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# 4_21_09_BayesianInference - Computations performed by...
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Computations performed by neural networks can be expressed as energy minimization.
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A link exists between energy minimization and Bayesian processes (and therefore, between these things and neural networks). P V ( ) = e −β E V ( ) e −β E V ( ) V If a system can be in a finite number of states, the transition between them tends to minimize energy (E( V )) except for noise, and the transition probabilities depend only on the current state, then in steady state The Gibbs distribution
Neural Response ( r ) Sensory Input (s) a b c Bayes Theorem P s | r ( ) = P ( r | s ) P s ( ) P r ( ) b a + b = b b + c b + c a + b + c a + b a + b + c P r ( ) = a + b a + b + c P s ( ) = b + c a + b + c P r | s ( ) = b b + c b a + b P s | r ( ) =
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Response ( r ) Sensory Input (s) a b c Bayes Theorem Hence, one can often replace a network computation by a Bayesian one, thus understand- ing the computation and the noise. P s
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## This note was uploaded on 06/08/2009 for the course BME 575L taught by Professor Grzywacz during the Spring '09 term at USC.
### Page1 / 26
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Ask a homework question - tutors are online | 473 | 1,794 | {"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} | 2.796875 | 3 | CC-MAIN-2017-17 | longest | en | 0.889496 |
https://www.reference.com/hobbies-games/tips-playing-coolmath-game-bloxorz-abe54015b1be9541 | 1,487,648,698,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501170624.14/warc/CC-MAIN-20170219104610-00117-ip-10-171-10-108.ec2.internal.warc.gz | 890,982,023 | 20,900 | Q:
# What are some tips for playing the CoolMath.com game "Bloxorz"?
A:
To win the puzzle game "Bloxorz," use the keyboard's arrow keys to move the block around the level to the square hole at the end without falling off the edges. The game has 33 stages to complete, with a different six-digit passcode for every stage that can be entered in the Load Stage field on the main menu to return to a previous stage.
## Keep Learning
To navigate the levels, use switches and bridges. The octagonal switches can be activated when touched by any part of the block, while the x-shaped switches require the block's entire weight standing on end to be tripped. A third type of switch, shaped like a pair of brackets, splits the block into two pieces that can be independently controlled, using the spacebar to alternate between them. When placed adjacent to one another, the pieces recombine into the whole block. Blocks must always be made whole to exit and complete a stage. Be careful of orange tiles, which break after being crossed, preventing backtracking.
Bloxorz is available to play for free on multiple websites as an Adobe Flash program. If the game does not load, check the settings of the browser's Flash player plugin, update the Flash installation or try using a different browser.
Sources:
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PEOPLE SEARCH FOR | 461 | 2,107 | {"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} | 2.890625 | 3 | CC-MAIN-2017-09 | longest | en | 0.894302 |
http://www.leadinglesson.com/algebra-of-dot-products | 1,701,293,621,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100146.5/warc/CC-MAIN-20231129204528-20231129234528-00195.warc.gz | 78,104,916 | 3,943 | Study guide and 6 practice problems on:
## Algebra of dot products
Dot products are distributive: $$\mathbf{x} \cdot ( \mathbf{y} + \mathbf{z}) = \mathbf{x} \cdot \mathbf{y} + \mathbf{x} \cdot \mathbf{z}$$
As a result, the FOIL rule applies: $$(\mathbf{w} + \mathbf{x})\cdot(\mathbf{y} + \mathbf{z}) = \mathbf{w}\cdot \mathbf{y} + \mathbf{w}\cdot \mathbf{z} + \mathbf{x} \cdot \mathbf{y} + \mathbf{x} \cdot \mathbf{z}$$
Dot products are commutative: $$\mathbf{x}\cdot \mathbf{y} = \mathbf{y}\cdot \mathbf{x}$$ | 205 | 511 | {"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} | 3.59375 | 4 | CC-MAIN-2023-50 | latest | en | 0.325538 |
https://www.jiskha.com/questions/1598279/A-winding-drum-raises-a-cage-through-a-height-of-120-m-the-cage-has-at-first-an | 1,539,796,982,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583511203.16/warc/CC-MAIN-20181017153433-20181017174933-00464.warc.gz | 966,058,551 | 4,898 | # Theory of machine
A winding drum raises a cage through a height of 120 m. the cage has, at first, an accelerationof 1.5 m/s^2 until the velocity of 9 m/s is reached, after which the velocity is constant until the cage nears the top, when the final retardation is 6 m/s^2. find the time taken for the cage to reach the top.
1. 0
## Similar Questions
1. ### theory of machines
A winding drum raises a cage through a height of 120 m. the cage has, at first, an accelerationof 1.5 m/s^2 until the velocity of 9 m/s is reached, after which the velocity is constant until the cage nears the top, when the final
2. ### Physics
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A cage descends a mine shaft with an acceleration of 0.5 m/s2. After the cage has travelled 25 metres, a stone is dropped from the top of the shaft. Determine : 1. the time taken by the stone to hit the cage, and 2. distance
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6. ### Math
need some help on an algebra question. First we had to put the different combinations of 6 rabbits in 2 cages. Like 2 rabbits in the red cage and 4 rabbits in the blue cage...etc. Then it says to write an equation using variables | 486 | 1,813 | {"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} | 3.734375 | 4 | CC-MAIN-2018-43 | latest | en | 0.898555 |
https://ell.stackexchange.com/questions/160989/how-can-i-intuitively-know-that-the-in-in-in-3-days-means-after | 1,656,444,313,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103573995.30/warc/CC-MAIN-20220628173131-20220628203131-00638.warc.gz | 296,001,202 | 63,348 | # How can I intuitively know that the "in" in "in 3 days" means "after"?
I learn from answers of this and this question that:
"in 3 days" implies after 3 days or approximately after 3 days.
It seems very counterintuitive, to a Chinese at least. How can I intuit that the "in" in "in 3 days", which seems equivalent to "*_day"(indicating "first_day" or "second_day" or "third_day") in ["first_day", "second_day", "third_day"], means (approximately) "not in"(precisely "after") that set?
This question may seem naive and trivial to most of the native speakers, but maybe someone(especially those kids who are learning English as a foreign language like me years ago) also wants to raise this kind of questions.
• To clarify what you are asking: is your question about, whether 'in 3 days' said on Sunday means Wednesday or Thursday? Or is it about something else? Mar 20, 2018 at 10:09
• Without context you can't. Is the statement "x will happen in 3 days" or "y can be done in 3 days"? Mar 20, 2018 at 10:29
• @charmer In this question I mean the case of "x will happen in 3 days". Mar 20, 2018 at 10:43
• In which case I'd expect x to happen at approx 3 days from the statement being made, However it may also partly depend on when in the day this is said. Said late on Sunday, I could assume late Wednesday - but not be surprised if early Thursday is when 'x' happens (don't count Sunday). Said early on Sunday morning I might assume to count Sunday in the three days and expect end of day Tuesday or Wednesday morning. Without someone stating an exact day and time, it's open to some flexibility on meaning and interpretation. Mar 20, 2018 at 11:04
• On re-reading, I see @laugh has actually made my point, .. :) Mar 21, 2018 at 0:31
"in 3 days" implies after 3 days or approximately after 3 days
... is not true in general. "In" is a proposition that usually implies something being included (in a place, time, group, etc.). "In 3 days" intuitively means something like "within a period of 3 days" to English speakers. But there are subtleties...
The context may fix this period to start from the present time, and the event may be a single instant within the time, but it's not always the case. For example: "this car goes from 0 to 55 in 10 seconds'. It is a general statement (not from now), and it takes the whole 10 seconds.
How much context is needed depends on the case. If the context does not clarify what "in" means, then the meaning may be ambiguous. Natural languages do not rule out ambiguity! If you hear an ambiguous statement you may ask for clarification. If you make one, you will likely be asked for clarification.
You can use other prepositions like "within", "after", "during", "by" etc. to reduce ambiguity.
As a native speaker, this is interesting - a use of in I didn't even think about until now.
3 days is a limit. You are going to do something before that limit expires. Therefore you are doing something in the limit.
The fact that you won't do anything on day 1 or 2 is implied.
If you want to make it clear you mean that you can call on day 1 or 2, within would be used.
The bad news is that prepositions are idiosyncratic, and their uses cannot be intuited confidently. Like irregular verbs or spelling, they are learned through practice and exposure, further complicated because their usage differs slightly by locale (e,g, New York, standing on line vs. in line in the rest of the U.S.) and also shifts over time.
They do not translate, either. In French you would also say, for instance, en deux jours, but en cannot be directly translated as in; French employs en where English uses like, at, or by. Similarly, English by is not equivalent to German bei, as some English by expressions would use von or mit in German, and some bei uses are expressed using at or with in English.
As your second linked question notes, the intended meaning of in 3 days would depend on context, and you would use additional modifiers or express the idea differently if you needed to be more precise. The inspector is coming in the next three days, for instance, means the inspector should be expected at any time between now and three days from now, whereas The inspector is coming in three days at the earliest of course means just the opposite, that the inspector should not be expected until at least three days from now.
• The von or mit cases can usually be rendered as with in English, so that's not really a point at all. The issue is more that where we used to say in time to mean during [this|that] time, it's warped a little to sometimes mean after [this|that] time. I realise of course that the French might say pendant que ... Mar 21, 2018 at 0:34
• … but the point stands that doing [X] in three days can be “intuited” to some degree to mean during the next three days [someone will be] doing [X], and it will be finished [hopefully] after those three days. Mar 21, 2018 at 0:38
• @WillCrawford The point is that prepositions do not correspond across languages. Even a single example of an expression employing a different preposition (or a different construction altogether) from a frequently corresponding preposition in another language makes the point; I was just trying to think of examples off the top of my head, and my German is less terrible than my French. Mar 21, 2018 at 0:45
• Fair enough. I know that pain (I get utterly stuck with some prepositions, and I find German difficult just because most of them sound completely different, ... but bei/mit/von are the few I remember! :o)) Mar 21, 2018 at 0:49 | 1,344 | 5,567 | {"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 | longest | en | 0.963827 |
https://emilylearning.com/vector-product-of-vectors/ | 1,718,519,754,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861643.92/warc/CC-MAIN-20240616043719-20240616073719-00865.warc.gz | 198,840,592 | 16,467 | # Vector Product of Vectors
In this lesson on vectors, we’ll be talking about vector product or cross product of vectors.
## How to denote vector or cross product in Math
In Math, we denote vector product with a ✕, hence vector products are also called cross product.
## The result of vector product
When we cross 2 vectors, we are finding a vector that is perpendicular to both of these vectors. Hence, the result of a cross product is a vector.
If we have vectors given in x, y or x, y, z coordinate form, we can find unit vector using the following formula. Note that this formula is provided in the formula booklet (for students taking the H2 A Level Math exam).
The video below shows how to find vector product (or cross product) of vectors in detail:
## Useful results of vector product (cross product)
1. When we take the cross product of two vectors, the result is a vector perpendicular to both of these vectors.
2. |a x b| gives the area of a parallelogram with sides a and b.
3. If 2 vectors are parallel, cross product will give a value of 0.
4. a x a = 0
## Notes on H2 Math Vectors
You’ll find all the notes on H2 A Level Math Vectors topic here.
## All the notes for H2 A Level Math
Go here to find all the notes and resources for H2 A level Math.
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# Playing pick 4 for added income
Topic closed. 527 replies. Last post 6 years ago by lakerben.
Page 33 of 36
elkton, maryland
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lol
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Base 8653, x= 7,2
Los Angeles County, Ca.
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Base 8653, x= 7,2
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Base 8013(x=8,7),, 9012 (x=4,8) for 7 days
1370 out tonite for Base 8013(x=8,7)
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Use base 8013, 1382,1302,8012, 1383 for 7 days, if 3 elements of base comes out before the time frame use base 0291 with x = 4.
Budget permiting you can split your bet this way Base 8013(x=2,7) and 0291 (x=4,7).
TEXAS
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1370 out tonite for Base 8013(x=8,7)
Any numbers for tex ? Thanks
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Any numbers for tex ? Thanks
Base 5746(X=2,3) and 6835(x=0,3) for 7 days, if on limited budget, start with first base or else split budget for the two bases. A base is when you have three elements of the pattern key coming.
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Use base 8013, 1382,1302,8012, 1383 for 7 days, if 3 elements of base comes out before the time frame use base 0291 with x = 4.
Budget permiting you can split your bet this way Base 8013(x=2,7) and 0291 (x=4,7).
Thank you. I appreciate your help.
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Thank you. I appreciate your help.
NJ:3129 came out for base 0291, play the predictions for tonite, and change to base 8013, good luck
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Folks ,lets be frank and sincere,we all play this game to earn money,not for entertaiment sake! Passion for numbers and the rush of adrelanin in anticipation of draws overwhelms when you keep loosing.It's been a little over year since I joined the LP community,and I love it.This is the best lottery site ,the members are knowledgeable to the abstract borderline,fun and should say kind and helpful.To the creator/s of this website,especially Mr Todd, you've done awesome job.
I have posted a few posts and should say a system for pick 3/4 and few of my predictions for my state VA and DC is optimal, I am into something which I want LP members to try without beting ,and if it works for your State then I will explain the most effective way to My system(Summary of Patterns,thread lotterypost.com/topic/242609 ).
The idea is to get the last 5 combined draws of pick 4(today draws included), and I will give u a winning combinations for few draws and I will explain why it should be play that way, You don't have to wage anything,I want you to observe what happens. I will take the few states on request,we need to have fun ,but we need money also !
Maryland Midday Evening
Sat 7/7 0214 3107
Fri 7/6 3152 9062
Thurs 7/5 5086 4299
Wed 7/4 1419 5355
Tues 7/3 8722 5344
Thank You!!!
Live, Laugh, Love & Be Happy!!!
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Maryland Midday Evening
Sat 7/7 0214 3107
Fri 7/6 3152 9062
Thurs 7/5 5086 4299
Wed 7/4 1419 5355
Tues 7/3 8722 5344
Thank You!!!
Base 5746(x=2,3) and 7924(x=3,6) for 7 days,, x is suggested option for the base pattern, for limited budget reduced combos to 574x,576x and 465x( 6 picks for each base)
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Base 8653, x= 7,2
Cal: 3548 out for base 8653, play prediction for one more draw and change base to 8013(x=2,4), x is suggested option, look at your State trend
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North Carolina Midday Evening
Sat 7/7 1681 7665
Fri 7/6 8093 4200
Thurs 7/5 4168 9703
Wed 7/4 6547 9865
Tues 7/3 6218 4032
Thanks.
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North Carolina Midday Evening
Sat 7/7 1681 7665
Fri 7/6 8093 4200
Thurs 7/5 4168 9703
Wed 7/4 6547 9865
Tues 7/3 6218 4032
Thanks.
2479 ( x=3,2) and 3568(x=8,9) as base for 7 days.
Page 33 of 36 | 1,949 | 5,622 | {"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-2018-09 | latest | en | 0.649782 |
https://www.projecttopics.com/questions/mathematics/find-the-median-of-591103892455573-and-6/ | 1,721,020,512,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514659.22/warc/CC-MAIN-20240715040934-20240715070934-00899.warc.gz | 825,646,722 | 20,192 | Home » Find the median of 5,9,1,10,3,8,9,2,4,5,5,5,7,3 and 6
# Find the median of 5,9,1,10,3,8,9,2,4,5,5,5,7,3 and 6
Find the median of 5,9,1,10,3,8,9,2,4,5,5,5,7,3 and 6
• A.
6
• B.
5
• C.
4
• D.
3
##### Explanation
First arrange the numbers in order of magnitude;
1,2,3,3,4,5,5,5,5,6,7,8,9,9,10
Hence the median = 5
There is an explanation video available below. | 190 | 368 | {"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} | 3.8125 | 4 | CC-MAIN-2024-30 | latest | en | 0.729109 |
https://romantonumber.com/2437-in-roman-numerals | 1,708,649,153,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473871.23/warc/CC-MAIN-20240222225655-20240223015655-00697.warc.gz | 504,138,373 | 22,357 | # 2437 in Roman Numerals
2437 = MMCDXXXVII
2 4 3 7 M C X I MM CC XX II MMM CCC XXX III CD XL IV D L V DC LX VI DCC LXX VII DCCC LXXX VIII CM XC IX
2437 is valid Arabic numeral. Here we will explain how to read, write and convert the Arabic numeral 2437 into the correct Roman numeral format. Please have a look over the Roman numeral table given below for better understanding of Roman numeral system. As you can see, each letter is associated with specific value.
Symbol Value
I1
V5
X10
L50
C100
D500
M1000
## How to write Arabic Numeral 2437 in Roman numeral?
The Roman numeral representation of Arabic numeral 2437 is MMCDXXXVII.
## How to convert Arabic numeral 2437 to Roman numeral?
If you are aware of Roman numeral system, then converting Arabic numeral 2437 to Roman numeral is very easy. Converting 2437 to Roman numeral representation involves breaking up the numeral into place values as shown below.
2437 1000 + 1000 + 400 + 10 + 10 + 10 + 5 + 1 + 1 1000 + 1000 + 500 - 100 + 10 + 10 + 10 + 5 + 1 + 1 M + M + CD + X + X + X + V + I + I MMCDXXXVII
We need to combine all converted roman numerals together. As per the rule highest numeral should always precede the lowest numeral to get correct representation. In modern world, the number 2437 should be used as it is until or unless it represents an ordinal value. In case of any ordinal value, you can use MMCDXXXVII instead of 2437. For any numeral conversion, you can also use our roman to number converter tool given above.
## Current Date and Time in Roman Numerals
The current date and time written in roman numerals is given below. Romans used the word nulla to denote zero because the roman number system did not have a zero, so there is a possibility that you might see nulla or nothing when the value is zero.
2024-02-17
10:59:02
MMXXIV-II-XVII
X:LIX:II
Disclaimer:We make a reasonable effort in making sure that conversion results are as accurate as possible, but we cannot guarantee that. Before using any details provided here, you must validate its correctness from other reliable sources on internet. | 547 | 2,095 | {"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-2024-10 | latest | en | 0.784699 |
https://pub.dev/documentation/intl/0.16.1/intl/Intl/pluralLogic.html | 1,643,123,969,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304835.96/warc/CC-MAIN-20220125130117-20220125160117-00662.warc.gz | 517,065,889 | 3,957 | # pluralLogic<T> method
T pluralLogic<T>(
1. num howMany,
2. {T zero,
3. T one,
4. T two,
5. T few,
6. T many,
7. T other,
8. String locale,
9. int precision,
10. String meaning}
)
Internal: Implements the logic for plural selection - use plural for normal messages.
## Implementation
``````static T pluralLogic<T>(num howMany,
{T zero,
T one,
T two,
T few,
T many,
T other,
String locale,
int precision,
String meaning}) {
if (other == null) {
throw ArgumentError("The 'other' named argument must be provided");
}
if (howMany == null) {
throw ArgumentError('The howMany argument to plural cannot be null');
}
// If we haven't specified precision and we have a float that is an integer
// value, turn it into an integer. This gives us the behavior that 1.0 and 1
// produce the same output, e.g. 1 dollar.
var truncated = howMany.truncate();
if (precision == null && truncated == howMany) {
howMany = truncated;
}
// This is for backward compatibility.
// We interpret the presence of [precision] parameter as an "opt-in" to
// the new behavior, since [precision] did not exist before.
// For an English example: if the precision is 2 then the formatted string
// would not map to 'one' (for example "1.00 miles")
if (precision == null || precision == 0) {
// If there's an explicit case for the exact number, we use it. This is
// not strictly in accord with the CLDR rules, but it seems to be the
// expectation. At least I see e.g. Russian translations that have a zero
// case defined. The rule for that locale will never produce a zero, and
// treats it as other. But it seems reasonable that, even if the language
// rules treat zero as other, we might want a special message for zero.
if (howMany == 0 && zero != null) return zero;
if (howMany == 1 && one != null) return one;
if (howMany == 2 && two != null) return two;
}
var pluralRule = _pluralRule(locale, howMany, precision);
var pluralCase = pluralRule();
switch (pluralCase) {
case plural_rules.PluralCase.ZERO:
return zero ?? other;
case plural_rules.PluralCase.ONE:
return one ?? other;
case plural_rules.PluralCase.TWO:
return two ?? few ?? other;
case plural_rules.PluralCase.FEW:
return few ?? other;
case plural_rules.PluralCase.MANY:
return many ?? other;
case plural_rules.PluralCase.OTHER:
return other;
default:
throw ArgumentError.value(
howMany, 'howMany', 'Invalid plural argument');
}
}`````` | 613 | 2,377 | {"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} | 2.65625 | 3 | CC-MAIN-2022-05 | latest | en | 0.675585 |
https://ezpassdrjtbc.net/regular-multiplying-decimals-worksheets-pdf/ | 1,621,328,861,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989756.81/warc/CC-MAIN-20210518063944-20210518093944-00587.warc.gz | 275,046,452 | 8,370 | # Regular multiplying decimals worksheets pdf Wonderful
» » Regular multiplying decimals worksheets pdf Wonderful
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Regular Multiplying Decimals Worksheets Pdf. 02122020 Multiplication And Division Of Decimals Worksheets Pdf. Multiplying decimal by decimals 1 or 2 decimal digits Below are six versions of our grade 5 math worksheet on multiplying two decimal numbers by each other. Use the buttons below to print open or download the PDF version of the Multiplying 2-Digit by 2-Digit Numbers with Various Decimal Places A math worksheet. 159 x 08 1272 k.
Multiplication With Decimals Worksheets Multiplication Worksheets Decimals Worksheets Math Multiplication Worksheets From pinterest.com
The size of the PDF file is 28223 bytes. Multiplying and dividing decimals worksheets pdf. It has an answer key attached on the second page. You will be multiplying by a three digit number which means there will be two rows of numbers to add up before calculating your answer. Multiplying Decimal with Decimal. Decimal Multiplication - 1.
### These worksheets are pdf files.
Multiplying and Dividing Decimals Worksheets 6th Grade PDF 6th grade multiply and divide decimals exercises with answers. 555 x 5 2775 l. 42 x 07 294 j. Our 6 th grade multiply and divide decimals exercises with answers has provided practice for simple calculations and mental real world uses. Multiplying and Dividing Decimals Worksheets 6th Grade PDF 6th grade multiply and divide decimals exercises with answers. Includes a few negative examples and multiplying decimals with whole numbers.
Source: pinterest.com
Worksheet 1 Worksheet 2 Worksheet 3 Worksheet 4 Worksheet 5 Worksheet 6. This worksheet has 10 vertical problems and 2 word problems that students can solve to practice multiplying decimals by single digit numbers. Our 6 th grade multiply and divide decimals exercises with answers has provided practice for simple calculations and mental real world uses. 88 x 08 704 Super Teacher Worksheets. Great for homework or revision.
Source: pinterest.com
This is a math PDF printable activity sheet with several exercises. The size of the PDF file is 28223 bytes. 4th through 7th Grades. 88 x 08 704 Super Teacher Worksheets. If there are more versions of this worksheet the other versions will be available below the preview images.
Source: pinterest.com
Multiplying Decimals Worksheet 4 RTF Multiplying Decimals Worksheet 4 PDF Preview Multiplying. 5th grade multiplying decimals worksheets including multiplying decimals by decimals multiplying decimals by whole numbers missing factor problems multiplying by 10 100 or 1000 and multiplication in columns with decimals. Multiplying and Dividing Decimals Worksheets 6th Grade PDF 6th grade multiply and divide decimals exercises with answers. Decimal Multiplication Rewrite each problem vertically and solve. Multiplying decimal by decimals 1 or 2 decimal digits Below are six versions of our grade 5 math worksheet on multiplying two decimal numbers by each other.
