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https://www.distancesto.com/fuel-cost/it/courmayeur-ao-to-6000-koper/history/337369.html
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# INR25.07 Total Cost of Fuel from Courmayeur AO to 6000 Koper
Your trip to 6000 Koper will consume a total of 10.03 gallons of fuel.
Trip start from Courmayeur AO, IT and ends at 6000 Koper, SI.
Trip (401.1 mi) Courmayeur AO » 6000 Koper
The map above shows you the route which was used to calculate fuel cost and consumption.
### Fuel Calculations Summary
Fuel calculations start from Courmayeur AO, Italy and end at 6000 Koper, Slovenia.
Fuel is costing you INR2.50 per gallon and your vehicle is consuming 40 MPG. The formula can be changed here.
The driving distance from Courmayeur AO to 6000 Koper plays a major role in the cost of your trip due to the amount of fuel that is being consumed. If you need to analyze the details of the distance for this trip, you may do so by viewing the distance from Courmayeur AO to 6000 Koper.
Or maybe you'd like to see why certain roads were chosen for the route. You can do so by zooming in on the map and choosing different views. Take a look now by viewing the road map from Courmayeur AO to 6000 Koper.
Of course, what good is it knowing the cost of the trip and seeing how to get there if you don't have exact directions? Well it is possible to get exact driving directions from Courmayeur AO to 6000 Koper.
Did you also know that how elevated the land is can have an impact on fuel consumption and cost? Well, if areas on the way to 6000 Koper are highly elevated, your vehicle may have to consume more gas because the engine would need to work harder to make it up there. In some cases, certain vehicles may not even be able to climb up the land. To find out, see route elevation from Courmayeur AO to 6000 Koper.
Travel time is of the essence when it comes to traveling which is why calculating the travel time is of the utmost importance. See the travel time from Courmayeur AO to 6000 Koper.
Speaking of travel time, a flight to 6000 Koper takes up a lot less. How much less? Flight time from Courmayeur AO to 6000 Koper.
Cost is of course why we are here... so is it worth flying? Well this depends on how far your trip is. Planes get to where they need to go faster due to the speed and shorter distance that they travel. They travel shorter distances due to their ability to fly straight to their destination rather than having to worry about roads and obstacles that are in a motor vehicle's way. You can see for yourself the flight route on a map by viewing the flight distance from Courmayeur AO to 6000 Koper.
*The cost above should be taken as an ESTIMATE due to factors which affect fuel consumption and cost of fuel.
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Q:
# How do I subtract percentages?
A:
To subtract a percentage, one starts by converting the percentage into its decimal equivalent and subtracting the result from one. Multiplying the answer by the original figure gives the value after subtracting the percentage. For instance, to subtract a tax of 27 percent from base pay, one coverts the percentage to 0.27 or (27/100). Basic pay x (1 - 0.27) gives the pay after the tax deduction.
Know More
## Keep Learning
Credit: Bart Broek E+ Getty Images
Alternatively, an employee may first calculate the actual tax (basic pay x 27/100) and then subtract it from the basic pay. If a buyer pays \$7.61 for an item that was priced at \$6.95, the applicable percentage sales tax (x) is calculated as follows: x = (7.61- 6.95)/ 6.95; x = 9.5 percent.
Mathematically, a percentage of any value means the value multiplied by the percentage in its decimal form. For example, 35 percent of 80 means 0.35 x 80. It’s also possible to calculate a percentage rate of a value if the actual fraction is known. For example, if x represents a percent of 20 that equals 30, the percent is calculated as follows: x = 100 x 30/20. One multiplies the rate by 100 to convert from the decimal to percentage form.
Sources:
## Related Questions
• A:
To subtract fractions, make sure the bottom numbers (denominators) are the same, then subtract the top numbers (numerators) from one another, retaining the denominator and simplifying if required. This process can be done by hand or by using a calculator.
Filed Under:
• A:
Subtracting decimals involves writing the two decimals one below the other and making sure that the decimal points line up before subtracting in the normal manner. However, the larger decimal should be placed above the smaller one. Similarly, if the two decimals do not have the same length, then add zeroes before subtracting them.
Filed Under:
• A:
Two-fifths is equivalent to 40 percent. Dividing the numerator, 2, by the denominator, 5, yields a decimal value of 0.40. Decimal values can be converted to percentages by multiplying by 100, which means that 0.40 is equal to 40 percent.
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http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=4117
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Welcome to ZOJ
Problem Sets Information Select Problem Runs Ranklist
ZOJ Problem Set - 4117
Time Limit: 1 Second Memory Limit: 65536 KB
BaoBao is a good student who loves reading, but compared with his huge bookshelf containing lots and lots of books, his reading desk, which can only hold at most $k$ books, is surprisingly small.
Today BaoBao decides to read some books for $n$ minutes by the desk. According to his reading plan, during the $i$-th minute, he is scheduled to read book $a_i$. The reading desk is initially empty and all the books are initially on the shelf. If the book BaoBao decides to read is not on the desk, BaoBao will have to fetch it from the shelf. Also, if the desk is full and BaoBao has to fetch another book from the shelf, he will have to put one book back from the desk to the shelf before fetching the new book.
Tired of deciding which book to put back, BaoBao searches the Internet and discovers an algorithm called the Least Recently Used (LRU) algorithm. According to the algorithm, when BaoBao has to put a book back from the desk to the shelf, he should put back the least recently read book.
For example, let's consider the reading plan {4, 3, 4, 2, 3, 1, 4} and assume that the capacity of the desk is 3. The following table explains what BaoBao should do according to the LRU algorithm. Note that in the following table, we use a pair of integer $(a, b)$ to represent a book, where $a$ is the index of the book, and $b$ is the last time when this book is read.
Minute Books on the Desk
Before This Minute
BaoBao's Action
1 {} Fetch book 4 from the shelf
2 {(4, 1)} Fetch book 3 from the shelf
3 {(4, 1), (3, 2)} Do nothing as book 4 is already on the desk
4 {(4, 3), (3, 2)} Fetch book 2 from the shelf
5 {(4, 3), (3, 2), (2, 4)} Do nothing as book 3 is already on the desk
6 {(4, 3), (3, 5), (2, 4)} Put book 4 back to the shelf as its the least recently read book,
and fetch book 1 from the shelf
7 {(3, 5), (2, 4), (1, 6)} Put book 2 back to the shelf as its the least recently read book,
and fetch book 4 from the shelf
Given the reading plan, what's the number of times BaoBao fetches a book from the shelf if the value of $k$ (the capacity of the desk) ranges from 1 to $n$ (both inclusive)?
#### Input
There are multiple test cases. The first line of the input contains an integer $T$, indicating the number of test cases. For each test case:
The first line contains an integer $n$ ($1 \le n \le 10^5$), indicating the length of the reading plan.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le n$), indicating the indices of the books to read.
It's guaranteed that the sum of $n$ of all test cases will not exceed $10^6$.
#### Output
For each test case output one line containing $n$ integers $f_1, f_2, \dots, f_n$ separated by a space, where $f_i$ indicates the number of times BaoBao fetches a book from the shelf when the capacity of the desk is $i$.
Please, DO NOT output extra spaces at the end of each line, or your solution may be considered incorrect!
#### Sample Input
1
7
4 3 4 2 3 1 4
#### Sample Output
7 6 5 4 4 4 4
Author: WANG, Yucheng
Source: The 10th Shandong Provincial Collegiate Programming Contest
Submit Status
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CFD Online Discussion Forums (https://www.cfd-online.com/Forums/)
- Main CFD Forum (https://www.cfd-online.com/Forums/main/)
- - SOR method (https://www.cfd-online.com/Forums/main/972-sor-method.html)
Yogesh Talekar June 30, 1999 23:52
SOR method
Can anybody tell me what is SOR method? IS it implicit or explicit? Is it diificult to programme than Crank-Nicolson method? Is there any book for this?
John C. Chien July 1, 1999 00:15
Re: SOR method
(1). Most numerical analysis books should cover SOR method. (2). There is no other method simpler than SOR method. (3). None of the methods you mentioned are difficult to program. These are very standard methods. They are all in the book. (4). Pick a book written for engineers and scientists. ( I had taken real analysis once and it is really hard to understand. Stay away from the books written by mathematicians for now.)
Yogesh Talekar July 1, 1999 00:59
What is long form of SOR method
Can you tell me what is long form of SOR
sudharsan natteri July 1, 1999 01:52
Re: What is long form of SOR method
successive over relaxation
Patrick Godon July 1, 1999 09:15
Re: SOR method
A good place to start is the book:
Numerical Recipes, by Press, Flannery, Teukolsky and Vetterling, 1989, Cambridge: Cambridge University Press. This book should be for sure in the Physics Library (in your Department I guess).
There you can find basic information, and it is written for Physicsits and Engineers (no mathematicians, John Chien).
Patrick
Phil Greenfield July 22, 1999 18:17
Re: SOR method
Successive Over-Relaxation : SOR (& SUR) Specifically, can dramatically accelerate convergance of the Gauss-Seidel iterative solution method using an extrapolation technique. I believe the matrix must be either diagonally dominant or symmetric positive definte. w>1; over-relaxation w=1; Gauss-Siedel 0<w<1: under-relaxation
All times are GMT -4. The time now is 08:04.
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# How Many Grams In A Tea Bag?
Depending on the variety of tea, a teabag that is designed to fit a single cup of tea typically contains between two and three grams of tea. Nevertheless, this quantity may be variable for a variety of brands. A double-cup teapot can produce tea that is quite bitter, in contrast to the flavorful cup of tea that can be produced using a single-cup teabag.
And how many cups of tea are included in each tea bag? The average weight of a regular tea bag is between 1.5 and 2 grams. That quantity is sufficient to brew an 8-ounce cup of steaming tea.
## How many grams are in an ounce of tea?
• There are roughly 28.35 grams in one ounce.
• Tea is normally sold in quantities of 1, 2, 4, or 8 ounces, however it can be purchased in greater quantities by the pound.
• There are sixteen ounces in a pound.
• A single ounce is a unit of measurement that is approximately equivalent to a big sample and provides enough material for you to boil and serve the tea eleven to eighteen times.
• 57 grams is equivalent to 2 ounces, which is an eighth of a pound.
## How much does a teaspoon of tea weigh?
• A teaspoon is a common unit of measurement that is equivalent to 1/6 of an ounce, or around 5 milliliters.
• It was common knowledge among certain individuals that 1 teaspoon equals 4 grams of tea leaves.
• Is that what you mean?
• After putting it to the test with a kitchen scale, I discovered that one teaspoon is equivalent to 2.5 grams.
• So would you say that a teaspoon of loose leaf tea weighs around 2.5 grams?
• Absolute no.
See also: How Many Cups Of Green Tea Should I Drink A Day?
Why?
## How many ML is a cup of tea?
• When filled to a level that is 1.5 centimeters below the rim, the capacity of the typical traditional teacup is 150 milliliters.
• A cup is equal to 250 milliliters when referring to culinary measurements (236 ml, or 8 ounces in US).
• When it is completely full, that is the volume that a cup considered to be ″standard″ can contain.
• Nevertheless, relatively few people really pour their tea in that manner into their cups.
• The actual tea cups themselves come in a wide variety of forms and sizes.
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Can someone help me solve this problem ?
label Mathematics
account_circle Unassigned
schedule 1 Day
account_balance_wallet \$5
Solve the given linear equation.
9 − x = 23 + x
x =
Oct 1st, 2015
Thank you for the opportunity to help you with your question!
9 - x = 23 + x
Put like terms together. Take 9 to the right side of the equal sign and the x on the right side to the left side of the equal sign. when numbers cross the equal sign, they change the mathematical sign before them. therefore:
-x - x = 23 - 9
-2x = 12
diving both side by (-2) gives
x = 12/-2
Hence x = -6
Please let me know if you need any clarification. I'm always happy to answer your questions.
Oct 1st, 2015
...
Oct 1st, 2015
...
Oct 1st, 2015
Oct 21st, 2017
check_circle
Mark as Final Answer
check_circle
Unmark as Final Answer
check_circle
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# Unit 9 - Real Numbers
## Lesson 1 - Introduction to Real Numbers
Topic 1 - Variables and Expressions
Topic 2 - Integers
Topic 3 - Rational and Real Numbers
## Lesson 2 - Operations with Real Numbers
Topic 2 - Adding Real Numbers
Topic 3 - Subtracting Real Numbers
Topic 4 - Multiplying and Dividing Real Numbers
## Lesson 3 - Properties of Real Numbers
Topic 1 - Associative, Commutative, and Distributive Properties
## Lesson 4 - Simplifying Expressions
Topic 1 - Order of Operations
# Unit 10 - Solving Equations and Inequalities
## Lesson 1 - Solving Equations
Topic 1 - Solving One-Step Equations Using Properties of Equality
Topic 2 - Solving Multi-Step Equations
Topic 3 - Special Cases and Applications
Topic 4 - Formulas
## Lesson 2 - Solving Inequalities
Topic 1 - Solving One-Step Inequalities
Topic 2 - Multi-Step Inequalities
## Lesson 3 - Compound Inequalities and Absolute Value
Topic 1 - Compound Inequalities
Topic 2 - Equations and Inequalities and Absolute Value
# Unit 11 - Exponents and Polynomials
## Lesson 1 - Integer Exponents
Topic 1 - Exponential Notation
Topic 2 - Simplify by Using the Product, Quotient, and Power Rules
Topic 3 - Products and Quotients Raised to Powers
Topic 4 - Scientific Notation
## Lesson 2 - Polynomials with Single Variables
Topic 1 - Introduction to Single Variable Polynomials
Topic 2 - Adding and Subtracting Polynomials
Topic 3 - Multiplying Polynomials
Topic 4 - Multiplying Special Cases
Topic 5 - Dividing by a Monomial
Topic 6 - Dividing by Binomials and Polynomials
## Lesson 3 - Polynomials with Several Variables
Topic 1 - Simplifying and Evaluating Polynomials with More than One Term
Topic 2 - Operations with Polynomials
# Unit 12 - Factoring
## Lesson 1 - Introduction to Factoring
Topic 1 - Greatest Common Factor
## Lesson 2 - Factoring Polynomials
Topic 1 - Factoring Trinomials
Topic 2 - Factoring: Special Cases
Topic 3 - Special Cases: Cubes
## Lesson 3 - Solving Quadratic Equations
Topic 1 - Solve Quadratic Equations by Factoring
# Unit 13 - Graphing
## Lesson 1 - Graphs and Applications
Topic 1 - The Coordinate Plane
Topic 2 - Graphing Linear Equations
## Lesson 2 - Slope and Writing the Equation of a Line
Topic 1 - Finding the Slope of a Line
Topic 2 - Writing the Equation of a Line
Topic 3 - Parallel and Perpendicular Lines
Topic 4 - Graphing Linear Inequalities
# Unit 14 - Systems of Equations and Inequalities
## Lesson 1 - Graphing Systems of Equations and Inequalities
Topic 1 - Graphing Systems of Linear Equations
Topic 2 - Graphing Systems of Inequalities
## Lesson 2 - Algebraic Methods to Solve Systems of Equations
Topic 1 - The Substitution Method
Topic 2 - The Elimination Method
## Lesson 3 - Systems of Equations in Three or More Variables
Topic 1 - Solving Systems of Three Variables
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Vous êtes sur la page 1sur 14
# 1
2
Single-factor measures
Output / (Single Input)
All-factors measure
Output / (Total Inputs)
Productivity =
Outputs
Inputs
Productivity
3
Single-factor Output Output Output Output
measures Labor Machine Capital Energy
All-factors Output
measure All inputs
Measures of Productivity
4
Single-factor Output Output Output Output
measures Labor Machine Capital Energy
If we produce only one product, the numerator can be either
the total units of the product or the total \$ value of the
product. If we produce several products, the numerator is
The denominator can be the units of input or the total \$
value of input.
Single Factor
5
10,000 Units Produced
Sold for \$10/unit
500 labor hours
Labor rate: \$9/hr
What is the
labor productivity?
Example: Single Factor Productivity
6
10,000 units / 500hrs = 20 units/hr
(10,000 units * \$10/unit) / 500hrs = \$200/hr
10,000 units / (500hrs * \$9/hr) = 2.2 unit/\$
(10,000 units * \$10/unit) / (500hrs * \$9/hr) = 22.22
The last one is unit-less
Example: Labor Productivity
7
Labor Productivity
Quantity (or value) of output / labor hrs
Quantity (or value) of output / shift
Machine Productivity
Quantity (or value) of output / machine hrs
Energy Productivity
Quantity (or value of output) / kwh
Capital Productivity
Quantity (or value) of output / value of input
Some Single Factor Measurements
8
All-factors Goods or Services produced
measure All inputs used to produce them
If we produce only one product, the numerator can be either
the total units of product or total \$ value of the product.
If we produce several products, the numerator is the total
Usually, the numerator is the total \$ value of all outputs.
The denominator is total \$ value of all inputs.
All Factors
9
10,000 Units Produced
Sold for \$10/unit
500 labor hours
Labor rate: \$9/hr
Cost of raw material: \$30,000
Example
10
Output
(10,000 units) * (\$10)
(500)*(\$9) + (\$30,000) + (\$15,500)
AFP = 2.0
Example : All-Factor Productivity
AFP =
AFP =
11
Training Methods
Technology Management
What are the factors that affect productivity?
12
1. A company that makes shopping carts for supermarkets recently purchased
new equipment, which reduced the labor content needed to produce the
carts. Information concerning the old system (before adding the new
equipment) and the new system (after adding the new machines) includes:
Old System New System
Output/hr 80 84
Workers 5 4
Wage \$/hr 10 10
Machine \$/hr 40 50
a) Compute labor productivity for both the Old System and the New System.
b) Compute AFP productivity for both the Old System and the New System.
c) Suppose production with old equipment was 30 units of cart A at a price of
\$100 per cart, and 50 units of cart B at a price of \$120. Also suppose that
production with new equipment is 50 units of cart A, at a price of \$100 per
cart, and 30 units of cart B at a price of \$120. Compare all-factor productivity
for the old and the new systems.
Assignment 1.1.. Due at the beginning of the next class
13
Assignment 1.2.. Due at the beginning of the next class
2. A company has introduced a process improvement that reduces the
processing time for each unit and increases output by 25% with less
Under the old process, five workers could produce 60 units per hour.
Labor costs are \$12/hour, and material input was \$16/unit.
For the new process, material input is now \$10/unit and overhead is
charged at 1.6 times direct labor cost. Finished units sell for \$31 each.
a) Compute single factor productivity of labor in the old system.
(Compute it in four possible ways.)
b) Compute all factor productivity for both old and new systems.
Factor Old System New System
Output 60 60(1.25) = 75
# of workers 5 6
Worker cost \$12/hr \$12/hr
Material \$16/unit \$10/unit
Price 31 31
14
Assignment 1.3 For your own practice
3. A milk factory seeks advice from an external consulting company concerning its
business and production processes. The final consulting report describes
several steps to increase productivity including implementation of cutting-edge
processing techniques through more powerful filtering systems.
a) Calculate the labor productivity for the existing as well as the proposed system.
b) Find the All-Factor Productivity for both systems.
c) Assume that current processing includes 700 gallons of Grade-A milk sold at
\$2.40/gallon and 300 gallons of Grade-B milk at \$1.90/gallon. Furthermore,
assume that under the proposed system, processing will include 600 gallons of
Grade-A milk at \$2.40/gallon and 400 gallons of Grade-B milk at \$1.90/gallon.
Compare all-factor productivity for both the existing and the new system.
Existing System Proposed System
Workers 12 9
Milk Output/hour 1,000 gallons 1,400 gallons
Wage Rate/hour \$12 \$12
Filtration Cost/hour \$120 \$170
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# cse331-week5 - CSE 331 Computer Organization and Design...
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CSE331 W05.1 KB-Fall 2008PSU CSE 331 Computer Organization and Design Fall 2008 Week 5 Course material on ANGEL: cms.psu.edu [Many thanks to Mary Jane Irwin adapted from D. Patterson slides ]
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CSE331 W05.2 KB-Fall 2008PSU Head’s Up Last week’s material Supporting procedure calls and returns; addressing modes This week’s material Assemblers, linkers and loaders - Reading assignment - PH: 2.10, A.1-A.5 Next week’s material (after Exam #1) Introduction to VHDL and basic VHDL constructs - Reading assignment – Y, Chapters 1 through 3 Reminders HW 4 is due October 3 (by 11:55pm) Quiz 3 is due (by 11:55pm)
CSE331 W05.3 KB-Fall 2008PSU Review: Addressing Modes Illustrated 1. Register addressing op rs rt rd funct Register word operand 3. Immediate addressing op rs rt operand 4. PC-relative addressing op rs rt offset Program Counter (PC) Memory branch destination instruction 5. Pseudo-direct addressing op jump address Program Counter (PC) Memory jump destination instruction || op rs rt offset 2. Base addressing base register Memory word or byte operand
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CSE331 W05.4 KB-Fall 2008PSU Review: MIPS Instructions, so far Category Instr OpC Example Meaning Arithmetic (R & I format) add add \$s1, \$s2, \$s3 \$s1 = \$s2 + \$s3 subtract sub \$s1, \$s2, \$s3 \$s1 = \$s2 - \$s3 add immediate 8 addi \$s1, \$s2, 4 \$s1 = \$s2 + 4 shift left logical sll \$s1, \$s2, 4 \$s1 = \$s2 << 4 shift right logical srl \$s1, \$s2, 4 \$s1 = \$s2 >> 4 (fill with zeros) shift right arithmetic sra \$s1, \$s2, 4 \$s1 = \$s2 >> 4 (fill with sign bit) and and \$s1, \$s2, \$s3 or or \$s1, \$s2, \$s3 \$s1 = \$s2 | \$s3 nor nor \$s1, \$s2, \$s3 \$s1 = not (\$s2 | \$s3) and immediate c andi \$s1, \$s2, ff00 or immediate d ori \$s1, \$s2, ff00 \$s1 = \$s2 | 0xff00 load upper immediate f lui \$s1, 0xffff \$s1 = 0xffff0000
CSE331 W05.5 KB-Fall 2008PSU Review: MIPS Instructions, so far Category Instr OpC Example Meaning Data transfer (I format) load word 23 lw \$s1, 100(\$s2) \$s1 = Memory(\$s2+100) store word 2b sw \$s1, 100(\$s2) Memory(\$s2+100) = \$s1 load byte 20 lb \$s1, 101(\$s2) \$s1 = Memory(\$s2+101) store byte 28 sb \$s1, 101(\$s2) Memory(\$s2+101) = \$s1 load half
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## This note was uploaded on 11/03/2008 for the course CMPEN 331 taught by Professor Bhat during the Fall '08 term at Pennsylvania State University, University Park.
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You are Here: Home
# How do you complete the square on this question? watch
1. Write y=2x^2 -5x +10 in the form
y=a(x+b)^2 + c
2. tip: put the rest of the equation into a bracket with 2 outside and try and solve from there
3. Your coefficient for your ‘a’ value will be placed outside the brackets. From here try to factorise in the form it asks it to be given in
4. Or just expand the completed square form to get and compare with to obtain the values of a,b,c
5. first step:
2(x^2-5/2x)+10
second step:
2(x^2-5/2x +5/4^2 -5/4^2) +10
third step:
2[(x-5/4)^2 -5/4^2]+10
fourth step:
2(x-5/4)^2 +13.125
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# Re: current A.6 draft
```On Fri, Nov 29, 2002 at 11:48:53AM -0500, Raul Miller wrote:
> On Fri, Nov 29, 2002 at 12:22:32AM -0500, Andrew Pimlott wrote:
> > What you did propose doesn't seem to do what you want. It says that
> > a defeat by the default option can't be weaker than another defeat.
> > It can still be stronger.
> I wanted to say that a defeat by the default option can't be weaker
> than another defeat.
> > b. A defeat (A,X) is weaker than a defeat (B,Y) if A is not
> > the default option and V(A,X) is less than V(B,Y). Also,
> > (A,X) is weaker than (B,Y) A is not the default option and if
> > V(A,X) is equal to V(B,Y) and V(X,A) is greater than V(Y,B).
You could spell it out:
A defeat (A,X) is weaker than a defeat (D,Y) if A is not the default
option and D is the default option.
A defeat (D,X) is weaker than a defeat (D,Y) if D is the default
option, and V(D,X) is less than V(D,Y).
A defeat (A,X) is weaker than a defeat (B,Y) if neither A nor B
are the default option, and V(A,X) is equal to V(B,Y) and V(X,A)
is greater than V(Y,B).
A defeat (A,X) is weaker than a defeat (B,Y) if neither A nor B are the
default option, and V(A,X) is less than V(B,Y).
A defeat (A,X) is weaker than a defeat (B,Y) if neither A nor B
are the default option, and V(A,X) is equal to V(B,Y) and V(X,A)
is greater than V(Y,B).
The catch is you should be looking at the V(...) stuff only if either
both A and B are the default option, or if neither of them are. I hate
writing these things in programming languages too :(
Maybe you could say:
The strength of defeats are determined as follows:
(1) defeats by non-default options are always weaker than defeats
by the default option; otherwise
(2) a defeat (A,X) is weaker than a defeat (B,Y) is V(A,X) < V(B,Y);
otherwise
(3) a defeat (A,X) is weaker than a defeat (B,Y) when V(X,A) > V(Y,B);
otherwise
(4) the defeats are equal
or something similar, with the idea that each rule is checked in both
directions before moving on...
> A has a 2:1 supermajority requirement, B has no special majority
> requirement, D is the default option, votes are
> 3 ABD
> 1 BDA
> 1 DBA
> A defeats B by 4:1
> B defeats D by 4:1
> D defeats A by 4:3
Actually, A defeats B by 3:2. You meant:
3 ABD
1 BDA
1 DAB
> Which means that A is not in the Schwartz set, but B and D are. So this
> is a tie between B and D, and the person with the casting vote chooses
> whether to settle for B or whether it's worth discussing this more and
> holding another election.
How interesting. So they're not as equivalent as I'd thought. Hrm.
Cheers,
aj
--
Anthony Towns <aj@humbug.org.au> <http://azure.humbug.org.au/~aj/>
I don't speak for anyone save myself. GPG signed mail preferred.
``If you don't do it now, you'll be one year older when you do.''
```
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## Welcome to the Treehouse Community
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# my if statement in else statement doesn't work
```first_name = input("What is your name? ")
print("hello,", first_name)
if first_name == "begana":
print(first_name, "is learning Python.")
elif first_name == "maximiliane":
print(first_name, "is learning Python with fellow students in the community")
else:
age = int(input("How old are you? "))
if age <= 6:
print("you should totally learn Python, {}!".format(first_name))
print("Have a great day, {}!".format(first_name))
```
after I give a condition of age, the message about age isn't printed out. can somebody point out what is my mistake??
If the first name inputted is either "begana" or "maximiliane" then the else statement for the age won't run.
If you want it to run then you need to remove the else statement and put the age variable under the elif or if statement. The else statement only works if the name is neither "begana" or "maximiliane"
Benjamin Boulter We're dealing with when the else statement does run here.
These 2 lines will run if the age is less than or equal to 6:
``` print("wow, you are {}! if you are confident with your reading already...".format(age))
print("you should totally learn Python, {}!".format(first_name))
```
if age is greater than 6, only this line will run within the else statement:
``` print("you should totally learn Python, {}!".format(first_name))
```
if you want different behavior than this, you'll have to change the indentation. Based on the video, seems like your code is working as it should.
Let me know if this helps!
Thank you !! :D but I didn't indent this line,
```print("you should totally learn Python, {}!".format(first_name))
```
Why does it print out when the age is less than 6?
Begana Choi The age doesn't matter with that print statement. Watch the short video I recorded about this.
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html, body, form { margin: 0; padding: 0; width: 100%; } #calculate { position: relative; width: 177px; height: 110px; background: transparent url(/images/alphabox/embed_functions_inside.gif) no-repeat scroll 0 0; } #i { position: relative; left: 18px; top: 44px; width: 133px; border: 0 none; outline: 0; font-size: 11px; } #eq { width: 9px; height: 10px; background: transparent; position: absolute; top: 47px; right: 18px; cursor: pointer; }
Cosh
http://functions.wolfram.com/01.20.21.4457.01
Input Form
Integrate[z^n Sinh[b z^2 + d z + e] Cosh[c z]^v, z] == 2^(-2 - v) (b^(-1 - n) E^(-(d^2/(4 b)) - e) Binomial[v, v/2] (-1 + Mod[v, 2]) (E^(2 e) Sum[2^(-n + q) (-d)^(n - q) (d + 2 b z)^(1 + q) (-((d + 2 b z)^2/b))^((1/2) (-1 - q)) Binomial[n, q] Gamma[(1 + q)/2, -((d + 2 b z)^2/(4 b))], {q, 0, n}] - E^(d^2/(2 b)) Sum[2^(-n + q) (-d)^(n - q) (d + 2 b z)^(1 + q) ((d + 2 b z)^2/b)^((1/2) (-1 - q)) Binomial[n, q] Gamma[(1 + q)/2, (d + 2 b z)^2/(4 b)], {q, 0, n}]) + b^(-1 - n) Sum[E^((d^2 - 4 b e + c^2 (-2 s + v)^2 - 2 c d (2 s + v))/ (4 b)) Binomial[v, s] ((-1)^(1 + n) E^((2 c d s)/b) Sum[2^(-n + q) (d + 2 c s - c v)^(n - q) (-d - 2 c s + c v - 2 b z)^ (1 + q) ((-d - 2 c s + c v - 2 b z)^2/b)^((1/2) (-1 - q)) Binomial[n, q] Gamma[(1 + q)/2, (-d - 2 c s + c v - 2 b z)^2/ (4 b)], {q, 0, n}] - E^((c d v)/b) (E^(-(d^2/(2 b)) + 2 e - (c^2 (-2 s + v)^2)/(2 b)) Sum[2^(-n + q) (-d - 2 c s + c v)^(n - q) (-((-d - 2 c s + c v - 2 b z)^2/b))^((1/2) (-1 - q)) (d + 2 c s - c v + 2 b z)^(1 + q) Binomial[n, q] Gamma[(1 + q)/2, -((-d - 2 c s + c v - 2 b z)^2/(4 b))], {q, 0, n}] + E^(2 e - (d - 2 c s + c v)^2/(2 b)) Sum[(-(d/2) + c s - (c v)/2)^(n - q) (-((d - 2 c s + c v + 2 b z)^ 2/b))^((1/2) (-1 - q)) (d + c (-2 s + v) + 2 b z)^(1 + q) Binomial[n, q] Gamma[(1 + q)/2, -((d - 2 c s + c v + 2 b z)^2/ (4 b))], {q, 0, n}] + (-1)^n Sum[2^(-n + q) (d - 2 c s + c v)^ (n - q) (-d + 2 c s - c v - 2 b z)^(1 + q) ((d + c (-2 s + v) + 2 b z)^2/b)^((1/2) (-1 - q)) Binomial[n, q] Gamma[(1 + q)/2, (d - 2 c s + c v + 2 b z)^2/(4 b)], {q, 0, n}])), {s, 0, Floor[(1/2) (-1 + v)]}]) /; Element[n, Integers] && n >= 0 && Element[v, Integers] && v > 0
Standard Form
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MathML Form
z n sinh ( b z 2 + d z + e ) cosh v ( c z ) z 2 - v - 2 ( - d 2 4 b - e ( v v 2 ) TagBox[RowBox[List["(", GridBox[List[List[TagBox["v", Identity]], List[TagBox[FractionBox["v", "2"], Identity]]]], ")"]], InterpretTemplate[Function[Binomial[Slot[1], Slot[2]]]]] ( v mod 2 \$CellContext`v 2 - 1 ) b - n - 1 ( 2 e q = 0 n 2 q - n ( - d ) n - q ( d + 2 b z ) q + 1 ( - ( d + 2 b z ) 2 b ) 1 2 ( - q - 1 ) ( n q ) TagBox[RowBox[List["(", GridBox[List[List[TagBox["n", Identity]], List[TagBox["q", Identity]]]], ")"]], InterpretTemplate[Function[Binomial[Slot[1], Slot[2]]]]] Γ ( q + 1 2 , - ( d + 2 b z ) 2 4 b ) - d 2 2 b q = 0 n 2 q - n ( - d ) n - q ( d + 2 b z ) q + 1 ( ( d + 2 b z ) 2 b ) 1 2 ( - q - 1 ) ( n q ) TagBox[RowBox[List["(", GridBox[List[List[TagBox["n", Identity]], List[TagBox["q", Identity]]]], ")"]], InterpretTemplate[Function[Binomial[Slot[1], Slot[2]]]]] Γ ( q + 1 2 , ( d + 2 b z ) 2 4 b ) ) + b - n - 1 s = 0 v - 1 2 d 2 - 2 c ( 2 s + v ) d + c 2 ( v - 2 s ) 2 - 4 b e 4 b ( v s ) TagBox[RowBox[List["(", GridBox[List[List[TagBox["v", Identity]], List[TagBox["s", Identity]]]], ")"]], InterpretTemplate[Function[Binomial[Slot[1], Slot[2]]]]] ( ( - 1 ) n + 1 2 c d s b q = 0 n 2 q - n ( d + 2 c s - c v ) n - q ( - d - 2 c s + c v - 2 b z ) q + 1 ( ( - d - 2 c s + c v - 2 b z ) 2 b ) 1 2 ( - q - 1 ) ( n q ) TagBox[RowBox[List["(", GridBox[List[List[TagBox["n", Identity]], List[TagBox["q", Identity]]]], ")"]], InterpretTemplate[Function[Binomial[Slot[1], Slot[2]]]]] Γ ( q + 1 2 , ( - d - 2 c s + c v - 2 b z ) 2 4 b ) - c d v b ( ( - 1 ) n q = 0 n 2 q - n ( d - 2 c s + c v ) n - q ( - d + 2 c s - c v - 2 b z ) q + 1 ( ( d + c ( v - 2 s ) + 2 b z ) 2 b ) 1 2 ( - q - 1 ) ( n q ) TagBox[RowBox[List["(", GridBox[List[List[TagBox["n", Identity]], List[TagBox["q", Identity]]]], ")"]], InterpretTemplate[Function[Binomial[Slot[1], Slot[2]]]]] Γ ( q + 1 2 , ( d - 2 c s + c v + 2 b z ) 2 4 b ) + 2 e - ( d - 2 c s + c v ) 2 2 b q = 0 n ( - d 2 + c s - c v 2 ) n - q ( - ( d - 2 c s + c v + 2 b z ) 2 b ) 1 2 ( - q - 1 ) ( d + c ( v - 2 s ) + 2 b z ) q + 1 ( n q ) TagBox[RowBox[List["(", GridBox[List[List[TagBox["n", Identity]], List[TagBox["q", Identity]]]], ")"]], InterpretTemplate[Function[Binomial[Slot[1], Slot[2]]]]] Γ ( q + 1 2 , - ( d - 2 c s + c v + 2 b z ) 2 4 b ) + - d 2 2 b + 2 e - c 2 ( v - 2 s ) 2 2 b q = 0 n 2 q - n ( - d - 2 c s + c v ) n - q ( - ( - d - 2 c s + c v - 2 b z ) 2 b ) 1 2 ( - q - 1 ) ( d + 2 c s - c v + 2 b z ) q + 1 ( n q ) TagBox[RowBox[List["(", GridBox[List[List[TagBox["n", Identity]], List[TagBox["q", Identity]]]], ")"]], InterpretTemplate[Function[Binomial[Slot[1], Slot[2]]]]] Γ ( q + 1 2 , - ( - d - 2 c s + c v - 2 b z ) 2 4 b ) ) ) ) /; n v + Condition z z n b z 2 d z e c z v 2 -1 v -2 -1 d 2 4 b -1 -1 e Binomial v v 2 -1 \$CellContext`v 2 -1 b -1 n -1 2 e q 0 n 2 q -1 n -1 d n -1 q d 2 b z q 1 -1 d 2 b z 2 b -1 1 2 -1 q -1 Binomial n q Gamma q 1 2 -1 -1 d 2 b z 2 4 b -1 -1 d 2 2 b -1 q 0 n 2 q -1 n -1 d n -1 q d 2 b z q 1 d 2 b z 2 b -1 1 2 -1 q -1 Binomial n q Gamma q 1 2 -1 d 2 b z 2 4 b -1 b -1 n -1 s 0 v -1 2 -1 d 2 -1 2 c 2 s v d c 2 v -1 2 s 2 -1 4 b e 4 b -1 Binomial v s -1 n 1 2 c d s b -1 q 0 n 2 q -1 n d 2 c s -1 c v n -1 q -1 d -1 2 c s c v -1 2 b z q 1 -1 d -1 2 c s c v -1 2 b z 2 b -1 1 2 -1 q -1 Binomial n q Gamma q 1 2 -1 -1 d -1 2 c s c v -1 2 b z 2 4 b -1 -1 c d v b -1 -1 n q 0 n 2 q -1 n d -1 2 c s c v n -1 q -1 d 2 c s -1 c v -1 2 b z q 1 d c v -1 2 s 2 b z 2 b -1 1 2 -1 q -1 Binomial n q Gamma q 1 2 -1 d -1 2 c s c v 2 b z 2 4 b -1 2 e -1 d -1 2 c s c v 2 2 b -1 q 0 n -1 d 2 -1 c s -1 c v 2 -1 n -1 q -1 d -1 2 c s c v 2 b z 2 b -1 1 2 -1 q -1 d c v -1 2 s 2 b z q 1 Binomial n q Gamma q 1 2 -1 -1 d -1 2 c s c v 2 b z 2 4 b -1 -1 d 2 2 b -1 2 e -1 c 2 v -1 2 s 2 2 b -1 q 0 n 2 q -1 n -1 d -1 2 c s c v n -1 q -1 -1 d -1 2 c s c v -1 2 b z 2 b -1 1 2 -1 q -1 d 2 c s -1 c v 2 b z q 1 Binomial n q Gamma q 1 2 -1 -1 -1 d -1 2 c s c v -1 2 b z 2 4 b -1 n v SuperPlus [/itex]
Rule Form
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# Solving polynomial equations by decomposition
I have a very little background on mathematics. I have a very basic question about solving polynomial equations. if we have $$P_n(x) = 0$$ where $$P_n(x)$$ is a polynomial of degree $$n$$.For example: $$x^5-3x^3+x^2-7x+3 = 0$$
Why we cannot just decompose the polynomial equation to a system of equations solving them separately and the solutions to the original polynomial will be the intersection of the roots of each equation: $$\begin{cases} x^5-3x^3 = 0 \\ \text{and} \\x^2-7x+3=0 \end{cases}$$ There is a lot of possibilities (infinity?) to choose this sub-equations. I know this seems very stupid but could you tell me which rule of logic I broke.
As far as I know, if $$a=0$$ and $$b=0$$ then we can add them to form $$a+b=0$$ we cannot go the other way, that is $$a+b=0\implies a=0 \quad\text{and}\, b=0$$ why?
• It does not follow from $a+b=0$ that $a=0$ and $b=0$. – J. W. Tanner Jan 21 at 22:40
• @J.W.Tanner $a=0$ and $b=0$ then we can write $a+b=0$ why we cannot go the other way? – IamNotaMathematician Jan 21 at 22:45
• It could be, for example, that $a=1$ and $b=-1$. – J. W. Tanner Jan 21 at 22:52
• You have 2 equations for 1 unknown, apart very special cases you will get an incompatible system (no solution). Moreover not all solutions of a+b=0 are of the form a=0 and b=0 – Picaud Vincent Jan 21 at 22:53
• @PicaudVincent but we accept only the values that satisfy both equations – IamNotaMathematician Jan 21 at 22:57
We have that: $$a +b =0 \to a=-b$$ Which means it holds for $$a=b=0$$, but also that you can't go the other way, as you said.
Your system of equations should be: $$\begin{cases} x^5-3x^3 = k \\ \text{and} \\x^2-7x+3=-k \end{cases}$$
We can solve $$x^2-7x+(3+k)=0$$ for $$x=\frac{7\pm\sqrt{37-4k}}{2}$$, and then insert than into the quintic for solutions, but this results in:
$$\frac{(7\pm\sqrt{37-4k})^5}{32}-\frac{3(7\pm\sqrt{37-4k})^3}{8}-k=0$$
One question springs to mind, is this really any simpler? To my mind, not at all.
We can substitute $$t=7+\sqrt{37-4k}$$, but this requires the solution of:
$$t^5-12t^3+8t^2-112t+96=0$$ which is more complicated than the polynomial we are trying to solve.
All in all: is this a valid method? Absolutely. Is it practical? Absolutely not in my opinion.
• I think that is what I was looking for thank you – IamNotaMathematician Jan 21 at 23:34
Note that this is not exactly the same question, for example what if $$x^5 - 3x^3 = 1$$ and $$x^2 - 7x + 3 = -1$$? Then the original equation would be satisfied, but neither of the original equations would be satisfied. However, they also take on these values at different values of $$x$$. Additionally, if you continued to break down the equation into systems of equations, you could eventually get to the case of a $$0th$$ order polynomial which could never be equal to $$0$$. In your particular example this would be looking for solutions of $$3 = 0$$ which clearly doesn't make sense.
One counter-example is enough to show that there are solutions not contained in the intersection of the zeros. As for the proving that the intersection of the zeros is also a solution to the original polynomial do as follows:
Let $$P_n(x) = P_k(x) + P_m(k).$$ Suppose $$P_k(a) = 0,$$ and $$P_m(a) = 0$$. Then, $$P_n(a) = P_k(a) + P_m(a) = 0 +0.$$
• Yes I know but why it doesn't work how to prove that – IamNotaMathematician Jan 21 at 22:49
• You are correct however, that if both equations are $0$ for the same value of $x$, then that would be a solution. Consider for example $x^5 - 3x^3+x^2-7x = 0$. You can break it into: $x^5 - 3x^3 = 0,$$and x^2-7x = 0.$ Both of which have solutions at $x = 0$, and so the original equation also has a solution at $x = 0$. – Jack Pfaffinger Jan 21 at 22:52
• but how could you prove that there is no such decomposition not necessarily 2 terms at a time. in other words if I say for a given polynomial equation there is an equivalent system of simple equations how can you disprove that – IamNotaMathematician Jan 21 at 23:03
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# The Black Chamber
When you read about cryptography (the science of secrecy), you will encounter lots of jargon, as well as different words that all seem to mean roughly the same thing. This short, rather boring section is supposed to clarify some of the jargon.
Definitions
Code: A system for hiding the meaning of a message by replacing each word or phrase in the original message with another character or set of characters. The list of replacements is contained within a so-called codebook. A code has no built in flexibility, other than re-writing the codebook. (An alternative definition of a code is any form of encryption which has no built in flexibility.) To protect a message in this way is called encoding.
Cipher: Any general system for hiding the meaning of a message by replacing each letter in the original message with another letter. To protect a message in this way is called enciphering. Each cipher can be split into two halves – the algorithm and the key. The key gives a cipher some built in flexibility.
Encrypt: A term that covers encoding and enciphering.
Encicode: The process of encoding a message followed by enciphering. The application of multiple layers of encryption is also known as superencipherment.
Key: The flexible component of a cipher. The cipher is a general algorithm that is specified by the key. For example, substitution is a general algorithm that is specified by a key, which is the substitution for each letter. Rival groups can use the same substitution cipher, but they will choose different keys so that they can not read each other’s messages.
The Enigma Cipher
To put some of the definitions into context, let's use the Enigma cipher machine (pictured above with Simon Singh) as an example. It is definitely a cipher, because it encrypts at the level of letters and the algorithm depends on a flexible key that is chosen by the sender.
The key is the set-up of the machine - rotor orientations, plug selections, etc. all determine the encipherment. The receiver must have an Enigma machine, but must also know the key in order to decipher the message.
The More Keys The Better
A secure cipher system must have a wide range of potential keys. For example, if the sender uses the pigpen cipher to encrypt a message, then encryption is very weak because there is only 1 potential key, given by the pigpen grid. From the enemy's point of view, if they intercept the message and suspect that the pigpen cipher is being used, then they merely have to use the grid to decode the message. Let us look at the number of keys used in other methods of encryption.
The Caesar Cipher: The alphabet can be shifted up to 25 places, but shifting a letter 26 places takes it back to its original position, and shifting it 27 places is the same as shifting it 1 place. So there are 25 keys.
The Kama-Sutra Cipher: This is a much stronger cipher. To construct the cipher alphabet, the letter A could be paired with any of the remaining 25 letters. Another letter could be paired with any of the remaining 23 letters. This continues until there is only 1 letter left. The number of keys is therefore 25 x 23 x ... x 1 = 7,905,853,580,625.
General Monoalphabetic Cipher:To construct the cipher alphabet, the letter A could be represented as any of the 26 letters. The letter B could be represented as any of the remaining 25 letters, C could be represented as any of the remaining 24 letters, and so on until the entire cipher alphabet has been formed. The total number of keys is therefore 26 x 25 x 24 x ... x 1 = 403,291,461,126,605,635,584,000,000.
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# Joule Conversion Calculator
Enter a value in joules below to convert to another unit of energy.
## What Is a Joule?
A joule is a unit used to measure energy.
The joule is the SI derived unit for energy in the metric system. Joules can be abbreviated as J; for example, 1 joule can be written as 1 J.
### Background and Origin
The joule is named for English physicist James Prescott Joule, who discovered the relationship between the current passing through a resistor and the heat it dissipates.
## How Much Is a Joule?
The joule is the energy equal to the force on an object of one newton at a distance of one meter.[1] One joule is equal to the heat energy dissipated by the current of one ampere through one ohm of resistance for one second.
One joule is also equal to the energy needed to move an electric charge of one coulomb through a potential difference of one volt. In addition, one joule is also equal to the one watt-second.
## How to Convert Joules
To convert joules to another unit of energy, you need to multiply the value by a conversion factor. A conversion factor is a numerical value used to change the units of a measurement without changing the value.[2]
You can find the conversion factors for joules in the conversion table below.
Then, multiply the energy measurement by the conversion factor to find the equivalent value in the desired unit of measurement.
joules × conversion factor = result
You can also use a calculator, such as one of the converters below, for the conversion.
## Joule Conversion Table
Common joule values and equivalent imperial and metric energy measurements
joules kilojoules megajoules gigajoules calories kilocalories megacalories watt-hours kilowatt-hours megawatt-hours watt-seconds british thermal units million BTU therms quads electronvolts kiloelectronvolts megaelectronvolts
1 J 0.001 kJ 0.000001 MJ 0.000000001 GJ 0.239006 cal 0.000239 kcal 0.00000023901 Mcal 0.000278 Wh 0.00000027778 kWh 0.00000000027778 MWh 1 Ws 0.000948 BTU 0.00000000094782 MMBTU 0.0000000094782 0.00000000000000000094782 6,241,500,000,000,000,000 eV 6,241,500,000,000,000 keV 6,241,509,343,260 MeV
2 J 0.002 kJ 0.000002 MJ 0.000000002 GJ 0.478011 cal 0.000478 kcal 0.00000047801 Mcal 0.000556 Wh 0.00000055556 kWh 0.00000000055556 MWh 2 Ws 0.001896 BTU 0.0000000018956 MMBTU 0.000000018956 0.0000000000000000018956 12,483,000,000,000,000,000 eV 12,483,000,000,000,000 keV 12,483,018,686,520 MeV
3 J 0.003 kJ 0.000003 MJ 0.000000003 GJ 0.717017 cal 0.000717 kcal 0.00000071702 Mcal 0.000833 Wh 0.00000083333 kWh 0.00000000083333 MWh 3 Ws 0.002843 BTU 0.0000000028435 MMBTU 0.000000028435 0.0000000000000000028435 18,725,000,000,000,000,000 eV 18,725,000,000,000,000 keV 18,724,528,029,781 MeV
4 J 0.004 kJ 0.000004 MJ 0.000000004 GJ 0.956023 cal 0.000956 kcal 0.00000095602 Mcal 0.001111 Wh 0.0000011111 kWh 0.0000000011111 MWh 4 Ws 0.003791 BTU 0.0000000037913 MMBTU 0.000000037913 0.0000000000000000037913 24,966,000,000,000,000,000 eV 24,966,000,000,000,000 keV 24,966,037,373,041 MeV
5 J 0.005 kJ 0.000005 MJ 0.000000005 GJ 1.195029 cal 0.001195 kcal 0.000001195 Mcal 0.001389 Wh 0.0000013889 kWh 0.0000000013889 MWh 5 Ws 0.004739 BTU 0.0000000047391 MMBTU 0.000000047391 0.0000000000000000047391 31,208,000,000,000,000,000 eV 31,208,000,000,000,000 keV 31,207,546,716,301 MeV
6 J 0.006 kJ 0.000006 MJ 0.000000006 GJ 1.434034 cal 0.001434 kcal 0.000001434 Mcal 0.001667 Wh 0.0000016667 kWh 0.0000000016667 MWh 6 Ws 0.005687 BTU 0.0000000056869 MMBTU 0.000000056869 0.0000000000000000056869 37,449,000,000,000,000,000 eV 37,449,000,000,000,000 keV 37,449,056,059,561 MeV
7 J 0.007 kJ 0.000007 MJ 0.000000007 GJ 1.67304 cal 0.001673 kcal 0.000001673 Mcal 0.001944 Wh 0.0000019444 kWh 0.0000000019444 MWh 7 Ws 0.006635 BTU 0.0000000066347 MMBTU 0.000000066347 0.0000000000000000066347 43,691,000,000,000,000,000 eV 43,691,000,000,000,000 keV 43,690,565,402,821 MeV
8 J 0.008 kJ 0.000008 MJ 0.000000008 GJ 1.912046 cal 0.001912 kcal 0.000001912 Mcal 0.002222 Wh 0.0000022222 kWh 0.0000000022222 MWh 8 Ws 0.007583 BTU 0.0000000075825 MMBTU 0.000000075825 0.0000000000000000075825 49,932,000,000,000,000,000 eV 49,932,000,000,000,000 keV 49,932,074,746,081 MeV
9 J 0.009 kJ 0.000009 MJ 0.000000009 GJ 2.151052 cal 0.002151 kcal 0.0000021511 Mcal 0.0025 Wh 0.0000025 kWh 0.0000000025 MWh 9 Ws 0.00853 BTU 0.0000000085304 MMBTU 0.000000085304 0.0000000000000000085304 56,174,000,000,000,000,000 eV 56,174,000,000,000,000 keV 56,173,584,089,342 MeV
10 J 0.01 kJ 0.00001 MJ 0.00000001 GJ 2.390057 cal 0.00239 kcal 0.0000023901 Mcal 0.002778 Wh 0.0000027778 kWh 0.0000000027778 MWh 10 Ws 0.009478 BTU 0.0000000094782 MMBTU 0.000000094782 0.0000000000000000094782 62,415,000,000,000,000,000 eV 62,415,000,000,000,000 keV 62,415,093,432,602 MeV
11 J 0.011 kJ 0.000011 MJ 0.000000011 GJ 2.629063 cal 0.002629 kcal 0.0000026291 Mcal 0.003056 Wh 0.0000030556 kWh 0.0000000030556 MWh 11 Ws 0.010426 BTU 0.000000010426 MMBTU 0.00000010426 0.000000000000000010426 68,657,000,000,000,000,000 eV 68,657,000,000,000,000 keV 68,656,602,775,862 MeV
12 J 0.012 kJ 0.000012 MJ 0.000000012 GJ 2.868069 cal 0.002868 kcal 0.0000028681 Mcal 0.003333 Wh 0.0000033333 kWh 0.0000000033333 MWh 12 Ws 0.011374 BTU 0.000000011374 MMBTU 0.00000011374 0.000000000000000011374 74,898,000,000,000,000,000 eV 74,898,000,000,000,000 keV 74,898,112,119,122 MeV
13 J 0.013 kJ 0.000013 MJ 0.000000013 GJ 3.107075 cal 0.003107 kcal 0.0000031071 Mcal 0.003611 Wh 0.0000036111 kWh 0.0000000036111 MWh 13 Ws 0.012322 BTU 0.000000012322 MMBTU 0.00000012322 0.000000000000000012322 81,140,000,000,000,000,000 eV 81,140,000,000,000,000 keV 81,139,621,462,382 MeV
14 J 0.014 kJ 0.000014 MJ 0.000000014 GJ 3.34608 cal 0.003346 kcal 0.0000033461 Mcal 0.003889 Wh 0.0000038889 kWh 0.0000000038889 MWh 14 Ws 0.013269 BTU 0.000000013269 MMBTU 0.00000013269 0.000000000000000013269 87,381,000,000,000,000,000 eV 87,381,000,000,000,000 keV 87,381,130,805,643 MeV
15 J 0.015 kJ 0.000015 MJ 0.000000015 GJ 3.585086 cal 0.003585 kcal 0.0000035851 Mcal 0.004167 Wh 0.0000041667 kWh 0.0000000041667 MWh 15 Ws 0.014217 BTU 0.000000014217 MMBTU 0.00000014217 0.000000000000000014217 93,623,000,000,000,000,000 eV 93,623,000,000,000,000 keV 93,622,640,148,903 MeV
16 J 0.016 kJ 0.000016 MJ 0.000000016 GJ 3.824092 cal 0.003824 kcal 0.0000038241 Mcal 0.004444 Wh 0.0000044444 kWh 0.0000000044444 MWh 16 Ws 0.015165 BTU 0.000000015165 MMBTU 0.00000015165 0.000000000000000015165 99,864,000,000,000,000,000 eV 99,864,000,000,000,000 keV 99,864,149,492,163 MeV
17 J 0.017 kJ 0.000017 MJ 0.000000017 GJ 4.063098 cal 0.004063 kcal 0.0000040631 Mcal 0.004722 Wh 0.0000047222 kWh 0.0000000047222 MWh 17 Ws 0.016113 BTU 0.000000016113 MMBTU 0.00000016113 0.000000000000000016113 106,110,000,000,000,000,000 eV 106,110,000,000,000,000 keV 106,110,000,000,000 MeV
18 J 0.018 kJ 0.000018 MJ 0.000000018 GJ 4.302103 cal 0.004302 kcal 0.0000043021 Mcal 0.005 Wh 0.000005 kWh 0.000000005 MWh 18 Ws 0.017061 BTU 0.000000017061 MMBTU 0.00000017061 0.000000000000000017061 112,350,000,000,000,000,000 eV 112,350,000,000,000,000 keV 112,350,000,000,000 MeV
19 J 0.019 kJ 0.000019 MJ 0.000000019 GJ 4.541109 cal 0.004541 kcal 0.0000045411 Mcal 0.005278 Wh 0.0000052778 kWh 0.0000000052778 MWh 19 Ws 0.018009 BTU 0.000000018009 MMBTU 0.00000018009 0.000000000000000018009 118,590,000,000,000,000,000 eV 118,590,000,000,000,000 keV 118,590,000,000,000 MeV
20 J 0.02 kJ 0.00002 MJ 0.00000002 GJ 4.780115 cal 0.00478 kcal 0.0000047801 Mcal 0.005556 Wh 0.0000055556 kWh 0.0000000055556 MWh 20 Ws 0.018956 BTU 0.000000018956 MMBTU 0.00000018956 0.000000000000000018956 124,830,000,000,000,000,000 eV 124,830,000,000,000,000 keV 124,830,000,000,000 MeV
## Joule Conversion Calculators
You can also convert energy using one of our joule converters below.
