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# The MMS Big Data Book: Where Data and Magnetosphere Come Together
Designed to help students learn about NASA's missions currently studying the sun and its effects on Earth, this second of two storybooks in the series focuses on the importance of collaboration in data acquisition and analysis. Through a series of... (View More)
Audience: Elementary school, Middle school
Materials Cost: Free
# Water in the Atmosphere
In this activity, participants learn about the atmosphere by making observations and taking measurements. They will go outside and use scientific equipment to collect atmospheric moisture data (temperature, relative humidity, precipitation and cloud... (View More)
# Water in the Geosphere
In this activity, participants learn about the geosphere by making observations and taking measurements. They will go outside and use scientific equipment to investigate water in the soil by measuring soil moisture, temperature, color and... (View More)
# Year of the Solar System Math Guide
This collection of 160 math problems covers the 20 science topic themes presented by the NASA/JPL Year of the Solar System (YOSS) website, covering the solar system, planets, the search for life, and robotics. Examples of topics included are: scale... (View More)
# MRC: How Do I Measure This? (Grades 3-5)
This is a lesson about measurement and cratering. Learners will read about the origin of the foot as a standardized unit of measure, work collaboratively to conduct an experiment about cratering, and collect and record data to draw logical and... (View More)
# Earth's Water
The total amount of water on Earth, the places in which it is found and the percentages of fresh vs. salt are examined in this lesson. A short demonstration allows students to visualize the percentage differences and a coloring exercise illustrates... (View More)
# Water in the Biosphere
Water is essential to Earth's living system, the biosphere. In this lesson, students determine the types of living things within an outdoor study site, use a taxonomic key to determine the predominant land cover type of their site, then predict the... (View More)
# Water in the Hydrosphere
In this activity, participants learn about the hydrosphere by making observations and taking measurements. They will go outside and use scientific equipment to investigate temperature, pH and transparency of a body of water. They will use this... (View More)
# Water Conservation
This activity was developed to get students thinking about the many ways that people use freshwater and how we can conserve this precious and fundamental natural resource. Students will watch a short documentary describing issues related to clean... (View More)
# Measuring Precipitation
Students will design, build and then test a rain gauge to measure precipitation. By sharing their results, they will recognize the need for standardization and precision in scientific tools. All background information, student worksheets and... (View More)
«Previous Page1234 Next Page» | 706 | 3,481 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2018-51 | latest | en | 0.889727 |
https://www.kopykitab.com/blog/rd-sharma-solutions-class-12-maths-chapter-19-exercise-19-26/ | 1,643,196,005,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304947.93/warc/CC-MAIN-20220126101419-20220126131419-00512.warc.gz | 873,184,447 | 28,374 | # RD Sharma Solutions Class 12 Maths Chapter 19 Exercise 19.26 (2021)
RD Sharma Solutions Class 12 Maths Chapter 19 Exercise 19.26: Solutions PDF is an important reference guide to help students score well in the Class 12 exam. By solving exercise problems using RD Sharma Solutions Class 12 Maths Chapter 19 on a daily basis, students improve their problem solving and logical thinking skills. This is important for getting better academic grades. The main goal is to help students self-analyze the areas that require the most practice from the exam point of view.
## Download RD Sharma Solutions Class 12 Maths Chapter 19 Exercise 19.26 PDF:
RD-Sharma-Solutions-Class-12-Maths-Chapter-19-Ex-19.26
## Access RD Sharma Solutions Class 12 Maths Chapter 19 Exercise 19.26
Indefinite Integrals Ex 19.26 Q1
Indefinite Integrals Ex 19.26 Q2
Indefinite Integrals Ex 19.26 Q3
Indefinite Integrals Ex 19.26 Q4
Indefinite Integrals Ex 19.26 Q5
Indefinite Integrals Ex 19.26 Q6
Indefinite Integrals Ex 19.26 Q7
Indefinite Integrals Ex 19.26 Q8
Indefinite Integrals Ex 19.26 Q9
Indefinite Integrals Ex 19.26 Q10
Indefinite Integrals Ex 19.26 Q11
Indefinite Integrals Ex 19.26 Q12
Indefinite Integrals Ex 19.26 Q13
Indefinite Integrals Ex 19.26 Q14
Indefinite Integrals Ex 19.26 Q15
Indefinite Integrals Ex 19.26 Q16
Indefinite Integrals Ex 19.26 Q17
Indefinite Integrals Ex 19.26 Q18
Indefinite Integrals Ex 19.26 Q19
Indefinite Integrals Ex 19.26 Q20
Indefinite Integrals Ex 19.26 Q21
Indefinite Integrals Ex 19.26 Q22
Indefinite Integrals Ex 19.26 Q23
Indefinite Integrals Ex 19.26 Q24
## RD Sharma Solutions Class 12 Maths Chapter 19 Exercise 19.26: Important Topics From The Chapter
Some of the essential topics covered in RD Sharma Solutions Class 12 Maths Chapter 19 Exercise 19.26 chapter are listed below.
• Definition of primitive or antiderivative
• Definition and meaning of indefinite integral
• Fundamental integration formulae
• Some standard results on integration along with the corollary
• Integration of trigonometric functions
• Integration of exponential functions
• Miscellaneous problems
• Geometrical interpretation of indefinite integral
• Comparison between differentiation and integration
• Methods of integration
• Integration by substitution
• Some standard results
• Evaluation of integrals by using trigonometric substitutions
• Some special integrals
• Integration by parts
• Some important integrals along with theorems
• Integration of rational algebraic functions by using partial fractions
• When the denominator is expressible as a product of distinct linear factors
• When the denominator contains some repeating linear factors
• The denominator contains irreducible quadratic factors
• Integration of some special irrational algebraic functions
We have included all the information regarding CBSE RD Sharma Solutions Class 12 Maths Chapter 19 Exercise 19.26. If you have any query feel free to ask in the comment section.
### What is meant by indefinite integral?
An indefinite integral is a function that takes the antiderivative of another function.
### Are Integrals tough?
Some students find integrals harder than derivatives.
### How can I get RD Sharma Solutions Class 12 Maths Chapter 19 Exercise 19.26 PDF free?
You can refer to the above article.
### Are RD Sharma’s questions tough?
Yes, some students find RD Sharma as the toughest book for Maths. But we have provided solutions for RD Sharma which makes it easier to solve questions.
### What is the important topics included in RD Sharma Solutions Class 12 Maths Chapter 19 Exercise 19.26?
You can refer to the above article to know about the important topics included in RD Sharma Solutions Class 12 Maths Chapter 19 Exercise 19.26. | 919 | 3,765 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2022-05 | latest | en | 0.620415 |
https://gateoverflow.in/234684/addressing-modes-in-coa | 1,575,740,968,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540500637.40/warc/CC-MAIN-20191207160050-20191207184050-00242.warc.gz | 383,051,460 | 17,627 | 254 views
A computer supports one address and two address instructions. All the Addresses are the memory addresses. Memory size is 1Mbyte. How many one address instructions are possible if it has 240 two addresses instructions? (Assume binary instructions code has 48 bits)
| 254 views
in this problem total size of instruction code = 48 bit
= 8 + 20 + 20
given that 240 is two address instruction
remaining = 2^8 - 240
= 256 - 240 = 16
hence one address instruction = 16* 220 = 224
by Loyal (6.3k points)
selected by
0
what is the meaning of tag and how is it 8?
+1
because total instruction 48 bit if u substract 2 address it means (48-20-20) = 8 bit
+1
sub a,b ----> in this instruction a and b are memory addresses and sub is opcode ( tag in this solution )
0
ok thank u sir.
0
sir but how did we get memory address=20?
+1
@ritus
Memory Size = 1 MB = 220 B ===> 20 bits required to identify one Byte in this architecture
0
but there will be 23 bit not 20 because it's 1mbyte | 302 | 1,007 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2019-51 | latest | en | 0.892561 |
https://whatisconvert.com/530-centimeters-in-inches | 1,675,267,429,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499946.80/warc/CC-MAIN-20230201144459-20230201174459-00860.warc.gz | 629,122,984 | 7,106 | ## Convert 530 Centimeters to Inches
To calculate 530 Centimeters to the corresponding value in Inches, multiply the quantity in Centimeters by 0.39370078740157 (conversion factor). In this case we should multiply 530 Centimeters by 0.39370078740157 to get the equivalent result in Inches:
530 Centimeters x 0.39370078740157 = 208.66141732283 Inches
530 Centimeters is equivalent to 208.66141732283 Inches.
## How to convert from Centimeters to Inches
The conversion factor from Centimeters to Inches is 0.39370078740157. To find out how many Centimeters in Inches, multiply by the conversion factor or use the Length converter above. Five hundred thirty Centimeters is equivalent to two hundred eight point six six one Inches.
## Definition of Centimeter
The centimeter (symbol: cm) is a unit of length in the metric system. It is also the base unit in the centimeter-gram-second system of units. The centimeter practical unit of length for many everyday measurements. A centimeter is equal to 0.01(or 1E-2) meter.
## Definition of Inch
An inch (symbol: in) is a unit of length. It is defined as 1⁄12 of a foot, also is 1⁄36 of a yard. Though traditional standards for the exact length of an inch have varied, it is equal to exactly 25.4 mm. The inch is a popularly used customary unit of length in the United States, Canada, and the United Kingdom.
## Using the Centimeters to Inches converter you can get answers to questions like the following:
• How many Inches are in 530 Centimeters?
• 530 Centimeters is equal to how many Inches?
• How to convert 530 Centimeters to Inches?
• How many is 530 Centimeters in Inches?
• What is 530 Centimeters in Inches?
• How much is 530 Centimeters in Inches?
• How many in are in 530 cm?
• 530 cm is equal to how many in?
• How to convert 530 cm to in?
• How many is 530 cm in in?
• What is 530 cm in in?
• How much is 530 cm in in? | 478 | 1,885 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2023-06 | latest | en | 0.893249 |
https://www.pascalgamedevelopment.com/showthread.php?22687-Regarding-a-loop-and-drawing-in-console-mode-(etc-)&s=9d7bae6004a9be634ef7c5d998a01a12&p=97200&mode=threaded | 1,591,508,683,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348523564.99/warc/CC-MAIN-20200607044626-20200607074626-00426.warc.gz | 811,341,546 | 13,191 | ## Regarding a loop and drawing in console mode (etc.)
Well, since I'm back after so long and have picked again up the process of learning and doing programming, it has been going pretty well, hah hah hah, and I was actually having some fun with it I thought "If I had only continued this little fun hobby after these 4, 5 years or so since I stopped that previous time, then by now I would've been a much better programmer." I still recalled things I learnt in the past and applied them creatively and I couldn't help but suddenly start with a clone of a DOS bowling game I played as a kid, just for the fun of it. I think that, after this is finished and it shouldn't be too long, I will move on to something more difficult and, of course, more pleasing to the eye. (I recall another member long ago telling me I should use more modern libraries and he is right, but first I want to finish this thing for the sake of completing a small project, quick practicing, getting used to seeing code again, trying to think like a programmer and solve programming questions, nostalgia, and fun) (I think (but don't know for sure) there was someone giving me an example how such a simple game could be made, but I could not find this on the forum if it were there.)
Anyway, unfortunately I've encountered a problem and I'll phrase it as follows: how can I keep on drawing a moving ball, have it stop only if the player hits the button? When hit, the ball will move to the left. So far, for example temporarily:
Code:
{ GAME LOOP }
repeat
case Key of
'=': moveBallUp(1, 1, 60); { the parameters are a bunch of bytes for the variables X, Y (for a for loop and coordinates) and SpeedMs for use with a delay }
'\': moveBallDown(1, 1, 60); { here too }
#8 : throwBall(1, 1);
end;
until Key = #27;
Maybe drawBallUp and drawBallDown are better descriptions for these procedures. moveBallUp and moveBallDown draw the ball, going on one fixed X coordinate and where the Y coordinate changes position of course.
While the program checks for user input, I want the ball to move. Then, upon hitting <Backspace>, the ball will stop its moving up or down and will go straight to the left right on the Y coordinate that the user chose.
As you know, the problem is that now one must press = or \ to move the ball. While I might also use that anyway to make it more difficult, I really want to clone that old game and have the ball move up and down without the user having to specify it. Maybe something like:
Code:
repeat
moveBallUp(1, 1, 60);
moveBallDown(1, 1, 60):
until (Key = #27) or (Key = #8);
...
...
...but of course the program waits for user input and doesn't continue to move the ball up and down.
I'm thinking at the moment and maybe something in the sense of the following might do:
Code:
repeat
moveBallUp(1, 1, 60);
moveBallDown(1, 1, 60);
until keyPressed;
case Key of
#27: goto ENDING;
#8 : throwBall(1, 1);
end;
...
...
?
The goto ENDING is just there for ease of quickly jumping to the "ENDING" label which is just before "end." of the main program. Hmmm, well, at least at the moment, to me, this loop seems much better. Then somehow get the user input while the key was pressed. throwBall would have to start exactly from the Y coordinate when the user pressed <Backspace>.
I know it's a simple thing. Even if so, I really want to finish this thing as I have spent some time on it, drawing a cheap start screen, writing a few simple algorithms to print characters of character arrays or strings at certain speed forwards and backwards and using it in this little program, etc., where I finally creatively applied basic programming knowledge. As one who knows another subject well, I can certainly appreciate creativity, so I can certainly appreciate computer programming creativity. I am eager to also apply knowledge in regard to programming creatively to produce programs and this is one of the experiences. When this little game is done, I will play it and have a laugh. I will also send it to certain others so we can have a laugh together and talk of the old DOS days Then, if possible time-wise and whatever-wise, move on to the next little programming project to increase experience, probably finally having a go at Pong or a version of Snake either in console mode or using something like SDL, or maybe to simply go and learn some more about classes and other things first. Please advise | 1,031 | 4,415 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2020-24 | latest | en | 0.963587 |
https://bugsfixing.com/solved-plotting-value-of-each-node-using-python/ | 1,675,363,057,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500035.14/warc/CC-MAIN-20230202165041-20230202195041-00199.warc.gz | 167,820,413 | 11,691 | # [SOLVED] Plotting value of each node using Python
## Issue
I would like to plot the values of matrix `A` on y-axis as a function of node number on x-axis. However, since I have a 5×5 matrix, I don’t wish to define the node numbers manually. For instance, node 1 corresponds to 2.53734572e-01, node 2 to -1.08940733e-01,…, node 6 to -5.02000098e-01 and so on.
``````import numpy as np
import matplotlib.pyplot as plt
Node=np.array([[1,2,3,4,5],[6,7,8,9,10]])
A=np.array([[ 2.53734572e-01, -1.08940733e-01, 3.26138649e-03,
-6.10246692e-03, -2.59115145e-02],
[-5.02000098e-01, 1.08933714e-01, -3.65540228e-02,
5.93536044e-03, 3.88767438e-02],
[-1.42775456e+00, 4.52103243e-01, -2.33067190e-02,
7.27554880e-03, 1.15638039e-01],
[ 4.81030592e-01, -8.91302226e-02, 1.40486724e-03,
2.28801066e-02, -3.83389182e-02],
[ 8.39965176e-01, -2.81589587e-01, 2.24843962e-01,
-8.47758268e-03, -6.84721033e-02]])
plt.scatter(Node, A)
plt.xlabel('Node')
plt.ylabel('Velocity')
``````
## Solution
We can reduce the matrix to one dimension and use numpy.arange on the length of the matrix:
``````import numpy as np
import matplotlib.pyplot as plt
ys=np.array([[ 2.53734572e-01, -1.08940733e-01, 3.26138649e-03,
-6.10246692e-03, -2.59115145e-02],
[-5.02000098e-01, 1.08933714e-01, -3.65540228e-02,
5.93536044e-03, 3.88767438e-02],
[-1.42775456e+00, 4.52103243e-01, -2.33067190e-02,
7.27554880e-03, 1.15638039e-01],
[ 4.81030592e-01, -8.91302226e-02, 1.40486724e-03,
2.28801066e-02, -3.83389182e-02],
[ 8.39965176e-01, -2.81589587e-01, 2.24843962e-01,
-8.47758268e-03, -6.84721033e-02]]).flatten()
nodes=np.arange(len(ys))
plt.scatter(nodes, ys)
plt.xlabel('Node')
plt.ylabel('Velocity')
plt.show()
``````
Answered By – Christian P
Answer Checked By – Pedro (BugsFixing Volunteer) | 783 | 1,786 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2023-06 | latest | en | 0.488324 |
http://shuklan.com/haskell/lec04.html | 1,618,104,281,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038060603.10/warc/CC-MAIN-20210411000036-20210411030036-00485.warc.gz | 97,299,967 | 6,641 | Lecture 4
#SyntaxSwag
### Using These Slides
#### Every slide has a secret note.
• On Chrome: press `F12`, then click Console
• On IE: press `F12`, then click Console
• On Firefox: `Ctrl+Shift+k`
### Shortcut Keys:
`↓`, `PgDn`, `n`, `j` next slide `↑`, `PgUp`, `p`, `k` prev slide `Esc` enables `ctrl+f` globally
## Review
Haskell is Purely functional Statically typed Lazy Data looks like Bool Int Char ...
Lists and Tuples are useful data structures
List `[1,2,3]` Tuple `(1, "one")`
### More Review
Everything in Haskell has a Type
Here are some Type declarations.
``````
head :: [a] -> a -- gets the first element of a list
tail :: [a] -> [a] -- gets everything but the first element
last :: [a] -> a -- gets the last element of a list
init :: [a] -> [a] -- gets everything but the last element
(++) :: [a] -> [a] -> [a] -- concatenates two lists together
(:) :: a -> [a] -> [a] -- prepends an element to a list
fst :: (a,b) -> a -- gets the first element of a tuple
snd :: (a,b) -> b -- gets the second element of a tuple
``````
### Review of Homework 3
Implement a Caesar Cipher
A B C D
``````
-- This example uses 'succ' to get next letter
cipher :: String -> Int-> String
cipher "" n = ""
cipher str n = rotate (head str) n : cipher (tail str) n
rotate :: Char -> Int -> Char
rotate c 0 = c
rotate c n = rotate (next c) (n-1)
next :: Char -> Char
next c = if c=='z' then 'a' else succ c
``````
## Pattern Matching
A function can have multiple patterns
``````
guess :: Int -> [Char]
guess 42 = "correct!"
guess x = "wrong guess!"
``````
Each pattern has the same type declaration
#### Pattern Matching
• Patterns are matched in order, top-down
• Only the first matched pattern is evaluated
• The patterns must exhaust the entire domain
### What's wrong with this code?
``````
fib :: Int -> Int
fib n = fib(n-1) + fib(n-2)
fib 0 = 1
fib 1 = 1
``````
The base case is never hit.
The first pattern eats up everything!
∞ loop
### More Pattern Matching
You can even match lists using Construct
``````
head (firstItem : everythingElse) = firstItem
``````
``````
tail (x:xs) = xs
``````
### More Pattern Matching
Write a function to detect if a list is a palindrome
``````
isPal :: Eq a => [a] -> Bool
``````
### Pattern matching is powerful
We can define `fst` with pattern matching
``````
fst :: (a,b) -> a
fst (x,y) = x
``````
Try defining `head` with pattern matching
``````
``````
### Wildcard in Pattern Matching
• We can specify when a value is unused.
• The "_" symbol is called a wildcard in Haskell.
• This is how it's used:
• ``````
tail (_:xs) = xs
``````
### Error Handling
When GHCi is angry, it produces error messages through the `error` function.
``````
error :: [Char] -> a
``````
The official implementation of `head` is
``````
``````
*
## Guards
• Guards are clean `if` statements.
• Just like with pattern matching, order matters.
• A guard is introduced by the `|` symbol.
• And it's followed by a Bool expression.
• Then followed by the function body
• ``````
guessMyNumber x
| x > 27 = "Too high!"
| x < 27 = "Too low!"
| otherwise = "Correct!"
``````
`otherwise` is just a fancy word for `True`
### Guards
Guards are very powerful.
Anything done with pattern matching can be done with guards.
``````
| null xs = error "list is empty"
| otherwise = xs !! 0
``````
`!!` is a function that gives an element at an index
### Variables
These are not like your typical Java variables
In Java or C++, you can redefine variables:
``````
x = 1;
...
x = 2;
``````
Mathematically, this makes no sense.
## Variables
Once defined, they can't change.
They can be used with the `let` keyword.
``````
slope (x1,y1) (x2,y2) = let dy = y2-y1
dx = x2-x1
in dy/dx
``````
Or with the `where` keyword.
``````
slope (x1,y1) (x2,y2) = dy/dx
where dy = y2-y1
dx = x2-x1
``````
## Whitespace
Code which is part of some expression should be indented further in than the beginning of that expression *
``````
Level-1
Level-2
Level-3
Level-3
Level-3
Level-2
Level-3
Level-3
Level-1
Level-2
Level-2
Level-1
``````
Don't use tab. Use spaces '` `'.
# Homework
## A Useful Tool
1. Fill out this week's form.
2. Convert between metric and imperial.
`convert :: (Double, [Char]) -> (Double, [Char])`
• m ↔ yd
• L ↔ gal
• kg ↔ lb
``````
Prelude> convert (1, "m")
(1.09361, "yd")
Prelude> convert (1, "L")
(0.264172, "gal")
Prelude> convert (1, "kg")
(2.20462, "lb")
`````` | 1,362 | 4,472 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2021-17 | latest | en | 0.745488 |
https://gateoverflow.in/232192/toc-rl | 1,550,934,624,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550249504746.91/warc/CC-MAIN-20190223142639-20190223164639-00412.warc.gz | 551,545,417 | 16,711 | 54 views
Consider the following language L = {w ∈ (a+b)* | w has atleast as many occurrences of (bba)’s as (abb)’s}. Which of the following statements is/are true?
S1: Language L is regular.
S2: Complement of L is CFL.
S3: Complement of L is CSL.
S4: Reversal of L is CFL.
+1 vote
there is no need to keep the number of bba’s in the memory because whenever two abb’s comes together (adjacent), then one bba’s always come between them. So L is regular. regular language is closed under complement, reversal and regular language are subset of CSL and CFL.
So, all the statements are correct.
Option A
Language is regular
1
+1 vote
2 | 175 | 635 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2019-09 | latest | en | 0.96526 |
https://electronics.stackexchange.com/questions/250665/calculate-total-impedance-in-rlc-circuit-with-l-and-c-parallel | 1,696,011,279,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510520.98/warc/CC-MAIN-20230929154432-20230929184432-00603.warc.gz | 253,521,953 | 40,062 | # Calculate total impedance in RLC circuit with L and C parallel
I am trying to find the total impedance of this circuit but im not getting the right answer. According to my notes the total impedance is 1002 Ω, -86.6°.
This is the method i tried: I got
Xc = 111.1, -90° => 0-j111.1
Xl= 125, 90° => 0+j125
R = 60+j0
then from here ive tried multiple different methods. ive done ((Xc*Xl)/(Xc+Xl))+R. Just adding Xc+Xl+R then converting them back to polar form. Nothing i try give me the right answer. What am i doing wrong here?
• So I get 1002 at -86.6 deg for ((Xc*Xl)/(Xc+Xl))+R. Is this not what you got, or is this what you have and it doesn't match the answer key? Aug 4, 2016 at 14:41
• 1002 is what it says in the answer key. Thats not what im getting. Aug 4, 2016 at 14:44
Your notes are correct and I think the mistake you may have made is adding 60 ohm (resistive) to the 1000 ohms (reactive) to get 1060 ohms. You need to add them as squares then take the square root hence $\sqrt{1000^2 + 60^2}$ = 1001.8 ohms. | 336 | 1,029 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2023-40 | longest | en | 0.921302 |
https://kr.mathworks.com/matlabcentral/cody/problems/183-pull-the-variable-y_correct-from-the-caller-s-workspace/solutions/1681896 | 1,568,852,492,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573385.29/warc/CC-MAIN-20190918234431-20190919020431-00007.warc.gz | 548,517,782 | 15,600 | Cody
Problem 183. Pull the variable y_correct from the Caller's Workspace
Solution 1681896
Submitted on 28 Nov 2018 by li haitao
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
y_correct = 1; assert(isequal(your_fcn_name,y_correct))
2 Pass
y_correct = 1:100; assert(isequal(your_fcn_name,y_correct))
3 Pass
y_correct = rand(100); assert(isequal(your_fcn_name,y_correct))
4 Pass
y_correct = randi(100); assert(isequal(your_fcn_name,y_correct))
5 Pass
y_correct = 'lol'; assert(isequal(your_fcn_name,y_correct)) | 187 | 640 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2019-39 | longest | en | 0.65587 |
http://www.wiziq.com/online-tests/13479-cbse-xii-phy-ch-2-current-electricity-paper-3 | 1,369,445,464,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368705310619/warc/CC-MAIN-20130516115510-00015-ip-10-60-113-184.ec2.internal.warc.gz | 811,385,272 | 29,191 | # CBSE-XII-Phy-Ch-2-Current Electricity-Paper-3 online Test
Zero
20 V
60 V
120 V
B will be zero
A will be zero
A and B will be 2V
A will be > 2V and B will be <2V
3 A
13 V
23 A
-3 A
5(
10(
15(
20(
How much work is required to carry a 6 (C charge from the negative terminal to the positive terminal of a 9 V battery?
54(10–3 j
54(10–6 j
54(10–9 j
54(10–12 j
In order to pass 10 % of main current through a moving coil galvanometer of 99 (, the resistance of the required shunt is:
9.9(
10(
11(
9(
320(
8(
20(
40(
In an experiment to measure the internal resistance of a cell by potentiometer, it is found that the balance point is at a length of 2m when the cell is shunted by a 5( resistance; and is at a length. Of 3 m when the cell is shunted by 10( resistance. The internal resistance of the cell is, then:
1.5(
10(
15(
1(
1.5 V
3.0 V
0.67 V
1.33 V
1(
3(
10(
5(
10/9
0.1
1.0
10.0
3 V
2 V
5 V
4 V
If R1 and R2 are respectively the filament resistance of a 200 W bulb and 100 W bulb designed to operator on the same voltage, then:
R1 is two times R2
R2 is two times R1
R2 is four times R 1
R1 is four times R2
Description:
15 Questions on Current Electricity
Tags:
By: Ashish
912 days 17 hours 54 minutes ago
I just came across practice questions on IIT-JEE on Current Electricity, it was fun and new questions kept coming even after I tried more than 30 questions, try it out at http://www.simplylearnt.com/practice/questions/Current-Electricity-1
By: vimanyu sahu
817 days 10 hours 45 minutes ago
good
By: kutta
770 days 11 hours 21 minutes ago
thanx heped mee get air 203 in iit jee 2010
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Connect | 748 | 2,278 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2013-20 | latest | en | 0.870136 |
https://www.mathworks.com/matlabcentral/cody/problems/2315-reverse-the-input/solutions/461984 | 1,496,046,691,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463612036.99/warc/CC-MAIN-20170529072631-20170529092631-00402.warc.gz | 1,142,343,085 | 11,573 | Cody
# Problem 2315. Reverse the input
Solution 461984
Submitted on 25 Jun 2014 by Zikobrelli
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 = 123; y_correct = 321; assert(isequal(reverse(x),y_correct));
``` ans = 321 ```
2 Pass
%% x = 96721; y_correct = 12769; assert(isequal(reverse(x),y_correct));
``` ans = 321 ans = 12769 ``` | 140 | 464 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2017-22 | longest | en | 0.62011 |
http://www.sbnation.com/2012/4/19/2961005/fangraphs-bartolo-colon-consecutive-strikes/in/2724833 | 1,406,380,517,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1405997901589.57/warc/CC-MAIN-20140722025821-00237-ip-10-33-131-23.ec2.internal.warc.gz | 1,104,652,522 | 21,956 | ## FanGraphs: The Odds Of Bartolo Colon Doing What He Did
Bartolo Colon threw 38 consecutive strikes Wednesday night. He also threw 70 other pitches, which were consecutive, but people don't care so much about those 70 pitches. Maybe you feel like you've heard enough about Colon's strike streak, but you're wrong. You can't hear enough about it. It's so unbelievable I still haven't grasped how unbelievable it is.
In an earlier feature, I linked to a FanGraphs post on the odds of Colon throwing 38 strikes in a row, but I figured I should give it its own entry. So here again is Dave Cameron, on Bartolo Colon:
Last night, we got 38 consecutive strikes without a ball. Binomial distribution tells us that the odds of that occurring, given what we know about Colon’s career strike percentage, is about 0.000000246. In other words, you’d expect to find one string of 38 consecutive strikes if you had a population of approximately 4.1 million strings of pitches thrown by Bartolo Colon. One in 4.1 million.
Dave issues an update that acknowledges that the math isn't that easy. And it isn't - the pitches aren't all independent of one another. But this works as an approximation, and if you just focus on the "million" part, you kind of get how crazy this was. Colon's achievement wasn't as visibly impressive or jaw-dropping as some other achievements, but it was statistically improbable.
By the way, here's Colon's first ball after the streak:
I would be steamed!
1. Apr 19
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We have our own Community Guidelines at SBNation.com. You should read them. | 462 | 2,016 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2014-23 | longest | en | 0.971345 |
https://thelaptopadviser.com/what-is-30-minutes-from-now-discover-the-surprising-answer/ | 1,718,983,551,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862125.45/warc/CC-MAIN-20240621125006-20240621155006-00303.warc.gz | 507,417,197 | 45,062 | # What Is 30 Minutes From Now? Discover The Surprising Answer!
## What Is 30 Minutes From Now
What is 30 minutes from now? It’s a question that often arises when we’re trying to plan our schedule or make arrangements for the near future. While it may seem like a simple query, the answer can vary depending on the context.
If you’re asking what will happen in 30 minutes from the current moment, it’s important to consider that time is constantly moving forward. So, in half an hour from now, you’ll find yourself in a new timeframe with different events unfolding around you.
However, if you’re looking for a more specific response regarding a particular event or activity happening exactly 30 minutes from now, then it would require knowing the exact time and location of that event. With this information at hand, you can determine what will be occurring at precisely half an hour into the future.
In conclusion, “what is 30 minutes from now?” doesn’t have a fixed answer as it depends on various factors such as your current situation and any scheduled events. To get a more precise response, it’s crucial to have specific details about the time and location in question.
## What Does ’30 Minutes From Now’ Mean?
When we talk about “30 minutes from now,” we are referring to a specific time that is half an hour in the future. It’s a way of indicating the time that will pass after the present moment.
To understand what “30 minutes from now” means, let’s break it down:
1. The Present Moment: This refers to the current point in time when you make the statement or consider a future event.
2. Minutes: Minutes are units of time used to measure small intervals. There are 60 minutes in an hour, and each minute represents 1/60th of an hour.
3. From Now: “From now” indicates that we’re measuring the time relative to this very moment, not any other reference point.
So, when we say “30 minutes from now,” it means adding 30 minutes to the current time or moment you’re considering. It’s like fast-forwarding half an hour into the future.
It’s important to note that this phrase is commonly used for scheduling or planning purposes. For example:
• If it’s currently 2:00 PM and you say, “Let’s meet in 30 minutes from now,” it would mean meeting at 2:30 PM.
• If you have a task deadline set for “30 minutes from now,” you need to complete it within half an hour.
Understanding what “30 minutes from now” means allows us to effectively manage our time and coordinate activities with others without any confusion about when something should happen.
In conclusion, when someone mentions “30 minutes from now,” they are referring to a point in time that is half an hour ahead of the present moment. So keep track of your clock or timer if you want things done on schedule!
## Time Conversions: From Now to Minutes
In this section, I’ll be discussing how to calculate what time will it be 30 minutes from now. It’s a common question that arises when we want to plan our activities or schedule our tasks efficiently.
To determine the time 30 minutes from now, you can follow these steps:
1. Start by looking at the current time on your clock or device.
2. Identify the number of minutes remaining until the next hour.
3. Add 30 minutes to that count.
4. If the resulting total is less than 60, simply add it to the current minute value.
5. If the total exceeds 60, subtract 60 from it and increment the hour value by one.
Let’s consider an example:
Assuming it’s currently 2:45 PM, we want to find out what time it will be exactly 30 minutes later.
1. The next hour is at 3:00 PM, which means there are still 15 minutes left in the current hour.
2. Adding 30 minutes to that count gives us a total of 45 (15 + 30).
3. Since this total falls within one hour, we can directly add it to the current minute value of 45.
4. Therefore, after adding those extra 30 minutes, we get a new minute value of 75, indicating that it will be quarter past three (3:15 PM).
By following these simple steps and calculations, you can determine what time it will be precisely 30 minutes from now for any given starting point. | 920 | 4,125 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2024-26 | latest | en | 0.935363 |
https://www.theessayworld.com/how-many-vertices-does-a-triangular-pyramid-have/ | 1,680,335,774,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949701.56/warc/CC-MAIN-20230401063607-20230401093607-00564.warc.gz | 1,114,875,630 | 14,074 | Question
1. # How Many Vertices Does A Triangular Pyramid Have
When it comes to geometry, there are a lot of different shapes and figures that you need to be aware of. One such figure is the triangular pyramid. This shape is pretty self-explanatory – it’s a pyramid with three sides. But what you may not know is how many vertices (or corners) this shape actually has. The answer may surprise you – a triangular pyramid actually has four vertices. This is because each of the three sides of the triangle creates a point, and then there’s one more at the top of the pyramid. If you’re interested in learning more about geometry and all of the different shapes out there, then this post is for you! We’ll explore the triangular pyramid in more depth and uncover some interesting facts about this geometric figure.
## Three Sides
A triangular pyramid has four sides. The three sides of the triangle are the base, the two lateral faces, and the apex.
## Four Vertices
A triangular pyramid has four vertices. The vertices are the points where the three sides of the pyramid meet.
## Five Faces
A triangular pyramid has five faces. The base is a triangle, and the other four faces are triangles that meet at a point (the apex).
## Six Edges
A triangular pyramid has six edges. The base of the pyramid is a triangle, and the three sides of the pyramid are triangles.
2. Hey there everyone!
If you’re looking for the answer to the question “how many vertices does a triangular pyramid have?”, you’ve come to the right place!
It may seem like a daunting question if you’re new to geometry, but the answer is actually quite simple. A triangular pyramid has four vertices, or corners.
Let’s take a closer look at how this works. A triangular pyramid is formed by connecting three triangular faces, with each of those faces having three vertices. In total, this means that the triangle has nine vertices.
However, when you bring the three faces together, some of those vertices overlap. Each face will overlap with two other faces, so in total, there will be six overlapping vertices. This leaves three non-overlapping vertices, making a total of four vertices for the triangular pyramid.
To sum it up, a triangular pyramid has four vertices. You can remember this by thinking of the four corners of the pyramid.
So the next time someone asks you “how many vertices does a triangular pyramid have?”, you’ll know the answer! | 506 | 2,428 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.84375 | 5 | CC-MAIN-2023-14 | latest | en | 0.948677 |
https://appliedmacro.com/2017/06/12/the-misuse-of-significance/ | 1,685,507,926,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224646257.46/warc/CC-MAIN-20230531022541-20230531052541-00484.warc.gz | 118,753,150 | 27,692 | The misuse of Significance
Definition
What does the word “significant” mean?
Dictionaries most often suggest a range of closely related definitions.
In a more everyday sense:
1. Importance e.g. this new discovery is a significant development
2. Meaningful e.g. the significance of the message was not lost on John
In mathematics, you get the example of:
1. Significant figures – e.g. 1.524658 is 1.5 to 2 sig fig
This use of the word is mathematical jargon with a precise meaning, but it also tallies with our general use of the word. We only want to look at the digits which are important and mean something.
In statistics:
1. “significant” means probably true (not due to chance)
Some issues arise from this
1. Something statistically significant may not be important
A result may be true and therefore significant when backed up by statistics, it doesn’t however mean it is important in the more standard English usage sense. I think this statistical interpretation can easily come into conflict with the everyday meaning and is fraught with danger.
When you jump out of a plane without a parachute it is likely that holding up an umbrella has a “significant” effect on your speed. I doubt you would think that this effect was important when you hit the ground.
I’m sure you can think of many things that are probably true but not important!
1. Statistical relationships are not transitive
An example from medicine, drugs for the most part are tested against a placebo rather than against each other. Drug A may perform better in tests against a placebo than Drug B. (ie has more significant results) However, that does not mean you know that Drug A will perform better in tests against Drug B. Unfortunately, current medical practice makes this implicit assumption when approving drugs.
This is a common misconception that you can use simple logic to infer other relationships. Unfortunately, this is not true. There is a similarly confused relationship with correlation. Statistical relationships like this are not transitive. https://iase-web.org/documents/papers/isi56/CPM80_CastroSotos.pdf
1. The 5% threshold for statistical significance is arbitrary
When you say that one result is significant and another is not because one has a 4.9% chance of being random and the other has 5.1%. This is the correct usage of the technical term but people ascribe more meaning to the word than that. One of the ideas is held to be “true” and the other is discarded.
1. A significant result may have happened by random chance
Saying that a certain outcome would only occur 1 time in 20 if it were random sounds good. But what if you ran 20 sets of analysis? By random chance you should expect one of them to pass the “significance” test.
Was the test constructed properly?
This relates to a supremely important point that often statistics are quoted in situations they are not supposed to be used or have been not properly applied
1. How many relationships did you test?
In finance, all analysts look at lots of different data sets, over different time periods in search of something “significant”.
1. Did you look at any of the data before choosing what test to run?
I cannot imagine how someone could not fall into this trap. We only run tests on things we think might work. But the reason we think they might work is that we have done some rough statistical work already e.g. looked at a picture or perhaps just subconsciously noted some signs of a relationship. This means that the data has been mined and your choice of test is not independent.
2. How many people are trying to find these relationships?
Let’s say that you are extremely careful in how you do your statistics. Let’s imagine that everyone else in the firm you work at is similarly careful. Then when you produce a “significant” result you may reasonably think it is meaningful. After all you only ran one test and it worked! You then show your boss. Should she be impressed? Maybe not.
1. How many failed tests are not shown?
In my experience, analysts do not show me large quantities of research they have done which they think is completely meaningless. Highly trained with great degrees, they want to show me “good” work with “good” results. This means that the 19 analysts that did not find anything today do not show me anything. From the perspective of the individual the result appears to be strongly non-random. From my perspective, it looks entirely consistent with being random.
Is it meaningless?
No. it just means exactly what the equation says it means. You should remain aware of the context if you want to use it. My interaction with professionals of all types is that they are enormously well trained in the complexity of statistical methods and woefully under trained in the limitations of them. In fact, their high proficiency with manipulating the data and the methods makes them even more prone to methodological error of this type as they have essentially been trained in the art of data-mining.
Conclusion
I am yet to read a research piece from a bank which presents data demonstrating that their hypothesis is has no statistical significance. We should remember that this is significant.
3 thoughts on “The misuse of Significance”
1. From another one of Castro Soto’s papers:
“[It] is proposed in Batanero et al. (2004) to drop the word significant from data analysis vocabulary and use it only in its everyday sense to describe something actually noteworthy or important.”
A related topic: I’ve recently been revisiting the controversy between Fischer and Neyman-Pearson. Putting aside the philosophy of science debate, I was interested to learn that I and most other users of statistical analysis have been applying an incoherent muddle between the two approaches. Here are two papers that address this, though I think they could have been clearer, perhaps by using some simple equations!
Click to access 03-26.pdf
Click to access Hubbard-Bayarri-2003.pdf
If I recall correctly, this was covered in How Not to be Wrong chapter 16. Does Lung Cancer Make You Smoke Cigarettes? However, I don’t think I really got the point when I read that treatment.
Like | 1,284 | 6,179 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2023-23 | latest | en | 0.953279 |
https://www.coursehero.com/tutors-problems/Computer-Science/10003726-This-is-java-programming-actually-I-dont-know-how-to-write-down-code/ | 1,560,674,841,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998084.36/warc/CC-MAIN-20190616082703-20190616104703-00005.warc.gz | 722,809,288 | 199,041 | View the step-by-step solution to:
# COM S 207 Triangle Pattern Write a program that reads in lines from the input. Each line has a single integer followed by a single character. The...
This is java programming, actually I dont know how to write down code
COM S 207 Triangle Pattern Write a program that reads in lines from the input. Each line has a single integer followed by a single character . The objective is to print a triangle pattern using the input character. The first integer represents the number of rows of the triangle. Input values will contain only integers greater than or equal to 1 and less than or equal to 50. Input The format of an input line is noRows char The first is an integer (and can be read using nextInt()). The last one is a single character (like *) and can be read using next() as a String and then using charAt(0) to get at the first character. Output For each line of input, a triangular pattern is printed consisting of several rows. If the no of rows is less than or equal to zero or more than 50, the program prints out an error statement as shown in the sample output. Sample Input 2 * -1 ' 10 = Sample Output * ** error in input = == === ==== ===== ====== ======= ======== ========= ========== HINT 1. use in.next().charAt(0) to get the character. 2. don’t forget to discard the rest of the input line using in.nextLine() if you are using in.hasNextLine() to check if there is a next line. 3. You will need two loops for this one as well.
COM S 207 ThreeNumbersM This problem is almost exactly like 006 ThreeNumbers problem, except that here you need to use a method to do the calculations. Declare a method named calcSumProdAvg as follows: private static void calcSumProdAvg(double d1, double d2, double d3){ // write your code for doing the calculations AND printing here. } In the main method, do all the reading of the lines and doubles. Call this method from the main method to calculate the sum, product, and average and to print them out. Write a program that reads in three integers and prints out their sum and product and average Input The input will be one or more lines of numbers that have been typed in by the user. Each line is to have exactly three real numbers. Output The output will have for each line, the numbers that had been read in separated by a single space followed by their sum, product, and average separated by a space. Each number is to be printed using "%.2e" format (see hint below). Sample Input 1 2 3 1.1 2.2 3.3 Sample Output 1.00e+00 2.00e+00 3.00e+00 6.00e+00 6.00e+00 2.00e+00 1.10e+00 2.20e+00 3.30e+00 6.60e+00 7.99e+00 2.20e+00
Page 2 of 2 pages HINT Here use in.nextDouble() to read a number. After reading three doubles, use in.nextLine() – so that you discard the rest of the "white spaces". To print two numbers num1 and num2 using %.2e format, I would use System.out. printf ("%.2e %.2e\n", num1, num2); The \n is used to make sure it prints a newline character. The printf method can be found in your textbook – section 2.5.5 Formatted Output
COM S 207 Rectangle Patterns Write a program that reads in lines from the input. Each line has two integer values followed by a single character . The objective is to print a rectangle pattern using the input pattern. The first integer represents the number of rows and the second integer represents the number of columns. Input The format of an input line is noRows noCols char The first two are integers (and can be read using nextInt()). The last one is a single character (like *) and can be read using next() as a String and then using charAt(0) to get at the first character. Output For each line of input, a rectangular pattern is printed consisting of several rows. If the no of rows or cols is less than or equal to zero or more than 50, the program prints out an error statement as shown in the sample output. Sample Input 2 2 , 2 51 + 10 10 = 0 11 * Sample Output ,, ,, error in input ========== ========== ========== ========== ========== ========== ========== ========== ========== ========== error in input HINT 1. use in.next().charAt(0) to get the character. 2. don’t forget to discard the rest of the input line using in.nextLine() if you are using in.hasNextLine() to check if there is a next line. 3. You will need TWO loops to print each rectangle.
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https://leanprover-community.github.io/mathlib4_docs/Mathlib/Algebra/Order/SMul.html | 1,680,240,053,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949573.84/warc/CC-MAIN-20230331051439-20230331081439-00606.warc.gz | 407,595,935 | 8,287 | Documentation
Mathlib.Algebra.Order.SMul
Ordered scalar product #
In this file we define
• OrderedSMul R M : an ordered additive commutative monoid M is an OrderedSMul over an OrderedSemiring R if the scalar product respects the order relation on the monoid and on the ring. There is a correspondence between this structure and convex cones, which is proven in Analysis/Convex/Cone.lean.
Implementation notes #
• We choose to define OrderedSMul as a Prop-valued mixin, so that it can be used for actions, modules, and algebras (the axioms for an "ordered algebra" are exactly that the algebra is ordered as a module).
• To get ordered modules and ordered vector spaces, it suffices to replace the OrderedAddCommMonoid and the OrderedSemiring as desired.
References #
• https://en.wikipedia.org/wiki/Ordered_module
Tags #
ordered module, ordered scalar, ordered smul, ordered action, ordered vector space
class OrderedSMul (R : Type u_1) (M : Type u_2) [inst : ] [inst : ] [inst : ] :
• Scalar multiplication by positive elements preserves the order.
smul_lt_smul_of_pos : ∀ {a b : M} {c : R}, a < b0 < cc a < c b
• If c • a < c • b for some positive c, then a < b.
lt_of_smul_lt_smul_of_pos : ∀ {a b : M} {c : R}, c a < c b0 < ca < b
The ordered scalar product property is when an ordered additive commutative monoid with a partial order has a scalar multiplication which is compatible with the order.
Instances
instance OrderDual.instSMulWithZeroOrderDualInstZeroOrderDualToZero {R : Type u_1} {M : Type u_2} [inst : Zero R] [inst : ] [inst : ] :
Equations
• OrderDual.instSMulWithZeroOrderDualInstZeroOrderDualToZero = let src := instSMulOrderDual; SMulWithZero.mk (_ : ∀ (m : Mᵒᵈ), 0 m = 0)
instance OrderDual.instMulActionOrderDual {R : Type u_1} {M : Type u_2} [inst : ] [inst : ] :
Equations
instance OrderDual.instMulActionWithZeroOrderDualInstZeroOrderDualToZero {R : Type u_1} {M : Type u_2} [inst : ] [inst : ] [inst : ] :
Equations
• One or more equations did not get rendered due to their size.
instance OrderDual.instDistribMulActionOrderDualToMonoidInstAddMonoidOrderDual {R : Type u_1} {M : Type u_2} [inst : ] [inst : ] [inst : ] :
Equations
Equations
• One or more equations did not get rendered due to their size.
theorem smul_lt_smul_of_pos {R : Type u_2} {M : Type u_1} [inst : ] [inst : ] [inst : ] [inst : ] {a : M} {b : M} {c : R} :
a < b0 < cc a < c b
theorem smul_le_smul_of_nonneg {R : Type u_2} {M : Type u_1} [inst : ] [inst : ] [inst : ] [inst : ] {a : M} {b : M} {c : R} (h₁ : a b) (h₂ : 0 c) :
c a c b
theorem smul_nonneg {R : Type u_1} {M : Type u_2} [inst : ] [inst : ] [inst : ] [inst : ] {a : M} {c : R} (hc : 0 c) (ha : 0 a) :
0 c a
theorem smul_nonpos_of_nonneg_of_nonpos {R : Type u_1} {M : Type u_2} [inst : ] [inst : ] [inst : ] [inst : ] {a : M} {c : R} (hc : 0 c) (ha : a 0) :
c a 0
theorem eq_of_smul_eq_smul_of_pos_of_le {R : Type u_2} {M : Type u_1} [inst : ] [inst : ] [inst : ] [inst : ] {a : M} {b : M} {c : R} (h₁ : c a = c b) (hc : 0 < c) (hle : a b) :
a = b
theorem lt_of_smul_lt_smul_of_nonneg {R : Type u_2} {M : Type u_1} [inst : ] [inst : ] [inst : ] [inst : ] {a : M} {b : M} {c : R} (h : c a < c b) (hc : 0 c) :
a < b
theorem smul_lt_smul_iff_of_pos {R : Type u_1} {M : Type u_2} [inst : ] [inst : ] [inst : ] [inst : ] {a : M} {b : M} {c : R} (hc : 0 < c) :
c a < c b a < b
theorem smul_pos_iff_of_pos {R : Type u_1} {M : Type u_2} [inst : ] [inst : ] [inst : ] [inst : ] {a : M} {c : R} (hc : 0 < c) :
0 < c a 0 < a
theorem smul_pos {R : Type u_1} {M : Type u_2} [inst : ] [inst : ] [inst : ] [inst : ] {a : M} {c : R} (hc : 0 < c) :
0 < a0 < c a
Alias of the reverse direction of smul_pos_iff_of_pos.
theorem monotone_smul_left {R : Type u_1} {M : Type u_2} [inst : ] [inst : ] [inst : ] [inst : ] {c : R} (hc : 0 c) :
theorem strictMono_smul_left {R : Type u_1} {M : Type u_2} [inst : ] [inst : ] [inst : ] [inst : ] {c : R} (hc : 0 < c) :
theorem smul_lowerBounds_subset_lowerBounds_smul {R : Type u_1} {M : Type u_2} [inst : ] [inst : ] [inst : ] [inst : ] {s : Set M} {c : R} (hc : 0 c) :
theorem smul_upperBounds_subset_upperBounds_smul {R : Type u_1} {M : Type u_2} [inst : ] [inst : ] [inst : ] [inst : ] {s : Set M} {c : R} (hc : 0 c) :
theorem BddBelow.smul_of_nonneg {R : Type u_2} {M : Type u_1} [inst : ] [inst : ] [inst : ] [inst : ] {s : Set M} {c : R} (hs : ) (hc : 0 c) :
theorem BddAbove.smul_of_nonneg {R : Type u_2} {M : Type u_1} [inst : ] [inst : ] [inst : ] [inst : ] {s : Set M} {c : R} (hs : ) (hc : 0 c) :
theorem OrderedSMul.mk'' {𝕜 : Type u_1} {M : Type u_2} [inst : ] [inst : ] [inst : ] (h : ∀ ⦃c : 𝕜⦄, 0 < cStrictMono fun a => c a) :
To prove that a linear ordered monoid is an ordered module, it suffices to verify only the first axiom of OrderedSMul.
instance Nat.orderedSMul {M : Type u_1} [inst : ] :
Equations
instance Int.orderedSMul {M : Type u_1} [inst : ] :
Equations
instance LinearOrderedSemiring.toOrderedSMul {R : Type u_1} [inst : ] :
Equations
theorem OrderedSMul.mk' {𝕜 : Type u_2} {M : Type u_1} [inst : ] [inst : ] [inst : ] (h : ∀ ⦃a b : M⦄ ⦃c : 𝕜⦄, a < b0 < cc a c b) :
To prove that a vector space over a linear ordered field is ordered, it suffices to verify only the first axiom of OrderedSMul.
instance Pi.orderedSMul {ι : Type u_1} {𝕜 : Type u_2} [inst : ] {M : ιType u_3} [inst : (i : ι) → OrderedAddCommMonoid (M i)] [inst : (i : ι) → MulActionWithZero 𝕜 (M i)] [inst : ∀ (i : ι), OrderedSMul 𝕜 (M i)] :
OrderedSMul 𝕜 ((i : ι) → M i)
Equations
instance Pi.orderedSMul' {ι : Type u_1} {𝕜 : Type u_2} {M : Type u_3} [inst : ] [inst : ] [inst : ] [inst : ] :
OrderedSMul 𝕜 (ιM)
Equations
instance Pi.orderedSMul'' {ι : Type u_1} {𝕜 : Type u_2} [inst : ] :
OrderedSMul 𝕜 (ι𝕜)
Equations
theorem smul_le_smul_iff_of_pos {𝕜 : Type u_1} {M : Type u_2} [inst : ] [inst : ] [inst : ] [inst : ] {a : M} {b : M} {c : 𝕜} (hc : 0 < c) :
c a c b a b
theorem inv_smul_le_iff {𝕜 : Type u_1} {M : Type u_2} [inst : ] [inst : ] [inst : ] [inst : ] {a : M} {b : M} {c : 𝕜} (h : 0 < c) :
c⁻¹ a b a c b
theorem inv_smul_lt_iff {𝕜 : Type u_1} {M : Type u_2} [inst : ] [inst : ] [inst : ] [inst : ] {a : M} {b : M} {c : 𝕜} (h : 0 < c) :
c⁻¹ a < b a < c b
theorem le_inv_smul_iff {𝕜 : Type u_1} {M : Type u_2} [inst : ] [inst : ] [inst : ] [inst : ] {a : M} {b : M} {c : 𝕜} (h : 0 < c) :
a c⁻¹ b c a b
theorem lt_inv_smul_iff {𝕜 : Type u_1} {M : Type u_2} [inst : ] [inst : ] [inst : ] [inst : ] {a : M} {b : M} {c : 𝕜} (h : 0 < c) :
a < c⁻¹ b c a < b
@[simp]
theorem OrderIso.smulLeft_symm_apply {𝕜 : Type u_1} (M : Type u_2) [inst : ] [inst : ] [inst : ] [inst : ] {c : 𝕜} (hc : 0 < c) (b : M) :
().toEmbedding b = c⁻¹ b
@[simp]
theorem OrderIso.smulLeft_apply {𝕜 : Type u_1} (M : Type u_2) [inst : ] [inst : ] [inst : ] [inst : ] {c : 𝕜} (hc : 0 < c) (b : M) :
().toEmbedding b = c b
def OrderIso.smulLeft {𝕜 : Type u_1} (M : Type u_2) [inst : ] [inst : ] [inst : ] [inst : ] {c : 𝕜} (hc : 0 < c) :
M ≃o M
Left scalar multiplication as an order isomorphism.
Equations
• One or more equations did not get rendered due to their size.
@[simp]
theorem lowerBounds_smul_of_pos {𝕜 : Type u_1} {M : Type u_2} [inst : ] [inst : ] [inst : ] [inst : ] {s : Set M} {c : 𝕜} (hc : 0 < c) :
lowerBounds (c s) = c
@[simp]
theorem upperBounds_smul_of_pos {𝕜 : Type u_1} {M : Type u_2} [inst : ] [inst : ] [inst : ] [inst : ] {s : Set M} {c : 𝕜} (hc : 0 < c) :
upperBounds (c s) = c
@[simp]
theorem bddBelow_smul_iff_of_pos {𝕜 : Type u_1} {M : Type u_2} [inst : ] [inst : ] [inst : ] [inst : ] {s : Set M} {c : 𝕜} (hc : 0 < c) :
@[simp]
theorem bddAbove_smul_iff_of_pos {𝕜 : Type u_1} {M : Type u_2} [inst : ] [inst : ] [inst : ] [inst : ] {s : Set M} {c : 𝕜} (hc : 0 < c) : | 3,045 | 7,675 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2023-14 | longest | en | 0.73462 |
https://www.knowpia.com/knowpedia/Optical_depth | 1,669,948,232,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710890.97/warc/CC-MAIN-20221202014312-20221202044312-00643.warc.gz | 894,826,786 | 22,649 | BREAKING NEWS
Optical depth
## Summary
In physics, optical depth or optical thickness is the natural logarithm of the ratio of incident to transmitted radiant power through a material. Thus, the larger the optical depth, the smaller the amount of transmitted radiant power through the material. Spectral optical depth or spectral optical thickness is the natural logarithm of the ratio of incident to transmitted spectral radiant power through a material.[1] Optical depth is dimensionless, and in particular is not a length, though it is a monotonically increasing function of optical path length, and approaches zero as the path length approaches zero. The use of the term "optical density" for optical depth is discouraged.[1]
Aerosol Optical Depth (AOD) at 830 nm measured with the same LED sun photometer from 1990 to 2016 at Geronimo Creek Observatory, Texas. Measurements made at or near solar noon when the sun is not obstructed by clouds. Peaks indicate smoke, dust and smog. Saharan dust events are measured each summer.
In chemistry, a closely related quantity called "absorbance" or "decadic absorbance" is used instead of optical depth: the common logarithm of the ratio of incident to transmitted radiant power through a material, that is the optical depth divided by ln 10.
## Mathematical definitions
### Optical depth
Optical depth of a material, denoted ${\textstyle \tau }$ , is given by:[2]
${\displaystyle \tau =\ln \!\left({\frac {\Phi _{\mathrm {e} }^{\mathrm {i} }}{\Phi _{\mathrm {e} }^{\mathrm {t} }}}\right)=-\ln T}$
where
• ${\textstyle \Phi _{\mathrm {e} }^{\mathrm {i} }}$ is the radiant flux received by that material;
• ${\textstyle \Phi _{\mathrm {e} }^{\mathrm {t} }}$ is the radiant flux transmitted by that material;
• ${\textstyle T}$ is the transmittance of that material.
The absorbance ${\textstyle A}$ is related to optical depth by:
${\displaystyle \tau =A\ln {10}}$
### Spectral optical depth
Spectral optical depth in frequency and spectral optical depth in wavelength of a material, denoted ${\displaystyle \tau _{\nu }}$ and ${\displaystyle \tau _{\lambda }}$ respectively, are given by:[1]
${\displaystyle \tau _{\nu }=\ln \!\left({\frac {\Phi _{\mathrm {e} ,\nu }^{\mathrm {i} }}{\Phi _{\mathrm {e} ,\nu }^{\mathrm {t} }}}\right)=-\ln T_{\nu }}$
${\displaystyle \tau _{\lambda }=\ln \!\left({\frac {\Phi _{\mathrm {e} ,\lambda }^{\mathrm {i} }}{\Phi _{\mathrm {e} ,\lambda }^{\mathrm {t} }}}\right)=-\ln T_{\lambda },}$
where
• ${\displaystyle \Phi _{\mathrm {e} ,\nu }^{\mathrm {t} }}$ is the spectral radiant flux in frequency transmitted by that material;
• ${\displaystyle \Phi _{\mathrm {e} ,\nu }^{\mathrm {i} }}$ is the spectral radiant flux in frequency received by that material;
• ${\displaystyle T_{\nu }}$ is the spectral transmittance in frequency of that material;
• ${\displaystyle \Phi _{\mathrm {e} ,\lambda }^{\mathrm {t} }}$ is the spectral radiant flux in wavelength transmitted by that material;
• ${\displaystyle \Phi _{\mathrm {e} ,\lambda }^{\mathrm {i} }}$ is the spectral radiant flux in wavelength received by that material;
• ${\displaystyle T_{\lambda }}$ is the spectral transmittance in wavelength of that material.
Spectral absorbance is related to spectral optical depth by:
${\displaystyle \tau _{\nu }=A_{\nu }\ln 10,}$
${\displaystyle \tau _{\lambda }=A_{\lambda }\ln 10,}$
where
• ${\displaystyle A_{\nu }}$ is the spectral absorbance in frequency;
• ${\displaystyle A_{\lambda }}$ is the spectral absorbance in wavelength.
## Relationship with attenuation
### Attenuation
Optical depth measures the attenuation of the transmitted radiant power in a material. Attenuation can be caused by absorption, but also reflection, scattering, and other physical processes. Optical depth of a material is approximately equal to its attenuation when both the absorbance is much less than 1 and the emittance of that material (not to be confused with radiant exitance or emissivity) is much less than the optical depth:
${\displaystyle \Phi _{\mathrm {e} }^{\mathrm {t} }+\Phi _{\mathrm {e} }^{\mathrm {att} }=\Phi _{\mathrm {e} }^{\mathrm {i} }+\Phi _{\mathrm {e} }^{\mathrm {e} },}$
${\displaystyle T+ATT=1+E,}$
where
• Φet is the radiant power transmitted by that material;
• Φeatt is the radiant power attenuated by that material;
• Φee is the radiant power emitted by that material;
• T = Φetei is the transmittance of that material;
• ATT = Φeattei is the attenuation of that material;
• E = Φeeei is the emittance of that material,
and according to the Beer–Lambert law,
${\displaystyle T=e^{-\tau },}$
so:
${\displaystyle ATT=1-e^{-\tau }+E\approx \tau +E\approx \tau ,\quad {\text{if}}\ \tau \ll 1\ {\text{and}}\ E\ll \tau .}$
### Attenuation coefficient
Optical depth of a material is also related to its attenuation coefficient by:
${\displaystyle \tau =\int _{0}^{l}\alpha (z)\,\mathrm {d} z,}$
where
• l is the thickness of that material through which the light travels;
• α(z) is the attenuation coefficient or Napierian attenuation coefficient of that material at z,
and if α(z) is uniform along the path, the attenuation is said to be a linear attenuation and the relation becomes:
${\displaystyle \tau =\alpha l}$
Sometimes the relation is given using the attenuation cross section of the material, that is its attenuation coefficient divided by its number density:
${\displaystyle \tau =\int _{0}^{l}\sigma n(z)\,\mathrm {d} z,}$
where
• σ is the attenuation cross section of that material;
• n(z) is the number density of that material at z,
and if ${\displaystyle n}$ is uniform along the path, i.e., ${\displaystyle n(z)\equiv N}$ , the relation becomes:
${\displaystyle \tau =\sigma Nl}$
## Applications
### Atomic physics
In atomic physics, the spectral optical depth of a cloud of atoms can be calculated from the quantum-mechanical properties of the atoms. It is given by
${\displaystyle \tau _{\nu }={\frac {d^{2}n\nu }{2\mathrm {c} \hbar \varepsilon _{0}\sigma \gamma }}}$
where
### Atmospheric sciences
In atmospheric sciences, one often refers to the optical depth of the atmosphere as corresponding to the vertical path from Earth's surface to outer space; at other times the optical path is from the observer's altitude to outer space. The optical depth for a slant path is τ = , where τ′ refers to a vertical path, m is called the relative airmass, and for a plane-parallel atmosphere it is determined as m = sec θ where θ is the zenith angle corresponding to the given path. Therefore,
${\displaystyle T=e^{-\tau }=e^{-m\tau '}}$
The optical depth of the atmosphere can be divided into several components, ascribed to Rayleigh scattering, aerosols, and gaseous absorption. The optical depth of the atmosphere can be measured with a sun photometer.