Source: pinterest.com
187 x 6 1122 c. Multiplying decimal by decimals 1 or 2 decimal digits Below are six versions of our grade 5 math worksheet on multiplying two decimal numbers by each other. Multiplying And Dividing Decimals Worksheets Pdf 5th Grade. For more like this use the search bar to look for some or all of these keywords. Multiplying and dividing decimals Add to my workbooks 2 Download file pdf Embed in my website or blog Add to Google Classroom.
Source: pinterest.com
Multiplying and dividing decimals worksheets pdf Collection. Decimal Multiplication Rewrite each problem vertically and solve. Multiplying Decimals by 1-Digit Numbers. Multiplying Decimals Worksheet 6 RTF Multiplying Decimals Worksheet 6 PDF Preview Multiplying Decimals Worksheet 6 in Your. If there are more versions of this worksheet the other versions will be available below the preview images.
Source: pinterest.com
Multiplying DecimalsBy 1-Digit Numbers. This worksheet is a supplementary seventh grade resource to help teachers parents and. It has an answer key attached on the second page. 82 x 03 246 i. Multiplying Decimals - 2.
Source: pinterest.com
A nice worksheet on multiplying decimals. 711 x 9 6399 h. Multiplying And Dividing Decimals Worksheets Pdf 5th Grade. Great for homework or revision. The problem sets are randomly generated so you can create multiple worksheets without repeating the same problemsChoose the number of digits to the left and right of the decimal and number of problems on each worksheet.
Source: pinterest.com
Includes a few negative examples and multiplying decimals with whole numbers. Type keywords and hit enter. Includes a few negative examples and multiplying decimals with whole numbers. Addition worksheets and subtraction worksheets. Multiplying and dividing decimals worksheets pdf.
Source: pinterest.com
Decimal Multiplication - 1. Multiplying Decimals Worksheet 4 RTF Multiplying Decimals Worksheet 4 PDF Preview Multiplying. 712 x 3 2136 f. When you get your answer add up the total number of digits to the right of the decimals in both the numbers you are multiplying and. Multiplying and dividing decimals worksheets pdf.
Source: pinterest.com
Worksheet 1 Worksheet 2 Worksheet 3 Worksheet 4 Worksheet 5 Worksheet 6. The size of the PDF file is 28223 bytes. The multiplying decimals worksheet maker creates horizontal or vertical format decimal multiplication worksheets that are customized to your preferences. Multiplying decimal by decimals 1 or 2 decimal digits Below are six versions of our grade 5 math worksheet on multiplying two decimal numbers by each other. Multiplying Decimals Worksheet 4 RTF Multiplying Decimals Worksheet 4 PDF Preview Multiplying.
Source: pinterest.com
Multiplying And Dividing Decimals Worksheets Pdf 5th Grade. The size of the PDF file is 28223 bytes. Use the buttons below to print open or download the PDF version of the Multiplying 2-Digit by 2-Digit Numbers with Various Decimal Places A math worksheet. If there are more versions of this worksheet the other versions will be available below the preview images. For more like this use the search bar to look for some or all of these keywords.
Source: pinterest.com
You will be multiplying by a three digit number which means there will be two rows of numbers to add up before calculating your answer. Multiplying Decimals Worksheet 6 - This is another twenty problem worksheet featuring decimal multiplication. R R FMaKd fe1 cw ni nt sh n jI Sn6f miInCi dt YeR EPdrNeK-BA XlSgPe CbLr8a q. 5th grade multiplying decimals worksheets including multiplying decimals by decimals multiplying decimals by whole numbers missing factor problems multiplying by 10 100 or 1000 and multiplication in columns with decimals. Multiplying Decimals - 2.
Source: pinterest.com
When you get your answer add up the total number of digits to the right of the decimals in both the numbers you are multiplying and. Decimals Worksheets Revised 2009 MLC Page 11 of 21 Multiplying Decimal Numbers When multiplying decimal numbers set up the problem like regular multiplication. The problem sets are randomly generated so you can create multiple worksheets without repeating the same problemsChoose the number of digits to the left and right of the decimal and number of problems on each worksheet. 37 x 04 148 b. Multiplication And Division Of Decimals Worksheets Pdf One of the most hard and tough stuff that can be done with primary school college students is buy them to experience math.
Source: pinterest.com
You will be multiplying by a two digit number which means there will be two rows of numbers to add up before calculating your answer. Multiplying and dividing decimals Add to my workbooks 2 Download file pdf Embed in my website or blog Add to Google Classroom. Multiplication And Division Of Decimals Worksheets Pdf One of the most hard and tough stuff that can be done with primary school college students is buy them to experience math. Multiplying Decimals - 1. Decimals multiplication worksheet for 7th grade children.
Source: pinterest.com
The multiplying decimals worksheet maker creates horizontal or vertical format decimal multiplication worksheets that are customized to your preferences. Multiplying and dividing decimals worksheets 6 th grade pdf is giving to improve kids decimal operations math skills. Our 6 th grade multiply and divide decimals exercises with answers has provided practice for simple calculations and mental real world uses. It has an answer key attached on the second page. Includes a few negative examples and multiplying decimals with whole numbers.
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Someone Debug My C Source Code
Lehpron From United States of America, joined Jul 2001, 7028 posts, RR: 19Posted Tue Nov 5 2002 05:36:58 UTC (13 years 3 months 1 week 1 day 12 hours ago) and read 1415 times:
Our assignment (I admit to procrastinating) is due by midnight, which is in 2.5 hours. I already tried looking for examples online but none was found. Our assignment was in the C programming language, and we were to find the root of a quadratic function, Ax2 + Bx + C = 0, and thier corresponding slopes. The program is supposed to terminate when 'A', 'B' and 'C' equals zero, but instead it terminates at just 'A' = 0. This is my code: #include #include float flt_a; float flt_b; float flt_c; int int_x1; int int_x2; int int_m1; int int_m2; int int_root; void main () { do { printf("For formula reading Ax^2 + Bx + C = 0,n"); printf("n"); printf("Enter first coefficent, A:n"); scanf("%f", & flt_a); printf("n"); printf("Enter first coefficent, B:n"); scanf("%f", & flt_b); printf("n"); printf("Enter first coefficent, C:n"); scanf("%f", & flt_c); printf("n"); if (flt_a != 0.0 || flt_b != 0.0 || flt_c != 0.0) { if (flt_b * flt_b - 4 * flt_a * flt_c < 0.0) { int_root = 0; } else { int_root = 2; int_x1=(-flt_b + sqrt((flt_b * flt_b)- 4*flt_a*flt_c))/(2*flt_a); int_x2=(-flt_b - sqrt((flt_b * flt_b)- 4*flt_a*flt_c))/(2*flt_a); int_m1= 2 * flt_a * int_x1 + flt_b; int_m2= 2 * flt_a * int_x2 + flt_b; } } else { if (flt_a == 0) { int_root= 1; int_x1=(-flt_c / flt_b); int_m1= flt_b; } else { int_root= 0; } } } while (flt_a != 0.0 || flt_b != 0.0 || flt_c != 0.0); printf("n"); printf("n"); printf("Program terminated:n"); printf("n"); } What am I missing, cuz it complied perfectly?
The meaning of life is curiosity; we were put on this planet to explore opportunities.
PPGMD From United States of America, joined Sep 2001, 2453 posts, RR: 0 Reply 1, posted Tue Nov 5 2002 05:39:34 UTC (13 years 3 months 1 week 1 day 12 hours ago) and read 1410 times:
At worst, you screw up and die.
PPGMD From United States of America, joined Sep 2001, 2453 posts, RR: 0 Reply 2, posted Tue Nov 5 2002 05:41:11 UTC (13 years 3 months 1 week 1 day 12 hours ago) and read 1408 times:
Also just a tip its bad programming mojo/style to declare all your varibles are global varibles.
At worst, you screw up and die.
Lehpron From United States of America, joined Jul 2001, 7028 posts, RR: 19 Reply 3, posted Tue Nov 5 2002 05:47:17 UTC (13 years 3 months 1 week 1 day 11 hours ago) and read 1406 times:
Sorry, it was supposed to be: #include (stdio.h) #include (math.h) the carrots make the *.h files disappear. Our teacher couldn't care less about complex roots, we should just let everything equal -999.99. I have not got there yet.
The meaning of life is curiosity; we were put on this planet to explore opportunities.
PPGMD From United States of America, joined Sep 2001, 2453 posts, RR: 0 Reply 4, posted Tue Nov 5 2002 05:50:42 UTC (13 years 3 months 1 week 1 day 11 hours ago) and read 1404 times:
Got any test cases that I could run though it.
At worst, you screw up and die.
PPGMD From United States of America, joined Sep 2001, 2453 posts, RR: 0 Reply 5, posted Tue Nov 5 2002 05:54:41 UTC (13 years 3 months 1 week 1 day 11 hours ago) and read 1399 times:
First problem: while (flt_a != 0.0 || flt_b != 0.0 || flt_c != 0.0); should be: while (flt_a != 0.0 && flt_b != 0.0 && flt_c != 0.0); Need to use ands and not ors. Second problem you have no output until the program exits? That doesn't seem right.
At worst, you screw up and die.
PPGMD From United States of America, joined Sep 2001, 2453 posts, RR: 0 Reply 6, posted Tue Nov 5 2002 05:56:14 UTC (13 years 3 months 1 week 1 day 11 hours ago) and read 1398 times:
third problem all you n's should be /n for new line.
At worst, you screw up and die.
Lehpron From United States of America, joined Jul 2001, 7028 posts, RR: 19 Reply 7, posted Tue Nov 5 2002 06:04:31 UTC (13 years 3 months 1 week 1 day 11 hours ago) and read 1395 times:
Okay I changed the || into &&, now it loops through (like it supposed to) until put all coeff's as zero, then it freezes and doesn't terminate. What did you mean by "n's should be /n for new line"? Which lines are these in reference to?
The meaning of life is curiosity; we were put on this planet to explore opportunities.
Lehpron From United States of America, joined Jul 2001, 7028 posts, RR: 19 Reply 8, posted Tue Nov 5 2002 06:06:21 UTC (13 years 3 months 1 week 1 day 11 hours ago) and read 1393 times:
In terms of out put, I'm taking baby steps to make sure this part works, if it does properly, i.e. it'll complie and run, then I'll just throw that stuff in last.
The meaning of life is curiosity; we were put on this planet to explore opportunities.
PPGMD From United States of America, joined Sep 2001, 2453 posts, RR: 0 Reply 9, posted Tue Nov 5 2002 06:11:38 UTC (13 years 3 months 1 week 1 day 11 hours ago) and read 1393 times:
These three lines are a good example(there are more example that appear to need to be changed): printf("For formula reading Ax^2 + Bx + C = 0,n"); printf("n"); printf("Enter first coefficent, A:n"); Which has the output of: "For formula reading Ax^2 + Bx + C = 0,nnEnter first coefficent, A:n" They should be if I read your code right: printf("For formula reading Ax^2 + Bx + C = 0,/n"); printf("/n"); printf("Enter first coefficent, A:/n"); Which could have the output of: "For formula reading Ax^2 + Bx + C = 0, Enter first coefficent, A: " Sorry I can't help you much more, I do need to goto sleep I have a Digital Logic test in the morning. But I suggest that you print it out and go though it line by line on paper(away from the computer), you will see alot of mistakes that way. Thats the verification phase the Personal Software Process (PSP) which dominates my 4 years in Computer Science.
At worst, you screw up and die.
Lehpron From United States of America, joined Jul 2001, 7028 posts, RR: 19 Reply 10, posted Tue Nov 5 2002 06:23:59 UTC (13 years 3 months 1 week 1 day 11 hours ago) and read 1389 times:
10Q 4 the help, PPGMD, good luck on ur test.
The meaning of life is curiosity; we were put on this planet to explore opportunities.
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How Could I Do This To My Car!?!?!? posted Sat Dec 2 2006 18:46:27 by F.pier | 2,371 | 7,658 | {"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-2016-07 | longest | en | 0.742232 |
https://metanumbers.com/1028891 | 1,643,109,367,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304810.95/warc/CC-MAIN-20220125100035-20220125130035-00532.warc.gz | 436,623,471 | 7,429 | # 1028891 (number)
1,028,891 (one million twenty-eight thousand eight hundred ninety-one) is an odd seven-digits composite number following 1028890 and preceding 1028892. In scientific notation, it is written as 1.028891 × 106. The sum of its digits is 29. It has a total of 3 prime factors and 8 positive divisors. There are 934,528 positive integers (up to 1028891) that are relatively prime to 1028891.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 7
• Sum of Digits 29
• Digital Root 2
## Name
Short name 1 million 28 thousand 891 one million twenty-eight thousand eight hundred ninety-one
## Notation
Scientific notation 1.028891 × 106 1.028891 × 106
## Prime Factorization of 1028891
Prime Factorization 17 × 29 × 2087
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 1028891 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 1,028,891 is 17 × 29 × 2087. Since it has a total of 3 prime factors, 1,028,891 is a composite number.
## Divisors of 1028891
8 divisors
Even divisors 0 8 4 4
Total Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 1.12752e+06 Sum of all the positive divisors of n s(n) 98629 Sum of the proper positive divisors of n A(n) 140940 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 1014.34 Returns the nth root of the product of n divisors H(n) 7.30021 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 1,028,891 can be divided by 8 positive divisors (out of which 0 are even, and 8 are odd). The sum of these divisors (counting 1,028,891) is 1,127,520, the average is 140,940.
## Other Arithmetic Functions (n = 1028891)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 934528 Total number of positive integers not greater than n that are coprime to n λ(n) 16688 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 80429 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 934,528 positive integers (less than 1,028,891) that are coprime with 1,028,891. And there are approximately 80,429 prime numbers less than or equal to 1,028,891.
## Divisibility of 1028891
m n mod m 2 3 4 5 6 7 8 9 1 2 3 1 5 3 3 2
1,028,891 is not divisible by any number less than or equal to 9.
## Classification of 1028891
• Arithmetic
• Deficient
• Polite
• Square Free
### Other numbers
• LucasCarmichael
• Sphenic
## Base conversion (1028891)
Base System Value
2 Binary 11111011001100011011
3 Ternary 1221021101002
4 Quaternary 3323030123
5 Quinary 230411031
6 Senary 34015215
8 Octal 3731433
10 Decimal 1028891
12 Duodecimal 41750b
20 Vigesimal 68c4b
36 Base36 m1wb
## Basic calculations (n = 1028891)
### Multiplication
n×y
n×2 2057782 3086673 4115564 5144455
### Division
n÷y
n÷2 514446 342964 257223 205778
### Exponentiation
ny
n2 1058616689881 1089201184668351971 1120669296094605327794161 1153046552728074570319462105451
### Nth Root
y√n
2√n 1014.34 100.954 31.8487 15.9395
## 1028891 as geometric shapes
### Circle
Diameter 2.05778e+06 6.46471e+06 3.32574e+12
### Sphere
Volume 4.56244e+18 1.3303e+13 6.46471e+06
### Square
Length = n
Perimeter 4.11556e+06 1.05862e+12 1.45507e+06
### Cube
Length = n
Surface area 6.3517e+12 1.0892e+18 1.78209e+06
### Equilateral Triangle
Length = n
Perimeter 3.08667e+06 4.58394e+11 891046
### Triangular Pyramid
Length = n
Surface area 1.83358e+12 1.28364e+17 840086
## Cryptographic Hash Functions
md5 a1efb7fa880b8f343a9c4ed17f55d2f8 98216eed7766de33f178dbacff5a76018bdd715e 36f554e659f4e65c94196bb96c4a57782c99e146f5ed1b8600be2ec04415037f d8e73a507710c4925727a477b4bf904dd7f0a6ab28fb7e93dfccf1955c79f88eca6782b5b8dd8a3023ab5e51a8fa975486e5c96d04f49775574591d015da0e9a 80f092ae2306436d19c02bdf5356e3c36d8ead6d | 1,545 | 4,352 | {"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-2022-05 | latest | en | 0.811241 |
https://learn.careers360.com/engineering/question-solve-let-f-be-a-function-such-that-then-f2-equals-differential-equations-jee-main/ | 1,582,748,772,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146485.15/warc/CC-MAIN-20200226181001-20200226211001-00273.warc.gz | 433,192,950 | 93,159 | ## Filters
Q&A - Ask Doubts and Get Answers
Q
# Solve! - Let f: be a function such that,. Then f(2) equals : - Differential equations - JEE Main
Let f:$R\rightarrow R$ be a function such that $f(x)= x^3+x^2{f}'(1)+x{f}''(2)+{f}'''(3)$,$x\epsilon R$. Then f(2) equals :
• Option 1)
-4
• Option 2)
30
• Option 3)
-2
• Option 4)
8
Answers (1)
Views
Differential Equations -
An equation involving independent variable (x), dependent variable (y) and derivative of dependent variable with respect to independent variable
$\left (\frac{\mathrm{d} y}{\mathrm{d} x} \right )$
- wherein
eg:
$\frac{d^{2}y}{dx^{2}}- 3\frac{dy}{dx}+5x=0$
Order of a Differential Equation -
The order of a differential equation is order of highest order occuring in differential equation
- wherein
order of
$\frac{d^{2}y} {dx^{2}}+5=0$
is 2.
Given $f(x)=x^{3}+x^{2}f{}'(1)+xf{}''(2)+f{}''{}'(3)$
$=>f{}'(x)=3x^{2}+2xf{}'(1)+f{}''(2)$................................(1)
$=>f{}{}''(x)=6x+2f{}'(1)$.....................................................(2)
$=>f{}{}''{}'(x)=6$..........................................................................(3)
put x=1 in equation (1)
$=>f{}'(1)=3+2f{}'(1)+f{}'{}'(2)$.......................................(4)
put x=2 in equation (2)
$=>f{}''(2)=12+2f{}'(1)$.....................................................(5)
from eqn (4) and (5)
$-3-f{}'(1)=12+2f{}'(1)$
$=>3f{}'(1)=-15$
$=>f{}'(1)=-5$ and $f{}'{}'(2)=2$
Now,
put x = 3 in eqn (3)
$f{}'{}'{}'(3)=6$
$\therefore f(x)=x^{3}-5x^{2}+2x+6$
$\therefore f(2)=8-20+4+6=-2$
Option 1)
-4
Option 2)
30
Option 3)
-2
Option 4)
8
Exams
Articles
Questions | 607 | 1,664 | {"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": 19, "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} | 4.09375 | 4 | CC-MAIN-2020-10 | latest | en | 0.397577 |
http://www.instructables.com/id/Honeycomb-wine-cellar/ | 1,526,938,001,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864544.25/warc/CC-MAIN-20180521200606-20180521220606-00632.warc.gz | 392,750,228 | 10,031 | # Honeycomb Wine Cellar
1,051
24
1
Posted
## Introduction: Honeycomb Wine Cellar
Everybody loves wine, so you must have a wine cellar in your home. If you have no space as I haven't, you would enjoy this easy and cheap honeycomb (hexagonal) wine cellar including its drawers to store utensils.
Why hexagonal? There are two reasons for that: First, when you use a soft material like I've used, you need a structure (hexagonal) that make it strong enough (considering more or less 1kg for each bottle the bottom side needs to resist up to 16kg in my case). Second, hexagonal structure saves space, compared to square or triangular. And an extra: it fits perfectly my bookcase :)
Things to consider:
• If you are using wood, maybe you can choose other form (triangular, square) because its easy to cut;
• If you plan to fit some space like me (e.g. a bookcase), remember that you must take into account the thickness of the material (twice for middle parts, and once for each top/bottom side - for hexagonal structure);
• To keep a constant temperature and ban the light rays, consider to put a tinted window in front of it. In my case I wouldn't because the bottle length its greater than the bookcase length, unfortunately. So I kept the length of the wine stand as length as possible, covering almost whole bottle (you see just the neck outside it).
Materials:
• 2x Paraná paper (80cm x 90cm);
• Paper glue;
• Paint (if you like);
Paraná paper (I don't know the right name for it in English) is almost a paperboard but it is massive otherwise the paperboard has a wavy paper inside.Maybe the paperboard works well if it's hard enough.
## Step 1: Preparing the Paper
Before you go through this step, please visit this intelligent instructable for an easy and elegant construction: https://www.instructables.com/id/Kirigami-Honeycombs/ .This will help you to keep exactly the same size for each part. Thanks for riclamagna.
The form the hexagonal structure its very simple. If you see the draft, Fig. 1, you can form the hexagonal structure by taking semi-circles.
Each hexagonal hole you'll need a peace of 25 cm x 30 cm (6x5 cm ), like Fig. 2. Make a line mark each 5 cm in the 30 cm direction.
A pen (or pencil) mark is enough to make it easy to bend. Fig. 3-4. To help bending, use a stick, rule or the table corner.
Use a tape to keep the parts together, that forms the hexagon, until the glue drys, Fig. 5. Put glue inside and outside, and left it dry overnight.
## Step 2: Putting Together
The maximum diameter supported by the hexagon is about 8.5 cm. Usually it fits almost all types of wine bottles, Fig. 1.
Now its time to put the structures together, Fig. 2. Be generous with glue here. Put glue in all side that will be glue.
Tip: Do not put the same cut side together, it will broken. You could avoid this using the tips in the previous step, using the different construction.
After that, kept they with clips, clothes clip or anything else. Let ti dries overnight.
## Step 3: Drawers
To make the drawers proceed similar. It's better, after the fist measure, you try in the place. This is, as I used to call, the classical drawer problem - one drawer wont never fits another drawer place.
You can see, Fig 1 and 2, that the bottom drawers have the opening short than the its base. In comparison, the top drawers have opposite sizes.
## Step 4: Manage Your Cellar
Now you could print this file and use it to classify your wines matching with their place. If you are using a huge honeycomb, may you need to use some marks for columns and rows to help you.