Joules
Calories
Watts x Time
Thermal Units
Electronvolts
Related Calculators
You might also find our other electrical calculators useful.
## References
1. International Bureau of Weights and Measures, The International System of Units, 9th Edition, 2019, https://www.bipm.org/documents/20126/41483022/SI-Brochure-9-EN.pdf
2. National Institute of Standards & Technology, Unit Conversion, https://www.nist.gov/pml/owm/metric-si/unit-conversion
Joules
Calories
Watts x Time
Thermal Units
Electronvolts
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# How do you measure distance in arcore 2
In this tutorial, we will cover how to measure distance in Arcore 2. This will allow you to find out how far two objects are from each other and to calculate the distance between any two points.
I have given my best possible advice on this topic if you want to know anything comment down below, our team is thriving hard to give you the best.
## How is augmented reality distance measured?
One way to measure distance in augmented reality is to use the camera on your device. This is the method that Google used when they first introduced their AR platform.
You can also use motion tracking to measure distance. This is the method that Apple and Microsoft use for their AR platforms. This method uses the camera on your device to track the movement of objects in the environment.
Another way to measure distance in augmented reality is to use lasers. This is the method that Facebook and Snapchat use. They scan a QR code or target location with a laser and then use that information to determine how far away the object is.
## How do you gauge the distance?
To measure the distance between two points in a 3D space, you need to use a coordinate system. A coordinate system is a set of axes and coordinates that allow you to specify the location of objects in space.
I would appreciate a thankyou in comments or a suggestion if you have any. Looking forward to your reaction if we were able to answer you
Usually, you would use a Cartesian coordinate system when measuring distances in the real world. This system uses three axes, called x, y, and z. The x-axis runs perpendicular to the plane of the page and measures east/west distance; y-axis measures up/down distance; and z-axis measures north/south distance.
When measuring distances in a 3D space, you need to convert these coordinates into another format. In this format, the x-axis is renamed as the local X-axis and points towards the object you are measuring; y-axis becomes the local Y-axis and points towards the object you are measuring; and z-axis becomes the local Z-axis and points towards the object you are measuring.
I should tell about the next thing that everyone is asking on social media and searching all over the web to find out the answer, well i have compiled answers further below
READ : How do you permanently delete snapchat messages on both sides
For example, if you want to measure the distance between points A and B in a 3D space, you would use these coordinates: (0,0), (1,1), (2,2). You would then convert these coordinates
## How do you measure height width and distance of an object using camera android github?
When measuring the height, width and distance of an object using a camera on an Android device, there are several methods that can be used.
Further answered questions are also very related but given separately because we can't put everything in one subheading let's check further
One method is to use the camera’s built-in sensors. These sensors can measure the height, width and distance of an object based on the light that is reflected off of it. This method is useful if you want to measure the height, width and distance of an object without having to use a physical measurement tool.
Another method is to use the camera’s flash. When the flash is activated, it will illuminate the object and allow the camera to measure its height, width and distance. This method is useful if you want to measure the height, width and distance of an object in a dark room or if you want to measure the height, width and distance of an object that is obscured by another object.
I would say this is the best explanation to the topic in a breif, however there are many questions that need thorrough reading
Finally, a third method is to use a physical measurement tool. When using a physical measurement tool, you will need to position the tool next to the object that you want to measure and take measurements using its built-in sensors or its flash.
## What can ARCore do?
ARCore can measure distance in a variety of ways. It can use the camera on your phone to measure distances between objects, or it can use the GPS in your phone to measure distances between points. ARCore can also use the sensors on your phone to measure distances between objects.
## What is AR world tracking?
AR world tracking is a technology that allows you to track and measure distances in real time. This technology is used in many different applications, including gaming and marketing.
READ : How to inverted filter on tiktok
AR world tracking works by tracking the movements of the user’s head and hands. This information is then used to calculate the distance between the user and any objects or virtual objects that they are interacting with.
This technology is incredibly useful for marketing and gaming applications. It allows brands to create realistic marketing campaigns without having to spend a lot of money on filming or photoshopping. It also allows gamers to measure their progress in games and make better decisions about how to optimize their playstyle.
## What is motion tracking in AR?
One of the most common uses for AR is to measure distance. This can be done in a number of ways, but one popular method is motion tracking. Motion tracking involves tracking the movements of an object in order to determine its location. It can be used in a number of applications, such as measurement, navigation, and 3D mapping.
One of the main advantages of using motion tracking is that it is very accurate. Even small movements can be tracked, which makes it ideal for measuring distance. This accuracy is also useful when accuracy is important, such as when measuring the size or shape of an object.
Another advantage of motion tracking is that it is relatively easy to set up. Most devices that are capable of motion tracking are relatively inexpensive and easy to use. This means that you can easily measure distance without having to invest a lot of money in equipment or hire a specialist.
## How do you find the distance between two objects in a video?
To measure the distance between two objects in a video, you first need to find the center of each object. To do this, you can use a method called centering.
To center an object in a video, you first need to find its center point. To do this, you can use a method called centering. This method involves locating the object’s middle point on the screen and then moving the object until it is centered in the window.
READ : How to see friends activity on instagram 2021
Once you have found the object’s center point, you can use simple math to calculate the distance between the object and any other points on the screen. To do this, you divide the distance between the center point and each other point by the size of the object in pixels.
## conclusion
When measuring distance in the air, it is important to use the correct units. One common unit for measuring distance in the air is meters. To measure a distance in meters, divide the length of the object being measured by 1,000. For example, if you want to measure a distance of 2 meters, you would divide 2 by 1000 to get 200 meters.
Another common unit for measuring distance in the air is feet. To measure a distance in feet, divide the length of the object being measured by 3.2808. For example, if you want to measure a distance of 1 foot, you would divide 1 by 3.2808 to get 0.3048 feet.
It is important to be aware of which unit you are using when measuring distances in the air. Using the wrong unit can result in inaccurate measurements.
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Next: Post-Processing for Eigenvectors Using Up: General Use of ARPACK Previous: Post-Processing for Eigenvectors Using
# Computational Modes for Complex Problems
This section describes the solution of eigenvalue problems in complex arithmetic. The reverse communication interface subroutine for the double precision complex eigenvalue problem is znaupd . This routine is to be used for both Hermitian and non-Hermitian problems. The routine is called as shown in Figure 3.8. It should be noted that the calling sequences for znaupd and zneupd differ slightly from those of dnaupd and dneupd. The main difference is that an additional work array rwork is required by znaupd that is not required by dnaupd.
Occasionally, when using znaupd on a complex Hermitian problem, eigenvalues will be returned with small but non-zero imaginary part due to unavoidable round-off errors. These should be ignored unless they are significant with respect to the eigenvalues of largest magnitude that have been computed. There is little computational penalty for using the non-Hermitian routines in this case. The only additional cost is to compute eigenvalues of a Hessenberg rather than a tridiagonal matrix. For the problem configurations this software is designed to solve, the size of these matrices are small enough that the differences in computational cost are negligible compared to the major work that is required.
call znaupd ( ido, bmat, n, which, nev, tol, resid, ncv, v, & ldv, iparam, ipntr, workd, workl, lworkl, rwork,info )
The integer ido is the reverse communication flag that specifies a requested action on return from dnaupd. The character*1 parameter bmat specifies if this is a standard bmat = 'I' or a generalized bmat = 'G' problem. The integer n specifies the dimension of the problem. The character*2 parameter which may take the same possible values listed for subroutine dnaupd in Table 3.4.
There are three shift-invert modes for complex problems. These modes are specified by setting the parameter entry iparam(7) = mode where mode = 1,2, or 3.
In the following list, the specification of OP and are given for the various modes. Also, the iparam(7) and bmat settings are listed along with the name of the sample driver for the given mode. Sample drivers for the following modes may be found in the EXAMPLES/COMPLEX subdirectory.
1.
Regular mode (iparam(7) = 1, bmat = 'I' ). Use driver zndrv1.
(a)
Solve in regular mode.
(b)
and
2.
Shift-invert mode (iparam(7) = 3, bmat = 'I'). Use driver zndrv2.
(a)
Solve in shift-invert mode.
(b)
and
3.
Regular inverse mode (iparam(7) = 2, bmat = 'G'). Use driver zndrv3.
(a)
Solve in regular inverse mode.
(b)
and
4.
Shift-invert mode (iparam(7) = 3, bmat = 'G'). Use driver zndrv4.
(a)
Solve in shift-invert mode.
(b)
and
Next: Post-Processing for Eigenvectors Using Up: General Use of ARPACK Previous: Post-Processing for Eigenvectors Using
Chao Yang
11/7/1997
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# Roku Developer Program
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Level 7
## Re: HSVA to RGBA function
Exactly what do you mean by the color stops changing?
What color do you expect at 2147483647 + 1? If I had to guess *tries to find Calc... oh you RMB on the "start" button! that makes sense. ah, I see the Run command...* that would be &h7FFFFFFF + 1, right? And wouldn't that be an Alpha of 0?
But beyond all that, where are you getting this 2147483647? I thought you were converting from HSV.
BTW, you left out, "s = 0 then rgb = [v, v, v]" in your port. Dunno if that plays a factor in what you're trying to do, but figured I should mention it somewhere.
Level 7
## Re: HSVA to RGBA function
OK. Fine! It's Baby Blue. *grumbles*
I think I figured it out though... what should &h7FFFFFFF + 1 = ?
[spoiler=tricky computer math:1zasyl27]BrightScript Debugger> ? 2147483647
2147483647
BrightScript Debugger> ? 2147483647+1
2147483648
BrightScript Debugger> ? type(2147483647)
Double
BrightScript Debugger> ? type(2147483647-1)
Double
BrightScript Debugger> i% = 2147483647
BrightScript Debugger> ? i%
2147483647
BrightScript Debugger> ? int(2147483647)
2147483647
BrightScript Debugger> ? type(int(2147483647))
Integer
BrightScript Debugger> ? type(int(2147483647+1))
Integer[/spoiler:1zasyl27]
I've actually learned this before. Go figure.
BrightScript Debugger> i% = 2147483647 : ? i% + 1
[spoiler=only for the strong of heart:1zasyl27]BrightScript Debugger> ? i%+1
-2147483648
Hex for clarity... The "7" in &h7F... means "0111" in binary. The "0" is the bit that tells the sign. "0" = positive number, "1" = negative. Since BrightScript can't read your mind... it is assuming you want to use Doubles. You need Integers.
There are at least a couple ways to tell BrightScript you want Integers...[/spoiler:1zasyl27]
Level 7
## Re: HSVA to RGBA function
Not out of the woods yet...
BrightScript Debugger> ? int(2147483647+42)
2147483647 <<<===== Say, Whaaaaaaat?
BrightScript Debugger> ? int(2147483647)
2147483647
BrightScript Debugger> ? int(2147483647)+42
-2147483607 <<< this GOOD.
BrightScript Debugger> ? int(2147483647+42)
______________
Now on the LEFT I get a REDish gradient thing with a hard edge in the middle-ish due to the alpha cycling.
`for hue = 0 to 360 scr.DrawLine( 450,100+hue, 550,100+hue, int(2147483647)+hue) scr.DrawLine( 650,100+hue, 750,100+hue, int(2147483647)+(hue<<8))end for`
______________
So, the changes to the HSVtoRGB func that fixed this ( by pure luck! ) are the assigning the return value to Integer ( ... As Integer ) and the use of "%" at the end of the retval.
Level 8
## Re: HSVA to RGBA function
@dev42
Take a look at the second to last post on this thread.
viewtopic.php?f=34&t=84652&p=484454
Probably should have mentioned the earlier but couldn't remember where it was, plus I'm posting from my phone which is difficult (not home right now). RokuKC explains in depth the difference between a hex value and an integer.
Cheers
Level 7
## Re: HSVA to RGBA function
@Romans -
Thanks for that link! Awesome stuff there.
Now, is there still a problem? Because, to me at least, once the signed/unsigned issue is understood, it should become clear that, when talking about signed integers:
&hFFFFFFFF != 4294967295
&hFFFFFFFF == -1
still working on how the 64 bit Double gets clipped to the 32 bit Signed Integer.
Edit: Double is to Float as int is to char!
4294967295 == 0x41EFFFFFFFE00000 ==
01000001 11101111 11111111 11111111
11111111 11100000 00000000 00000000
but it is getting clipped... possibly taking the least significant 32 bits. ( chime in any time gurus! )
________________________
Wikipedia: 2's Complement
Level 11
## Re: HSVA to RGBA function
Dev42 - note that library int() function is (mis)defined as taking Float. If you pass Double, it will be down-converted to Float first - 4 bytes, one of which is exponent, so only ~3 bytes make it (25 bits or thereabout) - and only then to Int. You can make your own convertor Double->Int though (magic!) and explore:
`BrightScript Debugger> toInt = function(x as Integer) as Integer: return x: end functionBrightScript Debugger> y = toInt(4294967295): ? y, type(y) 2147483647 IntegerBrightScript Debugger> for x= 2#^31-3 to 2#^31+2: ? x, toInt(x): next 2147483645 2147483645 2147483646 2147483646 2147483647 2147483647 2147483648 2147483647 2147483649 2147483647 2147483650 2147483647`
Level 11
## Re: HSVA to RGBA function
"Romans_I_XVI" wrote:
Glad you like it dev42 . How were you planning on implementing it though? I'm currently jumping through about 3 hoops. But it doesn't matter too much because I'm not running these functions often.
Hmm, the way you describe doing it is broken in principle. Not sure why/if you are getting the right results.
I start with that original HSVAtoRGBA function, which gives me a decimal. Then I have to pass it through this function.
It actually gives you a Double. There is no Decimal as such in B/S (Decimal is fairly rare type, i mostly know it in some DBs, also Java has BigDecimal).
`function decToHex (dec as integer) as string`
Here lies the problem. This function is declared to take Integer. So when you call it with a Double (from above) and that number is >2147483647#, it will get chopped to 2147483647 (MAXINT, 2^31-1). Which means when you call with R>=128, for any G, B, A you will be getting "7FFFFFFF" from that fn.
In my Brightscript I'm doing it like this.
` some_hex_color = val(decTohex(HSVAtoRGBA(hue,sat,100,255)),16)`
Maybe calling it always with V=100, A=255 has saved your bacon so far, IDK. But check for colors with strong R, they should all be washing to cyan (?).
Level 9
## Re: HSVA to RGBA function
"EnTerr" wrote:
Maybe calling it always with V=100, A=255 has saved your bacon so far, IDK. But check for colors with strong R, they should all be washing to cyan (?).
Thats what I found, passing it pure red returned a really pale cyan. I already had another solution so i didnt pursue it though.
Kinetics Screensavers
Level 7
## Re: HSVA to RGBA function
I haven't yet tried this with DrawObject and I'm not understanding the point of converting 2^32 to "decimal" and then use that as a color IF the HSVAtoRGBA() func is working as it should ...
The main tweak's to HSVAtoRGBA() are that it now uses Integers, Integer bit shifting which supposedly avoids the Double to Integer conversion issue, and returns an Integer. The function (latest ver) returned negative values and a spectrum from Red through all the hues back to Red... so, I was pleasantly content.
I did call out how fully I tested this.
I'll dig deeper later.
__________________
"squirreltown" wrote:
Thats what I found, passing it pure red returned a really pale cyan. I already had another solution so i didn't pursue it though.
I did make mention of the "baby blue" :roll: ... and again, my changes to the HSV...() fixed it. No more "pale cyan". Could some kind soul test it out?
Level 8
## Re: HSVA to RGBA function
Sorry, the dectohex function I posted earlier is not the same as what I'm using, I forgot I removed that "As Integer" so it is not getting chopped off and I'm getting back the correct hex as a string, I just now tested it.
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Friday
May 6, 2016
# Homework Help: Physics
Posted by Drew Meryll on Saturday, January 26, 2013 at 9:29pm.
A 2.1 kg ball is attached to a ceiling by a string. The distance from the ceiling to the center of the ball is 1.20 m, and the height of the room is 3.00 m.
(a) What is the gravitational potential energy associated with the ball relative to the ceiling?
(b) What is the gravitational potential energy associated with the ball relative to the floor?
(c) What is the gravitational potential energy associated with the ball relative to a point at the same elevation as the ball?
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Cody
# Problem 17. Find all elements less than 0 or greater than 10 and replace them with NaN
Solution 3031951
Submitted on 2 Oct 2020 by Abhijeet Deshmukh
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = [ 5 17 -20 99 3.4 2 8 -6 ]; y_correct = [ 5 NaN NaN NaN 3.4 2 8 NaN ]; assert(isequalwithequalnans(cleanUp(x),y_correct))
x = 5.0000 NaN -20.0000 99.0000 3.4000 2.0000 8.0000 -6.0000 x = 5.0000 NaN NaN 99.0000 3.4000 2.0000 8.0000 -6.0000 x = 5.0000 NaN NaN NaN 3.4000 2.0000 8.0000 -6.0000 x = 5.0000 NaN NaN NaN 3.4000 2.0000 8.0000 NaN y = 5.0000 NaN NaN NaN 3.4000 2.0000 8.0000 NaN
2 Pass
x = [ -2.80 -6.50 -12.60 4.00 2.20 0.20 -10.60 9.00]; y_correct = [ NaN NaN NaN 4.00 2.20 0.20 NaN 9.00] assert(isequalwithequalnans(cleanUp(x),y_correct))
y_correct = NaN NaN NaN 4.0000 2.2000 0.2000 NaN 9.0000 x = NaN -6.5000 -12.6000 4.0000 2.2000 0.2000 -10.6000 9.0000 x = NaN NaN -12.6000 4.0000 2.2000 0.2000 -10.6000 9.0000 x = NaN NaN NaN 4.0000 2.2000 0.2000 -10.6000 9.0000 x = NaN NaN NaN 4.0000 2.2000 0.2000 NaN 9.0000 y = NaN NaN NaN 4.0000 2.2000 0.2000 NaN 9.0000
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# A wooden cube floating in water supports a mass 0.2 kg on its top. When the mass is removed the cube rises by 2cm. What is the side legnth of the cube ? Density of water = 10^3 kg//m^3
Khareedo DN Pro and dekho sari videos bina kisi ad ki rukaavat ke!
Solution : Decrease in weight = decrease in upthrust <br> :. Mg = V rho_w g <br> or m = V rho_w <br> or 0.2 = (2xx10^(-2)Axx10^3)<br> where A is the area of the base of cube. Then, A = 10^(-2) m^2 = 100 cm^2 <br> Side of the cube, L = sqrtA = sqrt(100) = 10 cm
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## Convert Septet (Number Of Performers) to Hecto (h, Metric Prefixes)
This page features online conversion from septet to hecto. These units belong to different measurement systems. The first one is from Number Of Performers. The second one is from Metric Prefixes.
If you need to convert septet to another compatible unit, please pick the one you need on the page below. You can also switch to the converter for hecto to septet.
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## Quantity Units
Units: unit, point (1) / pair, brace, yoke / nest, hat trick / half-dozen / decade, dicker / dozen / baker's dozen / score / flock / shock / hundred / great hundred / gross / thousand / great gross
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## Percentages and Parts
septet to percent (%) septet to permille (‰)
septet to parts per million (ppm) septet to parts per billion (ppb)
Units: percent (%) / permille (‰) / parts per million (ppm) / parts per billion (ppb)
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## Fractions
This section answers a question like "How many one sevenths are there in 1 half?". To get an answer enter 1 under half and see the result under 1/7. See if you can use this section to find out how many one-sixths are there in 15 one-nineths.
septet to half or .5 (1/2) septet to one third or .(3) (1/3) septet to quart, one forth or .25 (1/4) septet to tithe, one fifth or .2 (1/5) septet to one sixth or .1(6) (1/6)
septet to one seventh or .142857 (1/7) septet to one eights or .125 (1/8) septet to one ninth or .(1) (1/9) septet to one tenth or .1 (1/10) septet to one sixteenth or .0625 (1/16) septet to one thirty-second or .03125 (1/32)
Units: half or .5 (1/2) / one third or .(3) (1/3) / quart, one forth or .25 (1/4) / tithe, one fifth or .2 (1/5) / one sixth or .1(6) (1/6) / one seventh or .142857 (1/7) / one eights or .125 (1/8) / one ninth or .(1) (1/9) / one tenth or .1 (1/10) / one sixteenth or .0625 (1/16) / one thirty-second or .03125 (1/32)
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## Metric Prefixes
These prefixes are widely used in Metric System. They apply to any unit, so if you ever see, e.g. kiloapple, you know it's 1000 apples.
septet to yocto (y) septet to zepto (z) septet to atto (a) septet to femto (f) septet to pico (p) septet to nano (n) septet to micro (µ, mc) septet to milli (m) septet to centi (c) septet to deci (d)
septet to deka (da) septet to hecto (h) septet to kilo (k) septet to mega (M) septet to giga (G) septet to tera (T) septet to peta (P) septet to exa (E) septet to zetta (Z) septet to yotta (Y)
Units: yocto (y) / zepto (z) / atto (a) / femto (f) / pico (p) / nano (n) / micro (µ, mc) / milli (m) / centi (c) / deci (d) / deka (da) / hecto (h) / kilo (k) / mega (M) / giga (G) / tera (T) / peta (P) / exa (E) / zetta (Z) / yotta (Y)
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## Number of Performers
Units: solo / duet / trio / quartet / quintet / sextet / septet / octet
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Select Page
Here is the Axiom:
If I would like to swim a Perfect 1500 meters, I first need to learn to swim a Perfect 500 meters. If I would like to swim a Perfect 500 meters, I first need to learn to swim a Perfect 100 meters. If I want to swim a Perfect 100 meters, I first need to learn to swim a Perfect 25 meters. If I want to swim a Perfect 25 meters, I first need to learn to take 16 Perfect Strokes. And if I want to take 16 Perfect Strokes, I first need to learn to take just one Perfect Stroke.
## The Pieces of The Perfect 25
No matter how short or how far we intend to swim, it starts one 25 meter (or yard) pool length at a time. The excellence of the whole swim is never greater than the excellence of each part of the swim- so why permit inferior parts? Excellence does not happen by accident, only by deliberate focus on important skills. Not all 10,000 hours of time spent in practice are equal. Only an excellence-oriented 10,000 hours will produce excellence.
So I’ve set out to build my Perfect 25. Here are the basic components of it:
• The push-off & underwater glide
• The break-out
• The stroke
• The finish
We must also eventually add the flip-turn (or open-turn) if we intend to swim for anything beyond 25. And we must add the block-start if that will be used for the first lap. I won’t. I intend to set my PR from an in-water wall start.
## The Concepts Behind It
Following the understanding that Speed is simply Stroke Length (SL) x Stroke Rate (SR) by ‘Perfect 25’ I mean the longest SL I can achieve at a certain SR, and then hold it consistently for every length of my chosen event, and better, to hold that SL as I increase SR, thus speeding up. In every component of my 25 meter length I am working to maximize my abilities and put them together in a finely-tuned choreography.
Just to clarify- I am using a Tempo Trainer (TT) to set my SR (1/tempo) constraints. Another way to view my goal is that I am trying to get to the other wall in as few beeps of the TT as possible. Since I get one stroke per one beep, I need to take as few strokes as possible and make each of those strokes as effective as possible- which means I am making each stroke as long as possible (long SL). Longer SL means less SPL (strokes per length of the pool). The Tempo Trainer fixes my SR, and counting strokes tells me SL.
I’ve been developing this Perfect 25 training concept for the goal I have set for achieving an new lifetime PR in the 100 meter sprint. Although I would designate myself as a marathon-distance OW swimmer I recognize that the kind of skills and the manner in which I am training for my 100 meter sprint will directly translate into strengths which will serve me on long swims. This is counter to the traditional muscular mindset and training methods which view sprint work as contrary to distance work. I am not limiting my muscles to a certain kind of movement pattern (e.g. fast-twitch or slow-twitch), I am expanding my ability to control them with precision in a wider range of EFFECTIVE patterns- to employ those patterns at will, as needed, like gearing on a bicycle or sports car. An open-water swim will often involve a wide range of patterns to negotiate the conditions and competition to keep best position.
For 100 meters I need to swim four Perfect 25 meter segments. Before I attempt to execute 4 of them (or 400 of them) I need to be able to make just one good enough that it can hold up under the inevitable exhaustion that comes with each consecutive length. I first build a perfect prototype then work on duplicating it.
The key element of this training strategy is that I am confronting my neurological threshold and expanding it- a concept that Terry Laughlin has coincidentally, just recently started a TI Forum thread about.
What I have exposed in this training process is that my ability to hold my targets for the Perfect 25 is much more about the strength of my focus than it is about the power in my muscles, though power is developing as a natural result of my neurologically-oriented training. When I notice that I lose focus, I lose SPL. And when I lose SPL it means I am slowing down. If I cannot preserve stroke length I have to increase stroke rate disproportionately (in terms of effort) to sustain my pace.
For instance, if I have been holding 15 SPL at .95 tempo to hold a 64.6 second 100m pace, but surrender one stroke to swim with 16 SPL, I have to increase my tempo to .89 seconds per stroke just to maintain the same pace. That’s a huge increase in tempo and a huge increase in power output just because I lost focus on stroke quality.
Once we start losing SL, we start slipping into a battle to minimize loss. This is why ALL swimmers slow down at the end of their sprint race although their stroke rate increases- it becomes a race for who loses the least amount of stroke length, not who spins the arms the fastest. This is why protecting SL is the paramount challenge for any swimmer in any event.
In the next essay I will talk about how I am training for the Perfect 25.
© 2012, Mediterra International, LLC. All rights reserved. Unauthorized use and/or duplication of this material without express and written permission from this site’s author and/or owner is strictly prohibited. Excerpts and links may be used, provided that full and clear credit is given to Mediterra International, LLC and Mediterraswim.com with appropriate and specific direction to the original content.
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Unit 4 linear equations & graphs. Unit 5 forms of linear equations. Graphing quadratic functions name_____ id: 1 ©g l2t0z1d5a dk[uxtqaa gsaoffvtmwgaxrseq. Your aos (in this chapter) will always. One important feature of the graph is that it. Web when finished with this set of worksheets, students will be able to solve linear and quadratic functions graphically.
1 ©g l2t0z1d5a dk[uxtqaa gsaoffvtmwgaxrseq. Unit 4 linear equations & graphs. Web assessment over graphing quadratic functions in standard form, vertex form and intercept form. In this article, we review how to graph quadratic. Graphing quadratic functions name_____ id:
## Web Benefits Of Graphing Quadratic Functions Worksheets.
Your aos (in this chapter) will always. Cuemath experts developed a set of graphing quadratic functions. ¥ the highest or lowest point of. Web assessment over graphing quadratic functions in standard form, vertex form and intercept form.
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### Calculate the price of the bond
Assignment Help Accounting Basics
##### Reference no: EM132280240
Question - A company issued a bond on January 1st, 2019. (Face value of \$500,000, 3 years maturity, and coupon rate of 8% per year, paid semiannually).
Suppose the market interest rate is 10%.
Calculate the Price of the bond and indicate whether it is discount bond or premium bond.
Calculate the interest expense when the company pays the second coupon payment.
#### Balance of interest payable for the loan
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How Does Lagrange's Theorem Prove a^(|G|+1) = a in Finite Groups?
• Aju
In summary, Prove that if G is a finite group with multiplicative notation, and 'a' is element of G then a^(|G|+1) = a. Prove that if G is a finite group with multiplicative notation, and 'a' is element of G then k must be a factor of |G|. What, then, is a|G|?
Aju
Prove that if G is a finite group with multiplicative notation, and 'a' is element of G then
a^ (|G|+1) = a
Aju said:
Prove that if G is a finite group with multiplicative notation, and 'a' is element of G then
a^ (|G|+1) = a
|G| is the number of elements in G. Look at a0= e (the group identity), a1, a2, a3, ..., aG. Since there are only |G| elements in G, there must be a duplicate in that list of |G|+1 values. Suppose an= am with n< m. Then am-n= e. Now you finish it: Show that the set of powers of a: a, a2, ..., ak=e, with k= m-n, foms a subgroup of G. It follows that k must be a factor of |G|. What, then is a|G|?
perhaps the op does not know "lagrange's" theorem? (due to gauss)
If he doesn't then he surely shouldn't be doing a problem like this!
I think Lagranges theorem is actually due to Camille Jordan. (Gauss discovered the special case of cyclic groups)
agreed, gauss discovered the special case of subgroups of Z/n, (and I believe of units in Z/n). what i refer to is the proof gauss gave, which is exactly the disjoint cosets method which proves the general case.
i.e. although in the special case a more special proof is available, gauss gave a general proof which works in the general case.
i am perhaps more strict than some in my method of attribution, but i decline to attribute a theorem to someone who merely observes that someone else's proof works again in anther case, without providing a new argument.
in teaching group theory this year i worked out the theory of abelian groups first, and was able to take advantage of commutativity in this proof. I.e. in an abelian group, given to elements, there is another element whose order is the lcm of their orders. this implies the lagrqnge theorem for abelian groups.
i was not able however to come up with a new argument for lagrange without using the idea, due to gauss, of disjoint cosets. this is a genuine idea, which one appreciate when trying to circumvent it. when i noticed gauss had introduced this diea in his proof, i became convinced the real idea of the argument for lagranges theorem was due to gauss. indeed i cannot see anyhting new at all in the general argument not present in gauss version.
so more accurately this theorem is perhaps first stated and proved in general by lagrange, but the argument given in books today for this theorem is identical that used by gauss earlier on a special case. (see his disquisitiones for this argument).
so if one were to share the credit, i would still give most of it to gauss.
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after further reflection, i think it is a contribution to realize that an existing argument has anew application. i have often done this myself, although niot always with knowledge of the prior argument.
but the later proof should ideally acknowledge the earlier instance of it. since gauss name is almost never mentioend in connection with algranges theorem, i emophasized gauss's priority.
but the reference to jordan is also interesting to me. i am curiious to read his traite des substitutioins and see for myself whetehr he mentiions gauss. and why is lagrange usually credited?
ah history is indeed afscinating and eye opening. a little research reveals that fermats little theorem is a precursor of lagrange's theorem and fermat died in 1665, over a century before gauss's birth.
so i should remember to be more careful in making attributions. someone always knows an earlier reference for a seminal idea on a topic. note too the encouraging side that even great innovators are only building a little on what went before.
so we should read the great mathematicians and try to understand what they have done so well that we see ourselves how to extend it.
indeed the concept of a gcd in euclid could be said to lie beneath all these concepts for cyclic groups. i.e. a cyclic group is a homomorphic image of Z, and subgroups of subgroups of Z correspond to divisors of generators.tht wasin fact my motive in giving credit to gauss for this proof. i.e. euclidfs method works on abelian groups, and gauss method works on non abelian groups. amd i know of no other methods, so to me gauss made the conceptual leap allowing the theorem to generalize to non abelian groups.
someone else actually noticed that however.
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I wonder if poor Aju is still with us!
Mr Aju, the result we are referring to says that, for an element a of a finite group, the first positive exponenent t such that a^t = 1, where 1 is the identity element, is a divisor of the number of elements of the group.
Gauss's proof was to observe that one can define a partition of the group into subsets of the same order as follows. one set is the distinct powers a,a^2, a^3,...,a^t = 1. another set is the collection of translates of this first set by any other element g of the group. I.e. ga,ga^2,ga^3,...,ga^t = g. one repeats this process until no more elements remain. this shows that thw group is a disjoint union of translates of the distinct powers of a. hence the number of these powers is a divisor of the order of the group.
this is called lagranges theorem, although it is predated by many other similar results, by fermat, gauss, jordan, euclid,...moses...
if you kniow lagranges theorem your problem follows immediately as follows: a^|G|+1 = a^|G| a = 1a = a. so lagrange is the whole matter.
1. How do I begin to prove this problem?
The first step in proving a problem is to clearly define and understand the problem. Make sure you understand all the given information and any restrictions. Then, brainstorm different approaches or methods that could potentially be used to prove the problem.
2. What type of evidence do I need to prove the problem?
The type of evidence needed to prove a problem will depend on the specific problem and the approach being used. This could include mathematical equations, logical reasoning, experimental data, or other forms of evidence. It is important to carefully consider what evidence will be most effective in proving the problem.
3. How can I verify the accuracy of my proof?
One way to verify the accuracy of a proof is to have someone else review and critique it. This could be a colleague, mentor, or even a peer reviewer if you plan to publish your proof. Another way is to double check all calculations and logical steps to ensure they are correct.
4. What if I encounter roadblocks or challenges while trying to prove the problem?
It is common to face challenges when trying to prove a problem. Some strategies for overcoming these roadblocks include taking a break and coming back to the problem with fresh eyes, seeking help or advice from colleagues or mentors, and breaking the problem down into smaller, more manageable parts.
5. What are some common mistakes to avoid when proving a problem?
Some common mistakes to avoid when proving a problem include making assumptions without justification, skipping steps or logical connections, and not thoroughly checking all calculations and reasoning. It is also important to clearly communicate your thoughts and reasoning throughout the proof.
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# Category Archives: Interview Questions – Problem Solving
## Chocolate Feast Problem
By | June 10, 2021
Little Bobby loves chocolate. He frequently goes to his favorite 5 & 10 store, Penny Auntie, to buy them. They are having a promotion at Penny Auntie. If Bobby saves enough wrappers, he can turn them in for a free chocolate. Example n = 15 c = 3 m = 2 Answer 5 + 2… Read More »
## Interview Problem Solving – Mars Exploration
By | April 26, 2019
Problem Sami’s spaceship crashed on Mars! She sends a series of SOS messages to Earth for help. Letters in some of the SOS messages are altered by cosmic radiation during transmission. Given the signal received by Earth as a string, , determine how many letters of Sami’s SOS have been changed by radiation. For example,… Read More »
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# How is the degree of membership calculated?
CRPFALKSD (crpfalksd@aol.com)
Sun, 26 Apr 1998 22:53:54 +0200 (MET DST)
Question: How is the degree of membership calculated?
Thus far, I understand the basic theories of Fuzzy Logic; however, I am
stumpted by exactly how the degree of membership is calculated. In Bart Kosko's
Book entitled, "Fuzzy Thinking: The New Science of Fuzzy Logic," (which I
thought was a good introductory book) there is an example of Fuzzy Logic
controlling an air conditioner.
In Bart's book, he has this graph (you may want to copy and paste it into
Cool Just Right Warm
1 | /\ /\ /\
| / \ O \ / \
| / \ / \ / \
| / X X \
| / / \ / \ \
| / / O / \ \
0 |/______/_____\/_____\______\_____
45 60 65 70 85
At 63 degrees (noted by the 'O' in the graph), the temperature is 80% (0.80 out
of 1.0) Just Right and 15% (0.15 out of 1.0) Cool. How are these percentages
calculated from 63 degrees? What is the formula for this type of calculation?
Is the Pythagorean formula used in calculating (so long as triangles are used)?
I'm sure it's really simple, and I'm just not seeing it yet. All help
appreciated. :-)
Thanks,
David
crpfalksd@aol.com
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# Math
posted by .
To find 62 - 24, you could first find 64 - 24 = 40. Then what should you do? Explain>
• Math -
62 - 24 = 38
• Math -
40-2= 38 because there's 2 extras so I subtract it.
• Math -
40-2= 38 because there's 2 extras so I subtract it. Is this reasonable
• Math -
Reasonable
• Math -
to find 62-24 you could first find 64-24 equals 40
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2. ### algebra
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# How to make the question about the relation between statistics and set theory clearer?
It is closed as lacking details or clarity. According to one mod, the question "might be answerable, but there are a lot of different possible answers, so it would help to rephrase that question into something more specific." However, I don't know what I should include more. Can you give some examples on how this question can be answered in different ways, so I can choose the one I'm interested most?
You should only ask practical, answerable questions based on actual problems that you face. Chatty, open-ended questions diminish the usefulness of our site and push other questions off the front page
(from our help on asking questions).
The set theory/correlation question strikes me as being of this type. I said as much in comments, but because the space for explanations is limited there, here I will elaborate a little.
Set theory is fundamental to mathematics, so it would not be unfair to draw some parallels between the subject question and hypothetical questions one can think of asking about how correlation (or some other statistical concept) might be "related" to, say, addition or logic or real numbers. These hypotheticals might help make the breadth and vagueness of such questions more apparent. In particular,
• What form of "correlation" do you have in mind? Pearson, Spearman, Kendall's Tau, etc.? Some generic concept of "linear relation"? Or even a broad concept of "statistically related"?
• What kind of "relation" are you considering and why might it be important? Many correlation coefficient formulas involve addition in some form, but in most cases it's doubtful this sheds any light either on the concept of correlation or of addition.
• Addition can be exploited in many different ways, according to the many different formulas for the many different forms of correlation. Which way is of interest and why?
Another problem with the original question is that if such questions were considered on-topic here, we would open ourselves to moderating and curating an endless number of vague, open-ended questions of the form "is there any relation between [broad mathematical concept X] and [broad statistical concept] Y?" No such question is sufficiently specific and focused to be appropriate here on CV.
For more guidance, please read the remainder of the help page I cited.
• I see. I'd thought that there is only one concept of correlation and one concept of intersection. It seems that I have to role up my sleeves to read more, because if I don't have the sufficient vocabulary then even experts can't see how to answer the question. That's sad because I expect experts can help me unclear it. Commented May 10, 2021 at 17:21
• I won't rise to that bait--you're coming awfully close to trolling us, now.
– whuber Mod
Commented May 10, 2021 at 17:22
• Trolling? Can you tell me which part I'm trolling? I just sharing my emotion for what I experience. I don't blame you, or beg you to help me. I just state a phenomenon happening many times in many Stack Exchange sites when a novice asking a question and the experts see it's unclear. You may read another example in Biology. Sharing experience and emotion isn't trolling. Commented May 10, 2021 at 17:33
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# BIO 500 Week 8 Rank Correlation Assignment Questions
BIO 500 Week 8 Rank Correlation Assignment BIO 500 Week 8 Rank Correlation Assignment Details: Question 1. Use Excel or SPSS to complete Exercise 1 in the Cumulative Review Exercises on page 485 in the textbook. Permalink: https://nursingpaperessays.com/ bio-500-week-8-r
ion-assignment / APA format is not required, but solid academic writing is expected. Due Tuesday NOV 29th Question 2. Chapter 11 of the textbook covers Analysis of Variance, however the authors state that Testing for Equality of Three or More Population Means might be a better chapter title. Due December 6th Question 3 Use Excel or SPSS to perform the one-way analysis of variance (ANOVA) for the data in From Data to Decision on page 565 of the textbook. APA format is not required, but solid academic writing is expected. Question 4 For the following assignment, use the Rank Correlation that was demonstrated in Chapter 12 of the textbook (page 566). Utilizing Excel or SPSS: Permalink: https://ulcius.com/ bio-500-week-8-r
ation-assignment / ? Use a rank correlation ato test for a correlation between two variables. Use a significance level of ?=0.05. The new health care program in the United States makes provisions for capitation programs where health care insurers work with clinical facilities to perform risk analysis of patients to determine the cost of providing care. The following assignment might be used to assess how much a person smokes. When nicotine is absorbed by the body, cotinine is produced. A measurement of cotinine in the body is therefore a good indicator of how much a person smokes. The reported number of cigarettes smoked per day and the measured amounts of cotinine (in ng/ml) are provided. (The values are from randomly selected subjects in a National Health Examination Survey.) Is there a significant linear correlation? How would you measure the cotinine level in the body? Explain the result. Refer to the Rank Correlation Table. APA format is not required, but solid academic writing is expected. You are not required to submit this assignment to Turnitin. BIO 500 Week 8 Rank Correlation Assignment Get a 10 % discount on an order above \$ 100 Use the following coupon code : NURSING10
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## Binary Search
Wow that is ancient technology you might think – can more be written about that? Probably yes π
So Binary Search has been around for as long as we’ve had sorted lists and the need to locate a specific element in the list. If you have studied Computer Science you most probably has come across binary search algorithms.
If you need to locate an element in a list and just start searching from one end you in average will do n/2 comparisons if there are n elements in the list. That is O(n) performance.
That is probably fine if the list always contains a limited number of elements or you don’t have performance as a requirement.
If the elements you are dealing with can be compared you can sort the list and use binary search to locate the element. The algorithm basically starts in the middle of the sorted list and if the element you are searching for is “smaller” than the middle element the algorithm repeats for the middle element in the first half of the list, and if it is “larger” it repeats in the middle for the last half of the list. And so on, until the element is located.
So with n elements in the list the algorithm does on average log2(n) comparisons. That is O(log n) performance. If there are 1 million elements in the sorted list the algorithm will do at most 19 comparisons before the element is found or it is determined that the element does not exist.
So why are we discussing Binary Search again? Doesn’t JAVA have binary search like built-in?
Absolutely for example Collections has a few variants and so does Arrays. But my use-case is a bit special so neither fits and I felt like rolling my own π
So in my case I have a number of ranges given and I need to relatively fast locate a given number in one of the ranges. For example 4455000000 belongs to the range 4450000000..4459999999. In the particular challenge these strings of digits are actually not treated as numbers and they might actually have different lengths, so 44550000 also would belong to the range 4450000000..4459999999…
So I have a list of ranges, they can be sorted, and I have a string of digits and need to know which range it belongs to. That of course could be solved by normalizing my single string of digits and turn it into a short range like 4455000000..4455000000 and the standard `Collections.binarySearch()` would probably do nicely.
But that is no fun. Also the range contains other stuff than just the low and high values of the range and it would feel unnatural to turn my string of digits into a range. In my “day-time job” implementation (which is not the same as the one presented here) the normalization takes place in the comparison so we never have to bother about which length of the string of digits is the right. It could be 16, it could be 18 or even 19 – we don’t care and do not have to make a decision about that π
So I decided to roll my own completely generic `BinarySearcher<T>` that can do binary searches on sorted lists locating anything that can be compared to the elements in the list. It uses a `BiComparator<T, K>` because it needs to be able to compare elements of potentially different types. As the `BinarySearcherSimpleIntegerTest` illustrates it can also be used with simple types.
In my setup the `BinarySearcher` owns the list of elements to be searched because it makes best sense in my setup. But it can easily be turned into a static method and be used in other setups.
Enjoy π
Posted in Java | Tagged , | Leave a comment
## Mutable
So you might wonder – “Mutable” – what is that all about?
In general we prefer things to be Immutable. With immutability comes improved performance because we do not have to copy things around and the JIT can do some optimizations. And there are less surprises, less WTFs because referenced things doesn’t suddenly change state.
One of my favorites is Optional. It either has a present value or is empty. Even after a few thousand lines of code a reference to an `Optional` is either empty or has the value it was initialized with. Love it! No surprises, no WTFs…
It also has a nice stream-like API with `map()` and `filter()` and whatnot.
Then in java 16 came `records`. And I love records. With a few lines of code you define immutable data carriers with sane `equals()`, `hashCode()` and `toString()` implementations and natural named accessors not prefixed with `get` as in the JavaBeans legacy. Love it! No surprises, no WTFs…
So whenever I’m going to introduce a class with the primary responsibility of moving some information around I try to use records. Sometimes annotated with lombok `@Builder` because it makes construction of these instances somewhat easier and easily split over multiple lines…
And as already mentioned one of the selling points of records is their immutability!
Then recently I was refactoring a rather complex piece of code that dealt with parsing some interesting external files into some rather complex hierarchies and at the bottom sometimes enhancing higher-level items with some information from internal systems. In other words, adding information at a later stage, information not available at construction time.
I believe we’ve all been there…
And I wanted to refactor the code to use records simply because there is much less code to maintain (less TCO) and there are less … surprises…
This phrase seems to come up often… “Less surprises”… Believe me I am getting old and with age comes some resistance towards surprises. I don’t need them and basically want less of them. And compared to how I coded like 25 years ago that mentality has changed my coding style A LOT! So, less surprises, less WTFs are important to me π
My younger co-workers might find me grumpy at times, but with less surprises comes increased maintainability and with that less TCO. So… Lets get on with it!
To solve my hierarchy files parsing issue I needed a `TopLevel` type and a `Child` type, they have things in common so they implement a `HierarchyElement` interface with reference to `parent()` which for `TopLevel` is it self and for `Child` is another child or a `TopLevel`. Both has a `children()` list and some other stuff. And `TopLevel` has a field that can only be set when parsing a deep child and some children might get values from another system later, before we’re done parsing… Got it so far? π
One way of doing this would be to – when we have the values for the “late fields” – replace the `Child` or the `TopLevel` with a copy of the existing with a field or two added…
With a complex hierarchy that is just not trivial. Doable but not trivial and the code would be… Complex and full of surprises. Some junior developer might come along and think WTF – not needed – DELETE… Hmmm…
So I decided to introduce `Mutable<T>`. I could have used `AtomicReference<T>` but I need no concurrency and has some other needs I might as well cater for.
One thing that really annoyed me in the original implementation was 9 lines of code that basically checked if the top- or the child already had a field set and if not, set the field to some new value. All those ifs-and-buts needs to be unit-tested and then you have some very complex unit-tests to cater for something extremely simple. Must be tested or it does not work but considering TCO…
``````if (someForeignThing != null) {
if (top.getSomeField() != null) {
top.setSomeField(someForeignThing.getValueForField());
}
if (child.getSomeField() != null) {
child.setSomeField(someForeignThing.getAnotherValue());
}
if (child.getAnotherField() != null) {
child.setAnotherField(someForeignThing.getCompletelyDifferentField());
}
}``````
So I came up with `Mutable<T>` to make it possible to set some rather important fields later. Both the `TopLevel` and `Child` types already has mutable collections (list of children) that are set to new `ArrayList<>()` so having one or two other mutable fields does not seem too arcane.
What should it look like… Well, it must have `get()` and `set()` like in `AtomicReference<T>`. Except I like methods that change a single state of an object to return the previous state so we have `T set(T newValue)`.
And since it can supply it’s current value and it can consume a `T`, why not have it implement Supplier<T> and Consumer<T>. Supplier uses the method `accept()` to set a new value so we delegate that to the `set()` method.
And to boil the 9 lines of code down to three we introduce a `setIfNull()` method that only set the mutable value if it is already `null` and we have:
``````if (someForeignThing != null) {
top.someField().setIfNull(someForeignThing.getValueForField());
child.someField().setIfNull(someForeignThing.getAnotherValue());
child.anotherField()
.setIfNull(someForeignThing.getCompletelyDifferentField());
}``````
Less is more! Love it!! And the `null` checking only needs to be unit-tested with the `Mutable<T>` tests. And I do believe even a novice developer can see what is going on π
So the `Mutable<T>` has a not-null concept. It looks very much like the `Optional` empty or not. But slightly different since the `Optional` is immutable so either it has a value present or it is empty. Even the API states this. My mutable has a null or not-null concept, empty is something slightly different and the not-null state can change to another not-null state and even back. So decided to go with a `isNull()` method simply because it describes the situation better. Also there is no `NoSuchElementException` involved when `get()` returns `null` π
We are not trying to solve the infamous `NullPointerException` thing here π
So the core of it is like the following:
``````public final class Mutable<T> implements Supplier<T>, Consumer<T> {
private T value;
public boolean isNull() {
return value == null;
}
public boolean setIfNull(T value) {
if (isNull()) {
set(value);
return true;
}
return false;
}
public T set(T value) {
final var old = this.value;
this.value = value;
return old;
}
@Override
public T get() {
return value;
}
@Override
public void accept(T value) {
set(value);
}
@Override
public String toString() {
return String.valueOf(value);
}
}``````
So `get()`, `set()` and `isNull()` are the very core part of the implementation – the other methods just delegates. Again this makes testing simpler and the JIT can definitely inline those.
The `setIfNull()` returns `true` if the `Mutable` was changed – that information might come in handy sometimes.
There is also a `toString()` but that is mostly for debugging purposes and for ensuring a relatively sane output in logs etc.
## Things to consider
First, this thing is not thread-safe and is not meant to be. It is for temporary eventually lazy settable items in same-thread situations. If you need thread-safety and concurrency use the AtomicReference<T> and similar. Period π
There are no `equals()` and `hashCode()` and why not?
Basically these methods are for things to be put in collections specifically in Sets and Maps. Since Mutables are … mutable … they should never be put into Sets or used as keys in Maps. Never ever. Remember surprises and WTFs.
One could consider implementing `equals()` – it is not difficult to come up with a reasonable implementation, eg:
``````@Override
public boolean equals(Object obj) {
if (this == obj) {
return true;
}
return obj instanceof Mutable<?> other
&& Objects.equals(value, other.value);
}``````
That would work perfectly fine for all trivial cases. But please do read the JavaDocs for Object.equals():
• It is reflexive: for any non-null reference value `x`, `x.equals(x)` should return `true`.
• It is symmetric: for any non-null reference values `x` and `y`, `x.equals(y)` should return `true` if and only if `y.equals(x)` returns `true`.
• It is transitive: for any non-null reference values `x`, `y`, and `z`, if `x.equals(y)` returns `true` and `y.equals(z)` returns `true`, then `x.equals(z)` should return `true`.
• It is consistent: for any non-null reference values `x` and `y`, multiple invocations of `x.equals(y)` consistently return `true` or consistently return `false`, provided no information used in `equals` comparisons on the objects is modified.
• For any non-null reference value `x`, `x.equals(null)` should return `false`.
The fourth bullet – it is consistent – we cannot ensure that with a `Mutable<T>` since after a call to `set()`, `setIfNull()` or `accept()` the method might return something different. The documentation actually allows that, but I hate surprises and things that are equal one moment should still be equal even a few lines of code later…
And another thing is if we implement `equals()` we really should also implement `hashCode()` (and really consider `compareTo()` too but that is another even longer story) with a similar contract and we’re basically in deep trouble. DO NOT. NO NO DO NOT. We hate nasty surprises π We like less WTFs.
By not implementing these methods they revert to the `Object` implementations which are based on object identity and are very deterministic. No surprises. We like π
Look for nasty surprises here: `SurprisesTest.java`
## Records in Sets and as keys in Maps
So since records are immutable they are perfectly safe to use in Sets and as keys in Maps?
Well not quite. If your implementation contains at least one mutable field (a `Mutable<T>`, an `AtomicSomething` or some mutable collection) then you are out for nasty surprises. Unless you specifically override `equals()` and `hashCode()` to be stable towards the mutable elements.
And then while you are at it, consider implementing `Comparable<T>`
So be careful out there, use the `Mutable<T>` with care, don’t ever overdo it and please do not ever be tempted to implement `equals()` and `hashCode()` or use instances in Set or as keys in Maps.
Posted in Java | Tagged , | Leave a comment
## Generating garbage…
At JCrete the last couple of years I’ve had the pleasure to be socializing with, among others, HFT people like Peter Lawrey and Martin Thompson.
Those HFT dudes really makes me thinking when I’m implementing stuff. Thinking about how much garbage I create.
In High Frequency Trading systems you cannot afford garbage collection and therefore they are doing lots of tricks to get around generating garbage.
With the stuff I’m doing a gc pause is generally not critical, so generating garbage is no big deal. Writing easy to read- and maintain code is more important.
But then again, it does not harm to think about garbage.
Posted in Fun, Java | Tagged , | 1 Comment
## SafeEquals
I’m currently designing a user-authentication OAuth2 based service.
I’m trying very hard never ever to reveal anything about users or passwords. Credential-lookup by userid is always done twice. If the user is not found a known dummy-user is looked up instead. If the user is found, a “by guarantee not existing” user is looked up. Just to try to make the timings of existing and not-existing users the same.
Internally a lot of hashing takes place with random and known stuff. Valid and invalid users and passwords all take exactly the same “route”.
At the end, byte arrays are compared, and if they contain the same binary information, the user is authenticated.
In order to ensure this check timing-wise is independent of the outcome, a special `equals(byte[], byte[])` method is implemented. Also, a rather special counterpart, the `neverEquals(byte[], byte[])` is implemented. This does the exact same comparisons as the equals method, just the outcome is always false, to be used when the user is invalid. Continue reading
Posted in Java, Open Source | Tagged , | Leave a comment
## YN
Yeah, I know, this isn’t rocket science. But it is rather nice π
So, I’m working on this JEE project, backed by an Oracle database. Some of my work involves refactoring CMP 2.1 entities into JPA entities.
Here I’m faced with the issue, that sometimes `true/false` columns are modeled using `CHAR(1)` as ‘Y’ vs ‘N’. And at other times they are modeled using `INTEGER` as 1 vs 0. Sometimes both versions are using in the same table.
The natural datatypes in JPA are `String` and `int`. So I’m having code doing `βYβ.equals(colum)` and `column == 1` etc. For the `String` I’ve generally been using an `enum` called `YN` having 2 constants, Y and N. So I can do `column == YN.Y`. Still a bit annoying though…
Just today I realized I could fix both issues with a simple version of my `YN` enum below. This works both for the `CHAR` ‘Y’ / ‘N’ case and the `INTEGER` 1 / 0 case.
And it’s damn simple containing 2 simple convenience methods that does make life simpler:
```/** * Being used in JPA entities as @Enumerated(EnumType.STRING) or @Enumerated(EnumType.ORDINAL) (N=0, Y=1) */ public enum YN { N, Y; public final boolean bool() { return this == Y; } public static final YN bool(boolean b) { return b ? Y : N; } }```
Now I can do `column.bool()` and I can do `YN.bool(something)` the other way. And the mapping in the JPA entity takes care of the rest.
Nice and simple, definitely not rocket science.
Posted in Databases, Java | Tagged , , , , , , | Leave a comment
## Unit testing really simple implementations
As an independent subcontractor (freelance consultant) I get to work in various very different organisations, with very different approaches towards testing in general and unit-testing in particular.
Some places unit-testing is mandatory and there might even be a code-coverage tool where you must strive to meet a specific threshold.
At other places unit-tests are more like integration tests, where the unit-testing is considered implicit in the integration-unit-tests.
And then there are places where unit-tests are of your own choice, as long as you do not spend too much (measurable) time on them.
In the latter cases you as an experienced developer might feel that unit-testing very simple stuff is superfluous.