The optical depth with respect to the height within the atmosphere is given by[3]
${\displaystyle \tau (z)=k_{a}w_{1}\rho _{0}He^{-z/H}}$
and it follows that the total atmospheric optical depth is given by[3]
${\displaystyle \tau (0)=k_{a}w_{1}\rho _{0}H}$
In both equations:
• ka is the absorption coefficient
• w1 is the mixing ratio
• ρ0 is the density of air at sea level
• H is the scale height of the atmosphere
• z is the height in question
The optical depth of a plane parallel cloud layer is given by[3]
${\displaystyle \tau =Q_{e}\left[{\frac {9\pi L^{2}HN}{16\rho _{l}^{2}}}\right]^{1/3}}$
where:
• Qe is the extinction efficiency
• L is the liquid water path
• H is the geometrical thickness
• N is the concentration of droplets
• ρl is the density of liquid water
So, with a fixed depth and total liquid water path, ${\textstyle \tau \propto N^{1/3}}$ .[3]
### Astronomy
In astronomy, the photosphere of a star is defined as the surface where its optical depth is 2/3. This means that each photon emitted at the photosphere suffers an average of less than one scattering before it reaches the observer. At the temperature at optical depth 2/3, the energy emitted by the star (the original derivation is for the Sun) matches the observed total energy emitted.[citation needed][clarification needed]
Note that the optical depth of a given medium will be different for different colors (wavelengths) of light.
For planetary rings, the optical depth is the (negative logarithm of the) proportion of light blocked by the ring when it lies between the source and the observer. This is usually obtained by observation of stellar occultations.
Mars dust storm – optical depth tau – May to September 2018
(Mars Climate Sounder; Mars Reconnaissance Orbiter)
(1:38; animation; 30 October 2018; file description) | 2,284 | 8,646 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 37, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2022-49 | longest | en | 0.850122 |
http://www.gradesaver.com/textbooks/math/calculus/calculus-early-transcendentals-2nd-edition/chapter-3-derivatives-3-7-the-chain-rule-3-7-exercises-page-192/62 | 1,524,305,360,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945111.79/warc/CC-MAIN-20180421090739-20180421110739-00556.warc.gz | 436,902,770 | 13,479 | ## Calculus: Early Transcendentals (2nd Edition)
$= \frac{{30\,{{\left( {3x} \right)}^4}}}{{\,{{\left( {4x + 2} \right)}^6}}}$
$\begin{gathered} y = \,{\left( {\frac{{3x}}{{4x + 2}}} \right)^5} \hfill \\ \hfill \\ differentiate \hfill \\ \hfill \\ {y^,} = \,{\left( {\frac{{3x}}{{4x + 2}}} \right)^5} \hfill \\ \hfill \\ use\,\,the\,Chain\,\,rule \hfill \\ \hfill \\ = 5\,{\left( {\frac{{3x}}{{4x + 2}}} \right)^4} \cdot \,\,{\left( {\frac{{3x}}{{4x + 2}}} \right)^,} \hfill \\ \hfill \\ Quotient\,\,rule \hfill \\ \hfill \\ = 5\,{\left( {\frac{{3x}}{{4x + 2}}} \right)^4} \cdot \,\frac{{3\,\left( {4x + 2} \right) - 4\,\left( {3x} \right)}}{{\,{{\left( {4x + 2} \right)}^2}}} \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ = 5\,{\left( {\frac{{3x}}{{4x + 2}}} \right)^4} \cdot \,\,\frac{6}{{\,{{\left( {4x + 2} \right)}^2}}} \hfill \\ \hfill \\ = \frac{{30\,{{\left( {3x} \right)}^4}}}{{\,{{\left( {4x + 2} \right)}^6}}} \hfill \\ \end{gathered}$ | 444 | 950 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2018-17 | latest | en | 0.402209 |
https://www.time24.news/2020/10/contest-2058-is-drawn-check-the-numbers.html | 1,603,563,138,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107884322.44/warc/CC-MAIN-20201024164841-20201024194841-00052.warc.gz | 925,738,982 | 19,053 | # contest 2058 is drawn; check the numbers
0
3
Caixa recently drew the 15 numbers of Lotofácil’s 2058 contest. The dozens that left the globe were: 01-05-07-08-10-12-13-14-15-17-20-21-22-23-24. The estimated value for those who put everything in the wheel is R \$ 1.5 million.
You will soon be able to follow on the Caixa website if there was a winner in this edition or if the prize will be accumulated and then increase if it is accumulated. In the contest held on Thursday (15), four winning bets took home the value of R \$ 803,932.83 each.
Earlier, the sixth edition of Super Sete was held, Caixa’s newest modality. The prize accumulated, according to the bank, and will have a prize of R \$ 1.5 million next Monday (19).
### How to participate in the next Lotofácil drawing?
You can place your bet on Lotofácil up to one hour before the draw in accredited lottery shops and on Caixa’s official website. That is, it is possible to register your game until 7 pm today. There are 25 numbers available, and bets are placed on games of 15 to 20 chosen tens. The smallest bet now costs R \$ 2.50, while the most expensive reaches more than R \$ 38.7 thousand. You can also use “Surprise” to let the system choose the numbers for you.
### What is the chance of winning the maximum Lotofácil prize?
With the minimum bet (R \$ 2.50), playing 15 tens, the chance of hitting them all is one in almost 3.3 million. Playing with a dozen more, the value of the game rises to R \$ 40, but the odds increase: they become one in just over 204 thousand. Whoever bets on 20 numbers increases the chance of winning to one in 211. Lotofácil also has prizes for those who get 14, 13, 12 and 11 points. With the lowest bet, the chance of winning at least the R \$ 5 prize is one in 11.
### How does Lotofácil draw?
Lotofácil also has a jackpot available for group bets. The minimum price charged by Caixa in this modality is R \$ 10, and the shares of each participant must start from R \$ 3. In bets with 15 numbers, the number of shares allowed varies between two and eight. With 20 tens, up to 100 quotas will be allowed. It is possible to place up to ten bets per pot between 15 and 18 tens. At 19, the number drops to six. Playing 20 numbers, only one bet is allowed. | 591 | 2,265 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2020-45 | latest | en | 0.932058 |
https://www.howmany.wiki/formula/calorie-%28thermochemical%29--to--calorie-%28IT%29 | 1,713,541,226,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817438.43/warc/CC-MAIN-20240419141145-20240419171145-00375.warc.gz | 724,628,976 | 29,189 | ## HowMany.wiki
Please get in touch with us if you:
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# Calories (thermochemical) to calories (IT) formula
Use the formula below to convert any value from calories (thermochemical) to calories (IT):
calories (IT) = calories (thermochemical) × 0.99933123148944
To convert from calories (thermochemical) to calorie (IT), you just need to multiply the value in calories (thermochemical) by 0.99933123148944. (It is called the conversion factor)
## Using the formula (some examples):
Convert full calorie (thermochemical) to calories (IT):
a calorie (thermochemical) = 1 × 0.99933123148944 = 0.99933123148944 calories (IT).
Convert two calorie (thermochemical) to calories (IT):
two calorie (thermochemical) = 2 × 0.99933123148944 = 1.9986624629789 calories (IT).
Convert five calories (thermochemical) to calories (IT):
5 calories (thermochemical) = 5 × 0.99933123148944 = 4.9966561574472 calories (IT).
## More Examples:
Convert ten calories (thermochemical) to calories (IT): 10 calories (thermochemical) = 10 × 0.99933123148944 = 9.9933123148944 calories (IT).
Convert twenty calories (thermochemical) to calories (IT): 20 calories (thermochemical) = 20 × 0.99933123148944 = 19.986624629789 calories (IT).
Convert fifty calories (thermochemical) to calories (IT): 50 calories (thermochemical) = 50 × 0.99933123148944 = 49.966561574472 calories (IT).
Convert a hundred calories (thermochemical) to calories (IT): 100 calories (thermochemical) = 100 × 0.99933123148944 = 99.933123148944 calories (IT).
Convert a thousand calories (thermochemical) to calories (IT): 1000 calories (thermochemical) = 1000 × 0.99933123148944 = 999.33123148944 calories (IT).
## More conversion Factors
### Disclaimer
Despite efforts to provide accurate information on this website, no guarantee of its accuracy is made. Therefore, the content should not be used for decisions regarding health, finances, or property. | 565 | 2,005 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2024-18 | latest | en | 0.761893 |
https://eepower.com/resistor-guide/resistor-fundamentals/electrical-resistivity/ | 1,719,294,194,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865560.33/warc/CC-MAIN-20240625041023-20240625071023-00468.warc.gz | 186,289,248 | 15,384 | # Electrical Resistivity
## What Is Electrical Resistivity?
Electrical resistivity is a measure of a material’s property to oppose the flow of electric current. This is expressed in Ohm-meters (Ω⋅m). The symbol of resistivity is usually the Greek letter ρ (rho). A high resistivity means that a material does not conduct electric charge well.
Electrical resistivity is defined as the relation between the electric field inside a material, and the electric current through it as a consequence:
$$\rho = \frac{E}{J}$$
in which ρ is the resistivity of the material (Ω⋅m),
E is the magnitude of the electrical field in the material (V/m),
J is the magnitude of the electric current density in the material (A/m2)
If the electric field (E) through a material is very large and the flow of current (J) is very small, it means that the material has a high resistivity.
Electrical conductivity is the inversion of resistivity and is a measure of how well a material conducts electric current:
$$\sigma = \frac{1}{\rho} = \frac{J}{E}$$
in which σ is the conductivity of the material expressed in siemens per meter (S/m). In electrical engineering, κ (kappa) is used often instead of σ.
## Electrical Resistance
Electrical resistance is expressed in ohms, and is not the same as resistivity. While resistivity is a material property, resistance is the property of an object. The electrical resistance of a resistor is determined by the combination of the shape and the resistivity of the material. For example, a wirewound resistor with a long, thin wire has a higher resistance than with a shorter and thicker wire. A wirewound resistor made from a material with high resistivity has a higher resistance value then one with a low resistivity. An analogy with a hydraulic system can be made, where water is pumped through a pipe. The longer and thinner the pipe, the higher the resistance will be. A pipe full with sand will resist the flow of water more than a pipe without sand (resistivity property).
Hydraulic analogy of electrical resistance
### Wire resistance
The resistance value of a wire depends on three parameters: resistivity, length and diameter. The formula to calculate wire resistance is as follows:
$$R = \rho (\frac{l}{A})$$
in which R is the resistance (Ω),
ρ is the resistivity of the material (Ω⋅m),
l is the length of the material (m),
A is the cross-sectional area of the material (m2)
The resistance value of a wire is dependent on three parameters; its resistivity, cross-sectional area and length.
As an example, consider a wirewound resistor with a wire of Nichrome with a resistivity of 1.10×10−6 Ω⋅m. The wire has a length of 1500 mm (1.5 m) and an area of 0.5 mm2 (0.5×10−6 m2). With these three parameters the resistance value is calculated:
$$R = \rho (\frac{l}{A}) = 1.1⋅10^{-6} ⋅ \frac{1.5}{0.5⋅10^{-6}} = 3.3 \Omega$$
Nichrome and Constantan are often used as resistance wire. Look in the table for material resistivity for commonly used materials.
### Sheet resistance
The resistance value for a sheet is calculated the exact same way as for wire resistance. The cross-sectional area can be written as the product of w and t:
$$R = \rho (\frac{l}{A}) = \rho (\frac{l}{wt})$$
For some applications like thin films, the ratio between resistivity and film thickness is called sheet resistance Rs:
$$R = \rho (\frac{l}{wt}) = R_s (\frac{l}{w})$$
in which Rs is in ohms/square (Ω/□). The film thickness needs to be uniform for this calculation.
Often resistor manufacturers increase resistance by cutting a pattern in the film to increase the path of the electric current.
Electrical resistance of a sheet depends on length, width, film thickness and resistivity. The resistance can be increased by cutting a pattern in the sheet.
## Resistive Properties of Materials
The resistivity of a material is dependent on the temperature and is normally given for room temperature (20°C). The change in resistivity as a result of temperature change is described by the temperature coefficient. For example, thermistors make use of this property to measure temperature. On the other hand, in precision electronics this is usually an unwanted effect. Metal foil resistors have excellent properties for temperature stability. This is achieved not only by the low resistivity of the material, but also by the mechanical design of the component.
Many different materials and alloys are used for resistors. Nichrome, an alloy of nickel and chromium, is often used as resistor wire material because of its high resistivity, and because it doesn’t oxidize at high temperatures. A disadvantage is that solder doesn’t adhere to it. Constantan, another popular material, is easily soldered and has a low temperature coefficient.
Material ρ (Ω⋅m) at 20°C σ (S/m) at 20°C Temperature coefficient (1/°C) x10-3 Silver 1.59×10−8 6.30×107 3.8 Copper 1.68×10−8 5.96×107 3.9 Gold 2.44×10−8 4.10×107 3.4 Aluminum 2.82×10−8 3.5×107 3.9 Tungsten 5.60×10−8 1.79×107 4.5 Zinc 5.90×10−8 1.69×107 3.7 Nickel 6.99×10−8 1.43×107 6 Lithium 9.28×10−8 1.08×107 6 Iron 1.0×10−7 1.00×107 5 Platinum 1.06×10−7 9.43×106 3.9 Tin 1.09×10−7 9.17×106 4.5 Lead 2.2×10−7 4.55×106 3.9 Manganin 4.82×10−7 2.07×106 0.002 Constantan 4.9×10−7 2.04×106 0.008 Mercury 9.8×10−7 1.02×106 0.9 Nichrome 1.10×10−6 9.09×105 0.4 Carbon (amorphous) 5×10−4 to 8×10−4 1.25 to 2×103 -0.5 | 1,471 | 5,380 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2024-26 | latest | en | 0.933248 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/5070/2/b/e/1351/2/ | 1,619,169,197,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039568689.89/warc/CC-MAIN-20210423070953-20210423100953-00276.warc.gz | 973,190,411 | 80,746 | # Properties
Label 5070.2.b.e.1351.2 Level $5070$ Weight $2$ Character 5070.1351 Analytic conductor $40.484$ Analytic rank $1$ Dimension $2$ CM no Inner twists $2$
# Related objects
## Newspace parameters
Level: $$N$$ $$=$$ $$5070 = 2 \cdot 3 \cdot 5 \cdot 13^{2}$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 5070.b (of order $$2$$, degree $$1$$, not minimal)
## Newform invariants
Self dual: no Analytic conductor: $$40.4841538248$$ Analytic rank: $$1$$ Dimension: $$2$$ Coefficient field: $$\Q(i)$$ Defining polynomial: $$x^{2} + 1$$ Coefficient ring: $$\Z[a_1, a_2]$$ Coefficient ring index: $$1$$ Twist minimal: no (minimal twist has level 390) Sato-Tate group: $\mathrm{SU}(2)[C_{2}]$
## Embedding invariants
Embedding label 1351.2 Root $$-1.00000i$$ of defining polynomial Character $$\chi$$ $$=$$ 5070.1351 Dual form 5070.2.b.e.1351.1
## $q$-expansion
$$f(q)$$ $$=$$ $$q+1.00000i q^{2} -1.00000 q^{3} -1.00000 q^{4} -1.00000i q^{5} -1.00000i q^{6} -2.00000i q^{7} -1.00000i q^{8} +1.00000 q^{9} +O(q^{10})$$ $$q+1.00000i q^{2} -1.00000 q^{3} -1.00000 q^{4} -1.00000i q^{5} -1.00000i q^{6} -2.00000i q^{7} -1.00000i q^{8} +1.00000 q^{9} +1.00000 q^{10} -4.00000i q^{11} +1.00000 q^{12} +2.00000 q^{14} +1.00000i q^{15} +1.00000 q^{16} -8.00000 q^{17} +1.00000i q^{18} -6.00000i q^{19} +1.00000i q^{20} +2.00000i q^{21} +4.00000 q^{22} -6.00000 q^{23} +1.00000i q^{24} -1.00000 q^{25} -1.00000 q^{27} +2.00000i q^{28} -4.00000 q^{29} -1.00000 q^{30} +1.00000i q^{32} +4.00000i q^{33} -8.00000i q^{34} -2.00000 q^{35} -1.00000 q^{36} +2.00000i q^{37} +6.00000 q^{38} -1.00000 q^{40} -2.00000i q^{41} -2.00000 q^{42} +4.00000 q^{43} +4.00000i q^{44} -1.00000i q^{45} -6.00000i q^{46} -1.00000 q^{48} +3.00000 q^{49} -1.00000i q^{50} +8.00000 q^{51} -10.0000 q^{53} -1.00000i q^{54} -4.00000 q^{55} -2.00000 q^{56} +6.00000i q^{57} -4.00000i q^{58} -4.00000i q^{59} -1.00000i q^{60} -10.0000 q^{61} -2.00000i q^{63} -1.00000 q^{64} -4.00000 q^{66} +12.0000i q^{67} +8.00000 q^{68} +6.00000 q^{69} -2.00000i q^{70} -8.00000i q^{71} -1.00000i q^{72} +8.00000i q^{73} -2.00000 q^{74} +1.00000 q^{75} +6.00000i q^{76} -8.00000 q^{77} +8.00000 q^{79} -1.00000i q^{80} +1.00000 q^{81} +2.00000 q^{82} +12.0000i q^{83} -2.00000i q^{84} +8.00000i q^{85} +4.00000i q^{86} +4.00000 q^{87} -4.00000 q^{88} +14.0000i q^{89} +1.00000 q^{90} +6.00000 q^{92} -6.00000 q^{95} -1.00000i q^{96} -16.0000i q^{97} +3.00000i q^{98} -4.00000i q^{99} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$2q - 2q^{3} - 2q^{4} + 2q^{9} + O(q^{10})$$ $$2q - 2q^{3} - 2q^{4} + 2q^{9} + 2q^{10} + 2q^{12} + 4q^{14} + 2q^{16} - 16q^{17} + 8q^{22} - 12q^{23} - 2q^{25} - 2q^{27} - 8q^{29} - 2q^{30} - 4q^{35} - 2q^{36} + 12q^{38} - 2q^{40} - 4q^{42} + 8q^{43} - 2q^{48} + 6q^{49} + 16q^{51} - 20q^{53} - 8q^{55} - 4q^{56} - 20q^{61} - 2q^{64} - 8q^{66} + 16q^{68} + 12q^{69} - 4q^{74} + 2q^{75} - 16q^{77} + 16q^{79} + 2q^{81} + 4q^{82} + 8q^{87} - 8q^{88} + 2q^{90} + 12q^{92} - 12q^{95} + O(q^{100})$$
## Character values
We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/5070\mathbb{Z}\right)^\times$$.
$$n$$ $$1691$$ $$1861$$ $$4057$$ $$\chi(n)$$ $$1$$ $$-1$$ $$1$$
## Coefficient data
For each $$n$$ we display the coefficients of the $$q$$-expansion $$a_n$$, the Satake parameters $$\alpha_p$$, and the Satake angles $$\theta_p = \textrm{Arg}(\alpha_p)$$.
Display $$a_p$$ with $$p$$ up to: 50 250 1000 Display $$a_n$$ with $$n$$ up to: 50 250 1000
$$n$$ $$a_n$$ $$a_n / n^{(k-1)/2}$$ $$\alpha_n$$ $$\theta_n$$
$$p$$ $$a_p$$ $$a_p / p^{(k-1)/2}$$ $$\alpha_p$$ $$\theta_p$$
$$2$$ 1.00000i 0.707107i
$$3$$ −1.00000 −0.577350
$$4$$ −1.00000 −0.500000
$$5$$ − 1.00000i − 0.447214i
$$6$$ − 1.00000i − 0.408248i
$$7$$ − 2.00000i − 0.755929i −0.925820 0.377964i $$-0.876624\pi$$
0.925820 0.377964i $$-0.123376\pi$$
$$8$$ − 1.00000i − 0.353553i
$$9$$ 1.00000 0.333333
$$10$$ 1.00000 0.316228
$$11$$ − 4.00000i − 1.20605i −0.797724 0.603023i $$-0.793963\pi$$
0.797724 0.603023i $$-0.206037\pi$$
$$12$$ 1.00000 0.288675
$$13$$ 0 0
$$14$$ 2.00000 0.534522
$$15$$ 1.00000i 0.258199i
$$16$$ 1.00000 0.250000
$$17$$ −8.00000 −1.94029 −0.970143 0.242536i $$-0.922021\pi$$
−0.970143 + 0.242536i $$0.922021\pi$$
$$18$$ 1.00000i 0.235702i
$$19$$ − 6.00000i − 1.37649i −0.725476 0.688247i $$-0.758380\pi$$
0.725476 0.688247i $$-0.241620\pi$$
$$20$$ 1.00000i 0.223607i
$$21$$ 2.00000i 0.436436i
$$22$$ 4.00000 0.852803
$$23$$ −6.00000 −1.25109 −0.625543 0.780189i $$-0.715123\pi$$
−0.625543 + 0.780189i $$0.715123\pi$$
$$24$$ 1.00000i 0.204124i
$$25$$ −1.00000 −0.200000
$$26$$ 0 0
$$27$$ −1.00000 −0.192450
$$28$$ 2.00000i 0.377964i
$$29$$ −4.00000 −0.742781 −0.371391 0.928477i $$-0.621119\pi$$
−0.371391 + 0.928477i $$0.621119\pi$$
$$30$$ −1.00000 −0.182574
$$31$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$32$$ 1.00000i 0.176777i
$$33$$ 4.00000i 0.696311i
$$34$$ − 8.00000i − 1.37199i
$$35$$ −2.00000 −0.338062
$$36$$ −1.00000 −0.166667
$$37$$ 2.00000i 0.328798i 0.986394 + 0.164399i $$0.0525685\pi$$
−0.986394 + 0.164399i $$0.947432\pi$$
$$38$$ 6.00000 0.973329
$$39$$ 0 0
$$40$$ −1.00000 −0.158114
$$41$$ − 2.00000i − 0.312348i −0.987730 0.156174i $$-0.950084\pi$$
0.987730 0.156174i $$-0.0499160\pi$$
$$42$$ −2.00000 −0.308607
$$43$$ 4.00000 0.609994 0.304997 0.952353i $$-0.401344\pi$$
0.304997 + 0.952353i $$0.401344\pi$$
$$44$$ 4.00000i 0.603023i
$$45$$ − 1.00000i − 0.149071i
$$46$$ − 6.00000i − 0.884652i
$$47$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$48$$ −1.00000 −0.144338
$$49$$ 3.00000 0.428571
$$50$$ − 1.00000i − 0.141421i
$$51$$ 8.00000 1.12022
$$52$$ 0 0
$$53$$ −10.0000 −1.37361 −0.686803 0.726844i $$-0.740986\pi$$
−0.686803 + 0.726844i $$0.740986\pi$$
$$54$$ − 1.00000i − 0.136083i
$$55$$ −4.00000 −0.539360
$$56$$ −2.00000 −0.267261
$$57$$ 6.00000i 0.794719i
$$58$$ − 4.00000i − 0.525226i
$$59$$ − 4.00000i − 0.520756i −0.965507 0.260378i $$-0.916153\pi$$
0.965507 0.260378i $$-0.0838471\pi$$
$$60$$ − 1.00000i − 0.129099i
$$61$$ −10.0000 −1.28037 −0.640184 0.768221i $$-0.721142\pi$$
−0.640184 + 0.768221i $$0.721142\pi$$
$$62$$ 0 0
$$63$$ − 2.00000i − 0.251976i
$$64$$ −1.00000 −0.125000
$$65$$ 0 0
$$66$$ −4.00000 −0.492366
$$67$$ 12.0000i 1.46603i 0.680211 + 0.733017i $$0.261888\pi$$
−0.680211 + 0.733017i $$0.738112\pi$$
$$68$$ 8.00000 0.970143
$$69$$ 6.00000 0.722315
$$70$$ − 2.00000i − 0.239046i
$$71$$ − 8.00000i − 0.949425i −0.880141 0.474713i $$-0.842552\pi$$
0.880141 0.474713i $$-0.157448\pi$$
$$72$$ − 1.00000i − 0.117851i
$$73$$ 8.00000i 0.936329i 0.883641 + 0.468165i $$0.155085\pi$$
−0.883641 + 0.468165i $$0.844915\pi$$
$$74$$ −2.00000 −0.232495
$$75$$ 1.00000 0.115470
$$76$$ 6.00000i 0.688247i
$$77$$ −8.00000 −0.911685
$$78$$ 0 0
$$79$$ 8.00000 0.900070 0.450035 0.893011i $$-0.351411\pi$$
0.450035 + 0.893011i $$0.351411\pi$$
$$80$$ − 1.00000i − 0.111803i
$$81$$ 1.00000 0.111111
$$82$$ 2.00000 0.220863
$$83$$ 12.0000i 1.31717i 0.752506 + 0.658586i $$0.228845\pi$$
−0.752506 + 0.658586i $$0.771155\pi$$
$$84$$ − 2.00000i − 0.218218i
$$85$$ 8.00000i 0.867722i
$$86$$ 4.00000i 0.431331i
$$87$$ 4.00000 0.428845
$$88$$ −4.00000 −0.426401
$$89$$ 14.0000i 1.48400i 0.670402 + 0.741999i $$0.266122\pi$$
−0.670402 + 0.741999i $$0.733878\pi$$
$$90$$ 1.00000 0.105409
$$91$$ 0 0
$$92$$ 6.00000 0.625543
$$93$$ 0 0
$$94$$ 0 0
$$95$$ −6.00000 −0.615587
$$96$$ − 1.00000i − 0.102062i
$$97$$ − 16.0000i − 1.62455i −0.583272 0.812277i $$-0.698228\pi$$
0.583272 0.812277i $$-0.301772\pi$$
$$98$$ 3.00000i 0.303046i
$$99$$ − 4.00000i − 0.402015i
$$100$$ 1.00000 0.100000
$$101$$ 16.0000 1.59206 0.796030 0.605257i $$-0.206930\pi$$
0.796030 + 0.605257i $$0.206930\pi$$
$$102$$ 8.00000i 0.792118i
$$103$$ 12.0000 1.18240 0.591198 0.806527i $$-0.298655\pi$$
0.591198 + 0.806527i $$0.298655\pi$$
$$104$$ 0 0
$$105$$ 2.00000 0.195180
$$106$$ − 10.0000i − 0.971286i
$$107$$ 12.0000 1.16008 0.580042 0.814587i $$-0.303036\pi$$
0.580042 + 0.814587i $$0.303036\pi$$
$$108$$ 1.00000 0.0962250
$$109$$ 12.0000i 1.14939i 0.818367 + 0.574696i $$0.194880\pi$$
−0.818367 + 0.574696i $$0.805120\pi$$
$$110$$ − 4.00000i − 0.381385i
$$111$$ − 2.00000i − 0.189832i
$$112$$ − 2.00000i − 0.188982i
$$113$$ 20.0000 1.88144 0.940721 0.339182i $$-0.110150\pi$$
0.940721 + 0.339182i $$0.110150\pi$$
$$114$$ −6.00000 −0.561951
$$115$$ 6.00000i 0.559503i
$$116$$ 4.00000 0.371391
$$117$$ 0 0
$$118$$ 4.00000 0.368230
$$119$$ 16.0000i 1.46672i
$$120$$ 1.00000 0.0912871
$$121$$ −5.00000 −0.454545
$$122$$ − 10.0000i − 0.905357i
$$123$$ 2.00000i 0.180334i
$$124$$ 0 0
$$125$$ 1.00000i 0.0894427i
$$126$$ 2.00000 0.178174
$$127$$ −4.00000 −0.354943 −0.177471 0.984126i $$-0.556792\pi$$
−0.177471 + 0.984126i $$0.556792\pi$$
$$128$$ − 1.00000i − 0.0883883i
$$129$$ −4.00000 −0.352180
$$130$$ 0 0
$$131$$ 10.0000 0.873704 0.436852 0.899533i $$-0.356093\pi$$
0.436852 + 0.899533i $$0.356093\pi$$
$$132$$ − 4.00000i − 0.348155i
$$133$$ −12.0000 −1.04053
$$134$$ −12.0000 −1.03664
$$135$$ 1.00000i 0.0860663i
$$136$$ 8.00000i 0.685994i
$$137$$ 6.00000i 0.512615i 0.966595 + 0.256307i $$0.0825059\pi$$
−0.966595 + 0.256307i $$0.917494\pi$$
$$138$$ 6.00000i 0.510754i
$$139$$ −8.00000 −0.678551 −0.339276 0.940687i $$-0.610182\pi$$
−0.339276 + 0.940687i $$0.610182\pi$$
$$140$$ 2.00000 0.169031
$$141$$ 0 0
$$142$$ 8.00000 0.671345
$$143$$ 0 0
$$144$$ 1.00000 0.0833333
$$145$$ 4.00000i 0.332182i
$$146$$ −8.00000 −0.662085
$$147$$ −3.00000 −0.247436
$$148$$ − 2.00000i − 0.164399i
$$149$$ − 10.0000i − 0.819232i −0.912258 0.409616i $$-0.865663\pi$$
0.912258 0.409616i $$-0.134337\pi$$
$$150$$ 1.00000i 0.0816497i
$$151$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$152$$ −6.00000 −0.486664
$$153$$ −8.00000 −0.646762
$$154$$ − 8.00000i − 0.644658i
$$155$$ 0 0
$$156$$ 0 0
$$157$$ 22.0000 1.75579 0.877896 0.478852i $$-0.158947\pi$$
0.877896 + 0.478852i $$0.158947\pi$$
$$158$$ 8.00000i 0.636446i
$$159$$ 10.0000 0.793052
$$160$$ 1.00000 0.0790569
$$161$$ 12.0000i 0.945732i
$$162$$ 1.00000i 0.0785674i
$$163$$ 16.0000i 1.25322i 0.779334 + 0.626608i $$0.215557\pi$$
−0.779334 + 0.626608i $$0.784443\pi$$
$$164$$ 2.00000i 0.156174i
$$165$$ 4.00000 0.311400
$$166$$ −12.0000 −0.931381
$$167$$ 4.00000i 0.309529i 0.987951 + 0.154765i $$0.0494619\pi$$
−0.987951 + 0.154765i $$0.950538\pi$$
$$168$$ 2.00000 0.154303
$$169$$ 0 0
$$170$$ −8.00000 −0.613572
$$171$$ − 6.00000i − 0.458831i
$$172$$ −4.00000 −0.304997
$$173$$ −22.0000 −1.67263 −0.836315 0.548250i $$-0.815294\pi$$
−0.836315 + 0.548250i $$0.815294\pi$$
$$174$$ 4.00000i 0.303239i
$$175$$ 2.00000i 0.151186i
$$176$$ − 4.00000i − 0.301511i
$$177$$ 4.00000i 0.300658i
$$178$$ −14.0000 −1.04934
$$179$$ 10.0000 0.747435 0.373718 0.927543i $$-0.378083\pi$$
0.373718 + 0.927543i $$0.378083\pi$$
$$180$$ 1.00000i 0.0745356i
$$181$$ −10.0000 −0.743294 −0.371647 0.928374i $$-0.621207\pi$$
−0.371647 + 0.928374i $$0.621207\pi$$
$$182$$ 0 0
$$183$$ 10.0000 0.739221
$$184$$ 6.00000i 0.442326i
$$185$$ 2.00000 0.147043
$$186$$ 0 0
$$187$$ 32.0000i 2.34007i
$$188$$ 0 0
$$189$$ 2.00000i 0.145479i
$$190$$ − 6.00000i − 0.435286i
$$191$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$192$$ 1.00000 0.0721688
$$193$$ 4.00000i 0.287926i 0.989583 + 0.143963i $$0.0459847\pi$$
−0.989583 + 0.143963i $$0.954015\pi$$
$$194$$ 16.0000 1.14873
$$195$$ 0 0
$$196$$ −3.00000 −0.214286
$$197$$ − 18.0000i − 1.28245i −0.767354 0.641223i $$-0.778427\pi$$
0.767354 0.641223i $$-0.221573\pi$$
$$198$$ 4.00000 0.284268
$$199$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$200$$ 1.00000i 0.0707107i
$$201$$ − 12.0000i − 0.846415i
$$202$$ 16.0000i 1.12576i
$$203$$ 8.00000i 0.561490i
$$204$$ −8.00000 −0.560112
$$205$$ −2.00000 −0.139686
$$206$$ 12.0000i 0.836080i
$$207$$ −6.00000 −0.417029
$$208$$ 0 0
$$209$$ −24.0000 −1.66011
$$210$$ 2.00000i 0.138013i
$$211$$ −20.0000 −1.37686 −0.688428 0.725304i $$-0.741699\pi$$
−0.688428 + 0.725304i $$0.741699\pi$$
$$212$$ 10.0000 0.686803
$$213$$ 8.00000i 0.548151i
$$214$$ 12.0000i 0.820303i
$$215$$ − 4.00000i − 0.272798i
$$216$$ 1.00000i 0.0680414i
$$217$$ 0 0
$$218$$ −12.0000 −0.812743
$$219$$ − 8.00000i − 0.540590i
$$220$$ 4.00000 0.269680
$$221$$ 0 0
$$222$$ 2.00000 0.134231
$$223$$ − 2.00000i − 0.133930i −0.997755 0.0669650i $$-0.978668\pi$$
0.997755 0.0669650i $$-0.0213316\pi$$
$$224$$ 2.00000 0.133631
$$225$$ −1.00000 −0.0666667
$$226$$ 20.0000i 1.33038i
$$227$$ − 4.00000i − 0.265489i −0.991150 0.132745i $$-0.957621\pi$$
0.991150 0.132745i $$-0.0423790\pi$$
$$228$$ − 6.00000i − 0.397360i
$$229$$ − 4.00000i − 0.264327i −0.991228 0.132164i $$-0.957808\pi$$
0.991228 0.132164i $$-0.0421925\pi$$
$$230$$ −6.00000 −0.395628
$$231$$ 8.00000 0.526361
$$232$$ 4.00000i 0.262613i
$$233$$ −24.0000 −1.57229 −0.786146 0.618041i $$-0.787927\pi$$
−0.786146 + 0.618041i $$0.787927\pi$$
$$234$$ 0 0
$$235$$ 0 0
$$236$$ 4.00000i 0.260378i
$$237$$ −8.00000 −0.519656
$$238$$ −16.0000 −1.03713
$$239$$ 16.0000i 1.03495i 0.855697 + 0.517477i $$0.173129\pi$$
−0.855697 + 0.517477i $$0.826871\pi$$
$$240$$ 1.00000i 0.0645497i
$$241$$ − 2.00000i − 0.128831i −0.997923 0.0644157i $$-0.979482\pi$$
0.997923 0.0644157i $$-0.0205183\pi$$
$$242$$ − 5.00000i − 0.321412i
$$243$$ −1.00000 −0.0641500
$$244$$ 10.0000 0.640184
$$245$$ − 3.00000i − 0.191663i
$$246$$ −2.00000 −0.127515
$$247$$ 0 0
$$248$$ 0 0
$$249$$ − 12.0000i − 0.760469i
$$250$$ −1.00000 −0.0632456
$$251$$ −6.00000 −0.378717 −0.189358 0.981908i $$-0.560641\pi$$
−0.189358 + 0.981908i $$0.560641\pi$$
$$252$$ 2.00000i 0.125988i
$$253$$ 24.0000i 1.50887i
$$254$$ − 4.00000i − 0.250982i
$$255$$ − 8.00000i − 0.500979i
$$256$$ 1.00000 0.0625000
$$257$$ −12.0000 −0.748539 −0.374270 0.927320i $$-0.622107\pi$$
−0.374270 + 0.927320i $$0.622107\pi$$
$$258$$ − 4.00000i − 0.249029i
$$259$$ 4.00000 0.248548
$$260$$ 0 0
$$261$$ −4.00000 −0.247594
$$262$$ 10.0000i 0.617802i
$$263$$ 2.00000 0.123325 0.0616626 0.998097i $$-0.480360\pi$$
0.0616626 + 0.998097i $$0.480360\pi$$
$$264$$ 4.00000 0.246183
$$265$$ 10.0000i 0.614295i
$$266$$ − 12.0000i − 0.735767i
$$267$$ − 14.0000i − 0.856786i
$$268$$ − 12.0000i − 0.733017i
$$269$$ −24.0000 −1.46331 −0.731653 0.681677i $$-0.761251\pi$$
−0.731653 + 0.681677i $$0.761251\pi$$
$$270$$ −1.00000 −0.0608581
$$271$$ − 16.0000i − 0.971931i −0.873978 0.485965i $$-0.838468\pi$$
0.873978 0.485965i $$-0.161532\pi$$
$$272$$ −8.00000 −0.485071
$$273$$ 0 0
$$274$$ −6.00000 −0.362473
$$275$$ 4.00000i 0.241209i
$$276$$ −6.00000 −0.361158
$$277$$ 10.0000 0.600842 0.300421 0.953807i $$-0.402873\pi$$
0.300421 + 0.953807i $$0.402873\pi$$
$$278$$ − 8.00000i − 0.479808i
$$279$$ 0 0
$$280$$ 2.00000i 0.119523i
$$281$$ 6.00000i 0.357930i 0.983855 + 0.178965i $$0.0572749\pi$$
−0.983855 + 0.178965i $$0.942725\pi$$
$$282$$ 0 0
$$283$$ 4.00000 0.237775 0.118888 0.992908i $$-0.462067\pi$$
0.118888 + 0.992908i $$0.462067\pi$$
$$284$$ 8.00000i 0.474713i
$$285$$ 6.00000 0.355409
$$286$$ 0 0
$$287$$ −4.00000 −0.236113
$$288$$ 1.00000i 0.0589256i
$$289$$ 47.0000 2.76471
$$290$$ −4.00000 −0.234888
$$291$$ 16.0000i 0.937937i
$$292$$ − 8.00000i − 0.468165i
$$293$$ 26.0000i 1.51894i 0.650545 + 0.759468i $$0.274541\pi$$
−0.650545 + 0.759468i $$0.725459\pi$$
$$294$$ − 3.00000i − 0.174964i
$$295$$ −4.00000 −0.232889
$$296$$ 2.00000 0.116248
$$297$$ 4.00000i 0.232104i
$$298$$ 10.0000 0.579284
$$299$$ 0 0
$$300$$ −1.00000 −0.0577350
$$301$$ − 8.00000i − 0.461112i
$$302$$ 0 0
$$303$$ −16.0000 −0.919176
$$304$$ − 6.00000i − 0.344124i
$$305$$ 10.0000i 0.572598i
$$306$$ − 8.00000i − 0.457330i
$$307$$ − 12.0000i − 0.684876i −0.939540 0.342438i $$-0.888747\pi$$
0.939540 0.342438i $$-0.111253\pi$$
$$308$$ 8.00000 0.455842
$$309$$ −12.0000 −0.682656
$$310$$ 0 0
$$311$$ −24.0000 −1.36092 −0.680458 0.732787i $$-0.738219\pi$$
−0.680458 + 0.732787i $$0.738219\pi$$
$$312$$ 0 0
$$313$$ 6.00000 0.339140 0.169570 0.985518i $$-0.445762\pi$$
0.169570 + 0.985518i $$0.445762\pi$$
$$314$$ 22.0000i 1.24153i
$$315$$ −2.00000 −0.112687
$$316$$ −8.00000 −0.450035
$$317$$ − 30.0000i − 1.68497i −0.538721 0.842484i $$-0.681092\pi$$
0.538721 0.842484i $$-0.318908\pi$$
$$318$$ 10.0000i 0.560772i
$$319$$ 16.0000i 0.895828i
$$320$$ 1.00000i 0.0559017i
$$321$$ −12.0000 −0.669775
$$322$$ −12.0000 −0.668734
$$323$$ 48.0000i 2.67079i
$$324$$ −1.00000 −0.0555556
$$325$$ 0 0
$$326$$ −16.0000 −0.886158
$$327$$ − 12.0000i − 0.663602i
$$328$$ −2.00000 −0.110432
$$329$$ 0 0
$$330$$ 4.00000i 0.220193i
$$331$$ 10.0000i 0.549650i 0.961494 + 0.274825i $$0.0886199\pi$$
−0.961494 + 0.274825i $$0.911380\pi$$
$$332$$ − 12.0000i − 0.658586i
$$333$$ 2.00000i 0.109599i
$$334$$ −4.00000 −0.218870
$$335$$ 12.0000 0.655630
$$336$$ 2.00000i 0.109109i
$$337$$ −14.0000 −0.762629 −0.381314 0.924445i $$-0.624528\pi$$
−0.381314 + 0.924445i $$0.624528\pi$$
$$338$$ 0 0
$$339$$ −20.0000 −1.08625
$$340$$ − 8.00000i − 0.433861i
$$341$$ 0 0
$$342$$ 6.00000 0.324443
$$343$$ − 20.0000i − 1.07990i
$$344$$ − 4.00000i − 0.215666i
$$345$$ − 6.00000i − 0.323029i
$$346$$ − 22.0000i − 1.18273i
$$347$$ −24.0000 −1.28839 −0.644194 0.764862i $$-0.722807\pi$$
−0.644194 + 0.764862i $$0.722807\pi$$
$$348$$ −4.00000 −0.214423
$$349$$ 28.0000i 1.49881i 0.662114 + 0.749403i $$0.269659\pi$$
−0.662114 + 0.749403i $$0.730341\pi$$
$$350$$ −2.00000 −0.106904
$$351$$ 0 0
$$352$$ 4.00000 0.213201
$$353$$ − 14.0000i − 0.745145i −0.928003 0.372572i $$-0.878476\pi$$
0.928003 0.372572i $$-0.121524\pi$$
$$354$$ −4.00000 −0.212598
$$355$$ −8.00000 −0.424596
$$356$$ − 14.0000i − 0.741999i
$$357$$ − 16.0000i − 0.846810i
$$358$$ 10.0000i 0.528516i
$$359$$ 24.0000i 1.26667i 0.773877 + 0.633336i $$0.218315\pi$$
−0.773877 + 0.633336i $$0.781685\pi$$
$$360$$ −1.00000 −0.0527046
$$361$$ −17.0000 −0.894737
$$362$$ − 10.0000i − 0.525588i
$$363$$ 5.00000 0.262432
$$364$$ 0 0
$$365$$ 8.00000 0.418739
$$366$$ 10.0000i 0.522708i
$$367$$ −36.0000 −1.87918 −0.939592 0.342296i $$-0.888796\pi$$
−0.939592 + 0.342296i $$0.888796\pi$$
$$368$$ −6.00000 −0.312772
$$369$$ − 2.00000i − 0.104116i
$$370$$ 2.00000i 0.103975i
$$371$$ 20.0000i 1.03835i
$$372$$ 0 0
$$373$$ −2.00000 −0.103556 −0.0517780 0.998659i $$-0.516489\pi$$
−0.0517780 + 0.998659i $$0.516489\pi$$
$$374$$ −32.0000 −1.65468
$$375$$ − 1.00000i − 0.0516398i
$$376$$ 0 0
$$377$$ 0 0
$$378$$ −2.00000 −0.102869
$$379$$ − 18.0000i − 0.924598i −0.886724 0.462299i $$-0.847025\pi$$
0.886724 0.462299i $$-0.152975\pi$$
$$380$$ 6.00000 0.307794
$$381$$ 4.00000 0.204926
$$382$$ 0 0
$$383$$ − 12.0000i − 0.613171i −0.951843 0.306586i $$-0.900813\pi$$
0.951843 0.306586i $$-0.0991866\pi$$
$$384$$ 1.00000i 0.0510310i
$$385$$ 8.00000i 0.407718i
$$386$$ −4.00000 −0.203595
$$387$$ 4.00000 0.203331
$$388$$ 16.0000i 0.812277i
$$389$$ −8.00000 −0.405616 −0.202808 0.979219i $$-0.565007\pi$$
−0.202808 + 0.979219i $$0.565007\pi$$
$$390$$ 0 0
$$391$$ 48.0000 2.42746
$$392$$ − 3.00000i − 0.151523i
$$393$$ −10.0000 −0.504433
$$394$$ 18.0000 0.906827
$$395$$ − 8.00000i − 0.402524i
$$396$$ 4.00000i 0.201008i
$$397$$ − 14.0000i − 0.702640i −0.936255 0.351320i $$-0.885733\pi$$
0.936255 0.351320i $$-0.114267\pi$$
$$398$$ 0 0
$$399$$ 12.0000 0.600751
$$400$$ −1.00000 −0.0500000
$$401$$ − 30.0000i − 1.49813i −0.662497 0.749064i $$-0.730503\pi$$
0.662497 0.749064i $$-0.269497\pi$$
$$402$$ 12.0000 0.598506
$$403$$ 0 0
$$404$$ −16.0000 −0.796030
$$405$$ − 1.00000i − 0.0496904i
$$406$$ −8.00000 −0.397033
$$407$$ 8.00000 0.396545
$$408$$ − 8.00000i − 0.396059i
$$409$$ − 26.0000i − 1.28562i −0.766027 0.642809i $$-0.777769\pi$$
0.766027 0.642809i $$-0.222231\pi$$
$$410$$ − 2.00000i − 0.0987730i
$$411$$ − 6.00000i − 0.295958i
$$412$$ −12.0000 −0.591198
$$413$$ −8.00000 −0.393654
$$414$$ − 6.00000i − 0.294884i
$$415$$ 12.0000 0.589057
$$416$$ 0 0
$$417$$ 8.00000 0.391762
$$418$$ − 24.0000i − 1.17388i
$$419$$ 10.0000 0.488532 0.244266 0.969708i $$-0.421453\pi$$
0.244266 + 0.969708i $$0.421453\pi$$
$$420$$ −2.00000 −0.0975900
$$421$$ − 8.00000i − 0.389896i −0.980814 0.194948i $$-0.937546\pi$$
0.980814 0.194948i $$-0.0624538\pi$$
$$422$$ − 20.0000i − 0.973585i
$$423$$ 0 0
$$424$$ 10.0000i 0.485643i
$$425$$ 8.00000 0.388057
$$426$$ −8.00000 −0.387601
$$427$$ 20.0000i 0.967868i
$$428$$ −12.0000 −0.580042
$$429$$ 0 0
$$430$$ 4.00000 0.192897
$$431$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$432$$ −1.00000 −0.0481125
$$433$$ 2.00000 0.0961139 0.0480569 0.998845i $$-0.484697\pi$$
0.0480569 + 0.998845i $$0.484697\pi$$
$$434$$ 0 0
$$435$$ − 4.00000i − 0.191785i
$$436$$ − 12.0000i − 0.574696i
$$437$$ 36.0000i 1.72211i
$$438$$ 8.00000 0.382255
$$439$$ 32.0000 1.52728 0.763638 0.645644i $$-0.223411\pi$$
0.763638 + 0.645644i $$0.223411\pi$$
$$440$$ 4.00000i 0.190693i
$$441$$ 3.00000 0.142857
$$442$$ 0 0
$$443$$ 16.0000 0.760183 0.380091 0.924949i $$-0.375893\pi$$
0.380091 + 0.924949i $$0.375893\pi$$
$$444$$ 2.00000i 0.0949158i
$$445$$ 14.0000 0.663664
$$446$$ 2.00000 0.0947027
$$447$$ 10.0000i 0.472984i
$$448$$ 2.00000i 0.0944911i
$$449$$ 6.00000i 0.283158i 0.989927 + 0.141579i $$0.0452178\pi$$
−0.989927 + 0.141579i $$0.954782\pi$$
$$450$$ − 1.00000i − 0.0471405i
$$451$$ −8.00000 −0.376705
$$452$$ −20.0000 −0.940721
$$453$$ 0 0
$$454$$ 4.00000 0.187729
$$455$$ 0 0
$$456$$ 6.00000 0.280976
$$457$$ 8.00000i 0.374224i 0.982339 + 0.187112i $$0.0599128\pi$$
−0.982339 + 0.187112i $$0.940087\pi$$
$$458$$ 4.00000 0.186908
$$459$$ 8.00000 0.373408
$$460$$ − 6.00000i − 0.279751i
$$461$$ − 6.00000i − 0.279448i −0.990190 0.139724i $$-0.955378\pi$$
0.990190 0.139724i $$-0.0446215\pi$$
$$462$$ 8.00000i 0.372194i
$$463$$ 26.0000i 1.20832i 0.796862 + 0.604161i $$0.206492\pi$$
−0.796862 + 0.604161i $$0.793508\pi$$
$$464$$ −4.00000 −0.185695
$$465$$ 0 0
$$466$$ − 24.0000i − 1.11178i
$$467$$ −28.0000 −1.29569 −0.647843 0.761774i $$-0.724329\pi$$
−0.647843 + 0.761774i $$0.724329\pi$$
$$468$$ 0 0
$$469$$ 24.0000 1.10822
$$470$$ 0 0
$$471$$ −22.0000 −1.01371
$$472$$ −4.00000 −0.184115
$$473$$ − 16.0000i − 0.735681i
$$474$$ − 8.00000i − 0.367452i
$$475$$ 6.00000i 0.275299i
$$476$$ − 16.0000i − 0.733359i
$$477$$ −10.0000 −0.457869
$$478$$ −16.0000 −0.731823
$$479$$ − 32.0000i − 1.46212i −0.682315 0.731059i $$-0.739027\pi$$
0.682315 0.731059i $$-0.260973\pi$$
$$480$$ −1.00000 −0.0456435
$$481$$ 0 0
$$482$$ 2.00000 0.0910975
$$483$$ − 12.0000i − 0.546019i
$$484$$ 5.00000 0.227273
$$485$$ −16.0000 −0.726523
$$486$$ − 1.00000i − 0.0453609i
$$487$$ 26.0000i 1.17817i 0.808070 + 0.589086i $$0.200512\pi$$
−0.808070 + 0.589086i $$0.799488\pi$$
$$488$$ 10.0000i 0.452679i
$$489$$ − 16.0000i − 0.723545i
$$490$$ 3.00000 0.135526
$$491$$ −42.0000 −1.89543 −0.947717 0.319113i $$-0.896615\pi$$
−0.947717 + 0.319113i $$0.896615\pi$$
$$492$$ − 2.00000i − 0.0901670i
$$493$$ 32.0000 1.44121
$$494$$ 0 0
$$495$$ −4.00000 −0.179787
$$496$$ 0 0
$$497$$ −16.0000 −0.717698
$$498$$ 12.0000 0.537733
$$499$$ 38.0000i 1.70111i 0.525883 + 0.850557i $$0.323735\pi$$
−0.525883 + 0.850557i $$0.676265\pi$$
$$500$$ − 1.00000i − 0.0447214i
$$501$$ − 4.00000i − 0.178707i
$$502$$ − 6.00000i − 0.267793i
$$503$$ 10.0000 0.445878 0.222939 0.974832i $$-0.428435\pi$$
0.222939 + 0.974832i $$0.428435\pi$$
$$504$$ −2.00000 −0.0890871
$$505$$ − 16.0000i − 0.711991i
$$506$$ −24.0000 −1.06693
$$507$$ 0 0
$$508$$ 4.00000 0.177471
$$509$$ 18.0000i 0.797836i 0.916987 + 0.398918i $$0.130614\pi$$
−0.916987 + 0.398918i $$0.869386\pi$$
$$510$$ 8.00000 0.354246
$$511$$ 16.0000 0.707798
$$512$$ 1.00000i 0.0441942i
$$513$$ 6.00000i 0.264906i
$$514$$ − 12.0000i − 0.529297i
$$515$$ − 12.0000i − 0.528783i
$$516$$ 4.00000 0.176090
$$517$$ 0 0
$$518$$ 4.00000i 0.175750i
$$519$$ 22.0000 0.965693
$$520$$ 0 0
$$521$$ −26.0000 −1.13908 −0.569540 0.821963i $$-0.692879\pi$$
−0.569540 + 0.821963i $$0.692879\pi$$
$$522$$ − 4.00000i − 0.175075i
$$523$$ 36.0000 1.57417 0.787085 0.616844i $$-0.211589\pi$$
0.787085 + 0.616844i $$0.211589\pi$$
$$524$$ −10.0000 −0.436852
$$525$$ − 2.00000i − 0.0872872i
$$526$$ 2.00000i 0.0872041i
$$527$$ 0 0
$$528$$ 4.00000i 0.174078i
$$529$$ 13.0000 0.565217
$$530$$ −10.0000 −0.434372
$$531$$ − 4.00000i − 0.173585i
$$532$$ 12.0000 0.520266
$$533$$ 0 0
$$534$$ 14.0000 0.605839
$$535$$ − 12.0000i − 0.518805i
$$536$$ 12.0000 0.518321
$$537$$ −10.0000 −0.431532
$$538$$ − 24.0000i − 1.03471i
$$539$$ − 12.0000i − 0.516877i
$$540$$ − 1.00000i − 0.0430331i
$$541$$ 8.00000i 0.343947i 0.985102 + 0.171973i $$0.0550143\pi$$
−0.985102 + 0.171973i $$0.944986\pi$$
$$542$$ 16.0000 0.687259
$$543$$ 10.0000 0.429141
$$544$$ − 8.00000i − 0.342997i
$$545$$ 12.0000 0.514024
$$546$$ 0 0
$$547$$ −20.0000 −0.855138 −0.427569 0.903983i $$-0.640630\pi$$
−0.427569 + 0.903983i $$0.640630\pi$$
$$548$$ − 6.00000i − 0.256307i
$$549$$ −10.0000 −0.426790
$$550$$ −4.00000 −0.170561
$$551$$ 24.0000i 1.02243i
$$552$$ − 6.00000i − 0.255377i
$$553$$ − 16.0000i − 0.680389i
$$554$$ 10.0000i 0.424859i
$$555$$ −2.00000 −0.0848953
$$556$$ 8.00000 0.339276
$$557$$ − 2.00000i − 0.0847427i −0.999102 0.0423714i $$-0.986509\pi$$
0.999102 0.0423714i $$-0.0134913\pi$$
$$558$$ 0 0
$$559$$ 0 0
$$560$$ −2.00000 −0.0845154
$$561$$ − 32.0000i − 1.35104i
$$562$$ −6.00000 −0.253095
$$563$$ −24.0000 −1.01148 −0.505740 0.862686i $$-0.668780\pi$$
−0.505740 + 0.862686i $$0.668780\pi$$
$$564$$ 0 0
$$565$$ − 20.0000i − 0.841406i
$$566$$ 4.00000i 0.168133i
$$567$$ − 2.00000i − 0.0839921i
$$568$$ −8.00000 −0.335673
$$569$$ −6.00000 −0.251533 −0.125767 0.992060i $$-0.540139\pi$$
−0.125767 + 0.992060i $$0.540139\pi$$
$$570$$ 6.00000i 0.251312i
$$571$$ −40.0000 −1.67395 −0.836974 0.547243i $$-0.815677\pi$$
−0.836974 + 0.547243i $$0.815677\pi$$
$$572$$ 0 0
$$573$$ 0 0
$$574$$ − 4.00000i − 0.166957i
$$575$$ 6.00000 0.250217
$$576$$ −1.00000 −0.0416667
$$577$$ 12.0000i 0.499567i 0.968302 + 0.249783i $$0.0803594\pi$$
−0.968302 + 0.249783i $$0.919641\pi$$
$$578$$ 47.0000i 1.95494i
$$579$$ − 4.00000i − 0.166234i
$$580$$ − 4.00000i − 0.166091i
$$581$$ 24.0000 0.995688
$$582$$ −16.0000 −0.663221
$$583$$ 40.0000i 1.65663i
$$584$$ 8.00000 0.331042
$$585$$ 0 0
$$586$$ −26.0000 −1.07405
$$587$$ 36.0000i 1.48588i 0.669359 + 0.742940i $$0.266569\pi$$
−0.669359 + 0.742940i $$0.733431\pi$$
$$588$$ 3.00000 0.123718
$$589$$ 0 0
$$590$$ − 4.00000i − 0.164677i
$$591$$ 18.0000i 0.740421i
$$592$$ 2.00000i 0.0821995i
$$593$$ − 22.0000i − 0.903432i −0.892162 0.451716i $$-0.850812\pi$$
0.892162 0.451716i $$-0.149188\pi$$
$$594$$ −4.00000 −0.164122
$$595$$ 16.0000 0.655936
$$596$$ 10.0000i 0.409616i
$$597$$ 0 0
$$598$$ 0 0
$$599$$ 24.0000 0.980613 0.490307 0.871550i $$-0.336885\pi$$
0.490307 + 0.871550i $$0.336885\pi$$
$$600$$ − 1.00000i − 0.0408248i
$$601$$ 22.0000 0.897399 0.448699 0.893683i $$-0.351887\pi$$
0.448699 + 0.893683i $$0.351887\pi$$
$$602$$ 8.00000 0.326056
$$603$$ 12.0000i 0.488678i
$$604$$ 0 0
$$605$$ 5.00000i 0.203279i
$$606$$ − 16.0000i − 0.649956i
$$607$$ −28.0000 −1.13648 −0.568242 0.822861i $$-0.692376\pi$$
−0.568242 + 0.822861i $$0.692376\pi$$
$$608$$ 6.00000 0.243332
$$609$$ − 8.00000i − 0.324176i
$$610$$ −10.0000 −0.404888
$$611$$ 0 0
$$612$$ 8.00000 0.323381
$$613$$ − 10.0000i − 0.403896i −0.979396 0.201948i $$-0.935273\pi$$
0.979396 0.201948i $$-0.0647272\pi$$
$$614$$ 12.0000 0.484281
$$615$$ 2.00000 0.0806478
$$616$$ 8.00000i 0.322329i
$$617$$ − 6.00000i − 0.241551i −0.992680 0.120775i $$-0.961462\pi$$
0.992680 0.120775i $$-0.0385381\pi$$
$$618$$ − 12.0000i − 0.482711i
$$619$$ 10.0000i 0.401934i 0.979598 + 0.200967i $$0.0644084\pi$$
−0.979598 + 0.200967i $$0.935592\pi$$
$$620$$ 0 0
$$621$$ 6.00000 0.240772
$$622$$ − 24.0000i − 0.962312i
$$623$$ 28.0000 1.12180
$$624$$ 0 0
$$625$$ 1.00000 0.0400000
$$626$$ 6.00000i 0.239808i
$$627$$ 24.0000 0.958468
$$628$$ −22.0000 −0.877896
$$629$$ − 16.0000i − 0.637962i
$$630$$ − 2.00000i − 0.0796819i
$$631$$ − 12.0000i − 0.477712i −0.971055 0.238856i $$-0.923228\pi$$
0.971055 0.238856i $$-0.0767725\pi$$
$$632$$ − 8.00000i − 0.318223i
$$633$$ 20.0000 0.794929
$$634$$ 30.0000 1.19145
$$635$$ 4.00000i 0.158735i
$$636$$ −10.0000 −0.396526
$$637$$ 0 0
$$638$$ −16.0000 −0.633446
$$639$$ − 8.00000i − 0.316475i
$$640$$ −1.00000 −0.0395285
$$641$$ −18.0000 −0.710957 −0.355479 0.934684i $$-0.615682\pi$$
−0.355479 + 0.934684i $$0.615682\pi$$
$$642$$ − 12.0000i − 0.473602i
$$643$$ 44.0000i 1.73519i 0.497271 + 0.867595i $$0.334335\pi$$
−0.497271 + 0.867595i $$0.665665\pi$$
$$644$$ − 12.0000i − 0.472866i
$$645$$ 4.00000i 0.157500i
$$646$$ −48.0000 −1.88853
$$647$$ 30.0000 1.17942 0.589711 0.807614i $$-0.299242\pi$$
0.589711 + 0.807614i $$0.299242\pi$$
$$648$$ − 1.00000i − 0.0392837i
$$649$$ −16.0000 −0.628055
$$650$$ 0 0
$$651$$ 0 0
$$652$$ − 16.0000i − 0.626608i
$$653$$ 42.0000 1.64359 0.821794 0.569785i $$-0.192974\pi$$
0.821794 + 0.569785i $$0.192974\pi$$
$$654$$ 12.0000 0.469237
$$655$$ − 10.0000i − 0.390732i
$$656$$ − 2.00000i − 0.0780869i
$$657$$ 8.00000i 0.312110i
$$658$$ 0 0
$$659$$ 2.00000 0.0779089 0.0389545 0.999241i $$-0.487597\pi$$
0.0389545 + 0.999241i $$0.487597\pi$$
$$660$$ −4.00000 −0.155700
$$661$$ − 48.0000i − 1.86698i −0.358599 0.933492i $$-0.616745\pi$$
0.358599 0.933492i $$-0.383255\pi$$
$$662$$ −10.0000 −0.388661
$$663$$ 0 0
$$664$$ 12.0000 0.465690
$$665$$ 12.0000i 0.465340i
$$666$$ −2.00000 −0.0774984
$$667$$ 24.0000 0.929284
$$668$$ − 4.00000i − 0.154765i
$$669$$ 2.00000i 0.0773245i
$$670$$ 12.0000i 0.463600i
$$671$$ 40.0000i 1.54418i
$$672$$ −2.00000 −0.0771517
$$673$$ −26.0000 −1.00223 −0.501113 0.865382i $$-0.667076\pi$$
−0.501113 + 0.865382i $$0.667076\pi$$
$$674$$ − 14.0000i − 0.539260i
$$675$$ 1.00000 0.0384900
$$676$$ 0 0
$$677$$ −30.0000 −1.15299 −0.576497 0.817099i $$-0.695581\pi$$
−0.576497 + 0.817099i $$0.695581\pi$$
$$678$$ − 20.0000i − 0.768095i
$$679$$ −32.0000 −1.22805
$$680$$ 8.00000 0.306786
$$681$$ 4.00000i 0.153280i
$$682$$ 0 0
$$683$$ 44.0000i 1.68361i 0.539779 + 0.841807i $$0.318508\pi$$
−0.539779 + 0.841807i $$0.681492\pi$$
$$684$$ 6.00000i 0.229416i
$$685$$ 6.00000 0.229248
$$686$$ 20.0000 0.763604
$$687$$ 4.00000i 0.152610i
$$688$$ 4.00000 0.152499
$$689$$ 0 0
$$690$$ 6.00000 0.228416
$$691$$ − 14.0000i − 0.532585i −0.963892 0.266293i $$-0.914201\pi$$
0.963892 0.266293i $$-0.0857987\pi$$
$$692$$ 22.0000 0.836315
$$693$$ −8.00000 −0.303895
$$694$$ − 24.0000i − 0.911028i
$$695$$ 8.00000i 0.303457i
$$696$$ − 4.00000i − 0.151620i
$$697$$ 16.0000i 0.606043i
$$698$$ −28.0000 −1.05982
$$699$$ 24.0000 0.907763
$$700$$ − 2.00000i − 0.0755929i
$$701$$ −32.0000 −1.20862 −0.604312 0.796748i $$-0.706552\pi$$
−0.604312 + 0.796748i $$0.706552\pi$$
$$702$$ 0 0
$$703$$ 12.0000 0.452589
$$704$$ 4.00000i 0.150756i
$$705$$ 0 0
$$706$$ 14.0000 0.526897
$$707$$ − 32.0000i − 1.20348i
$$708$$ − 4.00000i − 0.150329i
$$709$$ 4.00000i 0.150223i 0.997175 + 0.0751116i $$0.0239313\pi$$
−0.997175 + 0.0751116i $$0.976069\pi$$
$$710$$ − 8.00000i − 0.300235i
$$711$$ 8.00000 0.300023
$$712$$ 14.0000 0.524672
$$713$$ 0 0
$$714$$ 16.0000 0.598785
$$715$$ 0 0
$$716$$ −10.0000 −0.373718
$$717$$ − 16.0000i − 0.597531i
$$718$$ −24.0000 −0.895672
$$719$$ −36.0000 −1.34257 −0.671287 0.741198i $$-0.734258\pi$$
−0.671287 + 0.741198i $$0.734258\pi$$
$$720$$ − 1.00000i − 0.0372678i
$$721$$ − 24.0000i − 0.893807i
$$722$$ − 17.0000i − 0.632674i
$$723$$ 2.00000i 0.0743808i
$$724$$ 10.0000 0.371647
$$725$$ 4.00000 0.148556
$$726$$ 5.00000i 0.185567i
$$727$$ −16.0000 −0.593407 −0.296704 0.954970i $$-0.595887\pi$$
−0.296704 + 0.954970i $$0.595887\pi$$
$$728$$ 0 0
$$729$$ 1.00000 0.0370370
$$730$$ 8.00000i 0.296093i
$$731$$ −32.0000 −1.18356
$$732$$ −10.0000 −0.369611
$$733$$ 38.0000i 1.40356i 0.712393 + 0.701781i $$0.247612\pi$$
−0.712393 + 0.701781i $$0.752388\pi$$
$$734$$ − 36.0000i − 1.32878i
$$735$$ 3.00000i 0.110657i
$$736$$ − 6.00000i − 0.221163i
$$737$$ 48.0000 1.76810
$$738$$ 2.00000 0.0736210
$$739$$ − 30.0000i − 1.10357i −0.833987 0.551784i $$-0.813947\pi$$
0.833987 0.551784i $$-0.186053\pi$$
$$740$$ −2.00000 −0.0735215
$$741$$ 0 0
$$742$$ −20.0000 −0.734223
$$743$$ 12.0000i 0.440237i 0.975473 + 0.220119i $$0.0706445\pi$$
−0.975473 + 0.220119i $$0.929356\pi$$
$$744$$ 0 0
$$745$$ −10.0000 −0.366372
$$746$$ − 2.00000i − 0.0732252i
$$747$$ 12.0000i 0.439057i
$$748$$ − 32.0000i − 1.17004i
$$749$$ − 24.0000i − 0.876941i
$$750$$ 1.00000 0.0365148
$$751$$ 16.0000 0.583848 0.291924 0.956441i $$-0.405705\pi$$
0.291924 + 0.956441i $$0.405705\pi$$
$$752$$ 0 0
$$753$$ 6.00000 0.218652
$$754$$ 0 0
$$755$$ 0 0
$$756$$ − 2.00000i − 0.0727393i
$$757$$ −6.00000 −0.218074 −0.109037 0.994038i $$-0.534777\pi$$
−0.109037 + 0.994038i $$0.534777\pi$$
$$758$$ 18.0000 0.653789
$$759$$ − 24.0000i − 0.871145i
$$760$$ 6.00000i 0.217643i
$$761$$ − 22.0000i − 0.797499i −0.917060 0.398750i $$-0.869444\pi$$
0.917060 0.398750i $$-0.130556\pi$$
$$762$$ 4.00000i 0.144905i
$$763$$ 24.0000 0.868858
$$764$$ 0 0
$$765$$ 8.00000i 0.289241i
$$766$$ 12.0000 0.433578
$$767$$ 0 0
$$768$$ −1.00000 −0.0360844
$$769$$ 10.0000i 0.360609i 0.983611 + 0.180305i $$0.0577084\pi$$
−0.983611 + 0.180305i $$0.942292\pi$$
$$770$$ −8.00000 −0.288300
$$771$$ 12.0000 0.432169
$$772$$ − 4.00000i − 0.143963i
$$773$$ 38.0000i 1.36677i 0.730061 + 0.683383i $$0.239492\pi$$
−0.730061 + 0.683383i $$0.760508\pi$$
$$774$$ 4.00000i 0.143777i
$$775$$ 0 0
$$776$$ −16.0000 −0.574367
$$777$$ −4.00000 −0.143499
$$778$$ − 8.00000i − 0.286814i
$$779$$ −12.0000 −0.429945
$$780$$ 0 0
$$781$$ −32.0000 −1.14505
$$782$$ 48.0000i 1.71648i
$$783$$ 4.00000 0.142948
$$784$$ 3.00000 0.107143
$$785$$ − 22.0000i − 0.785214i
$$786$$ − 10.0000i − 0.356688i
$$787$$ 32.0000i 1.14068i 0.821410 + 0.570338i $$0.193188\pi$$
−0.821410 + 0.570338i $$0.806812\pi$$
$$788$$ 18.0000i 0.641223i
$$789$$ −2.00000 −0.0712019
$$790$$ 8.00000 0.284627
$$791$$ − 40.0000i − 1.42224i
$$792$$ −4.00000 −0.142134
$$793$$ 0 0
$$794$$ 14.0000 0.496841
$$795$$ − 10.0000i − 0.354663i
$$796$$ 0 0
$$797$$ 10.0000 0.354218 0.177109 0.984191i $$-0.443325\pi$$
0.177109 + 0.984191i $$0.443325\pi$$
$$798$$ 12.0000i 0.424795i
$$799$$ 0 0
$$800$$ − 1.00000i − 0.0353553i
$$801$$ 14.0000i 0.494666i
$$802$$ 30.0000 1.05934
$$803$$ 32.0000 1.12926
$$804$$ 12.0000i 0.423207i
$$805$$ 12.0000 0.422944
$$806$$ 0 0
$$807$$ 24.0000 0.844840
$$808$$ − 16.0000i − 0.562878i
$$809$$ 2.00000 0.0703163 0.0351581 0.999382i $$-0.488807\pi$$
0.0351581 + 0.999382i $$0.488807\pi$$
$$810$$ 1.00000 0.0351364
$$811$$ − 38.0000i − 1.33436i −0.744896 0.667180i $$-0.767501\pi$$
0.744896 0.667180i $$-0.232499\pi$$
$$812$$ − 8.00000i − 0.280745i
$$813$$ 16.0000i 0.561144i
$$814$$ 8.00000i 0.280400i
$$815$$ 16.0000 0.560456
$$816$$ 8.00000 0.280056
$$817$$ − 24.0000i − 0.839654i
$$818$$ 26.0000 0.909069
$$819$$ 0 0
$$820$$ 2.00000 0.0698430
$$821$$ − 10.0000i − 0.349002i −0.984657 0.174501i $$-0.944169\pi$$
0.984657 0.174501i $$-0.0558313\pi$$
$$822$$ 6.00000 0.209274
$$823$$ 52.0000 1.81261 0.906303 0.422628i $$-0.138892\pi$$
0.906303 + 0.422628i $$0.138892\pi$$
$$824$$ − 12.0000i − 0.418040i
$$825$$ − 4.00000i − 0.139262i
$$826$$ − 8.00000i − 0.278356i
$$827$$ 12.0000i 0.417281i 0.977992 + 0.208640i $$0.0669038\pi$$
−0.977992 + 0.208640i $$0.933096\pi$$
$$828$$ 6.00000 0.208514
$$829$$ 10.0000 0.347314 0.173657 0.984806i $$-0.444442\pi$$
0.173657 + 0.984806i $$0.444442\pi$$
$$830$$ 12.0000i 0.416526i
$$831$$ −10.0000 −0.346896
$$832$$ 0 0
$$833$$ −24.0000 −0.831551
$$834$$ 8.00000i 0.277017i
$$835$$ 4.00000 0.138426
$$836$$ 24.0000 0.830057
$$837$$ 0 0
$$838$$ 10.0000i 0.345444i
$$839$$ 40.0000i 1.38095i 0.723355 + 0.690477i $$0.242599\pi$$
−0.723355 + 0.690477i $$0.757401\pi$$
$$840$$ − 2.00000i − 0.0690066i
$$841$$ −13.0000 −0.448276
$$842$$ 8.00000 0.275698
$$843$$ − 6.00000i − 0.206651i
$$844$$ 20.0000 0.688428
$$845$$ 0 0
$$846$$ 0 0
$$847$$ 10.0000i 0.343604i
$$848$$ −10.0000 −0.343401
