including:
• Advanced Concepts of GD&T (ASME Y14.5M-1994) textbook by Alex Krulikowski
• A GD&T Ultimate Pocket Guide (1994)
• A Fundamentals of GD&T Exercise Workbook
• 90-day access to Fundamentals of GD&T 1994 web training course, (\$209 value)
Thousands of students have learned GD&T through Alex Krulikowski"s textbooks, self-study courses, computer based training, and online learning center. Students who attend courses like this one walk away with more than knowledge. They gain on-the-job skills because the learning materials are performance-based.
##### Learning Objectives
By attending this class, you will be able to:
• Explain the importance of product design and functional dimensioning
• Define the terms “feature” and “feature of size”
• Recognize which dimensioning standards apply to an engineering drawing
• Explain the fundamentals of drawing interpretation and how to handle substandard drawings
• Recognize the difference between a rigid and a flexible (non-rigid) part
• State the requirements for tolerancing parts measured in the restrained state
• Identify the two special considerations for datum usage on restrained (non-rigid parts
• Calculate advanced applications of form controls
• Describe uses, advantages, misconceptions, and common errors of the datum system
• List nine common datum feature types
• Describe advanced datum target concepts
• Explain how to specify / interpret specialized datum feature applications
• Describe modifier usage in tolerance of position applications
• Describe the effects of simultaneous and separate requirements with tolerance of position
• Explain composite position tolerancing and multiple single-segment position tolerancing
• Interpret tolerance of position applications with a conical tolerance zone
• Explain composite profile tolerancing and multiple single-segment profile tolerancing
• Describe profile applications
##### Who Should Attend
This course is valuable for individuals who create or interpret engineering drawings, product and gage designers; process, product, and manufacturing engineers; supplier quality engineers/professionals; CMM operators; buyers/purchasers; checkers; inspectors; technicians; and sales engineers/professionals.
##### Prerequisites
Please be aware that this is not an introductory course. In order to understand the course content, students should have completed ETI's GD&T Fundamentals course or equivalent.
GD&T Fundamentals Review
• GD&T skills survey
• GD&T fundamentals for further study
Importance of Product Design
• Product design effects on costs
• Consequences of drawing errors
Functional Dimensioning
• The purpose of tolerances
• The importance of specifying proper tolerances
• The importance of a common tolerancing approach
• Tolerancing principles and benefits
Interpretation of Feature
• The terms “element,” “gap,” and “interruption”
• Y14.5 definition of feature and types
• Regular, element, complex, and interrupted feature; sub-feature
Interpretation of Feature of Size
• The terms “opposed,” “fully opposed,” “partially opposed, “size dimension,” and “cylindrical”
• Importance of distinguishing between a feature and feature of size
• The definition of feature of size from Y14.5
• Requirements and categories of a feature of size
• Identifying and interpreting a complete, interrupted, partial, and bounded feature of size
Applicable Drawing Standards
• Determining on which standards an engineering drawing is based
• Clarifying a drawing when no dimensioning standard is referenced
• Reducing confusion on dimensioning standards
Drawing Interpretation
• Interpreting an engineering drawing
• Drawing title block, revision column, general drawing notes
• Fundamental rules that affect drawing interpretation
• Surface coating and heat treat
• Geometric controls and a valid datum system
• Misconceptions on measuring parts that use the datum system
• Controlling characteristics for each part feature
• Proper uses for coordinate tolerancing
• Specification / interpretation
Using Substandard Drawings
• Categories of substandard drawing specifications
• Steps for dealing with substandard drawings
• Things not to do when using a substandard drawing
Rigid/Non-Rigid Parts Definitions
• Free state
• Restrained state
• Rigid part
• Non-rigid part and part feature
Tolerancing Non-Rigid Parts
• Tolerancing a non-rigid (restrained) part
• Roles of a restraint note
• Determining restraining conditions on non-rigid parts
• Requirements that need to be addressed in a restraint note
• The difference between a general note and a local restraint note
• When a free state symbol should be used
• Areas that need special attention when inspecting a non-rigid part
Restrained Part Datum Usage
• How to use datum targets to support, orient, and locate a restrained part in the datum reference frame
• How datum shift occurs on a restrained part
Form Controls
• Calculating the flatness tolerance value for a gasketed joint application
• Calculating the cylindricity tolerance value in a support application
• Calculating the straightness tolerance value in an assembly application
• Overriding Rule #1 to limit flatness on a thin part
The Datum System
• When to use the datum system.
