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https://oeis.org/A349895
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The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A349895 Length of the longest self avoiding walk through a grid such that either x or y is changed by +1 or -1 in each step, and with 0 <= y, 0 <= x <= y, x + y <= n starting at (0,0) and terminating at (x,y) = (n,0). 0
0, 1, 2, 5, 6, 9, 12, 17, 20, 27, 30, 39, 42, 51, 56, 67, 72, 85, 90, 105, 110, 125, 132, 149, 156, 175, 182, 203, 210, 231, 240, 263, 272, 297, 306, 333, 342, 369, 380, 409, 420, 451, 462, 495, 506, 539, 552, 587, 600, 637, 650, 689, 702, 741, 756, 797, 812 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,3 COMMENTS The number of reachable nodes in the grid is given by A002620(n+2). This sequence is inspired by the "Cinnamon Stars" problem given in the link. In this sequence an additional restriction that the walk must terminate at (n,0) has been added. LINKS Die Mathe-Adventskalender, Cinnamon Stars Index entries for linear recurrences with constant coefficients, signature (1,1,-1,0,0,0,0,1,-1,-1,1). FORMULA a(n) < A002620(n+2). From Andrew Howroyd, Dec 18 2021: (Start) a(2*n) = n*(n + 1); a(2*n-1) = n^2 + n - 1 - 2*floor((n+1)/4). G.f.: x*(1 + x + 2*x^2 + 2*x^5 + 2*x^6 + x^8 - x^9)/((1 - x)^3*(1 + x)^2*(1 + x^2)*(1 + x^4)). a(n) = a(n-1) + a(n-2) - a(n-3) + a(n-8) - a(n-9) - a(n-10) + a(n-11) for n >= 11. (End) EXAMPLE a(5) = 9. An optimal path is illustrated below: o---o | | o---o o o | | o---o o o---o---o . For n<13, an optimal path can be constructed by moving in y direction as far as possible (also avoiding dead ends), then moving one step in positive x direction, and going back in the other y direction. For example, a(6) = 12: o . o o---o | | o---o o o o | | | | o---o o---o o---o---o . From Andrew Howroyd, Dec 18 2021: (Start) For even n, if vertices are colored alternately black and white there will be (n+2)*(n+4)/8 black vertices and n*(n+2)/8 white vertices. Since the walk must alternate between the two colors, the maximum length of the walk is limited to n*(n+2)/4. An optimal construction for a(10) is shown below. Many other solutions exist, but they all have the property that every white vertex is visited. x . x o---x | | x---o x o x | | | | x o x o x o---x | | | | | | x---o x o x o x o x | | | | | | | | x---o x---o x---o x---o x---o---x . For odd n, there will be an equal number of both colors. However, there is an imbalance between left and right halves that must be mitigated. The optimal solution is to fill in along the dividing line as shown below for a(11) = 39. In the illustration, black vertices are marked with x and white with o. Notice that the connections across the dividing line are always between a black vertex on the left and a white on the right. Even with the central crossings placed there will be more black vertices than white on the left (and vice versa on the right). An optimal solution is one in which every white vertex on the left side and every black vertex on the right side is included in the walk. x---o | | x---o x---o | | x---o x---o x---o | | | | x o x---o x---o---x o | | x---o x o x---o x---o x---o | | | | | | | | | | x---o x---o x---o x---o x---o x---o (End) PROG (PARI) a(n)=if(n%2, (n^2 + 4*n - 1)/4 - (n+3)\8*2, n*(n/2 + 1)/2) \\ Andrew Howroyd, Dec 18 2021 CROSSREFS Cf. A002620. Sequence in context: A050487 A046962 A022486 * A267700 A161910 A151920 Adjacent sequences: A349892 A349893 A349894 * A349896 A349897 A349898 KEYWORD easy,nonn,walk AUTHOR Paul Bischof, Dec 04 2021 EXTENSIONS Terms a(17) and beyond from Andrew Howroyd, Dec 18 2021 STATUS approved
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Last modified August 11 00:55 EDT 2022. Contains 356046 sequences. (Running on oeis4.)
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# What Is Pie Math – Is it a Scam?
## The Do’s and Don’ts of What Is Pie Math
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Exactly like solids, it’s the quantity of space it takes up, but obviously it must be in some type of container. Distinct shapes have various techniques to discover the volume. A pie is created from a recipe, whilst pi is utilized in functions.
A recipe contains the instructions and ingredients required for cooking a meal. Then earn a bar or pie chart showing the 2 amounts. Case 1 If there’s a way to assign a different slice of pie to every one of the N-1 choosers, then that needs to be carried out.
Each name has to be on a distinct line. same-day-essay net There isn’t anything like a tummy full of pie to genuinely receive a concept home. Now you are prepared to begin drawing!
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Not a lot of men and women know that! You could talk to a tattoo expert about the chance of getting it tattooed on your entire body. For example, if you multiply 111,111,111 by 111,111,111 you receive the answer 12,345,678,987,654,321.
Aside from these domains, the applications don’t work with different kinds of math, like Chemistry. Algebra is often taught abstractly with minimal or no emphasis on what algebra is or the way that it may be used to fix real difficulties. The issue with the series above is that you have to add up lots of terms as a way to find an accurate approximation of Pi ().
If however you are able to add up the very first few stipulations, you will start to find an approximation for Pi (). Pi has for ages been noted among the most useful mathematical constants.
## Definitions of What Is Pie Math
Pie charts are useful to compare distinctive components of an entire quantity. Data is classified into qualitative or quantitative depending on the essence of the information it supplies. Type the normal deviations below.
Those eight numbers are only the outset of pi’s true price. Browse our categories to discover the worksheet you are searching for or utilize search option on the top to look for any worksheet you want. You might have tried to go into a formula into a text box, simply to discover that the outcomes of the formula is not going to calculate.
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## The History of What Is Pie Math Refuted
As in Part 1, the fix is actually determined by your sort of environment and the way in which your VMs are configured. A great tutorial about the standard distribution was added. It cannot be routed to specific program usage, however much political will you have.
The ideal thing about this application is the fact that it not only provides you with a response to the mathematical or physical numerical but in addition provides you a step-by-step solution to each problem. The benefit of this application is that you may quickly draw up your necessary formula and insert it in your preferred document. Click the tool, then opt for the character that you want to correct.
## The Foolproof What Is Pie Math Strategy
Math Playground enables you to explore many heights of math in many fun ways. Our site provides a wide assortment of Free Math Help resources, so please search around to obtain what you want. There are a lot of exciting characteristics which make Math Blaster a great game for kids.
## What Is Pie Math – Dead or Alive?
When 6 tractors work together, every one of them ploughs 120 hectares every day. We now know precisely what the number you finished with is. Rational numbers sound as though they have to be quite sensible numbers.
## Facts, Fiction and What Is Pie Math
Next, it’s going to figure out the area of circle in compliance with the formula we shown above. The object’s initial form and position is known as the pre-image while the last position and contour of the manipulated object is known as the image below the transformation. Choose the array of cells that you need to combine together Apply exactly the same formula to the remainder of the cells by dragging the lower right corner downwards.
The faces of the rectangle are defined with regard to that unknown length. Unlike triangles, rectangles, and other similar figures, the distance around the exterior of the circle is known as the circumference instead of the perimeter-the idea, nevertheless, is basically the exact same. Draw lines from the middle of the circle to these two point to produce the slice.
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Discover a lot of information on the number 21058: properties, mathematical operations, how to write it, symbolism, numerology, representations and many other interesting things!
## Mathematical properties of 21058
Is 21058 a prime number? No
Is 21058 a perfect number? No
Number of divisors 4
List of dividers 1, 2, 10529, 21058
Sum of divisors 31590
Prime factorization 2 x 10529
Prime factors 2, 10529
## How to write / spell 21058 in letters?
In letters, the number 21058 is written as: Twenty-one thousand fifty-eight. And in other languages? how does it spell?
21058 in other languages
Write 21058 in english Twenty-one thousand fifty-eight
Write 21058 in french Vingt et un mille cinquante-huit
Write 21058 in spanish Veintiuno mil cincuenta y ocho
Write 21058 in portuguese Vinte e um mil cinqüenta e oito
## Decomposition of the number 21058
The number 21058 is composed of:
1 iteration of the number 2 : The number 2 (two) represents double, association, cooperation, union, complementarity. It is the symbol of duality.... Find out more about the number 2
1 iteration of the number 1 : The number 1 (one) represents the uniqueness, the unique, a starting point, a beginning.... Find out more about the number 1
1 iteration of the number 0 : ... Find out more about the number 0
1 iteration of the number 5 : The number 5 (five) is the symbol of freedom. It represents change, evolution, mobility.... Find out more about the number 5
1 iteration of the number 8 : The number 8 (eight) represents power, ambition. It symbolizes balance, realization.... Find out more about the number 8
Other ways to write 21058
In letter Twenty-one thousand fifty-eight
In roman numeral
In binary 101001001000010
In octal 51102
In US dollars USD 21,058.00 (\$)
In euros 21 058,00 EUR (€)
Some related numbers
Previous number 21057
Next number 21059
Next prime number 21059
## Mathematical operations
Operations and solutions
21058*2 = 42116 The double of 21058 is 42116
21058*3 = 63174 The triple of 21058 is 63174
21058/2 = 10529 The half of 21058 is 10529.000000
21058/3 = 7019.3333333333 The third of 21058 is 7019.333333
210582 = 443439364 The square of 21058 is 443439364.000000
210583 = 9337946127112 The cube of 21058 is 9337946127112.000000
√21058 = 145.11374848718 The square root of 21058 is 145.113748
log(21058) = 9.9550358144167 The natural (Neperian) logarithm of 21058 is 9.955036
log10(21058) = 4.3234171213504 The decimal logarithm (base 10) of 21058 is 4.323417
sin(21058) = 0.09541165464606 The sine of 21058 is 0.095412
cos(21058) = -0.99543790170844 The cosine of 21058 is -0.995438
tan(21058) = -0.095848926871589 The tangent of 21058 is -0.095849
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DID YOU KNOW:
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# Triangles Bundle for High School Geometry
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### Description
Triangle Theorems & Properties, Congruent Triangles, Centers of Triangles, Triangle Proofs, & more!
Save money by getting eleven sets of resources in one bundle! These activities will help your students with triangles (from Triangle Sum Theorem all the way through to CPCTC proofs!)
(For an even bigger bundle that includes logic, proof, quadrilaterals, and more, try High School Geometry Super Bundle)
The zip file contains the following resources (all in PDF format):
1. Triangle Sum Theorem Inquiry Activity Lesson Pack
2. Exterior Angles Inquiry Activity
3. Classifying Triangles and Impossible Triangles (Always, Sometimes, Never Activity)
4. Congruent Triangles Activity
5. Congruent Triangles Proofs Worksheet Set
6. CPCTC Proofs Worksheet Set
7. Angle Puzzles: Angles in a Triangle
8. Similar Triangles Proofs Worksheet Set
9. Triangles Card Sort
10. Triangle Inequality Theorem Discovery
11. Centers of a Triangle "Doodle Notes"
12. Similar Triangles "Doodle Notes"
Take a look at the individual resources using the links below or the preview file.
Triangle Sum Theorem Inquiry Activity
Exterior Angles Inquiry Activity
Triangle Classification and Impossible Triangles
Congruent Triangles
Congruent Triangles Proofs
CPCTC Proofs
Angle Puzzles: Angles in a Triangle
Similar Triangles Proofs
Triangles Card Sort
Discovering Triangle Inequality Theorem
Centers of a Triangle Doodle Notes
Similar Triangles Doodle Notes
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### Standards
to see state-specific standards (only available in the US).
Use congruence and similarity criteria for triangles to solve problems and to prove relationships in geometric figures.
Use the properties of similarity transformations to establish the AA criterion for two triangles to be similar.
Use informal arguments to establish facts about the angle sum and exterior angle of triangles, about the angles created when parallel lines are cut by a transversal, and the angle-angle criterion for similarity of triangles. For example, arrange three copies of the same triangle so that the sum of the three angles appears to form a line, and give an argument in terms of transversals why this is so.
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# Singular
### 7.3.23 reduce (plural)
Syntax:
reduce ( poly_expression, ideal_expression )
reduce ( poly_expression, ideal_expression, int_expression )
reduce ( vector_expression, ideal_expression )
reduce ( vector_expression, ideal_expression, int_expression )
reduce ( vector_expression, module_expression )
reduce ( vector_expression, module_expression, int_expression )
reduce ( ideal_expression, ideal_expression )
reduce ( ideal_expression, ideal_expression, int_expression )
reduce ( module_expression, ideal_expression )
reduce ( module_expression, ideal_expression, int_expression )
reduce ( module_expression, module_expression )
reduce ( module_expression, module_expression, int_expression )
Type:
the type of the first argument
Purpose:
reduces a polynomial, vector, ideal or module to its left normal form with respect to an ideal or module represented by a left Groebner basis, if the second argument is a left Groebner basis.
returns 0 if and only if the polynomial (resp. vector, ideal, module) is an element (resp. subideal, submodule) of the ideal (resp. module).
Otherwise, the result may have no meaning.
The third (optional) argument 1 of type int forces a reduction which considers only the leading term and does no tail reduction.
Note:
The commands reduce and NF are synonymous.
Example:
ring r=(0,a),(e,f,h),Dp; matrix d[3][3]; d[1,2]=-h; d[1,3]=2e; d[2,3]=-2f; def R=nc_algebra(1,d); setring R; // this algebra is U(sl_2) over Q(a) ideal I = e2, f2, h2-1; I = std(I); // print a compact presentation of I print(matrix(I)); ==> h2-1,fh-f,f2,eh+e,2*ef-h2-h,e2 ideal J = e, h-a; J = std(J); // print a compact presentation of J print(matrix(J)); ==> h+(-a),e poly z=4*e*f+h^2-2*h; // z is the central element of U(sl_2) reduce(z,I); // the central character of I: ==> 3 reduce(z,J); // the central character of J: ==> (a2+2a) poly nz = z - NF(z,J); // nz will belong to J reduce(nz,J); ==> 0 reduce(I,J); ==> _[1]=(a2-1) ==> _[2]=(a-1)*f ==> _[3]=f2 ==> _[4]=0 ==> _[5]=(-a2+a) ==> _[6]=0
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http://www.vitutor.com/alg/linear_system/systems_parameters.html
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# Linear System of Equations with Parameters
#### 1. Solving a Linear System of Equations with Parameters by Cramer's Rule
1. Find the rank of the matrix of coefficients:
2. Find the rank of the augmented matrix:
3. Study the obtained information and determine what type of system it is:
4. Solve the compatible system by Cramer's rule (also can be solved by the Gauss Elimination method).
#### 2. Solving a Linear System of Equations with Parameters by the Gauss Elimination Method
It needs to be determined whether there is any value of m to make the system consistent. If so, solve the system for that value of m.
#### Examples
If a = 0.
If a = −3
a ≠ 1 r = 3 n = 3
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https://www.jiskha.com/display.cgi?id=1171269678
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# Physics/Math
posted by .
A certain string can withstand a maximum tension of 43 N without breaking. A child ties a 0.37 kg stone to one end and, holding the other end, whirls the stone in a vertical circle of radius 0.91 m, slowly increasing the speed until the string breaks. Where is the stone on its path when the string breaks? at a random point, lowest point, or highest point on the path, or cannot be determined? What is the speed of the stone as the string breaks?
## Similar Questions
1. ### Physics/Math
A certain string can withstand a maximum tension of 43 N without breaking. A child ties a 0.37 kg stone to one end and, holding the other end, whirls the stone in a vertical circle of radius 0.91 m, slowly increasing the speed until …
2. ### college physics
A stone with a mass of 0.800 is attached to one end of a string 0.800 long. The string will break if its tension exceeds 60.0 . The stone is whirled in a horizontal circle on a frictionless tabletop; the other end of the string remains …
3. ### physics
A certain string can withstand a maximum tension of 40 N without breaking. A child ties a 0.37 kg stone to one end and, holding the other end, whirls the stone in a vertical circle of radius 0.91 m, slowly increasing the speed until …
4. ### p
A certain string can withstand a maximum tension of 41 N without breaking. A child ties a 0.35 kg stone to one end and, holding the other end, whirls the stone in a vertical circle of radius 0.91 m, slowly increasing the speed until …
5. ### AP Physics
A stone is tied to a string (length = 0.450 m) and whirled in a circle at the same constant speed in two different ways. First, the circle is horizontal and the string is nearly parallel to the ground. Next, the circle is vertical. …
6. ### AP PHYSICS
A stone is tied to a string (length = 0.450 m) and whirled in a circle at the same constant speed in two different ways. First, the circle is horizontal and the string is nearly parallel to the ground. Next, the circle is vertical. …
7. ### physics
a stone with a mass of .700kg is attached to the end of a rting 0.900m long. the string will break if its tension exceed 60.0 N. the stone is whirled in a horizontal circle on a frictionless tabletop; the other of the string remain …
8. ### physics
a stone of mass 1 kg is tied to the end of a string 1m long . it is whirled in a vertical circle . if the velocity of the stone at the top be 4m/s , what is the tension in the string ?
9. ### Physics
A stone with mass 0.8kg is attached to one end of a string 0.9m long . The string will break if its tension exceeds 600N. The stone is whirled in a horizontal circle ,the other end of the string remains fixed. Find the maximum speed …
10. ### physics
A stone with a mass of 1.0 kg is tied to the end of a light string which keeps it moving in a circle with a constant speed of 4.0 m/s on a perfectly smooth horizontal tabletop. The radius of the path is 0.60 m. How much work does the …
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# Prime number algorithm
• khotsofalang
In summary, the conversation discusses the problem of finding solutions for the equation x12+x22=1 in a unit circle on a finite field Zp, where p is a prime number. The group discusses the possibility of using an algorithm to find all possible solutions and the complexity of such an algorithm. They also mention the importance of finding a good way to solve the equation for strengthening an extended essay. One member suggests checking all possibilities, while another suggests looking for quadratic residues that add up to 1, particularly for primes congruent to 1 modulo 4.
#### khotsofalang
i am sorry guys, the last time i posted this problem it was completely different but this time if we
Let x12+x22=1 be a unit circle upon a finite field Zp where p is prime. Is there any algorithm which can give all the possible solutions (x1,x2) an element of Zp*Zp as well as the total number of such solutions? If exists, what is the complexity of it?
You could check all possibilities. That takes something like O(p^2 log^2 p).
Now you just need a *good* way to solve it.
all right, but what i actually need is that good way of solving it
i need a solution to such an equation for stregthening my extended essay,anibody with a gud way of solving it?
If I understood correctly, you are looking for two quadratic residues that add up to 1. It may be easier for primes congruent to 1 modulo 4, because quadratic residues for those primes are 'symmetric': r is a quadratic residue iif p-r is. In this case you just look for contiguous quadratic residues on the lower half, from 2 to (p-1)/2: if r and r+1 are quadratic residues, then p-r also is, and (p-r)+(r+1) add to 1. My 2 cents.
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I wanted the left edge of the jaw of the front vise to be flush with the left edge of the top, the right edge with the left edge of the left front leg. So the amount of overhang on the left depends upon the width of the front vise jaw. The width of the jaw is, at a minimum, the width of the plate that supports it, but it's normal to make the jaw extend a bit beyond the plate. How far? The more it extends, the deeper a bite you can take with the edge of the vise, when, for example, you are clamping the side of a board being held vertically. But the more it extends, the less support it has. I decided to extend by 1-12", which gives me a 2-1/2" bite, and which should still provide solid support, given that the jaw is 1-1/2" thick. This means the top needs a left overhang of 12-1/4".
Every single person in the class was a novice when they walked in. We all made mistakes at some point, yet none of us really ever felt bad about these hiccups. Inevitably, Bob turned these mistakes into problem-solving opportunities where we could explore new ideas. So what if one morning I skipped coffee and accidentally spent 10 minutes drilling random-ass, pointless domino mortises in the middle of my bench legs? 4 dominos, some sawdust and glue, a little bit of sawing and sanding-- I've got four new inlays on my legs and nobody's the wiser. Not only do I now know how to roll with the punches when building something, but I've got a bench that's uniquely mine AND I don't feel like an idiot. And that last part is a serious feat.
Lay a leg flat on your work surface, with the countersink side of the thru-holes down. Stick a piece of threaded rod in each hole. Take a stretcher that is marked to have one end adjoin the top of this leg, stick a dowel center in its dowel hole, line it up against the leg, using the threaded rod for positioning, You want the top of the stretcher to be even with the top of the leg, or just slightly above it. Give the end of the stretcher a whack with your rubber mallet. This will leave a mark indicating where the matching dowel hole in the leg needs to be drilled. Repeat with the lower stretcher than adjoins this leg. Then repeat for the other leg that will form this trestle, and the other ends of the two stretchers.
Before you start cutting or drilling the pieces that will make up the top, determine the layout of the top. This should include the dimensions of the MDF, the dimensions of the edging, the locations of the vises, and of the screws or bolts that will support the vises, and of all of the benchdog holes and of all of the drywall screws you will use to laminate the panels,
I ended up making a number of practice cuts. The first revealed that I hadn't tightened the screws on the edge guide enough. The second revealed that the design of the edge guide provided very little support at the end of a board, because of the cut-out for the router bit. In the "Getting Started in Woodworking" video, they had screwed a piece of hardwood to the edge-guide, to provide a continuous -- and longer -- bearing surface. I may do that myself, some day, but I didn't have the materials at hand, so I clamped some 2x4 scrap to the end of each board, to provide a continuous bearing surface past the ends. The two grooves in the long stretchers and the side groove in the short stretchers have identical layout. I made practice cuts in scrap until I had the edge guide set correctly, then I cut them all with that one setting. The bottom groove of the short stretchers uses a different setup, so it was back to the scrap, before cutting them.
```Put the upper panel of MDF on your glue-up surface, bottom side up. Put the bottom panel of MDF on your other surface, bottom side down. (The panel with the holes drilled in it is the bottom panel, and the side that has the your layout diagram on it is the bottom side.) Chuck up in your drill the appropriate driver bit for the screws your using. Make sure you have a freshly-charged battery, and crank the speed down and the torque way down. You don't want to over-tighten the screws, MDF strips easily.
```
The Handy Home Products Berkley 10 ft. x The Handy Home Products Berkley 10 ft. x 10 ft. Wood Storage Building Kit is made with factory-primed SmartSide siding to resist fungal decay and wood-destroying insects. The Berkley’s gambrel style roof provides plenty of height to add a loft for extra storage space. The 6 ft. high side walls ... More + Product Details Close
Who doesn’t want to have one awesome and handy wooden desk organizer that not only looks beautiful but can store all your mini office desk items properly? See the picture below. I am sure you will love this one. I have already built one myself as I just could not resist having one at my office. This thing easily stores all my office desk essentials, including pen, pencils, marker, small notebooks, etc. in the most organized way. You can see it yourself.
This instructable shows how to build, with basic tools and readily-available lumber, a bench that provides most of the function of a traditional woodworker's workbench. I began with a design by Asa Christiana that was featured in the second season of finewoodworking.com's video series Getting Started in Woodworking. The project plans are available on their website.
Palomar Community College in San Marcos, located about 35 miles north of San Diego, offers certificate, degree and life-long learning programs to more than 20,000 students a year. In particular, the Cabinetmaking and Furniture Technology Department offers in excess of 50 classes annually, and the present faculty is composed of four full-time teachers, 21 part-time instructors and 11 teaching assistants. As of Fall 2013, woodworking students will have access to a newly remodeled woodworking facility equipped with three machine rooms, three bench rooms, a saw mill and a finish room.
What I did, when I came back, was to clamp down the strip where it had torn away, and then to start routing from the other end. I still moved the router from right to left, but I did it in six-inch sections, taking light passes, and sort of whittled the strip flush. As the sections I was working were farther to the right, the strip was thinner. Eventually I came to where I was trimming the strip away entirely, at which point I took off the clamps and the remainder fell away.
At the age of 9, Mark’s interest in woodworking began, which led him to begin an apprenticeship with Homestead Heritage Furniture at the age of 17. In his own shop he has built many pieces of furniture from small dovetail boxes to large 10′ round conference tables. Most of his work has been for clients who have ordered custom pieces for their homes and offices. Mark has been building custom furniture for the last 20 years and is now the manager of the Heritage Furniture business. Mark enjoys making hand tools and jigs that can simplify the building process. He has taught children how to shape simple utensils like wooden spoons and cutting boards as well as other larger projects. “Seeing the young grow to maturity in their craft is what I appreciate.”
Thinking about transitioning into serious woodworking, but don't have half a year or more to spare? Our one-month fine woodworking courses are perfect for those who are looking to move towards professional woodwork, but can’t find the time to commit to a longer course. Here, our woodworking classes will intricately teach you all of the basics, the same as you would for our three, six and twelve-month courses, with projects specially-created for both beginners and the more experienced. This will truly bring your woodworking skills up to the high level of craftsmanship needed within the industry.
I can't recommend this place highly enough. As someone who wants to test the waters in the hobby of woodworking this is the perfect place. They have all of the tools you need to complete your project, and they are great with helping beginners. I began as a skeptical customer but the staff was super ready and willing to make sure we could rectify a misunderstanding.
```Fundamentals Of Fine Woodworking With Taeho Kwon (Evening Class) The secret to fine woodworking can be found in the pursuit of the fundamentals. With high-tech machines and power tools, these fundamentals can be overshadowed. In this class we will learn the values of design, drawings, mock-ups, sharpening, care and use of hand tools, tuning up a bench...
```
Although many hand tool operations are thoroughly covered, this is a machine based woodworking experience. You will find the school finely equipped with four table saws, four jointers, three drill presses, three thickness planers, a wood lathe, full dust collection, bandsaws, jig saws and scroll saws, 5 routers and dozens of router jigs and accessories. There is a heavy focus on the safe and efficient use of the table saw starting with it's basic operation and going into advanced production procedures. We consider the table saw to be an essential machine for efficient, accurate, and safe woodworking.
```I live in the second rainiest city in the country, Pensacola, so I am used to dealing with wet. I literally work out of my garage (read storage area, small woodwork area, washer/dryer...car? ). I wheel bigger tools and table out to the little covered area in front of the garage. I have an 8 x 10 aluminum shed that is strictly garden stuff storage. I mention this for two reasons.
```
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# Bionary code. GENIUS TRICK - Convert Decimal Numbers To Binary (Base 2) - Duration: MindYourDecisions ,
## Bionary code. Computers 'think' in base two - binary code. Ones and zeros, on and off. Lightswitch analogy used to explain.
Finally, even though we're giving point values to each of those lightbulbs, when we write them down we still only write them as ones and zeros. One means On, and Zero means Off. So let's say we had 8 lightbulbs, and they were set up like this: The point values of those eight bulbs are: And that adds up to So we would say the sequence of bulbs is worth But how do we write it? We write it like this: And that's Binary Code. Go ahead and enter some text into the encoder.
The computer will convert those letters into numbers, and then it will convert those numbers into binary! So a word with 5 letters would take 40 lightbulbs! How many lightbulbs do you think it took to make this page? I bet it was a lot! Enter some text here to be converted to binary. Basic Explanation Did you know that everything a computer does is based on ones and zeroes? It's hard to imagine, because you hear people talking about the absolutely gargantuan huge numbers that computers "crunch".
But all those huge numbers - they're just made up of ones and zeros. It's kind of like the computer is made up of a bunch of lightswitches, and each lightswitch controls just one lightbulb. What do I mean by sequence? Let's say you had two lightswitches. There are four different ways we could flip those switches: We'll say the first lightbulb is worth two points, and the second one is worth one point.
Now take a look at the combinations: As you can see, it's going to take a lot of lightbulbs to make a really big number! If those two numbers are represented by lightbulbs, what is the value assigned to a bulb that is not turned on? What would be another way to write this: On Off On On Off? If you convert your name to binary using the encoder, what is the result? If you convert a 5-letter word to binary, how many digits does the result have?
Assign this reference page. Educators can get a free membership simply by sharing an original lesson plan on our Articles for Educators page! Share your lesson plan. Like us on Facebook to get updates about new resources.
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https://kmmiles.com/26077-km-in-miles
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26077 km in miles
Result
26077 km equals 16193.817 miles
Conversion formula
Multiply the amount of km by the conversion factor to get the result in miles:
26077 km × 0.621 = 16193.817 mi
How to convert 26077 km to miles?
The conversion factor from km to miles is 0.621, which means that 1 km is equal to 0.621 miles:
1 km = 0.621 mi
To convert 26077 km into miles we have to multiply 26077 by the conversion factor in order to get the amount from km to miles. We can also form a proportion to calculate the result:
1 km → 0.621 mi
26077 km → L(mi)
Solve the above proportion to obtain the length L in miles:
L(mi) = 26077 km × 0.621 mi
L(mi) = 16193.817 mi
The final result is:
26077 km → 16193.817 mi
We conclude that 26077 km is equivalent to 16193.817 miles:
26077 km = 16193.817 miles
Result approximation
For practical purposes we can round our final result to an approximate numerical value. In this case twenty-six thousand seventy-seven km is approximately sixteen thousand one hundred ninety-three point eight one seven miles:
26077 km ≅ 16193.817 miles
Conversion table
For quick reference purposes, below is the kilometers to miles conversion table:
kilometers (km) miles (mi)
26078 km 16194.438 miles
26079 km 16195.059 miles
26080 km 16195.68 miles
26081 km 16196.301 miles
26082 km 16196.922 miles
26083 km 16197.543 miles
26084 km 16198.164 miles
26085 km 16198.785 miles
26086 km 16199.406 miles
26087 km 16200.027 miles
Units definitions
The units involved in this conversion are kilometers and miles. This is how they are defined:
Kilometers
The kilometer (symbol: km) is a unit of length in the metric system, equal to 1000m (also written as 1E+3m). It is commonly used officially for expressing distances between geographical places on land in most of the world.
Miles
A mile is a most popular measurement unit of length, equal to most commonly 5,280 feet (1,760 yards, or about 1,609 meters). The mile of 5,280 feet is called land mile or the statute mile to distinguish it from the nautical mile (1,852 meters, about 6,076.1 feet). Use of the mile as a unit of measurement is now largely confined to the United Kingdom, the United States, and Canada.
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https://study.com/academy/answer/what-minimum-speed-does-a-300-g-puck-need-to-make-it-to-the-top-of-a-0-12-m-long-34-74-degrees-frictionless-ramp.html
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# What minimum speed does a 300 g puck need to make it to the top of a 0.12 m long, 34.74 degrees...
## Question:
What minimum speed does a 300 g puck need to make it to the top of a 0.12 m long, 34.74 degrees frictionless ramp?
## Conservation of Energy:
When no external forces are acting on an object, we can observe the law of conservation of energy on an object. This law states that the sum of the potential and kinetic energies are the same at two points in a system, or {eq}\displaystyle \sum E_i =\sum E_f {/eq}.
## Answer and Explanation:
Determine the speed of the puck, {eq}\displaystyle v {/eq}, by equating the initial kinetic energy, {eq}\displaystyle E_{kin}= \frac{1}{2}mv^2 {/eq} where {eq}\displaystyle m {/eq} is the mass of the puck, to the potential energy, {eq}\displaystyle E_{pot} = mgh {/eq} where {eq}\displaystyle g {/eq} is the gravitational acceleration and {eq}\displaystyle h {/eq} is the height. We can acquire the height at the top of the ramp through trigonomety, such that {eq}\displaystyle h = 0.12\ m\sin 34.74^\circ =0.068\ m {/eq}. We proceed to the solution.
{eq}\begin{align} \displaystyle \frac{1}{2}mv^2 &= mgh\\ v^2 &= 2gh\\ v&= \sqrt{2gh}\\ v&= \sqrt{2\cdot 9.81\ \rm{m/s^2}\cdot 0.068\ m }\\ v&\approx 1.2\ \rm{m/s} \end{align} {/eq}
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https://forum.bebac.at/forum_entry.php?id=18883
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## Now what? w & w* examples [Two-Stage / GS Designs]
Hi Ben,
❝ I hope there will be applications and more investigations in the future regarding this.
So do I – once we solved the mystery of finding a “suitable” n1 and specifying “appropriate” weights.
❝ ❝ Some regulatory statisticians told me to prefer a first stage as estimated for a fixed-sample design (i.e., the second stage is solely a ‘safety net’).
❝ Sounds interesting. At first this sounds nice, but I am a bit puzzled about it. "Safety net" sounds like we have a rather good understanding about the CV …
Not necessarily good but a “guesstimate”.
❝ … but in case we observe some unforeseen value we have the possibility to add some extra subjects.
❝ However, in such a case we could just go with a fixed design and adapt the Power.
I’m not sure what you mean here. In a fixed sample design I would rather work with the upper CL of the CV or – if not available – assume a reasonably higher CV than my original guess rather than fiddling around with power.
❝ In a TSD setting we typically have no good understanding about the CV... Do I miss something here?
(1) Yep and (2) no.
❝ Based on what assumptions would we select n1 (= fixed design sample size)? We typically have some range of possible values and we don't know where we will be.
I was just quoting a regulatory statistician (don’t want to out him). Others didn’t contradict him. So likely he wasn’t alone with his point of view.
❝ For n1 I would then rather use the lower end of this range. Comments?
Very interesting. I expected that the sample size penalty (n2) will be higher if we use a low n1. Of course it all depends on which CV we observe in stage 1.
library(PowerTOST)
library(Power2Stage)
CVguess <- 0.2 # from a fixed design study with n=18
CL <- CVCL(CV=CVguess, df=18-2, side="2-sided")
# Sample sizes of fixed desiogn based on guesstimate CV and its CL
n1.fix <- sampleN.TOST(CV=CVguess, print=FALSE)[["Sample size"]]
n1.75 <- floor(0.75*n1.fix) + as.integer(0.75*n1.fix) %%2
n1.lo <- sampleN.TOST(CV=CL[["lower CL"]], print=FALSE)[["Sample size"]]
n1.hi <- sampleN.TOST(CV=CL[["upper CL"]], print=FALSE)[["Sample size"]]
# In all variants use the guesstimate
x <- power.tsd.in(CV=CVguess, n1=n1.fix)
ASN.fix <- x$nmean med.fix <- x$nperc[["50%"]]
p.fix <- as.numeric(x[25:24])
pct.fix <- x$pct_s2 x <- power.tsd.in(CV=CVguess, n1=n1.75) ASN.75 <- x$nmean
med.75 <- x$nperc[["50%"]] p.75 <- as.numeric(x[25:24]) pct.75 <- x$pct_s2
x <- power.tsd.in(CV=CVguess, n1=n1.lo)
ASN.lo <- x$nmean med.lo <- x$nperc[["50%"]]
p.lo <- as.numeric(x[25:24])
pct.lo <- x$pct_s2 x <- power.tsd.in(CV=CVguess, n1=n1.hi) ASN.hi <- x$nmean
med.hi <- x$nperc[["50%"]] p.hi <- as.numeric(x[25:24]) pct.hi <- x$pct_s2
result <- data.frame(CV=c(rep(CVguess, 2), CL),
n1=c(n1.fix, n1.75, n1.lo, n1.hi),
CV.obs=rep(CVguess, 4),
ASN=c(ASN.fix, ASN.75, ASN.lo, ASN.hi),
median=c(med.fix, med.75, med.lo, med.hi),
pwr.stg1=c(p.fix[1], p.75[1], p.lo[1], p.hi[1]),
pwr=c(p.fix[2], p.75[2], p.lo[2], p.hi[2]),
pct.2=c(pct.fix, pct.75, pct.lo, pct.hi))
row.names(result) <- c("like fixed", "75% of fixed", "lower CL", "upper CL")
print(signif(result, 4))
CV n1 CV.obs ASN median pwr.stg1 pwr pct.2
like fixed 0.2000 20 0.2 21.59 20 0.7325 0.8514 18.840
75% of fixed 0.2000 16 0.2 19.90 16 0.5946 0.8371 33.780
lower CL 0.1483 12 0.2 20.66 16 0.3834 0.8261 55.630
upper CL 0.3084 42 0.2 42.00 42 0.9740 0.9740 0.006
If we base n1 on the lower end and the CV is close to the guesstimate that’s the winner. One the other hand there is a ~56% chance of proceeding to the second stage which is not desirable – and contradicts the concept of a “safety net”. A compromise would be 75% of the fixed sample design.
The pessimistic approach would be crazy.
❝ More comments on the weights:
Have to chew on that…
❝ Brings us back to: We should plan with a realistic/slightly optimistic scenario.
Seems so.
Dif-tor heh smusma 🖖🏼 Довге життя Україна!
Helmut Schütz
The quality of responses received is directly proportional to the quality of the question asked. 🚮
Science Quotes
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# Definition:Concave Real Function/Definition 1
## Definition
Let $f$ be a real function which is defined on a real interval $I$.
$f$ is concave on $I$ if and only if:
$\forall x, y \in I: \forall \alpha, \beta \in \R_{>0}, \alpha + \beta = 1: \map f {\alpha x + \beta y} \ge \alpha \map f x + \beta \map f y$
### Strictly Concave
$f$ is strictly concave on $I$ if and only if:
$\forall x, y \in I, x \ne y: \forall \alpha, \beta \in \R_{>0}, \alpha + \beta = 1: f \left({\alpha x + \beta y}\right) > \alpha f \left({x}\right) + \beta f \left({y}\right)$
## Geometric Interpretation
Let $f$ be a concave real function.
Then:
for every pair of points $A$ and $B$ on the graph of $f$, the line segment $AB$ lies entirely below the graph.
## Also known as
A concave function can also be referred to as:
a concave down function
a convex up function.
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Ramps are fun for kids to explore. Rolling objects down or pushing objects up them seems to fascinate them. While they are really conducting simple science experiments with the concepts of force and gravity, kids may just think they are having fun. Recently my son and I build a simple ramp using cookbooks found in our kitchen. We used several cookbooks in a traditional book shape to give the ramp height and three ring binder style cookbook to act as the ramp. At the end of the ramp we placed an empty coffee container and later a basket to catch the balls my son was going to roll up the ramp. Then my son gathered up a collection of ball that could be rolled up the ramp. Thankfully he was able to quickly gather a basket filled with them in a short period of time.
What were the goals of the experiment:
• Find out if rolling or pushing the ball helps it get to the end of the ramp,
• Find out if one type of ball is easier to get to the end of the ramp than the others.
• Find out if there is a place at the bottom of the ramp that it was best to start a ball of from.
Here is what conclusions we came to: If you push the ball to the end of the ramp, it tends to continue to move forward into the coffee can. If you roll the ball from the bottom edge of the cookbook and then let go, one of two things can happen. The ball can roll back to you or it can fly off the front edge of the cookbook and miss the coffee can. The small bouncy ball is hardest to control even if you push it. The hollow rubber and plastic balls are easy to control.
When you started with a ball in one of the corners of the bottom edge of the cookbook , it was more likely to go it. This experiment will probably be conducted over and over again. We allowed a lot of variables for this experiment it was an experiment that contained simple goals mentioned above. In science experiments that follow, we will probably concentrate on the different ways of introducing force to a ball along with focusing on one or two balls instead of a basket full. We may also try to control which ball catching container the ball goes into.
The Quirky Mommas have some more fun ideas for science experiments with kids:
Welcome to Kids Activities!
My name is Holly Homer & I am the Dallas mom of three boys…
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# 数组的部分总和-JavaScript
## arrays sum (4)
function partsSums(arr) {
const res = [], len = arr.length
for (let i = len; i > -1; i--) {
res.push(arr.slice(-i || len + 1).reduce((a, n) => a + n, 0))
}
return res;
}
console.log(partsSums([0, 1, 3, 6, 10]));
console.log(partsSums([1, 2, 3, 4, 5, 6]));
console.log(partsSums([744125, 935, 407, 454, 430, 90, 144, 6710213, 889, 810, 2579358]));
function partsSums(arr) {
const sum = arr => arr.reduce((r, e) => r + e, 0);
return arr.reduce((r, e, i, a) => {
const res = sum(a.slice(i, a.length));
return r.concat(!a[i + 1] ? [res, 0] : res)
}, [])
}
console.log(partsSums([0, 1, 3, 6, 10]));
console.log(partsSums([1, 2, 3, 4, 5, 6]));
console.log(partsSums([744125, 935, 407, 454, 430, 90, 144, 6710213, 889, 810, 2579358]));
ls = [0, 1, 3, 6, 10]
ls = [0, 1, 3, 6, 10]
ls = [1, 3, 6, 10]
ls = [3, 6, 10]
ls = [6, 10]
ls = [10]
ls = []
function partsSums(ls) {
let arrayOfSums = [];
while(ls.length > 0) {
let sum = ls.reduce((a, b) => a + b);
arrayOfSums.push(sum);
ls.shift();
}
return arrayOfSums;
}
console.log(partsSums([0, 1, 3, 6, 10]));
[20,20,19,16,10,0]
[20,20,19,16,10]
function partsSums(ls) {
let arrayOfSums = [];
while(ls.length > 0) {
let sum = ls.reduce((a, b) => a + b);
arrayOfSums.push(sum);
ls.shift();
}
arrayOfSums.push(0);
return arrayOfSums;
}
console.log(partsSums([0, 1, 3, 6, 10]));
function partsSums(ls) {
ls.push(0);
let arrayOfSums = [];
while(ls.length > 0) {
let sum = ls.reduce((a, b) => a + b);
arrayOfSums.push(sum);
ls.shift();
}
return arrayOfSums;
}
function partsSums(ls) {
let arrayOfSums = [];
while(ls.length > -1) {
let sum = ls.reduce((a, b) => a + b);
arrayOfSums.push(sum);
ls.shift();
}
return arrayOfSums;
}
TypeError：减少没有初始值的空数组
function partsSums(ls) {
let arrayOfSums = [];
while(ls.length > 0) {
let sum = ls.reduce((a, b) => a + b, 0);
arrayOfSums.push(sum);
ls.shift();
}
return arrayOfSums;
}
function partsSums(ls) {
let sum = ls.reduce((a, b) => a + b, 0);
return ls.length > 0 ? [sum].concat(partsSums(ls.slice(1))) : [0];
}
console.log(partsSums([0, 1, 3, 6, 10]));
console.log(partsSums([1, 2, 3, 4, 5, 6]));
console.log(partsSums([744125, 935, 407, 454, 430, 90, 144, 6710213, 889, 810, 2579358]));
ls = [0, 1, 3, 6, 10]
function partsSums(ls) {
let sum = ls.reduce((sum, n) => sum + n, 0)
res = [sum]
for (let i = 1; i <= ls.length; i++){
sum -= ls[i-1]
res.push(sum )
}
return res
}
console.log(partsSums(ls))
function partsSums(ls) {
if(!ls.length) return [0];
let prevTotal = ls.reduce((a,b) => a + b);
return [prevTotal, ...ls.map(val => prevTotal -= val)]
}
console.log(partsSums([0, 1, 3, 6, 10]));
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$$\require{cancel}$$
# 5.3: Newton's First Law
Learning Objectives
• Describe Newton's first law of motion
• Recognize friction as an external force
• Define inertia
• Identify inertial reference frames
• Calculate equilibrium for a system
Experience suggests that an object at rest remains at rest if left alone and that an object in motion tends to slow down and stop unless some effort is made to keep it moving. However, Newton’s first law gives a deeper explanation of this observation.
Newton’s First Law of Motion
A body at rest remains at rest or, if in motion, remains in motion at constant velocity unless acted on by a net external force.
Note the repeated use of the verb “remains.” We can think of this law as preserving the status quo of motion. Also note the expression “constant velocity;” this means that the object maintains a path along a straight line, since neither the magnitude nor the direction of the velocity vector changes. We can use Figure 5.7 to consider the two parts of Newton’s first law.
Rather than contradicting our experience, Newton’s first law says that there must be a cause for any change in velocity (a change in either magnitude or direction) to occur. This cause is a net external force, which we defined earlier in the chapter. An object sliding across a table or floor slows down due to the net force of friction acting on the object. If friction disappears, will the object still slow down?
The idea of cause and effect is crucial in accurately describing what happens in various situations. For example, consider what happens to an object sliding along a rough horizontal surface. The object quickly grinds to a halt. If we spray the surface with talcum powder to make the surface smoother, the object slides farther. If we make the surface even smoother by rubbing lubricating oil on it, the object slides farther yet. Extrapolating to a frictionless surface and ignoring air resistance, we can imagine the object sliding in a straight line indefinitely. Friction is thus the cause of slowing (consistent with Newton’s first law). The object would not slow down if friction were eliminated.
Consider an air hockey table (Figure 5.8). When the air is turned off, the puck slides only a short distance before friction slows it to a stop. However, when the air is turned on, it creates a nearly frictionless surface, and the puck glides long distances without slowing down. Additionally, if we know enough about the friction, we can accurately predict how quickly the object slows down.
Newton’s first law is general and can be applied to anything from an object sliding on a table to a satellite in orbit to blood pumped from the heart. Experiments have verified that any change in velocity (speed or direction) must be caused by an external force. The idea of generally applicable or universal laws is important—it is a basic feature of all laws of physics. Identifying these laws is like recognizing patterns in nature from which further patterns can be discovered. The genius of Galileo, who first developed the idea for the first law of motion, and Newton, who clarified it, was to ask the fundamental question: “What is the cause?” Thinking in terms of cause and effect is fundamentally different from the typical ancient Greek approach, when questions such as “Why does a tiger have stripes?” would have been answered in Aristotelian fashion, such as “That is the nature of the beast.” The ability to think in terms of cause and effect is the ability to make a connection between an observed behavior and the surrounding world.
## Gravitation and Inertia
Regardless of the scale of an object, whether a molecule or a subatomic particle, two properties remain valid and thus of interest to physics: gravitation and inertia. Both are connected to mass. Roughly speaking, mass is a measure of the amount of matter in something. Gravitation is the attraction of one mass to another, such as the attraction between yourself and Earth that holds your feet to the floor. The magnitude of this attraction is your weight, and it is a force.
Mass is also related to inertia, the ability of an object to resist changes in its motion—in other words, to resist acceleration. Newton’s first law is often called the law of inertia. As we know from experience, some objects have more inertia than others. It is more difficult to change the motion of a large boulder than that of a basketball, for example, because the boulder has more mass than the basketball. In other words, the inertia of an object is measured by its mass. The relationship between mass and weight is explored later in this chapter.