$$849$$ −4.00000 −0.137280
$$850$$ 8.00000i 0.274398i
$$851$$ − 12.0000i − 0.411355i
$$852$$ − 8.00000i − 0.274075i
$$853$$ − 46.0000i − 1.57501i −0.616308 0.787505i $$-0.711372\pi$$
0.616308 0.787505i $$-0.288628\pi$$
$$854$$ −20.0000 −0.684386
$$855$$ −6.00000 −0.205196
$$856$$ − 12.0000i − 0.410152i
$$857$$ −4.00000 −0.136637 −0.0683187 0.997664i $$-0.521763\pi$$
−0.0683187 + 0.997664i $$0.521763\pi$$
$$858$$ 0 0
$$859$$ 36.0000 1.22830 0.614152 0.789188i $$-0.289498\pi$$
0.614152 + 0.789188i $$0.289498\pi$$
$$860$$ 4.00000i 0.136399i
$$861$$ 4.00000 0.136320
$$862$$ 0 0
$$863$$ − 12.0000i − 0.408485i −0.978920 0.204242i $$-0.934527\pi$$
0.978920 0.204242i $$-0.0654731\pi$$
$$864$$ − 1.00000i − 0.0340207i
$$865$$ 22.0000i 0.748022i
$$866$$ 2.00000i 0.0679628i
$$867$$ −47.0000 −1.59620
$$868$$ 0 0
$$869$$ − 32.0000i − 1.08553i
$$870$$ 4.00000 0.135613
$$871$$ 0 0
$$872$$ 12.0000 0.406371
$$873$$ − 16.0000i − 0.541518i
$$874$$ −36.0000 −1.21772
$$875$$ 2.00000 0.0676123
$$876$$ 8.00000i 0.270295i
$$877$$ − 14.0000i − 0.472746i −0.971662 0.236373i $$-0.924041\pi$$
0.971662 0.236373i $$-0.0759588\pi$$
$$878$$ 32.0000i 1.07995i
$$879$$ − 26.0000i − 0.876958i
$$880$$ −4.00000 −0.134840
$$881$$ 18.0000 0.606435 0.303218 0.952921i $$-0.401939\pi$$
0.303218 + 0.952921i $$0.401939\pi$$
$$882$$ 3.00000i 0.101015i
$$883$$ 28.0000 0.942275 0.471138 0.882060i $$-0.343844\pi$$
0.471138 + 0.882060i $$0.343844\pi$$
$$884$$ 0 0
$$885$$ 4.00000 0.134459
$$886$$ 16.0000i 0.537531i
$$887$$ −2.00000 −0.0671534 −0.0335767 0.999436i $$-0.510690\pi$$
−0.0335767 + 0.999436i $$0.510690\pi$$
$$888$$ −2.00000 −0.0671156
$$889$$ 8.00000i 0.268311i
$$890$$ 14.0000i 0.469281i
$$891$$ − 4.00000i − 0.134005i
$$892$$ 2.00000i 0.0669650i
$$893$$ 0 0
$$894$$ −10.0000 −0.334450
$$895$$ − 10.0000i − 0.334263i
$$896$$ −2.00000 −0.0668153
$$897$$ 0 0
$$898$$ −6.00000 −0.200223
$$899$$ 0 0
$$900$$ 1.00000 0.0333333
$$901$$ 80.0000 2.66519
$$902$$ − 8.00000i − 0.266371i
$$903$$ 8.00000i 0.266223i
$$904$$ − 20.0000i − 0.665190i
$$905$$ 10.0000i 0.332411i
$$906$$ 0 0
$$907$$ −52.0000 −1.72663 −0.863316 0.504664i $$-0.831616\pi$$
−0.863316 + 0.504664i $$0.831616\pi$$
$$908$$ 4.00000i 0.132745i
$$909$$ 16.0000 0.530687
$$910$$ 0 0
$$911$$ −20.0000 −0.662630 −0.331315 0.943520i $$-0.607492\pi$$
−0.331315 + 0.943520i $$0.607492\pi$$
$$912$$ 6.00000i 0.198680i
$$913$$ 48.0000 1.58857
$$914$$ −8.00000 −0.264616
$$915$$ − 10.0000i − 0.330590i
$$916$$ 4.00000i 0.132164i
$$917$$ − 20.0000i − 0.660458i
$$918$$ 8.00000i 0.264039i
$$919$$ 40.0000 1.31948 0.659739 0.751495i $$-0.270667\pi$$
0.659739 + 0.751495i $$0.270667\pi$$
$$920$$ 6.00000 0.197814
$$921$$ 12.0000i 0.395413i
$$922$$ 6.00000 0.197599
$$923$$ 0 0
$$924$$ −8.00000 −0.263181
$$925$$ − 2.00000i − 0.0657596i
$$926$$ −26.0000 −0.854413
$$927$$ 12.0000 0.394132
$$928$$ − 4.00000i − 0.131306i
$$929$$ − 30.0000i − 0.984268i −0.870519 0.492134i $$-0.836217\pi$$
0.870519 0.492134i $$-0.163783\pi$$
$$930$$ 0 0
$$931$$ − 18.0000i − 0.589926i
$$932$$ 24.0000 0.786146
$$933$$ 24.0000 0.785725
$$934$$ − 28.0000i − 0.916188i
$$935$$ 32.0000 1.04651
$$936$$ 0 0
$$937$$ 18.0000 0.588034 0.294017 0.955800i $$-0.405008\pi$$
0.294017 + 0.955800i $$0.405008\pi$$
$$938$$ 24.0000i 0.783628i
$$939$$ −6.00000 −0.195803
$$940$$ 0 0
$$941$$ − 34.0000i − 1.10837i −0.832394 0.554184i $$-0.813030\pi$$
0.832394 0.554184i $$-0.186970\pi$$
$$942$$ − 22.0000i − 0.716799i
$$943$$ 12.0000i 0.390774i
$$944$$ − 4.00000i − 0.130189i
$$945$$ 2.00000 0.0650600
$$946$$ 16.0000 0.520205
$$947$$ − 4.00000i − 0.129983i −0.997886 0.0649913i $$-0.979298\pi$$
0.997886 0.0649913i $$-0.0207020\pi$$
$$948$$ 8.00000 0.259828
$$949$$ 0 0
$$950$$ −6.00000 −0.194666
$$951$$ 30.0000i 0.972817i
$$952$$ 16.0000 0.518563
$$953$$ −24.0000 −0.777436 −0.388718 0.921357i $$-0.627082\pi$$
−0.388718 + 0.921357i $$0.627082\pi$$
$$954$$ − 10.0000i − 0.323762i
$$955$$ 0 0
$$956$$ − 16.0000i − 0.517477i
$$957$$ − 16.0000i − 0.517207i
$$958$$ 32.0000 1.03387
$$959$$ 12.0000 0.387500
$$960$$ − 1.00000i − 0.0322749i
$$961$$ 31.0000 1.00000
$$962$$ 0 0
$$963$$ 12.0000 0.386695
$$964$$ 2.00000i 0.0644157i
$$965$$ 4.00000 0.128765
$$966$$ 12.0000 0.386094
$$967$$ − 14.0000i − 0.450210i −0.974335 0.225105i $$-0.927728\pi$$
0.974335 0.225105i $$-0.0722725\pi$$
$$968$$ 5.00000i 0.160706i
$$969$$ − 48.0000i − 1.54198i
$$970$$ − 16.0000i − 0.513729i
$$971$$ 38.0000 1.21948 0.609739 0.792602i $$-0.291274\pi$$
0.609739 + 0.792602i $$0.291274\pi$$
$$972$$ 1.00000 0.0320750
$$973$$ 16.0000i 0.512936i
$$974$$ −26.0000 −0.833094
$$975$$ 0 0
$$976$$ −10.0000 −0.320092
$$977$$ 30.0000i 0.959785i 0.877327 + 0.479893i $$0.159324\pi$$
−0.877327 + 0.479893i $$0.840676\pi$$
$$978$$ 16.0000 0.511624
$$979$$ 56.0000 1.78977
$$980$$ 3.00000i 0.0958315i
$$981$$ 12.0000i 0.383131i
$$982$$ − 42.0000i − 1.34027i
$$983$$ 52.0000i 1.65854i 0.558846 + 0.829271i $$0.311244\pi$$
−0.558846 + 0.829271i $$0.688756\pi$$
$$984$$ 2.00000 0.0637577
$$985$$ −18.0000 −0.573528
$$986$$ 32.0000i 1.01909i
$$987$$ 0 0
$$988$$ 0 0
$$989$$ −24.0000 −0.763156
$$990$$ − 4.00000i − 0.127128i
$$991$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$992$$ 0 0
$$993$$ − 10.0000i − 0.317340i
$$994$$ − 16.0000i − 0.507489i
$$995$$ 0 0
$$996$$ 12.0000i 0.380235i
$$997$$ −26.0000 −0.823428 −0.411714 0.911313i $$-0.635070\pi$$
−0.411714 + 0.911313i $$0.635070\pi$$
$$998$$ −38.0000 −1.20287
$$999$$ − 2.00000i − 0.0632772i
Display $$a_p$$ with $$p$$ up to: 50 250 1000 Display $$a_n$$ with $$n$$ up to: 50 250 1000
## Twists
By twisting character
Char Parity Ord Type Twist Min Dim
1.1 even 1 trivial 5070.2.b.e.1351.2 2
13.5 odd 4 390.2.a.e.1.1 1
13.8 odd 4 5070.2.a.e.1.1 1
13.12 even 2 inner 5070.2.b.e.1351.1 2
39.5 even 4 1170.2.a.e.1.1 1
52.31 even 4 3120.2.a.o.1.1 1
65.18 even 4 1950.2.e.f.1249.1 2
65.44 odd 4 1950.2.a.h.1.1 1
65.57 even 4 1950.2.e.f.1249.2 2
156.83 odd 4 9360.2.a.bh.1.1 1
195.44 even 4 5850.2.a.bi.1.1 1
195.83 odd 4 5850.2.e.i.5149.2 2
195.122 odd 4 5850.2.e.i.5149.1 2
By twisted newform
Twist Min Dim Char Parity Ord Type
390.2.a.e.1.1 1 13.5 odd 4
1170.2.a.e.1.1 1 39.5 even 4
1950.2.a.h.1.1 1 65.44 odd 4
1950.2.e.f.1249.1 2 65.18 even 4
1950.2.e.f.1249.2 2 65.57 even 4
3120.2.a.o.1.1 1 52.31 even 4
5070.2.a.e.1.1 1 13.8 odd 4
5070.2.b.e.1351.1 2 13.12 even 2 inner
5070.2.b.e.1351.2 2 1.1 even 1 trivial
5850.2.a.bi.1.1 1 195.44 even 4
5850.2.e.i.5149.1 2 195.122 odd 4
5850.2.e.i.5149.2 2 195.83 odd 4
9360.2.a.bh.1.1 1 156.83 odd 4 | 23,516 | 41,692 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2021-17 | latest | en | 0.26539 |
https://gmatclub.com/forum/m12-183940.html?kudos=1 | 1,579,411,202,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250594209.12/warc/CC-MAIN-20200119035851-20200119063851-00057.warc.gz | 476,694,531 | 61,270 | GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video
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M12-21
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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 00:47
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55% (hard)
Question Stats:
61% (02:13) correct 39% (02:05) wrong based on 76 sessions
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How many integers in a group of 5 consecutive positive integers are divisible by 4?
(1) The median of these numbers is odd
(2) The average (arithmetic mean) of these numbers is a prime number
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Math Expert
Joined: 02 Sep 2009
Posts: 60480
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16 Sep 2014, 00:47
1
2
Official Solution:
How many integers in a group of 5 consecutive positive integers are divisible by 4?
(1) The median of these numbers is odd. The median of the set with odd number of terms is just a middle term, thus our set of 5 consecutive numbers is: {Odd, Even, Odd, Even, Odd}. Out of 2 consecutive even integers only one is a multiple of 4. Sufficient.
(2) The average (arithmetic mean) of these numbers is a prime number. In any evenly spaced set the arithmetic mean (average) is equal to the median, so we have that mean=median=prime. Since it's not possible that median=2=even, (in this case not all 5 numbers will be positive), then median=odd prime, and we have the same case as above. Sufficient.
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Intern
Joined: 22 Jun 2014
Posts: 21
Concentration: General Management, Finance
GMAT 1: 700 Q50 V34
GRE 1: Q800 V600
GPA: 3.68
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12 Jan 2016, 05:40
1
rhio wrote:
I understood solution, but I just wanted to know why exactly does 1 of the number will be divisible by 4? Is their any kind of rule here that i have missed. I took various different sets of positive consecutive numbers with median being odd and in all cases,one of the number was divisible by 4.
because the other no. will yield the reminder 2 and hence not divisible
Math Expert
Joined: 02 Aug 2009
Posts: 8343
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12 Jan 2016, 07:48
1
rhio wrote:
I understood solution, but I just wanted to know why exactly does 1 of the number will be divisible by 4? Is their any kind of rule here that i have missed. I took various different sets of positive consecutive numbers with median being odd and in all cases,one of the number was divisible by 4.
Hi,
In all cases, any number, n, will have only one number divisible by itself in 'n' consecutive numbers..
5 will have only one multiple in any 5 consecutive number because of the property of multiplication tables..
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05 Jan 2016, 14:07
I understood solution, but I just wanted to know why exactly does 1 of the number will be divisible by 4? Is their any kind of rule here that i have missed. I took various different sets of positive consecutive numbers with median being odd and in all cases,one of the number was divisible by 4.
Math Expert
Joined: 02 Sep 2009
Posts: 60480
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06 Jan 2016, 08:13
rhio wrote:
I understood solution, but I just wanted to know why exactly does 1 of the number will be divisible by 4? Is their any kind of rule here that i have missed. I took various different sets of positive consecutive numbers with median being odd and in all cases,one of the number was divisible by 4.
I think this is explained in the solution: out of 2 consecutive even integers only one is a multiple of 4.
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15 Jul 2016, 22:57
can someone explain me me how option 2 is right?
for eg 7 8 9 10 11
9 is a median and not prime so cannot determine . Ans should be just option A
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Posts: 60480
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15 Jul 2016, 23:13
r19 wrote:
can someone explain me me how option 2 is right?
for eg 7 8 9 10 11
9 is a median and not prime so cannot determine . Ans should be just option A
(2) says: The average (arithmetic mean) of these numbers is a prime number.
Why do you consider the case which contradicts the statement?
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GMAT 1: 600 Q47 V26
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16 Jul 2016, 01:29
I considered 2 cases 1 2 3 4 5 and 7 8 9 10 11, for one we can determine and for other cannot .I misunderstood the question. Thanks for the help
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Joined: 17 Feb 2017
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GMAT 1: 670 Q49 V32
GPA: 3.3
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18 Nov 2017, 07:27
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. Is 2 divisible by 4?
Ans would be 1/2. Does it satisfy our condition?
Shouldn't we look for a whole number as an answer?
In refrence Statement 1, considering the set : (1, 2, 3, 4, 5)
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18 Nov 2017, 07:47
jasanisanket24 wrote:
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. Is 2 divisible by 4?
Ans would be 1/2. Does it satisfy our condition?
Shouldn't we look for a whole number as an answer?
In refrence Statement 1, considering the set : (1, 2, 3, 4, 5)
Hi jasanisanket24
the question asks how many numbers in the set are divisible by 4. so 2 is not divisible by 4. Hence in the set (1,2,3,4,5) we have only 1 number that is divisible by 4. Hence the statement is sufficient.
Both statement 1 & 2 leads us to the conclusion that the consecutive set starts with an ODD integer. so the set will be of the form (odd,even,odd,even,odd)
hence there will be two even integers in the set and both will be consecutive even integers. out of any two consecutive even integers, only 1 is divisible by 4 (eg. 2,4; 4,6; 8,10....etc). SO we can conclude that there is only 1 number that is divisible by 4. Hence both the statements are sufficient
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Joined: 11 Sep 2019
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25 Nov 2019, 23:45
Bunuel wrote:
Official Solution:
How many integers in a group of 5 consecutive positive integers are divisible by 4?
(1) The median of these numbers is odd. The median of the set with odd number of terms is just a middle term, thus our set of 5 consecutive numbers is: {Odd, Even, Odd, Even, Odd}. Out of 2 consecutive even integers only one is a multiple of 4. Sufficient.
(2) The average (arithmetic mean) of these numbers is a prime number. In any evenly spaced set the arithmetic mean (average) is equal to the median, so we have that mean=median=prime. Since it's not possible that median=2=even, (in this case not all 5 numbers will be positive), then median=odd prime, and we have the same case as above. Sufficient.
For St1 What about -1,0,1,2,3; Median is odd but none is divisible by 4
Math Expert
Joined: 02 Sep 2009
Posts: 60480
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25 Nov 2019, 23:48
Nityanshu1990 wrote:
Bunuel wrote:
Official Solution:
How many integers in a group of 5 consecutive positive integers are divisible by 4?
(1) The median of these numbers is odd. The median of the set with odd number of terms is just a middle term, thus our set of 5 consecutive numbers is: {Odd, Even, Odd, Even, Odd}. Out of 2 consecutive even integers only one is a multiple of 4. Sufficient.
(2) The average (arithmetic mean) of these numbers is a prime number. In any evenly spaced set the arithmetic mean (average) is equal to the median, so we have that mean=median=prime. Since it's not possible that median=2=even, (in this case not all 5 numbers will be positive), then median=odd prime, and we have the same case as above. Sufficient.
For St1 What about -1,0,1,2,3; Median is odd but none is divisible by 4
0 is divisible by every integer except 0 itself: 0/integer = integer.
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26 Nov 2019, 08:09
This is an excellent question on statistics and number properties. From the question statement, we know that the integers are consecutive and that they are all positive integers.
In a group of n consecutive integers, one integer will always be a multiple of n. For example, in a set of 3 consecutive integers, there will always be a multiple of 3. This is the underlying concept that we can use to solve this question.
However, we need to be careful about the possibility that there could be two multiples of 4 in a set of 5 consecutive integers. For example, 4,5,6,7,8 contains two integers which are divisible by 4. This is where we use the statements to evaluate how many of them can be divisible by 4.
Let us represent the 5 consecutive integers as a, (a+1), (a+2), (a+3) and (a+4). Remember that we have already taken these in order since these are consecutive.
From statement I, the median of these numbers is odd. This means that (a+2) is odd, which means that there are 3 odd numbers and 2 even numbers. In this case, only one of the even numbers will be divisible by 4. A set of 5 numbers will have 2 multiples of 4 when the first number itself is a multiple of 4, because of the rule stated above.. Clearly, that’s not happening here.
Statement I is sufficient to say that the set of integers consists of ONE number that is divisible by 4. Possible answer options are A or D. Answer options B, C and E can be eliminated.
From statement II, the average of the given numbers is a prime number.
For equally spaced values in a data set, Mean = Median.
Consecutive integers definitely represent equally spaced values. So, for the numbers that we have considered, the average(mean) is going to be the middle value.
So, the average is (a+2).
If(a+2) = 2, a = 0 which is not possible because all the numbers in the set are positive integers.
Therefore, (a+2) has to be an odd prime number. This is equivalent to the data given in the first statement. Since statement I alone was sufficient, statement II alone will also be sufficient. This is a very simple and logical conclusion.
Answer option A can be eliminated. The correct answer option is D.
Hope that helps!
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26 Nov 2019, 08:14
Nityanshu1990 wrote:
Bunuel wrote:
Official Solution:
How many integers in a group of 5 consecutive positive integers are divisible by 4?
(1) The median of these numbers is odd. The median of the set with odd number of terms is just a middle term, thus our set of 5 consecutive numbers is: {Odd, Even, Odd, Even, Odd}. Out of 2 consecutive even integers only one is a multiple of 4. Sufficient.
(2) The average (arithmetic mean) of these numbers is a prime number. In any evenly spaced set the arithmetic mean (average) is equal to the median, so we have that mean=median=prime. Since it's not possible that median=2=even, (in this case not all 5 numbers will be positive), then median=odd prime, and we have the same case as above. Sufficient.
For St1 What about -1,0,1,2,3; Median is odd but none is divisible by 4
I think the question says that we have 5 consecutive positive integers. -1 is not a positive integer. Please be careful, do not miss out on such pieces of information given in the question. When you try out a case like this, please re-read the question just to ensure that you are not missing out on information.
Thanks
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28 Nov 2019, 01:43
Bunuel wrote:
Official Solution:
(2) The average (arithmetic mean) of these numbers is a prime number. In any evenly spaced set the arithmetic mean (average) is equal to the median, so we have that mean=median=prime. Since it's not possible that median=2=even, (in this case not all 5 numbers will be positive), then median=odd prime, and we have the same case as above. Sufficient.
So '0' is considered neither negative nor positive for all purposes on the GMAT?
Bcz [0,1,2,3,4] has a median=even=2 and contradicts St(2).
Math Expert
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Posts: 60480
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28 Nov 2019, 04:07
RaunaqSinghPunn wrote:
Bunuel wrote:
Official Solution:
(2) The average (arithmetic mean) of these numbers is a prime number. In any evenly spaced set the arithmetic mean (average) is equal to the median, so we have that mean=median=prime. Since it's not possible that median=2=even, (in this case not all 5 numbers will be positive), then median=odd prime, and we have the same case as above. Sufficient.
So '0' is considered neither negative nor positive for all purposes on the GMAT?
Bcz [0,1,2,3,4] has a median=even=2 and contradicts St(2).
0 is neither positive nor negative generally in math, not only for the GMAT.
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Re: M12-21 [#permalink] 28 Nov 2019, 04:07
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M12-21
Moderators: chetan2u, Bunuel | 3,588 | 13,189 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2020-05 | latest | en | 0.88673 |
https://serverfault.com/questions/51470/speed-calculator-for-raid-sets | 1,582,715,201,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146341.16/warc/CC-MAIN-20200226084902-20200226114902-00376.warc.gz | 523,977,170 | 37,155 | # Speed Calculator for Raid sets
First my apologies if this questions has been asked. I've googled… and googled some more and can't seem to find what I'm looking for.
Anyone know of a software or web based calculator that will let you plug in a RAID configuration (example below) and output expected R/W speeds, hopefully in MBs
Number of disk, size, spin, type , Raid type
EX. (8, 73Gb, 15k, SAS, Raid 1/0)
Or
EX. (6, 146Gb, 10k, FC, Raid 5)
I found severely that calculate available space. Some that give some speed info, but they can't be realistic because they don’t take spin or type in the consideration.
I don't know of a calcualtor that can tell you that, in part because there are so many other factors besides just the disk and connection type factors. The RAID controllers make a huge difference, as can the firmware on those controllers, the type of data, as does the ability of the motherboard to push data. Your best bet is benchmarking on your own. I can't even think of a way to write a calculator do do that sort of thing. Also I believe that probably for most operations the network will bottleneck before the RAID.
There are quite a few variables that can affect speed, but here's some basic ideas to get a feel for what a given raid set should be capable of.
Raw disk throughput
Assuming that a random seek completes an average of 1/2 of a rotation (180 degrees) away from the sector you want, the average random access time is one average seek plus the time the disk takes to rotate 180 degrees.
• On a 10K RPM disk 1/2 of a rotation takes approximately 3ms.
• On a 15K RPM disk 1/2 of a rotation takes approximately 2ms.
• Average seek time for a Seagate Cheetah 15K6 is quoted at 3.5ms for reads and 3.9ms for writes (I presume the writes include a period to align the head on the servo tracks). A 10K disk is slightly longer.
So, a raw estimate is an average of 5.5ms per random seek for a 15K drive and 7ms for a 10K drive. Tagged command queuing will optimise this slightly. Thus, for a 15k drive we have a theoretical random throughput of about 180 IOPS and 140 IOPS for a 10K drive.
RAID-1
On a non-striped RAID-1, reads can be split between the two disks, but writes must go to both drives. Random operations will give you twice the throughput of a single disk for reads and approximately the throughput of a single disk for writes. Sequential I/O tends to peak at the maximum throughput of a single disk. Interface cables may or may not present a bottleneck.
Striped RAID sets
RAID-5, RAID-10 or RAID-50 disks have the data split up into chunks spread in a round-robin fashion amongst the members of the RAID set. Assuming no read-ahead optimisation a disk can read at most one stripe per revolution of the disk. A 10K disk revolves about 170 times per second and a 15K disk revolves about 250 times per second.
For a 64K stripe this comes to approximately 10MB/sec per 10K disk or 15MB/sec per 15K disk. Larger stripe sizes give you better sequential throughput on the disks - for example a 256K stripe size on an array of 15K disks would give you 60MB/sec per disk. A heavily random access workload will reduce this by introducing more latency between seeks. Read-ahead on a controller might increase it.
Thus, an array with 14 15K disks using 64K stripes would have a theoretical streaming throughput of around 210MB/sec assuming no other constraints. If the controller is not fast enough the practical rate may be lower (for example, I could never get a dell PV660 (Mylex DAC-FFX) to get more than one read per two revolutions of the disks). A heavily random access workload would also be somewhat slower because the disk accesses will average less than one per revolution of the disk. Some reads will also be used on parity data so the actual application data throughput would be a bit slower.
Write bottlenecks
The fastest possible write on a RAID-5 involves two reads and two writes. The controller has to read the old block and corresponding parity block, XOR the old and new data with the parity block to recalculate the parity and write out the new block and parity. Caching can reduce the amount of disk activity if the old block and parity block are in cache. The same applies to a RAID-50.
A RAID-10 needs two disk accesses per write - one to the main and the other to the mirror. Read performance is roughly equivalent to a RAID-5.
Controller bottlenecks
In some cases (fibre channel is prone to this) the connections to the physical disk subsystem are of somewhat lower bandwidth than the disks are theoretically capable of delivering. Also, disk controllers can perform poorly. In many cases this is a more significant limitation than the disks themselves. High-end SAN hardware often has large multiprocessor machines as controllers - they may also have custom hardware for fast parity calculations. The controller for an EMC DMX takes up half a rack by itself - before you put any disks on it.
Tuning the disk itself
Caching and read-ahead parameters on the disks themselves may also affect performace for certain workloads. For example, disks using Seagate's 'V' firmware might be set up for fewer larger cache segments and agressive read-ahead to optimise for streaming throughput of media data. The same physical disk configured for use in a Clariion would be configured with more, smaller cache segments in order to support a larger number of smaller writes from many clients on a SAN.
• This is one damn fine piece of an answer. – sjas Sep 25 '14 at 11:22
• As always, @COTW, this is an excellent answer with great details that are hard to find elsewhere! Thanks! – Max Vernon May 17 '16 at 14:40
• @MaxVernon - Thanks. I spent an ungodly amount on fibre channel kit to find that out, right down to poking about with disk and controller firmware. Now it's all in landfill - the last of it went just a few weeks ago. R.I.P. – ConcernedOfTunbridgeWells May 17 '16 at 16:25
The sort of speed that you will see with vary a lot depending on the drives, the controller, and your workload, so you are not going to fine a nice easy calculator that will give good accurate+precise results.
You may already realise this, but..
Besides all of the drive characteristics, the speed is going to be largely governed by the performance of any given RAID card. Which will depend not only on obvious things like it's interface (eg PCI-X). But more dramatically the quality and performance of it's chipset routines.
As others have said, I don't think this can be done in the terms you've stated. I think the best you could do is work out the relative performance of different raid options i.e. treat the hardware as a constant. It would still be inaccurate, but may give some guidance.
But I think you also need to consider why there are different raid configurations. One usually chooses by judging the trade-offs between performance, capacity, data protection and cost.
If you're not familiar with the trade-offs, take a look at a comparison chart to see the relative merits.
It sounds like performance is your main criteria here, so you probably know what raid level you want; you just need to find the best performing hardware.
Here is an example : I made benchmarks with the same drives (7x750GB seagate barracuda ES2), same RAID configuration (stripe size, etc), same motherboard (Supermicro H8DMe), same CPU (dual Opteron 2214), same RAM (8GB ECC) and same operating system (Linux), same filesystem (XFS, nobarrier option) and different RAID controllers. Appreciate the results :
• Areca 1280 : 250MB/s write, 350MB/s read, 21000 file created/s
• Adaptec ASR52445 : 240MB/s write, 350 MB/s read, 18000 file created/s
• 3Ware 9550 : 310MB/s write, 410MB/s read, 6500 file created/s
• 3Ware 9650 : 440MB/s write, 410MB/s read, 4500 file created/s
Of course these are the optimal results after fine-setting all software parameters for each controller (read-ahead, caching options, request queue length, request size...) by doing long and repeated benchmarks while adjusting the various knobs.
One of the funny thing I discovered by careful benchmarking is that the settings are entirely different if you use Barracuda ES2 (32MB cache) and Barracuda ES (16MB cache) drives, though the top performance is about the same.
Unfortunately, storage and RAID is hard. That's why you won't find an easy-to-go performance calculator.
I found a calculator that will give you multipliers of speed.
It boils down to
• JBOD:
• Write: 1X
• Raid 0 (Striped Set)
• Write: [NumberOfVolumes]X
• Raid 1 (Mirror Set)
• Write: 1X
• Raid 5
• Write: N/A Dependent on the controller
• Raid 10: (Mirror of striped sets 4 drives)
• Write: [2]X
If those calculators exist, they will be on the vendor web-sites. So many things can affect throughput speeds that a simple calculator would be pretty worthless. Especially for any RAID that includes parity, as those tend to bottleneck more on the RAID Controller's CPU than anything else. The best you'll find is, "rule of thumb, your mileage may vary," type estimators.
There's a lot more involved to the speed than the underlying raid layout so I doubt you'll find such a calculator.
Things that make can make a difference:
Raid Type
• hardware
• software
• fakeraid (raid controller that offloads the xor checksum calcs to the cpu)
What type of bus is the controller in...what is it sharing this bus with. Most desktop class motherboards share PCI buses with multiple slots.
File system type, block size, and it's alignment with the chunksize on the underlaying raid also come into play.
Drive type, rotational speed, cache size
Finally the workload will also interact with all of these things. So the more important question is actually what disk and raid layout are a good match for your workload and data availability goals. | 2,297 | 9,839 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2020-10 | latest | en | 0.938198 |
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Jul 19th, 2015
Anonymous
Category:
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Price: \$5 USD
Question description
section 5.4
1.A bacteria population is 5000 at time
t = 0
and its rate of growth is
1000 · 8t
bacteria per hour after t hours. What is the population after one hour? (Round your answer to the nearest whole number.)
2.Water flows from the bottom of a storage tank at a rate of
r(t) = 200 − 4t
liters per minute, where 0 ≤ t ≤ 50. Find the amount of water that flows from the tank during the first 40 minutes.
3.The velocity function (in meters per second) is given for a particle moving along a line.
v(t) = 3t − 8, 0 ≤ t ≤ 5
(a) Find the displacement.
m
(b) Find the distance traveled by the particle during the given time interval.
m
4.Simplify the expression by using a Double-Angle Formula or a Half-Angle Formula.
(a)
2 sin 19° cos 19°
(b)
2 sin 4θ cos 4θ
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posted by .
Basic Estimating – Week 2
A security analyst specializing in the stocks of the motion picture industry the relation between the number of movie theater tickets sold in December and the annual level of earnings in the motion picture industry. Time-series data for the last 15 years are used to estimate the regression model. E = a + bN where E is total earnings of the motion picture industry measured in dollars per year and N is the number of tickets sold in December. The regression output is as follows:
DEPENDENT VARIABLE: E
R-SQUARE
F-RATIO
P-VALUE ON F
OBSERVATIONS: 15
0.8311
63.96
0.0001
VARIABLE
PARAMETER ESTIMATE
STANDARD ERROR
T-RATIO
P-VALUE
INTERCEPT
25042000.00
20131000.00
1.24
0.2369
N
32.31
8.54
3.78
0.0023
How well do movie ticket sales in December explain the level of earnings for the entire year? Present statistical evidence to support your answer. Also, sales of movie tickets in December are expected to be approximately 950,000. According to this regression analysis, what do you expect earnings for the year to be? Prior to this analysis, the estimate for annual earnings is \$48 million. Is this evidence strong enough for you to consider a improving the current recommendation for the motion picture industry? Explain. Respond to at least two of your fellow students’ postings
• Managerial Economics- need help -
Basic Estimating - Week 2
A security analyst specializing in the stocks of the motion picture industry the relation between the number of movie theater tickets sold in December and the annual level of earnings in the motion picture industry. Time-series data for the last 15 years are used to estimate the regression model. E = a + bN where E is total earnings of the motion picture industry measured in dollars per year and N is the number of tickets sold in December. The regression output is as follows:
DEPENDENT VARIABLE: E R-SQUARE F-RATIO P-VALUE ON F
OBSERVATIONS: 15 0.8311 63.96 0.0001
VARIABLE PARAMETER ESTIMATE STANDARD ERROR T-RATIO P-VALUE
INTERCEPT 25042000.00 20131000.00 1.24 0.2369
N 32.31 8.54 3.78 0.0023
How well do movie ticket sales in December explain the level of earnings for the entire year? Present statistical evidence to support your answer. Also, sales of movie tickets in December are expected to be approximately 950,000. According to this regression analysis, what do you expect earnings for the year to be? Prior to this analysis, the estimates for earnings in December are \$48 million. Is this evidence strong enough for you to consider a improving the current recommendation for the motion picture industry? Explain.
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# C3 Differentiation help Watch
1. Hi,
I've just been working through some questions and am not sure about a couple of them. I would be immensely grateful for someone to check my answers.
1. Differentiating e^(3x).(sinx +2cosx) = e^(3x).(cosx - 2 sinx) + 3e^(3x).(sinx +2cosx)
= e^(3x) (7cosx + sinx)?
2. y=(3xe^x) - 1
dy/dx = (3e^x).(x+1)
Turning point at x=-1 only as otherwise x=ln0 (which is impossible)
d2y/dx2 = (3e^x).(1+(x+1)) = (3e^x).(2+x)?
3. y=4sec5x.(sin2x)^3
dy/dx = 4(sinx)^2 . (6sec5xcos2x + 5sec5xtan5xsin2x)?
4. y=(x^3)ln(5x+2)
dy/dx = (x^2)(5x/5x+2 + 3ln(5x+2))? can this be simplified anymore?
5. y = x(1+x)^(1/2)
My answer for dy/dx = (x+3(1+x)^(1/2)) / 2(1+x)^(1/2)
Correct answer = dy/dx = (3x+2) / 2(1+x)^(1/2)
Turning point at x= -2/3
Correct answer for y = (-2.(3^1/2)) / 3.(3^1/2)
Finally, point P is (-1,0). The gradient is apparently infinite at this point - why is that?
dy/dx = -1/0 at P.
2. (Original post by PardonMyFrench96)
Hi,
I've just been working through some questions and am not sure about a couple of them. I would be immensely grateful for someone to check my answers.
1. Differentiating e^(3x).(sinx +2cosx) = e^(3x).(cosx - 2 sinx) + 3e^(3x).(sinx +2cosx)
= e^(3x) (7cosx + sinx)?
2. y=(3xe^x) - 1
dy/dx = (3e^x).(x+1)
Turning point at x=-1 only as otherwise x=ln0 (which is impossible)
d2y/dx2 = (3e^x).(1+(x+1)) = (3e^x).(2+x)?
3. y=4sec5x.(sin2x)^3
dy/dx = 4(sinx)^2 . (6sec5xcos2x + 5sec5xtan5xsin2x)?
4. y=(x^3)ln(5x+2)
dy/dx = (x^2)(5x/5x+2 + 3ln(5x+2))? can this be simplified anymore?
5. y = x(1+x)^(1/2)
My answer for dy/dx = (x+3(1+x)^(1/2)) / 2(1+x)^(1/2)
Correct answer = dy/dx = (3x+2) / 2(1+x)^(1/2)
Turning point at x= -2/3
Correct answer for y = (-2.(3^1/2)) / 3.(3^1/2)
Finally, point P is (-1,0). The gradient is apparently infinite at this point - why is that?
dy/dx = -1/0 at P.
Okay, 1-4 look completely correct to me (although obviously I may have made a mistake too). I don't think 4 can be meaningfully simplified. I'm not entirely sure how you got to your answer for dy/dx for 5. Just differentiate it normally using the product rule and then cross multiply when you're left with two terms which should lead you to the correct answer. There is indeed a turning point at x=-2/3 but I'm not entirely sure how you or the "correct answer" got to that value of y... if you sub that value of x into y, you should get -2root3/9, unless I've misunderstood you. The gradient is infinite at (-1,0). dy/dx = -1/0 which is undefined, but dx/dy = 0/-1 = 0. So if you flip the axes around, you've got a horizontal line so if you flip them back again, you've got a vertical line which therefore has an infinite gradient.
3. (Original post by Chlorophile)
Okay, 1-4 look completely correct to me (although obviously I may have made a mistake too). I don't think 4 can be meaningfully simplified. I'm not entirely sure how you got to your answer for dy/dx for 5. Just differentiate it normally using the product rule and then cross multiply when you're left with two terms which should lead you to the correct answer. There is indeed a turning point at x=-2/3 but I'm not entirely sure how you or the "correct answer" got to that value of y... if you sub that value of x into y, you should get -2root3/9, unless I've misunderstood you. The gradient is infinite at (-1,0). dy/dx = -1/0 which is undefined, but dx/dy = 0/-1 = 0. So if you flip the axes around, you've got a horizontal line so if you flip them back again, you've got a vertical line which therefore has an infinite gradient.
Thanks - for question 5 I got what you did, but the correct answer is apparently -2root3/3root3. Thanks for explaining that.
4. (Original post by PardonMyFrench96)
Thanks - for question 5 I got what you did, but the correct answer is apparently -2root3/3root3. Thanks for explaining that.
so that "answer" doesn't make much sense unless I'm misreading what you've written!
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# If PQRO is a square inside a Circle with centre at "O" and
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If PQRO is a square inside a Circle with centre at "O" and radius "a", what is the area of the shaded portion ?
A. a^2((3pi-8)/12)
B. a^2((pi-2)/4)
C. a^2((9pi-16)/12)
D. a((3pi-1)/12)
E. a^2/11
[Reveal] Spoiler: OA
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Re: If PQRO is a square inside a Circle with centre at "O" and [#permalink]
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Pansi wrote:
If PQRO is a square inside a Circle with centre at "O" and radius "a", what is the area of the shaded portion ?
A. a^2((3pi-8)/12)
B. a^2((pi-2)/4)
C.a^2((9pi-16)/12)
D.a((3pi-1)/12)
E.a^2/11
So the square should have a diagonal equal to length of radius of circle. Let x be the side of square.
Hence diagonal of a square with side x= x root2
=> x root2 = a (radius of circle)
=>x= a/root 2
Hence area of square = (a/root 2)^2 = a^2/2.
Now the area of circular quadrant is (pi * a^2)/4
So shaded area = (pi * a^2)/4 - a^2/2, by simplifying
=> a^2((pi-2)/4)
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Re: If PQRO is a square inside a Circle with centre at "O" and [#permalink]
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SOURH7WK wrote:
Pansi wrote:
If PQRO is a square inside a Circle with centre at "O" and radius "a", what is the area of the shaded portion ?
A. a^2((3pi-8)/12)
B. a^2((pi-2)/4)
C.a^2((9pi-16)/12)
D.a((3pi-1)/12)
E.a^2/11
So the square should have a diagonal equal to length of radius of circle. Let x be the side of square.
Hence diagonal of a square with side x= x root2
=> x root2 = a (radius of circle)
=>x= a/root 2
Hence area of square = (a/root 2)^2 = a^2/2.
Now the area of circular quadrant is (pi * a^2)/4
So shaded area = (pi * a^2)/4 - a^2/2, by simplifying
=> a^2((pi-2)/4)
Just a remark: For any quadrilateral with perpendicular diagonals (so obviously also for a square), the area is given by half the product of the diagonals.
(You can easily deduce it by expressing the areas of the triangles formed by the diagonals.)
So, when you know the diagonal of a square, you don't have to compute the side in order to find the area. You just have to square the diagonal and half it.
In the given question, the diagonal of the square is $$a$$ (the radius of the circle), so the area of the square is $$a^2/2.$$
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Re: If PQRO is a square inside a Circle with centre at "O" and [#permalink]
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13 Nov 2013, 14:57
SOURH7WK wrote:
Pansi wrote:
If PQRO is a square inside a Circle with centre at "O" and radius "a", what is the area of the shaded portion ?
A. a^2((3pi-8)/12)
B. a^2((pi-2)/4)
C.a^2((9pi-16)/12)
D.a((3pi-1)/12)
E.a^2/11
So the square should have a diagonal equal to length of radius of circle. Let x be the side of square.
Hence diagonal of a square with side x= x root2
=> x root2 = a (radius of circle)
=>x= a/root 2
Hence area of square = (a/root 2)^2 = a^2/2.
Now the area of circular quadrant is (pi * a^2)/4
So shaded area = (pi * a^2)/4 - a^2/2, by simplifying
=> a^2((pi-2)/4)
???? my answer is a square (a square (pi -1)/4) or asquare pi - 4 a square.!!!
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Re: If PQRO is a square inside a Circle with centre at "O" and [#permalink]
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14 Nov 2013, 07:45
Area of one quarter of the circle: (pi*a^2)/4
Area of the square:
Diagonal is equal to radius a. Therefore s(sqrt2) = a --> s = a/(sqrt2)
s^2 = (a^2)/2
Area of the shaded region is area of one quarter of the circle minus area of the square:
(pi*a^2)/4 - (a^2)/2 = [(pi*a^2) - 2(a^2)]/4
Factor out a^2:
a^2[(pi-2)/4]
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Re: If PQRO is a square inside a Circle with centre at "O" and [#permalink]
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09 May 2015, 02:36
Devon wrote:
Area of one quarter of the circle: (pi*a^2)/4
Area of the square:
Diagonal is equal to radius a. Therefore s(sqrt2) = a --> s = a/(sqrt2)
s^2 = (a^2)/2
Area of the shaded region is area of one quarter of the circle minus area of the square:
(pi*a^2)/4 - (a^2)/2 = [(pi*a^2) - 2(a^2)]/4
Factor out a^2:
a^2[(pi-2)/4]
You have reduced qone quadrant with area of Square, I am fine with it. But I have worked out complete Area of circle minus Area of Square which gives= a^2(Pi-1/2). What is wrong in this?
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Re: If PQRO is a square inside a Circle with centre at "O" and [#permalink]
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08 Jun 2016, 10:20
Pansi wrote:
Attachment:
Image.png
If PQRO is a square inside a Circle with centre at "O" and radius "a", what is the area of the shaded portion ?
A. a^2((3pi-8)/12)
B. a^2((pi-2)/4)
C. a^2((9pi-16)/12)
D. a((3pi-1)/12)
E. a^2/11
OQ=a,So OR=$$\frac{a}{\sqrt{2}}$$,So area of the Square=($$\frac{a}{\sqrt{2}}$$)^2=$$\frac{a^2}{2}$$
OQ=a,So area of the $$\frac{1}{4}$$th of the Circle=$$\pi$$$$a^2$$/4
So the area of the shaded portion=($$\pi$$$$a^2$$/4)-$$\frac{a^2}{2}$$=$$a^2$$($$\pi$$-2/4)
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Re: If PQRO is a square inside a Circle with centre at "O" and [#permalink] 08 Jun 2016, 10:20
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https://services.math.duke.edu/education/postcalc/predprey/pred3a.html | 1,723,100,509,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640723918.41/warc/CC-MAIN-20240808062406-20240808092406-00415.warc.gz | 402,135,074 | 2,357 | ### Systems of Differential Equations: Models of Species Interaction
Part 3.1: Competing Species I
Now we will look at a system of equations that is very similar to the predator-prey system. Indeed the only difference is a change in sign of some of the coefficients.
dx/dt = ax - bxy
dy/dt = cy - pxy
We still assume that a, b, c, and p are positive constants.
• (a) What are the sign changes in this system from the predator-prey system?
(b) Describe a biological scenario with two populations for which this system of equations might apply.
1. The new system again has an equilibrium point. Find formulas for the coordinates of this equilibrium point in terms of a, b, c, and p.
2. Let (x0,y0) be the equilibrium point for the general system
The lines x = x0 and y = y0 divide the first quadrant in the xy-plane into four subregions as indicated below.
For each subregion indicate whether x is increasing or decreasing and whether y is increasing or decreasing.
3. Using paper and pencil, sketch in what you think the pattern of the trajectories should look like.
4. Let a = 1, b = 0.1, c = 1, p = 0.2, and use your computer algebra system to draw the direction field and several trajectories for this system. Do they agree with your pencil and paper sketch?
• (a) What is the difference between the behavior of trajectories near the equilibrium point for this system and the behavior of trajectories near the equilibrium point for a predator-prey system?
(b) Does this seem reasonable in terms of the biological population description you gave in Step 1? Explain why or why not.
• (a) How do the trajectories of solutions to the system change as you vary the ratio b/p? Does this seem reasonable in terms of the biological population description you gave in Step 1? Explain why or why not.
(b) How do the trajectories of solutions to the system change as you vary the ratio a/c? Does this seem reasonable in terms of the biological population description you gave in Step 1? Explain why or why not. | 454 | 2,018 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2024-33 | latest | en | 0.921154 |
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BA 440 (0) | 541 | 1,540 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2019-51 | latest | en | 0.353284 |
http://blog.shriphani.com/2013/12/13/probabilistic-counting/ | 1,438,129,930,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042985140.15/warc/CC-MAIN-20150728002305-00092-ip-10-236-191-2.ec2.internal.warc.gz | 28,800,477 | 6,094 | ## Probabilistic Counting
I recently got my hands on the common-crawl URL dataset from here and wanted to compute some stats like the number of unique domains contained in the corpus. This was the perfect opportunity to whip up a script that implemented Durande et. al’s log-log paper.
The algorithm operates on a stream and produces an estimate for the cardinality of the input stream. Here’s the algorithm itself:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 (ns probabilistic-counting.log-log "The LogLog algorithnm" (:use [incanter.core]) (:import [org.apache.mahout.math MurmurHash] [org.apache.commons.lang3 SerializationUtils])) (defn rho "Number of leading zeros in the bit-representation Args: y : the number itself size : optional : the number of bytes used to represent the number. Default: 4 bytes/32 bits" ([y] (rho y 32)) ([y size] (int (- size (Math/ceil (/ (Math/log (+ y 1)) (Math/log 2))))))) (defn alpha [m] (if (< 64 m) 0.79402 (* 2 (Math/pow (* (gamma (- (/ 1 m))) (/ (- 1 (Math/pow 2 (/ 1 m))) (Math/log 2))) (- m))))) (defn log-log [xs k] (let [m (int (Math/pow 2 k)) buckets (make-array Integer/TYPE m)] (do (map #(aset buckets % 0) (range m)) (doseq [x xs] (let [h (int (MurmurHash/hash (SerializationUtils/serialize x) 1991)) idx (bit-and h (- m 1)) val (max (aget buckets idx) (rho h))] (aset buckets idx val))) (* (Math/pow 2 (/ (apply + buckets) m)) m (alpha m)))))
And here’s a test (actual cardinality = 1,000,000):
1 2 3 (defn demo-log-log [] (log-log (map #(mod % 1000000) (range 10000000)) 10))
When run, we get:
user> (demo-log-log)
1023513.5806580923
which is an error of 2.3%.
Now, let us deploy it on the URL dataset. There are 2,412,755,840 URLs in the format: host_reversed/path/scheme. The following piece of code constructs the host stream and estimates the cardinality.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 (ns probabilistic-counting.demo-urls "Estimate the number of unique domains in the Common crawl dataset" (:use [probabilistic-counting.log-log])) (defn urls-stream [url-file] (-> url-file clojure.java.io/reader line-seq)) (defn hosts-stream [url-stream] (map #(first (clojure.string/split % #"/")) url-stream)) (defn count-hosts [a-hosts-stream] (log-log a-hosts-stream 10)) (defn -main [& args] (let [path (first args) num-urls (when (second args) (-> args second Integer/parseInt))] (if num-urls (->> path urls-stream hosts-stream (take num-urls) count-hosts) (->> path urls-stream hosts-stream count-hosts))))
On a portion of the dataset, these are the results:
Using standard unix tools (uniq works here without a sort because the dataset is already sorted by hostname):
➜ probabilistic_counting git:(master) ✗ cat ~/common_crawl_index_urls | head -n 50000 | cut -d "/" -f 1 | uniq | wc -l
9205
➜ probabilistic_counting git:(master) ✗ cat ~/common_crawl_index_urls | head -n 500000 | cut -d "/" -f 1 | uniq | wc -l
22164
➜ probabilistic_counting git:(master) ✗ cat ~/common_crawl_index_urls | head -n 5000000 | cut -d "/" -f 1 | uniq | wc -l
196318
➜ probabilistic_counting git:(master) ✗ cat ~/common_crawl_index_urls | head -n 50000000 | cut -d "/" -f 1 | uniq | wc -l
1525445
And log-log produces:
➜ probabilistic_counting git:(master) ✗ lein trampoline run ~/common_crawl_index_urls 50000
9677.974613035705
➜ probabilistic_counting git:(master) ✗ lein trampoline run ~/common_crawl_index_urls 500000
22708.710878857888
➜ probabilistic_counting git:(master) ✗ lein trampoline run ~/common_crawl_index_urls 5000000
194919.74158794453
➜ probabilistic_counting git:(master) ✗ lein trampoline run ~/common_crawl_index_urls 5000000
1602155.9911824786
And the accuracy is really very good.