• Advantages of the datum system.
• Common misconceptions about the datum system.
• Common errors in datum usage
Datum Feature Types
• Common datum feature types
• When each datum feature type is typically used
• Degrees of freedom restrained when each datum feature type is used
• The datum feature simulator for the datum features referenced in a geometric tolerance
Datum Targets
• Reducing the impact that using datum targets has on functional dimensioning
• Application requirements
• Applications where datum targets should be used
• Specifying fixed and movable datum targets
• Special datum target types
• Dimensioning a simulated gage for datum target applications
Specialized Datum Applications
• Specifying a screw thread as a datum feature and interpreting application
• Specifying a gear or spline feature as a datum feature and interpreting application
• Temporary and permanent datum features
• Major disadvantage of temporary datum features
Tolerance of Position Usage
• When to use a tolerance of position control
• Loss function curve, customer robust dimension, and customer sensitive dimension
• Tolerance of position control and material condition used
Simultaneous and Separate Requirements
• Simultaneous and separate requirements, effects and where they apply
• Tolerance of position at MMC simultaneous requirement
• Tolerance of position controls as separate requirements
• One exception to the simultaneous requirement
Composite Position Tolerancing
• Rules, advantages, and when to use it
• “FRITZ” and “PLTZF”
• Tolerance of position composite application
Multiple Single-Segment Tolerance of Position Tolerancing
• Rules, advantages, and when to use it
• Tolerance of position vs. composite tolerance of position
Conical Tolerance Zones
• A conical tolerance zone and advantage of use
• Specifying a conical tolerance zone in a tolerance of position application
• When to use tolerance of position with a conical tolerance zone
Profile Tolerances
• When to use a profile control
• The four characteristics profile can control
• Converting coordinate tolerances into profile callouts
• The profile datum rule
Profile and Simultaneous Requirements
• Simultaneous requirement applied to profile
• Profile controls with separate requirements
Composite Profile Tolerancing
• Composite profile tolerancing, rules, and advantages
• Interpreting a composite profile application
Multiple Single-Segment Profile Tolerancing
• Rules, advantages, interpretation, when to use it
• Profile vs. a composite profile tolerance
Course summary, final learning assessment
All of ETI's instructors are industry professionals with years experience applying GD&T on the job. ETI trainers have:
• Expert knowledge of the Y14.5 Standard
• ASME certified and/or ASQ certified
• Current or recent industrial experience using GD&T
• At least five years experience using GD&T
• Experience and skill using ETI teaching materials
Our instructors use identical training materials and lesson plans, so you receive the same class presentation from every trainer.
Hotel & Travel Information
Fees: \$1280.00
SAE Members: \$1024.00 - \$1152.00
1.3 CEUs
You must complete all course contact hours and successfully pass the learning assessment to obtain CEUs.
For additional information, contact SAE Customer Service 1-877-606-7323 (724-776-4970 outside the U.S. and Canada) or at CustomerService@sae.org.
Duration: 2 Days
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If z - 4z > 5 then which of the following is always true : PS Archive
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# If z - 4z > 5 then which of the following is always true
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If z² - 4z > 5 then which of the following is always true
A) z > -5
B) z < 5
C) z > -1
D) z < 1
E) z < -1
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### Show Tags
14 Jul 2006, 18:17
u2lover wrote:
If z² - 4z > 5 then which of the following is always true
A) z > -5
B) z < 5
C) z > -1
D) z < 1
E) z < -1
(E) it is.
Its easy to see that inequality is false for z = 0, only (E) satisfies non-zero values of z..
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### Show Tags
22 Jul 2006, 12:19
E.