Not so! In my experience, it is when you do these simple no-brainer implementations that you make mistakes, simply because you do not have to think. Also, unit-testing simple utility stuff demonstrates corner-cases that might suddenly be used or mis-used in the code where changing even the simplest implementation might break important business logic
In this post I will go through a few very simple utility methods that I’ve worked on recently and talk about some of the finer details β and why unit-testing even the simplest is really important.
Posted in Java | Tagged , | Leave a comment
## Bad sample code…
Okay, I just HATE bad code.
I’m not religious about how you express your business logic in the code. I don’t mind βdifferentβ indentation or long methods although I try to keep my own code as short and simple as possible. I prefer lots of small classes over a few huge, and I prefer a bunch of short methods (βa screenfullβ) over a few huge ones. Complex IF-clauses formed as simple methods etc.
Whenever I’m maintaining code developed by others, I accept the previous developers different habits. I might rewrite (also known as refactoring) parts, especially if there are unit-tests. But only if I think it would benefit the quality and maintainability of the code.
But, when reading books or articles I get really REALLY annoyed when I encounter bad code and bad habits in samples etc. Whenever I come across these things, where the quality is as if the developer is not really a JAVA developer, absentminded or just plain drunk, I loose interest in the rest of the content. As if, if the code is really bad (and untrustworthy) how can I then trust the rest of the content?
Posted in Fun, Java | 1 Comment
## Handing over JAVA components to L3 support…
Java Specialist13:13
Hi
Udby, Jesper13:13
Hi
Java Specialist13:13
we are unable to build with existing set up
can you give us ear file to me
Udby, Jesper13:14
no, unfortunately not, I’m working from home today and do not have access to my dev env
Java Specialist13:14
i will put in weblogic server
Udby, Jesper13:14
I suggest you configure you’r maven setup to use the proper repo so you can build it. It has to be done anyways…
Java Specialist13:15
can you share your maven to me
_
folder
Udby, Jesper13:15
no… its on m dev env too…
the settings.xml linked from the document i sent you contains information about the repo – you should be able to use that?
which weblogic server btw?
Java Specialist13:18
weblogic server 11gR1
Udby, Jesper13:18
you have a local setup?
Java Specialist13:18
yes
Udby, Jesper13:18
nice:)
you got a local db setup too?
Java Specialist13:19
datasouce
Udby, Jesper13:20
yes, datasource. does it point to a local db or one of our dev environments?
Java Specialist13:20
yes
Udby, Jesper13:21
do you have a local db?
Java Specialist13:21
yes
Udby, Jesper13:21
nice:)
how’s the schema setup?
Java Specialist13:21
there
Udby, Jesper13:21
there?
Java Specialist13:21
yes
Udby, Jesper13:22
ok, i rephrase: how are the users/schemas, tables, sequences etc setup?
DDL
Java Specialist13:22
i will give schema name
yes
Udby, Jesper13:23
i don’t know if you are aware of it, but “we” are using what is called “database migration” scripts to do the DDL
Java Specialist13:24
1 min
Udby, Jesper13:24
1 min what?
Java Specialist13:25
wait
database migrationΒ is no there with me
Udby, Jesper13:27
well thats not important if you are able to setup the schema/users, tables etc yourself
I was just wondering where you got the details from
Java Specialist13:28
webservice setup document
Udby, Jesper13:29
ok
Java Specialist13:31
Do you have this type of document with you_
Udby, Jesper13:32
nope, not here, might have it in the office…
Java Specialist13:32
ok
is there anu body who will share information_
Udby, Jesper13:33
what information?
Java Specialist13:34
regadring configuartion of java compenent with oSB
Udby, Jesper13:34
what component and what kind of configuration?
Java Specialist13:36
two ear file ie two java componet and after that i will confuge jMS queue with SFDCΒ
and OSB
Udby, Jesper13:36
first, there are more than two components, you need to tell me which components you need information about
then I can tell you which datasouces and jms queues are necessary
but you can see that in the code, if you like?
take the xxx-feed-parser (from memory, you can see the proper name in the poms/mail/source etc)
Java Specialist13:38
ok
Udby, Jesper13:38
it needs a jms queue, can’t remember the jndi name, but it’s in the annotation in the MDB
and it needs a datasource, can’t remember the jndi name, but its in the persistence.xml file
Java Specialist13:39
i have JMSQUEue is there with me
Udby, Jesper13:39
what?
Java Specialist13:40
while i am doing yYYY project
Udby, Jesper13:40
ok, good, but its probably not the same queue:)
Java Specialist13:41
ok
Udby, Jesper13:41
you can see the jndi name of the jms queue in the annotation in the mdb – message driven bean
Java Specialist13:44
1min
wait
i am opening weblogic server
Java Specialist13:53
yes
Udby, Jesper13:54
yes?
Java Specialist13:55
i am able to jndi nameof the JMS queue
for YYYY
Udby, Jesper13:55
ok
I wasn’t referring to the jndi name of the jms queue in yyyy
you do know what an annotation is, right?
Java Specialist14:00
no
Udby, Jesper14:00
ok, you are in trouble
do you know what a message driven bean is?
Java Specialist14:01
yes
Udby, Jesper14:03
good
in the mdb (aka message driven bean) there is an annotation that tells it which jms queue it should connect to
i can’t remember the proper name of it, but somewhere in the xxx-feed-ejb (i think it is) there is the java source of a message driven bean – I even think it is called something like Xxx…Mdb or similar. In that source you can find the annotation containing the name of the jms queue
also, in that project, there is a persistence.xml – you do know what that is?
Java Specialist14:06
jesper please share me the ear file to deploy javacomponents in weblogic server
Udby, Jesper14:06
sorry, i cant
you should configure your maven to use the proper repo, then you can build it yourself
btw: there is more than one ear
one component = one ear
## Salesforce wsc (sfdc-wsc): new release 23 coming
The current situation is a mess. There are now 3 known versions of the sfdc-wsc tool:
I will try to merge relevant changes from the codebase at mvnrepository but please notice it has already forked. The code style at mvnrepository is slightly different to my style and I take the liberty to refactor stuff in order to keep classes clean and without too much responsibility (e.g. ConnectorConfig should not have the responsibility of creating connections, adding stream wrappers and what not).
Released – wsc-23.jar – with the following:
Posted in Java, Salesforce | Tagged , , , | Leave a comment
## Salesforce wsc hacking: yet another Open Source encounter?
As I wrote in an earlier post I’ve volunteered to become a committer to the wsc tool and are in the progress of making minor tweaks and enhancements for a coming release 23.
My plan was that this release should be no more than a “maintenance” release covering most of the relevant issues. And I do have a number of ideas for major changes for a possible release 24, e.g.:
• Fixing issue 51 – or at least provide a usable implementation hopefully inspired by input from Martin Haynes
• Reworking/reimplementing the wsdlc tool and templating mechanism
• Reworking infrastructure parts, possibly using JAXB/JAX-WS
• Mavenize the build, divide build tools and infrastructure, 2 perhaps 3 artifacts
• Adding unit tests
I stopped further developments on the “23 branch” when I realized that there apparently already was a “fork” on github.
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# Limit of a sequence
The sequence given by the perimeters of regular n-sided polygons that circumscribe the unit circle has a limit equal to the perimeter of the circle, i.e. $2\pi r$. The corresponding sequence for inscribed polygons has the same limit.
n n sin(1/n)
1 0.841471
2 0.958851
...
10 0.998334
...
100 0.999983
As the positive integer n becomes larger and larger, the value n sin(1/n) becomes arbitrarily close to 1. We say that "the limit of the sequence n sin(1/n) equals 1."
In mathematics, the limit of a sequence is the value that the terms of a sequence "tend to".[1] If such a limit exists, the sequence is called convergent. A sequence which does not converge is said to be divergent.[2] The limit of a sequence is said to be the fundamental notion on which the whole of analysis ultimately rests.[1]
Limits can be defined in any metric or topological space, but are usually first encountered in the real numbers.
## History
The Greek philosopher Zeno of Elea is famous for formulating paradoxes that involve limiting processes.
Leucippus, Democritus, Antiphon, Eudoxus and Archimedes developed the method of exhaustion, which uses an infinite sequence of approximations to determine an area or a volume. Archimedes succeeded in summing what is now called a geometric series.
Newton dealt with series in his works on Analysis with infinite series (written in 1669, circulated in manuscript, published in 1711), Method of fluxions and infinite series (written in 1671, published in English translation in 1736, Latin original published much later) and Tractatus de Quadratura Curvarum (written in 1693, published in 1704 as an Appendix to his Optiks). In the latter work, Newton considers the binomial expansion of (x+o)n which he then linearizes by taking limits (letting o→0).
In the 18th century, mathematicians such as Euler succeeded in summing some divergent series by stopping at the right moment; they did not much care whether a limit existed, as long as it could be calculated. At the end of the century, Lagrange in his Théorie des fonctions analytiques (1797) opined that the lack of rigour precluded further development in calculus. Gauss in his etude of hypergeometric series (1813) for the first time rigorously investigated under which conditions a series converged to a limit.
The modern definition of a limit (for any ε there exists an index N so that ...) was given by Bernhard Bolzano (Der binomische Lehrsatz, Prague 1816, little noticed at the time) and by Karl Weierstrass in the 1870s.
## Real numbers
The plot of a convergent sequence {an} is shown in blue. Visually we can see that the sequence is converging to the limit 0 as n increases.
In the real numbers, a number $L$ is the limit of the sequence $(x_n)$ if the numbers in the sequence become closer and closer to $L$ and not to any other number.
### Examples
• If $x_n = c$ for some constant c, then $x_n \to c$.[proof 1]
• If $x_n = 1/n$, then $x_n \to 0$.[proof 2]
• If $x_n = 1/n$ when $n$ is even, and $x_n = 1/n^2$ when $n$ is odd, then $x_n \to 0$. (The fact that $x_{n+1} > x_n$ whenever $n$ is odd is irrelevant.)
• Given any real number, one may easily construct a sequence that converges to that number by taking decimal approximations. For example, the sequence $0.3, 0.33, 0.333, 0.3333, ...$ converges to $1/3$. Note that the decimal representation $0.3333...$ is the limit of the previous sequence, defined by
$0.3333...\triangleq\lim_{n\to \infty} \sum_{i=1}^n \frac{3}{10^i}$.
• Finding the limit of a sequence is not always obvious. For instance, $\lim_{n\to\infty} \left( 1 + \frac{1}{n} \right)^n$, also known as the number e, or the Arithmetic–geometric mean. The squeeze theorem is often useful in such cases.
### Formal definition
We call $x$ the limit of the sequence $(x_n)$ if the following condition holds:
• For each real number $\epsilon > 0$, there exists a natural number $N$ such that, for every natural number $n > N$, we have $|x_n - x| < \epsilon$.
In other words, for every measure of closeness $\epsilon$, the sequence's terms are eventually that close to the limit. The sequence $(x_n)$ is said to converge to or tend to the limit $x$, written $x_n \to x$ or $\lim_{n \to \infty} x_n = x$.
If a sequence converges to some limit, then it is convergent; otherwise it is divergent.
### Properties
Limits of sequences behave well with respect to the usual arithmetic operations. If $a_n \to a$ and $b_n \to b$, then $a_n+b_n \to a+b$, $a_nb_n \to ab$ and, if neither b nor any $b_n$ is zero, $a_n/b_n \to a/b$.
For any continuous function f, if $x_n \to x$ then $f(x_n) \to f(x)$. In fact, any real-valued function f is continuous if and only if it preserves the limits of sequences (though this is not necessarily true when using more general notions of continuity).
Some other important properties of limits of real sequences include the following.
• The limit of a sequence is unique.
• $\lim_{n\to\infty} (a_n \pm b_n) = \lim_{n\to\infty} a_n \pm \lim_{n\to\infty} b_n$
• $\lim_{n\to\infty} c a_n = c \cdot \lim_{n\to\infty} a_n$
• $\lim_{n\to\infty} (a_n \cdot b_n) = (\lim_{n\to\infty} a_n)\cdot( \lim_{n\to\infty} b_n)$
• $\lim_{n\to\infty} \left(\frac{a_n}{b_n}\right) = \frac{\lim_{n\to\infty} a_n} {\lim_{n\to\infty} b_n}$ provided $\lim_{n\to\infty} b_n \ne 0$
• $\lim_{n\to\infty} a_n^p = \left[ \lim_{n\to\infty} a_n \right]^p$
• If $a_n \leq b_n$ for all $n$ greater than some $N$, then $\lim_{n\to\infty} a_n \leq \lim_{n\to\infty} b_n$
• (Squeeze theorem) If $a_n \leq c_n \leq b_n$ for all $n > N$, and $\lim_{n\to\infty} a_n = \lim_{n\to\infty} b_n = L$, then $\lim_{n\to\infty} c_n = L$.
• If a sequence is bounded and monotonic then it is convergent.
• A sequence is convergent if and only if every subsequence is convergent.
These properties are extensively used to prove limits without the need to directly use the cumbersome formal definition. Once proven that $\frac{1}{n} \to 0$ it becomes easy to show that $\frac{a}{b+\frac{c}{n}} \to \frac{a}{b}$, ($b \ne 0$), using the properties above.
### Infinite limits
A sequence $(x_n)$ is said to tend to infinity, written $x_n \to \infty$ or $\lim_{n\to\infty}x_n = \infty$ if, for every K, there is an N such that, for every $n \geq N$, $x_n > K$; that is, the sequence terms are eventually larger than any fixed K. Similarly, $x_n \to -\infty$ if, for every K, there is an N such that, for every $n \geq N$, $x_n < K$. If a sequence tends to infinity, or to minus infinity, then it is divergent (however, a divergent sequence need not tend to plus or minus infinity).
## Metric spaces
### Definition
A point x of the metric space (X, d) is the limit of the sequence (xn) if, for all ε > 0, there is an N such that, for every $n \geq N$, $d(x_n, x) < \epsilon$. This coincides with the definition given for real numbers when $X = \mathbb{R}$ and $d(x, y) = |x-y|$.
### Properties
For any continuous function f, if $x_n \to x$ then $f(x_n) \to f(x)$. In fact, a function f is continuous if and only if it preserves the limits of sequences.
Limits of sequences are unique when they exist, as distinct points are separated by some positive distance, so for $\epsilon$ less than half this distance, sequence terms cannot be within a distance $\epsilon$ of both points.
## Topological spaces
### Definition
A point x of the topological space (X, τ) is the limit of the sequence (xn) if, for every neighbourhood U of x, there is an N such that, for every $n \geq N$, $x_n \in U$. This coincides with the definition given for metric spaces if (X,d) is a metric space and $\tau$ is the topology generated by d.
The limit of a sequence of points $\left(x_n:n\in \mathbb{N}\right)\;$ in a topological space T is a special case of the limit of a function: the domain is $\mathbb{N}$ in the space $\mathbb{N} \cup \lbrace +\infty \rbrace$ with the induced topology of the affinely extended real number system, the range is T, and the function argument n tends to +∞, which in this space is a limit point of $\mathbb{N}$.
### Properties
If X is a Hausdorff space then limits of sequences are unique where they exist. Note that this need not be the case in general; in particular, if two points x and y are topologically indistinguishable, any sequence that converges to x must converge to y and vice versa.
## Cauchy sequences
The plot of a Cauchy sequence (xn), shown in blue, as xn versus n. Visually, we see that the sequence appears to be converging to a limit point as the terms in the sequence become closer together as n increases. In the real numbers every Cauchy sequence converges to some limit.
A Cauchy sequence is a sequence whose terms ultimately become arbitrarily close together, after sufficiently many initial terms have been discarded. The notion of a Cauchy sequence is important in the study of sequences in metric spaces, and, in particular, in real analysis. One particularly important result in real analysis is Cauchy characterization of convergence for sequences:
A sequence is convergent if and only if it is Cauchy.
## Definition in hyperreal numbers
The definition of the limit using the hyperreal numbers formalizes the intuition that for a "very large" value of the index, the corresponding term is "very close" to the limit. More precisely, a real sequence $(x_n)$ tends to L if for every infinite hypernatural H, the term xH is infinitely close to L, i.e., the difference xH - L is infinitesimal. Equivalently, L is the standard part of xH
$L = {\rm st}(x_H)\,$.
Thus, the limit can be defined by the formula
$\lim_{n \to \infty} x_n= {\rm st}(x_H),$
where the limit exists if and only if the righthand side is independent of the choice of an infinite H.
1. Proof: choose $N = 1$. For every $n > N$, $|x_n - c| = 0 < \epsilon$
2. Proof: choose $N = \left\lfloor\frac{1}{\epsilon}\right\rfloor$ (the floor function). For every $n > N$, $|x_n - 0| \le x_{N+1} = \frac{1}{\lfloor1/\epsilon\rfloor + 1} < \epsilon$.
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# Special Studies in Finance Prelims Questions 2012
0
PRELIMINARY EXAMINATION
00000012 / 3044 / 0005 / BMS / Special Studies in Finance (SSF) / V Semester / Marks 60 M / 2.00 Hrs. Seat ————
Â
Note:
1. Attempt both the sections on same answer sheet
2. All the questions are compulsory in Section I and attempt any three questions from Section II
3. Working notes should be part of the answers
4. Figures to indicate right will give you full Marks
5. Use of Simple Calculator is allowed
SECTION I
Q1 a. Concept Testing                                                                                                                                05
1)     Flash Report              2) ESOP         3) AS 31         4) Non – Monetary items as per AS 11
5) AS 17
Q1 b. Attempt any two                                                                                                                                10
1)Â Â Â Â Â Mahesh used his car of the value Rs. 89,730 for a lease rental of Rs. 30,000 payable at the end of each year of 5 years. Calculate the interest rate implicit in the lease.
2)     The following data pertain to division of Nidhi incorporated. The company’s required rate of return on invested capital is 8%.
Particulars Division Sales value (Rs.) ? Income (Rs.) ? Average investment (Rs.) ? Sales margin % 25% Capital turnover (Times ) ? ROI (%) 25% Residual income/ economic value added (Rs.) 1,20,000
3)Â Â Â Â Â Minal Ltd. acquired a machine on 1.4.2008 costing US \$ 40,000. The suppliers agreed following terms of payment:
Date Terms of payment 1.4.2008 1.4.2009 1.4.2010 Down payment 50% 25% 25%
The company depreciates machinery @ 10% on the straight line method. The rate of exchange is steady at US \$ 1= Rs.40 up to 30.9.2009. on 1.10.2009, due to an official revaluation of rates , the exchange rate is adjusted to US \$ 1=48. the extracts of the relevant entries in the profit & loss A/c. for the year ended 31st March 2010 and the balance sheet as on date, showing such workings as necessary as per AS-11.
Â
Q2. Case Study                                                                                                                                            15
You are approached by financial institution to appraise the following project:
Name of the borrowers: NIDHI Chemicals Private Limited
Proposed loan is taken to set up a chemical unit for processing industrial waste into a marketable product XYZ. The product has a demand for 50000 liters. The processing costs include variable cost of Rs. 5 per liter and fixes cost (excluding depreciation) Rs. 30000 per year. Advertising expenses are also expected to be Rs. 20000 per year.
XYZ can be sold at Rs.10 per liter. Raw material (Industrial Waste) is available at Re. 1 per liter. The capital cost of Chemical Unit is Rs. 750000.
The company has applied for a loan of Rs. 600000 for a term of 10 years and that is over the life of the asset. The promoters of the company are young, dynamic and highly qualified people but are doing the venture for the first time. The promoters are unable to provide any collateral security for the loan expect personal Guarantee for their parents.
They have thought of this project after market research. The said research has stated in the risk factors about invasion of Malaysia in chemical market and drastic reduction in selling price of similar products.
The above unit is a SSI unit and its average tax rate is 20%. Interest rate is 12% p.a.
Loan is repayable equally in 10 annual installments along with interest at the end of each year. You are required to:
1. Give the cash flow generated by the above project for the first 3 years only.
2. Calculate the Debt Service Coverage Ratio for the above 3 years.
3. 3.     Prepare the flash report presenting the above information to the financial institution.
Â
SECTION II (Any Three)
Q3. M/s. Sehwag and Co. purchased a machinery worth Rs. 7,92,500/- (cash pirce) from M/s Gambhir and Bros. on 1st January 2008. It was agreed by both the parties that eh payment of machinery will be done as under:
Down payment Rs. 1,58,500/- on the date of purchase and the balance will be discharge in four half-yearly installment of RS. 2 lakhs each, commencing from 30th June, 2008.
Your are required to prepare Machinery Account and M/s. Gambhir and Bros. account in the books of M/s. Sehwag and Co. for calendar years 2008 and 2008, considering that M/s. Shewag and co. closes its books of account on 31st December every year and charges depreciation on machinery @10% p.a. on written Down Value Method.
Q4. From the following information for NIDHI Ltd. For the year ended 31st March 2009 calculate the deferred tax asset/ liability as per AS-22.
 Accounting profit Rs. 1,00,000 Book profit as per MAT (minimum alternate tax ) Rs. 90,000 Profit as per income tax act Rs. 10,000 Tax rate 30% MAT rate 10%
Q5. PASS Journal Entries for the following transactions in foreign currency and in the books of nidhi Ltd. Nidhi Ltd. imported raw materials worth US \$ 40,000 on 12th December 2004. The Exchange rate for US \$ 1 as on 12-12-2004 was Rs. 46.50Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 10
        Date of Payment        Payment Exchange rate for 1US \$ 23-02-2005 US \$ 18,000 RS.48.75 21-03-2005 US \$ 12,000 RS. 47.25 10-04-2005 US \$ 10.000 Rs. 49.50
The accounting year of the company ended on 31st March .The exchange rate as on 31st March 2005 for US\$1was Rs. 46
Q6. Short Notes (Any Two)Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 10
a)Â Â Â Â Â Profit Maximization v/s Wealth Maximization
b)Â Â Â Â Â Steps involve in Term Loan
c)Â Â Â Â Â Distinguish Between Hire Purchase and Lease
d)Â Â Â Â Explain Merchant Banking
BEST OF LUCK
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# Description
Goal: Practice adding up, until all cards are gone.
MultipleData 1: 4 stars
Level 1: Find equivalent cards up to 11 with addition of 1.
MultipleData 2: 4 stars
Level 1: Find equivalent cards up to 12 with addition of 2.
MultipleData 3: 4 stars
Level 1: Find equivalent cards up to 13 with addition of 3.
MultipleData 4: 4 stars
Level 1: Find equivalent cards up to 14 with addition of 4.
MultipleData 5: 4 stars
Level 1: Find equivalent cards up to 15 with addition of 5.
MultipleData 6: 5 stars
Level 1: Find equivalent cards up to 16 with addition of 6.
MultipleData 7: 5 stars
Level 1: Find equivalent cards up to 17 with addition of 7.
MultipleData 8: 5 stars
Level 1: Find equivalent cards up to 18 with addition of 8.
MultipleData 9: 5 stars
Level 1: Find equivalent cards up to 19 with addition of 9.
MultipleData 10: 5 stars
Level 1: Find equivalent cards up to 20 with addition of 10.
# Details
Differential Revisions
dekumar updated the task description. (Show Details)May 9 2020, 12:22 PM
What means :
Goal: Find cards with equivalent quantities until 11.
Star 1 is 3 years old.
At this age they are supposed to quantify a maximum of 3.
I have done the job to describe this dataset in the general description, why did you not use it?
https://phabricator.kde.org/T13122
What means :
Goal: Find cards with equivalent quantities until 11.
Star 1 is 3 years old.
At this age they are supposed to quantify a maximum of 3.
I have done the job to describe this dataset in the general description, why did you not use it?
https://phabricator.kde.org/T13122
Hi, For each multiple datasets the table of 1, 2, 3 .. .10 will be used. As up to 11 means that the addition memory game with 1 as it would have maximum sum up to 11. This changes were suggested by @timotheegiet during the proposal review period. I would have a discussion again once and update the description if required.
Thanks!
@echarruau : you put the link to this task in your comment, I guess you meant https://phabricator.kde.org/T12428
Also, the dataset description you put for "Addition and subtraction memory game against Tux" only contains addition related data, and too much redundant content...
Addition-only and Subtraction-only memory games can not really have same content/description (as the formula is different...).
And the way the "Addition and subtraction memory game" works, it is relevant to use the same dataset description as for subtraction only.
So for the dataset content I suggested for this, as @dekumar described in the proposal, it is based on current levels we have.
I guess we should just adapt the number of stars of the datasets to fit the content.
Yes, we need to find multipledatasets that make sens.
Yesterday when I reviewed the descriptions I had the feeling that it was a
quick copy/paste job done in hurry because we asked to have descriptions! I
did not realise that it was the proposal content. My fault I should have
multipledataset that will be useful from the start. We are not doing
descriptions because we have to do them, we are doing the description to
define what we will do :)
Le dim. 10 mai 2020 à 20:17, Timothée Giet <noreply@phabricator.kde.org> a
écrit :
@echarruau https://phabricator.kde.org/p/echarruau/ : you put the link
https://phabricator.kde.org/T12428
Also, the dataset description you put for "Addition and subtraction memory
game against Tux" only contains addition related data, and too much
redundant content...
Addition-only and Subtraction-only memory games can not really have same
content/description (as the formula is different...).
And the way the "Addition and subtraction memory game" works, it is
relevant to use the same dataset description as for subtraction only.
So for the dataset content I suggested for this, as @dekumar
https://phabricator.kde.org/p/dekumar/ described in the proposal, it is
based on current levels we have.
I guess we should just adapt the number of stars of the datasets to fit
the content.
https://phabricator.kde.org/T13122
*To: *dekumar, timotheegiet
*Cc: *echarruau, amankumargupta, AkshayCHD, timotheegiet, jjazeix,
dekumar, sanjayshetty, parimalprasoon, harrymecwan, ganeshredcobra,
@echarruau @timotheegiet Can I implement the above described multiple datasets or is there any change which needs to be done?
to me the dataset description here is good, just the star numbers need to be changed to be better adapted to the actual difficulty of the levels.
dekumar updated the task description. (Show Details)Jun 7 2020, 5:58 PM
@echarruau Can you please review the difficulty level of all multiple datasets once?
dekumar closed this task as Resolved.Jun 21 2020, 4:28 PM
Description is not correct, again the stars are wrong. Please change them to follow the rules I gave you.
Description is not correct, again the stars are wrong. Please change them to follow the rules I gave you.
Sure, I will update it and make a commit to update it for the datasets content too which I have implemented.
dekumar reopened this task as Open.Jun 21 2020, 5:38 PM
dekumar updated the task description. (Show Details)Jun 21 2020, 6:22 PM
@echarruau I have updated the difficulty levels. Please review it and let me know if it's all good.
I agree with this dataset
I agree with this dataset
I will update my work. Thanks for the review!
dekumar updated the task description. (Show Details)Jun 23 2020, 10:00 AM
I think for the dataset description, it would be better and simpler to just say "addition table of 1", "addition table of 2", .... And it would be consistent with the other activities ( same as it is for subtraction memory, for example).
I think for the dataset description, it would be better and simpler to just say "addition table of 1", "addition table of 2", .... And it would be consistent with the other activities ( same as it is for subtraction memory, for example).
I will update the description.
Thanks!
dekumar updated the task description. (Show Details)Jun 25 2020, 5:42 PM
dekumar closed this task as Resolved.Jul 3 2020, 2:30 PM
jjazeix moved this task from Backlog to Done on the GCompris: Improvements board.Jul 5 2020, 1:58 PM
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# Modular Arithmetic
##### Citation preview
In computing, the modulo operation finds the remainder after division of one number by another (sometimes called modulus). Given two positive numbers, a (the dividend) and n (the divisor), amodulo n (abbreviated as a mod n) is the remainder of the Euclidean division of a by n.
Modular arithmetic is a special type of arithmetic that involves only integers. This goal of this article is to explain the basics of modular arithmetic while presenting a progression of more difficult and more interesting problems that are easily solved using modular arithmetic.
Motivation Let's use a clock as an example, except let's replace the
at the top of the clock with a .
Starting at noon, the hour hand points in order to the following:
This is the way in which we count in modulo 12. When we add to , we arrive back at . The same is true in any other modulus (modular arithmetic system). In modulo , we count
We can also count backwards in modulo 5. Any time we subtract 1 from 0, we get 4. So, the integers from to , when written in modulo 5, are
where , , , or
is the same as in modulo 5. Because all integers can be expressed as , in modulo 5, we give these integers their own name: the residue
classes modulo 5. In general, for a natural number that is greater than 1, the modulo residues are the integers that arewhole numbers less than :
This just relates each integer to its remainder from the Division Theorem. While this may not seem all that useful at first, counting in this way can help us solve an enormous array of number theory problems much more easily!
Residue We say that
is the modulo-
residue of
when
, and
.
Congruence There is a mathematical way of saying that all of the integers are the same as one of the modulo 5 residues. For instance, we say that 7 and 2 are congruent modulo 5. We write this using the symbol : In other words, this means in base 5, these integers have the same residue modulo 5:
The (mod 5) part just tells us that we are working with the integers modulo 5. In modulo 5, two integers are congruent when their difference is a multiple of 5. In general, two integers and are congruent modulo when is a multiple of . In other words, means that
and
when is an integer. Otherwise, are not congruent modulo .
, which
Examples
because
because
because
is a multiple of
.
, which is an integer.
, which is not a multiple of .
because
, which is not an integer.
Sample Problem Find the modulo
residue of
.
Solution: Since and
R , we know that is the modulo
residue of
.
Another Solution: Since
, we know that
We can now solve it easily and
is the modulo
residue of
Making Computation Easier We don't always need to perform tedious computations to discover solutions to interesting problems. If all we need to know about are remainders when integers are divided by , then we can work directly with those remainders in modulo . This can be more easily understood with a few examples.
Addition Problem Suppose we want to find the units digit of the following sum:
We could find their sum, which is , and note that the units digit is . However, we could find the units digit with far less calculation. Solution We can simply add the units digits of the addends:
The units digit of this sum is , which must be the same as the units digit of the four-digit sum we computed earlier. Why we only need to use remainders We can rewrite each of the integers in terms of multiples of
and remainders:
. When we add all four integers, we get
At this point, we already see the units digits grouped apart and added to a multiple of (which will not affect the units digit of the sum): . Solution using modular arithmetic
Fun With Modular Arithmetic Home›Math›
A reader recently suggested I write about modular arithmetic (aka “taking the remainder”). I hadn’t given it much thought, but realized the modulo is extremely powerful: it should be in our mental toolbox next to addition and multiplication. Instead of hitting you in the face with formulas, let’s explore an idea we’ve been subtly exposed to for years. There’s a nice article on modular arithmetic that inspired this post. Odd, Even And Threeven Shortly after discovering whole numbers (1, 2, 3, 4, 5…) we realized they fall into two groups:
Even: divisible by 2 (0, 2, 4, 6..) Odd: not divisible by 2 (1, 3, 5, 7…)
Why’s this distinction important? It’s the beginning of abstraction — we’re noticing the properties of a number (like being even or odd) and not just the number itself (“37”).
This is huge — it lets us explore math at a deeper level and find relationships between types of numbers, not specific ones. For example, we can make rules like this:
Even x Even = Even Odd x Odd = Odd Even x Odd = Even
These rules are general — they work at the property level. (Intuitively, I have a chemical analogy that “evenness” is a molecule some numbers have, and cannot be removed by multiplication.) But even/odd is a very specific property: division by 2. What about the number 3? How about this:
“Threeven” means a number is divisbile by 3 (0, 3, 6, 9…) “Throdd” means you are not divisible by 3 (1, 2, 4, 5, 7, 8…)
Weird, but workable. You’ll notice a few things: there’s two types of throdd. A number like “4” is 1 away from being threeven (remainder 1), while the number 5 is two away (remainder 2). Being “threeven” is just another property of a number. Perhaps not as immediately useful as even/odd, but it’s there: we can make rules like “threeven x threeven = threeven” and so on. But it’s getting crazy. We can’t make new words all the time. Enter The Modulo The modulo operation (abbreviated “mod”, or “%” in many programming languages) is the remainder when dividing. For example, “5 mod 3 = 2” which means 2 is the remainder when you divide 5 by 3. Converting everyday terms to math, an “even number” is one where it’s “0 mod 2” — that is, it has a remainder of 0 when divided by 2. An odd number is “1 mod 2” (has remainder 1). Why’s this cool? Well, our “odd/even” rules become this:
Even x Even = 0 x 0 = 0 [even] Odd x Odd = 1 x 1 = 1 [odd]
Even x Odd = 0 x 1 = 0 [even]
Cool, huh? Pretty easy to work out — we converted “properties” into actual equations and found some new facts. What’s even x even x odd x odd? Well, it’s 0 x 0 x 1 x 1 = 0. In fact, you can see if there’s an even being multiplied anywhere the entire result is going to be zero… I mean even :). Clock Math The sneaky thing about modular math is we’ve already been using it for keeping time — sometimes called “clock arithmetic”. For example: it’s 7:00 (am/pm doesn’t matter). Where will the hour hand be in 7 hours? Hrm. 7 + 7 = 14, but we can’t show “14:00” on a clock. So it must be 2. We do this reasoning intuitively, and in math terms:
(7 + 7) mod 12 = (14) mod 12 = 2 mod 12 [2 is the remainder when 14 is divided by 12]
The equation “14 mod 12 = 2 mod 12” means, “14 o’clock” and “2 o’clock” look the same on a 12-hour clock. They are congruent, indicated by a triple-equals sign: 14 ≡ 2 mod 12. Another example: it’s 8:00. Where will the big hand be in 25 hours? Instead of adding 25 to 8, you might realize that 25 hours is just “1 day + 1 hour”. So, the clock will end up 1 hour ahead, at 9:00.
(8 + 25) mod 12 ≡ (8) mod 12 + (25) mod 12 ≡ (8) mod 12 + (1) mod 12 ≡ 9 mod 12
You intuitively converted 25 to 1, and added that to 8. Fun Property: Math Just Works Using clocks as an analogy, we can figure out whether the rules of modular arithmetic “just work” (they do).
Addition/Subtraction Let’s say two times look the same on our clock (“2:00” and “14:00”). If we add the same “x” hours to both, what happens? Well, they change to the same amount on the clock! 2:00 + 5 hours ≡ 14:00 + 5 hours — both will show 7:00. Why? Well, we never cared about the excess “12:00” that the 14 was carrying around. We can just add 5 to the 2 remainder that both have, and they advance the same. For all congruent numbers (2 and 14), adding and subtracting has the same result. Multiplication It’s harder to see whether multiplication stays the same. If 14 ≡ 2 (mod 12), can we multiply both sides and get the same result? Let’s see — what happens when we multiply by 3? Well, 2:00 * 3 ≡ 6:00. But what’s “14:00” * 3? Remember, 14 = 12 + 2. So, we can say
14 * 3 = (12 + 2) * 3 = (12 * 3) + (2 * 3) mod 12
The first part (12 * 3) can be ignored! The “12 hour overflow” that 14 is carrying around just gets repeated a few times. But who cares? We ignore the overflow anyway. When multiplying, it’s only the remainder that matters, which is the same 2 hours for 14:00 and 2:00. Intuitively, this is how I see that multiplication doesn’t change relationships with modular math (you can multiply both sides of a modular relationship and get the same result). See the above link for more rigorous proofs — these are my intuitive pencil lines. Uses Of Modular Arithmetic Now the fun part — why is modular arithmetic useful? Simple time calculations
We do this intuitively, but it’s nice to give it a name. You have a flight arriving at 3pm. It’s getting delayed 14 hours. What time will it land? Well, 14 ≡ 2 mod 12. So I think of it as “2 hours and an am/pm switch”, so I know it will be “3 + 2 = 5am”. This is a bit more involved than a plain modulo operator, but the principle is the same. Putting Items In Random Groups Suppose you have people who bought movie tickets, with a confirmation number. You want to divide them into 2 groups. What do you do? “Odds over here, evens over there”. You don’t need to know how many tickets were issued (first half, second half), everyone can figure out their group instantly (without contacting a central authority), and the scheme works as more people buy tickets. Need 3 groups? Divide by 3 and take the remainder (aka mod 3). You’ll have groups “0”, “1” and “2”. In programming, taking the modulo is how you can fit items into a hash table: if your table has N entries, convert the item key to a number, do mod N, and put the item in that bucket (perhaps keeping a linked list there). As your hash table grows in size, you can recompute the modulo for the keys. Picking A Random Item I use the modulo in real life. Really. We have 4 people playing a game and need to pick someone to go first. Play the mod N mini-game! Give people numbers 0, 1, 2, and 3. Now everyone goes “one, two, three, shoot!” and puts out a random number of fingers. Add them up and divide by 4 — whoever gets the remainder exactly goes first. (For example: if the sum of fingers is 11, whoever had “3” gets to go first, since 11 mod 4 = 3). It’s fast and it works. Running Tasks On A Cycle Suppose tasks need to happen on a certain schedule:
How do you store this information and make a schedule? One way:
Have a timer running every minute (keep track of the minute as “n”) 3x / hour means once every 60/3 = 20 minutes. So task A runs whenever “n % 20 == 0” Task B runs whenever “n % 10 == 0” Task C runs whenever “n % 60 == 0”
Oh, you need task C1 which runs 1x per hour, but not the same time as task C? Sure, have it run when “n mod 60 == 1” (still once per hour, but not the same as C1). Mentally I see a cycle I want to “hit” at various intervals, so I insert a mod. The neat thing is that the hits can overlap independently. It’s a bit like XOR in that regard (each XOR can be layered — but that’s another article!). Similarly, when programming you can print every 100th log item by doing: if (n % 100 == 0){ print… }. It’s a very flexible, simple way to have items run on a schedule. In fact, it’s the way to answer the FizzBuzz sanity check. If you don’t have the modulo operation in your batbelt the question becomes much more tricky. Finding Properties Of Numbers Suppose I told you this:
a = (47 * 2 * 3)
What can you deduce quickly? Well, “a” must be even, since it’s equal to something which involves multiplication by 2. If I also told you:
a = (39 * 7)
You’d balk. Not because you “know” the two products are different, but because one is clearly even, and the other is odd. There’s a problem: a can’t be the same number in both since the properties don’t match up.
Things like “even”, “threeven” and “mod n” are properties that are more general than individual numbers, and which we can check for consistency. So we can use modulo to figure out whether numbers are consistent, without knowing what they are! If I tell you this:
3a + 5b = 8 3a + b = 2
Can these equations be solved with the integers? Let’s see:
3a + 5b = 8… let’s “mod 3 it”: 0 + 2b ≡ 2 mod 3, or b ≡ 1 mod 3 3a + b = 2… let’s “mod 3 it”: 0 + b ≡ 2 mod 3), or b ≡ 2 mod 3
A contradication, good fellows! B can’t be both “1 mod 3” and “2 mod 3” — it’s as absurd as being even and odd at the same time! But there’s one gotcha: numbers like “1.5” are neither even nor odd — they aren’t integers! The modular properties apply to integers, so what we can say is that b cannot be an integer. Because, in fact, we can solve that equation:
(3a + 5b) – (3a +b) = 8 – 2 4b = 6 b = 1.5 3a + 1.5 = 2, so 3a = 0.5, and a = 1/6
Don’t get seduced by the power of modulo! Know its limits: it applies to integers. Cryptography Playing with numbers has very important uses in cryptography. It’s too much to cover here, but modulo is used in Diffie-Hellman Key Exchange — used in setting up SSL connections to encrypt web traffic. Plain English Geeks love to use technical words in regular contexts. You might hear “X is the same as Y modulo Z” which means roughly “Ignoring Z, X and Y are the same.” For example:
b and B are identical, modulo capitalization The iTouch and iPad are identical, modulo size ;)
Onward And Upward It’s strange thinking about the “utility” of the modulo operator — it’s like someone asking why exponents are useful. In everyday life, not very, but it’s a tool to understand patterns in the world, and create your own. In general, I see a few general use cases:
Range reducer: take an input, mod N, and you have a number from 0 to N-1. Group assigner: take an input, mod N, and you have it tagged as a group from 0 to N-1. This group can be agreed upon by any number of parties — for example, different servers that know N = 20 can agree what group ID=57 belongs to. Property deducer: treat numbers according to properties (even, threeven, and so on) and work out principles derived at the property level
I’m sure there’s dozens more uses I’ve missed — feel free to comment below. Happy
Motivation Let's use a clock as an example, except let's replace the
at the top of the clock with a .
Starting at noon, the hour hand points in order to the following:
This is the way in which we count in modulo 12. When we add to , we arrive back at . The same is true in any other modulus (modular arithmetic system). In modulo , we count
We can also count backwards in modulo 5. Any time we subtract 1 from 0, we get 4. So, the integers from to , when written in modulo 5, are
where is the same as in modulo 5. Because all integers can be expressed as , , , , or in modulo 5, we give these integers their own name: the residue classes modulo 5. In general, for a natural number that is greater than 1, the modulo residues are the integers that arewhole numbers less than :
This just relates each integer to its remainder from the Division Theorem. While this may not seem all that useful at first, counting in this way can help us solve an enormous array of number theory problems much more easily!
Residue We say that
is the modulo-
residue of
when
, and
.
Congruence There is a mathematical way of saying that all of the integers are the same as one of the modulo 5 residues. For instance, we say that 7 and 2 are congruent modulo 5. We write this using the symbol : In other words, this means in base 5, these integers have the same residue modulo 5:
The (mod 5) part just tells us that we are working with the integers modulo 5. In modulo 5, two integers are congruent when their difference is a multiple of 5. In general, two integers and are congruent modulo when is a multiple of . In other words, means that
Examples
and
when is an integer. Otherwise, are not congruent modulo .
, which
because
because
is a multiple of
.
, which is an integer.
because
, which is not a multiple of .
because
, which is not an integer.
Sample Problem Find the modulo
residue of
.
Solution: Since
and
R , we know that
is the modulo
residue of
.
Another Solution: Since
, we know that
We can now solve it easily
and
is the modulo
residue of
Making Computation Easier We don't always need to perform tedious computations to discover solutions to interesting problems. If all we need to know about are remainders when integers are divided by , then we can work directly with those remainders in modulo . This can be more easily understood with a few examples.
Suppose we want to find the units digit of the following sum:
We could find their sum, which is , and note that the units digit is . However, we could find the units digit with far less calculation. Solution We can simply add the units digits of the addends:
The units digit of this sum is , which must be the same as the units digit of the four-digit sum we computed earlier. Why we only need to use remainders We can rewrite each of the integers in terms of multiples of
and remainders:
. When we add all four integers, we get
At this point, we already see the units digits grouped apart and added to a multiple of (which will not affect the units digit of the sum): . Solution using modular arithmetic Now let's look back at this solution, using modular arithmetic from the start. Note that
Because we only need the modulo summands:
so the units digit of the sum is just .
residue of the sum, we add just the residues of the
, and
are integers and
is a positive integer such that
the following is always true: . And as we did in the problem above, we can apply more pairs of equivalent integers to both sides, just repeating this simple principle. Proof of the addition rule Let
, and
where
and
equations we get: Which is equivalent to saying
Subtraction The same shortcut that works with addition of remainders works also with subtraction. Problem Find the remainder when the difference between
and
is divided by .
Solution Note that
and
. So,
Thus,
so 1 is the remainder when the difference is divided by . (Perform the subtraction yourself, divide by , and see!) Subtraction rule When
, and
are integers and
the following is always true:
.
is a positive integer such that
Multiplication Modular arithmetic provides an even larger advantage when multiplying than when adding or subtracting. Let's take a look at a problem that demonstrates the point. Problem Jerry has boxes of soda in his truck. The cans of soda in each box are packed oddly so that there are cans of soda in each box. Jerry plans to pack the sodas into cases of cans to sell. After making as many complete cases as possible, how many sodas will Jerry have leftover? Solution First, we note that this word problem is asking us to find the remainder when the product is divided by . Now, we can write each
and
in terms of multiples of
and remainders:
This gives us a nice way to view their product: Using FOIL, we get that this equals
We can already see that each part of the product is a multiple of , except the product of the remainders when each and are divided by 12. That part of the product is , which leaves a remainder of when divided by . So, Jerry has sodas leftover after making as many cases of as possible. Solution using modular arithmetic First, we note that
Thus,
meaning there are
sodas leftover. Yeah, that was much easier.
Multiplication rule When
, and
are integers and
is a positive integer such that
The following is always true: .
Exponentiation Since exponentiation is just repeated multiplication, it makes sense that modular arithmetic would make many problems involving exponents easier. In fact, the advantage in computation is even larger and we explore it a great deal more in the intermediate modular arithmetic article. Note to everybody: Exponentiation is very useful as in the following problem: Problem #1 What is the last digit of the middle?
if there are 1000 7s as exponents and only one 7 in
We could solve this problem using mods. This can also be stated as . After that, we see that 7 is congruent to -1 in mod 4, so we can use this fact to replace the 7s with -1s, because 7 has a pattern of repetitive period 4 for the units digit. is simply 1, so therefore , which really is the last digit. Problem #2 What are the tens and units digits of
?
We could (in theory) solve this problem by trying to compute , but this would be extremely time-consuming. Moreover, it would give us much more information than we need. Since we want only the tens and units digits of the number in question, it suffices to find the remainder when the number is divided by . In other words, all of the information we need can be found using arithmetic mod . We begin by writing down the first few powers of A pattern emerges! We see that we have
(mod (mod
(mod
mod
: ). So for any positive integer
,
). In particular, we can write
).
By the "multiplication" property above, then, it follows that (mod
).
Therefore, by the definition of congruence, differs from by a multiple of . Since both integers are positive, this means that they share the same tens and units digits. Those digits are and , respectively. Problem #3 Can you find a number that is both a multiple of square?
but not a multiple of
and a perfect
No, you cannot. Rewriting the question, we see that it asks us to find an integer satisfies .
that
Taking mod on both sides, we find that . Now, all we are missing is proof that no matter what is, will never be a multiple of plus , so we work with cases:
This assures us that it is impossible to find such a number.
Summary of Useful Facts Consider four integers and a positive integer such that and . In modular arithmetic, the following identities hold:
Addition: Subtraction: Multiplication: Division: . Exponentiation:
. . . , where where
is a positive integer that divides
and
is a positive integer.
Applications of Modular Arithmetic Modular arithmetic is an extremely flexible problem solving tool. The following topics are just a few applications and extensions of its use:
Divisibility rules Linear congruences
A few distributive properties of modulo are as follows: 1. ( a + b ) % c = ( ( a % c ) + ( b % c ) ) % c 2. ( a * b ) % c = ( ( a % c ) * ( b % c ) ) % c 3. ( a – b ) % c = ( ( a % c ) - ( b % c ) ) % c ( see notes at bottom ) 4. ( a / b ) % c NOT EQUAL TO ( ( a % c ) / ( b % c ) ) % c So, modulo is distributive over +, * and - but not / . One observation that I’d like to make here is that the result of ( a % b ) will always be less than b.
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# Joining strings in every third element in a list
How do I join a string to the first and every third element in the list?
Example:
list={1,2,3,4,5,6,7,8,9..}
={"string1",2,3,"string4",5,6,"string7",8,9..}
I have come to the conclusion that this would be some of the solution, but I don't know how to do it every third element.
"string"<>ToString@
There are a lot of ways to do this, but here is one:
Method1[list_] := Fold[
Function@ReplacePart[#1, #2 -> ("string" <> ToString[#1[[#2]]])],
list,
Range[1, Length[list], 3]];
Method1 @ Range[9]
{"string1", 2, 3, "string4", 5, 6, "string7", 8, 9}
For help understanding this, see the Fold function, and the ReplacePart function. This is a rather functional way of solving the problem; a more procedural method might look like this:
Method2[listArg_] := Module[
{list = listArg},
Do[
list[[i]] = "string" <> ToString[list[[i]]],
{i, 1, Length[list], 3}];
list];
Method2 @ Range[9]
{"string1", 2, 3, "string4", 5, 6, "string7", 8, 9}
You can also set the list directly, if list is not a function argument or a constant:
Method3[listArg_] := Module[
{list = listArg},
list[[1 ;; All ;; 3]] = ("string" <> ToString[#]) & /@ list[[1 ;; All ;; 3]];
list];
Method3 @ Range[9]
{"string1", 2, 3, "string4", 5, 6, "string7", 8, 9}
And, one final way:
Method4[list_] := MapAt[
Function["string" <> ToString[#]],
list,
List /@ Range[1, Length[list], 3]];
Method4 @ Range[9]
{"string1", 2, 3, "string4", 5, 6, "string7", 8, 9}
Edit:
As pointed out by Rorschach, some of these methods are slow for large lists (specifically, methods 1, 2, and 4 are fairly slow). For completeness, I've placed Rorschach's method just below. The Method3 above runs faster in my own tests than Rorschach's proposed method (due, probably, to the partitioning in Rorschach's method), while methods 1 and 4 can be rearranged slightly to run in comparable time; a new method 1 is below, and method 4 can be adapted to look like the solution given by Kuba.
RorschachsMethod[list_] := Flatten@Map[
{"string" <> ToString[#[[1]]], #[[2]], #[[3]]} &,
Partition[r, 3]];
KubaMethod[list_] := MapAt[
Function["string" <> ToString[#]],
list,
{1 ;; All ;; 3}];
NewMethod1[list_] := With[
{range = Range[1, Length[list], 3]},
ReplacePart[
list,
Thread[range -> (("string" <> ToString[#]) & /@ list[[range]])]]];
In case speed is an issue for anyone, and for the sake of completeness, the fastest version of this function I was able to come up with (without employing any parallel or compiling tools) is below:
FastMethod[listArg_] := Module[
{list = listArg},
list[[1 ;; All ;; 3]] = MapThread[
StringJoin,
{ConstantArray["string", Floor[(Length[list] + 2)/3]],
ToString /@ list[[1 ;; All ;; 3]]}];
list];
This last method runs slightly faster than the original Method3 and in about half the time (for argument Range[10^6]) of any other method I've seen:
With[
{r = Range[10^6]},
Table[
fn -> First@AbsoluteTiming[fn[r]],
{fn, {Method3, NewMethod1, RorschachMethod, KubaMethod, FastMethod}}]]
{Method3 -> 1.42383, NewMethod1 -> 2.59458, RorschachMethod -> 2.90768, KubaMethod -> 1.42317, FastMethod -> 1.26555}
• Very slow for large lists. Dec 9, 2015 at 8:33
• @Rorschach: The third method is not slow; the others are slower, but this is a beginner question, so they are intended to show the many tools for this kind of operation.
– nben
Dec 9, 2015 at 16:36
f = "string" <> ToString[#] &
or, if you have 10+:
f = StringTemplate["string"]
Can't find a duplicate, so:
list = Range@10;
list[[;; ;; 3]] = f /@ list[[;; ;; 3]];
list
{"string1", 2, 3, "string4", 5, 6, "string7", 8, 9, "string10"}
Or:
list = Range@10;
MapAt[
f,
list,
{;; ;; 3}
]
• Span + Part functionality is the best thing in MMA ;)
– Kuba
Dec 8, 2015 at 21:14
• I really should start thinking along these lines more! Dec 8, 2015 at 22:49
Code:
data = Range @ 12;
Table[{i}, {i, 1, Length @ data, 3}];
MapAt["string" <> ToString @ # &, data, %]
Output:
{"string1", 2, 3, "string4", 5, 6, "string7", 8, 9, "string10", 11, 12}
Alternative (as one line):
MapAt["string" <> ToString @ # &, #, Table[{i}, {i, 1, Length @ #, 3}]] &[Range @ 12]
Output:
{"string1", 2, 3, "string4", 5, 6, "string7", 8, 9, "string10", 11, 12}
Reference:
MapAt
ToString
Table
Length
Range
@|<>|#|&
• This is essentially the same as user21382's last solution. Dec 8, 2015 at 21:12
• @march I agree, all of the solutions are essentially the same since they achieve same objective. The above has been produced only out of curiousity since I am still learning MMA. :) Dec 8, 2015 at 21:19
• Very very slow for large lists. Dec 9, 2015 at 8:30
Rest@Table[If[Mod[i - 1, 3] == 0, "string" <> ToString@i, i], {i, 0, 9}]
{"string1", 2, 3, "string4", 5, 6, "string7", 8, 9}
I'm a bit sad nobody thought to use MapIndexed[]:
MapIndexed[With[{k = #2[[1]]}, If[Mod[k, 3] == 1, "string" <> ToString[k], #]] &,
Range[9]]
{"string1", 2, 3, "string4", 5, 6, "string7", 8, 9}
Partition into groups of three and just apply your operation on first of every sub list and flatten. Say for range of 10 elements, use patterns,
ReplaceRepeated[Partition[Range[10],3],
{s___, {a_, b_, d_}, k___} :> {s, "string" <> ToString[a], b, d, k},
MaxIterations -> Length[Ceiling[Range[10]/3]]]
=> {"string1", 2, 3, "string4", 5, 6, "string7", 8, 9}
But since we are using patterns, it slows down mostly with large data sets, Table version gives the fastest of all the answers.
AbsoluteTiming[
Flatten[Map[{"string" <> ToString[#[[1]]], #[[2]], #[[3]]} &,
Partition[Range[1000000], 3]]]]
giving 1.2649 seconds. I don't have V10+ so I couldn't check @Kuba solution.
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Search a number
33333331 is a prime number
BaseRepresentation
bin111111100101…
…0000001010011
32022201111201211
41333022001103
532013131311
63150240551
7553220503
oct177120123
968644654
1033333331
11178a7909
12b1b6157
136ba1285
1445d9a03
152dd6821
hex1fca053
33333331 has 2 divisors, whose sum is σ = 33333332. Its totient is φ = 33333330.
The previous prime is 33333329. The next prime is 33333347. The reversal of 33333331 is 13333333.
Adding to 33333331 its reverse (13333333), we get a palindrome (46666664).
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 33333331 - 21 = 33333329 is a prime.
It is a super-2 number, since 2×333333312 = 2222221911111122, which contains 22 as substring.
Together with 33333329, it forms a pair of twin primes.
It is a nialpdrome in base 10.
It is not a weakly prime, because it can be changed into another prime (33333031) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (13) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 16666665 + 16666666.
It is an arithmetic number, because the mean of its divisors is an integer number (16666666).
Almost surely, 233333331 is an apocalyptic number.
33333331 is a deficient number, since it is larger than the sum of its proper divisors (1).
33333331 is an equidigital number, since it uses as much as digits as its factorization.
33333331 is an odious number, because the sum of its binary digits is odd.
The product of its digits is 2187, while the sum is 22.
The square root of 33333331 is about 5773.5024898237. The cubic root of 33333331 is about 321.8297873592.
The spelling of 33333331 in words is "thirty-three million, three hundred thirty-three thousand, three hundred thirty-one".