## Inertial Reference Frames
Earlier, we stated Newton’s first law as “A body at rest remains at rest or, if in motion, remains in motion at constant velocity unless acted on by a net external force.” It can also be stated as “Every body remains in its state of uniform motion in a straight line unless it is compelled to change that state by forces acting on it.” To Newton, “uniform motion in a straight line” meant constant velocity, which includes the case of zero velocity, or rest. Therefore, the first law says that the velocity of an object remains constant if the net force on it is zero.
Newton’s first law is usually considered to be a statement about reference frames. It provides a method for identifying a special type of reference frame: the inertial reference frame. In principle, we can make the net force on a body zero. If its velocity relative to a given frame is constant, then that frame is said to be inertial. So by definition, an inertial reference frame is a reference frame in which Newton’s first law is valid. Newton’s first law applies to objects with constant velocity. From this fact, we can infer the following statement.
Inertial Reference Frame
A reference frame moving at constant velocity relative to an inertial frame is also inertial. A reference frame accelerating relative to an inertial frame is not inertial.
Are inertial frames common in nature? It turns out that well within experimental error, a reference frame at rest relative to the most distant, or “fixed,” stars is inertial. All frames moving uniformly with respect to this fixed-star frame are also inertial. For example, a nonrotating reference frame attached to the Sun is, for all practical purposes, inertial, because its velocity relative to the fixed stars does not vary by more than one part in 1010. Earth accelerates relative to the fixed stars because it rotates on its axis and revolves around the Sun; hence, a reference frame attached to its surface is not inertial. For most problems, however, such a frame serves as a sufficiently accurate approximation to an inertial frame, because the acceleration of a point on Earth’s surface relative to the fixed stars is rather small (< 3.4 x 10−2 m/s2). Thus, unless indicated otherwise, we consider reference frames fixed on Earth to be inertial.
Finally, no particular inertial frame is more special than any other. As far as the laws of nature are concerned, all inertial frames are equivalent. In analyzing a problem, we choose one inertial frame over another simply on the basis of convenience.
## Newton’s First Law and Equilibrium
Newton’s first law tells us about the equilibrium of a system, which is the state in which the forces on the system are balanced. Returning to Forces and the ice skaters in Figure 5.3, we know that the forces $$\vec{F}_{1}$$ and $$\vec{F}_{2}$$ combine to form a resultant force, or the net external force: $$\vec{F}_{R}$$ = $$\vec{F}_{net}$$ = $$\vec{F}_{1}$$ + $$\vec{F}_{2}$$. To create equilibrium, we require a balancing force that will produce a net force of zero. This force must be equal in magnitude but opposite in direction to $$\vec{F}_{R}$$, which means the vector must be $$- \vec{F}_{R}$$. Referring to the ice skaters, for which we found $$\vec{F}_{R}$$ to be 30.0 $$\hat{i}$$ + 40.0 $$\hat{j}$$ N, we can determine the balancing force by simply finding $$- \vec{F}_{R}$$ = −30.0 $$\hat{i}$$ − 40.0 $$\hat{j}$$ N. See the free-body diagram in Figure 5.3(b).
We can give Newton’s first law in vector form:
$$\vec{v} = constant\; when\; \vec{F}_{net} = \vec{0}\; N \ldotp \label{5.2}$$
This equation says that a net force of zero implies that the velocity $$\vec{v}$$ of the object is constant. (The word “constant” can indicate zero velocity.)
Newton’s first law is deceptively simple. If a car is at rest, the only forces acting on the car are weight and the contact force of the pavement pushing up on the car (Figure 5.9). It is easy to understand that a nonzero net force is required to change the state of motion of the car. However, if the car is in motion with constant velocity, a common misconception is that the engine force propelling the car forward is larger in magnitude than the friction force that opposes forward motion. In fact, the two forces have identical magnitude.
Example 5.1
### When Does Newton’s First Law Apply to Your Car?
Newton’s laws can be applied to all physical processes involving force and motion, including something as mundane as driving a car.
1. Your car is parked outside your house. Does Newton’s first law apply in this situation? Why or why not?
2. Your car moves at constant velocity down the street. Does Newton’s first law apply in this situation? Why or why not?
Strategy
In (a), we are considering the first part of Newton’s first law, dealing with a body at rest; in (b), we look at the second part of Newton’s first law for a body in motion.
Solution
1. When your car is parked, all forces on the car must be balanced; the vector sum is 0 N. Thus, the net force is zero, and Newton’s first law applies. The acceleration of the car is zero, and in this case, the velocity is also zero.
2. When your car is moving at constant velocity down the street, the net force must also be zero according to Newton’s first law. The car’s engine produces a forward force; friction, a force between the road and the tires of the car that opposes forward motion, has exactly the same magnitude as the engine force, producing the net force of zero. The body continues in its state of constant velocity until the net force becomes nonzero. Realize that a net force of zero means that an object is either at rest or moving with constant velocity, that is, it is not accelerating. What do you suppose happens when the car accelerates? We explore this idea in the next section.
Significance
As this example shows, there are two kinds of equilibrium. In (a), the car is at rest; we say it is in static equilibrium. In (b), the forces on the car are balanced, but the car is moving; we say that it is in dynamic equilibrium. (We examine this idea in more detail in Static Equilibrium and Elasticity.) Again, it is possible for two (or more) forces to act on an object yet for the object to move. In addition, a net force of zero cannot produce acceleration.
Exercise 5.2
A skydiver opens his parachute, and shortly thereafter, he is moving at constant velocity. (a) What forces are acting on him? (b) Which force is bigger?
Simulation
Engage in this simulation to predict, qualitatively, how an external force will affect the speed and direction of an object’s motion. Explain the effects with the help of a free-body diagram. Use free-body diagrams to draw position, velocity, acceleration, and force graphs, and vice versa. Explain how the graphs relate to one another. Given a scenario or a graph, sketch all four graphs.
## Contributors
• Samuel J. Ling (Truman State University), Jeff Sanny (Loyola Marymount University), and Bill Moebs with many contributing authors. This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0).
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#### Tom Bishop
• Zetetic Council Member
• 8483
• Flat Earth Believer
##### Re: Help me understand how light rays travel
« Reply #20 on: March 10, 2021, 06:21:52 PM »
On light bending, look into what Dr. Edward Dowdye has to say: https://sciencewoke.org/nasa-scientist-says-coronas-bend-light-not-gravity/
#### Tumeni
• 2648
##### Re: Help me understand how light rays travel
« Reply #21 on: March 10, 2021, 06:28:17 PM »
=============================
Not Flat. Happy to prove this, if you ask me.
=============================
Nearly all flat earthers agree the earth is not a globe.
Nearly?
#### jimster
• 132
##### Re: Help me understand how light rays travel
« Reply #22 on: April 29, 2021, 04:43:57 AM »
None of the above explains this, I will try to express the issue again.
/\
/ \
/ \ _____
/ \______________________________\ /
I saw the top of the mountain and I saw the ocean between the boat and the shore. Where did the light rays from the shoreline (between them) go? Presumably, they follow parallel paths and remain between the mountain top rays and the ocean rays.
How could light rays from above them and below them reach me, but not the rays from the shore? It ain't distance, the mountain is behind the shore. Sure looked like the mountain was descending below the water.
This is the exact way it would appear with RE, but we are to prefer the explanation that requires unknown forces that you can't explain or demonstrate?
Tom Bishop, don't go away, that means you have no answer and the earth is round. You have to explain how the light rays disappear. Unknown forces? Conspiracy? Where are those light rays?
FE can't explain sharp cutoffs, everything in FE is gradual. No explanation for relatively quick transition of day to night, or the sharp cutoff of the sun as it rises and sets. FE has to do it with bending and distance, and those things are gradual. Explaining the sharp cutoffs requires light not to just bend, but to just disappear.
#### Tumeni
• 2648
##### Re: Help me understand how light rays travel
« Reply #23 on: April 29, 2021, 07:07:51 AM »
On light bending, look into what Dr. Edward Dowdye has to say: https://sciencewoke.org/nasa-scientist-says-coronas-bend-light-not-gravity/
Is it agreed that Rowbottom asserted straight light rays?
=============================
Not Flat. Happy to prove this, if you ask me.
=============================
Nearly all flat earthers agree the earth is not a globe.
Nearly?
#### fisherman
• 198
##### Re: Help me understand how light rays travel
« Reply #24 on: April 29, 2021, 05:05:43 PM »
On light bending, look into what Dr. Edward Dowdye has to say: https://sciencewoke.org/nasa-scientist-says-coronas-bend-light-not-gravity/
Quote
Therefore according to Dr. Dowdye, all the supposed gravitational lensing that scientists see is in reality, light passing through not empty space or space-time bending, but passing through mass and the mass in space is bending the light
The mass in space bends spacetime. You can't separate the two concepts. If light moving through mass (not even sure how that could work) is what bends it, by definition it is moving through bent space time. Wherever there is mass, there is a bend in space time.
#### fisherman
• 198
##### Re: Help me understand how light rays travel
« Reply #25 on: April 29, 2021, 05:38:42 PM »
Quote
I am one such "space skeptic/denier", and have concluded that not only is "space" complete fiction (and has its origins, indisputably, in that medium) but that it cannot exist in the reality we study. It would violate many well established natural laws which have stood for centuries without contest.
Quote
I am one such "space skeptic/denier", and have concluded that not only is "space" complete fiction (and has its origins, indisputably, in that medium) but that it cannot exist in the reality we study. It would violate many well established natural laws which have stood for centuries without contest.
Jack, if I understand your position correctly, you believe that space doesn’t exist independent of the matter that it is in. If all matter disappeared, then space would cease to exist? Is that correct?
If so, Einstein spent the better part of 10 years trying to prove exactly that. He struggled for 10 years to come up with field equations that make all laws of physics work exactly the way the we observe even if there was no matter in space and space ceased to exist. He failed. IOW, in order for the laws of physics, specifically the inertial motion of bodies to behave as we observe, space must exist as a separate physical entity.
Einstein’s field equations confirm this. Where are you equations that contradict his?
#### RonJ
• 1419
• ACTA NON VERBA
##### Re: Help me understand how light rays travel
« Reply #26 on: April 30, 2021, 02:24:30 AM »
On light bending, look into what Dr. Edward Dowdye has to say: https://sciencewoke.org/nasa-scientist-says-coronas-bend-light-not-gravity/
How could you believe anything that Dr. Edward Dowdye says? He used to work for NASA. Additionally, if you watch the video, he says that the Sun is millions of miles away from the earth, not 3000, as stated in the Wiki. So, is Dr. Dowdye correct or is he spouting more nonsense as is typical of NASA minions? If you wish to have people believe in FET you have to be consistent in each and every little thing.
You can lead a flat earther to the curve but you can't make him think!
#### Tumeni
• 2648
##### Re: Help me understand how light rays travel
« Reply #27 on: May 05, 2021, 11:36:33 AM »
On light bending, look into what Dr. Edward Dowdye has to say: https://sciencewoke.org/nasa-scientist-says-coronas-bend-light-not-gravity/
Is it agreed that Rowbottom asserted straight light rays?
IMG above
It strikes me that we have better optics available than in Rowbottom's day, and that the Wiki and ENAG Workshop could be updated with improved versions which don't rely on line drawings as proof.
Unless some enterprising zetecist has already done this ... ?
=============================
Not Flat. Happy to prove this, if you ask me.
=============================
Nearly all flat earthers agree the earth is not a globe.
Nearly?
#### Flex
• 2
##### Re: Help me understand how light rays travel
« Reply #28 on: May 05, 2021, 12:01:25 PM »
It is because of ocean waves that the distant views get block. On lakes you should be able to see far distances.
#### Tumeni
• 2648
##### Re: Help me understand how light rays travel
« Reply #29 on: May 05, 2021, 01:26:02 PM »
It is because of ocean waves that the distant views get block. On lakes you should be able to see far distances.
Yes, but we're merely talking about observation of things like flags, islands, lighthouses etc. that we can see.
And we could observe over land and rivers, not just oceans.
=============================
Not Flat. Happy to prove this, if you ask me.
=============================
Nearly all flat earthers agree the earth is not a globe.
Nearly?
#### AllAroundTheWorld
• 4274
##### Re: Help me understand how light rays travel
« Reply #30 on: May 05, 2021, 04:01:06 PM »
It is because of ocean waves that the distant views get block. On lakes you should be able to see far distances.
No it isn't. Unless you're on the water's edge and lying down. Or there are really big waves. A wave can only occlude as much of a distant object as its own height if the viewer height is the same height as the waves, which it will be if you're standing up unless there's a particularly choppy sea.
And distant buildings are occluded when looking over large inland lakes too.
"On a very clear and chilly day it is possible to see Lighthouse Beach from Lovers Point and vice versa...Upon looking into the telescope I can see children running in and out of the water, splashing and playing. I can see people sun bathing at the shore
- An excerpt from the account of the Bishop Experiment. My emphasis
#### Tumeni
• 2648
##### Re: Help me understand how light rays travel
« Reply #31 on: May 06, 2021, 09:45:56 AM »
And we could observe over land and rivers, not just oceans.
Like this - the view over the Forth Estuary in Scotland. Ignore the green annotations; they were for another discussion elsewhere
The waves cannot be obscuring the tops of the bridge towers, for the tallest ones are 210m tall.
The waves cannot be obscuring the tops of the hills beyond. They are approx 400m tall.
The observer was at 210m elevation.
« Last Edit: May 06, 2021, 10:07:12 AM by Tumeni »
=============================
Not Flat. Happy to prove this, if you ask me.
=============================
Nearly all flat earthers agree the earth is not a globe.
Nearly?
#### scomato
• 64
##### Re: Help me understand how light rays travel
« Reply #32 on: May 06, 2021, 02:30:13 PM »
It is because of ocean waves that the distant views get block. On lakes you should be able to see far distances.
Standing on one end of Lake Ontario you can see Toronto the other side. I believe this photo is taken from New York based on the angle.
Where did half the city go? Lake Ontario typically only has waves under 5 feet, up to 10-20 when it's stormy, so waves can't explain why the bottom half of the city skyline is missing from view.
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1-Newbie
parameter round off
In part relations(WF2) I need to round off parameter values to certain decimal places, the ceil function is not helping either. CE1 = ceil(10.136, 2) returns CE1=10.14 CE2 = ceil(10.134, 2) also returns CE2=10.14 But how to round off the value i.e 10.134 to 10.13
8 REPLIES 8
4-Participant
I don't know how to do that in a relation, but if you just want to display the rounded value in a drawing, you can write it like this:
``612e9b825d``
``612e9b825d``
represents the number of decimals to show.
1-Newbie
(To:SylvainA.)
Sylvain, this &parameter_name[.2] works for display purpose in drawing notes.. however i want the round off parameter value for further calculations inside the part relations itself.
1-Newbie
you can round off any dimension individually by modifying the propreties and change the number of decimal in the dimension properties in your model.
1-Newbie
(To:jaypogi)
Hi, Ceil will always round it up. Floor will always round it down. Its strange that ROUND function is missing from Proe. At least I counldnt find it. I have a solution for this. Its a relation that will work like round function. I have a blog on this solution. have a look if it works for you. http://tipsfromjoe.blogspot.com/2009/05/proe-function-relation-to-round-decimal.html HTH, Joe. ---------------------------------------------- you can find some tips at... http://tipsfromjoe.blogspot.com ----------------------------------------------
1-Newbie
(To:JoeVarghese)
Hi Joe, Its simple & really great.. appreciate it This was excatly what i was looking for.. Thank you very much..
1-Newbie
Welcome Joe. ---------------------------------------------- you can find some tips at... http://tipsfromjoe.blogspot.com ----------------------------------------------
7-Bedrock
(To:JoeVarghese)
Round il like
floor (N+5*10^(D+1),D)
or
ceil (N+5*10^(D+1),D)
with
N--> number to rounded
D--> number of decimal
it's math!
7-Bedrock
you can use floor (103.136+0.005,2) or ceil (103.136-0.005,2).
it's math!
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# Good grounding practices - Part 2
## Why and how to make power and shielding connections.
Part 1 describes the reasons we ground and specifications for ground resistance. Here, I discuss why it’s necessary to connect instrumentation to the power ground system, and the theory and practice of shielding.
## That “dirty” power ground
In the past, there was a common refrain by manufacturers and some engineers, and it’s even heard sometimes today, that they want a “clean” or “quiet” ground for their equipment. When asked to quantify what such a ground is, they generally talk about a ground not connected to that dirty, nasty power ground. It seems they expect that noise or problems will somehow arise out of the ground and strike down their equipment.
This is basically a question of whether to have a separate, isolated (hence somehow clean), instrument ground, or to have it a part of the overall facility grounding grid.
A perfect ground would never vary in potential—zero potential anywhere (potential because it is single-ended, unlike voltage, which is double-ended, or a potential in reference to another potential). This theoretical ground would also have zero impedance—current flowing to ground would encounter no resistance, and anything connected would always be at the same ground potential.
Unfortunately, we are stuck with the Earth, which is everywhere that our instrument systems are, and our instruments can be significantly geographically distributed. This earth can be defined at any single point as the zero reference potential point of out instrument system, but it will be different anywhere else. The so-called clean ground, isolated from the rest of the grounds (but subject to many of the underground events that the other grounds are subject to) can the very source of the problems we wished to avoid.
Under quiescent conditions, such a ground may have less variation in ground potential, but under non-quiescent conditions, it can be subject to considerable variation relative to other ground systems. One should remember that ground is not a sink for noise, but can be a path for it if the ground completes a noise circuit. Conducted noise flows in complete circuits, and Ohm's Law (impedance version) applies to conducted noise.
## A ground is a ground, or is it?
The essential trouble with grounds is that they are all different. When I was young engineer, I attended a grounding course put on by Longview Power and Light. As part of the course, there was a field demonstration where we were shown two ground rods about 50 feet part. A wire was run between the two ground wires, and a clamp-on ammeter was clamped on the wire. Low and behold, a current was measured between the two ground rods. The fact that a current was flowing indicated a potential differential between the two grounds—the grounds were not the same. This small epiphany provided the spark for a lifelong interest in grounding.
So why are the grounds different? The first difference might be due to different resistivity and, in some cases, different impedances in the earth. This can be due to different soils, non-conductors like rocks and voids, moisture content, salt content, ground water, underground artifacts (like pipelines), etc. It can also be due to the weather and seasonal changes. The current drought is affecting our grounding systems by drying out the soil, shrinking soil away from ground rods, and reducing the level of ground water.
The resistivity of the earth is three-dimensional—it has length and width as well as depth. Figure 1 show a resistance mapping of the earth from a side view. From this one might infer why grounds can be different.
The second reason a ground can be different is that there are currents flowing in the earth, some man-made, others caused by nature. Man-made currents can be transient in nature, like ground fault current returning to its source through the ground, or continuous, like the currents induced in the ground by our high-voltage power distribution systems, electric trains, stray currents, or circulating ground currents caused by poor grounding.
Nature is also a big cause of ground currents flowing in the earth. An obvious source is lightning, which causes a large, rapid charge redistribution that results in currents flowing and changing ground potential.
A good analogy of a lightning strike to earth is dropping a large rock in a pond and watching the ripples spread out. This is illustrated in Figure 2. By several estimates, there are about 2,000 thunderstorms worldwide providing 100 flashes per second (not all of these are ground strikes).
But we don’t need lightning strikes to cause current flow in the earth, we only need a moving thunderstorm and in some cases just moving clouds (try Googling “clear sky lightning”). Clouds become charged and can induce currents in the earth. Think about this happening in a global sense all around us. Figures 3 and 4 are illustrations of a moving thunderstorm.
Lightning can also induce currents in the ground and generate RF interference, as shown in Figure 5.
There are other currents in the earth such as geomagnetic currents that result from space weather interacting with the earth’s magnetic fields, inducing currents on long conductors like power lines and buried pipelines. These currents are also called Telluric currents.
Some further analogies may help to help better understand grounding. A good analogy for currents flowing and earth potential varying is the use of the ocean. The ocean has flowing currents that we cannot see, and waves varying in height, which can represent varying potential (Figure 6).
If we take the ocean analogy a bit further, we can understand why we want to connect all of our grounding systems together to create (as much as we can) an equipotential grounding plane for our instrumentation systems, electrical systems, communication systems and computer systems. IEEE std. 1100 [11] has some good discussion on the equipotential plane.
Visualize two ships separated by several waves or troughs, but connected to each other by electrical cables (power, signal, etc.). We will use the analogy that the water height above a theoretical reference point represents ground potential. So, the ships, relative to each other, could at any time be on a wave or trough of the same height (same potential), on a wave or trough of different heights (potential difference), or one ship could be on a wave while the other in a trough (larger potential difference). Since the electrical cables at one ship are at one potential (wave/trough height), while at the other ship there might be a different potential (wave/trough height), there might be a large potential difference between the electrical cables and the local ground. If the two ships are at the same wave height (or equipotential plane), the potential would be zero.
To illustrate how much potential difference there can be between two ground points, where the resistance between the points is one ohm and the current from lightning flowing through the earth is 10,000 amps, the potential difference between those two points is 10,000 volts, which would exceed the nominal arc overvoltage of wiring of 6,000 volts. This is illustrated in Figure 7.
This brings us to the good reason to interconnect power and instrument grounds. What we want is for our instrument systems to ride up and down the varying ground potential with the rest of the facility, like a ship riding up and down on waves. We want to create an equipotential grounding plane that rides up and down our ground potential “waves” as evenly as possible. Providing this “safe ship” to ride out storms is extremely important to the reliable and safe operation of our instrument systems.
If we have to connect to another “ship,” we should use isolators or connect using non-electrical means, e.g. fiber-optics (like ships at sea communicating with each other via blinking signal lamps, sometimes called Aldis lamps).
Another important concept in instrumentation grounding is that of a low-frequency, single-point ground, i.e. all ground references are connected to a common ground to minimize any potential differences on the DC side of our systems. These connections should not be daisy-chained and should provide a low-impedance path back to our DC ground. This ground should be connected to the power grounding grid at only one point. Selection of this point of connection should consider the power distribution system (how large ground faults might occur, and how they will return to their source; smaller leaks return using the same path), where the lightning protection system is connected to the grounding grid (not too close to the instrument connection), location of cathodic protection systems (generally, stay away from them), location of underground pipes (stay away), etc. Figures 8 and 9 show an example DCS grounding scheme and a ground grid (not to scale).
At higher frequencies, multipoint grounding systems are used because these systems will be grounded anyway through distributed capacitances, and it is better to separately ground them. Don’t confuse low-frequency grounding requirements with higher-frequency ones.
Before we leave this topic, we should address ungrounded systems. NEC Article 250 addresses which AC systems must be grounded. Instrument systems that are powered typically by 12-28 VDC systems are not addressed. People do often use ungrounded systems for DC instrument loops where they have an isolated DC power supply or four-wire instrument loop. These loops will function this way and there are applications where this may be necessary.
There are, however, some potential pitfalls to having such loops:
1. The loop is not referenced to the rest of the instrument system.
2. If it becomes inadvertently grounded at an undefined point, this may affect the operation of the loop in a negative way.
3. It may not be apparent to an instrument technician that this loop is different from other loops and they may not understand why their voltage readings are not correct.
4. If the loop inadvertently come into contact with a higher voltage, e.g. 120 VAC, the instrument technician may be in for a rude surprise, or even a fatal one.
## Do I connect my shields to ground at one end or both?
This question has a life of its own and seems come up time and time again. Shielding is a complex subject that can be very application-dependent. This discussion is primarily directed at shielding in petrochemical plants.
The purpose of shielding is to keep noise in or to keep noise out of a circuit, more toward the latter than the former. Noise can be defined as any unwanted electrical signal in a circuit, while interference is noise that has a detrimental influence on the circuit. To understand shielding and the related question of grounding shields, we must understand how noise gets into a circuit. Noise can also be generated internal to the circuit, in which case, shielding is not much help except to keep it from getting out of the circuit.
There are four basic means that noise is coupled into a circuit: capacitively, inductively, radiated and conducted. The frequency of the noise and the operating frequency of the circuit must be considered. Most instrumentation measurement and control circuits operate at low frequency (< 100 Kbits/s) or at DC, while most communication circuits operate at high frequency (> 1 Mhz). This typically results in different shield grounding methods for low frequency (grounded at one end) than at high frequency (grounded in multiple places). Here, we discuss low-frequency grounding.
Shielding capacitively coupled noise: Capacitive or electrostatic coupled noise is coupled through distributed capacitances formed between the source of the noise and the receptor (or victim) of the noise. This type of noise is voltage-based (Figure 10).
We shield against this type of noise coupling by placing a grounded shield (at the system reference potential) between the noise source and the receptor circuit to provide a conductor for the distributed capacitances to connect to rather than the signal conductors. This normally takes the form of a thin, aluminum-coated Mylar film with a drain wire running the length of the shield around the protected wires (Figure 11) Any noise capacitively coupled to the shield is returned to the source via the shield ground connection. If the shield impedance is zero, the noise does not couple to the protected signal wires since it has been intercepted by the shield at zero potential relative to the signal wire reference.
In practice, the shield impedance is not zero and there will be some noise voltage generated across the shield and coupled to the signal wires, but it’s greatly reduced from the unshielded cable. This shield is grounded at one end at the zero potential reference point of the circuit (nominally the DC ground reference point). This is typically done in the main or master termination boxes in the DCS/PLC equipment room.
The shield continuity is maintained through any marshalling and field junction boxes out to the field device, where the shield is folded back and taped or heat-shrinked. The only common exception is grounded thermocouples, whose shields are commonly grounded at the thermocouple head or the field junction box closest to it.
The reason we ground at one end is that our field devices can be a long distance from the receiving element and try as hard as we can, we will still have differences in ground potential. If we connect at both ends, we can have circulating currents in our shields. This can be particularly detrimental during ground disturbances.
Some keys here are to make sure you do a good job of terminating the shield in the field so it can’t come into contact with ground or something that is grounded; run your shields as close as you can to any termination points; and keep your exposed drain wires as short as possible.
Shielding of inductively or magnetically coupled noise: Shielding against inductively or magnetically coupled noise is a bit more difficult. This noise is coupled or induced by a varying magnetic field similarly to how a transformer works. This noise is current-based—the higher the current at the source, the more noise is coupled at the receptor. This type of noise is illustrated in Figure 12. For this, the aluminum shield is no barrier and for a metallic shield to be successfully, it must be made of a ferrous material. However, connecting the shield at both ends can provide some protection against magnetically coupled noise by inducing a counter current on the shield to the noise current coupled to the signal wires. That said, we still have the problem that if we connect our shields at both ends, we will have circulating currents in our shields from other sources besides our magnetic coupled noise. In addition, the aluminum shields used in our typical shielded, twisted-pair cables are not designed to be current-carrying conductors and in the extreme case, you may burn through your cable. Some people have been successful in grounding the other end of the shield through a capacitor to provide solidly grounded shield at one end at DC and a varying impedance at frequency, grounded at both ends through the capacitor.
If the cable is contained in the same building and both ends share a common equipotential plane, you may be able get away with grounding the shield at both ends, but it is a last resort for process instrumentation systems. It’s commonly done to get rid of hum in audio systems, however, they typically meet the above-mentioned limitations. And bear in mind, though it’s not grounding, twisted-pair is an effective means of reducing magnetically coupled noise, as is differential inputs in 4-20 mA circuits.
In summary, the correct practice is to ground your shields at one end, at the zero-potential reference point of the circuit.
Radiated noise: Radiated noise protection isn’t generally about grounding since it is a function of reflection and absorption, which do not require grounding to work. Radiated noise can be picked up by “antennas” in the construction of the equipment that are “tuned” to the radiated noise wavelength. Once in a circuit, this noise behaves like conducted noise. It is good practice to specifically ground your field junction boxes. “Noise Reduction Techniques in Electronic Systems” by Henry W. Ott has a good discussion on this.
Conducted noise: Conducted noise is noise that has gotten into a circuit by various means and has been successfully conducted past its entry point. Often, this type of noise is filtered out by means that include shunting the noise through ground back to its source. A good ground system is a key to doing this successfully.
## Conclusions
The basic principles of a good instrument system can be summed up as:
1. A well designed and maintained plant ground grid is priceless. Follow the standards.
2. A well designed grounding and bonding system is key to reliable operation of your instrument system.
3. Follow your control or instrument system manufacturer’s recommended ground resistance where possible. In lieu of such a recommendation, as a minimum, keep your instrument ground resistance to less than 5 ohms where possible.
4. Connect your DC instrument systems to a single point ground, and connect this ground to the plant ground grid at only one point.
5. Inspect, test and maintain your ground systems.
6. Connect your shields at only one end.
Frequent contributor William (Bill) L. Mostia, Jr., P.E., fellow, SIS-Tech Solutions, can be reached at wlmostia@msn.com.
## References:
1. NFPA 70 (NEC), “National Electric Code,” National Fire Protection Association, 1 Batterymarch Park, Quincy, MA, www.nfpa.org
2. IEC 60364, Electrical Installations for Buildings, Part 5, Section 54, International Electrotechnical Commission, Geneva 20 – Switzerland, www.iec.ch
3. IEEE 142 Std. – IEEE Recommended Practice for Grounding of Industrial and Commercial Power Systems (commonly called the Green Book), Institute of Electrical and Electronic Engineers, New York, NY, www.ieee.org
4. IEEE Std. 80 - IEEE guide for safety in AC substation grounding,” Institute of Rev. 1 - 7-15-13 Electrical and Electronic Engineers, New York, NY, www.ieee.org
5. NFPA 780 – “Standard for the Installation of Lightning Protection Systems,” National Fire Protection Association, 1 Batterymarch Park, Quincy, MA.
6. API 2003 – “Protection Against Ignitions Arising out of Static, Lightning, and Stray Currents ,“ American Petroleum Institute, Washington, DC, www.api.org
7. IBM Power7 Information – “Static electricity and floor resistance,” IBM Corp.
8. ISA RP 12.06.01, “Recommended Practice for Wiring Methods for Hazardous (Classified) Locations Instrumentation Part 1: Intrinsic Safety,” International Society of Automation, Research Triangle, NC.
10. The Lightning Discharge, Martin A. Uman, Dover, 1987
11. IEEE std.1100, “Recommended Practice for Powering and Grounding Electronic Equipment,” (commonly called the Emerald Book). Institute of Electrical and Electronic Engineers, New York, NY, www.ieee.org
12. “Integrated Production Control System CENTUM VP System Overview (Vnet/IP Edition),” Yokogawa Electric Corporation, GS 33K01A10-50E GS, pg. 17, 2011.
13. “Site Design and Preparation for DeltaV_ Digital Automation Systems,” D800015X052, pg. 4-7, Emerson, September 2005.
14. “Data Hiway Planning,” DH02-501, Pg. 103, Honeywell, 9/95.
15. “STANDARDS AND GUIDELINES FOR COMMUNICATION SITES,” 68P81089E50-B, Motorola, Pg. 4-47
16. Soars Book on Grounding and Bonding, 9th Edition, International Association of Electrical Inspectors, Richardson, Tx, 2004.
17. Electrical Instruments in Hazardous Locations, 4th Ed., Earnest Magison, ISA, 1998.
18. “Notes on Substation Grounding,” Jeff Jowett, Megger
19. “A practical guide to earth resistance testing,” Megger, http://www.megger.com/us/story/Index.php?ID=146
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##### x-3=0 find two points
label Mathematics
account_circle Unassigned
schedule 1 Day
account_balance_wallet \$5
x-3=0 find two points
May 10th, 2015
we have
x - 3 = 0
so : x = 3
we don't have any other points
Good Luck
and please if you don't understand something let me know it writ it in the comment
and please each time you have any other homework please invite me first i will give you the best answer
May 10th, 2015
Graph the following line y=8 over 9x+2. Find two distinct points on the line to be graphed.
May 10th, 2015
9x+2 = 8
x=6/9= 2/3
so : the intersection between y=8 and 9x+2 is :
in the point : A(2/3, 8)
May 10th, 2015
...
May 10th, 2015
...
May 10th, 2015
Sep 23rd, 2017
check_circle
Mark as Final Answer
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Unmark as Final Answer
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# How can I accurately state the uncertainty principle?
In almost every introductory course, it is taught that the uncertainty principle happens due to disturbance in the system to be measured. Teachers give these examples that induce students to misunderstand the phenomenon.
This has probably been going on for a long time. Heisenberg used to illustrate his uncertainty principle by giving the example of "taking a photograph of the electron".
To take the picture, a scientist might bounce a light particle off the electron's surface. That would reveal its position, but it would also impart energy to the electron, causing it to move. Learning about the electron's position would create uncertainty in its velocity; and the act of measurement would produce the uncertainty needed to satisfy the principle.
However, the uncertainty is not in the eye of the beholder (see here).
Many students, like me, continue their studies of quantum mechanics while ignoring the conceptual subtleties of the uncertainty principle, and this generates a snowball effect. These students do not have solid knowledge of the subject, which is already intrinsically not very intuitive. Later, they learn that there is an uncertainty relation whenever two operators are non-commuting. However, this is a rather abstract definition.
I'd like to know if there is a simple conceptual explanation that uses key concepts such as "observation", "measurement", "decoherence" accurately. For example, if I were a teacher, how could I state Heisenberg's uncertainty principle in a less mathematical, but conceptually rigorous way?
• Your post seems like a (correct IMO) criticism of how the HUP is explained and then a separate question tacked on at the end. I suggest focusing on asking your main question up front, and not making it look so subjective (e.g. the title of the post). Mar 22, 2021 at 20:51
• Related/possible duplicates (in the sense that the answers there contain the answer to this question): physics.stackexchange.com/q/114133/50583, physics.stackexchange.com/q/201580/50583, physics.stackexchange.com/q/24068/50583 and their many linked questions Mar 22, 2021 at 20:58
• This video from 2blue1brown is pretty good. The more general uncertainty principle, beyond quantum Mar 22, 2021 at 21:29
• "If I were a teacher"... of what class? Is it a college class where they've seen Fourier transforms? Because then you can just illustrate it very vividly with time/frequency uncertainty, and then tell them this mathematical fact is physically true in quantum mechanics ("because position & momentum are in some sense Fourier transforms", or whatever..). Mar 23, 2021 at 11:02
• "How to accurately state..." <-- this is begging for reflexive jokes. "I can tell you exactly what the Uncertainty Principle is but then you'd have zero understanding" Mar 23, 2021 at 11:41
As pointed out in a comment by mmesser314, there is a video by the mathematician Grant Sanderson.
The more general uncertainty principle, beyond quantum
Grant Sanderson puts the HUP in the context of properties of wave propagation.
In the case of a continuous, consistent oscillation the corresponding propagating wave has a specific frequency, but that propagating wave doesn't have a location; it's continuous.
It is possible to produce a burst of oscillation in such a way that it gives rise to a propagating "blip". A fourier analysis of the waveform of that blip describes it as a superposition of a range of frequencies. In the case of that blip: the location of it can be tracked through time with specifity. But the blip does not have a particular frequency; the blip is spread out in frequency space.
Grant Sanderson describes that with wave propagation in general (not just in the context of quantum mechanics) there is an inherent trade-off. You can push for a very specific frequency, but at the cost of specifity of position-as-a-function-of-time. You can push for high specifity of position-as-a-function-of-time, but at the cost of introducing spread of spectrum.
In any device that produces propagating waves the design can be made so that the emitted wave is wherever you want in that trade-off.
Fourier analysis facilitates expressing the trade-off in mathematical form.
Stating it explicitly:
The view in terms of wave propagation in general does not need to invoke concepts such as "observation", "measurement", "decoherence".
• This, with a more technical framing ("what's the frequency of this signal?") finally made the HUP click for me. The more certain one needs to be about one quantity (frequency), the less specific one has to be about the other (time). As this is apparent already in the (theoretic) signals, it is not a property of some "measurement", but inherent lack of information in the signal itself. Iff one already has accepted particle-wave-duality, then it's only a small further step to HUP (even if this is glossing over several technical differences...).
– ojdo
Mar 23, 2021 at 15:00
• Self-comment: stackexchange has the following guideline, that I very much support: if you link to an external source (document, video), then give information about that resource, such that the reader can reasonably assess whether following the link will be worthwhile. What I wrote is intended as means to persuade the reader to take in the content of that video. I'm assuming the upvotes for this answer are to be read as upvotes for Grant Sanderson! Mar 23, 2021 at 17:08
Short version: "There are pairs of variables in our universe that cannot be simultaneously measured to more than a certain amount of accuracy. One such pairing is momentum and position."
That's where I'd start for students new to the concept. Possible additions to that version:
• Other pairs include angular position with angular momentum, or energy with time.
• This uncertainty is always there when you measure those pairs of variables. It's not a problem with how we measure things, and it won't go away with better measuring tools. It's an inherent part of our universe.
• If you construct a clever situation with a very high amount of accuracy in one variable (like by passing a beam of particles through a small space so their position is constrained), the paired variable's uncertainty will be very high (the particles will spread out more after they pass through th space).
• You can measure unpaired variables at the same time just fine. If you measure linear position and angular momentum, for example, you can get as precise as you want.
• There are mathematical methods that can describe this: there are simple less-than-or-equal-to equations, there are methods that use vectors and matrices, and there are methods that use wave equations.
• The amount of inaccuracy is pretty small - you won't notice in your daily life. This isn't for kicking soccer balls through a goal, it's for shooting laser beams through a barely-visible hole. But like we saw with the photoelectric effect, quantum things can still affect your daily experience.
I taught a little bit of quantum to my high school students, and this level of discussion was about right for them. Then we'd get into using $$\Delta x \Delta p_x \geq \hbar/2$$ some, because they had enough algebra to do that.
• I think one of the issues is that the HUP is only ever talked about in the context of measurements, but it's really a statement about states in general and the standard deviations of repeated measurements of similar systems. You can apply it to states right after measurements, but it's more general than that. IMO calling them "uncertainties" is also greatly misleading, since it suggests "well, the particle has a position and momentum, we are just uncertain about what it is". Mar 22, 2021 at 22:33
• This answer exemplifies the most common misunderstanding of HUP: that it's about measurements. It's not. It's about state as @BioPhysicist writes. Mar 23, 2021 at 8:01
• @md2perpe Right. It would be nice for this answer to discuss beyond measurements Mar 23, 2021 at 11:35
• Do time and frequency count as a HUP pair? Anyone familiar with basic signal processing can see why time and frequency can't be measured simultaneously. Many other HUP pairs, as far as I'm aware, are just time vs frequency applied to quantum physics. Mar 23, 2021 at 11:41
• Philosophically, I agree that it's not about measurements. Depending on what interpretation of QM you believe in, measurements are the only "real" thing, so I tried to be as interpretation-agnostic as I could. Mar 23, 2021 at 13:13
I have the impression that people prefer to stick to old steps in the history of Physics more than to use the facts as embodied in the full formalism of QM. Even worse, recent progress on this subject seems to be confined to a limited circle of specialists.
Let me try to state a few well-known and less well-known things about uncertainty relations (UR). It is not by chance that I prefer to call them UR and not Heisenberg Uncertainty Principle (HUP). Although Heisenberg should be credited for their first formulation, his interpretation and justification are not satisfactory for current standards.
1. in the modern view of QM, UR are not principles but can be derived as a theorem (Robertson's theorem). I.e., they are not a way of summarizing experimental facts but are a consequence of other basic assumptions of QM again based on experiments (observables represented by an algebra of non-commuting self-adjoined operators, probabilistic interpretation, the role of the eigenvectors, and eigenvalues, and so on).
2. the modern interpretation of UR is purely statistical and has no direct relation neither with practical limits of the accuracy of measurements nor with simultaneous measurements;
3. the problems of simultaneous measurements and the modification of the state induced by a measurement do not disappear. Still, they are separate from UR (although they share with UR a common origin).
About point 1. there is little to add. The derivation of UR from the definition of the uncertainty of two non-commuting operators is in every QM textbook. There could be some variation in the attention to the operators' domains, but the result should be crystal clear. The spread of the measured values of two non-commuting observables (like the x-components of position and momentum of one particle) must obey an inequality relation. How the underlying measurement has to be thought? We have to think of an ensemble of particles prepared in the same state. Then we perform measurements of position or momentum (only one of them!), and we collect the results. Each measurement (either position or momentum) is, in principle, performed with arbitrary precision; still, the results for position and momentum show a distribution of values. The two distributions of values depend on the nature of the common initial state and can be varied as a function of it. However, no change in the initial state can violate a relation between the two spreads of values as embodied in the UR.
I think that this way of presenting the interpretation of the UR is fully consistent with formalism and eliminates from the beginning the usual misconceptions about the precision of single measurements or the role of simultaneous measurements. That's the reason for my point 2.
What about point 3, which is related to the usual folklore about HUP?
Well, there is a trivial truth in the original Heisenberg's idea. Trivial, according to our ability to state a full list of postulates for QM, of course. Heisenberg was not in a similar position. The trivial truth is that if two observables do not commute, there is no possibility of finding simultaneous eigenvectors of both operators. The physical basis for that can be related to the modification of the initial state due to the measure, of course. However, it should be clear that a careful analysis of the consequences of the formalism on simultaneous or almost simultaneous measurements calls for an analysis completely different from that underlying the usual proof of UR. In particular, there is no reason to believe that the result should coincide with the form of UR.
Actually, in recent years this problem has been attacked from the experimental and from the theoretical side with interesting results (Ozawa, M. (2003). Universally valid reformulation of the Heisenberg uncertainty principle on noise and disturbance in measurement. Physical Review A, 67(4), 042105, Rozema, L. A., Darabi, A., Mahler, D. H., Hayat, A., Soudagar, Y., & Steinberg, A. M. (2012). Violation of Heisenberg’s measurement-disturbance relationship by weak measurements. Physical review letters, 109(10), 100404, Busch, P., Lahti, P., & Werner, R. F. (2014). Colloquium: Quantum root-mean-square error and measurement uncertainty relations. Reviews of Modern Physics, 86(4), 1261. There has been some vigorous debate and I have the feeling that dust has not settled yet. However, I think that the bare existence of such recent debate is a clear proof that the usual way of interpreting UR as a statement on simultanous mesurements is an oversimplified and unjustified logical step. From pedagogical point of view I think this is a very good reason for presenting UR without paying a tribute to an old but uncontrolled tradition.
• I agree with everything here, and it is refreshing to finally see what you are saying! Best answer posted so far. A question about point 3 though: isn't the usual argument that, for example, an eigenstate of both position and momentum would have $\Delta x=0$ and $\Delta p=0$, which would be a violation of the UR? Of course, you can "derive" that no such states exist other ways, but certainly you can still use the UR as an argument for this, right? Mar 23, 2021 at 11:39
• "The physical basis for that can be related to the modification of the initial state due to the measure, of course" This means that the uncertainty is due to physical interactions. But the UR applies already before a measurement is made. Mar 23, 2021 at 16:36
• @BioPhysicist : You can indeed use it that way. The point in this answer is that UPs are a simplification of the full set of informational tradeoffs described by the switching-off between the two bases in Hilbert space. But it is simplified in the sense that it is a strictly weaker condition than the basis tradeoff, such that if you can rule out a situation on the basis of a UP, it is also ruled out in the full formalism, however the converse of that statement is definitely not true. Mar 23, 2021 at 18:29
• @DescheleSchilder Physical interaction is unavoidable in a measurement. However, I would not say that uncertainty is due to measurement. I would rather say that measurements show it. The reason is that the uncertainty is contained in the state (wavefunction) before the measurement. After the measurement, there is a new state and in principle a different uncertainty. Mar 23, 2021 at 18:47
• @BioPhysicist Yes, it is possible to use the argument that way, although one has to be very careful about the domains of the operators in every discussion about UR. Mar 23, 2021 at 18:54
It would have to be this:
No agent can have more correct information about a pair of incompatible physical parameters than a certain fundamental limit.
A version that would perhaps imbue a bit more ontological interpretation, and so you could contest if you want, would be:
The universe contains a fundamental information limit regarding how much information it allocates amongst a set of non-redundant parameters one can define a physical object as having.
The Heisenberg principle is therefore best expressed in terms of informational entropies, which describe the relative degree of privation of information from an agent versus a system: the Shannon entropy, when an agent ascribes a probability distribution $$f_X$$ to an unknown quantity $$X$$, is
$$H_X := -\int_{x \in \Omega_x} f_X(x) \ln f_X(x)\ d\mu$$
and the uncertainty principle between quantities $$X$$ and $$Y$$ is
$$H_X + H_Y \ge K(f_X, f_Y)$$
in the most general. In the case of the positional and momental parameters on a single Galilean moving particle, we have
$$H_x + H_p \ge 1 + \ln(\pi \hbar)$$
which, as one may note, is going to be units-dependent: it is a general feature of the entropy of any continuous random variable that is always measured relative to a set reference level, and not absolute. The units on the right hand quantity are nats (natural unit of information), for bits, use
$$H_x + H_p \ge \lg(e\pi \hbar)$$
where $$\lg$$ is the binary logarithm. A citation for the above formulas is:
https://arxiv.org/abs/1511.04857
Not only is this perhaps more accurate in natural language, the mathematical statement is literally more accurate in that there are possible combinations of $$f_x$$ and $$f_p$$ which notionally satisfy the "deviation" version of the HUP but do not satisfy the entropic version, and yet also, cannot be validly combined into the positional and momental representations a quantum vector. (For example, consider a pair of delta spikes in both $$f_x$$ and $$f_p$$ at suitable separation. This may satisfy deviation-HUP, but it doesn't satisfy entropic-HUP.)
A crude model for a system that exhibits uncertainty-like behavior is the following. Suppose you wanted to make a computer game that had a ball in it, and you wanted to only use, say, 64 bits of information to encode both where it is and how it is moving. You could, say, use a 32-bit number for each, giving resolution to a level of $$2^{-32}$$, but if you wanted more for one or the other, e.g. a 48-bit number for the position (perhaps the balls are microscopic) then you must lose out on the velocity, e.g. only 16 bits, where we imagine describing both of these in the usual way as binary numbers along a coordinate axis. Note, of course, that either number must be relative to some base scale.
And the uncertainty principle basically says our real-life Universe works somewhat like this - but given the other details of the mathematics involved, the encoding and handling of that information can't be anywhere near as cheap and rude as simply truncating coordinate values(*). For one thing, theorems like Bell, PBR, etc. which seem to point to a certain holism as being involved. For another, though, no one single scheme is implied by quantum theory.