Z > 5 can also be the answer but its not in the choices.
C could have been the answer but when z = 0 then the equality doesn't hold true.
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22 Jul 2006, 12:19
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Sathyabama University Previous Years Question Papers
Transmission And Distribution BE Fifth Sem
1. Define single line diagram.
2. State Kelvin’s law.
3. Define GMD.
4. Mentions the factors on which the corona effect depends.
6. What do you mean by load compensation?
7. Draw the structure of a typical cable.
8. Define String Efficiency.
9. Define BIL.
10. What are arcing horns?
PART – B (5 x 12 = 60)
transmission.
(or)
12. Mention the various types of bus bar arrangements. Explain the
double bus bar scheme with sectionalisation with the neat sketch.
What is the use of bus coupler?
13. (a) Derive the inductance of single phase two wire transmission
line. (4)
(b) Calculate the inductance/phase of the 3 phase transmission
line with conductors A, B, C placed in an equilateral triangle 1 m
of its sides. The radius of the conductor is 20mm. (8)
(or)
14. (a) Explain the phenomena of corona and the conditions affecting
corona loss. (5)
(b) A 3 phase OH line is being supported by 4 disc insulators.
The self capacitance is equal to 10 times the mutual capacitance.
Find (i) the voltage distribution across various units expressed as
a percentage of total voltage across the string and
(ii) String efficiency. (7)
15. A 200km long, 3 phase overhead line has a resistance of 48.7
ohm/phase, inductive reactance of 80.20 ohm/phase and
capacitance (line to neutral) 8.42nF/km. It supplies a load of 13.5
MW at a voltage of 132kV and power factor 0.86 lagging. Use
rigorous method and hence find the sending end voltage, current,
regulation and power angle.
(or)
16. Explain the procedural steps for constructing the receiving end
power circle diagram. Also explain how to determine the capacity
of phase modifier from the circle diagram.
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# FIND THE MISSING SIDES AND ANGLES USING SIN LAW
Find the Missing Sides and Angle Using Sin Law :
The sine law is a relationship between the sides and angles in any triangle.
Let ABC be any triangle, where a, b, and c represent the measures of the sides opposite ∠A, ∠B, and ∠C, respectively. Then
a/sin A = b/sin B = c/sin C
(or)
sin A/a = sin B/b = sin C/c
## Find the Missing Sides and Angle Using Sin Law - Examples
Question 1 :
Determine the value of the marked unknown angle in each.
Solution :
(a) From the picture given above, we know that we need to find the angle C.
BC = a, AC = b = 31 m and AB = c = 28 m
<B = 62
a/sin A = b/sin B = c/sin C
31/sin 62 = 28/sin C
31/0.8829 = 28/sin C
35.11 = 28/sin C
sin C = 28/35.11
sin C = 0.7974
C = 52.88°
Hence the required angle is 53° (approximately).
(b) BC = a = 15, AC = b = 17.5 m and AB = c
<B = 98, <A = ?
a/sin A = b/sin B = c/sin C
15/sin A = 17.5/sin 98
15/sin A = 17.5/0.9902
15/sin A = 17.67
sin A = 15/17.67
sin A = 0.8488
Hence the required angle is 58° (approximately).
Question 2 :
Determining the lengths of all three sides and the measures of all three angles is called solving a triangle. Solve each triangle
Solution :
(a) AB = c = 13 m, BC = a, AC = b = 12 m
<B = 67
a/sin A = b/sin B = c/sin C ----(1)
a/sin A = 12/sin 67 = 13/sin C
12/sin 67 = 13/sin C
12/0.9205 = 13/sin C
13.03 = 13/sin C
sin C = 13/13.03
sin C = 0.9976
<C = 86
In triangle ABC,
<A + <B + <C = 180
<A + 67 + 86 = 180
<A + 153 = 180
<A = 180 - 153
<A = 27
a/sin 27 = 12/sin 67 = 13/sin 86
a/sin 27 = 12/sin 67
a/0.4539 = 13.03
a = 13.03(0.4539)
a = 5.91 approximately 6 m
Hence the missing side and missing angles are 6 m and 86 respectively.