This algorithm works since:
• $\rho(x)$ computes the position of the LSB in $x$.
• The probability that $\rho(x)$ is $k$ is $\frac{1}{2^k}$ (the probability of obtaining a sequence of $k – 1$ zeroes and a one).
• Say, the real cardinality is $n$. Thus, on $\frac{n}{2^k}$ members of this set, $\rho(x)$ will yield a value of $k$.
• As a result, if you can drive $2^k$ as close to $n$ as possible, you have a good estimation of cardinality. This is achieved by maximizing $k$.
• And thus, $\arg\max_{x \in M} \rho(x)$ is an estimator (albeit biased) for $\log(n)$.
This is one of those extremely sweet papers where the idea (minus the details) fits on a business card and the resulting algorithm has immense practical value.
The full source code is available in this github repository.
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(c) Shriphani Palakodety 2013-2014 | 1,481 | 4,665 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2015-32 | longest | en | 0.582622 |
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### Get Complete Lecture Notes for Statistical methods for data science on Cynohub APP
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### Statistical methods for data science Unit One
#### Basic Concepts
Population, sample, parameter and statistic; characteristics of a good estimator; Consistency –Invariance property of Consistent estimator, Sufficient condition for consistency; Unbiasedness; Sufficiency –Factorization Theorem –Minimal sufficiency; Efficiency –Most efficient estimator,likelihood equivalence, Uniformly minimum variance unbiased estimator,applications of Lehmann-Scheffe’s Theorem, Rao -Blackwell Theorem and applications
### Statistical methods for data science Unit Two
#### Point Estimation
Point Estimation-Estimator, Estimate, Methods of point estimation –Maximum likelihood method (the asymptotic properties of ML estimators are not included), Large sample properties of ML estimator(without proof)-applications , Method of moments, method of least squares, method of minimum chi-square and modified minimum chi-square-Asymptotic Maximum Likelihood Estimation and applications.
### Statistical methods for data science Unit Three
#### Interval Estimation
Confidence limits and confidence coefficient; Duality between acceptance region of a test and a confidence interval; Construction of confidence intervals for population proportion (small and large samples) and between two population proportions(large samples); Confidence intervals for mean and variance of a normal population; Difference between the mean and ratio of two normal populations.
### Statistical methods for data science Unit Four
#### Testing of hypotheses
Types of errors, power of a test, most powerful tests; Neyman-Pearson Fundamental Lemma and its applications; Notion of Uniformly most powerful tests; Likelihood Ratio tests: Description and property of LR tests -Application to standard distributions.
### Statistical methods for data science Unit Five
#### Small sample tests
Student’s t-test, test for a population mean, equality of two population means, paired t-test, F-test for equality of two population variances,CRD,RBD,LSD; Chi-square test for goodness of fit and test for independence of attributes, χ2 test for testing variance of a normal distributionSign test, Signed rank test, Median test, Mann-Whitney test, Run test and One sample Kolmogorov –Smirnov test ,Kruskal –Wallis H test(Description, properties and applications only).
### Statistical methods for data science Course Objectives
This course aims at providing knowledge on basic concepts of Statistics, Estimation and testing of hypotheses for large and small samples.
### Statistical methods for data science Course Outcomes
•Understand the basic concepts of Statistics
•Analyze data and draw conclusion about collection of data under study using Point estimation
•Analyze data and draw conclusion about collection of data under study using Interval estimation.
•Analyzing the tests and types of errors for large samples
•Apply testing of hypothesis for small samples.
### Statistical methods for data science Text Books
1.Manoj Kumar Srivastava and Namita Srivastava, Statistical Inference –Testing ofHypotheses, Prentice Hall of India, 2014.
2.Robert V Hogg, Elliot A Tannis and Dale L.Zimmerman, Probability and Statistical Inference,9th edition,Pearson publishers,2013.
### Statistical methods for data science Reference Books
1.S.P.Gupta, Statistical Methods, 33rd Edition, Sultan Chand & Sons.
2.Miller and John E Freund, Probability and Statistics for Engineers, 5th Edition.
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In Six Sigma, We always take decision based on data. We are using Minitab for six sigma projects because most of the statistical and analytical tools we use in Six Sigma are available in Minitab. Anyone who is having the basic knowledge of spreadsheet can use Minitab. You don't have to be a statistician to understand analysis of data and get the insight from your data. Minitab software does the work for you.
COURSE CONTENT:
Module 1 : Introduction and Interoperability to Minitab
1. How to install Minitab
2. Minitab Screen Layout
3. Worksheet Format and Structure
4. Exporting Analysis data from Minitab to Word and PPT What is Six Sigma
Module 2 : Pull Down Menus
Module 3 : Graphical Analysis
1. Bar Chart
2. Pareto Chart
3. Histogram
4. Graphical Summary
5. Pie Chart
6. Probability Plot
7. Box Plot
8. Scatter Plot
9. Matrix Plot
Module 4 : Statistical Analysis
1. Descriptive Statistics
2. Overview of Control Charts
3. Variable Control Charts
4. Attribute Control Charts
5. Overview of Gage R & R (MSA)
6. Attribute Gage R & R
7. Variable Gage R & R
8. Process Capability
9. Simple Linear Regression
10. Multiple Linear Regression
11. Overview of Hypothesis Testing
12. 2 sample t-test
13. Paired t-test
14. Chi-Square Test
15. ANOVA
16. Test of Equal Variances
17. Proportion Test
Who this course is for:
• Quality Managers
• Supervisors, Managers and professionals who want to implement Six Sigma process in the organization.
• Six Sigma Green / Black Belt Certified Professionals
• Beginner Six Sigma Practitioner
• Beginner to Data Analysis
Course content
Expand all 52 lectures 04:20:33
+ Introduction and Interoperability to Minitab
8 lectures 23:21
Welcome to the course
00:50
Installation of Minitab
02:31
Preview 03:44
UPDATE: What' New in Minitab 19 Version
00:17
UPDATE: What' New in Minitab 19 Version
09:47
Worksheet format and Structure
03:31
Export Analysis to Word and PPT
00:53
+ Pull Down Menus of Minitab
4 lectures 06:26
Preview 05:06
00:50
00:11
UPDATE: Practice data in Minitab 19 version
00:19
+ Graphical Tools
9 lectures 51:29
Bar Chart
05:16
Preview 05:36
Histogram
06:24
Preview 06:13
Pie Chart
04:15
Probability Plot
03:47
Box Plot
06:49
Scatter Plot
05:11
Matrix Plot
07:58
This quiz is related to Graphical tools that we have seen till now.
Graphical Tools Quiz
5 questions
+ Statistical Tools
31 lectures 02:59:17
Descriptive Statistics Part 1
10:21
Descriptive Statistics Part 2
05:10
Overview of Control Chart Part 1
04:36
Overview of Control Chart Part 2
04:42
Overview of Control Chart Part 3
11:37
I-MR Chart
05:25
X-bar R Chart
06:11
X-bar s Chart
03:17
p Chart
05:20
NP Chart
03:42
c Chart
02:29
u Chart
03:45
Overview of Gage R & R (MSA)
03:03
Attribute Gage R & R
09:36
Variable Gage R & R Part 1
07:48
Variable Gage R & R Part 2
04:45
Process Capability
07:45
Simple Linear Regression Part 1
04:17
Simple Linear Regression Part 2
06:19
Multiple Regression Part 1
09:15
Multiple Regression Part 2
06:22
Overview of Hypothesis Testing Part 1
04:47
Overview of Hypothesis Testing Part 2
05:56
2-Sample t-test Part 1
06:42
2-Sample t-test Part 2
02:28
2-Sample t-test Part 3
03:44
Paired T-Test
06:47
Contingency Table (Chi-square Test)
05:26
ANOVA (Analysis of Variance)
07:47
Test of Equal Variances
05:13
Proportion Test
04:42
This test is related to the statistical tools which we have seen in above section. All the best!!!
Statistical Tools Practice Test
10 questions
This is the final practice test in which questions from entire course will be there.
So All the best!!!
Final Practice Test
25 questions | 1,387 | 4,877 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2020-24 | latest | en | 0.775316 |
http://wordaligned.org/docs/oulipo/ | 1,632,521,468,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057580.39/warc/CC-MAIN-20210924201616-20210924231616-00441.warc.gz | 66,619,311 | 12,509 | # Lessons from the OuLiPo
Computer programmers and OuLiPians write constrained texts. This talk follows the links.
{ code ~ queneau ~ puzzle ~ perec ~ knight ~ knuth ~ tour ~ tag ~ easter ~ egg ~ bug }
# OuLiPo & Me
## X20
#### Richard Beard, 1996
{365 × 20 × 10} + {2 × 20} (leap years). Equals exactly 73,040. Plus 17 irregulars. Not give or take, not approximately, but exactly seventy-three thousand and fifty-seven. All the same, it’s difficult to prove.
## X20
{365 × 20 × 10} + {2 × 20} (leap years). Equals exactly 73,040. Plus 17 irregulars.
``````
cigarettes_in_pack = x20 = 20
days_in_year = 365
irregulars = 17
cigarettes_smoked = days_in_decade * x20 + irregulars
assert chapters == x20
assert cigarettes_smoked == word_count
assert sections[chapter] + chapter == x20 + 1
``````
## Life A User’s Manual
### Georges Perec, 1978
Grégoire Simpson
Gregory Simpson
Gregor Samsa
Perhaps the greatest 20th century novel — Donald Knuth
## Surreal Numbers
#### Donald Knuth, 1974
How two ex-students turned on to pure mathematics and found total happiness: a novelette
EODERMDROME
SHOES ON HENS
figs lizards snakes heat light figs snakes light lizards heat figs
# OuLiPo & We
## OuLiPo
Ouvroir de littérature potentielle
Founded by poet, Raymond Queneau, in Paris, 1960
Primarily mathematicians and writers
Analytic OuLiPo – investigate formal devices used by writers
Synthetic OuLiPo – apply patterns from formal languages such as maths, logic, computer science ... chess!
Deceased members are excused from meetings but are not allowed to withdraw.
{ queneau ~ berge ~ duchamps ~ calvino ~ perec ~ fournel }
Rats who build the labyrinth from which they will try to escape
## Why make it harder?
1. Structure
2. Comedy
3. Accessibility
## Why make it harder?
1. Structure
2. Comedy
3. Accessibility
4. Profit
## What’s happening?
### (Answers limited to 140 characters)
• no error in the system
• no reror in the system
• no rreor in the system
• no rroer in the system
• no rrore in the system
• no rror ein the system
• no rror ien the system
• no rror ine the system
• no rror in ethe system
• no rror in tehe system
• no rror in thee system
• no rror in the esystem
• no rror in the seystem
• no rror in the syestem
• no rror in the sysetem
• no rror in the systeem
• no rror in the systeem
• no rror in the systeme
• eno rror in the system
• neo rror in the system
• noe rror in the system
• no error in the system
Eugen Gomringer 1969
# The Rules
## integer literals
First, we want to establish the idea that a computer language is not just a way of getting a computer to perform operations but rather that it is a novel formal medium for expressing ideas about methodology. Thus, programs must be written for people to read, and only incidentally for machines to execute.
## Why make it harder?
1. ... for machines to execute
# 2147483647
A Magic Number
Proved a prime by Leonhard Euler
# ☠
### Integers must not exceed this value
We never thought a video would be watched in numbers greater than a 32-bit integer (=2,147,483,647 views), but that was before we met PSY. Gangnam Style has been viewed so many times we had to upgrade to a 64-bit integer (9,223,372,036,854,775,808)!
### The UNIX Time-Sharing System
There have always been fairly severe size constraints on the system and its software. Given the partially antagonistic desires for reasonable efficiency and expressive power, the size constraint has encouraged not only economy but a certain elegance of design.
— Dennis Ritchie and Ken Thompson
## Choose any three.
### !amanaP – lanac a ,nalp a ,nam A
A man, a plan, a canoe, pasta, heros, rajahs, a coloratura, maps, snipe, percale, macaroni, a gag, a banana bag, a tan, a tag, a banana bag again (or a camel), a crepe, pins, Spam, a rut, a Rolo, cash, a jar, sore hats, a peon, a canal – Panama!
## Le Grand Palindrome
### Georges Perec, 1969
Trace l’inégal palindrome. Neige. Bagatelle, dira Hercule. Le brut repentir, cet écrit né Perec. L’arc lu pèse trop, lis à vice-versa. Perte. Cerise ... Désire ce trépas rêvé : Ci va ! S’il porte, sépulcral, ce repentir, cet écrit ne perturbe le lucre : Haridelle, ta gabegie ne mord ni la plage ni l’écart
word count = 1247
## Why make it harder?
1. Structure
2. Comedy
3. Accessibility
4. Profit
5. Challenge
[…] Perec’s palindrome barely made sense to the readers. Some teachers took it for the work of an incompetent student, while others suspected that they had been treated to a surrealist text produced by “automatic writing”. Those with psychiatric interests identified the author as an adolescent in a dangerously paranoid state; those who had not forgotten the swinging sixties wondered whether it was LSD or marijuana that had generated the disconnected images of the text. Readers seem to project their own positive and negative fantasies onto Perec’s palindrome, as they do onto other difficult, obscure and unattributed works.
— David Bellos, Georges Perec: a Life in Words
## 20:02 20/02 2002
#### Peter Norvig
A man, a plan, a caddy, Ore, Lee, tsuba, Thaine, a lair, ..., Hell, a burial, Aeniah, Tabu, Steele, Roydd, a canal, Panama
word count = 15139
### Norvig’s Palindrome Finder
``````def consider_candidates(self):
"""Push a new state with a set of candidate words onto stack."""
if self.diff > 0: # Left is longer, consider adding on right
dir = 'right'
substr = self.left[-1][-self.diff:]
candidates = self.dict.endswith(substr)
elif self.diff < 0: # Right is longer, consider adding on left
dir = 'left'
substr = reversestr(self.right[-1][0:-self.diff])
candidates = self.dict.startswith(substr)
else: # Both sides are same size
dir = 'left'
if not self.used_reversibles:
substr = ''
candidates = self.dict.startswith('')
if substr == reversestr(substr):
self.report()
self.stack.append(('trying', dir, substr, candidates))
``````
## Why make it harder?
1. ... So machines can execute them
2. Challenge
3. Education
### Algol Poetry – Noël Arnaud, 1968
``` Table
Begin: to make format,
go down to comment
while channel false
(if not true). End.
```
### `Black Perl`
```BEFOREHAND: close door, each window & exit; wait until time.
open spellbook, study, read (scan, select, tell us);
write it, print the hex while each watches,
reverse its length, write again;
kill spiders, pop them, chop, split, kill them.
unlink arms, shift, wait & listen (listening, wait),
sort the flock (then, warn the "goats" & kill the "sheep");
kill them, dump qualms, shift moralities,
values aside, each one;
die sheep! die to reverse the system
you accept (reject, respect);
next step,
kill the next sacrifice, each sacrifice,
wait, redo ritual until "all the spirits are pleased";
do it ("as they say").
do it(*everyone***must***participate***in***forbidden**s*e*x*).
return last victim; package body;
exit crypt (time, times & "half a time") & close it,
select (quickly) & warn your next victim;
AFTERWARDS: tell nobody.
wait, wait until time;
wait until next year, next decade;
sleep, sleep, die yourself,
die at last
# Larry Wall, 1990
```
#### ⸮⸮ International Obfuscated C Code Contest ??
```/*(c) 2001 Thad */
#include<string.h>
#include <stdio.h>
#define abc stdout
int main(int a,ch\
ar*b){char*c="??="
"??(??/??/??)??'{"
"??!??>??-";while(
!((a=fgetc(stdin))
==EOF))fputc((b=s\
trchr(c,a))?fputc(
fputc(077,abc),abc
),"=(/)'<!>""-"??(
b-c??):a, abc);??>
```
# Perec
## Georges Perec, 1936 – 1982
parents died in World War II
won Prix Renaudot in 1965 for his first novel, Les Choses
archivist, 1961 – 1978
chess player
cruciverbist
genius
OuLiPian 1967 –
## Perec in the Lab
Designed and and managed a scientific database for the Laboratoire de Physiologie at the Centre Hospitalier et Universitaire Saint-Antoine.
## La Disparition
### 1969
This book avoids using a particular symbol. How crazy is that? Author GP must adapt sounds and jump through hoops to finish this book. Such a constraint constructs an unusual story about a group of humans looking for a missing companion, Anton Vowl. I could try and go on ...
✂ ✂ ✂ ✂ ✂ ✂ ✂
`8<--8<--8<--8<--8<--8<--8<--8<--`
## Why make it harder?
1. Structure
2. Comedy
3. Accessibility
4. Profit
5. Challenge
6. Inspiration
# Life A User’s Manual
## Knight’s Tour
On an 8 × 8 board, there are 26,534,728,821,064 directed closed tours.
## Easter Egg
The woman carries in her right hand a bulky set of keys, no doubt the keys of all the flats she has visited that day; some are fixed to novelty key rings: a miniature bottle of Marie Brizard apéritif, a golf tee and a wasp, a double-six domino, and a plastic octagonal token in which is set a tuberose flower.
## Easter Egg
When writing a specialization, be careful about its location; or to make it compile will be such a trial as to kindle its self-immolation.
# Clinamen
## On the Nature of Things
### Lucretius, 99-55 BC
When atoms move straight down through the void by their own weight, they deflect a bit in space at a quite uncertain time and in uncertain places, just enough that you could say that their motion has changed. But if they were not in the habit of swerving, they would all fall straight down through the depths of the void, like drops of rain, and no collision would occur, nor would any blow be produced among the atoms. In that case, nature would never have produced anything.
The clinamen can be used freely anywhere it isn’t required. According to Calvino it is the only device that “can make the text a true work of art”.
## References
• Life A User’s Manual, Georges Perec
• Georges Perec: A Life in Words, David Bellos
• X20, Richard Beard
• Oulipo Compendium, Harry Matthews
{ END ~ CODE ~ ONCE } | 2,698 | 9,743 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2021-39 | latest | en | 0.79205 |
https://labs.tib.eu/arxiv/?author=Sara%20A.%20Van%20De%20Geer | 1,586,149,271,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371618784.58/warc/CC-MAIN-20200406035448-20200406065948-00233.warc.gz | 532,535,704 | 8,189 | • ### The Smooth-Lasso and other $\ell_1+\ell_2$-penalized methods(1003.4885)
Oct. 7, 2011 math.ST, stat.TH
We consider a linear regression problem in a high dimensional setting where the number of covariates $p$ can be much larger than the sample size $n$. In such a situation, one often assumes sparsity of the regression vector, \textit i.e., the regression vector contains many zero components. We propose a Lasso-type estimator $\hat{\beta}^{Quad}$ (where '$Quad$' stands for quadratic) which is based on two penalty terms. The first one is the $\ell_1$ norm of the regression coefficients used to exploit the sparsity of the regression as done by the Lasso estimator, whereas the second is a quadratic penalty term introduced to capture some additional information on the setting of the problem. We detail two special cases: the Elastic-Net $\hat{\beta}^{EN}$, which deals with sparse problems where correlations between variables may exist; and the Smooth-Lasso $\hat{\beta}^{SL}$, which responds to sparse problems where successive regression coefficients are known to vary slowly (in some situations, this can also be interpreted in terms of correlations between successive variables). From a theoretical point of view, we establish variable selection consistency results and show that $\hat{\beta}^{Quad}$ achieves a Sparsity Inequality, \textit i.e., a bound in terms of the number of non-zero components of the 'true' regression vector. These results are provided under a weaker assumption on the Gram matrix than the one used by the Lasso. In some situations this guarantees a significant improvement over the Lasso. Furthermore, a simulation study is conducted and shows that the S-Lasso $\hat{\beta}^{SL}$ performs better than known methods as the Lasso, the Elastic-Net $\hat{\beta}^{EN}$, and the Fused-Lasso with respect to the estimation accuracy. This is especially the case when the regression vector is 'smooth', \textit i.e., when the variations between successive coefficients of the unknown parameter of the regression are small. The study also reveals that the theoretical calibration of the tuning parameters and the one based on 10 fold cross validation imply two S-Lasso solutions with close performance.
• ### On the conditions used to prove oracle results for the Lasso(0910.0722)
Oct. 5, 2009 math.ST, stat.TH, stat.ML
Oracle inequalities and variable selection properties for the Lasso in linear models have been established under a variety of different assumptions on the design matrix. We show in this paper how the different conditions and concepts relate to each other. The restricted eigenvalue condition (Bickel et al., 2009) or the slightly weaker compatibility condition (van de Geer, 2007) are sufficient for oracle results. We argue that both these conditions allow for a fairly general class of design matrices. Hence, optimality of the Lasso for prediction and estimation holds for more general situations than what it appears from coherence (Bunea et al, 2007b,c) or restricted isometry (Candes and Tao, 2005) assumptions.
• ### High-dimensional generalized linear models and the lasso(0804.0703)
April 4, 2008 math.ST, stat.TH
We consider high-dimensional generalized linear models with Lipschitz loss functions, and prove a nonasymptotic oracle inequality for the empirical risk minimizer with Lasso penalty. The penalty is based on the coefficients in the linear predictor, after normalization with the empirical norm. The examples include logistic regression, density estimation and classification with hinge loss. Least squares regression is also discussed.
• ### On non-asymptotic bounds for estimation in generalized linear models with highly correlated design(0709.0844)
Sept. 6, 2007 math.ST, stat.TH
We study a high-dimensional generalized linear model and penalized empirical risk minimization with $\ell_1$ penalty. Our aim is to provide a non-trivial illustration that non-asymptotic bounds for the estimator can be obtained without relying on the chaining technique and/or the peeling device. | 876 | 4,056 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2020-16 | latest | en | 0.889728 |
https://www.examfriend.in/questions-and-answers/TS-EAMCET-2017/Mathematics/General-questions/0.html | 1,726,146,348,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651457.35/warc/CC-MAIN-20240912110742-20240912140742-00642.warc.gz | 715,792,028 | 9,591 | 12345>>
1.
If A and B are events having probabilities P(A)=0.6 , P(B)=0.4 and $P(A\cap B)=0$ , then probability that neither A nor B occurs is
A) $\frac{1}{4}$
B) 1
C) $\frac{1}{2}$
D) 0
2.
If the line x-y=-4K is a tangent to the parabola $y^{2}=8x$ at P, then the perpendicular distance of normal at P from (K,2K) is
A) $\frac{5}{2\sqrt{2}}$
B) $\frac{7}{2\sqrt{2}}$
C) $\frac{9}{2\sqrt{2}}$
D) $\frac{1}{2\sqrt{2}}$
3.
The lengths of the sides of a triangle are13,14 and 15. If R and r repectively denote circumradius and inradius of that triangle , then 8R+r=
A) 84
B) $\frac{65}{8}$
C) 4
D) 69
4.
The number of solutions of $\cos 2 \theta=\sin \theta$ in $(0,2\pi)$ is
A) 4
B) 3
C) 2
D) 5
5.
If the pair of straight lines xy-x-y+1=0 and the line x+ay-3=0 are concurrent, then the acute angle between the pair of lines
$ax^{2}-13xy-7y^{2}+x+23y-6=0$ is
A) $\cos^{-1}\left(\frac{5}{\sqrt{218}}\right)$
B) $\cos^{-1}\left(\frac{1}{\sqrt{10}}\right)$
C) $\cos^{-1}\left(\frac{5}{\sqrt{173}}\right)$
D) $\cos^{-1}\left(\frac{1}{\sqrt{5}}\right)$
12345>> | 459 | 1,094 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2024-38 | latest | en | 0.434404 |
https://www.orchidsinternationalschool.com/calculators/math-calculator/antilog-calculator | 1,726,682,998,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651931.60/warc/CC-MAIN-20240918165253-20240918195253-00470.warc.gz | 833,467,984 | 8,968 | # Antilog Calculator
The antilogarithm, often referred to as "antilog," is the inverse operation of finding the exponent to which a given base must be raised to produce a specific number. In simpler terms, it is the inverse function of the logarithm. Calculating antilogs is particularly useful in various mathematical and scientific calculations, especially when dealing with exponential growth or decay.
### How Does the Antilog Calculator Work?
The calculator raises the given base to the power of the entered logarithm value to compute the antilog.
### When Should I Use an Antilog Calculator ?
An antilog calculator is beneficial in scenarios where you need to find the original value from its logarithm, aiding in exponential calculations and solving exponential equations.
### What is the Formula for Calculating Antilog?
`Antilog(x)=Basex`
### Examples
Example 1:
Calculate 10 3.5 (antilog of 3.5 with base 10)
Calculation:
Antilog (3.5) = 103.5
Antilog (3.5) = 3162.2776
Example 2:
Find e 2.7 (antilog of 2.7 with base e)
Calculation:
Antilog (2.7) = e2.7
Antilog (3.5) = 14.8797
Example 3:
Determine 2 4.2 (antilog of 4.2 with base 2)
Calculation:
Antilog (4.2) = 24.2
Antilog (4.2) = 18.3797
Example 4:
Calculate the antilog of 2.5 with a base of 3
Calculation:
Antilog (2.5) = 32.5
Antilog (2.5) = 15.59
Example 5:
Determine the antilog of -1 with a base of 2
Calculation:
Antilog (-1) = 3-1
Antilog (-1) = 0.5
Yes, finding the antilog is akin to determining a number raised to a specific power.
The antilog can be a positive real number, but not negative or zero, as it represents the inverse of a logarithm.
Antilogs can be calculated for different bases, such as base 10 (common logarithm), base e (natural logarithm), or any other specified base.
Yes, you can calculate the antilog of any number using a specific base in the formula Basex
If the logarithmic value is negative, the antilog would result in a fraction or decimal depending on the base used.
Yes, the antilog is the inverse operation of finding the number whose logarithm is known. It's the reverse process of exponentiation to find the original number from its logarithm.
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ORCHIDS - The International School | Terms | Privacy Policy | Cancellation | 613 | 2,307 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2024-38 | latest | en | 0.81211 |
http://umdberg.pbworks.com/w/page/47130472/Bulk%20modulus%20--%20solids | 1,718,640,123,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861733.59/warc/CC-MAIN-20240617153913-20240617183913-00628.warc.gz | 31,896,061 | 9,503 | • If you are citizen of an European Union member nation, you may not use this service unless you are at least 16 years old.
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# Bulk modulus -- solids
last edited by 12 years, 7 months ago
Working content > MacroModels > Solids
Prerequisites
From our discussions of phenomenological macroscopic forces like normal forces, tension, and friction, we expect that solids will deform a little if we exert a force on it. (We also expect this from our understanding of interatomic forces.) Let's consider how we might describe how a solid would respond to being squeezed.
Conceptually, the simplest situation for seeing how a solid might deform in response to forces is to consider a rectangular block of some solid material and push on it on all sides. How should we think about such forces? Should we try to exert equal forces on each side of the block? Even if the sides are not the same size?
We are interested in extracting a parameter that tells how a solid object responds to forces that is a property of the material in general -- not that of the specific object. Sort of like the way density tells us about the mass of a material generically -- whereas the mass (density times volume) tells us about a specific object. How can we extract the object independent property of the material?
A more natural way to think about what might be an appropriate way to exert forces on a solid it to think of a few molecules in the surface of the solid.
If the pressure pushing on all of the surfaces of a block increases, the block will compress and take up less volume (Fig 2). This change in volume might be very small if the atoms or molecules making up the block push against each other and resist the change. We describe the change in volume in response to a uniform increase in pressure as the bulk modulus, B:
Here, ΔP is the change in pressure necessary to cause a fractional change in volume ΔV/V, where ΔV is the volume change, and V is the original volume. The more resistant a solid is to compression, the more pressure it takes to change the volume, and the larger the bulk modulus is. Since ΔV/V has units of volume / volume, the denominator is unitless. Therefore, B has units of pressure, e.g. Pa. Since the air pressure is pushing on each of the surfaces, the block will shrink along all three dimensions.
This looks like the Young's modulus equation except we are looking at the fractional change in volume instead of in length:
The bulk modulus of solid materials can be quite large. For steel, B= 1.60x1011 Pa. This means that to get a volume change of 0.1% requires a pressure increase of 1.6x108 Pa which is 1580 times atmospheric pressure (1 atm ~ 105 Pa). Obviously, not much happens under typical atmospheric conditions.
The bulk modulus is important for another reason, as it tells how easy it is for a pressure wave to move through the solid. Since sound is essentially a pressure wave (alternating higher and lower pressure) B will be important in how well sound is transmitted through different materials.
The bulk modulus for biological materials is generally relatively high, though not as high as engineering materials such as steel (160 x 109 Pa). For bone, the bulk modulus is 15 x 109 Pa. The bulk modulus will determine how much a body changes volume in response to a change in pressure. Because the size of a body is mostly determined by the size of it’s bones, we can consider whether bones change size in response to different pressures. Organisms do encounter a range of pressures on earth. For example, air pressure changes with altitude. In Denver (the Mile high city) air pressures is 17% less than at sea level by 17%. Air pressure is 49% lower on the top of Denali in Alaska, and 65% on the top of Mt Everest. Animals that fly very high will also encounter lower pressures. These pressure decreases could result in a small increase in the volume of bones in the body. While these are unlikely to be large, they are not insignificant.
Even larger pressure changes are encountered when animals swim to depth in the ocean. For example, sperm whales dive to 400 m depths where the absolution pressure is 40 times greater than at the surface. Elephant seals may dive up to a mile down (1600 m) where the pressure is 160 times that at the surface. It is possible that bone volume shrinks a bit in response to this increase in pressure.
Karen Carleton and Joe Redish 10/27/11 | 1,017 | 4,637 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2024-26 | latest | en | 0.905792 |
https://www.stumblingrobot.com/2015/07/06/prove-by-induction-a-property-of-the-alternating-sum-of-odd-integers/ | 1,718,883,426,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861940.83/warc/CC-MAIN-20240620105805-20240620135805-00285.warc.gz | 887,248,487 | 15,563 | Home » Blog » Prove by induction a property of the alternating sum of odd integers
# Prove by induction a property of the alternating sum of odd integers
Prove that
This implies the sum is proportional to with constant of proportionality 2.
Proof. The proof is by induction. For the case we have, on the left,
On the right we have . Hence, the formula holds for this case.
Assume then that the formula holds for some . Then,
Thus, if the statement is true for then it is true for . Hence, we have established the statement is true for all | 123 | 544 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2024-26 | latest | en | 0.917332 |
http://lbartman.com/worksheet/free-math-worksheets-grade-4.php | 1,606,859,187,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141681524.75/warc/CC-MAIN-20201201200611-20201201230611-00067.warc.gz | 60,096,783 | 11,289 | ## lbartman.com - the pro math teacher
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# Free Math Worksheets Grade 4
Public on 17 Oct, 2016 by Cyun Lee
### 1000 images about maths worksheets on pinterest math worksheets
Name : __________________
Seat Num. : __________________
Date : __________________
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show printable version !!!hide the show
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fun kindergarten math worksheets | 1,406 | 3,898 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2020-50 | latest | en | 0.187006 |
https://practicaldev-herokuapp-com.global.ssl.fastly.net/akhilpokle/convert-number-to-hexadecimal-solving-a-short-facebook-interview-question-3427 | 1,642,828,667,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303747.41/warc/CC-MAIN-20220122043216-20220122073216-00153.warc.gz | 518,604,052 | 28,957 | ## DEV Community is a community of 788,395 amazing developers
We're a place where coders share, stay up-to-date and grow their careers.
Akhil
Posted on
Question : Given an integer, write an algorithm to convert it to hexadecimal.
Lets tart with what's hexadecimal numbers ?
Hexadecimal number are the number represented in base 16, it consists of 16 symbols
+--------------------------------------------------------------------+
| Decimal : 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 |
| Hexadecimal : 0 1 2 3 4 5 6 7 8 9 A B C D E F |
+--------------------------------------------------------------------+
To make our lives easier, let's create an array that will store the hexadecimal values to its corresponding decimal index.
let map = ['0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'a', 'b', 'c', 'd', 'e', 'f'];
### Handling Positive Numbers
Handling positive numbers is easy. It's a 3 step operation and it's intuitive :
Step 1 : store the result of num%16
Step 2 : perfrom num/16
Step 3 : perform step 1,2 till num > 0
let res = "";
while(num > 0) {
let digit = num % 16;
res = arr[digit] + res;
num = Math.floor(num / 16);
}
### Handling Negative Integers
Handling negative integers becomes a bit tricky, since we can't write -#3AC to represent negative hexadecimal numbers, so let's a dive deeper and represent numbers in their binary forms.
And since any number is bolied down to binary 0's and 1's, we face the same issue of representing negative numbers in binary format since a computer won't understand -0010.
So to solve this issue, negative numbers are represented by setting Most Significant Bit to 1.
So how can we use this two key information to solve our problem ?
On closer look we see this :
Since an integer is 32-bit, which is further broken down to parts of 4-bit, and binary representation of numbers 8 - 16 have 1 set as their Most Significant Bit and 8 - F represent those numbers, so we could say that 8 - F range could be used to represent negative numbers.
So a hex number #FFFFF63C represents a negative number.
Whenever we come across a number < 0, we add 2^32, to it to convert into a format which could be mapped with hexadecimal mappings.
var toHex = function(num) {
let map = ['0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'a', 'b', 'c', 'd', 'e', 'f'];
if (num == 0) return "0";
if (num < 0) {
num += Math.pow(2,32);
}
let res = "";
while(num > 0) {
let digit = num % 16;
res = map[digit] + res;
num = Math.floor(num / 16);
}
return res;
};
This was a normal way to do it, now let's see an even smarter way of achieving the same which will definitely impress your interviewer.
### Smarter way
For this we need to understand two basic concepts of bit manipulation.
& operator
1 & 1 = 1
0 & 0 = 0
0 & 1 = 0
1 & 0 = 0
>>> right shit operator
shifts bit's to right
5 >>> 1 = 101 >>> 1 = 10 = 2.
Here the number is being shifted right once.
So if we perform -14&15 , -14&15 we get, &15 because we want to convert it to hex and F equals 15 :
Based on this we can say that &15 will convert negative decimal in relevant hexadecimal negative value while preserving positive decimal value.
Now all the basics out of the way, this technique consists of two steps.
Step 1 > res += map[num&15]
Step 2 > num>>>4.
Step 3 Repeat step 1 & 2 till num != 0
We performing step 2 is similar to diving num/16. Since 15 is 1111 ie 4 bits in binary form, we preform the "&" operation and remove those 4 bits.
Converting it to code :
var toHex = function(num) {
if (num == 0) return '0';
let map = ['0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'a', 'b', 'c', 'd', 'e', 'f'];
let result = '';
while (num != 0) {
let c = map[num & 15]; // map last 4 bits to a hex digit
result = c + result;
num = num >> 4;
}
return result;
};
I hope you liked my article :) | 1,127 | 3,856 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2022-05 | longest | en | 0.803831 |
https://outschool.com/ko/classes/math-kindergarten-drills-nsdmQLCc | 1,716,126,585,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057788.73/warc/CC-MAIN-20240519132049-20240519162049-00450.warc.gz | 392,567,718 | 108,179 | 한국어
수학 유치원 훈련
Mrs. Mathis
평균 평점:4.7수강 후기 수:(59)
유치원 수학 숙달 수업에 오신 것을 환영합니다! 이 8주 프로그램에서 귀하의 자녀는 특별히 유치원생을 위해 맞춤화된 흥미진진한 수학 여행을 시작하게 됩니다.
보고계신 지문은 자동 번역 되었습니다
수업 경험
미국 등급 Kindergarten
`이 수업은 English로 진행됩니다.`
```Throughout the Math Kindergarten Drills class, students will embark on an exciting mathematical journey, gaining proficiency in various key areas:
Number Recognition and Counting:
Identifying and writing numbers from 1 to 20.
Counting forwards and backwards within 100.
Understanding the concept of odd and even numbers.
Basic Operations (Addition and Subtraction within 10):
Mastering addition and subtraction facts within 10 through hands-on activities and games.
Learning different strategies for solving simple math problems.```
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`I have the privilege of teaching a diverse range of classes. In the mornings, I bring the magic of learning to our enthusiastic kindergarteners, guiding them through their first steps in education. I embark on adventures with the inquisitive minds... .css-gw2jo8{position:relative;display:inline-block;font-family:'Ginto Normal',sans-serif;font-style:normal;font-weight:500;font-size:1.6rem;text-align:center;text-transform:none;height:auto;max-width:100%;white-space:nowrap;cursor:pointer;-webkit-user-select:none;-moz-user-select:none;-ms-user-select:none;user-select:none;outline:none;border:none;background:none;padding:0;-webkit-transition:all ease-in-out 0.05s,outline 0s,;transition:all ease-in-out 0.05s,outline 0s,;line-height:1;color:#380596;}.css-gw2jo8:hover:not(:disabled),.css-gw2jo8:focus:not(:disabled){color:#380596;-webkit-text-decoration:underline;text-decoration:underline;}.css-gw2jo8:active:not(:disabled){color:#380596;}.css-gw2jo8:disabled{color:#C2C2C2;cursor:default;}.css-gw2jo8:focus-visible{outline-width:2px;outline-style:solid;outline-color:#4B01D4;outline-offset:2px;}`
₩100
매주 또는 ₩264 24 개 수업의 경우
주당 3회, 8주
25분
실시간 화상 수업
연령: 5-7
수업당 학습자 3-3 명
고객 지원
안전 정책개인 정보CA 개인 정보어린이 개인정보이용 약관 | 1,198 | 3,632 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-22 | latest | en | 0.303957 |
https://www.physicsforums.com/threads/energy-work-pendulum-problem.346827/ | 1,532,058,392,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591481.75/warc/CC-MAIN-20180720022026-20180720042026-00140.warc.gz | 945,351,518 | 13,109 | # Homework Help: Energy/Work pendulum problem
1. Oct 18, 2009
### Leesh09
1. The problem statement, all variables and given/known data
Imagine a space station has been built on Venus, and a pendulum is taken outside to determine the acceleration of gravity. The pendulum is a ball having mass m is connected by a strong string of length L to a pivot point and held in place in a vertical position. A wind exerting constant force of magnitude F is blowing from left to right.
Venus has an extremely dense atmosphere, which consists mainly of carbon dioxide and a small amount of nitrogen. The winds near the surface of Venus are much slower than that on Earth. They actually move at only a few kilometers per hour (generally less than 2 m/s and with an average of 0.3 to 1.0 m/s), but due to the high density of the atmosphere at the surface, this is still enough to transport dust and small stones across the surface.
(a) If the ball is released from rest, what is the maximum height H reached by the ball, as measured from its initial height? Check if your result is valid both for cases when 0 H L, and for L H 2L.
(b) Compute the value of H using the values m = 2.00 kg, L = 2.00 m, and F = 14.7 N. The gravitational acceleration on Venus is measured to be m/s2
(c) Using these same values, determine the equilibrium height of the ball.
(d) Could the equilibrium height ever be larger than L? Explain.
2. Relevant equations
3. The attempt at a solution
I honestly have no idea where to even start this. My only thought was you take the maximum velocity that the wind could be and somehow use this to calculate height?
2. Oct 18, 2009
### rl.bhat
Identify the forces acting on the ball in its defected position. From that find the angle of deflection with vertical by resolving the forces along x and y axis. | 434 | 1,821 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2018-30 | latest | en | 0.948004 |
https://www.gradesaver.com/textbooks/math/algebra/algebra-a-combined-approach-4th-edition/chapter-11-section-11-1-solving-quadratic-equations-by-completing-the-square-practice-page-758/4 | 1,531,801,160,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589557.39/warc/CC-MAIN-20180717031623-20180717051623-00329.warc.gz | 887,691,957 | 13,974 | ## Algebra: A Combined Approach (4th Edition)
$x = (\frac{1+2i}{3}, \frac{1-2i}{3})$
$(3x-1)^2 = -4$ $3x-1 = ±\sqrt -4$ (Use Square root Property) $3x-1= ± 2i$ (Simplify the radical) $3x = 1 ± 2i$ (Add 1 to both sides) $x=\frac{1±2i}{3}$ (Divide by 3 on both sides) The solution set is $(\frac{1 + 2i}{3}, \frac{1-2i}{3})$ Note : Root of a negative number is a complex number and is represented by 'i' | 167 | 402 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2018-30 | latest | en | 0.767822 |
https://cs.stackexchange.com/questions/30297/how-to-verify-permutation-generated-in-constant-amortized-time | 1,716,031,553,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057379.11/warc/CC-MAIN-20240518085641-20240518115641-00411.warc.gz | 158,565,171 | 40,812 | how to verify permutation generated in constant amortized time?
Here is an algorithm that generates the next permutation in lexicographic order, changing the given permutation in-place:
1. Find the largest index k such that a[k] < a[k+1]. If no such index exists, exit (the permutation is the last permutation).
2. Find the largest index l such that a[k] < a[l].
3. Swap a[k] with a[l].
4. Reverse the sequence from a[k+1] up to and including the final element a[n].
Is the next permutation generated in constant amortized time and if yes, how to verify it?
• I suggest you run this algorithm for small values of $n$. This will allow you to check whether it indeed generates the next permutation in lexicographic order, and will allow you to measure the average number of operations per step. If this average doesn't increase with $n$, then probably the algorithm takes constant amortized time. Looking at the pattern of the running times for small values of $n$ will probably help with the proof. Sep 25, 2014 at 17:44
• More concretely, the running time of each step is $O(n-k)$, so I suggest you understand how the quantity $n-k$ changes as we go over all permutations for small $n$. Sep 25, 2014 at 17:45
For a number $m$, let $\alpha(m) = \max \{ k : k! | m \}$. This sequence starts at $\infty,1,2,1,2,1,3,\ldots$. I claim that in iteration $t$ of the algorithm (where $1 \leq t \leq n!-1$), we have $n-1-k = \alpha(t)$. The average value of $\alpha(m)$ is roughly $$\left(1 - \frac{1}{2}\right) 1 + \left(\frac{1}{2} - \frac{1}{6}\right) 2 + \left(\frac{1}{6} - \frac{1}{24}\right) 3 + \cdots = \\ 1 + \frac{1}{2} + \frac{1}{6} + \cdots = e-1.$$ Since the running time of each step of the algorithm is proportional to $n-1-k$, we get that the algorithm runs in constant amortized time.
It remains to prove the claim $n-1-k = \alpha(t)$, which I leave to the OP.
• finding $$l$$ costs $$n-k$$
• swap costs $$3$$
• reversing costs $$2(n-k)$$
Thus, having found $$k$$ we do $$3(n-k+1)$$ steps.
In order to find $$k$$, we need to do $$n-k$$ comparisons which costs $$2(n-k)$$. Observe that after reversing, the tail of our permutation is increasing, so we only need to fix the first $$k$$ elements at most. This gives us that each $$k$$ is looked up at most $$\frac{n!}{(n-k)!}$$ times. Now we can get an upper bound on the total cost: \begin{align*} \sum_{k=1}^{n-1} \frac{n!}{(n-k)!} \cdot 3(n-k+1) &\leq 3\sum_{k=1}^{n-1} \frac{n!}{(n-k)!} \cdot (n-k+1) \\ &\leq 3\cdot\sum_{k=1}^{n-1} \frac{n!}{(n-k-1)!} \cdot \frac{n-k+1}{n-k} \\ &\leq 3\cdot\sum_{k=2}^{n-1} \frac{n!}{(n-k)!} \cdot 2 \\ &\leq 6 \cdot n!\cdot\sum_{k=2}^{n-1} \frac{1}{(n-k)!} \\ &< 6 \cdot n!\cdot e \end{align*}
Thus, we obtain $$O(n!)$$ after all. Since there are $$n!$$ permutations, the amortized time to produce the next one is constant. | 916 | 2,825 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 15, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2024-22 | latest | en | 0.859412 |
https://de.mathworks.com/matlabcentral/profile/authors/81066?page=2 | 1,685,994,361,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652161.52/warc/CC-MAIN-20230605185809-20230605215809-00095.warc.gz | 242,924,579 | 3,348 | Gelöst
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https://math.stackexchange.com/questions/637728/splitting-a-sandwich-and-not-feeling-deceived/637744 | 1,716,039,833,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057422.43/warc/CC-MAIN-20240518121005-20240518151005-00226.warc.gz | 337,322,931 | 77,988 | # Splitting a sandwich and not feeling deceived
This is a problem that has haunted me for more than a decade. Not all the time - but from time to time, and always on windy or rainy days, it suddenly reappears in my mind, stares at me for half an hour to an hour, and then just grins at me, and whispers whole day: "You will never solve me..."
Please save me from this torturer.
Here it is:
Let's say there are two people and a sandwich. They want to share the sandwich, but they don't trust each other. However, they found the way how both of them will have a lunch without feeling deceived: One of them will cut the sandwich in two halves, and another will choose which half will be his. Fair, right?
The problem is:
Is there such mechanism for three people and a sandwich?
EDIT: This was roller-coaster for me. Now, it turns out that there are at least two books devoted exclusively on this problem and its variations:
Fair Division
Cake Cutting Algorithms
Yesterday, I was in a coffee shop in a small company. We ordered coffee and some chocolate cakes. As I was cutting my cake for my first bite, I felt sweat on my forehead. I thought, 'What if some of my buddies just interrupt me and say: Stop! You are not cutting the cake in a fair manner!' My hands started shaking in fear of that. But, no, nothing happened, fortunately.
• Yes, see Wikipedia: Category:Fair division and in particular Selfridge–Conway Jan 14, 2014 at 5:20
• I can't believe this. My salvation was in Wikipedia? Jan 14, 2014 at 5:23
• Since we are in mathematics, we might want to apply the Banach–Tarski paradox N-1 times to solve the problem for N people... Jan 14, 2014 at 10:57
• This seems trivial. Each person rolls dice. The low score cuts the sandwhich into portions. Then each participant chooses a single portion according to highest dice order. Now since the cutter chooses last, he will obtain the smallest portion, as in the case with 2 participants, but it is clear that the number of participants do not matter. Therefore, he will maximize the size of the smallest portion. But this must mean that all portions are equal. Am I missing something here?
– Paul
Jan 14, 2014 at 11:01
• @paul your algorithm assumes that the players agree on which pieces are bigger than each other. The problem becomes nontrivial when you remove that assumption. If A cuts the cake, then A will believe all three pieces are equal. Then B takes a piece-but if C disagrees with A and believes B's piece is bigger than the others, the solution isn't envy-free. That's the beauty of the Selfridge-Conway procedure: all three players are guaranteed to believe they got a fair slice. Jan 14, 2014 at 22:00
For more than two, the moving knife is a nice solution. Somebody takes a knife and moves it slowly across the sandwich. Any player may say "cut". At that moment, the sandwich is cut and the piece given to the one who said "cut". As he has said that is an acceptable piece, he believes he has at least $\frac 1n$ of the sandwich. The rest have asserted (by not saying "cut") that is it at most $\frac 1n$ of the sandwich, so the average available is now at least their share. Recurse.
• Note that this is not an envy-free division but just called a proportional division. In fact there is an easy solution for that that doesn't require a moving knife (en.wikipedia.org/wiki/Proportional_%28fair_division%29). Some people will not think a moving knife algorithm is fair because the one moving the knife may move slightly after someone says "Cut!". Jan 14, 2014 at 5:43
• Nice! You used the invisible hand of the market to cut the sandwich.
– user
Jan 14, 2014 at 10:29
• @user One could build a machine and everyone gets a button. Not exactly ideal for a student flat, unless they are engineering students... Jan 14, 2014 at 11:29
• @user1729 Engineering students would design the machine to ensure accurate proportioning to start with, you'd only need one button to start the machine (and to stop it, for safety reasons).
– JAB
Jan 14, 2014 at 12:55
• @user1729 then your boss is not an engineer. He just thinks he is.
– John
Jan 14, 2014 at 18:14
Just for the record, here's the Selfridge–Conway discrete procedure mentioned in the comments. The Wikipedia article also contains some commentary on its origin and why it works.
This procedure was the first envy-free discrete procedure devised for three players. The maximal number of cuts in the procedure is five. The pieces are not always contiguous. Solutions for n players were also found later.
Suppose we have three players P1, P2 and P3. Where the procedure gives a criterion for a decision it means that criterion gives an optimum choice for the player.
Step 1. P1 divides the cake into three pieces he considers of equal size.
Step 2. Let's call A the largest piece according to P2.
Step 3. P2 cuts off a bit of A to make it the same size as the second largest. Now A is divided into:
• the trimmed piece A1
• the trimmings A2.
Leave the trimmings A2 to one side. If P2 thinks that the two largest parts are equal, then each player chooses a part in this order: P3, P2 and finally P1.
Step 4. P3 chooses a piece among A1 and the two other pieces.
Step 5. P2 chooses a piece with the limitation that if P3 didn't choose A1, P2 must choose it.
Step 6. P1 chooses the last piece leaving just the trimmings A2 to be divided.
Now, the cake minus the trimmings A2 has been divided in an envy free manner. The trimmed piece A1 has been chosen by either P2 or P3. Let's call the player who chose it PA and the other one Player PB.
Step 7. PB cuts A2 into three equal pieces.
Step 8. PA chooses a piece of A2 - we name it A21.
Step 9. P1 chooses a piece of A2 - we name it A22.
Step 10. PB chooses the last remaining piece of A2 - we name it A23.
Wikipedia on the origins this procedure:
This procedure is named after John Selfridge and John Horton Conway. Selfridge discovered it in 1960, and told it to Richard Guy, who told it to many people, but Selfridge did not publish it. John Conway discovered it independently in 1993, and also never published it, but the result is attributed to them in a number of books.
• this method is envy-free, @Chris Culter, but is it coalition-resistant? Jan 14, 2014 at 18:10
• @VividD, yes it is. It's easy to see if you read the analysis in the end of the wikipedia articles. Jan 14, 2014 at 20:28
• This answer is no good. It refers to cutting up a cake, whereas the question is clearly about cutting sandwiches.
– AviD
Jan 14, 2014 at 22:16
• There is a difference between cutting a cake and cutting a pie. From Science News: Mathematicians simplify cake cutting by assuming that all slices are perpendicular to one particular side of a rectangular cake, with no crosswise cuts. Following that rule, there is only one way to split a rectangular cake into two equal-size pieces. For a pie, however, there are an infinite number of ways, because there an infinite number of lines going through the center of a circle.
– JRN
Jan 14, 2014 at 23:37
• @Mottie It handles it well. Each player divides the sandwich into equally valued pieces from their perspective. If they value meat more, pieces with meat should be smaller than pieces with less meat. If they don't care about bread at all, then they can just cut the meat up evenly, and ignore the amount of bread in each piece. The big downside is incompetence: in the two-player game, it is almost always a better idea to be the 2nd player than the 1st, because it is easier to determine which of two pieces are bigger than it is to cut something into two even pieces.
– Yakk
Jan 15, 2014 at 14:52
For dividing a cake to $n$ people, there is an algorithm that guarantees that:
• Each of the $n$ people gets a piece that he considers at least as good as each of the other pieces (i.e., the division is envy free);
• All pieces are connected (actually, all pieces are rectangles).
This algorithm was invented by Forest Simmons and published by Francis Su at 1999.
The only problem with this algorithm is that its runtime is not bounded, i.e., it might take forever (As proved by Stromquist at 2008, no bounded algorithm can find an envy-free division when we want the pieces to be connected).
But, you can stop at any time and get a division that is approximately envy free.
• Great answer, explains high-level facts about the algorithm. Jan 14, 2014 at 17:35
• Still, this is basically a link-only answer, since it doesn't even give an hint on how the actual algorithm works.
– o0'.
Jan 15, 2014 at 15:54
• What about the auction-style algorithm ? One player moves the knife over the surface. When it appears to have covered 1/n of the cake, he continues moving until a person calls out stop. When they do, he cuts the cake at that point. The player who called out stop takes the cut portion. This continues till the cake is divided. Mar 6, 2018 at 21:11
• @William Grannis: This solution is the one Ross Millilan referenced, the problem with (implementing) it in practice is that it is continuous-time based. May 3, 2020 at 18:09
Here's a variation on the accepted answer.
A cuts the sandwich into 3 parts
If B thinks the top 2 pieces are equal
C chooses a piece
B chooses a piece
A gets the remaining piece
Else
B re-balances the 2 biggest pieces.
C chooses a piece
If only one of B's re-balanced pieces remain
B gets that piece
A gets the remaining piece
else
A chooses a piece
B gets the remaining piece
Summary:
• A will get one of their original "equal" pieces or more, and is thus happy.
• B will get one of the pieces they adjusted, and are thus happy.
• C will get first choice, and is thus happy.
• Although each person should get at least a third in their mind, this doesn't qualify as an envy-free solution. For example, if B re-balances two pieces and C chooses the piece B increased, then A would envy C because he got more than 1/3 in A's mind. However, A still gets 1/3 in their mind.
I think you can generalize this pattern for $N$ people. Simplified, you give up your place in the picking order if you rebalance, but the trade-off is that you are guaranteed to get a piece you rebalanced or a bigger one. Here's a first take at the priority rules for the picking procedure, though I'm not sure it handles all cases:
1. If N people each have N pieces they rebalanced remaining, the first person who rebalanced from that group gets to choose from their rebalanced pieces. For example, if there is one piece left from the pieces you re-balanced, you get that piece.
2. If you did not re-balance, the highest number gets to pick first
3. If you did rebalance, the lowest cutter gets to pick first
Also, here's an example of how it would work for 4 people:
A cuts the sandwich into 4 equal pieces
If B thinks the top 3 pieces are equal:
If C thinks the top 2 pieces are equal:
D chooses a piece
C chooses a piece
B chooses a piece
A gets the last piece
Else:
C rebalances the top 2 pieces
D chooses a piece
If only one of C's rebalanced pieces remain:
C gets the other rebalanced piece
B chooses a piece
A gets the last piece
Else:
B chooses a piece
If only one of C's rebalanced pieces remain
C gets the other rebalanced piece
A gets the remaining piece
Else
A chooses
C gets the remaining piece
Else:
B rebalances the top 3 pieces
If C thinks the top 2 pieces are equal:
D chooses a piece
C chooses a piece
If only one of B's rebalanced pieces remain:
B takes that piece
A gets the final piece
Else:
A chooses a piece
B gets the final piece
Else:
C rebalances the top 2 pieces
D chooses a piece
If only one of C's rebalanced pieces remain:
C gets that piece
If only one of B's rebalanced pieces remain:
B gets that piece
A gets the last piece
Else:
A chooses a piece
B gets the other piece
Else if only 2 pieces rebalanced by both C & B are left:
B chooses one of the rebalanced pieces
C gets the other one
A gets the last piece
Else:
A chooses a piece
If only one of C's rebalanced pieces remain:
C gets that piece
B gets the final piece
Else:
B gets to choose
C gets the final piece
• A gets screwed if B and C are in cahoots together. Jan 14, 2014 at 23:32
• @AndrewLarsson: No, A will always get one of his fair pieces or more. Jan 15, 2014 at 2:42
• @AndrewLarsson: Thanks :) Jan 15, 2014 at 14:45
• One problem is the rebalancing, I personally wouldn't want a sandwich that is almost 1/3 + several small pieces of sandwich Jan 16, 2014 at 23:06
Okay, so this is just a small comment about the "general science'' behind all these. You can understand a sandwich as a measure space with $n$ given measures. Also, these measures are non-atomic, otherwise you can not solve the problem even for two people. These measures generate a vector measure on the sandwich''. Now you can apply Lyapunov theorem and say, that the range of this measure is closed and convex. But the points $(1,1,...,1)$ and $(0,0,...,0)$ lie in the range, thus the point $(1/n,..., 1/n)$ does. This means, that there is a piece of sandwich such that each person thinks that it is exactly $1/n$ of the whole sandwich. Now you can give it to anyone and repeat the procedure. Cool thing is that you will divide the sandwich in such a way that each person will think, that the sandwich is divided into equal parts. So one can solve the problem in the following form "There is a way to divide a sandwich such that each person thinks that no one have more sandwich than him". I have no idea how to make an explicit algorithm for that one.
• Sounds a lot like the ham sandwich theorem, in fact. Jan 14, 2014 at 7:34
• That result is originally due to Dubins and Spanier. Feb 18, 2014 at 13:02
• To elaborate on the unsolvability of the problem for atomic measures: If all measures assign measure $1$ to a given point, there is only one winner. In other words, if there is an indivisible "cherry" on the cake that everyone wants, only one person can get it and the rest will be unsatisfied. Jul 8, 2016 at 0:52
According to Wikipedia, the Brams-Taylor procedure is
the first finite procedure to produce an envy-free division of an infinitely divisible good among any positive integer number of players.
A nice discussion about it can be found here.
• Ironically it is patented instead of infinitely copied and shared in an envy-free way with the n players of the internet... Jan 14, 2014 at 13:32
• @ChrisWesseling I'm going to patent the method for taking into account the overhead costs of paying a royalty every time you use this method to determining whether you would be better off using it versus luck or fend for yourself. Jan 14, 2014 at 15:26
Complete EDIT (too lazy to formulate mathematically): Imagine X as a sandwich just for simplicity. The first person cuts X into thirds "horizontally" and they have to cut across the whole sandwich in a straight line with each cut. The second person cuts each third into thirds by cutting the sandwich "vertically" with each cut going through all of the original thirds in a straight line. Now there are 9 pieces. Third player picks one piece and then first person picks, then second and third again in a circle.... This operation with the horizontal and vertical constraints (and having to go right across the sandwich) should avoid anyone feeling deceived.
• Of course if it was an actual sandwich it would be destroyed probably by cutting it into 27 pieces :) Jan 14, 2014 at 5:17
• I am upvoting this solution because of its 3D nature. :) Jan 14, 2014 at 5:37
• So, for four people we would need sandwich in 4-dimensional space? ;) Jan 14, 2014 at 5:38
• Well imposing the ratios is just as difficult (if not more, what with all the crumbs) than just requiring one of the three to divide fairly. Furthermore even imposing ratios doesn't solve the problem because some parts of the sandwich may be mouldy... Jan 14, 2014 at 5:52
• You are assuming much too much about the 'uniformity' of the value ascribed to areas of the sandwich.
– jwg
Jan 14, 2014 at 13:14
Nobody said it has to be a discrete process. Just blend the sandwich into infinitestimally small pieces, everyone eats small pieces at the same speed until there is nothing left. Lucky person gets to lick the blades.
• "until they ate a third of the mess." How do you decide when that happens? Jan 14, 2014 at 14:19
• I believe it will naturally happen when the sandwich ends, if each person eats 1 piece at a time. Jan 14, 2014 at 15:07
• @DanHulme fixed Dec 16, 2014 at 17:51
• the trust problem is not solved. it will end as a speed eating competition. plus, the sandwich will be eaten infinitesimally fast, losing the benifit of taste (texture was already eliminated by blending). but,i have to admit, the solution adds much fun and adrenalin. Aug 12, 2016 at 3:37
From You cut I choose:
An interesting solution is outlined in "Mutiny on the H.M.S. Bounty", called the Who-shall-have-this Method. Say there are $N$ people to divide between, they choose from amongst themselves two people - $A$ and $B$. $A$ stands where $B$ cannot observe him, and divides the cake. B then points to someone and that someone comes to claim his share of the cake. $B$ may at any time point to $A$ or to himself, but since he never can see which segment of cake he's giving out, and since $B$ and $A$ are both chosen by the crowd, there is virtually no opportunity for intentional collusion.
This method still won't guarantee everyone a satisfactory cut of the cake, but it gives $A$ the correct motivation, since only by dividing the cake evenly can he hope to get a large piece for himself.