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# Basis question
• Oct 11th 2009, 09:48 AM
grandunification
Basis question
For the given matrix H in M2(R) and for any real number r let Vsubr denote the solution space in M2(R) of the equation
[xy]
[zw]H =
rH[xy]
[zw]
Find a basis for Vsub2 given that
H = [01]
[-10]
Okay. So I had some difficulty typing this question so it looks really awkward. But I think you can figure out what I'm doing. H is a 2X2 matrix.
[xy]
[zw] is supposed to be a 2x2 matrix also.
thanks.
• Oct 11th 2009, 11:36 AM
HallsofIvy
I think what you are saying is this: Let $V_r$ be the subspace of 2 by 2 matrices satisfying
$\begin{bmatrix}x & y \\ z & w\end{bmatrix}\begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix}= r\begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix}\begin{bmatrix}x & y \\ z & w\end{bmatrix}$.
Find a basis for that space.
Okay, go ahead and do the computation: this says
$\begin{bmatrix}-y & x \\ -w & z\end{bmatrix}= \begin{bmatrix}rz & rw \\ -rx & -rw\end{bmatrix}$.
And that reduces to 4 equations: -y= rz, x= rw, -w= -rx, and z= -ry. Since two involve only y and z and two only x and w, that is really two sets of twe equations. Divide the first equation by the last to get -y/z= -z/y which is equivalent to $y^2= z^2$. That means we must have either y= z or y= -z. But what about r? If y= z, then the first equation above becomes -y= rz so r must be -1. If y= -z, then r must be 1.
Divide the second equation the third to get -x/w= -w/x or $x^2= w^2$. We must have either x= w or x= -w. Again, if x= w then r must be -1 and if x= -w, r= 1.
That is, r must be either 1 or -1. Then matrices in $V_{1}$ are of the form $\begin{bmatrix}x & y \\ -y & -x\end{bmatrix}$ $= x\begin{bmatrix}1 & 0 \\ 0 & -1\end{bmatrix}+ y\begin{bmatrix}0 & 1 \\-1 & 0\end{bmatrix}$ so a basis for this space is the two matrices $\begin{bmatrix}1 & 0 \\ 0 & -1\end{bmatrix}$ and $\begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix}$.
Matrices in $V_{-1}$ are of the form $\begin{bmatrix} x & y \\ y & x\end{bmatrix}$ $= x\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}+ y\begin{bmatrix}0 & 1 \\ 1& 0\end{bmatrix}$ so that a basis for $V_{-1}$ is given by the two matrices $\begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}$ and $\begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}$
• Oct 11th 2009, 12:02 PM
grandunification
Yes, but here's the problem. He wants me to find a basis for Vsub2. But you're saying that's impossible. Maybe there is a problem with his question?
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# Public key cryptography with the RSA algorithm
An introduction to the original mathematics of asymmetric cryptography, February 2018
Last year I spent a few months working on a system which made extensive use of public key cryptography to sign and encrypt messages. It was my job to explain to new starters how the system all fitted together and how cryptography protected the confidentiality and integrity of messages.
At some point I started saying things like, "you can do RSA on the back of a napkin if you use small primes". Which in retrospect is a fairly bold claim (you definitely benefit from using a calculator), but the point I was trying to make is that the maths to generate secure messages is not that complex. You don't have to be a mathematician to understand this stuff.
In an attempt to prove that point and refresh my knowledge I thought I'd write up the method. So I'll step through the RSA algorithm now, creating a public key pair which I'll use to encrypt and decrypt a message. The terminology gets a bit difficult-sounding, but the underlying concepts and mathematical operations are simple enough. I hope to make this as easy to follow as possible.
## Step 1: Choose two primes and compute their product
To start with you need two distinct primes, which we'll refer to as `p` and `q`. In real world applications the primes selected need to be very big for RSA to be secure, but to make it possible to do the maths with mental arithmetic I've chosen very small ones.
`p = 3` `q = 7`
Once you've picked some primes you need to multiply them together. We'll call the product `n`.
`n = pq` `n = 21`
## Step 2: Compute Euler's totient of `n`, `λ(n)`
Euler's what? What's a `λ(n)`?
Euler's totient of `n` is the lowest common multiple of `p-1` and `q-1`. That number is represented by `λ(n)`.
To calculate it you multiply `p-1` and `q-1` together and divide by their greatest common divisor, which I'll abbreviate to `gcd`. That is the largest number that divides into both.
`p-1 = 2` `q-1 = 6` `gcd(6, 2) = 2` `λ(n) = (2 x 6) / 2` `λ(n) = 6`
## Step 3: Find an encryption key
The encryption key needs to be greater than `1` but smaller than the totient, `λ(n)`. It must not share any factors with `λ(n)` except `1`, which means it must be coprime.
`λ(n)` is `6` in this example, so the encryption key must be a number between `1` and `6` which does not share any common factors with `6`. There is only one possible encryption key, `5`.
`e = 5`
Checking for common factors is intensive work and in real world applications, where you don't want to wait all day to generate a key, the short cut is to select a prime number which is not a divisor of `λ(n)`. That gives you a coprime by default. So although `e` doesn't technically need to be prime, a prime will almost always be used.
## Step 4: Find a decryption key
The decryption key must be the modular multiplicative inverse of `e mod λ(n)`, which is actually a tricky one to explain, but the maths isn't particularly complex once you know what to do. The idea is to find the value of `d` here.
`(d x e) mod λ(n) = 1`
We can substitute the numbers we already know, so it's easier to read.
`(d x 5) mod 6 = 1`
`mod` stands for modulo, and refers to the remainder after division. So we're looking for a number where when you multiply it by `5` and then divide it by `6` the remainder is `1`.
To calculate this properly you can use the Extended Euclidean Algorithm, but `5` and `6` are small enough numbers to work out that `85` fits the bill without getting into that.
`d = 85 / 5` `d = 17`
Incidentally `25` would also fit the bill, but then we'd end up with `d = 5` and `e = 5` which would be a case of accidental symmetric encryption and just be confusing. What's you may not have realised before is that there are a whole load of possible encryption and decryption keys for any value of `n`, but that's the way it works.
## Step 5: Send secret messages
We now have a public and private key pair, `e` to encrypt and `d` to decrypt. Let's encrypt something!
`e = 5` `d = 17`
The encryption algorithm is.
`m ^ e mod n = c`
Let's start with something super small, the number `2`.
`2 ^ 5 mod 21 = 11`
This gives us the ciphertext `c = 11`.
Decryption takes place with the same function, but with the private key used as the exponent.
`c ^ d mod n = m`
This bit is not easy to do in your head, given that `11 ^ 17 = 505447028499293771`, so feel free to use a calculator to check the following.
`11 ^ 17 mod 21 = 2`
And that's it. A public key pair created on the digital equivalent of the back of a napkin.
If you're interested in encrypting a larger message than `2` then have a go. If you try something ambitious then you'll find the system breaks down if your message is larger than the modulus `n = 21`. That's a limitation of RSA. In real world applications a message tends to be encrypted with a symmetrical algorithm like AES, and then the sender will encrypt the key for that message using RSA.
Anyway, if you got this far reading this then I hope it's been interesting. If you are baying for more then try Prime Number Hide and Seek: How the RSA Cipher Works by Brian Raiter, it's an absolute gem of a webpage.
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Scilab Function
cdft - cumulative distribution function Student's T distribution
### Calling Sequence
[P,Q]=cdft("PQ",T,Df)
[T]=cdft("T",Df,P,Q)
[Df]=cdft("Df",P,Q,T)
### Parameters
• P,Q,T,Df : six real vectors of the same size.
• P,Q (Q=1-P) : The integral from -infinity to t of the t-density. Input range: (0,1].
• T : Upper limit of integration of the t-density. Input range: ( -infinity, +infinity). Search range: [ -1E150, 1E150 ]
• DF: Degrees of freedom of the t-distribution. Input range: (0 , +infinity). Search range: [1e-300, 1E10]
### Description
Calculates any one parameter of the T distribution given values for the others.
Formula 26.5.27 of Abramowitz and Stegun, Handbook of Mathematical Functions (1966) is used to reduce the computation of the cumulative distribution function to that of an incomplete beta.
Computation of other parameters involve a seach for a value that produces the desired value of P. The search relies on the monotinicity of P with the other parameter.
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Intuitionistic Logic Explorer < Previous Next > Nearby theorems Mirrors > Home > ILE Home > Th. List > mpbir3an GIF version
Theorem mpbir3an 1121
Description: Detach a conjunction of truths in a biconditional. (Contributed by NM, 16-Sep-2011.) (Revised by NM, 9-Jan-2015.)
Hypotheses
Ref Expression
mpbir3an.1 𝜓
mpbir3an.2 𝜒
mpbir3an.3 𝜃
mpbir3an.4 (𝜑 ↔ (𝜓𝜒𝜃))
Assertion
Ref Expression
mpbir3an 𝜑
Proof of Theorem mpbir3an
StepHypRef Expression
1 mpbir3an.1 . . 3 𝜓
2 mpbir3an.2 . . 3 𝜒
3 mpbir3an.3 . . 3 𝜃
41, 2, 33pm3.2i 1117 . 2 (𝜓𝜒𝜃)
5 mpbir3an.4 . 2 (𝜑 ↔ (𝜓𝜒𝜃))
64, 5mpbir 144 1 𝜑
Colors of variables: wff set class Syntax hints: ↔ wb 103 ∧ w3a 920 This theorem was proved from axioms: ax-1 5 ax-2 6 ax-mp 7 ax-ia1 104 ax-ia2 105 ax-ia3 106 This theorem depends on definitions: df-bi 115 df-3an 922 This theorem is referenced by: limon 4265 limom 4362 issmo 5937 1eluzge0 8743 2eluzge1 8745 0elunit 9084 1elunit 9085 4fvwrd4 9227 fzo0to42pr 9306 resqrexlemga 10047
Copyright terms: Public domain W3C validator
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# Ihp 525 module three problem set
IHP 525 Module Three Problem Set
1. What is the probability of being born on:
a) February 28?
b) February 29?
c) February 28 or February 29?
2. A patient newly diagnosed with a serious ailment is told he has a 60% probability of surviving 5 or more years. Let us assume this statement is accurate. Explain the meaning of this statement to someone with no statistical background in terms he or she will understand.
3. A lottery offers a grand prize of \$10 million. The probability of winning this grand prize is 1 in 55 million (about 1.8×10-8). There are no other prizes, so the probability of winning nothing = 1 – (1.8×10-8) = 0.999999982. The probability model is:
Winnings (X) 0 \$10 x 106 P(X = xi) 1 1.8 x 10-8
a) What is the expected value of a lottery ticket?
b) Fifty-five million lottery tickets will be sold. How much does the proprietor of the lottery need to charge per ticket to make a profit?
4. Suppose a population has 26 members identified with the letters A through Z.
a) You select one individual at random from this population. What is the probability of selecting individual A?
b) Assume person A gets selected on an initial draw, you replace person A into the sampling frame, and then take a second random draw. What is the probability of drawing person A on the second draw?
c) Assume person A gets selected on the initial draw and you sample again without replacement. What is the probability of drawing person A on the second draw?
5. Let A represent cat ownership and B represent dog ownership. Suppose 35% of households in a population own cats, 30% own dogs, and 15% own both a cat and a dog. Suppose you know that a household owns a cat. What is the probability that it also owns a dog?
6. What is the complement of an event?
7. Accidents occur along a 5-mile stretch of highway at a uniform rate. The following “curve” depicts the probability density function for accidents along this stretch:
a) What is the probability that an accident occurred in the first mile along this stretch of highway?
b) What is the probability that an accident did not occur in the first mile?
c) What is the probability that an accident occurred between miles 2.5 and 4?
8. Suppose there were 4,065,014 births in a given year. Of those births, 2,081,287 were boys and 1,983,727 were girls.
a) If we randomly select two women from the population who then become pregnant, what is the probability both children will be boys?
b) If we randomly select two women from the population who then become pregnant, what is the probability that the first woman’s child will be a boy and the second woman’s child will be a boy?
c) If we randomly select two women from the population who then become pregnant, what is the probability that both children will be boys given that at least one child is a boy?
9. Explain the difference between mutually exclusive and independent events.
10. Suppose a screening test has a sensitivity of 0.80 and a false-positive rate of 0.02. The test is used on a population that has a disease prevalence of 0.007. Find the probability of having the disease given a positive test result.
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# Statements with rare counter-examples [duplicate]
This question already has an answer here:
This is a soft question. I'm searching for examples of mathmatical statements (preferably in number theory, but other topics are also fine), that seem to be true, but are actually not. Statements where observing some examples would let you think it is always true, but then there is a well hidden counter-example.
If Riemann's $\zeta$ function had a zero beside the critical line, this would be such an example I'm looking for. Or if Fermat's Last Theorem would be false.
Do you know any such surprising counter-exmaples?
## marked as duplicate by Jack M, user147263, Michael Albanese, Hakim, user85798May 27 '14 at 23:30
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
• See the prime generating polynomials: mathworld.wolfram.com/Prime-GeneratingPolynomial.html – Amzoti May 26 '14 at 12:56
• The number of primes $\le x$ of the form $4k+3$ is never greater than the number of primes $\le x$ of the form $4k+1$. Please see Prime Number Races. Counterexamples are actually not rare, but the first one is big. – André Nicolas May 26 '14 at 13:15
• Fermat's Last Theorem is proven true... – Platonix May 26 '14 at 13:21
• There are some nice examples in this thread. – André 3000 May 26 '14 at 14:05
• All prime numbers are odd :-) – gnasher729 May 26 '14 at 16:32
## 8 Answers
The following spikedmath is cute.
The books "Counterexamples in Topology" by Steen and Seebach as well as "Counterexamples in Analysis" by Gelbaum and Olmstead have some that are surprising when you first see them.
For every $n$ $$\int_0^\infty 2 \cos(x) \prod_{i=0}^n\frac{\sin\frac{x}{2i+1}}{\frac{x}{2i+1}}\,dx=\pi/2$$
Is it true? Well, for every $n$ less than 56, but after...
An example of Borwein integral.
• I.e. provided $\prod_{i=0}^n \dfrac1{2i+1}<3$. – Rosie F Jan 29 '17 at 20:26
Pierre de Fermat conjectured that all Fermat numbers were prime, and a similar mistaken conjecture can be made for most pseudoprimes (Catalan, Fibonacci, Euler, Wieferich, etc.). Also see Euler's sum of Powers conjecture , and the Polya conjecture.
• Oh, and Merten's conjecture. – Platonix May 26 '14 at 13:49
Fermat's ‘little’ theorem states that if $n$ is prime, then $$a^n\equiv a\pmod n\tag{\ast}$$ holds for all $a$. The converse, which is false, states that if $(\ast)$ holds for all $a$, then $n$ is prime.
Counterexamples to this converse are uncommon; the smallest is $n=561$.
The largest odd value of $n$ such that regular $n$-gons are constructable by compass and straight edge is $n=1~431~655~765$.
There is one known exception: $n=4~294~967~295$.
All in all, there are currently only $31$ known constructable regular polygons with an odd number of sides. Prior to Gauss, the largest odd-sided constructable regular polygon was just the pentagon. No doubt much of the numerological and mystical lore surrounding the pentagon and pentagram can be traced back to observations by the ancients that this five sided shape represented some fundamental boundary between the world of finite imperfect man on one side, and the perfect and infinite on the other.
• I'd expect the ancients were smart enough to construct the regular $15$-gon from the triangle and pentagon. – Daniel Fischer May 26 '14 at 17:51
• @DanielFischer You're right. Book IV, Proposition 16, of Elements in fact gives the construction. Now editing. – David H May 26 '14 at 17:56
• Is there something missing in the first paragraphs, like "for all $n\leq N$"? The first sentence seems to say that the largest odd $n$ for which $P_n$ is true is $n=1\,431\,655\,765$. What is the exception? That $P_m$ is true for $m=4\,294\,967\,295$? If so, then what is the point of the first sentence? – JiK May 27 '14 at 13:43
• @JiK Exactly. If the first number was $15$ or something it would make more sense, but the fact that the first number itself is so large makes the point of the first example pointless. All it's saying is that the largest known example of a constructable regular polygon with odd sides is $n = 4294967295$, and the second largest example is $n = 1431655765$.. – MCT May 28 '14 at 4:22
$$a_1^k+\ldots+a_n^k=b^k\qquad\text{with }k>n>1$$
has no solutions in positive integers.
The smallest counterexample is
$$95800^4 + 217519^4 + 414560^4 = 422481^4$$
• smallest in what sense? – Bennett Gardiner May 27 '14 at 23:39
• @BennettGardiner Smallest in lexicographic order $(k,a_1,a_2,\dots,a_n)$. – Toscho May 28 '14 at 14:09
If $\ n\$ that $\ ◎(n)=\frac{n+1}{2^x}$ or $\ ◎(n)=\frac{n-1}{2^x}, \ n \in \mathbb{Z^+},\ x \in \mathbb{N}_{\gt 0},\$then $\ n\$ is prime.
First counter-example is $\ 92673$.
$◎(n)\$ is called the cycle length of $n$ , detail see: How to prove these two ways give the same numbers?
• If change 'then $\ n\$ is prime' to 'then $\ n\$ is prime or semiprime' ,then $\ 92673\$ is the only known counter-example. – miket May 28 '14 at 1:33
For every natural number $n$, the following number is prime: $$n^2-n+41$$
The smallest $n$ for which this conjecture is wrong, is 41 -- naturally.
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# Space station-nature of orbit?
1. Jan 18, 2012
### humanist rho
1. The problem statement, all variables and given/known data
A space station moving in a circular orbit around the earth goes into a new bound orbit by firing its engine radially outaward.This orbit is.......
Choices are;
(a) A larger circle
(b)A smaller circle
(c)An ellipse
(d)a parabola
2. Relevant equations
3. The attempt at a solution
I think it'll be a larger circle or parabola.
But donno how to work it out.
Last edited: Jan 18, 2012
2. Jan 18, 2012
### Simon Bridge
What is the free-body diagram of an object in a circular orbit?
The engine applies thrust radially outward - what does that do to the free body diagram?
Do the engines fire continuously or for a short while?
The change in velocity is important to this.
What can you say about the kind of velocity needed for a circular orbit?
(velocity is a vector - talk in terms of tangential and radial components)
What can you say about the kind of velocity needed for a parabolic orbit?
... an elliptical one?
3. Jan 18, 2012
### humanist rho
mv2/r = GMm/r2
v = Sqrt[GM/r] for circular orbit.
How can i modify these equations for elliptical orbit.
4. Jan 18, 2012
### Simon Bridge
Hint:
1. velocity is a vector - represent as radial and tangential.
2. free body diagram
the v in your equation is purely tangential.
when the thrust is applied for a short time it changes the velocity vector ...
what happens to the radial component
what happens to the tangential component
is it possible for an object in a circular orbit to have a radial component to it's velocity?
5. Jan 18, 2012
### Simon Bridge
btw: you equation gets modified as follows...
$$\underbrace{ma = F_g+F_{thrust}}_{\text{from FBD}} \Rightarrow \frac{mv_\perp^2}{R} = mg_R -\frac{m}{T}\Delta v_r$$... the thrust is applied for a fixed time T, so creates a change in speed Δv. You'll have heard of thrust being referred to as a "delta-vee"? R is the radius of the orbit so gR is the local acceleration of gravity at R. The minus sign is because the thrust is radially outwards, making the positive direction to be radially inwards. The instantanious velocity after the thrust has been applied is given by
$\vec{v}=v_r\hat{r}+v_\perp\hat{\small \perp}$
...which has magnitude $v =\sqrt{v_r^2 + v_\perp^2}$
...and the angle this makes to the radius is $\tan^{-1}(v_\perp / v_r)$.
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0
suppose the parabola has a vertex (4,-6) and also passes through the point (5,-4). write the equation of the parabola in vertex form
• y = a(x – 4)² – 6
=> -4 = a(5 – 4)² – 6
=> -4 = a – 6
i.e. a = 2
so, y = 2(x – 4)² – 6
:)>
• You are not dumb. Dumb means you cannot talk.
You are bright enough to ask for help. I call that SMART.
Y = A(X – H)² + K
Vertex = (H , K) = (4 , -6)…………H = 4 and K = – 6
Point is (X , Y) = (5 , -4)…………X = 5 and Y = – 4
Now solve for A by substituting in the values of H , K , X , Y
– 4 = A(5 – 4)² – 6
+6……………….+6
2 = A(1)²
A = 2
For the final answer replace A , H , K into the original equation
Y = 2(X – 4)² – 6………………ANSWER
Let me know if you have questions on this problem or any other math problems.
I am an excellent math teacher/tutor specializing in making students
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For convenience in the modeling, Pang and Sharma (1994) divide the entire filtration process into two phases:
1 the initial internal cake filtration, and
2 the later external cake filtration.
They separate these two filtration phases by a "transition time" after which the particle migration into porous formation becomes negligible and an external filter cake begins forming over the injection well formation face. Sharma and Pang (1997) assumed that transition from the internal to the external cake filtration occurs when the porosity, ɸ, of the formation face decreases to a minimum critical value, ɸ, by particle deposition, below which particle invasion into porous media is not possible. Their models apply for single phase water flow in the near wellbore region. Hence, the effect of the oil-water two-phase flow during the initial water injection period is neglected because this initial period is relatively short.
## Transition Time
Wennberg and Sharma (1997) estimate the transition time based on the expressions given by Iwasaki (1937) for particle deposition rate and the filtration coefficient, respectively, as:
where b is an empirical constant and X,0 is the filtration coefficient without particle deposition. Although more sophisticated expressions are available in the literature (See Chapter 8), they used Eq. 19-15 for simplicity. Thus, invoking Eq. 19-15 into Eq. 19-14 yields the following expression, similar to the rate equation for particle deposition in pluggable pathways given by Gruesbeck and Collins (1982):
They obtain the analytic solution of Eq. 19-16 for constant flow rate and suspension particle concentration for two cases as the following:
At the transition time, the porosity attains the minimum critical porosity, ɸ , and the maximum critical volume fraction of the deposited particles becomes a* = ɸ0 - ɸ* . Under these conditions, Eqs. 19-17 and 18 can be used to obtain the following expressions, respectively, for the transition time:
## Internal Filtration Models
Pang and Sharma (1994, 1995) and Sharma et al. (1997) have pursued their derivations in terms of the following variables:
Here, v represents the interstitial velocity of the fluid phase and n denotes the fraction of the pore space occupied by the particle deposits in porous media. X' is the product of the deposition rate constant, K, and the interstitial velocity of the flowing suspension. In the following, their formulations are presented in a manner consistent with the rest of the presentation of this chapter. The damage of a core plug by the injection of a dilute particulate suspension can be described by means of the volumetric balance equations of the suspended and deposited particles in porous media, given, respectively, by (Wennberg and Sharma, 1997):
subject to the initial and boundary conditions given by (Pang and Sharma, 1994):
The analytical solution of Eqs. 19-24 through 27 used by Pang and Sharma (1994) implies some simplifications. It applies for the injection of dilute suspension of particles. Therefore, the effect of small amount of particle deposition, a, compared to the initial porosity can be neglected, the deposition rate coefficient is assumed constant, and a constant rate injection is considered. Thus, the analytic solution for constant ɸ~ɸ0, X,~X0 , and u = u0 can be adopted from Rhee et al. (1986) as:
where the term inside the square brackets expresses that cf is a function of (t-ɸx/u). Considering that cf(t) = cf is constant and c0(x) = Q and a0(x) = 0 in the laboratory core flow tests, Pang and Sharma (1994) simplify Eqs. 19-29 and 30, respectively, as:
Pang and Sharma (1995) assumed that the permeability reduction primarily occurs by pore throat plugging. Therefore, they estimated the permeability of the porous formation as a harmonic average permeability of the combined plugged and unplugged regions as:
where fp represents the volume fraction of the deposited particles contributing to pore throat plugging, Kp denotes the permeability of the plugged region near the pore throats, and Km is the permeability of the formation matrix, assumed to remain constant, which is equal to the initial nondamaged permeability, K0 (i.e., Km =K0}. Therefore, Pang and Sharma (1995) rearranged Eq. 19-35 for the relative or fractional retained permeability of the porous formation undergoing particle deposition from dilute suspensions as, inferred by Payatakes et al. (1974):
Thus, they calculate the harmonic average permeability of the damaged portion of the core by:
Substituting Eqs. 19-33, 34, and 36 into Eq. 19-38, and then integrating, they derive the following expression:
Note that Eq. 19-39 was previously derived by Wojtanowicz et al. The injectivity ratio for the linear cases, applicable to laboratory core plugs without external cake formation (/?c = 0), is given by:
Pang and Sharma (1994) simplify Eq. 19-41 by considering that the injection front reaches the outlet end of the core rapidly. Therefore, neglecting the damage during the short period of time until the front reaches the core outlet, Eqs. 19-39 and 41 yield for xf = L the following equation indicating that the reciprocal injectivity ratio is a linear function of time:
## External Filtration Models
Considering the formation of an incompressible external cake without any particle invasion into the core plug, Pang and Sharma (1994) expressed the harmonic average permeability of the cake and the core system as:
Sharma et al. (1997) determine the thickness of the external filter cake by means of a volumetric balance of the particles in the cake as:
where A is the cross-sectional area of the core plug and cf is the volume fraction of the fine particles in the water injected at a flow rate of q. For constant cf and q, Eq. 19-47 simplifies as:
Thus, substituting Eq. 19-48 into 45 and considering that the filter cake thickness is much smaller than the length of the core plug (i.e., hc « L), they obtained the following expression indicating that the reciprocal injectivity ratio is a linear function of time:
## Filtration Coefficient
Wennberg and Sharma (1997) point out that the filtration coefficient varies by particle deposition according to:
Their review of the various expressions available for prediction of the filtration coefficient is summarized and presented in the following. Ives (1967) proposed a general expression as:
in which x, v, z, and p are some empirical parameters and aM is the maximum of the volume fraction of the deposited particles necessary to make the filtration coefficient of porous media zero. This equation indicates that the filtration coefficient is equal to one when there is no deposited particles in porous media, and the filtration coefficient becomes zero when the volume fraction of deposited particles reaches a certain characteristic value of maximum aM. Chiang and Tien (1985) developed an empirical correlation as:
in which the dimensionless groups are defined as following. As is Happel's dimensionless geometric parameter. A^ is the London parameter given by
v is velocity, ILL is viscosity, and pp and pe denote the particle and fluid densities, respectively. (3 is a packing parameter given by
where ɸ is the porosity in fraction. D is the coefficient of diffusion for the Brownian motion of particles. Wennberg and Sharma (1997) analyzed the measurements of the filtration coefficient reported by various investigators and determined that these data mostly indicate power law-type relationships to the volumetric flux, the suspended particle size, and the porous media grain size as:
## Diagnostic-Type Curves for Water Injectivity Tests
Pang and Sharma (1994, 1997) identified four distinct type curves that can be used for interpretation of the water-quality tests. They justified these type curves with experimental data obtained from the literature as shown in Figure 19-3. Type curve 1 is a straight line indicating the formation of an incompressible external filter cake or a thin internal cake near the injection face of the core plug according to Eq. 19-49. The slope remains constant. Type curve 2 is for the similar case, but applies for compressible cakes. In this case, the porosity and permeability of the cake decrease by increasing filtration pressures. As a result, the slope of the curve increases with the filtration time or pore volume injected. Type curve 3 refers to a deep particle invasion and pore filling in the core plug, leading to a slower gradual permeability decrease. As a result, the slope of the curve decreases with the filtration time. Type curve 4 may be an S-shaped or other types of curves indicating a shift of the dominance of the different damage mechanisms during the filtration process.
## References
Barkman, J. H., & Davidson, D. H., "Measuring Water Quality and Predicting Well Impairment," JPT, Trans., AIME, Vol. 253, July 1972, pp. 865-73.
Chiang, H. W., & Tien, C., "Transient Behavior of Deep-Bed Filters," Symp. Adv. in Solids-Liquid Separ., University College, London, 1983.
Chiang, H. W., & Tien, C., "Dynamics of Deep Bed Filtration, Part I and II," AIChE J., Vol. 31, August 1985, p. 1349.
Civan, F, "Quadrature Solution for Waterflooding of Naturally Fractured Reservoirs," SPE Reservoir Evaluation and Engineering, April 1998, pp. 141-147.
Davidson, D. H., "Invasion and Impairment of Formations by Particulates," SPE Paper 8210, Presented at the 54th Annual Fall Meeting of the SPE, Las Vegas, Nevada, September 23-26, 1979.
Donaldson, E. C., Baker, B. A., & Carroll, H. B., Jr., "Particle Transport in Sandstone," SPE Paper 6905, Presented at the 52nd Annual Technical Conference of SPE, Denver, Colorado, October 9-12, 1977.
Folger, H. S., Elements of Chemical Reaction Engineering, Prentice-Hall, Englewood Cliffs, New Jersey, 1986, pp. 197-198.
Gruesbeck, C., & Collins, R. E., "Entrainment and Deposition of Fine Particles in Porous Media," SPEJ, December 1982, pp. 847-856.
Hofsaess, T., & Kleinitz, W., "Injectivity Decline in Wells with Nonuniform Perforation Properties," SPE 39586 paper, Proceedings of the 1998 SPE International Symposium on Formation Damage Control, February 18-19, 1998, Lafayette, Louisiana, pp. 631-640.
Ives, K. J., Deep Filters, Filtr. Sep., March/April 1967, pp. 125-135.
Iwasaki, T, "Some Notes on Sand Filtration," J. Am. Water Works Ass., Vol. 29, 1937, pp. 1591-1602.
Liu, X., & Civan, F., "A Multi-Phase Mud Fluid Infiltration and Filter Cake Formation Model," SPE 25215 paper, Proceedings, SPE International Symposium on Oilfield Chemistry, February 28-March 3, 1993, New Orleans, Louisiana, pp. 607-621.
Liu, X., & Civan, F., "Characterization and Prediction of Formation Damage in Two-Phase Flow Systems, SPE 25429 paper, proceedings of the SPE Production Operations Symposium, March 21-23, 1993, Oklahoma City, Oklahoma, pp. 231-248.
Liu, X., & Civan, F, "Formation Damage and Skin Factors Due to Filter Cake Formation and Fines Migration in the Near-Wellbore Region," SPE 27364 paper, Proceedings of the 1994 SPE Formation Damage Control Symposium, Feb. 9-10, 1994, Lafayette, Louisiana, pp. 259-274.
Liu, X., & Civan, F., "Formation Damage by Fines Migration Including Effects of Filter Cake, Pore Compressibility and Non-Darcy Flow—A Modeling Approach to Scaling from Core to Field," SPE Paper No. 28980, SPE International Symposium on Oilfield Chemistry, February 14-17, 1995, San Antonio, TX.
Liu, X., & Civan, F., "Formation Damage and Filter Cake Buildup in Laboratory Core Tests: Modeling and Model-Assisted Analysis," SPE Formation Evaluation J. (March 1996) Vol. 11, No. 1, pp. 26-30.
Pang, S., & Sharma, M. M., "A Model for Predicting Injectivity Decline in Water Injection Wells," SPE Paper 28489, presented at the SPE 69th Annual Technical Conference and Exhibition in New Orleans, September 25-28, 1994, pp. 275-284.
Pang, S., & Sharma, M. M., "Evaluating the Performance of Open-Hole, Perforated and Fractured Water Injection Wells," SPE Paper 30127, presented at the SPE European Formation Damage Control Conference, the Hague, May 15-16, 1995.
Pang, S., & Sharma, M. M., "A Model for Predicting Injectivity Decline in Water Injection Wells," SPE Formation Evaluation, September 1997, pp. 194-201.
Pautz, J. F., et al., "Relating Water Quality and Formation Permeability to Loss of Injectivity," SPE Paper 18888, presented at the SPE Production Operations Symposium, Oklahoma City, Oklahoma, March 13-14, 1989.
Payatakes, A. C., Rajagopalan, R., & Tien, C., "Application of Porous Media Models to the Study of Deep Bed Filtration," The Canadian Journal of Chemical Engineering, Vol. 52, December 1974.
Rajagopalan, R., & Tien, C., "Trajectory Analysis of Deep-Bed Filtration with the Sphere-in-cell Porous Media Model," AIChE J., Vol. 22, May 1976, pp. 523-533.
Rhee, H. K., Aris, R., & Amundson, N. R., First-order Partial Differential Equations: Volume I, Prentice-Hall, 1986.
Sharma, M. M., Pang, S., & Wennberg, K. E., "Injectivity Decline in Water Injection Wells: An Offshore Gulf of Mexico Case Study," SPE Paper 38180, Proceedings of the 1997 SPE European Formation Damage Conference held in the Hague, The Netherlands, June 2-3, 1997, pp. 341-351.
Todd, A. C., et al., "Review of Permeability Damage Studies and Related North Sea Water Injection," SPE Paper 7883, presented at the SPE International Symposium on Oilfield and Geothermal Chemistry, Dallas, Texas, January 22-24, 1979.
Todd, A. C., et al., "The Application of Depth of Formation Damage Measurements in Predicting Water Injectivity Decline," SPE Paper 12498, presented at the Formation Damage Control Symposium held in Bakersfield, California, February 13-14, 1984.
Todd, A. C., et al., "The Value and Analysis of Core-Based Water Quality Experiments as Related to Water Injection Schemes," SPE Paper 17148, presented at the SPE Formation Damage Control Symposium held in Bakersfield, California, February 8-9, 1988.
van Oort, E., van Velzen, J. F. G., & Leerlooijer, K., "Impairment by Suspended Solids Invasion: Testing and Prediction," SPE Production. & Facilities, August 1993, pp. 178-184.
Vetter, O. J., et al., "Particle Invasion into Porous Medium and Related Injectivity Problems," SPE Paper 16625, presented at the SPE International Symposium on Oilfield and Geothermal Chemistry in San Antonio, Texas, February 4-6, 1987.
Wennberg, K. E., & Sharma, M. M., "Determination of the Filtration Coefficient and the Transition Time for Water Injection Wells," SPE Paper 38181, Proceedings of the 1997 SPE European Formation Damage Conference held in the Hague, The Netherlands, June 2-3, 1997, pp. 353-364.
Wojtanowicz, A. K., Krilov, Z., & Langlinais, J. P., "Study on the Effect of Pore Blocking Mechanisms on Formation Damage," SPE 16233 paper, presented at Society of Petroleum Engineers Symposium, Oklahoma City, Oklahoma, March 8-10, 1987, pp. 449-463.
Wojtanowicz, A. K., Krilov, Z., & Langlinais, J. P., "Experimental Determination of Formation Damage Pore Blocking Mechanisms," Trans, of the ASME, Journal of Energy Resources Technology, Vol. 110, 1988, pp. 34-42.
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Analytical Mech Homework Solutions 61
Analytical Mech Homework Solutions 61 - Collecting terms...
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Collecting terms and neglecting terms in 2 ω : cos sin 0 g g x x t y y t l l ω ω + + + ±± ±± = sin cos 0 g g x x t y y t l l ω ω + + ±± ±± = 5.21 24 sin T λ = hours 24 73.7 sin19 T = = D hours 5.22 Choose a coordinate system with the origin at the center of the wheel, the x and axes pointing toward fixed points on the rim of the wheel, and the axis pointing toward the center of curvature of the track. Take the initial position of the y z x axis to be horizontal in the V D G direction, so the initial position of the y axis is vertical. The bicycle wheel is rotating with angular velocity V about its axis, so … b D ˆ l V k b ω = D G A unit vector in the vertical direction is: ˆ ˆ ˆ sin cos V t V t n i j b b = + D D At the instant a point on the rim of the wheel reaches its highest point: ˆ ˆ ˆ sin cos V t V t r bn b i j b b = = + D D G Since the coordinate system is moving with the wheel, every point on the rim is
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# Nuclear Decay Equations Worksheet Answers
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# Calculated Column or Dax or combination
I’m new to PowerBi and still learning, and need some guidance. I have the following challenge
In the example data below. I have a table of invoices, mixed into that are what we call a credit rebills. One column contains service orders (column 4) that are generated and eventually get converted to invoices with unique numbers shown in (column 1). Periodically we have to credit the customer and create a new invoice (column 1) with a new invoice price. When a credit is created from the original SVO number a new transaction is created, it keeps the same SVO number but a new SVI or invoice is created with a new price. If I total up all the transactions in the table its not a true number until I pull out the transactions that fall under a credit rebill situation.
What I have to do is to identify any SVO that is a duplicate. Finding any that are duplicate means it was a credit rebill. Then I have to go back to column 1 and select the larger of the SVI numbers because it was the latest created. It will have the new price in Column 5. My thought was to create a calculated column that could capture all credit rebills. Any help on an approach would be greatly appreciated.
@lomorris Can you please share sample data and required output screen shot(How you want to display on report)
Hi @lomorris !
There might be are plenty of ways to solve this, but the fastest way that came to my mind is using DAX.
Firstly, I am not sure if you should rely only on dates to identify the most recent SVO since it is possible to get credits on the same date. So my solution is based on the invoice number. You could also use a combination of Invoice number and dates.
The solution only requires 2 steps
1. Identify the largest invoice number per each SVO, you could create a calculated column such as:
LargestInvoice = MAXX(FILTER(Orders,Orders[SVO]=EARLIER(Orders[SVO])),Orders[Invoice Number])
1. Filter the most recent invoices using in a new calculated table
SalesTable = FILTER(Orders,Orders[Invoice Number]=Orders[LargestInvoice])
That’s all
I have attached a PBI file with the solution. I appreciate if you can mark the problem as solved. Any question don’t hesitate!
Finance Solution Duplicated Values.pbix (91.2 KB)
Regards,
Diego
1 Like
Since this is more of a data modelling issue, I would favor handling this in Power Query. Is using PowerQuery an option here? If so, please attach some dummy data and a quick mock up of what the ideal output would be.
Thanks for posting your question @lomorris. To receive a resolution in a timely manner please make sure that you provide all the necessary details on this thread.
Here is a potential list of additional information to include in this thread; demo pbix file, images of the entire scenario you are dealing with, screenshot of the data model, details of how you want to visualize a result, and any other supporting links and details.
Including all of the above will likely enable a quick solution to your question.
Thanks Diego, You are correct! I totally left out that in the case of an SVO the related SVI to use would the one with the highest number. I am ignoring dates. The larger number would the the latest invoice associated with the SVO.
Diego,
Thanks again, it works as expected. I will take some time now to review the Dax formula to get a more thorough understanding of how it works. Thanks again, exactly what I was looking for.
Happy to help and thanks for marking the solution, you can also solve it using M as the expert said
@lomorris
As @diego mentioned there are many approaches for your requirement .
Here is another simple solution
Just create Rank calculated column… That’s it
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SITE-C teams: Lesson 3
Use the CASIO CFX-9850G to find the area of the region bounded between the functions:
Turn on the CASIO CFX-9850G by pressing AC/ON.
Use the cursor arrows to highlight the GRAPH menu and press EXE or just press 5 when at the main menu screen.
Enter into the Y1 slot the expression (x + 6) ¸ Ö ( x + 6) and press EXE to store it in memory.
Enter into the Y2 slot the expression ( 2 x2 + 2 x + 1 ) ¸ 5 and press EXE to store it in memory.
Press SHIFT F3 (V-Window) to enter the viewing window screen, and press F1 (INIT) to select the initial viewing window. Change the parameters for the y-values to:
Ymin = -0.7 Ymax = 5.5 Yscale = 1
Press EXIT to return to the function list, and press F6 (DRAW) to see the graph.
Press SHIFT F5 (G-Solv) F5 (ISCT) to find the first point of intersection of the two functions. This is (-2.5896800891, 1.8467051499). Press the right cursor arrow once to find the second point of intersection. This is (2.1219785028, 2.8499085078).
The area bounded by the two functions can be expressed mathematically by the following:
where the bounds of integration were determined by the x-values of the points of intersection of the functions.
To numerically compute the value of the area in this region press EXIT MENU 1 to access the RUN menu.
Press OPTN F4 (CALC)
F4 (ò dx) VARS F4 (GRPH) F1 (Y1) , -2.5896800891, 2.1219785028) -
OPTN F4 (CALC)
F4 (ò dx) VARS F4 (GRPH) F1 (Y2) , -2.5896800891, 2.1219785028)
These keystrokes will cause the following to appear on the screen
ò (Y1, -2.5896800891, 2.1219785028) - ò (Y2, -2.5896800891, 2.1219785028)
Press EXE and the area of the region will appear; the result is 7.141441971 square units.
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Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °
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#1 2017-07-13 14:23:34
Bhavya2
Guest
Area of triangle
In right triangle BAC , A is a right angle. AD is perpendicular to BC.
If BD = 9 cm and DC = 4 cm , can you find the area of triangle ABC.
If yes what is the area
If no why can't we find it out?
#2 2017-07-13 15:59:18
Monox D. I-Fly
Member
Registered: 2015-12-02
Posts: 798
Re: Area of triangle
The area is 39 cm² because AD is 6 cm.
Offline
#3 2017-07-13 19:20:52
bob bundy
Registered: 2010-06-20
Posts: 8,054
Re: Area of triangle
hi Bhavya2
Welcome to the forum.
I have deleted my answer because it was wrong. Monox D. I-Fly has the right result.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Offline
#4 2017-07-13 21:02:23
Monox D. I-Fly
Member
Registered: 2015-12-02
Posts: 798
Re: Area of triangle
bob bundy wrote:
Monox D. I-Fly: Where did you get 6 from?
Offline
#5 2017-07-13 22:04:40
bob bundy
Registered: 2010-06-20
Posts: 8,054
Re: Area of triangle
hi Monox D. I-Fly
Thanks for the explanation. My answer is completely wrong so I have deleted it. Whoops.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Offline
#6 2017-07-14 02:52:14
Monox D. I-Fly
Member
Registered: 2015-12-02
Posts: 798
You are welcome.
Offline
#7 2017-07-15 19:24:51
bob bundy
Registered: 2010-06-20
Posts: 8,054
Re: Area of triangle
hi NakulG
19.5? What, the area? I think not.
Monox D. I-Fly wrote:
The area is 39 cm² because AD is 6 cm.
This is correct. half base times height = 0.5 x (9+4) x 6 = 13 x 3 = 39.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Offline
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# Taking vector transposes seriously
## Introduction
The thesis of this paper is as follows:
Users want to write, using compositions of the transpose and product operators, linear algebra expressions that involve vectors and matrices, and the results make sense, i.e. the results of u'*v behaves like a scalar. Examples of linear algebraic expressions users expect to write include:
Inner product $$u^{T}v$$, a scalar,
Outer product $$uv^{T}$$, a matrix,
Quadratic form $$u^{T}Au$$, a scalar,
Bilinear form $$u^{T}Av$$, a scalar,
The description of these quantities employ Householder notation (Householder 1953, Householder 1955), a convention that is familiar to most, if not all, practitioners of numerical linear algebra today.
### Introduction (v1)
Humans, to a fault, are so good at special casing abstractions by context that it would be easy to conclude that a discussion of how linear algebra fits into computer languages would hardly seem necessary. Decades after APL made multidimensional arrays first class objects, and MATLAB made matrix laboratory syntax popular, we recently reached an inescapable conclusion, one we preferred not to believe, that vectors, in the sense of computer languages, and linear algebra vectors have serious coexistence issues.
To set the stage, in nearly every computer language (other than MATLAB (!!)) a “vector” is a one dimensional container. In concrete linear algebra there are column vectors and row vectors. In both concrete and abstract linear algebra, vectors are quantities that are subject to addition, multiplication by scalars, linear transformations, and play a role in scalar and outer products. As we shall see, the seemingly innocuous idea of transposing a vector turns out to be a highly contentious subject.
Consider the following familiar statements
• A matrix times a vector is a vector
• Matrix Multiplication of $$A$$ and $$B$$ is defined by taking the scalar product of a row of $$A$$ with a column of $$B$$.
• The scalar product is defined on a pair of vectors. The scalar product of $$v$$ with itself is a non-negative number known as $$\|v\|^{2}$$.
and consider common notations in math or software:
• A[i,:]*B[:,j] or dot(A[i,:],B[:,j])
• $$v^{T}\!v=(v,v)=$$ v'*v =dot(v,v)
(more coming)
### Introduction (v0)
Matrices and vectors are fundamental concepts for both computer science (Knuth 1967, Pratt 2001) and computational science (Strang 2003, Trefethen 1997). Nevertheless, the terms “vector” and “matrix” mean different things in the contexts of data structures and linear algebra. As data structures, they are simply arrays, whereas as linear algebraic objects, vectors are elements of a vector space and matrices are linear transformations. When we represent linear algebra vectors and linear algebra transformation in a basis, we obtain the familiar containers we know simply as vectors and matrices.
Computer science focuses primarily on the homogeneous container semantics of vectors and matrices. Oftentimes they are considered synonymous with arrays of rank 1 and 2 respectively. The seminal work of (Iliffe 1961) says
“Depending on the organization of the array it may be treated as a set, a vector, or a matrix.”
The classic (Knuth 1967) focuses only on indexing semantics; the index entry for “two-dimensional array” cross-references the entry for “matrix”. Even today, the conflation persists. A modern textbook on programming language design writes (p. 215 Pratt 2001):
A vector is a one-dimensional array; a matrix composed of rows and columns of components is a two-dimensional array[.]
In contrast, vectors and matrices in linear algebra are defined by their algebraic properties, such as inner products and matrix-vector products.
The aim of this paper is to identify if and when the semantics of array indexing and linear algebra may conflict.
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https://documen.tv/question/let-f-62-5-and-g-3-5-find-g-f-0-19800623-80/
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## Let f(x) = 6×2 + 5 and g(x) = 3x-5. Find g(f(0))
Question
Let f(x) = 6×2 + 5 and g(x) = 3x-5. Find g(f(0))
in progress 0
7 months 2021-08-02T05:48:52+00:00 1 Answers 1 views 0
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RBI Production by Lineup Spot: NL West
A player’s spot in the lineup is crucial to the amount of RBI they accumulate. What’s even more important is the production of the player hitting in front of that player. Have I stated the obvious? Probably. But, it’s easier to write down analytical laws in an agreeable way than it is to actually take a look at the data. In this post, I’m going to investigate the RBI production of players who had at least 100 plate appearances on teams in the NL West according to their most common lineup spot. Let’s see if hitting in the three spot is more productive from an RBI standpoint than the one spot, as they say it is.
The first step was to go to each team’s Roster Resource Lineup Tracker page and download the excel workbook. Next, I wrote some code in Python that would skip across each column and output the mode, or most common, batting order position for each player. Next, I merged in the player’s 2021 stats and put it all together in a table. A quick thing to note; the production listed in these tables is not isolated to one specific lineup spot, it is that player’s full 2021 stats. So, when you see Manny Machado with 106 RBI but a batting position of three, that does not mean that all 106 RBI came from hitting in the three spot. The batting position just tells us where that player batted the most in 2021. Lastly, I’ve included PAs so you can see the reason for RBI fluctuation among these hitters.
2021 Padres Production by Lineup Spot
Name PA RBI xwOBA Most Frequent Lineup Spot
Manny Machado 640 106 0.380 3
Fernando Tatis Jr. 546 97 0.406 2
Jake Cronenworth 643 71 0.344 3
Eric Hosmer 565 65 0.325 5
Trent Grisham 527 62 0.310 1
Tommy Pham 561 49 0.354 1
Adam Frazier 211 11 NaN 2
*Players with at least 100 PAs.
Results
Adam Frazier’s production above is only reflected by what he did post-trade. Otherwise, we see that the players with the most RBI hit from the three, two, and five spots most often. It was good to be Manny Machado. He had the second-most plate appearances with men on base (285), behind Jake Cronenworth (296) who lead the team in that category. Eric Hosmer followed with 258.
Fantasy Perspective
A new manager in southern California could swap players around, but it’s a good bet that Machado will continue to get ABs in productive lineup spots. Whoever is batting behind Tatis is sure to get a little boost in RBI production, but the Padres need Grisham to get on base more often if he is going to hold down the leadoff spot.
2021 Rockies Production by Lineup Spot
Name PA RBI xwOBA Most Frequent Lineup Spot
C.J. Cron 547 92 0.362 5
Ryan McMahon 596 86 0.324 5
Charlie Blackmon 582 78 0.357 3
Trevor Story 595 75 0.333 3
Brendan Rodgers 415 51 0.309 2
Raimel Tapia 533 50 0.279 1
Connor Joe 211 35 0.371 1
Garrett Hampson 494 33 0.290 1
Yonathan Daza 331 30 0.292 2
*Players with at least 100 PAs.
Results
We see the same results; players hitting more frequently in the three spot top the list. However, in this case, C.J. Cron tops the list, hitting from the five spot most frequently. With Tapia and Hampson leading off, McMahon (282 PAs), Blackmon (281), and Story (274) still had the most opportunities with men on base. But the production out of those three helped Cron, who came in fourth on the team with 251 plate appearances with men on base.
Fantasy Perspective
While Steamer has McMahon projected for 10 less RBI (proj. 76) in 2022, it is more in line with 2021 for Blackmon (74) and actually gives Story (88) more, but that could be due to the unknown of where he’ll play. If you’re a fantasy manager who streams Colorado hitters, pay close attention to where they are linning up in the order to be sure to get more production.
Arizona Diamondbacks
2021 Diamondbacks Production by Lineup Spot
Name PA RBI xwOBA Most Frequent Lineup Spot
Eduardo Escobar 400 65 NaN 3
David Peralta 538 63 0.293 4
Ketel Marte 374 50 0.371 2
Pavin Smith 545 49 0.316 1
Carson Kelly 359 46 0.343 4
Christian Walker 445 46 0.311 4
Josh Rojas 550 44 0.298 1
Josh VanMeter 310 36 0.275 2
Kole Calhoun 182 17 0.292 2
Tim Locastro 133 5 NaN 1
*Players with at least 100 PAs.
Results
Once again the hitters who lined up at three and four the most often were the most productive in the RBI department. Speed at the leadoff doesn’t translate to xwOBA at the leadoff and some of these managers might want to consider who get’s the leadoff spot. I think I may have just found my next article…
Fantasy Perspective
When you see that Salvador Perez hit 121 RBI in 2021, quick math will tell you that almost all Diamondback players recorded less than half of that. Perez did accumulate 665 PAs, so that’s an unfair comparison. But, David Peralta just barely makes the top 100 among hitters with at least 500 PAs at 98th. There’s not much to look at here. But, I think we can agree that had Ketel Marte and Carson Kelly been healthy this year, it would be a different outcome.
San Francisco Giants
2021 Giants Production by Lineup Spot
Name PA RBI xwOBA Most Frequent Lineup Spot
Brandon Crawford 549 90 0.356 5
Mike Yastrzemski 532 71 0.316 2
Brandon Belt 381 59 0.374 4
Buster Posey 454 56 0.361 3
LaMonte Wade Jr. 381 56 0.347 1
Wilmer Flores 436 53 0.312 5
Evan Longoria 291 46 0.352 5
Darin Ruf 러프 312 43 0.379 4
Alex Dickerson 312 38 0.335 4
Austin Slater 306 32 0.335 1
Tommy La Stella 242 27 0.339 1
Kris Bryant 212 22 NaN 5
Mike Tauchman 175 15 NaN 1
*Players with at least 100 PAs.
Results
A lot of crazy things happened in San Francisco in 2021. But, the RBI/Most Frequent Lineup Spot results seem to fit the narrative. You do see that the Giants really moved their players around in the order, playing more matchups and remaining unpredictable. They also had a lot of depth that would come in at different stages, leaving their opponents to game plan accordingly.
Fantasy Perspective
It’s going to be much harder to predict with any accuracy who will get the most three spot PAs in 2022. Expect Gabe Kapler to continue doing what he does and moving his players up and down the order. But, don’t count out the production that can come from veteran hitters who defy the aging curve.
Los Angeles Dodgers
2021 Dodgers Production by Lineup Spot
Name PA RBI xwOBA mode
Max Muncy 592 94 0.406 2
Justin Turner 612 87 0.359 3
Will Smith 501 76 0.363 5
Chris Taylor 582 73 0.329 6
Mookie Betts 550 58 0.356 1
Corey Seager 409 57 0.393 2
Albert Pujols 204 38 NaN 4
Cody Bellinger 350 36 0.281 4
Trea Turner 226 28 NaN 3
*Players with at least 100 PAs.
Results
Muncy, Turner and Smith; hitters are gonna hit. For most of the year it was Mookie Betts or Chris Taylor batting leadoff. But once Trea Turner came around, that changed. The utility man, Taylor, was a great producer in the bottom half. I would assume the biggest issue with batting from the three spot in LA would be the risk that the one and two hitters may have already scored.
Fantasy Perspective
Draft Dodgers.
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https://docs.nvidia.com/deeplearning/tensorrt/operators/docs/Unary.html
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# Unary¶
Computes a unary operation for each element of the input tensor.
## Attributes¶
operation Unary operation can be one of:
• EXP $$output=e^{input}$$
• LOG $$output=log_e(input)$$
• SQRT $$output=\sqrt{input}$$
• RECIP $$output=\frac{1}{input}$$
• ABS $$output=|input|$$
• NEG $$output=-input$$
• SIN $$output=sin(input)$$
• COS $$output=cos(input)$$
• TAN $$output=tan(input)$$
• SINH $$output=sinh(input)$$
• COSH $$output=cosh(input)$$
• ASIN $$output=asin(input)$$
• ACOS $$output=acos(input)$$
• ATAN $$output=atan(input)$$
• ASINH $$output=asinh(input)$$
• ACOSH $$output=acosh(input)$$
• ATANH $$output=atanh(input)$$
• CEIL $$output=\lceil input \rceil$$
• FLOOR $$output=\lfloor input \rfloor$$
• ERF $$output=gef(input) \, \text{, gef is a Gaussian Error Function}$$
• NOT $$output=~input$$
• SIGN $$output=sign(input)$$
• ROUND $$output=round(input)$$
• ISINF $$output=isinf(input)$$
## Inputs¶
input: tensor of type T
## Outputs¶
output: tensor of type T. ISINF only outputs bool type.
## Data Types¶
Operation
T
ABS
int8, int32, float16, float32
ACOS
int8, float16, float32
ACOSH
int8, float16, float32
ASIN
int8, float16, float32
ASINH
int8, float16, float32
ATAN
int8, float16, float32
ATANH
int8, float16, float32
CEIL
int8, float16, float32
ABS
int8, float16, float32
COS
int8, float16, float32
COSH
int8, float16, float32
ERF
int8, float16, float32
EXP
int8, float16, float32
FLOOR
int8, float16, float32
LOG
int8, float16, float32
NEG
int8, int32, float16, float32
RECIP
int8, float16, float32
ROUND
int8, float16, float32
SIN
int8, float16, float32
SINH
int8, float16, float32
SQRT
int8, float16, float32
ATAN
int8, float16, float32
NOT
bool
SIGN
int8, int32, float16, float32
ISINF
float16, float32
## Shape Information¶
The output has the same shape as the input.