(b) AB = c, BC = a, AC = b = 50 m
<A = 42, <B = 84
a/sin A = b/sin B = c/sin C
a/sin 42 = 50/sin 84 = c/sin C
a/sin 42 = 50/sin 84
a/0.6691 = 50/0.9945
a = 50.27 (0.6691)
a = 33.63 approximately 33.6 m
In triangle ABC,
<A + <B + <C = 180
42 + 84 + <C = 180
<C = 180 - 126
<C = 54
Hence the missing side and missing angle are 33.6 m and 54 degree.
After having gone through the stuff given above, we hope that the students would have understood "Find the Missing Sides and Angles Using Sin Law".
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# TIME Units Conversionpetaseconds to yoctoseconds
1 Petaseconds
= 1.0E+39 Yoctoseconds
Category: time
Conversion: Petaseconds to Yoctoseconds
The base unit for time is seconds (SI Unit)
[Petaseconds] symbol/abbrevation: (Ps)
[Yoctoseconds] symbol/abbrevation: (ys)
How to convert Petaseconds to Yoctoseconds (Ps to ys)?
1 Ps = 1.0E+39 ys.
1 x 1.0E+39 ys = 1.0E+39 Yoctoseconds.
Always check the results; rounding errors may occur.
Definition:
In relation to the base unit of [time] => (seconds), 1 Petaseconds (Ps) is equal to 1.0E+15 seconds, while 1 Yoctoseconds (ys) = 1.0E-24 seconds.
1 Petaseconds to common time units
1 Ps = 1.0E+15 seconds (s)
1 Ps = 16666666666667 minutes (min)
1 Ps = 277777777777.78 hours (hr)
1 Ps = 11574074074.074 days (day)
1 Ps = 1653439153.4392 weeks (wk)
1 Ps = 31709791.983765 years (yr)
1 Ps = 380517503.80518 months (mo)
1 Ps = 3170577.0450222 decades (dec)
1 Ps = 317057.70450222 centuries (cent)
1 Ps = 31705.770450222 millenniums (mill)
Petasecondsto Yoctoseconds (table conversion)
1 Ps = 1.0E+39 ys
2 Ps = 2.0E+39 ys
3 Ps = 3.0E+39 ys
4 Ps = 4.0E+39 ys
5 Ps = 5.0E+39 ys
6 Ps = 6.0E+39 ys
7 Ps = 7.0E+39 ys
8 Ps = 8.0E+39 ys
9 Ps = 9.0E+39 ys
10 Ps = 1.0E+40 ys
20 Ps = 2.0E+40 ys
30 Ps = 3.0E+40 ys
40 Ps = 4.0E+40 ys
50 Ps = 5.0E+40 ys
60 Ps = 6.0E+40 ys
70 Ps = 7.0E+40 ys
80 Ps = 8.0E+40 ys
90 Ps = 9.0E+40 ys
100 Ps = 1.0E+41 ys
200 Ps = 2.0E+41 ys
300 Ps = 3.0E+41 ys
400 Ps = 4.0E+41 ys
500 Ps = 5.0E+41 ys
600 Ps = 6.0E+41 ys
700 Ps = 7.0E+41 ys
800 Ps = 8.0E+41 ys
900 Ps = 9.0E+41 ys
1000 Ps = 1.0E+42 ys
2000 Ps = 2.0E+42 ys
4000 Ps = 4.0E+42 ys
5000 Ps = 5.0E+42 ys
7500 Ps = 7.5E+42 ys
10000 Ps = 1.0E+43 ys
25000 Ps = 2.5E+43 ys
50000 Ps = 5.0E+43 ys
100000 Ps = 1.0E+44 ys
1000000 Ps = 1.0E+45 ys
1000000000 Ps = 1.0E+48 ys
(Petaseconds) to (Yoctoseconds) conversions
:)
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# Adjusted Present Value Calculator | APV Calculator
The Adjusted Present Value Calculator (APV Calculator) allows you to calculate the APV based on Net Present Value (NPV) or investment adjusted for the interest and tax advantages of leveraging debt provided that equity is the only source of financing
🖹 Normal View🗖 Full Page View Calculator Precision (Decimal Places)012345678910 Project Cost (pc) Cash Flow (cf) Interest Expense (e) Risk Rate (r) % Asset Beta (a) Market Return (mr) % Cost of Debt (cod) % Tax Rate (t) %
Present Value of Cash Flows formula and calculations The Present Value of Cash Flows ( pvcf ) is The Present Value of Tax Shield ( pvts ) is The Adjusted Present Value ( apv ) is pvcf = cf/(r + a) × (mr - r)/100 - pc/( + ) × ( - )/100 - / × /100 - //100 - / - - pvcf = pvts = e × t/codpvts = × %/%pvts = × /pvts = × pvts = apv = pvcf + pvtsapv = + apv = Project Cost (pc) Cash Flow (cf) Interest Expense (e) Risk Rate (r) % Asset Beta (a) Market Return (mr) % Cost of Debt (cod) % Tax Rate (t) %