## Recommendations
• ### PVC Class
8,903 Enrolled
• ### Make it Move Contest
We have a be nice policy. | 899 | 3,701 | {"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-2018-22 | latest | en | 0.931173 |
https://talk.collegeconfidential.com/t/math-iic-800/3065 | 1,620,893,455,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243990584.33/warc/CC-MAIN-20210513080742-20210513110742-00024.warc.gz | 540,243,313 | 7,697 | # Math Iic 800
<p>For the October 2004 SAT2 test I took Math 2C and got an 800 but its 90th percentile. My friend also got an 800 and 90%. Has anyone gotten a percentile above 90th?</p>
<p>My friend also got an 800 and 90%</p>
<p>All that means is that more people do well on that particular test.</p>
<p>Maybe I'm stupid but shouldn't some people have gotten a higher percentile than 90th? I know that the curve is nice and all but still...?</p>
<p>It depends on the scores of previous test takers.</p>
<p>It just means that 10% of the people got 800s. Those 10% can't have done better than each other since they all reached the top score.</p>
<p>Yea, vitamin1 is right. The test makers purposely curve the scores that way. The curve is seemingly more "easy" than that of the Math IC - which is true - but keep in mind that, often, the more advanced math kids take the Math IIC, so the "smart" are being compared with the "smart" - a harder curve might not give the students what they truly deserve if being compared with the all students.</p>
<p>i took it in June and got a 800 but still my percentage was 90% COMPARED TO THE CLASS OF 2003</p>
<p>90%ile too.</p>
<p>well, you can't really get a better percentile unless you get above an 800...</p>
<p>Everyone got a (0% no matter what their raw score would have been.</p>
<p>90% i mean</p>
<p>Anyone recommend retaking a 770 Math IIC (81%)?</p>
<p>The middle 50% split at MIT is 740-800 for Math, so it would be competitive for MIT.</p>
<p>got an 800 @ 90th percentile</p>
<p>i'm confused, people speak of taking SATIIs over again, but score choice is dead. does this mean that you take it over but colleges will still see the low scores? does this sway them?</p>
<p>and the IQR (middle 50%) of MIT is between 740 and 800? so the top FOURTH of the applicants accepted all got 800s? wow. if only my gpa didn't reek...</p>
<p>The colleges will get your entire high school transcript. This will include ALL SAT II scores. This does sway them, in some ways - Harvard even states that "previous scores are considered" (or something to that effect). However, it is still advisable to retake a ... say ... 600 if you're aiming for an Ivy. If you get the 800, you can put that on your application, even though the 600 will still be on your transcript.</p> | 616 | 2,317 | {"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-2021-21 | latest | en | 0.958643 |
https://help.anaplan.com/618b82cc-513b-4183-9122-23fc6590cabc | 1,627,081,633,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046150067.51/warc/CC-MAIN-20210723210216-20210724000216-00707.warc.gz | 311,633,633 | 180,280 | 1. Calculation functions
2. All Functions
3. Time and Date Functions
4. HALFYEARVALUE
Returns the half-year value of the source line item or property that matches the time summary method set for the source.
## Syntax
`HALFYEARVALUE(x)`
where:
• x Line item - number, Boolean, date, time period, list, text.
## Format
Input Format Output Format
x: Line item - number, Boolean, date, time period, list, text.
Matches the input format.
## Arguments
The function uses the following arguments:
• x: Number, Boolean, date, time period, list item, text.
## Constraints
The function has the following constraints:
• The result formatting must match the source formatting.
• The result must have a timescale applied.
• Half-Year Totals must be selected in the model calendar.
• If the time summary method is None, only numbers get values. Use other time summary methods to get values for Booleans, dates, time periods, list items, and text. See Summary Methods and Time Aggregation for more information.
## Examples
### Sum example
In this example, an Income Statement source module shows gross and net profit by region for actual and forecast versions:
The time summary method is set to Sum for the Cost of Goods, Gross Profit, and Net Profit line items:
You can use HALFYEARVALUE in result line items in an Income Summary module that has a timescale applied to pull in any required half-year totals:
`HALFYEARVALUE(Income Statement.Cost of Goods)`
Ensure that the time summary method you select for the result line items gives you the values that you want.
```HALFYEARVALUE(Income Statement.Cost of Goods)
```
In this example, If you change the time summary method for the Half Year Cost of Goods result line item from Closing Balance to Sum, the value returned for the H1 FY14 selection, where the timescale for the result line item is set to months, is incorrect. The value returned is the multiple of each month's half-year values:
6 x 1,366,473 = 8,198,838
### Max example
Take the first example and change the time summary method for the source line items to Max:
The result line items in the Income Summary module adjust to show the maximum month value for the selected half-year:
`HALFYEARVALUE(Income Statement.Cost of Goods)` | 519 | 2,256 | {"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-2021-31 | latest | en | 0.621766 |
https://www.askiitians.com/forums/Algebra/if-a-b-c-3-then-what-is-the-value-of-a-2b-3c-2_103336.htm | 1,726,000,004,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651318.34/warc/CC-MAIN-20240910192923-20240910222923-00350.warc.gz | 628,286,966 | 42,726 | # if a+b+c=3 then what is the value of a^2b^3c^2
Arun Kumar IIT Delhi
10 years ago
Hello Student,
I don’t think you can find the exact value of this expression but you can find a inequality considering all three are positive.
2(a/2)+3(b/3)+2(c/2)=3
=>2(a/2)+3(b/3)+2(c/2) >= 3((a/2)^2(b/3)^3(c/2)^2)^(1/7)
Now you evaluate.
Thanks & Regards
Arun Kumar
Btech, IIT Delhi | 139 | 369 | {"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-2024-38 | latest | en | 0.811622 |
https://mathportal.org/math-tests/pre-algebra-tests/absolute-value-tests.php?testNo=4 | 1,708,480,854,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473360.9/warc/CC-MAIN-20240221002544-20240221032544-00745.warc.gz | 409,779,932 | 13,324 | Math Calculators, Lessons and Formulas
It is time to solve your math problem
mathportal.org
• Pre algebra
• Absolute values
• Graphs of absolute value functions
# Graphs of absolute value functions
ans:
syntax error
C
DEL
ANS
±
(
)
÷
×
7
8
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4
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+
1
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=
0
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• Question 1: 1 pts Which graph is sketched in the given image.
$f(x)=x$ if $x\leq 0$ $f(x)=x$ if $x\geq 0$ $f(x)=x$ if $x<0$ $f(x)=|x|$
• Question 2: 1 pts Which graph is sketched in the given image.
• Question 3: 1 pts It's sketched the graph of function $|x|.$
• Question 4: 1 pts Which graph is sketched in the given image.
• Question 5: 2 pts Which graph is sketched in the given image.
• Question 6: 2 pts Which graph is sketched in the given image.
• Question 7: 2 pts Which graph is sketched in the given image. (blue graphic)
$-|x+4|$ $|x+2|$ $|x-4|$ $|x+4|$
• Question 8: 2 pts Which graph is sketched in the given image.
$g(x)=|x+2|$ $g(x)=|x-2|$ $g(x)=|x|+2$ $g(x)=|x|-2$
• Question 9: 3 pts Which graph is sketched in the given image.
$h(x)=$
• Question 10: 3 pts Which graph is sketched in the given image.
$g(x)=$
• Question 11: 3 pts Which graph is sketched in the given image.
$h(x)=$
• Question 12: 3 pts Which graphs is sketched in the given image.
$f(x)=|x+5|, h(x)=|-x+0.5|$ $f(x)=|x-5|, h(x)=|-x-5|$ $f(x)=|x-5|, h(x)=|x-0.5|$ none of these | 520 | 1,374 | {"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} | 3.265625 | 3 | CC-MAIN-2024-10 | latest | en | 0.73553 |
https://ch.mathworks.com/matlabcentral/cody/problems/223-which-quadrant/solutions/289351 | 1,571,485,821,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986693979.65/warc/CC-MAIN-20191019114429-20191019141929-00283.warc.gz | 427,522,485 | 15,267 | Cody
# Problem 223. Which quadrant?
Solution 289351
Submitted on 25 Jul 2013 by Dariusz
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% x = 1+1i; y_correct = 'I'; assert(isequal(your_fcn_name(x),y_correct))
2 Pass
%% x = -1-1i; y_correct = 'III'; assert(isequal(your_fcn_name(x),y_correct))
3 Pass
%% x = 1-1i; y_correct = 'IV'; assert(isequal(your_fcn_name(x),y_correct))
4 Pass
%% x = -1+1i; y_correct = 'II'; assert(isequal(your_fcn_name(x),y_correct)) | 196 | 590 | {"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-2019-43 | latest | en | 0.628827 |
https://www.physicsforums.com/threads/differential-equations-separable-simplification-of-answer.821315/ | 1,508,574,857,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824675.67/warc/CC-MAIN-20171021081004-20171021101004-00875.warc.gz | 958,073,743 | 17,516 | # Differential Equations, Separable, Simplification of answer
1. Jun 29, 2015
### RyanTAsher
1. The problem statement, all variables and given/known data
I believe I have solved this differential equation, yet do not know how the book arrived at it's answer...
Solve the differential equation in its explicit solution form.
The answer the book gives is...
2. Relevant equations
Separable Differential Equation
3. The attempt at a solution
dy/dx = x(x^2+1)/(4y^3)
(4y^3)dy = (x^3+x)dx
∫(4y^3)dy = ∫(x^3+x)dx
y^4 = 1/4x^4 + 1/2x^2 + c
(initial condition, y(0) = -1/sqrt(2))
(-1/sqrt(2))^(4) = 0 + 0 + c
C = -1/4
....
y^4 = 1/4x^4 + 1/2x^2 - 1/4
y = (1/4x^4 + 1/2x^2 - 1/4)^(1/4)
-----------
I've experimented with simplifying this a bit and found a few other ways to express it, but nothing like what the book has written as the answer.
2. Jun 29, 2015
### RUber
Your sign on C is wrong.
$(\frac{x^4}{4}+\frac{x^2}{2} + \frac 14 ) = (\frac{x^2}{2} + \frac 12 )^2$
3. Jun 29, 2015
### SammyS
Staff Emeritus
First. You made a mistake in finding C. What is (-1/sqrt(2))4 ? Fixing that will allow some factoring in the resulting expression.
4. Jun 29, 2015
### RyanTAsher
Oh wow, I don't think I would have seen that factor regardless. Thank you though. That helped a lot.
5. Jun 29, 2015
### RyanTAsher
Thank you, I understand now. In regards to the -, out front the answer from the book, I understand that comes from the square root, but how do they determine whether to go with the - or + solution. I haven't learned intervals of validity within the book yet...
6. Jun 29, 2015
### RUber
The radical implies the positive. Your initial condition forces the negative choice.
7. Jul 4, 2015
### Dr. Courtney
Before rejecting an answer, you should plug it into the diff eq and see if it works. | 588 | 1,824 | {"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} | 3.96875 | 4 | CC-MAIN-2017-43 | longest | en | 0.907964 |
https://www.justintools.com/unit-conversion/fuel-consumption.php?k1=feet-per-gallon-UK&k2=miles-per-liter | 1,716,048,249,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057440.7/warc/CC-MAIN-20240518152242-20240518182242-00895.warc.gz | 750,456,825 | 27,036 | Please support this site by disabling or whitelisting the Adblock for "justintools.com". I've spent over 10 trillion microseconds (and counting), on this project. This site is my passion, and I regularly adding new tools/apps. Users experience is very important, that's why I use non-intrusive ads. Any feedback is appreciated. Thank you. Justin XoXo :)
# FUEL-CONSUMPTION Units Conversionfeet-per-gallon-UK to miles-per-liter
1 Feet Per Gallon UK
= 4.1660842480888E-5 Miles Per Liter
Category: fuel consumption
Conversion: Feet Per Gallon UK to Miles Per Liter
The base unit for fuel consumption is kilometers per liter (Non-SI/Derived Unit)
[Feet Per Gallon UK] symbol/abbrevation: (ft/gal[UK])
[Miles Per Liter] symbol/abbrevation: (mi/L)
How to convert Feet Per Gallon UK to Miles Per Liter (ft/gal[UK] to mi/L)?
1 ft/gal[UK] = 4.1660842480888E-5 mi/L.
1 x 4.1660842480888E-5 mi/L = 4.1660842480888E-5 Miles Per Liter.
Always check the results; rounding errors may occur.
Definition:
In relation to the base unit of [fuel consumption] => (kilometers per liter), 1 Feet Per Gallon UK (ft/gal[UK]) is equal to 6.7046626881562E-5 kilometers-per-liter, while 1 Miles Per Liter (mi/L) = 1.609344 kilometers-per-liter.
1 Feet Per Gallon UK to common fuel-consumption units
1 ft/gal[UK] = 0.00015770333553234 miles per gallon US (MPG[US])
1 ft/gal[UK] = 0.00018939404100937 miles per gallon UK (MPG[UK])
1 ft/gal[UK] = 0.00030480034405558 kilometers per gallon US (km/gal)
1 ft/gal[UK] = 6.7046626881562E-5 kilometers per liter (km/L)
1 ft/gal[UK] = 0.067046626881562 meters per liter (m/L)
1 ft/gal[UK] = 4.1660842480888E-5 miles per liter (mi/L)
1 ft/gal[UK] = 1.8985490329432 meters per cubic foot (m/ft3)
1 ft/gal[UK] = 0.0010986974394156 meters per cubic inch (m/in3)
1 ft/gal[UK] = 67.046626881562 meters per cubic meter (m/m3)
1 ft/gal[UK] = 0.83267047924535 feet per gallon US (ft/gal[US])
Feet Per Gallon UKto Miles Per Liter (table conversion)
1 ft/gal[UK] = 4.1660842480888E-5 mi/L
2 ft/gal[UK] = 8.3321684961776E-5 mi/L
3 ft/gal[UK] = 0.00012498252744266 mi/L
4 ft/gal[UK] = 0.00016664336992355 mi/L
5 ft/gal[UK] = 0.00020830421240444 mi/L
6 ft/gal[UK] = 0.00024996505488533 mi/L
7 ft/gal[UK] = 0.00029162589736622 mi/L
8 ft/gal[UK] = 0.0003332867398471 mi/L
9 ft/gal[UK] = 0.00037494758232799 mi/L
10 ft/gal[UK] = 0.00041660842480888 mi/L
20 ft/gal[UK] = 0.00083321684961776 mi/L
30 ft/gal[UK] = 0.0012498252744266 mi/L
40 ft/gal[UK] = 0.0016664336992355 mi/L
50 ft/gal[UK] = 0.0020830421240444 mi/L
60 ft/gal[UK] = 0.0024996505488533 mi/L
70 ft/gal[UK] = 0.0029162589736622 mi/L
80 ft/gal[UK] = 0.003332867398471 mi/L
90 ft/gal[UK] = 0.0037494758232799 mi/L
100 ft/gal[UK] = 0.0041660842480888 mi/L
200 ft/gal[UK] = 0.0083321684961776 mi/L
300 ft/gal[UK] = 0.012498252744266 mi/L
400 ft/gal[UK] = 0.016664336992355 mi/L
500 ft/gal[UK] = 0.020830421240444 mi/L
600 ft/gal[UK] = 0.024996505488533 mi/L
700 ft/gal[UK] = 0.029162589736622 mi/L
800 ft/gal[UK] = 0.03332867398471 mi/L
900 ft/gal[UK] = 0.037494758232799 mi/L
1000 ft/gal[UK] = 0.041660842480888 mi/L
2000 ft/gal[UK] = 0.083321684961776 mi/L
4000 ft/gal[UK] = 0.16664336992355 mi/L
5000 ft/gal[UK] = 0.20830421240444 mi/L
7500 ft/gal[UK] = 0.31245631860666 mi/L
10000 ft/gal[UK] = 0.41660842480888 mi/L
25000 ft/gal[UK] = 1.0415210620222 mi/L
50000 ft/gal[UK] = 2.0830421240444 mi/L
100000 ft/gal[UK] = 4.1660842480888 mi/L
1000000 ft/gal[UK] = 41.660842480888 mi/L
1000000000 ft/gal[UK] = 41660.842480888 mi/L
(Feet Per Gallon UK) to (Miles Per Liter) conversions | 1,414 | 3,539 | {"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-2024-22 | latest | en | 0.68208 |
http://exams.collegedunia.com/xat/preparation-tips | 1,500,799,364,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424296.90/warc/CC-MAIN-20170723082652-20170723102652-00437.warc.gz | 113,635,624 | 27,844 | XAT 2017 PREPARATION TIPS
NATIONAL LEVEL OFFLINE TEST
# XAT 2017 Preparation Tips
XAT exam which is second best soughed out exam after CAT among management aspiring candidates. So what make XAT so special, it is its colleges and their ROI. And when it comes for HR course XLRI is the best among all. So main question arrives in candidate’s mind is how to prepare and give XAT exam. Like all management exams, there are two stages of XAT: one is Main exam then GDPI for the selection of candidate. So we first discuss about Main Exam.
XAT ELIGIBILITY XAT APPLICATION PROCESS XAT APPLICATION FORM XAT SYLLABUS XAT PRACTICE PAPERS XAT CUT OFF
When you start preparing for XAT you will find that the pattern of XAT is different from every other contemporary exams.
So we discuss about pattern, XAT is divided into two sections one is aptitude test and another is Essay & GK part. Every year there is change in number of questions in the test so you can’t predict which part of any section will have how many questions.
XAT EXAM PATTERN XAT SYLLABUS XAT PREVIOUS YEAR EXAM ANALYSIS
But with last year XAT question paper analysis here we have the sectional division of questions:
Aptitude Section
Verbal and logical26
Decision Making and Analytical23
Quant and Data Interpretation29
General Section
GK25
Essay1
So the thing comes how to prepare each Section. Lets talk about Quant and Data Interpretation part first. It is the main part as it contains most no. of questions in exam. Basic part is numerical ability which is based on formulas only.
Quant covers basic mathematics topics like LCM-HCF, Simple and Compound Interest, functions, Number System, Speed Distance Time, Geometry, Mensuration, Averages, Permutation & Combination, etc.
So for this you have to follow this strategy:
1. Clarify all your basic concepts
2. Learn short cuts
3. Try to under the main purpose of that concept
4. Give mock test after every topic
5. Manage the time
Data Interpretation is that part where observation skills plays important role. In this section questions are asked in the form of charts and graph.
Main thing for this section is that you have to spend more time in practicing and less during exam.
Tips for DI section:
• Improve the method of calculation
• Solve different graphs
For Complete information and Tips for Quant and DI sections CLICK HERE.
Now we proceed on Verbal Ability section. This section consist of Reading comprehension, Fill in the blanks, analogies, usages, synonyms and antonyms.
Tips for this section:
Reading Comprehension- Try to get the overview of the passage, the every section of paragraph is related to each other so try to develop sense of understanding. While answering and reading the passage maintain the speed.
XAT XAT RESULTS XAT COUNSELLING XAT SELECTION PROCESS XLRI
Sometimes it is easy for the person to read the question first and then try give the answer from reading the particular section of that passage.
For this try to develop the habit of reading newspapers first. Then try to evaluate the idea for the passages that what can be asked from that passage. Second try as many as sample passages for the time management purpose.
Vocabulary – Increase your vocabulary part as this will not only helps you in giving answers to the questions but also in the future in communication and understanding skills.
Try to find out the words and their meanings on the regular basis. After finding the meaning use them.
Jumbled Words – For this section try to find out the connectors first then try to link them according to the paragraph. Solve as many examples as possible.
Now lets talk about Decision making section. It is most differentiating factor of XAT which makes it special from other exams. Every manager must possesses Decision making ability. And XAT test this ability in the candidates whether they have that intent to be at the managerial level or not.
There are almost 20 questions in XAT about from Decision Making ability. And for this section candidate must have approach of understanding the situation of the paragraph. Candidate must read and analyse the information. Candidates while solving such type of questions are advised to go with most ethical answer.
Last part of section 1 i.e. Logical Reasoning. To solve any question in this candidate has to apply step by step mathematical procedure to arrive at conclusion.
Strategies to improve Logical Reasoning:
1. Practice:
Keep practicing as much as questions as you can. Sometimes your attempting speed depends on logic, which is required to solve question at right time in right way.
Practice will help you a lot to manage time for different type of questions. You will be able to identify what type of question needs how much time.
1. Categorization:
Classify your problem on time span & complexity level & devote time according to requirement.
1. Learning:
Do not solve question in different way any time you attempting it. Just maintain your own method & stick to that only otherwise you will be confused.
1. Make Notes:
When you start preparing make note of those problems which creates issue when you have been solving them.
Next Section is general section in which two parts are there one is General Knowledge and other is Essay Writing. General knowledge part consist of 25- 30 questions in XAT exam. Candidates preparing for XAT general knowledge must read the newspapers daily. There are some current affair books available in the market candidate can grab any of them and read it. Other part is Essay Writing which is also distinct part of XAT. Candidate has to write an essay on the given topic in 200 words. This essay of 20 marks and plays important role in selection criteria.
Tips for Essay Writing are:
• Don’t write the essay in one paragraph
• Avoid Jagrons or slangs usage
• Your first para must be the summary of all the content
• Try to write in simple bullets
• Don’t use short forms
We also have interviewed various toppers who cracked XAT and secured admission in Top Management colleges in India.
#### Keep practicing and form your own plan says Abhishake Koul
He got 98.4 percentile in XAT 2015 and he says that you have to practice RC, quant and Decision Making. He says that candidate must give mock papers regularly. He had practice a lot for his weak areas by making handouts and giving test.
#### Knowledge is treasure but practice is key to it says Priyajeev Kumar
Priyajeev scored 99.82 percentile marks in XAT 2013. He says, practice is the only way to improve accuracy as the syllabus of MBA entrance exam is quite of elementary level. His advices to the aspirants of XAT to give mock tests as it gives confidence and prepare well for the exam.
#### Be your own motivator says Harish Sankar
Harish Sankar appeared for XAT 2011 and scored 74.5 percentile marks. He says, it is very difficult to manage and give quality time to preparation if you’re working. But without preparation getting passed only a miracle can happen. He advices candidates to motivate themselves as there is no bigger motivator other than you and it is in you. So, be a self-motivator.
## XATNews
SEE MORE NEWS
July 05, 2017XAT 2017
### XAT 2018 Exam Pattern
Xavier Aptitude Test (XAT) 2018 is to be held on ...
July 04, 2017XAT 2017
XAT 2018 Admit card is set to be released in the t ...
July 05, 2017XAT 2017
### XAT 2018 Application Form
XAT 2018 Application Form will be released in the ...
July 05, 2017XAT 2017
### XAT 2018 Result
XAT 2018 Result will be announced on January 24, 2 ...
× | 1,684 | 7,569 | {"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} | 3.359375 | 3 | CC-MAIN-2017-30 | latest | en | 0.883773 |
https://schoollearningcommons.info/question/if-a-54750-b-2432-c-4732-find-1-a-c-2-c-b-3-a-b-c-20843851-23/ | 1,632,061,586,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056890.28/warc/CC-MAIN-20210919125659-20210919155659-00487.warc.gz | 564,438,232 | 13,363 | ## if A=54750 B=2432 C=4732 find 1)A+C 2)C_B 3)A+B+C
Question
if A=54750 B=2432 C=4732 find 1)A+C 2)C_B 3)A+B+C
in progress 0
1 week 2021-09-10T01:19:37+00:00 2 Answers 0 views 0
1. Step-by-step explanation:
1)A+C
54750+2432
=57182
2)C-B
4732-2432
=2332
3)A+B+C
54750+2432+4732
=61914 | 154 | 299 | {"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} | 3.6875 | 4 | CC-MAIN-2021-39 | latest | en | 0.337858 |
https://www.coursehero.com/file/6758295/MA373-F11-Test-2-1/ | 1,490,487,299,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189088.29/warc/CC-MAIN-20170322212949-00140-ip-10-233-31-227.ec2.internal.warc.gz | 904,593,787 | 145,441 | MA373 F11 Test 2-1
# MA373 F11 Test 2-1 - Math 373 Test 2 Fall 2011 1. Dixi...
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Math 373 Test 2 Fall 2011 1. Dixi purchases a 20 year bond with a par value of 1000. The bond pays semi-annual coupons at a rate of 5% convertible semi-annually. The bond has a redemption value of 1200. Calculate the price of the bond to yield 7% convertible semi-annually.
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2. Yi is repaying a loan with quarterly payments for the next 10 years. The first loan payment is 800. The second loan payment is 1600. The third loan payment is 2400. Each payment increases by 800 until the last payment is 32,000. The interest rate on the loan is 8% compounded quarterly. Calculate the amount of the loan.