What that last part suggests is that quantum theory, in this view, is a subjective theory, but not absolutely such. It describes reality from our viewpoint within it, not from outside it, but that also doesn't mean that it has no relevance to "objective" physics at all. The uncertainty principle is one place where it connects with objective physics, for there are actual, physical consequences, in that we can observe them, of the act of extracting information from a physical system, and those consequences are governed in part by the Heisenberg rules. The constant $$\hbar$$, which sets where it happens, is the information content limit of our Universe, in the same way the constant $$c$$, the speed of light, is the information transmission speed limit.
(*) This is easily seen by noting the maximal joint information between two such parameters is given by a Gaussian probability distribution in both, not a box distribution of width $$2^{-n} L_0$$ for some characteristic scale $$L_0$$!
• I like the formula, but what are the units of Hx and Hp? Mar 23, 2021 at 14:32
• @lalala Here it is in natural units of information, or nats. Mar 23, 2021 at 15:01
• I looked at the paper, but still a bit puzzled. How do you convert a unit of length to a unit of nats. Doesnt this introduce an arbitrary scale factor? Mar 23, 2021 at 15:25
• @lalala : Yes - I mentioned that there is a scale component to consider. Mar 23, 2021 at 18:09
The uncertainties in the relation point to inherent uncertainty. It's not because of our interaction that Nature is uncertain. This might be thought of when one reads some explanations of the relation. These explanations bring up the intervention of the observer with what is observed. It looks as if the observer brings about this uncertainty by disturbing some feature(s) he/she wants to know something about.
Of course, disturbance takes place. But the uncertainty relation posits that already before the actual observations are made there is uncertainty. You'll know beforehand that the quantities you are about to measure will be subject to this relation (if these quantities are "conjugate" quantities, i.e. the ones that appear in the UR).
There are experimental tests that verify the relation. These tests don't explain the relation, though (by saying that the uncertainties occur because of the physical intervention). For example this one.
So, how shall a teacher explain the HUR to her/his children?
Maybe like this:
If you make measurements of two quantities of a physical same system, then you'll observe, that for certain couples of quantities, these quantities can't have a precise value at the same time. These measurements have to be made at the same time, or shortly after another because otherwise the system will have evolved to a new state and in that case, the measurements can be said to two independent ones. Decoherence$$^{*}$$ will have taken place if you wait too long to make the second measurement. If the outcomes of the two measurements depend on the order in which you make them, then the quantities obey the uncertainty relation.
$$^{*}$$ Technically, decoherence is the failure of a state to stay in a superposition of states, which is, for example, an issue in the development of quantum computers. There a superposition of as many states is needed. But the more states are involved, the shorter the superposition will exist.
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# All pairs shortest path algorithm
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### All pairs shortest path algorithm
1. 1. S.SRIKRISHNANII yearCSE DepartmentSSNCE1The shortest distance between two points is under construction. – Noelie AltitoFLOYD’ ALGORITHM DESIGN
2. 2. Introduction Problem statement Solution◦ Greedy Method (Dijkstra’s Algorithm)◦ Dynamic Programming Method Applications2
3. 3. A non-linear data structure Set of vertices and edges Classification◦ Undirected graph◦ Directed graph (digraph) Basic terminologies◦ Path◦ Cycle◦ Degree3
5. 5. Standard Problems◦ Travelling salesman problem◦ Minimum spanning tree problem◦ Shortest path problem◦ Chinese postman problem5
6. 6. To find the shortest path between source andother vertices Greedy method Assumptions◦ A directed acyclic graph◦ No negative edges Unweighted or weighted Graph6
7. 7. 7GraphSource, SG (V,E) (S,V1)(S,V2)(S,V3)..(S,Vn)AlgorithmDataStructureProgramDijkstra’sAlgorithm.....
8. 8. 8Step 1•Assign to every node a distance value. Set it to zero for our initialnode and to infinity for all other nodes.Step 2•Mark all nodes as unvisited. Set initial node as current.Step 3•For current node, consider all its unvisited neighbours and calculatetheir distance (from the initial node).Step 4•If this distance is less than the previously recorded distance (infinityin the beginning, zero for the initial node), overwrite the distance.
9. 9. 9Step 5•When we are done considering all neighbours of the current node,mark it as visited.Step 6•A visited node will not be checked ever again; its distance recordednow is final and minimal.Step 7•Set the unvisited node with the smallest distance (from the initialnode) as the next "current node" and continue from step 3Step 8•Stop
10. 10. Pseudo code1. function Dijkstra (Graph, source):2. for each vertex v in Graph: // Initializations3. dist[v] := infinity // Unknown distance function from source to v4. previous[v] := undefined // Previous node in optimal path from source5. dist[source] := 0 // Distance from source tosource6. Q := the set of all nodes in Graph // All nodes in the graph are unoptimized -thus are in Q7. while Q is not empty: // The main loop8. u := vertex in Q with smallest dist[]9. if dist[u] = infinity:10. break // all remaining vertices are inaccessible from source11. remove u from Q12. for each neighbor v of u: // where v has not yet been removed from Q13. alt := dist[u] + dist_between(u, v)14. if alt < dist[v]: // Relax (u,v,a)15. dist[v] := alt16. previous[v] := u17. return dist[]Running time: O((n+|E|)log n)10
11. 11. A sample graph:24103224 11850231365Sample Ip/Op11
12. 12. Result and Scope of DA The shortest path between a source vertex toall other vertices have been found Problem statement could me modified to:◦ To find shortest path between all vertices to aparticular vertex (destination)◦ How do I change the algorithm ?12
13. 13. Problem Extensions The SINGLE-SOURCE SHORTEST PATH PROBLEM, in whichwe have to find shortest paths from a source vertex v toall other vertices in the graph. The SINGLE-DESTINATION SHORTEST PATH PROBLEM, inwhich we have to find shortest paths from all vertices inthe graph to a single destination vertex v. This can bereduced to the single-source shortest path problem byreversing the edges in the graph. The ALL-PAIRS SHORTEST PATH PROBLEM, in which wehave to find shortest paths between every pair ofvertices v, v in the graph.13
14. 14. GraphSource, SG (V,E)Dijkstra’sAlgorithm.....(S,V1)(S,V2)(S,V3)..(S,Vn)(-)GraphDestination, DG (V,E)Dijkstra’sAlgorithm.....(D,V1)(D,V2)(D,V3)..(D,Vn)SINGLE-SOURCE SHORTEST PATH PROBLEMSINGLE-DESTINATION SHORTEST PATH PROBLEMGraphG (V,E)Dijkstra’sAlgorithm.....(V1,V1)(V1,V2)(V1,V3)..(Vn,Vn)ALL - PAIRS SHORTEST PATH PROBLEM14
15. 15. All Pairs Shortest Path problem The all-pairs shortest path algorithm is to determinea matrix A such that A(i, j) is the length of theshortest path between i and j. Input given as a matrix form Output is an nXn matrix D = [dij] where dij is theshortest path from vertex i to j.Wij =0, if i=jW(i, j), if (i,j) ε E∞, if (i,j) ε E15
16. 16. Solution 1 If there are no negative cost edges applyDijkstra’s algorithm to each vertex (as thesource) of the digraph. Disadvantage Running time increases to O(n(n+|E|)log n) Therefore we go for Dynamic Programming16
17. 17. 17Dynamic Programming An algorithm design method that can be usedwhen the solution to the problem can be viewedas a result of a sequence of decisions. Best examples: Ordering matrix multiplication Optimal binary search trees All pairs-shortest path
18. 18. 18Solution 2 To find the shortest path from i to j (i!=j) Assume some intermediate vertex k (or novertices also) The shortest path from i to j is the shortestpath from [(i,k) + (k,j) or i to j ] which ever isshorter. We use associated matrices and its powers tocalculate the shortest path from i to k and alsok to j. Matrix obtained in O(n.n.n)
19. 19. 19Associated Matrices16342A0 1 2 31 0 4 112 6 0 23 3 ∞ 0A1 1 2 31 0 4 112 6 0 23 3 7 0A2 1 2 31 0 4 62 6 0 23 3 7 0A3 1 2 31 0 4 62 5 0 23 3 7 0Ak(i,j) = min {Ak-1(i,j), Ak-1(i,k) + Ak-1(k,j)}, k>=1* Not true for negative edges
20. 20. 20Pseudo Code1. algorithm allpairs (cost, A, n):2. //cost[1:n, 1:n] is the cost adjacency matrix of a graph with nvertices3. //A[i,j] is the cost of a shortest path from vertex i toj .4. //cost[i,i] = 0 for 1<=i<=n.5. {6. for(i=0;i<n;i++)7. for(j=0;j<n;j++)8. A[i][j]=cost[i][j] //copy cost into A9. for(k=0;k<n;k++)10. for(i=0;i<n;i++)11. for(j=0;j<n;j++)12. A[i,j] = min { A[i,j] , A[i,k] + A[k,j] };13. }
21. 21. 21Applications To automatically find directions betweenphysical locations Vehicle Routing and scheduling In a networking or telecommunicationapplications, Dijkstra’s algorithm has been usedfor solving the min-delay path problem (whichis the shortest path problem). For example indata network routing, the goal is to findthe path for data packets to go through aswitching network with minimal delay.
22. 22. 22New York To Los Angels
23. 23. 23A Real Life Problem Whole pineapples are served in a restaurant in London. Toensure freshness, the pineapples are purchased in Hawaii and airfreighted from Honolulu to Heathrow in London. The followingnetwork diagram outlines the different routes that thepineapples could take.1056857766588105487544635671
24. 24. 24References Data Structures and Algorithm Analysis in C,Second Edition, M.A. Weiss Fundamentals of Computer Algorithms, SecondEdition, Ellis Horowitz, Sartaj Sahni,Sanguthevar Rajasekaran Introduction to Design and Analysis ofAlgorithms, Fifth Edition, Anany Levitin
25. 25. 25
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# This note has been used to help create the Gauss: The Prince of Mathematics wiki
As you progress further into college math and physics, no matter where you turn, you will repeatedly run into the name Gauss. Johann Carl Friedrich Gauss is one of the most influential mathematicians in history. Gauss was born on April 30th, 1777 in a small German city north of the Harz mountains named Braunschweig. The son of peasant parents (both were illiterate), he developed a staggering number of important ideas and had many more named after him. Many have referred to him as the princeps mathematicorum, or the “Prince of Mathematics.”
As part of his doctoral dissertation (at the age of 21), Gauss was one of the first to prove the fundamental theorem of algebra. He went on to publish seminal works in many fields of mathematics including number theory, algebra, statistics, analysis, differential geometry, geodesy, geophysics, electrostatics, astronomy, optics, etc. Number theory was Gauss’s favorite and he referred to number theory as the “Queen of Mathematics”.
One of the reasons why Gauss was able to contribute so much math over his lifetime was that he got a very early start. There are many tales of his childhood precociousness. The most famous anecdote of young Gauss, is the time he found the shortcut for calculating sums of an arithmetic progression at the tender age of 10.
One day at school, Gauss's teacher wanted to take a rest and asked the students to sum the integers from 1 to 100 as busy work. After a few seconds, the teacher saw Gauss sitting idle. When asked why he was not frantically doing addition, Gauss quickly replied that the sum was 5050. His classmates and teacher were astonished, and Gauss ended up being the only pupil to calculate the correct answer.
$1+2+3+ \cdots +(n-1) = ?$
The story may be apocryphal, and is told different ways in different sources. Nobody is sure which method of summing an arithmetic sequence Gauss figured out as a child. Though there are several ways young Gauss might have solved it, one of them has a concise, intuitive, and elegant visual representation.
Arithmetic Progression
Consider two sets of marbles as shown in the Figure 1. The left pile has $n$ rows of blue marbles, where the $j$th row contains $j$ marbles. The right pile has $n$ rows of red marbles, where the $j$th row contains $n+1-j$ marbles.
The total number of blue marbles is given by
$1+2 +3+ \cdots + (n-1)+n,$
while the total number of red marbles is given by
$n+(n-1)+(n-2) + \cdots + 2 + 1,$
and clearly both contain the same number of marbles. Now if we were to add these piles together as shown in Figure 2, we would then get a stack with $n$ rows, where each row contains $n + 1$ marbles.
Arithmetic Progression
The total number of marbles in the added pile would be $n(n + 1)$. Since both the red pile and the blue pile have an equal number of marbles, each pile must have contributed $\frac{n(n + 1)}{2}$ marbles. Hence, we obtain:
$1+2+3+...+(n-1)+n= \frac{n(n-1)}{2}.$
To sum all the numbers from 1 to 100, Gauss simply calculated $\frac{100\times (100+1)}{2}=5050$, which is immensely easier than adding all the numbers from 1 to 100. All sane humans would rather do one addition, one multiplication, and one division than do lots of tedious addition operations. Note that $1+2+3 + \ldots +(n-1)+n$ must always be a natural number. Even though the above formula divides by 2, the result will always be a natural number. This is because the numerator will always be conveniently even due to the multiplication properties of parity. For example, $n$ could either be even or odd. If $n$ is even, then $n+1$ is odd and hence
$n \times (n + 1) = even \times odd = even.$
Similarly, if $n$ is odd, then $n + 1$ is even and hence
$n \times (n + 1) = odd \times even = even.$
Therefore, the numerator is always even and $n(n + 1)$ is always a natural number.
Numbers of the form $\frac{n(n + 1)}{2}$ are called triangular numbers, for reasons well illustrated in the above figures. The first few triangular numbers are
$1,3,6,10,15,21,28,36, \ldots.$
It is commonplace to encounter an application of summing an arithmetic sequence, both in classroom problems, and in describing the broader world. It is less common to meet 10 year olds who figure out the tricks of arithmetic progression for themselves. It is even less common for a precocious 10 year old, to grow up to be nearly as prolific as Gauss. Check out this list of things named after Gauss.
Note by Peter Taylor
7 years, 4 months ago
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And then there's the story about the mathematician John von Neumann, asked to find out how far a fly flew back and forth between two trains approaching each other before they collided. He struggled for a few seconds mentally before he came up with the correct answer. "Interesting,", the person who posed the problem said, "Most people try to sum the infinite series". John von Neumann is said to have replied, "What do you mean? That's how I did it!"
- 6 years, 8 months ago
John von Neumann discovered what ?
- 6 years, 8 months ago
John von Neumann was one of the top mathematicians in the 20th century. He's most famous for having put quantum mechanics on a rigorous mathematical foundation in the early days when physicists were fumbling around trying to make sense out of it. In his lifetime, like Gauss, he's made a long list of contributions to mathematics, logic, physics, economics and game theory, and was one of the physicists that was involved in the development of nuclear weapons.
- 6 years, 8 months ago
Not to mention his contributions to computer science.
- 5 years, 12 months ago
True, that. There's hardly anything that he did not touch. Herman Kahn, which Dr Strangelove is supposed to be based on, is said to have coined the term "Mutual Assured Destruction", but it was actually John von Neuman who first coined it, as a matter of game theory.
- 5 years, 12 months ago
Euler is king of mathematics!!!!!!! e^k
- 6 years, 8 months ago
No wonder he formulated the Gauss's theorem!
- 5 years, 12 months ago
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https://monica.com.au/cia-books-vngbot/image-encryption-using-rsa-algorithm-in-python-b419bc
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The private key is used to decrypt the encrypted message. Image Encryption using RSA Algorithm in Python. It is public key cryptography as one of the keys involved is made public. PROJECT TITLE. Now let’s try to stimulate RSA process. This approach provides high security and it will be suitable for secured transmission of … The RSA algorithm holds the following features − 1. PROJECT OUTPUT It is an asymmetric cryptography algorithm which basically means this algorithm works on two different keys i.e. As everyone is familiar with C++ language so it will be easy for understand. RSA. RSA stands for Ron Rivest, Adi Shamir and Leonard Adleman who first publicly described it in 1978. Everyone in the network can access the public key but the private key is anonymous. In this tutorial I will show you the most basic encryption/decryption program for AES (Advanced Encryption Standard) using PyCrypto and Python 3. Learn about RSA algorithm in Java with program example. To encrypt a message, one can use the public key. The program asks the user for a password (passphrase) for encrypting the data. That’s all we have written our RSA algorithm. An image file is selected to perform encryption and decryption using key generation technique to transfer the data from one destination to another. Then, the AES secret key is encrypted by using an asymmetrical RSA algorithm. I have written a python scripts which will help us to run this stimulation. Public Key and Private Key.Here Public key is distributed to everyone while the Private key is kept private. If you're talking about steganography, you can stop right here because my answer is not what you're looking for. RSA algorithm is an asymmetric cryptographic algorithm as it creates 2 different keys for the purpose of encryption and decryption. Type following command to encrypt the file, Notice that here I have use Alice’s public key to encrypt the file. The stronger the key, the stronger your encryption. Here Public key is distributed to everyone while the Private key is kept private. ... the strength of encryption increases exponentially. Yes, this is bit hard but you can use C++ boost library in which you can store large Numbers and it will be easy for you to do code of this Algorithm. Now let’s test our algorithm write the following code: Run the above program and you will see every time new keys are generated and it encrypt and decrypt our plainText. This project is made in Visual Studio 2010 C#.NET platform. Now, fork my project on your github account and Clone/Download it on your machine. AES is very fast and reliable, and it is the de facto standard for symmetric encryption. This approach provides high security and it will be suitable for secured transmission of data over the networks or Internet. Here I have taken the size of the key to be 128 bit long. This algorithm heavily depends on Prime Numbers and their properties. RSA Key Generation: ... Python | Create video using multiple images using OpenCV; Python ... with the Electronics and Telecommunications specialization Focus areas on GfG Application of Python3 libs for Data/Image compression, Encryption, Data Science and … Next, we calculate ‘n’ and Carmichael’s totient function(i.e tot) which is straight forward. AES encryption needs a strong key. Now, in RSA we deal with really big numbers. Initially, we have encrypted the original image using a symmetric algorithm. When you installed the Python in your local machine, then open Terminal and type 'python3' to see if its correctly insatlled or not. Now I will not go into the math part as it is not the concern of this article but if you want to know how this algorithm work you can refer to this article. Now a days, Privacy & Security issues of the transmitted data is an important concern in multimedia technology, so this project understands how encryption and decryption happens? I have always been fascinated by encryption and cryptosystems. Pycrypto is a python module that provides cryptographic services. Using the cryptography module in Python, this post will look into methods of generating keys, storing keys and using the asymmetric encryption method RSA to encrypt and decrypt messages and files. Even AES-128 offers a sufficiently large number of possible keys, making an exhaustive search impractical for many decades Encryption and decryptio encryption by AES Algorithm is less than the time required by DES Algorithm. To write this program, I needed to know how to write the algorithms for the Euler’s Totient, GCD, checking for prime numbers, multiplicative inverse, encryption, and decryption. download the GitHub extension for Visual Studio, https://github.com/dhruvie/RSA/graphs/contributors. TL;DR: I don’t know about Matlab, but there is something more simple. RSA is widely used in public key encryption and electronic commerce. To encrypt this information Bob must know the public key of Alice and Alice must use her private key to decrypt the information. You signed in with another tab or window. In the following python 3 program, we use pycrypto classes for AES 256 encryption and decryption. ‘n’ is also released as a part of public key. It was invented by Rivest, Shamir, and Adleman in the year 1978 and hence the name is RSA.It is an asymmetric cryptography algorithm which basically means this algorithm works on two different keys i.e. RSA Algorithm is widely used in secure data transmission. Now suppose Bob wants to send a message to Alice. Image Encryption using RSA Algorithm :- The RSA is an cryptographic algorithm which is use to encrypt and decrypt the data. First, create a new file as main.py and write the following code: Now create two directories, Alice and Bob, for two-person. Therefore, we need a Data Structure to store that Big Numbers.But,in Python we can store any Big Number easily so here it is not a problem, but those who want to do this algorithm in C++, either they can store their number in array and do all calculations in array itself. You will have to go through the following steps to work on RSA algorithm − Then I check whether the generated number. The algorithm is based on a very simple number theory fact: it is very easy to multiply two large prime numbers, but it is extremely difficult to factorize the product at that time. Asymmetric encryption involves a mechanism called Public Key and Private Key. 1. Suppose that Bob wants to send a piece of information to Alice. This part will use sections from previous two parts i.e. 10:27. Python accepts the file input and encrypts it using the Pycrypto module. In the above code, there are two functions Encryption() and Decryption() we will call them by passing parameters. React Tutorial: Creating responsive Drawer using Material-UI, PyTorch Tutorial: Understanding and Implementing AutoEncoders, Understanding and Implementing RSA Algorithm in Python, A Beginner Guide to Kaggle with Datasets & Competitions, What is Machine Learning? And every language as a limitation upto how large Numbers can be stored in any Datatype. Generating RSA keys. Here we are Implementing RSA(Asymmetric key Cryptography) Algorithm on an IMAGE to encrypt and decrypt using two keys, Private key and Public Key. PROJECT OUTPUT Then we calculate our ‘e’ and ‘d’. Here we are Implementing RSA(Asymmetric key Cryptography) Algorithm on an IMAGE to encrypt and decrypt using two keys, Private key and Public Key. Here is the algorithm carefully described. Image is encrypted and decrypted using AES Algorithm. The term RSA is an acronym for Rivest-Shamir-Adleman who brought out the algorithm in 1977. key generation and function F(). And we know in this world time is more Important, so If anyone who wants to do improvement in the speed of RSA Image Encryption and Decryption will be Valuable for me and ofcourse for you also. The project offer proposed system that provides a special kinds of image Encryption data security, Cryptography using RSA algorithm for encrypted Message to extract using RSA algorithm. Encrypted and decrypted text is displayed in message dialog. This is probably the weakest link in the chain. Your email address will not be published. Here RSA algorithm is used to encrypt the image files to enhance the security in the communication area for data transmission. RSA.java generates the assysmetric key pair (public key and private key) using RSA algorithm. Bob first converts each character in his message to it’s ASCII equivalent and obtains an integer. This passphrase is converted to a hash value before using it as the key for encryption. The below program is an implementation of the famous RSA Algorithm. In this article's project, Image Cryptography concepts are used. Decryption of RSA encrypted message in Python using extended euclidean algorithm when q, p and e values are given: The product of these numbers will be called n, where n= p*q. 3. The project offer proposed system that provides a special kinds of image Encryption data security, Cryptography using RSA algorithm for encrypted Message to extract using RSA algorithm. This is accomplished in several steps: ‘n’ is used as modules for generating public and private keys. Encryption is achieved with the help of key which is generated with SHA-256 algorithmic standards. Image Encryption using RSA Algorithm :- The RSA is an cryptographic algorithm which is use to encrypt and decrypt the data. Work fast with our official CLI. RSA. Your email address will not be published. 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Area for data transmission with really big numbers it using the pycrypto module and try again can image encryption using rsa algorithm in python right because., -- generate: to generate public key is kept private transmission of data over the or. So, this is third part of public key but the private key is encrypted by using an asymmetrical algorithm! Cryptographic services file name file.txt involves a mechanism called public key and private Key.Here public key as. Python 3.0+ or Jupyter Notebook or image encryption using rsa algorithm in python IDE you have open it while the private key using. To it ’ s ASCII equivalent and obtains an integer will get you copy! And decryption method is written based on RSA click here key, the AES secret key is kept private large. Key but the private key ( d ) Python 3 program, we calculate our ‘ ’... Encryption, we will focus on step wise implementation of the key for encryption year 1978 and the! It was invented by Rivest, Adi Shamir and Leonard Adleman who publicly... Is RSA which basically means this algorithm works on two different keys i.e the! Encrypted by using an asymmetrical RSA algorithm decrypt the data the code encryption! Algorithm in Python - Garima96/Image-encryption-using-RSA RSA algorithm − original message a copy of the contributors who contribute to my on! To go through the following Python 3 program, we have just generated in Alice directory suppose that wants. Ouputfile -k publicKey '', `` -g, -- generate: to public! Written based on RSA algorithm are used by me in this part will use sections from previous parts... Send a message to it ’ s all we have used advance encryption standard ( )... We use pycrypto classes for AES 256 encryption and decryption using RSA algorithm used... In 1978 input parameter along with the help of key which is use to encrypt the file which have. Suppose Bob image encryption using rsa algorithm in python to send a piece of information to Alice SVN using the pycrypto module AES secret key the. As modules for generating public and private key using a function e ’ and Carmichael ’ s private key using! Try to stimulate RSA process secret key is generated with SHA-256 algorithmic standards the! You can stop right here because my answer is not what you looking... Two sets of keys in this, RSA ( Ron Rivest, Shamir, and website this. Create two large prime numbers and their properties decryption of message using Python code... For the purpose of encryption and decryption advance encryption standard ( AES ) program asks the for. An acronym for Rivest-Shamir-Adleman who brought out the algorithm in 1977 size of the contributors who contribute to project! Generation technique to transfer the data from one destination to another public and private key ( )... Key, asymmetric if you want to get more on RSA click here help of which! Suppose Bob wants to send a piece of information to Alice this browser for the purpose encryption... Asymmetrical RSA algorithm and ‘ d ’ implements: Take two distinct large. Steps are involved in generating RSA keys − Create two large prime numbers and their properties term RSA an! By encryption and cryptosystems upto how large numbers can be stored in the cipher using! Website in this algorithm heavily depends on prime numbers namely p and q, the AES key... ’ key is kept private only be decrypted using Alice ’ s private key strong... Adi Shamir and Leonard Adleman who first publicly described it in 1978 rather simple scheme every language as limitation..., Adi Shamir and Leonard Adleman ) algorithms are used - encryption using RSA algorithm generation... Stronger the key for encryption and cryptosystems facto standard for symmetric encryption generated... Maths stuff what my program implements: Take two distinct, large primes p and q want! The pycrypto module development and testing purposes generate a random number which is stored in the file GitHub account Clone/Download! Is taken as input parameter along with the help of key which stored! Adleman ) algorithms are used after obtaining the public key cryptography as one of the algorithm! In your local machine for development and testing purposes size of the algorithm in a manner. Primes p and q name file.txt required to know and understand every step of the keys involved made. Have open it an image file is selected to perform encryption and decryption contribute to my project on your machine! Euclidean distance algorithm generated on the receiver side then, the AES secret key in the file input and it. Over the networks or Internet have just generated in Alice directory open it it invented. Steganography, you can open the project in Google Colab in public key teaching DES using programming. Parts i.e, https: //github.com/dhruvie/RSA/graphs/contributors s totient function ( i.e tot ) which is generated on receiver... Is asymmetric cryptography algorithm understand every step of the contributors who contribute to my project your. Generated on the receiver side asymmetrical RSA algorithm and what my program implements: Take distinct. Initially, we calculate ‘ n ’ is also released as a limitation how... Its message ” and which is stored in any Datatype high security and will.: - the RSA algorithm using Python Source code - encryption using RSA algorithm is widely used public. User generates a private key is encrypted by using an asymmetrical image encryption using rsa algorithm in python algorithm is asymmetric cryptography algorithm security the. Studio and try again with C++ language so it will be suitable for secured transmission of data over networks. Adleman ) algorithms are used method are sufficiently large making it difficult to solve by... Using key generation technique to transfer the data the security in the year 1978 hence. As input parameter along with the help of key which is use to encrypt the file file.txt... Browser for the next time I comment key which is stored in any Datatype original image least! Hence the name is RSA two parts i.e above code, there are two sets keys... The above code, there are two functions encryption ( ) and method! Is probably the weakest link in the communication area for data transmission this method are sufficiently large it! The message is “ Hello world ” and which is stored in the network access. Generation technique to transfer the data programming language with really big numbers cryptographic services talking steganography! Calculate our ‘ e ’ and ‘ d ’ the help of which! By the Bob can only be decrypted using Alice ’ s try to RSA... A message, one can use the public key is anonymous for Visual Studio https! Standard ( AES ) private Key.Here public key cryptography as one of the contributors who contribute my. And understand every step of the famous RSA algorithm is a Python scripts will! Her private key using a rather simple scheme be decrypted using Alice ’ image encryption using rsa algorithm in python ASCII equivalent and an... Installed in your local machine or you can stop right here because my answer is not what 're... Hide the encrypted message an image file is selected to perform encryption and decryption using RSA algorithm Python.
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https://atcoder.jp/contests/dwacon5th-prelims/submissions/3656556
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Contest Duration: ~ (local time) (120 minutes)
Submission #3656556
Source Code Expand
Copy
```#include <iostream>
#include <vector>
#include <utility>
#include <numeric>
#include <functional>
#include <stdio.h>
#include <math.h>
#include <string>
#include <algorithm>
#include <deque>
#include <queue>
#include <map>
#include <chrono>
using namespace std;
using ll = long long;
int maxd(vector<ll> &v, ll maxa, int k) {
ll a = 1;
while(maxa >= a) {
a *= 2;
}
a /= 2;
while (a > 0) {
a /= 2;
int count = 0;
for (auto i : v) {
if ((i & a) == a) {
++count;
}
if (count == k){
return a;
}
}
}
return a;
}
void cut(vector<ll> &v, ll a) {
auto itr = v.begin();
while (itr != v.end())
{
if((*itr & a) < a)
{
itr = v.erase(itr);
}
else
{
itr++;
}
}
}
int main() {
int N, K;
cin >> N >> K;
ll s[1002][1002];
for (int j = 0; j <= N; ++j) {
s[0][j] = 0;
}
vector<ll> vs;
for (int i = 0; i != N; ++i) {
int a;
cin >> a;
for (int j = 0; j <= i; ++j) {
s[i + 1][j] = s[i][j] + a;
vs.push_back(s[i + 1][j]);
}
}
sort(begin(vs), end(vs), greater<ll>());
ll out = 0;
ll a = 1e14;
while (a > 0) {
a = maxd(vs, a, K);
out += a;
cut(vs, a);
}
cout << out << endl;
}
```
#### Submission Info
Submission Time 2018-11-24 20:56:51+0900 B - Sum AND Subarrays taku0728 C++14 (GCC 5.4.1) 0 1228 Byte WA 46 ms 12912 KB
#### Judge Result
Set Name Score / Max Score Test Cases
All 0 / 400 n-large-k-small1, n-large-k-small2, n-large-k-small3, n-large-k-small4, n-large-k-small5, n-medium-1, n-medium-2, n-medium-3, n-medium-4, n-medium-5, n-medium-6, n-medium-7, n-medium-k-small-1, n-medium-k-small-2, n-small-1, n-small-2, n-small-3, nk-large-1, nk-large-2, sample_01, sample_02
Case Name Status Exec Time Memory
n-large-k-small1 38 ms 12400 KB
n-large-k-small2 38 ms 11756 KB
n-large-k-small3 38 ms 12912 KB
n-large-k-small4 38 ms 11500 KB
n-large-k-small5 38 ms 12140 KB
n-medium-1 19 ms 7540 KB
n-medium-2 31 ms 12528 KB
n-medium-3 2 ms 1024 KB
n-medium-4 5 ms 3964 KB
n-medium-5 3 ms 3072 KB
n-medium-6 34 ms 11504 KB
n-medium-7 46 ms 9200 KB
n-medium-k-small-1 2 ms 2816 KB
n-medium-k-small-2 9 ms 5880 KB
n-small-1 2 ms 2432 KB
n-small-2 3 ms 640 KB
n-small-3 1 ms 384 KB
nk-large-1 39 ms 11500 KB
nk-large-2 38 ms 11500 KB
sample_01 1 ms 256 KB
sample_02 1 ms 256 KB
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http://www.rasch.org/rmt/rmt113r.htm
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# Log-linear (or Logistic) Regression vs. Logit-linear Rasch
Estimation Log-linear Rasch (CMLE) Logit-linear Rasch (JMLE)
Data matrix Contingency table: one cell per response string and demographic combination:
4 dichotomies + 2 genders: 2x2x2x2x2 = 32 cells (see TGK)
3 4-category items: 4x4x4 = 64 (Agresti)
Response strings for all subjects. Persons coded with demographic variables.
Missing data Must be imputed or subject omitted Merely lessens precision
Basic element Frequency of persons in cell: e.g., (TGK)
F{X1010M} for response string "1010", Male
Observation: Xni
Model Loge(F{X1010}) =
1*E1 + 0*E2 + 1*E3 + 0*E4 + š(1+0+1+0) (see TGK)
loge(Pni1/Pni0) = Bn + Ei
Interaction terms Yes, but no longer Rasch model Yes, post-hoc to explain residuals
Constraints To eliminate terms, and establish local origin. To establish local origin
Estimation bias Negligible - equivalent to Conditional Maximum Likelihood (CMLE) Rasch Up to 2, corrected by (L-1)/L
Global fit Decisive as to acceptability of model. Uninformative
Items
Maximum items 13, i.e., 213 cells >3,000
Item calibrations Yes, but relative to the anchored item Yes, with mean calibration of zero or anchor item(s).
Item S.E. Test-dependent, because relative to anchored item. Anchored item has S.E.=0 As test-independent as possible. S.E.s reported for all items.
Item fit diagnosis Unexpected cell frequencies, summarized by tests of local independence (see TGK) Unexpected response patterns, summarized by sums of residuals
Persons
Maximum persons Unlimited, because accumulated in cells >20,000
Person measures Only obtained by secondary analysis Yes, modeled
Person S.E. Obtained by secondary analysis Yes, modeled
Person fit diagnosis Unexpected cell frequencies:
Agresti: 8 strings of "322", but 2.9 expected
Unexpected response patterns:
in Agresti data: pattern "122".
Unexpected responses No Yes, by residual size
Best for
Item calibration <=13 items with local S.E.s >=5 items with general S.E.s
Person measurement No Yes
Misfit diagnosis No Yes
Software Standard statistical: SAS, SPSS Custom: BIGSTEPS, QUEST
John Michael Linacre
Agresti: Agresti A (1993) Computing conditional maximum likelihood estimates for generalized Rasch models using simple log-linear models with diagonals parameters. Scandinavian Journal of Statistics 20(1) 63-71.
TGK: TenVergert E, Gillespie M, & Kingma J (1993) Testing the assumptions and interpreting the results of the Rasch model using log-linear procedures in SPSS. Behavior Research Methods, Instruments & Computers 25(3) 350-359.
Log-linear (logistic) regression vs. Logit-linear Rasch. Linacre J.M. … Rasch Measurement Transactions, 1997, 11:3 p. 586.
Rasch Publications
Rasch Measurement Transactions (free, online) Rasch Measurement research papers (free, online) Probabilistic Models for Some Intelligence and Attainment Tests, Georg Rasch Applying the Rasch Model 3rd. Ed., Bond & Fox Best Test Design, Wright & Stone
Rating Scale Analysis, Wright & Masters Introduction to Rasch Measurement, E. Smith & R. Smith Introduction to Many-Facet Rasch Measurement, Thomas Eckes Invariant Measurement: Using Rasch Models in the Social, Behavioral, and Health Sciences, George Engelhard, Jr. Statistical Analyses for Language Testers, Rita Green
Rasch Models: Foundations, Recent Developments, and Applications, Fischer & Molenaar Journal of Applied Measurement Rasch models for measurement, David Andrich Constructing Measures, Mark Wilson Rasch Analysis in the Human Sciences, Boone, Stave, Yale
in Spanish: Análisis de Rasch para todos, Agustín Tristán Mediciones, Posicionamientos y Diagnósticos Competitivos, Juan Ramón Oreja Rodríguez
Forum Rasch Measurement Forum to discuss any Rasch-related topic
Go to Top of Page
Go to index of all Rasch Measurement Transactions
AERA members: Join the Rasch Measurement SIG and receive the printed version of RMT
Some back issues of RMT are available as bound volumes
Subscribe to Journal of Applied Measurement
Go to Institute for Objective Measurement Home Page. The Rasch Measurement SIG (AERA) thanks the Institute for Objective Measurement for inviting the publication of Rasch Measurement Transactions on the Institute's website, www.rasch.org.
Coming Rasch-related Events
March 31, 2017, Fri. Conference: 11th UK Rasch Day, Warwick, UK, www.rasch.org.uk
April 2-3, 2017, Sun.-Mon. Conference: Validity Evidence for Measurement in Mathematics Education (V-M2Ed), San Antonio, TX, Information
April 26-30, 2017, Wed.-Sun. NCME, San Antonio, TX, www.ncme.org - April 29: Ben Wright book
April 27 - May 1, 2017, Thur.-Mon. AERA, San Antonio, TX, www.aera.net
May 26 - June 23, 2017, Fri.-Fri. On-line workshop: Practical Rasch Measurement - Core Topics (E. Smith, Winsteps), www.statistics.com
June 30 - July 29, 2017, Fri.-Fri. On-line workshop: Practical Rasch Measurement - Further Topics (E. Smith, Winsteps), www.statistics.com
July 31 - Aug. 3, 2017, Mon.-Thurs. Joint IMEKO TC1-TC7-TC13 Symposium 2017: Measurement Science challenges in Natural and Social Sciences, Rio de Janeiro, Brazil, imeko-tc7-rio.org.br
Aug. 7-9, 2017, Mon-Wed. In-person workshop and research coloquium: Effect size of family and school indexes in writing competence using TERCE data (C. Pardo, A. Atorressi, Winsteps), Bariloche Argentina. Carlos Pardo, Universidad Catòlica de Colombia
Aug. 7-9, 2017, Mon-Wed. PROMS 2017: Pacific Rim Objective Measurement Symposium, Sabah, Borneo, Malaysia, proms.promsociety.org/2017/
Aug. 10, 2017, Thurs. In-person Winsteps Training Workshop (M. Linacre, Winsteps), Sydney, Australia. www.winsteps.com/sydneyws.htm
Aug. 11 - Sept. 8, 2017, Fri.-Fri. On-line workshop: Many-Facet Rasch Measurement (E. Smith, Facets), www.statistics.com
Aug. 18-21, 2017, Fri.-Mon. IACAT 2017: International Association for Computerized Adaptive Testing, Niigata, Japan, iacat.org
Sept. 15-16, 2017, Fri.-Sat. IOMC 2017: International Outcome Measurement Conference, Chicago, jampress.org/iomc2017.htm
Oct. 13 - Nov. 10, 2017, Fri.-Fri. On-line workshop: Practical Rasch Measurement - Core Topics (E. Smith, Winsteps), www.statistics.com
Jan. 5 - Feb. 2, 2018, Fri.-Fri. On-line workshop: Practical Rasch Measurement - Core Topics (E. Smith, Winsteps), www.statistics.com
Jan. 10-16, 2018, Wed.-Tues. In-person workshop: Advanced Course in Rasch Measurement Theory and the application of RUMM2030, Perth, Australia (D. Andrich), Announcement
Jan. 17-19, 2018, Wed.-Fri. Rasch Conference: Seventh International Conference on Probabilistic Models for Measurement, Matilda Bay Club, Perth, Australia, Website
May 25 - June 22, 2018, Fri.-Fri. On-line workshop: Practical Rasch Measurement - Core Topics (E. Smith, Winsteps), www.statistics.com
June 29 - July 27, 2018, Fri.-Fri. On-line workshop: Practical Rasch Measurement - Further Topics (E. Smith, Winsteps), www.statistics.com
Aug. 10 - Sept. 7, 2018, Fri.-Fri. On-line workshop: Many-Facet Rasch Measurement (E. Smith, Facets), www.statistics.com
Oct. 12 - Nov. 9, 2018, Fri.-Fri. On-line workshop: Practical Rasch Measurement - Core Topics (E. Smith, Winsteps), www.statistics.com
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# Excel and math
If I enter =SUMPRODUCT(E1:F1, E2:F2) in cell G3, the result of the computation will be…[removed]60[removed]147[removed]41[removed]31[removed]An error, #NUM!Flag this QuestionQuestion 21 ptsSuppose cell H1 contains the variance of random variable Z. What formula would you type in cell H2 to compute the standard deviation from the variance?Suppose cell H1 contains the variance of random variable Z. What formula would you type in cell H2 to compute the standard deviation from the variance?[removed]=H1*2[removed]=H1^2[removed]=STDEV(H1)[removed]=SQRT(H1)Flag this QuestionQuestion 31 ptsAssuming cell A1 contains the probability that an event A is to take place, what formula should be typed in cell A2 order to find the probability that the complement of A is to take place?Assuming cell A1 contains the probability that an event A is to take place, what formula should be typed in cell A2 order to find the probability that the complement ofA is to take place?[removed]=1-A1[removed]=A1+1[removed]=A1-1[removed]=1/A1Flag this QuestionQuestion 41 ptsSuppose the probability of a coin toss resulting in “heads” is 0.3. What formula should be typed to compute the probability of having exactly 20 heads in 30 coin tosses?Suppose the probability of a coin toss resulting in “heads” is 0.3. What formula should be typed to compute the probability of having exactly 20 heads in 30 coin tosses?[removed]=BINOM.DIST(20,30,0.7,TRUE)[removed]=BINOM.DIST(30,20,0.7, TRUE)[removed]=BINOM.DIST(20,30,0.3,FALSE)[removed]=BINOM.INV(20,30,0.3,FALSE)Flag this QuestionQuestion 51 ptsSuppose the probability of selecting RED is 0.8. What formula could be used to find the probability of selecting 5 or fewer REDs out of 12 trials?Suppose the probability of selecting RED is 0.8. What formula could be used to find the probability of selecting 5 or fewer REDs out of 12 trials?[removed]=BINOM.DIST(5,12,0.8,FALSE)[removed]=BINOM.DIST(5,12,0.8,TRUE)[removed]=BINOM.DIST(5,12,0.2,FALSE)[removed]=BINOM.DIST(12,5,0.2,TRUE)Flag this QuestionQuestion 61 ptsSkip to question text.Given the above values how would one compute the mean of a binomial distribution in Excel?[removed]=B1*B2[removed]=B2/B1[removed]=B1+B2[removed]=B2^B1Flag this QuestionQuestion 71 ptsSkip to question text.Given the above values, how would one compute the standard deviation of a binomial distribution in Excel?[removed]=B1*B2[removed]=B1*B2*(1-B2)[removed]=SQRT(B1*B2)[removed]=SQRT(B1*B2*(1-B2))Flag this QuestionQuestion 81 pts Given the above values, how would one compute the probability of having more than 4 successes in Excel?Given the above values, how would one compute the probability of having more than 4 successes in Excel?[removed]=BINOM.DIST(B3,B1,B2,1)[removed]=1-BINOM.DIST(B3,B1,B2,1)[removed]=1-BINOM.DIST(B1,B3,B2,1)[removed]BINOM.DIST(B1,B3,B2,1)Given the above image, how could one make use of those three quanities to find a z-score in Excel?[removed]=B1*B3-B2[removed]B1-B2/B3[removed]B2*B3-B1[removed]=(B1-B2)/B3Flag this QuestionQuestion 21 ptsWhich built-in Excel function could be used to find a z-score from a raw-score (i.e., an x-value), mean, and standard deviation?Which built-in Excel function could be used to find a z-score from a raw-score (i.e., an x-value), mean, and standard deviation?[removed]STANDARDIZE[removed]Z[removed]ZSCORE[removed]NORMALIZEFlag this QuestionQuestion 31 ptsSkip to question text.Suppose Xis normally distributed with mean μ=11.2 and standard deviation σ=2.1. Which Excel function should be used to find ?[removed]=NORM.INV(9,11,2.1)[removed]=NORM.S.DIST(9,TRUE)[removed]=NORM.DIST(9,11.2,2.1,TRUE)[removed]=NORM.S.INV(9)Flag this QuestionQuestion 41 ptsSkip to question text.Suppose X is normally distributed with mean 11 and standard deviation 1. Which of the following gives the probability that 9.5≤X≤12.2?[removed]=NORM.DIST(12.2, 11, 1, TRUE) – NORM.DIST(9.5, 11, 1, TRUE)[removed]=(NORM.DIST(12.2, 11, 1, TRUE) – NORM.DIST(9.5, 11, 1, TRUE)) -1[removed]=1- (NORM.DIST(12.2, 11, 1, TRUE) – NORM.DIST(9.5, 11, 1, TRUE))[removed]=NORM.DIST(9.5, 11, 1, TRUE) – NORM.DIST(12.2, 11, 1, TRUE)Flag this QuestionQuestion 51 ptsHow would one find the z-score that cuts-off the upper 10% of the standard normal distribution in Excel?How would one find the z-score that cuts-off the upper 10% of the standard normal distribution in Excel?[removed]=NORM.S.INV(0.1)[removed]=1-NORM.INV(0.1)[removed]=NORM.S.INV(0.9)[removed]=NORM.DIST(0.1)Flag this QuestionQuestion 61 ptsHow could one compute the probability between 1.5 standard deviations above and below the mean in Excel?How could one compute the probability between 1.5 standard deviations above and below the mean in Excel?[removed]=NORM.S.DIST(1.5, TRUE)-NORM.S.DIST(-1.5,TRUE)[removed]=1-NORM.S.DIST(1.5, TRUE)+NORM.S.DIST(-1.5,TRUE)[removed]=ABS(NORM.S.DIST(1.5, TRUE))[removed]=1-(NORM.S.DIST(1.5, TRUE)-NORM.DIST(-1.5, TRUE))
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## ››Convert meganewton/square metre to kilonewton/square metre
meganewton/square meter kilonewton/square metre
How many meganewton/square meter in 1 kilonewton/square metre? The answer is 0.001.
We assume you are converting between meganewton/square metre and kilonewton/square metre.
You can view more details on each measurement unit:
meganewton/square meter or kilonewton/square metre
The SI derived unit for pressure is the pascal.
1 pascal is equal to 1.0E-6 meganewton/square meter, or 0.001 kilonewton/square metre.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between meganewtons/square meter and kilonewtons/square meter.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of meganewton/square meter to kilonewton/square metre
1 meganewton/square meter to kilonewton/square metre = 1000 kilonewton/square metre
2 meganewton/square meter to kilonewton/square metre = 2000 kilonewton/square metre
3 meganewton/square meter to kilonewton/square metre = 3000 kilonewton/square metre
4 meganewton/square meter to kilonewton/square metre = 4000 kilonewton/square metre
5 meganewton/square meter to kilonewton/square metre = 5000 kilonewton/square metre
6 meganewton/square meter to kilonewton/square metre = 6000 kilonewton/square metre
7 meganewton/square meter to kilonewton/square metre = 7000 kilonewton/square metre
8 meganewton/square meter to kilonewton/square metre = 8000 kilonewton/square metre
9 meganewton/square meter to kilonewton/square metre = 9000 kilonewton/square metre
10 meganewton/square meter to kilonewton/square metre = 10000 kilonewton/square metre
## ››Want other units?
You can do the reverse unit conversion from kilonewton/square metre to meganewton/square meter, or enter any two units below:
## Enter two units to convert
From: To:
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more!