## Why do companies use Adjusted Present Value?
A company can finance a project/investment using only shareholder equity without leverage/borrowed, cash flows. This provides unleveraged cash flow from shareholder equity
The company then repays associated debts using the unleveraged cash flow from shareholder equity.
This benefits the company by leveraging tax deductions on the interest part of these payments.
This helps to increase the project/investment bottom line which in turn puts the company at an advantage as far as the project/investment profitability.
The company can analyse the project/investment profitability using the adjusted present value (APV) including adjusting tax benefits from interest obligations on associated outstanding debts and subsequently comparing these against the return on alternate cash flows for funding the project/investment.
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The first balloon passengers were a sheep, a duck, and a : GMAT Sentence Correction (SC)
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# The first balloon passengers were a sheep, a duck, and a
Author Message
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Manager
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The first balloon passengers were a sheep, a duck, and a [#permalink]
### Show Tags
16 May 2007, 04:32
00:00
Difficulty:
(N/A)
Question Stats:
0% (00:00) correct 0% (00:00) wrong based on 1 sessions
### HideShow timer Statistics
The first balloon passengers were a sheep, a duck, and a rooster that were ascended from Versailles, France, in a basket beneath a hot-air balloon in September 1783.
A.
B. The first balloon passengers were a sheep, a duck, and a rooster that ascended
C. A sheep, a duck, and a rooster were the first balloon passengers that ascended
D. A sheep, a duck, and a rooster that were the first balloon passengers to ascend
E. A sheep, a duck, and a rooster were the first balloon passengers ascending
a tough one
If you have any questions
New!
Manager
Joined: 14 May 2007
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### Show Tags
16 May 2007, 04:47
Yup its a bit tough !!
List is ending properly with an "and" in all options
E - wrong ( ascending - wrong tense )
D - wrong ( "that were the first balloon passengers" dont need "that"
can directly say "were the first balloon passengers" )
its tough between B & C ( there is no option A )
I will go with B
My Reason:
In B the subject is "The first balloon passengers" which is OK !!
but
In C the subject is a compound subject that needs "and"
(e.g. x and y and z ....)
thats what i think ....
Senior Manager
Joined: 27 Mar 2007
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### Show Tags
16 May 2007, 05:47
(B) seems correct!
(C), (D) and ( E) changes meaning! these answers sounds like the animals were the first passanger that ascended from Versailles, but in fact they werde the first passangers ever in a baloon! (A) has passive voice and sounds awkward.
cheers
Manager
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### Show Tags
16 May 2007, 06:46
i disagree with u fellas.....
b says 'a rooster that'...... was it the rooster only who ascended?
b is definitly out.....C remains.....
thats wot I think...... pls correct me.....
_________________
Eat like a Pig, Lift like a Demon & Sleep like Dead.............
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### Show Tags
21 May 2007, 17:55
Even i don't see any reason why B is right and c is not.
"that" can refer only immediate precedence, so in B it does not refer to all of them
VP
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### Show Tags
21 May 2007, 19:17
C for the same reason as pointed out by devansh_god.