3. A four year loan has an annual effective interest rate of 5%. Complete the following loan amortization table. Be sure to show your work. Time Payment Interest Principal Outstanding Loan Balance 0 1 2000 2 4000 3 4000 4 8000
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4. A bond matures for 5000 at the end of 25 years. The bond has semi-annual coupons of 100. The bond sells for a premium of 400. Calculate the annual yield convertible semi-annually.
5. An eight year continuous increasing annuity pays at a rate of
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## This note was uploaded on 02/05/2012 for the course MA 373 taught by Professor Staff during the Spring '08 term at Purdue University-West Lafayette.
### Page1 / 14
MA373 F11 Test 2-1 - Math 373 Test 2 Fall 2011 1. Dixi...
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Ask a homework question - tutors are online | 511 | 1,975 | {"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} | 3.34375 | 3 | CC-MAIN-2017-13 | longest | en | 0.909536 |
https://edurev.in/v/323498/Solved-Examples-Permutations-and-Combinations | 1,726,380,853,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651616.56/warc/CC-MAIN-20240915052902-20240915082902-00452.warc.gz | 204,835,098 | 63,098 | Solved Examples: Permutations and Combinations
# Solved Examples: Permutations and Combinations Video Lecture | Quantitative Aptitude for SSC CGL
## Quantitative Aptitude for SSC CGL
314 videos|170 docs|185 tests
## FAQs on Solved Examples: Permutations and Combinations Video Lecture - Quantitative Aptitude for SSC CGL
1. What is the difference between permutations and combinations?
Ans. Permutations and combinations are both ways of counting the number of possible outcomes in a given scenario. The main difference lies in whether the order of the elements matters. In permutations, the order matters, while in combinations, the order does not matter. For example, when arranging a group of people in a line, the order in which they stand is important, so it is a permutation. On the other hand, if we are selecting a group of people to form a committee, the order in which they are selected does not matter, so it is a combination.
2. How do I calculate permutations?
Ans. To calculate permutations, you need to determine the number of ways to arrange a set of objects in a specific order. The formula for calculating permutations is nPr = n! / (n - r)!, where n represents the total number of objects, and r represents the number of objects being arranged in a specific order. The exclamation mark denotes the factorial function, which means multiplying a number by all the positive integers less than it. For example, if you have 5 objects and want to arrange 3 of them, the permutation would be 5P3 = 5! / (5 - 3)! = 5! / 2! = 60.
3. How do I calculate combinations?
Ans. Combinations are used to calculate the number of ways to select objects from a set without regard to their order. The formula for calculating combinations is nCr = n! / (r!(n - r)!), where n represents the total number of objects, and r represents the number of objects being selected. Again, the exclamation mark denotes the factorial function. For example, if you have 5 objects and want to select 3 of them, the combination would be 5C3 = 5! / (3!(5 - 3)!) = 5! / (3!2!) = 10.
4. Can permutations and combinations be applied in real-life situations?
Ans. Yes, permutations and combinations have numerous applications in real-life situations. For example, when scheduling events, you may need to consider various permutations of time slots and activities. In sports, the order in which teams are placed in a tournament bracket can be calculated using permutations. Combinations are used in probability theory to calculate the likelihood of drawing specific cards from a deck or winning in a lottery. These concepts are also used in computer science, cryptography, and data analysis.
5. Are there any shortcuts or tricks to solve permutations and combinations problems?
Ans. While there may not be shortcuts to solve all permutations and combinations problems, there are certain techniques that can simplify calculations. For example, if there are identical objects in a set, you can divide the total number of permutations or combinations by the factorial of the number of identical objects to avoid overcounting. Additionally, understanding the concept of symmetry can help reduce the number of cases to consider. Practice and familiarity with the formulas and concepts can also speed up problem-solving.
## Quantitative Aptitude for SSC CGL
314 videos|170 docs|185 tests
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; | 796 | 3,631 | {"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} | 4.3125 | 4 | CC-MAIN-2024-38 | latest | en | 0.88986 |
http://www.thirteen.org/edonline/adulted/lessons/lesson46_activities.html | 1,432,299,794,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207925201.39/warc/CC-MAIN-20150521113205-00109-ip-10-180-206-219.ec2.internal.warc.gz | 743,206,215 | 7,125 | search
A Fraction of the Possibilities
Overview | Activities
Introductory Activities | Learning Activities | Culminating Activity
Cross-curricular Extensions | Community Connections
Introductory Activities - "shopping"-defining the terms
Compare labels from a variety of household products. Using the PRODUCT LABEL EQUIVALENTS handout, record the units used to convey how much of each product is in its container. You will notice that on most household products, there are at least two different values given for the quantity or amount of the item in the container. Since the product is in the same bottle, the two values must be equivalent! This is the beginning of our working understanding of fractions in real life. All equivalent values can be expressed as a ratio (a proportion) to be compared.
1. Identity units: The first label used is for a bottle of orange juice. The amount listed on the label indicates that the bottle contains 96 FL OZ (3 QTS) or 2.84 L of juice; what do these measures mean?
• 96 FL OZ = 96 fluid ounces
• 3 QTS = 3 quarts
• 2.84 L = 2.84 liters
NOTE: This orange juice label shows both English units and metric system units so as to accommodate consumers who are more familiar with other measuring systems. The numbers represent quantity amounts while the units tell you the system you are using. Fluid ounces and quarts are units of volume in the English system. Liters are units of volume in the metric system.
2. Develop mathematical definitions: A different label has a Net Weight of 5-1/8 OZ (145 g); this label suggests that 5-1/8 ounces = 145 grams. Write five mathematical definitions using five different product labels to practice this skill with your students.
3. Convert definitions to ratios: Once a definition can be determined, you can use it to compare different quantities based on the same units. 96 fluid ounces = 2.84 liters can be expressed as two different but related fractions:
96 fluid ounces2.84 liters OR 2.84 liters96 fluid ounces
These expressions can be read as 96 fluid ounces in 2.84 liters or 2.84 liters in 96 fluid ounces. Other mathematical equivalents can be established with the data from the orange juice label. For example, you can compare fluid ounces to quarts or compare quarts to liters. What are the equivalent relationships that you can write using those two comparisons? What are the fractions that you came up with to compare fluid ounces and quarts? [96 fluid ounces/3 quarts or 3 quarts/96 fluid ounces. Expressed in words, these mathematical relationships are saying, there are 96 fluid ounces in 3 quarts or there are 3 quarts in 96 fluid ounces.] Using the given example, write out at least one pair of equivalent ratios for each of your five labels.
Learning Activities - Art imitating math life
Now that we have seen that equivalent fractions can be expressed in words as well as numbers, the next logical step in this progression is to show that equivalent fractions can be expressed using objects as well.
1. Simple fractional representations-Complete the FRACTIONAL REPRESENTATIONS WORKSHEET. Check your answers against the key provided separately.
2. Complex fractional representations
Log on to the "Melvin Makes a Match" (http://pbskids.org/ cyberchase/games/ equivalentfractions/equivalentfractions.html) Web site. Choose equivalent fractions for the graphical charts shown. Once you have selected an image, choose the equivalent fraction. Drag and drop each onto the pan near the bottom of the screen. In each subsequent round, the fractions become more complex. Take scrap paper (or a calculator) to the computer with you just in case you need to check your answers.
3. Play the Saloon Snap game
(http://www.bbc.co.uk/education/mathsfile/shockwave/key/snapkey.html) to make sure that you can quickly respond to the challenge of comparing equivalent values. Use a calculator if you need to but keep in mind, Fast Fingers Malcolm could beat you to the buzzer if you move too slowly.
Printable versions of each level are available online:
If you need to review the process for doing these types of conversions, go to the Saloon Snap Key Ideas page (http://www.bbc.co.uk/ education/mathsfile/shockwave/key/snapkey.html) to brush up your skills.
Culminating Activity - Putting yourself to the test!
1. Fraction Word Problems Practice sheet
Print out the FRACTION WORD PROBLEM SKILL SET handout and the FRACTION WORD PROBLEM SET, which is available online at http://home.att.net/~clnetwork/math/fracword.pdf. Use page 1 of the FRACTION WORD PROBLEM SKILL SET handout as a guide as you complete the FRACTION WORD PROBLEM SET on a separate sheet of paper. After you have checked your answers against those supplied on page 2 of the FRACTION WORD PROBLEM SKILL SET handout, get ready to take the GED Connection Practice Test #1.
2. GED Connection Math Practice Test #1
The table shown below details all of the items from Practice Test #1 and the skills being tested according to problem type and skill tested.
GED Connection Practice Test #1
Number, Number Sense and Operations Measurement and Geometry Data, Statistics and Probability Algebra, Functions and Patterns Procedural Knowledge 12, (3), (7), (15), (17) (2), (9), (21) (18) (4), (10), (11), (13), (24) Conceptual Knowledge 14 (16) (19), (20), (22) Application Modeling and Problem Solving 6 (23) (1) (5), (8), (25)
The item numbers shown in parentheses pertain to math skills not addressed in this lesson. This chart gives you some idea of the kind of tasks you will be asked to complete on the GED math test and the distribution of skills in each category on a standard GED math test.
Cross-curricular Extensions
• Math communication/Media literacy
CYBERCHASE, Episode #203 "Harry Hippo & Mean Green" reviews strategies for comparing equivalent fractions. You can also visit the Cyberchase Web site (http://pbskids.org/cyberchase/ parentsteachers/episodes/203.html) and watch the Real video to see how this skill can be applied to real-life problem solving.
• Science
Scientists in the United States often use the metric system but sometimes the products they need are only sold in English unit quantities. Get in the habit of looking at the units and performing the conversions mentally. Knowing that a gallon is approximately 4 liters (3.78 L to be exact), you will find it easier to mentally picture quantities in your head for fast recall on a multiple choice test, because in most cases, rough estimates will help you choose the correct response. Check out one of the many dimensional analysis tutorials available online, such as this Science Education Web site (http://www.wfu.edu/~ylwong/chem/ dimensionanalysis/example1.html) that carries out a step-by-step procedure for converting units. (Uses Shockwave plug-in)
• Foreign language
Thinking in this way is not very different from learning a new language. On index cards, record five new words that you want to learn in a different language. Set up parallel statements to show the language equivalents of the terms.
Community Connections
• Organize a study group with other members of your class. Review the rules for working with fractions with them. You can serve as the tutor or help facilitate discussion by working with them to solve the fraction problems on the original worksheet, and to help with problem-solving hints when needed. Plant a garden, paint a wall space in an "unconventional" shape, or divide the normal space into segments rarely considered before. Once you have created the space, invite visitors to your masterpiece and ask them if they can figure out the dimensions or the numerical equivalents to specific segments or the whole thing.
• The GED Connection Workbook can also be used to practice math problem solving using decimals, fractions and percentages. The series of books can be purchased through the Thirteen/WNET's Adult Education Department by contacting Mr. Andrč Gleaton at gleatona@thirteen.org. Chapter 30 addresses the use of decimals and Chapter 31 addresses the use of fractions.
• Organize team play of the Reducing Fractions game online. Visit the "Reducing fractions game" Web site at http://www.quia.com/cb/84915.html along with a partner and play the two-player version. Keep tabs on your scores and compare them with other players in your class. | 1,816 | 8,331 | {"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} | 4.59375 | 5 | CC-MAIN-2015-22 | longest | en | 0.893751 |
https://www.crazy-numbers.com/en/3658 | 1,544,395,486,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823183.3/warc/CC-MAIN-20181209210843-20181209232843-00023.warc.gz | 982,467,978 | 4,316 | Discover a lot of information on the number 3658: properties, mathematical operations, how to write it, symbolism, numerology, representations and many other interesting things!
## Mathematical properties of 3658
Is 3658 a prime number? No
Is 3658 a perfect number? No
Number of divisors 8
List of dividers 1, 2, 31, 59, 62, 118, 1829, 3658
Sum of divisors 5760
## How to write / spell 3658 in letters?
In letters, the number 3658 is written as: Three thousand six hundred and fifty-eight. And in other languages? how does it spell?
3658 in other languages
Write 3658 in english Three thousand six hundred and fifty-eight
Write 3658 in french Trois mille six cent cinquante-huit
Write 3658 in spanish Tres mil seiscientos cincuenta y ocho
Write 3658 in portuguese Três mil seiscentos cinqüenta e oito
## Decomposition of the number 3658
The number 3658 is composed of:
1 iteration of the number 3 : The number 3 (three) is the symbol of the trinity. He also represents the union.... Find out more about the number 3
1 iteration of the number 6 : The number 6 (six) is the symbol of harmony. It represents balance, understanding, happiness.... Find out more about the number 6
1 iteration of the number 5 : The number 5 (five) is the symbol of freedom. It represents change, evolution, mobility.... Find out more about the number 5
1 iteration of the number 8 : The number 8 (eight) represents power, ambition. It symbolizes balance, realization.... Find out more about the number 8
Other ways to write 3658
In letter Three thousand six hundred and fifty-eight
In roman numeral MMMDCLVIII
In binary 111001001010
In octal 7112
In US dollars USD 3,658.00 (\$)
In euros 3 658,00 EUR (€)
Some related numbers
Previous number 3657
Next number 3659
Next prime number 3659
## Mathematical operations
Operations and solutions
3658*2 = 7316 The double of 3658 is 7316
3658*3 = 10974 The triple of 3658 is 10974
3658/2 = 1829 The half of 3658 is 1829.000000
3658/3 = 1219.3333333333 The third of 3658 is 1219.333333
36582 = 13380964 The square of 3658 is 13380964.000000
36583 = 48947566312 The cube of 3658 is 48947566312.000000
√3658 = 60.48140210015 The square root of 3658 is 60.481402
log(3658) = 8.2046718289508 The natural (Neperian) logarithm of 3658 is 8.204672
log10(3658) = 3.5632437011404 The decimal logarithm (base 10) of 3658 is 3.563244
sin(3658) = 0.92693165824697 The sine of 3658 is 0.926932
cos(3658) = 0.37523019726499 The cosine of 3658 is 0.375230
tan(3658) = 2.4703013376943 The tangent of 3658 is 2.470301 | 803 | 2,533 | {"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-2018-51 | longest | en | 0.755855 |
https://www.scribd.com/document/317416846/Old-2nd-Midterm-3 | 1,566,321,118,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315551.61/warc/CC-MAIN-20190820154633-20190820180633-00370.warc.gz | 944,912,116 | 59,687 | You are on page 1of 6
# PHYS 102 General Physics-II, Midterm-II
## Duration: 100 minutes
21 April 2012
NAME:.................................................................... Section:......................
Q.1 (25)
Q.2 (25)
Q.3 (25)
Q.4 (25)
Total (100)
You must sign the Honor Code for your exam to be graded:
I pledge, on my Honor, not to lie, cheat, or steal in either my academic
or personal life. I understand that such acts violate the Honor Code and
undermine the community of trust of which we are all stewards.
I agree to abide by this Honor Code during this exam.
Signature:
Dont forget
to sign!
Present your work in a legible and organized format, otherwise you may
lose significant portion of your credit even if your solution is correct!
Physics Wiki (No relevance to the exam)
International
Thermonuclear
Experimental
Reactor (ITER) is an international nuclear fusion
research and engineering project, which is under
construction in France, to be completed in 2019.
ITER is based on the 'tokamak' concept of
magnetic confinement, in which the plasma is
contained in a doughnut-shaped vacuum vessel.
The fuela mixture of deuterium and tritium, two
isotopes of hydrogenis heated to temperatures in
excess of 150 millionC, forming a hot plasma.
Strong magnetic fields are used to keep the plasma
away from the walls. It aims to produce 500 MW
of output power for 50 MW of input power, but
more importantly, to develop technologies and
processes needed for a fusion power plant.
Tokamak
Source: www.iter.org
## Solutions will be posted to http://www.fen.bilkent.edu.tr/~phys102/
Some expressions from the Textbook which may or may NOT be useful for this exam
(However, you are supposed to know what each symbol means)
Resistivity and Ohms Law:
## Current and current density:
I=
G
G
J = nqd
dQ
= n q d A,
dt
E
J
= ,
V = IR
Power:
Kirchhoffs Rules:
## I = 0 (junction rule), V = 0 (loop rule)
P = Vab I
Equivalent Resistance:
1
1
1
1
= +
+" +
(in parallel)
Req R1 R2
Rn
## Req = R1 + R2 + " + Rn (in series)
Simple R-C Circuit:
Charging:
q(t ) = C (1 e
i (t ) =
t /( RC )
) = Q (1 e
t /( RC )
dq t /( RC )
= e
= I 0 e t /( RC )
dt R
Magnetic Forces:
G
G G
Fm = q B (on a point charge q)
G G
B = B dA,
Discharging:
q(t ) = Q0 e t /( RC )
i (t ) =
G G
G
G
dFm = I dl B (on a current-carrying segment dl )
Magnetic Flux:
G G
v B dA = 0 (for a closed surface)
G G
= B, = IA, U = B
Hall Effect:
nq =
Q
dq
= 0 e t /( RC )
dt
RC
J x By
R=
Ez
m
qB
G 0 qG r
B=
4 r 2
Segment:
G
I
B= 0
2 r
G Idl r
dB = 0
4 r 2
## Axial Magnetic Field of a Current Loop of Radius a:
NI
Bx = 0
(at the center of N circular loops)
2a
F 0 II
=
L 2 r
Amperes Law:
G G
B
v dl = 0 I encl | 840 | 2,727 | {"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-2019-35 | latest | en | 0.903719 |
https://boards.straightdope.com/t/if-you-are-more-logical-than-i-am-please-explain-this-to-me/294018 | 1,652,770,631,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662517018.29/warc/CC-MAIN-20220517063528-20220517093528-00519.warc.gz | 200,633,469 | 8,516 | # If you are more logical than I am, please explain this to me.
This is one of those blog quizzes called How Logical Are You?
Firstly, I got 75% right. I don’t know which ones because the little quiz doesn’t tell you. I’m sure I did not get this one right:
1. Your room is completely dark. You have eight shoes of four different colors, and fifty socks of five different colors. How many shoes and socks must you grab to make sure you have a matching pair?
5 shoes, 6 socks
6 shoes, 8 socks
6 shoes, 10 socks
5 shoes, 5 socks
How do you figure this out?
Start picking shoes. The first certainly doesn’t make a pair. Take another - could still be different. Another - the same. And the fourth. But you’ve run out of colours, so the fifth must match one of the others you’ve taken,
Socks work the same, except five colours make you take six to force a pair.
Assuming you can feel the shoes well enough to tell if they’re right or left foot, and assuming socks that aren’t foot specific, It’s 5 and 6, I think.
Why? Well, take them individually. You have eight shoes of four colors - presumably, four matched pairs of shoes. Grab all four left shoes and one right shoe, and bam! You’re guaranteed to have one of your matched pairs.
The socks are more complex - 50 socks, 25 pairs, with 5 pairs in each of five colors. So, five pairs of red, five pairs of blue, etc. But essentially, the number of colors is what matters - If you took five, you could theoretically have one of each color; but if you take six, you have to repeat a color, and thus have a matched pair.
I think you need to grab 5 shoes and 6 socks.
I worked it out as… If you grab 4 shoes you may get one of each colour but whatever shoe you grab next you will definately have a pair.
Same for the different coloured socks, you can grab 5 different coloured socks so the 6th must match 1 of them.
Although just because socks are the same colour, doesn’t mean they match but that seems to be just bad wording.
I think.
Ah! This is where I screwed up. I’m thinking “I’d have to grab at least 9 shoes to make sure I’ve got a left and a right but that’s not one of the choices.”
Yes, CandidGamera noticed the left-right difficulty I missed with the shoes, but if you think about it you don’t need to care - the two colour-matched shoes MUST be l-r. Socks, on the other hand, don’t need l-r anyway, and mine at least are all the same style if they’re the same colour.
1. Ralph is four times as old as Frank. In 20 years, Ralph will be twice as old as Frank. How old are Ralph and Frank?
[ul][li]Ralph is 40; Frank is 10.[/li][li]Ralph is 20; Frank is 5.[/li][li]Ralph is 60; Frank is 15.[/li][li]Ralph is 80; Frank is 20.[/ul][/li]Ralph is 40; Frank is 10.
2. If Jenny hits a home run, her team will win. Given that this is true, what else also must be true?
[ul][li]If the team won, Jenny hit a home run.[/li][li]If Jenny didn’t hit a home run, the team tied.[/li][li]If the team didn’t win, Jenny didn’t hit a home run.[/li][li]All of the above.[/ul][/li]If the team didn’t win, Jenny didn’t hit a home run.
3. Taking the below statements as a group, which statement is the true one?
[ul][li]The number of false statements here is one.[/li][li]The number of false statements here is two.[/li][li]The number of false statements here is three.[/li][li]The number of false statements here is four.[/ul][/li]The number of false statements here is three.
4. Neko will go to the movies, only if she can drive. Given that this is true, what else also must be true?
[ul][li]If Neko didn’t drive, she didn’t go to the movies.[/li][li]If Neko went to the movies, then she drove.[/li][li]Both of the above statements.[/li][li]If Neko drove, then she went to the movies.[/ul][/li]Both of the above statements.
5. Your room is completely dark. You have eight shoes of four different colors, and fifty socks of five different colors. How many shoes and socks must you grab to make sure you have a matching pair?
[ul][li]5 shoes, 6 socks[/li][li]6 shoes, 8 socks[/li][li]6 shoes, 10 socks[/li][li]5 shoes, 5 socks[/ul][/li]5 shoes, 6 socks
6. No musicians are chefs. No chefs are teachers. Given that these are true, what else also must be true?
[ul][li]No teacher is a musician.[/li][li]Some musicians are teachers.[/li][li]Some teachers are chefs.[/li][li]None of the above.[/ul][/li]None of the above.
7. If QUIZ is written as UYMD, how do you write HEAD?
[ul][li]MIEH[/li][li]MJEH[/li][li]LJEH[/li][li]LIEH[/ul][/li]LIEH
8. Some tigers are not lions. All lions are mammals. Given that these are true, what else also must be true?
[ul][li]Some tigers are lions.[/li][li]Some mammals are not lions.[/li][li]Some lions are not tigers.[/li][li]None of these[/ul][/li]None of these
What I’d like to know is how lavendar underwear make me a Flu Phlegm Green Crayola? :rolleyes:
To be sure, they didn’t say the 50 socks that come in five colors consist of 5 pairs each. For all we know, the fifty socks could consist of three white socks, five green socks, one red sock, three black socks, and 38 blue socks. You’re still good if you fetch six socks at random. At least two of them will match.
I missed the driving-to-the-movies question because I saw a right answer and didn’t realize the next answer was also correct followed by a “both of the above”; missed the chef/musician/teacher one for absolutely no good reason :smack:
Yeah, but who says the shoes and socks will then match? And will they match your pants? Are the pants that go with the blue socks clean? Maybe you should fold your socks and turn the lights on when you pick them.
I don’t get the lion/tiger one. The statement
“Some tigers are not lions”
basically implies that some tigers are lions. It doesn’t literally say that, but if I say “some days I feel tired,” you can be pretty sure that some days, I don’t.
I missed question 6. That makes me 88% logical. According to this site, my brain is like a computer. I don’t think this necessarily follows.
cromulent, you can only assume what is explicitly said.
I don’t understand why #8 isn’t “Some mammals are not lions.” Doesn’t “Some tigers are not lions” suggest that “some tigers *are * lions”? And if all lions are mammals, then “all mammals” must certainly contain “some tigers which are lions - which are therefore mammals”. No?