• I think I like this method the best so far, even better than the solution I proposed. Jan 14, 2014 at 21:20
• This method really has some appeal. @Nancy Hastings-Trew : I maybe don't fully understand it. Does this mean that A cuts the cake in N peaces? And that B repeatedly points to next person who in turn chooses his/her share? Jan 15, 2014 at 17:30
• @VividD Yes, I believe so. A cuts all pieces the same so that if he is pointed to last, he still gets 1/n of the cake. However, this still has the same flaw as the "You Cut I Choose" method, in which A cannot plan to receive more than 1/n of the cake, while the first person chosen is guaranteed more than 1/n (because no one can cut cake that evenly). Jan 15, 2014 at 22:08
• @Timtech - "... while the first person chosen is guaranteed more than 1/n (because no one can cut cake that evenly". It's assumed that A will attempt to cut the cake as evenly as possible (of course, there will be some variation). But, the first piece A chooses to distribute will not necessarily be the biggest. A will just start at some "random" point for the first piece, so there is no guarantee that the first person chosen will get a piece that is larger than 1/n. If this was somehow guaranteed, then B would always choose himself first to be guaranteed a bigger piece. Jan 16, 2014 at 20:08
• @VividD - your interpretation is close... A first divides the cake to allow for a piece for everyone, and then selects a piece to be distributed. B selects a person to receive the selected piece, but B can't see the piece that A selected. The selected person must accept the piece that A selected. A then selects the next piece, and so on. Jan 16, 2014 at 20:29
This was covered in the first class I ever TAed in graduate school. Naturally, the reaction of most of the students was, "why don't they just eat the cake?"
Person A cuts it into two pieces she views as "of equal value." Person B then chooses her favorite half. A and B then each divide their halves into thirds that they views as "of equal value." Person C then chooses her 1 favorite piece from A and one favorite piece from B. This ensures a fair (but not envy-free) division.
• Yes this is the inefficient solution in the Wikipedia article I gave. The other solution there takes a polynomial number of cuts which comes up to only 3 cuts for 3 people. Jan 14, 2014 at 5:55
• I like this version for its simplicity. The general recursive algorithm: if $n = 1$ then take the whole cake; if $n > 1$ then recursively divide the cake among $n-1$ people and then have each of those people divide their cake into $n$ portions they believe are equal. The $n$-th person now chooses one portion from each of the $n-1$ other people.
– ntc2
Jan 16, 2014 at 21:51
Divide the sandwich into three portions and weigh each one on a digital scale. Adjust until the portions are the same. If equal distribution of condiments is a point of contention, the sandwich must be pureed prior to portioning. No one said it had to taste good.
• Amusing to say the least. Jan 14, 2014 at 21:07
• Perhaps we need an entropy measure of the resulting condiment distributions to quantify the tolerance of our eaters to condiment envy risk ;) Jan 15, 2014 at 7:05
My professor for Game Theory asked me this question,and I gave him an answer that he said that is the simplest he had ever heard (also Steaven Brhams said it's nice):
P1 divides the cake to 3.
P2 notes the smallest,in his opinion; let it be A.
If P3 wants A, let him have it.
Else P1 takes it, and we have only 2 which we know how to solve.
• Sorry for the super late comment, but suppose that P3 does not want the piece. Then P1 takes it. So P1 is out of consideration for future pieces. THen how do P2 and P3 decide who gets which of the remaining pieces? Sep 18, 2019 at 14:52
• @JoeSjoberg I fixed that in my edit Jan 6, 2020 at 19:01
• @JoeSjoberg either they agree, or just cut each of them to 2, as mentioned in the question. Jul 5, 2023 at 5:45
Okay, I've got a weird idea that I think might work.
Viewed from above, the sandwich is a square, right? Since a square is technically a disc in $L_1$ space, I think you can use the pizza theorem to cut it into 12 pieces and divide it evenly with way less work than Selfridge-Conway.
I think this should work because 12 divides into 4 (a pizza theorem requirement) but also divides into 3. I started with the following image from Wikipedia:
My algorithm was, first divide the pizza along the cardinal directions. Then, enumerating over each such section, color each first wedge, second wedge, third wedge, and fourth wedge (modulo 3) black. Offset that sequence by 1, and repeat twice more. Here is the result:
I have a feeling that this is at least a good approximation. The color permutation might not be so simple, though--the claim is that there exists such a permutation for any set of cuts that satisfy the requirements of the theorem.
The algorithm for this one is much more intuitive:
1. Observe that there are 3 "opposite pairs" of slices of each color.
2. Select one opposite pair of each color, and recolor it.
The claim is that each pair of opposites is a third of half of the pizza sandwich.
• Disclaimer: This may only work on square circles in a Manhattan world ;) Jan 14, 2014 at 7:08
• Alternatively, you may perform a certain transformation on the sandwich to get a circle before attempting to use this algorithm. Jan 14, 2014 at 7:20
• +1 for introducing the Pizza theorem to me, something I always thought to be true but had never completely worked out.
– SQB
Jan 14, 2014 at 7:30
• You can subdivide the (square) sandwich in an infinite number of circles, on which you can apply this theorem. Though, lunch time might be over by then.. Jan 14, 2014 at 12:39
• It's a cool idea, but ultimately doesn't solve the sandwich-dividing problem, because you can't guarantee straight cuts and you can't guarantee precise angles. Slightly rotate the vertical cut clockwise, and green gets ripped off to black's benefit. Jan 14, 2014 at 21:41
The "one person cuts, the other chooses" algorithm generalises to any number of people, but the generalisation is (a) not envy-free; (b) vulnerable to collusion in a subset of at least two of the participants. It does work for cases where one or more participants want a piece of size $< \frac{1}{n}$, but it splits their share unfairly among the remaining participants.
The participants sit in a circle. Person 0 cuts a portion from the cake. (They don't cut the cake all the way through, but just cut one slice from it.) The resulting portion is offered to person 1. If he accepts it, he takes it and is removed from the circle; otherwise, it's offered to person 2, and so on around the circle. If no one takes it, the portion is returned to person 0, who must take it and is removed from the circle.
This procedure is repeated with the remaining participants (starting with person 1 if they're still present, otherwise person 2), until there is only one person left. This person gets the left-over slice of cake which has never been offered to anyone.
It's clearly very gameable, but if the participants are basically honest, it works well. At least it runs in bounded time and delivers connected portions. Its main practical problem is that one participant can cut a slice that's much too large, advantaging one participant at the expense of all the others. But unlike the "one person cuts all slices" method, the resentment gets to be spread among all the participants who slice, not just focused on one person.
• Collusion, though possible, would not be subgame perfect. There are N people. I cut the sandwich into a large piece and N-1 tiny pieces, thinking that the recipient of the large piece will tip me a portion > (1/N). Problem is, they now have their meal, and they don't have to share. Such a bargain could only be enforced because (a) I have a knife, or (b) there is a repeated context, or (c) some other enforcement mechanism outside the game
– Paul
Jan 14, 2014 at 11:13
• That's true, but realistically, context outside the game is likely. Sitting a husband far from his wife would be an obvious precaution, for instance. Jan 14, 2014 at 11:22
• I think it's just too complex. If player 1 doesn't accept the first portion, and player 2 does, there is something wrong, unless you count other factors (getting a smaller portion instead of waiting longer).
– G B
Jan 14, 2014 at 12:30
• @GB - "If player 1 doesn't accept the first portion..." the piece may be a "fair" sized piece (1/n) and so player 2 accepts it. Player 1 may just be waiting for a piece that is larger or smaller (than 1/n), or a piece with a cherry (decoration) on it, or more/less frosting, or just one that "looks" better. Jan 16, 2014 at 20:50
• @KevinFegan There goes the "not feeling deceived" part. If different pieces have different (subjective) values, there is no solution.
– G B
Jan 17, 2014 at 9:14
Here's my solution...
Let's say it's a pizza and there are 3 people, but it would work for other items or numbers of people.
The pizza is divided into 3 pieces. It's assumed that the pieces are cut into equal sizes, but it's not important... because the "value" of a piece may not be solely dependent on it's size. For example, a smaller piece may have more of a particular topping, or it may be more/less burned, or thicker/thinner, etc.
Each person identifies the piece they want. If each piece is chosen, then everyone is happy... we're done.
Otherwise, each person "bids" an amount of money they are willing to pay for the piece they want. The piece with the highest bid is set aside for that person.
Bidding continues for the remaining pieces until the last piece (and 1 person) is left.
Everyone pays the amount that they bid, and the last person pays an amount that is the average of all the bids, less the amount that the highest bid exceeded the average (which could be $0 or even less, but that should not be typical). From this money, the pizza is paid for, and the remaining money is divided equally among all the people. In the case where the money collected is not enough to pay for the pizza, then everyone pays an equal amount to cover the shortage. Also in this case, you should consider 1: choose another place for your pizza, 2: choose different people to share pizza with. Call the players A, B, C: A cuts the sandwich in two pieces, and calls it "one third" and "two thirds". B can either • accept the "one third" or • choose to cut the "two thirds" in two halves. In the first case, C gets to cut the "two thirds", and A will choose one of the two halves. In the second case, C chooses one of the two halves, B gets the other half, and A gets the "one third" he cut at the beginning. I think it's the easiest possible solution, assuming the players are honest. • Sounds like C loses if A cuts poorly. Jan 14, 2014 at 12:35 • Actually, C has to wait for A and for B to cut/choose, but A "loses" just as much as C. – G B Jan 14, 2014 at 12:51 • This isn't a solution. It only works if the participants are in agreement over the value of each piece. – jwg Jan 14, 2014 at 13:13 • @GB There are solutions in that case - that's what "envy-free" means. In the case of 2 players, for instance, the person who cut gets what he thinks is an equal share, while the person who chose gets what he thinks is the bigger share. No one feels cheated. Jan 15, 2014 at 0:16 • @GB: The problem is that this is not coalition-resistant. If A and B are friends, then A can cut the sandwich into two equal pieces, so B gets a ½ piece and C is limited to a ¼ piece. Then A and B can get together later and share ¾ of the sandwich. Worse, what if A cuts the sandwich into a huge piece and a tiny piece (e.g., 90% and 10%) and then points to the huge piece and says, “That’s the ⅓ piece.” What recourse does C have then? Jan 15, 2014 at 1:20 I think I have a more practical solution than the one selected: 1. A cuts the cake in three. 2. B cuts each piece in three. 3. C cuts each piece in three. We now have 27 pieces, each takes one piece in the same order they have cut the cake then reverse the sequence: ABC-CBA-ABC-CBA-... No matter how A and B cut the cake, C can make sure that every piece is very close to the same size AND it is his best interest to do so. Similarly B, by cutting each piece properly in 3 equal parts, minimizes C's ability to screw up. A has least incentive to cut fairly, but his cuts have the least influence on the final outcome. He can however set a baseline: if his cuts are fair, no screw-up can be bigger than one third of the cake. The assumption here is that none of them is capable of making perfect cuts, but that the errors will cancel themselves out. The important issue is to make it in their best interest to attempt the fairest cuts possible. By reverting the picking order on each turn, A & C will alternatively chose the biggest remaining piece and B will get the medium piece of each set, so even if there is a discrepancy in the portion sizes the picking order should minimize its impact. There is also another aspect which is that the greater the number of portions of somewhat similar size, the harder it becomes for each person to keep track of the total amount of cake gotten. With 1 piece each, it's fairly easy for each one to compare portions. But if each person must chose 9 pieces, it's a lot harder to compare the total volume gotten by each. • A likes cake and B doesn't like cake, and A and B dislike C, so A cuts one huge and two small pieces, and B cuts each piece into one large and two tiny pieces. A goes first and takes the single huge piece, and B discards whatever crumbs he gets, but this eventually leaves C very sad and hungry. Jan 17, 2014 at 9:37 • Read again and that time do try to understand the solution. A and B cuts the cake into a large one and 2 tiny pieces. Now we have 2 big pieces and a bunch of crumbs. What you didn't understand is that C cuts EACH piece in 3 EQUAL pieces. Nothing A and B can do about it. After C, you have 6 equal pieces and a bunch of crumbs. If C cuts properly, he will get exactly 1 third of the cake. Jan 18, 2014 at 2:57 • I think I responded to the wrong answer, but your method is still wrong. A likes cake and C doesn't like cake, and A and C dislike B, so A cuts equally. If B cuts equally, C cuts all equally except one which he cuts into one large and two tiny, so A goes first and takes that large piece, and B cannot get his fair share. If B does not cut equally, then C cuts equally and I am pretty certain that B also will not get his fair share if both A and C play perfectly. Jan 18, 2014 at 4:52 • You are outside the scope of the question: the original problem was sharing a cake betwen 2 people who want their fair share but don't trust the other. Expanding it to 3 people, we have 3 people who want their fair share. The failure can only occur if C does not actually wants his fair share, in which case most answers above fail completely, whereas my solution limits the failure to a single 1/9th piece. B gets less, but only if C agrees to take even less than B, which is outside the scope of the question asked. Jan 18, 2014 at 22:20 • Ah you interpret the problem differently. The problem never explicitly stated that each of them wants his fair share. It only says that they don't trust each other, and implies that we are to find a procedure by which any player can obtain at least his fair share by playing correctly. But if you assume that every player wants his fair share, then it is enough to let one cut into 3 pieces and the rest choose, since the one cutting will then cut evenly. And I have indeed responded to many answers that fail, not just yours. Jan 23, 2014 at 1:33 I think I might have an algorithm that a) Is fair b) Can be generalized for n players c) Only uses a bounded number of cuts (a different bound for each n) odds are that I am wrong =P The procedure for 3 players: 1. P1 divides the cake in 3 parts 2. P2 and P3 choose each a 'smallest piece', such that they dont want part of it. 2.1 If P2 and P3 agree,then they take the 2 pieces they like, and divide each amongst themselves. P3 gets the remaining 2.2 If P2 and P3 disagree, we have: P2 'likes' two pieces P3 'likes' two pieces P1 will be assumed to like the two pieces that are not liked by both P2 and P3 Now, each piece has 2 people that like it. These 2 divide the piece amongst themselves P1 is happy (if he divided correctly): he either got 1 piece, or half of 2 pieces P2 and P3 are happy: they each got half of the two pieces they think are bigger The procedure for n (using the procedure for n-1) 1. P1 divides in N 2. Each P2, ..., PN chooses a piece they dislike (and therefore, n-1 they like). They get a 'stake' in each piece they like. 3. P1 gets 'stakes' in pieces such that each piece is 'divided' in n-1 stakes 4. recurse: each piece is divided amonsgt the people with stakes on them. If P1 has more then 1 stake, he can 'play as many players' in the division. As far as I can tell, that is not a problem • Yeah, apealing novelty is "don't like" method. I like it. :) Jan 16, 2014 at 22:45 It's simple: one person divides the sandwich into 3 pieces, and that person is the last one to choose (the other two can decide using odds and evens, for instance). The dividing person will not want to make no piece larger than another, because he/she would obviously be left with the smaller one. • If the first person makes one large piece and two small ones, then the person who picks the large piece will be envied by the person who picks the smaller. Jan 14, 2014 at 14:19 • By accident. Cutting three equal pieces of something is not that easy. Jan 14, 2014 at 14:26 • The accident could happen cutting in two, too, even being an easier division. Anyway, @VividD asked for a "such mechanism", not for a "better one". Jan 14, 2014 at 14:31 • The point is that when cutting in two, only the cutter loses out if they screw up. In your algorithm, the cutter and one other person lose out, so that other person will "feel deceived". Jan 14, 2014 at 14:39 • Yes, you're right. But considering the practicality of it, I still recommend my solution. Have you memorized the "Selfridge–Conway discrete procedure" to apply it in the real world in a moment of hunger? :) Jan 14, 2014 at 15:11 It might help what is in the sandwitch. If it's a BLT for example then one of the people might be a vegetarian, but might feel cheated if the other gets more tomato. On the otherhand the bacon will be up for grabs. The lettuce is a wild card of course if they all want it. I think what you have to do is negotiate sandwitch parts with the others. Perhaps one person will be happy letting the other eat all the bacon if they can eat all the tomato and lettuce. Perhaps one is gluten intolerant so they don't mind the 3rd person having all the bread. The other alternative is for each person to bring their own sandwitches. We had a similar problem like this at work once with pizza and we basically decided not to eat shared pizza anymore. • Yeah, what if pizza is quattro stagione? Jan 14, 2014 at 10:41 • We had a similar problem like this at work once with pizza and we basically decided not to eat shared pizza anymore. — You must not be mathematicians at work. Jan 14, 2014 at 12:32 Person 1 cut out a 1/3 piece. Person 2 and 3 decide what piece they want. If they pick different pieces, the 1/3 piece is given, and other play the 2 people game with the 2/3 piece and players 1. If they both pick the 2/3 piece, player 1 receives the 1/3 piece and they play the 2 people game with the 2/3 piece. If they both pick the 1/3 piece, person 2 cut off a part of the 1/3 piece, that part is then added to the 2/3 piece. Person 3 decide if he takes the remaining 1/3 piece, or plays the 2 people game with player 1. If you do not want to cut the sandwich in more than 3 pieces, go as follow: Person 1 cut out his piece (1/3); Person 2 cut out his piece (1/3); Person 3 can now switch with person 1 or 2 or keep the remaining piece himself. • Your first solution is equivalent to the (not envy-free) proportional division given on Wikipedia. Your second solution is incorrect. Person 1 does not like cake and Person 3 likes cake and Persons 1,3 dislike Person 2, so Person 1 cuts a huge piece out and Person 3 will definitely take it, leaving Person 2 forlorn. Jan 17, 2014 at 10:33 • The second solution is obvious less work and therefore less fair; however if you add the assumption that when person 1 cut out 5/6 cake, person 2 and 3 might argue he meant to take only 1/6 part himself. Even when person 1 and 3 work together, this method still makes it hard to give person 2 less than 1/4th of the cake. If person 2 expect treason, person 1 can give him 1/2 cake and he can cut it into 1/4th parts. It works of course better when all players try to maximize there share. Jan 20, 2014 at 14:57 • Also i did not read wikipedia but nobody else here mentioned that specific way so i tought it was fair to add it. Jan 20, 2014 at 14:59 • I can understand, but we would like to have correct solutions listed here, so if there are like twenty solutions every reader will have to take a long time to think through every single solution, and in that case I would rather go to Wikipedia than come here to find the answer to this question. That said, I commented only to demonstrate that it is not as easy as one might think at first glance, and if you try to prove any of the correct solutions you will find that it is even harder to prove than to devise them. Jan 23, 2014 at 1:23 • Also, if players always maximize their share, then it suffices to have one player cut the cake into 3 pieces and the rest choose one for each of themselves. Jan 23, 2014 at 1:24 Take the sandwich and divide it into 4 even squares, take 1 of the squares and cut it evenly into 3 rectangular pieces. Everyone now gets a big square and a small rectangular piece. • This solution is vaguely reminiscent of Egyptian fractions and becomes very appropriate if you wanted to divide two sandwiches between three people (each gets 2/3). The Egyptian fraction for 2/3 is: 1/2 + 1/6. Thus, you would divide the sandwich into halves and the remaining half into sixths. (See the "practical use of Egyptian fractions" in the link). – user49763 Jan 26, 2014 at 23:23 This may be silly, but I think it's fairly straightforward for as many people as you want. If you have$n$people, order person number$1$to cut the sandwich into$n$pieces. All the other$n-1$people then pick their own piece. The person who was cutting gets the piece that was left. If he cut all pieces equally, he will have$1/n\$ of the sandwich. If any one piece was bigger, this means one piece had to be smaller, and the smaller piece will not get selected, so the cutter has motivation to cut equally.
• What if... Person number 1 doesn't really care that much if their piece is a little smaller and that lack of care is reflected in their cutting accuracy. So person 2 selects the biggest piece and person 3-n are left to choose from smaller pieces, leaving the smallest piece for person 1. The only person happy with this is person 2 (and perhaps person 1 who didn't care). The others are left to wonder why person 2 was able to choose first. Even if person 1 cut carefully, some may view person 2's ability to choose first as being unfair. Jan 14, 2014 at 20:14
• This comment can also be applied to the original case, so I don't think it's all that relevant. I think the task only makes sense if all the people present have the same measure of fairness and are equally interested in it. With those assumptions, my method works.
– 5xum
Jan 14, 2014 at 20:52
• It is different. In the original case, with 2 people, if the cutter didn't care and cut poorly (for any reason), the chooser would be happy, and the cutter would be happy (didn't care). The same is the case for the first-chooser and the cutter with 3 or more people (or if the cutter can't cut accurately). The difference is, the remaining "middle" people have no control of the outcome if they are not happy with their piece, compared to pieces choosen before they were able to choose. Jan 14, 2014 at 21:17
• Point taken. I admit my method needs slighlty stronger assumptions if there are 3 od more people. I still kinda like it, though.
– 5xum
Jan 16, 2014 at 10:20
• It is a "classic" method... it's hard to beat the classics. Jan 16, 2014 at 19:54
This presentation contains background and main results in resource allocation and fair division areas. "A cake" is formally defined as (0,1) interval. The author outlines and compares eight procedures for fair division of a cake. For the sake of comparison, following properties of fair division procedures are defined: proportionality, envy-freeness, equability, Paretto-efficiency. Some of the procedures require a referee, and some do not. Some produce contiguous pieces, some do not. All procedures are evaluated against the set of properties above.
Following table from the presentation summarizes features of described procedures:
Regarding avoiding envy, the authors even mention and define degree of envy-freeness for some of presented procedures.
The key conclusion of the presentation is:
The problem becomes non-trivial for more than two players, and there are many open problems relating to finding procedures with “good” properties for larger numbers of players.
Hmm, if you cut it into an approximate 3rd and 2/3rds portions, then the cutter would get a slice that both other people think is about 1/3rd. To attempt to subvert this, player A could cheat and make 2 equal halves. In the real world, such a player would never get invited to party games again. ;) But let's just say that then one of the other players would have to divide each half into 1/6th and 1/3rd slices. Repeat ad nauseum?
If you can 'uncut' the sandwich, then it's easier: Just keep having a different player cut the sandwich until the other 2 agree. This is similar to the moving-knife solution but where the cut is not done unless all 3 agree, in actual practice since 'uncutting' really means just not cutting yet.
Once you get the bigger piece being roughly 2/3rds, then you can apply the normal 2-person method. To go to 4 people, vote in pairs? You could also apply this idea with 1/4th & 3/4ths slices. This is generalizable to 5,...,dozens of people. | 11,000 | 44,000 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2024-22 | latest | en | 0.969868 |
https://slideplayer.com/slide/6429160/ | 1,624,054,673,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487641593.43/warc/CC-MAIN-20210618200114-20210618230114-00065.warc.gz | 474,815,796 | 20,483 | # Scientific Measurements
## Presentation on theme: "Scientific Measurements"— Presentation transcript:
Scientific Measurements
Scientific Measurements
Measurement: a means of describing observations using numbers and a UNIT of measure (ex: 11 inches) - makes a comparison with a known standard - ALL measurements contain at least one of the four basic dimensional quantities (length, mass, time, temperature) Systems of measurement: 1) Metric System (aka International System of Units (SI) 2) English System * See handout reference)
Basic Measurement Units
Unit Metric English Instrument Time: instant something happens OR period of change Length: distance between 2 points Mass: amount of matter in an object Temperature: measurement of how hot or cold something is (average KE = average Kinetic Energy) Second (s), minute (min), hour (hr), year (yr) Meter (m), centimeter (cm), millimeter (mm), kilometer (km) Gram (g), kilogram (kg) Celsius (0C) Kelvin (0K) Clock, stopwatch, sundial, hour glass Same as Metric Mile (mi), yard (yd), foot (ft), inch (in) Meter stick, yardstick, ruler, tape measure Ounce (oz), pound (lb), ton (t) Balance, scale Fahrenheit (0F) Thermometer
Derived Units Examples of Derived Units: Volume; Density
** Some properties are best described by using some mathematical combination of the basic quantities. These mathematical combination of basic quantities are referred to as DERIVED UNITS. Examples of Derived Units: Volume; Density Volume: amount of space an object takes up Density: amount of material contained in a space Regular Shape: Volume L . W . H cm . cm . cm = cm3 Density mass (g) volume (cm3) Irregular Shape: displacement method (1 ml = 1 cm3) Drop object in known volume of water and record the difference Density = mass (g) volume (cm3)
Let’s measure some irregular shapes!
What is the volume of the dinosaur in ml? 0.8 ml
Working With Numbers Every number has a decimal version (4 = 4.0)
Rounding off numbers: most of the time to the nearest 10th or 100th Every number has a decimal version (4 = 4.0) Rules: you always check the number to the right to decide > 5 Ex: < 4 Leave it alone thousandths tenths hundredths (5.363) (5.4) (5.36) The “rules” Five or more, raise the score Four or less, leave it rest
Rounding Practice Time
Working With Numbers Rounding Practice Time Round to the nearest tenths place: 3.73= 2. 65= 4.324= 18.957= = Round to the nearest Hundredths place: 6.546= 16.342= = 12.997= = 3.7 6.55 2.7 16.34 4.3 124.86 19.0 13.00 2345.6
Converting numbers percents (%)
Number % - move the decimal point to the right 2 places and add % sign (** multiply by 100 = same thing) 54% 0.54= % Number - move the decimal point to the left 2 places and lose % sign (** divide by 100 = same thing) 47%= 0.47
Converting numbers percents (%)
Numbers / Percents Practice Time Convert the following percents to numbers: 75%= 83%= 58.8%= 73.94%= Convert the following numbers to percents: 0.76= 0.43= 0.435= 0.683= 0.75 76% 0.83 43% 43.5% 0.588 0.7394 68.3%
Scientific Notation (aka math shorthand)
Takes very large & very small numbers and makes them shorter. HOW? Move the decimal point until there is only one number to the left of the decimal point and count the numbers – this is the EXPONENT If you need to count to the RIGHT to get back to the original number then the EXPONENT is POSITIVE. If you need to count to the LEFT to get back to the original number then the EXPONENT is NEGATIVE. Example: 1,750, x x 10-5 Positive # See pg. 1 of ESRTs Negative #
Sci Notation Practice Time
Move the decimal point until there is only one number to the left of the decimal point and count the spaces – this is the EXPONENT Practice: ** do these on your own, then we’ll check them 76,000 = = 130 = = 7.6 x x x x 10-2 8.4 x 103 = 1.2 x 10-4 = 5.5 x 10-5 = x 107 = 8, ,504,000
Unit Conversions X = 65.6 miles
1) Set up the problem; KNOW the conversion factor 2) Make into a fraction and multiply by what you know (set up the conversion factor in order to cancel units) Example: 1) 105 km = how many miles? 2) 105 km mi km X Conversion factor Unwanted units cancel 3) = 65.6 miles
Unit Conversions Ctd X = 96 km
1) Set up the problem; KNOW the conversion factor 2) Make into a fraction and multiply by what you know (set up the conversion factor in order to cancel units) Example: 1) 60 miles = how many km? 2) 60 mi km mi X Conversion factor Unwanted units cancel 3) 60 x 1.6 = 96 km
Who wants more practice : Try, “Working With Numbers worksheet”
Rounding Percents Scientific notation Unit Conversions Try, “Working With Numbers worksheet” | 1,256 | 4,598 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2021-25 | latest | en | 0.831909 |
https://www.cut-the-knot.org/do_you_know/5-6-GoldenRatioIn.shtml | 1,542,415,155,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743247.22/warc/CC-MAIN-20181116235534-20181117021534-00341.warc.gz | 847,526,180 | 9,826 | # Golden Ratio in 5-gon and 6-gon
The following has been posted by Tran Quang Hung at the CutTheKnotMath facebook page, with a proof by Leo Giugiuc.
Indeed, due to symmetry, suffice it to prove that $\displaystyle\frac{DR}{ER}=\varphi.\;$
Choose, WLOG, $A=0,\;$ $F=2,\;$ $E=3+i\sqrt{3},\;$ and $D=2+2i\sqrt{3}.\;$ We have,$\displaystyle\frac{A-F}{P-F}=\cos\frac{3\pi}{5}+i\sin\frac{3\pi}{5},\;$ implying
\displaystyle\begin{align} P&=2-2\cos\frac{3\pi}{5}+2i\sin\frac{3\pi}{5}\\ &=2+2\sin\frac{2\pi}{5}+2i\sin\frac{3\pi}{5}\\ &=4\cos^2\frac{\pi}{5}+2i\sin\frac{3\pi}{5}\\ &=\varphi^2+2i\sin\frac{3\pi}{5}\\ &=1+\varphi+2i\sin\frac{3\pi}{5}. \end{align}
From here, the line $PR\;$ is $x=1+\varphi,\;$ so that $R=1+\varphi+ki\;$ (we are not interested in the value of k.) It follows that $\displaystyle\overrightarrow{DR}=\varphi-1+mi=\frac{1}{\varphi}+mi\;$ and $\displaystyle\overrightarrow{RE}=2-\varphi+ni=\frac{1}{\varphi^2}+ni\;$ and, finally, $\displaystyle\frac{DR}{ER}=\frac{\displaystyle\frac{1}{\varphi}}{\displaystyle\frac{1}{\varphi^2}}=\varphi.$
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# » Nesting OR and AND Functions to Meet Multiple Criteria
Problem:
We want to check whether or not each value in List1 (column A) meets both of the following criteria:
1. The number must be even.
2. The number must be smaller than 3 or larger than 8.
Solution:
Use the AND, MOD, and OR functions as shown in the following formula:
=AND(MOD(A2,2)=0,OR(A2<3,A2>8))
Example:
List1____Result
2________TRUE
5________FALSE
4________FALSE
10_______TRUE
Screenshot // Nesting OR and AND Functions to Meet Multiple Criteria
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XP Math - Forums This is hard
11-08-2006 #11 Sillysidley Join Date: Oct 2006 Posts: 822 no, 1/2 __________________ .
11-09-2006 #12 Mr. Hui Join Date: Mar 2005 Posts: 10,609 (1+1)^-1 = 1/2 <- Correct __________________ Do Math and you can do Anything!
11-10-2006 #13 Alex a.k.a whiteboy Guest Posts: n/a mann mann then what is ^?
11-10-2006 #14
Sillysidley
Join Date: Oct 2006
Posts: 822
Quote:
Originally Posted by Alex a.k.a whiteboy mann then what is ^?
^ is " to the power of"
__________________
.
11-10-2006 #15 Mr. Hui Join Date: Mar 2005 Posts: 10,609 To display superscripts on the forum, use [sup][/sup] For example, 2[sup]3[/sup] will give 23 __________________ Do Math and you can do Anything! Last edited by Mr. Hui; 11-25-2006 at 11:28 AM..
11-10-2006 #16
Sillysidley
Join Date: Oct 2006
Posts: 822
Quote:
Originally Posted by Mr. Hui To display superscripts on the forum, use [sup][/sup] For example, 2[sup]3[/sup] will give 23
Mr. Hui, what is this?
__________________
.
Last edited by Mr. Hui; 11-25-2006 at 11:28 AM..
11-11-2006 #17
Sillysidley
Join Date: Oct 2006
Posts: 822
Quote:
Originally Posted by somedude 1. What is the arithmetic mean, expressed as a common fraction, of the solutions to|10x|-sqrt(10x)=0 2. Triangle ABC is such that AB=8, AC=9, and BC=10 cm. The bisector of the interior angle BAC intersects BC in a point M. What is the length of the segment BM? Give you r answer as a common fraction in lowest term. These are tricky. I have worked on them for hours, and don't know what they are. If anyone wants interesting problems tell me. ( I kow the answers to those )
1. The answer is 1/20, because 10x must be 0, or 1.
__________________
.
11-12-2006 #18 Alex a.k.a whiteboy Guest Posts: n/a umm umm is number two 11?
11-13-2006 #19 Sillysidley Join Date: Oct 2006 Posts: 822 I don't know the answer... but thats wrong anyway because it's not a common fraction __________________ .
11-13-2006 #20 Sillysidley Join Date: Oct 2006 Posts: 822 i bet u can't even figure out what the negative reciprocal of 500 is... __________________ .
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Presentation on theme: "34.1 Flow of Charge Heat flows through a conductor when a temperature difference exists. Heat flows from higher temperature to lower temperature. When."— Presentation transcript:
Electric current is related to the voltage that produces it, and the resistance that opposes it.
34.1 Flow of Charge Heat flows through a conductor when a temperature difference exists. Heat flows from higher temperature to lower temperature. When temperature is at equilibrium, the flow of heat ceases.
34.1 Flow of Charge Charge flows in a similar way. Charge flows when there is a potential difference, or difference in potential (voltage), between the ends of a conductor. The flow continues until both ends reach the same potential. When there is no potential difference, there is no longer a flow of charge through the conductor. To attain a sustained flow of charge in a conductor, one end must remain at a higher potential than the other. The situation is analogous to the flow of water.
34.1 Flow of Charge Water flows from higher pressure to lower pressure. The flow will cease when the difference in pressure ceases.
34.1 Flow of Charge Water flows from higher pressure to lower pressure. The flow will cease when the difference in pressure ceases. Water continues to flow because a difference in pressure is maintained with the pump. The same is true of electric current.
34.2 Electric Current Measuring Current
Electric current is measured in amperes, symbol A. An ampere is the flow of 1 coulomb of charge per second. When the flow of charge past any cross section is 1 coulomb (6.24 billion billion electrons) per second, the current is 1 ampere.
34.3 Voltage Sources Voltage sources such as batteries and generators supply energy that allows charges to move steadily.
34.3 Voltage Sources Charges do not flow unless there is a potential difference. Something that provides a potential difference is known as a voltage source. Batteries and generators are capable of maintaining a continuous flow of electrons.
34.3 Voltage Sources The potential energy per coulomb of charge available to electrons moving between terminals is the voltage. The voltage provides the “electric pressure” to move electrons between the terminals in a circuit.
34.3 Voltage Sources Power utilities use electric generators to provide the 120 volts delivered to home outlets. The alternating potential difference between the two holes in the outlet averages 120 volts. When the prongs of a plug are inserted into the outlet, an average electric “pressure” of 120 volts is placed across the circuit. This means that 120 joules of energy is supplied to each coulomb of charge that is made to flow in the circuit.
34.3 Voltage Sources Consider a long pipe filled with water.
Water will flow through the pipe if there is a difference in pressure across the pipe or between its ends. Water flows from high pressure to low pressure. Similarly, charges flow through a circuit because of an applied voltage across the circuit. You don’t say that voltage flows through a circuit. Voltage doesn’t go anywhere, for it is the charges that move. Voltage causes current.
34.4 Electric Resistance The amount of charge that flows in a circuit depends on the voltage provided by the voltage source. The current also depends on the resistance that the conductor offers to the flow of charge—the electric resistance. This is similar to the rate of water flow in a pipe, which depends on the pressure difference and on the resistance of the pipe.
34.4 Electric Resistance For a given pressure, more water passes through a large pipe than a small one. Similarly, for a given voltage, more electric current passes through a large-diameter wire than a small-diameter one.
34.4 Electric Resistance A simple hydraulic circuit is analogous to an electric circuit.
34.5 Ohm’s Law Ohm’s law states that the current in a circuit is directly proportional to the voltage impressed across the circuit, and is inversely proportional to the resistance of the circuit.
34.5 Ohm’s Law Electric resistance is measured in units called ohms.
Georg Simon Ohm, a German physicist, tested wires in circuits to see what effect the resistance of the wire had on the current. The relationship among voltage, current, and resistance is called Ohm’s law.
34.5 Ohm’s Law For a given circuit of constant resistance, current and voltage are proportional. Twice the current flows through a circuit for twice the voltage across the circuit. The greater the voltage, the greater the current. If the resistance is doubled for a circuit, the current will be half what it would be otherwise.
34.5 Ohm’s Law The relationship among the units of measurement is:
A potential difference of 1 volt impressed across a circuit that has a resistance of 1 ohm will produce a current of 1 ampere. If a voltage of 12 volts is impressed across the same circuit, the current will be 12 amperes.
34.5 Ohm’s Law The resistance of a typical lamp cord is much less than 1 ohm, while a typical light bulb has a resistance of about 100 ohms. An iron or electric toaster has a resistance of 15 to 20 ohms. The low resistance permits a large current, which produces considerable heat.
34.5 Ohm’s Law Current inside electric devices is regulated by circuit elements called resistors. The stripes on these resistors are color coded to indicate the resistance in ohms.
34.5 Ohm’s Law think! How much current is drawn by a lamp that has a resistance of 100 ohms when a voltage of 50 volts is impressed across it?
34.5 Ohm’s Law think! How much current is drawn by a lamp that has a resistance of 100 ohms when a voltage of 50 volts is impressed across it? Answer:
34.6 Ohm’s Law and Electric Shock
think! If your skin were very moist, so that your resistance was only 1000 ohms, and you touched the terminals of a 24-volt battery, how much current would you draw?
34.11 Electric Power Electric power is equal to the product of current and voltage.
34.11 Electric Power Unless it is in a superconductor, a charge moving in a circuit expends energy. This may result in heating the circuit or in turning a motor. Electric power is the rate at which electrical energy is converted into another form such as mechanical energy, heat, or light.
34.11 Electric Power Electric power is equal to the product of current and voltage. electric power = current × voltage If the voltage is expressed in volts and the current in amperes, then the power is expressed in watts. 1 watt = (1 ampere) × (1 volt)
34.11 Electric Power The power and voltage on the light bulb read “60 W 120 V.” The current that would flow through the bulb is: I = P/V = (60 W)/(120 V) = 0.5 A.
34.11 Electric Power A lamp rated at 120 watts operated on a 120-volt line will draw a current of 1 ampere: 120 watts = (1 ampere) × (120 volts). A 60-watt lamp draws 0.5 ampere on a 120-volt line.
34.11 Electric Power A kilowatt is 1000 watts, and a kilowatt-hour represents the amount of energy consumed in 1 hour at the rate of 1 kilowatt. Where electrical energy costs 10 cents per kilowatt-hour, a 100-watt light bulb burns for 10 hours for 10 cents. A toaster or iron, which draws more current and therefore more power, costs several times as much to operate for the same time.
34.11 Electric Power think! How much power is used by a calculator that operates on 8 volts and 0.1 ampere? If it is used for one hour, how much energy does it use?
34.11 Electric Power think! How much power is used by a calculator that operates on 8 volts and 0.1 ampere? If it is used for one hour, how much energy does it use? Answer: Power = current × voltage = (0.1 A) × (8 V) = 0.8 W. Energy = power × time = (0.8 W) × (1 h) = 0.8 watt-hour, or kilowatt-hour.
34.11 Electric Power think! Will a 1200-watt hair dryer operate on a 120-volt line if the current is limited to 15 amperes by a safety fuse? Can two hair dryers operate on this line?
34.11 Electric Power think! Will a 1200-watt hair dryer operate on a 120-volt line if the current is limited to 15 amperes by a safety fuse? Can two hair dryers operate on this line? Answer: One 1200-W hair dryer can be operated because the circuit can provide (15 A) × (120 V) = 1800 W. But there is inadequate power to operate two hair dryers of combined power 2400 W. In terms of current, (1200 W)/(120 V) = 10 A; so the hair dryer will operate when connected to the circuit. But two hair dryers will require 20 A and will blow the 15-A fuse.
34.11 Electric Power How can you express electric power in terms of current and voltage?
Assessment Questions Electric charge will flow in an electric circuit when electrical resistance is low enough. a potential difference exists. the circuit is grounded. electrical devices in the circuit are not defective.
Assessment Questions Electric charge will flow in an electric circuit when electrical resistance is low enough. a potential difference exists. the circuit is grounded. electrical devices in the circuit are not defective. Answer: B
Assessment Questions The electric current in a copper wire is normally composed of electrons. protons. ions. amperes.
Assessment Questions The electric current in a copper wire is normally composed of electrons. protons. ions. amperes. Answer: A
Assessment Questions Which statement is correct?
Voltage flows in a circuit. Charge flows in a circuit. A battery is the source of electrons in a circuit. A generator is the source of electrons in a circuit.
Assessment Questions Which statement is correct?
Voltage flows in a circuit. Charge flows in a circuit. A battery is the source of electrons in a circuit. A generator is the source of electrons in a circuit. Answer: B
Assessment Questions Which of the following type of copper wire would you expect to have the least electric resistance? a thick long wire a thick short wire a thin long wire a thin short wire
Assessment Questions Which of the following type of copper wire would you expect to have the least electric resistance? a thick long wire a thick short wire a thin long wire a thin short wire Answer: D
Assessment Questions When you double the voltage in a simple electric circuit, you double the current. resistance. ohms. resistors.
Assessment Questions When you double the voltage in a simple electric circuit, you double the current. resistance. ohms. resistors. Answer: A
Assessment Questions To receive an electric shock there must be
current in one direction. moisture in an electrical device being used. high voltage and low body resistance. a difference in potential across part or all of your body.
Assessment Questions To receive an electric shock there must be
current in one direction. moisture in an electrical device being used. high voltage and low body resistance. a difference in potential across part or all of your body. Answer: D
Assessment Questions The difference between DC and AC in electrical circuits is that in DC charges flow steadily in one direction only. charges flow in one direction. charges steadily flow to and fro. charges flow to and fro.
Assessment Questions The difference between DC and AC in electrical circuits is that in DC charges flow steadily in one direction only. charges flow in one direction. charges steadily flow to and fro. charges flow to and fro. Answer: B
Assessment Questions To convert AC to a fairly steady DC, which devices are used? diodes and batteries capacitors and diodes capacitors and batteries resistors and batteries
Assessment Questions To convert AC to a fairly steady DC, which devices are used? diodes and batteries capacitors and diodes capacitors and batteries resistors and batteries Answer: B
Assessment Questions What is it that travels at about the speed of light in an electric circuit? charges current electric field voltage
Assessment Questions What is it that travels at about the speed of light in an electric circuit? charges current electric field voltage Answer: C
Assessment Questions When you buy a water pipe in a hardware store, the water isn’t included. When you buy copper wire, electrons must be supplied by you, just as water must be supplied for a water pipe. are already in the wire. may fall out, which is why wires are insulated. enter it from the electric outlet.
Assessment Questions When you buy a water pipe in a hardware store, the water isn’t included. When you buy copper wire, electrons must be supplied by you, just as water must be supplied for a water pipe. are already in the wire. may fall out, which is why wires are insulated. enter it from the electric outlet. Answer: B
Assessment Questions If you double both the current and the voltage in a circuit, the power remains unchanged if resistance remains constant. halves. doubles. quadruples.
Assessment Questions If you double both the current and the voltage in a circuit, the power remains unchanged if resistance remains constant. halves. doubles. quadruples. Answer: D
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https://www.allquestionandanswer.com/questions/tags/neometrix-engineering/ | 1,675,005,445,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499744.74/warc/CC-MAIN-20230129144110-20230129174110-00537.warc.gz | 674,178,012 | 17,970 | # neometrix engineering Interview Questions
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418 views | 210 | 611 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2023-06 | latest | en | 0.860603 |
https://pythongdb.com/dice-roller-python-code/ | 1,696,160,662,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510888.64/warc/CC-MAIN-20231001105617-20231001135617-00329.warc.gz | 502,777,766 | 23,993 | # Dice Roller Python Code
You can run this code and each time it executes, it will simulate rolling a dice and display the corresponding value on the screen.
## In this version of code, there are three main functions:
1. The roll_dice function allows you to specify the number of dice and the number of sides on each die. It returns a list of rolled dice values.
2. The print function prints the rolled dice values in a user-friendly format.
3. Th`s `unction calculates the sum of the rolled dice values.
## The role with trick function introduces additional tricks for rolling the dice:
• If you add ‘!’ to the number of dice, it will roll the specified number of dice and display their sum.
• If you add ‘d’ to the number of dice, it will roll the specified number of dice with the specified number of sides on each die.
• If you enter just a number, it will roll that number of 6-sided dice.
Code
``````import random
def roll_dice(num_dice=1, num_sides=6):
"""
Simulates rolling multiple dice with a specified number of sides.
Parameters:
- num_dice (int): Number of dice to roll (default: 1)
- num_sides (int): Number of sides on each die (default: 6)
Returns:
- List[int]: List of rolled dice values
"""
dice_rolls = []
for _ in range(num_dice):
dice_value = random.randint(1, num_sides)
dice_rolls.append(dice_value)
return dice_rolls
def print_dice(dice_rolls):
"""
Prints the rolled dice values.
Parameters:
- dice_rolls (List[int]): List of rolled dice values
"""
for i, dice_value in enumerate(dice_rolls):
print(f"Die {i + 1}: {dice_value}")
def sum_dice(dice_rolls):
"""
Calculates the sum of the rolled dice values.
Parameters:
- dice_rolls (List[int]): List of rolled dice values
Returns:
- int: Sum of the rolled dice values
"""
return sum(dice_rolls)
def roll_with_tricks():
"""
Simulates rolling dice with additional tricks.
Tricks:
- Add '!' to the number of dice to roll to get the sum of all dice.
- Add 'd' to the number of dice to roll to specify the number of sides on each die.
Examples:
- '2' will roll two 6-sided dice.
- '3d10' will roll three 10-sided dice.
- '4!' will roll four 6-sided dice and display the sum.
Returns:
- List[int] or int: List of rolled dice values or sum of the rolled dice values
"""
trick_input = input("Enter the number of dice to roll (or use tricks): ").strip().lower()
if trick_input.endswith('!'):
num_dice = int(trick_input[:-1])
dice_rolls = roll_dice(num_dice)
print_dice(dice_rolls)
return sum_dice(dice_rolls)
if 'd' in trick_input:
num_dice, num_sides = map(int, trick_input.split('d'))
dice_rolls = roll_dice(num_dice, num_sides)
print_dice(dice_rolls)
return dice_rolls
num_dice = int(trick_input)
dice_rolls = roll_dice(num_dice)
print_dice(dice_rolls)
# Roll the dice with tricks
roll_with_tricks()
``````
Output
Note:- You can run this code and experiment with different inputs to roll dice, display the values, and calculate sums. | 740 | 2,924 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2023-40 | latest | en | 0.519386 |
https://www.vikschaatcorner.com/what-did-archimedes-discover-in-eureka/ | 1,685,418,391,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224645089.3/warc/CC-MAIN-20230530032334-20230530062334-00457.warc.gz | 1,175,480,961 | 18,658 | # What did Archimedes discover in Eureka?
## What did Archimedes discover in Eureka?
In Physics, this Archimedes’ Eureka moment is called the Archimedes Principle, which states that when a body is immersed in a liquid, it experiences an upward buoyant force, which is equal to the weight of the liquid displaced by the body. In fact, buoyancy explains why some objects float, and others don’t.
## Why did Archimedes jump out of the bath shouting Eureka?
Therefore, it would displace more water than a crown of pure gold. Archimedes, allegedly, became so excited by this discovery that he jumped out of the bath and ran naked through the city streets, shouting: “Eureka! Eureka!” meaning, “I have found it! I have found it!”
How did Archimedes solve the gold Crown problem?
According to legend, Archimedes weighed the king’s crown. Then he got a piece of pure gold that weighed the same amount as the crown. He placed the gold into a bowl of water, measured how much it made the water rise, and took the gold out.
Did Archimedes cry Eureka?
Realizing he has hit upon a solution, the young Greek math whiz leaps out of the bath and rushes home naked crying “Eureka! Eureka!” Or, translated: “I’ve found it! I’ve found it!”
### Who discovered water displacement?
Archimedes
Archimedes took one mass of gold and one of silver, both equal in weight to the crown. He filled a vessel to the brim with water, put the silver in, and found how much water the silver displaced.
### When did Archimedes discover buoyancy?
ARCHIMEDES’S PRINCIPLE: 287-212 b.c.), who first identified it.
Why did Einstein say Eureka?
He reportedly proclaimed “Eureka! Eureka!” after he had stepped into a bath and noticed that the water level rose, whereupon he suddenly understood that the volume of water displaced must be equal to the volume of the part of his body he had submerged.
How did he prove that the goldsmith cheated on the king’s crown?
Archimedes found that the crown did, in fact displace more water than the lump of gold of equal weight. Thus he came to the conclusion that the crown was not pure gold, and that the goldsmith had indeed mixed some silver (or other, lighter metal) into the gold in an attempt to cheat the king.
#### Who invented density?
Archimedes of Syracuse
Archimedes
Archimedes of Syracuse
Born c. 287 BC Syracuse, Sicily
Died c. 212 BC (aged approximately 75) Syracuse, Sicily
Known for List Archimedes’ principle Archimedes’ screw Center of gravity Statics Hydrostatics Law of the lever Indivisibles Neuseis constructions List of other things named after him
Scientific career
#### Did Archimedes Discover displacement?
Legend says that Archimedes discovered the principle of displacement while stepping into a full bath. He realized that the water that ran over equaled in volume the submerged part of his body. Through further experiments, he deduced the above mentioned Archimedes’ principle.
How did Archimedes estimated pi?
Archimedes’ method finds an approximation of pi by determining the length of the perimeter of a polygon inscribed within a circle (which is less than the circumference of the circle) and the perimeter of a polygon circumscribed outside a circle (which is greater than the circumference).
What is Archimedes Eureka moment?
It was Archimedes Eureka moment, a 2300 years old event which changed the history. Who was Archimedes? Archimedes, born in 287 BC, is a well-known name to all of us, we all have studied his principle of buoyancy in high school physics. He was a Greek physicist, mathematician, astronomer and engineer.
## What is the Archimedes’box?
Also known as Loculus of Archimedes or Archimedes’ Box, this is a dissection puzzle similar to a Tangram, and the treatise describing it was found in more complete form in the Archimedes Palimpsest. Archimedes calculates the areas of the 14 pieces which can be assembled to form a square.
## What did Archimedes say when he ran in Syracuse?
Eureka!’ Archimedes has gone down in history as the guy who ran naked through the streets of Syracuse shouting “Eureka!” — or “I have it!”
How much food do you need in Archimedes Eureka?
Archimedes: Eureka! Collector’s Edition There are many screenshots in this walkthrough, you can choose to download the full walkthrough now. You can see the goals of the current level on top right corner. Build nut need 25 foods and 25 woods. | 1,014 | 4,396 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2023-23 | latest | en | 0.959165 |
https://justaaa.com/statistics-and-probability/12006-1-a-simple-random-sample-from-a-population-with-a | 1,713,240,986,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817043.36/warc/CC-MAIN-20240416031446-20240416061446-00060.warc.gz | 310,527,660 | 10,195 | Question
# 1. A simple random sample from a population with a normal distribution of 97 body temperatures...
1. A simple random sample from a population with a normal distribution of 97 body temperatures has x=98.90°F and s=0.62°F. Construct a 95% confidence interval estimate of the standard deviation of body temperature of all healthy humans.
?°F.< σ < ?°F
(Round to two decimal places as needed.)
2. The test statistic of z=2.45 is obtained when testing the claim that p>0.3.
a. Identify the hypothesis test as being two-tailed, left-tailed, or right-tailed.
b. Find the P-value.
c. Using a significance level of f a=0.05, should we reject H0 or should we fail to reject H0?
a. This is a ▼ test.
b. P-value=. (Round to three decimal places as needed.)
c. Choose the correct conclusion below.
A. Fail to rejectFail to reject H0. There is notis not sufficient evidence to support the claim that p>0.3.
B. Fail to rejectFail to reject H0. There is sufficient evidence to support the claim that p>0.3.
C. RejectReject Upper H0. There is notis not sufficient evidence to support the claim that p>0.3.
D. RejectReject H0. There Is sufficient evidence to support the claim that p>0.3.
#### Homework Answers
Answer #1
1)
here n = 97 s2= 0.384
Critical value of chi square distribution for n-1=96 df and 95 % CI Lower critical value χ2L= 70.783 from excel: chiinv(0.975,96) Upper critical valueχ2U= 125.000 from excel: chiinv(0.025,96)
for Confidence interval of standard deviation: Lower bound =√((n-1)s2/χ2U)= 0.543 Upper bound =√((n-1)s2/χ2L)= 0.722
from above 95% confidence interval for population standard deviation =(0.54<σ<0.72)
2)
a) since we are checking if proportion is greater than 0.3
this is a right tailed test
b)
p value from excel: =1-normsdist(2.45)=0.007
c)
since p value <0.05 ; option D is correct
Reject H0. There Is sufficient evidence to support the claim that p>0.3.
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Total # Posts: 3,530
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Post a New Question | 3,187 | 10,200 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2017-26 | longest | en | 0.903073 |
http://pidanic.com/en/blog/hackerrank-sequence-equation/ | 1,726,514,067,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651710.86/warc/CC-MAIN-20240916180320-20240916210320-00740.warc.gz | 21,517,499 | 24,342 | # Hackerrank – Sequence Equation
## Hackerrank – Problem Statement
A description of the problem can be found on Hackerrank.
## Solution
I have given count of numbers n.
I will iterate over all numbers from `1` to `n`
I will find index of each iterated number in all elements and then index of index within all elements
I created solution in:
All solutions are also available on my GitHub profile.
## 1 thoughts on “Hackerrank – Sequence Equation”
1. Jared says:
Here is a solution for Python:
def permutationEquation(p):
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# Four extra-large sandwiches of exactly the same size were ordered for
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Intern
Joined: 07 May 2015
Posts: 35
Re: Four extra-large sandwiches of exactly the same size were ordered for [#permalink]
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15 Jun 2018, 08:38
Hi all,
I don't mean to bump an old post but this leads me to think I am really over-thinking problems. I took the same approach as many of you, picking numbers, but I got lost in the question stem. I kept thinking "is Carol included in M? is she an outsider that eats them after they've been divided evenly among the M students?" This seems to be a common error I make. Any tips for getting over this?
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Posts: 522
Location: India
Re: Four extra-large sandwiches of exactly the same size were ordered for [#permalink]
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17 Jun 2018, 01:17
Bunuel wrote:
Four extra-large sandwiches of exactly the same size were ordered for m students, where m > 4. Three of the sandwiches were evenly divided among the students. Since 4 students did not want any of the fourth sandwich, it was evenly divided among the remaining students. If Carol ate one piece from each of the four sandwiches, the amount of sandwich that she ate would be what fraction of a whole extra-large sandwich?
A) (m+4)/[m(m-4)]
B) (2m-4)/[m(m-4)]
C) (4m-4)/[m(m-4)]
D) (4m-8)/[m(m-4)]
E) (4m-12)/[m(m-4)]
Kudos for a correct solution.
Let X be the area of each sandwich.
Given 3 sandwiches are evenly divided among m students. Hence each student received one piece of area 3X/m
Now the 4th sandwich is refused by 4 students & is evenly divided among the rest of the students. Hence each of these students received a X/(m-4) area a piece of this sandwich.
Carol ate one piece from each sandwich, hence total area of sandwich consumed by carol = 3X/m + X/(m-4) = X(4m-12)/[m(m-4)]
Hence Carol ate (4m-12)/[m(m-4)] fraction of the extra large sandwich.
Thanks,
GyM
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Re: Four extra-large sandwiches of exactly the same size were ordered for [#permalink]
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17 Jun 2018, 03:13
Bunuel wrote:
Four extra-large sandwiches of exactly the same size were ordered for m students, where m > 4. Three of the sandwiches were evenly divided among the students. Since 4 students did not want any of the fourth sandwich, it was evenly divided among the remaining students. If Carol ate one piece from each of the four sandwiches, the amount of sandwich that she ate would be what fraction of a whole extra-large sandwich?
A) (m+4)/[m(m-4)]
B) (2m-4)/[m(m-4)]
C) (4m-4)/[m(m-4)]
D) (4m-8)/[m(m-4)]
E) (4m-12)/[m(m-4)]
Let m=5, implying that there are 5 students.
Let each sandwich = 5 units, implying that 3 sandwiches = 3*5 = 15 units.
Since 3 sandwiches are distributed among all 5 students -- including Carol -- the number of units received by Carol from these 3 sandwiches = 15/5 = 3 units.
Since 4 of the 5 students do not share in the last sandwich, and Carol eats a portion of EVERY sandwich, all 5 units of the last sandwich must be given to Carol.
Thus, total units for Carol = 3+5 = 8 units.
Resulting fraction:
(Carol's units)/(units per sandwich) = 8/5. This is our target.
Now plug m=5 into the answer choices to see which yields our target of 8/5.
Every answer choice has the same denominator:
m(m-4) = 5(5-4) = 5.
To yield our target of 8/5, the correct answer choice must have a numerator of 8.
Only E works:
4m-12 = (4*5) - 12 = 8.
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Re: Four extra-large sandwiches of exactly the same size were ordered for [#permalink]
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27 Jun 2018, 10:53
I feel that this question could be best solved by substituting some value to m.
Let m = 12.
Scenario 1:
3 sandwiches --- 12 students
? --- 1 student
$$= \frac{3}{12} = \frac{1}{4}$$
Scenario 2:
1 sandwich --- 8 students
? --- 1 student
$$= \frac{1}{8}$$
$$Total : \frac{1}{4}+ \frac{1}{8} = \frac{3}{8}$$
When substituted m=12 in all of the options, only E gives the correct value.
Intern
Joined: 15 Sep 2018
Posts: 30
Re: Four extra-large sandwiches of exactly the same size were ordered for [#permalink]
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25 Sep 2018, 06:07
If the $$4$$ sandwiches are to be divided among $$m$$ students, this means that each of student will have $$\frac{4}{m}$$ of the entire portion.
Given that $$4$$ of the students didn’t want their share on the fourth sandwich, these were then divided among the total students (minus those $$4$$); that’s $$1$$ divided by $$(m-4)$$ students.
This means each of the students has received an additional $$\frac{1}{(m-4)}$$ of the sandwiches.
Carol ate $$\frac{3}{m}$$ and $$\frac{1}{(m-4)}$$ of the portions. That adds up to :
$$\frac{3}{m}+\frac{1}{(m-4)}$$
$$=\frac{3(m-4)}{m(m-4)} +\frac{m}{m(m-4)}$$
$$=\frac{(3m-12+m)}{(m(m-4))}$$
$$=\frac{(4m-12)}{(m(m-4))}$$
Manager
Joined: 13 Aug 2018
Posts: 63
Re: Four extra-large sandwiches of exactly the same size were ordered for [#permalink]
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13 Oct 2018, 21:02
Hi, I assumed m = 6 and what I got was 3/6 (3 sandwiches divided among 6) + 1/2 (6-4= 2 people equally sharing the one sandwich) and got 1 and selected E, which was correct.
However, I had a little problem with the language of the question when it said Carol had each of the four sandwiches. I was confused because I thought it said that Carol should have one piece of four sandwiches which would mean that she should have 1/2 (the share equally divided for all the sandwiches) * 4 which is 2.
Any help will be appreciated to how to translate this part properly
Re: Four extra-large sandwiches of exactly the same size were ordered for [#permalink] 13 Oct 2018, 21:02
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Display posts from previous: Sort by | 1,904 | 6,709 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2019-22 | latest | en | 0.956192 |
http://mathhelpforum.com/pre-calculus/36675-performance-final-done-today-problem-there-i-could-not-figure-out.html | 1,529,481,770,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863489.85/warc/CC-MAIN-20180620065936-20180620085936-00082.warc.gz | 202,280,489 | 10,024 | # Thread: Performance Final Done today. Problem on there I could not figure out.
1. ## Performance Final Done today. Problem on there I could not figure out.
I got done with it about 2 hours ago, and there was one problem I could not figure out.
I had to prove that:
$\displaystyle cot x+\frac{sinx}{1+cosx}=cscx$
Could not figure it out to save my life. Could someone show me how to do it?
2. $\displaystyle \cot (x) + \frac{{\sin (x)}}{{1 + \cos (x)}} = \frac{{\cos (x)}}{{\sin (x)}} + \frac{{\sin (x)}}{{1 + \cos (x)}}$
$\displaystyle \frac{{\cos (x)}}{{\sin (x)}} + \frac{{\sin (x)}}{{1 + \cos (x)}} = \frac{{\cos (x) + \cos ^2 (x) + \sin ^2 (x)}}{{\sin (x)\left( {1 + \cos (x)} \right)}} = \frac{1}{{\sin (x)}}$
3. Hello, mathgeek777!
Another approach . . .