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### Show Tags
21 May 2007, 19:29
Jamesk486 wrote:
The first balloon passengers were a sheep, a duck, and a rooster that were ascended from Versailles, France, in a basket beneath a hot-air balloon in September 1783.
A.
B. The first balloon passengers were a sheep, a duck, and a rooster that ascended
C. A sheep, a duck, and a rooster were the first balloon passengers that ascended
D. A sheep, a duck, and a rooster that were the first balloon passengers to ascend
E. A sheep, a duck, and a rooster were the first balloon passengers ascending
a tough one
C for me.
In B), 'that' refers to the rooster, not all the passengers.
Manager
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### Show Tags
21 May 2007, 21:02
B is my choice.
As mentioned in one of the earlier posts, that refers "a sheep, a duck, and a rooster".
Problem with C is that "that" refers to "passengers".
Manager
Status: Post MBA, working in the area of Development Finance
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### Show Tags
21 May 2007, 22:15
that refers to "passengers" which comprises the three beings.
So, B seems to be correct.
What's the OA?
devansh_god wrote:
i disagree with u fellas.....
b says 'a rooster that'...... was it the rooster only who ascended?
b is definitly out.....C remains.....
thats wot I think...... pls correct me.....
Manager
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### Show Tags
22 May 2007, 09:01
Artemov wrote:
that refers to "passengers" which comprises the three beings.
So, B seems to be correct.
What's the OA?
devansh_god wrote:
i disagree with u fellas.....
b says 'a rooster that'...... was it the rooster only who ascended?
b is definitly out.....C remains.....
thats wot I think...... pls correct me.....
Doesnt A fit here...Actually we need passive voice here as the animals did not ascend themselves....Probably thinking out of the box
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### Show Tags
22 May 2007, 17:46
It would be C if there was 'who' instead of 'that'.
Manager
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### Show Tags
22 May 2007, 21:06
The first balloon passengers were a sheep, a duck, and a rooster that were ascended from Versailles, France, in a basket beneath a hot-air balloon in September 1783.
A.
B. The first balloon passengers were a sheep, a duck, and a rooster that ascended
A is def. out because of the use of "were" before ascended. If the sentence is read as
The first balloon passengers....that ascended..
Choice B seemed appropriate.
[It must be noted, though, that the omitted portion is not an independent clause]
vij101 wrote:
Artemov wrote:
that refers to "passengers" which comprises the three beings.
So, B seems to be correct.
What's the OA?
devansh_god wrote:
i disagree with u fellas.....
b says 'a rooster that'...... was it the rooster only who ascended?
b is definitly out.....C remains.....
thats wot I think...... pls correct me.....
Doesnt A fit here...Actually we need passive voice here as the animals did not ascend themselves....Probably thinking out of the box
Director
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### Show Tags
22 May 2007, 21:10
Jamesk486 wrote:
The first balloon passengers were a sheep, a duck, and a rooster that were ascended from Versailles, France, in a basket beneath a hot-air balloon in September 1783.
A.
B. The first balloon passengers were a sheep, a duck, and a rooster that ascended
C. A sheep, a duck, and a rooster were the first balloon passengers that ascended
D. A sheep, a duck, and a rooster that were the first balloon passengers to ascend
E. A sheep, a duck, and a rooster were the first balloon passengers ascending
a tough one
C. only has clear pronoun "that" thatr refers to passangers. So C should be the answer.
Intern
Joined: 07 Jan 2007
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22 May 2007, 21:12
This is a good one! My choice is C.
Can someone please post the OA?
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24 May 2007, 09:46
I think the ans is B...
except A and B, all other options distort the meaning, implying that sheep, duct and rooster were the first passengers to ascend from Versailles.
Sheep, duct and rooster were the first balloon passengers and it was the balloon that ascended from Versailles
between A and B, A is gramatically incorrect.
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24 May 2007, 10:17
A. The first balloon passengers were a sheep, a duck, and a rooster that were ascended – sentence construction – too long.