## DLA Restrictions¶
When using DLA, only operation of type ABS is supported.
## Examples¶
Unary
in1 = network.add_input("input1", dtype=trt.float32, shape=(2, 3))
layer = network.add_unary(in1, op=trt.UnaryOperation.ABS)
network.mark_output(layer.get_output(0))
inputs[in1.name] = np.array([[-3.0, -2.0, -1.0], [0.0, 1.0, 2.0]])
outputs[layer.get_output(0).name] = layer.get_output(0).shape
expected[layer.get_output(0).name] = np.array([[3.0, 2.0, 1.0], [0.0, 1.0, 2.0]])
## C++ API¶
For more information about the C++ IUnaryLayer operator, refer to the C++ IUnaryLayer documentation.
## Python API¶
For more information about the Python IUnaryLayer operator, refer to the Python IUnaryLayer documentation.
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https://sat.magoosh.com/flashcards/vocabulary/SAT_math-algebra-exponents-roots/dividing-powers-with-different-bases
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← Math: Algebra, Exponents, Roots Cards you don't know will reappear later
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dividing powers with different bases
To divide exponents with different bases, divide the bases and keep the exponent the same.
Examples: y4 ÷ z4 = (y/z)4 ; 64 ÷ 34 = (6/3)4 = 24 = 16
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https://setscholars.net/binary-classification-using-gaussiannb-multinomialnb-bernoullinb-classifiers/
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# Binary Classification using GaussianNB, MultinomialNB, BernoulliNB classifiers
Binary classification is a type of supervised learning where the goal is to predict one of two possible outcomes, such as “positive” or “negative”. The Naive Bayes Classifier is a popular algorithm for binary classification, and it is implemented in the scikit-learn library (sklearn) in three different forms: GaussianNB, MultinomialNB and BernoulliNB.
GaussianNB is used when the data is continuous and follows a normal distribution. It is based on the assumption that the likelihood of a feature is a Gaussian distribution. This implementation is particularly useful when the features of the data are continuous, such as in problems related to image processing, speech recognition, and natural language processing.
MultinomialNB is used when the data is discrete, and the features represent the occurrence of events. It is based on the assumption that the features are generated by a multinomial distribution. This implementation is particularly useful when the features of the data are counts, such as in problems related to text classification, document classification, and sentiment analysis.
BernoulliNB is also used when the data is discrete, but the features represent binary events. It is based on the assumption that the features are generated by a Bernoulli distribution. This implementation is particularly useful when the features of the data are binary, such as in problems related to spam detection, fraud detection, and customer churn prediction.
The first step in using these classifiers is to prepare the dataset, this dataset should contain labeled examples of the two possible outcomes. The dataset should also be split into a training set and a test set. The training set is used to train the algorithm, and the test set is used to evaluate its performance.
After selecting the appropriate classifier and preparing the data, we can train the classifier on the training set using the `fit()` function. The classifier will learn the patterns in the data and use them to make predictions about the outcomes of new examples. After the classifier is trained, we can evaluate its performance on the test set using the `predict()` function. This function will take in the test set and return an array of predicted labels for each example in the test set. We can then compare the predicted labels with the actual labels to calculate the accuracy of the classifier.
Finally, we can use the trained classifier to make predictions on new, unseen examples. This can be done by calling the `predict()` function on the trained classifier and passing in the new examples.
In summary, using Naive Bayes Classifier for binary classification in scikit-learn (sklearn) involves selecting the appropriate classifier from the GaussianNB, MultinomialNB and BernoulliNB based on the nature of data. Preparing a labeled dataset, splitting it into a training set and a test set. Training the classifier on the training set using the `fit()` function and evaluating its performance on the test set using the `predict()` function. Finally, using the trained classifier to make predictions on new unseen examples by calling the `predict()` function on the trained classifier and passing in the new examples. Each of these implementations is suited for different types of data, and the choice of implementation will depend on the nature of the data.
In this Applied Machine Learning & Data Science Recipe (Jupyter Notebook), the reader will find the practical use of applied machine learning and data science in Python programming: Binary Classification using GaussianNB, MultinomialNB, BernoulliNB classifiers.
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`Disclaimer: The information and code presented within this recipe/tutorial is only for educational and coaching purposes for beginners and developers. Anyone can practice and apply the recipe/tutorial presented here, but the reader is taking full responsibility for his/her actions. The author (content curator) of this recipe (code / program) has made every effort to ensure the accuracy of the information was correct at time of publication. The author (content curator) does not assume and hereby disclaims any liability to any party for any loss, damage, or disruption caused by errors or omissions, whether such errors or omissions result from accident, negligence, or any other cause. The information presented here could also be found in public knowledge domains.`
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How to compare boosting ensemble Classifiers in Multiclass Classification
How to compare boosting ensemble Classifiers in Python
How to tune Hyperparameters in Gradient boosting Classifiers in Python
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## Topic 14.6: Stiff Differential Equations (HOWTO)
Introduction Notes Theory HOWTO Examples Engineering Error Questions Matlab Maple
# Problem
Given the IVP
y(1)(t) = f( t, y(t) )
y(t0) = y0
approximate y(t1).
# Assumptions
The function f(t, y) should be continuous in both variables.
# Tools
We will use the backward divided-difference formula.
# Initial Conditions
Set h = t1t0. Let y1 be the approximation of y(t1).
# Process
Solve the equation
for &upsilon and set y1 = υ.
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# If $Q=\left\{x:x=\frac{1}{y},\text{where}\phantom{\rule{1ex}{0ex}}y\in N\right\}$, then
A
$0\in Q$
B
$1\in Q$
C
$2\in Q$
D
$-\frac{2}{3}\in Q$
Text Solution
Verified by Experts
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Updated on:21/07/2023
### Knowledge Check
• Question 1 - Select One
## If x+y=k, where x,yεN then xy is maximum when (x,y)≡..
A(k3,2k3)
B(k4,3k4)
C(k2,k2)
D(2k3,2k3)
• Question 2 - Select One
## If xy=k, where x,yεN, then x+y is minimum when (x,y)≡…
A(k,1)
B(k,k)
C(k1/4,k3/4)
D(k2/3,k2/3)
• Question 3 - Select One
## If xy+yx=−1, where x,y≠0 then the value of (x3−y3) is
A1
B-1
C0
D12
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# Modelling tips and tricks: Using OFFSET for scenario analysis
istock_savushkin_ss
Tags
## Welcome to a new series of applied tips and tricks in Excel for accountants and financial analysts wherever you may be. As a professional modeller, FCA and Excel MVP Liam Bastick highlights some of the more useful functions for financial modelling/Excel spreadsheeting.
In the first part of a two-part discussion on OFFSET, Liam explains how OFFSET can assist with “what if?” analysis.
## OFFSET at the outset
Here is a useful function for modelling that is often maligned: OFFSET. This function considers disposition or displacement and has the following syntax:
OFFSET(Reference,Rows,Columns,[Height],[Width])
The arguments in square brackets (Height and Width) can be omitted from the formula (they both have a default value of 1 which will be explained further in my next article).
In its most basic form, OFFSET(Reference, Rows, Columns) will select a reference Rows rows down (-Rows would be Rows rows up) and Columns rows to the right (-Columns would be Columns rows to the left) of the Reference.
For example, consider the following grid:
## Content series
### Modelling tips and tricks: Flagging IF
OFFSET(A1,2,3) would take us two rows down and three columns across to cell D3. Therefore, OFFSET(A1,2,3) = 16, viz.
OFFSET(D4,-1,-2) would take us one row up and two rows to the left to cell B3. Therefore, OFFSET(D4,-1,-2) = 14, viz.
It is this displacement technique that can create a scenario table:
This example is included in the attached Excel file. Essentially, the assumptions used in the model are linked from cells L17:L24 (mainly in cyan). These values are drawn from the scenario table to the right of the highlighted yellow range (e.g. cells N17:N24 constitute Scenario 1. The “Base” case, cells O17:O24 constitute Scenario 2).
The Scenario Selector is located in cell H12. Using OFFSET we can retain all scenarios and select as we see fit. For example, the formula in cell L18 (highlighted) is simply =OFFSET(M18,,\$H\$12), that is, start at cell M18 and displace zero rows and the value in H12 columns across. In the illustration above, the formula locates the cell one column to the right, which is Scenario 1.
The advantage of OFFSET over other functions such as INDEX, CHOOSE and LOOKUP functions (we’ll cover these another time!) is that the range of data can be added to. The image below shows Scenarios 6 and 7 added in an instant. Whilst the other functions require a specified range whereas we can keep adding scenarios without changing the formula / making the model inefficient.
It should be noted that OFFSET is a volatile function, i.e. a function that causes recalculation of a formula in the cell where it resides every time Excel recalculates. This occurs regardless of whether precedent cells/calculations have changed, or whether the formula also contains non-volatile functions. One test to check whether your workbook is volatile is close a file after saving and see if Excel prompts you to save it a second time (this is an indicative test only).
OFFSET is also what is known as a non-auditable function in that it does not recognise dependent cells that are linked via an OFFSET function. For example, in my illustration above, the \$3.70 in cell N18 is clearly used. However, if you were to select this cell and trace dependents (ALT + M + D), you would get the following message:
This should not put you off using OFFSET; it is a function that frequently calculates much quicker than the alternative options and its advantages may often outweigh the potential pitfalls.
It’s such a versatile and useful function I shall be continuing with further examples next time.
### About Liam Bastick
Recognised by Microsoft as one of 104 Most Valuable Professionals (MVPs) in Excel worldwide by Microsoft, Liam has over 30 years’ experience in financial model development/auditing, valuations, M&A, strategy, training and consultancy. He has headed Ernst & Young’s modelling team in Melbourne and was an Assistant Director in their strategic valuations team in London. He was also a senior member of the UK Post Office’s M&A and strategy teams and has worked for / assisted various other Australian modelling companies including BPM, Corality, Navigator Project Finance, PKF and SumProduct.
He has worked in the UK, Australia, Belgium, Denmark, France, Germany, Hong Kong, Indonesia, Malaysia, New Zealand, United States, Switzerland and Vietnam, with many internationally recognised clients, constructing and reviewing strategic, operational and valuation models for many high profile IPOs, LBOs and strategic assignments. Liam is a Fellow of the Institute of Chartered Accountants (ICAEW), a Fellow of the Institute of Chartered Management Accountants (CIMA) and is a professional mathematician.
### Please login or register to join the discussion.
There are currently no replies, be the first to post a reply.
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# Problem solver program
Problem solver program can be a useful tool for these scholars. We can help me with math work.
## The Best Problem solver program
One tool that can be used is Problem solver program. In other words, if you know the lengths of two sides of a right triangle, you can use this theorem to find the length of the third side. For example, if you know that one leg is 3 inches long and the other leg is 4 inches long, you can use the Pythagorean theorem to find that the length of the hypotenuse is 5 inches. In general, solving for a side in a right triangle is a matter of applying simple algebra to the Pythagorean theorem. With a little practice, you will be able to solve for sides in right triangles with ease.
A series solver is a program that solves mathematical series. Series are mathematical expressions that can be represented in the form of an infinite summation. Series solvers are used to find the value of a particular series at a certain point. Series solvers can be used to solve Series for Convergence, Series for Divergence, and Series for Alternating Series. Series solvers have a wide range of applications in mathematics and physics. Series solvers are essential in solving complex mathematical problems that cannot be solved by hand. Series solvers can be used to solve problems in physics, engineering, and other sciences. Series solvers are also used in financial analysis and in business decision-making.
Calculus can be a difficult subject for many students. In addition to mastering a new set of concepts, students must also learn how to apply those concepts to solve complex problems. While some students may be able to do this on their own, others may find it helpful to use a calculus solver with steps. A calculus solver with steps can show students how to work through a problem from start to finish, allowing them to see the thought process behind each step. This can be a valuable tool for students who are struggling to understand the material or for those who simply want to check their work. Calculus solvers with steps are available online and in many textbooks. With a little bit of searching, students should be able to find a calculator that meets their needs.
Solving by square roots is a mathematical process for finding the value of a number that, when squared, equals a given number. For example, the square root of 9 is 3, because 3 squared (3 x 3) equals 9. In general, the square root of x is equal to the number that, when multiplied by itself, equals x. Solving by square roots can be done by hand or with the help of a calculator. The process involves finding the value of one number that, when multiplied by itself, equals the given number. This value is then used to determine the answer to the original problem. Solving by square roots is a useful tool for solving many mathematical problems.
There are a variety of methods that can be used to solve differential equations, depending on the specific equation and the desired results. The most common method is known as separation of variables, which involves breaking the equation down into simpler pieces that can be solved individually. Other methods include integrating factors and substitution, among others. With practice, solving differential equations can become second nature.
## Help with math
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# 285/65r17 In Inches – Tire Size, Rim Fitment & Best Tires
The 285/65R17 tire has an overall diameter of approximately 31.6 inches, a section width of roughly 11.2 inches, and is designed to be mounted on a 17-inch diameter rim. The equivalent tire size in the high flotation system is 31.6×11.2R17.
## What does 285/65r17 Tire mean?
Ever looked at your tire and thought, “285/65r17, what in the rubber world does that mean?” Each of these numbers and letters have specific meanings which might surprise you. Here’s a simple breakdown in a two-column table to clear the air of confusion:
So, a 285/65r17 tire is a radial tire with a width of 285mm, an aspect ratio of 65%, designed to fit a 17-inch rim.
## What is 285/65r17 Tire in inches?
Now that we’ve decoded the hieroglyphs of tire measurement, let’s get down to converting our tire size from the metric system (millimeters) to inches, because, let’s face it, in some parts of the world, millimeters might as well be light years.
Let’s prepare a table with six rows and three columns showing the conversion of 285/65r17 into inches and millimeters:
For the math nerds among you, here’s how we got these numbers:
• Tire Diameter = (2 x Sidewall Height) + Rim Diameter
• Sidewall Height = (Width x Aspect Ratio / 100) / 25.4 (to convert from mm to inches)
• Circumference = Diameter x Pi (3.14)
The “Revolutions per Mile” is important to know as it impacts the accuracy of your speedometer. A larger tire will have fewer revolutions per mile, and a smaller tire will have more. Consequently, if you change your tire size, it might affect your speedometer reading.
## A Deeper Dive into the 285/65r17 Tire Specifications
So, you want to know more about the magical 285/65r17 tire, huh? Buckle up, my tire enthusiast friend, because we’re about to take a wild ride through the nitty-gritty details of this rubber marvel. Get ready to be blown away by facts, figures, and some good old tire-related banter!
### Tire width of 285/65r17
Let’s talk about width, baby! The 285 in 285/65r17 represents the width of the tire in millimeters. That’s right, we’re measuring rubber with metric precision here. This bad boy spans a robust 285mm from sidewall to sidewall, ensuring a wide and stable footprint on the road. It’s like having a pair of monster truck tires, ready to conquer any terrain that stands in your way.
### Tire height of 285/65r17
We all want some height in our lives, and our tires are no exception! The 65 in 285/65r17 reveals the aspect ratio of the tire. Now, hold onto your socks because we’re about to do some tire height calculations! The aspect ratio is presented as a percentage, representing the height of the tire sidewall compared to its width. In this case, the tire stands tall at 65% of its width. It’s like the tire is saying, “Hey, I’m not just wide, I’ve got some serious height too!”
### Sidewall height of 285/65r17
Ah, the sidewall height. It’s like the unsung hero of tire specifications, quietly providing stability and protection. For the 285/65r17 tire, the sidewall height measures around 7.3 inches. This measurement is crucial as it affects the tire’s overall performance and handling. A taller sidewall can offer a smoother ride, absorbing bumps and imperfections like a pro. It’s like having your own personal shock absorber attached to each wheel.
### Rim diameter for 285/65r17
Now, let’s talk about rims, those shiny circles that hold our precious tires in place. For the 285/65r17 tire, you’ll need a rim with a diameter of 17 inches. This is the size of the wheel that the tire is specifically designed to fit. So, make sure you have the right-sized rims to keep those tires hugging the road like a long-lost lover.
### Tire circumference of 285/65r17
Time to get a little circular here! The tire circumference is like the tire’s own personal marathon track. It’s the distance the tire covers in one complete revolution. For the 285/65r17 tire, the circumference measures around 102.2 inches. That’s a whole lot of rubber rotating per revolution! Just imagine the tire sprinting around at high speeds, ready to conquer any mile you throw at it.
## What are the best 285/65r17 tires?
When it comes to finding the best 285/65r17 tires, you’re in luck because I’ve done the research for you. These top-notch picks will have you cruising in style and confidence, whether you’re tackling off-road adventures or dominating the city streets. So, without further ado, let’s dive into the top 3 contenders:
### 1 – Yokohama Geolandar AT G015 Performance Tire 285/65R17 116H
Description: The Yokohama Geolandar AT G015 is a performance tire that delivers the perfect balance between on-road comfort and off-road capability. Its aggressive tread design and advanced technology ensure excellent traction in various conditions, from muddy trails to dry highways. With enhanced durability and exceptional handling, this tire is ready to conquer any terrain.
Key Features:
• Triple 3D sipes for enhanced stability and improved grip
• Dual sidewall protectors for added resistance against cuts and impacts
• Geo-Shield technology for excellent all-season performance
• Long-lasting tread life for extended mileage
Pros:
• Impressive off-road traction and grip
• Smooth and comfortable ride on highways
• Resistant to wear and tear
• Low road noise for a quiet driving experience
Cons:
• Slightly limited snow performance compared to dedicated winter tires
• Not the most fuel-efficient option in its class
Final Recommendation: If you’re seeking a tire that excels in both on-road comfort and off-road prowess, the Yokohama Geolandar AT G015 is an excellent choice. Its impressive performance and durability make it a reliable companion for any adventure.
### 2 – Cooper Discoverer AT3 LT All-Season LT285/65R17 121/118S Tire
Description: The Cooper Discoverer AT3 LT is an all-season tire that combines rugged performance with refined on-road capabilities. Designed for light trucks and SUVs, this tire offers reliable traction on any terrain, from muddy trails to dry pavement. With its aggressive tread pattern and strong construction, the Discoverer AT3 LT is built to handle all your off-road escapades.
Key Features:
• Adaptive-Traction technology for superior grip in various conditions
• Whisper Grooves for reduced road noise and a quiet ride
• Stone ejector ribs to prevent stone drilling and enhance tire life
Pros:
• Excellent off-road traction and durability
• Smooth and comfortable on-road performance
• Low road noise for a peaceful driving experience
• Impressive wet and dry traction
Cons:
• Slightly reduced snow and ice performance compared to dedicated winter tires
• Fuel efficiency could be improved
Final Recommendation: The Cooper Discoverer AT3 LT is a solid choice if you’re seeking a tire that can handle both off-road adventures and daily commuting. Its rugged performance and versatility make it a reliable companion for any terrain.
### 3 – Antares COMFORT A5 All-Season Radial Tire – 285/65R17 116T
Description: The Antares COMFORT A5 is an all-season radial tire designed for SUVs and trucks. With its stylish appearance and reliable performance, this tire delivers a smooth and comfortable ride on both city streets and highways. It offers excellent traction and stability, ensuring a confident driving experience in various weather conditions.
Key Features:
• Four wide circumferential grooves for efficient water evacuation
• Optimized tread pattern for enhanced grip and handling
• Long-lasting tread life for extended mileage
Pros:
• Smooth and comfortable ride
• Good traction in wet and dry conditions
• Affordable price point
• Low road noise for a quiet driving experience
Cons:
• Snow and ice performance could be improved
Final Recommendation: If you’re looking for a budget-friendly option that offers decent on-road performance and comfort, the Antares COMFORT A5 is worth considering. It’s a reliable tire for
Got some burning questions about the magnificent 285/65r17 tire? Don’t worry, I’ve got you covered. From vehicle compatibility to air pressure and more, I’m here to shed light on your curiosities. So, grab a seat, and let’s dive into these frequently asked questions!
### Q: What vehicles use 285/65r17 Tires?
A: Ah, the 285/65r17 tire—a popular choice among off-road enthusiasts and truck lovers alike! This tire size is often found on a variety of vehicles, including:
• Light trucks such as the Ford F-150, Chevrolet Silverado, and Toyota Tacoma.
• SUVs like the Jeep Wrangler, Toyota 4Runner, and Ford Explorer.
• Some heavy-duty trucks, such as the Ford F-250 and Ram 2500.
These vehicles embrace the 285/65r17 tire with open arms, ready to tackle both on and off-road adventures.
### Q: How many revolutions per mile does a 285/65r17 have?
A: Revolutions per mile—quite the mind-boggling concept, isn’t it? The 285/65r17 tire typically completes around 621 revolutions per mile. That’s a whole lot of rotations, my friend! Keep in mind that this number may vary slightly depending on factors such as tire wear, tread depth, and inflation.
### Q: What is the price of a 285/65r17 tire?
A: Ah, the price question—a favorite among budget-conscious tire shoppers. While prices can vary based on brand, model, and retailer, you can expect to find 285/65r17 tires ranging from around \$150 to \$300 per tire. Remember, investing in quality tires is like pampering your vehicle’s feet, ensuring optimal performance and safety.
### Q: What kind of rim does a 285/65r17 tire fit on?
A: The 285/65r17 tire is designed to fit on a 17-inch rim. So, make sure you have a set of snazzy 17-inch rims to match your tire choice. It’s like finding the perfect dance partner—tire and rim twirling harmoniously together.
### Q: How much air should be in a 285/65r17 tire?
A: Ah, the crucial question of air pressure—the life force of any tire! For the 285/65r17 tire, the recommended air pressure usually falls in the range of 35 to 40 PSI (pounds per square inch). However, it’s essential to consult your vehicle’s owner’s manual or the tire manufacturer’s guidelines for the specific air pressure recommendation. Keeping your tires properly inflated ensures optimal performance, handling, and tread life.
### Q: What is a 285/65r17 equivalent to?
A: Oh, you’re seeking some tire size equivalency knowledge—impressive! The 285/65r17 tire is similar in overall diameter and circumference to other tire sizes such as 33×11.50r17 or 305/60r17. These sizes offer a comparable overall diameter, ensuring a similar ride height and fitment on your vehicle. It’s like finding different entrances to the same fabulous tire party!
## Comparison with similar tires
When it comes to tire sizes, it’s always intriguing to compare and contrast. So, let’s dive into some exciting tire matchups and see how the 285/65r17 tire stacks up against its tire brethren.
### 285/65r17 vs 285/70r17
In the battle of the 285/65r17 vs 285/70r17, we have two formidable contenders. The main difference lies in the aspect ratio—the 285/70r17 has a higher aspect ratio, meaning a taller sidewall compared to the 285/65r17. This translates to a slightly higher ride height and potential for increased off-road capability. However, the 285/65r17 offers a wider tread, providing better traction in certain conditions. So, it boils down to your personal preference and the specific demands of your driving adventures.
### 285/65r17 vs 265/70r17
Ah, the clash between the 285/65r17 and the 265/70r17—two tire sizes vying for supremacy. The key distinction here is the width—the 285/65r17 boasts a wider footprint compared to the 265/70r17. This wider tread can enhance stability and grip, especially during off-road endeavors. On the other hand, the 265/70r17 might offer a smoother ride due to its slightly higher sidewall. So, it’s a trade-off between width and comfort, depending on your driving priorities.
### 285/65r17 vs 265/65r17
Let’s pit the 285/65r17 against the 265/65r17 in an epic tire showdown. Once again, the difference lies in the width—the 285/65r17 takes the lead with its broader shoulder and increased traction capabilities. However, the 265/65r17 may offer a slightly quieter and more comfortable ride due to its narrower profile. It’s a balancing act between grip and refinement, depending on your driving style and preferences.
### 285/65r17 vs 275/70r17
The showdown between the 285/65r17 and the 275/70r17 is about to commence. In this matchup, the 285/65r17 takes the crown for width, providing a wider surface area for improved grip and stability. On the other hand, the 275/70r17 offers a taller sidewall, potentially providing a bit more cushioning and impact resistance. It’s a tug-of-war between width and sidewall height, depending on your desired balance of performance and comfort.
### Conclusion
After delving into the intricate world of the 285/65r17 tire, exploring its measurements, best tire options, and comparisons, it’s time to draw a conclusion. If you’re seeking a tire that combines versatility, performance, and reliability, the 285/65r17 is an excellent choice. Its width and aspect ratio strike a balance between on-road comfort and off-road capabilities, making it a favorite among truck and SUV enthusiasts.
Remember, tire selection is a personal decision, influenced by your driving style, vehicle requirements, and desired performance attributes. Consider factors such as traction, tread life, road noise, and budget to find the perfect match for your needs.
So, my fellow tire aficionados, armed with the knowledge gained from this comprehensive guide, go forth and conquer the roads with confidence. Embrace the power of the 285/65r17 tire and let it take you on extraordinary journeys, wherever your adventurous spirit leads you. Safe travels!
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# Question on validity of Heisenberg's Uncertainty principle
1. Feb 27, 2015
### Shan K
Hi,
I have a question about the validity of Hiesenberg's principle when relativity is in action.
Hiesenberg's principle tells us that simultaneous measurement of position and momentum can not be done accurately . But relativity tells us that simultaneity is relative , so simultaneous measurement in my frame is not remain simultaneous in others' frame . So can they accurately measure the position and momentum?
If they can, then we can transform the data in our frame with the help of lorentz transformation, and we will have the simultaneous position and momentum.
CAN IT BE DONE ?
2. Feb 27, 2015
### laudas
What is position ? If something is moving how can it have a absolute position ?
3. Feb 27, 2015
### Shan K
I am not talking about absolute position but the position in my reference frame. And in classical physics we can define the position of a particle at each instant of time whether it is moving or not.
4. Feb 27, 2015
### atyy
In non-relativistic quantum mechanics, observables (in the Heisenberg picture) are labelled by time. In relativistic quantum mechanics position is no longer an observable but a label for an observable. So in relativistic quantum mechanics, observables are labelled by position and time. It remains the case that only commuting observables can be measured simultaneously and accurately.
So the short answer is that one has to adjust things a bit to make quantum mechanics work in relativity, but the basic principles restricting simultaneous measurement remain the same.
5. Feb 27, 2015
### laudas
Position is always relative to something, (excuse my use of term absolute position), and how to you get object's position ?
You use a photon ? that photo must have energy which must affect the object your trying to find out about,
As a photon is not of zero size(the one you using to do the measure with), there must be a uncertainty.
So called classical physics dose not take this into account.
The other way of looking at it, all information is energy, the closer the energy of the measurement is to the energy of the particle your trying
to measure, the more affect it will have on that particle.
6. Feb 27, 2015
### bhobba
That's not what it says. You will find many threads on this forum explaining it.
First there there is no observation that simultaneously measures position and momentum.
Secondly QM is based on the Galilean transformations in which simultaneity is absolute.
Thanks
Bill
7. Feb 27, 2015
### atyy
In classical special relativity, position and time are only labels - it is the spacetime event that is absolute. In quantum special relativity, position and time are labels for events at which observations occur.
8. Feb 27, 2015
### bhobba
Good point. Regardless of relativity the output of an observation is a space-time event. QM says that output cant tell you both position and momentum.
Thanks
Bill
9. Feb 27, 2015
### tijana
What do you mean by photon is not zero size, that it have mass?
And I was puzzled with the thing that if we say that photons are particles then uncertainty as for other particles are correct. But what does it mean then? Does it mean that E is transporting with some probability?
10. Feb 27, 2015
### Staff: Mentor
Relativity tells us that simultaneity is relative for events happening at different points (to be precise, spacelike-separated events). There is no trouble establishing simultaneity for things happening at the same point and thus no difficulty applying the uncertainty principle here.
11. Mar 2, 2015
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# The ball was thrown vertically upward at a speed of 15 m / s. what movement will he make in 2 s?
At the top of the ball trajectory V = 0:
V = V0 – gt = 0,
where V0 is the initial speed of the ball, V0 = 15 m / s;
g – acceleration due to gravity, g = 9.8 m / s2.
Ball lifting time: t1 = V0 / g = 15 / 9.8 = 1.53 s.
Maximum lifting height of the ball:
h1 = V0 t1 – gt1 ^ 2/2 = 15 × 1.53 – 9.8 × 1.53 ^ 2/2 = 11.48 m.
Ball drop time: t2 = t – t1 = 2 – 1.53 = 0.47 s.
Lowering the ball in time t2:
h2 = gt2 ^ 2/2 = 9.8 × 0.47 ^ 2/2 = 1.08 m.
Ball movement in time t = 2 s:
s = h1 + h2 = 11.48 + 1.08 = 12.56 m.
One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities.
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# trajectory density plot
I want to make a plot that shows the density of trajectories (x-y plots). In other words I want a plot that's analogous to this but with curves instead of points as data.
Here's what I'm doing right now (with a simplified function) which gives okay results for 100 curves but doesn't scale to 10,000:
ta = RandomVariate[NormalDistribution[0, 0.3], {100}];
p = Map[Cos[t - #] &, ta];
Plot[p, {t, 0, 10}, PlotStyle -> {{Black, Opacity[0.05]}}]
I'd also prefer to not have to quantize the x-y space. But if that's the only solution I'll accept it.
• Have you seen DensityPlot, could you use that to display the number of trajectories that pass through the X-Y space? – image_doctor Jun 27 '15 at 0:07
• I have. I'm not sure how i'd compute a density function from 10,000 stochastic trajectories though... – user1816847 Jun 27 '15 at 5:17
• At each of may constant-x surfaces, compute the distribution (in y) of lines that pass through it. Of course, this will require discretizing x-y space. – bbgodfrey Jun 27 '15 at 6:29
fs = Function[{s, t}, (1 + Cos[t - #]) (1 + Sin[s - #])] & /@
DensityPlot[Plus @@ (#[x, y] & /@ fs), {x, 0, 2 π}, {y, 0, 2 π}]
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# Is normality temperature dependent?
Asked by: Kathryn Terry Jr.
Score: 4.4/5 (32 votes)
Normality depends upon the volume of the solution. Thus, it will vary with temperature changes.
## What does normality depend on?
Normality is the only unit of chemical concentration that depends on the chemical reaction being studied.
## Which is independent of temperature?
Molality and mole fraction are independent of temperature as both are w/w relations.
## Which of the following are temperature dependent?
Molarity is dependent on temperature. As the molarity includes the volume of a solution, that can change with a change in temperature.
## Is molarity temperature dependent?
Molarity is the function of the mole of solute and volume of the solution so, molarity will be affected by the change in temperature.
## Testing For Normality - Clearly Explained
43 related questions found
### Is molarity directly proportional to temperature?
The volume of a solution is directly proportional to temperature and is known to increase as the temperature does. Therefore, molarity is inversely proportional to the temperature. ... Thus, the molarity is known to decrease as the temperature is increased.
### Why is molarity dependent on temperature?
Molarity is solute moles per litre of solution. As the temperature increases, water expands, so the solution's volume therefore increases. In more litres, we have the same number of moles, but the molarity is smaller at higher temperatures. The temperature determines molarity.
### Which of the following is not temperature dependent?
Molality is not dependent upon any quantity that changes with temperature. So, it is temperature independent. Normality depends upon the volume of the solution.
### What is the molality of pure water?
Molality of pure water is 55.55 m.
### Which of the following concentration of solution depends on temperature?
The concentration of solution that depends on temperature is molarity.
### Which of the following is independent of temperature for a gas?
The temperature of a gas depends on volume, rates of diffusion and pressure. But vapour density is the density of a gas relative to that of hydrogen at the same temperature is known as the vapour density of the gas. Thus, vapour density is independent of temperature.
### Which concentration of temperature is independent?
The concentration terms that are "independent of temperature" are Molality and Mole Fraction.
### Which one is correct molality changes with temperature?
Molarity and normality change with temperature because of expansion or contraction of the liquid with temperature. However, molality does not change with temperature because mass of the solvent does not change with temperature.
### Why is molarity better than normality?
While Molarity refers to the concentration of a compound or ion in a solution, normality refers to the molar concentration only of the acid component or only of the base component of the solution. Thus, normality offers a more in-depth understanding of the solution's concentration in acid-base reactions.
### What is the relationship between molarity and normality?
If you know the Molarity of an acid or base solution, you can easily convert it to Normality by multiplying Molarity by the number of hydrogen (or hydroxide) ions in the acid (or base). For example, a 2 M H2SO4 solution will have a Normality of 4N (2 M x 2 hydrogen ions).
### Why do we calculate normality?
Normality is used in precipitation reactions to measure the number of ions which are likely to precipitate in a specific reaction. It is used in redox reactions to determine the number of electrons that a reducing or an oxidizing agent can donate or accept.
### What is the normality of pure water?
The sub-atomic weight of water is 18. 1000/18 = mole for each L, or 55.6M. Since n = 1 for water, water is 55.6N as well.
### What is density of pure water?
A common unit of measurement for water's density is gram per milliliter (1 g/ml) or 1 gram per cubic centimeter (1 g/cm3). ... Actually, the exact density of water is not really 1 g/ml, but rather a bit less (very, very little less), at 0.9998395 g/ml at 4.0° Celsius (39.2° Fahrenheit).
### What is the molarity of 1 liter of pure water?
The standard state for a liquid is the pure liquid, so the standard state of water is pure water, whose concentration is 55.5 M (in a liter, there are 55.5 moles of water, so its concentration is 55.5 mol/L).
### Which one of the following concentration unit does not depend on temperature?
Explanation: Molality is a quality of a solution and can be explained as the number of moles of solute per 1000 gm of solvent. The molality is represented by mol/kg. Molality does not show dependence on the temperature, neither moles of solute nor mass solvent will be affected by the deviation of temperature.
### Which one of the following will change with change in temperature?
Molarity changes with temperature.
### What is the effect of temperature on molarity?
Molarity is dependent on temperature. As the temperature increases, water expands, so the solution's volume therefore increases. In more liters, we have the same number of moles, but the molarity is smaller at higher temperatures. The temperature determines molarity.
### Why molality does not change with temperature?
At every temperature, mass is the same but with variations in temperature, molality will not change. ... Since molality is defined in terms of the solvent's mass, not its volume, the temperature does not change the molality of a solution. With temperature, molarity varies. Molarity is solute moles per litre of solution.
### Is molarity or molality dependent on temperature?
Molality (m) of a solution is defined as the number of moles of the solute present in one kilogram of the solvent. Molarity is temperature dependent while molality is independent of the temperature.
### When temperature of a solution is increased What happens to molarity?
Molarity decreases because Volume of Solution increases with increase in temperature, but the number of moles of the solute remains the same.
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Purchase Solution
# What is the horizontal force required to move a crate?
Not what you're looking for?
If the coefficient of kinetic friction between a 12.0 kg crate and the floor is .30, what horizontal force is required to move the crate at a steady speed across the floor? What horizontal force is required if the force of friction is zero?
##### Solution Summary
In the first case, the solution is found by applying Newton's first law in both vertical and horizontal directions.
##### Solution Preview
Newton's first law states that the total force applied to a body must be zero if the body is at rest or moving with constant velocity.
In this case we have to apply this law in both vertical and horizontal directions.
The forces applied on the obect are ...
##### The Moon
Test your knowledge of moon phases and movement.
##### Introduction to Nanotechnology/Nanomaterials
This quiz is for any area of science. Test yourself to see what knowledge of nanotechnology you have. This content will also make you familiar with basic concepts of nanotechnology.
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Cody
# Problem 1068. Guess the Coefficients!
Solution 250503
Submitted on 29 May 2013
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Fail
%% assert(isequal(guess_the_coefficients(@(x)x^2+2*x+15),[1 2 15]))
```Error: Inputs must be a scalar and a square matrix. To compute elementwise POWER, use POWER (.^) instead. ```
2 Pass
%% assert(isequal(guess_the_coefficients(@(x)polyval(53,x)),53))
``` ```
3 Pass
%% assert(isequal(guess_the_coefficients(@(x)polyval([54 87],x)),[54 87]))
``` ```
4 Pass
%% assert(isequal(guess_the_coefficients(@(x)polyval([49 40 68],x)),[49 40 68]))
``` ```
5 Pass
%% assert(isequal(guess_the_coefficients(@(x)polyval([75 53 35 15],x)),[75 53 35 15]))
``` ```
6 Pass
%% assert(isequal(guess_the_coefficients(@(x)polyval([59 27 5 76 25],x)),[59 27 5 76 25]))
``` ```
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Re: List plot with random size for each point
• To: mathgroup at smc.vnet.net
• Subject: [mg122444] Re: List plot with random size for each point
• From: Ray Koopman <koopman at sfu.ca>
• Date: Sat, 29 Oct 2011 07:07:45 -0400 (EDT)
• Delivered-to: l-mathgroup@mail-archive0.wolfram.com
• References: <j8dtjl\$kgs\$1@smc.vnet.net>
```On Oct 28, 2:45 am, Rudresh <rudr... at email.unc.edu> wrote:
> I want to plot an array of numbers on a 2d plot, however each point has a random number associated with it which should be its size on the plot. Can this be done in mathematica and if so how. My code as of now is:
>
> ===================================
>
> particle = Array[f, {200, 4}];
> position = Array[g, {200, 2}];
> n = 1;
> While [n < 201,
> particle[[n, 1]] = n;
> particle[[n, 2]] = RandomInteger[{15, 25}];
> particle[[n, 3]] = RandomInteger[{0, 200}];
> particle[[n, 4]] = RandomInteger[{0, 200}];
>
> position[[n, 1]] = particle[[n, 3]];
> position[[n, 2]] = particle[[n, 4]];
> n++
> ];
>
> ListPlot[position, PlotStyle -> PointSize[0.01],
> PlotRange -> {{0, 200}, {0, 200}}]
>
> ==========================================
>
> I want particle[[n,2]] to be the size of position[[n,1],[n,2]]
n = 200
particle = Transpose@{Range[n],
RandomInteger[{15,25},n],
RandomInteger[{0,200},n],
RandomInteger[{0,200},n]};
Show[Graphics[{PointSize[#2*.001],Point@{#3,#4}}& @@@ particle],
Frame->True, PlotRange->{{-5,205},{-5,205}}, AspectRatio->Automatic]
```
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# Physics 247 homework help speed velocity acceleration
### "Word Problems" - Physics: "Speed, Velocity, and Acceleration
The total distance traveled is directly proportional to the square of the time.IB Physics notes on 2.1 Kinematics. Tweet. IB Guides why fail.When an object is speeding up, the acceleration is in the same direction as the velocity.Our rules are designed to help you get a useful answer in the fewest number of posts.
It is found in the Physics Interactive section and allows a learner to apply concepts of speed, velocity and acceleration.AP Physics Practice Test: Motion in One-Dimension. including speed, velocity, acceleration,.
Given these average velocity values during each consecutive 1-second time interval, we could say that the object would fall 5 meters in the first second, 15 meters in the second second (for a total distance of 20 meters), 25 meters in the third second (for a total distance of 45 meters), 35 meters in the fourth second (for a total distance of 80 meters after four seconds).Velocity-time graphs help to cement and. velocity, and acceleration are.Both velocity and acceleration are vector quantities and a full description of the quantity demands the use of a directional adjective.
### Physics 247 Homework Help Speed Velocity Acceleration
Negative accelerations do not refer acceleration values that are less than 0.Consistent with the mathematical convention used on number lines and graphs, positive often means to the right or up and negative often means to the left or down.If an object is not changing its velocity, then the object is not accelerating.ANSWERS: Worksheet 2D - Velocity, Displacement and Acceleration. Worksheet 2D - Velocity, Displacement and.Physics homework: acceleration, velocity and speed. help. Physics homework: acceleration, velocity and speed. help please.
This subreddit is for help, pushes in the right direction, not answers.Come here for homework help in. velocity and acceleration as.In Example A, the object is moving in the positive direction (i.e., has a positive velocity) and is speeding up.
### Formal Homework Assignment 2 | Velocity | Acceleration
Now you can use this time t to, again with a related formula giving you the height of the sandbag at a time, get the maximum height and so on for the following tasks.Practice Science Questions on the subject of Physics Velocity and Acceleration.
### Velocity and acceleration, Physics - Experts Mind
About the Physics Interactives Kinematics Usage Policy Newtons Laws Vectors and Projectiles Momentum and Collisions Work and Energy Circular and Satellite Motion Balance and Rotation Electric Circuits Static Electricity Magnetism Light and Color Waves and Sound Reflection and Mirrors Refraction and Lenses.If an object is changing its velocity -whether by a constant amount or a varying amount - then it is an accelerating object.The data tables below depict motions of objects with a constant acceleration and a changing acceleration.In physics, the use of positive and negative always has a physical meaning.
### Speed, Velocity and Acceleration Unit by Masfar - Teaching
Acceleration has to do with changing how fast an object is moving.
### Maze Game - Position | Velocity | Acceleration - PhET
Have some respect for people who take time to answer your question and follow the posting rules.Banner Ads Infographics Interactive Physics Simulations Our Standard Flyer Pinterest Pages Skyscraper Ads Small Classroom Posters Teacher Toolkits Test Reviews What Can Students Do.
This discussion illustrates that a free-falling object that is accelerating at a constant rate will cover different distances in each consecutive second.A person can be moving very fast and still not be accelerating.Sometimes an accelerating object will change its velocity by the same amount each second.About notebook philosophy purpose reporting Teacher Guide Using Lab Notebooks Share.Been out in pdf file, text file physics library. 5m s epsilnn density of mass.
### Physics - Science - Brightstorm
In Example B, the object is moving in the negative direction (i.e., has a negative velocity) and is slowing down.
### physics homework: Velocity - Course Hero
A worksheet to help learners visualise the difference between.
### One-dimensional motion | Physics | Science | Khan Academy
You may need to add four spaces before or put backticks around math fragments.
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# Difference between arithmetic and mathematics
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Mar 03, 2020
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# Probability Sample Test
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| By Doriarg
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Quizzes Created: 4 | Total Attempts: 8,933
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This is a sample test to gauge your knowledge about statistics and probability. Probability Sample Test quiz will help you revise for exams and increase you know how. All the best.
• 1.
### AN event has probability 1.0. Which one of the following statements best describes this event in along sequence of trials?
• A.
This event is certain. It will definitely occur on every trial.
• B.
This event is extremely likely and will occur almost all the time.
• C.
This event is likely and will occur more often than not in the long run.
• D.
This event is likely; however it is more likely to NOT occur than to occur.
• E.
This event is very unlikely, but it will occur once in a while.
A. This event is certain. It will definitely occur on every trial.
Explanation
1. Prob=1, this event is certain. It will definitely occur on every trial.
2. .8
Rate this question:
• 2.
### AN event has probability 0.95. Which one of the following statements best describes this event in along sequence of trials?
• A.
This event is certain. It will definitely occur on every trial.
• B.
This event is extremely likely and will occur almost all the time.
• C.
This event is likely and will occur more often than not in the long run.
• D.
This event is likely; however it is more likely to NOT occur than to occur.
• E.
This event is very unlikely, but it will occur once in a while.
B. This event is extremely likely and will occur almost all the time.
Explanation
1. Prob=1, this event is certain. It will definitely occur on every trial.
2. .8
Rate this question:
• 3.
### AN event has probability 0.5. Which one of the following statements best describes this event in along sequence of trials?
• A.
This event is certain. It will definitely occur on every trial.
• B.
This event is extremely likely and will occur almost all the time.
• C.
This event is likely and will occur more often than not in the long run.
• D.
This event is likely; however it is more likely to NOT occur than to occur.
• E.
This event will occur half of the time in the long run
E. This event will occur half of the time in the long run
Explanation
1. Prob=1, this event is certain. It will definitely occur on every trial.
2. .8
Rate this question:
• 4.
### AN event has probability 0.35. Which one of the following statements best describes this event in along sequence of trials?
• A.
This event is certain. It will definitely occur on every trial.
• B.
This event is extremely likely and will occur almost all the time.
• C.
This event is likely and will occur more often than not in the long run.
• D.
This event is likely; however it is more likely to NOT occur than to occur.
• E.
This event will occur half of the time in the long run
D. This event is likely; however it is more likely to NOT occur than to occur.
Explanation
1. Prob=1, this event is certain. It will definitely occur on every trial.
2. .8
Rate this question:
• 5.
### AN event has probability 0.05. Which one of the following statements best describes this event in along sequence of trials?
• A.
This event is extremely likely and will occur almost all the time.
• B.
This event is likely and will occur more often than not in the long run.
• C.
This event is likely; however it is more likely to NOT occur than to occur.
• D.
This event is very unlikely, but it will occur once in a while.
• E.
This event is impossible. It can never occur.
D. This event is very unlikely, but it will occur once in a while.
Explanation
1. Prob=1, this event is certain. It will definitely occur on every trial.
2. .8
Rate this question:
• 6.
### AN event has probability 0.0. Which one of the following statements best describes this event in along sequence of trials?
• A.
This event is extremely likely and will occur almost all the time.
• B.
This event is likely and will occur more often than not in the long run.
• C.
This event is likely; however it is more likely to NOT occur than to occur.
• D.
This event is very unlikely, but it will occur once in a while.
• E.
This event is impossible. It can never occur.
E. This event is impossible. It can never occur.
Explanation
1. Prob=1, this event is certain. It will definitely occur on every trial.
2. .8
Rate this question:
• 7.
### Of the following statements about standard deviation, three statements True and one is False. Which statement is FALSE?
• A.
Standard deviation has a unit of measure.
• B.
Standard deviation is only positive.
• C.
Standard deviation is inflated by outliers.
• D.
Standard deviation should be used to measure spread even when the mean is not an appropriate measure of center.
D. Standard deviation should be used to measure spread even when the mean is not an appropriate measure of center.
Explanation
Properties of Standard deviation:
1. Has a unit of measure
2. Can only be positive values.
3. Is affected by outliers. Outliers increases standard deviation.
4. It is appropriate measure of spread if the data is symmetric. When the data is symmetric, the mean is the appropriate measure of center.
Rate this question:
• 8.
### A survey conducted by the Statistics Department to assess the study habits of the Stats221 students and obtained data from a sample of 1000 Stats221 students. The average number of hours per week studying for Stats221 class was found to be 15 with a standard deviation of 5. Is 15 a parameter or a statistic? Why?
• A.
A parameter because the average is for all Stats221 students.
• B.
A parameter because is the average for all Stats221 students at BYU.
• C.
A statistic because it summarizes the results of the entire population of Stats221 students at BYU.
• D.
A statistic because it summarizes the results of this sample of 1000 Stats221 students.
D. A statistic because it summarizes the results of this sample of 1000 Stats221 students.
Explanation
Parameter refers to population mean or population standard deviation or population proportion.
Statistics refers to sample mean or sample standard deviation or sample proportion.
Rate this question:
• 9.
### A survey conducted by the BYU-Salt Lake Center to assess the reasons students from BYU-Provo take classes at the Salt Lake Center instead of in Provo and obtained data from a sample of 1500 BYU-Provo students. The administration believes that BYU-Provo students take an average of 6 credits per semester at the BYU-SLC. The average number of credits taken by BYU-Provo students was found to be 11 with a standard deviation of 3. Is 6 a parameter or a statistic? Why?
• A.
A parameter because the average is for all students.
• B.
A parameter because it is the average for all students at BYU-Provo.
• C.
A statistic because it summarizes the results of the entire population of students at BYU-Provo.
• D.
A statistic because it summarizes the results of this sample of 1500 BYU-Provo students.
B. A parameter because it is the average for all students at BYU-Provo.
Explanation
Parameter refers to population mean or population standard deviation or population proportion.
Statistics refers to sample mean or sample standard deviation or sample proportion.
Rate this question:
• 10.
### Which one of the following concepts says that if sample mean is based on a large simple random sample from a non-normal population, we can calculate approximate probabilities on sample mean using the standard Normal table?
• A.
The Law of Large Numbers.
• B.
The Central Limit Theorem.
• C.
The principle of least squares.
• D.
The sampling distribution of the mean.
• E.
There is no such result.
B. The Central Limit Theorem.
Explanation
The Central Limit theorem states that if sample mean is based on a large simple random sample from a non-normal population, we can calculate approximate probabilities on sample mean using the standard Normal table
Rate this question:
• 11.
### Theoretically, the "sampling distribution of sample mean" is which of the following?
• A.
The distribution of values of x in the population of interest.
• B.
The distribution of values of sample mean, but only in a very large samples.
• C.
The distribution of values of x in a specified sample from the population.
• D.
The distribution of values of sample mean from all possible samples of a specified size.
• E.
The histogram of values of x in a random sample of size n.
D. The distribution of values of sample mean from all possible samples of a specified size.
Explanation
The "sampling distribution of sample mean" is the distribution of values of sample mean from all possible samples of a specified size.
Rate this question:
• 12.
### The price of a pound of grapes in the grocery stores along the Wasatch Front is distributed normally with a mean of \$1.40 and a standard deviation of \$0.15. A store selling a pound of grapes for \$1.60 is considered expensive. What is the probability that a randomly selected store is selling expensive grapes?
• A.
0.9082
• B.
0.9099
• C.
0.0918
• D.
0.0901
• E.
0.1600
C. 0.0918
Explanation
z=(1.60-1.4)/0.15=1.33
From the Normal Table:
z Percentage to the left of z
1.33 .9082
The percentage to the right of z: 1-.9082=0.0918
Rate this question:
• 13.
### The price of a pound of grapes in the grocery stores along the Wasatch Front is distributed normally with a mean of \$1.40 and a standard deviation of \$0.15. What is the probability that the mean price of a pound of grapes of four randomly selected grocery stores will exceed \$1.60?
• A.
0.0038
• B.
0.0918
• C.
0.9962
• D.
0.9082
A. 0.0038
Explanation
z=(1.6-1.4)/(0.15/sqrt(4))=2.67
Percentage to the right of z: 1-0.9962=0.0038
Rate this question:
• 14.
### What does Central Limit Theorem allow us to do?
• A.
Calculate probabilities of sample mean from a large random sample taken from a non-Normal population.
• B.
Determine whether the data are sampled from a population which is Normally distributed.
• C.
Specify the probability of obtaining each possible random sample of size n.
• D.
Know exactly what the value of the sample mean will be.
A. Calculate probabilities of sample mean from a large random sample taken from a non-Normal population.
Explanation
The Central Limit Theorem allows us to calculate probabilities of sample mean from a large random sample taken from a non-Normal population.
Rate this question:
• 15.
### The histogram below represents the distribution of a population. Suppose a sampling distribution of is constructed from samples of size 60 from this population. Which one of the following figures best represents the sampling distribution?
• A.
A
• B.
B
• C.
C
• D.
D
A. A
Explanation
The sampling distribution is Normal with less spread and taller when n>=30.
Rate this question:
• 16.
### The histogram below represents the distribution of a population. Suppose a sampling distribution of is constructed from samples of size 60 from this population. Which one of the following figures best represents the sampling distribution?
• A.
A
• B.
B
• C.
C
• D.
D
A. A
Explanation
The sampling distribution is Normal with less spread and taller when n>=30.
Rate this question:
• 17.
### The histogram below represents the distribution of a population. Suppose a sampling distribution of is constructed from samples of size 60 from this population. Which one of the following figures best represents the sampling distribution?
• A.
A
• B.
B
• C.
C
• D.
D
A. A
Explanation
The sampling distribution is Normal with less spread and taller regardless of size when the population is already Normal.
Rate this question:
• 18.
### Use the counts of the following table to answer the following questions:Which of the following is the conditional distribution for Professions for those whose ethnicity is Asian?
• A.
A
• B.
B
• C.
C
• D.
D
C. C
Explanation
50/110=45.4; 20/110=18.2; 10/110=9.1; 30/110=27.3
Rate this question:
• 19.
### Use the counts of the following table to answer the following questions: What is the marginal percent of respondents who are lawyers?
• A.
18.18%
• B.
15.38%
• C.
42.86%
• D.
30%
• E.
27.08%
E. 27.08%
Explanation
130/480=27.08
Rate this question:
• 20.
### Use the counts of the following table to answer the following questions: Of the Asians, what percent are lawyers?
• A.
45.45%
• B.
18.18%
• C.
10.42%
• D.
81.81%
B. 18.18%
Explanation
20/110=18.18
Rate this question:
• 21.
### Use the counts of the following table to answer the following questions: Of the lawyers, what percent are Asians?
• A.
18.18%
• B.
15.38%
• C.
27.08%
• D.
30.76%
B. 15.38%
Explanation
20/130=15.38
Rate this question:
• 22.
### Consider the statement " The survey finds differences between Asians and Caucasians over professions. Just 7% of the Caucasians say that they are engineers; by contrast, more than 45% of the Asians say that they are engineers." Which of the following agrees with this statement:
• A.
There is a curved relationship between ethnicity and profession.
• B.
There is a linear relationship between ethnicity and profession.
• C.
There is an association between ethnicity and profession.
• D.
There is no association between ethnicity and profession.
C. There is an association between ethnicity and profession.
Explanation
When the percentages are not equal then there is association between the two categorical variables.
Rate this question:
• 23.
### Students who have taken Stats221 were asked in a survey about their Stats221 Final exam scores and how much time they spent studying for the class.The correlation coefficient between these two variables is 0.90. On the basis of this information, approximately what percentage of the variation in Stats221 Final exam score can be explained by how time students spent studying for the class?
• A.
90%
• B.
9%
• C.
81%
• D.
None of the above
C. 81%
Explanation
(.90)^2=.81 or 81%
Rate this question:
• 24.
### Consider these plots: the plot on the left is a scatterplot and the plot on the right is the corresponding residual plot from fitting a regression line. On the basis of these plots, is fitting the data with least-squares regression appropriate? Why or why not?
• A.
No, because the relationship between X and Y is non-linear.
• B.
No, because Y has more spread for larger value of X.
• C.
No, because there is no relationship between X and Y.
• D.
Yes, because X and Y has a linear relationship.
• E.
Yes, because there is a relationship between X and Y.
A. No, because the relationship between X and Y is non-linear.
Explanation
fitting the data with least-squares regression is NOT appropriate if the scatter plot shows non-linear pattern.
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• 25.
### Consider these plots: the plot on the left is a scatterplot and the plot on the right is the corresponding residual plot from fitting a regression line. On the basis of these plots, is fitting the data with least-squares regression appropriate? Why or why not?
• A.
No, because the relationship between X and Y is non-linear.
• B.
No, because Y has more spread for larger value of X.
• C.
No, because there is no relationship between X and Y.
• D.
Yes, because X and Y has a linear relationship.
• E.
Yes, because there is a relationship between X and Y.
D. Yes, because X and Y has a linear relationship.
Explanation
fitting the data with least-squares regression is appropriate when the scatter plot shows a linear pattern.
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• 26.
### In order to apply the Central Limit Theorem when sampling from a non-Normal population, which one of the following states the necessary conditions?
• A.