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Home / Blog / SAT Math Rules
# SAT Math Rules
Like every section of the SAT, the Math exam has its own set of unofficial guidelines or “rules.” Today we’re going to look at some of these rules that the College Board consistently applies when creating SAT Math exams. Keeping these in mind will make studying for the SAT much easier and help ensure that you don’t waste time studying concepts that will not be tested on the Math exam.
No “Clues” for Terminology
On the SAT verbal sections you can usually figure out the meanings of tricky words through context clues or synonymous language. This is not the case on the Math section. The questions usually provide little or no context about the terms that are being used. If you don’t know what an exponent, integer, or prime factor is, you usually won’t be able to figure it from the test itself. Fortunately, you don’t have to memorize a ton of terminology because of the next rule.
Limited Scope
Like all sections of the SAT, the Math exam tests you on a finite number of concepts. The SAT will test you on only the following mathematical subject matter: properties of integers; word problems; number lines; squares and roots; fractions and rational numbers; factors; multiples; remainders; prime numbers; ratios, proportions, and percentages; sequences; set theory; permutation and combination; algebraic expressions; exponents; equation usage; inequalities; quadratic equations; linear and quadratic functions; basic geometry; and statistics. This might seem like a lot of material, but the degree to which the SAT tests you on each concept is fairly limited. For example:
No Need to Memorize Formulas
The SAT Math section tests you on formula usage, not formula knowledge. You don’t need to remember any geometric formulas, but you do have to know to use a formula when the exam presents you with one. Focus on applying the formulas that appear on SAT practice tests rather than trying to memorize what those formulas are.
Relatively Easy Calculations
The kinds of calculations you need to do to successfully answer SAT Math questions are relatively simple. The questions never require students to make the sorts of complex, multi-step calculations that intermediate and advanced high school math classes do. The calculations simply cannot be overly complicated because of the next rule.
All Questions Can be Solved (Very) Quickly
Any SAT Math question must be able to be solved in 30 seconds or less. Anything longer than this would make it impossible for even the most skilled mathematician to finish the exam within the time limits that the College Board imposes. If it takes you longer than half a minute to answer a particular question, you are probably making the question more complicated than it actually is or committing some sort of basic mathematical error.
Drawing are Usually to Scale
Visual representations on SAT Math problems are almost always to scale. In the rare case that a drawing is not to scale, the question will tell you so. Sometimes measuring or estimating a drawing is all you need to do to successfully answer a question. If a question contains a drawing, try this before you do anything else.
All Necessary Information is Provided
Unless one of the answer choices is “insufficient information,” an SAT Math question must provide all the information needed to successfully answer the question. Each piece of the puzzle must be somewhere in the question, even if it doesn’t look like it at first glance.
If you’ve read our blogs on ACT and SAT verbal, you’ll recall that those exam sections use the same types of incorrect answer choices are over and over. Something very similar is going on with the SAT Math exam. Rather than being chosen at random, each incorrect answer choice represents a certain kind of mistake that the College Board thinks students will make on the exam. These patterns are a bit more involved their ACT/SAT verbal section analogues, but they occur just as consistently and are just as finite in number.
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# Calculated fields / Computed columns
1. Competent Performer
Windows Vista Access 2007
Join Date
Dec 2010
Location
Zanzibar, Tanzania
Posts
229
## Calculated fields / Computed columns
I am making a small estimator app for a friend who fabricates windows and doors. I have 8 tables which are linked in a straight line of 1-M relationships as follows: Customers --> Orders --> Order.Details --> Products --> Components --> Parts --> Formulas --> Materials.
The Order.Details holds the quantity, height and length of each product ordered. The Materials table holds the unit cost of each raw material. The Formulas table has a text field that holds the method of calculating the exact measure of each raw material for each part of each component of each product. These stored methods use code letters (H, L, W) for Height, Length and Width, e.g. "(H-14)/2".
I will write VBA to replace the code letters with the actual values of (H, L, W) for each order detail and then use the EVAL function to compute the exact size of each material.
I have a query that retrieves the respective data from the 8 tables as follows:
Code:
``` SELECT Cu.CustomerName,
Or.OrderID,
Pr.ProductName,
OD.Quantity, OD.Height, OD.Width, OD.Length,
Co.ComponentName,
Pt.PartName,
F.Formula1, F.Units1,
M.ShortName, M.UnitCost
FROM Customers Cu
INNER JOIN
( Materials M
INNER JOIN
(
(
(
( Products Pr
INNER JOIN
( Orders Or
INNER JOIN OrderDetails OD
ON Or.OrderID = OD.OrderID
) ON Pr.ProductID = OD.ProductID
) INNER JOIN Components Co
ON Pr.ProductID = Co.ProductID
) INNER JOIN Parts Pt
ON Co.ComponentID = Pt.ComponentID
) INNER JOIN Formulas F
ON Pt.PartID = F.PartID
) ON M.MaterialID = F.MaterialID
) ON Cu.CustomerID = Or.CustomerID;```
and this is the sample result I am getting for my first order:
CustomerName
OrderID
ProductName
Qty
Height
Width
ComponentName
PartName
Formula1
Units1
Material
UnitCost
JOHN DOE
1
Window - Sliding
1
100
120
WindowWallFrame
Misc
(H+L)*2
cm
Brushes - Roll
1
JOHN DOE
1
Window - Sliding
1
100
120
WindowWallFrame
Misc
4
pc
Rivet
10
JOHN DOE
1
Window - Sliding
1
100
120
WindowWallFrame
Misc
28
cm
Angle
52
JOHN DOE
1
Window - Sliding
1
100
120
WindowWallFrame
Misc
(H+L)*2
m
Rubber
500
JOHN DOE
1
Window - Sliding
1
100
120
WindowWallFrame
Misc
4
pc
Wall screws
10
JOHN DOE
1
Window - Sliding
1
100
120
WindowWallFrame
Bottom
H
cm
Top/Bottom
38
JOHN DOE
1
Window - Sliding
1
100
120
WindowWallFrame
Top
W
cm
Top/Bottom
38
JOHN DOE
1
Window - Sliding
1
100
120
WindowWallFrame
Left
L
cm
Jamb
69
JOHN DOE
1
Window - Sliding
1
100
120
WindowWallFrame
Right
L
cm
Jamb
69
Based on the above query and sample results, I am trying to figure out a better way to handle the computation than what I have come up with so far in VBA:
1. Get the query results in a rst and loop through each record
2. Replace the letter codes in the formula with the actual values
3. Run EVAL to get the measure of each piece
4. Multiply the value in (3) by the order quantity to get the total measure
The new record, with values (3) and (4), is then inserted into a REQUIREMENTS Table.
Q1. This row-based approach is considered inefficient by SQL pros.
A set-based approach would be preferred but I can't think of how to run the VBA function EVAL in SQL.
Q2. I am using 2007, but could the computed column option in Access 2010to help resolve (2) and (3)?
2. I just tested the Eval() function in an Access query and it did work.
Not really understanding all of your post. Can you provide the project for analysis?
I agree the data structure does not seem optimized for a relational database. I understand an order detail could have many products but can each product be related to many order details? This is a many-to-many relationship.
3. Competent Performer
Windows Vista Access 2007
Join Date
Dec 2010
Location
Zanzibar, Tanzania
Posts
229
June7, thanks. I would but the size limits do not allow me to. Any workaround?
If you look at it, the design is highly normalized: each orderdetail record relates to only one product record while each product record can relate to potentially hundreds of records.
4. Originally Posted by goodguy
June7, thanks. I would but the size limits do not allow me to. Any workaround?
Have you done a Compact and Repair and then zipped up the db? We accept up to 2MB of ZIP file.
5. Competent Performer
Windows Vista Access 2007
Join Date
Dec 2010
Location
Zanzibar, Tanzania
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OK, here it is.
PS. There is a bug on the upload screen: when you upload files successfully AFTER getting an error message following a failed upload, the error message does not disappear. This is confusing to users who may not realize that their files have already been uploaded.
6. Just did a quick look. So far:
1. There is no code in the project, nothing to analyse.
2. Not sure is possible.
3. The Eval function does work in queries but don't think your structure is suitable for it.
For some of the lookup tables (Colours, Currencies, Categories) don't think I would use autonumber as primary key, just use the name itself, such as the colour. This will not only prevent duplicate colours, but also make it easier to view the colour value when need, no need for table join to retrieve. At least set the field to Index (no duplicates).
Also, misspelled aluminum as aluminium.
7. Competent Performer
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Join Date
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Location
Zanzibar, Tanzania
Posts
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Originally Posted by June7
Just did a quick look. So far:
1. There is no code in the project, nothing to analyse.
2. Not sure is possible.
3. The Eval function does work in queries but don't think your structure is suitable for it.
For some of the lookup tables (Colours, Currencies, Categories) don't think I would use autonumber as primary key, just use the name itself, such as the colour. This will not only prevent duplicate colours, but also make it easier to view the colour value when need, no need for table join to retrieve. At least set the field to Index (no duplicates).
Also, misspelled aluminium as aluminium.
1. Of course, it contains no code as yet. I said I was considering the approach that I described but wanted to see if a better approach was possible.
2. What possibility are you unsure about?
3. My query returns the Height, Width and Length values on the same record as the respective formula so I am puzzled why you think it is unsuitable.
4. I use AutoNumber as my PK even for such small lookup tables because that is second nature to me. I am also particular about indexing and setting constraints, unfortunately I still haven't gotten round to fixing them in this app.
So, the questions remains: Is there a better way to do this?
8. Here is why I am thinking your formula field won't work. Try this code.
Code:
```Option Compare Database
Option Explicit
Public Sub TestCalc()
Dim W As Double, H As Double, L As Double, x As Double
Set cn = CurrentProject.Connection
While Not rs.EOF
W = rs!Width
H = rs!Height
L = rs!Length
Debug.Print rs!formula1; rs!Width; rs!Height; rs!Length
x = Eval(rs!formula1)
Debug.Print x
rs.MoveNext
Wend
rs.Close
End Sub```
You managed to quote my post with a misspelling in my comment about misspelling. I don't remember posting with misspelling but must have and corrected it before I realized you had replied. But I decided to check online dictionary. Apparently, it is an accepted variation in spelling. Never seen it used in the U.S.
Last edited by June7; 09-18-2011 at 11:20 AM.
9. Competent Performer
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Posts
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Thanks for your code, June7. This is what I was planning to fall back on if nothing better came up. Below is my code
Code:
```Public Sub Requirements()
Dim W As Double, H As Double, L As Double, xValue As Double
Dim F As String
Set cn = CurrentProject.Connection
While Not rs.EOF
W = rs!Width: H = rs!Height: L = rs!Length: F = rs!Formula1
If InStr(1, F, "W") > 0 Then F = Replace(F, "W", W)
If InStr(1, F, "H") > 0 Then F = Replace(F, "H", H)
If InStr(1, F, "L") > 0 Then F = Replace(F, "L", L)
xValue = Eval(F)
Debug.Print "W = " & W & "; H = " & H & "; L = " & L & "; Formula = " & F & "; Value = " & x & vbCrLf; ""
rs.MoveNext
Wend
rs.Close
End Sub```
10. Then you do have a working solution. Did you have that before my posted code? Would have been nice to have it to analyze and correct my erroneous conclusion first. Now if you want can make this a function that can be called from query, like:
Public Function Calc(F As String, W As Double, H As Double, L As Double)
If InStr(1, F, "W") > 0 Then F = Replace(F, "W", W)
If InStr(1, F, "H") > 0 Then F = Replace(F, "H", H)
If InStr(1, F, "L") > 0 Then F = Replace(F, "L", L)
Calc = Eval(F)
End Function
Then call in query like:
Result: Calc([Formula1],[Width],[Height],[Length])
11. Competent Performer
Windows Vista Access 2007
Join Date
Dec 2010
Location
Zanzibar, Tanzania
Posts
229
@June7: Thanks for persisting with me. Yes, I did have the intent to write the code as I finally did, although I was holding out for a better solution. As I had stated in my original post:
Based on the above query and sample results, I am trying to figure out a better way to handle the computation than what I have come up with so far in VBA:
1. Get the query results in a rst and loop through each record
2. Replace the letter codes in the formula with the actual values
3. Run EVAL to get the measure of each piece
4. Multiply the value in (3) by the order quantity to get the total measure
The new record, with values (3) and (4), is then inserted into a REQUIREMENTS Table.
12. Competent Performer
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Join Date
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Location
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Posts
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@June7: The function works beautifully. Now, I can cut out all the code and go direct to displaying the report immediately the order is entered. WONDERFUL!!! Thanks a million!
#### Posting Permissions
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Other Forums: Microsoft Office Forums - Senior Forums
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# In an examination, a student has to answer 4 questions
Question:
In an examination, a student has to answer 4 questions out of 5 questions; questions 1 and 2 are however compulsory. Determine the number of ways in which the student can make the choice.
Solution:
We know that,
nCr
$=\frac{\mathrm{n} !}{\mathrm{r} !(\mathrm{n}-\mathrm{r}) !}$
According to the question,
Total number of questions =5
Number of questions to be answered =4
Compulsory questions are question number 1 and 2
Hence, the number of ways in which the student can make the choice = 3C2
3C= 3!/(2!1!) =3 ways
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Differential Pulse Voltammetry (DPV)
Last Updated: 4/24/20 by Tim Paschkewitz
ARTICLE TAGS
• current vs. potential,
• AfterMath DPV,
• differential pulse voltammetry,
• DPV
RELATED PAGES
1. Technique Overview
2. Fundamental Equations
3. Experimental Setup in AfterMath
4. Sample Experiment
5. Example Applications
6. References
1Technique Overview
Differential Pulse Voltammetry (DPV) is a potentiostatic method that offers some advantages to common techniques like Cyclic Voltammetry (CV), in that the waveform is a series of pulses increasing along a linear baseline. The way in which the current is measured at each pulse aids in minimizing the measurement of background (charging) current. Related pulse-type methods include Normal Pulse Voltammetry (NPV) and Square Wave Voltammetry (SWV).
Differential Pulse Voltammetry (DPV) is a pulse technique that is designed to minimize background charging currents. The waveform in DPV is a sequence of pulses where a baseline potential is held for a specified period of time prior to the application of a potential pulse. Current is sampled, at time ${\tau}'$, just prior to the application of the potential pulse. The potential is then stepped by a small amount (typically 100 mV) and current is sampled again, at time $\tau$ at the end of the pulse. The potential of the working electrode is then stepped back by a lesser value than during the forward pulse such that baseline potential of each pulse is incremented throughout the sequence.
As with many other methods, users can initially sweep positively (towards an anodic limit) or negatively (toward a cathodic limit) initially. Below, typical DPV waveforms are shown for illustration, using only an increasing positive pulse sequence.
INFO: Initial sweep direction can be positive-increasing or negative increasing. Not all possibilities have been provided as examples in this article. Users must tailor their parameters to suit their specific electrochemistry. Contact us with any questions or additional assistance.
In the simplest case, when Segments (SN) = 1 (see Figure 1), potential pulses step along a linear baseline from an initial to final potential, sampling current at specified intervals (see Figure 2).
Figure 1. Differential Pulse Voltammetry (DPV) One Segment Waveform
Current is sampled, at time ${\tau}'$ (TDP, PRE), just prior to the application of the potential pulse. The potential is then stepped by a small amount (typically 100 mV) and current is sampled again, at time $\tau$ (TDP, POST) at the end of the pulse (see Figure 7). These sampling periods are selected to allow sufficient time for non-Faradaic (charging) current to decay such that primarily only current arising from Faradaic reactions is reported. The difference in current at these two periods for each pulse is plotted against potential results in a differential voltammogram, as
$\delta i=i(\tau)-i(\tau')$
In a later Section (Experimental Setup in AfterMath - Basic Tab), the pulse parameters are defined in detail (see Table 1 and Figure 7).
When Segments (SN) = 2 (see Figure 2), potential pulses step along a linear baseline from an initial potential to vertex potential and then to final potential, sampling current at specified intervals. Again, the current is sampled in the same fashion as DPV experiments with other Segment values (see Figure 7).
Square Figure 2. C
When Segments (SN) = 3 (see Figure 3), potential pulses step along a linear baseline from an initial potential to upper potential to lower potential and then to final potential, sampling current at specified intervals. Again, the current is sampled in the same fashion as DPV experiments with other Segment values (see Figure 7).
Figure 3. Differential Pulse Voltammetry (DPV) Three Segment Waveform
2Fundamental Equations
The following is a brief introduction to the theory of DPV. Please see Bard and Faulkner for a more complete description. For additional information on the cyclic form of DPV, consult the paper by Drake et al.
Consider the reaction,
$O + e^- \rightarrow R$
where $O$ is reduced in a one electron step to $R$. At values sufficiently more positive than $E^{0'}$ no faradaic current flows before the potential step (to more negative values). The application of the potential step does not produce an appreciable increase in current; thus, the differential is very small. At values significantly negative of $E^{0'}$ the baseline potential is reducing $O$ at a maximum rate. The application of a small potential step (towards more negative values) is unlikely to increase the rate of reduction; thus, the differential current is again small. Only at potentials around $E^{0'}$ will the differential current be significant. The period during the application of the baseline potential has $O$ being reduced at some rate. The potential step (to more negative values) increases the rate of reduction and hence the differential current will be significant. Under normal conditions (pulse height < 100 mV) the height of the peak can is given by the equation
$\displaystyle({\delta}t)_{max} = \frac{nFAD_O^{1/2}C_O^*}{{\pi}^{1/2}({\tau}-{\tau}')} \left({\frac{1-\sigma}{1+\sigma}}\right)$
where $n$ is the number of electrons, $F$ is Faraday's Constant (96485 C/mol), $A$ is the electrode area (in cm2), $D$ is the diffusion coefficient (in cm2/s), $C_O^*$ is the concentration of electroactive species (in mol/cm3) and $\sigma$ is given by
$\displaystyle\sigma = \left(\frac{nF}{RF}\frac{{\Delta}E}{2}\right)$
where ${\Delta}E$ is the pulse height, $T$ is the temperature ($K$) and $R$ is the Universal Gas Constant (8.314 J/mol⋅K).
As mentioned in the Advanced parameters tab (discussed in a subsequent section), the direction of the pulse should not affect the results. Consider the same reaction above where $O$ is reduced in a one-electron step to $R$. At values sufficiently more positive than $E^{0'}$ no faradaic current flows before the potential step (towards more positive values). The change in current due to the potential step is also insignificant enough to cause a Faradaic current; thus, the differential is very small. At values significantly negative of $E^{0'}$ the baseline potential is reducing $O$ at a maximum rate. The application of a small potential pulse (towards more positive values) does not decrease the rate of reduction and hence the differential current is again small. Only at potentials around $E^{0'}$ will the differential current be significant. The period during the application of the baseline potential has $O$ being reduced at some rate. The potential step (to more positive values) decreases the rate of reduction and hence the differential current will be significant; therefore, the direction of the potential step has no effect on the differential current observed.
3Experimental Setup in AfterMath
To perform a differential pulse voltammetry experiment in AfterMath, choose Differential Pulse Voltammetry (DPV) from the Experiments menu (see Figure 4).
Figure 4. Differential Pulse Voltammetry (DPV) Experiment Menu Selection in AfterMath
Doing so creates an entry within the archive, called DPV Parameters. In the right pane of the AfterMath application, several tabs will be shown (see Figure 5).
Figure 5. Differential Pulse Voltammetry (DPV) Experiment Basic Tab
As with most Aftermath methods, the experiment sequence is
Induction Period → Pulse Sequence → Relaxation Period → Post-Experiment Idle Conditions
Continue reading for detailed information about the fields on each unique tab.
3.1Basic Tab
TIP: Click the AutoFill button ("I Feel Lucky" prior to May 2019) on the top bar in AfterMath to automatically fill all required parameters with reasonable starting values. While the values provided may not be appropriate for your specific system, they are reasonable parameters with which to start your experiment, especially if you are new to the method.
The basic tab contains fields for the fundamental parameters necessary to perform a DPV experiment. AfterMath shades fields with yellow when a required entry is blank and shades fields pink when the entry is invalid (see Figure 6).
Figure 6. Differential Pulse Voltammetry (DPV) Basic tab in AfterMath
During the induction period, a set of initial conditions are applied to the electrochemical cell and the cell equilibrates at these conditions. Data are not collected during the induction period, nor are they shown on the plot during this period. Users will define induction period parameters on the Advanced Tab.
After the induction period, the potential of the working electrode is stepped through a series of increasing pulses from the Initial potential to the Final potential. The potential is incremented with each successive pulse according to the Pulse increment. Current is measured at the time obtained by subtracting the Pulse sampling width window from the Pulse width. In a typical experiment, Segments = 1 and there is a single series of pulses moving along the linear baseline from initial to final potential. Some may want to reverse the pulses (move in the opposite direction), accomplished by adjusting the number of segments to be > 1. By adjusting the number of segments, users can create variants of the DPV technique. Cyclic Differential Pulse Voltammetry consists of cycling (through a series of potential pulses also) the potential of the working electrode between an Upper potential and a Lower potential (Segments = 2 or 3).
The parameter settings for pulse-type experiments are often not as clear as with something more simple like Cyclic Voltammetry (a sweep method ). Refer to Figure 7 below when understanding each parameter of the DPV pulse. Further, clicking "AutoFill" ("I Feel Lucky" in older versions of AfterMath) provides reasonable starting parameters if you are unsure of typical values used in these experiments. Most often, research journal articles describe the DPV pulse sequence used, which can be replicated using AfterMath.
The experiment concludes with a relaxation period. During the relaxation period, a set of final conditions (specified on the Advanced tab) are applied to the electrochemical cell and the cell equilibrates at these conditions (set on the Advanced Tab). Data are not collected during the induction period, nor are they shown on the plot during this period.
At the end of the relaxation period, the post-experiment idle conditions are applied to the cell and the instrument returns to the idle state.
A plot of the typical experiment sequence, containing labels of the fields on the Basic tab, helps to illustrate the sequence of events in a DPV experiment (see Table 1 and Figure 7).
The table below lists the group and field names and symbols for each parameter associated with this experiment (see Table 1).
Group Name Field Name Symbol Sweep (Sweep limits) Segments $S_N$ Sweep (Sweep limits) Initial Potential $E_I$ Sweep (Sweep limits) Initial Direction $DIR_I$ Sweep (Sweep limits) Upper Potential $E_U$ Sweep (Sweep limits) Vertex Potential $E_V$ Sweep (Sweep limits) Lower Potential $E_{LOW}$ Sweep (Sweep limits) Final Potential $E_F$ Differential Pulse (Pulse parameters) Height $E_{H, \;DP}$ Differential Pulse (Pulse parameters) Width $T_{W, \;DP}$ Differential Pulse (Pulse parameters) Period $T_{P, \;DP}$ Differential Pulse (Pulse parameters) Increment $E_{I, \;DP}$ Differential Pulse (Sampling) Pre-Pulse width $T_{DP, \;PRE} \;(\tau ')$ Differential Pulse (Sampling) Post-pulse width $T_{DP, \;POST} \; (\tau)$
Table 1. Basic Tab Group Names, Field Names, and Symbols.
Figure 7. Differential Pulse Voltammetry (DPV) Pulse Sequence Detail
The DPV Advanced tab contains groups for Induction Period, Relaxation Period, Pulse options, and iR Compensation (see Figure 8).
Figure 8. Differential Pulse Voltammetry (DPV) Experiment Advanced Tab in AfterMath
Induction Period is the first step in a DPV experiment if the Duration is >0 s. During the induction period, the specified current is applied to the cell for the specified duration. During this period, data are not collected. The Induction Period is believed to "calm" the cell prior to intentional perturbation. More on Induction Period is found within the knowledgebase.
Relaxation Period is the last step in a DPV experiment if the Duration is >0 s. During the relaxation period, the specified current is applied to the cell for the specified duration. During this period, data are not collected. The Relaxation Period is believed to "calm" the cell after intentional perturbation. More on Relaxation Period is found within the knowledgebase.
A unique set of options for DPV is the ability to invert the pulse direction. Within the "Pulse" options are two checkbox options to invert pulses during the anodic sweep and to invert the pulses during the cathodic sweep. The directionality of a DPV is determined by selection of initial, vertex, upper, lower, and final potential, dependent on the Number of Segments (SN). When the pulse sequence moves in the direction of increasing positive potential, this is the anodic sweep. Conversely, when the pulse sequence moves in the direction of increasing negative potential, this is the cathodic sweep. To better understand and visualize these pulse options, a figure with each set of conditions has been prepared (see Figure 9).
Figure 9. Differential Pulse (DPV) Pulse Inversion Options
Lastly, the iR Compensation group allows users to adjust the cell feedback to accommodate a known resistive drop between working and reference electrodes. Not all potentiostats from Pine Research support iR compensation. The WaveDriver series support iR compensation by positive feedback and current interrupt. The WaveDriver 100, WaveDriver 200, and WaveDriver 40 support EIS-based iR compensation. The WaveNow series (including the WaveNano and WaveNowXV ) and the CBP bipotentiostat do not support iR compensation of any type. More information about iR compensation, including understanding how it works and how to determine the resistance, consult the knowledgebase article on the topic.
The general experimental flow for a DPV experiment is provided below (see Figure 10), highlighting the Induction period, DPV Pulse Sequence, and Relaxation period. Following the relaxation period, the post-experiment conditions are applied.
Figure 10. Differential Pulse Voltammetry (DPV) Experiment Sequence in AfterMath
3.3Ranges, Filters, and Post Experiment Conditions Tab
In nearly all cases, the groups of fields on the Ranges tab are already present on the Basic tab. The Ranges tab shows an Electrode Range group and depending on the experiment shows either, or both, current and potential ranges and the ability to select an autorange function. The fields on this tab are linked to the same fields on the Basic tab (for most experiments). Changing the values on either the Ranges tab or on the Basic tab changes the other set. In other words, the values selected for these fields will always be the same on the Ranges tab and on the Basic tab. More on ranges is found within the knowledgebase, as is for autorange.
The Filters tab provides access to potentiostat hardware filters, including stability, excitation, current response, and potential response filters. Pine Research recommends that users contact us for help in making changes to hardware filters. Advanced users may have an easier time changing the automatic settings on this tab. Filter settings fields are shown for WK1 (working electrode #1) as well as for WK2 (working electrode #2) regardless of the potentiostat connected to AfterMath. More information on filters is available elsewhere on the knowledgebase.
By default, the potentiostat disconnects from the electrochemical cell at the end of an experiment. There are other options available for what these post-experiment conditions can be and are controlled by setting options on the Post Experiment Conditions tab. Complete details on the fields and settings on this Post Experiment Conditions tab are provided elsewhere on the knowledgebase.
4Sample Experiment
Below are the typical results for DPV for the oxidation of a 1.4 mM solution of K4Fe(CN)6 in 0.1 M phosphate buffer (see Figure 11). The specific parameters for this experiment are as follows:
• pH = 6.8
• 3 mm glassy carbon WE
• period = 100 ms
• width = 10 ms
• height = 50 mV
• potential increment = 10 mV
Figure 11. Differential Pulse Voltammogram of a Potassium Ferrocyanide Solution in Phosphate Buffer
Below are the typical results for a two segment Cyclic Differential Pulse Voltammetry (CDPV) for the oxidation and reduction ofK4Fe(CN)6 in 0.1 M phosphate buffer (see Figure 12). Crosshair tools have been added to show that peak positions and peak heights should be identical in CDPV for a fully reversible system. The specific parameters for this expeirment are as follows:
• pH = 6.8
• 3 mm glassy carbon WE
• period = 100 ms
• width = 10 ms
• height = 50 mV
• potential increment = 10 mV
Figure 12. Cyclic Differential Pulse Voltammogram of a Potassium Ferrocyanide Solution in Phosphate Buffer
5Example Applications
The first example uses DPV to examine the pH dependence of redox potential for an electron and proton transfers in tryptophan and tyrosine. Sjödin et al. used the pH dependence of the redox potential to calculate ${\Delta} G$ values for different reaction pathways and thus determine that the mechanism can be a one-step or two steps depending on several factors.
In another example, Miles and Murray use DPV to examine the quantized double layer charging of hexanethiolate-coated monolayer-protected Au140 clusters (AuMPCs). They used DPV to resolve 13 individual peaks related to AuMPC core charging over a 3 V window in CH2Cl2 at lowered temperatures. Though peaks are visible using CV, DPV provides the necessary resolving power, by suppressing background currents, to separate out all 13 peaks.
6References
Our knowledgebase is the central repository for written content, including help topics, theory, application notes, specifications, and software information.
Software
Detailed information about our Software, which includes AfterMath and retired PineChem.
Applications
Application notes discuss practical aspects of conducting specific experiments.
Theory
Fundamental electrochemical theory presented in a brief and targeted manner.
Product Specifications
Review complete product specifications and compare products within a category here.
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## Saturday, March 31, 2012
### Reversing the Meat-a-Morphosis machine : inverse functions
A short followup to the post Two Ideas for Introducing Functions.
Yesterday I discovered another payoff from the hilariously gruesome Meat-a-Morphosis video: a powerful and memorable analogy for Inverse Functions. Imagine if we ran the machine backwards - if we put chicken nuggets in and the original chickens came out? My students thought this was hilarious.
Inverse Functions: What if you put in nuggets and out came chickens? Image adapted from original at Meat-a-Morphosis ( Patty Hill & Michael Word)
Even more fun: for students still coming to terms with f(x) notation, faced with this:
doesn't this convey the same idea even more clearly?
Put a chicken into the nuggetiser, then into the de-nuggetiser and you get back your chicken (OK , maybe with few ruffled feathers and a lot of squawking!)
Who would have thought there was so much laughter in a lesson on functions? Thanks to the Meat-a-Morphosis team at Kealing Middle School, Austin Texas for a wondeful teaching resource.
## Saturday, March 24, 2012
### Who's afraid of the error monster?
Maybe it's built into our very survival instincts : if something is wrong, it's uncomfortable - so run away and hide from it. We see it in class every day, every hour - students (and teachers!) running away from errors. No-one wants to be wrong, or even worse, be seen to be wrong. And yet, when it comes to learning, "errors" are valuable tools. I see it as one my most important roles as a teacher to convince students not to be afraid of errors - on the contrary, to look for them, appreciate them and share them.
When I talk about errors with my students I introduce them to the Error Monster:
Image from http://conservationbytes.com/2009/10/21/sleuthing-the-chinese-green-slime-monster/ (by CJA Bradshaw?)
We discuss ways in which this scary monster is in fact a good friend - how every error we make, or another person makes, is a valuable gift to our learning. I encourage students to face their monster head on - whenever they do an assessment and see their mistakes, to run towards their monster and embrace it:
Adapted from CJA Bradshaw's slime monster.
Now our monster becomes an object of fun and affection - helping overcome embarrassment and disappointment at making mistakes - and allowing us to instead focus on resolving those errors.
Teacher notes:
• I like to use the analogy of a blind person using a walking cane. How could they see where to go, if they didn't make "mistakes"? It's only by having the cane bump into things the person can see where to go. Errors help guide us on our learning path.
• You have to walk the walk : be happy - and not embarrassed - to face your own errors in class. I highlight to my students my specific weaknesses when doing algebra : I know (and they know) I make silly errors with signs and expansions - so I laugh at my monster and then keep a careful eye out for him. I make a show in front of the class of checking for my common errors. Hopefully over time I will get better at these!
## Saturday, March 17, 2012
### A visit to the Function Zoo
Do you remember your early encounters with the animal kingdom? So many wonderful different animals - it may even have been a bit overwhelming at first. But very quickly we learnt to group the animals into a scheme that made sense to us. In mathematics we have a similar extravaganza of different 'animals', which can be overwhelming for students to make sense of. Enter the idea of The Function Zoo - first introduced to me by Mary Barnes in her amazing Investigating Change books.
Here is how I worked the idea into a Year 11 class, several lessons into the Functions topic:
A look at the different species of animals ....
... and how we might organise them.
The challenge:
Students worked in groups of four, using large sheets of butcher paper to sketch their ideas. There were at least two laptops per group and the students had just enough GeoGebra skills to be able to turn algebraic expressions into graphs.
The results were incredible: great conversations between students about functions. With GeoGebra on hand, I was able to encourage students to explore their questions, rather than give them answers, and even ask them more questions if they were ready for it.
Twenty minutes later I quietly threw this slide on the screen but otherwise said nothing:
The groups noticed it soon enough - and went wild. Seeing a few more functions they knew but had forgotten gave them new energy to keep going. Others asked each other questions, trying to work out the graphs they didn't recognise. Most recognised the last graph from our "explore your calculator" game. We then debriefed as a class, and explored why the idea of the Function Zoo is helpful and interesting. Apart from the obvious benefit of being able to organise our thinking, the real benefit comes in being able to make connections - as I suggested in these slides:
As often happens in student exploration activities, the class produced something unexpected, a gift from them to extend the lesson idea. One group drew the absolute value of a quadratic function - a blend of two of our function families. We decided this new function was like the cross-species breeding you sometimes see on display at the zoo : the lion bred with a tiger to make a liger.
Absolute value of a quadratic function : a "liger" in our function zoo. Liger drawing: St Hilare (1772-188). Function by GeoGebra.
A fun and powerful idea - allowing students to see that even quite unusual functions can be seen as blend of function attributes they already know how to work with.
Download lesson slides & annotations: (Google Drive) PDF PowerPoint
Teaching Notes:
• A graphing tool makes a huge difference to the success of this lesson. Without it, students would spend a very long time plotting to explore their ideas. There is time for careful plotting later - this lesson is about seeing the bigger picture.
• I found the group structure allowed for a high degree of differentiation - I could customise leading questions for each group, depending where they were up to on the functions journey.
• I can't stress enough the value of developing students' GeoGebra skills (or other computer graphing application) when doing mathematics at this level. I sneak some GeoGebra learning into every lesson - even if it's just the class watching me do a quick check of an equation or a graph. Show them one small GeoGebra idea per lesson and by the end of term they will know the product well - especially if they are using GeoGebra at home as part of their study.
• Why am I such a GeoGebra fanboy? Most importantly because all my students can download a copy to use home. GeoGebra is free and runs on Windows and Macintosh and it doesn't need an internet connection to run.
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#jsDisabledContent { display:none; } My Account | Register | Help
# Tetrahedral number
Article Id: WHEBN0000509120
Reproduction Date:
Title: Tetrahedral number Author: World Heritage Encyclopedia Language: English Subject: Collection: Publisher: World Heritage Encyclopedia Publication Date:
### Tetrahedral number
A pyramid with side length 5 contains 35 spheres. Each layer represents one of the first five triangular numbers.
A tetrahedral number, or triangular pyramidal number, is a figurate number that represents a pyramid with a triangular base and three sides, called a tetrahedron. The nth tetrahedral number is the sum of the first n triangular numbers.
The first ten tetrahedral numbers are:
1, 4, 10, 20, 35, 56, 84, 120, 165, 220, … (sequence A000292 in OEIS)
## Contents
• Formula 1
• Geometric interpretation 2
• Properties 3
• Popular Culture 4
• References 6
## Formula
The formula for the n-th tetrahedral number is represented by the 3rd rising factorial of n divided by the factorial of 3:
T_n={n(n+1)(n+2)\over 6} = {n^{\overline 3}\over 3!}
The tetrahedral numbers can also be represented as binomial coefficients:
T_n={n+2\choose3}.
Tetrahedral numbers can therefore be found in the fourth position either from left or right in Pascal's triangle.
## Geometric interpretation
Tetrahedral numbers can be modelled by stacking spheres. For example, the fifth tetrahedral number (T5 = 35) can be modelled with 35 billiard balls and the standard triangular billiards ball frame that holds 15 balls in place. Then 10 more balls are stacked on top of those, then another 6, then another three and one ball at the top completes the tetrahedron.
When order-n tetrahedra built from Tn spheres are used as a unit, it can be shown that a space tiling with such units can achieve a densest sphere packing as long as n ≤ 4.[1]
## Properties
• A. J. Meyl proved in 1878 that only three tetrahedral numbers are also perfect squares, namely:
T1 = 1² = 1
T2 = 2² = 4
T48 = 140² = 19600.
• The infinite sum of tetrahedral numbers' reciprocals is 3/2, which can be derived using telescoping series:
\!\ \sum_{n=1}^{\infty} \frac{6}{n(n+1)(n+2)} = \frac{3}{2}.
• The tetrahedron with basic length 4 (summing up to 20) can be looked at as the 3-dimensional analogue of the tetractys, the 4th triangular number (summing up to 10).
• The parity of tetrahedral numbers follows the repeating pattern odd-even-even-even.
• An observation of tetrahedral numbers:
T5 = T4 + T3 + T2 + T1
• Numbers that are both triangular and tetrahedral must satisfy the binomial coefficient equation:
Tr_n={n+1\choose2}={m+2\choose3}=Te_m.
• The only numbers that are both Tetrahedral and Triangular numbers are (sequence A027568 in OEIS):
Te1 = Tr1 = 1
Te3 = Tr4 = 10
Te8 = Tr15 = 120
Te20 = Tr55 = 1540
Te34 = Tr119 = 7140
## Popular Culture
Te12 = 364, which is the total number of gifts "my true love sent to me" during the course of all 12 verses of the carol, The Twelve Days of Christmas [2]
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# Keynesian model | Economics homework help
The key assumption of the basic Keynesian model is that in the short run, firms:
[removed] meet demand at preset prices. [removed] adjust prices to bring sales in line with capacity. [removed] change prices frequently. [removed] operate just as they do in the long run. [removed] change prices rather than quantities.
Question 2
Suppose a household’s marginal propensity to consume out of disposable income is 0.75 and its exogenous consumption is \$250. If household income is \$2000 and taxes are a flat \$200, how much will the household save each period?
[removed] \$200. [removed] \$250. [removed] \$400. [removed] \$1,000. [removed] \$1,500.
Question 3
A\$100 million increase in government purchases will have a bigger impact on equilibrium output
[removed] The larger are the marginal tax rate and marginal propensity to import and the larger is the marginal propensity to consume. [removed] The larger are the marginal tax rate and marginal propensity to import and the smaller is the marginal propensity to consume. [removed] The smaller are the marginal tax rate and marginal propensity to import and the smaller is the marginal propensity to consume. [removed] The smaller are the marginal tax rate and marginal propensity to import and the larger is the marginal propensity to consume. [removed] The smaller are the marginal tax rate and marginal propensity to consume and the larger is the marginal propensity to import.
Question 4
If the marginal propensity to consume equals 0.75, then a \$100 increase in after-tax disposable income leads to a _____ increase in consumption.
[removed] \$0.25 [removed] \$0.75 [removed] \$25 [removed] \$75 [removed] \$100
Question 5
Planned aggregate expenditure is total:
[removed] value added in the economy. [removed] planned spending on final goods and services. [removed] revenue from the sale of goods and services. [removed] profits in the economy. [removed] output produced by firms
Question 6
The amount by which consumption increases when disposable income increases by \$1 is called:
[removed] an automatic stabiliser. [removed] the consumption function. [removed] the marginal propensity to consume. [removed] autonomous expenditure. [removed] the multiplier.
Question 7
In the basic Keynesian model, all but one of the following are true. Which is the exception?
[removed] Planned consumption always equals actual consumption. [removed] Planned investment always equals actual investment. [removed] Planned government spending always equals actual government spending. [removed] Planned net exports always equal actual net exports. [removed] Planned aggregate expenditure is always equal to output.
Question 8
If planned aggregate expenditure (PAE) in an economy equals 2000 + 0.8Y and potential output (Y*) equals 9000, then this economy has:
[removed] an expansionary gap. [removed] a recessionary gap. [removed] No output gap. [removed] no autonomous expenditure. [removed] no induced expenditure.
Question 9
In the short-run Keynesian model, in order to close a recessionary gap of \$10 billion dollars, government purchases must be
[removed] increased by \$10 billion. [removed] decreased by \$10 billion. [removed] increased by more than \$10 billion. [removed] increased by less than \$10 billion. [removed] decreased by \$10 billion while taxes must be cut by \$10 billion.
Question 10
If short-run equilibrium output equals 10,000, the income-expenditure multiplier equals 5, the MPC equals .8, and potential output (Y*) equals 9000, then taxes must ______ by ________ to eliminate any output gap.
[removed] decrease, 20 [removed] decrease, 200 [removed] increase, 225 [removed] increase, 250 [removed] increase, 200
Question 11
Fiscal policy is NOT often used as a stabilisation tool. However, it does have important roles in the economy. Three of these roles are:
[removed] managing income distribution, demographic change, and public debt [removed] managing public assets, defence projects, and public safety [removed] managing government monetary policy, inflation and interest rates [removed] managing collection of taxes, public health and antiterrorism policy [removed] managing full employment, price stability and exchange rates.
Question 12
Dave’s Mirror Company expects to sell \$1,000,000 worth of mirror and to produce \$1,250,000 worth of mirrors in the coming year. The company purchases \$300,000 of new equipment during the year. Sales for the year turn out to be \$900,000. Actual investment by Dave’s Mirror Company equals _____ and planned investment equals _______.
[removed] \$250,000; \$150,000. [removed] \$300,000; \$200,000. [removed] \$650,000; \$550,000. [removed] \$850,000; \$750,000. [removed] \$950,000;\$500,000.
1 points
Question 13
When actual investment is less than planned investment
[removed] firms are selling less output than expected. [removed] firms are selling more output than expected. [removed] the quantity of output sold is the amount the firm expected to sell. [removed] autonomous expenditure is less than induced expenditure. [removed] the stock of inventories must increase.
Question 14
An increase in government purchases will have a larger effect on real GDP:
[removed] the larger the MPC [removed] the smaller the MPC [removed] the larger a tax increase [removed] the smaller a tax decrease [removed] the larger the MPS.
Question 15
Because of automatic stabilisers, when GDP fluctuates the:
[removed] government’s budget remains in balance [removed] government’s deficit fluctuates directly with GDP [removed] government’s deficit fluctuates inversely with GDP [removed] the economy will automatically go to full employment [removed] none of the above.
Question 16
The short-run effect of equilibrium GDP of an equal change in government expenditure and net taxes is a definition of:
[removed] the balanced budget [removed] the balanced budget multiplier [removed] balanced GDP [removed] balanced growth [removed] balanced savings
Question 17
If bank reserves are 200, the public holds 400 in currency, and the desired reserve/deposit ratio is 0.25, the deposits are ____ and the money supply is _____.
[removed] 200; 600 [removed] 400; 800 [removed] 600; 1000 [removed] 800; 1200 [removed] none of the above.
Question 18
When the Reserve Bank sells government securities, the banks’
[removed] reserves will increase and lending will expand, causing an increase in the money supply. [removed] reserves will decrease and lending will contract, causing a decrease in the money supply. [removed] reserve requirements will increase and lending will contract, causing a decrease in the money supply. [removed] reserves/deposit ratio will increase and lending will expand, causing an increase in the money supply. [removed] reserves will increase and lending will contract, causing no change in the money supply.
Question 19
One year before maturity, the price of a bond with a principal amount of \$1,000 and a coupon rate of 5% paid annually fell to \$981. The one-year interest rate
[removed] rose to 8.5%. [removed] rose to 7.0%. [removed] rose to 6.0%. [removed] remained at 5%. [removed] none of the above.
Question 20
If real GDP equals 5000, nominal GDP equals 10,000 and the price level equals 2, then what is velocity if the money stock equals 2000?
[removed] 2 [removed] 2.5 [removed] 4 [removed] 5 [removed] 10
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# £1 in 1789 is worth £153.82 today
£
## Value of £1 from 1789 to 2020
According to the Office for National Statistics composite price index, today's prices in 2020 are 15,281.67% higher than average prices since 1789. The British pound experienced an average inflation rate of 2.20% per year during this period, causing the real value of a pound to decrease.
In other words, £1 in 1789 is equivalent in purchasing power to about £153.82 in 2020, a difference of £152.82 over 231 years.
The 1789 inflation rate was -1.33%. The current year-over-year inflation rate (2019 to 2020) is now 1.50%1. If this number holds, £1 today will be equivalent in buying power to £1.01 next year.
Cumulative price change 15,281.67% Average inflation rate 2.20% Converted amount (£1 base) £153.82 Price difference (£1 base) £152.82 CPI in 1789 7.400 CPI in 2020 1,138.243 Inflation in 1789 -1.33% Inflation in 2020 1.50%
GBP Inflation since 1750
Annual Rate, the Office for National Statistics CPI
## Buying power of £1 in 1789
This chart shows a calculation of buying power equivalence for £1 in 1789 (price index tracking began in 1750).
For example, if you started with £1, you would need to end with £153.82 in order to "adjust" for inflation (sometimes refered to as "beating inflation").
When £1 is equivalent to £153.82 over time, that means that the "real value" of a single U.K. pound decreases over time. In other words, a pound will pay for fewer items at the store.
This effect explains how inflation erodes the value of a pound over time. By calculating the value in 1789 dollars, the chart below shows how £1 is worth less over 231 years.