B. The first balloon passengers were a sheep, a duck, and a rooster that ascended
C. A sheep, a duck, and a rooster were the first balloon passengers that ascended
D. A sheep, a duck, and a rooster that were the first balloon passengers to ascend – Misplaced Pronoun
E. A sheep, a duck, and a rooster were the first balloon passengers ascending – what does that mean? Verb Tense.
Ok so A, D & E can be eliminated straight away.
left with B & C
B - Changes the context slightly. It seems to say that the first passengers a sheep and a duck but rooster was the one who ascended!!!
So I am left with C which says it quite correctly. With the pronouns in the right place.
Hence C is the correct answer.
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24 May 2007, 10:17
Jamesk486 wrote:
The first balloon passengers were a sheep, a duck, and a rooster that were ascended from Versailles, France, in a basket beneath a hot-air balloon in September 1783.
A.
B. The first balloon passengers were a sheep, a duck, and a rooster that ascended
C. A sheep, a duck, and a rooster were the first balloon passengers that ascended
D. A sheep, a duck, and a rooster that were the first balloon passengers to ascend
E. A sheep, a duck, and a rooster were the first balloon passengers ascending
a tough one
I will go with C.
B) Akward construction
C) Perfectly manages the tense.
D) Changes the tense and meaning.
E) Changes the tense and meaning.
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29 May 2007, 19:21
In this sentence, subject(noun) is passengers ,and that refers to passegers. Option C should be correct choice.
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30 May 2007, 10:07
A real good one. Between B and C , I am still not able to digest ".. rooster that ...' in B.
Still need more logic to negate C as an option.
30 May 2007, 10:07
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# Date differences in MS Access
LillyC used Ask the Experts™
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How to calculate the number of years and months between [Start Date] and [End Date] within a query.
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EE Solution Guide - Technical Dept Head
Most Valuable Expert 2017
Commented:
You can use DATEDIFF to calculate the difference between 2 dates.
https://www.experts-exchange.com/questions/27601269/Access-2007-How-to-calculate-YEARS-of-Service-in-company.html
Database Developer
Commented:
Thanks for your help but I'm trying to find a solution via a query not by coding if possible.
Most Valuable Expert 2012
Top Expert 2014
Commented:
You can use DateDiff in a query:
SELECT DateDiff("m", DateField1, DateField2) AS DateDifference FROM SomeTable
That would give you the number of Months between the two. You can divide by 12 to get the number of Years, and then do some subtraction to get any remainder months:
SELECT (DateDiff("m", DateField1, DateField2))\12 AS YearDiff, (((DateDiff("m", DateField1, DateField2))/12) - ((DateDiff("m", DateField1, DateField2))\12)) AS MonthDiff FROM SomeTable
You may also be able to use the MOD function to get the Months, but that could perform rounding (which you wouldn't want).
DateDiff: https://support.office.com/en-us/article/datediff-function-e6dd7ee6-3d01-4531-905c-e24fc238f85f
Owner, Dev-Soln LLC
Most Valuable Expert 2014
Top Expert 2010
Commented:
keep in mind that the difference between 12/31/17 and 1/1/18 is one month:
?datediff("m", #12/31/17#, #1/1/18#)
so as long as you are good with that then Scott's answer would be a good start.
Another method would be to create a function you could call from you query and do the processing there.
You could pass in the start and end date, then keep adding a month to the start date and count the number of increments until the
DateAdd("m", intLoop, StartDate) > EndDate
Then compute years and months as Scott describes above.
You could also try:
DateDiff("m", StartDate, EndDate) + (Day(EndDate) < Day(StartDate))
which would subtract 1 from the number of months if the day of the month of the EndDate is less than the day of the month of the start date.
Most Valuable Expert 2015
Distinguished Expert 2018
Commented:
I'm trying to find a solution via a query not by coding if possible.
You can't. You will have to use a function like this to get it right. And, after all, it's nothing more than a copy-n-paste into a module:
``````Public Function Months( _
ByVal datDate1 As Date, _
ByVal datDate2 As Date, _
Optional ByVal booLinear As Boolean) _
As Integer
' Returns the difference in full months between datDate1 and datDate2.