Prove: .$\displaystyle \cot x+\frac{\sin x}{1+\cos x}\:=\:\csc x$
Multiply the fraction by .$\displaystyle \frac{1-\cos x}{1 - \cos x}$
. . $\displaystyle \cot x \:+ \:\frac{\sin x}{1 + \cos x}\cdot{\color{blue}\frac{1-\cos x}{1-\cos x}} \;\;=\; \;\cot x + \frac{\sin x(1-\cos x)}{1-\cos^2\!x}$
. . $\displaystyle =\;\;\cot x + \frac{\sin x(1-\cos x)}{\sin^2\!x} \;\;=\;\;\cot x + \frac{1-\cos x}{\sin x}$
. . $\displaystyle = \;\;\frac{\cos x}{\sin x} + \frac{1-\cos x}{\sin x} \;\;=\;\;\frac{1}{\sin x} \;\;=\;\;\csc x$ | 499 | 1,297 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2018-26 | latest | en | 0.670467 |
https://www.jiskha.com/questions/1654304/write-the-equation-in-standard-form-and-then-graph-the-equation-y-2x-2-8x-11 | 1,534,905,552,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221219242.93/warc/CC-MAIN-20180822010128-20180822030128-00024.warc.gz | 894,628,983 | 5,424 | # Algebra
write the equation in standard form and then graph the equation.
y=2x^2-8x+11
1. You will have to use "completing the square".
y = 2x^2-8x+11
= 2(x^2 - 4x+ .... ) + 11 <---- took out the coefficient of the square term from the first two terms
= 2(x^2 - 4x + 4 - 4) <--- took half of the coefficient of the x term, then squared it, added it and immediately subtracted it again, the net result was adding zero.
= 2( (x-2)^2 - 4) + 11 <----- formed the perfect square plus the left over constant
= 2(x-2)^2 - 8 + 11 <---- dististributed the 2 over the perfect square and the left-over constant
= 2(x-2)^2 + 3 <---- added the stuff at the end
all done, now I can read the vertex
which is .......
and if I let x = 0 in the original (or this last) equation, I get
y = 11
So plot the vertex, plot the y-intercept and form a neat parabola from the vertex to hit (0,11). Complete the other half of the parabola.
It should look like this:
http://www.wolframalpha.com/input/?i=plot+y+%3D+2x%5E2-8x%2B11,+y+%3D+2(x-2)%5E2+%2B+3
notice I graphed both forms of the equation and the graphs coincide. This was a neat way to check my work.
posted by Reiny
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https://askrose.org/student-resources/science-resources/chemistry/behavior-of-gases/gaseous-reactions/ | 1,702,204,063,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679101779.95/warc/CC-MAIN-20231210092457-20231210122457-00123.warc.gz | 128,540,209 | 14,993 | Gaseous Reactions
20 mols gaseous ammonia and 15 mols oxygen gas react to form nitrogen monoxide gas and water vapor.
If the reaction occurs at and atmospheric conditions of pressure what will be the final volume of the unreacted component and products.
Step 1: Determine the limiting reactant
• Assuming all reactants react to determine which reactant produces less product
20 mol mols NO produced
15 mol mols NO produced
Step 2: Determine # of mols present after reaction
15 mol produced
15 mol react
20 mol
8mol
Step 3: Ideal gas law Assumption
Assume water vapor, NH_3 and NO behave as IG
PV=nRT
• Solve for Volume
• V=
• Substitution
• =945.3 L | 169 | 664 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2023-50 | longest | en | 0.81847 |
http://www.fixya.com/support/t21743631-much_cubic_ft_kenmore_model_106_9552810 | 1,513,372,191,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948579564.61/warc/CC-MAIN-20171215192327-20171215214327-00669.warc.gz | 371,121,514 | 33,200 | # How much cubic ft is the Kenmore model 106.9552810
Posted by Anonymous on
Hi,
a 6ya expert can help you resolve that issue over the phone in a minute or two.
Best thing about this new service is that you are never placed on hold and get to talk to real repairmen in the US.
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Goodluck!
Posted on Jan 02, 2017
SOURCE: cubic feet?
Most refrigerator models with two digits, the rest letters, the digits stand for the cubic foot capacity of the refrigerator. In you case, 23 cubic foot model.
Posted on Sep 16, 2008
If you can provide it, then it can be determined and given to you.
Posted on Jan 15, 2009
i have this problem .whats the fix
Posted on May 29, 2009
SOURCE: Would like to know specs
i believe this model is a standard size unit,
and should be very easily repalced by almost any manufacturer today
this unit has a capacity of 3.5 cubic feet on the drum
Posted on Nov 03, 2009
I tried downloading a manual for your Kenmore Top-mount refrigerator-freezer but it is no-longer available. the next best way is to measure the inside dimensions of your refrigerator. Measure the depth, height and width in inches. Times them together and divide by 1728.
example: if your inside of your refigerator is 30 inches wide by 20 inched deep by 50 inches tall you get 30 x 50 x 20 = 30000 cuin / 1728 = 17.36 cuft. Do the freezer section seperate to know thier individual measurements and add them together to get the total. I hope this helps. - subdriver
Posted on Mar 25, 2010
Testimonial: "Thanks for taking the time to read and respond to my question"
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## Related Questions:
### What is the cubic feet of Kenmore refrigarator model #253.562824on and also need mearsurements of outside.
You seem to be missing a few digits in the model number. Sears Parts Direct lists 11 refrigerator models that start 253.562824. (The full models numbers have 2 extra characters, including 253.56282400 and 253.5628240N.) See http://www.searspartsdirect.com/model/search.html?q=253.562824 for the list of models.
Kenmore uses the 3rd digit after the decimal point to indicate the cubic footage of their refrigerators. With a 2 in that position, this model has a 22 cubic feet capacity (probably 21.6 or 22.1 cubic feet before rounding). Using the size of other models, the external dimensions will be between 30" W x 31-3/4" D x 69" H (the size of a 21 cu. ft. model) and 35-5/8" W x 28-1/4" D x 69-3/8" H (23 cu. ft. model). One current Frigidaire 22 cubic foot side by side model has the external dimensions: 33" W x 32" D x 69-7/8" H. (Note: the 253 in the model number indicates a product made by Frigidaire for the Kenmore line.) Unfortunately, I can't find the dimensions for any of 253.562824 models.
I hope this helps.
Cindy Wells
Aug 30, 2016 | Kenmore Refrigerators
### How many cubic ft. in a Kenmore microwave over stove model # 665-61612100
Just measure the inside dimensions of the unit in inches. Get the length, width and height then use this formula: (L x W x H)/12 = Cubic Feet
Feb 26, 2015 | Kenmore Microwave Ovens
### Kenmore model 795.7500401 need to know cu sq ft
Hi,
Here is a tip that I wrote that tells you how to figure out how big your refrigerator and/or your freezer is...
New Refrigerator Size or Refrigerator Cubic Feet
Heatman101
Aug 27, 2010 | Kenmore 71279 Bottom Freezer Refrigerator
### How many cubic feet are in the Kenmore Washer Model 110 80 series
The cubic feet of the Kenmore washer model 110 heavy duty is 2.8 cubic feet. Measured to the top water line 14.5 inches and using the radius of the tub of 10.375 inches.
Oct 26, 2009 | Kenmore 3.2 cu. ft. Plus Top Load Washer
### Kenmore refrigerator and unknown capacity specs
It should be 25 cubic ft. It is usally the first few numbers in the model number.
Phil
Apr 10, 2009 | Kenmore Refrigerators
### Cubic Feet of Washer
3.2 cu ft.
Thanks for using FixYa - a 4 THUMBS rating is appreciated for answering your FREE question.
Jan 15, 2009 | Kenmore 22422 Top Load Washer
## Open Questions:
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55 people viewed this question
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Level 3 Expert | 1,143 | 4,246 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2017-51 | latest | en | 0.926487 |
https://www.geeksforgeeks.org/pairs-of-positive-negative-values-in-an-array/ | 1,708,985,396,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474663.47/warc/CC-MAIN-20240226194006-20240226224006-00382.warc.gz | 764,704,119 | 65,526 | # Find pairs of Positive and Negative values present in given array
Given an array of distinct integers, print all the pairs having both positive and negative values of a number that exists in the array. The pairs can be printed in any order.
Examples:
Input: arr[] = {1, -3, 2, 3, 6, -1}
Output: -1 1 -3 3
Input: arr[] = {4, 8, 9, -4, 1, -1, -8, -9}
Output: -4 4 -8 8 -9 9 -1 1
Recommended Practice
Naive Approach: To solve the problem follow the below idea:
The idea is to use two nested loops. For each element arr[i], find negative of arr[i] from index i + 1 to n – 1 and store it in another array
Below is the implementation of this approach:
## C++
`// Simple CPP program to find pairs of positive``// and negative values present in an array.``#include ``using` `namespace` `std;` `// Print pair with negative and positive value``void` `printPairs(``int` `arr[], ``int` `n)``{`` ``vector<``int``> v;` ` ``// For each element of array.`` ``for` `(``int` `i = 0; i < n; i++)` ` ``// Try to find the negative value of`` ``// arr[i] from i + 1 to n`` ``for` `(``int` `j = i + 1; j < n; j++)` ` ``// If absolute values are equal print pair.`` ``if` `(``abs``(arr[i]) == ``abs``(arr[j]))`` ``v.push_back(``abs``(arr[i]));` ` ``// If size of vector is 0, therefore there is no`` ``// element with positive negative value, print "0"`` ``if` `(v.size() == 0)`` ``return``;` ` ``// Print the pair with negative positive value.`` ``for` `(``int` `i = 0; i < v.size(); i++)`` ``cout << -v[i] << ``" "` `<< v[i] << ``" "``;``}` `// Driver code``int` `main()``{`` ``int` `arr[] = { 4, 8, 9, -4, 1, -1, -8, -9 };`` ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);` ` ``// Function call`` ``printPairs(arr, n);`` ``return` `0;``}`
## Java
`// Java program to find pairs of positive``// and negative values present in an array.``import` `java.lang.*;``import` `java.util.*;` `class` `GFG {` ` ``// Print pair with negative and positive value`` ``public` `static` `void` `printPairs(``int` `arr[], ``int` `n)`` ``{`` ``Vector v = ``new` `Vector();`` ``// For each element of array.`` ``for` `(``int` `i = ``0``; i < n; i++)` ` ``// Try to find the negative value of`` ``// arr[i] from i + 1 to n`` ``for` `(``int` `j = i + ``1``; j < n; j++)` ` ``// If absolute values are equal`` ``// print pair.`` ``if` `(Math.abs(arr[i]) == Math.abs(arr[j]))`` ``v.add(Math.abs(arr[i]));` ` ``// If size of vector is 0, therefore there`` ``// is no element with positive negative`` ``// value, print "0"`` ``if` `(v.size() == ``0``)`` ``return``;` ` ``// Print the pair with negative positive`` ``// value.`` ``for` `(``int` `i = ``0``; i < v.size(); i++)`` ``System.out.print(-v.get(i) + ``" "` `+ v.get(i)`` ``+ ``" "``);`` ``}` ` ``// Driver code`` ``public` `static` `void` `main(String[] args)`` ``{`` ``int` `arr[] = { ``4``, ``8``, ``9``, -``4``, ``1``, -``1``, -``8``, -``9` `};`` ``int` `n = arr.length;` ` ``// Function call`` ``printPairs(arr, n);`` ``}``}` `// This code is contributed by Prasad Kshirsagar.`
## Python 3
`# Simple Python 3 program to find``# pairs of positive and negative``# values present in an array.` `# Print pair with negative and``# positive value` `def` `printPairs(arr, n):`` ``v ``=` `[]` ` ``# For each element of array.`` ``for` `i ``in` `range``(n):` ` ``# Try to find the negative value`` ``# of arr[i] from i + 1 to n`` ``for` `j ``in` `range``(i ``+` `1``, n):` ` ``# If absolute values are`` ``# equal print pair.`` ``if` `(``abs``(arr[i]) ``=``=` `abs``(arr[j])):`` ``v.append(``abs``(arr[i]))` ` ``# If size of vector is 0, therefore`` ``# there is no element with positive`` ``# negative value, print "0"`` ``if` `(``len``(v) ``=``=` `0``):`` ``return` ` ``# Print the pair with negative`` ``# positive value.`` ``for` `i ``in` `range``(``len``(v)):`` ``print``(``-``v[i], "``", v[i], end="` `")` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:`` ``arr ``=` `[``4``, ``8``, ``9``, ``-``4``, ``1``, ``-``1``, ``-``8``, ``-``9``]`` ``n ``=` `len``(arr)` ` ``# Function call`` ``printPairs(arr, n)` `# This code is contributed``# by ChitraNayal`
## C#
`// C# program to find pairs of positive``// and negative values present in an array.``using` `System;``using` `System.Collections.Generic;` `class` `GFG {` ` ``// Print pair with negative and positive value`` ``public` `static` `void` `printPairs(``int``[] arr, ``int` `n)`` ``{`` ``List<``int``> v = ``new` `List<``int``>();` ` ``// For each element of array.`` ``for` `(``int` `i = 0; i < n; i++)` ` ``// Try to find the negative value of`` ``// arr[i] from i + 1 to n`` ``for` `(``int` `j = i + 1; j < n; j++)` ` ``// If absolute values are equal`` ``// print pair.`` ``if` `(Math.Abs(arr[i]) == Math.Abs(arr[j]))`` ``v.Add(Math.Abs(arr[i]));` ` ``// If size of vector is 0, therefore there`` ``// is no element with positive negative`` ``// value, print "0"`` ``if` `(v.Count == 0)`` ``return``;` ` ``// Print the pair with negative positive`` ``// value.`` ``for` `(``int` `i = 0; i < v.Count; i++)`` ``Console.Write(-v[i] + ``" "` `+ v[i] + ``" "``);`` ``}` ` ``// Driver code`` ``public` `static` `void` `Main(String[] args)`` ``{`` ``int``[] arr = { 4, 8, 9, -4, 1, -1, -8, -9 };`` ``int` `n = arr.Length;` ` ``// Function call`` ``printPairs(arr, n);`` ``}``}` `// This code has been contributed by 29AjayKumar`
## Javascript
``
Output
`-4 4 -8 8 -9 9 -1 1 `
Time Complexity: O(N2)
Auxiliary Space: O(N)
## Find pairs of Positive and Negative values present in given array using hashing:
To solve the problem follow the below idea:
The idea is to use hashing to store count of absolute value of every element present in the array. If the count of any element is equal to 2, then it means that a pair has been found
Follow the given steps to solve the problem:
• Traverse the given array, and increase the count at the absolute value of the hash table.
• If the count becomes 2, store its absolute value in another vector.
• If the size of the vector is 0, print “0”,
• else for each term in the vector print first its negative value and the positive value
Below is the implementation of this approach:
## C++
`// CPP program to find pairs of``// positive and negative values present in``// an array.``#include ``using` `namespace` `std;` `// Print pair with negative and positive value``void` `printPairs(``int` `arr[], ``int` `n)``{`` ``vector<``int``> v;`` ``unordered_map<``int``, ``bool``> cnt;` ` ``// For each element of array.`` ``for` `(``int` `i = 0; i < n; i++) {` ` ``// If element has not encounter early,`` ``// mark it on cnt array.`` ``if` `(cnt[``abs``(arr[i])] == 0)`` ``cnt[``abs``(arr[i])] = 1;` ` ``// If seen before, push it in vector (`` ``// given that elements are distinct)`` ``else` `{`` ``v.push_back(``abs``(arr[i]));`` ``cnt[``abs``(arr[i])] = 0;`` ``}`` ``}` ` ``if` `(v.size() == 0)`` ``return``;` ` ``for` `(``int` `i = 0; i < v.size(); i++)`` ``cout << ``"-"` `<< v[i] << ``" "` `<< v[i] << ``" "``;``}` `// Driver code``int` `main()``{`` ``int` `arr[] = { 4, 8, 9, -4, 1, -1, -8, -9 };`` ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);` ` ``// Function call`` ``printPairs(arr, n);` ` ``return` `0;``}`
## Java
`// Java program to find pairs of``// positive and negative values present in``// an array.``import` `java.util.*;``class` `GFG {` ` ``// Print pair with negative`` ``// and positive value`` ``static` `void` `printPairs(``int` `arr[], ``int` `n)`` ``{`` ``ArrayList v = ``new` `ArrayList();`` ``HashMap cnt`` ``= ``new` `HashMap();` ` ``// For each element of array.`` ``for` `(``int` `i = ``0``; i < n; i++) {` ` ``// If element has encounter early,`` ``// then increment its count`` ``if` `(cnt.containsKey(Math.abs(arr[i])))`` ``cnt.put(Math.abs(arr[i]),`` ``cnt.get(Math.abs(arr[i])) + ``1``);` ` ``// If element has not seen before,`` ``// then initialize its count to 1`` ``else` `{`` ``cnt.put(Math.abs(arr[i]), ``1``);`` ``}`` ``if` `(cnt.get(Math.abs(arr[i])) == ``2``) {`` ``v.add(Math.abs(arr[i]));`` ``}`` ``}`` ``if` `(v.size() == ``0``)`` ``return``;` ` ``for` `(``int` `i = ``0``; i < v.size(); i++)`` ``System.out.print(``"-"` `+ v.get(i) + ``" "` `+ v.get(i)`` ``+ ``" "``);`` ``}` ` ``// Driver Code`` ``public` `static` `void` `main(String[] args)`` ``{`` ``int` `arr[] = { ``4``, ``8``, ``9``, -``4``, ``1``, -``1``, -``8``, -``9` `};`` ``int` `n = arr.length;` ` ``// Function call`` ``printPairs(arr, n);`` ``}``}``// This code is contributed by Prerna Saini`
## Python3
`# Python3 program to find pairs of``# positive and negative values present in``# an array.``# Print pair with negative and``# positive value` `def` `printPairs(arr, n):` ` ``s ``=` `set``()`` ``ret ``=` `[]` ` ``# For each element of array.`` ``for` `i ``in` `arr:`` ``if` `abs``(i) ``in` `s:`` ``ret.append(``abs``(i))`` ``else``:`` ``s.add(``abs``(i))` ` ``ret.sort()` ` ``for` `i ``in` `range``(``0``, ``len``(ret)):`` ``print``(``-``ret[i], "``", ret[i], end="` `")` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:`` ``arr ``=` `[``4``, ``8``, ``9``, ``-``4``, ``1``, ``-``1``, ``-``8``, ``-``9``]`` ``n ``=` `len``(arr)` ` ``# Function call`` ``printPairs(arr, n)``# This code is contributed by RohitOberoi`
## C#
`// C# program to find pairs of``// positive and negative values present in``// an array.``using` `System;``using` `System.Collections.Generic;``class` `GFG {` ` ``// Print pair with negative and positive value`` ``static` `void` `printPairs(``int``[] arr, ``int` `n)`` ``{`` ``List<``int``> v = ``new` `List<``int``>();`` ``Dictionary<``int``, ``bool``> cnt`` ``= ``new` `Dictionary<``int``, ``bool``>();` ` ``// For each element of array.`` ``for` `(``int` `i = 0; i < n; i++) {` ` ``// If element has not encounter early,`` ``// mark it on cnt array.`` ``int` `absVal = Math.Abs(arr[i]);`` ``if` `(!cnt.ContainsKey(absVal)) {`` ``cnt[absVal] = ``true``;`` ``}`` ``else` `if` `(cnt[absVal] == ``false``) {`` ``cnt[absVal] = ``true``;`` ``}`` ``else` `{`` ``v.Add(Math.Abs(arr[i]));`` ``cnt[absVal] = ``false``;`` ``}`` ``}` ` ``if` `(v.Count == 0)`` ``return``;` ` ``v.Sort();`` ``for` `(``int` `i = 0; i < v.Count; i++)`` ``Console.Write(-v[i] + ``" "` `+ v[i] + ``" "``);`` ``}` ` ``// Driver code`` ``static` `void` `Main()`` ``{`` ``int``[] arr = { 4, 8, 9, -4, 1, -1, -8, -9 };`` ``int` `n = arr.Length;` ` ``// Function call`` ``printPairs(arr, n);`` ``}``}``// This code is contributed by divyeshrabadiya07`
## Javascript
``
Output
`-4 4 -1 1 -8 8 -9 9 `
Time Complexity: O(N)
Auxiliary Space: O(N)
## Find pairs of Positive and Negative values present in given array using set:
To solve the problem follow the below idea:
The idea is to use a set. Find the negative of the number in the set. If it exits then print both the numbers and if it does not exist then add it to the set
Below is the implementation of this approach:
## C++
`// CPP program to find pairs of``// positive and negative values present in``// an array.``#include ``using` `namespace` `std;` `// Print pair with negative and positive value``void` `printPairs(``int` `arr[], ``int` `n)``{`` ``unordered_set<``int``> hs;`` ``vector<``int``> ans;`` ``for` `(``int` `i = 0; i < n; i++) {`` ``if` `(hs.find((arr[i]) * -1) != hs.end()) {`` ``if` `(arr[i] < 0) {`` ``cout << arr[i] << ``" "``;`` ``cout << (arr[i] * -1) << ``" "``;`` ``}`` ``else` `{`` ``cout << (arr[i] * -1) << ``" "``;`` ``cout << arr[i] << ``" "``;`` ``}`` ``}`` ``hs.insert(arr[i]);`` ``}` ` ``return``;``}` `// Driver code``int` `main()``{`` ``int` `arr[] = { 4, 8, 9, -4, 1, -1, -8, -9 };`` ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);` ` ``// Function call`` ``printPairs(arr, n);`` ``return` `0;``}`
## Java
`// Java program to find pairs of``// positive and negative values present in``// an array.``import` `java.util.*;` `public` `class` `GFG {` ` ``// Print pair with negative and positive value`` ``public` `static` `void` `printPairs(``int``[] arr, ``int` `n)`` ``{`` ``HashSet hs = ``new` `HashSet();`` ``ArrayList ans = ``new` `ArrayList();`` ``for` `(``int` `i = ``0``; i < n; i++) {`` ``if` `(hs.contains((arr[i]) * -``1``)) {`` ``if` `(arr[i] < ``0``) {`` ``System.out.print(arr[i]);`` ``System.out.print(``" "``);`` ``System.out.print((arr[i] * -``1``));`` ``System.out.print(``" "``);`` ``}`` ``else` `{`` ``System.out.print((arr[i] * -``1``));`` ``System.out.print(``" "``);`` ``System.out.print(arr[i]);`` ``System.out.print(``" "``);`` ``}`` ``}` ` ``hs.add(arr[i]);`` ``}` ` ``return``;`` ``}` ` ``// Driver code`` ``public` `static` `void` `main(String[] args)`` ``{`` ``int``[] arr = { ``4``, ``8``, ``9``, -``4``, ``1``, -``1``, -``8``, -``9` `};`` ``int` `n = arr.length;` ` ``// Function call`` ``printPairs(arr, n);`` ``}``}` `// This code is contributed by Aarti_Rathi`
## Python3
`# Python3 program to find pairs of``# positive and negative values present in``# an array` `def` `printPairs(arr, n):` ` ``hs ``=` `set``()`` ``ans ``=` `[]`` ``for` `i ``in` `range``(n):`` ``if` `(arr[i] ``*` `-``1``) ``in` `hs:` ` ``if` `(arr[i] < ``0``):`` ``print``(arr[i], end``=``" "``)`` ``print``((arr[i] ``*` `-``1``), end``=``" "``)` ` ``else``:`` ``print``((arr[i] ``*` `-``1``), end``=``" "``)`` ``print``(arr[i], end``=``" "``)` ` ``hs.add(arr[i])` ` ``return` `# Driver code``arr ``=` `[``4``, ``8``, ``9``, ``-``4``, ``1``, ``-``1``, ``-``8``, ``-``9``]``n ``=` `len``(arr)` `# Function call``printPairs(arr, n)` `# This code is contributed by shinjanpatra.`
## C#
`// C# program to find pairs of positive and``// negative values present in an array.``using` `System;``using` `System.Collections;``using` `System.Collections.Generic;` `public` `class` `GFG {` ` ``// Print pair with negative and positive value`` ``public` `static` `void` `printPairs(``int``[] arr, ``int` `n)`` ``{`` ``HashSet<``int``> hs = ``new` `HashSet<``int``>();`` ``for` `(``int` `i = 0; i < n; i++) {`` ``if` `(hs.Contains((arr[i]) * -1)) {`` ``if` `(arr[i] < 0) {`` ``Console.Write(arr[i]);`` ``Console.Write(``" "``);`` ``Console.Write((arr[i] * -1));`` ``Console.Write(``" "``);`` ``}`` ``else` `{`` ``Console.Write((arr[i] * -1));`` ``Console.Write(``" "``);`` ``Console.Write(arr[i]);`` ``Console.Write(``" "``);`` ``}`` ``}`` ``hs.Add(arr[i]);`` ``}` ` ``return``;`` ``}` ` ``// Driver code`` ``static` `public` `void` `Main()`` ``{`` ``int``[] arr = { 4, 8, 9, -4, 1, -1, -8, -9 };`` ``int` `n = arr.Length;` ` ``// Function call`` ``printPairs(arr, n);`` ``}``}` `// This code is contributed by lokeshmvs21.`
## Javascript
``
Output
`-4 4 -1 1 -8 8 -9 9 `
Time Complexity: O(N)
Auxiliary Space: O(N)
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https://www.unitconverters.net/power/joule-minute-to-dekajoule-second.htm | 1,721,222,667,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514771.72/warc/CC-MAIN-20240717120911-20240717150911-00239.warc.gz | 913,908,129 | 3,639 | Home / Power Conversion / Convert Joule/minute to Dekajoule/second
Convert Joule/minute to Dekajoule/second
Please provide values below to convert joule/minute [J/min] to dekajoule/second [daJ/s], or vice versa.
From: joule/minute To: dekajoule/second
Joule/minute to Dekajoule/second Conversion Table
Joule/minute [J/min]Dekajoule/second [daJ/s]
0.01 J/min1.66667E-5 daJ/s
0.1 J/min0.0001666667 daJ/s
1 J/min0.0016666667 daJ/s
2 J/min0.0033333333 daJ/s
3 J/min0.005 daJ/s
5 J/min0.0083333333 daJ/s
10 J/min0.0166666667 daJ/s
20 J/min0.0333333333 daJ/s
50 J/min0.0833333333 daJ/s
100 J/min0.1666666667 daJ/s
1000 J/min1.6666666667 daJ/s
How to Convert Joule/minute to Dekajoule/second
1 J/min = 0.0016666667 daJ/s
1 daJ/s = 600 J/min
Example: convert 15 J/min to daJ/s:
15 J/min = 15 × 0.0016666667 daJ/s = 0.025 daJ/s | 336 | 827 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2024-30 | latest | en | 0.088612 |
http://www.finedictionary.com/nor-q-d.html | 1,601,134,207,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400244231.61/warc/CC-MAIN-20200926134026-20200926164026-00369.warc.gz | 180,563,647 | 4,611 | # Nor-Q-D
## Definitions
• WordNet 3.6
• n Nor-Q-D trade name for an oral contraceptive containing norethindrone
• ***
## Usage
### In poetry:
Q. What were the things they promis'd to fulfil
For thee — 'till thou to proper age shou'dst grow,
When of thyself thou hadst no pow'r, nor will?
Tell me aloud, if thou the same dost know.
"The Catechism " by Rees Prichard
### In science:
Given Q > 0 and writing Ac = Ω \ A, we let P0 (f , Q) = {∃∆ ∈ D(f , ω ), ∆ ⊆ [−Q/ǫ, Q/ǫ]}c , be the set of realizations of the random field that neither give rise to a positive nor to a negative elongation in the interval [−Q/ǫ, Q/ǫ].
One-dimensional random field Kac's model: localization of the phases
The power transformation φ(D) = Dq with 0 < q < 1 is not rectifiable nor bounded.
Robust Estimation through Schoenberg transformations
Note also that P ∪ R[w , a] ∪ T ∪ S [b, x] and Q ∪ R[y , c] ∪ U ∪ S [d, z ] are odd circuits A and B , respectively, for neither G[Hi−1 ∪ P ] nor G[Hi−1 ∪ Q] is an even-circuit-connected graph.
Even circuits of prescribed clockwise parity
As a consequence, neither ¯Q{γ · D, B+ }Q nor ¯Q[γ · D, E− ]Q is redundant.
New lattice action for heavy quarks
Any choice of p, 0 < p < (q − 1)/q now gives a proof of the existence of µ(k , d) such that if n ≥ µ(d, k) and C is an [n, k , d] code, then neither C nor its dual C ⊥ are good for error detection.
Codes for error detection, good or not good
*** | 466 | 1,422 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2020-40 | latest | en | 0.813716 |
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# Personal Finance and accounting Problem
This content was STOLEN from BrainMass.com - View the original, and get the already-completed solution here!
PROBLEMS 6.38
Nonconstant Growth: A company will pay a \$2 per share dividend in 1 year. The dividend in 2 years will be \$4 per share, and it is expected that dividends will grow at 5 percent per year thereafter. The expected rate of return on the stock is 12 percent.
A. What is the current price of the stock?
B. What is the expected price of the stock in a year?
C. Show that the expected return, 12 percent, equals dividend yield plus capital appreciation
PROBLEM 15.14 (Chapter 15 problem #14)
Earnings and Leverage: Reliable Gearing currently is all-equity financed. It has 10,000 shares of equity outstanding, selling at \$100 a share. The firm is considering a capital-restructuring. The low-debt plan calls for a debt issue of \$200,000 with the proceeds used to buy back stock. The high-debt plan would exchange \$400,000 of debt for equity. The debt will pay an interest rate of 10 percent. The firm pays no taxes.
A. What will be the debt-to-equity ratio after each possible restructuring?
B. If earnings before interest and tax (EBIT ) will be either \$90,000 or \$130,000, what will be earnings per share for each financing mix for both possible values of EBIT? If both scenarios are equally likely, what is expected(i.e., average) EPS under each financing mix? Is the high-debt mix preferable?
C. Suppose that EBIT is \$100,000. What is EPS under each financing mix? Why are they the same in this particular case?
Problem 16.21 (Chapter 16, Problem #21)
Dividend Policy. For each of the following four groups of companies, state whether you would expect them to distribute a relatively high or low proportion of current earnings and whether you would expect them to have a relatively high or low price-earnings ratio.
A. High-risk companies.
B. Companies that have recently experienced a temporary decline in profits.
C. Companies that expect to experience a decline in profits.
D."Growth" companies with valuable future investment opportunities.
#### Solution Preview
PROBLEMS 6.38
Non constant Growth: A company will pay a \$2 per share dividend in 1 year. The dividend in 2 years will be \$4 per share, and it is expected that dividends will grow at 5 percent per year thereafter. The expected rate of return on the stock is 12 percent.
A. What is the current price of the stock?
The current price of the stock is the sum of the discounted dividends that would be received. When dividends grow at a constant rate, we can use the dividend discount model to get the price. The growth rate will be constant fro m year 3 onward. We can get the price in year 2 using the dividend discount model. The price would be
Price = Div1/(ke-g), where Div1 would be dividend in year 3, ke is the discounting rate and g is the growth rate.
Price = 4X(1+5%)/(0.12-0.05) = 4.2/0.07 = \$60.
The price today would the discounted sum of dividend in year 1, dividend in year 2 and the price is year 2.
Price today = 2/1.12 + 4/(1.12)^2 + 60/(1.12)^2
Price today = \$52.80
B. What is the expected price of the stock in a year?
The expected price of the stock in a year would be the discounted cash flows. After a year there would be \$4 of dividend and \$60 of price in the next year.
The price in a year would be 4/1.12+60/1.12 = \$57.14
C. Show that the expected return, 12 percent, equals dividend yield plus capital appreciation
In one year the dividend in \$2 and the capital appreciation is \$57.14-\$52.80=\$4.34. The dividend yield is 2/52.80 = 3.79%. The capital ...
#### Solution Summary
The solution has 3 questions in financial management relating to stock valuation, financing mix and dividend policy.
\$2.19
## Principles of Accounting: E1-1, E1-5, E1-12, E1-13, E2-2, E2-3, E7-3
Prepare written answers to the following assignments from the text, Financial Accounting 6th ed:
a. Chapter 1:
1) Exercise E1-1
2) Exercise E1-5
3) Exercise E1-12
4) Exercise E1-13
b. Chapter 2:
1) Exercise E2-2
2) Exercise E2-3
c. Chapter 7:
Exercise E7-3
View Full Posting Details | 1,077 | 4,173 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2018-51 | longest | en | 0.926614 |
http://scifi.stackexchange.com/questions/5489/short-story-where-a-character-moves-through-dimensions-via-mental-exercises | 1,448,482,749,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398445679.7/warc/CC-MAIN-20151124205405-00182-ip-10-71-132-137.ec2.internal.warc.gz | 211,399,105 | 19,489 | # Short story where a character moves through dimensions via mental exercises
It's been about 25 years since I read a story in an anthology. The only details I remember from it are that a guy is told that if, while walking down a road (or something similar), he mentally forces himself to think of dimensions opposite of what they are (down is up, left is right, etc.), he will be able to move into another dimension. He succeeds, but that's all I remember of the story.
-
"The Gostak and the Doshes" by Miles J. Breuer, M.D.. There's a review at Alex Kasman's Mathematical Fiction site. The full text is available at Wikisource.
If you read it in an anthology around 1985, that could have been The Arbor House Treasury of Science Fiction Masterpieces (which was republished in slightly abridged form as Great Tales of Science Fiction), or perhaps Amazing Stories: 60 Years of the Best Science Fiction.
"According to fiction writers, to switch into the t dimension, some sort of apparatus with an electrical field ought to be necessary. It is not. You need nothing more to rotate into the t dimension than you do to stop the moon and make the trees move as you ride down the road; or than you do to turn the cubes upside down. It is a matter of relativity."
I had ceased trying to wonder or to understand.
"Show me!" was all I could gasp.
"The success of this experiment in changing from the z to the t co-ordinate has depended largely on my lucky discovery of a favourable location. It is just as, when you want the moon to ride the tree-tops successfully, there have to be favourable features in the topography or it won't work. The edge of this building and that little walk between the two rows of Norway poplars seems to be an angle between planes in the z and t dimensions. It seems to slope downward, does it not?—Now walk from here to the end and imagine yourself going upward. That is all. Instead of feeling this building behind and above you, conceive it as behind and below. Just as on your ride by moonlight, you must tell yourself that the moon is not moving while the trees ride by. Can you do that? Go ahead, then." He spoke in a confident tone, as though he knew exactly what would happen.
The title is explained in the Wikipedia article on Gostak:
Gostak is a meaningless noun that is used in the phrase "the gostak distims the doshes", which is an example of how it is possible to derive meaning from the syntax of a sentence even if the referents of the terms are entirely unknown.
[. . . .]
Coined in 1903 by Andrew Ingraham, the sentence became more widely known through its quotation in 1923 by C. K. Ogden and I. A. Richards in their book The Meaning of Meaning (p. 46).
Ogden and Richards refer to Ingraham as an "able but little known writer", and quote his following dialogue:
"Suppose someone to assert: The gostak distims the doshes. You do not know what this means; nor do I. But if we assume that it is English, we know that the doshes are distimmed by the gostak. We know too that one distimmer of doshes is a gostak. If, moreover, the doshes are galloons, we know that some galloons are distimmed by the gostak. And so we may go on, and so we often do go on."
[. . . .]
In Amazing Stories, Dr. Miles Breuer wrote a story, now considered a classic, titled "The Gostak and the Doshes" whose protagonist pops into an alternate world in which the phrase is a political slogan that induces sufficient umbrage throughout the populace to declare justified, righteous war. Other writers have picked up on the reference, notably David Gerrold.
-
HOLY CRAP THAT'S IT!! (sorry for the caps, but I've been trying to figure this out for years!) – Larry G. Wapnitsky Jun 8 at 18:39
This sounds to me like a Roger Zelazny story, but I'm not quite sure from your description. It could be one of the short stories he wrote in the Amber universe, but it actually makes me think of "Roadmarks", which would be a short book rather than a short story. Hope that helps.
-
just looked at the summary...doesn't ring a bell. thanks, though – Larry G. Wapnitsky Sep 15 '11 at 14:45
Roadmarks has walking down a road, but no “down is up” business that I can remember. Amber has interdimensional travel, but no “up is right” either. – Gilles Sep 15 '11 at 20:07
Sounds a bit like Elsewhen from Robert Heinlein, but there are probably too many differences for it to be the one you're looking for.
-
I'm a big Heinlein fan, and when I started reading that story a few years ago I thought it was it. Unfortunately not. – Larry G. Wapnitsky Oct 13 '11 at 13:50 | 1,112 | 4,582 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2015-48 | longest | en | 0.976229 |
https://rhettallain.com/tag/random/ | 1,679,629,802,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945242.64/warc/CC-MAIN-20230324020038-20230324050038-00415.warc.gz | 542,232,990 | 17,421 | # People in and out of a bar
So, I am still in Alabama because of lack of electricity in Louisiana (although we got power back last night – YAY!). My wife convinced me to go out and hear this band since we are staying at my parents and they said they would watch the kids. In general, I am way too old and crotchety to go out – but it appears I had no choice. Overall, it was not too bad except for staying out too late. (the band was actually pretty good – I think they were called [Fly By Radio](http://www.fly-byradio.com/) and they played 80s musics) The one thing I thought about while in the bar was the long line of people waiting to go in. At some point, the number of people inside reaches its maximum capacity. People still come in because some people leave. Here is my rough sketch of this: (I call it a bar graph – get it?)
My first idea was to think of the bar as a capacitor and maybe this works. Ok, now I made something positive out of my trip.
# How long do you wait for a question to be answered?
I teach classes. I ask questions in class. I wait for answers. All faculty do this, so who cares. If you are in a class or teaching a class, how long do you wait for someone to answer your question? Well, I asked two questions of my class this week.
1. Estimate how long I wait when I ask you questions.
2. How long should you (ideally) wait in a class for someone to answer?
Here is the data I gathered: (and I will tell you how long I actually wait)
# ARRRGGGGG RapidWeaver, why do you hate me?
Dear RapidWeaver,
What did I do to you? I like you, I really do – but this is why we broke up. It’s not you, it’s me. No wait, it’s you. I am looking back through my old dotphys posts and most of my introductions are GONE. I was starting to think I was crazy when I wrote these. Take for instance [My analysis of Kobe Bryant’s Jump](http://dotphys.net/files/kobe.html). Isn’t it odd how this starts? That is because it is MISSING the first paragraph. The only reason I know I am not insane is because of Apple’s Time Machine. I went back and looked and the older file has an intro. RapidWeaver, see how you are!
# This is a test post
I have been using RapidWeaver to make my blog at http://dotphys.net.
I wonder how hard it is to put in an image.
this is a chair | 570 | 2,285 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2023-14 | latest | en | 0.977629 |
http://www.asmcommunity.net/forums/topic/?id=15000 | 1,680,040,087,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948871.42/warc/CC-MAIN-20230328201715-20230328231715-00214.warc.gz | 59,814,448 | 2,081 | Since shifting an unsigned integer value to the right one position is equivalent to dividing that value by two, you can also use the shift right instruction for division by powers of two:
shr ax, 1 ;Equivalent to AX/2
shr ax, 2 ;Equivalent to AX/4
shr ax, 3 ;Equivalent to AX/8
shr ax, 4 ;Equivalent to AX/16
shr ax, 5 ;Equivlaent to AX/32
shr ax, 6 ;Equivalent to AX/64
shr ax, 7 ;Equivalent to AX/128
shr ax, 8 ;Equivalent to AX/256
if ax contains 0000000000000010, shr ax, 1 produces 0000000000000001, which is 2/2 = 1, quite simple.
But what if we do shr ax, 3? the results is 0, which is not ax/3. How to explain???
Thanks
Posted on 2003-09-01 02:07:16 by catman123ok
If ax = 2h,
shr ax, 3 will result in 0 (2/8 = .25 or rounded down = 0). This is because the shift is just shifting of the bits. For accuracy, you would would not be using shifts, but fpu. But of course does the decimal value matters? Oh yes, one more thing the last bit shifted out is stored in the carry flag. That could be of some uses of course.
Posted on 2003-09-01 02:14:14 by roticv
catman, here's a li'l toy you might like: | 351 | 1,106 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2023-14 | latest | en | 0.912324 |
https://fr.mathworks.com/matlabcentral/cody/problems/42519-sum-my-indices/solutions/1498404 | 1,571,788,007,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987826436.88/warc/CC-MAIN-20191022232751-20191023020251-00112.warc.gz | 475,831,766 | 15,382 | Cody
# Problem 42519. Sum my indices
Solution 1498404
Submitted on 20 Apr 2018 by bainhome
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
M = 1; N = 1; y_correct = [2]; assert(isequal(sumMyIndices(M,N),y_correct))
2 Pass
M = 1; N = 2; y_correct = [2,3]; assert(isequal(sumMyIndices(M,N),y_correct))
3 Pass
M = 3; N = 2; y_correct = [2,3; 3,4; 4,5]; assert(isequal(sumMyIndices(M,N),y_correct))
4 Pass
M = 3; N = 3; y_correct = [2,3,4; 3,4,5; 4,5,6]; assert(isequal(sumMyIndices(M,N),y_correct))
5 Pass
M = 4; N = 5; y_correct = [ 2 3 4 5 6 3 4 5 6 7 4 5 6 7 8 5 6 7 8 9 ]; assert(isequal(sumMyIndices(M,N),y_correct))
6 Pass
M = 5; N = 8; y_correct = [ 2 3 4 5 6 7 8 9 3 4 5 6 7 8 9 10 4 5 6 7 8 9 10 11 5 6 7 8 9 10 11 12 6 7 8 9 10 11 12 13 ]; assert(isequal(sumMyIndices(M,N),y_correct)) | 415 | 929 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2019-43 | latest | en | 0.553697 |
https://www.hindawi.com/journals/amse/2013/810508/ | 1,713,312,061,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817112.71/warc/CC-MAIN-20240416222403-20240417012403-00607.warc.gz | 714,539,603 | 104,690 | #### Abstract
The thermal elastic-plastic stresses have been investigated for a rotating functionally graded stainless steel composite cylinder under internal and external pressure with general nonlinear strain hardening law and von Mises’ yield criterion using finite difference method. The modulus of elasticity in the rotating cylinder varies radially according to power law and the temperature distribution satisfies Laplace heat equation in radial direction. From the analysis, we can conclude that cylinder made of functionally graded stainless steel composite material with variable thickness and variable density under thermal loading for Swift’s strain hardening measure is better choice of the design as compared to homogeneous cylinder. This is because of the reason that circumferential stress is less for functionally graded stainless steel composite cylinder as compared to homogeneous cylinder for Swift’s strain hardening measure under internal and external pressure. This leads to the idea of “stress saving” which minimizes the possibility of fracture of cylinder.
#### 1. Introduction
In this paper, thermal elastic-plastic stresses have been calculated for cylinder made up of functionally graded stainless steel composite material under internal and external pressure using finite difference method. In this problem, a general nonlinear strain hardening law with von Mises’ yield criterion has been considered. Results have been discussed numerically with the help of graphs and tables.
#### 2. Objective
For a rotating cylinder with varying material properties, circumferential stress at the hub does not exceed the allowable value which tells the designers little more than that the design of the cylinder is safe at the given pressure. Thus our prime objective is to calculate allowable thermal elastic-plastic stresses in an open ended functionally graded stainless steel composite rotating cylinder under pressure for varying expansion to incorporate a “safety factor” that prevents the cylinder from bursting under pressure and thermal loading.
#### 3. Mathematical Formulation
##### 3.1. Distribution of Material Properties
Consider a long open ended axisymmetric cylinder made up of functionally graded stainless steel composite material with inner and outer radii and , respectively, and the cylinder is subjected to internal pressure and external pressure as shown in Figure 1. The cylindrical polar coordinates under plane stress condition have been considered in this problem.
In this study, Poisson’s ratio and thermal expansion coefficient () are assumed to be constants. The other properties, that is, Young’s modulus which is defined by power law, temperature distribution that follows Laplace heat equation in radial direction, thickness, and density, are varying radially and are expressed as where , is the radius of the cylinder, , , , and are material constants, and , , and are the geometric parameters.
##### 3.2. Basic Equations
The equilibrium equation for the cylinder in the absence of body forces is where and are radial and circumferential stresses, respectively.
Using infinitesimal theory of elasticity, the relations between strains and radial displacements are where and are radial and circumferential strains, respectively, and is the radial displacement.
The equation of compatibility can be derived from (3) as
From infinitesimal theory of elasticity, the stress-strain relations are where , , and are the elastic radial, circumferential, and axial strains, respectively.
Due to geometric symmetry of the cylinder, circumferential displacement, shear stresses, and strains are assumed to be zero.
Using deformation theory of plasticity, the relation between the stresses and plastic strains can be determined as where is the equivalent stress, is the equivalent plastic strain, and , , and are the plastic radial, circumferential, and axial strains, respectively.
von Mises’ yield criterion is given by
The total radial, circumferential, and axial strains in thick-walled rotating cylinder are
The temperature field satisfying Laplace heat equation is with at and at , where is a constant, given by .
We define the stress function for thick-walled rotating cylinder which is related to radial and hoop stresses as
Since it has been assumed that the cylinder is long and open ended and there is plane stress condition, therefore axial stress is zero; that is, .
Substituting (10) and (5) into (8), we have
Substituting of (11) into compatibility (4), we have where , , and .
The relation between the yield stress and the equivalent plastic strain for Swift’s hardening law can be expressed as where , , , , and are hardening parameter, material parameter, yield limit, equivalent total strain, and yield strain, respectively.
Substituting from (13) into (6) results in
Substituting (14) into (12), we have
Equation (15) is the differential equation of the functionally graded stainless steel composite rotating cylinder with nonlinear strain hardening subjected to thermal loading in the plastic region in terms of stresses and stress function.
Equation (15) can be described in the general form in terms of stress function as
Equation (16) is a nonlinear two point boundary value problem and can be solved numerically, subjected to the boundary conditions where and are the inner and outer radii of the cylinder and and are internal and external pressures, respectively.
Using finite difference method with central difference in (16), we get the following system of equations:
Equation (18) consists of algebraic system of equations with the boundary conditions and . After solving (18) with boundary conditions we get a stress function . Then, the radial and circumferential stresses can be obtained from (10) after substituting the value of stress function .
#### 4. Numerical Discussion
The properties of a functionally graded stainless steel composite thick-walled rotating cylinder under internal and external pressure and MPa, respectively, subjected to thermal loading () are defined as follows: the radii of the cylinder are taken as m and m, Poisson’s ratio , Young’s modulus GPa, and thermal expansion coefficient . The geometric parameters of the cylinder are taken as in Young’s modulus function and is nonlinear strain hardening measure.
To show the effect of internal and external pressure on a functionally graded stainless steel composite rotating cylinder with strain hardening measure having constant thickness and constant density, Tables 1 and 2 show the circumferential stresses with different parameters of Young’s modulus .
It has been observed from Table 1 that, when external pressure is greater than the internal pressure, circumferential stresses approaches tensile to compressible. Also, these stresses are maximum at external surface for homogeneous as well as functionally graded stainless steel composite cylinder. These stresses are less for functionally graded stainless steel composite cylinder as compared to homogeneous cylinder. As nonhomogeneity changes from to , circumferential stresses decrease significantly. With the introduction of thermal effects circumferential stresses increase for homogeneous as well as for functionally graded stainless steel composite cylinder () but decrease for functionally graded stainless steel composite cylinder with . It has also been noticed from Table 1 that with the increase in thermal effects these stresses decrease significantly for homogeneous as well as for functionally graded stainless steel composite cylinder and are less for the cylinder with nonhomogeneity parameter . With the increase in angular speed, circumferential stresses increase significantly. When external pressure is less than the internal pressure, circumferential stresses are maximum at internal surface for homogeneous cylinder while maximum at external surface for functionally graded stainless steel composite cylinder and these stresses decreased with the change in nonhomogeneity measure from to . With the introduction of thermal effects circumferential stresses increase for homogenous as well as for functionally graded stainless steel composite measure while decrease for functionally graded stainless steel composite cylinder with . With the increase in thermal effects these circumferential stresses decrease significantly while increase with the increase in angular speed. It has been observed from Table 2 that the behavior of homogeneous and functionally graded stainless steel composite cylinder same as discussed in Table 1 but it has been observed that, with increase in strain hardening measure from to , these stresses decrease significantly for functionally graded stainless steel composite cylinder.
Tables 3 and 4 have been made for circumferential stresses in rotating cylinders with variable thickness and variable density with different parameters of Young’s modulus and strain hardening measure .
It has been observed from Table 3 that for cylinder with varying thickness and density, whose external pressure is greater than the internal pressure, circumferential stresses approach from tensile to compressible and are maximum at external surface for homogeneous as well as functionally graded stainless steel composite cylinder. These stresses are less for functionally graded stainless steel composite cylinder as compared to homogeneous cylinder with varying thickness and density as well as cylinder with constant thickness and density. With increase in strain hardening measure from to these circumferential stresses decrease significantly for functionally graded stainless steel composite cylinder with varying thickness and density as can be seen from Table 4.
Figures 24 have been drawn to discuss the effect of internal and external pressure on stresses in rotating cylinder made of functionally graded stainless steel composite material with constant thickness and constant density with nonlinear strain hardening measure.
It has been observed from Figure 2 that circumferential stress approaches towards compressive from tensile. It has also been observed that when external pressure is greater than the internal pressure, these stresses are maximum at external surface for homogeneous as well as functionally graded stainless steel composite cylinder. Also, it has been observed that circumferential stress is maximum at internal surface for homogeneous cylinder while maximum at external surface for functionally graded stainless steel composite cylinder when external pressure is less than the internal pressure. Also, with the increase in angular speed, circumferential stresses increase significantly. From Figure 3 it can be seen that, as circumferential stresses increase for homogeneous cylinder as well as for nonhomogenous cylinder but with the change in nonhomogeneity from to , circumferential stresses decrease when external pressure is greater than the internal pressure with thermal effects. Also, these stresses increase for homogenous cylinder as well as for functionally graded stainless steel composite cylinder () while decrease for functionally graded stainless steel composite cylinder with . With the increase in temperature these stresses decrease significantly for homogeneous as well as for functionally graded stainless steel composite cylinder as can be seen from Figure 4.
Figures 57 have been drawn to discuss the effect of internal and external pressure on stresses in rotating cylinder made of functionally graded stainless steel composite material with variable thickness and variable density with nonlinear strain hardening measure.
It has been observed from Figure 5 that circumferential stresses are maximum at external surface for homogeneous as well as functionally graded stainless steel composite cylinder. It has also been observed that with the increase in angular speed circumferential stresses increase significantly. With the introduction of thermal effects, circumferential stresses increase for homogeneous cylinder as well as for functionally graded stainless steel composite cylinder (), but with the change in nonhomogeneity from to , circumferential stresses decrease when external pressure is greater than the internal pressure as can be seen from Figure 6. Circumferential stresses decrease for homogeneous cylinder while increase for functionally graded stainless steel composite cylinder when external pressure is less than the internal pressure. It has also been observed from Figure 7 that with the increase in temperature these stresses increase significantly for homogeneous cylinder as well for functionally graded stainless steel composite cylinder with while decrease for functionally graded stainless steel composite cylinder with when external pressure is greater than the internal pressure, while these stresses decrease significantly for homogeneous cylinder as well for functionally graded stainless steel composite cylinder when external pressure is less than the internal pressure.
#### 5. Conclusion
From the analysis, we can conclude that rotating cylinder made of functionally graded stainless steel composite material having variable thickness and variable density with Swift’s strain hardening measure and thermal loading is better choice for designers as compared to rotating cylinder with constant thickness and constant density. This is because of the reason that circumferential stress is less for functionally graded stainless steel composite cylinder with variable thickness and variable density as compared to other cases, which leads to the idea of stress saving that minimizes the possibility of fracture of cylinder.
#### Conflict of Interests
The authors declare that they have no conflict of interest. | 2,441 | 13,789 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-18 | latest | en | 0.895839 |
https://math.stackexchange.com/questions/4505802/graph-theory-maximal-graph-without-cycles-and-maximal-planar-graph-without-tri | 1,675,327,817,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499967.46/warc/CC-MAIN-20230202070522-20230202100522-00161.warc.gz | 402,590,420 | 31,715 | # graph theory - maximal graph without cycles and maximal planar graph without triangles
Let there be $$G = (V,E_1\cup E_2)$$ such that $$(V,E_1)$$ is a planar graph without triangles, and $$(V,E_2)$$ is without circles, show that $$G$$ is $$6$$-colorable
Can I prove here that $$E_1$$ and $$E_2$$ has at least one edge that is inside both of them when the vertices are $$v_1,v_2,v_3,\dots,v_n$$, I couldn't find why this is wrong, can someone please show it using the pigeon hole principle.
• Comments are not for extended discussion; this conversation has been moved to chat.
– Pedro
Aug 5, 2022 at 9:12 | 179 | 608 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 8, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2023-06 | latest | en | 0.935442 |
http://lists.openscenegraph.org/htdig.cgi/osg-users-openscenegraph.org/2008-July/147811.html | 1,657,171,705,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104683683.99/warc/CC-MAIN-20220707033101-20220707063101-00588.warc.gz | 35,294,455 | 2,414 | # [osg-users] How to scale rotations ?
Janusz janusz at 3dcalc.pl
Fri Jul 25 12:13:32 PDT 2008
```Dear all:
Appreciate any answer on the topic:
I am reading an orientation / rotation matrix info from a sensor, then
visualizing the rotation of an object on screen with OSG. Given the
matrix or the resulting quaternion I need to be able to scale the
rotations independently, i.e.
attitude_scale*attitude_angle
bank_scale*bank_angle
in order to be able to control them on its own or restrict the rotation
to one angle only.
I have written 2 functions. The first one converts a quaternion to the
Euler angles, the other one does the opposite. In between the calls to
the functions I am trying to scale the rotations or switch them on/off
as needed.
The results are not quite good, however, and I often get trapped in the
famous "gimbal lock" as probably expected. :-)
Q: Is there any other possibility to handle rotations on its own? Is
there a way to do that without running the conversion, scaling, then
converting back to quat ?
All I have is the rotation matrix.
Best regards,
Janusz Goldasz
--
void getEulerFromQuat(osg::Quat q, double& heading, double& attitude,
double& bank)
{
double sqx = q.x()*q.x();
double sqy = q.y()*q.y();
double sqz = q.z()*q.z();
double t = q.x()*q.y() + q.z()*q.w();
if (t>0.49999)
{
attitude = osg::PI_2;
bank = 0;
}
else if (t<-0.49999)
{
attitude = - osg::PI_2;
bank = 0;
}
else
{
heading = atan2(2*q.y()*q.w()-2*q.x()*q.z() , 1 - 2*sqy - 2*sqz);
attitude = asin(2*t);
bank = atan2(2*q.x()*q.w()-2*q.y()*q.z() , 1 - 2*sqx - 2*sqz);
}
}
void getQuatFromEuler(double heading, double attitude, double bank,
osg::Quat& q)
{
double c2 = cos(attitude/2);
double s2 = sin(attitude/2);
double c3 = cos(bank/2);
double s3 = sin(bank/2);
double c1c2 = c1*c2;
double s1s2 = s1*s2;
double w =c1c2*c3 - s1s2*s3;
double x =c1c2*s3 + s1s2*c3;
double y =s1*c2*c3 + c1*s2*s3;
double z =c1*s2*c3 - s1*c2*s3;
q[0] = x; q[1] = y;
q[2] = z; q[3] = w;
}
``` | 666 | 1,986 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2022-27 | latest | en | 0.785104 |
https://nrich.maths.org/public/leg.php?code=164&cl=3&cldcmpid=4873 | 1,474,751,008,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738659496.36/warc/CC-MAIN-20160924173739-00082-ip-10-143-35-109.ec2.internal.warc.gz | 877,295,670 | 8,999 | # Search by Topic
#### Resources tagged with Speed similar to Motion Capture:
Filter by: Content type:
Stage:
Challenge level:
### There are 37 results
Broad Topics > Measures and Mensuration > Speed
### Motion Capture
##### Stage: 3 and 4 Challenge Level:
Explore displacement/time and velocity/time graphs with this mouse motion sensor.
### An Average Average Speed
##### Stage: 4 Challenge Level:
My average speed for a journey was 50 mph, my return average speed of 70 mph. Why wasn't my average speed for the round trip 60mph ?
### How Do You React?
##### Stage: 4 Challenge Level:
To investigate the relationship between the distance the ruler drops and the time taken, we need to do some mathematical modelling...
### Nutrition and Cycling
##### Stage: 3 and 4 Challenge Level:
Andy wants to cycle from Land's End to John o'Groats. Will he be able to eat enough to keep him going?
### In Constantly Passing
##### Stage: 4 Challenge Level:
A car is travelling along a dual carriageway at constant speed. Every 3 minutes a bus passes going in the opposite direction, while every 6 minutes a bus passes the car travelling in the same. . . .
### Escalator
##### Stage: 4 Challenge Level:
At Holborn underground station there is a very long escalator. Two people are in a hurry and so climb the escalator as it is moving upwards, thus adding their speed to that of the moving steps. . . .
### Speedy Sidney
##### Stage: 3 Challenge Level:
Two trains set off at the same time from each end of a single straight railway line. A very fast bee starts off in front of the first train and flies continuously back and forth between the. . . .
### Triathlon and Fitness
##### Stage: 3 Challenge Level:
The triathlon is a physically gruelling challenge. Can you work out which athlete burnt the most calories?
### Circuit Training
##### Stage: 4 Challenge Level:
Mike and Monisha meet at the race track, which is 400m round. Just to make a point, Mike runs anticlockwise whilst Monisha runs clockwise. Where will they meet on their way around and will they ever. . . .
### Alternative Record Book
##### Stage: 4 and 5 Challenge Level:
In which Olympic event does a human travel fastest? Decide which events to include in your Alternative Record Book.
### Walk and Ride
##### Stage: 2 and 3 Challenge Level:
How far have these students walked by the time the teacher's car reaches them after their bus broke down?
### Lap Times
##### Stage: 4 Challenge Level:
Two cyclists, practising on a track, pass each other at the starting line and go at constant speeds... Can you find lap times that are such that the cyclists will meet exactly half way round the. . . .
### An Unhappy End
##### Stage: 3 Challenge Level:
Two engines, at opposite ends of a single track railway line, set off towards one another just as a fly, sitting on the front of one of the engines, sets off flying along the railway line...
### Graphical Interpretation
##### Stage: 4 Challenge Level:
This set of resources for teachers offers interactive environments to support work on graphical interpretation at Key Stage 4.
### John's Train Is on Time
##### Stage: 3 Challenge Level:
A train leaves on time. After it has gone 8 miles (at 33mph) the driver looks at his watch and sees that the hour hand is exactly over the minute hand. When did the train leave the station?
### Illusion
##### Stage: 3 and 4 Challenge Level:
A security camera, taking pictures each half a second, films a cyclist going by. In the film, the cyclist appears to go forward while the wheels appear to go backwards. Why?
### How Far Does it Move?
##### Stage: 3 Challenge Level:
Experiment with the interactivity of "rolling" regular polygons, and explore how the different positions of the red dot affects the distance it travels at each stage.
### Jumping Gerbils
##### Stage: 4 Challenge Level:
A conveyor belt, with tins placed at regular intervals, is moving at a steady rate towards a labelling machine. A gerbil starts from the beginning of the belt and jumps from tin to tin.
### Speed-time Problems at the Olympics
##### Stage: 4 Challenge Level:
Have you ever wondered what it would be like to race against Usain Bolt?
### Up and Across
##### Stage: 3 Challenge Level:
Experiment with the interactivity of "rolling" regular polygons, and explore how the different positions of the red dot affects its vertical and horizontal movement at each stage.
### Bus Stop
##### Stage: 4 Challenge Level:
Two buses leave at the same time from two towns Shipton and Veston on the same long road, travelling towards each other. At each mile along the road are milestones. The buses' speeds are constant. . . .
### Speeding Up, Slowing Down
##### Stage: 3 Challenge Level:
Experiment with the interactivity of "rolling" regular polygons, and explore how the different positions of the red dot affects its speed at each stage.
##### Stage: 4 Challenge Level:
Four vehicles travel along a road one afternoon. Can you make sense of the graphs showing their motion?
##### Stage: 3 Challenge Level:
Can you rank these sets of quantities in order, from smallest to largest? Can you provide convincing evidence for your rankings?
### Speeding Boats
##### Stage: 4 Challenge Level:
Two boats travel up and down a lake. Can you picture where they will cross if you know how fast each boat is travelling?