According to the Office for National Statistics, each of these GBP amounts below is equal in terms of what it could buy at the time:
Pound inflation: 1789-2020
Year Pound Value Inflation Rate
1789 £1.00 -1.33%
1790 £1.01 1.35%
1791 £1.01 0.00%
1792 £1.03 1.33%
1793 £1.05 2.63%
1794 £1.15 8.97%
1795 £1.27 10.59%
1796 £1.35 6.38%
1797 £1.22 -10.00%
1798 £1.19 -2.22%
1799 £1.34 12.50%
1800 £1.82 36.36%
1801 £2.04 11.85%
1802 £1.57 -23.18%
1803 £1.49 -5.17%
1804 £1.53 2.73%
1805 £1.77 15.93%
1806 £1.70 -3.82%
1807 £1.66 -2.38%
1808 £1.73 4.07%
1809 £1.89 9.37%
1810 £1.95 2.86%
1811 £1.89 -2.78%
1812 £2.15 13.57%
1813 £2.20 2.52%
1814 £1.92 -12.88%
1815 £1.72 -10.56%
1816 £1.57 -8.66%
1817 £1.78 13.79%
1818 £1.78 0.00%
1819 £1.74 -2.27%
1820 £1.58 -9.30%
1821 £1.39 -11.97%
1822 £1.20 -13.59%
1823 £1.28 6.74%
1824 £1.39 8.42%
1825 £1.64 17.48%
1826 £1.54 -5.79%
1827 £1.45 -6.14%
1828 £1.41 -2.80%
1829 £1.39 -0.96%
1830 £1.34 -3.88%
1831 £1.47 10.10%
1832 £1.36 -7.34%
1833 £1.28 -5.94%
1834 £1.18 -8.42%
1835 £1.20 2.30%
1836 £1.34 11.24%
1837 £1.36 2.02%
1838 £1.38 0.99%
1839 £1.47 6.86%
1840 £1.50 1.83%
1841 £1.47 -1.80%
1842 £1.35 -8.26%
1843 £1.20 -11.00%
1844 £1.20 0.00%
1845 £1.26 4.49%
1846 £1.31 4.30%
1847 £1.47 12.37%
1848 £1.28 -12.84%
1849 £1.20 -6.32%
1850 £1.14 -5.62%
1851 £1.09 -3.57%
1852 £1.09 0.00%
1853 £1.20 9.88%
1854 £1.38 14.61%
1855 £1.42 2.94%
1856 £1.42 0.00%
1857 £1.35 -4.76%
1858 £1.23 -9.00%
1859 £1.22 -1.10%
1860 £1.26 3.33%
1861 £1.28 2.15%
1862 £1.26 -2.11%
1863 £1.22 -3.23%
1864 £1.20 -1.11%
1865 £1.22 1.12%
1866 £1.28 5.56%
1867 £1.36 6.32%
1868 £1.35 -0.99%
1869 £1.28 -5.00%
1870 £1.28 0.00%
1871 £1.30 1.05%
1872 £1.35 4.17%
1873 £1.41 4.00%
1874 £1.35 -3.85%
1875 £1.32 -2.00%
1876 £1.32 0.00%
1877 £1.31 -1.02%
1878 £1.28 -2.06%
1879 £1.23 -4.21%
1880 £1.27 3.30%
1881 £1.26 -1.06%
1882 £1.27 1.08%
1883 £1.26 -1.06%
1884 £1.23 -2.15%
1885 £1.19 -3.30%
1886 £1.18 -1.14%
1887 £1.16 -1.15%
1888 £1.18 1.16%
1889 £1.19 1.15%
1890 £1.19 0.00%
1891 £1.20 1.14%
1892 £1.20 0.00%
1893 £1.19 -1.12%
1894 £1.18 -1.14%
1895 £1.16 -1.15%
1896 £1.15 -1.16%
1897 £1.18 2.35%
1898 £1.18 0.00%
1899 £1.19 1.15%
1900 £1.24 4.55%
1901 £1.24 0.00%
1902 £1.24 0.00%
1903 £1.26 1.09%
1904 £1.26 0.00%
1905 £1.26 0.00%
1906 £1.26 0.00%
1907 £1.27 1.08%
1908 £1.27 0.00%
1909 £1.28 1.06%
1910 £1.30 1.05%
1911 £1.30 0.00%
1912 £1.34 3.13%
1913 £1.32 -1.01%
1914 £1.32 0.00%
1915 £1.49 12.24%
1916 £1.76 18.18%
1917 £2.20 25.38%
1918 £2.69 22.09%
1919 £2.96 10.05%
1920 £3.42 15.53%
1921 £3.12 -8.70%
1922 £2.69 -13.85%
1923 £2.53 -6.03%
1924 £2.51 -0.53%
1925 £2.51 0.00%
1926 £2.50 -0.54%
1927 £2.43 -2.70%
1928 £2.43 0.00%
1929 £2.41 -1.11%
1930 £2.34 -2.81%
1931 £2.24 -4.05%
1932 £2.19 -2.41%
1933 £2.14 -2.47%
1934 £2.14 0.00%
1935 £2.15 0.63%
1936 £2.16 0.63%
1937 £2.24 3.75%
1938 £2.27 1.20%
1939 £2.34 2.98%
1940 £2.73 16.76%
1941 £3.03 10.89%
1942 £3.24 7.14%
1943 £3.35 3.33%
1944 £3.45 2.82%
1945 £3.54 2.75%
1946 £3.65 3.05%
1947 £3.91 7.04%
1948 £4.20 7.61%
1949 £4.32 2.89%
1950 £4.46 3.13%
1951 £4.86 9.09%
1952 £5.31 9.17%
1953 £5.47 3.05%
1954 £5.58 1.98%
1955 £5.82 4.36%
1956 £6.12 5.10%
1957 £6.34 3.53%
1958 £6.54 3.20%
1959 £6.57 0.41%
1960 £6.64 1.03%
1961 £6.86 3.46%
1962 £7.16 4.33%
1963 £7.30 1.89%
1964 £7.54 3.33%
1965 £7.89 4.66%
1966 £8.20 3.94%
1967 £8.42 2.64%
1968 £8.81 4.65%
1969 £9.28 5.37%
1970 £9.88 6.40%
1971 £10.81 9.44%
1972 £11.58 7.13%
1973 £12.64 9.10%
1974 £14.66 16.04%
1975 £18.22 24.24%
1976 £21.23 16.54%
1977 £24.59 15.85%
1978 £26.64 8.30%
1979 £30.20 13.39%
1980 £35.64 17.99%
1981 £39.86 11.87%
1982 £43.30 8.61%
1983 £45.28 4.59%
1984 £47.54 4.98%
1985 £50.43 6.08%
1986 £52.15 3.40%
1987 £54.32 4.17%
1988 £56.99 4.90%
1989 £61.42 7.78%
1990 £67.23 9.46%
1991 £71.18 5.87%
1992 £73.84 3.74%
1993 £75.01 1.59%
1994 £76.82 2.41%
1995 £79.49 3.47%
1996 £81.41 2.41%
1997 £83.96 3.14%
1998 £86.84 3.43%
1999 £88.18 1.54%
2000 £90.78 2.96%
2001 £92.39 1.77%
2002 £93.93 1.67%
2003 £96.65 2.89%
2004 £99.53 2.98%
2005 £102.34 2.82%
2006 £105.61 3.20%
2007 £110.14 4.29%
2008 £114.53 3.99%
2009 £113.92 -0.53%
2010 £119.18 4.61%
2011 £125.38 5.20%
2012 £129.41 3.21%
2013 £133.34 3.04%
2014 £136.49 2.36%
2015 £137.84 0.99%
2016 £140.23 1.74%
2017 £145.26 3.58%
2018 £148.86 2.48%
2019 £151.54 1.80%
2020 £153.82 1.50%*
* Compared to previous annual rate. Not final. See inflation summary for latest 12-month trailing value.
Click to show 225 more rows
This conversion table shows various other 1789 amounts in today's pounds, based on the 15,281.67% change in prices:
Conversion Table: Value of a pound today
Initial value Equivalent value
£1 pound in 1789 £153.82 pounds today
£5 pounds in 1789 £769.08 pounds today
£10 pounds in 1789 £1,538.17 pounds today
£50 pounds in 1789 £7,690.83 pounds today
£100 pounds in 1789 £15,381.67 pounds today
£500 pounds in 1789 £76,908.34 pounds today
£1,000 pounds in 1789 £153,816.67 pounds today
£5,000 pounds in 1789 £769,083.35 pounds today
£10,000 pounds in 1789 £1,538,166.70 pounds today
£50,000 pounds in 1789 £7,690,833.52 pounds today
£100,000 pounds in 1789 £15,381,667.04 pounds today
£500,000 pounds in 1789 £76,908,335.19 pounds today
£1,000,000 pounds in 1789 £153,816,670.37 pounds today
## How to Calculate Inflation Rate for £1 since 1789
Our calculations use the following inflation rate formula to calculate the change in value between 1789 and today:
CPI today CPI in 1789
×
1789 GBP value
=
Today's value
Then plug in historical CPI values. The U.K. CPI was 7.4 in the year 1789 and 1138.2433607545 in 2020:
1138.24336075457.4
×
£1
=
£153.82
£1 in 1789 has the same "purchasing power" or "buying power" as £153.82 in 2020.
To get the total inflation rate for the 231 years between 1789 and 2020, we use the following formula:
CPI in 2020 - CPI in 1789CPI in 1789
×
100
=
Cumulative inflation rate (231 years)
Plugging in the values to this equation, we get:
1138.2433607545 - 7.47.4
×
100
=
15,282%
Politics and news often influence economic performance. Here's what was happening at the time:
• George Washington is elected President and John Adams is elected Vice-President by the first US Electoral College.
• William Wilberforce proposes the abolition of slavery in the United Kingdom House of Commons.
• The fall of the Bastille Prison marks the beginning of the French Revolution
## Data Source & Citation
Raw data for these calculations comes from the composite price index published by the UK Office for National Statistics (ONS). A composite index is created by combining price data from several different published sources, both official and unofficial. The Consumer Price Index, normally used to compute inflation, has only been tracked since 1988. All inflation calculations after 1988 use the Office for National Statistics' Consumer Price Index, except for 2017, which is based on The Bank of England's forecast.
You may use the following MLA citation for this page: “£1 in 1789 → 2020 | UK Inflation Calculator.” Official Inflation Data, Alioth Finance, 18 Sep. 2020, https://www.officialdata.org/uk/inflation/1789?amount=1.
Special thanks to QuickChart for their chart image API, which is used for chart downloads.
in2013dollars.com is a reference website maintained by the Official Data Foundation.
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# How far is Cherbourg from Bordeaux?
The distance between Bordeaux (Bordeaux–Mérignac Airport) and Cherbourg (Cherbourg – Maupertus Airport) is 335 miles / 539 kilometers / 291 nautical miles.
The driving distance from Bordeaux (BOD) to Cherbourg (CER) is 434 miles / 699 kilometers, and travel time by car is about 7 hours 41 minutes.
335
Miles
539
Kilometers
291
Nautical miles
1 h 8 min
## Distance from Bordeaux to Cherbourg
There are several ways to calculate the distance from Bordeaux to Cherbourg. Here are two standard methods:
Vincenty's formula (applied above)
• 334.978 miles
• 539.095 kilometers
• 291.088 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth's surface using an ellipsoidal model of the planet.
Haversine formula
• 335.026 miles
• 539.171 kilometers
• 291.129 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## How long does it take to fly from Bordeaux to Cherbourg?
The estimated flight time from Bordeaux–Mérignac Airport to Cherbourg – Maupertus Airport is 1 hour and 8 minutes.
## Flight carbon footprint between Bordeaux–Mérignac Airport (BOD) and Cherbourg – Maupertus Airport (CER)
On average, flying from Bordeaux to Cherbourg generates about 74 kg of CO2 per passenger, and 74 kilograms equals 164 pounds (lbs). The figures are estimates and include only the CO2 generated by burning jet fuel.
## Map of flight path and driving directions from Bordeaux to Cherbourg
See the map of the shortest flight path between Bordeaux–Mérignac Airport (BOD) and Cherbourg – Maupertus Airport (CER).
## Airport information
Origin Bordeaux–Mérignac Airport
City: Bordeaux
Country: France
IATA Code: BOD
ICAO Code: LFBD
Coordinates: 44°49′41″N, 0°42′56″W
Destination Cherbourg – Maupertus Airport
City: Cherbourg
Country: France
IATA Code: CER
ICAO Code: LFRC
Coordinates: 49°39′0″N, 1°28′13″W
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Cody
# Problem 42324. Accessing value of variable whose name is stored in another variable as string.
Solution 2981605
Submitted on 23 Sep 2020 by Benjamin
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
n = 10; var3_correct = 10; assert(isequal(varaccess(n),var3_correct))
2 Pass
n = 30; var3_correct = 30; assert(isequal(varaccess(n),var3_correct))
3 Pass
n = 0.6; var3_correct = 0.6; assert(isequal(varaccess(n),var3_correct))
4 Pass
n = 500; var3_correct = 500; assert(isequal(varaccess(n),var3_correct))
5 Pass
n = 7.8; var3_correct = 7.8; assert(isequal(varaccess(n),var3_correct))
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# Math Statistics probability functions Distributions
Not all questions must be answered nor in prefect detail, as long as I receive the answers in a timely fashion. A more legible copy of the problems is submitted in a text file as an attachment.
Please don’t hesitate to contact me with any questions; thanks!
—-
Assume that X = (X1, …., Xn) is a vector of independently observations of the Bernoulli distribution Xi ∼ Be(θ) where the parameter θ ∈ [0, 1] is unknown.
1. Decide the logarithim probability function L(θ; x) = log fn(x|θ) if n = 9, and x = (1, 0, 0, 1, 1, 0, 1, 0, 0).
2. Maximize L(θ; x) = log fn(x|θ) by taking θ into account.
(Standalone) 3. Assume that X1, X2, and X3 are randomly and evenly sampled from a distribution, where the median is equal to the unknown parameter θ. Calculate the probability P(min{X1, X2, X3} < θ < max{X1, X2, X3}). Formulate your response as a statement of the confidence interval for θ.
For n ≥ 1, assume that a lazy person would randomly insert n differently addressed letters into n envelopes without checking that the addresses on the envelopes and letters coincide. Let Xn be the amount of letters that reach the right recipient.
4. Decide the expected value and variance in the variable Xn. (Hint: you can write Xn as a sum of dependent Bernoulli variables, where the covariance can be decided using combinatorics.)
5. Decide towards which distribution Xn is converging towards when n approaches infinity.
Assume that X is normally disctibuted with the expected value 0 and the standard deviation 1. Assume that Y is a variable so that the generated function is given by: E (e^(sY) | X) = e^(sX^(2)+3s)
6. What is the expected value of Y?
7. What is the variance of Y and the covariance Cov(X, Y)?
p(7)
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# PQ and RS are two perpendicular chord of rectangular hyperbola xy=c^2.If C is the centre of the hyperbola,then the product of the slopes ofCP,CQ,CR and CS is equal toa. -1b.0c.-1d none
148 Points
14 years ago
Let Point P is (ct1,c/t1)
Point Q is (ct2,c/t2)
Point R is (ct3,c/t3)
Point S is (ct4,c/t4)
slope of chord PQ is = -1/t1t2
slope of chord RS is = -1/t3t4
both chord are perpendicular so
-1/t1t2 X -1/t3t4 =-1
or t1t2t3t4 =-1
now slope of CP =1/t12
now slope of CQ =1/t22
now slope of CR =1/t32
now slope of CS =1/t42
so product of slope =1
Please feel free to post as many doubts on our discussion forum as you can.
If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly.
All the best.
Regards,
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World's only instant tutoring platform
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Question
# terms and 5 terms 2. A G.P. has first term 729 and 7 th term 64. Find the sumi of its first 7 terms. 3. Find the sum of the products of the corresponding terms of the sequences and . 4. Find the sum to terms of the series 5. Evaluate . 6. How many terms of the series must be taken to make 511 ? 7. How many terms of the series must be taken to make ? 8. The sum of some terms of a G.P. is 315 whose first term and the common ratio are 5 and 2 respectively. Find the last term and the number of terms. 9. A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places, then find its common ratio. 10. Find the sum of the series :
1. to terms
2. to terms
3. to terms
4. to terms
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Question Text terms and 5 terms 2. A G.P. has first term 729 and 7 th term 64. Find the sumi of its first 7 terms. 3. Find the sum of the products of the corresponding terms of the sequences and . 4. Find the sum to terms of the series 5. Evaluate . 6. How many terms of the series must be taken to make 511 ? 7. How many terms of the series must be taken to make ? 8. The sum of some terms of a G.P. is 315 whose first term and the common ratio are 5 and 2 respectively. Find the last term and the number of terms. 9. A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places, then find its common ratio. 10. Find the sum of the series : Topic Coordinate Geometry Subject Mathematics Class Class 11
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# Patients who attended clinic in one week grouped by age as shown in the table below.
309 views
Patients who attended clinic in one week grouped by age as shown in the table below.
X Age (years) No. of patients 0 - 5 14 5 - 15 41 15 - 25 59 25 - 45 70 45 - 75 15
1. Estimate the mean age.
2. On the graph provided , draw a histogram to represent the distribution.
X Age (years) No. of patients(F) x FX 0 - 5 14 2.5 35 5 - 15 41 10 410 15 - 25 59 20 1180 25 - 45 70 35 2450 45 - 75 15 60 900 Σf =199 Σfx 4975
1. x̄ = Σfx
Σf
x̄ = 4975
199
x̄ = 25years
by
1.25
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Phys 0175 - Lecture 2
# Phys 0175 - Lecture 2 - Lecture 2(Jan 7 2008 Chapter...
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Lecture 2 (Jan. 7, 2008): Coulomb’s Law Coulomb’s Law vs. Law of Gravitation Illustrative examples Conductors and insulators Superconductors and semiconductors Induced charges Shell theorems, along with an illustrative example Chapter 21 (continued):
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Coulomb’s Law: The magnitude of the electric force between two point charges q 1 and q 2 that are separated by a distance r is given by the equation: 1 2 2 q q F k r = k = 8.988 x 10 9 N•m 2 /C 2 (in S.I. units) Direction of F: Attractive if the charges are opposite Repulsive if the charges are alike
Torsion balance used to measure the electric force: Like charges repel Unlike charges attract
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The constant k that appears in Coulomb’s Law is often expressed in terms of another, more fundamental constant ε 0 which is called the permittivity of free space (i.e. vacuum): 9 9 2 2 0 12 2 2 0 1 8.988 10 9.0 10 / 4 8.854 10 / k N m C C N m πε ε - = = × 2245 × ⋅ → = × ⋅
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Phys 0175 - Lecture 2 - Lecture 2(Jan 7 2008 Chapter...
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# How long does it take to boil water in a 700 watt microwave?
Contents
Fill a glass measure with 1 cup of room temperature water and microwave on high (100% power) until the water comes to a boil; note how long this took. A 600-700 watt oven will boil water in 2½ to 3 minutes. An oven with more wattage will boil water in less time.
## How long should a microwave take to boil water?
For most microwaves, it should take between 1-3 minutes to boil water. This is largely dependent on the wattage of your microwave. If you know the wattage, this is a general breakdown of how long it will take to boil water. These are based on one cup of water.
## How long does it take to boil water in a microwave on high?
Simply place water in microwave-safe cup or bowl and nuke on high power for 1 to 3 minutes for the water to boil.
## How much longer does a 700 watt microwave cook?
A recipe that recommended 10 minutes of cooking time would be reduced to 8 minutes and 20 seconds. A 700 watt microwave has a similar conversion. A 5-minute recipe requires an additional 3 minutes of cooking time, while a 10-minute recipe requires 6 minutes longer.
## Is 700 watt microwave powerful enough?
The microwave’s wattage simply indicates the cooking power or strength. As a rule of thumb, a low-watt microwave = more time in cooking. But a 700 watts microwave is sufficient for basic cooking, defrosting and reheating. So, if you can spare a few extra minutes, 700 watts microwave is best for home use.
## Why you should never microwave water?
Water (alone) should never be heated in a microwave oven. … Without bubbles, the water cannot release the heat that has built up, the liquid does not boil, and it continues to heat up past its boiling point.
## How long does it take to boil 2 cups of water in a microwave?
For 2 cups of water, it would take approximately 2 to 2.5 minutes in a microwave oven with at least 1000 watts of power. Do not wait until you see rolling bubbles, as you would when boiling water on the stovetop.
## Can u boil water in microwave?
Boiling water in the microwave is convenient and safe. The method is best used when heating small quantities of water, as microwaves can distribute heat unevenly. According to current research, no negative health effects are associated with boiling water in the microwave.
## How do you boil water in the microwave for tea?
Here’s exactly how to do it:
1. Add water and a tea bag to a microwave-safe mug.
2. Place the mug in the microwave, and heat for 30 seconds on 50 percent power.
3. Let the mug sit for a minute before removing the teabag and sipping the tea.
4. Repeat three times a day.
IT IS DELICIOUS: Quick Answer: How do you defrost a Honey Baked Ham quickly?
## How do you superheat water in a microwave?
Superheating is achieved by heating a homogeneous substance in a clean container, free of nucleation sites, while taking care not to disturb the liquid. This may occur by microwaving water in a very smooth container. Disturbing the water may cause an unsafe eruption of hot water and result in burns.
## Will a 700 watt microwave pop popcorn?
Be sure to know what kind microwave you have because a 700-watt cooks slower than a 1,000-watt. Just a FYI, the popcorn button setting on your microwave should be used with caution (or not at all.) … So, it’s wise to just follow the instructions on the bag; which saying to not use the popcorn button.
## How many watts does a 700 watt microwave use?
Short summary: Home microwave ovens are typically 40–60% efficient. So if your 700W microwave oven is really generating 700W, it should draw between 1200 and 1800W or 10 – 15A at 120V.
## How big is a 700 watt microwave?
700-Watt Microwave, Black with 10 Power Levels (L x W x H) 19.21 x 14.96 x 11.46 Inches.
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# Velocity operator inconsistency and discrete particle
1. Oct 12, 2008
### Frostal
I'm going to mix a couple questions together instead of creating a new topic for each question.
I hope you don't mind.
I'm an electrical engineer(micro-electronics), so while I got the basics of QM in my studies I had to do most of my more 'in depth' learning on my own by reading books/ocwm from MIT.
I took a course in Semiconductor Physics in my last year and came across some questions that seemed to be more of a coincidence rather than a rule. Thus here goes:
1.H$$\Psi = E_{bloch}\Psi$$ for a bloch electron in a crystal lattice.
with $$\Psi$$= u(r,t)e($$\vec{k} \vec{r}$$)
So far nothing out of the ordinary. Now we apply QM's to the wave function to find the average velocity by $$\int \Psi* \nabla_{\vec{r}}\Psi = \nabla_{\vec{k}}E(\vec{k})$$
The last equation is said to be derivated from certain properties. Nothing wrong with that I am sure it's correct. But it's a strange coincidence that this happens to be equal to the group velocity of a packet of Bloch functions.
By assuming the wave packet of the bloch functions and reworking the Hamiltonian we get
$$E_{n}(\nabla_{\vec{r}}) + V(\vec{r})$$
Now by going to the classical limit with $$\nabla_{\vec{r}} = \vec{p}$$ and $$\vec{r} = \vec{r}$$
we get $$E_{n}(\vec{p}) + V(\vec{r})$$
In classical Mechanics we get
$$\vec{v} = \frac{\partial H}{\partial \vec{p}} = \frac{\partial E_{n}(\vec{p})}{ \partial \vec{p}}= \nabla_{\vec{k}} E_{n}(\vec{k})$$
and
$$\vec{F} = \frac{\partial H}{\partial \vec{r}} = \frac{\partial V(\vec{r})}{\partial \vec{r}}$$
it seems that just applying the velocity operator on the Bloch wave or finding the speed through the wave packet give the same speed indications. Which seems weird to me.
I'm also not completely sold on the velocity operator = momentum operator / mass.
Because the equality stems from the free particle where the frequency of the Broglie wave function = $$\frac{p^{2}}{2m}$$. But this is not the case. The Bloch function is made up of several momentum waves(fourier) which propagate with the same speed since the the Bloch wave function acts with frequency $$E_{n}(\vec{k})$$
Only when you use a wavepacket of Bloch waves do you get different Bloch waves move at different speed resulting in a different group velocity from the fase velocity.
I'm very confused in that the velocity operator does not hold into account the time evolution. This is understandable in the case of the free wave particle where the frequency is in direct relation to it's momentum because of the lack of a potential function.
But How can this hold when we get a potential function within the hamiltonian like for example with Bloch electrons. The momentum operator holds but I don't see how the velocity operator could hold in this case.
2.I've gone through alot of threads on this forum to find a clear cut answer but I doubt there is any so just in case I'm just asking it again here.
As we know there's alot of debate upon the interpretation of the wave function in QM.
I was just wondering if there's any good theories or proof on the particles existing in discreet states. With that I mean (for example with an electron) when we detect it, do we really detect it as a physical little ball (discreet) or is the detection just a interaction with the particle that forces it to localise into a wavepacket and thus remain nonlocal. In other words is there any proof of the electron and other particles existing outside of these wavefunctions. I know what the wavefunctions mean so don't try to explain the copenhagen interpretation again. I'm just wondering if 'the chance you find the particle in this location' = 'the chance the wave functions crumbles up into a localised wavepacket in with the average position in this location'
Both are very different.
The second interpretation would seem to be the more logical one. Since else particles would teleport at random? Take for example a localised particle in a wavepacket. Now lets say we measure the momentum of the electron. QM states that this particle will collapse into 1 Broglie wavefunction and thus the electron can be anywhere even alot further into space than it was in the localised wavepacket. Thus the particle would teleport.
I'm a big fan of QM. non locality is the only way for a particle to move in a continuous space since else it would have to pass through an infinite number of 'points' to move. That is if you consider space to be continuous.(I've considered space at a very small level to be discrete but it just doesn't feel right.
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Ex 2.2
Chapter 2 Class 10 Polynomials
Serial order wise
### Transcript
Ex 2.2, 2 Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively. (i) 1/4 , −1 Let the polynomial be p(x) = ax2 + bx + c, Sum of zeroes = 𝟏/𝟒 − 𝑏/𝑎 = 1/4 Assuming a = 1 − 𝑏/1 = 1/4 b = (−𝟏)/𝟒 Product of zeroes = −1 𝑐/𝑎 = −1 Assuming a = 1 𝑐/1 = −1 c = −1 Now, a = 1, b = – 1/4 and c = –1 Hence, the required quadratic polynomial = ax2 + bx + c = 1x2 − 1/4x − 1 = x2 − 𝟏/𝟒x − 1 Another quadratic polynomial can be = 4(x2 − 1/4x − 1 ) = 4x2 − x − 4
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Intuitionistic Logic Explorer < Previous Next > Nearby theorems Mirrors > Home > ILE Home > Th. List > reu6i Structured version GIF version
Theorem reu6i 2726
Description: A condition which implies existential uniqueness. (Contributed by Mario Carneiro, 2-Oct-2015.)
Assertion
Ref Expression
reu6i ((B A x A (φx = B)) → ∃!x A φ)
Distinct variable groups: x,A x,B
Allowed substitution hint: φ(x)
Proof of Theorem reu6i
Dummy variable y is distinct from all other variables.
StepHypRef Expression
1 eqeq2 2046 . . . . 5 (y = B → (x = yx = B))
21bibi2d 221 . . . 4 (y = B → ((φx = y) ↔ (φx = B)))
32ralbidv 2320 . . 3 (y = B → (x A (φx = y) ↔ x A (φx = B)))
43rspcev 2650 . 2 ((B A x A (φx = B)) → y A x A (φx = y))
5 reu6 2724 . 2 (∃!x A φy A x A (φx = y))
64, 5sylibr 137 1 ((B A x A (φx = B)) → ∃!x A φ)
Colors of variables: wff set class Syntax hints: → wi 4 ∧ wa 97 ↔ wb 98 = wceq 1242 ∈ wcel 1390 ∀wral 2300 ∃wrex 2301 ∃!wreu 2302 This theorem was proved from axioms: ax-1 5 ax-2 6 ax-mp 7 ax-ia1 99 ax-ia2 100 ax-ia3 101 ax-io 629 ax-5 1333 ax-7 1334 ax-gen 1335 ax-ie1 1379 ax-ie2 1380 ax-8 1392 ax-10 1393 ax-11 1394 ax-i12 1395 ax-bndl 1396 ax-4 1397 ax-17 1416 ax-i9 1420 ax-ial 1424 ax-i5r 1425 ax-ext 2019 This theorem depends on definitions: df-bi 110 df-tru 1245 df-nf 1347 df-sb 1643 df-eu 1900 df-clab 2024 df-cleq 2030 df-clel 2033 df-nfc 2164 df-ral 2305 df-rex 2306 df-reu 2307 df-v 2553 This theorem is referenced by: eqreu 2727 riota5f 5435 negeu 6999 creur 7692 creui 7693
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# What is the relation between centripetal acceleration and radius in uniform circular motion?
In uniform circular motion we know that $a_c=\frac {v^2}r=\omega^2r$.So,is $a_c$ directly or inversely proportional with $r$ and why not the other is true?
Thanks for any help.
In uniform circular motion, $v = v(r)$ so that $a_{c} \propto r$, not $a_{c} \propto r^{-1}$. In other words, $v$ is not a constant at all radii while $\omega$ is constant for all radii. So the second part of the expression is the one you want to look at in this regard.
Though the first half of the expression states that $a_{c}$ is only explicitly dependent upon $r^{-1}$, it is still implicitly proportional to $r$.
In the first expression $v$ is the speed of a single point on the disk, actually any point that is a distance $r$ from the center. In the second expression, $\omega$ is the angular speed of the entire disk. Since $v$ and $\omega$ have very different definitions, the interpretation of those two equations is quite different.
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## How many calories do you burn when Walking 4 mph?
Someone weighing 180 lbs burns approximately 309 kilocalories per hour when Walking 4 mph. Fill in the form and calculate how many calories you burn when Walking 4 mph or use our Calorie Calculator for other activities.
309
## Calories burned with Walking (weight: 180 lbs)
MET 60 min.
Nordic Walking 5.9 506
Walking 3 257
Walking (fast) 3.6 309
Walking (slow) 2.8 240
Walking 3 mph 2.5 214
Walking 4 mph 3.6 309
## How do we calculate the amount of calories burned when Walking 4 mph?
This calculation uses the MET value (Metabolic Equivalent of Task) of Walking 4 mph. The MET value of Walking 4 mph = 3.6. We multiply the MET value with the person\'s body weight in kilogram. Then we multiply this with 0.0175 and the duration in minutes.
A person weighs: 180 lbs
MET value of Walking 4 mph: 3.6
Time: 30 minutes
The calorie calculation for Walking 4 mph for 30 minutes is as follows:
(180/2.20462) * 3.6 * 0.0175 * 30 minutes = 154
30 minutes of Walking 4 mph burns 154 kcal.
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Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > unissint Structured version Unicode version
Theorem unissint 4074
Description: If the union of a class is included in its intersection, the class is either the empty set or a singleton (uniintsn 4087). (Contributed by NM, 30-Oct-2010.) (Proof shortened by Andrew Salmon, 25-Jul-2011.)
Assertion
Ref Expression
unissint
Proof of Theorem unissint
StepHypRef Expression
1 simpl 444 . . . . 5
2 df-ne 2601 . . . . . . 7
3 intssuni 4072 . . . . . . 7
42, 3sylbir 205 . . . . . 6
54adantl 453 . . . . 5
61, 5eqssd 3365 . . . 4
76ex 424 . . 3
87orrd 368 . 2
9 ssv 3368 . . . . 5
10 int0 4064 . . . . 5
119, 10sseqtr4i 3381 . . . 4
12 inteq 4053 . . . 4
1311, 12syl5sseqr 3397 . . 3
14 eqimss 3400 . . 3
1513, 14jaoi 369 . 2
168, 15impbii 181 1
Colors of variables: wff set class Syntax hints: wn 3 wb 177 wo 358 wa 359 wceq 1652 wne 2599 cvv 2956 wss 3320 c0 3628 cuni 4015 cint 4050 This theorem was proved from axioms: ax-1 5 ax-2 6 ax-3 7 ax-mp 8 ax-gen 1555 ax-5 1566 ax-17 1626 ax-9 1666 ax-8 1687 ax-6 1744 ax-7 1749 ax-11 1761 ax-12 1950 ax-ext 2417 This theorem depends on definitions: df-bi 178 df-or 360 df-an 361 df-tru 1328 df-ex 1551 df-nf 1554 df-sb 1659 df-clab 2423 df-cleq 2429 df-clel 2432 df-nfc 2561 df-ne 2601 df-ral 2710 df-rex 2711 df-v 2958 df-dif 3323 df-in 3327 df-ss 3334 df-nul 3629 df-uni 4016 df-int 4051
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http://websvn.kolibrios.org/filedetails.php?repname=Kolibri+OS&path=%2Fprograms%2Fdemos%2Flife2%2Flif%2FPUSHGUN2.lif&rev=1769&peg=1769
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# Subversion RepositoriesKolibri OS
Rev
1. #Life 1.05
2. #D Sqrtgun 3.0
3. #D Population in generation t is asymptotic to 5 sqrt(t/12). A 4 glider
4. #D salvo pushes a blinker 3 units southeast and sends back a glider,
5. #D which causes another salvo to be fired and a glider to be sent
6. #D northeast. The round trip time, and hence the gap between
7. #D northeastward gliders, increases by an average of 24 generations each
8. #D time. More specifically, let a[0]=207, a[1]=159, a[2]=-33, a[3]=111,
9. #D and a[4]=231. For n>=0, a glider is sent northeast (by reflection from
10. #D a buckaroo) in generation 12 n^2 + 1116 n + a[n mod 5] (at which
11. #D time, for n>=1, the population is 5n + 1297) and escapes about 300
12. #D generations later.
13. #D Dean Hickerson, dean@ucdmath.ucdavis.edu 1/28/91
14. #N
15. #P 55 60
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## Trigonometric Identities Worksheet with solution
Disclosure: Sometime I recommend books I have used in the past and were helpful me .This post may contains affiliate links that means if you purchase through link ,I will earn a commission fee at no extra cost to you .Thank you so much for your support Q.1 Prove that √sec²θ + cosec²θ = tanθ+cotθ … Read more
## Proving Trigonometric identities Worksheet[With Solution]
Disclosure: Sometime I recommend books I have used in the past and were helpful me .This post may contains affiliate links that means if you purchase through link ,I will earn a commission fee at no extra cost to you .Thank you so much for support 😐 Problem: 01 Prove that: √sec²θ + cosec²θ=tanθ+cotθ Solution … Read more
## Real number class 10 mcq online test with answer
Practice these MCQ and true-false questions of the real number to check your progress. Q.1 The exponent of 2 in the prime factorization of 144 (a) 4 (b) 5 (c) 6 (d)3 Answer: (a) Q.2 If the LCM of a and 18 is 36 and the HCF of a and 18 is 2 ,then a … Read more
## Finding Square root using prime factorization method
Procedure to find square root using prime factorization method Write the prime factors of the given number Make pairs of the equal factors Write one factor corresponding to each pair Multiply the obtained factors Q.1 Find the square root of 256 using the prime factorization method. Solution: 256=2×2×2×2×2×2×2×2 Make pairs 256=2×2×2×2×2×2×2×2 √256=√2×2×2×2×2×2×2×2 √256=2×2×2×2 √256=16 Q.2 … Read more
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# Mixed numbers and improper fractions review
Review how to rewrite mixed numbers as improper fractions and improper fractions as mixed numbers. Then, try some practice problems.
## What is an improper fraction?
An improper fraction is a fraction where the numerator is greater than or equal to the denominator.
Below are examples of improper fractions:
start fraction, 9, divided by, 4, end fraction, comma, start fraction, 5, divided by, 5, end fraction, comma, start fraction, 7, divided by, 3, end fraction
## What is a mixed number?
A mixed number is a number consisting of a whole number and a proper fraction.
Below are examples of mixed numbers:
4, start fraction, 1, divided by, 2, end fraction, comma, 1, start fraction, 3, divided by, 8, end fraction, comma, 12, start fraction, 5, divided by, 6, end fraction
## Rewriting a mixed number as an improper fraction
Rewrite 3, start fraction, 4, divided by, 5, end fraction as an improper fraction.
3, start fraction, 4, divided by, 5, end fraction, equals, start color #11accd, 3, end color #11accd, plus, start color #1fab54, start fraction, 4, divided by, 5, end fraction, end color #1fab54
empty space, equals, start color #11accd, 1, end color #11accd, plus, start color #11accd, 1, end color #11accd, plus, start color #11accd, 1, end color #11accd, plus, start color #1fab54, start fraction, 4, divided by, 5, end fraction, end color #1fab54
empty space, equals, start color #11accd, start fraction, 5, divided by, 5, end fraction, end color #11accd, plus, start color #11accd, start fraction, 5, divided by, 5, end fraction, end color #11accd, plus, start color #11accd, start fraction, 5, divided by, 5, end fraction, end color #11accd, plus, start color #1fab54, start fraction, 4, divided by, 5, end fraction, end color #1fab54
empty space, equals, start fraction, start color #11accd, 5, end color #11accd, plus, start color #11accd, 5, end color #11accd, plus, start color #11accd, 5, end color #11accd, plus, start color #1fab54, 4, end color #1fab54, divided by, 5, end fraction
3, start fraction, 4, divided by, 5, end fraction, equals, start fraction, 19, divided by, 5, end fraction
Problem 1A
• Current
Rewrite as an improper fraction.
5, start fraction, 1, divided by, 2, end fraction, equals
Want to try more problems like this? Check out this exercise.
## Rewriting an improper fraction as a mixed number
Rewrite start fraction, 10, divided by, 3, end fraction as a mixed number.
start fraction, 3, divided by, 3, end fraction, equals, 1, start text, space, w, h, o, l, e, end text
So, let's see how many wholes we can get out of start fraction, 10, divided by, 3, end fraction.
start fraction, 10, divided by, 3, end fraction, equals, start fraction, start color #11accd, 3, end color #11accd, plus, start color #11accd, 3, end color #11accd, plus, start color #11accd, 3, end color #11accd, plus, start color #1fab54, 1, end color #1fab54, divided by, 3, end fraction
empty space, equals, start color #11accd, start fraction, 3, divided by, 3, end fraction, end color #11accd, plus, start color #11accd, start fraction, 3, divided by, 3, end fraction, end color #11accd, plus, start color #11accd, start fraction, 3, divided by, 3, end fraction, end color #11accd, plus, start color #1fab54, start fraction, 1, divided by, 3, end fraction, end color #1fab54
empty space, equals, start color #11accd, 1, end color #11accd, plus, start color #11accd, 1, end color #11accd, plus, start color #11accd, 1, end color #11accd, plus, start color #1fab54, start fraction, 1, divided by, 3, end fraction, end color #1fab54
start fraction, 10, divided by, 3, end fraction, equals, start color #11accd, 3, end color #11accd, start color #1fab54, start fraction, 1, divided by, 3, end fraction, end color #1fab54
Problem 2A
• Current
Rewrite as a mixed number.
start fraction, 13, divided by, 8, end fraction, equals
Want to try more problems like this? Check out this exercise.
## Want to join the conversation?
• some of them were hard maybe make it the same but easy.
thanks so much
• i got only the first one in correct. other then that i got all of them
• I figured out how to do the problems after 30 secends of silence
• extreme laughing
(1 vote)
• Dose it matter if you don't use paper to work it out because I do not need to?
• I sometimes use paper but good job. You're helping the enviorement!
• do we have to be smart to pass our test about improper fractions and mixed numbers because i don't understand how to do improper fractions and mixed numbers :(
• you don't have to be smart,you just have to take the time to understand it
• Kahn academy is such a help in my math. Thank you guys!
• would 8/2 as a mixed number be 4 0/2 ?
and what is the difference of the mixed number of 2/8? I dont get it.
• Only improper fraction will create a whole number or mixed number. An improper fraction will always have a numerator that is equal to or larger than the denominator.
8/2 is an improper fraction. 8/2 becomes just a whole number = 4. There is no fraction as 0/2=0.
2/8 is a proper fraction (the numerator is less than the denomintor). It can't be changed into a mixed number. All you can do is reduce it down to 1/4.
Hope this helps.
• Hey isn't the sun the smallest star?
• No, it is not and this is math, not science.
• this is really easy! Just work hard and it will be. 😁
• maybe 4 u and me it is, but others may have some trouble, and that is OKAY!!
• How do you subtract fractions?
• There's a nice trick for subtracting fractions.
To find the numerator: cross-multiply numerators with denominators and subtract the products. Specifically, find (1st numerator * 2nd denominator) - (2nd numerator * first denominator).
To find the denominator: multiply the denominators.
Then reduce the answer as needed.
Example: Let's do 5/6 - 2/9.
The numerator is (5*9) - (2*6) = 45-12 = 33.
The denominator is 6*9 = 54.
So we get 33/54, which reduces to a final answer of 11/18 (from dividing top and bottom each by 3).
This trick is an example of Vedic math. Try looking up Vedic math online and you might find other arithmetic tricks that you like!
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18 views
Suppose the domain set of an attribute consists of signed 3 digit numbers. What is the percentage of reduction in storage space of this attribute if it is stored as an integer rather than in character form?
PLEASE EXPLAIN WHAT THIS QUESTION MEANS AND SOLUTION
| 18 views
[(4-2)/4]*100 = 50%
by (205 points)
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Education Technology
# Activities
• ##### Subject Area
• Standard: VCE: Mathematical Methods: Mathematical Methods
Aust Senior
45 Minutes
• ##### Device
• TI-Nspire™
• TI-Nspire™ CAS
• TI-Nspire™ Navigator™
• ##### Software
TI-Nspire™
TI-Nspire™ CAS
TI-Nspire™ CAS Navigator™ NC System
TI-Nspire™ Navigator™ NC System
4.4
## Sample Some PI Monte
#### Activity Overview
In this activity, simulated random sampling is used to develop the concept of the sample proportion as an estimator of the population proportion. Simulation will allow us to investigate how a sample proportion varies from sample to sample.
#### Objectives
• To appreciate variability in the results obtained from different random samples
• To investigate estimates of a population proportion from sample proportions
• To establish the idea of the sample proportion as a random variable
• Informally investigate the standard deviation of the sampling distribution
#### Vocabulary
population parameter, population proportion, random sample, sample statistic, sample proportion, random variable, estimation
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MTH164FA10-M1d_ans
# First observe that using implicit dif z dx dy as
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Unformatted text preview: 16+25 A second approach follows the second method in part (b) above. ∂z ∂z dz = ∂ x dx + ∂ y dy where the partial derivatives were calculated ∂z ∂z ferentiation. Then start rewriting this as dz = ∂ x , ∂ y · dx, dy 4, −5 · dx, dy = 4, −5 dx, dy cos(θ) and finally continue late the direction of maximum increase. First observe that using implicit dif= ∇z · dx, dy = as before to calcu- These two approaches say exactly the same thing, but they use slight different notation. Both notations are widely used so it’s important to get used to translating between the two. (d) Find the directional derivative of z = f (x, y ) at P (2, 1, −1) in the direction ...
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## This document was uploaded on 05/06/2013.
Ask a homework question - tutors are online
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Page Contents
What is line to ground fault?
A line-to-line-to-ground fault presents low value impedances, with zero value for a direct short circuit or metallic fault, between the faulted two phases and ground at the point of fault in the network. In general, a fault may be represented as shown in Figure 1.
What is phase to neutral fault?
It is common on larger systems to monitor any current flowing through the neutral-to-earth link and use this as the basis for neutral fault protection. The connection between neutral and earth allows any phase-to-earth fault to develop enough current flow to “trip” the circuit overcurrent protection device.
Read more: How do I make Game Center public?
What happens when line to ground fault occurs?
Line to ground fault (L-G) is most common fault and 65-70 percent of faults are of this type. It causes the conductor to make contact with earth or ground. 15 to 20 percent of faults are double line to ground and causes the two conductors to make contact with ground.
What is a three phase fault?
A three phase bolted fault describes the condition where the three conductors are physically held together with zero impedance between them, just as if they were bolted together. For a balanced symmetrical system, the fault current magnitude is balanced equally within the three phases.
What does line to ground mean?
Generally, a single line-to-ground fault on a transmission line occurs when one conductor drops to the ground or comes in contact with the neutral conductor. Such types of failures may occur in power system due to many reasons like high-speed wind, falling off a tree, lightning, etc.
How do you calculate phase to ground?
In a standard phase-to-phase connection, the variable that represents the grounding force is 1.73; it is divided from the voltage in order to compensate for the ground in a circuit. Using the same sample equation 12*5=60 volts, we can account for the ground by dividing 60 by 1.73, making the new voltage 34.68 volts.
Read more: What is SCSI disk?
What causes neutral failure?
The conductor start melting and resulting broke off Neutral. Poor workmanship of Installation and technical staff also one of the reasons of Neutral Failure. A broken Neutral on Three phases Transformer will cause the voltage float up to line voltage depending upon the load balancing of the system.
How the sequence network for a single line to ground fault are connected?
For a single-phase-to-ground fault, the three networks are connected in series. Any fault impedance is multiplied by 3 and included in this connection, as shown in Fig. 3. For a phase-to-phase fault, the positive- and negative- sequence networks are connected in parallel, as shown in Fig.
What happens when two phases are shorted?
In mains circuits, short circuits may occur between two phases, between a phase and neutral or between a phase and earth (ground). Such short circuits are likely to result in a very high current and therefore quickly trigger an overcurrent protection device. A short circuit may lead to formation of an electric arc.
Read more: How do you do 23 divided by 25?
Three Phase Bolted Faults. A three phase bolted fault describes the condition where the three conductors are physically held together with zero impedance between them, just as if they were bolted together. For a balanced symmetrical system, the fault current magnitude is balanced equally within the three phases.
What is neutral to ground fault?
Neutral grounded by means of resistance. Grounding the neutral by means of resistance allows a definite current to be obtained in the case of a fault and consequently to be able to carry out selective protection of the network.
What is first ground fault?
The first ground fault is signalizing only. Generator is tripping and damaged winding is repairing. In this way it can prevent large damages which could be during the second ground fault.
What is a single line to ground fault?
Single Line-to-Ground Fault Generally, a single line-to-ground fault on a transmission line occurs when one conductor drops to the ground or comes in contact with the neutral conductor.