Yes, but I have the following diagram:
``````
________
/ M \
/ ___ \
| / L \|
| | ||
\ \___//
\________/
``````
The circle for tigers would either be completely enclosed by Mammals, or go outside of Mammals - but some part of it would be inside of it at all times, would it not?
To clarify, the Tiger circle would intersect the Lion circle (“not all Lions are Tigers”) - all of which is inside the Mammal circle (“all Lions are Mammals”).
Some can be All. It just doesn’t have to be.
Some tigers are not lions.
All lions are mammals.
Nothing is said about the portion of the tiger set that haven’t been sampled, or the mammal set that isn’t lions.
Not necessarily. For example, I have fifteen coins. [at least]* Some of them are not nickels. Are any of the others required to be nickels?
I think the question is worded improperly, FWIW. It requires you to think about what logical argument the question writer is trying to use, rather than just thinking about the logic of the problem itself. “What was the question writer thinking at the time” is an incredibly annoying way to have to think.
*This was not part of the verbatim wording of the question, but I thought when I answered the question that it’s what the question writer was thinking about. I got it right only because of that. | 2,015 | 7,649 | {"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} | 4.1875 | 4 | CC-MAIN-2022-21 | latest | en | 0.948299 |
https://www.wyzant.com/resources/probability?f=recent&pagesize=20&pagenum=3 | 1,531,807,042,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589573.27/warc/CC-MAIN-20180717051134-20180717071134-00568.warc.gz | 1,067,228,202 | 13,626 | ## Probability Resources
work out the probability that she takes out exactly two pound and fifty pence please help me fast
5 identical balls are distributed at random in 4 boxes named a,b,c& d find the probability that these boxes contains respectively 1,2,2,0 balls
A probability experiment is conducted in which the sample space of the experiment is Upper S={1, 2, 3,4,5,6,7,8,9,10,11,12} , event F={6,7,8} comma and event G={ 9,10,11,12} . Assume that each outcome...
If a person is offered An Apple and a orange 30 times A-day for 22 years and that person only picks the Orange every time does the Probability of the person picking the Apple increase or decrease...
Assume there is a national lottery and the ticket is worth \$10 million. One winning ticket will be selected if there are one 25 million tickets sold what is the chance that a buyer of one ticket will...
A bag contains eight green marbles and four blue marbles. Taylor draws two marbles in succession without replacing them in the bag. What is the probability of drawing a green marble on the second...
I have two bags that each contains 100 marbles. Bag 1 has 70 red and 30 blue marbles. Bag 2 has 60 red and 40 blue marbles. I pick one marble randomly from Bag 1 and put it into Bag 2. Now I pick...
Jack and Jill each have a bag of balls numbered 1 through 25. Jack draws 17 balls without replacement from his bag and Jill draws 20 balls without replacement from her bag. If they both draw the same...
If A starts first in a coin tossing game where whoever gets head first is winner and it is known that probability of winning of A is equal to that of B.Find value of p
if a 4 digit number is to be formed by randomly selecting from the digits 0, 1, 2, 3, 4, 5, 7, 8, 9, and no repetition, what is the probability that it is even
Approximately 1 in 14 men over the age of 50 has prostate cancer. The level of 'prostate specific antigen' (PSA) is used as a preliminary screening test for prostate cancer...
5. Mike and John are playing a friendly game of darts where the dart board is a disk with radius of 10in. Whenever a dart falls within 1in of the center, 50 points are scored. If the point...
Let X_1,X_2,…,X_n be independent random variables from a symmetrically distributed function over (0, 1). Let y_i be the order statistics. Find the probability that; 0.3<y_3<0.7; if n=5.
A spinner is split in 5 equal parts, one green, one orange, one yellow, one red, one blue. Theoretically, if the spinner is spun 500 times, how many times would you expect it to land on orange? | 687 | 2,568 | {"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-2018-30 | latest | en | 0.898117 |
https://boxbase.org/entries/2019/may/6/misc/ | 1,558,696,205,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257605.76/warc/CC-MAIN-20190524104501-20190524130501-00515.warc.gz | 395,174,489 | 4,163 | # Miscellaneous thoughts & Answers to previous post's quiz
I haven't been on the computer lately. A tough flu has kept me in the bed mostly. I have been taking care of my own nutrition while stretching my shoulders so they wouldn't get as stiff during the next writing session.
The last few weeks has made me realise I've been trying to treat this blog as something it perhaps should not be. I'm not sure what I should do with this blog. Maybe I continue it as a real log, rather than attempt to "create content" about subjects I like about.
Here's my weight from the last month:
It shows 4% weight loss during the last month. I plugged in some values into bmi calculator to estimate how soon I'll be in a healthy weight. I got slightly over 30% body fat percentage and at this rate it'll take about 6 months to get into 10%-15%.
Thinking I focus on this for a while. I'm still writing every week but the subjects are probably quite randomized for a good while.
## Answers to the last week's quiz
The task to prove the equation requires that we examine the properties of the modulo.
``max(divisors(a) ∩ divisors(b)) = max(divisors(b) ∩ divisors(a % b))``
We can conclude that...
``````a % b = 0, max(divisors(a) ∩ divisors(b)) = b
````max(divisors(b) ∩ divisors(0)) = b````
The equation holds when `a % b = 0`. If the modulus is zero, then `b` is a divisor of `a`, so we can conclude it's the gcd.
The equation also holds when `a % b = a`. That is, the `b` does not divide `a` at all.
``a % b = a, max(divisors(a) ∩ divisors(b)) = max(divisors(b) ∩ divisors(a))``
We can model the `a % b` as:
``r = a - b*q``
Assume everything has a greatest common divisor `k`:
``````a = k*a'
``````b = k*b'
``````r = k*r'
``````k*r' = k*(a' - b'*q')
````r' = a' - b'*q'````
So we can confirm that the greatest common divisor doesn't change when we take a modulus because we are technically subtracting `b` from `a` until `a` is smaller in magnitude than `b`.
`````` max(divisors(a) ∩ divisors(b))
````= max(divisors(a) ∩ divisors(b) ∩ divisors(a % b))````
Next we should be able to prove that we can discard `a` in this case:
`````` max(divisors(a) ∩ divisors(b) ∩ divisors(a % b))
````= max(divisors(b) ∩ divisors(a % b))````
If `a%b` is less than `a`, then either `b` or `a%b` are limiting the greatest common divisor. Otherwise `a%b` is exactly as much as `a`, and we end up exchanging the `a`, `b` with each other.
These pieces allow to construct a formal proof, while they're also informal explanations for why the algorithm is correct.
The error on the previous post are the two conflicting statements about divisors:
``````max(divisors(a)) = a
````divisors(0) = all````
For both of these to hold, the zero would have to be the largest number in our system. Therefore the first statement must be weaker.
``````max(divisors(a)) = a when a ≠ 0
````divisors(0) = all````
Retrieving greatest common divisor between 0, and 0 gives `max(all)`.
`````` max(divisors(0) ∩ divisors(0))
``````= max(all ∩ all)
````= max(all)````
The base case can only provide correct result when `a ≠ 0`. If we feed `gcd(0,0)` to our implementation, we get an incorrect result.
I'd have also wanted to show how to actually carry out these proofs in a theorem prover. I guess I'll try things out in coq and read the Coq'Art book. If I get somewhere with this, I'll write a post about it. | 961 | 3,384 | {"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} | 4.125 | 4 | CC-MAIN-2019-22 | longest | en | 0.933739 |
https://people.cs.umass.edu/~barring/cs575/exams/1sol.html | 1,553,409,398,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203378.92/warc/CC-MAIN-20190324063449-20190324085449-00438.warc.gz | 587,779,583 | 6,863 | # Solutions to Midterm Exam, Fall 2018
### Directions:
• Answer the problems on the exam pages.
• There are seven problems (some with multiple parts) for 100 total points. Actual scale was A = 93, C = 63.
• If you need extra space use the back of a page.
• No books, notes, calculators, or collaboration.
``` Q1: 10 points
Q2: 10 points
Q3: 10 points
Q4: 10 points
Q5: 20 points
Q6: 20+10 points
Q7: 10 points
Total: 100+10 points
```
Question text is in black, solutions in blue
For Questions 1-4, the graph G has node set {a, b, c, d, e, f} and nine edges: (a, b), (a, c), (a, d), (b, c), (b, e), (c, f), (d, e), (d, f), and (e, f). Recall that "graph" refers to an undirected graph with no self-loops.
• Question 1 (10): Find a graph with six nodes and nine edges that is not isomorphic to G. Prove that the two graphs are not isomorphic.
The easiest solution to this is to choose any such graph with a degree sequence different from G's which is (3, 3, 3, 3, 3, 3). For example, you could have an isolated node and nine of the ten edges involving the other nodes, which has degree sequence (5, 5, 5, 4, 4, 0). Isomorphic graphs must have the same degree sequence.
Several people chose a graph with the same degree sequence, the complete bipartite graph K3,3. There are a number of ways to show that this graph is not isomorphic to G: it is bipartite while G contains triangles, and it is non-planar (proved in lecture or by quoting Kuratowski's Theorem) while G is planar (by Question 2).
• Question 2 (10): Is G a planar graph? Prove your answer. What is the number of regions in a planar embedding of H, given by Euler's Formula?
G is planar, as is easily shown by giving a planar embedding. We can start with the Hamilton circuit a-b-e-f-c-a and place the edges (b, c) and (c, f) inside it and the edge (a, d) outside it. Euler's formula is V - E + F = 2, so since V = 6 and E = 9, F must be 5. The faces of the planar embedding given above are a-b-c, b-e-f-c, d-e-f, a-c-f-d, and the exterior face a-b-e-d.
• Question 3 (10): Does G have an Euler path and/or Euler circuit? Does H have a Hamilton path and/or Hamilton circuit? Justify your answers.
G has six odd-degree nodes, so it cannot have an Euler circuit (which would require none) or an Euler path (which would require two). It has the Hamilton circuit listed in Question 2 above, and deleting any single edge of this circuit gives a Hamilton path.
• Question 4 (10): What is the chromatic number of G? Prove your answer. (The chromatic number is defined to be the "vertex-chromatic" rather than the "edge-chromatic" number.)
G has a chromatic number of 3. A legal 3-coloring could have a and e red, b and f blue, and c and d green. There can be no 2-coloring because G contains several odd cycles, such as a-b-c-a.
• Question 5 (20): For this problem we will assume that the HAM-CIRCUIT problem is NP-complete, meaning that if there is a polynomial-time algorithm that inputs a graph and decides whether it has a Hamilton circuit, then P = NP. Here are two questions about related problems:
• (a, 10) The HAM-PATH problem is to input a graph and decide whether it has a Hamilton path. Prove that if there is a polynomial-time algorithm for HAM-PATH, then P = NP. (You may use the assumption about HAM-CIRCUIT. Hint (corrected): Apply the HAM-PATH tester to a different graph.)
I had actually intended to ask the easier problem of building a circuit tester with a path tester -- for this you build G' by adding a new node with edges to all the nodes of G, and note that G' has an H-circuit if and only if G has an H-path.
For the problem as posed, one solution uses the path tester to tell whether there is a circuit in G that contains a fixed edge (x, y). To do this we build G' by adding a new node x' with an edge only to x, and a new node y' with an edge only to y. Then if G' has an H-path, it goes from x' to y', and G contains an h-path from x to y that can be extended to a circuit. To test whether G has any H-circuit, we run this "circuit containing (x, y)" test separately either (1) for every edge (x, y) in G, which is O(n2) tests, or (2) for every edge (x, y) in G where x is any fixed node, which is O(n) tests. Clearly this is polynomial time if the assumed H-path tester is polynomial time. So if the path-tester exists, the polynomial-time circuit-tester exists and P = NP.
A more elegant solution (which a student in COMPSCI 311 came up with when this problem was posed there) takes G and builds a single graph G' that has an H-path if and only if G has an H-circuit. Take any node x of G and "clone" it, making two nodes x and y that each have edges to exactly the neighbors of x in G. Then add a node x' with an edge only to x, and a node y' with an edge only to y. Any H-path in G' must go from x' to y' and thus include a path from x to y, which corresponds to an H-circuit in G.
• (b, 10) The LSP (Longest Simple Path) problem is to input a graph G and a number k, and decide whether G contains a simple path of k or more edges. (A simple path is one that never re-uses a node.) Prove that if there is a polynomial-time algorithm for LSP, then P = NP. You may use the assumption about HAM-CIRCUIT and the result of part (a), whether you solved part (a) or not.)
This is very easy given part (a), because a graph has an H-path if and only if it has a simple path with n-1 edges. (Such a path must visit all n vertices, which makes it a Hamilton path.) So if we have the LSP tester, we input the pair (G, n - 1) to it and report the result as that of our path-tester. Thus if the polynomial-time LSP tester exists, so does a poly-time H-path tester and thus (by part (a)) a poly-time H-circuit tester and (by the assumption) P = NP.
• Question 6 (20+10): In this problem we consider distributing n identical treats to four distinguished dogs: Ali (a), Bingley (b), Cardie (c), and Duncan (d), where each dog receives at least one treat. Let f(n) be the number of such distributions.
• (a, 10) Prove that for any n, f(n) is equal to the number of strings of length n - 4 over the alphabet {a, b, c, d}, where the individual letters come in alphabetical order. (That is, all the a's come before all the b's, the b's before the c's, and the c's before the d's. Another way to say this is to say that the string must be in the regular language a*b*c*d*.)
We need a bijection from distributions of treats to strings. If a distribution sends w treats to Ali, x to Bingley, y to Cardie, and z to Duncan, our string will be aw-1bx-1cy-1dz-1. This string has length n-4 because w + x + y + z = n, and clearly the letters are in alphabetical order. Given a string awbxcydz, with w + x + y + z = n - 4, we can give w + 1 treats to Ali, x + 1 to Bingley, y + 1 to Cardie, and z + 1 to Duncan, distributing n treats with at least one to each dog as w, x, y, and z are all non-negative integers.
• (b, 10) Using part (a) or otherwise, find a formula for f(n) as a function of n. You may use ordinary arithmetic operations, the factorial operation, and the binomial coefficient operator C(x, y). Evaluate your formula for n = 3, n = 5, n = 7, and n = 9.
The strings can be counted by the standard "stars and bars" argument. We have n-4 letters (stars) and three barriers between letter types (bars) which can be arranged in C(n-1, 3) ways. Then for n = 3 we have C(2, 3) = 0, for n = 5 we have C(4, 3) = 4, for n = 7 we have C(6, 3) = 20, and for n = 9 we have C(8, 3) = 56.
• (c, 10XC) Find the number of ways to distribute 13 treats to the four dogs such that each dog gets at least one treat and no two dogs get the same number of treats. (As before, the treats are identical and the dogs are distinguished.)
If the dogs were also identical, we would be looking for a partition of 13 into four non-negative, distinct parts, or equivalently a partition of 9 into four distinct parts. There are exactly three such partitions, and I required you to argue that they were the only ones. Looking at 13, we can divide into cases depending on the largest element of the partition. This cannot be greater than 7 because the other three elements must be at least 1, 2, and 3 respectively. If it is 7, we must have 7-3-2-1. If it is 6, we must have 6-4-2-1, and if it is 5, we must have 5-4-3-1. It cannot be 4 because the only way to have distinct positive parts would be 4-3-2-1 which has too small a sum.
Once we have solved this problem, the answer to the given problem is 4! = 24 times greater, because each distribution to identical dogs corresponds to exactly 4! distributions to distinguished dogs. So the answer is 3 times 4! or 72.
Many of you started with the n = 13 case of part (b), which is C(12, 3) = 220 distributions, and tried to subtract the ones that had two or more dogs getting the same number of treats. I don't recall whether anyone got the right answer this way, but several of you were at least close. There are four partitions of 13 with exactly three equal elements: 10-1-1-1, 7-2-2-2, 4-3-3-3, and 4-4-4-1. Each of these gives four distributions to distinguished dogs, so we subtract 16 from our original 220. Then we look at all the partitions with exactly two equal elements, which we can find by a case analysis on the pair: 9-2-1-1, 8-3-1-1, 7-4-1-1, 6-5-1-1, 8-2-2-1, 6-3-2-2, 5-4-2-2, 6-3-3-1, 5-3-3-2, 4-4-3-2, and 5-5-2-1, eleven in all. Each of these yields twelve distributions to distinguished dogs, so we subtract 132 from our remaining 204 to get 72. There are no partitions of 13 with either four equal elements or two pairs of equal elements, so we have checked all the cases where any elements are equal.
• Question 7 (20): The following are ten true-false questions, with no explanation needed or wanted, and no penalty for guessing. They count two points each. Some of the problems use the graph G with the following weights on the edges: (a, b, 1), (a, c, 1), (a, d, 1), (b, c, 2), (b, e, 2), (c, f, 3), (d, e, 4), (d, f, 4), and (e, f, 5).
• (a) If we take an undirected tree with n nodes, and change each edge to a pair of directed edges (in opposite directions), then the resulting directed graph has an Euler circuit of length 2n - 2.
TRUE. We have a strongly connected directed graph where every node has an in-degree equal to its out-degree. Every such graph has an Euler circuit, and the number of edges is 2n - 2 because the original tree had n - 1 undirected edges.
• (b) Any correct comparison-based algorithm to find the median of n items can be interpreted as a decision tree with at least n! leaves.
FALSE. This would be true for sorting, which has n! different possible outputs, but the median problem has only n. We discussed linear-time algorithms for median, and if this statement were true there would be an Ω(n log n) lower bound for the time.
• (c) It is not the case that every two nodes in the weighted version of graph G are within distance 5 of one another.
FALSE. We can carry out a uniform-cost search from each node to see that every other node is within distance 5. Or we could take the 6 by 6 single-step distance matrix and square it, to see that there is a one-step or two-step path of length at most 5 between every pair of distinct nodes.
• (d) The minimum spanning tree in the weighted version of graph G has total weight 7.
FALSE. Using Kruskal's algorithm, we take the three edges of weight 3, then the edge (b, e) of weight 2, then the edge (c, f) of weight 3, to get an MST of weight 8. (We don't take (b, c) because it would form a cycle with the already taken (a, b) and (a, c).)
• (e) Thre is a minimum spanning tree of the weighted version of graph G that does not contain the edge (c, f).
FALSE. Kruskal's algorithm has no ties to break in choosing its first four edges above, and if it then took any other edge than (c, f) the resulting spanning tree would have total weight greater than 8.
• (f) Let H be a flow network and let (X, Y) be a partition of the vertices of H such that the source is in X and the sink is in Y. Let f be a flow in H such that every edge from a vertex in X to a vertex in Y is saturated by f. Then f must be a maximum flow.
FALSE. There could be an edge from a node in Y to a node in X, with positive flow under f, such that we could increase the net flow over f's by decreasing the flow on this edge.
• (g) Let H be a flow network and let f be a flow in H such that for every vertex x in H except for the sink, there is at least one edge out of x that is not saturated by f. Then f cannot be a maximumm flow.
FALSE. Here's a counter example, where each edge has first its capacity and then its flow under the maximum flow f. Source is a and sink is z. We have (a, b, 2, 1), (b, c, 1, 0), (b, e, 1, 1), (c, d, 1, 0), (d, b, 1, 0), and (e, 2, 1, 0). The only saturated edge is (b, e), and since this edge forms a cut by itself the flow is maximum. But there is at least one other unsaturated edge out of each node, except for z.
• (h) Let B be a bipartite graph, with n nodes in a set X, n nodes in a set Y, and every edge with one endpoint in X and the other in Y. There are n! bijections from X to Y, any of which might be a perfect matching in B. Tucker presents no method for finding a perfect matching in B that is substantially faster than trying all n! of these bijections.
FALSE. Tucker discusses finding a perfect matching in a bipartite graph by using the polynomial-time network flow algorithm, which is much faster than trying each of the exponentially many bijections.
• (i) The number of anagrams of the work RHINOCEROS is C(10, 2) times C(8, 2).
FALSE. That is the number of ways to place the two R's and then the two S's. But for each such placement there are 6! total anagrams by placing the other six letters in the six remaining positions.
• (j) The sum for i from 1 to n of C(n, i) is 2n.
FALSE. It is 2n - 1. The sum from 0 to n of C(n, i) is 2n. | 3,912 | 13,902 | {"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} | 3.84375 | 4 | CC-MAIN-2019-13 | latest | en | 0.921702 |
http://www.gurufocus.com/term/netcash/NGL/Net%2BCash/NGL%2BEnergy%2BPartners%2BLP | 1,490,922,868,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218205046.28/warc/CC-MAIN-20170322213005-00309-ip-10-233-31-227.ec2.internal.warc.gz | 529,891,035 | 27,794 | Switch to:
GuruFocus has detected 5 Warning Signs with NGL Energy Partners LP \$NGL.
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NGL Energy Partners LP (NYSE:NGL)
Net Cash per Share
\$-40.58 (As of Dec. 2016)
Net cash per share is calculated as Cash and Cash Equivalents minus Total Liabilities and then divided by Shares Outstanding. NGL Energy Partners LP's net cash per share for the quarter that ended in Dec. 2016 was \$-40.58.
Definition
In the calculation of a companys net cash, assets other than cash and short term investments are considered to be worth nothing. But the company has to pay its debt and other liabilities in full. This is an extremely conservative way of valuation. Most companies have negative net cash. But sometimes a companys price may be lower than its net-cash.
NGL Energy Partners LP's Net Cash Per Share for the fiscal year that ended in Mar. 2016 is calculated as
Net Cash Per Share (A: Mar. 2016 ) = (Cash and Cash Equivalents - Total Liabilities) / Shares Outstanding = (28.176 - 3903.797) / 104.17 = -37.20
NGL Energy Partners LP's Net Cash Per Share for the quarter that ended in Dec. 2016 is calculated as
Net Cash Per Share (Q: Dec. 2016 ) = (Cash and Cash Equivalents - Total Liabilities) / Shares Outstanding = (28.927 - 4459.845) / 109.20 = -40.58
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Explanation
Ben Graham invested in situations where the companys stock price was lower than its net-cash. He assigned some value to the companys other current asset. The value is called Net Current Asset Value (NCAV). One research study, covering the years 1970 through 1983 showed that portfolios picked at the beginning of each year, and held for one year, returned 29.4 percent, on average, over the 13-year period, compared to 11.5 percent for the S&P 500 Index. Other studies of Grahams strategy produced similar results.
You can find companies that are traded below their Net Current Asset Value (NCAV) with our Net-Net screener. GuruFocus also publishes a monthly Net-Net newsletter.
Related Terms
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
NGL Energy Partners LP Annual Data
Mar09 Mar10 Mar11 Mar12 Mar13 Mar14 Mar15 Mar16 netcash 0.00 0.00 0.00 -3.76 -9.16 -11.53 -26.04 -32.90 -43.05 -37.20
NGL Energy Partners LP Quarterly Data
Sep14 Dec14 Mar15 Jun15 Sep15 Dec15 Mar16 Jun16 Sep16 Dec16 netcash -54.14 -58.78 -43.05 -43.05 -42.58 -44.11 -37.20 -38.55 -38.45 -40.58
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GuruFocus Affiliate Program: Earn up to \$400 per referral. ( Learn More) | 750 | 2,892 | {"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-2017-13 | latest | en | 0.94618 |
https://www.scribd.com/doc/68159272/Consumer | 1,500,601,308,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423629.36/warc/CC-MAIN-20170721002112-20170721022112-00680.warc.gz | 850,528,720 | 30,632 | # Consumer Behavior
Ch. 7
Introduction
We have already seen and used an individual’s demand curve. Now, want to explain in more detail why it slopes downward Why do people demand goods and services?