The population must not be skewed.
• B.
The population distribution from which the data are sampled must be Normal.
• C.
A large simple random sample must be taken.
• D.
The data must have a bell-shaped distribution.
C. A large simple random sample must be taken.
Explanation
When sampling from a non-Normal population, a large (n>=30) simple random sample must be taken for the Central Limit to hold that the shape of the sampling distribution is Normal.
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• 27.
### Consider a population that is non-Normally distributed with mean 70 and a standard deviation of 10. A sample size of 25 is taken from this population. Which of the following statements about the sampling distribution of sample means is true?
• A.
The sampling distribution of sample means is Normally distributed with standard deviation less than 10.
• B.
The sampling distribution of sample means is Normally distributed with standard deviation of 10.
• C.
The sampling distribution of sample means is Normally distributed with standard deviation greater than 10.
• D.
The sampling distribution of sample means is not Normally distributed with standard deviation less than 10.
• E.
The sampling distribution of sample means is not Normally distributed with standard deviation of 10.
D. The sampling distribution of sample means is not Normally distributed with standard deviation less than 10.
Explanation
When sampling from a non-Normal population, a large (n>=30) simple random sample must be taken for the Central Limit to hold that the shape of the sampling distribution is Normal.
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• 28.
### Home prices follow an extremely right skewed distribution with mean \$200k and a standard deviation of \$30k. If we select an SRS of 16 homes, can we use the standard Normal table to validly calculate the probability that their average home price is is greater than \$250k?
• A.
No, because the population standard deviation is not known.
• B.
No, because the Central Limit Theorem does not apply.
• C.
Yes, because of the sampling distribution of sample means is Normal.
• D.
Yes, because of the Law of Large numbers.
B. No, because the Central Limit Theorem does not apply.
Explanation
When sampling from a non-Normal population, a large (n>=30) simple random sample must be taken for the Central Limit to hold that the shape of the sampling distribution is Normal.
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• 29.
### Which one of the following control charts show a process that is in statistical control?
• A.
A
• B.
B
• C.
C
• D.
D
C. C
Explanation
Out of control charts when points are outside the UCL or LCL, and 9 consecutive points below or above the mean.
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• 30.
### The mean amount of money spent by BYU single students on dates this semester was \$380, with a standard deviation of \$100. The distribution was skewed to the right. Suppose you took 20,000 simple random samples of 49 students, and computed the mean amount for each sample. Between what two values would you expect approximately 95% of the mean amounts for the samples to be?
• A.
\$80 to \$680
• B.
\$351.4 to 408.6
• C.
\$280 to \$480
• D.
\$100 to \$380
B. \$351.4 to 408.6
Explanation
stdev for x-bar=100/sqrt(49)=14.29
95% covers 2 stdev above and below the mean, therefore, 380-3(14.29)=351.4, and 380+3(14.29)=408.6
Rate this question:
• 31.
### Cost of dates in a semester of single students at BYU are Normally distributed with mean \$380. A study is being planned to determine whether cost of dates in a semester for single students who are returned missionaries differ from the BYU mean. Proposed sample #1 has a sample size of 50 and proposed sample #2 has a sample size of 25. Which proposed sample is more likely to have an unusually high sample mean (i.e., a sample mean exceeding \$500 on dates)?
• A.
Proposal sample #1
• B.
Proposal sample #2
• C.
Both proposed samples have equal chance of having an unusually high sample mean.
B. Proposal sample #2
Explanation
The smaller the sample size, the larger the stdev of x-bar.
The larger the sample size, the smaller the stdev of x-bar.
stdev of x-bar=sigma/sqrt(n)
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• 32.
### The scatter plot below depicts the relationship between calories and sodium content of several brands of meat hotdogs. The least squares line has been drawn on the plot. What does the point labeled A in the scatterplot indicate?
• A.
The point labeled A does not have a residual.
• B.
The point labeled A has a positive residual.
• C.
The point labeled A has a negative residual.
• D.
The point labeled A is not an outlier.
B. The point labeled A has a positive residual.
Explanation
Points above the line have positive residuals.
Points below the line have negative residuals.
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• 33.
### A study was done of the accuracy of three-point shooting using Doria’s method and the traditional method. Dan Bentley took sixty shots using Doria’s method and 50 shots using the traditional method. The results are in the following table:Based on the information above, which of the following best describes Dan’s three-point shooting?
• A.
Dan shoots better using Doria method.
• B.
Dan shoots better using Traditional method.
• C.
There is no association between Dan’s shooting percentage and method of shooting.
• D.
There is a strong association between Dan’s shooting percentage and method of shooting.
C. There is no association between Dan’s shooting percentage and method of shooting.
Explanation
36/60=.6; 30/50=.6; 66/110=.6
If the conditional and marginal percentages are equal then there is NO association between the two categorical variables.
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• 34.
### Fill in the blank: If sampling distributions of sample means are examined for samples of size 9, 16, 25, and 36 from a non-Normal population, you will notice that as n increases in size, the shape of the sampling distribution appears more like that of the _____________________.
• A.
Sample distribution
• B.
Uniform distribution
• C.
Population distribution
• D.
Normal distribution.
D. Normal distribution.
Explanation
from a non-Normal population, you will notice that as n increases in size, the shape of the sampling distribution appears more like that of the Normal distribution.
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• 35.
### Consider the theoretical sampling distribution of sample means based on samples of size 50 from a uniform (flat) population where the population mean=70 and population standard deviation =10.The mean of the theoretical sampling distribution of sample means
• A.
Is greater than 70 because of the skewness.
• B.
Is close to but not equal to 70.
• C.
Is exactly equal to 70.
• D.
Cannot be determined.
C. Is exactly equal to 70.
Explanation
The mean of all the sample means=the population mean.
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• 36.
### Consider the theretical smpling distribution of sample means based on samples of size 50 from a uniform (flat) population where the population mean=70 and population standard deviation =10.The standard deviation of the theoretical sampling distribution of sample means
• A.
Is less than 10.
• B.
Is exactly equal to 10.
• C.
Is greater than 10.
• D.
Cannot be determined.
A. Is less than 10.
Explanation
stdev of x-bar=sigma/sqrt(n)=10/sqrt(50)=1.41 < 10
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• 37.
### Consider the theoretical sampling distribution of sample means based on samples of size 50 from a uniform (flat) population where the population mean=70 and population standard deviation =10.The shape of the theoretical sampling distribution of sample means is
• A.
Approximately Normal.
• B.
Right skewed.
• C.
Left skewed.
• D.
Flat or uniform.
A. Approximately Normal.
Explanation
If n>=30 and the population in non-normal, the shape of the sampling distribution for x-bar is normal.
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• 38.
### Andy, Samantha, Britt, and Christine have each taken a random sample of single students from BYU to estimate the variability in amount spent on dates this semester. Andy asked 15 people, Samantha 25, Britt 40, and Christine 50. Whose sample standard deviation probably differs most from the population standard deviation?
• A.
Andy
• B.
Samantha
• C.
Britt
• D.
Christine
A. Andy
Explanation
From the law of large numbers, the sample stdev will be closer to the population stdev the larger the sample size, just like the sample mean will be closer to the population mean as the sample size increases.
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• 39.
### Andy, Samantha, Britt, and Christine have each taken a random sample of single students from BYU to estimate the variability in amount spent on dates this semester. Andy asked 15 people, Samantha 25, Britt 40, and Christine 50. Whose sample standard deviation probably differs the least from the population standard deviation?
• A.
Andy
• B.
Samantha
• C.
Britt
• D.
Christine
D. Christine
Explanation
From the law of large numbers, the sample stdev will be closer to the population stdev the larger the sample size, just like the sample mean will be closer to the population mean as the sample size increases.
Please note that we are not talking about the standard deviation of the sampling distribution.
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• 40.
### Assuming the scales on the x and y axis are the same for all graphs, which of the following shows a weak positive relationship.Edit Quiz / Quiz School - Create Free Quizzes
• A.
A
• B.
B
• C.
C
• D.
D
C. C
Explanation
The more scattered the points from an imaginary line, the weaker the correlation.
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• 41.
### In the Casino Royale movie, James Bond played a card game called Baccarat. He wins by playing in a specified manner. Which of the following is true about the probability of winning the game?
• A.
It can be computed from the number of ways the 52 cards can be dealt.
• B.
It is impossible to compute.
• C.
It is approximated by playing lots of times and dividing the number of times you win by the number of times you play.
• D.
It is ,007 because this is Bond's number.
C. It is approximated by playing lots of times and dividing the number of times you win by the number of times you play.
Explanation
Probability is the number of success divided by the total possibilities.
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• 42.
### The data in the table below represent distance a shipment must travel and the length of time, in days, it takes the shipment to arrive . The line in the scatterplot is the least squares line: where y-hat is the predicted length of time and x=distance of shipment, in miles.Interpret the slope of the regression line in context.
• A.
For each additional mile, time in days to deliver the shipment increases by .02 day on average.
• B.
For each additional mile, time in days to deliver the shipment increases by 6.9 days on average
• C.
For each additional day, there is a 0.02 miles a shipment traveled on the average.
• D.
For each additional day, there is a 6.9 miles a shipment traveled on the average.
A. For each additional mile, time in days to deliver the shipment increases by .02 day on average.
Explanation
slope is the average increase/decrease in y per one unit change in x.
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• 43.
### The data in the table below represent distance a shipment must travel and the length of time, in days, it takes the shipment to arrive at month. The line in the scatterplot is the least squares line: where y-hat is the predicted length of time and x=distance of shipment, in miles. For these data r2 = 83.7%. Interpret r2in context.
• A.
Eighty-three point seven percent of delivery time is based on distance.
• B.
Distance explains 83.7% of the variation in delivery time.
• C.
Distance can be used to predict delivery time about 83.7% of the time.
• D.
Delivery time can be used to predict distance about 83.7% of the time.
B. Distance explains 83.7% of the variation in delivery time.
Explanation
r^2 is the percent variation in y that can be explained by x.
Rate this question:
• 44.
### The data in the table below represent distance a shipment must travel and the length of time, in days, it takes the shipment to arrive at month. The line in the scatterplot is the least squares line: where y-hat is the predicted length of time and x=distance of shipment, in miles. Why is the line on the scatter plot the least squares line?
• A.
The distances of the points to the line are as small as possible.
• B.
The sum of the squared residuals is as small as possible.
• C.
The predicted shipping time are identical to the actual shipping time used to determine the line.
• D.
The sum of the deviations of the shipping days from the mean shipping days is minimized
• E.
The predicted shipping days is as small as possible.
B. The sum of the squared residuals is as small as possible.
Explanation
the least squares line is The sum of the squared residuals is as small as possible.
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• 45.
### The data in the table below represent distance a shipment must travel and the length of time, in days, it takes the shipment to arrive at month. The line in the scatterplot is the least squares line: where y-hat is the predicted length of time and x=distance of shipment, in miles. The distance of a package to be shipped is 500 miles. What is the predicted delivery time in days?
• A.
6.9 days
• B.
.02 day
• C.
3.1 days
• D.
Cannot be computed.
C. 3.1 days
Explanation
y=-6.9+.02(500) = 3.1
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• 46.
### The data in the table below represent distance a shipment must travel and the length of time, in days, it takes the shipment to arrive at month. The line in the scatterplot is the least squares line: where y-hat is the predicted length of time and x=distance of shipment, in miles. Seattle is 2984 shipping miles from Cleveland, Ohio. Should you use this least squares equation to predict the shipping days?
• A.
No, because we do not know whether the deviation will be positive or negative.
• B.
No, because Seattle is much farther away than any other observed city in the data set.
• C.
Yes, because the predicted value is reasonable.
• D.
Yes, because the increase in shipping days is the same for each shipping mile.
B. No, because Seattle is much farther away than any other observed city in the data set.
Explanation
Do not use the least squares equation to predict y when x is outside the range of the data. This is called extrapolation.
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• 47.
### Researchers recorded facts about 100 dates, including cost spent on date and number of hugs they gave each other on a date. The linear model relating cost spend on date and the number of hugs the couple gave each other is predicted number of hugs = 3.5 + 1.5cost. Based only on the given information, what can be said about the correlation between cost spend on date and the number of hugs the couple gave each other?
• A.
The correlation is 0.
• B.
The correlation is 1.5
• C.
The correlation 3.5
• D.
The correlation is positive, but we cannot say what the exact value is.
• E.
It is impossible to say anything about the correlation from the information given.
D. The correlation is positive, but we cannot say what the exact value is.
Explanation
The slope has the same sign as the correlation.
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• 48.
### Referring to the scatterplot below, what is the best guess for the value of the correlation coefficient r between the explanatory and response variables?
• A.
0.85
• B.
0.45
• C.
-0.85
• D.
-0.45
• E.
0.35
C. -0.85
Explanation
This has a negative correlation.
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• 49.
### Which one of the following is a FALSE statement about the correlation coefficient r?
• A.
R is affected by outliers.
• B.
Values of r range from -1 to +1.
• C.
The correlation r has no unit of measure.
• D.
If the sign of r is negative, then increases in one variable are associated with decreases in the other variable.
• E.
The designations of which is the explanatory variable and which is the response variable cannot be interchanged.
E. The designations of which is the explanatory variable and which is the response variable cannot be interchanged.
Explanation
We get the same correlation even if we interchange the explanatory and the response variables.
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• 50.
### How does the correlation coefficient r for the data in Plot A compares with the correlation coefficient for the data in Plot B?
• A.
R in Plot A is less than r in Plot B.
• B.
R in Plot A is greater than r in Plot B.
• C.
R in Plot A is equal to r in Plot B.
• D.
R in Plot A is not comparable with r in Plot B.
• E.
Not enough information exists to compare the two r's.
A. R in Plot A is less than r in Plot B.
Explanation
Outliers lower the correlation.
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## Finding work by linearizing pressure and volume
The below table lists pressure and volume measurements for air in a cylindrical piston of an air compressor. The variables may be assumed related by: PV^k=C, where k and C are constants. Pressure (mmHg) Volume(cm3) 760. 48.3 1140 37.4 1520 31.3 2280 24.1 3040 20.0 3800 17.4 a.) First linearize the model equation and then determine the best fit values for k and C (include units and use correct significant figures). b.) What work is done by the compressor to the gas from the first measurement to the last? (for part a, I simply plugged in the values to a scatter plot and set a linear trend line, but unsure how to find values for C and k, short of assuming k=1, and then solving for C) (for part b, I took the equation of the trend line and integrated it from 48.3..17.4, is this on the right track or completely bogus?) Thanks in advance for help.
• Anonymous commented
the following pressure values correspond to the volume respectively, 760.- 48.3, 1140 -37.4, 1520-31.3, 2280-24.1, 3040-20, 3800-17.4
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are viewing an older version of this Concept. Go to the latest version.
# Conservation of Energy
## The amount of energy in a closed system never changes.
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The World's Fastest Roller Coaster
### The World’s Fastest Roller Coaster
Credit: Sarah Ackerman
Source: http://www.flickr.com/photos/sackerman519/8638541154/
Known as the fastest roller coaster in the world, Formula Rossa accelerates its riders to 149 mph. Not only does this 92 second ride have a total length of 2200 m and a vertical drop of 51.5 m, but its passengers will experience almost 5.0Gs!
#### Amazing But True
Credit: Jeremy Thomson
Source: http://www.flickr.com/photos/rollercoasterphilosophy/3228818023/
The energy equation lets engineers determine how fast a roller coaster can travel [Figure2]
• While almost every roller coaster is different, they all follow the same basic principle. The system's potential energy is increased and then gravity is allowed to take over, causing most of the potential energy to be converted into kinetic energy. Any of the energy that is not converted into kinetic energy by the end of the ride is energy that has been lost due to friction.
• The energy at any given point along the ride is given as:
Kinitial+Uinitial+Wext=Kfinal+Ufinal\begin{align*}K_{initial}+U_{initial}+W_{ext}=K_{final}+U_{final}\end{align*}
Where K\begin{align*}K\end{align*} and U\begin{align*}U\end{align*} represent the kinetic and potential energies respectively. W\begin{align*}W\end{align*} represents any work that is done on the ride by external forces (friction/air resistance). Using this equation and the principle that energy must be conserved, engineers are able to determine how fast they can make the ride as well as how sharp each turn can be.
#### What Do You Think?
Using the information provided above, answer the following questions.
1. Why do most roller coasters start by bringing the passengers up a really large incline?
2. Does the Wext\begin{align*}W_{\text{ext}}\end{align*} in the above equation serve to increase or decrease the speed at which the riders on a roller coaster can reach?
3. How is the energy in the system distributed when one of the carts on the roller coaster go through an upside down loop?
### Notes/Highlights Having trouble? Report an issue.
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# Posted by: Student138538z4 – (3+6i)z2 – 8 + 6i =0Slove the value of Z . It is a simple calculus to solve the value of ZFormulas are to be used are(a+b)2=a2+b2+2ab(�+�)2=�2+�2+2��and i2=−1�2=−1Carefully follow
Posted by: Student138538
z4 – (3+6i)z2 – 8 + 6i =0
Slove the value of Z . It is a simple calculus to solve the value of Z
Formulas are to be used are
(a+b)2=a2+b2+2ab(�+�)2=�2+�2+2��
and i2=1�2=−1
Carefully follow through the attached answer and kindly check the basis of complex numbers and if you any doubts regarding further kindly ask me without any hesitation
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# aircraft rotation code problem
This topic is 3813 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
## Recommended Posts
here i got an very complicated thing with some source, its about airplane rotation. my plan is. getting the camera matrix(ok) calculate a target point in the direction of the camera a bit away with a litte y++ ->that point is after rotation the nose of the airplane the new point the plane and the camera should look at i try to recalculate the new point (angle yaw,angle pitch)..and thats all for the theory i dont now if rotation of the plane(pulling the stick back) change the roll angle.. all i get are strange randomly look like roations.i am very confused i only want to rotate the planes nose up (using the roll angle)..
..eclipse java + openGL(jogl)
//calculate angle
public float winkel(float wtx,float wty)
{
float e;
wtx+=0.00000001f;
wty+=0.00000001f;
wiii=(float) ((Math.atan2( wtx ,wty)*180)/3.141593)-90;
if (wtx<0 && wty<0.0001) wiii+=360;
if (wty>0) wiii=360+wiii;
return wiii;
}
//Calculate distance
public float entf(float wtx,float wty)
{
float e=(float) (Math.sqrt( (wtx*wtx)+(wty*wty) ));
return e;
}
// get the Matrix of the Camera
...
gl.glTranslatef(0.0f,0.0f,0.0f);
gl.glRotatef(-view_r3, 0, 0, 1); //roll
gl.glRotatef(-view_r2, 1, 0, 0); //pitch
gl.glRotatef(-view_r1, 0, 1, 0); //yaw
gl.glGetFloatv(GL.GL_MODELVIEW_MATRIX, mvm, 0);
...
// view_r1 = yaw // view_r2= pitch // view_r3=roll
float tx=0.0f; float ty=0.1f; float tz=10.0f;
float tx1,ty1,tz1;
tx1= (mvm[0]*tx)+(mvm[1]*ty)+(mvm[2]*tz);
ty1= (mvm[4]*tx)+(mvm[5]*ty)+(mvm[6]*tz);
tz1= (mvm[8]*tx)+(mvm[9]*ty)+(mvm[10]*tz);
//New Angle view_r1
new_view_r1=winkel(-tx1,-tz1)-90; // the result looks ok ...
float en=entf(tx1,tz1); //getting distance
new_view_r2=winkel(en,ty1); //totally bug
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Did you write that code? Are you looking for advice on a bug, or are you interested in understanding the theory of rotations?
##### Share on other sites
hello grhodes_at_work,
yes, i wrote the code and its open for everybody
if somebody need it.
'Are you looking for advice on a bug, or are you interested in understanding the theory of rotations?'
i am intressting in both.
a direct advice on the code would be perfect,
but its very important to understand the rotation itself.
i was trying a lot of things, but i know there
is an logical error where i need help
:)
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# Max subarray sum using dynamic programming in C
The maximum subarray problem is to find the subarray within an array of integers with the largest sum. There is a linear time dynamic programming algorithm for solving it invented by Joseph Kadane.
Here it is in C:
int max(int a, int b)
{
return a > b ? a : b;
}
int max_subarray(int *array, size_t n)
{
int max_ending_here = 0;
int max_so_far = 0;
unsigned int i;
for (i = 0; i < n; i++) {
max_ending_here = max(0, max_ending_here + array[i]);
max_so_far = max(max_so_far, max_ending_here);
}
return max_so_far;
}
Example program:
int main(void)
{
int array[] = {-2, 1, -3, 4, -1, 2, 1, -5, 4};
const size_t n = sizeof(array) / sizeof(int);
printf("Max subarray sum is %d\n", max_subarray(array, n));
return 0;
}
Output:
6
# Partition function using dynamic programming in C
A partition of an integer is a way of writing it as a sum of positive integers. For example, partitions of 5 are:
5
4 + 1
3 + 2
3 + 1 + 1
2 + 2 + 1
2 + 1 + 1 + 1
1 + 1 + 1 + 1 + 1
The order of numbers in a partition doesn’t matter. The number of partitions of a number $$n$$ is given by the partition function $$p(n)$$.
The first few values of $$p(n)$$ are 1, 1, 2, 3, 5, 7, 11, 15, 22, 30, 42, 56, 77, 101, 135, 176, 231, 297, 385, 490, 627, 792, 1002, 1255, 1575, 1958, 2436, 3010, 3718, 4565, 5604, … (A000041).
The partition function can be computed using a dynamic programming table-filling algorithm. Here it is in C:
#include <stdlib.h>
unsigned int partition(unsigned int n)
{
unsigned int i, j;
unsigned int result;
unsigned int **table;
if (n == 0) {
return 1;
}
table = malloc((n + 1) * sizeof(unsigned int *));
for (i = 0; i <= n; i++) {
table[i] = malloc((n + 1) * sizeof(unsigned int));
}
for (i = 0;i <= n; i++) {
table[i][0]=0;
}
for (i = 1; i <= n; i++) {
table[0][i] = 1;
}
for (i = 1; i <= n; i++) {
for (j = 1; j <= n; j++) {
if (j > i) {
table[i][j] = table[i][j - 1];
}
else {
table[i][j] = table[i][j - 1] + table[i - j][j];
}
}
}
result = table[n][n];
for (i = 0; i <= n; i++) {
free(table[i]);
}
free(table);
return result;
}
Example program:
#include <stdio.h>
int main(void)
{
unsigned int i;
for (i = 0; i < 31; i++) {
printf("%d\n",partition(i));
}
return 0;
}
Output:
1
1
2
3
5
7
11
15
22
30
42
56
77
101
135
176
231
297
385
490
627
792
1002
1255
1575
1958
2436
3010
3718
4565
5604
# Shortest Common Supersequence using dynamic programming in C
The Shortest Common Supersequence (SCS) problem is to find the shortest sequence that is a supersequence of two or more sequences. It is closely related to the Longest Common Subsequence problem, in that the SCS of two strings is precisely their LCS, with those letters from each that are not in the LCS spliced into the string in their proper place relative to the SCS letters. For example, given the strings “pAqBrCs” and “wAxByCz”, their LCS is “ABC”, and an SCS is “wpAxqByrCzs”. Note that the order in which the letters occurring between the LCS letters is chosen from the two strings doesn’t matter, as long as the letters from each string are in the same order as they occur in it, so “pwAqxByrCsz” is also an SCS, and in general the SCS is not unique.
One way of computing an SCS is to use a dynamic programming algorithm similar to the Longest Common Subsequence one, but in which the letters occurring between the LCS letters are added to the solution. This is what I have implemented in C here. Rather than using pointers in the table to retrieve the solution as I have done in other DP algorithms, this time I have used enumerated values that indicate in which direction the next cell in the sequence is to be found – DOWN, RIGHT, or DIAG (diagonally down and to the right).
typedef enum {
DOWN,
RIGHT,
DIAG,
STOP
} direction;
typedef struct {
size_t len;
char letter;
direction dir;
} cell;
char *shortest_common_supersequence(const char *str1, const char *str2)
{
size_t lstr1 = strlen(str1);
size_t lstr2 = strlen(str2);
cell **matrix;
int i;
int j;
char *result;
unsigned int r;
matrix = malloc((lstr2 + 1) * sizeof(cell*));
for (i = 0; i <= lstr2; i++) {
matrix[i] = malloc((lstr1 + 1) * sizeof(cell));
}
for (i = 0; i < lstr2; i++) {
matrix[i][lstr1].len = lstr2 - i;
matrix[i][lstr1].letter = str2[i];
matrix[i][lstr1].dir = DOWN;
}
for (j = 0; j < lstr1; j++) {
matrix[lstr2][j].len = lstr1 - j;
matrix[lstr2][j].letter = str1[j];
matrix[lstr2][j].dir = RIGHT;
}
matrix[lstr2][lstr1].len = 0;
matrix[lstr2][lstr1].letter = 0;
matrix[lstr2][lstr1].dir = STOP;
for (i = lstr2 - 1; i >= 0; i--) {
for (j = lstr1 - 1; j >= 0; j--) {
cell *pcell = &matrix[i][j];
if (str2[i] == str1[j]) {
pcell->dir = DIAG;
pcell->letter = str1[j];
pcell->len = matrix[i + 1][j + 1].len + 1;
}
else if (matrix[i][j + 1].len < matrix[i + 1][j].len) {
pcell->dir = RIGHT;
pcell->letter = str1[j];
pcell->len = matrix[i][j + 1].len + 1;
}
else {
pcell->dir = DOWN;
pcell->letter = str2[i];
pcell->len = matrix[i + 1][j].len + 1;
}
}
}
result = malloc(matrix[0][0].len + 1);
i = 0;
j = 0;
r = 0;
while (i <= lstr2 && j <= lstr1) {
result[r] = matrix[i][j].letter;
r++;
if (matrix[i][j].dir == DOWN) {
i += 1;
}
else if (matrix[i][j].dir == RIGHT) {
j += 1;
}
else {
i += 1;
j += 1;
}
}
for (i = 0; i <= lstr2; i++) {
free(matrix[i]);
}
free(matrix);
return result;
}
Example program:
int main(void)
{
char a[] = "pAqBrCs";
char b[] = "wAxByCz";
char *scs = shortest_common_supersequence(a, b);
printf("%s\n", scs);
free(scs);
return 0;
}
Output:
wpAxqByrCzs
# Unbounded Knapsack using dynamic programming in C
The unbounded knapsack problem is to try to fill a knapsack of a given capacity with items of given weight and value in such a way as to maximise the value of the knapsack. There is no limit on the number of instances of each item, hence the “unbounded” label. The decision problem of whether the knapsack can be filled to greater than or equal to a value is NP-complete. The maximisation problem can be solved in pseudo-polynomial time using dynamic programming.
The algorithm works by filling an array of 0, …, the capacity of the knapsack from the bottom up with the value most valuable knapsack for each capacity. In order to work out at each capacity what the most profitable knapsack is, it considers each item in turn and finds out what the value of the knapsack would be if this was the last item added. This value is the value of the earlier knapsack that is lighter by the weight of the current item, plus the value of the current item. Since the knapsack values are calculated from lower to higher capacities, the value of this earlier, lighter knapsack can always be found without needing to recompute it. Once each of the items has been considered as the most recent addition, the one that gives rise to the most valuable knapsack is added, and this value is recorded in the current cell of the array. The algorithm then proceeds to the next cell of the array. When the algorithm reaches the top of the array, and fills that cell in, it has found the most valuable knapsack for the specified capacity.
There is an implicit link between each knapsack in the array and the previous lighter knapsack from which it was derived by adding an item. These links form multiple chains through the array. The chain that begins at the top cell is the one that traces the path of additions that have led to the most profitable knapsack. In order to realise this link, the array cells contain pointers to the earlier knapsack from which they were constructed, and these pointers are set as the algorithm progresses. Once the array is full, the values of the items can be retrieved by following this linked list of pointers.
struct knapsack {
unsigned int profit;
struct knapsack *prev;
};
typedef struct knapsack knapsack;
/* Find the minimum weight item with this profit */
int min_weight_item(unsigned int profit, const unsigned int *weights,
const unsigned int *profits, size_t len)
{
int item = -1;
unsigned int i;
for (i = 0; i < len; i++) {
if (profits[i] == profit) {
if (item == -1 || weights[i] < weights[item]) {
item = i;
}
}
}
return item;
}
unsigned int unbounded_knapsack(unsigned int capacity, unsigned int *weights,
unsigned int *profits, unsigned int *counts, size_t len)
{
knapsack *z = malloc((capacity + 1) * sizeof(knapsack));
unsigned int c, i;
unsigned int solution, profit;
z[0].profit = 0;
z[0].prev = NULL;
knapsack *current;
/* Fill in the array */
for (c = 1; c <= capacity; c++) {
z.profit = z.profit;
z.prev = &(z);
for (i = 0; i < len; i++) {
if (weights[i] <= c) {
/* prev is the best knapsack without adding this item */
knapsack *prev = z + (c - weights[i]);
if (prev->profit + profits[i] > z.profit) {
z.profit = prev->profit + profits[i];
z.prev = prev;
}
}
}
}
/* Read back the best solution */
for (profit = z[capacity].profit, current = z[capacity].prev;
current != NULL;
profit = current->profit, current = current->prev) {
counts[min_weight_item(profit - current->profit, weights, profits, len)]++;
}
solution = z[capacity].profit;
free(z);
return solution;
}
An example program:
static void print_knapsack(const unsigned int *counts, const unsigned int *profits, size_t len)
{
unsigned int i;
for (i = 0; i < len; i++) {
if (counts[i] > 0) {
printf("%d x %d\n", counts[i], profits[i]);
}
}
}
int main(void)
{
unsigned int weights[] = {4, 3, 5, 7, 11};
unsigned int profits[] = {5, 3, 6, 2, 7};
unsigned int counts[5] = {0};
const size_t len = sizeof(weights) / sizeof(unsigned int);
const unsigned int capacity = 17;
printf("The maximum profit is %u\n",
unbounded_knapsack(capacity, weights, profits, counts, len));
print_knapsack(counts, profits, len);
return 0;
}
Output:
The maximum profit is 21
3 x 5
1 x 6
# Longest Repeated Subsequence using dynamic programming in C
This problem is very like Longest Common Subsequence, except that we are looking for the longest subsequence occurring twice in the same string, rather than once in different strings. Once again, this is a dynamic programming solution using chains of pointers in the table to keep track of the subsequences under consideration.
struct lrs_entry {
unsigned int score;
char letter;
struct lrs_entry *prev;
};
typedef struct lrs_entry lrs_entry;
unsigned int longest_repeated_subsequence(const char *str, char **result)
{
const size_t len = strlen(str);
lrs_entry **lrstab;
unsigned int i, j;
unsigned int max;
/* Allocate the table */
lrstab = malloc((len + 1) * sizeof(lrs_entry *));
for (i = 0; i <= len; i++) {
lrstab[i] = malloc((len + 1) * sizeof(lrs_entry));
}
/* Calculate the scores and build the chains */
for (i = 0; i <= len; i++) {
for (j = 0; j <= len; j++) {
if (i == 0 || j == 0) {
/* Initialising the first row or column */
lrstab[i][j].score = 0;
lrstab[i][j].letter = 0;
lrstab[i][j].prev = NULL;
}
else if (str[i - 1] == str[j - 1] && i != j) {
/* Match */
lrstab[i][j].score = lrstab[i - 1][j - 1].score + 1;
lrstab[i][j].letter = str[i - 1];
lrstab[i][j].prev = &lrstab[i - 1][j - 1];
}
else {
/* Mismatch */
if (lrstab[i][j-1].score > lrstab[i-1][j].score) {
lrstab[i][j].score = lrstab[i][j-1].score;
lrstab[i][j].letter = lrstab[i][j-1].letter;
lrstab[i][j].prev = lrstab[i][j-1].prev;
}
else {
lrstab[i][j].score = lrstab[i - 1][j].score;
lrstab[i][j].letter = lrstab[i - 1][j].letter;
lrstab[i][j].prev = lrstab[i - 1][j].prev;
}
}
}
}
max = lrstab[len][len].score;
/* Allocate the output array and copy the LRS into it */
*result = malloc(max + 1);
for (i = max - 1, head = &lrstab[len][len];
}
(*result)[max] = '\0';
for (i = 0; i <= len; i++) {
free(lrstab[i]);
}
free(lrstab);
return max;
}
Example program:
int main()
{
char str[] = "abcXdYeXZfgYhZijk";
char *result;
printf("Longest Repeated Subsequence: %u\n", longest_repeated_subsequence(str, &result));
printf("%s\n", result);
free(result);
return 0;
}
Output:
Longest Repeated Subsequence: 3
XYZ
# Longest Common Substring using dynamic programming in C
The Longest Common Substring problem is similar to Longest Common Subsequence, except that the subsequence must be made of adjacent characters – a substring. The dynamic programming algorithm is also very similar. For this one I have used chains of pointers in the table to retrieve the solution, as I did with Longest Increasing Subsequence,
#include <string.h>
#include <stdlib.h>
struct lcs_entry {
unsigned int score;
char letter;
struct lcs_entry* prev;
};
typedef struct lcs_entry lcs_entry;
unsigned int longest_common_substring(const char *str1, const char *str2,
char **result)
{
unsigned int i, j;
unsigned int max = 0;
const size_t m = strlen(str1);
const size_t n = strlen(str2);
lcs_entry **lcstab;
/* Allocate the table */
lcstab = malloc((m + 1) * sizeof(lcs_entry*));
for (i = 0; i < m + 1; i++) {
lcstab[i] = malloc((n + 1) * sizeof(lcs_entry));
}
/* Calculate scores for each substring and build chains */
for (i = 0; i <= m; i++) {
for (j = 0; j <= n; j++) {
if (i == 0 || j == 0) {
lcstab[i][j].score = 0;
lcstab[i][j].letter = 0;
lcstab[i][j].prev = NULL;
}
else if (str1[i - 1] == str2[j - 1]) {
lcstab[i][j].score = lcstab[i - 1][j - 1].score + 1;
lcstab[i][j].letter = str1[i - 1];
lcstab[i][j].prev = &lcstab[i - 1][j - 1];
if (lcstab[i][j].score > max) {
max = lcstab[i][j].score;
}
}
else {
lcstab[i][j].score = 0;
lcstab[i][j].letter = 0;
lcstab[i][j].prev = NULL;
}
}
}
/* Allocate the output array and copy the LC substring */
*result = malloc(max + 1);
(*result)[max] = '\0';
for (i = max - 1; head->prev != NULL; i--) {
}
for (i = 0; i < m + 1; i++) {
free(lcstab[i]);
}
free(lcstab);
return max;
}
An example program:
int main(void)
{
char str1[] = "magnetohydrodynamics";
char str2[] = "hydrogen";
char *result;
printf("Length of Longest Common Substring is %u\n",
longest_common_substring(str1, str2, &result));
printf("%s\n", result);
free(result);
return 0;
}
The output:
Length of Longest Common Substring is 5
hydro
# Longest Increasing Subsequence using dynamic programming in C
A longest increasing subsequence (LIS) of a sequence is a maximal subsequence which is not necessarily contiguous, and is monotonically increasing across its length, and is as long as the longest such subsequence for that sequence. A given sequence may have more than one longest increasing subsequence.
This is a dynamic programming algorithm for finding an LIS of a sequence of integers. It works by building chains of increasing subsequences as it scans the main sequence, and keeping track of a maximum one. At the end of the scan the maximum subsequence chain is read back and copied to an output array.
struct lis_entry {
int value;
int score;
struct lis_entry *prev;
};
typedef struct lis_entry lis_entry;
unsigned int longest_increasing_subsequence(const unsigned int *arr, size_t len,
unsigned int **result)
{
lis_entry *lis;
unsigned int i, j, max = 0;
lis = malloc(len * sizeof(lis_entry));
if (lis == NULL) {
return 0;
}
/* Initialise entries */
for (i = 0; i < len; i++ ) {
lis[i].value = arr[i];
lis[i].score = 1;
lis[i].prev = NULL;
}
/* Calculate LIS scores and build chains */
for (i = 1; i < len; i++ ) {
for (j = 0; j < i; j++ ) {
if (arr[i] > arr[j] && lis[i].score < lis[j].score + 1) {
lis[i].score = lis[j].score + 1;
lis[i].prev = &lis[j];
if (lis[i].score > max) {
max = lis[i].score;
}
}
}
}
/* Allocate the output array and copy the max chain */
*result = malloc(max * sizeof(int));
if (*result != NULL) {
for (i = max - 1; head != NULL; i--) {
}
}
else {
max = 0;
}
free(lis);
return max;
}
An example program using the first few terms of the Van der Corput Sequence:
int main(void)
{
unsigned int arr[] = {0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15};
const size_t len = sizeof(arr)/sizeof(arr[0]);
unsigned int *result;
unsigned int i;
unsigned int lis_length = longest_increasing_subsequence(arr, len, &result);
printf("Length of LIS is %d\n", lis_length);
for (i = 0; i < lis_length; i++) {
printf("%u\n", result[i]);
}
free(result);
return 0;
}
Output:
Length of LIS is 6
0
4
6
9
13
15
# Longest Common Subsequence in C
This Longest Common Subsequence (LCS) algorithm is a classic dynamic programming table-filling algorithm.
static size_t longest_common_subsequence_internal(const char *str1, const char *str2, matrix **mat)
{
int i, j;
const size_t alen = strlen(str1);
const size_t blen = strlen(str2);
*mat = matrix_create(alen, blen);
if (*mat == NULL) {
return 0;
}
for (i = alen - 1; i >= 0; i--) {
for (j = blen - 1; j >= 0; j--) {
if (str1[i] == str2[j]) {
matrix_set(*mat, i, j, 1 + matrix_get(*mat, i + 1, j + 1));
}
else {
matrix_set(*mat, i, j, max(matrix_get(*mat, i + 1, j), matrix_get(*mat, i, j + 1)));
}
}
}
return matrix_get(*mat, 0, 0);
}
size_t longest_common_subsequence_length(const char *str1, const char *str2)
{
matrix *mat;
size_t result = longest_common_subsequence_internal(str1, str2, &mat);
matrix_delete(mat);
return result;
}
char *longest_common_subsequence(const char *str1, const char *str2)
{
matrix *mat;
size_t lcs_len = longest_common_subsequence_internal(str1, str2, &mat);
unsigned int i = 0;
unsigned int j = 0;
unsigned int k = 0;
const unsigned int alen = strlen(str1);
const unsigned int blen = strlen(str2);
char *lcs = malloc(lcs_len + 1);
while (i < alen && j < blen) {
if (str1[i] == str2[j]) {
lcs[k] = str1[i];
i++;
j++;
k++;
}
else if (matrix_get(mat, i + 1, j) >= matrix_get(mat, i ,j + 1)) {
i++;
}
else {
j++;
}
}
lcs[k] = '\0';
matrix_delete(mat);
return lcs;
}
Example program:
int main(void)
{
char str1[] = "pAqBrCs";
char str2[] = "wAxByCz";
char *lcs;
lcs = longest_common_subsequence(str1, str2);
printf("The longest common subsequence is \"%s\"\n", lcs);
free(lcs);
return 0;
}
The longest common subsequence is "ABC"
# xkcd 287
I’m not sure what this problem is called – I’m going to call it "multicombination sum" – but I don’t doubt that it is NP-complete, as it’s a variety of knapsack problem in which the values of the items are the same as their weight.
Below are three methods of solving it: a brute force method, using backtracking, and using dynamic programming.
## Brute force method
The brute force method is just to construct all of the possible orders that might total \$15.05.
The combinatorial algorithm we want is combinations with duplicates, or multicombinations.
Since 3 of the most expensive appetizer – SAMPLER PLATE – exceeds the target, and 7 of the cheapest appetizer – MIXED FRUIT – equals the target (so that’s one of the solutions), we want to iterate over all multicombinations with k ranging from 3 to 7.
unsigned int multiset_sum(const unsigned int *multiset, const unsigned int *values, unsigned int k)
{
unsigned int i;
unsigned int sum = 0;
for (i = 0; i < k; i++) {
sum += values[multiset[i]];
}
return sum;
}
typedef void (*multiset1fn)(const unsigned int *, unsigned int);
void multicombination_sum(const unsigned int *items, size_t len, unsigned int target,
multiset1fn fun)
{
unsigned int first = target / array_max(items, len);
unsigned int last = target / array_min(items, len);
unsigned int *multiset = calloc(last + 1, sizeof(unsigned int));
unsigned int k;
if (!multiset) {
return;
}
for (k = first; k <= last; k++) {
do {
if (multiset_sum(multiset, items, k) == target) {
fun(multiset, k);
}
} while (next_multicombination(multiset, len, k));
}
free(multiset);
}
void order_print(const unsigned int *numbers, unsigned int k)
{
const char *appetizers[] = {"MIXED FRUIT", "FRENCH FRIES", "SIDE SALAD",
"HOT WINGS", "MOZARELLA STICKS", "SAMPLER PLATE"};
unsigned int i, item, count;
for (i = 0; i < k; i++) {
if (i == 0 || numbers[i] != item) {
if (i > 0) {
printf("%s (x%d) ", appetizers[item], count);
}
count = 1;
item = numbers[i];
}
else {
count++;
}
}
printf("%s (x%d)\n", appetizers[item], count);
}
int main(void)
{
unsigned int prices[] = {215, 275, 335, 355, 420, 580};
const size_t n = sizeof(prices) / sizeof(prices[0]);
const unsigned int target = 1505;
multicombination_sum(prices, n, target, (multiset1fn)order_print);
return 0;
}
Output:
MIXED FRUIT (x1) HOT WINGS (x2) SAMPLER PLATE (x1)
MIXED FRUIT (x7)
This took 1709 iterations to come up with the answer.
## Backtracking
We can drastically reduce the search space by building the orders up one item at a time, and backtracking if the target is exceeded.
typedef void(*multiset2fn)(unsigned int *, const unsigned int *, size_t);
static void multicombination_sum_recursive(int i, unsigned int *combination,
const unsigned int *items, size_t len, unsigned int target, multiset2fn fun,
unsigned int sum)
{
if (i == (int)len - 1) {
if (sum == target) {
fun(combination, items, i);
}
}
else {
unsigned int quantity;
unsigned int max_quantity = (target - sum) / items[i + 1];
for (quantity = 0; quantity <= max_quantity; quantity++) {
combination[i + 1] = quantity;
multicombination_sum_recursive(i + 1, combination, items, len, target, fun,
sum + quantity * items[i + 1]);
}
}
}
void multicombination_sum(const unsigned int *items, size_t len, unsigned int target,
multiset2fn fun)
{
unsigned int *combination = malloc(len * sizeof(unsigned int));
multicombination_sum_recursive(-1, combination, items, len, target, fun, 0);
free(combination);
}
void order_print(const unsigned int *combination, const unsigned int *items, size_t len)
{
const char *appetizers[] = {"MIXED FRUIT", "FRENCH FRIES", "SIDE SALAD",
"HOT WINGS", "MOZARELLA STICKS", "SAMPLER PLATE"};
unsigned int i;
for (i = 0; i <= len; i++) {
if (combination[i] > 0) {
printf("%s (x%d) ", appetizers[i], combination[i]);
}
}
putchar('\n');
}
int main(void)
{
unsigned int prices[] = {215, 275, 335, 355, 420, 580};
const unsigned int len = sizeof(prices) / sizeof(unsigned int);
const unsigned int target = 1505;
multicombination_sum(prices, len, target, (multiset2fn)order_print);
return 0;
}
Output:
MIXED FRUIT (x1) HOT WINGS (x2) SAMPLER PLATE (x1)
MIXED FRUIT (x7)
This took just 211 iterations.
## Dynamic Programming
As I said at the beginning, this problem is a special case of the knapsack problem in which the values of the items are the same as their weight. This means that we can use the classic dynamic programming algorithm for knapsack to solve it. The algorithm works by calculating the most profitable knapsack for each capacity up to the target capacity.
Here is an implementation of the algorithm:
struct knapsack {
unsigned int profit;
struct knapsack *prev;
};
typedef struct knapsack knapsack;
/* Find the minimum weight item with this profit */
int min_weight_item(unsigned int profit, const unsigned int *weights, const unsigned int *profits,
size_t len)
{
int item = -1;
unsigned int i;
for (i = 0; i < len; i++) {
if (profits[i] == profit) {
if (item == -1 || weights[i] < weights[item]) {
item = i;
}
}
}
return item;
}
unsigned int unbounded_knapsack(unsigned int capacity, unsigned int *weights,
unsigned int *profits, unsigned int *counts, size_t len)
{
knapsack *z = malloc((capacity + 1) * sizeof(knapsack));
unsigned int c, i;
unsigned int solution, profit;
z[0].profit = 0;
z[0].prev = NULL;
knapsack *current;
/* Fill in the array */
for (c = 1; c <= capacity; c++) {
z.profit = z.profit;
z.prev = &(z);
for (i = 0; i < len; i++) {
if (weights[i] <= c) {
knapsack *prev = z + (c - weights[i]);
if (prev->profit + profits[i] > z.profit) {
z.profit = prev->profit + profits[i];
z.prev = prev;
}
}
}
}
/* Read back the best solution */
for (profit = z[capacity].profit, current = z[capacity].prev;
current != NULL;
profit = current->profit, current = current->prev) {
counts[min_weight_item(profit - current->profit, weights, profits, len)]++;
}
solution = z[capacity].profit;
free(z);
return solution;
}
We just need to use this algorithm and pass the prices of the menu items for both weights and profits:
int main(void)
{
unsigned int values[] = {215, 275, 335, 355, 420, 580};
const size_t len = sizeof(values) / sizeof(values[0]);
unsigned int counts[6] = {0};
const unsigned int target = 1505;
unbounded_knapsack(target, values, values, counts, len);
order_print(counts, len);
return 0;
}
This produces one of the solutions:
MIXED FRUIT (x7)
We could modify the algorithm to produce all of them.
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# Functions in Formulas
Contents
[ Hide Show ]
Aspose.Tasks for C++ API supports evaluating functions defined as formula expression in Extended Attribute. These include calculation of Math, General, Text, and DateTime functions.
## Calculation of Math Expressions
1. Abs( number )
2. Atn( number )
3. Cos( number )
4. Exp( number )
5. Fix( number )
6. Int( number )
7. Log( number )
8. Rnd( number )
9. Sgn( number )
10. Sin( number )
11. Sqr( number )
12. Tan( number )
`````` 1void CalculateMathExpressions::EvaluateSine()
2{
3 System::SharedPtr<Project> project = CreateTestProjectWithCustomField();
4
5 // Set formula Sin(pi/2)
6 project->get_ExtendedAttributes()->idx_get(0)->set_Formula(u"Sin(3.1415926/2)");
7
8 // Print Calculated value
11}
12
13System::SharedPtr<Project> CalculateMathExpressions::CreateTestProjectWithCustomField()
14{
15 System::SharedPtr<Project> project = System::MakeObject<Project>();
18
20
21 System::SharedPtr<ExtendedAttribute> a = attr->CreateExtendedAttribute();
23 return project;
24}``````
## Calculation of General Functions
The following General functions can be calculated by the API.
1. Choose( index, choice-1, choice-2, … , choice-n|, choice-n)
2. IIf( expr, truepart, falsepart )
3. IsNumeric( expression)
4. IsNull( expression )
5. Switch( expr-1, value-1, expr-2, value-2, … , expr-n,value-n|, expr-n,value-n )
`````` 1void CalculateGeneralFunctions::EvaluateChoose()
2{
3 System::SharedPtr<Project> project = CreateTestProjectWithCustomField();
4
5 // Set Formula
6 project->get_ExtendedAttributes()->idx_get(0)->set_Formula(u"Choose(3, \"This is a\", \"right\", \"choice\")");
7
8 // Print extended attribute value
11}
12
13void CalculateGeneralFunctions::EvaluateIsNumeric()
14{
15 System::ArrayPtr<System::String> numericFormulas = System::MakeArray<System::String>({u"IsNumeric('AAA')", u"IsNUmeric(1)", u"IsNumeric(1<0)", u"IsNumeric(\"1.1\")", u"IsNumeric(Choose((2 + Sgn(2^-3)), 123, \"one two three\"))"});
16
17 System::SharedPtr<Project> project = CreateTestProjectWithCustomField();
18
19 {
20 for (System::String numericFormula : numericFormulas)
21 {
22 // Set Formula
23 project->get_ExtendedAttributes()->idx_get(0)->set_Formula(numericFormula);
24
25 // Print extended attribute value
28 }
29
30 }
31}
32
33void CalculateGeneralFunctions::EvaluateSwitch()
34{
35 System::SharedPtr<Project> project = CreateTestProjectWithCustomField();
36
37 // Set Formula
38 project->get_ExtendedAttributes()->idx_get(0)->set_Formula(u"Switch( 0 < 1, \"0 is lesser than 1\", 0 > 1, \"0 is greater than 1\")");
39
40 // Print extended attribute value
43}
44
45System::SharedPtr<Project> CalculateGeneralFunctions::CreateTestProjectWithCustomField()
46{
47 System::SharedPtr<Project> project = System::MakeObject<Project>();
50
52
53 System::SharedPtr<ExtendedAttribute> a = attr->CreateExtendedAttribute();
55 return project;
56}``````
## Calculation of Text Functions
1. Asc( string )
2. Chr( charcode )
3. Format( expression, format, firstdayofweek, firstweekofyear)
4. Instr( start,string1, string2, compare )
5. LCase( string )
6. Left( string, length )
7. Len( string )
8. LTrim( string )
9. Mid( string, start, length )
10. Right( string, length )
11. RTrim( string )
12. Space( number )
13. StrComp( string1, string2, compare )
14. StrConv( string, conversion, LCID )
15. String( number, character )
16. Trim( string )
17. UCase( string )
`````` 1void CalculateTextFunctions::EvaluateStrConv()
2{
3 System::SharedPtr<Project> project = CreateTestProjectWithCustomField();
5
6 // Set formulas and print extended attribute value
7 project->get_ExtendedAttributes()->idx_get(0)->set_Formula(u"StrConv(\"sTring and sTRINg\",3)");
9 project->get_ExtendedAttributes()->idx_get(0)->set_Formula(u"StrConv(\"sTring and sTRINg\",1)");
11 project->get_ExtendedAttributes()->idx_get(0)->set_Formula(u"StrConv(\"sTring and sTRINg\",2)");
13}
14
15void CalculateTextFunctions::EvaluateStringFunction()
16{
17 System::SharedPtr<Project> project = CreateTestProjectWithCustomField();
19
20 // Set formulas and print extended attribute value
21 project->get_ExtendedAttributes()->idx_get(0)->set_Formula(u"String(5, 40)");
23 project->get_ExtendedAttributes()->idx_get(0)->set_Formula(u"String(5, \"A\")");
25 project->get_ExtendedAttributes()->idx_get(0)->set_Formula(u"String(-5, \"A\")");
26 // #Error
28}
29
30System::SharedPtr<Project> CalculateTextFunctions::CreateTestProjectWithCustomField()
31{
32 System::SharedPtr<Project> project = System::MakeObject<Project>();
35
37
38 System::SharedPtr<ExtendedAttribute> a = attr->CreateExtendedAttribute();
40 return project;
41}``````
## Calculation of Date/Time Functions
1. CDate( expression )
2. Date ()
3. DateAdd( interval, number, date )
4. DateDiff( interval, date1, date2, firstdayofweek, firstweekofyear )
5. DatePart( interval, date, firstdayofweek, firstweekofyear)
6. DateSerial( year, month, day )
7. DateValue( date)
8. Day( date)
9. Hour( time )
10. IsDate( expression )
11. Minute( time)
12. Month( date)
13. Now ()
14. ProjDateAdd( date, duration, calendar )
15. ProjDateConv( expression, dateformat )
16. ProjDateDiff( date1, date2, calendar )
17. ProjDateSub( date, duration, calendar )
18. ProjDateValue( expression )
19. ProjDurConv( expression, durationunits )
20. ProjDurValue( duration )
21. Second( time )
22. Time ()
23. Timer ()
24. TimeSerial( hour, minute, second)
25. TimeValue( time)
26. Weekday( date, firstdayofweek )
27. Year( date)
`````` 1System::SharedPtr<Project> project = CreateTestProject();
3
6
7System::SharedPtr<ExtendedAttribute> numberAttribute = numberDefinition->CreateExtendedAttribute();
9
10// Set ProjDateDiff formula and print extended attribute value
11numberDefinition->set_Formula(u"ProjDateDiff(\"03/23/2015\",\"03/18/2015\")");
12System::Console::WriteLine(numberAttribute->get_NumericValue());
13numberDefinition->set_Formula(u"ProjDateDiff(\"03/23/2015\",\"03/25/2015\")");
14System::Console::WriteLine(numberAttribute->get_NumericValue());
15
18System::SharedPtr<ExtendedAttribute> dateAttribute = dateDefinition->CreateExtendedAttribute();
20
23System::SharedPtr<ExtendedAttribute> durationAttribute = durationDefinition->CreateExtendedAttribute();
25
28System::SharedPtr<ExtendedAttribute> textAttribute = textDefinition->CreateExtendedAttribute();
30
31// Set ProjDateSub formula and print extended attribute value
32dateDefinition->set_Formula(u"ProjDateSub(\"3/19/2015\", \"1d\")");
33System::Console::WriteLine(System::ObjectExt::Box<System::DateTime>(dateAttribute->get_DateValue()));
34
35// We can set ProjDurConv formula to duration-valued attribute as well as to text-valued attribute.
36
37// Set ProjDurConv formula to duration-valued extended attribute and print its value.
38durationDefinition->set_Formula(u"ProjDurConv([Duration], pjHours)");
39System::Console::WriteLine(System::ObjectExt::Box<Duration>(durationAttribute->get_DurationValue()));
40
41// Set ProjDurConv formula to text-valued extended attribute and print its value.
42textDefinition->set_Formula(u"ProjDurConv([Duration], pjHours)");
43System::Console::WriteLine(textAttribute->get_TextValue());
44
45textDefinition->set_Formula(u"ProjDurConv([Duration], pjWeeks)");
46System::Console::WriteLine(textAttribute->get_TextValue());
47
48// Set Second formula and print entended attribute value
49numberDefinition->set_Formula(u"Second(\"4/21/2015 2:53:41 AM\")");
50System::Console::WriteLine(numberAttribute->get_NumericValue());
51
52// Set Weekday formula and print entended attribute value
53numberDefinition->set_Formula(u"Weekday(\"24/3/2015\", 1)");
54System::Console::WriteLine(numberAttribute->get_NumericValue());
55numberDefinition->set_Formula(u"Weekday(\"24/3/2015\", 2)");
56System::Console::WriteLine(numberAttribute->get_NumericValue());
57numberDefinition->set_Formula(u"Weekday(\"24/3/2015\", 3)");
58System::Console::WriteLine(numberAttribute->get_NumericValue());``````
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### Subject: What is the pressure of water at 'x' depth?