'
' Calculates correctly for:
' negative differences
' leap years
' dates of 29. February
' date/time values with embedded time values
' negative date/time values (prior to 1899-12-29)
'
' Optionally returns negative counts rounded down to provide a
' linear sequence of month counts.
' For a given datDate1, if datDate2 is decreased stepwise one month from
' returning a positive count to returning a negative count, one or two
' occurrences of count zero will be returned.
' If booLinear is False, the sequence will be:
' 3, 2, 1, 0, 0, -1, -2
' If booLinear is True, the sequence will be:
' 3, 2, 1, 0, -1, -2, -3
'
' If booLinear is False, reversing datDate1 and datDate2 will return
' results of same absolute Value, only the sign will change.
' This behaviour mimics that of Fix().
' If booLinear is True, reversing datDate1 and datDate2 will return
' results where the negative count is offset by -1.
' This behaviour mimics that of Int().
' DateAdd() is used for check for month end of February as it correctly
' returns Feb. 28. when adding a count of months to dates of Feb. 29.
' when the resulting year is a common year.
'
' 2010-03-30. Cactus Data ApS, CPH.
Dim intDiff As Integer
Dim intSign As Integer
Dim intMonths As Integer
' Find difference in calendar months.
intMonths = DateDiff("m", datDate1, datDate2)
' For positive resp. negative intervals, check if the second date
' falls before, on, or after the crossing date for a 1 month period
' while at the same time correcting for February 29. of leap years.
If DateDiff("d", datDate1, datDate2) > 0 Then
intSign = Sgn(DateDiff("d", DateAdd("m", intMonths, datDate1), datDate2))
intDiff = Abs(intSign < 0)
Else
intSign = Sgn(DateDiff("d", DateAdd("m", -intMonths, datDate2), datDate1))
If intSign <> 0 Then
' Offset negative count of months to continuous sequence if requested.
intDiff = Abs(booLinear)
End If
intDiff = intDiff - Abs(intSign < 0)
End If
' Return count of months as count of full 1 month periods.
Months = intMonths - intDiff
End Function
``````
That will returns the exact full count of months. To obtain years and months, do:
``````TotalMonths = 37 ' example
Years = TotalMonths \ 12
Months = TotalMonths Mod 12
``````
Thus, your query could look like:
``````Select *, CStr(Months([Start Date], [End Date]) \ 12) & " years, " & CStr(Months([Start Date], [End Date]) Mod 12) & " months" As YearMonth
From YourTable
``````
Top Expert 2014
Commented:
This problem is also related to the age calculation problem. If you are born on a date (DOB), what is your age, in years, on a later date? I normally code a VBA function for this.
@LillyC
What do you need the year_count/month_count answer to be for the following:
1. 12/31/17, 1/1/18 - Dale's example
2. 12/15/17, 1/14/18
3. 12/15/17, 1/15/18
4. 12/15/17, 1/16/18
5. 12/15/17, 2/14/19
6. 12/15/17, 2/15/19
7. 12/15/17, 2/16/19
Retired IT Professional
Commented:
upload a sample database.
Include a table with a few records.
Include a table showing manually the required output.
Most Valuable Expert 2015
Distinguished Expert 2018
Commented:
The results for those dates will be:
``````1: 0 years, 0 months
2: 0 years, 0 months
3: 0 years, 1 months
4: 0 years, 1 months
5: 0 years, 1 months
5: 0 years, 2 months
6: 0 years, 2 months
``````
Top Expert 2014
Commented:
Gustav
I think you should double-check those results :-)
Most Valuable Expert 2015
Distinguished Expert 2018
Commented:
Ah, missed the 19:
``````1: 0 years, 0 months
2: 0 years, 0 months
3: 0 years, 1 months
4: 0 years, 1 months
5: 1 years, 1 months
5: 1 years, 2 months
6: 1 years, 2 months
``````
Thanks!
Database Developer
Commented:
Thanks Scott your solution works well for me. Thanks everyone else for you input.
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