### Swimmers
##### Stage: 4 Challenge Level:
Swimmers in opposite directions cross at 20m and at 30m from each end of a swimming pool. How long is the pool ?
### All in a Jumble
##### Stage: 3 Challenge Level:
My measurements have got all jumbled up! Swap them around and see if you can find a combination where every measurement is valid.
### There and Back
##### Stage: 4 Challenge Level:
Brian swims at twice the speed that a river is flowing, downstream from one moored boat to another and back again, taking 12 minutes altogether. How long would it have taken him in still water?
##### Stage: 4 Challenge Level:
Four vehicles travelled on a road with constant velocities. The car overtook the scooter at 12 o'clock, then met the bike at 14.00 and the motorcycle at 16.00. The motorcycle met the scooter at. . . .
##### Stage: 4 Challenge Level:
Can you adjust the curve so the bead drops with near constant vertical velocity?
### Motion Sensor
##### Stage: 4 Challenge Level:
Looking at the graph - when was the person moving fastest? Slowest?
##### Stage: 2 and 3
A paradox is a statement that seems to be both untrue and true at the same time. This article looks at a few examples and challenges you to investigate them for yourself.
### Take a Message Soldier
##### Stage: 4 and 5 Challenge Level:
A messenger runs from the rear to the head of a marching column and back. When he gets back, the rear is where the head was when he set off. What is the ratio of his speed to that of the column?
### Around and Back
##### Stage: 4 Challenge Level:
A cyclist and a runner start off simultaneously around a race track each going at a constant speed. The cyclist goes all the way around and then catches up with the runner. He then instantly turns. . . .
### Hike and Hitch
##### Stage: 4 Challenge Level:
Fifteen students had to travel 60 miles. They could use a car, which could only carry 5 students. As the car left with the first 5 (at 40 miles per hour), the remaining 10 commenced hiking along the. . . .
### One and Three
##### Stage: 4 Challenge Level:
Two motorboats travelling up and down a lake at constant speeds leave opposite ends A and B at the same instant, passing each other, for the first time 600 metres from A, and on their return, 400. . . .
### N Is a Number
##### Stage: 3 Challenge Level:
N people visit their friends staying N kilometres along the coast. Some walk along the cliff path at N km an hour, the rest go by car. How long is the road? | 1,772 | 7,912 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2016-40 | longest | en | 0.90743 |
https://www.civilengineeringhandbook.tk/structural-engineering/r1-12r1-x-%5Bp21-r2-12r2.html | 1,553,362,988,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202889.30/warc/CC-MAIN-20190323161556-20190323183556-00436.warc.gz | 734,578,829 | 8,273 | ## R1 12r1 X [p21 r2 12r2
Equation 23.19 yields K factor values having a maximum error of less than 4% compared to corresponding theoretically correct values.
The effective-length factor concept is considered to be an essential part of many analysis procedures, and a corresponding alignment chart approach is recommended by almost all current design standards [1-4]. The idealizations and assumptions associated with the approach are as follows:
1. All beams and columns are purely elastic.
2. All members have constant cross-section.
### 3. All joints are rigid.
4. For braced (sidesway inhibited or sway prevented) frames, rotations at opposite ends of the restraining beams are equal in magnitude, producing single-curvature bending.
5. For unbraced (sidesway uninhibited or sway permitted) frames, rotations at opposite ends of the restraining beams are equal in magnitude, producing reverse-curvature bending.
6. The stiffness parameters L^/P/EI of all columns are equal.
7. Joint restraint is distributed to the column above and below the joint in proportion to the I/L values for the two columns.
8. All columns buckle simultaneously.
9. No significant axial compression force exists in the beams.
Based on the foregoing assumptions, the effective-length factor K of a column in a braced frame is determined from the following equation:
For an unbraced frame, the current Canadian standard CSA S16 [3] takes sway effects into account directly by using K = 1.0 for columns and performing a second-order elastic analysis under actual and notional loads (i.e., in lieu of using the traditional effective-length factor to account for frame stability, as is permitted by the American design specifications [1,2]). Following the alignment chart approach, the K factor for a column in an unbraced frame is determined from the following equation:
In Equations 23.20 and 23.21, GU and GL are the stiffness factors for the upper and lower ends of the column, respectively. A stiffness factor G is defined as
where the summation is taken over all members connected to the joint, Ic is the moment of inertia of the column section corresponding to the plane of buckling, Lc is the unsupported length of the column, Ib is the moment of inertia of the beam/girder section corresponding to the plane of buckling, and Lb is the unsupported length of the beam/girder. Having the stiffness factors GU and GL evaluated in accordance with Equation 23.22, the effective-length factor K of a column in a braced or unbraced frame can be obtained from the applicable alignment charts developed from Equations 23.20 and 23.21.
The evaluation of the effective-length factor K for columns in semirigid frames can be conveniently achieved by pursuing the foregoing approach for rigid frames with a necessary modification to account for connection flexibility when calculating a stiffness factor as
where ab is a modification factor applied to the moment of inertia Ib of the restraining beams that can be expressed as a function of the end-fixity factors associated with each restraining beam depending on whether the frame is of braced or unbraced frames. For braced semirigid frames
while for unbraced semirigid frames r1(2 + r2)
where the end-fixity factors r1 and r2 in Equations 23.24 and 23.25 correspond to the so called ''near end'' and ''far end,'' respectively, of the beam. In the case where the ''far end'' of the restraining beam is connected to a rotation-restrained support instead of a column, the corresponding modification factors for braced and unbraced semirigid frames are ab = ——— (23.26)
As the modification factor ab is expressed by the end-fixity factors, beams with pinned, semirigid, and rigid connections can all be considered. For instance, Equation 23.24 will yield a value of 1.5 for the modification factor ab for a beam in a braced frame with a rigid connection at the ''near end'' (r1 = 1) and a pinned connection at the ''far-end'' (r2 = 0). If the ''near end'' connection is a pinned connection (r1 = 0), the corresponding modification factor ab becomes zero, which indicates the beam does not provide any rotational restraint to the connected column.
Having the modification factor value ab based on the end-fixity of the restraining beams, the stiffness factors for the upper and lower ends of the column can be calculated according to Equation 23.23. The corresponding column effective-length factor is then obtained from the applicable alignment chart. However, unlike that for the rigid frame case, in which the effective-length factor of a column is ab —
evaluated regardless of the applied loads, the rotational stiffnesses of beam-to-column connections in semirigid frames are interrelated to the loads and, hence, so are the end-fixity factors rj (j = 1, 2) and modification factors ab. As a frame is loaded, the connection stiffness and, thus, the restraint provided to the columns gradually decreases, causing the column effective-length factor to increase. Consequently, the effective-length factor of the column in a semirigid frame must be evaluated for each applied load case.
0 0 | 1,097 | 5,133 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2019-13 | latest | en | 0.900962 |
http://mail.ivoa.net/pipermail/dal/2017-September/007838.html | 1,656,937,684,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104375714.75/warc/CC-MAIN-20220704111005-20220704141005-00283.warc.gz | 34,392,852 | 2,884 | # Arrays of geometries?
Fri Sep 15 18:22:03 CEST 2017
```An array of 4 points includes those 4 points while a polygon with 4
vertices incudes *all* the points inside or on the boundary. So the
polygon xtype says more than just an array of points. It is a sequence
of points that enclose a portion of the coordinate system (with great
circles) in a specific winding direction that defines the "inside" and
"outside". It is a set of all those points - not just the 4 listed.
Pat
On 15 September 2017 at 08:49, Dave Morris <dave.morris at metagrid.co.uk> wrote:
> Hi DAL,
>
> Are POLYGONs arrays of POINTs ?
>
> The current email thread is considering this:
>
> <!-- Four points, each made up of two floats -->
> <FIELD datatype="float" arraysize="2x4" xtype="point"...>
>
> to mean an array of four POINTs. Where the arraysize is split, with the
> first number, 2, applied to the 'float' part and the 4 applied to the
> 'point' part.
>
> What does that mean for POLYGONs ?
>
> At the moment POLYGONs are handled as 1D arrays of floats:
>
> <!-- A single polygon made up from 8 floats -->
> <FIELD datatype="float" arraysize="8" xtype="polygon"...>
>
> where the first part of the arraysize, 8, applies to the 'float' part and
> the 'polygon' part is unmodified.
>
> Are these equivalent ?
>
> <!-- Four points, each made up of two floats -->
> <FIELD datatype="float" arraysize="2x4" xtype="point"...>
>
> <!-- A single polygon made up from 8 floats -->
> <FIELD datatype="float" arraysize="8" xtype="polygon"...>
>
> What happens if we push the model to handle multiple shapes ?
>
> <!-- A single polygon made up from 8 floats -->
> <FIELD datatype="float" arraysize="8" xtype="polygon"...>
>
> <!-- Three polygons, each made up from eight floats -->
> <FIELD datatype="float" arraysize="8x3" xtype="polygon"...>
>
> <!-- Four points, each made up of two floats -->
> <FIELD datatype="float" arraysize="2x4" xtype="point"...>
>
> <!-- Three shapes, each made up of four points, each made of two floats
> -->
> <FIELD datatype="float" arraysize="2x4x3" xtype="point"...>
>
> I *think* this works.
>
> If we serialize polygons as arrays of points, do we need the polygon xtype?
>
> It might help avoid the ambiguity caused by an incomplete polygon:
>
> <!-- An incomplete polygon made up from 3 floats -->
> <FIELD datatype="float" arraysize="3" xtype="polygon"...>
>
>
> Any thoughts ?
>
> -- Dave
>
> --------
> Dave Morris
> Research Software Engineer
> Wide Field Astronomy Unit
> Institute for Astronomy
> University of Edinburgh
> --------
>
> On 2017-09-05 12:24, Markus Demleitner wrote:
>>
>> Dear DAL,
>>
>> While implementing DALI 1.1 with a view to TAP 1.1, one of my test
>> cases lead to the encoding of a 2-array of points to a
>>
>> <FIELD datatype="float" arraysize="2x2" xtype="point"...>
>>
>> This made me realise that, as far as I can see, it's unclear what
>> that means by DALI. And it leads to the more fundamental question of
>> whether xtype amends the datatype -- which would support such usage
>> -- or whether it amends (datatype, arraysize) -- which probably would
>> outlaw things like these.
>>
>> So:
>>
>> Is anyone strongly in favour of such usage? Note that another,
>> perhaps more common, use case is arrays of timestamps, which *could*
>> play a role in certain time series serialisations?
>>
>> ...or is anyone strongly opposed to it?
>>
>> Either way, I think it's one of these corner cases where we should
>> somehow agree upon a policy. Should xtype+array be allowed?
>> Forbidden? Undefined at this point?
>>
>> -- markus
--
Patrick Dowler | 978 | 3,673 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2022-27 | latest | en | 0.782782 |
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GUIDE: Mathematics of the Discrete Fourier Transform (DFT) - Julius O. Smith III. LTI Filters and the Convolution Theorem
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NOTE: THIS DOCUMENT IS OBSOLETE, PLEASE CHECK THE NEW VERSION: "Mathematics of the Discrete Fourier Transform (DFT), with Audio Applications --- Second Edition", by Julius O. Smith III, W3K Publishing, 2007, ISBN 978-0-9745607-4-8. - Copyright © 2017-09-28 by Julius O. Smith III - Center for Computer Research in Music and Acoustics (CCRMA), Stanford University
<< Previous page TOC INDEX Next page >>
LTI Filters and the Convolution Theorem
Definition: The frequency response of an LTI filter is defined as the Fourier transform of its impulse response. In particular, for finite, discrete-time signals , the sampled frequency response is defined as
The complete frequency response is defined using the DTFT, i.e.,
where we used the fact that is zero for and to truncate the summation limits. Thus, the infinitely zero-padded DTFT can be obtained from the DFT by simply replacing by . In principle, the continuous frequency response is being obtained using ''time-limited interpolation in the frequency domain'' based on the samples . This interpolation is possible only when the frequency samples are sufficiently dense: for a length finite-impulse-response (FIR) filter , we require at least samples around the unit circle (length DFT) in order that be sufficiently well sampled in the frequency domain. This is of course the dual of the usual sampling rate requirement in the time domain.8.10
Definition: The amplitude response of a filter is defined as the magnitude of the frequency response
From the convolution theorem, we can see that the amplitude response is the gain of the filter at frequency , since
Definition: The phase response of a filter is defined as the phase of the frequency response
From the convolution theorem, we can see that the phase response is the phase-shift added by the filter to an input sinusoidal component at frequency , since
The subject of this section is developed in detail in [1].
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January 15, 2003, 17:55 projection method #1 ilker basol Guest Posts: n/a hi, I am trying to implement projection 2 algorithm of Gresho using finite element method. I would like to discuss some points with anybody who already implemented this scheme. I would be very happy if you can contact me.
January 16, 2003, 20:24 Re: projection method #2 Tony Guest Posts: n/a Do you mean the incremental scheme? What are your questions?
January 17, 2003, 06:22 Re: projection method #3 ilker basol Guest Posts: n/a I am talking about the projection 2 that gresho presents in his paper " on the theory of semi-implicit projection methods for viscous incompressible flow and its implementation via a finite element method that also introduces a nearly consistent mass matrix Part 2: implementation " I try to implement the scheme using a Q2Q1 element (quadratic shape functions for velocities and linear shape functions for pressure). I am trying to debug it using a test problem. It is the classical channel flow with parabolic velocity profile. but i don't get logical results for the velocities at the first time step. the pressures look quite logical because pressure is linearly decreasing starting from inlet to outlet as it is in the analytical results. but I wonder why it doesn't give logical results for the velocities. my first question is do you any boundary condition for the solution of laplace equation (CT*ML-1*C)*PHI=CT*U~n+1-gn+1 (in discrete form as it appears in his paper) second question is about the term (CT*ML-1*C). How does it look like for one element, or if you already implemented this scheme is it possible to verify it for one element from the results of your code. In fact I still have some tiny questions in my mind but i keep them for my self.
January 17, 2003, 15:07 Re: projection method #4 Tony Guest Posts: n/a I will try to answer your questions, though I have not personally used Q2Q1. (CT*ML-1*C) is the so-called consistent Laplace matrix with M usually being properly diagnosed or lumped, since otherwise it is computationally unfeasible. Even using lumped M, the entries of this matrix are larger than the traditional Laplace matrix. I am not sure that it can be assembled in the element level. For example, for a given node, its support for the traditional Laplace is the intermediate neighbor nodes though element connectivity as we all know. But for the consistent matrix, the support involves one more level of the elements. This probably answers your second question. The implied boundary condition for the pressure is dPHI/dn=0 where n the unit vector normal to the boundary. Since PHI is the increment of pressure (probably scaled by dt), this condition means dP^n+1/dn = dP^n/dn. Hope this helps. Tony
January 18, 2003, 09:13 Re: projection method #5 ilker basol Guest Posts: n/a thanks for your suggestions and directions... what i have done up to now is to consider no boundary condition for the pressure poisson equation (CT*ML-1*C)*PHI = CT*U~n+1 - g_n+1 because in gresho's and chorin's paper and also in gresho book, it says that the boundary conditions are automatically built in this equation, therefore i don't implement any boundary condition for this equation. and the laplace operator i compute it globally as it is mentioned in the appendix of gresho's book not in element level... another thing that i wonder about is the g term. it say that g term is the complete gradient matrix tranposed times the u_n+1 g_n+1 = CT * u_n+1. i would like to know why it appears in all of the gresho's papers but not in others. and what does it physically represent.. and did you also implement this term in your code.. and lastly i would like to know which discretization did you implemeted in your code, i guess Q1Q0 ?? thanks a lot ilker
January 20, 2003, 15:50 Re: projection method #6 Tony Guest Posts: n/a Ilker, My experience with this issue was using Taylor-Hood type elements (P2P1). The consistent matrix was tested as the pressure preconditioner in Uzawa iteration. It worked fine. I bet the ¡°g¡± term is related to some BC¡¯s, and you may ignore it in your case. As you mentioned earlier that the pressure seems correct but the velocity does not. My guess is that something might be wrong with the gradient operator in the projection step. Also do not forget to check if the prescribed velocity (no-slip?) BC is enforced. Good luck.
January 22, 2003, 11:29 Re: projection method #7 ilker basol Guest Posts: n/a Probably, i found the major error... up to now i was calculating the A matrix (CT*ML-1*C) after reducing the dirichlet boundary conditions from the C and lumped mass matrix (the contributions of the boundary nodes to gradient operator and lumped mass matrix are not taken into account ). But this should be done just before reducing the dirichlet boundary conditions from the equations.. at the end we get the same dimension for A either i calculate without reducing the C or after reducing the C. but the final matrices for A in both cases are totaly different. now for the first time step the velocity values look better than before .. thanks for your suggestions ilker basol
January 28, 2003, 11:51 Re: projection method #8 ilker basol Guest Posts: n/a hi Tony, Now i have some improvements in the results but i still have problems and questions. My question is about the K (viscous term). I checked Mark A.Christon's paper named "Dealing with pressure:FEM solution strategies for the pressure in the time-dependent Navier-Stokes equations" and he says that he uses K which is augmented by balancing tensor diffusivity (BTD) and uses trapezoidal rule for this term as below; (M+dt/2*K)U~n+1 = (M-dt/2*K)Un + dt*(.......) did you also augmented the K term and also implemented the trapezoidal rule. and the last question is whether you can solve with any reynolds number. Did you ever try reynolds number of one for example? best regards, ilker
January 28, 2003, 15:51 Re: projection method #9 Tony Guest Posts: n/a Yes, I did try very low Reynolds numbers. It is very stable as long as you treat the viscous term implicitly, like the Crank Nicolson as you mentioned above. I believe the trapezoidal rule is good enough for integrating the viscous term. Good luck.
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Given two dependent random samples with the following results: Population 1 41 33 18 34 42...
Given two dependent random samples with the following results:
Population 1 Population 2 41 33 18 34 42 39 50 50 29 29 28 47 24 44
Use this data to find the 95% confidence interval for the true difference between the population means. Assume that both populations are normally distributed.
Step 1 of 4:
Find the point estimate for the population mean of the paired differences. Let x1 be the value from Population 1 and x2 be the value from Population 2 and use the formula d=x2−x1 to calculate the paired differences. Round your answer to one decimal place.
Step 2 of 4:
Calculate the sample standard deviation of the paired differences. Round your answer to six decimal places.
Step 3 of 4:
Calculate the margin of error to be used in constructing the confidence interval. Round your answer to six decimal places.
Step 4 of 4:
Construct the 95% confidence interval. Round your answers to one decimal place.
Answer #1
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https://numberworld.info/726143512 | 1,632,443,793,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057479.26/warc/CC-MAIN-20210923225758-20210924015758-00277.warc.gz | 468,223,596 | 4,112 | # Number 726143512
### Properties of number 726143512
Cross Sum:
Factorization:
2 * 2 * 2 * 61 * 797 * 1867
Divisors:
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
2b481218
Base 32:
lkg4go
sin(726143512)
0.87958841173385
cos(726143512)
-0.47573545794224
tan(726143512)
-1.8489023617
ln(726143512)
20.403258228185
lg(726143512)
8.8610224613479
sqrt(726143512)
26947.050153959
Square(726143512)
### Number Look Up
Look Up
726143512 which is pronounced (seven hundred twenty-six million one hundred forty-three thousand five hundred twelve) is a very great number. The cross sum of 726143512 is 31. If you factorisate the number 726143512 you will get these result 2 * 2 * 2 * 61 * 797 * 1867. The number 726143512 has 32 divisors ( 1, 2, 4, 8, 61, 122, 244, 488, 797, 1594, 1867, 3188, 3734, 6376, 7468, 14936, 48617, 97234, 113887, 194468, 227774, 388936, 455548, 911096, 1487999, 2975998, 5951996, 11903992, 90767939, 181535878, 363071756, 726143512 ) whith a sum of 1386317520. The number 726143512 is not a prime number. The figure 726143512 is not a fibonacci number. The figure 726143512 is not a Bell Number. The number 726143512 is not a Catalan Number. The convertion of 726143512 to base 2 (Binary) is 101011010010000001001000011000. The convertion of 726143512 to base 3 (Ternary) is 1212121100220122011. The convertion of 726143512 to base 4 (Quaternary) is 223102001020120. The convertion of 726143512 to base 5 (Quintal) is 2441343043022. The convertion of 726143512 to base 8 (Octal) is 5322011030. The convertion of 726143512 to base 16 (Hexadecimal) is 2b481218. The convertion of 726143512 to base 32 is lkg4go. The sine of 726143512 is 0.87958841173385. The cosine of the number 726143512 is -0.47573545794224. The tangent of 726143512 is -1.8489023617. The root of 726143512 is 26947.050153959.
If you square 726143512 you will get the following result 527284400019694144. The natural logarithm of 726143512 is 20.403258228185 and the decimal logarithm is 8.8610224613479. that 726143512 is very impressive number! | 792 | 2,173 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2021-39 | latest | en | 0.621874 |
https://grandyang.com/leetcode/688/ | 1,680,263,776,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949642.35/warc/CC-MAIN-20230331113819-20230331143819-00468.warc.gz | 341,693,304 | 34,210 | # 688. Knight Probability in Chessboard
On an `N`x`N` chessboard, a knight starts at the `r`-th row and `c`-th column and attempts to make exactly `K` moves. The rows and columns are 0 indexed, so the top-left square is `(0, 0)`, and the bottom-right square is `(N-1, N-1)`.
A chess knight has 8 possible moves it can make, as illustrated below. Each move is two squares in a cardinal direction, then one square in an orthogonal direction.
Each time the knight is to move, it chooses one of eight possible moves uniformly at random (even if the piece would go off the chessboard) and moves there.
The knight continues moving until it has made exactly `K` moves or has moved off the chessboard. Return the probability that the knight remains on the board after it has stopped moving.
Example:
``````Input: 3, 2, 0, 0
Output: 0.0625
Explanation: There are two moves (to (1,2), (2,1)) that will keep the knight on the board.
From each of those positions, there are also two moves that will keep the knight on the board.
The total probability the knight stays on the board is 0.0625.
``````
Note:
• `N` will be between 1 and 25.
• `K` will be between 0 and 100.
• The knight always initially starts on the board.
``````class Solution {
public:
double knightProbability(int N, int K, int r, int c) {
if (K == 0) return 1;
vector<vector<double>> dp(N, vector<double>(N, 1));
vector<vector<int>> dirs{{-1,-2},{-2,-1},{-2,1},{-1,2},{1,2},{2,1},{2,-1},{1,-2}};
for (int m = 0; m < K; ++m) {
vector<vector<double>> t(N, vector<double>(N, 0));
for (int i = 0; i < N; ++i) {
for (int j = 0; j < N; ++j) {
for (auto dir : dirs) {
int x = i + dir[0], y = j + dir[1];
if (x < 0 || x >= N || y < 0 || y >= N) continue;
t[i][j] += dp[x][y];
}
}
}
dp = t;
}
return dp[r][c] / pow(8, K);
}
};
``````
``````class Solution {
public:
vector<vector<int>> dirs{{-1,-2},{-2,-1},{-2,1},{-1,2},{1,2},{2,1},{2,-1},{1,-2}};
double knightProbability(int N, int K, int r, int c) {
vector<vector<vector<double>>> memo(K + 1, vector<vector<double>>(N, vector<double>(N, 0.0)));
return helper(memo, N, K, r, c) / pow(8, K);
}
double helper(vector<vector<vector<double>>>& memo, int N, int k, int r, int c) {
if (k == 0) return 1.0;
if (memo[k][r][c] != 0.0) return memo[k][r][c];
for (auto dir : dirs) {
int x = r + dir[0], y = c + dir[1];
if (x < 0 || x >= N || y < 0 || y >= N) continue;
memo[k][r][c] += helper(memo, N, k - 1, x, y);
}
return memo[k][r][c];
}
};
``````
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https://discuss.leetcode.com/topic/105571/my-accepted-dp-solution
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Help us with donation | 959 | 2,739 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2023-14 | latest | en | 0.781046 |
https://wahoosam.net/rules/what-effect-might-the-stitches-on-a-baseball-have-on-the-path-of-a-baseball-quizlet.html | 1,643,385,548,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320306301.52/warc/CC-MAIN-20220128152530-20220128182530-00444.warc.gz | 665,075,358 | 19,222 | # What effect might the stitches on a baseball have on the path of a baseball quizlet?
Contents
First, they provide a place to get a better grip on the ball. Secondly, they interact with the air as the ball moves. The stitches slightly affect the air drag, but they strongly affect the Magnus force.
## How do the stitches on baseballs affect their flight?
Depending on the orientation of the ball in flight, the drag changes as the flow is disturbed by the stitches. … For a baseball, the drag can be determined experimentally by throwing the ball and accurately measuring the change in velocity as the ball passes between two points of known distance.
## Why does a baseball have 108 stitches?
Why are There 108 Stitches on a Baseball? How many stitches on a baseball is determined by dimensions of the baseball. The size, as well as the shape of the cowhide used both contribute to how many stitches on a baseball are needed. The 108 stitches are double stitched, meaning the ball actually contains 216 stitches.
## What do stitches do for a baseball?
The purpose of having stitches on a baseball helps pitchers throw different pitches to hitters. By gripping the ball differently on or across the baseball seams, they can change their pitch trajectory. The spin of the ball against the air can cause a pitch to break in a specific way, or drop as it comes to a batter.
## How does lift affect a baseball?
When the baseball spins while it is moving toward home plate it generates aerodynamic lift due to the Magnus effect. In the case of a fastball this lift is upward and will help give the pitch a flatter trajectory that will break later.
## How are baseballs stitched?
They are stitched by hand using 108 stitches taking about 10 minutes. Once stitched, the ROMLB’s are machine rolled for 15 seconds to flatten the stitching. Then the Rawlings trademark, MLB logo, and commissioner’s signature are stamped on the balls and allowed to dry for one week.
## How heavy is a baseball?
The ball has a cork-and-rubber core, around which yarn is tightly wrapped; the cover consists of two snugly fitted pieces of white leather sewn together. The circumference is 9 to 9.25 inches (23 to 23.5 cm) and the weight between 5 and 5.25 ounces (142 and…
## What does a red K mean in baseball?
The backward K can be seen in ballparks all across the world. It’s a symbol for the fans to remind the pitcher and the batter how many strikeouts the pitcher has. It’s often seen hanging in ballparks, in the outfield in big red letters.
## How do the stitches on a baseball affect the speed at which it travels?
The raised red cotton stitching that holds the cowhide covering of the ball together serves more than just an ornamental function. Without it, the ball wouldn’t travel as far or as fast. When the ball is airborne, the stitching disturbs the boundary layer, the paper-thin layer of air closest to the surface of the ball.
THIS IS INTERESTING: How big is a 12 inch baseball glove?
## What are the stitches on a baseball made of?
The ball starts as a round cushioned cork center called a “pill,” then is wrapped tightly in windings of wool and polyester/cotton yarn, and then covered by stitched cowhide.
## What happens baseball?
Baseball is a bat-and-ball game played between two opposing teams, typically of nine players each, that take turns batting and fielding. The game proceeds when a player on the fielding team, called the pitcher, throws a ball which a player on the batting team tries to hit with a bat.
## What causes a baseball to move?
The rotation of the baseball in combination with the way the seam is spinning through the air causes the ball to move in different patterns. A few common pitches are the fastball, curve ball, slider, and the screwball. Each of these pitches has a different spin because of how the player grips the ball.
## What is lift force in baseball?
The lift force generated by a spinning baseball is the origin of the curve ball thrown by big league pitchers. … For a curve ball, the direction of the lift force is used to move the ball parallel to the surface of the earth, from side to side.
## What is drag force on a baseball?
The drag on a ball is being generated by the boundary layer separating from the back of the ball. As the flow separates, it creates a viscous wake behind the ball. A large, wide wake generates a large amount of drag; a thin wake produces less drag. | 951 | 4,444 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2022-05 | latest | en | 0.961924 |
http://slidegur.com/doc/1138135/document | 1,537,509,962,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267156857.16/warc/CC-MAIN-20180921053049-20180921073449-00283.warc.gz | 230,275,967 | 8,915 | ### Document
```St Columba’s high School
Electricity and Electronics
•Electric Fields and Resistors in
Circuits
•Alternating Current and
Voltage
Which of the following statements
is/are correct?
A. There is a force on a charge in an E-field.
B. When an E-field is applied to a conductor, free charges
move through it.
C. Work is done when a charge is moved in an E-field.
ly
A
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an
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an
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C
C
d
0%
C
0%
on
ly
on
B
on
0%
ly
0%
ly
0%
an
A only
B only
A and C only
B and C only
A, B and C
A
1.
2.
3.
4.
5.
30
One Volt is…….
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ne
C
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ul
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ou
lo
lo
m
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m
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b
pe
Jo
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pe
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nd
m
b
pe
rJ
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0% 0% 0% 0% 0%
O
1. One Joule Coulomb
2. One Coulomb per
Second
3. One Joule per
Coulomb
4. One Joule per
Second
5. One Coulomb per
Joule
30
Metal parallel plates X and Y are
connected to a 3.0 V d.c. supply as
shown. What is the gain in electrical
potential energy of an electron moving
between the plates?
0%
0%
0%
0%
0%
1.
2.
3.
4.
5.
1.6 x 10 –19 J
4.8 x 10 –19 J
9.6 x 10 –19 J
1.9 x 10 19 J
5.3 x 10 –20 J
3.0V
Plate X
Plate Y
90
Two parallel metal plates X and Y in a
vacuum have a potential difference V
across them. What is the speed of an
electron of charge e and mass m as it
reaches Y from X?
0%
0%
0%
0%
0%
1. 2eV/m
2. √2eV/m
3. √2V/em
4. 2V/em
5. 2mV/e
X
Y
-
+
V
90
In the circuit below, when the switch is
open V = 2.6V. With the switch closed,
V= 2.2V. This decrease of 0.4 V is
called…
0%
0%
0%
0%
0%
1.
2.
3.
4.
5.
V
Peak voltage
e.m.f.
r.m.s voltage
Lost volts
Terminal potential difference
0%
30
0%
0%
0%
0%
0%
The battery in the circuit below has an
emf of 6.0V and an internal resistance
of 0.20 ohms. What is the resistance
of R if the reading on the ammeter is
4A?
6.0V
1. 1.5 ohms
0.20
2. 2.2 ohms
A
3. 1.3 ohms
R
4. 0.7 0hms
5. 6.4 ohms
90
What is the p.d. across the 10
resistor?
0%
0%
0%
0%
0%
1.
2.
3.
4.
5.
1.0V
1.2V
4.8V
5.0V
6.0V
6.0V
2
A
10
90
The p.d. across the 16 ohm resistor is
40V when the switch is open. What is
the p.d. across the 16 ohm resistor
when S is closed?
0%
0%
0%
0%
0%
1.
2.
3.
4.
5.
12V
15V
30V
45V
48V
60V
8
S
16
8
90
0%
0%
0%
0%
0%
In the circuit below, each resistor has
a resistance of 20 ohms. What is the
voltage across XY?
1. 3.5V
3.5V
2. 0.2V
3. 2.0V
4. 0.5V
X
Y
5. 1V
90
In the circuit shown, the p.d. between
points P and Q is 12 V. What is the
0%
0%
0%
0%
0%
1.
2.
3.
4.
5.
0V
2V
4V
6V
8V
P
2
4
Q
V
4
8
30
In the Wheatstone bridge shown, there
is a small reading on the voltmeter.
What should be done to balance the
bridge?
6
0%
0%
0%
0%
0%
1.
2.
3.
4.
5.
12
Increase P by 6
Q
P
Increase Q by 6
R
S
4
12
Increase R by 6
Increase S by 6
Insert a 6 resistor in series with
the voltmeter.
V
90
What is the total power developed in
the circuit below?
0%
0%
0%
0%
0%
1.
2.
3.
4.
5.
12W
24W
48W
300W
1200W
12V
6
6
90
In the circuit below, the frequency of
the a.c. supply output is varied in
frequency from 50Hz to 5kHz. What is
0%
0%
0%
0%
0%
1.
2.
3.
4.
5.
Remains constant
Rises then falls
Falls then rises
A
1K
30
A supply with alternating output 6V
r.m.s. is connected as shown. What is
the peak voltage across the resistor
and the peak current in the circuit?
0%
0%
0%
0%
0%
1.
2.
3.
4.
5.
Vp=6√2
Vp=6√2
Vp=6
Vp=6√2
Vp=6
Ip=2√2
Ip=2
Ip=2
Ip=1/(2√2)
Ip=2√2
6V
3
90
The resistance of the LDR decreases when the light
intensity increases. Which row shows the correct
voltmeter readings when the lights are off?
1.
2.
3.
4.
5.
V1= increases V2 = increases
V1= decreases V3 = increases
V2 = increases V3 = decreases
V2 = no change V3 = decreases
V1= decreases V2 = no change
-
V1
V
+
V2
V3
The peak voltage of a supply is 70V.
What is the r.m.s. value?
99.0V
35.0V
17.5V
49.5V
50.3V
.3
50
.5
49
V
0%
V
0%
V
.5
17
.0
35
.0
0%
V
0%
V
0%
99
1.
2.
3.
4.
5.
90
A student measures the frequency of
an a.c. supply and obtains the trace
below. If the grid has 1cm squares,
what is the frequency of the supply?
0%
0%
0%
0%
0%
1.
2.
3.
4.
5.
0.06Hz
0.17Hz
41.7Hz
62.5Hz
125Hz
Y-gain = 3mV/cm
Timebase = 2ms/cm
90
``` | 1,849 | 4,062 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2018-39 | latest | en | 0.800166 |
https://www.varsitytutors.com/gre_math-help/how-to-simplify-an-expression | 1,721,220,185,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514771.72/warc/CC-MAIN-20240717120911-20240717150911-00707.warc.gz | 896,942,469 | 43,548 | # GRE Math : How to simplify an expression
## Example Questions
### Example Question #401 : Algebra
a # b = (a * b) + a
What is 3 # (4 # 1)?
27
8
20
12
15
27
Explanation:
Work from the "inside" outward. Therefore, first solve 4 # 1 by replacing a with 4 and b with 1:
4 # 1 = (4 * 1) + 4 = 4 + 4 = 8
That means: 3 # (4 # 1) = 3 # 8. Solve this now:
3 # 8 = (3 * 8) + 3 = 24 + 3 = 27
### Example Question #1 : Simplifying Expressions
Simplify the result of the following steps, to be completed in order:
2. Multiply the sum by 4
3. Add x to the product
4. Subtract x – y from the sum
28x + 13y
28x – 13y
29x + 13y
28x + 12y
28x + 11y
28x + 13y
Explanation:
Step 1: 7x + 3y
Step 2: 4 * (7x + 3y) = 28x + 12y
Step 3: 28x + 12y + x = 29x + 12y
Step 4: 29x + 12y – (x – y) = 29x + 12y – x + y = 28x + 13y
### Example Question #1 : Simplifying Expressions
Quantitative Comparison
x and y are non-zero integers.
Quantity A: (x – y)2
Quantity B: (x + y)2
Quantity A is greater.
The answer cannot be determined from the information given.
The two quantities are equal.
Quantity B is greater.
The answer cannot be determined from the information given.
Explanation:
Quantity A: (x – y)2 = x2 – 2xy + y2
Quantity B: (x + y)2 = x2 + 2xy + y2
Both have x2 + y2 so cancel those from both columns and just compare –2xy in Quantity A to 2xy in Quantity B. If x = 1 and y = 1, –2xy = –2 and 2xy = 2, so Quantity B is greater. But if x = 1 and y = –1, –2xy = 2 and 2xy = –2, so Quantity A is greater. The contradiction means the answer cannot be determined.
### Example Question #1 : How To Simplify An Expression
Which is the greater quantity: the median of 5 positive sequential integers or the mean of 5 positive sequential integers?
The relationship cannot be determined
The median is greater
The quantities are equal
The mean is greater
The quantities are equal
Explanation:
If the first integer is , then
This is the same as the median.
### Example Question #121 : Expressions
You are told that can be determined from the expression:
Determine whether the absolute value of is greater than or less than 2.
The relationship cannot be determined from the information given.
The quantities are equal
Explanation:
The expression is simplified as follows:
Since the value of must be slightly greater for it to be 17 when raised to the 4th power.
### Example Question #402 : Gre Quantitative Reasoning
Quantitative Comparison
Quantity A:
Quantity B:
Quantity B is greater.
The two quantities are equal.
Quantity A is greater.
The relationship cannot be determined from the information given.
The relationship cannot be determined from the information given.
Explanation:
(x + y)2 = x2 + 2xy + y2
Now, since there are no specifications on what x and y can equal, one or both of them could be 0, making the two columns equal. Any value other than 0 will make the columns unequal because of the additional 2xy term, so the answer cannot be determined.
### Example Question #1 : How To Simplify An Expression
Which best describes the relationship between and if ?
The relationship cannot be determined from the information given.
The relationship cannot be determined from the information given.
Explanation:
Use substitution to determine the relationship.
For example, we could plug in and .
So far it looks like the first expression is greater, but it's a good idea to try other values of x and y to be sure. This time, we'll try some negative values, say, and .
This time the first quantity is smaller. Therefore the relationship cannot be determined from the information given.
### Example Question #3 : How To Simplify An Expression
If and , then
Cannot be determined
Explanation:
We have three variables and only two equations, so we will not be able to solve for each independent variable. We need to think of another solution.
Notice what happens if we line up the two equations and add them together.
(x + y) + (3x – y + z) = 4x + z
and 5 + 3 = 8
Lets take this equation and multiply the whole thing by 3:
3(4x + z = 8)
Thus, 12x + 3z = 24.
### Example Question #402 : Gre Quantitative Reasoning
Quantitative Comparison
is an integer.
Quantity A:
Quantity B:
Quantity A is greater.
The relationship cannot be determined from the information given.
Quantity B is greater.
The two quantities are equal. | 1,208 | 4,401 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.875 | 5 | CC-MAIN-2024-30 | latest | en | 0.797814 |
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# How Many Tablespoons of Butter in 1/4 Cup: A Simple Guide
Published
on
I enjoy cooking, yet at times, determining the right measurements can be somewhat challenging. With butter, knowing the number of tablespoons in 1/4 cup is crucial for the outcome of a recipe.
In this article, I’ll break down the conversion for you, making butter measurements easy to grasp. By the end, you’ll be able to confidently convert cup measurements to tablespoons and never second-guess your butter amounts again.
Let’s dive in!
## Key Takeaways
• Accurate measurements of butter ensure successful recipes
• 1/4 cup of butter is equivalent to 4 tablespoons
• Converting measurements from cups to tablespoons is important for precise measurement
• Understanding common kitchen measurements, such as tablespoons and cups, is crucial in cooking and baking.
## Conversion of 1/4 Cup to Tablespoons
To convert 1/4 cup to tablespoons, you’ll need to multiply it by 4.
When it comes to measuring butter, it’s important to know the correct conversions to ensure accurate and delicious results in your recipes. Butter is commonly measured in tablespoons, and having a butter conversion chart can be a helpful tool in the kitchen.
By converting recipes to tablespoons, you can easily determine the amount of butter needed for a specific dish. For example, if a recipe calls for 1/4 cup of butter, you would multiply this measurement by 4 to get the equivalent amount in tablespoons, which is 4 tablespoons.
This conversion ensures that you use the right amount of butter and achieve the desired taste and texture in your culinary creations.
## Understanding Butter Measurements
Understanding butter measurements can be confusing, but did you know that 1/4 cup of butter is equal to 4 tablespoons?
When it comes to culinary measurement conversions, it’s important to know the butter measurement equivalents. In the culinary world, butter is often measured in cups and tablespoons. A cup is a larger unit of measurement compared to a tablespoon.
One cup contains 16 tablespoons of butter. Therefore, when you have 1/4 cup of butter, you can easily convert it to 4 tablespoons. This knowledge is handy when following recipes or adjusting ingredient quantities.
## The Ratio of Tablespoons to Cups
Did you know that a tablespoon is equal to 1/16 of a cup? This culinary conversion is essential for accurately measuring butter in recipes. To help you understand the ratio of tablespoons to cups, I have created a table below:
Tablespoons Cups
1 1/16
2 1/8
3 3/16
4 1/4
## Tablespoons in a Quarter Cup of Butter
When it comes to cooking and baking, precise measurements are crucial for successful results. One common measurement conversion that often arises is the conversion of butter from fractions to tablespoons.
Understanding the ratio of tablespoons to a quarter cup of butter can be useful in many recipes. Additionally, knowing other culinary measurement equivalents can help streamline the cooking process and ensure accuracy in the kitchen.
### Butter Measurement Conversions
To convert 1/4 cup of butter to tablespoons, you’ll need 4 tablespoons. When it comes to butter, measurements can sometimes be confusing, especially when following recipes or looking for substitutes. It’s important to understand the conversion ratios to ensure accurate measurements in your butter recipes. Here is a table that shows the conversion of common butter measurements to tablespoons:
Butter Measurement Tablespoons
1/4 cup 4
1/2 cup 8
1 cup 16
1 pound 32
1 stick 8
Knowing these conversions will help you navigate butter recipes and make informed decisions when using butter substitutes. Now that we have established the conversion ratio for 1/4 cup of butter to tablespoons, let’s explore other fraction to tablespoon ratios in the next section.
### Fraction to Tablespoon Ratio
Now that we’ve discussed butter measurement conversions, let’s dive into the topic of converting fractions to tablespoons accurately.
When it comes to measuring butter in tablespoons, it’s important to be precise to ensure accurate results in your recipes. To convert fractions to tablespoons, you need to know that there are 16 tablespoons in 1 cup of butter. So, if you have 1/4 cup of butter, you can simply divide 16 by 4 to find that it is equal to 4 tablespoons.
However, there are common mistakes that people make when measuring butter in tablespoons. One common mistake is not properly leveling off the butter when using a tablespoon. This can lead to inaccurate measurements and affect the outcome of your dish. So, always make sure to level off the butter to get the right amount in tablespoons.
### Culinary Measurement Equivalents
If you’re cooking and need to convert culinary measurements, it’s helpful to know some common equivalents.
One common measurement technique is converting butter measurements. When a recipe calls for a specific amount of butter, it’s important to know how much that is in tablespoons.
To convert butter measurements, you can use the following conversion: 1/4 cup of butter is equal to 4 tablespoons. This means that if a recipe calls for 1/4 cup of butter, you will need to use 4 tablespoons instead.
Knowing these culinary measurement equivalents can be extremely useful when following recipes or creating your own culinary masterpieces.
## Calculating Butter Amounts in Tablespoons
You can easily convert 1/4 cup of butter into tablespoons. Butter measurement conversions are essential in the culinary world. It’s important to know how to convert between different units of measurement for butter. To help you out, here’s a table that shows the equivalent measurements of butter in tablespoons:
Butter Measurement Tablespoons
1/4 cup 4 tablespoons
1/2 cup 8 tablespoons
1 cup 16 tablespoons
2 cups 32 tablespoons
Butter substitutes are also worth exploring, especially if you’re looking for alternative options. Here are some common substitutes and their equivalent measurements in tablespoons:
Butter Substitute Tablespoons
Margarine Same as butter measurement
Coconut oil Same as butter measurement
Olive oil Same as butter measurement
Applesauce 1/4 cup = 4 tablespoons
## Equivalent Measurements for 1/4 Cup of Butter
To determine the equivalent measurements for 1/4 cup of butter, multiply the measurement by 4. Therefore, 1/4 cup of butter is equal to 4 tablespoons of butter.
This conversion is important when following recipes that require precise measurements. Converting cups to tablespoons allows for more accuracy when measuring ingredients. Knowing the equivalent measurements for 1/4 cup of butter can also be helpful in adjusting recipes or scaling up or down the amount of butter needed.
It is a simple and straightforward conversion that can be easily applied in the kitchen. So, the next time you come across a recipe that calls for 1/4 cup of butter, remember that it is equivalent to 4 tablespoons of butter.
## How Many Tablespoons Are in 1/4 Cup
There’s a total of 4 tablespoons in 1/4 cup.
Accurate butter measurements are crucial in baking recipes to ensure the perfect texture and taste. When converting tablespoons to grams for precise butter measurement, it is important to understand the density and consistency of butter.
Butter is typically measured by volume, but for more accurate results, it is recommended to weigh it in grams. This is especially important when it comes to baking, where even small variations in ingredient quantities can greatly affect the final outcome.
To convert tablespoons to grams, you can refer to a conversion chart or use an online calculator. By taking the time to measure butter accurately, you can achieve consistent and delicious results in your baked goods.
Accurate measurements are crucial in baking recipes to ensure the perfect texture and taste. It’s especially important to understand the density and consistency of butter when converting measurements.
When it comes to butter substitutes for baking, it’s important to choose options that have a similar fat content and consistency.
Here are some tips for measuring butter accurately:
1. Softened Butter: Before measuring, make sure the butter is softened to room temperature. This allows for easier and more accurate measurement.
2. Use a Scale: For precise measurements, using a kitchen scale is recommended. This ensures accuracy and eliminates any guesswork.
3. Marked Butter Wrappers: Many butter brands have measurements marked on their packaging. These markings help you measure the desired amount without the need for additional tools.
## Converting Cup Measurements to Tablespoons
When measuring ingredients for baking, it’s helpful to know that 1 cup is equivalent to 16 tablespoons. This conversion formula is essential in the kitchen, as it allows you to accurately measure ingredients.
Understanding common kitchen measurements is crucial for successful baking. For example, if a recipe calls for 1/4 cup of butter, you can easily convert it to tablespoons. Since there are 16 tablespoons in a cup, you can divide 1/4 cup by 1 and multiply it by 16. Doing the math, 1/4 cup of butter is equal to 4 tablespoons.
This simple conversion formula saves time and ensures that your measurements are precise. By mastering these common kitchen measurements, you can confidently tackle any baking recipe.
### Can I Use Margarine Instead of Butter in Recipes That Require 1/4 Cup?
I can use margarine instead of butter in recipes that require 1/4 cup. However, it’s important to note that margarine is not as healthy as butter. Vegan butter can also be used as a substitute.
### How Many Ounces Are in 1/4 Cup of Butter?
When it comes to cooking conversions, measuring butter can be tricky. To answer the current question, 1/4 cup of butter is equivalent to 4 tablespoons.
### What Is the Weight of 1/4 Cup of Butter in Grams?
To convert 1/4 cup of butter to tablespoons, you’ll need 4 tablespoons. When it comes to butter measurement equivalents, 1/4 cup is approximately 57 grams.
### How Much Does 1/4 Cup of Melted Butter Weigh?
To measure melted butter accurately, 1/4 cup weighs 4 tablespoons. When comparing butter and margarine, butter is generally considered healthier due to its natural composition and higher vitamin content.
### Can I Substitute Oil for Butter in Recipes That Call for 1/4 Cup?
Substituting butter with oil in recipes that call for 1/4 cup has its pros and cons. While oil can add moisture and enhance flavor, it may affect the texture and rise of baked goods.
## Conclusion
After conducting a thorough investigation, it has been determined that there are indeed 4 tablespoons in 1/4 cup of butter.
This theory has been proven to be true through precise measurements and calculations.
Understanding the conversion of cup measurements to tablespoons is essential for accurate and efficient cooking.
By knowing the exact measurements, you can confidently follow recipes and create delicious dishes.
Mastering butter measurements will undoubtedly make your culinary adventures much easier and more enjoyable.
# Discover Who Owns Butter Restaurant: A Culinary Insight
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Stepping into the cozy and welcoming ambiance of Butter Restaurant, I can’t help but ponder: who is the owner of this culinary treasure?
In this article, we will delve into the fascinating history and ownership changes of Butter Restaurant. From its humble beginnings to its current renowned status, we will explore the notable figures behind its success and the plans for its future.
So, grab a seat and get ready to uncover the secrets of who truly owns Butter Restaurant.
## Key Takeaways
• Chef Alex Guarnaschelli was the original owner of Butter Restaurant, founding it in 2002.
• Chef Richie Notar took over as the owner in 2010.
• In 2019, Butter was sold to a group of investors led by Sam Nazarian.
• Each new owner has contributed to Butter’s legacy and success.
## History of Butter Restaurant
You’ll be interested to know that the history of Butter Restaurant dates back to its opening in 2002.
Over the years, the restaurant has undergone significant development to become the renowned establishment it is today.
It all began when Chef Alex Guarnaschelli, a well-known figure in the culinary world, decided to open her own restaurant in the heart of New York City.
With a vision of creating a space that combines exquisite food with a comfortable atmosphere, Butter Restaurant quickly gained popularity among food enthusiasts and celebrities alike.
The history of Butter Restaurant is marked by a commitment to culinary excellence, with Chef Guarnaschelli constantly pushing the boundaries of flavor and technique.
The restaurant’s development has been shaped by the dedication and passion of its team, resulting in a dining experience that is both memorable and delightful.
## Founding of Butter Restaurant
The ownership history of Butter Restaurant is a fascinating tale of passion, dedication, and perseverance.
From its humble beginnings as a small food truck to its current status as a renowned culinary establishment, Butter has seen a remarkable journey.
Founded by renowned chef Alex Guarnaschelli, the vision behind Butter was to create a dining experience that combined exquisite flavors with a warm and inviting atmosphere.
### Ownership History of Butter
Don’t miss out on the fascinating ownership history of Butter restaurant.
Over the years, Butter has seen several ownership changes, with each transition bringing new energy and notable figures to the establishment.
The journey began with Chef Alex Guarnaschelli, who founded Butter in 2002 and built a strong foundation for the restaurant.
In 2010, Chef Richie Notar took over as the owner, bringing his experience and vision to the table. Notar’s tenure saw the restaurant soar to new heights, becoming a popular destination for food enthusiasts.
In 2019, Butter changed hands once again, with Notar selling the restaurant to a group of investors led by renowned restaurateur Sam Nazarian.
With each new owner, Butter’s legacy grows, ensuring its continued success in the ever-evolving culinary landscape.
### Butter’s Founding Vision
Founded in 2002 by Chef Alex Guarnaschelli, Butter has always aimed to embody a unique culinary vision. Our founding principles revolve around creating exceptional dishes using the finest ingredients, providing impeccable service, and creating a warm and inviting atmosphere for our guests.
Butter has had a significant impact on the culinary scene, not only in terms of the delectable food we serve but also as a pioneer in modern American cuisine. We achieve this by uniting traditional techniques with innovative flavors, elevating comfort food to new heights, celebrating local and seasonal ingredients, and fostering a sense of community through food.
Our commitment to these principles has helped us create a dining experience that is both memorable and satisfying. As we delve into the ownership changes at Butter Restaurant, we remain dedicated to upholding our founding principles and continuing our impact on the culinary scene.
## Ownership Changes at Butter Restaurant
So, there’s been a change in management at Butter Restaurant. The new management team has taken over and it’s expected that there will be some implications for the staff.
It’s important to discuss what these implications might be and how they could potentially affect the employees.
### New Management at Butter
The new management at Butter has brought a fresh perspective to the restaurant. They have faced several challenges in their quest to improve the customer experience.
• Innovative Menu: The new management has introduced exciting and unique dishes that have delighted customers, making dining at Butter a truly memorable experience.
• Improved Service: The new team has implemented rigorous training programs for the staff, resulting in exceptional service and a warm, welcoming atmosphere.
• Updated Decor: The restaurant has undergone a stunning makeover, with new furniture, lighting, and artwork, creating a modern and stylish ambiance that enhances the overall dining experience.
• Enhanced Communication: The management has implemented effective communication channels, ensuring that customer feedback is taken seriously and promptly addressed, leading to a more personalized and satisfying experience for all.
Overall, the impact of the new management can be felt in every aspect of the restaurant, from the enticing menu to the attentive service, providing customers with a truly exceptional dining experience at Butter.
### Implications for Staff?
With the new management, staff members at Butter have experienced positive changes in their training programs and communication channels. The new management team has implemented a more structured and comprehensive training program, providing employees with the necessary skills and knowledge to excel in their roles. Additionally, they have introduced new communication channels, such as regular team meetings and an online platform for sharing information and updates. These changes have greatly improved staff morale and productivity. In terms of staff implications, the new management has emphasized job security and has made efforts to retain and develop existing talent within the organization. They have implemented performance evaluation systems and career development plans to ensure that employees have opportunities for growth and advancement. The table below highlights some of the key changes and their impact on staff members:
Changes Staff Implications
Improved training programs Enhanced skills and performance
Enhanced communication channels Better coordination and collaboration
Emphasis on job security Increased employee satisfaction and loyalty
Performance evaluation systems Opportunities for growth and advancement
Career development plans Increased motivation and engagement
Overall, the new management at Butter has made significant positive changes that have had a profound impact on staff members, ensuring job security and providing opportunities for professional growth.
## Notable Figures Behind Butter Restaurant
Did you know that Butter Restaurant is owned by renowned chef Alex Guarnaschelli? As one of the notable figures in the culinary world, Guarnaschelli has brought her expertise and passion to Butter, creating a dining experience like no other.
The ownership history of Butter Restaurant has seen Guarnaschelli’s influence and dedication since she took over in 2003. Here are some reasons why Alex Guarnaschelli’s ownership of Butter is significant:
• Culinary expertise: With Guarnaschelli’s extensive knowledge and experience, guests can expect exceptional food and innovative flavors.
• Celebrity status: As a well-known chef and TV personality, Guarnaschelli’s ownership adds a touch of glamour and excitement to the restaurant.
• Commitment to quality: Guarnaschelli’s ownership ensures that Butter maintains its reputation for using the finest ingredients and delivering top-notch service.
• Legacy of excellence: With Guarnaschelli at the helm, Butter continues to be a destination for food lovers seeking a memorable dining experience.
## Expansion and Growth of Butter Restaurant
Despite its initial location, Butter has expanded and grown to become a renowned culinary destination.
As the owner of Butter Restaurant, I have always been on the lookout for expansion opportunities. With the success and popularity of our original location, we saw a demand for more Butter experiences.
We carefully researched and identified potential markets where we could replicate our winning formula. However, we also had to consider the market competition in these new locations. We analyzed the local culinary scene, identified gaps in the market, and adjusted our menu and concept accordingly.
## Current Ownership of Butter Restaurant
As the current owner, I have embraced innovation and kept Butter Restaurant at the forefront of the culinary industry. Under my leadership, the restaurant has thrived and grown, bringing in new customers and expanding our reach.
The current ownership of Butter Restaurant has significant implications for the staff, as I have implemented various changes to improve efficiency and enhance the dining experience. These changes include:
• Upgrading kitchen equipment to streamline cooking processes and reduce wait times
• Implementing a comprehensive training program to ensure consistent high-quality service
• Introducing new menu items and seasonal specials to keep things fresh and exciting for both staff and customers
• Prioritizing employee well-being by offering competitive compensation packages and opportunities for career growth.
## Future Plans for Butter Restaurant
Looking ahead, you can expect exciting changes and expansions at Butter that will elevate your dining experience even further.
We are thrilled to announce our future expansion plans, as we aim to bring the Butter experience to more locations. Our team is actively searching for new opportunities to bring our unique concept to different cities and neighborhoods.
In addition to our expansion efforts, we are also working on exciting menu changes to keep our offerings fresh and innovative. Our talented chefs are constantly experimenting with new flavors and techniques to create dishes that will surprise and delight our guests.
We are committed to providing you with an exceptional dining experience, and we can’t wait to share these exciting updates with you in the near future.
### What Is the Average Price Range for Menu Items at Butter Restaurant?
The average price range for menu items at Butter Restaurant varies depending on the dish, but generally falls between \$20 and \$40. They offer a diverse selection of options to suit different tastes and dietary preferences.
### Can I Make a Reservation at Butter Restaurant?
Sure, you can make a reservation at Butter Restaurant. They offer both indoor and outdoor seating, with availability for special dietary accommodations. It’s always best to book in advance to secure your spot.
### Are There Any Vegetarian or Vegan Options Available at Butter Restaurant?
There are several vegetarian and vegan options available at Butter Restaurant. They offer a variety of dishes that cater to those with dietary restrictions, ensuring everyone can enjoy their meal.
### Does Butter Restaurant Offer Catering Services for Private Events?
Yes, Butter Restaurant offers catering services for private events. From intimate gatherings to grand celebrations, they provide delicious food and exceptional service to make your event truly memorable.
### What Are Some Signature Dishes or Must-Try Items at Butter Restaurant?
Some signature dishes at Butter Restaurant include the herb-roasted chicken and the truffle mac and cheese. These chef-recommended dishes are known for their rich flavors and are a must-try for any food lover.
## Conclusion
In conclusion, Butter Restaurant has had a rich history, starting from its founding to its current ownership. Throughout the years, the restaurant has seen changes in ownership, but it has remained a staple in the culinary world.
Notable figures have contributed to the success and growth of Butter Restaurant, making it a renowned establishment. As for the future, the restaurant plans to continue expanding and providing exceptional dining experiences.
For instance, imagine a food critic raving about Butter Restaurant’s innovative menu and impeccable service, solidifying its place as a top dining destination.
# Which Attachment to Cream Butter and Sugar: A Comprehensive Guide
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I have always been curious about which attachment is the most effective for creaming butter and sugar. After thorough research and testing in my own kitchen, I am happy to share the results.
In this article, I will share my findings on the various attachments available for stand mixers and hand mixers, including the whisk, paddle, beater, blender, and food processor attachments.
Get ready to elevate your baking game and achieve the perfect creamy texture with the right attachment.
## Key Takeaways
• The paddle attachment in a stand mixer is the best for creaming butter and sugar, while the whisk attachment in a hand mixer is also effective.
• Creaming butter and sugar at room temperature helps with easier mixing and achieving a light and fluffy texture.
• Stand mixers offer better control over mixing speed and consistent mixing for thorough creaming.
• The creaming technique is crucial for achieving the desired texture in baked goods, and the beater attachment is highly effective for creaming butter and sugar.
## Stand Mixer Attachments
You can use a stand mixer attachment to easily and efficiently cream butter and sugar together. When it comes to choosing the best stand mixer attachment for this task, the paddle attachment is the most suitable. This attachment has a flat, paddle-shaped head that works well for creaming butter and sugar. It effectively combines the two ingredients, creating a light and fluffy mixture.
To achieve the best results using the creaming method, ensure that the butter is at room temperature. This allows for easier mixing and better incorporation of the sugar. Start by beating the butter alone until it becomes smooth and creamy. Then gradually add the sugar, beating until the mixture is light and fluffy. This process helps to incorporate air into the mixture, resulting in a lighter texture for your baked goods.
Using a stand mixer attachment with the creaming method can save you time and effort, giving you consistent and professional results every time.
## Hand Mixer Attachments
When it comes to hand mixer attachments, finding the best one can make all the difference in your baking experience.
From wire beaters to dough hooks, each attachment serves a specific purpose and can greatly affect the outcome of your recipes.
Additionally, understanding the mixing time differences between various attachments is crucial in achieving the desired consistency and texture in your batter or dough.
### Best Hand Mixer Attachment
The best hand mixer attachment for creaming butter and sugar is the whisk attachment. Whisking techniques, such as using a whisk attachment, are essential for achieving a light and fluffy texture when creaming butter and sugar.
The whisk attachment helps to incorporate air into the mixture, resulting in a smoother and creamier consistency. It is important to consider the creaming temperature as well. The butter should be at room temperature, around 68-70°F (20-21°C), to ensure proper creaming.
Using a whisk attachment at this temperature allows the butter and sugar to blend together more efficiently, creating a homogenous mixture. By understanding the importance of whisking techniques and creaming temperature, you can achieve the perfect creamed butter and sugar mixture for your recipes.
Now let’s explore the differences in mixing time.
### Mixing Time Differences
To achieve the perfect texture, it’s important to understand the differences in mixing times.
When it comes to the effective creaming technique, using a stand mixer offers several benefits.
Firstly, a stand mixer provides consistent and even mixing, ensuring that the butter and sugar are thoroughly combined. This helps to create a light and fluffy texture in your baked goods.
Additionally, a stand mixer allows for better control over the mixing speed, allowing you to gradually incorporate air into the mixture. This air is essential for leavening and creating a tender crumb in your baked goods.
Lastly, using a stand mixer reduces the amount of manual labor required, saving you time and effort.
## Whisk Attachment
Using the whisk attachment will help incorporate air into the butter and sugar mixture. This whisking technique is essential for achieving a light and fluffy texture in baked goods.