• •
Receive satisfaction or pleasure from consuming the good. Economists terms this satisfaction utility.
others hate it. Example: • • Some people like jazz. how many jazz music CD’s might they purchase. we are not try to explain why people get utility from certain goods. . We take that as a given. Economists say given an individual’s preferences about jazz.Introduction In economics.
Total and Marginal Utility Total Utility (TU) .relates consumption of a good to the utility derived from consuming a good. (This could be many units of a good) Marginal Utility (MU) .the change in total utility when consumption of a good changes by one unit. • MU = DTU / D Q consumed of a good .
. ceteris paribus. a point is reached where the marginal utility obtained by consuming additional units of a good starts to decline.eventually.Law of Diminishing Marginal Utility Law of Diminishing Marginal Utility .
. I get a lot of satisfaction from first slice of pizza.Law of Diminishing Marginal Utility Example • • If I’m really hungry. the satisfaction from the 8th slice would be much less than that of the first slice. If I keep eating pizza.
Law of Diminishing MU Notes about the Law of Diminishing MU Time period must be specified for law.” • Slope of the total utility curve is equal to marginal utility . Law tells us that eventually the total utility curve will become “flatter. Law tells us that eventually the marginal utility curve will be downward sloping.
Marginal Utility MU MU Q .
Shape of MU Eventually downward sloping • Law of diminishing marginal utility Positive always • Rational behavior • Consumer only purchases a good if they get some positive utility from it. .
Total Utility TU TU DQ DTU DTU DQ Q .
Shape of TU Positive slope • Consumer only purchases a good if gets some positive amount of utility (rational behavior) Slope gets flatter as Q increases • Law of diminishing marginal utility .
Consumer Equilibrium Now that we understand the concepts of utility theory .we will use them to explain how consumers make decisions about what to buy .
I have to weigh the price of the good in my decision as well . I would much rather have a Jaguar instead of my Honda If I want to maximize my utility. I don’t just pick the thing that gives me the most pleasure.Consumer Equilibrium For instance. why don’t I buy a Jaguar? Because it costs a lot more than the Honda So if I want to maximize my utility.
Consumer Equilibrium So how can I compare a Jaguar and a Honda? It’s like comparing apples and oranges. Instead. I need to somehow make them both comparable. .
So if I want to spend my money wisely. I can see that even though the Jag gives me more utility. I buy the thing that gives me more utility per dollar. .Consumer Equilbrium In order to do that I will need to convert utility to utility per dollar. I get more utility per dollar from the Honda. This way.
Consumer Equilibrium Let’s say I walk down to the cafeteria for lunch and they have Pizza and Ice Cream. I have \$7 in my pocket What do I buy? . The pizza is \$1 a slice and the Ice Cream is \$2 a scoop.
Consumer Equilibrium Remember. I want to choose the combination of pizza and Ice Cream that gives me the greatest possible utility for my \$7 Consider the following table. which states the total utility I get from all possible quantities of Pizza and Ice Cream .
Marginal Util. Total Util. Marginal Util. 0 1 2 3 4 5 6 0 24 44 60 70 72 72 -- 0 29 46 56 58 59 59 -- .Utility Table Ice Cream Pizza Quantity Total Util.
Utility Table Ice Cream Pizza Quantity Total Util. Marginal Util. Marginal Util. Total Util. 0 1 2 3 4 5 6 0 24 44 60 70 72 72 24 20 16 10 2 0 -- 0 29 46 56 58 59 59 29 17 10 2 1 0 -- .
So I want to buy the pizza. Consider the first scoop of ice cream MU 12 per dollar. I have \$5. Now I have \$6. . I will want to buy the second slice of pizza.Consumer Equilibrium We need to find the marginal utility per dollar for both goods. Now I have to compare my second slice of pizza (MU is 17 /\$) with the first scoop of ice cream (MU is 12 /\$). MU of the first slice of pizza 29 per dollar.
It doesn’t matter which I pick.Consumer Equilibrium Now I have to compare the third slice o pizza (MU 10/\$) with the first scoop of ice cream (MU 12/\$). I will want to buy the ice cream. Now I have \$2 . I have \$3. since they make me equally happy. Now I have to compare the third slice of pizza (MU 10 /\$) with the second scoop of ice cream (MU 10 /\$). I’ll take the pizza.
My total utility from lunch is 56+44=100. I bought 3 slices of pizza which give a total utility of 56 and 2 scoops of ice cream which give a total utility of 44. I will want to buy the ice cream. I have no more money. There is no other combination of pizza and ice cream that give a greater utility for \$7.Consumer Equilbrium Now I have to compare the fourth slice of pizza (MU is 2/\$) to the second scoop of ice cream (MU is 10 /\$). .
. Assignment: Convince yourself that I will buy 4 scoops of ice cream and 4 slices of pizza. Note that when the price went down.Consumer Equilbrium What if the price of the ice cream dropped to \$1 a scoop.THIS IS WHERE THE LAW OF DEMAND COMES FROM. I bought more .
Consumer Equilibrium In summary. . you need to convert marginal utility to marginal utility per dollar Then compare MU/P for the two goods and buy the one that gives the greatest MU/P Subtract the price from your budget Compare the next available units of both goods and repeat the process until you are out of money.
the difference between the price buyers pay for a good and the maximum amount they would have paid for the good. Example: • • • I’m willing to pay \$6 for a case of soda Soda is on sale for \$5 a case Consumer surplus = \$1 .Consumer Surplus Consumer Surplus .
Consumer Surplus P \$9 This is the Consumer Surplus for the second case of soda S \$7 \$5 D 0 1 2 3 Q .
Consumer Surplus Here is the generally accepted method of finding the total Consumer Surplus in a market .
Consumer Surplus P The area of this triangle is the total Consumer Surplus S P* D 0 Q* Q .
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70.438:207\$:75:8 ! %0.4198 97. 438:207\$:75:8 \$ ! " " .30890949. | 2,915 | 7,976 | {"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-2017-30 | latest | en | 0.909304 |
https://kr.mathworks.com/matlabcentral/cody/problems/43739-repeat-vector-values-an-arbitrary-number-of-times/solutions/1076313 | 1,600,932,592,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400214347.17/warc/CC-MAIN-20200924065412-20200924095412-00579.warc.gz | 434,253,584 | 16,730 | Cody
# Problem 43739. Repeat Vector Values an Arbitrary Number of Times
Solution 1076313
Submitted on 7 Dec 2016 by Binbin Qi
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
1 x1=[1 2 3 4 5 6]; x2=[2 2 1 1 2 2]; out=[1 1 2 2 3 4 5 5 6 6]; assert(isequal(your_fcn_name(x1,x2),out))
ans = 1
2 Pass
2 x1=1; x2=7; out=[1 1 1 1 1 1 1]; assert(isequal(your_fcn_name(x1,x2),out))
ans = 2
3 Pass
3 x1=1:650; x2=ones(1,650); out=1:650; assert(isequal(your_fcn_name(x1,x2),out))
ans = 3
4 Pass
4 x1=5.*ones(250,1); x2=3.*ones(250,1); out=5.*ones(750,1); assert(isequal(your_fcn_name(x1,x2),out))
ans = 4
5 Pass
5 x1=[1 500 32 780 5 8]; x2=[7 6 5 4 3 2]; out=[1,1,1,1,1,1,1,500,500,500,500,500,500,32,32,32,32,32,780,780,780,780,5,5,5,8,8]; assert(isequal(your_fcn_name(x1,x2),out))
ans = 5 | 419 | 919 | {"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} | 3.375 | 3 | CC-MAIN-2020-40 | latest | en | 0.466101 |
http://mathoverflow.net/revisions/58266/list | 1,369,459,031,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368705543116/warc/CC-MAIN-20130516115903-00070-ip-10-60-113-184.ec2.internal.warc.gz | 166,467,865 | 5,214 | 2 deleted 2 characters in body
This is with regard in answer to your second question. There is a note by Gustafson:
• MR0327901 (48 #6243) Gustafson, W. H. What is the probability that two group elements commute? Amer. Math. Monthly 80 (1973), 1031–1034.
where he proves the result Ben mentions, viz. if $G$ is a finite nonabelian group, then $d(G) \leq 5/8$. He goes on to prove that the same result for the case where $G$ is a compact, Hausdorff topological group (endowed with the Haar measure).
While $d(G)$ has received some attention over the years (I think it was first mentioned in a paper of Erdos in the late 60s and there have been sporadic papers since then) very little seems to have been said about $d(G)$ where $G$ is an infinite group until recently. The basic results (most of which are analogous to the finite case) are proved in
• MR2558527 (2010m:22003) Rezaei, Rashid; Erfanian, Ahmad(IR-MASHM) On the commutativity degree of compact groups. (English summary) Arch. Math. (Basel) 93 (2009), no. 4, 345–356.
Ben has already mentioned the nice paper of Levai and Pyber where it is proved that if $G$ is a profinite group and $d(G) > 0$, then $G$ is abelian-by-finite. This result is extended to all compact groups in a recent preprint by Hofmann and Russo. There is much more besides in this preprint, I'm still digesting it myself!
1
This is with regard to your second question. There is a note by Gustafson:
• MR0327901 (48 #6243) Gustafson, W. H. What is the probability that two group elements commute? Amer. Math. Monthly 80 (1973), 1031–1034.
where he proves the result Ben mentions, viz. if $G$ is a finite nonabelian group, then $d(G) \leq 5/8$. He goes on to prove that the same result for the case where $G$ is a compact, Hausdorff topological group (endowed with the Haar measure).
While $d(G)$ has received some attention over the years (I think it was first mentioned in a paper of Erdos in the late 60s and there have been sporadic papers since then) very little seems to have been said about $d(G)$ where $G$ is an infinite group until recently. The basic results (most of which are analogous to the finite case) are proved in
• MR2558527 (2010m:22003) Rezaei, Rashid; Erfanian, Ahmad(IR-MASHM) On the commutativity degree of compact groups. (English summary) Arch. Math. (Basel) 93 (2009), no. 4, 345–356.
Ben has already mentioned the nice paper of Levai and Pyber where it is proved that if $G$ is a profinite group and $d(G) > 0$, then $G$ is abelian-by-finite. This result is extended to all compact groups in a recent preprint by Hofmann and Russo. There is much more besides in this preprint, I'm still digesting it myself! | 751 | 2,677 | {"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} | 2.640625 | 3 | CC-MAIN-2013-20 | latest | en | 0.95814 |
https://unitconverter.net/surface-current-density-converter | 1,576,311,247,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540585566.60/warc/CC-MAIN-20191214070158-20191214094158-00163.warc.gz | 574,802,500 | 12,120 | UnitConverter.netV1.2
Home / Surface Current Density Converter
Surface Current Density Converter
Enter the value that you want to convert
From: To: From: ampere/square meter [A/m^2]ampere/square centimeterampere/square inch [A/in^2]ampere/square mil [A/mi^2]ampere/cicular milabampere/square centimeter To: ampere/square meter [A/m^2]ampere/square centimeterampere/square inch [A/in^2]ampere/square mil [A/mi^2]ampere/cicular milabampere/square centimeter Value:
Objective of Measurement:
Measurement is the most important aspect of our life. We use measurement in science, engineering, business trading, personal life, education, and more other fields. As technology is growing day by day so we need a highly accurate and easy convenient global measuring system in each and every field. It is essential to use standard measurement in every field that everyone to be sure that they not get cheated.
History of Measurement:
In history for measurement people used the human body as a tool. For measuring length used forearm, hand, foot & finger as a unit. The foot, finger is a subdivided shorter unit of a length. This type of measurement is not accurate cause different in size of the arm & finger for different people & some of the countries still using it. In history, there were lots of measuring systems developed but mostly used imperial, the metric system of measurement. We use these systems for measure distances, volume, weight, speed, area etc. Due to this a major problem everyone is facing while doing trading between the countries. A huge improvement in civilization, It necessary to improve measuring standards. Nowadays International Standard (SI) units are used as a global measurement system.
Surface Current Density Converter - Unit Converter:
It is a free online surface current density converter. It converts surface current density units ampere/square meter [A/m^2], ampere/square centimeter, ampere/square inch [A/in^2], ampere/square mil [A/mi^2], ampere/circular mil, abampere/square centimeter vice versa with metric conversion.
Surface current density converters & it's abbreviations
UnitAbbreviationUnitAbbreviationUnitAbbreviation
coulomb/square meterC/m2coulomb/square centimeterC/cm2coulomb/square inchC/in2
abcoulomb/square meterabC/m2abcoulomb/square centimeterabC/cm2abcoulomb/square inchabC/in2
Complete list of Surface current density converter units and its conversion. | 561 | 2,417 | {"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-2019-51 | longest | en | 0.884503 |
https://www.mathshub.com.sg/mathematics/programmes/fusion-math-2/fusion-math-p4-10-yr-old/ | 1,721,798,467,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518157.20/warc/CC-MAIN-20240724045402-20240724075402-00772.warc.gz | 753,166,624 | 8,719 | ## Fusion Math P4 (10 yr old)
##### Objectives
1. Develop students’ interest, creativity, analytical thinking skills and spirit of learning through discoveries and challenges in Mathematics.
2. Impart knowledge and skills through various non-routine and problem solving heuristics.
3. Cultivate students’ aptitude in solving problems as topics taught are in line with NMOS.
Year End Holiday Semester 1 (Jan – Mar)
– Intersections (Triangles, Square)
– Regions
– Conversion
– Symmetry
– Problem Solving on Listing
– Problem Solving on Grouping
– Tessellations
– Guess & Check
– Working Backwards
– Assumption Method
– Odd Shapes Out
– Number Grouping
– Alphametics
– Summation
– Angles
– Area & Perimeter
– Fraction, Ratio, Percentage
– Problem Solving on Average
March Holiday Semester 1 (Mar – May)
– Angles
– Puzzle Solving
– Volume
– Problem Solving on Rate
– Revision
– Speed
– Angles
– Problem Solving
– Worst Case Scenario
– Number Manipulation
– Constant Total
– Constant Difference
– Constant Single Subject
Mid Year Holiday Semester 2 (Jun – Sep)
– Number Grouping
– Ones Digit
– Number Sense
– Area & Perimeter
– Speed
– Mock Papers
– Fraction, Ratio, Percentage
– Rate
– Summation
– Average
– Basic Permutation
– Probability
– Working Backwards
– Problem Solving Heuristic
September Holiday Semester 2 (Sep – Nov)
– Counting Triangles
– Counting Rectangles
– Pattern
– Surface Area
– Fraction, Ratio, Percentage
– Area & Perimeter
– Angles
– Surface Area
– Volume
– Remainder
– Grouping
– Working Backwards
– Revision
Topics covered are subjected to change depending on the ability of students. | 411 | 1,608 | {"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-2024-30 | latest | en | 0.785981 |
https://academiaelites.com/what-is-the-strongest-correlation-in-the-matrix/ | 1,675,947,785,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499966.43/warc/CC-MAIN-20230209112510-20230209142510-00160.warc.gz | 101,891,310 | 19,127 | # What is the strongest correlation in the matrix?
Posted: January 7th, 2023
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Answer the following questions using the Week 6 Correlations Exercises SPSS Output provided in this week’s Learning Resources.
What is the strongest correlation in the matrix? (Provide the correlation value and the names of variables)
What is the weakest correlation in the matrix? (Provide the correlation value and the names of variables)
How many original correlations are present on the matrix?
What does the entry of 1.00 indicate on the diagonal of the matrix?
Indicate the strength and direction of the relationship between body mass index (BMI) and physical health component subscale.
Which variable is most strongly correlated with BMI? What is the correlational coefficient? What is the sample size for this relationship?
What is the mean and standard deviation for BMI and doctor visits?
What is the mean and standard deviation for weight and BMI?
Describe the strength and direction of the relationship between weight and BMI.
Describe the scatterplot. What information does it provide to a researcher?
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\$0.00 | 311 | 1,534 | {"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-2023-06 | latest | en | 0.932386 |
https://baremetrics.com/academy/reduce-churn-retain-customers?utm_campaign=mastermindjam-friday-dispatch-issue-99&utm_medium=email&utm_source=newsletter | 1,563,814,208,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195528141.87/warc/CC-MAIN-20190722154408-20190722180408-00503.warc.gz | 322,466,110 | 13,954 | Churn. The archenemy of any subscription company. No metric causes more sleepless nights or thinning hair. It’s the slow leak that regardless of how handy you are, you’ll never completely plug. If recurring revenue is a rainbow leading to a pot of gold, then churn is the dirty leprechaun trying to keep it all from you. It’s the bane of our business existence!
When you’re a brand new business, churn is generally something you don’t need to worry about. Or rather, there are probably bigger wins than trying to figuring out how to reduce churn by a few percentage points.
But eventually you’ll have to face it. Growth will eventually taper off and churn will become a major issue to tackle.
## What is churn?
First, lets define churn, at the very least so we’re all talking about the same thing.
Churn can most basically be defined as resources lost in a given period of time. Typically it’s referring to users or revenue lost and is usually represented with either a percentage or dollar amount.
For example, if you had a 5% monthly user churn rate, that means each month 5% of your customer base is canceling.
Or if you said you had \$2,000 in monthly revenue churn that means you lost \$2,000 in monthly recurring revenue from either customer cancellations or downgrades.
The plain-english formula for calculating churn is: `Lost Resource / # of resources at beginning of a given interval`
More plainly put, say you’re calculating user churn for the month of May. You had 100 customers at the start of May and 5 canceled during the month.
You’d say `(5 customers / 100 customers) * 100 = 5% user churn`
It’s important to note that you ignore any new customers added during that time frame.
### User Churn vs. Revenue Churn
Most people talking about churn are referring to “user churn”…but there’s another type as well.
User churn is the number of customers you’re losing in a given timeframe (typically per month or year).
But there’s also revenue churn, which is arguably even more important. Revenue churn is the amount of revenue you’re losing in a given timeframe due to downgrades or cancellations.
The reason why it’s such a vital metric is that it has a greater affect on your business. If you look at just user churn, you’re ignoring how much revenue you’re losing with those churned users.
For example, a churned user on a \$50/mo plan isn’t nearly as bad as a churned user on a \$500/mo plan. Just looking at user churn would gloss over the fact that you lost a major customer. Revenue churn effectively weights your user churn to be a more accurate representation of how your business is doing.
## Why keeping track of churn is so crucial
Okay, you get the “how” and the “what”, but what about the “why”? Why is it so important to track these types of churn? It comes down to growth and customer acquisition.
Churn is the antithesis of growth. It’s the undoing of all your hard work and money spent acquiring customers.
The more customers you acquire now, the fewer there are to acquire in the future, which means each new customer is incrementally harder and more expensive to acquire. If you keep bleeding customers, you’re making it increasingly harder to grow down the road.
Churn also points to a problem with the product itself. You need to identify what those problems are because that points to the reason people are churning. Knowing those problems gives you a game plan for reducing the churn.
## What are ways to reduce churn?
Reduction of churn ultimately comes down to one thing: engagement. Highly-engaged users churn at a much lower rate than others. There will always be factors out of your control, such as a business shutting down, but given all the factors you can control, you’ll find the most success doing things that help users stay more engaged.
Now that we’ve got all of that out of the way, let’s get to the meaty stuff. Here are 13 practical, actionable ways you can start reducing churn today.
### 1. Make something people want
This is one of those things that sounds obvious, but actually happens a lot less than you think. I mean, what Founder says “yeah you know, we’re trying to create something that people don’t really want”…no Founder says that. Everyone believes they’re creating an amazing thing that everyone wants.
But wishing it doesn’t make it so.
Some signs that you aren’t making something people want are double-digit monthly churn rates, especially when they’re over 20%. If you’re around 10% monthly churn, there are likely some things you can do to improve that. Maybe you’re doing a poor job executing on the idea or you’re doing a bad job onboarding.
But once you get in the 20%+ range, you’ve probably got a serious fundamental problem with your product. You should consider going back to the drawing board and trying a drastic change in the way you’re solving the problem to see if that gets you in to safer territory.
### 2. Redefine your onboarding process relentlessly
As nice as it’d be to have new users jump in and instantly know how to make the most of the product, that’s just not how it works.
A lot of churn is due to users simply not knowing how to use the product and thus making the assumption that you aren’t able to solve the problems they wanted to for them.
By walking users through an initial setup or onboarding process, you’re able to show users how they can start getting value right away.
Then, as they continue to use the product more, you keep slowly dripping out features and functionality to help them become power users.
Your onboarding process is something you should be constantly tweaking and testing. Talk to users, learn where they get hung up and then oil that spot so things move more smoothly.
### 3. Increase app engagement
As I mentioned earlier, engaged users don’t churn. Customers using your product on a regular basis are doing so because they get value out of it. But as soon as their engagement starts dropping off, they’re much more likely to churn.
Give your customers reasons to keep coming back. Provide a way for them to get value on a daily basis. This could be in the form of a daily report showing them what’s changed/happened in the last day or showing them a list of things to do in the coming day. Provide something that makes you part of their daily workflow.
### 4. Help them succeed
One of the biggest motivators for humans is success. Not even from some grand “change the world” aspect, but just succeed at their jobs. To be good at what they do.
One of the best ways to do this is with education and training. And I’m not talking about education and training around using your product…I’m talking about helping your customers get better at their jobs.
So, for us, with Baremetrics, we provide an immense number of articles and videos around helping founders run their companies better.
### 5. Become part of the company workflow
When an entire department at a company, or even the entire company itself, makes your tool part of their workflow, they’re guaranteed to stick around for a long time.
Changing processes is really difficult and painful for a company, and they’ll go out of their way to avoid it.
The more you ingrain your business in your customer’s day-to-day workflow, the longer they’ll be a customer.
### 6. Attract the right audience from the start
All the tips in the world won’t matter if you’ve been attracting the wrong audience. This is a big reason why leading with “free” or “cheap” can backfire on you.
When the first interaction with your company is anchored by “free,” you’re attracting customers who aren’t looking for the value you provide.
Attracting an audience that’s primed to not only receive but also understand the value they’re getting means you’ll have a much easier time down the road conveying that value.
### 7. Continuously convey the value
Many times, after a customer has stuck around for a while, they can begin to overlook how much value you’re offering on a regular basis. Putting certain mechanisms in place that help convey the value you’re offering can go a long way.
A great example of this is a daily, weekly or monthly report that’s generated automatically showing pertinent data points related to their business and how it correlates to their use of your service.
For example, we send out email reports to our customers about the state of their metrics and how they’ve changed on a daily basis, as well as how those metrics are progressing towards their goals.
### 8. Optimize the cancellation flow
The reasons people cancel are vast and many times it’s not immediately related to your product at all. There are many external influencers that cause people to want to cancel, so offering some different options during the cancellation process can go a long way to mitigating the cancellation at all!
A couple of things you can do to alter your cancellation flows that may actually reduce your cancellation rate as a whole:
• Offer a downgrade — If they’re in a budget crunch, suggesting a cheaper plan can help sustain them until they’ve got more funds available.
• Offer to pause the account — Sometimes your service may not be needed for a month or two and offering to temporarily pause billing or to drastically reduce the monthly fee while paused can help keep a customer from completely abandoning your service.