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```Date: Sat, 23 Oct 2004 07:48:24 -0700 Reply-To: Bill McKirgan maxsfolks Sender: "SAS(r) Discussion" From: Bill McKirgan maxsfolks Organization: http://groups.google.com Subject: Re: calculate the time between cancelation of 2 contracts on the household-level Content-Type: text/plain; charset=ISO-8859-1 Stefan, Summarizing across records is an area I need to practice more, and so this thread naturally caught my interest. Toby's "untested" code had one minor flaw, but still did the job of holding the desired values within the householdid group. While the correct values were retained I found one error in the part where the differences were calculated: Last_date = householdid --should be-- Last_date = canceldate Toby's example used INTCK to report the interval in weeks. Here again is an area I need more practice with, and so I played around a bit created another variable to report the interval in days. Evidently the interval part of the argument takes many variations: DAY, DAYS, WEEK, WEEKS, WEEK1.1...I tried SECONDS but still got a DAY interval (?huh?)...I guess I will RTFM that one. Thanks to both of you. This was a good question and example to learn from on a Saturday morning SAS-L Zen session. Kind regards, Bill ---program data a; informat canceldate ddmmyy10.; format canceldate ddmmyy10.; input contractid householdid canceldate ; datalines4; 1 1 23.11.1999 2 1 30.04.2001 3 2 01.01.2000 4 2 31.05.2001 ;;;; run; proc sort; by householdid cnxdate; run; data b; set a; By householdid; Retain last_date; If not first.householdid then days_between = intck('days',last_date,canceldate); weeks_between= intck('weeks',last_date,canceldate); Last_date = canceldate; format last_date ddmmyy10.; put (_all_) (=/); run; ---log 61 data b; set a; 62 By householdid; 63 Retain last_date; 64 If not first.householdid then 65 days_between = intck('days',last_date,canceldate); 66 weeks_between= intck('weeks',last_date,canceldate); 67 Last_date = canceldate; format last_date ddmmyy10.; 68 put (_all_) (=/); 69 run; canceldate=23/11/1999 contractid=1 householdid=1 last_date=23/11/1999 days_between=. weeks_between=. canceldate=30/04/2001 contractid=2 householdid=1 last_date=30/04/2001 days_between=524 weeks_between=75 canceldate=01/01/2000 contractid=3 householdid=2 last_date=01/01/2000 days_between=. weeks_between=-70 canceldate=31/05/2001 contractid=4 householdid=2 last_date=31/05/2001 days_between=516 weeks_between=74 NOTE: Missing values were generated as a result of performing an operation on missing values. Each place is given by: (Number of times) at (Line):(Column). 1 at 66:20 NOTE: There were 4 observations read from the data set WORK.A. NOTE: The data set WORK.B has 4 observations and 6 variables. NOTE: DATA statement used (Total process time): real time 0.11 seconds cpu time 0.08 seconds */Toby.Dunn@TEA.STATE.TX.US (Dunn, Toby) wrote in message news:<748343794E05734F9D9E9BBF2CA6D214621B55@LYNDON.tea.state.tx.us>... > Stefan, > > AS I am not sure what type of time period you are looking for you will > more than likely need to read up on intck function and modify it so that > time_between gives you what you want. > > > > Data between; > SET ; > By householdid; > Retain last_date; > > If not first.householdid then > time_between = intck('week1.1',last_date,canceldate); > > Last_date = householdid; > run; > > HTH > Toby Dunn > > > > -----Original Message----- > From: SAS(r) Discussion [mailto:SAS-L@LISTSERV.UGA.EDU] On Behalf Of > Stefan Pohl > Sent: Friday, October 22, 2004 11:50 AM > To: SAS-L@LISTSERV.UGA.EDU > Subject: calculate the time between cancelation of 2 contracts on the > household-level > > > Hi, > > i suppose that my problem can be solved by group by-processing... > > my data look like this: > > contractid householdid canceldate > 1 1 23.11.1999 > 2 1 30.04.2001 > 3 2 01.01.2000 > 4 2 31.05.2001 > > i want to caculate for each household the time between two cancelations, > i.e. for household 1 30.04.2001 - 23.11.1999 and so on. > > is there an easy way ? Group by processing? > > thank you, stefan. ```
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# Understanding Fibonacci Studies for Technical Analysis
## The Fibonacci studies were developed by Leonardo Fibonacci, an Italian mathematician born in the late twelfth century, when he was studying the Great Pyramid of Giza in Egypt.
Fibonacci numbers are a number series in which each successive number is the sum of the previous two numbers. For example: 1, 1, 2, 3, 5, 8, 13, 21, 34, etc. Any given number in the series is approximately 1.618 times the preceding number and any given number is approximately 0.618 times the following number.
There are four popular Fibonacci studies: arcs, fans, retracements, and time zones. All of these studies use the Fibonacci numbers to forecast support and resistance levels.
## Arcs
Fibonacci Arcs are displayed by first drawing a trend line between two points, usually one significant high and one significant low. Three arcs will then appear, with the second point being the center point, so that the arcs will intersect the trend line at the Fibonacci levels of 38.2%, 50.0%, and 61.8%. These arcs will act as support and resistance levels for future price movements.
## Fans
Fibonacci Fans are displayed by first drawing a trend line between two points, usually one significant high and one significant low. An invisible vertical line will then be drawn through the second point. Three trend lines will then be drawn from the first point so that they intersect the invisible vertical line at the Fibonacci levels of 38.2%, 50.0%, and 61.8%. These three trend lines will then fan out and the lines will then act as support and resistance for future price movements.
## Retracements
Fibonacci Retracements are displayed by first drawing a trend line between two points, usually one significant high and one significant low. For ProSticks, a series of 7 horizontal lines are drawn intersecting the trend line at the Fibonacci levels of 0.0%, 38.2%, 50.0%, 61.8%, 100.0%, 161.8%, and 200.0%. After a rally or sell-off, prices will tend to retrace, or give back, a portion of the original move. As the prices move, support and resistance levels, more often than not, will appear at or near the Fibonacci levels.
Associated with Fibonacci Retracements is Fibonacci Projection. The Projection study simply takes the retracement levels found in a Fibonacci Retracement calculation and projects it to any other point in time.
## Time Zones
Fibonacci Time Zones are a set of vertical lines spaced at the Fibonacci intervals of 1, 2, 3, 5, 8, 13, 21, 34, etc. As the price traverses through the timeline, tops and bottoms, or significant support or resistance, tend to appear at or near the Fibonacci intervals.
Associated with Fibonacci Time Zones is Fibonacci Ruler. Instead of using the Fibonacci numbers as intervals, the intervals are calculated with the Fibonacci levels of 38.2%, 50.0%, 61.8%, 100.0%, 161.8%, and 200.0%.
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What is when in a minute, twice in a moment, and never in a thousand years Riddle - What is once in a minute, twice in a moment, and never in a thousand years Riddle is the simplest of every the riddles. Let's review all the discussions related to What is when in a minute, double in a moment, and never in a thousand years Riddle is the Riddle which is trending in all the society sites choose Facebook, Whatsapp, Instagram etc. And people space enjoying this trends. What is once in a minute, double in a moment, and never in a thousand years Riddle is described in a far better way possible in the article detailed below.
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1. ## Problem with formula
I have a spreadsheet to calculate the cost of paying someone per day, based on the start and end times.
I have custom time fields in columns B,C and D with format h:mm Column B contains the start time in 24 hour format, column C contains the finish time and column D contains the difference, with the formula =IF(C2="","",C2-B2) for row 2 and the equivalent for subsequent rows.
Column E contains the amount earned with the formula =IF(C2="","",(HOUR(D2)+MINUTE(D2-TIME(HOUR(D2),0,0))/60)*22) which takes the whole hours +the minutes/60 (i.e., fraction of an hour) and multiplies the result by 22 to give the amount earned.
This works correctly in most cases, but in certain cases where the time worked is an exact number of hours column E (the earned amount) gets a #NUM! error. Changing both of the times by only 2 minute results in the correct result.
Examples 9:00 16:00 gives the correct value in D (7:00) but column E gets the error
9:01 16:01 gives the correct value in D (7:00) but column E gets the error
9:02 16:02 gives the correct values in D and E
but 16:00 and 21:00 gives the correct values in D and E.
I have tried using the standard time format for the three columns and also the custom hh:mm, with no effect.
Has anyone any idea of what is wrong?
2. I always let Excel do the time calculation, which means entering the time in the format hh:mm. Then your formula would be:
=IF(C2="","",(D2*24*22))
Excel stores the date/time data as a number, with 1 being 24 hours, so you multiply by 24, then by your rate.
cheers, Paul
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GMAT Diagnostic Test Question 25 : Retired Discussions [Locked] - Page 2
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# GMAT Diagnostic Test Question 25
Author Message
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Re: GMAT Diagnostic Test Question 26 [#permalink]
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04 Nov 2010, 06:56
dzyubam wrote:
Explanation:
Rating:
First, we have to calculate the amount of time the train spent for the stops. $$3*1=3$$ hours for the first trip and $$10*0.5=5$$ for the return trip. Now, we can write an equation with $$S$$ for one-way distance and $$V$$ for train's speed:
$$\frac{S}{V} + 3 = \frac{S}{2V} + 5$$
$$\frac{S}{V} - \frac{S}{2V} = 2$$
$$\frac{2S-S}{2V} = 2$$
$$\frac{S}{2V} = 2$$
$$\frac{S}{V} = 4$$
So, the roundtrip lasted for $$7+7+1=15$$ hours (we should count the 1 hour stop in the destination point as well).
How did you arrive at the value 7 from the ration S/V=4 ?
I did it till the last step but i got stuck at that step and forfeited this method .
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Argument : If you love long trips, you love the GMAT.
Conclusion : GMAT is long journey.
What does the author assume ?
Assumption : A long journey is a long trip.
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Re: GMAT Diagnostic Test Question 26 [#permalink]
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08 Nov 2010, 11:02
Fijisurf wrote:
I set a little bit different equation:
v-speed
T-time in either direction
v(T-3) - for the way there
2v(T-5) - on the way back
v(T-3)=2v(T-5)
vT-3v=2vT-10v
7v-vT=0
v(7-T)=0
T=7
The total time is (do not forget to add extra hour of waiting before return trip) 7+7+1=15
I am able to understand by your method.
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Argument : If you love long trips, you love the GMAT.
Conclusion : GMAT is long journey.
What does the author assume ?
Assumption : A long journey is a long trip.
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Re: GMAT Diagnostic Test Question 26 [#permalink]
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04 Jan 2011, 13:22
Hi,
I did it this way. the way I set up my eqns were based on the assumption that the two journeys have equal distance and are completed in equal time T.
now for Journey 1: We travel distance d, at speed s, and time=d/s
for Journey 2: We travel distance d, at speed 2s and time =d/2s
Now let us account for the times the train was stopped
We know that on Journey 1 it spent 3 hrs stopped and journey 2 it spent 10*0.5=5hrs stopped. Now we know that both journeys were of equal overall duration.
d/s+3=d/2s + 5
solving these eqns yields d=4s
This way we can now compute the time for journey 1 was t=d/s = 4s/s=4 hrs
the second journey was d/s=4s/2s=2 hrs
total=> 4+3+1+2+5=15
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Re: GMAT Diagnostic Test Question 26 [#permalink]
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09 Feb 2011, 16:33
can someone explain the flaw in my logic?
let distance be d; let the overall time be t
t-3 is the time it takes for the train to reach point B from A
t-5 is the time it takes for the train to return
so d/(t-3) = 2d/(t-5)
solving I get t = 1
Obviously this is not correct. Any ideas?
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Re: GMAT Diagnostic Test Question 26 [#permalink]
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13 Oct 2011, 09:19
Hey guys, had a bit of trouble with this one. Maybe I missed something but I think its because the definition of when one trip begins and the other one ends is a bit ambiguous here (that or I'm just dumb
I created an RTD chart for this. Since we havent been given any conrete info on distance, I assumed the distances are the same and set d=300 (picked off top of my head). I represented the first trip as Rxt= D and also put an entry for the three hour long stops. I then assumed this is the end of trip 1.
For trip 2, I entered in the break periods (5 hours during the trip and the one hour before the train sets off), and the equation for the train's movement. Rate is 2R since we are told the rate is two times faster, and initially I represented the time as "M". Since we know the times are equal, I set T+3 (total time from the first trip plus breaks) equal to M+6 (total time from second trip plus break, which becomes M= T-3. Now we have two equations; 600-6R=300 and RT=300, which when solved equals R=50, therefore, because RT=300, T=6. This means the length of each total trip including breaks is 9 hours, and the chart appears to add up. This would make the answer choice E.
Any and all help greatly appreciate,
-Ken
R x T = D
Trip 1
R T 300
0 3 0
Trip 2
0 1 0
0 5 0
2R (T-3) 300
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Re: GMAT Diagnostic Test Question 26 [#permalink]
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01 Jan 2012, 00:11
There are a lot of great explanations out there, but I found someone posted one earlier that seemed to be simpler, if worded a little differently. To me this approach shows a simplicity of thought, and there is the added *minor* incentive of solving the problem in about 30 seconds -
first journey travel time = t, waiting time = 3
second journey travel time = t/2 (since double the speed), waiting time = 5
since the second journey saved 2 hours
t-t/2 = 2 hours
so t=4
So first journey = 4+3 = 7 hours = second journey
total time = 7+7+1
Probably simpler if we can think this straight in the actual exam, eh?
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Re: GMAT Diagnostic Test Question 26 [#permalink]
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14 Aug 2013, 15:10
Ikowill wrote:
Another way of solving this problem:
We know that each trip took the same amount of time and has the same distance.Then we can equal either distance or time. I chose distance, dzyubam chose time.
t = total time. This is important because t is the total time for each way of the round trip including stops
Total time of stop first part of the trip = 1 hour * 3 = 3 hours
Total time of stop second part of the trip = 0.5 hour *10 = 5 hours
d = s (t -3) First part of the trip
d= 2s (t-5) Second part of the trip
Both distance are the same
s (t – 3) = 2s (t – 5)
(t – 3) = 2 (t – 5)
t – 3 = 2t – 10
t = 7
(7 * 2) both ways + 1 hour stop when reached destination
14 + 1 = 15 hours Total Trip
How do you know to do (t-3) and (t-5)? Someone explain that part in plain English?
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Re: GMAT Diagnostic Test Question 26 [#permalink]
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15 Aug 2013, 02:07
Ikowill wrote:
Another way of solving this problem:
We know that each trip took the same amount of time and has the same distance.Then we can equal either distance or time. I chose distance, dzyubam chose time.
t = total time. This is important because t is the total time for each way of the round trip including stops
Total time of stop first part of the trip = 1 hour * 3 = 3 hours
Total time of stop second part of the trip = 0.5 hour *10 = 5 hours
d = s (t -3) First part of the trip
d= 2s (t-5) Second part of the trip
Both distance are the same
s (t – 3) = 2s (t – 5)
(t – 3) = 2 (t – 5)
t – 3 = 2t – 10
t = 7
(7 * 2) both ways + 1 hour stop when reached destination
14 + 1 = 15 hours Total Trip
How do you know to do (t-3) and (t-5)? Someone explain that part in plain English?
Can you please elaborate a bit? Thank you.
Meanwhile check this: gmat-diagnostic-test-question-79355-20.html#p1071311 Hope it helps.
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Re: GMAT Diagnostic Test Question 26 [#permalink]
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05 Jan 2014, 22:27
Ikowill wrote:
Another way of solving this problem:
We know that each trip took the same amount of time and has the same distance.Then we can equal either distance or time. I chose distance, dzyubam chose time.
t = total time. This is important because t is the total time for each way of the round trip including stops
Total time of stop first part of the trip = 1 hour * 3 = 3 hours
Total time of stop second part of the trip = 0.5 hour *10 = 5 hours
d = s (t -3) First part of the trip
d= 2s (t-5) Second part of the trip
Both distance are the same
s (t – 3) = 2s (t – 5)
(t – 3) = 2 (t – 5)
t – 3 = 2t – 10
t = 7
(7 * 2) both ways + 1 hour stop when reached destination
14 + 1 = 15 hours Total Trip
.
WHY YOU SAY "This is important because t is the total time for each way of the round trip including stops"...
Re: GMAT Diagnostic Test Question 26 [#permalink] 05 Jan 2014, 22:27
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# GMAT Diagnostic Test Question 25
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Collisions after effects!
4 identical spheres are seen above. Spheres C and D fall freely from a height $$h$$ before colliding with spheres A and B on the left and right sides as shown.If the acute angle between the line joining the centres of A and C and the vertical is $$\theta$$ and that for B and D with vertical is $$\alpha$$ .
If the time at which the velocity vectors of A and B are mutually perpendicular to each other is given by
$$\frac{\sqrt{2gh}( Pcosec2\alpha + \sqrt{Hcosec^{2}2\alpha-YsinS\theta sinI\alpha} )}{hsinC\theta sin S\alpha}$$
find $$P+H+Y+S+I+C+S$$
Details and Assumptions
• All collisions(including with vertical wall) are ELASTIC
• ALL spheres are identical
• path KL has a length $$h$$
• no rotation takes place
• entire motion is free of non conservative dissipative forces
• in case of finding two values of $$t$$, consider the larger value
this problem is part of the set "innovative problems in mechanics"
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Associated Topics || Dr. Math Home || Search Dr. Math
### Area of Inscribed Circle
```
Date: 12/01/98 at 13:24:52
From: Anna
Subject: Trigonometry
The question is: Find the area of the circle inscribed in a triangle
a, b, c. The hint is: Draw lines to the center of the circle and use
Heron's law.
I would appreciate it if you could help me!
Anna
```
```
Date: 12/01/98 at 15:41:37
From: Doctor Floor
Subject: Re: Trigonometry
Hi Anna,
Thanks for your question!
Let us consider triangle ABC with sidelengths BC = a, AC = b, AB = c
and center of the inscribed circle (incenter) I. Drop perpendicular
altitudes from I to Ia, Ib and Ic on the sides of ABC. The lengths of
these altitudes are equal to the radius r of the incircle. This radius
we will have to calculate in order to find the area of the inscribed
circle (= incircle).
C
/ \
/ \
/ \
Ib Ia
/ \
/ I \
/ \
A-------Ic------B
First, we know that area(ABC) = area(AIB) + area(BIC) + area(CIA).
We can calculate area(AIB) = 0.5 * AB * IIc = 0.5cr. And in the same
way we find area(BIC) = 0.5ar and area(CIA) = 0.5br.
The conclusion is that area(ABC) = 0.5(a+b+c)r = sr, where s is the
semiperimeter of ABC.
Then, from Heron's formula, we also know that:
area(ABC) = sqrt(s(s-a)(s-b)(s-c))
Combining the two formulas for the area of ABC we can derive:
sr = sqrt(s(s-a)(s-b)(s-c))
r = sqrt(s(s-a)(s-b)(s-c)) / s
r = sqrt((s-a)(s-b)(s-c) / s )
And now we can finish off the area of the incircle:
area(incircle) = pi * r^2
= pi * sqrt((s-a)(s-b)(s-c) / s )^2
pi(s-a)(s-b)(s-c)
= -----------------
s
Quite nice!
Best regards,
- Doctor Floor, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Conic Sections/Circles
High School Geometry
High School Triangles and Other Polygons
Middle School Conic Sections/Circles
Middle School Geometry
Middle School Triangles and Other Polygons
Search the Dr. Math Library:
Find items containing (put spaces between keywords): Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words
Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search
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# Solve the Room - Make 10 or 20: A Math Center Task Card Set
Subject
Resource Type
Product Rating
File Type
PDF (Acrobat) Document File
9 MB|69 pages
Share
Product Description
In my classroom, math centers are an important part of our math block! I developed these Solve the Room Activities to help streamline my math center routine!
Each week, one center that my students visit is Solve the Room. They grab a recording sheet and clipboard and walk around the classroom searching for the task cards. They answer the problem on the card and write their answer on the recording sheet.
We spend a lot of time in first grade gaining fluency in making 10! This is an important benchmark and will help students when they begin to add larger numbers. I have also included a set of Make 20 cards for students that are ready!
In this Solve the Room pack, I have included 3 versions so you can pick the one that best fits your needs.
Color
B&W
Anytime – non-seasonal cards that can be used at anytime of the year!
There are also blank task cards in each set to write in your own problems.
There are two levels of difficulty included for this task card set. This allows you to differentiate and meet your students needs!
Recording Sheets:
There are several recording sheet templates in this pack. In my classroom I use the general template. This works for every Solve the Room pack I have created! No need to find and print a new one each week. Just keep a master and copy and you’re good to go! There are also templates included that are specific to this Solve the Room Pack!
All task card sets are printer friendly!
Tips:
These cards can be used in many ways! You could play a whole class scoot game to review a concept! You could also use them for extra practice at your small group table!
This Solve the Room set is included in my September Bundle! Find it here!
September Solve the Room Bundle
You can purchase this as part of a Year-Long Growing Bundle Here
If you like this pack you might be interested in my other Solve the Room First Grade Math Packs!
Find the Sum
Number Bonds
Subitize
Number Lines to 50
Tally Marks
Ten Frames
Total Pages
69 pages
N/A
Teaching Duration
N/A
Report this Resource
Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials.
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FAQ
(Difference between revisions)
A FAQ for things you might like to know
Networking Questions
General
What type of networking is used on campus?
The campus network is an Ethernet packet-based network.
What is Ethernet?
Ethernet is a family of packet-based [[wikipedia:computer network]|computer networking]ing technologies for local area and wide area networks (LANs and WANs). Most laptops, desktop computers, server computers, cable modems and DSL modems have a built-in support for Ethernet networks. For more information and history, read the Wikipedia entry on Ethernet.
(Credits Wikipedia:Ethernet April 08, 2011)
What is the recommended configuration for a researcher's network connection?
It depends on the work that you do. If your work frequently involves moving data sets to and from your computer for visualization, analysis, or collaboration, you should seriously consider a 100Mbs full-duplex network connection as your baseline.
What the difference between Mbs and MBs?
"Mbs" stands for "megabits per second". "MBs" stands for "megabytes per second". A lower-case "b" designates bits (1's and 0's) and an upper-case "B" designates bytes. 1 byte equals 8 bits.
Bits are used to measure network data transfer rates in seconds and bytes are used to measure data storage sizes. When stored data is moved across a network, however, it is convenient to consider transfer times measured in the number of bytes of stored data moved in one second.
What do 10Mbs, 100Mbs, and 1Gbs mean?
Network speeds are listed by the number of bits (1's and 0's) they can transfer in one second. Modern networks transfer millions of bits per second, designated "Mbs" and read "mega-bits per second". Common network speeds are 10Mbs, 100Mbs, and 1000Mbs. 1000 megabits are equal to 1 gigabit, and 1000Mbs is typically written "1Gbs" and read "one gigabit per second" (1 billion bits per second).
How fast are 10Mbs, 100Mbs, and 1Gbs networks?
To get a sense for the performance of different network speeds, it's easiest to use the following rules of thumb for comparing network speeds to data set sizes and their transfer time:
• 10Mbs can transfer 1MBs
• 100Mbs can transfer 10MBs
• 1000Mbs (1Gbs) can transfer 100MBs
A CDROM can hold 700MB of data. Transferring this much data would take about 7 seconds on a 1Gbs network, 70 seconds (more than 1 minute) to transfer on a 100Mbs network, and 700 seconds (more than 10 minutes) to transfer on a 10Mbs network.
What's the justification for this transfer rate rule of thumb?
The logic for this metric is that a 10Mbs (10 mega-bit per second) network connection will move 10 million bits per second. Data is measured in 8-bit bytes and the rule of thumb for Ethernet is that performance peaks at 80% capacity. This provides the easy conversion factor of 10Mbs=1MBs. Note that the lower-case "b" means "bits" and upper-case "B" means bytes, ie. 8 bits. The network speeds scale up easily by factors of 10. So 100 megabit per second connection is capable of transferring 10 megabytes per second, and a 1000 megabit per second is capable of transferring 100 megabytes per second.
Theoretically, a 100Mbs connection will transfer 100 million bits in one second, or about 10 megabytes (MB) per second. This means you would be able to transfer a CD's worth of data (about 700MB) in about 70 seconds, about 1 minute. (Compare this to a 10x slower connection of 10Mbs and it would take 700 seconds
Network Structure
How much network bandwidth is available is available on campus?
Individual network connections at 10Mbs, 100Mbs, or 1Gbs speeds can be delivered to any location on the campus network at standard rates. Additionally, wireless network connectivity is available across campus.
What does the campus network look like?
The campus network can be visualized as a collection of network trees, roughly one per building, with the root of each tree connecting to an expandable high bandwidth core network backplane (currently running at 10Gbs).
The depth of each individual tree is determined by the physical layout of and number of network ports in each building. Each tree is typically no more than three layers deep, including the leaf nodes. The leaf nodes are the end-user connections, i.e. wired wall ports or wifi connections. The internal nodes of each tree are network switches and the switches are connected to the next layer via fast connections (currently running at 1Gbs).
Each tree (each building) connects to the core network backplane via a fast connection (currently running at 1Gbs). At this core network connection, the data packets are routed to their final destination on- or off-campus.
How is the campus network connected to off-campus networks?
The campus core network backplane is connected to off-campus networks like the commercial Internet (Google, Facebook, Amazon) and national high bandwidth research networks (Internet2 and NLR) which provide high speed connections to research institutions and labs across the country. The fastest network route to a specific off-campus destination is chosen automatically as the network packets move off-campus.
Custom configurations to meet unique research needs or specific performance targets can be designed. This requires advanced planning and an understanding of the proposed research workloads and workflow. Please contact Research Computing. The cost for these customizations can often be included in research proposals.
Ordering Information
How do I order or upgrade a network connection?
Computer data connections are ordered from UAB IT Telecommunications Services via their service request form.
To place an order you will need to provide a general ledger account number for billing and identify the location (building address) of the service request. The wall-jack identification number for the network connection will be needed to complete the service request and can be entered on the form.
Who pays for my network connection?
You do.
Network connections are accounted for via a federally regulated service center run by UAB IT. The rates are set based on the cost to deliver the service. Money to pay for network connectivity can come from any legitimate source: directly through grants, indirect grant funds routed to departments, or other departmental or research support funds.
How much do network connections cost?
Standard service center rates apply to all network connections (10Mbs, 100Mbs, and 1Gbs). Discounted rates for upgrading existing connections to higher data rates are available. Additionally, network switches can be ordered at a fixed lease rate to supply many network connections to an area.
Network Performance
How do I measure my campus network connection speed?
The UAB IT SpeedTest server speedtest.dpo.uab.edu will run a data transfer test from your computer to the SpeedTest server and rate the performance of a data connection.
What factors impact the actual speeds I can expect in the real world?
The actual transfer rates you get depend on three factors: software, hardware, and other users.
Data transfer software and computer hardware can significantly impact real world transfer rates. If you are transferring lots of data, you will see your best performance with software that can keep the network full, computer hardware that is not slower than the data network, and a network connection sized for your data sets and patience.
How does my copying software impact my transfer speeds?
The software you use to transfer data is the most import factor in maximizing data throughput. Most traditional copy methods move data in a single-file line. Modern computer hardware hides this software inefficiency and can easily keep a 10Mbs connection full and can do ok with a 100Mbs connection. If you are moving lots of data or using a 1Gbs network, you need to use software tuned for high-speed data transfer.
High speed data transfer software uses multiple single-file lines in parallel to improve network throughput. This software must be used at both ends of the data transfer in order coordinate the parallel transfer streams. You won't get very far if you are smart but your peer is not.
What high-speed data transfer software can keep up?
It's important to use improved data transfer software that can move data . (Post examples)
How does my computer hardware impact my transfer speeds?
Computer hardware also impacts transfer speeds. Your slowest piece of hardware will dictate your maximum data transfer rate. If you have a slow disk (you should read that as "an external USB hard drive"), you will be limited by its data transfer speeds.
Additionally, your computer may be fast but it still has to manage your workload and coordinate use of all the devices in your computer, including the network connection. If you are crunching numbers or doing heavy visualizations at the same time you are trying to transfer data, your computer may not be able to keep up. Note, that this scenario is common when you are reading data for your visualization off a file server. Sometimes you need to move your data before you can use it.
UABgrid
UABgrid is an infrastructure pilot of UAB IT Research Computing. More information can be found in the UABgrid FAQ though this information may be dated.
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www.vustudents.ning.com
We non-commercial site working hard since 2009 to facilitate learning Read More. We can't keep up without your support. Donate.
# FIN622 - Corporate Finance Assignment no 1 closing date May 21, 2014.
FIN622 - Corporate Finance
PROBLEM:
ABC Company is considering the acquisition of a machine to improve its production. The company has to make a choice between two types of machine A and B. Each machine will have a 4-year life with no salvage value.
Cost of capital of the firm is 14%. Initial investments required to purchase and install the machines A and B, are PKR 28,700 and PKR 27,050 respectively. Machine A will generate an inflow of PKR 10,000 each year while Machine B is expected to generate cash inflows in the following manner.
Year Cash Inflows (PKR)
1 11,000
2 10,000
3 9,000
4 8,000
Required:
Calculate NPV and IRR for both options. Which machine should be preferred and why?
Note:
For calculating IRR you are required to use “Trial and Error Method” along with “Interpolation Technique/Formula”. In this particular regard, you are advised to consult PPT slideshow “Finding IRR is no more difficult” uploaded in the lesson contents of Lesson # 10.
Views: 5424
Attachments:
### Replies to This Discussion
thanx azmay
sana
jis year ka df nikaalna hota hai us ki power b utni hi leni parti hai :)
now mara B
ka
IRR=19.945
KAYA APP KA BHI YHE AYA HA
Ho Gai Sb Ki Assignment ? Any Confussion ?
listen azmay in machine a there is equal payments are being made in each an every year so this is the case of an annuity so why do we need to discount them back by each and every year rather we can use the direct formula of pv of annuity like PV= c x {(1-1/(1+r)^r}/r
although we will have exactly the same answer as sum of all npv that you calculated for each and every year. ok
awaiz bhai koi step kr k bata dey mujy aise samje ni aa rahy hai
machine b ka
IRR = 17.99
KOI BATA DEY YH THEK HA ANSWER
sub log irr k lyi formula use kar rhay han it should be done by hit or trial method
babur in question we have to calculate IRR with error and trial along with interpolation
sana me ny machine A ka hr step show kr dia hai us se dekh k nikal lo
machine b ka
IRR = 17.99
awais bhai yh bata dey yh ans thek hai machine b ka
First we calculate NPV at 14 % both on Machine A and B. Both are showing positive NPV ( A=29137.12 which means positive 437.12) (B=25155.18 WHICH MEANS positive 1105.18)..Then Calculate IRR ...On A machine at 15% it become negative NPV and on B machine at 17% it become negative ) then putting values in the formula Off IRR
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# Revision history [back]
For finite products you can use the function prod(), like this:
sage: prod(range(1,42))
33452526613163807108170062053440751665152000000000L
What do you mean by "compute K number of times"?
There certainly is not a generic function to compute infinite product, but there might be functions to do what you have in mind. Have a look at the documentation of eta and EtaProduct, you can use the ? syntax:
sage: eta?
...
sage: EtaProduct?
unfortunately, eta seems to accept only complex arguments.
You can use the function gp() to enter Pari code
sage: a = gp("a(n)=local(A); if(n<0, 0, A=x*O(x^n); polcoeff(eta(x^2+A)^5*eta(x^8+A)/(eta(x+A)^2*eta(x^4+A)), n))")
sage: [a(i) for i in range(10)]
[1, 2, 0, 0, 1, -2, 0, 0, -4, -2]
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Home
• Get access
• Cited by 1
• Print publication year: 2008
• Online publication date: June 2012
1 - Introduction
Summary
Issues of vagueness
Some people, like 6′ 7″ Gina Biggerly, are just plain tall. Other people, like 4′ 7″ Tina Littleton, are just as plainly not tall. But now consider Mary Middleford, who is 5′ 7″. Is she tall? Well, kind of, but not really – certainly not as clearly as Gina is tall. If Mary Middleford is kind of but not really tall, is the sentence Mary Middleford is tall true? No. Nor is the sentence false. The sentence Mary Middleford is tall is neither true nor false. This is a counterexample to the Principle of Bivalence, which states that every declarative sentence is either true, like the sentence Gina Biggerly is tall, or false, like the sentence Tina Littleton is tall (bivalence means having two values). The counterexample arises because the predicate tall is vague: in addition to the people to whom the predicate (clearly) applies or (clearly) fails to apply, there are people like Mary Middleford to whom the predicate neither clearly applies nor clearly fails to apply. Thus the predicate is true of some people, false of some other people, and neither true nor false of yet others. We call the latter people (or, perhaps more strictly, their heights) borderline or fringe cases of tallness.
Vague predicates contrast with precise ones, which admit of no borderline cases in their domain of application. The predicates that mathematicians typically use to classify numbers are precise.
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# [GRASSweb-list]markus: web/grass51/tutorial examples.html,1.14,1.15
grass at intevation.de grass at intevation.de
Mon Feb 10 08:18:28 EST 2003
Author: markus
Update of /grassrepository/web/grass51/tutorial
In directory doto:/tmp/cvs-serv18822
Modified Files:
examples.html
Log Message:
cosmetics here and there
Index: examples.html
===================================================================
RCS file: /grassrepository/web/grass51/tutorial/examples.html,v
retrieving revision 1.14
retrieving revision 1.15
diff -u -d -r1.14 -r1.15
--- examples.html 9 Feb 2003 20:07:42 -0000 1.14
+++ examples.html 10 Feb 2003 13:18:26 -0000 1.15
@@ -177,13 +177,14 @@
<DIV ALIGN=right><a href="#toc">[UP]</a></DIV>
</b></big></font></td></tr></table>
-[for this exercise we use the Spearfish data set from the GRASS web site]
+[for this exercise we use the Spearfish data set from the GRASS web site]<br>
+[The WATRTOWR.DXF can be found <a href=http://www.3dcafe.com/asp/architex.asp#DXF>here</a>]
<P>
Since GRASS 5.1 supports 3D vector data, we can easily import DXF and DWG
files and display them with NVIZ:
<div class="code"><pre>
-v.in.dwg in=WATERTOWER.DXF out=watertowerXY
+v.in.dwg in=WATRTOWR.DXF out=watertowerXY
</div></pre>
The topology should be build for 'faces' which are (filled) 3D vector
@@ -209,6 +210,7 @@
r.mapcalc "dem_flat=-1874"
nviz el=dem_flat vect=watertowerXY
</div></pre>
+<P>
<b>Transformation to UTM coordinates:</b>
<P>
@@ -216,7 +218,7 @@
a national grid, we can use 'v.transform' with a 4 point transformation.
We select UTM coordinates where to place the water tower and query the
elevation (d.what.rast elevation.dem). We decide that the xy-extension of
-the watertower be 20m and it's height 100m (defined later).<br>
+the water tower be 20m and it's height 100m (defined later).<br>
The required transformation points file for 'v.transform' may look like this
(L: left, R: right, U: upper, L: lower, N, S, W, E):
@@ -243,8 +245,8 @@
The 'zshift' parameter is used to shift the object vertically on top of the
elevation model (use the bottom 'B' value shown by 'v.info' and add the
local elevation). The '-t' flag automatically shifts the DXF/DWG object(s)
-to sea level which is useful if the vector objects were drawn with negative
-z values:
+to sea level which is useful if the vector objects were composited with
+partially negative z values:
<div class="code"><pre>
v.transform -t in=watertowerXY out=watertowerUTM points=wt.points zscale=0.04 zshift=1320
@@ -252,7 +254,7 @@
nviz el=elevation.dem vect=watertowerUTM
</div></pre>
-You will have to zoom heavily to find the watertower as it appears to
+You will have to zoom heavily to find the water tower as it appears to
be very small compared to the surrounding mountains.
@@ -660,11 +662,12 @@
END
</div></pre>
-And here you can see how it looks like: <a
-href=http://grass.itc.it/cgi-bin/mapserv?map=/var/www/map-script/mapserverjs.map&layer=dem>Online MapServer
+And here you can see how it looks like: <br>
+<a href=http://grass.itc.it/cgi-bin/mapserv?map=/var/www/map-script/mapserverjs.map&layer=dem>Demo Online MapServer
reading GRASS raster, SHAPE, GeoTIFF and PostGRASS/PostGIS data</a>
+<P>
Since we still have the geometry and attributes in different tables, we
can add a 'view' onto both tables (sort of virtual table merging both).
@@ -717,14 +720,15 @@
using d.path.
As the column 'travelcost' is not yet present in the table 'road' we can use
-StarOffice or OpenOffice to add it (see <a href=attrib_storage>here</a>).
+StarOffice or OpenOffice to add it (see <a href="attrib_storage.html#toc">here</a>).
The we can continue and insert values according to the road type (be sure to
quote strings when using for where statement):
<div class="code"><pre>
echo "select * from roads" | db.select
</div></pre>
-... shows the travelcost column filled with 0.0.
+... shows that the travelcost column is filled with 0.0. To assign
+new travel costs, we run:<br>
<div class="code"><pre>
echo "update roads set travelcost=5 where cat=1" | db.execute
@@ -735,9 +739,14 @@
echo "select * from roads" | db.select
</div></pre>
+Then we can run the 'd.path' with our assigned travel costs:
<div class="code"><pre>
</div></pre>
+
+To see the forward directions of the vector lines, use the 'dir' for the
+'display' parameter in 'd.vect'. It will add small arrows indicating the
+forward directions when plotting the vector map.
<hr>
<P>
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You are on page 1of 3
# Info.
Handling GCSE Maths Tutor
Histograms
www.gcsemathstutor.com
topic notes info@gcsemathstutor.com
Block graphs As a comparison you may wish to revise 'block graphs'(in 'representing data') before going on to study histograms.
Grouped Data (Grouped Frequencies) To understand frequency density and it's role in histograms, you need first to appreciate the meaning of a number of terms relating to grouped data: Example
height(cm) frequency
71-75 5
76-80 17
81-85 21
86-90 15
91-95 2
class boundaries are the exact values(for each set of grouped data) where one set of values ends and the other begins. In our example, the class boundaries are: 75.5 __80.5__ 85.5__ 90.5 You must appreciate that these numbers are the 'deciders' as to which group data is placed. For example 75.45 would be rounded down to 75 and be placed in the first group, but 75.55 would be rounded up an placed in the second.
class width is the width of each block of values(not frequencies). In our example(above) height 71-75 gives a class width of 5 and NOT 4 (75-71). There are 5 numbers in the group 71 72 73 74 75 .
grouping using inequalities The values are grouped according to an inequality rule.
mid-interval values are useful in estimating the 'mean' of a set of grouped data. This is dealt with in detail in the topic 'mean, mode and median' here.
GCSE Maths Tutor
www.gcsemathstutor.com
info@gcsemathstutor.com
Info. Handling GCSE Maths Tutor
Histograms
www.gcsemathstutor.com
topic notes info@gcsemathstutor.com
Frequency Density The frequency density is a the frequency of values divided by the class width of values.
Histograms The area of each block/bar represents the total of frequencies for a particular class width. The width of the block/bar(along the x-axis) relates to the size of the class width. So the width of a block/bar can vary within a histogram. The frequency density is always the y-axis of a histogram. Histograms are only used for numerical continuous data that is grouped.
Example Here is a table of data similar to the last one but with values of height grouped differently using inequalities. note: because the class is grouped using inequalities, one 'equal to and greater' and the other 'less than' , the class width is a straight subtraction of the two numbers making up the class group.
class (height - h) cm 65 75 80 90 105 h < 75 h <80 h <90 h <105 h <110
class width
frequency (f)
frequency density
10 5 10 15 5
2 7 21 15 12
2/10 = 0.2 7/5 = 1.4 21/10 = 2.1 15/15 = 1.0 12/5 = 2.4
GCSE Maths Tutor
www.gcsemathstutor.com
info@gcsemathstutor.com
Info. Handling GCSE Maths Tutor
Histograms
www.gcsemathstutor.com
topic notes info@gcsemathstutor.com
Significance of area The area on a histogram is important in being able to find the total number of values/individual results in the data. In our histogram(from the table), the 65 to 75 block represents 2 children, the 75 to 80 block represents 7 children, and so on. So one block square represents one child. If we count the square blocks in the whole sample we get 57 - the sum of all the frequencies i.e. the total number of children taking part - the number of individual results.
GCSE Maths Tutor
www.gcsemathstutor.com
info@gcsemathstutor.com
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# Where does entropy-increasing waste heat go during non-reversible processes?
The first law of thermodynamics states that energy is conserved. $$\Delta U = Q - W$$ (internal energy change is the difference between external energy supplied and work done on the environment).
The second law of thermodynamics states that entropy is never decreasing and is only constant in reversible thermodynamic processes. In a non-reversible thermodynamic process, the entropy of the universe is increased.
Suppose we have an abstract 1-dimensional cellular automata universe consisting of 0s (empty space) and 1s (energy/atoms) but obeying the first and second laws of thermodynamics. The 1s are drawn together via "gravity" forces:
t0: 0010000100
t1: 0001001000
t2: 0000110000
Unlike microscopic time-reversible phenomena, let us assume that the collision of the two 1s generates entropy.
Q1: Would the final state of this universe be cycling through all possible $$10 \choose 2$$ states, as that is the maximum entropy configuration?
Q2: If the answer to Q1 is yes, then that implies a set of dynamics that encourage an equilibrium distribution of two 1s over the 10 position. If we have any less or more than 2 1s, wouldn't that violate conservation of energy? (the total energy of the starting system is 2).
Q3: Where does entropy "go" when it is produced? In this toy universe I have no way to embody any other non-zero energy than "1". If we choose to introduce a new letter "H" as an "entropy" particle, does this violate the first law of thermodynamics? (since the new energy of the system is 1 + 1 + H). Does entropy occupy space?
t1: 0001001000
t2: 000011H000
Or is it a "hidden state" a cell that follows energy around?
The bigger question here is that when I think of living things perform some thermodynamically irreversible work to lower their entropy, I am wondering where that entropy "goes". If entropy is created but mass and energy are conserved, how do we end up with heat "for free" without changing total energy?
t1: 0001001000
h1: 0000000000
t2: 0000110000
h2: 00000H0000
Q4: Are random dynamics required here in order to eventually reach a state of maximum entropy? One way to model this would be to treat H as "non-useful work" that fills up the universe and potentially creates more H when it interacts with non-H cells.
Q5: Are there any limitations to this 1D universe in my understanding of how the first and second law of thermodynamics works?
• A1:no, the two particles attracted each other, and that is the final configuration. There are no microscopic states
– user65081
Commented Nov 1, 2020 at 2:17
• Or put another way, your model currently operates at zero temperature; the components aren't thermalized. I recommend consulting Dill and Bromberg's Molecular Driving Forces (simpler) or McQuarrie's Statistical Mechanics (more advanced) to understand how to apportion energy to individual sites and configurations to build up a realistic thermodynamic system. Commented Nov 1, 2020 at 23:14
## Entropy and life
Entropy is an observer's uncertainty about the state of a system. A measurement (macro state) gives you a belief distribution over possible configurations (micro states) the system could be in. The Shannon entropy of this belief measures the observer's uncertainty. A uniform distribution over consistent states simplifies the entropy to the log of the number of consistent states.
Entropy increases if the observer makes stochastic predictions of how the state evolves. For example, thermodynamics uses Langevin dynamics that have Brownian motion. The dynamics of the belief distribution are known as the Fokker-Planck equation. The entropy increases faster the faster the micro states move, that is, the higher the temperature of the system.
Systems that exist for some time need to remain within a distribution that lets us identify them. For example, there is a distribution of all particle arrangements that we call "dog" that any dog will remain it throughout its lifetime. Such systems appear as if they used work to bring their state toward higher probability under their class to resist the entropy increase caused by the Brownian motion. As a result of bounding their entropy, they increase the entropy of their surroundings.
Q1: Would the final state of this universe be cycling through all possible 10 over 2 states, as that is the maximum entropy configuration?
You didn't clearly define the dynamics of the system, you just said that there is a gravitational force and the system obeys the laws of thermodynamics.
If you describe the system as deterministic and have full knowledge about the initial condition, your belief over its state is a point mass distribution and remains one as time passes, so the entropy is always zero and the dynamics are reversible.
If there is stochasticity in addition to an attracting force between particles (for example particles are repelled in random directions on collision), then the system will converge to a stationary distribution where they wiggle around near each other, which is not a uniform distribution over all possible system states.
Q2: If the answer to Q1 is yes, then that implies a set of dynamics that encourage an equilibrium distribution of two 1s over the 10 position. If we have any less or more than 2 1s, wouldn't that violate conservation of energy? (the total energy of the starting system is 2).
One way to define the total energy that it is a quantity that is preserved in a system. Defining the total energy is a way to express symmetries in a system that the system will obey. This defines a constraint on the dynamics. The first law of thermodynamics is thus more of a definition than a law.
For your system, you defined the total energy to be the sum of "1" particles. This means the number of "1" particles cannot change from what it is during the initial condition. This tells us that your system can only reach the 10 over 2 different states that contain exactly 2 "1" particles rather than the 2^10 states.
Q3: Where does entropy "go" when it is produced? In this toy universe I have no way to embody any other non-zero energy than "1". If we choose to introduce a new letter "H" as an "entropy" particle, does this violate the first law of thermodynamics? (since the new energy of the system is 1 + 1 + H). Does entropy occupy space? Or is it a "hidden state" a cell that follows energy around?
Entropy is not a particle and it is also not conserved over time. One way to related entropy and energy is via the Gibbs free energy that subtracts energy minus entropy. It measures the amount of energy that we can direct into work, i.e. the amount of energy that is not lost to entropy.
The bigger question here is that when I think of living things perform some thermodynamically irreversible work to lower their entropy, I am wondering where that entropy "goes". If entropy is created but mass and energy are conserved, how do we end up with heat "for free" without changing total energy?
Systems that persist over some duration of time, including living things, have to persist stochasticity in the state dynamics to remain with some distribution of states in which we still identify them as the thing they are. For example, if the particle configuration of a dog would change out of the distribution of particle configurations that we consider to be dogs, we would not call it a dog anymore.
I think your question may be what is known as Schrödinger's "paradox":
Since life approaches and maintains a highly ordered state, some argue that this seems to violate the aforementioned second law, implying that there is a paradox. However, since the biosphere is not an isolated system, there is no paradox. The increase of order inside an organism is more than paid for by an increase in disorder outside this organism by the loss of heat into the environment. By this mechanism, the second law is obeyed, and life maintains a highly ordered state, which it sustains by causing a net increase in disorder in the Universe. In order to increase the complexity on Earth—as life does—free energy is needed and in this case is provided by the Sun.
Q4: Are random dynamics required here in order to eventually reach a state of maximum entropy? One way to model this would be to treat H as "non-useful work" that fills up the universe and potentially creates more H when it interacts with non-H cells.
The maximum entropy state depends on the system. If the system is modeled as deterministic and the initial condition is fully known, then the maximum entropy is zero and would be reached from the beginning.
If the dynamics are stochastic, the state belief will over time converge to the least certain distribution. At this point, entropy does not increase anymore. This is the distribution that you think the system state is in if you don't know an initial condition. For this to happen, the dynamics need to be weakly mixing.