The whisk attachment, with its wire loops, creates small air pockets by rapidly beating the butter and sugar together. As the whisk spins, the loops push through the mixture, incorporating air and creating a lighter consistency. This method is particularly effective when making cakes, cookies, and other delicate desserts.
Alternative methods, such as using a paddle attachment or hand mixing, may not achieve the same level of aeration. The whisk attachment’s rapid spinning action ensures a thorough incorporation of air, resulting in a superior texture and rise in your baked goods.
When it comes to the creaming method, using the paddle attachment is an effective option. It helps to create a smooth and fluffy texture by thoroughly incorporating the butter and sugar.
However, it’s important to note that there are other attachment options available, such as the whisk or beater attachments, which may yield different results.
The mixing time required with the paddle attachment will depend on the recipe, but generally, it should be mixed until the butter and sugar are light and creamy.
### Effective Creaming Method
Creaming butter and sugar is essential for achieving a light and fluffy texture in baked goods. When it comes to the creaming method, there are two main techniques: creaming and beating. Creaming involves combining softened butter and sugar together until the mixture is light and creamy. This process incorporates air into the mixture, resulting in a tender and airy baked good. On the other hand, beating involves melting the butter and then beating it with the sugar. While this method can still produce a delicious treat, it often results in a denser texture. To illustrate the differences between creaming and beating, here is a table outlining their characteristics:
Creaming Method Beating Method
Airiness Light and fluffy Denser
Texture Tender Dense
Ease Requires softened butter Requires melted butter
Now that we understand the effectiveness of the creaming method, let’s explore other attachment options for achieving the perfect texture in baked goods.
### Other Attachment Options
Let’s explore different ways to achieve the perfect texture in baked goods.
When it comes to creaming butter and sugar, most people rely on the traditional paddle attachment. However, there are alternative methods and attachments that can yield equally fantastic results.
One such option is using the whisk attachment. This attachment incorporates more air into the mixture, resulting in a lighter and fluffier texture.
Another alternative is the dough hook attachment, typically used for kneading bread dough. Surprisingly, this attachment can also be used for creaming butter and sugar. The dough hook gently combines the ingredients without incorporating too much air, resulting in a denser and more tender texture.
### Mixing Time Required
To achieve the perfect texture in your baked goods, make sure you don’t overmix the ingredients. Overmixing can lead to tough and dense results.
When it comes to creaming butter and sugar without a mixer, it’s important to consider the impact of mixing time on the final baked goods. Here are some key points to keep in mind:
• Creaming butter and sugar by hand requires more time and effort compared to using a mixer.
• The longer you mix, the more air is incorporated into the mixture, resulting in a lighter texture.
• However, too much mixing can cause the butter to melt, affecting the overall structure.
• It’s crucial to pay attention to the consistency and appearance of the mixture, stopping when it reaches a light and fluffy state.
• The final texture of your baked goods will greatly depend on the mixing time, so be mindful and adjust accordingly.
## Beater Attachment
The beater attachment mixes the butter and sugar together until creamy. This attachment is highly effective in achieving the desired texture for the creaming technique. The beater’s rotating action combines the butter and sugar, creating a smooth and fluffy mixture. It evenly distributes the sugar particles throughout the butter, incorporating air into the mixture. This aeration process helps to create a light and tender baked good.
To further illustrate the beater attachment’s effectiveness, let’s compare it to other attachments commonly used for creaming.
Attachment Effectiveness
Whisk attachment Less effective
Dough hook attachment Not recommended for creaming
As you can see from the table, the beater attachment outperforms other attachments when it comes to creaming butter and sugar. Its design and rotating action make it the ideal choice for achieving a creamy and light texture in your baked goods.
## Blender Attachment
If you’re looking for a versatile attachment, try the blender attachment for a variety of blending and pureeing tasks. The blender attachment is a powerful tool that can handle a wide range of ingredients and tasks.
Here are some key features of the blender attachment:
• Variable blender speed: The blender attachment comes with different speed settings, allowing you to control the blending process and achieve the desired texture.
• High blender capacity: With a large capacity, the blender attachment can handle larger batches of ingredients, making it perfect for making smoothies or pureeing soups.
• Sharp blender blades: The blender attachment is equipped with sharp blades that can easily blend through tough ingredients, ensuring a smooth and consistent blend.
• Easy to clean: The blender attachment is designed for easy cleaning, with detachable parts that can be easily washed and dried.
• Versatile use: Apart from blending and pureeing tasks, the blender attachment can also be used for making sauces, dressings, and even grinding nuts or coffee beans.
Overall, the blender attachment is a must-have for any kitchen, providing you with the convenience and efficiency you need for a variety of blending tasks.
## Food Processor Attachment
With its multiple functions, the food processor attachment is a versatile tool for various food preparation tasks.
When it comes to creaming butter and sugar, the food processor offers several benefits. Firstly, it saves time and effort by quickly and efficiently combining the ingredients. The powerful motor and sharp blades ensure a smooth and creamy consistency.
Additionally, the food processor attachment allows for precise control over the creaming process, enabling you to achieve the desired texture for your recipe. Whether you’re making cookies, cakes, or frosting, the food processor attachment can handle the task with ease. Its ability to cream butter and sugar evenly ensures a consistent result every time.
### Can I Use a Blender Attachment to Cream Butter and Sugar?
Yes, you can use a blender attachment to cream butter and sugar, but it may not yield the best results. A stand mixer with a paddle attachment or a hand mixer would be better alternatives.
### What Is the Difference Between a Hand Mixer Attachment and a Stand Mixer Attachment?
The difference between a hand mixer and a stand mixer attachment is that a hand mixer is portable and easier to store, while a stand mixer is more powerful and has more attachments available. Each has its pros and cons.
### Can I Use a Beater Attachment Instead of a Paddle Attachment?
Using a beater attachment instead of a paddle attachment may affect the creaming process. The paddle attachment’s flat shape helps incorporate air into the butter and sugar, resulting in a lighter texture.
### Is It Necessary to Use a Whisk Attachment for Creaming Butter and Sugar?
Yes, there is an alternative method to creaming butter and sugar without using a whisk attachment. The pros of using a paddle attachment are it’s faster and more efficient, but the cons are it may not incorporate as much air.
### Can a Food Processor Attachment Be Used to Cream Butter and Sugar Effectively?
A food processor attachment or a blender attachment can be effectively used to cream butter and sugar. It provides a convenient alternative to using a whisk attachment, resulting in a smooth and well-mixed mixture.
## Conclusion
In conclusion, after examining the various attachments available for creaming butter and sugar, it is clear that the paddle attachment is the most efficient and effective choice.
With its flat surface and flexible edges, it effortlessly combines the ingredients, resulting in a smooth and creamy texture.
Just like a conductor leading an orchestra, the paddle attachment harmonizes the butter and sugar, creating a symphony of flavor.
So, next time you’re in the kitchen, grab your paddle attachment and let it work its magic!
# Why is the Maker of Bread and Butter called in the Family?
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In my family, I am the one responsible for making bread and butter, providing warmth and nourishment at our table. This simple yet essential staple is deeply grounded in our traditions and has developed over the years.
Despite the gender stereotypes associated with bread and butter making, I see myself as a symbol of nourishment and care for my loved ones. In this article, we will explore the importance of the bread and butter maker’s role in family health, celebrate their contributions, and appreciate their unsung heroism.
## Key Takeaways
• Bread and butter have cultural significance, representing nourishment, comfort, and togetherness in the family.
• Traditional gender roles have historically seen women as the primary bread makers, but these roles are shifting with changing family dynamics and cultural influences.
• Modern households are seeing a more equal distribution of responsibilities, with both partners actively involved in household chores and childcare.
• The role of the bread and butter maker has become more egalitarian in dual-income households, with younger generations being more flexible and adaptable to changing perspectives on gender roles.
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## The Importance of Bread and Butter in the Family
You know how important bread and butter are in my family. They hold a significant cultural significance and are deeply rooted in our family traditions. Bread and butter have been a staple in our meals for generations, and they symbolize nourishment, comfort, and togetherness.
In my family, bread is not just a simple food item; it is a symbol of unity and generosity. We believe that breaking bread with loved ones brings people closer and strengthens the bond between family members. It is a tradition that has been passed down through generations, and we hold it dear to our hearts.
Butter, on the other hand, adds richness and flavor to our meals. It is a versatile ingredient that can be used in various dishes, from spreading it on warm bread to using it in baking and cooking. The creamy texture and distinct taste of butter elevate the overall taste of our family recipes, making them even more special.
Together, bread and butter create a harmonious combination that represents the core values of our family. They remind us of the importance of gathering around the table, sharing stories, and creating lasting memories. They are the building blocks of our family traditions, keeping us connected to our cultural heritage and bringing us closer as a family.
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When discussing traditional roles in bread and butter making, it is important to delve into the gendered bread-making roles that have been prevalent in many cultures throughout history.
In many societies, women have traditionally been seen as the primary bread makers, responsible for kneading the dough, baking the bread, and providing nourishment for the family.
However, with changing family dynamics and cultural influences, these roles have started to shift, allowing for more flexibility and equality in bread-making responsibilities.
As the family’s bread-maker, it’s important for me to challenge gendered roles and encourage everyone to participate in the process. Traditionally, bread-making has been associated with women, with the belief that it is their responsibility to provide nourishment for the family.
However, this gendered division of labor is influenced by cultural norms and expectations. In many societies, women are expected to fulfill domestic duties, including bread-making, while men are seen as the primary breadwinners. These cultural influences shape our perceptions of gender roles and can limit opportunities for both men and women to explore different aspects of cooking and food preparation.
### Changing Family Dynamics
In today’s modern households, it’s common to see shifting dynamics and roles within the family. As families evolve and adapt to changing societal norms and expectations, the traditional division of labor is being challenged.
Here are four ways in which family dynamics are changing:
1. Shared responsibilities: Both partners are actively involved in household chores and childcare, blurring the lines of traditional gender roles.
2. Dual-income households: With more women entering the workforce, families are becoming reliant on dual incomes, leading to a redistribution of responsibilities.
3. Stay-at-home dads: Men are increasingly taking on the role of primary caregivers, challenging the notion that women are solely responsible for childcare.
4. Single-parent households: With the rise in single-parent households, individuals are taking on multiple roles within the family, fulfilling both parental and breadwinner duties.
As family dynamics continue to evolve, it is important to consider the cultural influences that shape these changing roles.
### Cultural Influences on Roles?
One key factor influencing the shifting dynamics and roles within households today is the impact of cultural norms and expectations.
Cultural influences play a significant role in shaping the gender dynamics within families. In many cultures, there is a traditional expectation that men are the breadwinners and women are responsible for household duties. These cultural norms can have a profound effect on the division of labor within the family.
However, it is important to note that cultural influences are not static and are subject to change over time. As societies evolve and become more progressive, there is a growing recognition of the need for gender equality within households.
This shift in cultural norms has led to a reevaluation of traditional gender roles and a more equal distribution of responsibilities within the family unit.
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## Evolution of the Bread and Butter Maker
You’re the one who has become the master of making bread and butter in the family. It’s fascinating to see how this role has evolved over time, influenced by both cultural factors and personal preferences. Here are four key points that paint a picture of the evolution of the bread and butter maker:
1. Cultural Influences: Throughout history, different cultures have shaped the role of the bread and butter maker. In some societies, it was primarily a woman’s responsibility, tied to notions of domesticity and nurturing. In others, it was a shared duty between men and women, reflecting a more egalitarian approach to household chores.
2. Technological Advancements: The evolution of kitchen appliances and techniques has greatly influenced the bread and butter maker’s role. From hand-kneading dough to the invention of the stand mixer, these advancements have made the process more efficient and accessible to everyone.
3. Changing Gender Roles: As societies have progressed, traditional gender roles have shifted, impacting the bread and butter maker’s identity. Today, both men and women take pride in their ability to provide homemade bread and butter for their families, breaking free from outdated stereotypes.
4. Personalization and Creativity: In recent years, there has been a resurgence of homemade bread and butter making, driven by a desire for healthier, artisanal options. This has allowed individuals to experiment with different flours, flavors, and techniques, adding a personal touch to their creations.
As the bread and butter maker in my family, I find it fascinating to trace the evolution of this role and how it has been shaped by cultural influences, technological advancements, changing gender roles, and the desire for personalization and creativity.
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## Gender Stereotypes in Bread and Butter Making
When it comes to gender roles in cooking, traditional stereotypes often dictate that women should be the ones doing the cooking, while men take on more ‘masculine’ roles. However, breaking these stereotypes can lead to a more inclusive and diverse culinary world.
### Gender Roles in Cooking
In many households, it’s common for women to be considered the makers of bread and butter. However, this gender role in cooking comes with its own set of challenges. Breaking stereotypes in this area can be difficult, but it’s not impossible.
Here are four key challenges faced by women in breaking gender roles in cooking:
1. Social expectations: Society often expects women to excel in cooking and homemaking, which can create pressure and limit their opportunities in other areas.
2. Lack of support: Women may face resistance and criticism when trying to break free from traditional gender roles, making it harder to pursue their passions and ambitions.
3. Stereotyping: Society tends to stereotype women as the primary caregivers and assumes they should be responsible for all household chores, including cooking.
4. Work-life balance: Balancing cooking responsibilities with other commitments, such as work and family, can be challenging for women, leading to feelings of overwhelm and exhaustion.
Despite these challenges, women are breaking stereotypes in cooking and proving that anyone can be the maker of bread and butter, regardless of gender.
Breaking traditional gender roles can be challenging, but it’s important to challenge societal expectations and stereotypes.
In today’s society, many individuals are redefining family dynamics by breaking away from traditional gender roles. This shift is evident in various aspects of family life, including parenting, household chores, and career choices.
Men are taking on more active roles in raising children and performing household tasks traditionally associated with women, such as cooking and cleaning. Similarly, women are pursuing careers and leadership positions, challenging the notion that their primary role is to be homemakers.
This breaking of traditional gender roles has led to more egalitarian and balanced family dynamics, where responsibilities and decision-making are shared. By redefining family dynamics, individuals are challenging and dismantling societal expectations, fostering greater equality and empowerment for all.
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## The Bread and Butter Maker as a Symbol of Family Nourishment
The bread and butter maker is often seen as the heart of the family, providing nourishment and sustenance for all. In many cultures, bread making holds deep cultural significance and has been passed down through generations.
Here are four ways in which the cultural influences on bread making and the role of bread and butter in family bonding are evident:
1. Tradition: Bread making techniques and recipes are often passed down from one generation to another, creating a sense of continuity and connection to one’s cultural roots.
2. Ritual: Bread making can be a sacred act, involving specific rituals or ceremonies that bring the family together and strengthen bonds.
3. Communication: The process of making bread involves shared tasks and responsibilities, fostering communication and cooperation among family members.
4. Sharing and Nourishment: Bread and butter are often shared during family meals, symbolizing love, care, and sustenance. This act of sharing strengthens family bonds and promotes a sense of togetherness.
As the bread and butter maker takes on the role of providing nourishment and facilitating family bonding, they also face certain challenges.
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## Challenges Faced by the Bread and Butter Maker
You face various obstacles as you strive to provide nourishment and foster family bonding through the process of bread and butter making. The challenges of being the bread and butter maker in the family are numerous, but the importance of this role cannot be overstated.
Challenges Importance
Time constraints The bread and butter maker often has to juggle multiple responsibilities, including work and family obligations. Finding the time to bake fresh bread and churn homemade butter can be a challenge. However, the effort is worth it, as freshly made bread and butter not only taste better, but also provide essential nutrients for the family.
Skill development Mastering the art of bread making and butter churning takes time and practice. It requires knowledge of different types of yeast, flour, and techniques. The bread and butter maker must constantly improve their skills to ensure the best results.
Ingredient availability Sourcing high-quality ingredients can be a challenge, especially if you prefer to use organic or locally sourced products. The bread and butter maker must be resourceful in finding the best ingredients for their creations.
Financial considerations Making bread and butter from scratch can be more expensive than buying store-bought alternatives. The bread and butter maker must carefully budget and plan their expenses to ensure they can afford the necessary ingredients.
Despite these challenges, the role of the bread and butter maker in the family is of utmost importance. They provide nourishment and create a sense of togetherness through the process of making and sharing food. Balancing bread and butter making with other responsibilities requires careful time management and dedication, but the rewards are well worth it.
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## Balancing Bread and Butter Making With Other Responsibilities
Finding a balance between my passion for baking and my other obligations can be a challenge. Balancing work life and time management is crucial to ensure that I can pursue my love for bread and butter making while fulfilling my responsibilities.
Here are four strategies I have found helpful in achieving this balance:
1. Prioritization: I prioritize my tasks and responsibilities by categorizing them into urgent, important, and non-essential. This helps me allocate my time effectively and ensures that my bread and butter making does not compromise my other obligations.
2. Time blocking: I create a schedule that includes dedicated time for bread and butter making as well as other tasks. By setting aside specific blocks of time for each activity, I can ensure that I give my full attention to both my passion and my other responsibilities.
3. Delegation: I delegate tasks that can be done by others, such as grocery shopping or meal preparation, to free up more time for my bread and butter making. This allows me to focus on what I truly love while still managing my other commitments.
4. Self-care: Taking care of myself is essential for maintaining a healthy work-life balance. I prioritize self-care activities such as exercise, relaxation, and spending time with loved ones. This rejuvenates me and helps me approach both my passion and other responsibilities with renewed energy and focus.
Finding this balance allows me to fully enjoy bread and butter making as a labor of love, knowing that I am managing my time effectively and fulfilling all my obligations.
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## Bread and Butter Making as a Labor of Love
In my quest to balance bread and butter making with my other responsibilities, I discovered that this culinary art goes beyond mere sustenance. It became a form of self-expression and a labor of love.
Bread and butter making allowed me to explore my creativity and showcase my personality. From experimenting with different types of bread, like sourdough or whole wheat, to crafting unique butter spreads infused with herbs or spices, I found joy in the process of creation. The act of kneading the dough, watching it rise, and shaping it into a beautiful loaf became a therapeutic ritual that allowed me to escape from the stress of daily life.
There is something inherently satisfying about the simplicity of bread and butter making. As I immersed myself in the task, I found solace in the rhythmic motions and the aroma that wafted through the kitchen. The repetitive actions of kneading and shaping the dough provided a sense of calm and focus, relieving my mind of any worries or anxieties.
Research has shown that engaging in activities like bread and butter making can have significant therapeutic benefits. It can reduce stress, promote mindfulness, and enhance overall well-being. The tactile nature of the process, from feeling the softness of the dough to the satisfaction of slicing into a warm loaf, can provide a sense of grounding and connection to the present moment.
Bread and butter making became more than just a task on my to-do list. It became a way for me to express myself, find inner peace, and nourish both my body and soul.
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## Bread and Butter Making as a Source of Family Unity
When sharing the process, involving your loved ones in the bread and butter making experience can strengthen the bond between all members of your household. There is something special about working together as a family to create something delicious and comforting. Here are four ways in which bread and butter making can serve as a bonding activity:
1. Collaboration: Each family member can contribute to different aspects of the process, whether it’s kneading the dough, shaping the bread, or churning the butter. This collaboration fosters a sense of teamwork and togetherness.
2. Traditions: The bread and butter maker plays a crucial role in family traditions. Passing down recipes and techniques from generation to generation creates a sense of continuity and heritage within the family.
3. Quality time: Bread and butter making requires time and patience. Spending this time together allows for meaningful conversations, laughter, and the opportunity to create lasting memories.
4. Sense of accomplishment: When the bread is finally out of the oven and the butter is ready to be spread, the sense of accomplishment is shared by all. This shared achievement strengthens the bond between family members.
Involving your loved ones in the bread and butter making process not only creates delicious food but also nurtures the connection between family members.
Now, let’s dive into the bread and butter maker’s secret recipes.
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## The Bread and Butter Maker’s Secret Recipes
To unlock the secrets of these delicious recipes, all you need is a pinch of curiosity and a dash of adventure. The bread and butter maker in the family holds the key to these mouthwatering creations. They have perfected the art of bread and butter making through years of practice and experimentation. The secret ingredients and rituals they follow are what make their recipes truly special.
One of the key secrets to their bread and butter making is the use of high-quality ingredients. From selecting the finest flour to using fresh and creamy butter, every ingredient plays a crucial role in the final product. The bread and butter maker knows the importance of using the best ingredients and sources them meticulously.
In addition to the ingredients, the bread and butter maker also follows specific rituals that enhance the flavor and texture of their creations. These rituals may include kneading the dough for a specific duration, allowing it to rise at the perfect temperature, or using a particular technique to spread the butter on the bread. These small details make a significant difference in the final result.
To give you a glimpse into the world of the bread and butter maker, here’s a table showcasing some of their secret recipes:
Recipe Name Secret Ingredient Ritual
Garlic Bread Roasted garlic Double rising
Cinnamon Swirl Cinnamon sugar Butter brushing
These recipes are just a taste of the bread and butter maker’s expertise. By incorporating their secret ingredients and following their rituals, you can elevate your own bread and butter making skills to new heights. So, grab your apron and embark on a culinary adventure with these delightful recipes.
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## Passing Down the Bread and Butter Making Legacy
You can continue the rich tradition of bread and butter making by learning and practicing the secret recipes passed down through generations. There is something truly special about preserving family recipes and passing down traditions from one generation to the next.
Here are four reasons why it is important to continue the legacy of bread and butter making in your family:
1. Cultural Heritage: By learning and practicing these secret recipes, you are keeping alive a part of your family’s cultural heritage. Bread and butter making techniques have been perfected over time, and by continuing this tradition, you are honoring your ancestors and their culinary expertise.
2. Family Bonding: Making bread and butter together can be a wonderful bonding experience for the whole family. It provides an opportunity to spend quality time together, share stories, and create lasting memories.
3. Sense of Identity: Passing down these recipes helps to establish a sense of identity within your family. It gives future generations a connection to their roots and a sense of belonging.
4. Legacy Preservation: By continuing the tradition of bread and butter making, you are preserving a part of your family’s history for future generations to enjoy. It ensures that the knowledge and skills are not lost and can be passed on to your children and grandchildren.
Preserving these family recipes and traditions goes beyond just making delicious bread and butter. It has a profound impact on family health and well-being, which we will explore in the next section.
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## The Bread and Butter Maker’s Impact on Family Health
When it comes to the impact of dietary choices on family health, it is important to consider the nutritional aspect. Our food choices play a significant role in our overall well-being, and as the bread and butter maker in my family, I am responsible for ensuring that we have a balanced and nutritious diet.
### Dietary Choices and Health
In terms of dietary choices and health, it’s important to consider the impact of consuming bread and butter regularly. As someone who strives for a healthy lifestyle, I have done extensive research on the topic. Here are four key points to consider:
1. Nutritional value: Bread provides carbohydrates for energy while butter adds fat and flavor. It’s important to choose whole grain breads for added fiber and nutrients.
2. Portion control: Bread and butter can be high in calories, so it’s crucial to watch serving sizes. Moderation is key to maintaining a balanced diet.
3. Health effects: Consuming too much bread and butter can contribute to weight gain, increased risk of heart disease, and elevated cholesterol levels.
4. Alternatives: There are healthier options available, such as using spreads made from avocado or nut butters, along with whole grain or gluten-free breads.
Considering the impact of our dietary choices on our health is essential for maintaining a healthy lifestyle. Transitioning into the next section, let’s explore the nutritional impact on our families.
### Nutritional Impact on Family
As parents, it’s important to prioritize the nutritional well-being of our loved ones and be mindful of the impact our dietary choices have on their health. The nutritional implications of our family’s diet can have a profound effect on their overall well-being.
Research has shown that a well-balanced diet, rich in fruits, vegetables, whole grains, and lean proteins, can help prevent chronic diseases, maintain a healthy weight, and support optimal growth and development. On the other hand, a diet high in processed foods, sugar, and unhealthy fats can increase the risk of obesity, heart disease, and other health problems.
### Role in Preventing Diseases
You can play a significant role in preventing diseases by making mindful dietary choices and prioritizing the health of your loved ones. Here are four key ways in which your choices can have a positive impact on family health:
1. Opt for a balanced diet: Ensure that your family’s meals include a variety of fruits, vegetables, whole grains, lean proteins, and healthy fats. This will provide essential nutrients and help prevent chronic diseases like heart disease and diabetes.
2. Limit processed foods: Processed foods often contain high levels of unhealthy fats, added sugars, and sodium. By reducing their consumption, you can decrease the risk of obesity, high blood pressure, and other health issues.
3. Stay hydrated: Encourage your family members to drink plenty of water throughout the day. Proper hydration supports overall health and helps prevent conditions like kidney stones and urinary tract infections.
4. Practice portion control: Be mindful of serving sizes and avoid overeating. This can help maintain a healthy weight and reduce the risk of obesity-related diseases.
By making these dietary choices, you can significantly contribute to preventing diseases and safeguarding the health of your loved ones.
Now, let’s explore the importance of celebrating the bread and butter maker’s contributions.
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## Celebrating the Bread and Butter Maker’s Contributions
The family is really grateful for all the hard work the bread and butter maker has put in. In today’s society, changing gender roles have allowed individuals to pursue their passions and talents regardless of traditional stereotypes. The art of bread making, once considered a predominantly female domain, has now become a shared responsibility in many households.
As the bread and butter maker in my family, I take pride in this ancient craft that combines science, artistry, and tradition. I have spent countless hours perfecting my skills, experimenting with different recipes, and learning the intricacies of the fermentation process. From kneading the dough to shaping and baking it to golden perfection, I cherish every step of the journey.
Bread making is a labor of love, requiring patience, precision, and attention to detail. It is a skill that has been passed down through generations, and I feel honored to carry on this tradition. The satisfaction of seeing my loved ones enjoy a warm slice of freshly baked bread, slathered with homemade butter, is immeasurable.
I am grateful for the opportunity to contribute to the family’s well-being and nourishment through my bread making. It brings us together, creating a sense of warmth, comfort, and connection. The aroma of freshly baked bread fills our home, creating a welcoming atmosphere that brings joy to all. I am humbled by the appreciation and love my family has shown for my efforts, and I am grateful to be the bread and butter maker in our household.
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## Bread and Butter Making in Different Cultural Contexts
When it comes to bread and butter making, the variations across different cultures are truly fascinating. From the crusty baguettes of France to the fluffy naan of India, each culture has its own unique take on this staple food.
Additionally, the techniques used to make butter vary worldwide, with some cultures favoring churning by hand while others rely on modern machinery.
Lastly, bread and butter hold deep symbolic meaning in many cultures, representing nourishment, hospitality, and even prosperity.
Exploring these cultural nuances is not only a culinary adventure but also a window into the rich traditions and customs of different societies.
If you’re a fan of bread, you’ll love how different cultures have their own unique variations. Here are four fascinating cultural bread variations that showcase the diversity of this staple food:
1. Naan: This popular bread in Indian cuisine is traditionally cooked in a tandoor oven. It is made with yogurt, flour, and yeast, resulting in a soft and fluffy texture.
2. Baguette: A symbol of French culinary heritage, the baguette is a long, thin loaf with a crispy crust and a soft interior. It is made with just four ingredients: flour, water, yeast, and salt.
3. Pita: Commonly found in Middle Eastern cuisine, pita bread is versatile and perfect for dipping or stuffing. It is made by baking dough balls at high temperatures, causing them to puff up and create a pocket.
4. Tortilla: A staple in Mexican cuisine, tortillas are thin, flatbreads made primarily with corn or wheat flour. They can be used as a base for tacos, burritos, or enchiladas.
These cultural bread variations showcase the diversity of flavors, textures, and techniques used in breadmaking around the world.
Additionally, many cultures have unique butter making techniques that add even more depth to the bread experience.
### Butter-Making Techniques Worldwide
Explore the rich diversity of butter-making techniques around the world and discover how different cultures infuse unique flavors and textures into this beloved spread.
Butter-making techniques vary greatly from country to country, reflecting the cultural influences and traditions of each region. In India, for example, butter is made by churning yogurt, resulting in a tangy and creamy spread called ghee.
In France, the traditional method involves using raw cream and a wooden churn, creating a rich and velvety butter. In Scandinavia, butter is often made from cultured cream, giving it a slightly tangy and complex flavor.
Other countries, such as Ireland and the United States, rely on modern industrial techniques to produce butter on a large scale. Whether it’s the method of churning, the type of cream used, or the addition of herbs and spices, butter-making techniques around the world showcase the cultural diversity and culinary creativity of different societies.
Symbolism is a powerful tool used to represent various ideas and concepts, and bread and butter have long been symbolic of basic sustenance and nourishment. Here are four ways in which bread and butter symbolism is influenced by cultural influences on bread making:
1. Cultural traditions: Different cultures have their own unique bread making techniques and rituals. For example, in France, bread is often seen as a symbol of national pride and is deeply rooted in their culinary heritage. Butter, on the other hand, is considered a symbol of indulgence and luxury.
2. Social status: In many societies, bread and butter can represent social status. For instance, in ancient Egypt, only the wealthy could afford to eat bread made from fine flour and enjoy butter as a spread.
3. Religious significance: Bread and butter also hold religious significance in many cultures. In Christianity, bread is often associated with the body of Christ, and butter is seen as a symbol of richness and abundance.
4. Family and home: Bread and butter symbolism extends to the idea of family and home. The phrase ‘breadwinner’ refers to the person who provides for the family, while ‘bread and butter’ is used to describe the basic necessities of life. This reflects the importance of bread and butter as essential elements for sustaining a household.
Overall, the symbolism of bread and butter is deeply ingrained in cultural influences on bread making, reflecting traditions, social status, religion, and the concept of family and home.
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## Appreciating the Unsung Hero of the Family: The Bread and Butter Maker
Let’s take a moment to appreciate the unsung hero of the family, the bread and butter maker. In today’s society, the role of the bread and butter maker has evolved and so have the perspectives surrounding it. Generational differences have played a significant role in shaping these changing perspectives.
Traditionally, the bread and butter maker was seen as the provider, the one who brought home the bacon. This role was often taken on by the patriarch of the family, who worked tirelessly to ensure that the family had enough to eat and a roof over their heads. However, with changing times and the rise of dual-income households, the responsibility of being the bread and butter maker has become more egalitarian.
Nowadays, the bread and butter maker can be anyone in the family, irrespective of gender or age. It could be the stay-at-home dad who takes care of the household chores and prepares meals, or the working mom who juggles a career and still manages to put food on the table. This shift in perspective reflects a more inclusive and progressive outlook on gender roles and responsibilities.
Generational differences also come into play when it comes to the bread and butter maker. Older generations may have a more traditional view of this role, while younger generations may have a more flexible and adaptable approach. With advancements in technology and changing societal norms, the bread and butter maker of today may rely on online grocery shopping, meal delivery services, and convenient recipes that cater to busy lifestyles.
Traditional roles in bread and butter making include kneading the dough, churning butter, and baking the bread. Cultural influences shape these techniques, with different regions having their own unique methods and recipes.
### How Has the Bread and Butter Maker Evolved Over Time?
Over time, the bread and butter maker has evolved due to advancements in technology and changing societal roles. New techniques in bread making have emerged, while innovations in butter production have improved the process.
### What Are Some Gender Stereotypes Associated With Bread and Butter Making?
Gender stereotypes associated with bread and butter making include the belief that it is a woman’s role. However, research shows that men can also be skilled in this area. Promoting equality in bread making challenges these stereotypes.
### What Are Some Challenges Faced by the Bread and Butter Maker?
As the bread and butter maker in the family, I face various challenges and responsibilities. From managing time to ensuring quality ingredients, it’s a constant juggling act. But seeing my loved ones enjoy my creations makes it all worth it.
### How Does Bread and Butter Making Impact Family Health?
When it comes to family nutrition, the impact of bread and butter making techniques cannot be overlooked. From kneading to spreading, each step contributes to the health and well-being of our loved ones.
## Conclusion
In conclusion, as the bread and butter maker in my family, I take pride in the role I play in providing nourishment and sustenance for my loved ones.
While some may argue that this responsibility is outdated or mundane, I believe that it is a vital part of maintaining a strong and healthy family unit.
By creating delicious and wholesome bread and butter, I not only satisfy their hunger but also create a sense of comfort and unity.
So next time you enjoy a warm slice of bread with butter, remember the unsung hero behind it all – the bread and butter maker. | 13,613 | 71,620 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2024-22 | longest | en | 0.914751 |
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# TEXTBOOK SOLUTIONS FOR Analysis With an Introduction to Proof 4th Edition
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Chapter: Problem:
Suppose that (an) is a sequence of positive numbers. For each n ∈ ℕ, let bn = (a1 + a2 + … + an)/n. Prove that Σbn diverges to + ∞.
STEP-BY-STEP SOLUTION:
Chapter: Problem:
• Step 1 of 4
Consider the sequence is a series of positive numbers.
For each, let.
Now we have to prove that diverse to pluse infinite as follows:
• Chapter , Problem is solved.
Corresponding Textbook
Analysis With an Introduction to Proof | 4th Edition
9780131481015ISBN-13: 0131481010ISBN: Steven R LayAuthors: | 219 | 804 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2016-30 | latest | en | 0.811671 |
http://soscholar.com/domain/detail?domain_id=dd58bbb5-62a1-b4e7-1a2d-c1e2504f21dc | 1,553,493,016,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203755.18/warc/CC-MAIN-20190325051359-20190325073359-00367.warc.gz | 190,455,009 | 15,022 | Rational Curve 205 浏览 0关注
In mathematics, algebraic curves are the simplest objects of Euclidean geometry that can not be defined by linear properties. Specifically, in Euclidean geometry, a plane algebraic curve is the set (mathematics)|set of the points of the Euclidean plane whose coordinates are zero of a function|zeros of some polynomial in two variables.For example, the unit circle is an algebraic curve, being the set of zeros of the polynomial x2 + y2 − 1 Various technical considerations have led to consider that the complex number|complex zeros of a polynomial belong to the curve. Also, the notion of algebraic curve has been generalized to allow the coefficients of the defining polynomial and the coordinates of the points of the curve to belong to any field (mathematics)|field, leading to the following definition. In algebraic geometry, a plane affine algebraic curve defined over a field k is the set of points of K2 whose coordinates are zeros of some polynomial#Number of variables|bivariate polynomial ...
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主要的会议/期刊 CAGD TOG JSC ISSAC GMP CVGIP TIT DCC
Rational Curve文章数量变化趋势 | 254 | 1,107 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2019-13 | latest | en | 0.878252 |
https://www.convertunits.com/from/petagram/to/grain | 1,632,760,834,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058456.86/warc/CC-MAIN-20210927151238-20210927181238-00081.warc.gz | 732,388,858 | 12,933 | ## ››Convert petagram to grain
petagram grain
How many petagram in 1 grain? The answer is 6.479891E-17.
We assume you are converting between petagram and grain.
You can view more details on each measurement unit:
petagram or grain
The SI base unit for mass is the kilogram.
1 kilogram is equal to 1.0E-12 petagram, or 15432.358352941 grain.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between petagrams and grains.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of petagram to grain
1 petagram to grain = 1.5432358352941E+16 grain
2 petagram to grain = 3.0864716705883E+16 grain
3 petagram to grain = 4.6297075058824E+16 grain
4 petagram to grain = 6.1729433411766E+16 grain
5 petagram to grain = 7.7161791764707E+16 grain
6 petagram to grain = 9.2594150117649E+16 grain
7 petagram to grain = 1.0802650847059E+17 grain
8 petagram to grain = 1.2345886682353E+17 grain
9 petagram to grain = 1.3889122517647E+17 grain
10 petagram to grain = 1.5432358352941E+17 grain
## ››Want other units?
You can do the reverse unit conversion from grain to petagram, or enter any two units below:
## Enter two units to convert
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## ››Definition: Petagram
The SI prefix "peta" represents a factor of 1015, or in exponential notation, 1E15.
So 1 petagram = 1015 grams-force.
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 538 | 1,870 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2021-39 | longest | en | 0.777635 |
https://h2g2.com/edited_entry/A23428541 | 1,624,505,262,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488550571.96/warc/CC-MAIN-20210624015641-20210624045641-00158.warc.gz | 269,865,274 | 10,377 | Cycling into the Wind
The headwind is the natural enemy of the cyclist. However, it will also be an unavoidable fact of your cycling life and you might as well learn to cope. Think of the wind as a bully - do you recall how a larger kid could simply put their hand on your forehead and keep you at arm's length while you punched away at the empty space between you with tears of frustration streaming down your cheeks? Cycling into the wind is often like that, but it needn't always be. Let's start with some basics.
The Wind is Not your Friend
A friend will let you win sometimes. The wind never will. Consider this fact - there is no aerodynamic difference between a 35mph wind blowing past a stationary cyclist, and a cyclist sprinting along at 35mph in calm air. Over 100 years ago this was a matter of some dispute among early aerodynamicists including the Wright Brothers, but it has been settled - the two bicycle-mechanics were correct1. So as soon as you begin to move forward, which is really the point of cycling after all, you begin to experience a headwind. If you're in really, really good shape, you might be able to achieve a speed of 45mph, but it's the consequent 45mph headwind you are generating that will ultimately defeat you. One obvious exception would be if there was a tailwind, which is a very big if.
The Wind is a Vector
Based on its direction, wind speed can be added or subtracted. Here's how it works. Say that in each of the following cases you are cycling along at 20mph. If the wind is not blowing at all, you will feel an apparent 20mph headwind. If a 20mph wind is blowing directly against you, you will feel an apparent 40mph headwind. If a 20mph wind is blowing from directly behind, you would feel no wind at all (but of course this scenario has never occurred; it is merely a theoretical possibility). Crosswinds help or hinder to the extent that their direction is aligned with yours. A 20mph wind directly at your side will not slow you down at all, unless of course it causes you to fall over. Now, if you are travelling north, say, then a 20mph wind from the north-east will feel like a combined 14mph apparent increase to your headwind and a 14mph side-wind that merely tries to knock you over. This doesn't seem fair because if the wind is divided equally like that, it ought to be that only half (10mph) is in your face and the other half is at your side - and it would be if the wind was your friend, but it's not - it's a vector.
Then Where is it Pointing?
When you're riding along and the wind seems always in your face anyway, how can you tell what direction it is actually blowing? This may seem like a stupid question, but if your trip is more than even a few miles, you will have plenty of time to ponder such things. Here is a subtle clue to look for; it's one that pilots have used. Notice the trees - obviously if there is a gale, then they'll be leaning hard and so will you - however, the more practical thing to notice about them is their leaves. The wind always pushes against them so that the backs of the leaves, which are lighter in colour, are windward and the tops, which are darker, are leeward. This is a difference that you can discern from a pretty fair distance. So, look around in all directions and where you see the dark-coloured leaves is where the wind is coming from; where you see the light-coloured leaves is the opposite direction to where the wind is heading. (This is also the direction you would like to ride in if at all possible.) The problem is in the winter when there are no leaves, but at that time of year your ride will not be a long or leisurely one anyway and you'll be be thinking of other more immediate things to do with survival.
So What Can You Do?
Why not? It could work. The key is to be wind-aware. Winds tend to be calmer in the early and later hours of the day. Winds tend to blow in from large bodies of water in the afternoon and blow back out in the morning. Winds roar through canyons and ravines all the time. Windswept plains are just that. In the spring and autumn winds will tend to be more changeable, and will be more constant in the summer. Don't worry; as you cycle more your awareness will naturally increase and you'll find yourself making choices that result in fewer beatings.
Find an Ally
So, the wind is a bully. Then who or what can help you? Slipstreaming may be an option. If you are riding with someone, you can quietly steer in just behind them and allow them to take the brunt of the wind. This is rather a win-lose situation that is inherently unfair - a fact sure to be noticed by your companion sooner or later.
Riding a tandem bike, with another person of course, is actually a much better option as it is win-win. From your point of view, you have nearly twice the pedal-power and from the wind's point of view, there is only one of you. It does require that you get along well with your co-rider as you can't just ride away in a huff and expect to leave them behind.
Hunker Down
If there's only you, then you'll just have to tough it out. Try to make yourself as small as possible. On most bikes, this will mean keeping your knees and elbows in close while leaning over so that your forehead is touching the handlebars. It's not particularly comfortable.
Some good news then, hunkering-down-wise, is that stretching out on a recumbent bike gives a much lower profile with respect to the wind than folding up like a bat and is also much more comfortable.
Sailboats can manage in a full-on headwind by tacking (also known as 'zigzagging'), but they aren't constrained to a narrow strip of pavement as you are. This alarming tactic is to be avoided. Speaking of sailboats, if wearing a raincoat you can stop pedalling, lean back, outstretch the arms and make a sail that will (literally) foil the wind. Even though it will actually work, this hazardous manoeuvre should be avoided too if you are riding in a group, or in traffic - the reasons should be obvious.
No, in case of a headwind, you want to change your direction not just slightly, but completely. This involves having a Plan B with a different destination altogether and in the opposite direction of Plan A. For a day trip, this just means having considered the options of cycling out to a beach for a nice swim and lunch, or cycling inland to a pond, lake or river for the same. If your trip will be longer by days, weeks or months, then this option may not be for you and you'll just have to hunker down.
Treat it as an Uphill Climb
Too many people have fallen by the wayside with injured knees because they thought they could fight the wind. Don't be one of these. Cycling against the wind is really no different than cycling against gravity. Mountains are implacable, and so is the wind. Shift down to a lower gear and keep your rpm's up. You'll go slower, but you'll go the distance with knees intact.
Get Into It
Consider all the benefits of having a great big wind roaring past.
Ok, here's one: flies can't tolerate high winds so there won't be many of those about. Consider too, that there is actually no such thing as a constant wind. If you really pay attention you will notice brief lulls, helpful gusts, and slight changes of direction and you will come to appreciate these. Also, use visualisation techniques to motivate yourself - athletes do this all the time. Think of yourself as a salmon swimming upstream, there to spawn briefly and then die of exhaustion.
Admittedly that seems like a stupid visualisation, but if your trip is more than even a few miles, you will have plenty of time to ponder such things. You'll think of something.
1There's glory for you. | 1,731 | 7,693 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2021-25 | longest | en | 0.975708 |
https://solvedlib.com/n/to-calibrate-standard-curve-for-assaying-protein-concentrations,9772563 | 1,685,432,553,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224645417.33/warc/CC-MAIN-20230530063958-20230530093958-00634.warc.gz | 612,397,075 | 24,902 | # To calibrate standard curve for assaying protein concentrations plant pathologist used spectrophotometer measure the absombance of light (wavelength 500 nm)
###### Question:
To calibrate standard curve for assaying protein concentrations plant pathologist used spectrophotometer measure the absombance of light (wavelength 500 nm) by protein solution_ The results of 27 replicate assays of standard solution containing 60 ug protein per ml water were as follows: 0.111 0.115 0.115 0.110 0.099 0.121 0.107 0.107 0.100 0.110 0.106 0.116 0.098 0.116 0.108 0.098 0.120 0.123 0.124 0.122 0.116 0.130 0.114 0.100 0.123 0.119 0.107 Detemine the mean; the median; and the mode_ Calculate the range, Ine variance, the standard deviation, and the coeficient of varation. Detemine the values of the three quartiles and the interquartile range Calculate the values of the 3r decile, the 45t percentile; and the 95* percentile. Find the percentile rank of 75. Give brief description of this percentile rank Calculate the skewness and kurtosis_ Prepare box-and-whisker plot
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##### Name:Unit 3 Test Gas Laws Part 2Date:/10/10Total:120Short Answer Problem Solving 1. Aluminum metal reacts with hydrochloric acid HCl to produce aluminum chloride and hydrogen gas 2 Alts) HClaq) 2 AICIs (aq) 3 Hztg) How many grams of Aluminum metal must be used with excess HCl to produce 16.8 L of hydrogen gas? Assume STP conditions [T 14]
Name: Unit 3 Test Gas Laws Part 2 Date: /10 /10 Total: 120 Short Answer Problem Solving 1. Aluminum metal reacts with hydrochloric acid HCl to produce aluminum chloride and hydrogen gas 2 Alts) HClaq) 2 AICIs (aq) 3 Hztg) How many grams of Aluminum metal must be used with excess HCl to produce 16.8 ... | 4,202 | 15,099 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2023-23 | latest | en | 0.743788 |
http://community.hpe.com/t5/Application-Perf-Mgmt-BAC-BSM/SLA-Time-Based-Calculation-Month-vs-Quarter/m-p/6334429/highlight/true | 1,472,545,275,000,000,000 | text/html | crawl-data/CC-MAIN-2016-36/segments/1471982974951.92/warc/CC-MAIN-20160823200934-00187-ip-10-153-172-175.ec2.internal.warc.gz | 59,226,157 | 33,635 | Application Perf Mgmt (BAC / BSM) Practitioners Forum
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# SLA Time Based Calculation - Month vs. Quarter
[ Edited ]
Hi All,
Maybe I am missing the point but is it better to ask.
One of our time based SLA shows following situation (same for all SLAs, this is only one of them):
Availability Month October:
1) 99,967%
2) 99,944%
Group Average: 99,956% - correct
Availability Month November:
1) 100%
2) 99,519%
Group Average: 99,760% - correct
Availability Month December:
1) 100
2) 100
Group Average: 100 - correct
Average Availability Quarter Oct-Dec (CI Summary/CI Status):
1) 100%
2) 99,998%
Group Average: 99,999% - seems to be correct but...
...how can we have an average availability for 1) of 100% if the 3 Months were. 99,967%, 100% and 100%?
Same for 2). How can be the average availability 99,998% if we have for each month 99,944%, 99,519% and 100% ?
Changing the calculation method to sample based, we have a value for the average quarter availability that is near to the truth:
1) 99,989%
2) 99,827%
Group Average: 99,908%
Here a screenshot (SLA V2 is Time Based - SLA-V3 is exactly the same but Sample Based).
Is there an explanation?
Thanks for the help!
4 REPLIES
## Re: SLA Time Based Calculation - Month vs. Quarter
[ Edited ]
Here the screenshot again, in the post is too small. Thanks.
Highlighted
HPE Expert
## Re: SLA Time Based Calculation - Month vs. Quarter
it is not easy to answer w/o looking at the system.
I also see that you use custom Calendar, which can have impact on the calculation.
The 1st point to check is why 1 location gives 100% availability in time-based and 99.x% in the sample based. make sure both of your SLAs start at the same time and you recalculate them from the same time.
## Re: SLA Time Based Calculation - Month vs. Quarter
To check the start point and the recalculation point of the SLA was a good idea.
The V3 SLA (Sample Based) was a Clone but started not at the same time as the V2 SLA (Time Based).
What I did:
Cloned the V3 SLA (Sample Based) into V4 SLA (Time Based) with the same starting point and recalculate them.
V2 SLA - Time based - old SLA
V3 SLA - Sample based - new sla starting from a later date
V4 SLA - Time based - new sla starting from the same date of V3
Results:
The monthly CI reports of V4 SLA displayed other values for the monthly availability then the V2 SLA. The quarter report of the time based V4 SLA was also correct.
Some hours before the clone and recalc., I installed the patch for BSM 9.23 and I do not know is this solved some SLA issues of BSM 9.22
All SLAs (V2, V3 and V4) had the same amount of samples (before and after the installation of the patch)
The final check was to recalculate the V2 SLA after the patch installation and it suddenly showed the same values of V4 SLA. So maybe the patch solved really some SLA issues of BSM 9.22
Thanks again and kind regards!
## Re: SLA Time Based Calculation - Month vs. Quarter
Hi ,
I am also facing the same issue where in if I run the CI summary report for each day for same calender I get 100% SLA for each day whereas if I run the same report for month , it is showing 97% SLA.
Could anyone please let me know what can be the issue for this
Regards,
Ravi | 889 | 3,279 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2016-36 | latest | en | 0.892694 |
http://www.actuarialoutpost.com/actuarial_discussion_forum/showthread.php?s=8b25755229e878bf5ca24b4e8a9b1a16&p=9203673 | 1,531,820,709,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589634.23/warc/CC-MAIN-20180717090227-20180717110227-00551.warc.gz | 398,873,728 | 9,589 | Actuarial Outpost SOA practice problems, Question 44
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#1
01-03-2018, 11:23 AM
colt484 SOA Join Date: Aug 2014 Posts: 14
SOA practice problems, Question 44
You are given:
(i) Losses follow an exponential distribution with mean (Theta.)
(ii) A random sample of 20 losses is distributed as follows:
Loss Range Frequency
[0, 1000] 7
(1000, 2000] 6
(2000, (Infinity) 7
Calculate the maximum likelihood estimate of (Theta.)
When I work this out, I get Theta = -1000ln(20/33). Looking at the solutions for the SOA this is the same answer, however, I evaluate this at 500. This is not listed as the correct answer and is way outside the range of possible answers. Have I missed the mark here?
#2
01-03-2018, 11:40 AM
daaaave David Revelle Join Date: Feb 2006 Posts: 2,968
I don't have the current SOA solutions handy, but the answer is -1000/ln(20/33), not -1000*ln(20/33).
__________________
#3
01-03-2018, 11:55 AM
colt484 SOA Join Date: Aug 2014 Posts: 14
Quote:
Originally Posted by daaaave I don't have the current SOA solutions handy, but the answer is -1000/ln(20/33), not -1000*ln(20/33).
Algebra Mistake. Thank you. | 461 | 1,499 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2018-30 | longest | en | 0.849087 |
https://howmanykit.com/units/convert-153-5-cm-to-inches | 1,701,363,846,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100229.44/warc/CC-MAIN-20231130161920-20231130191920-00200.warc.gz | 362,860,822 | 19,419 | # Convert 153.5 CM to Inches
## How many inches is in a centimeter?
Do you are looking to convert 153.5 centimeters to a number in inches? first, you should be aware of how many inches 1 cm is equal to.
This is how I will give you a direct indication that one centimeter is equivalent to 0.3937 inches.
## Centimeter
A centimeter is a common unit of length in the metric system.
It equals to 0.01 meter.
This unit is used in CGS system, maps, home repaire and all areas in our life.
A single centimeter is roughly equivalent to 39.37 inches.
## Definition of Inch
The inch is a unit of length in the UK and the US customary systems of measurement. An inch is equal to 1/12 of a foot or 1/36 yard.
## How do I convert 1 cm to inches?
To convert inches from 1cm, multiply 1cm times 0.3937.
This makes it much easier to convert 153.5 cm to inches.
So 1 cm to inches = 1 times 0.3937 = 0.3937 inches.
This allows you to answer this question with ease and simplicity.
• What is one centimeter into inches?
• What is the cm to inch conversion?
• How many inches is equal to 1 cm?
• What does 1 cm equal in inches?
### What is 153.5 cm converted to inches?
You have fully understood cm to inches by the above.
This is the formula:
Value in inches = value in cm × 0.3937
So, 153.5 cm to inches = 153.5 cm × 0.3937 = 60.43295 inches
This formula can be used to answer the related questions:
• What’s the formula to convert inches from 153.5 cm?
• How do I convert cm to inches?
• How do you change cm to inches?
• What is cm to inch ratio?
• What size are 153.5 cm into inches?
cm inches 152.7 cm 60.11799 inches 152.8 cm 60.15736 inches 152.9 cm 60.19673 inches 153 cm 60.2361 inches 153.1 cm 60.27547 inches 153.2 cm 60.31484 inches 153.3 cm 60.35421 inches 153.4 cm 60.39358 inches 153.5 cm 60.43295 inches 153.6 cm 60.47232 inches 153.7 cm 60.51169 inches 153.8 cm 60.55106 inches 153.9 cm 60.59043 inches 154 cm 60.6298 inches 154.1 cm 60.66917 inches 154.2 cm 60.70854 inches 154.3 cm 60.74791 inches | 623 | 2,019 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2023-50 | latest | en | 0.861039 |
http://casuallivingresourceguide.com/molar-mass-of-butane-gas.html | 1,561,625,410,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628001014.85/warc/CC-MAIN-20190627075525-20190627101525-00324.warc.gz | 30,373,107 | 3,953 | # Molar mass of butane gas. Lab: Molar Mass of Butane 2019-01-31
Molar mass of butane gas Rating: 6,1/10 228 reviews
## Molar Mass of Butane
Any help would be appreciated. Three moles of H2 per one mole of N2. We recorded the initial volume of water, then equalized the pressure inside and outside the beaker until the water levels inside and outside the beaker were the same. What effect would this have on the experimental molar mass? I just need to know how to set up each. The gas is an alkane contains only carbon and hydrogen with single bonds.
Next
## Butane Lab
Under identical conditions, you have 12. What conclusion can you draw from these data? Record the volume of Butane in cylinder. . Wet lighter would have affected the weight of the lighter. The molarity of the solution is 0. As the stopper was removed, the lighter was held under the cylinder.
Next
## Molecular weight of Butane
By adding a fractionating column between the boiling flask and the condenser… 2105 Words 9 Pages Abstract: This experiment utilizes skills of titration in order to find the percent mass of actual aspirin acetylsalicylic acid in store-bought aspirin. Some research has shown me that ethylene glycol is an electrolyte, but I don't know how it dissociates in water. The mass of the disposable lighter was weighed and a tub was filled with water. The measurements in the lab are reliable taking into consideration the equipment windows of error and there are few to no flaws and weaknesses in the procedure itself. The mass of the gas was found to be 0. We concluded that the molar mass of butane can be determined through the average of repeated trials of the experiment were no systematic or random errors occur.
Next
## what is the molar mass of Butane,C7H10?
Determine the molar mass the formula and the name of the monomer. The temperature of the snow is initially at -20 °C, and we need 80 °C water to brew tea. Another systematic error was that if the beaker was even slightly cracked at the top, allowing air to enter the beaker, the value of molar mass would change because the gas collected at the top of the beaker would no longer be strictly butane. Page 9 6 — Bibliography …………………………………………. The shift from ice to water solid to a liquid is called the heat of fusion.
Next
## Determining the Molar Mass of Butane
Data Weight of lighter before releasing gas- 17. The volume of the glass bottle was 262 mL and the volume of water after adding gas mixture was 187 mL. Please read for more information about how you can control adserving and the information collected. If the molar mass of oxygen is 32. Repetition does not solve systematic errors, and the value is referred to as a bias or the deviation from the true value.
Next
## Lab: Molar Mass of Butane
The butane gas was aired out while the lighter dried. Using the chemical formula of the compound and the periodic table of elements, we can add up the atomic weights and calculate molecular weight of the substance. What is the molar mass of the gas, based on your calculations? Walter Snellings in Pittsburg and he gas is used for cigarette lighters, heaters, stove fuels, and other heating appliances. Press the lighter so that the Butane goes through the tube and into the cylinder. The rest of the molar mass of the mixture was fitted by molar mass of hydrogen. What is the minimum possible molar mass of hemoglobin? Then we recorded the final volume of water, and repeated the experiment until we had three trials.
Next
## Molecular Weight
Add the Engineering ToolBox extension to your SketchUp from the Sketchup Extension Warehouse! Also, what is the mass in grams of 2. The equivalence point is reached after adding 12. I found individual molar mass of: Ca-40. The calculated molar mass of the gas should have the average molar mass of the entire gas, including butane. The temperature of the freezing point was lowered when the solid was added and it decreased as the concentration of the solid increased. Remove the tube from the cylinder. A reasonably accurate atmospheric pressure value can be obtained using a google search of atmospheric pressures in your area or by using a phone app.
Next
## Molecular weight of Butane
To find the mass of the gas mixture, the lighter's mass was weighted for both before and after releasing of the gas. Using the molar mass, combinations of carbon and hydrogen were made to propose plausible gaseous compounds in the lighter. Determine percentage error for your data. This site explains how to find molar mass. Only emails and answers are saved in our archive. More number of moles results in lower molar mass number, because mole is on the denominator side.
Next | 1,090 | 4,735 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2019-26 | latest | en | 0.952613 |
https://teacherscoutmom.com/2017/08/20/multiplication-strategies-math-anchor-chart-kahoot-and-game/ | 1,685,625,312,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224647810.28/warc/CC-MAIN-20230601110845-20230601140845-00395.warc.gz | 622,843,157 | 41,803 | Posted in Kahoots, Math Anchor Charts, Math is Awesome!
# Multiplication Strategies – Math Anchor Chart, Kahoot and Game
Last year I noticed that I had several of my 5th grade students struggling with multiplication facts. It wasn’t just that they didn’t know them, but several of them couldn’t even skip count. Some who were still drawing out dots and groups to count, would get so tied up with figuring out what 8 x 7 was that they couldn’t remember what step they were on in long division or fraction computation. It was getting to be a little crazy.
I started researching strategies and found this handy anchor chart. (This year, it was the first anchor chart hung in my room!) I tried researching where the original idea came from but there were at least 25 variations on Pinterest and the web. I took out some of the facts that I felt my kids should know like 0s, 1s, and 10s and left the rest. I’m hoping that if I get this out early enough this year and teach from it, that it might help those students who are struggling. Chart Paper and Sharpie Markers are the best- I usually laminate all of my charts so it is super easy to pull out each year.
I also found this great little Kahoot called Multiplciation Facts by Jordan Manning that I love to use. I started using that about midyear last year as a warm up to get our brains thinking about math. I found that the kids loved competing against each other and for those who kept missing the same facts, I had them make flash cards on index cards to help them study. I plan on starting this a lot earlier in the year than last.
If you’ve never used a Kahoot before check out my Kahoot Introduction. I have several with different topics that I made under menu, Kahoots or just click here.
Another game, I like to use to help with fact fluency, is a game called Multiplication War. The only item needed is a deck of cards. (Amazon has 12 packs and 2 packs) In this game, students deal out the entire deck of cards, then turn over two cards. Each player multiplies the two cards together while the highest product wins the cards. In case of a tie, players flip again – this time winner takes all the cards turned over. Aces are worth 1, Jokers are 11, and Jacks, Queens and Kings are 10. The students use the cards that they won to continue playing until one person has all the cards or time is up. This year, I played it the first week of school and gave my students a multiplication table to look up the facts if they didn’t know them. As the year progress, I plan to slowly remove that handy chart.
http://www.multiplication.com also has some amazing games that the students love to play!
Hopefully, starting all of these games and strategies at the beginning of the year will help solidify some of those crazy facts that the kids need to survive secondary math! | 672 | 2,850 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2023-23 | latest | en | 0.974042 |
https://www.spanning-boundaries.eu/forums/topic/torah-and-gematria-3/ | 1,685,854,294,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649439.65/warc/CC-MAIN-20230604025306-20230604055306-00608.warc.gz | 1,106,059,087 | 21,754 | Home Forums BB Form Torah and Gematria
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However, gematria is essential to Kabbalah, the Jewish mystical tradition. The very basis of the kabbalistic cosmological system rests on the belief that God created the universe through the power of the Hebrew letters along with their numerical values. Indeed the many names of God and their permutations in Kabbalah have numerical values that are believed to contain potent power. The Babylonians preceded the Hebrew practice with a form of gematria that used tables with a logographic writing script. The gematria calculator is a more interactive way to find the numerical values in the most famous Gematria against the alphabet provided. Use or study of hidden meanings through numbers, especially the numerical equivalence of letters. Gematria was much employed by the kabbalists and was used to prove the messiahship of Shabbetai Zevi. Despite criticism, its use was widespread in both Sephardi and Ashkenazi circles. Gematria is the process by which numerical values are assigned to the letters of the Hebrew alphabet. And so many gematrias in widespread use are as trite as that joke. When you have any questions concerning in which and the way to use homepage, you are able to email us with the site. Over the centuries, scholars have created numerous sophisticated systems of gematria for interpreting Jewish texts and traditions. For example, the numerical value of each letter in a word may be deciphered separately to indicate something explicit. And each letter in a word really has a hidden meaning or secret behind its mathematical quantity. Many individuals believe there are an unlimited number of secrets in the Torah that can be unlocked utilizing gematria. The first numerical cipher that is known to be assigned to the English Alphabet was by Cornelius Agrippa in 1533, in his work De Occulta Philosopha. Agrippa gave value to the English letters without trying to transliterate them from Hebrew or Greek, so L is 20, rather than 30 , M is 30, rather than 40 and N is 40, rather than 50 . This cipher is sometimes erroneously labelled as “Jewish” or “Hebrew” by popular numerology calculators, such as Gematrix and Gematrinator. Mispar ha-Akhor – The value of each letter is its standard value multiplied by the position of the letter in a word or a phrase in either ascending or descending order. This method is particularly interesting, because the result is sensitive to the order of letters. It assigns the values 1–9, 10–90, 100–400 to the 22 Hebrew letters in order.
Viewing 1 post (of 1 total) | 589 | 2,648 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2023-23 | longest | en | 0.942203 |
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