Date: Mon Nov 16 13:20:20 1998
Posted by Sean Reid
School: Leysin American School in Switzerland
City: Leysin State/Province: Vaud
Country: Switzerland
Area of science: Physics
ID: 911244020.Ph
Message:
```
I have asked many of my teachers gone through many books,and have emailed
some
professors at some of the top universities and either they didn't know the
answer or they thought I was out of my mind and that I watched too much T.V.
So, you are my last hope!
What I need to find out is a formula(s). Let's say I were in a large "box"
with a hole of five square meters in the bottom, and I submerged the box (with
me in it) to a certain depth. What I need is, is a formula that would tell me
what the air pressure needs to be inside the "box" to keep the water from
coming through the hole. I need to know this at different depths and with
different size holes. I could really use any information that you could give
me on this matter and if there is anything you need clarification on just let
me know and I'll try to do my best! I just find it a bit hard to try and
explain my idea! Thank you!
Sean
```
Re: What is the pressure of water at 'x' depth?
Current Queue | Current Queue for Physics | Physics archives
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# There are infinitely many primes congruent to 9 mod 10
I want to show that there are infinitely many primes $p$ such that $p = 9 \pmod {10}$.
First, I can see that 19 is one of them. Assume there are finitely many, i.e., 19, $p_1, p_2 , \cdots , p_k$. Let $P = 19p_1 p_2 \cdots p_k$ and $N =4P^2 -5.$ I want to show that all prime divisors of $N$ are congruent modulo 1 or 4 mod 5 and $N=9 \pmod{10}$.
Thank you so much.
Any help will be appreciated.
• Step 1: $p=9\pmod{10}$ if and only if $p=9\pmod{2}$ and $p=9\pmod{5}$. This simplifies the problem, as all primes except 2 satisfy the first condition. Apr 19, 2016 at 4:18
• @vadim123: Please tell why all prime divisors of $N$ are congruent to 1 or 4 mod5 Apr 19, 2016 at 4:21
• math.stackexchange.com/questions/373750/… Apr 19, 2016 at 4:37
It is enough to show that there are infinitely many primes of the form $5k-1$.
Let $n\ge 2$, and let $N=5(n!)^2-1$. We first show that every prime divisor of $N$ is of the form $5k\pm 1$.
Suppose that $p$ is a prime divisor of $N$. Then $5$ is a quadratic residue of $p$. A simple Legendre symbol calculation shows that this forces $p\equiv \pm 1\pmod{5}$.
Finally, the prime divisors of $N$ cannot be all of the form $5k+1$, else their product $N$ would be, but it isn't. So $N$ has a prime divisor of the form $5k-1$.
Any prime divisor of $5(n!)^2-1$ must be greater than $n$. So we have shown that for any $n$ there is a prime of the form $5k-1$ which is greater than $n$.
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## 27309 - Statistics I
### Teaching Plan Information
2019/20
Subject:
27309 - Statistics I
Faculty / School:
109 - Facultad de Economía y Empresa
228 - Facultad de Empresa y Gestión Pública
301 - Facultad de Ciencias Sociales y Humanas
Degree:
ECTS:
6.0
Year:
1
Semester:
Second semester
Subject Type:
Basic Education
Module:
---
### 1.3. Recommendations to take this course
There are no previous requirements to take this course. To achieve greater progress, it is recommended to attend and to participate actively in the classes.
In the first session of the course, the contents of the course, the teaching methodology and the assessment criteria are explained in detail. Through the e-learning platform the teachers will inform the students about the readings, practice cases or relevant news to be employed in the activities of the course.
### 2.2. Learning goals
Passing this course will enable the student to...
1. Understand and situate the statistical description of a data set within the stages of the statistical study of an economic phenomenon.
2. Be able to handle statistical information sources in the Business and Economics areas.
3. Define, calculate and deduce the properties of the basic descriptive statistical measures in order to synthesise the location, the dispersion and the shape of the frequency distribution of a univariate data set.
4. Analyse the relationship between two statistical variables depending on the type of the variable (qualitative/quantitative).
5. Be able to handle index numbers employed in the economy and interpret the results that are obtained.
6. Define basic concepts of probability and apply the fundamental theorems to solve simple problems of Probability Calculus.
7. Be able to solve discrete decision problems in an environment of uncertainty.
8. Implement, using a spreadsheet, the statistical measures and the graphical techniques studied in the course.
9. Be able to write statistical reports formulating the conclusions that are derived from the study of a data set.
### 3.1. Assessment tasks (description of tasks, marking system and assessment criteria)
Students must show they have attained the learning results foreseen through the following assessment methods. The assessment is GLOBAL and the proposed assessment activities are of two types:
• Computer tests (CT) in which the students should apply the descriptive techniques presented in the first part of the course (Lessons 1 to 6) to a set of real data using a spreadsheet. In these computer tests, the evaluation criteria will take into account the use of the Excel functions related with the statistical analysis of data, the numerical results obtained and their concordance and suitability with the situation analysed in the context of socio-economic data, as well as their interpretation and the conclusions.
• Written test (WT) in which the students will have to solve several practice exercises referring to the application of the statistical techniques presented in the two last blocks of the course (Lessons 7 to 9). In each problem, several questions will be posed, and the following issues will be evaluated: the statement of the problem in statistical terms, the correct use of the statistical notation and terminology, the correct numerical resolution and the interpretation/comparison of the results obtained.
Each test will be scored from 0 to 10 points.
The part of the course evaluated by the computer tests (CT) will account for 60% of the total score and the part evaluated by the written test (WT) will account for the other 40%. In order to pass the course, two conditions will have to be met: (1) obtain a minimum of 4 points in each of the two parts (CT and WT) and (2) obtain a minimum of 5 out of 10 points in the total score. The total score will be obtained as:
TOTAL SCORE = 0.6*CT + 0.4*WT
For students who do not obtain the minimum of 4 points in each of the two parts, the total score will be obtained as:
TOTAL SCORE = Minimum {0.6*CT + 0.4*WT; 4}
The part evaluated by computer tests (CT) may be passed in two ways: (1) taking two intermediate computer tests, CT1
(Lessons 1 to 4) and CT2 (Lessons 5 and 6) which will be organized during the period of regular classes, or (2) taking a single global computer test (GCT) which will be organized in the official exam period established by the faculty for each call.
The written test (WT) will take place only in the official exam period established by the faculty for each call.
In order to pass the part evaluated by computer tests using the first option (two intermediate tests), the student should obtain at least 3 points in each of the intermediate computer tests, and the average score of the two tests (CT = 0.5*CT1
+ 0.5*CT2) should be greater than or equal to 4 points. The score for the students who have not obtained these minimums will be CT = 0. Students who have obtained these minimum scores in the intermediate computer tests but would like to improve their score in the part evaluated by the computer test, will be able to take the global computer test in the first call and will maintain the better of the two scores.
Second call
The students who have obtained at least 5 points in one of the two parts in the first call, but who have not passed the course, will be allowed to take only the part they did not pass in the second call. The tests in this call will have the same format as those of the first call.
### 4.1. Methodological overview
The methodology followed in this course is oriented towards achievement of the learning objectives. A wide range of teaching and learning tasks are implemented, such as such as lectures, practice sessions, computer practice sessions, and tutorials.
Classroom materials will be available via Moodle. These include a repository of the slides and lecture notes used in class, the course syllabus, as well as other learning resources such as leaning exercises, data files and outlines of the computer practices sessions.
The course is worth 6 ECTS implying a workload for the student of 150 hours divided between the classroom and private study hours. This workload is distributed in the following way:
Activities Hours in the classroom Private study hours Total student hours Lectures (whole group) 30 30 60 Computer practice sessions (Two subgroups if the POD allows it) 14 26 40 Practice sessions (Two subgroups if the POD allows it) 12 22 34 Additional practice sessions (P6) (Carrying out these practices and organise them in two subgroups will depend on the final POD) 4 7 11 Intermediate tests (Four subgroups) 2 2 Written exam 3 3 TOTAL 65 85 150
Lectures: The professors will present the main contents of the course and try to motivate participation and discussion in the classroom. Slides will be employed in these sessions to help the students to understand the topics. It is recommended to attend the lectures and make notes to complement and clarify the slides.
Practice sessions: In these sessions, the students will learn how to manage and solve practical problems. Before each practical session, the students will have at their disposal the set of problems that will be solved.
Computer practice sessions: During the semester, the students will do several computer practice sessions. In these sessions, they will solve some problems applying the methods and techniques studied in class by using a spreadsheet. Each practice session will consist of two parts. In the first one, the students will be guided to learn the main theoretical concepts; in the second, these concepts will be employed to solve real problems.
### 4.3. Syllabus
Lesson 1: Statistical Methods in Business and Economics Lesson 2: Scales of Measurement and Information Sources
Lesson 3: Describing Univariate Data: Frequency Tables and Distributions, and Graphic Presentation
Lesson 4: Describing Univariate Data: Numerical Measures
Lesson 5: Frequency Tables and Distributions and Graphic Presentation of Bivariate Data
Lesson 6: Correlation and Simple Linear Regression
Lesson 7: Index Numbers
Lesson 8: Probability Concepts
Lesson 9: Statistical Decision Theory
### 4.4. Course planning and calendar
For further details concerning the timetable, classroom and further information regarding this course please refer to the course website (moodle).
### Información del Plan Docente
2019/20
Asignatura:
109 - Facultad de Economía y Empresa
228 - Facultad de Empresa y Gestión Pública
301 - Facultad de Ciencias Sociales y Humanas
Titulación:
Créditos:
6.0
Curso:
1
Periodo de impartición:
Segundo semestre
Clase de asignatura:
Formación básica
Materia:
### 1.1. Objetivos de la asignatura
La asignatura y sus resultados previstos responden a los siguientes planteamientos y objetivos:
La asignatura pretende que el alumno conozca algunas de las principales fuentes de información utilizadas en el mundo de la Estadística Económica así como que sea capaz de realizar un análisis inicial de un conjunto de datos uni y bivariante. Se pretende además que el alumno disponga del conocimiento necesario para su actividad profesional sobre todo lo relativo a la elaboración de indicadores como medidas comparativas de la evolución de una magnitud.
Finalmente, y con el fin de proporcionar los fundamentos básicos de asignaturas como Estadística II, Econometría, ... las cuales se estudian en cursos posteriores, se realiza una introducción al Cálculo de Probabilidades enfatizando su aplicación a la resolución de problemas de decisión discretos en un ambiente de incertidumbre.
Todos los contenidos de la asignatura se desarrollan con una orientación marcadamente práctica, promoviendo la participación del alumno y el debate en el aula.
### 1.2. Contexto y sentido de la asignatura en la titulación
La asignatura Estadística I pertenece al módulo de Métodos Cuantitativos para la Empresa del plan de estudios, junto a las asignaturas de Estadística II, Investigación Operativa y Las TICs en la empresa. Asimismo, es evidente su relación con las asignaturas del módulo de Fundamentos del Análisis Económico e Instrumentos (Matemáticas I, Matemáticas II, Econometría y Aplicaciones econométricas en la empresa).
Tiene como objetivo dotar al estudiante de las herramientas básicas para el tratamiento de la información y la cuantificación de la economía o de la empresa y constituye una herramienta de apoyo a la toma de decisiones en dicho ámbito.
La asignatura está ubicada en el principio del bloque formativo y es el inicio de las asignaturas que abordan el tratamiento de datos en ambiente de incertidumbre (las materias de estadística, las materias de econometría y el resto de optativas).
### 1.3. Recomendaciones para cursar la asignatura
No existen requisitos previos para cursar esta asignatura. Para su mayor aprovechamiento se recomienda la presencialidad y la participación activa en las sesiones de trabajo.
### 2.1. Competencias
Al superar la asignatura, el estudiante será más competente para...
1. Valorar la situación y la evolución previsible de empresas y organizaciones, tomar decisiones y extraer el conocimiento relevante con referencia a la responsabilidad social.
2. Comprender y aplicar criterios profesionales y rigor científico a la resolución de los problemas económicos, empresariales y organizacionales.
3. Elaborar y redactar proyectos.
4. Capacidad para la resolución de problemas
5. Capacidad de organización y planificación
6. Habilidad para analizar y buscar información proveniente de fuentes diversas.
8. Motivación por la calidad y la excelencia
10. Capacidad de aplicar los conocimientos en la práctica
11. Capacidad para usar las herramientas e instrumentos tecnológicos necesarios en su desempeño profesional.
El estudiante, para superar esta asignatura, deberá demostrar los siguientes resultados...
1. Comprende y sitúa la descripción estadística de un conjunto de datos en las etapas de la investigación estadística de un fenómeno de naturaleza económica.
2. Es capaz de manejar fuentes de información estadística en el ámbito económico-empresarial.
3. Define, calcula y deduce sus propiedades, de las medidas estadísticas descriptivas básicas para sintetizar la posición, la dispersión y la forma de la distribución de frecuencias de un conjunto de datos univariantes.
4. Analiza la relación entre dos variables estadísticas distinguiendo por el tipo de variable (cualitativa/cuantitativa).
5. Es capaz de manejar los números índices más utilizados en Economía e interpretar los resultados obtenidos.
6. Define conceptos básicos de probabilidad y aplica los teoremas fundamentales para la resolución de problemas sencillos del Cálculo de Probabilidades.
7. Es capaz de resolver problemas de decisión discretos en ambiente de incertidumbre.
8. Implementa mediante una hoja de cálculo las medidas estadísticas y representaciones gráficas presentadas a lo largo de la asignatura.
9. Es capaz de elaborar informes estadísticos formulando las conclusiones que se desprenden del estudio.
### 2.3. Importancia de los resultados de aprendizaje
Un primer interés de la materia viene dado por el carácter instrumental de la materia que pretende dotar al estudiante de las herramientas básicas para resumir, interpretar y comprender una realidad económica que se desarrollará en otras materias de los estudios universitarios. Otro interés es la propia formación intrínseca del futuro profesional que le permita discernir y comprender la validez de los resultados de cualquier estudio empírico.
En la mayoría de las Ciencias Sociales y, en concreto, en el mundo económico-empresarial es necesario observar la realidad e intentar comprender y resumir dicha información, detectando cuando sea posible los modelos o patrones que siguen. Estos dos intereses sitúan a la estadística como una materia transversal en la mayoría de los estudios universitarios (ingenierías, medicina, veterinaria, economía, ciencias sociales, matemáticas, biología, sociología, ...).
Es claro que cada estudio universitario realiza una selección tanto de las técnicas como de la profundidad de éstas según los objetivos y fines perseguidos en la capacitación de sus futuros profesionales.
### 3.1. Tipo de pruebas y su valor sobre la nota final y criterios de evaluación para cada prueba
El estudiante deberá demostrar que ha logrado los resultados de aprendizaje mediante las siguientes actividades de evaluación
El sistema de evaluación será GLOBAL, y las actividades de evaluación previstas son de dos tipos:
Pruebas informáticas (PI) a realizar en el aula de informática, en las que los alumnos deberán aplicar las herramientas descriptivas presentadas en la primera parte del curso (Temas 1 a 6) a un conjunto de datos reales utilizando una hoja de cálculo. En las pruebas informáticas se valorará el uso de las funciones de Excel relacionadas con el Análisis Estadístico de datos, los resultados numéricos obtenidos y su concordancia y adecuación con la situación analizada en el contexto de datos de carácter socio-económico, así como su interpretación y conclusiones.
Prueba escrita (PE) en la que los alumnos deberán resolver diversos problemas prácticos referentes a la aplicación de las técnicas estadísticas presentadas en los 2 últimos bloques temáticos de la asignatura (Temas 7 a 9). En cada problema se plantearán diversos apartados en cuya resolución se valorará tanto el planteamiento estadístico del problema, el uso de la notación y terminología estadística, la correcta resolución numérica y la interpretación/comparación de los resultados obtenidos.
Cada prueba se calificará en una escala de 0 a 10 puntos.
La parte de la asignatura evaluada mediante pruebas informáticas (PI) tendrá un peso del 60% en la calificación global, mientras que la parte evaluada mediante prueba escrita (PE) tendrá el 40% restante. Para superar la asignatura se exigirá un mínimo de 4 puntos en cada una de las partes (PI y PE) y obtener una puntuación superior o igual a 5 puntos sobre 10 en la nota final. La calificación final se obtendrá como:
NOTA_FINAL = 0.6*PI + 0.4*PE
Para los estudiantes que no alcancen el mínimo de 4 puntos en cada una de las partes, la calificación final se obtendrá como:
NOTA_FINAL = Mínimo {0.6*PI + 0.4*PE; 4}
La parte informática podrá ser superada por los alumnos mediante dos pruebas informáticas parciales, PI1 (Temas 1 a 4) y PI2 (Temas 5 y 6) que se realizarán durante el periodo de clases o mediante una única prueba informática global (PIG) que se realizará en las fechas de las convocatorias oficiales.
La prueba escrita se realizará únicamente en las fechas de las convocatorias oficiales.
Para superar la parte informática mediante las pruebas informáticas parciales el alumno deberá obtener al menos 3 puntos en cada una de las pruebas, y la nota media de las dos pruebas (PI = 0.5*PI1 + 0.5*PI2) deberá ser igual o superior a 4 puntos. Para los estudiantes que no hayan obtenido estos mínimos, la puntuación será PI=0. Los estudiantes que, aun habiendo obtenido estas puntuaciones mínimas en las pruebas informáticas parciales, quieran mejorar su calificación de la parte informática para la primera convocatoria podrán realizar la prueba informática global (PIG), manteniendo la mejor de las dos calificaciones.
Segunda convocatoria
Los estudiantes que en la primera convocatoria hubieran obtenido al menos 5 puntos en alguna de las partes pero no hubieran superado finalmente la asignatura, podrán presentarse únicamente a la parte no superada en la segunda convocatoria, que tendrán un formato análogo al de la primera convocatoria.
### 4.1. Presentación metodológica general
En el proceso de enseñanza-aprendizaje de "Estadística I" se utilizarán diferentes métodos docentes. Se hará uso de técnicas didácticas expositivas para el caso de las clases teóricas, aunque se podrán utilizar otros métodos docentes
-formas didácticas de participación- que buscan la implicación del alumno en el proceso de enseñanza-aprendizaje. Así, este segundo tipo de métodos favorece la interacción tanto entre el profesor y el estudiante como entre los propios estudiantes, y son métodos adecuados para el desarrollo de las denominadas competencias genéricas.
Como apoyo se colgará en el ADD, información básica sobre la asignatura: el Programa de la asignatura, la Guía docente, las presentaciones utilizadas por el profesor en la exposición de los temas, el material complementario para la ampliación de las explicaciones realizadas en las clases, el material práctico de problemas a resolver en las sesiones de prácticas y otros propuestos para el trabajo personal del estudiante, los guiones de las prácticas de informática y las plantillas para agilizar su elaboración.
La asignatura consta de 6 créditos ECTS, lo que supone una carga de trabajo para el estudiante de 150 horas, entre presenciales y trabajo individual. Esta carga se reparte de la siguiente forma:
Actividades Horas presenciales Horas trabajo autónomo Total carga estudiante Clases teóricas (Grupo grande) 30 30 60 P Prácticas de problemas en aula informática (Se realizará un desdoble en dos subgrupos si el POD lo permite) 22 43 65 P Prácticas de problemas en aula convencional (Se realizará un desdoble en dos subgrupos si el POD lo permite) 4 6 10 Prácticas problemas adicionales (P6) (La impartición de estas prácticas así como su posible desdoble dependerá del POD final) 4 6 10 Pruebas informáticas (Cuatro subgrupos) 2 2 Examen escrito 3 3 TOTAL 65 85 150
Clases Teóricas: Se desarrollarán principalmente mediante clases magistrales expositivas motivando la participación y discusión en clase. En ellas se utilizarán las transparencias elaboradas para la parte teórica de la materia y su misión es apoyar al alumno en el seguimiento de las explicaciones, pero nunca como sustitución de la clase. Se recomienda la asistencia a clase y la toma de notas o aclaraciones a dichas transparencias.
Clases prácticas de problemas: Esta actividad pretende mostrar al estudiante como abordar y resolver problemas. Previo a cada sesión de prácticas, se anunciarán los problemas a abordar para que el alumno individualmente pueda valorar su comprensión y su posible resolución. Con este fin, el estudiante tendrá publicado con anterioridad la resolución de problemas tipo que le sirvan de base para resolver situaciones similares. Se alternarán las clases en pizarra y en aula de informática.
Clases prácticas de informática: Esta actividad se desarrollará en el aula de informática, trabajando una o dos personas por equipo. Cada sesión durará dos horas organizando una primera parte de trabajo guiado por el profesor y el resto para trabajo por parte de los estudiantes.
Tutorías Presenciales: Junto con las tutorías convencionales, se realizarán tutorías en el aula para resolver dudas, realizar aclaraciones y supervisar el desarrollo de la asignatura y de los trabajos.
### 4.3. Programa
Tema 1: Los métodos estadísticos en el ámbito económico-empresarial
Tema 2: Escalas de Medida y Fuentes de Información
Introducción. Fuentes de datos estadísticos. Conceptos básicos. Tipos de datos y variables. Medición y escalas de medida.
Tema 3: Tabulación y Representación gráfica de datos univariantes Tabulación de datos. Representación gráfica de datos.
Tema 4: Descripción numérica
Introducción. Medidas de posición. Medidas de dispersión. Medidas de forma. Otras medidas.
Tema 5: Tabulación y Representación gráfica de datos bivariantes
Introducción. Distribución de frecuencias conjunta. Distribuciones marginales. Distribuciones condicionadas. Representaciones gráficas. Independencia estadística.
Tema 6: Correlación y Regresión lineal simple
Introducción. Covarianza. Regresión lineal simple: criterio de los mínimos cuadrados. Bondad de ajuste y correlación. Predicción. Regresión no lineal.
Tema 7: Números índices
Introducción. Índices simples. Índices complejos. Deflación de series económicas. Enlace y cambio de base. Repercusión. Algunos índices notables.
Tema 9: Análisis Estadístico de Decisiones
Introducción. Conceptos básicos. Decisión en ambiente de riesgo. Decisión en ambiente de incertidumbre. Toma de decisiones con experimentación. Valor y eficiencia de la información.
### 4.4. Planificación de las actividades de aprendizaje y calendario de fechas clave
Fecha Contenido Método Docente Sesión 1 Presentación. Tema 1. Clase expositiva Sesión 2 Tema 2 y Tema 3. Tabulación y repres. Gráfica Clase expositiva Sesión 3 Práctica Informática 1. Tabulación y repres. gráfica de datos Prácticas en aula informática Sesión 4 Tema 4. Medidas de posición I Clase expositiva Sesión 5 Práctica 2. Medidas de posición Prácticas en aula informática Sesión 6 Tema 4. Medidas posición II. Medidas dispersión y forma Clase expositiva Sesión 7 Práctica 3. Medidas dispersión y forma Prácticas en aula informática Sesión 8 Tema 4. Medidas de desigualdad y repaso Clase expositiva Sesión 9 Práctica 4. Medidas de desigualdad y repaso Prácticas en aula informática Sesión 10 Prueba 1: Descriptiva Prueba en aula informática Sesión 11 Tema 5. Distribuciones bidimensionales Clase expositiva Sesión 12 Práctica 5. Distribuciones bidimensionales Prácticas en aula informática Sesión 13 Tema 6. Regresión y correlación (1) Clase expositiva Sesión 14 Práctica 6. Regresión lineal Prácticas en aula informática Sesión 15 Tema 6. Regresión y correlación (2) Clase expositiva Sesión 16 Práctica 7. Regresión no lineal Prácticas en aula informática Sesión 17 Prueba 2: Bidimensional, regresión y correlación Prueba en aula informática Sesión 18 Tema 7. Números índices (1). Teoría y problemas Clase expositiva Sesión 19 Tema 7. Números índices (2). Teoría y problemas Clase expositiva/problemas Sesión 20 Tema 7. Números índices (3). Teoría y problemas Clase expositiva/problemas Sesión 21 Tema 7. Números índices (4). Teoría y problemas Clase expositiva/problemas Sesión 22 Tema 7. Números índices (5). Teoría y problemas Prácticas de problemas Sesión 23 Tema 8. Probabilidad (1). Teoría y problemas Clase expositiva Sesión 24 Tema 8. Probabilidad (2). Teoría y problemas Clase expositiva/problemas Sesión 25 Tema 8. Probabilidad (3). Teoría y problemas Clase expositiva/problemas Sesión 26 Tema 8. Probabilidad (4). Teoría y problemas Clase expositiva/problemas Sesión 27 Tema 8. Probabilidad (5). Teoría y problemas Prácticas de problemas Sesión 28 Tema 9: Análisis de Decisiones (1). Teoría y problemas Clase expositiva Sesión 29 Tema 9: Análisis de Decisiones (2). Problemas Prácticas de problemas Sesión 30 Repaso Prácticas de problemas Prueba Global
Presentación de la asignatura: En la primera sesión del curso se explican de forma detallada los contenidos de la asignatura, se plantea la metodología docente utilizada en las clases y se exponen los criterios de evaluación con nitidez.
Prácticas informáticas: A lo largo del curso se realizarán prácticas de informática que consistirán en la resolución de problemas aplicando las técnicas y métodos analizados en clase mediante una hoja de cálculo. Cada práctica tiene una primera parte docente, en la que se ayuda al estudiante a comprender los conceptos teóricos fundamentales, y en la segunda parte, se utilizan esos conceptos para resolver problemas.
Pruebas intermedias: Se realizarán dos pruebas intermedias que consistirán en resolver problemas utilizando una hoja de cálculo. Las fechas concretas de dichas pruebas se fijarán de acuerdo al calendario académico y al horario establecido por el Centro, informándose de ello a los estudiantes a través del programa de la asignatura.
Prueba global: De acuerdo al calendario establecido por el centro, en el periodo de exámenes, el estudiante realizará una prueba global que consistirá en un examen escrito donde se evaluarán las competencias y destrezas adquiridas con un peso del 40% y un examen práctico, mediante la resolución de problemas con una hoja de cálculo, que tendrá un peso del 60%. Este examen práctico no será necesario para los alumnos que hayan obtenido en las pruebas intermedias estas dos condiciones:
• un mínimo de 3 puntos en cada una de ellas
• una nota promedio de ambas no inferior a 4.
A través del Anillo Digital Docente (ADD) el profesor irá informando puntualmente a los alumnos sobre la disponibilidad en el ADD de lecturas, casos prácticos, ejercicios, noticias relevantes sobre la materia para la realización de diferentes tareas.
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# Scale Model of the Solar System
Carl Sagan once claimed that the most important lesson we learn from studying the stars is perspective. To address this concept, this activity offers a scale model of the solar system to be evaluated. There are many versions of solar system scale... (View More)
# MISSIONMakers: OSIRIS-REx Mission Design
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Interested in becoming a citizen scientist? Join Dr. Michelle Thaller as she explains how the general public, using scientific protocols, careful observations and accurate measurements, can help NASA make exciting new discoveries. NASA eClips™ are... (View More)
# Playdough Planets
Learners create scale models of Earth, the moon, and Mars out of playdough. Based on the size of the models, they must determine the relative distance between them and then display them at that scale. This activity was designed for use in a library... (View More)
# ICESat-2 Bouncy Ball Photon Collection Challenge
Acting as the ICESat-2 satellite, students investigate the reflection of light photons off Earth's surface by catching and recording a number of photons. Using bouncy balls to represent the photons, students drop, let bounce once and try to catch in... (View More)
# MY NASA DATA: Using Hovmuller Plots to Better Understand Temperature and Salinity
A Hovmuller plot is a diagram that visibly displays data patterns from a selected latitude or longitude over a time period. Through a storyline and several samples, students are introduced to a Hovmuller plot of temperature data along a longitude in... (View More)
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# NASA eClips™ Real World: History of Winter - Abiotic Conditions
In this video clip, join scientists and teachers as they learn how to measure some of the abiotic conditions of winter. Find out about latent heat, how thermochrons can be used to collect data points and the importance of snow:water equivalents.... (View More)
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# NASA eClips™ Real World: Lightning Protection System for Launch Complex 39
See how NASA is using a rolling spheres lightning protection system to expand the cone of safety currently used on Launch Complex 39. NASA eClips™ are short, relevant educational video segments. These videos inspire and engage students, helping... (View More)
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# NASA eClips™ Real World: Global Cloud Observation Day
In this video clip, learn about precipitation and how clouds are formed. Find out why scientists study clouds and how students can help NASA collect cloud observation data. NASA eClips™ are short, relevant educational video segments. These videos... (View More)
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# NASA eClips™ Real World: What is Soil Moisture?
In this video clip, see how NASA measures soil moisture from space with the Soil Moisture Active Passive Mission (SMAP). Learn to calculate soil moisture in your own backyard and discover the real world applications for this data. NASA eClips™ are... (View More)
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# Note for Computer Programming - CP By ANNA SUPERKINGS
• Computer Programming - CP
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SCAD ENGINEERING COLLEGE DEPARTMENTOF COMPUTERSCIENCEAND ENGINEERING COURSE MATERIAL GE6151COMPUTER PROGRAMMING Page 1
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SYLLABUS GE6151COMPUTER PROGRAMMING UNITI INTRODUCTION 8 Generation and Classification of Computers- Basic Organization of a Computer –Number System – Binary – Decimal – Conversion – Problems. Need for logical analysis and thinking – Algorithm – Pseudo code – Flow Chart. UNITII C PROGRAMMING BASICS 10 Problemformulation – Problem Solving - Introduction to ‘ C’ programming –fundamentals – structureof a ‘C’ program – compilation and linking processes – Constants, Variables – Data Types –Expressions using operators in ‘C’ – Managing Input and Output operations – Decision Making andBranching – Looping statements – solving simple scientific and statistical problems. UNITIII ARRAYS AND STRINGS 9 Arrays – Initialization – Declaration – One dimensional and Two dimensional arrays. StringStringoperations – String Arrays. Simple programs- sorting- searching – matrix operations. UNITIV FUNCTIONS AND POINTERS 9 Function – definition of function – Declaration of function – Pass by value – Pass by reference – Recursion – Pointers - Definition – Initialization – Pointers arithmetic – Pointers and arrays- ExampleProblems. UNITV STRUCTURES AND UNIONS 9 Introduction – need for structure data type – structure definition – Structure declaration – Structurewithin a structure - Union - Programs using structures and Unions – Storage classes, Pre-processor directives. TOTAL:45 PERIODS TEXTBOOKS: 1. Anita Goel and Ajay Mittal, “Computer Fundamentals and Programming in C”, Dorling Kindersley(India) Pvt. Ltd., Pearson Education in South Asia, 2011. Page 2
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2. Pradip Dey, Manas Ghosh, “Fundamentals of Computing and Programming in C”, First Edition,OxfordUniversityPress, 2009 3. Yashavant P. Kanetkar. “ Let Us C”, BPB Publications, 2011. REFERENCES: 1. Byron S Gottfried, “Programming with C”, Schaum’s Outlines, Second Edition, Tata McGrawHill,2006. 2. Dromey R.G., “How to Solve it by Computer”, Pearson Education, Fourth Reprint, 2007. 3. Kernighan,B.W and Ritchie,D.M, “The C Programming language”, Second Edition, PearsonEducation, 2006. UNITI INTRODUCTION Generation and Classification of Computers- Basic Organization of a Computer –Number System – Binary – Decimal – Conversion – Problems. Need for logical analysis and thinking – Algorithm – Pseudo code – Flow Chart. GENERATIONS OF COMPUTERS The Zeroth Generation The term Zeroth generation is used to refer to the period of development of computing, which predated the commercial production and sale of computer equipment. The period might be dated as extending from the mid-1800s. In particular, this period witnessed the emergence of the first electronics digital computers on the ABC, since it was the first to fullyimplement the idea of the stored program and serial execution of instructions. The development of EDVAC set the stage for the evolution of commercial computing and operating system software. The hardware component technology of this period was electronic vacuum tubes. The actual operation of these early computers took place without be benefit of an operating system. Early programs were written in machine language and each contained code for initiating operation of the computer itself. This system was clearly inefficient and depended on the varying competencies of the individual programmer as operators. The First Generation, 1951-1956 The first generation marked the beginning of commercial computing. The first generation was Page 3
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characterized by high-speed vacuum tube as the active component technology. Operation continued without the benefit of an operating system for a time. The mode was called "closed shop" and was characterized by the appearance of hired operators who would select the job to be run, initial program load the system, run the user’s program, and then select another job, and so forth. Programs began to be written in higher level, procedure-oriented languages, and thus the operator’s routine expanded. The operator now selected a job, ran the translation program to assemble or compile the source program, and combined the translated object program along with any existing library programs that the program might need for input to the linking program, loaded and ran the composite linked program, and then handled the next job in a similar fashion. Application programs were run one at a time, and were translated with absolute computer addresses. There was no provision for moving a program to different location in storage for any reason. Similarly, a program bound to specific devices could not be run at all if any of these devices were busy or broken. At the same time, the development of programming languages was moving away from the basic machine languages; first to assembly language, and later to procedure oriented languages, the most significant being the development of FORTRAN The Second Generation, 1956-1964 The second generation of computer hardware was most notably characterized by transistors replacing vacuum tubes as the hardware component technology. In addition, some very important changes in hardware and software architectures occurred during this period. For the most part, computer systems remained card and tape-oriented systems. Significant use of random access devices, that is, disks, did not appear until towards the end of the second generation. Program processing was, for the most part, provided by large centralized computers operated under monoprogrammed batch processing operating systems. Themost significant innovations addressed the problem of excessive central processor delay due to waiting for input/output operations. Recall that programs were executed by processing the machine instructions in a strictly sequential order. As a result, the CPU, with its high speed electronic component, was often forced to wait for completion of I/O operations which involved mechanical devices (card readers and tape drives) that were order of magnitude slower. Thesehardware developments led to enhancements of the operating system. I/O and data channel communication and control became functions of the operating system, both to relieve the application programmer from the difficult details of I/O programming and to protect the integrity of the system to provide improved service to users by segmenting jobs and running shorter jobs first (during "prime time") and relegating longer jobs to lower priority or night time runs. System libraries became more widely available and more comprehensive as new utilities and application software components were available to programmers. The second generation was a period of intense operating system development. Also it was the period for sequential batch processing. Researchers began to experiment with multiprogramming and multiprocessing. The Third Generation, 1964-1979 , The Third generation officially began in April 1964 with IBM’s announcement of its System/360 page4
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# Distance between Sioux City, IA (SUX) and Newark, NJ (EWR)
Flight distance from Sioux City to Newark (Sioux Gateway Airport – New York Newark Liberty International Airport) is 1154 miles / 1858 kilometers / 1003 nautical miles. Estimated flight time is 2 hours 41 minutes.
Driving distance from Sioux City (SUX) to Newark (EWR) is 1292 miles / 2079 kilometers and travel time by car is about 22 hours 22 minutes.
## Map of flight path and driving directions from Sioux City to Newark.
Shortest flight path between Sioux Gateway Airport (SUX) and New York Newark Liberty International Airport (EWR).
## How far is Newark from Sioux City?
There are several ways to calculate distances between Sioux City and Newark. Here are two common methods:
Vincenty's formula (applied above)
• 1154.460 miles
• 1857.923 kilometers
• 1003.198 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 1151.516 miles
• 1853.186 kilometers
• 1000.640 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).
## Airport information
A Sioux Gateway Airport
City: Sioux City, IA
Country: United States
IATA Code: SUX
ICAO Code: KSUX
Coordinates: 42°24′9″N, 96°23′3″W
B New York Newark Liberty International Airport
City: Newark, NJ
Country: United States
IATA Code: EWR
ICAO Code: KEWR
Coordinates: 40°41′33″N, 74°10′7″W
## Time difference and current local times
The time difference between Sioux City and Newark is 1 hour. Newark is 1 hour ahead of Sioux City.
CDT
EDT
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 159 kg (352 pounds).
## Frequent Flyer Miles Calculator
Sioux City (SUX) → Newark (EWR).
Distance:
1154
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
1154
Round trip?
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# Excel MATCH Function
When we want to get the exact position of a given value in a column, row or table, we can use the Excel MATCH function. To fully understand how the function works, please read this post to the end.
Figure 1: Using Excel MATCH function to find relative positions
## General syntax of the formula
`=MATCH (look_up, look_array, [match_type])`
Where;
• Lookup_value– value to match in look_up array
• Lookup_array– refers to the range of cells or array reference
• Match_type– shows how you want to match. Default is 1
## How the MATCH function works
We use the MATCH function when we want to find the relative position of an item in an array. This function is very flexible given that it has different matching modes.
When used together with INDEX, MATCH can also help to retrieve a value at the matched position.
## Different match types in MATCH function
• Match_type 1: this shows that the MATCH function will find the largest value which is less or equal to the lookup_value. Here, you will have to sort the lookup_array in ascending order.
• Match_type 0: here, the MATCH function will find the first value that is exactly equal to the lookup_value. You don’t have to sort the lookup_array when zero is your preferred match_type.
• Match_type -1: the MATCH function will find the smallest value which is greater than or equal to the lookup_value. You will have to sort the lookup_array in a descending order.
## Note
When the match_type is omitted, then it will be assumed to be 1.
Also, all these match_types will return the exact match of what you are seeking.
## Example
In this example, we want to find the relative positions of Dennis and Juliet. To do this, proceed as follows:
Step 1: Prepare your data sheet, complete with the names, value column as well as Result column
Step 2: Click in the cell where you want to get the result
Step 3: Write the formula in the active cell, i.e. `=MATCH (C2,A1:A5,0)`
Step 4: Press Enter
Step 5: Do the same for the other items.
Most of the time, the problem you will need to solve will be more complex than a simple application of a formula or function. If you want to save hours of research and frustration, try our live Excelchat service! Our Excel Experts are available 24/7 to answer any Excel question you may have. We guarantee a connection within 30 seconds and a customized solution within 20 minutes.
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Solution examples
index and match with duplicates. I need to use another column as a reference, so my return value has two match the value of two things for it to return
Solved by E. H. in 60 mins
Hello, I have a big spreadsheet in which I need to know how many patients came in each month based on provider. I am using the following formula but I still get the N/A error. =INDEX('No Show Appts Data'!D:E, MATCH(1, 'No Show Appts Data'!D:D='No Shows Data'!M2)*('No Shows Data'!E:E='No Shows Data'!N1),0)) No Show Appts Data is the name of the sheet where Column D is Month Year of date in question and Column E is the provider. No Shows Data is the sheet where I am making all the formula calculation where Column M is Month and Year and Column N is the provider in question and therefore M2 is the month in question and N1 the provider in question. How do I fix this error? Thanks
Solved by F. H. in 40 mins
I need to find an INDEX function that will convert the Call Day (a number 1-7) to the actual weekday found in row 1 of the DayofWeek named range. I don't know if I'm supposed to use the MATCH function as well or not.
Solved by D. D. in 11 mins
I have two columns with names and a third one with email addresses. For every name that matches, I need to copy the email address to an empty column next to the matching name. .
Solved by E. U. in 15 mins
help with INDEX(IndexArea,MATCH(\$M\$5,MatchRows,0),MATCH(\$M\$6,MatchColumns,0))
Solved by Z. Y. in 12 mins
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# Properties
Base field $$\Q(\sqrt{-11})$$ Label 2.0.11.1-25344.2-z4 Conductor $$(-96 a + 48)$$ Conductor norm $$25344$$ CM no base-change yes: 528.i1,5808.be1 Q-curve yes Torsion order $$4$$ Rank not available
# Related objects
Show commands for: Magma / SageMath / Pari/GP
## Base field $$\Q(\sqrt{-11})$$
Generator $$a$$, with minimal polynomial $$x^{2} - x + 3$$; class number $$1$$.
magma: R<x> := PolynomialRing(Rationals()); K<a> := NumberField(R![3, -1, 1]);
sage: x = polygen(QQ); K.<a> = NumberField(x^2 - x + 3)
gp: K = nfinit(a^2 - a + 3);
## Weierstrass equation
$$y^2 = x^{3} + x^{2} - 472 x - 4108$$
magma: E := ChangeRing(EllipticCurve([0, 1, 0, -472, -4108]),K);
sage: E = EllipticCurve(K, [0, 1, 0, -472, -4108])
gp: E = ellinit([0, 1, 0, -472, -4108],K)
This is a global minimal model.
sage: E.is_global_minimal_model()
## Invariants
$$\mathfrak{N}$$ = $$(-96 a + 48)$$ = $$\left(2\right)^{4} \cdot \left(-a\right) \cdot \left(a - 1\right) \cdot \left(-2 a + 1\right)$$ magma: Conductor(E); sage: E.conductor() $$N(\mathfrak{N})$$ = $$25344$$ = $$3^{2} \cdot 4^{4} \cdot 11$$ magma: Norm(Conductor(E)); sage: E.conductor().norm() $$\mathfrak{D}$$ = $$(1824768)$$ = $$\left(2\right)^{11} \cdot \left(-a\right)^{4} \cdot \left(a - 1\right)^{4} \cdot \left(-2 a + 1\right)^{2}$$ magma: Discriminant(E); sage: E.discriminant() gp: E.disc $$N(\mathfrak{D})$$ = $$3329778253824$$ = $$3^{8} \cdot 4^{11} \cdot 11^{2}$$ magma: Norm(Discriminant(E)); sage: E.discriminant().norm() gp: norm(E.disc) $$j$$ = $$\frac{5690357426}{891}$$ magma: jInvariant(E); sage: E.j_invariant() gp: E.j $$\text{End} (E)$$ = $$\Z$$ (no Complex Multiplication ) magma: HasComplexMultiplication(E); sage: E.has_cm(), E.cm_discriminant() $$\text{ST} (E)$$ = $\mathrm{SU}(2)$
## Mordell-Weil group
Rank not available.
magma: Rank(E);
sage: E.rank()
Regulator: not available
magma: gens := [P:P in Generators(E)|Order(P) eq 0]; gens;
sage: gens = E.gens(); gens
magma: Regulator(gens);
sage: E.regulator_of_points(gens)
## Torsion subgroup
Structure: $$\Z/4\Z$$ magma: T,piT := TorsionSubgroup(E); Invariants(T); sage: T = E.torsion_subgroup(); T.invariants() gp: T = elltors(E); T[2] $\left(-16 : -12 a + 6 : 1\right)$ magma: [piT(P) : P in Generators(T)]; sage: T.gens() gp: T[3]
## Local data at primes of bad reduction
magma: LocalInformation(E);
sage: E.local_data()
prime Norm Tamagawa number Kodaira symbol Reduction type Root number ord($$\mathfrak{N}$$) ord($$\mathfrak{D}$$) ord$$(j)_{-}$$
$$\left(-a\right)$$ $$3$$ $$4$$ $$I_{4}$$ Split multiplicative $$-1$$ $$1$$ $$4$$ $$4$$
$$\left(a - 1\right)$$ $$3$$ $$4$$ $$I_{4}$$ Split multiplicative $$-1$$ $$1$$ $$4$$ $$4$$
$$\left(-2 a + 1\right)$$ $$11$$ $$2$$ $$I_{2}$$ Non-split multiplicative $$1$$ $$1$$ $$2$$ $$2$$
$$\left(2\right)$$ $$4$$ $$4$$ $$I_{3}^*$$ Additive $$1$$ $$4$$ $$11$$ $$0$$
## Galois Representations
The mod $$p$$ Galois Representation has maximal image for all primes $$p$$ except those listed.
prime Image of Galois Representation
$$2$$ 2B
## Isogenies and isogeny class
This curve has non-trivial cyclic isogenies of degree $$d$$ for $$d=$$ 2 and 4.
Its isogeny class 25344.2-z consists of curves linked by isogenies of degrees dividing 4.
## Base change
This curve is the base-change of elliptic curves 528.i1, 5808.be1, defined over $$\Q$$, so it is also a $$\Q$$-curve.
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# KiBps to Pibps - 1 KiBps to Pibps Conversion
expand_more
S = Second, M = Minute, H = Hour, D = Day
Sec
Min
Hr
Day
Sec
Min
Hr
Day
KiBps
label_important RESULT close
1 KiBps =0.0000000000072759576141834259033203125 Pibps
( Equal to 7.2759576141834259033203125E-12 Pibps )
content_copy
Calculated as → 1 x 8 ÷ 10244 smart_display Show Stepsexpand_more
Below chart table shows the amount of data that can be transferred at a constant speed of 1 KiBps in various time frames.
Transfer RateAmount of Data can be transferred
@ 1 KiBpsin 1 Second0.0000000000072759576141834259033203125 Pebibits
in 1 Minute0.00000000043655745685100555419921875 Pebibits
in 1 Hour0.000000026193447411060333251953125 Pebibits
in 1 Day0.000000628642737865447998046875 Pebibits
## Kibibytes per Second (KiBps) to Pebibits per Second (Pibps) Conversion - Formula & Steps
The KiBps to Pibps Calculator Tool provides a convenient solution for effortlessly converting data rates from Kibibytes per Second (KiBps) to Pebibits per Second (Pibps). 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 (Kibibyte) and target (Pebibit) data units.
Source Data Unit Target Data Unit
Equal to 1024 bytes
(Binary Unit)
Equal to 1024^5 bits
(Binary Unit)
The conversion diagram provided below offers a visual representation to help you better grasp the steps involved in calculating Kibibyte to Pebibit in a simplified manner.
÷ 1024
÷ 1024
÷ 1024
÷ 1024
÷ 1024
x 1024
x 1024
x 1024
x 1024
x 1024
Based on the provided diagram and steps outlined earlier, the formula for converting the Kibibytes per Second (KiBps) to Pebibits per Second (Pibps) can be expressed as follows:
diamond CONVERSION FORMULA Pibps = KiBps x 8 ÷ 10244
Now, let's apply the aforementioned formula and explore the manual conversion process from Kibibytes per Second (KiBps) to Pebibits per Second (Pibps). To streamline the calculation further, we can simplify the formula for added convenience.
FORMULA
Pebibits per Second = Kibibytes per Second x 8 ÷ 10244
STEP 1
Pebibits per Second = Kibibytes per Second x 8 ÷ (1024x1024x1024x1024)
STEP 2
Pebibits per Second = Kibibytes per Second x 8 ÷ 1099511627776
STEP 3
Pebibits per Second = Kibibytes per Second x 0.0000000000072759576141834259033203125
By applying the previously mentioned formula and steps, the conversion from 1 Kibibytes per Second (KiBps) to Pebibits per Second (Pibps) can be processed as outlined below.
1. = 1 x 8 ÷ 10244
2. = 1 x 8 ÷ (1024x1024x1024x1024)
3. = 1 x 8 ÷ 1099511627776
4. = 1 x 0.0000000000072759576141834259033203125
5. = 0.0000000000072759576141834259033203125
6. i.e. 1 KiBps is equal to 0.0000000000072759576141834259033203125 Pibps.
Note : Result rounded off to 40 decimal positions.
You can employ the formula and steps mentioned above to convert Kibibytes per Second to Pebibits per Second using any of the programming language such as Java, Python, or Powershell.
### Unit Definitions
#### What is Kibibyte ?
A Kibibyte (KiB) is a binary unit of digital information that is equal to 1024 bytes (or 8,192 bits) and is defined by the International Electro technical Commission(IEC). The prefix 'kibi' is derived from the binary number system and it is used to distinguish it from the decimal-based 'kilobyte' (KB). 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.
arrow_downward
#### What is Pebibit ?
A Pebibit (Pib or Pibit) is a binary unit of digital information that is equal to 1,125,899,906,842,624 bits and is defined by the International Electro technical Commission(IEC). The prefix 'pebi' is derived from the binary number system and it is used to distinguish it from the decimal-based 'petabit' (Pb). 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 Kibibytes per Second (KiBps) to Pebibits per Second (Pibps)
Apply the formula as shown below to convert from 1 Kibibytes per Second (KiBps) to Pebibits per Second (Pibps).
A B C
1 Kibibytes per Second (KiBps) Pebibits per Second (Pibps)
2 1 =A2 * 0.0000000000072759576141834259033203125
3
If you want to perform bulk conversion locally in your system, then download and make use of above Excel template.
## Python Code for Kibibytes per Second (KiBps) to Pebibits per Second (Pibps) Conversion
You can use below code to convert any value in Kibibytes per Second (KiBps) to Kibibytes per Second (KiBps) in Python.
kibibytesperSecond = int(input("Enter Kibibytes per Second: "))
pebibitsperSecond = kibibytesperSecond * 8 / (1024*1024*1024*1024)
print("{} Kibibytes per Second = {} Pebibits per Second".format(kibibytesperSecond,pebibitsperSecond))
The first line of code will prompt the user to enter the Kibibytes per Second (KiBps) as an input. The value of Pebibits per Second (Pibps) is calculated on the next line, and the code in third line will display the result.
## Similar Conversions & Calculators
All below conversions basically referring to the same calculation.
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Education Technology
# Activities
• ##### Subject Area
• Math: Algebra I: Exponential Functions
6-8
90 Minutes
• ##### Device
• TI-84 Plus
• TI-84 Plus Silver Edition
• TI-Navigator™
## Compound Interest: Show Me the Money
#### Activity Overview
This is an activity at the conclusion of the exponential relationship unit where students have experience with equations in the form y=a*b^x. Students use the random integer function on the TI-83 Plus to generate a rate of return for the investment profile they choose.
#### Before the Activity
Students choose from four different investment profiles: conservative, growth, growth/income, or aggressive.
#### During the Activity
Students use the random integer function on the TI-84 Plus to generate a rate of return for the investment profile they choose. For an aggressive investor the range is from -35% to 35%; a conservative portfolio ranges from 2%-5%.
After a 10 year investment total is determined, students submit an equation in "Activity Center." Each equation is color-coated based on portfolio type.
#### After the Activity
A follow-up activity could allow the students to use the random integer function for each year and do a 25 year analysis the investor profile they chose.
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# Mechanics 1 answer below »
A boy pulls a 5.0 Kg sled 10.0 m along a horizontal surface at a constant speed. The coefficient of kinetic friction is 0.20 and his pull is 37 degrees above the horizontal.
A.) what work does he do pulling the sled? (that is, what is the work done on the sled by the boy?)
B.) would the boy do more, l...
Mark B
A) since the speed is constant => a = 0 hence Fcos37 = Fr Fr = μk (mg - Fsin37) = 0.2 (5x9.8 - 0.6F) so F cos37 = 9.8 - 0.12F => F = 9.8 / 0.92 = 10.65 N <<<<<<<< so W = Fd cos37 =...
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https://www.freelancer.com/projects/visual-basic-statistics/fortran-conversion-vba-excel/
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# Fortran conversion to VBA (Excel)
I would like the gslib [url removed, login to view] source code translated to Excel macro modules.
It's all there in Fortran, most of the syntax is the same as VBA, just a few small changes.
Will be useful if you know something about statistics and mathematics to be able to check your work.
The routines will have to work together, as they call each other.
No changes or "improvements" must be made, just porting of the same code to Excel.
For example:
subroutine gauinv(p,xp,ierr)
c-----------------------------------------------------------------------
c
c Computes the inverse of the standard normal cumulative distribution
c function with a numerical approximation from : Statistical Computing,
c by W.J. Kennedy, Jr. and James E. Gentle, 1980, p. 95.
c
c
c
c INPUT/OUTPUT:
c
c p = double precision cumulative probability value: dble(psingle)
c xp = G^-1 (p) in single precision
c ierr = 1 - then error situation (p out of range), 0 - OK
c
c
c-----------------------------------------------------------------------
real*8 p0,p1,p2,p3,p4,q0,q1,q2,q3,q4,y,pp,lim,p
save p0,p1,p2,p3,p4,q0,q1,q2,q3,q4,lim
c
c Coefficients of approximation:
c
data lim/1.0e-10/
data p0/-0.322232431088/,p1/-1.0/,p2/-0.342242088547/,
+ p3/-0.0204231210245/,p4/-0.0000453642210148/
data q0/0.0993484626060/,q1/0.588581570495/,q2/0.531103462366/,
+ q3/0.103537752850/,q4/0.0038560700634/
c
c Check for an error situation:
c
ierr = 1
if([url removed, login to view]) then
xp = -1.0e10
return
end if
if(p.gt.(1.0-lim)) then
xp = 1.0e10
return
end if
ierr = 0
c
c Get k for an error situation:
c
pp = p
if(p.gt.0.5) pp = 1 - pp
xp = 0.0
if(p.eq.0.5) return
c
c Approximate the function:
c
y = dsqrt(dlog(1.0/(pp*pp)))
xp = real( y + ((((y*p4+p3)*y+p2)*y+p1)*y+p0) /
+ ((((y*q4+q3)*y+q2)*y+q1)*y+q0) )
if(real(p).[url removed, login to view](pp)) xp = -xp
c
c Return with G^-1(p):
c
return
end
Skills: Fortran, Mathematics, Statistics, Visual Basic
( 0 reviews ) San Isidro, Peru
Project ID: #1100655
## 3 freelancers are bidding on average \$200 for this job
PerfectSquare
\$200 USD in 5 days
(73 Reviews)
5.6
bharanisrihari
Hello, Excel VBA expert here. Please check PM for details. Thanks!
\$200 USD in 5 days
(32 Reviews)
5.6
am3ndoim
When do we start? I think you don't go use some routines, for example the function strlen in vba has feature for exchange it. You need just function in macro format or you need that use system information about the e More
\$200 USD in 15 days
(0 Reviews)
0.0
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https://nucleoniitjeekota.com/topic-notes.php?topic=ElectroMagnetic%20Induction
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ElectroMagnetic Induction Revision Notes - IIT JEE/NEET Preparation | Nucleon
ElectroMagnetic Induction
• FARADAYS LAWS OF ELECTROMAGNETIC INDUCTION
• When magnetic flux passing through a loop changes with time or magnetic lines of
force are cut by a conducting wire then an emf is produced in the loop or in that
wire. This emf is called induced emf.
Read more
• LENZS LAW (CONSERVATION OF ENERGY PRINCIPLE)
• According to this law, emf will be induced in such a way that it will oppose the cause which has
produced it. Figure shows a magnet approaching a ring with its north pole towards the ring
Read more
• MOTIONAL EMF
• We can find emf induced in a moving rod by considering the number
of lines cut by it per sec assuming there are ‘B’ lines per unit
area. Thus when a rod of length moves with velocity v in a magnetic
field B, as shown, it will sweep area per unit time equal to v
and hence it
Read more
• EXPLANATION OF EMF INDUCED IN ROD ON THE BASIS OF MAGNETIC FORCE
• EXPLANATION OF EMF INDUCED IN ROD ON THE BASIS OF MAGNETIC FORCE
Read more
• INDUCED EMF DUE TO ROTATION
• ROTATION OF THE ROD
Consider a conducting rod of length l rotating in a uniform magnetic field.