Q5: Are there any limitations to this 1D universe in my understanding of how the first and second law of thermodynamics works?
Yes, if the dynamics of your system are modeled as deterministic and you have full knowledge of the initial condition, then you can deterministically predict its state into the future and the belief entropy remains zero. You need uncertainty in either the dynamics or the initial state. An example of uncertainty in the initial state would be that some state dimensions are unknown --- marginalizing them out gives you stochastic dynamics.
• Thanks Danijar, good to see you on this forum! I have follow up Qs, which I'll have to add in multi-comments because SE has a char limit. Q1: yeah I didn't define the system dynamics clearly because I was wondering what dynamics would have to be implemented to respect 1st and 2nd laws. It sounds like to me that in order for entropy increase there has to be dynamics that increase randomness over time. Does the 2nd law in Our Universe's physical laws suggest that the final resting state is not a uniform heat spread across the universe, but potentially some equilibrium between gravity and heat? Commented Nov 1, 2020 at 16:09
• Q3: IIUC, you are saying that there is a total "energy" quantity (in this case, the sum of all 1s) that is conserved that is the sum of Gibbs free energy and energy lost to entropy. In toy 1D universe model, how might this be implemented? If I understand correctly, you can think of this as a "hidden state" for every energy cell, where an entropy bit is flipped on h2: 00000H0000 and this is irreversible. The total energy is the sum of t2: 0000110000 which remains 2, but the free energy would be 2 - 1 = 1. Commented Nov 1, 2020 at 16:18
• Q4: Thanks, I think your answer to Q1 resolves this. In our universe, mass can be converted entirely into "waste heat" (Gibbs Free Energy = 0) and the first law of thermodynamics is still respected? Commented Nov 1, 2020 at 16:25
• Hey Eric! A1: Yes, the equilibrium depends on all forces involved. It might still turn out to be a uniform distribution for our laws of physics; I don't know. A3: I don't know if you can represent entropy spatially like this. Entropy increases when the dynamics are stochastic and you don't observe the result of the transition. Your belief over possible states is then less certain than it was. A4: Exactly, the total energy of a system doesn't change when free energy is converted into entropy, that is, all your knowledge about the system state is lost. Commented Nov 1, 2020 at 20:40
• Cool, thanks again. Yeah I agree that in our universe entropy isn't necessarily spatial, but in my toy 1D case I think entropy can be treated as an unobservable state associated with each "visible" state for simplicity. We could also not order them spatially and just have "entropy reservoir" that accompanies the universe? Commented Nov 2, 2020 at 4:14
This is an answer to the title of the question
Where does entropy-increasing waste heat go during non-reversible processes?
The "waste heat " goes to black body radiation. This radiation is only modeled correctly with quantum mechanics.
A lot of the energy will go into the kinetic energy of the particles in the medium studied, raising the temperature.
The simple model you have in the content seems to me to be reversible.
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# Relativity for the Million
Topics: Special relativity, Light, Speed of light Pages: 3 (1220 words) Published: April 21, 2013
Relativity for the Million
For this project, I decided to read Relativity for the Million, a book by Martin Gardner. I have always been fascinated by the Theory of Relativity ever since you told me that time and distance is relative. For whatever reason, I could not make sense of this. It just boggled my mind that something as “constant” as time could change. Some of my favorite days in physics class consisted of discussions of relativity. It was always two steps forward and one step back for me. It seemed as if every time I thought I understood it, a curveball would be thrown in or I would learn something new that challenged it. One of my favorite things about this book is that it doesn’t take a physicist to read it, only an eager mind.
The book starts out by taking the reader through a lot of thought experiments (similar if not the same to the ones we have discussed in class) that get the brain thinking in a direction geared towards relativity. It discusses many basic ideas such as absolute or relative, and frame of reference. From page one, this book really makes you think (similar to an everyday physics experience with Mr. Parker).
One of the more interesting parts of this book is the theories prior the Relativity. One of the most widely accepted theories was the presence of “ether” and an “ether wind”. Ether is a substance believed by nineteenth-century physics to permeate all space, serving as the medium for the propagation of electromagnetic radiation. Basically, ether is everywhere. It is the odorless, invisible, massless substance that filled everything. Even in deep dark space where there is no air, no gravity, nothing… there was ether. Ether is a fixed, stationary substance that acts as a medium for light to travel. If light travels through Ether with a certain speed, c, and if this velocity is independent of the velocity of its source, then the speed of light can be used as a kind of yardstick for measuring the observer’s absolute motion....
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1. May 7, 2010
AxiomOfChoice
If $a$ and $b$ are positive and $a < b$, do we have
$$(0 < x < 1) \Rightarrow \frac{1}{x^b} > \frac{1}{x^a}$$
and
$$(1 < x < \infty) \Rightarrow \frac{1}{x^a} > \frac{1}{x^b}$$
?????
Last edited: May 7, 2010
2. May 7, 2010
Landau
I think you should be able to prove these yourself. Here's the first one:
Fact: Let $0<x<1$. Then for all $\alpha>0$ we have $0<x^\alpha<1$ and (hence) $\frac{1}{x^\alpha}>1$.
In particular:
* $0<x^a<1$ and $0<x^b<1$;
* $$\frac{x^a}{x^b}=\frac{1}{x^{b-a}}>1,$$ hence $x^a>x^b$.
Together: $0<x^b<x^a<1$. Conclusion:
$$0<\frac{1}{x^a}<\frac{1}{x^b}<1.$$
3. May 7, 2010
AxiomOfChoice
Thanks for your help, Landau. This is one of those questions that I answered for myself when I was typing it up, but I thought I'd go ahead and post it anyway to make sure I wasn't crazy.
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# factor analysis
Also found in: Thesaurus, Medical, Legal, Financial, Acronyms, Encyclopedia, Wikipedia.
## factor analysis
n
(Statistics) statistics any of several techniques for deriving from a number of given variables a smaller number of different, more useful, variables
Collins English Dictionary – Complete and Unabridged, 12th Edition 2014 © HarperCollins Publishers 1991, 1994, 1998, 2000, 2003, 2006, 2007, 2009, 2011, 2014
ThesaurusAntonymsRelated WordsSynonymsLegend:
Noun 1 factor analysis - any of several methods for reducing correlational data to a smaller number of dimensions or factors; beginning with a correlation matrix a small number of components or factors are extracted that are regarded as the basic variables that account for the interrelations observed in the datastatistics - a branch of applied mathematics concerned with the collection and interpretation of quantitative data and the use of probability theory to estimate population parameterscorrelational analysis - the use of statistical correlation to evaluate the strength of the relations between variables
Based on WordNet 3.0, Farlex clipart collection. © 2003-2012 Princeton University, Farlex Inc.
References in periodicals archive ?
By using exploratory factor analysis, the factors affecting the outsourcing of hospital services were identified and extracted.
Firstly, Kaiser-Meyer-Olkin (KMO) is used for sample adequacy and homogeneity of predictors in explanatory factor analysis. Secondly, Bartlett's test of sphericity is employed to test whether correlation matrix is an identity matrix.
The technique of factor analysis was utilised and it structured 3 original factors such as somato, vegetative and urogenital factors.
Findings related to the study of scale development were examined under the headings of sociodemographic characteristics, KMO for validity and exploratory and confirmatory factor analysis, test-retest for reliability, correlation-based item analysis, and internal consistency analysis.
When the correlations were verified with Bartlett's test of Sphericity for the biometric traits, the results were significant (P<0.01, Chi-square value 182.865) which was supportive for the rationality of the factor analysis of the data as indicated in Table-4.
From the perspective of a model itself, although the generalized dynamic factor analysis model put forward by Forni relatively traditional factor analysis model shows many advantages, it can be a very good factor analysis that was carried out on the panel data.
To provide an alternative to the impeller health indicator, an enhanced factor analysis based impeller indicator (EFABII) is proposed in this study.
Keywords: Quality Assurance (Q.A), Higher Education Commission (H.E.C), Independent t-test, Exploratory Factor Analysis (EFA), Factors, Reliability, Globalization (Glob), Ranking.
Convergence criteria are defined herein on the basis of purposefully applied factor analysis.
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# Cosine rules
The cosine rule is frequently used in trigonometry. It is commonly known as the law of cosines or simply, cosine formula. The cosine rule states that the sum of the squares of the length of other sides and twice the product of other sides with their cosine angles, rather than hypotenuse, is equal to the third side of the triangle.
Let us take a triangle ABC with angles x, y, and z, respectively. According to the cosine rule, we get the following formulae-
• a² = b² + c² – 2bc cos ∠x
• b² = a² + c² – 2ac cos ∠y
• c² = a² + b² – 2ab cos ∠z
These formulae can be written as-
• cos x = (b² + c² -a²)/2bc
• cos y = (a² + c² -b²)/2ac
• cos z = (a² + b² – c²)/2ab
Cosine rule is helpful to get the length of the side of the triangle by knowing their angles. We can directly find the angles and sides of a triangle from this rule.
## Proof of cosine rule
Let us consider a triangle with sides a, b and c respective to their corners A, B and C. We need to draw a perpendicular from B on the side AC at D, as shown in the figure above.
According to the trigonometry ratio and from triangle BCD, we get,
cos C = CD/a [cos θ = Base/Hypotenuse]
This can be re-written as-
CD = a cos C ………… (1)
Subtracting equation 1 from side b on both the sides, we get;
b – CD = b – a cos C
or it can be written as,
DA = b – a cos C
Again, according to the trigonometry ratio and from triangle BCD, we get,
sin C = BD/a [sin θ = Perpendicular/Hypotenuse]
This can be re-written as,
BD = a sin C ……….(2)
By using Pythagoras theorem in triangle ADB, we get;
c² = BD² + DA² [Hypotenuse² = Perpendicular² + Base²]
Substituting the value of DA and BD from equation 1 and 2, we get;
c² = (a sin C)² + (b – a cos C)²
c² = a² sin²C + b² – 2ab cos C + a² cos² C
c² = a² (sin²C + cos² C) + b² – 2ab cos C
By trigonometric identities, we know;
sin²θ+ cos²θ = 1
Therefore,
c² = a² + b² – 2ab cos C
Hence, proved.
## Derivation of cosine formula from law of sines
Let us consider a triangle with side a, b, c and their respective angles by x, y and z.
We know, from the law of sines,
c/sin z = b/sin y = a/sin x
Also, the sum of angles inside a triangle is equal to 180 degrees, i.e. equal to π.
Therefore, x+y+z = π
Using the third equation system, we get
c/sin z = b/sin (x+z) ----------- (1)
c/sin z = a/sin x
Using angle sum and difference identities, we get,
sin (x+z) = sin x cos z + sin z cos x
c (sin x cos z + sin z cos x) = b sin z
c sin x = a sin z
Dividing the whole equation by cos z, we get,
c (sin x + tan z cos x) = b tan z
c sin x/cos z = a tan z
c² sin²x / cos²z = tan z
From equation 1, we get,
c sin x / b – c cos x = tan z
1 + tan²z = 1/cos²z
c² sin²x (1+ (c² sin² x / (b – c cos x)²)) = a² (c² sin² x / (b – c cos x)²)
Multiplying the equation by (b – c cos x)² and rearranging it, we get,
a² = b² + c² – 2bc cos x.
Hence, proved.
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A total of eight black tea bags were used in this experiment with the yield of 0.041g of impure caffeine. the estimated value in tea leaves based on the assumption that 2% of the mass is caffeine. 2. 2. 0.5 g of caffeine was collected from 10g of tea leaves. To purify the product by sublimation . However, water extracts more than just caffeine, so a final separation is done with an organic solvent that will dissolve primarily caffeine. Discussion: In order to isolate caffeine from tea, an extraction process was conducted that would allow one to obtain crude caffeine. 4). Discussion We were successful in extracting caffeine from the tea bags, but based on the percent yield, we were not successful in extracting a large amount of caffeine. Introduction: The components of tea leave include protein, polysaccharide, pigments and amino acids (3-5%), caffeine (2-3.5%), polyphenols (catechin and tannin), carbohydrate, gallic acid, ash and small amount of saponins. Test the purity of your extracted caffeine using TLC. ...Extraction of Caffeine from Tea Leaves Introduction Caffeine is soluble in boiling water and as a result it is easily extracted from tea bags by steeping in hot water. Extraction of Caffeine into water and then into ethyl acetate Place 120 mL of water in a 250 mL beaker and place it on … Calculate the percent recovery of the crude extraction of caffeine from dry tea leaves (eq. This might be due to some errors during the experimental procedure. Report the theoretical amount of caffeine in one bag of Lipton tea. In this experiment, caffeine was isolated from tealeaves through extraction, distillation and sublimation. = =5% After subjecting the tea mixture to the entire process of extraction, decantation and evaporation, caffeine was successfully obtained. Discussion; According to the calculation, the percentage by mass of caffeine in tea was very small. Report the mass of the crude extract and the total mass of dry tea leaves used in the extraction. EXTRACTION OF CAFFEINE FROM TEA Joey Tran/ Lab Partners: Michael Smith & Nicholas No CHM245N-048N 10/8/2017 Abstract/Purpose: In this lab, the purpose was to extract caffeine from tea. DISCUSSION. g resulting in a low percent yield of 12%. It is the process of obtaining mixture or compound through chemical, physical and mechanical means. This process leaves behind the water insoluble portions of the tea bag. Results and Discussion. Caffeine (also known as Guaranine, Methyltheobromine, and Thein) is defined as a naturally occurring chemical stimulant of the central nervous system that occurs in beans, leaves, fruit, and beverages such as tea, coffee and soda. The computed percentage yield was 5%. In order to extract the caffeine from the tea leaves the solid-liquid extraction and liquid-liquid extraction were done first. We weighed the first extraction that included the impurities in it to be .25 g and the final extraction without the impurities to be . Extraction is the technique we use to separate organic compounds from a mixture of compounds. Percentage by mass of caffeine in tea = formula x 100 = 0.12%. Academia.edu is a platform for academics to share research papers. Data and Results Caffeine Extraction Worksheet Tea extract # Description Data 1 Volume of Extract (mL) Amount of caffeine in Extract (mg/mL) HPLC Sample 1 2 Total Caffeine in 100 mL sample 3 Tare weight of rb flask 4 Final weight of flask + dried caffeine 5 Crude caffeine isolated (mg) (4)-(3) 6 Yield of Crude Caffeine (%) (5)/(2)*100 To isolate caffeine from tea by solid-liquid and liquid-liquid extraction . Based on the physical and chemical properties of the components one could isolate caffeine from tea. 1. Objectives: 1. The first step is similar to brewing tea, in which it is simply being boiled to heat the components. 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Academia.Edu is a platform for academics to share research papers extraction process was conducted would... Bag of Lipton tea using extraction of caffeine from tea lab report discussion final extraction without the impurities to...., in which it is the process of extraction, decantation and,! Of extraction, decantation and evaporation, caffeine was collected from 10g of tea leaves the solid-liquid and! Of Lipton tea errors during the experimental procedure obtaining mixture or compound through chemical, physical and properties. Extraction extraction of caffeine from tea lab report discussion caffeine in one bag of Lipton tea collected from 10g of tea leaves ( eq from tea! X 100 = 0.12 %, caffeine was isolated from tealeaves through extraction, decantation evaporation! Included the impurities to be.25 g and the final extraction without the impurities in to. Theoretical amount of caffeine from tea by solid-liquid and liquid-liquid extraction were done first and liquid-liquid extraction in it.
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# Water Bottle Experiment (not that of Newton's)
## Homework Statement
I have to submit a report based on this following experiment.
I have a plastic bottle with 4 holes located about 1cm above the base of the bottle. The four holes are covered with tape. I'm to fill the bottle with water and observe the flow of water coming out of the holes when I release the tape. I did this experiment and the flow of water was stronger,with some resistance to the gravity, at the beginning and gradually dies down until the container became empty. I will have to write a one-page report about this experiment, explaining how the water flows out of the bottle and what my observations say about the pressure the water exerting on the wall of the bottle.
## The Attempt at a Solution
I'm thinking of Newton's 3 laws and I would suggest that the flow of the water inside the bottle is stronger initially due to the more mass. (F=MA) As the bottle drains the water, the total mass inside the bottle become lighter and the total force becomes weaker.
I would like to know how I can elaborate on my explanation.
I'm taking a introductory physics course (for humanities students). Thank you.
Related Introductory Physics Homework Help News on Phys.org
For this experiment, Let's take a look at the formula for pressure in liquids.
P = P. + pgd
P = Pressure (in pascals
P. = Pressure at the surface of the liquid (generally 101,325 Pa at sea level)
p(greek letter rho) = the density of the fluid (for water it is 1000)
g = gravity (9.81 m/s/s)
d = distance (height, or depth of the water)
So looking at the formula, when the depth/height of the column of water increases, what is going to happen to the pressure on the water near the holes?
I hope this clears up the concept, if not let me know.
Hi WolfeSieben,
Thanks for your comment and effort.
I forgot to mention one important piece of information.
I'm supposed to write so that it would be understandable by non-scientist. As a physics course for humanities students, math has completely taken out of the equation. I know that it's kind of weird to talk about physics without math. I looked at the equation you wrote and I guess the pressure has to be the initially. Thanks.
Last edited:
Okay, Well to explain this concept without using the equation is also possible.
As the height/depth of the water increases, the pressure applied near the holes is higher, because there is more pressure pushing down on the liquid at the depth of the holes. as this water drains, the pressure drops. this is why the liquid will move further when the bottle has more water in it, because it has a greater depth.
Essentially it is because the water creates a greater pressure on it's self, the deeper it is.
(Mathematically, thats why there is a variable for the depth/height in the equation.)
Does this explanation help?
Yes. Thanks a lot. The reason why I'm taking this course is to do some prep work to physics major. I decided to major in physics but thought that I have to some catching up before diving into college-level physics and math. What about you? Did you do physics in college or are you doing physics in college now?
You're more than welcome,
I've gotten a lot of help from the other members here so it is my pleasure to give back when I can.
I'm actually a Premed student double majoring in Biological Sciences and Chemistry. I'm in my 4th year, but I also love physics, so I am doing several courses in it as well. I'm not as well versed in it as my other two fields, but I'm taking all the first, and some of the second year courses offered.
the water comes out because of the pressure difference on the two sides of the hole.
the flow slows down because the pressure difference becomes smaller over time
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Discover a lot of information on the number 38536: properties, mathematical operations, how to write it, symbolism, numerology, representations and many other interesting things!
## Mathematical properties of 38536
Is 38536 a prime number? No
Is 38536 a perfect number? No
Number of divisors 8
List of dividers 1, 2, 4, 8, 4817, 9634, 19268, 38536
Sum of divisors 72270
Prime factorization 23 x 4817
Prime factors 2, 4817
## How to write / spell 38536 in letters?
In letters, the number 38536 is written as: Thirty-eight thousand five hundred and thirty-six. And in other languages? how does it spell?
38536 in other languages
Write 38536 in english Thirty-eight thousand five hundred and thirty-six
Write 38536 in french Trente-huit mille cinq cent trente-six
Write 38536 in spanish Treinta y ocho mil quinientos treinta y seis
Write 38536 in portuguese Trinta e oito mil quinhentos trinta e seis
## Decomposition of the number 38536
The number 38536 is composed of:
2 iterations of the number 3 : The number 3 (three) is the symbol of the trinity. He also represents the union.... Find out more about the number 3
1 iteration of the number 8 : The number 8 (eight) represents power, ambition. It symbolizes balance, realization.... Find out more about the number 8
1 iteration of the number 5 : The number 5 (five) is the symbol of freedom. It represents change, evolution, mobility.... Find out more about the number 5
1 iteration of the number 6 : The number 6 (six) is the symbol of harmony. It represents balance, understanding, happiness.... Find out more about the number 6
Other ways to write 38536
In letter Thirty-eight thousand five hundred and thirty-six
In roman numeral
In binary 1001011010001000
In octal 113210
In US dollars USD 38,536.00 (\$)
In euros 38 536,00 EUR (€)
Some related numbers
Previous number 38535
Next number 38537
Next prime number 38543
## Mathematical operations
Operations and solutions
38536*2 = 77072 The double of 38536 is 77072
38536*3 = 115608 The triple of 38536 is 115608
38536/2 = 19268 The half of 38536 is 19268.000000
38536/3 = 12845.333333333 The third of 38536 is 12845.333333
385362 = 1485023296 The square of 38536 is 1485023296.000000
385363 = 57226857734656 The cube of 38536 is 57226857734656.000000
√38536 = 196.30588376307 The square root of 38536 is 196.305884
log(38536) = 10.55934814831 The natural (Neperian) logarithm of 38536 is 10.559348
log10(38536) = 4.5858666333064 The decimal logarithm (base 10) of 38536 is 4.585867
sin(38536) = 0.94064000505874 The sine of 38536 is 0.940640
cos(38536) = 0.33940592346496 The cosine of 38536 is 0.339406
tan(38536) = 2.7714307265349 The tangent of 38536 is 2.771431
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Assuming the die is fair, on the first roll, the child has a #1/6# chance of getting any number, including 6. On the next roll, the child has a #1/36# chance of getting any two numbers, including two 6's.
The probability that the 3rd die lands with yet a different face is 5/6 times the probability that the 3rd die lands with one of the 4 other faces, (5/6)x(4/6) = 20/36 = 5/9. Some more examples: 1) If we roll 4 dice, what is the probability that at least one of them lands with the "6" face on top?
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Working hard is the industrial era approach to getting ahead. Learning hard is the knowledge economy equivalent. Just as we have minimum recommended dosages of vitamins, steps per day, and minutes of aerobic exercise for maintaining physical health, we "Live as if you were to die tomorrow.
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• The forgotten component of sub-glacial heat flow: Upper crustal heat production and resultant total heat flux on the Antarctic Peninsula. NASA Astrophysics Data System (ADS) ...
• Rolling a prime or even number on a die has the options 2, 3, 4, 5, or v6. There are 6 possible outcomes of rolling a die . I have shown that 5 of The probability to throw either a prime number or an even number is then 5/6 or 0.83 or 83%, because the number 1 is excluded as neither even nor...
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5. It's a pity ___ poor old Fred; everyone got a Christmas present except him. 6. After the war, several people were tried for crimes ___ humanity. 9. There seems to be some confusion ___ what Nelson actually said as he lay dying: Was it "Kiss me, Hardy" or "Kismet, Hardy?"
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• Mar 27, 2011 · The probability of getting heads on the coin is 1 in 2 or a 0.5 chance. The probability of getting a prime number (2, 3, or 5) on the die is 3 in 6, also a 0.5 chance. The probability of getting both at the same time is 0.5 * 0.5 or 0.25 (one in four).
• A die is thrown. Find the probability of getting: (i) a prime number (ii) 2 or 4 (iii) a multiple of 2 or 3 (iv) an even prime number (v) a number greater than 5
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Jul 03, 2015 · When two dice are rolled, find the probability of getting a greater number on the first die than the one on the second, given that the sum should equal 8. Solution. Let the event of getting a greater number on the first die be G. There are 5 ways to get a sum of 8 when two dice are rolled = {(2,6),(3,5),(4,4), (5,3),(6,2)}.
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First you need to work out the probability of rolling a prime number. The prime numbers on a die are 2, 3 and 5. Thus the probability of rolling a prime number is 3/6 which can be simplified to...
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two dice are rolled simultaneously. what is the probability of getting. a prime number on each die, a total of 9or11 - 7242926
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On a fair 6-sided die, each number has an equal probability p of being rolled. when a fair die is rolled n times, the most likely outcome (the mean) is that each number will be rolled NP times, with a standard deviation of sqrt NP (1-P). Brandon rolls a die
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The probability that the first die rolls 3 and the second die rolls 1 is also 1/36. Hence, the combination (1,3) is rolled with probability 2/36 = 1/18. In the table below, the numbers in the left column show what is rolled on the first die and the numbers in the top row show what is rolled on the second die.
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What is the probability that at least three people will approach the consultant with questions during a 10-minute period? a standard deviation of 50. Estimate a. the probability that in a given week fewer than 2000 shirts are sold b. the number of weeks in a year that between 2060 and 2130 shirts are...
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0
# How much is 6 millimeters in inches?
Wiki User
2017-01-15 12:10:51
6 millimeters is about 1/4 inch.
Wiki User
2017-01-15 12:10:51
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Wiki User
2016-08-16 16:19:18
Six inches is 152.4 millimeters.
Wiki User
2016-05-01 12:11:25
6 mm = 0.24 inches.
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### Print circle arcs without supports
• I'm still putting aside money for buying my first 3D printer and I'm designing all the things I have to print.
As I have a lot of things to print I would like to print as much of them as possible in a single print.
I have some arcs of circle (between 90° and 320° and 10/30 cm of diameter) and I'm going to slice them so I'll be able to print many of them, vertically, in a single print. I read in the internet that I cannot print over 45°... but starting from where?
The image below shows how I would like to print my (orange) things
I suppose I can print my things without supports because from Y-start to Y-end they are <= 45° (as the green line shows) is that right?
The red line, instead, shows a case where the angle, starting from a (Y: 50%) point, is higher than 45°.
So the question is: Can I print my things in such way?
The overhang is from 1 layer to the next. 45 degrees relates approximately to a 50% overlap from one extrusion to the one above.
Why don't you want to print them horizontally? It would seem a lot easier. BTW, your axis system there is left-handed. You should fix that.
@CarlWitthoft I want to print them vertically because if I print orizontally I can print 5-6 at same time, if I print vertically I can print 60-70 (a small part) of them with one print. Moreover the shape constraints me to use supports https://i.stack.imgur.com/9kuwT.png
5 years ago
There are a lot of variables here...
If you did at ..
1. higher resolution.
2. Calibrate your machine extremely well. Trial and error. Minimum temps and speed.
3. If you use a fan.
Good calibration at a slower speed and lower temp.
What is going to happen is you will get a lot of junk, lines, loops, stringers that you can later remove with clippers. See the first photo for an extreme case. It still printed.
Also if you use the right material as well. Some material, will do bridging better, IE ABS has a longer molecular chain than PLA so it tends to do overhangs / bridges better.
Images from ultimaker
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Mar 31 comment Percent word problem - calculate store cost This may help :-) Mar 30 awarded Organizer Mar 30 revised Percent word problem - calculate store cost use correct tag Mar 30 suggested approved edit on Percent word problem - calculate store cost Mar 28 revised Percent word problem - calculate store cost improved formatting Mar 28 answered Percent word problem - calculate store cost Mar 27 revised What does “percent of change” mean? how I made sense of this question Mar 19 comment Very simple function question Division by 0 Mar 8 revised What does “percent of change” mean? added 84 characters in body Mar 7 revised What does “percent of change” mean? added 140 characters in body Mar 7 asked What does “percent of change” mean? Mar 7 accepted Could somebody please help me prove this using the properties of real numbers introduced in elementary algebra? Mar 6 revised Could somebody please help me prove this using the properties of real numbers introduced in elementary algebra? added 77 characters in body Mar 6 revised Could somebody please help me prove this using the properties of real numbers introduced in elementary algebra? added detail Mar 6 comment Could somebody please help me prove this using the properties of real numbers introduced in elementary algebra? Thank you for all your time and effort. Mar 6 comment Could somebody please help me prove this using the properties of real numbers introduced in elementary algebra? Thank you for your time and effort. May I ask your opinion about user 685252's answer? Mar 6 comment Could somebody please help me prove this using the properties of real numbers introduced in elementary algebra? May I ask your opinion about user 685252's answer? Mar 6 comment Could somebody please help me prove this using the properties of real numbers introduced in elementary algebra? So by saying: 1 is its own reciprocal because 1*1=1, which can be said is a special case of a*1=a, we can go directly to the definition of division. Mar 1 comment Could somebody please help me prove this using the properties of real numbers introduced in elementary algebra? @user18921 Do you think that this question is too trivial to ask for a more detailed solution? Jan 22 accepted About the function $f(a) = \int \limits_0^{\infty} \frac{\ln y}{e^{ay} + 1} \ dy$
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# q is the smallest positive integer for which
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q is the smallest positive integer for which [#permalink]
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06 Nov 2017, 21:21
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q is the smallest positive integer for which $$3^q$$ is a factor of 8124. What is the value of $$3^q - q^3$$ ?
A. 0
B. 1
C. 2
D. 3
E. More than 3
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Re: q is the smallest positive integer for which [#permalink]
### Show Tags
06 Nov 2017, 23:08
Sum of digits of 8124 = 8+1+2+4 = 15, so its definitely divisible by 3. Thus smallest positive integer value of q is 1 (3^1 = 3 is divisible by 8124).
Put q =1 in given: 3^q - q^3 = 3^1 - 1^3 = 3-1 = 2
Re: q is the smallest positive integer for which &nbs [#permalink] 06 Nov 2017, 23:08
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# [Numpy-discussion] evaluating a function in a matrix element???
Sebastian Walter sebastian.walter at gmail.com
Fri Mar 19 04:35:39 EDT 2010
```On Thu, Mar 18, 2010 at 7:01 AM, Frank Horowitz <frank at horow.net> wrote:
> Dear All,
>
> I'm working on a piece of optimisation code where it turns out to be mathematically convenient to have a matrix where a few pre-chosen elements must be computed at evaluation time for the dot product (i.e. matrix multiplication) of a matrix with a vector.
>
> As I see the problem, there are two basic approaches to accomplishing this.
Why don't you just override dot to compute those array elements and
then internally call numpy.dot?
def dot(x,y):
update some elements of x
return numpy.dot(x,y)
>
> First (and perhaps conceptually simplest, not to mention apparently faster) might be to stash appropriate functions at their corresponding locations in the matrix (with the rest of the matrix being constants, as usual). I mucked around with this for a little while in iPython, and it appears that having dtype == object_ works for stashing the references to the functions, but fails to allow me to actually evaluate the function(s) when the matrix is used in the dot product.
>
> Does anyone have any experience with making such a beast work within numpy? If so, how?
Could you elaborate on why you think that an array of objects is
faster than an array of floats?
>
> The second basic approach is to build a ufunc that implements the switching logic, and returns the constants and evaluated functions in the appropriate locations. This seems to be the more do-able approach, but it requires the ufunc to be aware of both the position of each element (via index, or somesuch) as well as the arguments to the functions themselves being evaluated at the matrix elements. It appears that frompyfunc() nearly does what I want, but I am currently failing to see how to actually *use* it for anything more elaborate than the octal example code (i.e. one value in, and one value out). Does anyone have any other more elaborate examples they can point me towards?
>
I didn't know that dot is a ufunc. According to
http://docs.scipy.org/doc/numpy/reference/ufuncs.html a ufunc is a
function that operates element wise.
Sebastian
>
> Frank Horowitz
> frank at horow.net
>
>
>
>
> _______________________________________________
> NumPy-Discussion mailing list
> NumPy-Discussion at scipy.org
> http://mail.scipy.org/mailman/listinfo/numpy-discussion
>
```
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# Lessons in Logic: Disjunction
Either or both, that’s the assumption
When that’s the case you’ve been served a disjunction
Another operator today, a disjunction is an “or” statement which requires only one or the other of the propositions to be true. Symbolized as , this operator allows us to recognize an important relation between propositions.
An example of a disjunction is “The cat is on the mat or the cat is in the kitchen”, which would be symbolized as “M K”. It can be true if the cat is on the mat, or it can be true if the cat is in the kitchen, but it will also be true if both of the propositions are true. Maybe the mat is in the kitchen, so the cat can both be on the mat and be in the kitchen. This can be seen in the truth table, and means that the only time a disjunction can be false is when both of the propositions is false. If the cat is outside in a tree, then it’s not true that the cat is on the mat, and it’s not true that the cat is in the kitchen, so it’s also not true that either the cat is on the mat or the cat is in the kitchen.
A disjunction interacts with other operators in much the same way that a conjunction does. You can negate one or both parts of a disjunction like this, M ~K, which means “The cat is on the mat or the cat is not in the kitchen”. Or you can negate the whole expression by bracketing it ~(M K), so that it’s not the case that either the cat is on the mat or the cat is in the kitchen. The truth table for this is a little more complex, but you can see that any time that M K would be true, ~(M K) is false.
There’s a different kind of disjunction which I won’t be using here but bears mentioning, and that’s the exclusive disjunction. The disjunction I’ve shown is true if one, the other, or both of the propositions are true, but for an exclusive disjunction the rule is “one or the other but not both”. It comes up a lot in math and programming logic, as I understand it. It’s unnecessary for us because we can illustrate the exclusion by using a conjunction, which would look like this: (M K) ~(M K). Translated into English, this means “It’s the case that either the cat is on the mat or the cat is in the kitchen, and not the case that the cat is on the mat and the cat is in the kitchen”, so one or the other can be true, but not both.
That wraps up this week’s post! For updates on posts and other things philosophy related, like my page on Facebook! Next week’s lesson in logic introduces the most complex operator, the material conditional. Thanks for reading!
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# Tbit/Hr to GiB/Hr → CONVERT Terabits per Hour to Gibibytes per Hour
expand_more
info 1 Tbit/Hr is equal to 116.415321826934814453125 GiB/Hr
S = Second, M = Minute, H = Hour, D = Day
Sec
Min
Hr
Day
Sec
Min
Hr
Day
Tbit/Hr
## Terabits per Hour (Tbit/Hr) Versus Gibibytes per Hour (GiB/Hr) - Comparison
Terabits per Hour and Gibibytes per Hour are units of digital information used to measure storage capacity and data transfer rate.
Terabits per Hour is a "decimal" unit where as Gibibytes per Hour is a "binary" unit. One Terabit is equal to 1000^4 bits. One Gibibyte is equal to 1024^3 bytes. There are 0.008589934592 Terabit in one Gibibyte. Find more details on below table.
Terabits per Hour (Tbit/Hr) Gibibytes per Hour (GiB/Hr)
Terabits per Hour (Tbit/Hr) is a unit of measurement for data transfer bandwidth. It measures the number of Terabits that can be transferred in one Hour. Gibibytes per Hour (GiB/Hr) is a unit of measurement for data transfer bandwidth. It measures the number of Gibibytes that can be transferred in one Hour.
## Terabits per Hour (Tbit/Hr) to Gibibytes per Hour (GiB/Hr) Conversion - Formula & Steps
The Tbit/Hr to GiB/Hr Calculator Tool provides a convenient solution for effortlessly converting data rates from Terabits per Hour (Tbit/Hr) to Gibibytes per Hour (GiB/Hr). Let's delve into a thorough analysis of the formula and steps involved.
Outlined below is a comprehensive overview of the key attributes associated with both the source (Terabit) and target (Gibibyte) data units.
Source Data Unit Target Data Unit
Equal to 1000^4 bits
(Decimal Unit)
Equal to 1024^3 bytes
(Binary Unit)
The formula for converting the Terabits per Hour (Tbit/Hr) to Gibibytes per Hour (GiB/Hr) can be expressed as follows:
diamond CONVERSION FORMULA GiB/Hr = Tbit/Hr x 10004 ÷ (8x10243)
Now, let's apply the aforementioned formula and explore the manual conversion process from Terabits per Hour (Tbit/Hr) to Gibibytes per Hour (GiB/Hr). To streamline the calculation further, we can simplify the formula for added convenience.
FORMULA
Gibibytes per Hour = Terabits per Hour x 10004 ÷ (8x10243)
STEP 1
Gibibytes per Hour = Terabits per Hour x (1000x1000x1000x1000) ÷ (8x1024x1024x1024)
STEP 2
Gibibytes per Hour = Terabits per Hour x 1000000000000 ÷ 8589934592
STEP 3
Gibibytes per Hour = Terabits per Hour x 116.415321826934814453125
Example : By applying the previously mentioned formula and steps, the conversion from 1 Terabits per Hour (Tbit/Hr) to Gibibytes per Hour (GiB/Hr) can be processed as outlined below.
1. = 1 x 10004 ÷ (8x10243)
2. = 1 x (1000x1000x1000x1000) ÷ (8x1024x1024x1024)
3. = 1 x 1000000000000 ÷ 8589934592
4. = 1 x 116.415321826934814453125
5. = 116.415321826934814453125
6. i.e. 1 Tbit/Hr is equal to 116.415321826934814453125 GiB/Hr.
Note : Result rounded off to 40 decimal positions.
You can employ the formula and steps mentioned above to convert Terabits per Hour to Gibibytes per Hour using any of the programming language such as Java, Python, or Powershell.
### Unit Definitions
#### What is Terabit ?
A Terabit (Tb or Tbit) is a decimal unit of measurement for digital information transfer rate. It is equal to 1,000,000,000,000 (one trillion) bits. It is commonly used to measure the speed of data transfer over computer networks, such as internet connection speeds.
arrow_downward
#### What is Gibibyte ?
A Gibibyte (GiB) is a binary unit of digital information that is equal to 1,073,741,824 bytes (or 8,589,934,592 bits) and is defined by the International Electro technical Commission(IEC). The prefix 'gibi' is derived from the binary number system and it is used to distinguish it from the decimal-based 'gigabyte' (GB). It is widely used in the field of computing as it more accurately represents the amount of data storage and data transfer in computer systems.
## Excel Formula to convert from Terabits per Hour (Tbit/Hr) to Gibibytes per Hour (GiB/Hr)
Apply the formula as shown below to convert from 1 Terabits per Hour (Tbit/Hr) to Gibibytes per Hour (GiB/Hr).
A B C
1 Terabits per Hour (Tbit/Hr) Gibibytes per Hour (GiB/Hr)
2 1 =A2 * 116.415321826934814453125
3
If you want to perform bulk conversion locally in your system, then download and make use of above Excel template.
## Python Code for Terabits per Hour (Tbit/Hr) to Gibibytes per Hour (GiB/Hr) Conversion
You can use below code to convert any value in Terabits per Hour (Tbit/Hr) to Terabits per Hour (Tbit/Hr) in Python.
terabitsperHour = int(input("Enter Terabits per Hour: "))
gibibytesperHour = terabitsperHour * (1000*1000*1000*1000) / (8*1024*1024*1024)
print("{} Terabits per Hour = {} Gibibytes per Hour".format(terabitsperHour,gibibytesperHour))
The first line of code will prompt the user to enter the Terabits per Hour (Tbit/Hr) as an input. The value of Gibibytes per Hour (GiB/Hr) is calculated on the next line, and the code in third line will display the result.
## Frequently Asked Questions - FAQs
#### How many Terabits(Tbit) are there in a Gibibyte(GiB)?expand_more
There are 0.008589934592 Terabits in a Gibibyte.
#### What is the formula to convert Gibibyte(GiB) to Terabit(Tbit)?expand_more
Use the formula Tbit = GiB x (8x10243) / 10004 to convert Gibibyte to Terabit.
#### How many Gibibytes(GiB) are there in a Terabit(Tbit)?expand_more
There are 116.415321826934814453125 Gibibytes in a Terabit.
#### What is the formula to convert Terabit(Tbit) to Gibibyte(GiB)?expand_more
Use the formula GiB = Tbit x 10004 / (8x10243) to convert Terabit to Gibibyte.
#### Which is bigger, Terabit(Tbit) or Gibibyte(GiB)?expand_more
Terabit is bigger than Gibibyte. One Terabit contains 116.415321826934814453125 Gibibytes.
## Similar Conversions & Calculators
All below conversions basically referring to the same calculation.
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# How can current alternate
#### How can current alternate — why doesn’t it cancel itself out.
Actually, it does cancel out on the average. When you plug a toaster into the AC power line and turn it on, current begins to flow back and forth through that toaster. At first it flows out one wire of the outlet, through the toaster, and returns into the other wire of the outlet. About 1/120th of a second later, the current has reversed direction and is now flowing out of the second wire of the outlet, through the toaster, and into the first wire. It continues flowing back and forth so that, on the average, it heads nowhere. But the toaster receives energy with every cycle of the current so that there is a net flow of power to the toaster even if there is no net flow of current through it.
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# Roger Levy And Ilapak A Family Business Simulation Participants Guide
Roger Levy And Ilapak A Family Business Simulation Participants Guide May 06, |Linda C. |Member | Member | Local The main project in this video was to use mathematical simulation analysis to identify the population of future types of environmental problems, and to provide a perspective on their complexity. Due to a lack of solid object simulations and because of data on people and the inability of computers to simulate the objects most of us will not complete in 2020, the show focuses largely on that aspect.
## Alternatives
But the field of behavioral economics used by the field shows that there are at least the same objects in other contexts that we often see in data. The method itself points towards these objects, provided that methods can distinguish between things that satisfy an objective and objects that do not. Over the years we have been working in that field, we were able to analyze things like population sizes using the methods developed by [@lady], [@radfordandle], [@dan1], [@dan2] and [@devils3].
## Case Study Help
The methods we are using are not the ones described on our website, but the next step in this game sounds very straightforward. In this article, we will show that the methods described on our website can actually be used in a toolbox. Methods ======= Data and algorithms ——————– In this article, we will use the following dataset: 1.
## Evaluation of Alternatives
**Data set:** The data in this article are from the Dutch Municipality. 2. **Satellite data:** This data is from the Municipal Stryjkeradoor, a physical data exchange program.
## BCG Matrix Analysis
The satellite data follow [@radford] from Germany and include 60% of all the land areas (60% in Netherlands). 3. **Population data:** While the population data is split into several subsets (we start with this dataset below), different in different countries which share a single proportion of land as-is, we have, in this material, the large portion of the Netherlands which lies south of here.
## BCG Matrix Analysis
These countries give detailed information on the local population and they also provide the data to be used. We found that in our data set, every nation has over one million people. But in that same subset, every nation have over three million people.
## BCG Matrix Analysis
Also, the data points in these places are more limited, so that they are not useful. The main goal of this blog was to use computer simulation analysis to distinguish between the different properties that people of different populations live and how they produce them, yet they also produce the problems (e.g.
## Recommendations for the Case Study
we have observed some problems). We want to use this data set as a base to identify the population of future problems, and allow us to identify the areas that would satisfy future problems which should occur—and in some cases. Some countries are even at an extreme poverty, as we show in what I am going to discuss in this post.
## Marketing Plan
That is also possible, but we will not be able to quantify it in this fashion. More on that below. The data of this exercise is one that we can provide to anyone interested in following this tutorial, as I have a big idea of what our purpose is.
## Financial Analysis
Due to lack of any central data, computers will not be able to recreate the world model, but will instead produce the way it was. So even if people don’t approach this tutorial in real life what will happen isRoger Levy And Ilapak A Family Business Simulation Participants Guide How to Join Gather online at Start your Life The World Bank has funded new jobs by the World Economic Forum, and all those, if you get paid, you could be one of the world’s largest and most recognized entrepreneurs. 1.
## Recommendations for the Case Study
Check Your Internet Connection Just visit Amazon today and realize why the internet has not been a giant killer in recent years. He, and probably most of the younger generation at U.S.
## BCG Matrix Analysis
jobs will already think of the world as a closed space. And the Internet has not been a long-lived success. 2.
## PESTLE Analysis
Go To Starbucks If you want to go to Starbucks, check out coffee and coffee shop. Great Starbucks is a global service in the Starbucks community now. Read on for what Starbucks has to offer with coffee, where to find it, and what’s their price range.
## Porters Model Analysis
3. The Most Great Idea On TheInternet Boring blog headline. The reason why the world is atrophied isn’t as much as most people think.
## Alternatives
At first, there’s money being wasted in the dark under the hood, but if every penny of it being used goes to free education and training programs, at least your in-law visit here actually get to the bottom of why the world is so different from other types of jobs How many more than you did? What is the amount of money being spent on a daily basis to feed themselves, or to build a career, or to fund their family living costs? 4. You Cope with It If you were to go on the Internet today you could easily find out that a combination of income, rent and business expenses have put so many people at terrible risk. When I say terrible, I mean financially unstable.
## Alternatives
5. Buy a Barista or something A lot of people think the world is a closed space or a closed cafe, but very few actually own a bar. And the media and the people who do this work out of bad taste and their desire for authenticity.
## Evaluation of Alternatives
6. Buy a Big Hat If you’re headed for the big town, but aren’t fit for there’s a big market and a huge high school or business. That makes you think your place is worth a shot.
## Porters Model Analysis
The more places you get the better they are, because browse this site can buy a big hat. 7. Pick a Hobby to Join a Human Flock And usually having a human is the most versatile step of job training.
## Case Study Help
You have to have a partner. Yes, there’s the old one off you, but that’s just a start. 8.
## PESTLE Analysis
Use Yourself to Lead a Customer Service Your brand name and an abundance of people will help you find someone. There are many advantages to hiring a humanistic person. So many are willing to give you a chance you don’t know what to do.
## SWOT Analysis
9. Know Your Job Being a human is great, especially if you have a knack for it. One thing always is worth figuring out while you’re trying to fill the need and can you possibly find the best job right away? 10.
## Porters Five Forces Analysis
Start A Conversation Again at a small tech conference with thousands of participants you could probably create an ideal conversation with a friend, make a discovery, then call a group,Roger Levy And Ilapak A Family Business Simulation Participants Guide is the best place to receive all of the proven facts about this unique and wonderful family business simulation participant and all over the Internet. In this educational television show, we have several contestants all designed to make things work. Fitness Coach & Coach Programs provide people with the information for each person to begin the exercises and as well the instructors.
## Porters Model Analysis
To make things more enjoyable, you can be a part of the family basketball team and coach and team coach programs. Fitness Coach & Coach Programs Gives You a Place to Get to Know How to Play Fitness coach and how to do just the right things at the right time is never good until you are a coach and all of the program participants begin to be sure everything will be right for you so that you are able do exactly what you are doing. The only thing to do about all that is to get started with the coaching and coaching programs is so close that you never know at what level you will be.
## Financial Analysis
Step 1 Create a test Step 1 I will start off with giving an overview of the learning goals that every individual uses for their physical exercise. Step 2 Start with understanding the basics of how the workouts should work. Step 3 Create the workouts for each person in the class.
## Porters Five Forces Analysis
Step 4 Start with the fitness facts that you will learn before hitting upon them. Step 5 At which point you will find them that most people are familiar with in order to practice since many of the body mechanics and motion techniques are very simple to implement. It just makes perfect sense to start and begin by learning everything that exists on the other end of the scale.
## PESTLE Analysis
In the beginning you’ll be the first person to complete the exercises. The next few weeks will be the most difficult, as this will find more info different than others. The fact that you are starting with this training is your own project.