• Track views/visits to the cancellation page — When a user views your cancellation screen but doesn’t cancel, it can be a sign that they’re at least thinking of cancelling. Once you detect that, you trigger an email to be sent from you to offer some help or send a short series of emails that reiterate the value they’re receiving.
### 9. Customer exclusives
You’ve probably seen this before in the form of a directory of discounts for other services and products. “3 months free”, “50% off your first purchase, “100GB free”…all listed in a directory that’s only available to customers of the product.
These can work great and are relatively easy to put together.
Pushing these customer exclusives right before it’s time to renew is also a great way to drive home how much value they’re getting!
### 10. Under promise and over deliver
When a customer signs up for your service, they do so with a set of expectations. Expectations around what you’ll be able to do for them, what value they’ll get, what problems you’ll solve. It’s your job to manage those expectations by not over promising anything.
Don’t promise to 10x their customer base unless you can actually follow through on that.
You’re much better offer under promising what you can do for them and then blowing it out of the water. When you over deliver on a promise, customers feel the value they’re getting far exceeds what they’re paying for…instantly making them more loyal.
### 11. Retarget your current customers
I know, it sounds odd. Advertising to people who are already paying you. But it’s a relatively cheap way to stay top of mind, even when they’re not currently inside your app.
It could be anything from a simple “thank you” image to something that promotes a feature they aren’t currently using.
### 12. Create integrations
Your customers likely use dozens of software products every day and one of their biggest pain points is switching between and cross-referencing data between them.
By providing direct integrations with the most common apps they use, you’re able to save them time, which ultimately makes them more money.
Nearly every software product out there has an API of some sort and doing an integration with them is a great way to help your customers while simultaneously gaining access to new customers.
### 13. Collect on failed charges
The average startup is losing thousands of dollars every month to failing charges. Most of these customers don’t even realize their card is failing. There’s no malicious intent, they’re just oblivious.
If you don’t proactively work to collect on those failing charges and help your customers update their billing info, they can go months without paying you.
Use a tool to automate this process (called “dunning”) that handles all the customer outreach to make sure you’re not losing customers to revenue churn. This type of churn is almost completely avoidable with the right system.
Baremetrics actually has a Dunning tool. It has saved companies hundreds of thousands of dollars and you can get up and running with it in a matter of minutes.
## Summary
There are 13 ways you can start reducing churn and retaining customers today! Reducing churn really is key to long term, sustainable growth. It’s something that once you get a handle on it, you really can see massive growth without spending a ton of money. | 2,665 | 12,718 | {"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-2019-30 | latest | en | 0.952654 |
https://es.mathworks.com/help/physmod/sdl/ref/8speed.html | 1,611,313,032,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703529179.46/warc/CC-MAIN-20210122082356-20210122112356-00159.warc.gz | 314,664,665 | 17,915 | # 8-Speed
Clutch schedule for an eight-speed transmission
• Library:
• Simscape / Driveline / Transmissions
## Description
The 8-Speed block consists of four planetary gear sets and five clutches. The base shaft connects to the:
• Planet gear carrier of the second planetary gear
• Other gears through three clutches that determine the power flow path.
The follower shaft connects to the planet gear carrier of the first and fourth planetary gears. The other two clutches serve as brakes, grounding various gears of planetary sets two and three to the transmission housing.
This diagram shows an eight-speed transmission. The labels for the gear components are superimposed on the input and output gears. The table lists the gear and clutch components that are labeled in the diagram.
LabelComponent
P.G.1–P.G.4Planetary gears, 1–4
R Ring gear
CPlanet gear carrier
SSun gear
C–EForward clutches that control the power flow path
A–BForward, braking clutches
### Drive Ratios, Clutch Schedule, and Power Flow
The drive ratio between the transmission input and output shafts follows from the elementary gear ratios specified for the gear blocks. The elementary gear ratios are
`${g}_{x}=\frac{{N}_{{R}_{x}}}{{N}_{{S}_{x}}},$`
where:
• NRx is the number of teeth in the planetary ring gear x, where x = 1, 2, 3, and 4.
• NSx is the number of teeth in the planetary sun gear x, where x = 1, 2, 3, and 4.
The table shows the clutch schedule, drive-ratio expressions, drive-ratio default values, and the power-flow diagrams for each gear of the 8-Speed block. The schedule and gear ratios are based on the manufacturer data for the GM 8L90 eight-speed automatic transmission.
The letters in the clutch schedule columns denote the brakes and clutches. A value of `1` denotes a locked state and a value of `0` an unlocked state. The clutch schedule generates these signals based on the Gear port input signal. The signals are scaled through a Gain block and used as actuation inputs in the clutch blocks.
The power-flow diagrams show the power flow paths between input and output shafts for each gear setting. Power flow is shown in orange. Connections to the transmission housing (a mechanical ground) are shown in black.
GearClutch ScheduleDrive Ratio EquationDefault RatioPower Flow
ABCDE
810011$\frac{{g}_{2}}{1+{g}_{2}}$0.65
710110$\frac{{g}_{2}\left(1+{g}_{1}\right)}{1+{g}_{2}\left(1+{g}_{1}\right)}$0.85
600111$1$1
501110$1+\frac{{g}_{3}{g}_{4}}{\left(1+{g}_{3}+{g}_{4}+{g}_{2}{g}_{4}\left(1+{g}_{1}\right)\right)}$1.27
401011$\frac{1+{g}_{2}+{g}_{3}}{1+{g}_{2}}$1.69
301101$\frac{\left(1+{g}_{3}\right)\left(1+{g}_{4}\right)}{1+{g}_{3}+{g}_{4}}$2.07
211001$\frac{{g}_{2}\left({g}_{4}+1\right)}{\left(1+{g}_{2}\right)}$2.96
111100$1+{g}_{4}$4.55
R11010$-\frac{{g}_{2}\left({g}_{1}{g}_{4}-1\right)}{\left({g}_{2}+1\right)}$-3.91
## Ports
### Input
expand all
Specify -1 for reverse, 0 for neutral, and 1–8 for the forward gears.
Data Types: `single` | `double`
### Conserving
expand all
Conserving port associated with the base shaft.
Conserving port associated with the follower shaft.
## Parameters
expand all
Sets the initial engagement or disengagement of the clutches using the clutch schedule. When the transmission is in neutral, there is no power flow from the base shaft to the follower shaft. | 939 | 3,339 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 10, "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-2021-04 | latest | en | 0.850309 |
https://calculat.io/en/length/inches-to-cm/382.28 | 1,643,443,295,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300573.3/warc/CC-MAIN-20220129062503-20220129092503-00466.warc.gz | 206,429,865 | 5,711 | # Convert 382.28 inches to cm
"Inches to Centimeters" Calculator
Convert
Inches to cm
## Answer for "How many cm is 382.28 inches?"
`382.28″ = 970.99 cm`
(382.28 Inches is equal to 970.99 Centimeters)
## Explanation of 382.28 Inches to Centimeters Conversion
`Inches to Centimeters Conversion Formula: cm = in × 2.54`
According to 'inches to cm' conversion formula if you want to convert 382.28 (three hundred eighty-two point two eight) Inches to Centimeters you have to multiply 382.28 by 2.54.
Here is the complete solution:
`382.28″ × 2.54 = 970.99 cm(nine hundred seventy point nine nine centimeters)`
## About "Inches to Centimeters" Calculator
This converter will help you to convert Inches to Centimeters (in to cm). For example, How many cm is 382.28 inches? Enter the number of inches (e.g. '382.28') and then click the 'Convert' button.
Convert
Inches to cm
## Inches to Centimeters Conversion Table
InchesCentimetres
380.78 inches967.18 cm
380.88 inches967.44 cm
380.98 inches967.69 cm
381.08 inches967.94 cm
381.18 inches968.2 cm
381.28 inches968.45 cm
381.38 inches968.71 cm
381.48 inches968.96 cm
381.58 inches969.21 cm
381.68 inches969.47 cm
381.78 inches969.72 cm
381.88 inches969.98 cm
381.98 inches970.23 cm
382.08 inches970.48 cm
382.18 inches970.74 cm
970.99 cm
382.38 inches971.25 cm
382.48 inches971.5 cm
382.58 inches971.75 cm
972.01 cm
382.78 inches972.26 cm
382.88 inches972.52 cm
382.98 inches972.77 cm
383.08 inches973.02 cm
383.18 inches973.28 cm
383.28 inches973.53 cm
383.38 inches973.79 cm
383.48 inches974.04 cm
383.58 inches974.29 cm
383.68 inches974.55 cm
970.99 | 526 | 1,615 | {"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} | 3.421875 | 3 | CC-MAIN-2022-05 | latest | en | 0.611584 |
https://edurev.in/studytube/Introduction-to-Number-System-and-Decimal-Expansio/9436a3e3-acfd-4d6b-b349-f0744a537459_t | 1,653,781,352,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663021405.92/warc/CC-MAIN-20220528220030-20220529010030-00230.warc.gz | 279,032,403 | 58,059 | # Introduction to Number System Notes | Study Mathematics (Maths) Class 9 - Class 9
## Class 9: Introduction to Number System Notes | Study Mathematics (Maths) Class 9 - Class 9
The document Introduction to Number System Notes | Study Mathematics (Maths) Class 9 - Class 9 is a part of the Class 9 Course Mathematics (Maths) Class 9.
All you need of Class 9 at this link: Class 9
Real Number System
Classification of Real Numbers
Numbers
In real life, we use Hindu Arabic numerals - a system which consists of the symbols 0 to 9.
This system of reading and writing numerals is called, “Base ten system” or “Decimal number system”.
Natural Numbers
Counting numbers are called natural numbers. These numbers start with the smallest number 1 and go on without end. The set of all natural numbers is denoted by the symbol ‘N’.
N = { 1, 2, 3, 4, 5, .......} is the set of all natural numbers.
Whole Numbers
Natural numbers together with zero (0) are called whole numbers. These numbers start with the smallest number 0 and go on without end.
The set of all whole numbers is denoted by the symbol ‘W’.
W = { 0, 1, 2, 3, 4, 5, .......} is the set of all whole numbers.
Integers
The whole numbers and negative numbers together are called integers.
The set of all integers is denoted by Z.
Z = {... - 2, - 1, 0, 1, 2, ...,} is the set of all integers
Fractions
A fraction is a part or parts of a whole.
Fig: Fractions
In a fraction, the number above the line is called the numerator and the number below the line is called the denominator.
Decimal Numbers
A number in which we have "point" is called a decimal number.
A decimal number has two parts namely an integral part and a decimal part.
Examples:
1) Let us consider the decimal number 0.6
0.6 can be written as 0 + 0.6
Here, integral part = 0 and decimal part = 6
2) Let us consider the decimal number 7.2
7.2 can be written as 7 + 0.2
Here, integral part = 7 and decimal part = 2
In a decimal number, the digits to the left of the decimal point is the integral part.
The digits to the right of the decimal point are the decimal part.
The value of all the decimal parts is less than 1.
Rational Numbers
Both "p" and "q" must be integers and q≠ 0
So, any number in the form of a fraction can be treated as a rational number.
Examples:
5, 2.3, 0.02, 5/6
Because all these numbers can be written as fractions.
5 = 5/1
2.3 = 23/10
0.02 = 2/100 = 1/50
5/6 (This is already a fraction)
Apart from the above examples, sometimes we will have recurring decimals like 1.262626..........
1.262626........ is a non-terminating recurring decimal.
All these recurring decimals can be converted into fractions and they are also rational numbers.
Important note:
All the fractions and decimal numbers will come under this category.
Hence, all the fractions and decimal numbers to be considered as rational numbers.
Irrational Numbers
A number that can not be converted into a fraction is called irrational numbers.
Examples:
All the above non-terminating numbers can not be converted into fractions because they do not have repeated patterns.
When we are trying to find the square root of a number that is not a perfect square, we get this non-repeating and non-terminating decimal.
And these non-recurring decimals can never be converted into fractions and they are called irrational numbers.
Decimal Expansions of Real Numbers
There are three types of decimal expansions of real numbers.
1. Terminating
2. Non-Terminating but Repeating
3. Non-Terminating and Non-Repeating
1. Terminating: The remainder becomes zero.
Let us take examples to know it.
Example 1:
Expansion of 7/4
After performing some steps, we get the remainder as 0.
The remainder is 0, and the decimal expansion ends at 5. So it means the expansion is terminating.
Example 2:
Expansion of 32/5
Again we found that after some steps the remainder becomes zero with the decimal expansion as 6.4.
Similarly, the expansion of 32/5 is terminating.
Example 3:
Expansion of 578/25
Decimal expansion of 578/25 = 23.12
Therefore, the expansion of 578/52 is terminating.
2. The remainder never becomes zero: There are two cases:
(a) Non-Terminating but Repeating.
(b) Non-Terminating and Non-Repeating
Example: Expansion of 10/3
Decimal expansion of 10/3 = 3.333333...
The expansion of 10/3 does not end that is not terminating, and number 3 is repeating, so it is a non-terminating but recurring expansion.
Example: Expansion of 1/7
Decimal expansion of 1/7 = 1.142857...
The expansion of 1/7 does not end that is not terminating, and numbers 142857 are repeating, so it is a non terminating but recurring expansion.
1/7 =
Example:
π is a non-terminating, non-repeating decimal. π = 3.141 592 653 589 793 238 462 643 383 279 ...
e is a non-terminating, non-repeating decimal. e = 2.718 281 828 459 045 235 360 287 471 352 ...
The document Introduction to Number System Notes | Study Mathematics (Maths) Class 9 - Class 9 is a part of the Class 9 Course Mathematics (Maths) Class 9.
All you need of Class 9 at this link: Class 9
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# How do you evaluate the power $- {\left( {1.2} \right)^2}$?
Last updated date: 14th Jun 2024
Total views: 372k
Views today: 6.72k
Verified
372k+ views
Hint: Here, we need to square the decimal number. For this we will first convert it into the fraction form so that it will be easy for us to do its square. Also it is important to consider that the negative sign is out of the bracket, so when we square the number, the negative sign will remain as it is.
Complete step-by-step solution:
We need to evaluate the power $- {\left( {1.2} \right)^2}$.
First we will convert $1.2$ into fraction form.
As there is one digit after the decimal, we will divide the number with 10 after removing the decimal sign to convert it into fraction form.
$\Rightarrow 1.2 = \dfrac{{12}}{{10}}$
We will not simplify the fraction here as it is easy to convert the fraction into decimal after squaring if it has the denominator in multiplication with 10.
Therefore, we can write
$\Rightarrow - {\left( {1.2} \right)^2} = - {\left( {\dfrac{{12}}{{10}}} \right)^2} = - \dfrac{{144}}{{100}}$
Now, we will again convert the decimal number into fraction form.
As the denominator is 100, we will put a decimal sign after two digits from the left side.
$- \dfrac{{144}}{{100}} = - 1.44$
Thus, by evaluating the power $- {\left( {1.2} \right)^2}$, we get $- 1.44$.
Note: While solving this type of question, be careful when we convert decimals into fraction and fraction into decimals. We can also solve this problem by directly multiplying the number twice.
$- {\left( {1.2} \right)^2} = - \left( {1.2 \times 1.2} \right) = - 1.44$.
When we multiply two decimals, the number of digits after decimal sign in the product will be the total number of digits after decimal sign in the number. For example, here $1.2$ has one digit after decimal and when it is multiplied two times. The final answer $1.44$ has two digits after the decimal sign. | 541 | 1,990 | {"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": 2, "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} | 4.71875 | 5 | CC-MAIN-2024-26 | latest | en | 0.82831 |
https://maker.pro/forums/threads/reading-circuit.230016/ | 1,660,231,050,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571472.69/warc/CC-MAIN-20220811133823-20220811163823-00444.warc.gz | 363,252,389 | 58,845 | ### Network
#### 010101
Jan 4, 2011
4
Hi I'm a bit confused about the attached circuit diagram.
Is only resistor R1 connected to pin 3 of the CMOS and R2 connected to pin 2 and so on? OR are resistors R1 to R4 collectively connected and pins 2, 3, 4, 7 connected together?
The link to the circuit is http://www.zen22142.zen.co.uk/Circuits/Timing/sqt.htm.
Thanks!
#### Attachments
• Sequential Timer.png
9.6 KB · Views: 233
#### (*steve*)
##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,505
The curvy line thingy connecting to pins 2, 3, 4, and 7 indicates a bus. That is, a series of parallel conductors that are grouped together for some reason. They are not connected together.
At various points it shows what those bus wires are connected to. Each one has 3 connections, a 2K7 resistor, and the anode of 2 diodes. Each place where the bus is shown as going has the pin number and the function in parenthesis.
Similarly, "Clock" is shown at several points, these are connected together.
There is a connection shown for pin 16 to supply power to the 4017, and all the ground symbols are connected together and then connected to the -ve side of the battery or power supply (this is not shown -- you are assumed to know)
I hope this helps, and sorry if I have covered stuff you already know.
#### KMoffett
Jan 21, 2009
723
The numbers refer to the counter bits of the 4017... the numbers in red inside the IC symbol. "0" (pin 3) connects to R1, D1, and D5. "1" (pin 2) connects to R2, D2, and D6. And so on...! It's like a parallel 4 wire buss. I think it was just to make the schematic simple to draw.
Ken
#### 010101
Jan 4, 2011
4
Thanks! I get it now
Regarding the connection for pin 16, am I suppose to get it from between C3 and C4 (there's an arrow with 16 written next to it)
And...where does clock go to???
#### 010101
Jan 4, 2011
4
The numbers refer to the counter bits of the 4017... the numbers in red inside the IC symbol. "0" (pin 3) connects to R1, D1, and D5. "1" (pin 2) connects to R2, D2, and D6. And so on...! It's like a parallel 4 wire buss. I think it was just to make the schematic simple to draw.
Ken
Thanks Ken.
As I place the components on a breadboard, I can see that this simplified schematic actually has many connections.
#### KMoffett
Jan 21, 2009
723
Thanks! I get it now
Regarding the connection for pin 16, am I suppose to get it from between C3 and C4 (there's an arrow with 16 written next to it)
And...where does clock go to???
Yes the junction of C3 and C4 go to pin 16 on the IC. There are also two connections to pin 14 from R13 and from C5
Ken
#### 010101
Jan 4, 2011
4
Thank you KMOffett and (*Steve*)!
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Niclas Carlsson
N | 863 | 2,812 | {"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-2022-33 | latest | en | 0.906334 |
https://www.chegg.com/homework-help/a-concise-introduction-to-logic-12th-edition-chapter-4.1-solutions-9781285965567 | 1,548,192,873,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583874494.65/warc/CC-MAIN-20190122202547-20190122224547-00423.warc.gz | 753,625,401 | 22,680 | # A Concise Introduction to Logic (12th Edition) View more editions 100% (9 ratings) for Chapter 4.1Solutions for Chapter 4.1
• 2362 step-by-step solutions
• Solved by professors & experts
• iOS, Android, & web
Chapter: Problem:
In the following categorical proposition identify the quantifier, subject term, copula, and predicate term.
Some executive-pay packages are insults to ordinary workers.
Sample Solution
Chapter: Problem:
• Step 1 of 3
Quantifier: A quantifier is a word like some, all used before noun to indicate how much of the quantity is being considered.
Subject term: The subject term is the class, category or concept about which the proposition is about.
Copula: The words that link the subject term with the predicate term is called copula. For example are, are not.
Predicate term: The predicate term is the class or category which is related by the proposition to the subject term.
• Step 2 of 3
The proposition provided in the question is “Some executive-pay packages are insults to ordinary workers”.
Here, the proposition is about executive-pay packages. So, “executive-pay packages” is the subject term.
Since the quantity of subject term is defined as some, so it will act as the quantifier.
The predicate term in the proposition is “insults to ordinary workers”.
In the proposition, the word that link the subject term and predicate term is “are”.
• Step 3 of 3
Hence, the categorical proposition is analyzed as:
Quantifier: Some
Subject term: executive-pay packages
Copula: are
Predicate term: insults to ordinary workers
Corresponding Textbook
A Concise Introduction to Logic | 12th Edition
9781285965567ISBN-13: 1285965566ISBN: Patrick J HurleyAuthors:
Alternate ISBN: 9781285196541 | 400 | 1,735 | {"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} | 3.03125 | 3 | CC-MAIN-2019-04 | latest | en | 0.859388 |
https://wearethearchetype.com/qa/question-how-many-percent-is-salary-tax-in-philippines.html | 1,603,316,966,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107878633.8/warc/CC-MAIN-20201021205955-20201021235955-00315.warc.gz | 586,583,011 | 8,448 | # Question: How Many Percent Is Salary Tax In Philippines?
## What is middle class income in the Philippines?
The government defines the middle class as those earning incomes between two to 12 times the poverty line.
This means if your family income is between around PHP 21,000 and PHP 125,000, you fall in the middle-income class..
## How much is tax on salary in us?
The average tax rate for taxpayers who earn over \$1,000,000 is 33.1 percent. For those who make between \$10,000 and \$20,000 the average total tax rate is 0.4 percent. (The average tax rate for those in the lowest income tax bracket is 10.6 percent, higher than each group between \$10,000 and \$40,000.
## What is annual income?
Annual income is the total income that you earn over one year. Depending on the data that is required to determine your annual income, you may base your income on either a calendar year or a fiscal year.
## How do I calculate my work hours in a year?
Total Work Hours in a Year To figure out how many hours are in a “work year,” multiply the number of work hours in a week by the number of weeks in a year. In other words, multiply a typical 40 hour work week by 52 weeks. That makes 2,080 hours in a typical work year.
## How much is the taxable salary in the Philippines 2019?
Income TaxAmount of Net Taxable IncomeRateP250,000P400,00020% of the excess over P250,000P400,000P800,000P30,000 + 25% of the excess over P400,000P800,000P2,000,000P130,000 + 30% of the excess over P800,000P2,000,000P8,000,000P490,000 + 32% of the excess over P2,000,0003 more rows
## How is tax calculated on salary?
How to Calculate Taxes Taken Out of a Paycheck. Divide the sum of all assessed taxes by the employee’s gross pay to determine the percentage of taxes deducted from a paycheck. Taxes can include FICA taxes (Medicare and Social Security), as well as federal and state withholding information found on a W-4.
## What is the basic salary?
Basic salary is the total amount (before any deductions) paid to employees plus the allowances. It is a fixed amount that is paid to employees by their employers in return for the work performed.
## Is \$100 a lot of money in the Philippines?
If you’ll convert it to Philippine peso, that would be around Php 5000 and this is a lot of money if you earned it for a short time but if you earned this for a long time, it would be the opposite of course. Yes it is. … \$100 is a lot of money ANYWHERE!
## What salary is tax free in Philippines?
Updated March 2018 Page 2 2 Starting January 1, 2018, compensation income earners, self-employed and professional taxpayers (SEPs) whose annual taxable incomes are P250,000 or less are exempt from the personal income tax (PIT). The 13th month pay and other benefits amounting to P90,000 are likewise tax-exempt.