Read more
• EMF INDUCED IN A ROTATING DISC
• Consider a disc of radius r rotating in a magnetic field B.
Consider an element dx at a distance x form the centre. This element is moving with
Read more
• FIXED LOOP IN A VARYING MAGNETIC FIELD
• Now consider a circular loop, at rest in a varying magnetic field. Suppose the magnetic field is
directed inside the page and it is increasing in magnitude. The emf induced in the loop will be
Read more
• SELF INDUCTION
• Self induction is induction of emf in a coil due to its own current change.Total flux N?
passing through a coil due to its own current is proportional to the current and is given as N
L i where L is called coefficient of self induction or inductance.The inductance L is purely
Read more
• SELF INDUCTANCE OF SOLENOID
• Let the volume of the solenoid be V , the number of turns per unit length be n.
Let a current I be flowing in the solenoid.Magnetic field in the solenoid is given as
Read more
• INDUCTOR
• If current i through the inductor is increasing the induced emf will oppose the increase in
current and hence will be opposite to the current.If current i through the inductor is decreasing
the induced emf will oppose the decrease in current and hence will be in the direction of the
current.
Read more
• ENERGY STORED IN AN INDUCTOR
• If current in an inductor at an instant is i and is increasing at the rate di/dt,the induced emf
will oppose the current . Its behaviour is shown in the figure.
Read more
• GROWTH OF CURRENT IN SERIES R–L CIRCUIT
• Figure shows a circuit consisting of a cell, an inductor L and a resistor R ,connected in series. Let the
switch S be closed at t=0.Suppose at an instant current in the circuit be i which is increasing at the rate
di/dt.
Read more
• DECAY OF CURRENT IN THE CIRCUIT CONTAINING RESISTOR AND INDUCTOR
• DECAY OF CURRENT IN THE CIRCUIT CONTAINING
RESISTOR AND INDUCTOR
Read more
• MUTUAL INDUCTANCE
• Consider two arbitrary conducting loops 1 and 2. Suppose that ?1 is the instantaneous current flowing around
loop 1. This current generates a magnetic field B1 which links the second circuit, giving rise to a magnetic flux ?2
through that circuit. If the current ?1doubles, then the magnetic field B1 doubles in strength at all points in space,
so the magnetic flux ?2 through the second circuit also doubles. Furthermore, it is obvious that the flux through
the second circuit is zero whenever the current flowing around the first circuit is zero. It follows that the flux ?2
Read more
• TRANSFORMER
• A transformer changes an alternating potential difference from one value to another of greater or
smaller value using the principle of mutual induction .Two coils called the primary and secondary
windings,which are not connected to one another in any way , are wound on a complete soft iron
core.When an alternating voltag
Read more
• ENERGY LOSSES IN TRANSFORMER
• Although transformers are very efficient devices, small energy losses do occur in them due to four main causes.
1. RESISTANCE OF THE WINDINGS
Read more
Online Live Session JEE/NEET
Online live session that help you visualize each concept making it easier to understand Efficient way to Clear your doubts from your place.
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Notice: Only variables should be passed by reference in /home/nucleon479/public_html/nucleoniitjeekota.com/include/footer-script.php on line 1
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## Problem Solving
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CC-MAIN-2023-06
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latest
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https://patents.justia.com/patent/20240184956
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crawl-data/CC-MAIN-2024-30/segments/1720763518029.81/warc/CC-MAIN-20240723072353-20240723102353-00675.warc.gz
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PREDICTION METHOD OF CROWN OF STEEL PLATES AND STRIPS BASED ON DATA DRIVING AND MECHANISM MODEL FUSION
The invention belongs to the technical field of quality control of steel plates and strips products, and relates to a prediction method of crown of steel plates and strips based on data driving and mechanism model fusion. By establishing an outlet crown mechanism model of a hot continuous rolling, the mechanism model and a DNN model are combined to establish a DNN model for predicting crown of steel plates and strips, and the calculated value of the mechanism model is taken as a benchmark value of the outlet crown. The deviation amount between the benchmark value and the actual values of the outlet crown is taken as output of the DNN model for predicting crown of the steel plates and strips, and then sum of the predicted value and the benchmark value based on the DNN model for predicting the crown of the steel plates is taken as the final predicted value of the crown of the steel plates and strips.
Description
BACKGROUND OF THE INVENTION 1. Field of the Invention
The invention belongs to the technical field of quality control of steel plates and strips products, and relates to a prediction method of crown of steel plates and strips based on data driving and mechanism model fusion.
2. The Prior Arts
Hot rolled strips occupy an important position in the entire industrial system, wherein plate shape is a key indicator to measure whether the quality of hot rolled strips is qualified, and plate shape control has become an important technique in steel plates and strips production. In recent years, a lot of scientific research has been carried out at home and abroad on the rolling process of hot continuous rolling steel plates and strips, such as derivation and establishment of mathematical models. However, the actual rolling process is more complex, and has strong coupling, nonlinear, multi-variable characteristics and other characteristics, and there are uncertain unknown factors, so that it is difficult to establish an accurate mathematical model. Therefore, it is necessary to use an artificial intelligence method based on industrial data drive in combination with a mathematical model to predict the crown of the steel plates and strips and improve prediction precision, so that the site can be controlled more accurately.
In the prediction process of outlet crown of traditional hot continuous rolling, the crown of the steel plates and strips is directly taken as the output value of a neural network. Besides, the benchmark value of the crown of the steel plates and strips is not set, and parameter prediction is performed only by the neural network. The prediction error range is large, and the prediction precision of the model is reduced.
SUMMARY OF THE INVENTION
In order to solve the technical problems, the invention aims to provide a prediction method of crown of steel plates and strips based on data driving and mechanism model fusion. By establishing a DNN model for predicting crown of steel plates and strips based on data driving and mechanism model fusion, the deviation amount between the calculated value of the mechanism model and the actual values of the outlet crown is taken as the output of the DNN model for predicting crown of steel plates and strips, and the prediction error range can be narrowed.
The invention provides a prediction method of crown of steel plates and strips based on data driving and mechanism model fusion, and the method comprises the following steps:
Step 1: acquiring actual values of an outlet crown, actual measured data related to outlet crown of a hot continuous rolling production line, and calculated data of a process automation level, and using the actual measured data and the calculated data as input data to establish a DNN model for predicting a crown of the steel plates and strips;
Step 2: establishing an outlet crown mechanism model of a hot continuous rolling, performing calculating to obtain the calculated value of the outlet crown of the steel plates and strips as a benchmark value of the outlet crown, and calculating deviation amount of the benchmark value of the outlet crown and the actual values of the outlet crown as output data to establish the DNN model for predicting the crown of the steel plates and strips;
Step 3: randomly dividing modeling data consisting of the input data and the output data into training set data and test set data;
Step 4: based on the training set data, constructing the DNN model for predicting the crown of the steel plates and strips, selecting model parameters, and training the DNN model for predicting the crown of the steel plates and strips;
Step 5: inputting the test set data into the trained DNN model for predicting the crown of the steel plates and strips to predict parameters, and obtaining a predicted value of the deviation amount of the outlet crown;
Step 6: adding up the predicted value of the deviation amount of the outlet crown and the benchmark value of the outlet crown to obtain a final predicted value of the crown, evaluating predicted results by using a mean square error (MSE), a root mean square error (RMSE), a mean absolute error (MAE) of performance indexes and a correlation coefficient R, and analysing a prediction precision.
According to the prediction method of crown of steel plates and strips based on data driving and mechanism model fusion, the Step 1 specially comprises the steps:
Step 1.1: selecting an eight-stand continuous rolling production line for finish rolling, and determining following influencing factors based on a crown mechanism and combined with a hot continuous rolling technology: an outlet width of a rolled piece, an inlet temperature of the rolled piece, an outlet temperature of the rolled piece, a rolling force of stands, a roll-bending force of the stands, a roll wear amount of the stands, an outlet speed of the rolled piece, an outlet thickness of the rolled piece, a thermal expansion of the rolled piece, and a deformation resistance of the rolled piece; and
Step 1.2: according to the influencing factors, extracting the actual measured data and the calculated data of the process automation level from a site, wherein the actual measured data comprises the outlet width of the rolled piece of a finish rolling F8 stand, the inlet temperature of the rolled piece of a finish rolling F1 stand, the outlet temperature of the rolled piece of the finish rolling F8 stand, the rolling force of finish rolling F1-F8 stands, the roll-bending force of the finish rolling F1-F8 stands, the outlet thickness of the rolled piece of the finish rolling F8 stand, the outlet speed of the rolled piece of the finish rolling F1-F8 stands, and the outlet crown of the rolled piece of the finish rolling F8 stand; and the calculated data of the process automation level comprises the deformation resistance of the rolled piece of the finish rolling F1-F8 stands, the outlet thickness of the rolled piece of finish rolling F1-F7 stands, rolling kilometers of the finish rolling F1-F8 stands, and the thermal expansion of the rolled piece during the finish rolling process.
According to the prediction method of crown of steel plates and strips based on data driving and mechanism model fusion, the Step 2 specially comprises the steps:
Step 2.1: establishing an outlet crown mechanism model of the hot continuous rolling, wherein a mathematical equation is as follows:
$C = P K P + F K F + E C ω C + E ∑ ( ω H + ω W + ω O ) + E 0 Δ ; ( 1 )$
Wherein C represents the crown of the steel plates and strips; P and F respectively represent a rolling force of stands and a roll-bending force of the stands for enabling roll systems to bend and deform; KP and KF respectively represent a transverse stiffness of a rolling mill and a transverse stiffness of a bending roll; ωC represents a controllable roll crown; ωH represents a hot crown of the rolls caused by a thermal expansion of the rolls; ωW represents a wear crown of the rolls, caused by a wear of the rolls; ωO represents an initial roll crown of the rolls; Δ represents an inlet crown of the steel plates and strips; E0 represents inlet crown coefficients, EC represents controllable roll crown coefficients, and EΣ represents comprehensive crown coefficients;
Step 2.2: calculating the hot crown of the rolls caused by the thermal expansion of the rolls according to the following equation:
$ω H = 2 ( 1 + v ) β t R ∫ R 0 r [ T ( r , z ) - T 0 ( r , z ) ] dr ; ( 2 )$
Wherein Bt represents thermal expansion coefficients of the rolls and is calculated according to the equation below; v represents a Poisson coefficient of the rolls; T(r,z) represents a temperature at (r,z) where a coordinate is located, r represents a variable along a radius direction of the rolls, and z represents a variable along a length direction of the rolls; T0(r,z) represents an initial temperature of the rolls; a model is simplified and a temperature of the rolls is regarded as uniform distribution:
$β t = Δ L L × Δ T ; ( 3 )$
Wherein ΔL represents a thermal expansion of the steel plates and strips when the temperature changes by ΔT; L represents a length before expansion;
Step 2.3: calculating a wear amount of the rolls according to the following equation:
$wear n = k × ∑ P in × l in ( 1 + α X 4 ) w ; ( 4 )$
Wherein wearn represents the wear amount of the rolls; k represents coefficients related to roll materials and steel plates and strips materials, and Pin represents the rolling force of the nth rolling mill during rolling an ith steel coil; lin represents a length of the ith steel coil after being rolled by the nth rolling mill and is calculated according to the equation below; α represents wear coefficients of the rolls; X represents a position of the wear amount; w represents a width of the steel plates and strips:
$l in = L n × B n × H n b in × h in ; ( 5 )$
Wherein lin, bin and lin respectively represent a length, a width and a thickness of the ith steel coil after being rolled by the nth rolling mill, and Ln, Bn and Hn respectively represent a length, a width and a thickness of the steel plates and strips before being rolled;
Step 2.4: calculating the wear crown of the rolls caused by wear of the rolls according to the following equation:
ωw=wearn0−wearn1 (6);
Wherein ωw represents the wear crown of the rolls, wearn0 represents the wear amount of the rolls when a position X of the wear amount is equal to 0, and wearn1 represents the wear amount of the rolls when the position X of the wear amount is equal to ±1;
When X=0, at a center line of the corresponding steel plates and strips:
$wear n 0 = k × ∑ P in × l in w ; ( 7 )$
When X=±1, at an edge of the corresponding steel plates and strips:
$wear n 1 = k × ∑ P in × l in ( 1 + α ) w ; ( 8 )$
Step 2.5: taking remaining variables except for the rolling force of the stands, the roll-bending force of the stands, the hot crown of the rolls and the wear crown of the rolls, in the outlet crown mechanism model of the hot continuous rolling, as fixed values, calculating the outlet crown of the steel plates and strips, and taking the outlet crown of the steel plates and strips as the benchmark value of outlet crown.
According to the prediction method of crown of steel plates and strips based on data driving and mechanism model fusion, the Step 4 specially comprises the steps:
Step 4.1: designing a forward propagation algorithm of the DNN model for predicting the crown of the steel plates and strips and determining an activation function according to the equations below:
a1=x (9),
al=σ(dl)=σ(Wlal−1+bl) (10);
Wherein a1 represents an output of a first layer, expressed by a matrix method; d represents an output of a lth layer, expressed by the matrix method, wherein 2≤l≤L, L is a total number of layers of a neural network; Wl represents a matrix of the lth layer and bl represents a bias vector of the lth layer; x represents an input vector; σ (d) represents the activation function;
The activation function is specifically a Sigmoid activation function:
$σ ( d ) = 1 e - d ; ( 11 )$
Wherein d is an input of the activation function;
Step 4.2: designing a loss function in a backward propagation algorithm of the DNN model for predicting the crown of the steel plates and strips;
A mean square function is used to measure an output loss of the training set data:
$J ( W , b , x , y ) = 1 2 a L - y 2 2 = 1 2 σ ( W L × a L - 1 + b L ) - y 2 2 ; ( 12 )$
Wherein y is a target output of the DNN model for predicting the crown of the steel plates and strips;
Step 4.3: adopting an Adam optimization algorithm and updating and calculating the model parameters to minimize the loss function;
Step 4.4: adopting a Cosine annealing algorithm based on an unequal interval annealing strategy to adjust a learning rate of the DNN model for predicting the crown of the steel plates and strips;
Step 4.5: adopting a variable controlling method to select a number of hidden layers of the network, selecting a number of hidden layer nodes and a number of data groups used during each training, and completing training of the DNN model for predicting the crown of the steel plates and strips.
According to the prediction method of crown of steel plates and strips based on data driving and mechanism model fusion, the number of hidden layers of the constructed DNN model for predicting the crown of the steel plates and strips is 3, the number of hidden layer nodes is 50, and the number of the data groups selected from each training is 128.
According to the prediction method of crown of steel plates and strips based on data driving and mechanism model fusion, the Step 6 specially comprises the steps:
Step 6.1: adding up the predicted value of the deviation amount of the outlet crown and the benchmark value of the outlet crown to obtain the predicted value of the crown of the DNN model for predicting crown of the steel plates and strips;
Step 6.2: directly taking the outlet crown as the output of the DNN model and performing predicting to obtain the predicted value of the crown based on the DNN model;
Step 6.3: performing calculating according to the outlet crown mechanism model of the hot continuous rolling to obtain the calculated value of the outlet crown;
Step 6.4: evaluating the predicted results of the Steps 6.1-6.3 by using the mean square error (MSE), the root mean square error (RMSE), the mean absolute error (MAE) of performance indexes and the correlation coefficient R, and analyzing prediction precision.
According to the prediction method of crown of steel plates and strips based on data driving and mechanism model fusion, in the Step 6.4:
The mean square error (MSE) is calculated according to the following equation:
$M S E = 1 n ∑ j = 1 n ( y j - y j ′ ) 2 ; ( 13 )$
The root mean square error (RMSE) is calculated according to the following equation:
$R M S E = 1 n ∑ j = 1 n ( y j - y j ′ ) 2 ; ( 14 )$
The mean absolute error (MAE) of performance indexes is calculated according to the following equation:
$MAE = 1 n ∑ j = 1 n ❘ "\[LeftBracketingBar]" y j - y j ′ ❘ "\[RightBracketingBar]" ; ( 15 )$
The correlation coefficient R is calculated according to the following equation:
$R = 1 - ∑ j = 1 n ( y j - y j ′ ) 2 ∑ j = 1 n ( y j - y _ ) 2 ; ( 16 )$
Wherein yi represents the actual values of the outlet crown, y′i represents the predicted value obtained through the corresponding model, y represents a mean value of the actual values of the outlet crown, and n represents a total number of data groups in the test set data.
The prediction method of crown of steel plates and strips based on data driving and mechanism model fusion, disclosed by the invention, at least has the following beneficial effects:
According to the method, the variable controlling method is used to determine appropriate parameters of the DNN model for predicting crown of steel plates and strips, and an appropriate optimizer algorithm and a learning rate adjustment algorithm are selected to make the DNN model for predicting crown of steel plates and strips more accurately predict the deviation amount. Then, the deviation amount between the calculated value and the actual values of the mechanism model of crown of the steel plates and strips is taken as the predicted value of the output of the DNN model for predicting crown of steel plates and strips. On one hand, because the introduced benchmark value is close to the actual values in magnitude, the fluctuation range of deviation between the two values is smaller in comparison. Therefore, the deviation between the benchmark value and the actual values is taken as the output of the DNN model for predicting crown of steel plates and strips, which can further narrow the error prediction range, and is closer to the actual value, thereby making the prediction precision of the model higher; and on the other hand, by combining the mechanism model with the DNN model, the whole model can be more consistent with an actual physical process and more persuasive and interpretable. At the present stage, the hot continuous rolling production line is relatively perfect in the collection and storage of industrial data, so that the method disclosed by the invention has strong generalization ability, and a new method is provided to improve the accuracy of outlet crown of the steel plates and strips.
BRIEF DESCRIPTION OF DRAWINGS
FIG. 1 shows a flow chart of a prediction method of crown of steel plates and strips based on data driving and mechanism model fusion;
FIG. 2 shows an effect chart comparing the predicted value of crown based on a DNN model for predicting crown of steel plates and strips and the predicted value of crown based on the DNN model;
FIG. 3 shows an effect chart comparing the predicted value of the crown based on the DNN model for predicting crown of steel plates and strips and the calculated value of the outlet crown based on the outlet crown mechanism model of hot continuous rolling; and
FIG. 4 shows an effect chart comparing the calculated value of the outlet crown based on the outlet crown mechanism model of hot continuous rolling with the predicted value of the crown based on the DNN model.
DETAILED DESCRIPTION OF THE PREFERRED EMBODIMENT
The method of the invention will be further described in detail below in combination with the drawings and examples.
The example is based on a domestic hot continuous rolling production line, and takes the relevant data of the outlet crown of the steel plates and strips as data to establish a model. The overall flow is shown as FIG. 1, and specially comprises the following steps:
Step 1: acquiring actual values of an outlet crown, actual measured data related to outlet crown of a hot continuous rolling production line, and calculated data of a process automation level, and using the actual measured data and the calculated data as input data to establish a DNN model for predicting a crown of steel plates and strips, wherein the Step 1 specially comprises:
Step 1.1: selecting an eight-stand continuous rolling production line for finish rolling, and determining following influencing factors based on a crown mechanism and combined with a hot continuous rolling technology: an outlet width of a rolled piece, an inlet temperature of the rolled piece, an outlet temperature of the rolled piece, a rolling force of stands, a roll-bending force of the stands, a roll wear amount of the stands, an outlet speed of the rolled piece, an outlet thickness of the rolled piece, a thermal expansion of the rolled piece, and a deformation resistance of the rolled piece;
Step 1.2: according to the influencing factors, extracting the actual measured data and the calculated data of the process automation level from a site, wherein the actual measured data comprises the outlet width of the rolled piece of a finish rolling F8 stand, the inlet temperature of the rolled piece of a finish rolling F1 stand, the outlet temperature of the rolled piece of the finish rolling F8 stand, the rolling force of finish rolling F1-F8 stands, the roll-bending force of the finish rolling F1-F8 stands, the outlet thickness of the rolled piece of the finish rolling F8 stand, the outlet speed of the rolled piece of the finish rolling F1-F8 stands, and the outlet crown of the rolled piece of the finish rolling F8 stand; and the calculated data of the process automation level comprises the deformation resistance of the rolled piece of the finish rolling F1-F8 stands, the outlet thickness of the rolled piece of finish rolling F1-F7 stands, rolling kilometers of the finish rolling F1-F8 stands, and the thermal expansion of the rolled piece during the finish rolling process.
A certain roll change cycle is taken as an example, 180 pieces of steel are rolled during the period, and some specific data is shown as Table 1.
TABLE 1 Part specific data in roll change cycle roll-bending rolling force/kN force/kN rolling kilometers/m S/N F1 F2-F8 F1 F2-F8 F1 F2-F8 . . . crown/mm 1 3111.55 . . . 190.42 . . . 414.556434 . . . . . . 0.021 2 4724.57 . . . 189.16 . . . 825.3701278 . . . . . . 0.012 3 4666.17 . . . 195.45 . . . 1238.017136 . . . . . . 0.022 4 5060.44 . . . 217.45 . . . 1721.493268 . . . . . . 0.014 5 4740 . . . 220.7 . . . 2203.676473 . . . . . . 0.012 6 4878.29 . . . 226.94 . . . 2685.907602 . . . . . . 0.007 7 4739.87 . . . 228.39 . . . 3169.359711 . . . . . . 0.006 8 4745.19 . . . 229.42 . . . 3651.788625 . . . . . . 0.005 9 5412.33 . . . 251.02 . . . 4245.185755 . . . . . . 0.013 10 5137.93 . . . 253.62 . . . 4838.863531 . . . . . . 0.011 11 4994.29 . . . 254.02 . . . 5432.440532 . . . . . . 0.012 12 4697.31 . . . 253.25 . . . 6024.171825 . . . . . . 0.01 13 4677.23 . . . 249.45 . . . 6617.543757 . . . . . . 0.011 14 4633.8 . . . 253.44 . . . 7210.230985 . . . . . . 0.008 15 4659.15 . . . 252.46 . . . 7802.127369 . . . . . . 0.006 16 4520 . . . 254.83 . . . 8394.685041 . . . . . . 0.006 17 4587.33 . . . 253.13 . . . 8987.026914 . . . . . . 0.011 18 4384.08 . . . 252.91 . . . 9580.326697 . . . . . . 0.005 19 4564.4 . . . 253.12 . . . 10172.57511 . . . . . . 0.007 20 4247.43 . . . 251.12 . . . 10771.9406 . . . . . . 0.01 21-180 . . . . . . . . . . . . . . . . . . . . . . . .
Step 2: an outlet crown mechanism model of hot continuous rolling is established, calculating is performed to obtain the calculated value of the outlet crown of the steel plates and strips as a benchmark value of the outlet crown, and deviation amount of the benchmark value of the outlet crown and the actual values of the outlet crown is calculated as output data to establish the DNN model for predicting the crown of the steel plates and strips, wherein the Step 2 specially comprises:
Step 2.1: an outlet crown mechanism model of the hot continuous rolling is established, wherein a mathematical equation is as follows:
$C = P K P + F K F + E C ω C + E ∑ ( ω H + ω W + ω O ) + E 0 Δ ; ( 1 )$
Wherein C represents the crown of the steel plates and strips; P and F respectively represent a rolling force of stands and a roll-bending force of the stands for enabling roll systems to bend and deform; KP and KF respectively represent a transverse stiffness of a rolling mill and a transverse stiffness of a bending roll; ωC a represents controllable roll crown; ωH represents a hot crown of the rolls caused by a thermal expansion of the rolls; ωW represents a wear crown of the rolls, caused by a wear of the rolls; ωO represents an initial roll crown of the rolls; Δ represents an inlet crown of the steel plates and strips; E0 represents inlet crown coefficients, EC represents controllable roll crown coefficients, and EΣ represents comprehensive crown coefficients;
Step 2.2: the hot crown of the rolls caused by the thermal expansion of the rolls is calculated according to the following equation:
$ω H = 2 ( 1 + v ) β t R ∫ 0 R r [ T ( r , z ) - T 0 ( r , z ) ] dr ; ( 2 )$
Wherein βt represents thermal expansion coefficients of the rolls and is calculated according to the equation below; v represents a Poisson coefficient of the rolls; T(r,z) represents a temperature at (r,z) where a coordinate is located, r represents a variable along a radius direction of the rolls, and z represents a variable along a length direction of the rolls; T0(r,z) represents an initial temperature of the rolls; a model is simplified and a temperature of the rolls is regarded as uniform distribution:
$β t = Δ L L × Δ T ; ( 3 )$
Wherein ΔL represents a thermal expansion of the steel plates and strips when the temperature changes by ΔT; L represents a length before expansion;
Step 2.3: a wear amount of the rolls is calculated according to the following equation:
$wear n = k × ∑ P in × l in ( 1 + α X 4 ) w ; ( 4 )$
Wherein wearn represents the wear amount of the rolls; k represents coefficients related to roll materials and steel plates and strips materials, and Pin represents the rolling force of the nth rolling mill during rolling an ith steel coil; lin represents a length of the ith steel coil after being rolled by the nth rolling mill and is calculated according to the equation below; X is a position of the wear amount; w represents a width of the steel plates and strips; α represents wear coefficients of the rolls, and is related to the accumulated length of the steel plates and strips (one rolling cycle), rolling force of the stands and the roll materials, and can be manually set in the range of [0.0004-0.0006], in the example, 0.006 is used as the wear coefficients of the rolls, and the value of each roll change cycle is obtained by regression fitting with a least square method:
$l in = L n × B n × H n b in × h in ( 5 )$
Wherein lin, bin and hin respectively represent a length, a width and a thickness of the ith steel coil after being rolled by the nth rolling mill, and Ln, Bn and Hn respectively represent a length, a width and a thickness of the steel plates and strips before being rolled;
Step 2.4: the wear crown of the rolls caused by wear of the rolls is calculated according to the following equation:
ωw=wearn0−wear n1 (6);
Wherein ωw represents the wear crown of the rolls, wearn0 represents the wear amount of the rolls when a position X of the wear amount is equal to 0, and wearn1 represents the wear amount of the rolls when the position X of the wear amount is equal to ±1;
When X=0, at a center line of the corresponding steel plates and strips:
$wear n 0 = k × ∑ P in × l in w ; ( 7 )$
When X=±1, at an edge of the corresponding steel plates and strips:
$wear n 1 = k × ∑ P in × l in ( 1 + α ) w ; ( 8 )$
Step 2.5: remaining variables except for the rolling force of the stands, the roll-bending force of the stands, the hot crown of the rolls and the wear crown of the rolls, in the outlet crown mechanism model of the hot continuous rolling, are taken as fixed values, the outlet crown of the steel plates and strips is calculated, and the outlet crown of the steel plates and strips is taken as the benchmark value of outlet crown.
In the specific implementation, since the crown is greatly affected by the rolling force P of the stands, the roll-bending force F of the stands, thermal deformation of the rolls and the wear deformation of the rolls, but least affected by the remaining variables, the remaining variables are approximately regarded as the fixed values; and hot crown and wear crown of the rolls are obtained through calculation, the rolling force of the stands and the roll-bending force of the stands are extracted from the actual rolling in site, then the crown model is simplified, the calculated value of outlet crown of the steel plates and strips is calculated and the calculated value is taken as the benchmark value of outlet crown.
Step 3: modeling data consisting of the input data and the output data is randomly divided into training set data and test set data in a certain proportion, wherein in the example, the modeling data is divided into training set data and test set data in a ratio of 7:3. If the roll change cycle is taken as an example, 180 groups exist, including 126 groups of the training set data and 54 groups of the test set data.
Step 4: based on the training set data, the DNN model for predicting the crown of the steel plates and strips is constructed, model parameters are selected, and the DNN model for predicting the crown of the steel plates and strips is trained, wherein the Step 4 specially comprises:
Step 4.1: a forward propagation algorithm of the DNN model for predicting the crown of steel plates and strips is designed and an activation function is determined according to the equations below:
a1=x (9),
al=σ(dl)=σ(Wlal−1+bl) (10);
Wherein a1 represents an output of a first layer, expressed by a matrix method; d represents an output of a lth layer, expressed by the matrix method, wherein 2≤l≤L, L is a total number of layers of a neural network; Wl represents a matrix of the lth layer and bl represents a bias vector of the lth layer; x represents an input vector; σ(d) represents the activation function;
The activation function is specifically a Sigmoid activation function:
$σ ( d ) = 1 e - d ; ( 11 )$
Wherein d is an input of the activation function;
Step 4.2: a loss function in a backward propagation algorithm of the DNN model for predicting the crown of the steel plates and strips is designed:
Wherein in order to select appropriate parameters to make the output calculated from all training data inputs be equal to or closer to the actual values as much as possible, it is necessary to select the appropriate loss function to measure the output loss of training samples, further update W and b until stop iteration threshold is reached, and output a linear relationship coefficient matrix W and a bias vector b of the hidden layers and the output layers.
In the embodiment, the mean square function is used to measure an output loss of the training set data:
$J ( W , b , x , y ) = 1 2 a L - y 2 2 = 1 2 σ ( W L × a L - 1 + b L ) - y 2 2 ; ( 12 )$
Wherein y is a target output of the DNN model for predicting the crown of the steel plates and strips;
Step 4.3: the optimizer algorithm selected for the model is determined, in this way, network parameters affecting model training and model output are updated and calculated to approximate or reach the optimal value so that the loss function is minimized or maximized; and in the example, an Adam optimization algorithm is adopted, and model parameters are updated and calculated, so that the loss function is minimized.
Step 4.4: the selected learning rate adjustment algorithm and related parameters thereof are determined to prevent the condition that due to excessive learning rate, the network fails to converge, wanders around the optimal value and fails to reach the position of the optimal value, and also prevent the condition that due to too low learning rate, the network converges very slowly, the optimization time is greatly prolonged, the network converges easily as soon as entering the local extreme value point, and the optimal solution is not really found. In the embodiment, the Cosine annealing algorithm based on an unequal interval annealing strategy is adopted to adjust the learning rate of the DNN model for predicting crown of steel plates and strips.
Step 4.5: a parameter selection method lies in that the variable controlling method is used to select a number of corresponding hidden layers of the network according to the different effects of the number of different hidden layers on generalization performance, and then the appropriate number of hidden layer nodes is selected according to different errors generated by the number of different hidden layer nodes; similarly, the most appropriate number of training data groups is selected according to the influence of the number of different data groups selected from each training on the degree and speed of model optimization, training of the DNN model for predicting crown of steel plates and strips is completed. The number of hidden layers of the constructed DNN model for predicting the crown of steel plates and strips is 3, the number of hidden layer nodes is 50, and the number of the training data groups selected from each training is 128.
Step 5: the test set data is input into the trained DNN model for predicting the crown of the steel plates and strips to predict parameters, and a predicted value of the deviation amount of the outlet crown is obtained.
Step 6: the predicted value of the deviation amount of the outlet crown and the benchmark value of the outlet crown are added up to obtain a final predicted value of the crown, predicted results are evaluated by using a mean square error (MSE), a root mean square error (RMSE), a mean absolute error (MAE) of performance indexes and a correlation coefficient R, and a prediction precision is analyzed, wherein the Step 6 specially comprises:
Step 6.1: adding up the predicted value of the deviation amount of the outlet crown and the benchmark value of the outlet crown to obtain the predicted value of the crown of the DNN model for predicting crown of steel plates and strips;
Step 6.2: directly taking the outlet crown as the output of the DNN model and performing predicting to obtain the predicted value of the crown based on the DNN model;
Step 6.3: performing calculating according to the outlet crown mechanism model of the hot continuous rolling to obtain the calculated value of the outlet crown; and
Step 6.4: evaluating the predicted results of the Steps 6.1-6.3 by using the mean square error (MSE), the root mean square error (RMSE), the mean absolute error (MAE) of performance indexes and the correlation coefficient R, and analyzing prediction precision.
During the implementation, the mean square error (MSE) is calculated according to the following equation:
$M S E = 1 n ∑ j = 1 n ( y j - y j ′ ) 2 ; ( 13 )$
The root mean square error (RMSE) is calculated according to the following equation:
$R M S E = 1 n ∑ j = 1 n ( y j - y j ′ ) 2 ; ( 14 )$
The mean absolute error (MAE) of performance indexes is calculated according to the following equation:
$MAE = 1 n ∑ j = 1 n ❘ "\[LeftBracketingBar]" y j - y j ′ ❘ "\[RightBracketingBar]" ; ( 15 )$
The correlation coefficient R is calculated according to the following equation:
$R = 1 - ∑ j = 1 n ( y j - y j ′ ) 2 ∑ j = 1 n ( y j - y _ ) 2 ; ( 16 )$
Wherein yi represents the actual values of the outlet crown, y′i represents the predicted value obtained through the corresponding model, y represents a mean value of the actual values of the outlet crown, and n represents a total number of data groups in the test set data.
The predicted results are shown as Table 2.
TABLE 2 predicted parameter model type result type MSE RMSE MAE SMAPE R outlet application (deviation + 9.81E−6 0.0062 0.0047 53.25% 0.9002 crown model benchmark value) predicted value DNN predicted value 3.59E−4 0.0189 0.0150 71.19% 0.8730 model of crown mechanism calculated value 5.51E−5 0.0074 0.0057 68.25% 0.9091 model of crown
See FIGS. 2-4 for comparison of predicted effects. FIG. 2 shows an effect chart comparing the predicted value of crown based on a DNN model for predicting crown of steel plates and strips and the predicted value of crown based on the DNN model; FIG. 3 shows an effect chart comparing the predicted value of the crown based on the DNN model for predicting crown of steel plates and strips and the calculated value of the outlet crown based on the outlet crown mechanism model of hot continuous rolling; FIG. 4 shows an effect chart comparing the calculated value of the outlet crown based on the outlet crown mechanism model of hot continuous rolling with the predicted value of the crown based on the DNN model.
It can be seen from FIGS. 2-4 that data points are clearly and regularly distributed. Data distribution of the DNN model for predicting crown of steel plates and strips combined with the mechanism model and the DNN model of the invention is more consistent with a straight line y=x in the figs., that is, the predicted value is closer to the actual values. The DNN model for predicting crown of steel plates and strips of the invention has the best performance, and the values of MSE, MAE and RMSE are significantly reduced, which are 9.81E-6, 0.0047 and 0.0062 respectively.
The above is only preferred embodiments of the invention, and does not limit the idea of the invention. Any modification, equivalent replacement, improvement, and the like made within the spirit and principle of the invention shall be included in the protection scope of the invention.
Claims
1. A prediction method of crown of steel plates and strips based on data driving and mechanism model fusion, comprising the following steps:
Step 1: acquiring actual values of an outlet crown, actual measured data related to the outlet crown of a hot continuous rolling production line and calculated data of a process automation level, and using the actual measured data and the calculated data as input data to establish a DNN model for predicting a crown of steel plates and strips;
Step 2: establishing an outlet crown mechanism model of a hot continuous rolling, performing calculating to obtain a calculated value of the outlet crown of the steel plates and strips as a benchmark value of the outlet crown, and calculating a deviation amount of the benchmark value of the outlet crown and the actual values of the outlet crown as output data to establish the DNN model for predicting the crown of the steel plates and strips;
Step 3: randomly dividing modeling data consisting of the input data and the output data into training set data and test set data;
Step 4: based on the training set data, constructing the DNN model for predicting the crown of the steel plates and strips, selecting model parameters, and training the DNN model for predicting the crown of the steel plates and strips;
Step 5: inputting the test set data into the trained DNN model for predicting the crown of the steel plates and strips to predict parameters, and obtaining a predicted value of the deviation amount of the outlet crown; and
Step 6: adding up the predicted value of the deviation amount of the outlet crown and the benchmark value of the outlet crown to obtain a final predicted value of the crown, evaluating predicted results by using a mean square error (MSE), a root mean square error (RMSE), a mean absolute error (MAE) of performance indexes and a correlation coefficient R, and analyzing a prediction precision.
2. The prediction method of claim 1, wherein the Step 1 further comprises the steps of:
Step 1.1: selecting an eight-stand continuous rolling production line for finish rolling, and determining following influencing factors based on a crown mechanism and combined with a hot continuous rolling technology: an outlet width of a rolled piece, an inlet temperature of the rolled piece, an outlet temperature of the rolled piece, a rolling force of stands, a roll-bending force of the stands, a roll wear amount of the stands, an outlet speed of the rolled piece, an outlet thickness of the rolled piece, a thermal expansion of the rolled piece, and a deformation resistance of the rolled piece; and
Step 1.2: according to the influencing factors, extracting the actual measured data and the calculated data of the process automation level from a site, wherein the actual measured data comprises the outlet width of the rolled piece of a finish rolling F8 stand, the inlet temperature of the rolled piece of a finish rolling F1 stand, the outlet temperature of the rolled piece of the finish rolling F8 stand, the rolling force of finish rolling F1-F8 stands, the roll-bending force of the finish rolling F1-F8 stands, the outlet thickness of the rolled piece of the finish rolling F8 stand, the outlet speed of the rolled piece of the finish rolling F1-F8 stands, and the outlet crown of the rolled piece of the finish rolling F8 stand; and the calculated data of the process automation level comprises the deformation resistance of the rolled piece of the finish rolling F1-F8 stands, the outlet thickness of the rolled piece of finish rolling F1-F7 stands, rolling kilometers of the finish rolling F1-F8 stands, and the thermal expansion of the rolled piece during the finish rolling process.
3. The prediction method of claim 1, wherein the Step 2 further comprises the steps of: C = P K P + F K F + E C ω C + E ∑ ( ω H + ω W + ω O ) + E 0 Δ; ( 1 ) ω H = 2 ( 1 + v ) β t R ∫ 0 R r [ T ( r, z ) - T 0 ( r, z ) ] dr; ( 2 ) β t = Δ L L × Δ T; ( 3 ) wear n = k × ∑ P in × l in ( 1 + α X 4 ) w; ( 4 ) l in = L n × B n × H n b in × h in; ( 5 ) wear n 0 = k × ∑ P in × l in w; ( 7 ) wear n 1 = k × ∑ P in × l in ( 1 + α ) w; ( 8 )
Step 2.1: establishing the outlet crown mechanism model of the hot continuous rolling, wherein a mathematical equation is as follows:
wherein, C represents the crown of the steel plates and strips; P and F respectively represent a rolling force of stands and a roll-bending force of the stands for enabling roll systems to bend and deform; KP and KF respectively represent the a transverse stiffness of a rolling mill and a transverse stiffness of a bending roll; ωC represents a controllable roll crown; ωH represents a hot crown of the rolls caused by a thermal expansion of the rolls; ωW represents a wear crown of the rolls, caused by a wear of the rolls; ωO represents an initial roll crown of the rolls; Δ represents an inlet crown of the steel plates and strips; E0 represents inlet crown coefficients, EC represents controllable roll crown coefficients, and EΣ represents comprehensive crown coefficients;
Step 2.2: calculating the hot crown of the rolls caused by the thermal expansion of the rolls according to the following equation:
wherein Bt represents thermal expansion coefficients of the rolls and is calculated according to the equation below; v represents a Poisson coefficient of the rolls; T(r,z) represents a temperature at (r,z) where a coordinate is located, r represents a variable along a radius direction of the rolls, and z represents a variable along a length direction of the rolls; T0(r,z) represents an initial temperature of the rolls; a model is simplified and a temperature of the rolls is regarded as uniform distribution:
wherein ΔL represents a thermal expansion of the steel plates and strips when the temperature changes by ΔT; L represents a length before expansion;
Step 2.3: calculating a wear amount of the rolls according to the following equation:
wherein wearn represents the wear amount of the rolls; k represents coefficients related to roll materials and steel plates and strips materials, and Pin represents the rolling force of the nth rolling mill during rolling an ith steel coil; lin represents a length of the ith steel coil after being rolled by the nth rolling mill and is calculated according to the equation below; α represents wear coefficients of the rolls; X represents a position of the wear amount; W represents a width of the steel plates and strips:
wherein lin, bin and hin respectively represent a length, a width and a thickness of the ith steel coil after being rolled by the nih rolling mill, and Ln, Bn and Hn respectively represent a length, a width and a thickness of the steel plates and strips before being rolled;
Step 2.4: calculating the wear crown of rolls caused by wear of the rolls according to the following equation: ωw=wearn0−wearn1 (6);
wherein ωw represents the wear crown of the rolls, wearn0 represents the wear amount of the rolls when a position X of the wear amount is equal to 0, and wearn1 represents the wear amount of the rolls when the position X of the wear amount is equal to ±1;
when X=0, at a center line of the corresponding steel plates and strips:
when X=±1, at an edge of the corresponding steel plates and strips:
Step 2.5: taking remaining variables except for the rolling force of the stands, the roll-bending force of the stands, the hot crown of the rolls and the wear crown of the rolls, in the outlet crown mechanism model of the hot continuous rolling, as fixed values, calculating the outlet crown of the steel plates and strips, and taking the outlet crown of the steel plates and strips as the benchmark value of the outlet crown.
4. The prediction method of claim 1, wherein the Step 4 further comprises the steps of: σ ( d ) = 1 e - d; ( 11 ) J ( W, b, x, y ) = 1 2 a L - y 2 2 = 1 2 σ ( W L × a L - 1 + b L ) - y 2 2; ( 12 )
Step 4.1: designing a forward propagation algorithm of the DNN model for predicting the crown of the steel plates and strips and determining an activation function according to the equations below: a1=x (9), al=σ(dl)=σ(Wlal−130 bl) (10);
wherein a1 represents an output of a first layer, expressed by a matrix method; d represents an output of a lth layer, expressed by the matrix method, wherein 2≤l≤L, L is a total number of layers of a neural network; Wl represents a matrix of the lth layer and bl represents a bias vector of the lth layer; x represents an input vector; σ(d) represents the activation function;
the activation function is specifically a Sigmoid activation function:
wherein d is an input of the activation function;
Step 4.2: designing a loss function in a backward propagation algorithm of the DNN model for predicting the crown of the steel plates and strips:
a mean square function is used to measure an output loss of the training set data:
wherein y is a target output of the DNN model for predicting the crown of the steel plates and strips;
Step 4.3: adopting an Adam optimization algorithm and updating and calculating the model parameters to minimize the loss function;
Step 4.4: adopting a Cosine annealing algorithm based on an unequal interval annealing strategy to adjust a learning rate of the DNN model for predicting the crown of the steel plates and strips; and
Step 4.5: adopting a variable controlling method to select a number of hidden layers of the network, selecting a number of hidden layer nodes and a number of data groups used during each training, and completing training of the DNN model for predicting the crown of the steel plates and strips.
5. The prediction method of claim 4, wherein the number of the hidden layers of the constructed DNN model for predicting the crown of the steel plates and strips is 3, the number of the hidden layer nodes is 50, and the number of the data groups selected from each training is 128.
6. The prediction method of claim 1, wherein the Step 6 further comprises the steps of:
Step 6.1: adding up the predicted value of the deviation amount of the outlet crown and the benchmark value of the outlet crown to obtain the predicted value of the crown of the DNN model for predicting crown of the steel plates and strips;
Step 6.2: directly taking the outlet crown as the output of the DNN model and performing predicting to obtain the predicted value of the crown based on the DNN model;
Step 6.3: performing calculating according to the outlet crown mechanism model of the hot continuous rolling to obtain the calculated value of the outlet crown; and
Step 6.4: evaluating the predicted results of the Steps 6.1-6.3 by using the mean square error (MSE), the root mean square error (RMSE), the mean absolute error (MAE) of performance indexes and the correlation coefficient R, and analyzing the prediction precision.
7. The prediction method of claim 6, wherein in the Step 6.4: MSE = 1 n ∑ j = 1 n ( y j - y j ′ ) 2; ( 13 ) RMSE = 1 n ∑ j = 1 n ( y j - y j ′ ) 2; ( 14 ) MAE = 1 n ∑ j = 1 n ❘ "\[LeftBracketingBar]" y j - y j ′ ❘ "\[RightBracketingBar]"; ( 15 ) R = 1 - ∑ j = 1 n ( y j - y j ′ ) 2 ∑ j = 1 n ( y j - y _ ) 2; ( 16 )
the mean square error (MSE) is calculated according to the following equation:
the root mean square error (RMSE) is calculated according to the following equation:
the mean absolute error (MAE) of performance indexes is calculated according to the following equation:
the correlation coefficient R is calculated according to the following equation:
wherein yi represents the actual values of the outlet crown, y′i represents the predicted value obtained through the corresponding model, y represents a mean value of the actual values of the outlet crown, and n represents a total number of data groups in the test set data.
Patent History
Publication number: 20240184956
Type: Application
Filed: Jun 8, 2022
Publication Date: Jun 6, 2024
Inventors: Xu LI (Shenyang City), Nan CHEN (Shenyang City), Jingguo DING (Shenyang City), Feng LUAN (Shenyang City), Yan WU (Shenyang City), Bingbing MA (Shenyang City), Kun GAO (Shenyang City), Lifeng HUO (Shenyang City), Dianhua ZHANG (Shenyang City)
Application Number: 18/014,594
Classifications
International Classification: G06F 30/27 (20060101); G06F 30/17 (20060101); G06F 119/08 (20060101); G06F 119/14 (20060101);
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0
# What is 6 millimeters in thousandths?
Updated: 11/10/2022
Wiki User
15y ago
1 inch = 25.4 mm 1/1000 in = .0254 mm 6/.0254 = 236 thou
Wiki User
15y ago
Wiki User
13y ago
6 thousandths of an inch = 0.1524 millimeter (rounded)
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# Is there a way to answer this problem without using arithmetic at each step?
I was running through this mock high school math graduation test (try it out!) http://www.minnpost.com/data/2012/12/can-you-pass-mathematics-grad-test-high-school-students
Question 6 gave me pause:
Dan bought a new computer for $900. Each year, the value of the computer decreased by 25% of the previous year's value. At this rate, what can Dan expect the approximate value of the computer to be after 8 years? I can solve this easily enough by looping through the calculation for each year's deprecation. Programming has made me lazy, I let loops handle the arithmetic for me. But when solving by hand it was tedious. There must be a better way! Here's what I know: there is a rate of change which is constant at 25% / year, and we know the initial value at$900, and how long the change will be applied, 8 years, but I'm fuzzy on how to put it all together to get a formula.
My college calculus knowledge has imploded on itself, leaving little trace. The Wikipedia page on derivatives is causing my vision to blur. Where do I start to reason my way towards an intelligently-calculated answer?
-
This is a little off-topic, but I began reading some of the questions in that Minn. GRAD test: I weep for the mathematical level of those high school kids. Hopefully that was an exam for the lowest level and there are higher ones... – DonAntonio Jan 31 '13 at 2:56
Each year the value of the computer is $\frac34$ of its value the previous year, so after $8$ years the value is $$900\cdot\left(\frac34\right)^8$$ dollars, or $\$90.10$to the nearest cent. - I'm almost certain it is not 60.07 – Calvin Lin Jan 30 '13 at 23:51 @Calvin: True. I wonder what on earth I punched into the calculator. – Brian M. Scott Jan 30 '13 at 23:52 If$a$is the value this year, then the value next year is$(0.75)a$. So the value after$2$years is$(0.75)(0.75)a$. After$3$years it is$(0.75)(0.75)(0.75)a$, and so on. So the value after$8$years is$(0.75)^8 a$. To evaluate$(0.75)^8$on a simple calculator, square$0.75$, square the result, then square again, and it's done. Or else one can use the$y^x$key. It's even practical to do it by hand. For$0.75=\frac{3}{4}$. Note that$3^8=(3^4)^2=81^2=6561$. Also,$4^8=2^{16}=65536$. So$(0.75)^8=\frac{6561}{65536}$. - This is almost magic (i.e. unique to the values). We're looking for$0.75^8$. The trick here is to know that$\log 0.75 \approx -0.125$. As such,$\log (0.75)^8 = 8 \times \log 0.75 \approx -1$. Thus, it will be worth about 1/10 of it's original price, i.e.$90.
So, to obtain $\log 0.75$, you can use $\log 0.75 = \log \frac {3}{2^2} \approx 0.477 - 2 \times 0.301$. It is useful to know what $\log 2, \log 3$ is approximately equal to.
After the first year, the value is $\frac{3}{4}\times 900$. After another year, it is $\frac{3}{4}\times\frac{3}{4}\times 900$. Continuing this way, one sees that after eight years, the value is $(\frac{3}{4})^8\times 900$.
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It is my understanding that OAS takes into account liquidity and credit risk. Z-spread adds option premium to OAS. Here is my confusion: If you are valuing a bond with embedded options (e.g. callable bond), you add same OAS to forward rates to find an arbitrage-free present value. Why are we callin this additional spread OAS and not Z-spread? Second Confusion: Similarly, I am somewhat confused where we do spread analysis for Fixed Income securities. For example, when looking at Mortgage Back Securities (with prepayment option), why are we using OAS and not Z-spread? Thank you for your comments.
1rst: we add OAS because in the binomial tree is taking into account the fact that the callable bond will follow the ‘call rule’ and its value at certain nodes will reflect the fact that it will be called at the given interest rates… if there was no possibility that the bond will be called then z-spread would be appropriate because the cash flows would be the same and OAS = Z spread 2nd: we use OAS so that different securities (with different option costs) can be compared, and the relative credit and liquidity can be evaluated. If we use z-spread a security with an z-spread of 250 bps will appear to a security (with similar credit and liq) with a z-spread of 200 bps, however is the first security’s has a prepayment option worth 150 bps and the second security has a prepayment option worth only 50 bps then the first security will actually be overvalued relative to the second
Thanks, Marty3 but I am still confused. In several places in CFAI and kaplan the OAS as defined as the “spread with the option cost removed” But if the OAS is calculated using the tree while accounting for the call rules, how is does this NOT include the option cost?! Its certainly not the same spread as you’d get on an otherwise identical bond WITHOUT the options. If you calculate the spread on the tree and fully incorporate the call rules, how does this NOT reflect the option cost? This has been REALLY bothering me. The two treatments of the OAS seem to conflict: if z spread = OAS + cost of call option but OAS is calculated in a way that reflects the option, how does this work!?
Just as an example CMBS valuation is done with z-spread. Virtually no prepayment risks. Plain MBS valuation done with OAS using a binomial i-r- model because prepayment risk is heavy
z spread = OAS + cost of call option cost of option could be positive or negative, so the above is correct.
But if the OAS is derived using a tree that reflects the call rules, how can we say that the OAS does NOT reflect the option cost??
Say you buy a security with a coupon of 40 paid every 6 months for four years callable at 100 after two years and there is a downward sloping yield curve, the yield on a treasury with same maturity is 3% so the nominal spread is 5% and the z-spread will be greater than the nominal because of the downward sloping yield curve so lets say the z-spread is 6% However since the bond will probably be called in two years because of the the downward sloping yield curve you will receive your principal in two years and no interest. These forgone payments of interest and early return of principal have a value representing the cost of the option. When this cost is deducted from the z-spread you have the OAS spread. People say that the OAS spread “does not reflect the option cost” because it is the spread on the payments that you are expected to receive given the structure of interest rates therefore it can be used to compare a security with an option to securities with different or no options. OAS essentially represents the spread on the payments you are expected to receive while z-spread is the spread if all payments are received.
Ahh… Thanks, Marty. Seriously, this was bugging the \$!#@# out of me. Is this related to what you do for a living, by chance? In other words, the OAS very much reflects the cost that the option imposes on a security. It only “does not reflect the option cost” when it is used viz benchmark that also has similar option. This way, if you compare OAS between to callable securities, any differences will likely be to non option related factors. And so, if you use the OAS viz benchmark for bond of same issuer, coupon and option characteristics, you can judge whether your particular bond is under- or overvalued. In the sense of the difference between the OAS and the z spread, the difference IS the option cost. So OAS does in fact fully reflect the fact that the cashflows from a a bond that can be called are different from an otherwise identical noncallable security. Phew.
zoya Wrote: ------------------------------------------------------- > In the sense of the difference between the OAS and > the z spread, the difference IS the option cost. > So OAS does in fact fully reflect the fact that > the cashflows from a a bond that can be called are > different from an otherwise identical noncallable > security. thats right. Think of two rate trees with all the same attributes except one is callable and one is not. The price for the non callable, less the price for the callable is the option cost. The “spread” or percentage over the callable options interest rates at every point in the tree that would make the value equal to that of the Z-Spread (market price of the non callable bond)
I got another one for you all. Question number 35, reading 54: Jokinen states: “If interest rates rise and volatility is unchanged, the value of a callable bond should decrease” Correct / incorrect? Isnt this a bit iffy tho? Interest rates rise, so bond values fall, ok. But, as interest rates rise, the (negative!) value of the call decreases, right? As the likelihood of the call being exercised decreases, yes? So how can we conclusively say that this statement is correct, as the book tells us.
zoya Wrote: ------------------------------------------------------- > I got another one for you all. > Question number 35, reading 54: > > Jokinen states: > “If interest rates rise and volatility is > unchanged, the value of a callable bond should > decrease” > > Correct / incorrect? > > Isnt this a bit iffy tho? Interest rates rise, so > bond values fall, ok. > But, as interest rates rise, the (negative!) value > of the call decreases, right? As the likelihood of > the call being exercised decreases, yes? > > So how can we conclusively say that this statement > is correct, as the book tells us. it specifically says that volatility is unchanged so it will be treated like a non-callable( even though it inst) its just coz interest rate movement will not effect the price of the call option and keeping the call option price constant the bond price will decrease… that’s how i put it together correct me if i am wrong
yep, I think YAH got that right.
I also think the call option value does decrease, but the decrease in the bond price is more than the small gain from the reduction in the call option.
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Power of Ten
To enroll in any of our courses, click here
"Powers of 10" are the number 10 raised to any power, including negative powers. The metric system is based on powers of 10. Examples: Powers of 10 102 , 103 , 108 , 10-2 , 10-14 Positive powers of ten equal 10 times itself that many times. Examples: Positive Powers of 10 102 = 10 times itself 2 times = 10 x 10 103 = 10 times itself 3 times = 10 x 10 x 10 107 = 10 times itself 7 times = 10 x 10 x 10 x 10 x 10 x 10 x 10
Negative powers of ten equal 1 divided by ten times itself that many times. (Don't worry about why.) Examples: Negative Powers of 10 10-2 = 1 divided by 10 times itself 2 times = 1/(10 x 10 ) 10-3 = 1 divided by 10 times itself 2 times = 1/(10 x 10 x 10 ) 10-7 = 1 divided by 10 times itself 2 times = 1/(10 x 10 x 10 x 10 10 x 10 x 10 ) Any number to the zero power equals 1. So 100 = 1 Don't worry about why, unless you like worrying. :)
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