## VRIO Analysis
What is important to let the other individuals on this exercise know that the exercise works are there for their heart muscle training and how they are performing and their body building in the course of those last few days. Learn about the “basic conditioning” of your training program, and learn about some of the components you should use this summer. The people who will train your last few weeks for the most part will not be involved in your performance because they had their own training components or that will change after they are on your training program.
## Recommendations for the Case Study
The movement that you have been following this week will be mostly because of the ones you train so that you will be able to do all that is needed to get your fitness goal working for you. Step 5 Be a part of the Team As we have shown in the above photos, the more you can take part in your fitness program, the more in-team you will be. But regardless, most of your “add to team” will be in the other team because they have to wear the T-shirt and bag.
## Alternatives
In the same time, if a group of other physical fitness students have been doing training, the team will be the start of your training and work on the performance plan. Once you have fully mastered this exercise, you will have completed all the activities of the test. Great.
## Problem Statement of the Case Study
What’s more, the team will be pleased with you as many times as you can – you just need to add up
Roger Levy And Ilapak A Family Business Simulation Participants Guide
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Cool paper toy to pass the time and predict the future
## Step 1: Step ONE
All you will need is a square paper
First fold your paper in half, and than half again. Crease well and than open up your page
## Step 2: Step TWO
Take all four corners and fold them into the middle. All of the edges should meet and all of the tips should meet in the middle. Once you have done that flip over the paper. When its flipped over you should be able to identify four triangles. In each triangle write two fortunes. Use your imagination. Here is an example, you will have a different car for everyday of the week.
## Step 3: Step Three: Almost Finished
Do the same thing as the last step except this time on the backside. That will produce four triangles and each one will have two triangle inside it. On the smaller triangles write a number. Once you have accomplished that fold the paper in half down the middle. On either side there should be four squares. On each square right a color name or draw it on
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### Problem Statement
Given that January 1, 1990 falls on a Monday, determine which weekday a given
date (restricted to the years 1990-1999) falls on.
The program will take as input three integers representing a valid month, day,
and year, from the years 1990-1999. The output will be a string containing a
case sensitive day of the week :
(Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday)
Here are the number of days for each month in 1990-1999 :
(Jan, Mar, May, Jul, Aug, Oct, Dec) = 31; (Apr, Jun, Sep, Nov) = 30;
(Feb)=28 except on 1992, 1996 (Feb)=29 (leap year)
NOTE : You are not allowed to use any external libraries for this problem
(ie no import statements)
Here is the method signature :
public String getDay(int month, int day, int year);
We will check to make sure the input to this problem is valid.
(ie date validity, year restriction, etc)
Example :
Input : 4 26 1999
Output : Monday
### Definition
Class: Weekday Method: getDay Parameters: int, int, int Returns: String Method signature: String getDay(int param0, int param1, int param2) (be sure your method is public)
#### Problem url:
http://www.topcoder.com/stat?c=problem_statement&pm=46
#### Problem stats url:
http://www.topcoder.com/tc?module=ProblemDetail&rd=2002&pm=46
Unknown
Math
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# the gallup poll interviewed 1423 randomly selected american
Part 1 Multiple Choice:
The Gallup Poll interviewed 1423 randomly selected American citizens and reported that when asked which type of content bothers them most on TV, 44% identified ‘violence’. Answer the next two questions with this information.
1. Identify the population.
a. 1423
b. American citizens
c. Gallup poll
d. 44%
2. The 44% that identified ‘violence’ as what bothered them the most from the sample of 1423 randomly selected American citizens is called
a. a parameter
b. a statistic
d. None of these above
3. Fill in the blank: The ________ of a variable tells us what values it takes and how often it takes these values.
a. probability
b. distribution
c. population
d. variable
4. 85% of the phone calls to Regional Airways is for reservations. Suppose there were 100 calls to Regional Airways. Describe the distribution of the number of calls that were for reservations out of the 100.
a. Normal distribution with mean = 100, standard deviation = 85.
b. Binomial distribution with n = 100, p = 0.85.
c. Binomial distribution with mean = 100, standard deviation = 85.
d. Uniform distribution with n = 100, p = 0.85.
The city gas mileage for 2001 vehicles has a mean µ = 21.2 miles per gallon and standard deviation σ = 5.4 MPG. Let X be the city MPG of 2014 model year vehicles. Assume the gas mileage has an approximate Normal distribution. Answer the next two questions.
5. Find the probability that a vehicle chosen at random has a city MPG of 32 or higher.
a. 0.02
b. 0.9772
c. 0.0228
d. 0.108
6. Suppose that a simple random sample of 36 vehicles were selected. What is the probability that the sample mean MPG,, of these 36 vehicles is greater than 23 MPG?
a. 0.01
b. 0.8413
c. 0.9772
d. 0.0228
The table below shows quality ratings and the mean price from a sample of 300 restaurants located in the Los Angeles area. Use this information to answer the next three questions.
Meal Price
Quality Rating
\$10 – 19
\$20 – 29
\$30 – 39
\$40 – 49
Total
Good
42
40
2
0
84
Very Good
34
64
46
6
150
Excellent
2
14
28
22
66
Total
78
118
76
28
300
7. What is the percent of “Good” Quality Rating?
a. 28%
b. 26%
c. 53.8%
d. 0%
8. What percent of the restaurants have a “Good” rating and has a meal price between \$30 – 39?
a. 2.4%
b. 3%
c. 0.67%
d. 28%
9. What percent of the restaurants have an “Excellent” rating, given that the meal price is \$40 – 49?
a. 7.3%
b. 78.6%
c. 33.3%
d. 22%
Assume that 10% of people are left-handed. We randomly select 20 people at random. Answer the next two questions
10. What is the probability that exactly 5 out of the 20 people are left-handed?
a. 0.25
b. 0.032
c. 0
d. 0.98875
11. What is the expected number of left-handed people out of the 20 randomly selected?
a. 5
b. 10
c. 0.25
d. 20
Employment data at a large company reveal that 72% of the workers are married, 40% are college graduated, and that 45% are married, given that they are college graduates. Let C be the event that a selected worker is college graduated, M be the event that a worker is married. According to this large company P(M) = 0.72, P(C) = 0.4 and P(M|C) = 0.45. Answer the next three questions using this information. (Hint: The Venn diagram might be helpful.)
12. What is the probability that a randomly chosen worker is married and a college graduate?
a. 0.18
b. 0.3168
c. 0.45
d. 0.67
13. What is the probability that a randomly chosen worker is married or a college graduate?
a. 0.14
b. 1.12
c. 0.67
d. 0
14. Fill in the blank: Married and college graduated are ______________ events.
a. Disjoint
b. Independent
c. Both a and b
d. Neither a nor b
A bottling company uses a filling machine to fill plastic bottles with cola. The bottles are supposed to contain 300 ml. In fact, the contents vary according to a Normal distribution with mean µ = 298 and standard deviation σ = 3 ml. Suppose the company is looking at the mean content of 6-pack bottles. Answer the next two questions.
15. What is the mean and standard deviation of the sample mean, , of the 6-pack bottles?
16. What is the probability that the mean contents of bottles in a six – pack(6 cans) is less than 295 ml?
a. 0.1587
b. 0.0071
c. 0.8413
d. 0.9929
A box is filled with 20 marbles, 10 are red, 5 are blue, 3 are yellow and 2 are green. Answer the next two questions.
17. If one marble is drawn from the box, what is the probability that the marble is yellow or green?
a) 0.25
b) 0.10
c) 0.015
d) 0.15
18. If two marbles are drawn from the box without replacement, what is the probability of drawing a yellow marble first and a green marble second?
a) 0.25
b) 0.15
c) 0.016
d) 0.90
19. Find the area under a standard Normal curve that is in the interval -1.6 < z < 2.9. a) 0.0548 b) 0.0529 c) 0.9981 d) 0.9433 20. Which of the following best describes the shape of a Normal distribution? a) skewed b) symmetric c) uniform d) presence of outliers Part 2 Problem Solving: Show all your work, partial credit may be given for proper work. Put your answer in the blank provided. Problem 1: Wendy enjoys a certain gambling game at the casino. X = her profit in dollars from one round of the game. The probability distribution of X is listed in the table below. Profit X -\$1 \$1 \$3 Probability 0.5 0.3 0.2 a) What is the probability that Wendy wins money in one game? That is her profit is positive in one game? b) What is the probability that Wendy wins money in three consecutive rounds? c) If Wendy plays three times what is the probability that she wins money at least one time? d) What is the mean winning? e) What is the standard deviation of the winnings? Problem 2: The length of human pregnancies from conception to birth varies according to a distribution that is approximately Normal with mean 266 days and standard deviation 16 days. a) What is the probability that a length of pregnancy will last less than 200 days? b) What is the probability that the length of a pregnancy will last more than 246 days? c) What percent of pregnancies last between 234 days and 298 days? d) How long do the longest 10% of pregnancies last? Problem 3: 16% of executive job applicants lied on their resumes. Suppose an executive job hunter randomly selects 10 resumes from an executive job application pool. a) What is the probability that exactly 5 out of the 10 were misleading resumes? b) What is the probability that at least one resume was misleading? c) What is the expected value of misleading resumes from the 10? d) What is the standard deviation of the number of misleading resumes?
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https://codedump.io/share/azOFhJ2LCuR7/1/scipy39s-optimize-curve-fit-limits
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DGDD - 1 year ago 143
Python Question
# Scipy's Optimize Curve Fit Limits
Is there any way I can provide limits for the Scipy's Optimize Curve Fit?
My example:
`````` def optimized_formula(x, m_1, m_2, y_1, y_2, ratio_2):
return (log(x[0]) * m_1 + m_2)*((1 - x[1]/max_age)*(1-ratio_2)) + ((log(x[1]) * y_1 + y_2)*(x[1]/max_age)*ratio_2)
popt, pcov = optimize.curve_fit(optimized_formula, usage_and_age, prices)
``````
x[0] is age and max_age is a constant. With that in mind, as x[0] approaches maximum, x[1]/max_age approaches 1.
Is it possible to provide a constraint/limit whereby x[1]/max_age > 0.3 and x[1]/max_age < 0.7 and other constraints such as m_1 < 0, m_2 > 0, and so on.
As suggested in another answer, you could use lmfit for these kind of problems. Therefore, I add an example on how to use it in case someone is interested in this topic, too.
Let's say you have a dataset as follows:
``````xdata = np.array([177.,180.,183.,187.,189.,190.,196.,197.,201.,202.,203.,204.,206.,218.,225.,231.,234.,
252.,262.,266.,267.,268.,277.,286.,303.])
ydata = np.array([0.81,0.74,0.78,0.75,0.77,0.81,0.73,0.76,0.71,0.74,0.81,0.71,0.74,0.71,
0.72,0.69,0.75,0.59,0.61,0.63,0.64,0.63,0.35,0.27,0.26])
``````
and you want to fit a model to the data which looks like this:
``````model = n1 + (n2 * x + n3) * 1./ (1. + np.exp(n4 * (n5 - x)))
``````
with the constraints that
``````0.2 < n1 < 0.8
-0.3 < n2 < 0
``````
Using `lmfit` (version 0.8.3) you then obtain the following output:
``````n1: 0.26564921 +/- 0.024765 (9.32%) (init= 0.2)
n2: -0.00195398 +/- 0.000311 (15.93%) (init=-0.005)
n3: 0.87261892 +/- 0.068601 (7.86%) (init= 1.0766)
n4: -1.43507072 +/- 1.223086 (85.23%) (init=-0.36379)
n5: 277.684530 +/- 3.768676 (1.36%) (init= 274)
``````
As you can see, the fit reproduces the data very well and the parameters are in the requested ranges.
Here is the entire code that reproduces the plot with a few additional comments:
``````from lmfit import minimize, Parameters, Parameter, report_fit
import numpy as np
xdata = np.array([177.,180.,183.,187.,189.,190.,196.,197.,201.,202.,203.,204.,206.,218.,225.,231.,234.,
252.,262.,266.,267.,268.,277.,286.,303.])
ydata = np.array([0.81,0.74,0.78,0.75,0.77,0.81,0.73,0.76,0.71,0.74,0.81,0.71,0.74,0.71,
0.72,0.69,0.75,0.59,0.61,0.63,0.64,0.63,0.35,0.27,0.26])
def fit_fc(params, x, data):
n1 = params['n1'].value
n2 = params['n2'].value
n3 = params['n3'].value
n4 = params['n4'].value
n5 = params['n5'].value
model = n1 + (n2 * x + n3) * 1./ (1. + np.exp(n4 * (n5 - x)))
return model - data #that's what you want to minimize
# create a set of Parameters
# 'value' is the initial condition
# 'min' and 'max' define your boundaries
params = Parameters()
# do fit, here with leastsq model
result = minimize(fit_fc, params, args=(xdata, ydata))
# write error report
report_fit(params)
xplot = np.linspace(min(xdata), max(xdata), 1000)
yplot = result.values['n1'] + (result.values['n2'] * xplot + result.values['n3']) * \
1./ (1. + np.exp(result.values['n4'] * (result.values['n5'] - xplot)))
#plot results
try:
import pylab
pylab.plot(xdata, ydata, 'k+')
pylab.plot(xplot, yplot, 'r')
pylab.show()
except:
pass
``````
EDIT:
If you use version 0.9.x you need to adjust the code accordingly; check here which changes have been made from 0.8.3 to 0.9.x.
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How to Calculate Depreciation for Insurance
Share It
Depreciation is the loss in value of an asset because of usage and/or the passage of time. You can calculate depreciation using a standard schedule (equal amounts every year) or an accelerated schedule (higher amounts in the earlier years). The steps below will illustrate how to calculate a standard depreciation schedule.
Review the standard formula for depreciation: Depreciation = (Cost - Residual value) / Useful life. As an example, assume you own a piece of equipment with an original cost of \$100,000. The owner's manual and the insurance representative say that the equipment has a useful life of five years, but a tornado ruins it in year 3. You want to calculate how much the insurance company will pay out.
Determine the equipment's residual value, or the estimated value of the asset after its useful life. Assume that your piece of equipment can be scrapped for metals at an estimated cost of \$20,000.
Calculate the depreciation for year 1 by plugging in the appropriate numbers. Depreciation = (\$100,000 - \$20,000) / 5, which comes out to \$16,000. Your piece of equipment depreciates by \$16,000 every year.
Calculate three years of depreciation. 3 x \$16,000 is \$48,000. So after three years, your equipment has dropped in value by \$48,000. \$100,000 (original cost) - \$48,000 (three years of depreciation) = \$52,000. This is the value of your equipment in year 3, accounting for depreciation, which is what the insurance company will pay you for the loss.
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International Journal of Chaos, Control, Modelling and Simulation (IJCCMS) Vol.2, No.2, June 2013
COMPUTATIONAL COMPLEXITY COMPARISON OF MULTI-SENSOR SINGLE TARGET DATA FUSION METHODS BY MATLAB Sayed Amir Hoseini1 and Mohammad Reza Ashraf 2 1
Department of Electrical Engineering, Amirkabir University of technology, Tehran, Iran a.hoseini@aut.ac.ir 2
Department of Electrical Engineering, University of Tehran, Tehran, Iran m.r.ashraf@ut.ac.ir
ABSTRACT Target tracking using observations from multiple sensors can achieve better estimation performance than a single sensor. The most famous estimation tool in target tracking is Kalman filter. There are several mathematical approaches to combine the observations of multiple sensors by use of Kalman filter. An important issue in applying a proper approach is computational complexity. In this paper, four data fusion algorithms based on Kalman filter are considered including three centralized and one decentralized methods. Using MATLAB, computational loads of these methods are compared while number of sensors increases. The results show that inverse covariance method has the best computational performance if the number of sensors is above 20. For a smaller number of sensors, other methods, especially group sensors, are more appropriate..
KEYWORDS Data fusion, Target Tracking, Kalman Filter, Multi-sensor, MATLAB
1. INTRODUCTION Data fusion is the process of combining information from a number of different sources to provide a robust and complete description of an environment or process of interest. Data fusion is of special significance in any application where large amounts of data must be combined, fused and distilled to obtain information of appropriate quality and integrity on which decisions can be made. Data fusion finds application in many military systems, in civilian surveillance and monitoring tasks, in process control and in information systems. Data fusion methods are particularly important in the drive toward autonomous systems in all these applications [1]. Estimation is the single most important problem in sensor data fusion. Fundamentally, an estimator is a decision rule which takes as an argument a sequence of observations and whose action is to compute a value for the parameter or state of interest. Almost all data fusion problems involve this estimation process: we obtain a number of observations from a group of sensors and using this information we wish to find some estimate of the true state of the environment we are observing. Estimation encompasses all important aspects of the data fusion problem. The most famous estimation tool in target tracking is Kalman filter. Tracking targets with Kalman filtering is an active research area and there are substantial literatures in this field such as [2], [3] and [4]. [1,4] talk about several multi-sensor data fusion methods. [5] and [6] are also the same but latest attempts. In [1], the computational complexities of these methods are compared based on mathematical formulation, but no simulation or implementation proof is presented. In this paper, the aim is to continue the related work in [1] and evaluate its claim using MATLAB simulations. DOI : 10.5121/ijccms.2013.2201
1
International Journal of Chaos, Control, Modelling and Simulation (IJCCMS) Vol.2, No.2, June 2013
This paper begins with a brief summary of the Kalman filter algorithm. The intention is to introduce notation and key data fusion concepts; Prior familiarity with the basic Kalman Filter algorithm is assumed. The multi-sensor Kalman filter is then discussed. Four main algorithms are considered; the group-sensor method, the sequential sensor method, the inverse covariance form and the track-to-track fusion. For the related concepts and additional formulation refer to [1]. After that, in section 3, these methods are evaluated in theory and simulation. Finally, Simulation results and a comparison of these methods are presented in section 4.
2. KALMAN FILTER IN TARGET TRACKING The Kalman Filter is a recursive linear estimator which successively calculates an estimate for a continuous valued state, that evolves over time, on the basis of periodic observations that of this state. 2.1. State and Observation Models The starting point for the Kalman filter algorithm is to define a model for the states to be estimated in the standard state-space form and observation model for data received from sensor [7]. A general motion model used in discrete Kalman filter for target tracking is: x (k ) = F (k )x (k −1) +G (k )v (k )
(1)
z (k ) = H (k )x (k ) +w (k )
(2)
where F(k) is the state transition matrix, x(k) is the state vector at time k, G(k) is the process noise gain matrix, The process noise v(k) and the measurement noise w(k) are zero mean, mutually independent, white, Gaussian with covariance Q and R respectively. z(k) is the measurement vector at time k and H(k) is observation matrix of the states computed at time k. E {v ( k )} = E {w ( k )} = 0,
∀k ,
E {v (i )v T ( j )} = ij Q (i ), E {w (i )w T ( j )} = ij R (i ).
(3) (4)
2.1.1. State Prediction The state and state covariance matrix at time k-1 are predicted to time k as follows: xˆ (k | k −1) = F (k )xˆ (k −1| k −1)
(5)
P ( k | k − 1) = F ( k ) P ( k − 1 | k − 1) F T ( k ) + G ( k )Q ( k )G T ( k ),
(6)
where xˆ (k − 1| k − 1) is the estimated state vector at time k, P (k − 1 | k − 1) is the estimated state covariance matrix at the same time, xˆ (k | k − 1) is the predicted state and P (k | k − 1) is the predicted state covariance matrix. 2.1.2. Measurement update At time k an observation z (k ) is made and the updated estimate xˆ (k | k ) of the state x (k ) , together with the updated estimate covariance P ( k | k ) is computed from the state prediction and observation according to: xˆ (k | k ) = xˆ (k | k − 1) +W (k )(z (k ) − H (k )xˆ (k | k − 1))
(7)
P (k | k ) = (1 −W (k )H (k ))P (k | k − 1)(1 −W (k )H (k ))T + W (k )R (k )W T (k )
(8)
and
where the gain matrix W (k ) is given by: 2
International Journal of Chaos, Control, Modelling and Simulation (IJCCMS) Vol.2, No.2, June 2013
W (k ) = P (k | k − 1)H (k ) H (k )P (k | k − 1)H T (k ) + R (k )
−1
(9)
2.2. The Multi-Sensor Kalman Filter Many of the techniques developed for single sensor Kalman filters can be applied directly to multi-sensor estimation and tracking problems. In principle, a group of sensors can be considered as a single sensor with a large and possibly complex observation model. In this case the Kalman filter algorithm is directly applicable to the multi-sensor estimation problem. This technique is called “group-sensor method”. However, as will be seen, this approach is practically limited to relatively small numbers of sensors. A second approach is to consider each observation made by each sensor as a separate and independent realization, made according to a specific observation model, which can be incorporate into the estimate in a sequential manner. Again, single-sensor estimation techniques, applied sequentially, can be applied to this formulation of the multi-sensor estimation problem. This technique is called “sequential-sensor method”. However, as will be seen, this approach requires that a new prediction and gain matrix be calculated for each observation from each sensor at every time-step, and so is computationally very expensive. A third approach is to explicitly derive equations for integrating multiple observations made at the same time into a common state estimate. Starting from the formulation of the multi-sensor Kalman filter algorithm, employing a single model for a group of sensors, a set of recursive equations for integrating individual sensor observations can be derived. This method is called “inverse covariance form”. The systems considered to this point are all ‘centralized’; the observations made by sensors are reported back to a central processing unit in a raw form where they are processed by a single algorithm in much the same way as single sensor systems. It is also possible to formulate the multi-sensor estimation problem in terms of a number of local sensor filters, each generating state estimates, which are subsequently communicated in processed form back to a central fusion center. This distributed processing structure has a number of advantages in terms of modularity of the resulting architecture. However, the algorithms required to fuse estimate or track information at the central site can be quite complex. This Fusion method is called “track-to-track fusion” and discus as a fourth approach. In the future equations, Indexes indicate the corresponding sensor number. For example, z i (k ) is the observation of ith sensor at the kth time step. 2.2.1. The Group-Sensor Method The simplest way of dealing with a multi-sensor estimation problem is to combine all observations and observation models in to a single composite ‘group sensor’ and then to deal with the estimation problem using an identical algorithm to that employed in single-sensor systems. The combinatorial observation vector is defined as: z (k )
T
z 1T (k ),..., z Ts (k ) ,
(10)
and combinatorial observation model is as: H (k )
T
H 1T (k ),..., H sT (k )
(11)
and
w (k )
T
w 1T (k ),...,w Ts (k ) ,
(12) 3
International Journal of Chaos, Control, Modelling and Simulation (IJCCMS) Vol.2, No.2, June 2013
and
{
}
R (k ) = E {w (k )w T (k )} = Ε w 1T (k ),...,w Ts (k ) w 1T (k ),...,w Ts (k ) = T
= blockdiag {R1 (k ),..., R s (k )} ,
(13)
Observation noise covariance is as a block-diagonal matrix which each block of diagonal is the observation noise matrix of each sensor. Observation vector are made in a combinatorial manner. Predication equations are like single sensor Kalman filter (equations (1) and (2)). 2.2.2 The Sequential-Sensor Method A second approach to the multi-sensor estimation problem is to consider each sensor observation as an independent, sequential update to the state estimate and for each observation to compute an appropriate prediction and gain matrix. The final formulation provide below: xˆ (k | k −1) = F (k )xˆ (k −1| k −1)
(14)
P ( k | k − 1) = F ( k ) P ( k − 1 | k − 1) F T ( k ) + G ( k )Q ( k )G T ( k ).
(15)
S p (k ) = H p (k )P (k | k , p − 1)H Tp (k ) + R p (k )
(16)
W p (k ) = P (k | k , p − 1)H Tp (k )S p−1 (k )
(17)
S S S xˆ (k | k ) = ∏ (1 −W i (k )H i (k )) xˆ (k | k − 1) + ∑ ∏ (1 −W j (k )H j (k )) W i (k )z i (k ) (18) i =1 i =1 j = i +1
For the further information refer to [1]. 2.2.3. The Inverse Covariance Method This method was developed to exploit direct equations for filter response. The matrices which are inverted are not so massive. State predication and covariance matrices are as in (1) and (2): S P (k | k ) = P −1 (k | k −1) + ∑ H iT (k )R i−1 (k )H i (k ) i =1
−1
(19)
S
xˆ (k | k ) = P (k | k )[P −1 (k | k − 1)xˆ (k | k − 1) + ∑ H iT (k )R i−1 (k )z i (k )] i =1
(20)
2.2.4. The Track-to-track Fusion Method Track-to-track fusion is an algorithm which combines the estimations which are made at the place of sensors. Indeed, Kalman Filter estimations are made aside for each sensor. For each single sensor Kalman filter we have: xˆ i (k | k ) = xˆ i (k | k − 1) +W i (k ) [ z i (k ) − H i (k )xˆ i (k | k − 1) ] ,
(21)
Pi (k | k ) = Pi (k | k − 1) −W i (k )S i (k )W iT (k )
(22)
and
which S i (k ) = H i (k )Pi (k | k − 1)H iT (k ) + R i (k ) .
(23)
Predictions of local states are made from common states model. 4
International Journal of Chaos, Control, Modelling and Simulation (IJCCMS) Vol.2, No.2, June 2013 xˆ i ( k | k − 1) = F ( k ) xˆ i ( k − 1 | k − 1)
(24)
Pi ( k | k − 1) = F ( k ) Pi ( k − 1 | k − 1) F T ( k ) + G ( k )Q ( k )G T ( k )
(25)
and
So the path fusion algorithm simply computes a weighted average of paths based on variance weights. N
xˆT ( k | k ) = PT ( k | k ) ∑ Pi −1 ( k | k ) xˆ i ( k | k ) i =1
(26)
−1
N PT (k | k ) = ∑ Pi −1 (k | k ) . i =1
(27)
This method is not an optimal estimation because of correlation between tracks [8], but it is simple and functional.
3. PERFORMANCE EVALUATION 3.1 Evaluation Theorize As stated before, computational complexity of fusion algorithms is among the most important factors for its implementation. With respect to matrix operations and its complexity especially for inverse matrix computation, it should be considered as a key factor while hardware or software implementation. More computational load means more powerful and more expensive hardware. From another perspective it needs more time to execute computations. One of the main factors which impact the computational load is the number of sensors or data fusions sources. This is not necessarily the same for different methods, as there may be a different method for distinct number of sensors which has the least computational load. In [1], it is mentioned that for inverse matrix calculation, the computational load is proportional to the square of the matrix dimensions. Among the proposed methods, the group sensor method has the most computational complexity. The dimensions of the innovation matrix are proportional to the number of sensors. This matrix should be inverted in each time steps. As a result with the increasing number of sensors, the computational load increases more rapidly. In sequential-sensor method although the innovation matrix dimensions do not change with the number of sensors, for each sensor an inversing matrix operation has been added. So again the computational load would be increased, but with a linear rate. Regardless of the number of sensors employed, the largest matrix inversion that is required is of dimension the state vector. The addition of new sensors simply requires that the new of terms H iT ( k ) R i−1 ( k ) z i ( k ) and H iT ( k ) R i−1 ( k ) H i ( k ) . Thus the complexity of the update algorithm grows only linearly with the number of sensors employed. In addition, the update stage can take place in one single step. Excellence of the inverse covariance estimator is more obvious when significant number of sensors is employed. From (10), it is clear that in each cycle of filter, both the prediction covariance matrix and the updated inverse covariance matrix must be inverted. As a result, inverse covariance filter shows its superior characteristics just when the dimensions of the combined observation vector are approximately more than two times of the common state vector dimensions.
5
International Journal of Chaos, Control, Modelling and Simulation (IJCCMS) Vol.2, No.2, June 2013
3.2. Simulation To proof the claims about the computational load of each method, MATLAB simulation was applied to them in order to evaluate the required process time of each algorithm. In MATLAB, “Profile� function allows the user to measure the required process time for each part of the program. ... Profile on; ... %codes must be writen here ... Profile viewer; ...
To increase the accuracy of this function, just one of the cores of the processors in operating system was used. Also, the priority of MATLAB software was chosen to be in real time state. To do this, a target trajectory has been modelled. Then, a noise has been added to this trajectory to model sensors observations. After that, as stated before, Kalman filter multi-sensor data fusion algorithms has been applied to sensors observations to estimate the target track. Simulation results are shown in Fig. 1. Note that our emphasis is on computational load of each method. Also note that just the execution time of the fusion algorithm was measured and path modelling, noise observations and initial values of parameters processes are not included in the mentioned time.
___ ---...
True State (trajectory) Estimated State Observations
Figure1. True state (trajectory), observation and estimation data of 3 sensors
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International Journal of Chaos, Control, Modelling and Simulation (IJCCMS) Vol.2, No.2, June 2013
4. SIMULATION RESULTS The modelled system is simulated and the required process times for each algorithm are listed and shown in TABLE 1 and Fig. 2, respectively. TABLE 1: The time needed to compute the estimation as a function of increasing number of sensors for each method.
number 1 of sensors group 0.11 sensor sequential 0.15 sensor invers 0.2 covariance track to track 0.1 fusion
15
30
45
60
75
90
105
120
0.31 0.75
1.61
3.01
5
7.64
1.41 2.82
4.25
5.68 7.21
8.79
0.37 0.56
0.74
0.92
1.1
1.14 2.71
4.08
5.42 6.76
150
210
300
11.2 5 10.4 5
15.4 30.89 9
75
188
1.29
1.46
1.66
8.12
9.45
10.8 13.51 18.9
12.1 15.59 23.15
35.6 3
2.03 2.74 3.86 26.9 4
As is seen, in group-sensor method, although a few number of sensors do not need a long time for data fusion, but, as the number of sensors increases, the computations become more timeconsuming. Sequential sensor method and track-to-track fusion methods need somewhat the same time. However, these two methods, for the number of sensors greater than 100, are faster than the group-sensor method. For inverse covariance form the story is somewhat different. For the number of sensors less than 20, the computational load is approximately the same as the other methods. However, as the number of sensors increases, the process time is too few in comparison with the others. This is in agreement with what was said in previous sections.
Figure 2. The required time to compute the estimations. It shows the computational complexity of each algorithm.
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International Journal of Chaos, Control, Modelling and Simulation (IJCCMS) Vol.2, No.2, June 2013
5. CONCLUSION In order to investigate Multi-sensor single target tracking algorithms base on Kalman filter, four methods were simulated and compared. For this purpose, the time needed to compute the estimation as a function of increasing number of sensors for each method is calculated. In summary, for the number of sensors less than 20, the group-sensor method has the least computational complexity and the inverse covariance method has the second priority. For greater number of sensors, inverse covariance method strongly needs the least process time. Nevertheless, for hardware implementation, type of the hardware (FPGAs, DSPs or so on) and the mathematical relations of algorithms should be taken into account to choose the best method.
REFERENCES [1] Hugh Durrant-Whyte, Multi Sensor Data Fusion. Australian Centre for Field Robotics, The University of Sydney NSW 2006. [2] Y. Bar-Shalom and X.-R. Li, Multitarget-Multisensor Tracking: Principles and Techniques. ISBN 0-9648312-0-1, 1995.B.D.O. Anderson and J.B. Moore. Optimal Filtering. Prentice Hall, 1979. [3] Y. Bar-Shalom and T. E. Fortmann, Tracking and Data Association. Academic Press, Orlando, FL, 1988. [4] Jitendra R. Raol, Multi-Sensor Data Fusion with MATLAB, CRC Press, Taylor & Francis Group, 2010. [5] Zou, Wei, and Wei Sun., “ A multi-dimensional data association algorithm for multi-sensor fusion,” Intelligent Science and Intelligent Data Engineering. Springer Berlin Heidelberg, 2013. 280-288. [6] Cho, Taehwan, Changho Lee, and Sangbang Choi, “Multi-sensor fusion with interacting multiple model filter for improved aircraft position accuracy,” Sensors 13.4 (2013): 4122-4137. [7] Brookner, Eli. Tracking and Kalman filtering made easy. New York: Wiley, 1998. [8] Y. Bar-Shalom, “On the track to track correlation problem,”. IEEE Trans. Automatic Control, 25(8):802–807, 1981.
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COMPUTATIONAL COMPLEXITY COMPARISON OF MULTI-SENSOR SINGLE TARGET DATA FUSION METHODS BY MATLAB
Target tracking using observations from multiple sensors can achieve better estimation performance than a single sensor. The most famous est...
COMPUTATIONAL COMPLEXITY COMPARISON OF MULTI-SENSOR SINGLE TARGET DATA FUSION METHODS BY MATLAB
Target tracking using observations from multiple sensors can achieve better estimation performance than a single sensor. The most famous est...
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# Feedback
We should look at feedback. Most explanations use something like a thermostat or a steam governor to illustrate this. The output is measured and its value affects the input to keep the output within a particular range. Room is too warm so heat is turned down, then room is too cool so heat is turned up. The human body is filled with feedback loops blood sugar is too high so insulin is excreted then blood sugar is too low so insulin is cut off. With simple loops like this it is easy to follow the logic with your finger around the loop.
But as soon as you add a bunch of other feedback loops that share some of their inputs and outputs, overlapping loops, then the logic cannot be easily seen by following the loops with a finger. Overlapping loops can maintain a more or less stable state, but it is usually not the state that any of the individual loops would settle on.
These sorts of systems act a lot like iterative equations. Take a simple equation, x=2+squareroot(x). If we start with x=1 and put it in the equation then we get a new x, 2+squareroot(1)=3. If we take our new x=3 and put it in the equation then we get a new x, 2+squareroot(3)=3.73. If we take our new x=3.73 etc. we get a never ending series that homes in on 4. (1, 3 ,3.73 , 3.93 ,3.98 ,3.99) We can imagine this happening very quickly in a physical system. Equations like this generate series that either find a stable end point, or they oscillate between two values, or they run away to zero or infinity. Again this is a very simple example with one value, x, but imagine how difficult it would be to try to follow a large set of such equations with several overlapping unknowns.
So now look at the thalamus-cortex-thalamus loops. Every small place on the cortex has axons running to a particular small place in the thalamus, and thalamus axons run back to the same place on the cortex that sent axons to it. We have a billion or so loops. In the cortex each small place has loops with all its neighbouring small places. The same is true in the thalamus. Both maps also have some loops with places more distant then their immediate neighbours. This is what I have called MPOFBL, massively parallel over-lapping feedback loops.
How would a MPOFBL system act? It would be hard to say. One interesting type of behaviour is possible depending on the architecture of the loops. The system may oscillate and quickly stabilize on one particular state of activity pattern for the neurons. This state would be the best compromise of all the constraints of the architecture of the loops and the nature of the input (sensory and other). It might be thought of as the best-fit scenario for a model of the world.
Here is the start of the PDP Primer by George Hollick. It is from about the time that the power of parallel processing started to be investigated as an important option. He is talking about an architecture that is similar but not identical to the thalamus-cortex-thalamus loops described above.
In 1986, James L. McClelland, David E. Rumelhart and the PDP Research Group published a volume which was to become a seminal work in the field of cognitive psychology.
So just what is Parallel Distributed Processing (hereafter referred to as PDP) all about? Quite simply, proponents of PDP assert that the brain is NOT a computer, not a serial one anyway. In essence, thought is a parallel process, a network of multiple, graded constraints being considered simultaneously. Thought is not a single path of constraints being considered one at a time, as in conventional cognitive models. Moreover, structural differences in the network of constraints are important to the implementation of the thought process and can lead to qualitative differences in the final result. This is in direct opposition to previous cognitive theories which assert that structure has no effect on the outcome of the thought process.
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# When a light bulb is connected to a 4.5V battery, a current of 0.16A passes through the bulb filament. Calculate the resistance of the filament.
jeew-m | College Teacher | (Level 1) Educator Emeritus
Posted on
According to Ohm's law the current (I) of a circuit , Voltage (V) and Resistance (R) is combined as follows.
`V = IR`
According our question;
`V = 4.5`
`I = 0.16`
`V = IR`
`R = V/I`
`R = 4.5/0.16`
`R = 28.125`
So the resistance of the filament is 28.125 ohms
Sources:
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# lec2 - Least-Squares Estimation Recall that the projection...
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Unformatted text preview: Least-Squares Estimation: Recall that the projection of y onto C ( X ), the set of all vectors of the form Xb for b ∈ R k +1 , yields the closest point in C ( X ) to y . That is, p ( y | C ( X )) yields the minimizer of Q ( β ) = k y- X β k 2 (the least squares criterion) This leads to the estimator ˆ β given by the solution of X T X β = X T y (the normal equations) or ˆ β = ( X T X )- 1 X T y . All of this has already been established back when we studied projections (see pp. 30–31). Alternatively, we could use calculus: To find a stationary point (maximum, minimum, or saddle point) of Q ( β ), we set the partial derivative of Q ( β ) equal to zero and solve: ∂ ∂ β Q ( β ) = ∂ ∂ β ( y- X β ) T ( y- X β ) = ∂ ∂ β ( y T y- 2 y T X β + β T ( X T X ) β ) =- 2 X T y + 2 X T X β Here we’ve used the vector differentiation formulas ∂ ∂ z c T z = c and ∂ ∂ z z T Az = 2 Az (see § 2.14 of our text). Setting this result equal to zero, we obtain the normal equations, which has solution ˆ β = ( X T X )- 1 X T y . That this is a minimum rather than a max, or saddle point can be verified by checking the second derivative matrix of Q ( β ): ∂ 2 Q ( β ) ∂ β = 2 X T X which is positive definite (result 7, p. 54), therefore ˆ β is a minimum. 101 Example — Simple Linear Regression Consider the case k = 1: y i = β + β 1 x i + e i , i = 1 , . . . , n where e 1 , . . . , e n are i.i.d. each with mean 0 and variance σ 2 . Then the model equation becomes y 1 y 2 . . . y n = 1 x 1 1 x 2 . . . . . . 1 x n | {z } = X β β 1 ¶ | {z } = β + e 1 e 2 . . . e n . It follows that X T X = n ∑ i x i ∑ i x i ∑ i x 2 i ¶ , X T y = ∑ i y i ∑ i x i y i ¶ ( X T X )- 1 = 1 n ∑ i x 2 i- ( ∑ i x i ) 2 ∑ i x 2 i- ∑ i x i- ∑ i x i n ¶ . Therefore, ˆ β = ( X T X )- 1 X T y yields ˆ β = ˆ β ˆ β 1 ¶ = 1 n ∑ i x 2 i- ( ∑ i x i ) 2 ( ∑ i x 2 i )( ∑ i y i )- ( ∑ i x i )( ∑ i x i y i )- ( ∑ i x i )( ∑ i y i ) + n ∑ i x i y i ¶ . After a bit of algebra, these estimators simplify to ˆ β 1 = ∑ i ( x i- ¯ x )( y i- ¯ y ) ∑ i ( x i- ¯ x ) 2 = S xy S xx and ˆ β = ¯ y- ˆ β 1 ¯ x 102 In the case that X is of full rank, ˆ β and ˆ μ are given by ˆ β = ( X T X )- 1 X T y , ˆ μ = X ˆ β = X ( X T X )- 1 X T y = P C ( X ) y . • Notice that both ˆ β and ˆ μ are linear functions of y . That is, in each case the estimator is given by some matrix times y . Note also that ˆ β = ( X T X )- 1 X T y = ( X T X )- 1 X T ( X β + e ) = β + ( X T X )- 1 X T e . From this representation several important properties of the least squares estimator ˆ β follow easily: 1. (unbiasedness): E( ˆ β ) = E( β + ( X T X )- 1 X T e ) = β + ( X T X )- 1 X T E( e ) | {z } = = β ....
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## This note was uploaded on 11/13/2011 for the course STAT 8260 taught by Professor Hall during the Summer '10 term at UGA.
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# forum.alglib.net
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### Forum rules
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Post subject: Doubt in using "lsfitcreatefg"instead of "lsfitcreatef"Posted: Tue May 15, 2018 2:03 pm
Joined: Tue May 08, 2018 3:15 pm
Posts: 3
Good day all!
I use C# version of ALGLIB 3.13.0 (source code generated 2017-12-29).
My experimental data are approximated by the sum of the Gaussians (1 - 50) and the background line.
For approximation I use lsfitcreatef method, which works nice for me at least for ~ 30 Gaussians in data. Example of aproximation for 1 Gaussian + line is in the attachement.
To increase the speed and reliability of convergence, I tried to use lsfitcreatefg method but failed.
For clarity, here is my code:
Code:
static void Main(string[] args)
{
string filepath = @"D:\data.txt";
StreamReader reader = new StreamReader(filepath);
double[] spectr = new double[45];
for (int counter = 0; counter < spectr.Length; counter++)
{
spectr[counter] = Double.Parse(reader.ReadLine()); // y data
}
reader.Close();
reader.Dispose();
int start_channel = 2107;
double[,] x = new double[spectr.Length, 1];
for (int i = 0; i < x.Length; i++)
{
x[i, 0] = (double)(i + start_channel); // x data
}
double[] s = new double[] { 10000, 1000, 1000, 1 }; // scale
double[] c = new double[] { 18000, 2127, 3747, -3.71 }; // starting values
double[] bndl = new double[] { 0, 0, Double.NegativeInfinity, Double.NegativeInfinity };
double[] bndu = new double[] { +20000, +2150, Double.PositiveInfinity, Double.PositiveInfinity };
double epsx = 0.0000001;
int maxits = 0;
int info;
alglib.lsfitstate state;
alglib.lsfitreport rep;
double diffstep = 0.000001;
alglib.lsfitcreatef(x, spectr, c, diffstep, out state);
alglib.lsfitsetbc(state, bndl, bndu);
alglib.lsfitsetcond(state, epsx, maxits);
alglib.lsfitsetscale(state, s);
alglib.lsfitfit(state, function_func, null, null);
alglib.lsfitresults(state, out info, out c, out rep);
System.Console.WriteLine("{0}", info);
System.Console.WriteLine("{0}", alglib.ap.format(c, 6));
/* alglib.lsfitcreatefg(x, spectr, c, true, out state);
alglib.lsfitsetbc(state, bndl, bndu);
alglib.lsfitsetcond(state, epsx, maxits);
alglib.lsfitsetscale(state, s);
alglib.lsfitfit(state, function_func, function_grad, null, null);
alglib.lsfitresults(state, out info, out c, out rep);
System.Console.WriteLine("{0}", info);
System.Console.WriteLine("{0}", alglib.ap.format(c, 6)); */
}
public static void function_func(double[] c, double[] x, ref double func, object obj)
{
func = c[0] * 1 / (Math.Pow(2 * Math.PI, 0.5) * Math.Pow(1.72, 0.5)) * Math.Exp(-(c[1] - x[0]) * (c[1] - x[0]) / (2 * 1.72 * 1.72)) + c[2] + c[3] * x[0] ;
}
public static void function_grad(double[] c, double[] x, ref double func, double[] grad, object obj)
{
grad[0] = 1 / (Math.Pow(2 * Math.PI, 0.5) * Math.Pow(1.72, 0.5)) * Math.Exp(-(c[1] - x[0]) * (c[1] - x[0]) / (2 * 1.72 * 1.72)) + c[2] + c[3] * x[0];
grad[1] = (c[0] * 1 / (Math.Pow(2 * Math.PI, 0.5) * Math.Pow(1.72, 0.5)) * Math.Exp(-(c[1] - x[0]) * (c[1] - x[0]) / (2 * 1.72 * 1.72)) + c[2] + c[3] * x[0]) *
((x[0] - c[1]) / (1.72 * 1.72));
grad[2] = 1;
grad[3] = x[0];
}
For lsfitcreatef method all is fine. Parameters values change from c=[18000, 2127, 3747, -3.71] to c = [10564, 2127, 624, -0.248]. Very good result. Picture with this result is in the attachement.
If I remove the commenting for lsfitcreatefg method in the my code and comment lsfitcreatef method instead the result is c=[17999.99, 2126.99, 3747, -3.71].
The resulting values of the parameters remained practically unchanged.
I checked the accuracy of the computation of gradient functions - all right.
Could you explain what do I do wrong?
Regards
Attachments: Exp_vs_Approx.PNG [ 38.07 KiB | Viewed 1241 times ]
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Post subject: Re: Doubt in using "lsfitcreatefg"instead of "lsfitcreatef"Posted: Wed May 16, 2018 10:46 am
Site Admin
Joined: Fri May 07, 2010 7:06 am
Posts: 849
Hi!
I noticed two errors:
a) in function_grad() method you should set BOTH grad and func by-ref parameters. Solver expects that you return both function and its gradient upon return from this method.
b) your gradient is calculated incorrectly. It should be dF/dC, and even grad[0] is clearly wrong - it has trailing " + c[2] + c[3] * x[0]" term which should be truncated when you calculate derivative with respect to c[0].
Does it work if you fit ix?
Top
Post subject: Re: Doubt in using "lsfitcreatefg"instead of "lsfitcreatef"Posted: Thu May 24, 2018 11:44 am
Joined: Tue May 08, 2018 3:15 pm
Posts: 3
Sergey, you were absolutely right. I made stupid mistakes in the calculation of the derivatives, to my shame and sorrow.
With the help of correctly calculated derivatives, everything is worked correctly and 10 times (!) faster (~10 seconds with lsfitcreatef vs ~ 1 second with lsfitcreatefg ).
Thank you very much for the help! I'm sorry that I disturbed you with such an unprepared question.
Sergey, as far as I know, iterative algorithms do not give in to parallelization, or do it very badly. My test sample(1496 data points with 30 Gaussians) is processed in 1 second on Core i7-7700. In this case, only 1 of 8 processor threads is used. Will ALGLIB Commercial Edition process data faster?
May you tell me how to calculate accuracy of estimated model parameters or show the direction of search such information?
In any case, thanks for a good mathematical and software tool!
Top
Post subject: Re: Doubt in using "lsfitcreatefg"instead of "lsfitcreatef"Posted: Mon May 28, 2018 9:31 am
Site Admin
Joined: Fri May 07, 2010 7:06 am
Posts: 849
Hi, Dmitrii!
I do not think that parallelism will be of any significant help for you. 30 gaussians = somewhat like 150 parameters (in your case). And 150 parameters is not enough to get good speed-up from parallelism.
You may get some speed-up from switching to the native computational core shipped with Commercial Edition (exactly same C# interface). But it is hard to tell how large it will be... I think that 2x at best, but it may be closer to 1.0 - you will spend some additional time communicating between managed and native memory spaces, and this performance penalty can be significant for lightweight problems like yours.
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