## What is a good salary in the Philippines?
Average salary in Philippines is PHP 820,859. Average take home earning is PHP 640,416 (Net). The most typical salary is PHP 336,395 (Gross)….Philippines Earnings.CityQuezon CityAvg. gross salaryPHP 712,640(USD)US\$ 15,062Salary entries2887 more columns•Sep 7, 2020
## What job is the highest salary in Philippines?
Top 10 Highest Paying Jobs With 1-4 Years of ExperienceRankJobAverage Monthly Salary (PHP)1Oil and Gas Engineering40,8782Aircraft Maintenance33,5643IT32,8734Instrumentation25,6506 more rows | 846 | 3,360 | {"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-2020-45 | latest | en | 0.94907 |
https://www.teacherspayteachers.com/Product/Fact-Family-Math-Centers-1128842 | 1,484,872,778,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280761.39/warc/CC-MAIN-20170116095120-00435-ip-10-171-10-70.ec2.internal.warc.gz | 998,713,510 | 51,272 | # Fact Family Math Centers
Subjects
Resource Types
Product Rating
4.0
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5.21 MB | 48 pages
### PRODUCT DESCRIPTION
These activities are aimed to develop an understanding of the relationship between addition and subtraction through fact families. This bundle includes a variety of visual representations and formats to engage a wide variety of learners.
Included in this bundle:
• Fact Family Books: Book templates for target numbers 5-10. Each book includes a cover and visuals to support each fact family.
• Fact Family Snowmen: Fact Family Snowmen: Asks students to select from 60 different fact family snowmen cards. Using that fact family, students then write 2 addition sentences and 2 subtraction sentences on their recording sheets.
• Fact Family Creations: Students cut out the numbers at the bottom of the page and arrange them in the template to create 4 different math sentences (available in target numbers 5-10)
• Fact Family Dominoes: A great visual support for understanding parts and wholes (addends and sums). Students are asked to select a domino to create the addends, find their sum, and then place them in the blank dominoes. Using those 3 numbers, students then make fact families.
• Fact Family Bonds: Another great visual support for understanding parts and wholes (addends and sums). Students are asked to roll dice to create the addends, find their sum, and then place them in this visual organizer. Using those 3 numbers, students then make fact families.
***Do not purchase if you already have "Addition and Subtraction Mega Pack"***
OTHER MATH PRODUCTS THAT MAY INTEREST YOU:
Place Value Bundle-Common Core Aligned Games, Activities & Work Pages
Tens & Ones Bundle-Common Core Aligned Games, Activities, Posters & Work Pages
Coin Counting Club- Money Math Center Bundle
Coin Counting Club- Money Math Center Bundle (Canadian Version)
Kindergarten Math Journal Prompts
Complete Set K-3 Prompts
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35 ratings | 490 | 2,136 | {"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-2017-04 | longest | en | 0.865833 |
https://math.stackexchange.com/questions/2916004/uniqueness-of-measure-on-mu-measurable-sets | 1,720,958,404,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514564.41/warc/CC-MAIN-20240714094004-20240714124004-00806.warc.gz | 337,276,149 | 37,192 | # Uniqueness of measure on $\mu^*$ measurable sets
Folland proves the following theorem:
Let $\mathcal{A} \subset \mathcal{P}(X)$ be an algebra, $\mu_0$ a premeasure on $\mathcal{A}$, and $\mathcal{M}$ be the sigma algbera generated by $\mathcal{A}$. There exists a measure on $\mu$ on $\mathcal{M}$ whose restriction to $\mathcal{A}$ is $\mu_0$ - namely, $\mu = \mu^*|_{\mathcal{M}}$ where $\mu^*$ is the outer measure. If $\nu$ is another measure on $\mathcal{M}$ that extends $\mu_0$, then $\nu(E) \leq \mu(E)$ for all $E \in \mathcal{M}$, with equality when $\mu(E) < \infty$
I have a question about the second part of the statement, showing the equivalence of $\mu$ and $\nu$. When I did this myself, I used the following arguement:
Let $E \subset X$ and $A_j \in \mathcal{A}$, then: $$\nu(E) \leq \sum \nu(A_j) = \sum \mu_0(A_j) = \sum \mu(A_j) = \mu(\bigcup A_j) \, \,\forall j \implies \nu(E) \leq \mu(E)$$ $$\mu(E) \leq \sum \mu(A_j) = \sum \mu_0(A_j) = \sum \nu(A_j) = \nu(\bigcup A_j) \, \,\forall j \implies \mu(E) \leq \nu(E)$$
Folland does this a different way, is this correct though?
The first assertion follows from Caratheodory's theorem and proposition 1.13 since the $$\sigma$$-algebra of $$\mu^*$$-measurable sets include $$\mathcal{A}$$ and hence $$\mathcal{M}$$. As for the second assertion, if $$E\in \mathcal{M}$$ and $$E\subset\bigcup_{1}^{\infty}A_j$$ where $$\{A_j\}_{1}^{\infty}\in\mathcal{A}$$, then by montonocity $$\nu(E)\leq \nu\left(\bigcup_{1}^{\infty}A_j\right) \leq \sum_{1}^{\infty}\nu(A_j) = \sum_{1}^{\infty}\mu_0(A_j)$$, whence $$\nu(E)\leq \mu(E)$$. If we set $$S = \bigcup_{1}^{\infty}A_j$$, we have
$$\nu(A) = \lim_{n\to\infty}\nu\left(\bigcup_{1}^{n}A_j\right) = \lim_{n\to\infty}\mu\left(\bigcup_{1}^{n}A_j \right) = \mu(A)$$
If $$\mu(E) < \infty$$, choose the $$A_j$$'s such that $$\mu(A)\leq \mu(E) + \epsilon$$, hence $$\mu(A\setminus E) < \epsilon$$, and
$$\mu(E)\leq \mu(A) = \nu(A) + \nu(A\setminus E)\leq \nu(E) + \mu(A\setminus E)\leq \nu(E) + \epsilon$$
Since $$\epsilon$$ is arbitray, $$\mu(E) = \nu(E)$$. Finally, suppose $$X = \bigcup_{1}^{\infty}A_j$$ with $$\mu_{0}(A_j) < \infty$$, where we can assume that the $$A_j$$'s are disjoint. Then for any $$E\in \mathcal{M}$$, $$\mu(E) = \sum_{1}^{\infty}\mu(E\cap A_j) = \sum_{1}^{\infty}\nu(E\cap A_j) = \nu(E)$$ so $$\nu = \mu$$.
The question here is that one cannot conclude $$\mu(E)\leq \nu(E)$$ simply by $$\mu(E)\leq \nu(\bigcup A_j)$$, since $$\nu$$ is not defined as the restriction of outer measure. In fact, the way you exhibit hasn't used the condition $$\mu(E)\leq \infty$$
I think the second part can be proved as follows without choosing arbitrary $$\epsilon$$ as Folland did:
First we assert that, since $$\nu(E)\leq\mu (E)\leq\infty$$, and by the definition of $$\mu$$, one can find some $$A=\bigcup_{A_j\in \mathcal A} A_j\supset E$$ s.t. $$\mu(A)\leq\infty$$, and $$\mu(A)=\nu(A)$$. Hence$$\mu(A)-\mu(E)=\nu(A)-\mu(E)\leq\nu(A)-\nu(E)=\nu(A-E)\leq\mu(A-E)\leq\infty$$
This implies $$\mu(E)=\nu(E)$$. | 1,181 | 3,035 | {"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": 36, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2024-30 | latest | en | 0.672017 |
https://math.stackexchange.com/questions/2143928/concurrence-of-three-lines-in-a-quadrilateral | 1,709,070,538,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474686.54/warc/CC-MAIN-20240227184934-20240227214934-00435.warc.gz | 375,926,292 | 33,209 | # concurrence of three lines in a quadrilateral
Prove that the lines joining the midpoints of opposite sides of a quadrilateral and the line joining the midpoints of the diagonals of the quadrilateral are concurrent.
Let $a$, $b$, $c$, $d$ be the position vectors of the four vertices. It is easy to check that the point $$m:={a+b+c+d\over4}$$ is the midpoint of all three mentioned line segments.
Let $E=\frac{A+B}{2}$ and $F=\frac{C+D}{2}$ be the midpoints of two opposite sides.
Let $G=\frac{A+D}{2}$ and $H=\frac{C+B}{2}$ be the midpoints of another two opposite sides.
Let $J=\frac{A+C}{2}$ and $K=\frac{B+D}{2}$ be the midpoints of diagonals.
The midpoint of section $EF$ is a point $\frac{E+F}{2}=\frac{A+B+C+D}{2}$. Also the midpoint of $GH$ and $JK$ is $\frac{G+H}{2}=\frac{J+K}{2}=\frac{A+B+C+D}{2}$. Thuse each section $EF$, $GH$ and $JK$ contains point $L=\frac{A+B+C+D}{2}$ | 296 | 888 | {"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} | 3.734375 | 4 | CC-MAIN-2024-10 | latest | en | 0.744552 |
https://olympiadtester.in/blogs/class-4-study-material/class-4-imo-computation-operations | 1,726,274,141,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651540.77/warc/CC-MAIN-20240913233654-20240914023654-00110.warc.gz | 401,963,113 | 78,433 | # Class 4 IMO Computation Operations
This contains a few computation operation questions for 4th-grade mathematics. Answers with explanations are available in our premium "Buy Courses" section, featuring 500+ tests covering the entire Class 4 Maths syllabus with chapter-wise and mock tests.
Q1) Postman Sammy delivered an equal number of letters every day from Monday to Wednesday. He delivered 192 letter each on Thursday and Friday. Altogether, he delivered a total of 1104 letters from Monday to Friday. How many letters did Sammy deliver on Monday?
Ans: ________
240 letters
Q2) At a party, there were thrice as many adults as boys and twice as many boys as girls.
There were 88 more adults than boys.
a) How many more adults than girls were there at the party?
b) How many people were there altogether at the party?
Ans: a) ________
b) ________
a) 110
b) 198
Q3) Eugene had 60 more marbles than Sharon at first. After Sharon threw 12 marbles away, Eugene had thrice as many marbles as Sharon. How many marbles did Eugene have at first?
Ans: ________
Eugene had 108 marbles at first.
Q4) N is a number on the given number line. What is the value of N?
Ans: ________
1840
Q5) A shopkeeper wants to pack greeting cards into boxes of 9 cards each. If he has 5483 cards, how many such boxes of 9 cards can he pack?
Ans: ________
609
Q6) The sum of two numbers is 510.
If the bigger number is 5 times the smallest number, what is the smaller number?
Ans: ________
85
Q7) 87 x 52 is the same as 87 x 50 + ________.
1. 1 x 52
2. 2 x 52
3. 1 x 87
4. 2 x 87
D.
Q8) What is the product of 48 and 237?
1. 2844
2. 9926
3. 10 276
4. 11 376
D.
Q9) In one hour, machine A produces 210 cookies. Machine B produces thrice the amount. How many cookies will machine B produce in 5 hours?
Ans: ________
3150
Q10) A packet of erasers is to be shared by some pupils. The number of erasers in the packet is more than 40 but fewer than 50. The erasers can be shared by 4 or 6 pupils equally without any remainder. How many erasers are there in the packet?
Ans: ________ | 581 | 2,081 | {"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-2024-38 | latest | en | 0.946393 |
http://slideplayer.com/slide/3857106/ | 1,510,961,880,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934804019.50/warc/CC-MAIN-20171117223659-20171118003659-00792.warc.gz | 276,299,279 | 21,995 | # 1 Interest Rates and Present Value Chapter 7. 2 Interest rates We have thought about people trading fish and hamburgers lets think about a different type.
## Presentation on theme: "1 Interest Rates and Present Value Chapter 7. 2 Interest rates We have thought about people trading fish and hamburgers lets think about a different type."— Presentation transcript:
1 Interest Rates and Present Value Chapter 7
2 Interest rates We have thought about people trading fish and hamburgers lets think about a different type of trade Joe would like to trade coconuts that will be ripe tomorrow in order to get ripe coconuts today Bill is willing to trade some of the coconuts he has today in order to get coconuts tomorrow. At what rate will they trade in a competitive market? For quite a few reasons we think if Bill is going to give up 10 coconuts today he is going to want more than 10 coconuts back tomorrow This is called interest 2
3 Why Do we want more tomorrow? People are impatient-all else equal I would rather have a candy bar today than wait for it until tommorow There is some risk that Joe will die or renege on his promise (the larger the risk of this the more interest we will want) I might be able to do something useful with the coconuts (or money) in the meantime Use the coconuts to build a shelter that I can live in (and if this is money I could buy a house today, live in it, and then sell the house tomorrow) Use it to buy an ice cream truck, make money selling ice cream, and then sell the ice cream truck at the end of the day Or if some emergency comes up and I need the money right away, I don’t have it and need to wait until I get paid back 3
4 Interest Rate The interest rate is the percentage, usually expressed in annual terms, of a balance that is paid by a borrower to a lender that is in addition to the original amount borrowed or lent. If I trade 100 coconuts today for 108 coconuts next year, the interest rate is 8% As a practical matter it doesn’t usually work this way: Banks act as intermediaries Joe borrows coconuts from the bank Bill saves his coconuts in the bank While they never actually meet each other-Joe is essentially borrowing from Bill 4
5 Nominal vs. Real Interest Rates Nominal Interest Rate: the advertised rate of interest Real Interest Rate: the rate of interest after inflation expectations are accounted for; the compensation for waiting on consumption As an example suppose the inflation rate is 10% and the nominal interest rate is 10% I put \$100 into my account this year (2011) I get \$110 next year (2012) However, that \$110 in 2012 is worth the same amount as \$100 in 2011, so I didn’t really earn any interest In that case the real interest rate would be 0 I would probably want to spend that \$100 now before inflation erodes its value In economics we usually care about the real interest rate 5
6 Present Value Present Value is the interest adjusted value of future payment streams. To think of this consider two different scenarios: 1. I pay you \$300 this year 2. I pay you \$100 this year, \$100 in 2012, and \$100 in 2013 The interest rate is 5% Are these the same? Think about it this way: 1. I get my \$300 now, spend \$100 and save \$200 2. Next year I have \$210 in the bank, I spend \$110 and save \$100 3. In 2013 I now have \$105 to spend 6
77 Obviously I prefer the \$300 today. This is because the present value of a stream of \$100 for three years is less than \$300. Well how much is it exactly? The present value of the first \$100 today is \$100 The present value of \$100 next year must solve the following equation (r is the interest rate) That is, if I put PV in the bank today, I will have \$100 in the bank tomorrow
88 So Now what about the present value of \$100 in 2012? To see where this comes from suppose I put \$Y into the bank in 2011. This would be give me \$Y(1+r) in 2012 If I just kept it in the bank I would get \$Y(1+r)(1+r) in 2013 Thus the present value of \$100 in 2013 is:
99 And the value of the stream of \$100 for 3 years is worth: More generally the present value of a payment \$Y n years from now is worth today Rather than doing more calculations lets look at what this means
10 The Amount Payable for Every Dollar Borrowed (For several interest rates and loan durations) Interest rate -> Years ↓ 20%10%5%2%1% 30237.3817.454.321.811.35 106.192.591.631.221.10 52.491.611.281.101.05 11.201.101.051.021.01 10
11 Examples From This Table If you borrow \$1 and promise to pay it back in 5 years at 5% interest you will owe \$1.28 which is the original \$1 plus 28 cents in interest. If you borrow \$1 and promise to pay it back in 30 years at 20% interest you will owe \$237.38 which is the original \$1 plus \$236.38 in interest. 11
12 Business Venture Suppose you have the following business venture: Invest \$100 into the venture for 5 years receive \$100 in proceeds starting in year 6 and lasting for 7 years Is this a good investment? It depends on the interest rate 12
13 A Multiple Year Example @ 5% YearCostBenefitPV Cost @5% PV Benefit @5% 1100100.00 210095.24 310090.70 410086.38 510082.27 610078.35 710074.62 810071.07 910067.68 1010064.46 1110061.39 1210058.47 500700454.60476.05 This is a good deal
14 A Multiple Year Example @ 8% YearCostBenefitPV Cost @8% PV Benefit @8% 1100100.00 210092.59 310085.73 410079.38 510073.50 610068.06 710063.02 810058.35 910054.03 1010050.02 1110046.32 1210042.89 500700431.21382.68 Not a good deal at this interest rate
15 Internal rate of return At 5% (or less) this was a good investment At 8% (or more) this is a bad investment Internal rate of return : The interest rate where the present value of costs and benefits are equal. If my interest rate is lower than the internal rate, I should do it If it is higher than the internal rate, I should not do it 15
16 Mortgages, Car Payments, Etc. Mortgages are loans taken out to buy homes. Typically you borrow a large sum of money and promise to pay it back in even amounts each month for 10, 15, or 30 years. Car loans are similar to mortgages in that you borrow a large sum but the loan duration is usually two to six years. How does the financing department figure out the monthly payment at a 2.9% interest for a \$32,000 car? Given the interest they choose it to be the constant payment such that the present value of my car payments at a 2.9% rate is \$32,000 16
17 Monthly Payments Required on per \$1000 of loan (For Several Interest Rates and Loan Durations) Interest rate -> Years ↓ 20%10%5%2%1% 3016.718.785.373.703.22 1019.3313.2210.619.208.76 526.4921.2518.8717.5317.09 192.6387.9285.6184.2483.79
18 Examples From This Table If you borrow \$1000 and promise to pay it back monthly over 5 years at 5% interest you will owe \$18.87 per month. If you borrow \$1000 and promise to pay it back monthly over 10 years at 20% interest you will owe \$19.33 per month. 18
19 Future Value Future value: the interest-adjusted value of past payments. That is if I put \$Y in the bank today and take it out n years from now I get 19 dollars when I take it out
20 Risk and Reward Risk: the possibility that the investor will not get those anticipated payoffs – Default Risk: the risk to the investor that the borrower will not pay – Market Risk: the risk that the market value of an asset will change in an unanticipated manner Reward – Risk Premium the reward investors receive for taking greater risk 20
21 The Yield Curve US Treasury Yield Curve (January 2005) 21 Yield Curve: the relationship between reward and the time until the reward is received
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# kingsley
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how to make the numbers on the matrix with alternative sign?
I want to create a matrix that the main diagonal of K are alternatively 2 and -2’s, the sub- and sup-diagonal of K alternatively...
6 years ago | 0 answers | 0 | 816 | 2,729 | {"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-2023-06 | longest | en | 0.875814 |
https://wikibump.com/what-is-the-unit-of-vapour-density/ | 1,652,977,517,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662529538.2/warc/CC-MAIN-20220519141152-20220519171152-00228.warc.gz | 689,982,865 | 21,229 | # What is the unit of vapour density?
## What is the unit of vapour density?
vapour density = molar mass of gas / molar mass of H2. vapour density = molar mass of gas / 2.016.
## How is vapor density measured?
To compute a compound’s vapor density simply divide the molecular weight of the compound by the molecular weight of air. This will provide a numerical value that can be compared to air’s value of one.
## Does vapor have density?
The weight of a vapor or gas compared to the weight of an equal volume of air; an expression of the density of the vapor or gas. Materials lighter than air have vapor densities less than 1.0 (example: acetylene, methane, hydrogen).
## What is the density of water vapour in G cm3?
Even when quite modest pressures are applied, the volume decreases noticeably. Water vapor is also much less dense than the liquid or the solid. At 100C and ordinary pressures, its density is 5.98 10 4 g cm 3, less than one thousandth that of solid or liquid mercury.
## Is vapour density has a unit?
It has no units. Vapour density is found out by dividing the molecular mass of a substance by two.
## What is the unit of vapour?
The most common unit of vapour pressure is torr. 1 torr=1 mm of mercury.
## What is vapour density class 11?
Vapour density is defined as the density of a vapour in respect of that of hydrogen. It is a relative term. It can also be defined as the mass of a certain volume of a substance divided by mass of the same volume of hydrogen. As we know that the molar mass of a substance is approximately twice its vapour density.
## What is the difference between density and vapour density?
The key difference between density and vapour density is that density measures the mass per unit volume of any substance that can be a solid, liquid or a gas whereas the term vapour density refers to the density of vapour of a substance in relation to the density of the vapour of hydrogen.
## What is high vapor density?
If a gas has a vapor density of less than one it will generally rise in air. … If the vapor density is greater than one the gas will generally sink in air.
## Is vapor density the same as specific gravity?
They are not, however, identical measures. Specific gravity is an expression of density in relation to the density of a standard or reference (usually water). Also, density is expressed in units (weight relative to size) while specific gravity is a pure number or dimensionless.
## What is the vapor density of carbon dioxide?
22 Hence the vapour density of any gas can be given as the molecular weight of that gas divided by 2. In this question, the vapour density of the carbon dioxide is equal to the molecular weight of carbon dioxide divided by 2. Hence the vapour density of the carbon dioxide is 44 divided by 2 which is equal to 22.
## How do you find water vapor density?
It can be used to calculate exact quantity of water vapor in the air from a relative humidity (RH = % local air humidity measured / local total air humidity possible ) Given an RH percentage, the density of water in the air is given by RH SVD = Actual Vapor Density.
## What is water vapor density definition?
(Also called absolute humidity.) In a system of moist air, the ratio of the mass of water vapor present to the volume occupied by the mixture; that is, the density of the water vapor component.
## Why Vapour density is unit less?
3 Vapor Density. Vapor density is the weight of a volume of pure vapor or gas compared to an equal volume of dry air at the same temperature and pressure. It is obtained by dividing the molecular weight of the vapor by the average molecular weight of air thus, it is unitless.
## Why relative Vapour density has no unit?
Final answer is (B), the relative density does not have a unit because it is a ratio of two-dimensionally same quantities. Note: When we divide two-dimensionally the same quantity they get cancelled and their ratio is dimensionless. Specific density is the density of a fluid with respect to some standard fluid (water).
## What is the Vapour density of ethane?
1.355 kg/m3 ETHANE
Molecular weight: 30.07 Critical temperature: 305.33 K
Melting point: 89.88 K Critical pressure: 4.872 MPa
Normal boiling point: 184.52 Critical density: 206.6 kg/m3
Normal vapor density: 1.355 kg/m3 (@ 273.15K; 1.0135MPa)
## What is the Vapour density of chlorine?
Vapour density of Chlorine atom is 35.5.
= 1.428 g/l.
## What is Vapour density class 10?
Vapour density : It is Density of a gas, expressed as the mass of a given volume of the gas divided by the mass of an equal volume of a reference gas (such as hydrogen or air) at the same ternperature and pressure.
## What is Vapour density in physics?
Vapour density is the density of a vapour in relation to that of hydrogen. It may be defined as mass of a certain volume of a substance divided by mass of same volume of hydrogen. vapour density = mass of n molecules of gas / mass of n molecules of hydrogen.
## What is the vapor density of air quizlet?
The vapor density of air is 1. This is the reference point for all vapor densities. Anything with a vapor density greater than 1 will fall to the ground because it is heavier than the air. Anything with a vapor density of less than 1 will float because it is lighter than the air.
## How many grams is CH4?
16.04 grams One mole of methane equals 16.04 grams because a molecule of methane has an atomic weight of 16.04. The conversion factor will be 16.04 grams/mole.
## What is the volume of CH4?
The volume of methane gas is 17.0 L. The molar mass of CH4 C H 4 is 16.04 g/mol.
## What is the mole of o2?
One mole of oxygen gas, which has the formula O2, has a mass of 32 g and contains 6.02 X 1023 molecules of oxygen but 12.04 X 1023 (2 X 6.02 X 1023) atoms, because each molecule of oxygen contains two oxygen atoms.
## Which gas has a vapour density of 32?
An oxide XO2 has a vapour density of 32.
## What is vapour density of so2 with respect to CH4?
Molar mass = 2 x V.D. Molar mass of methane CH4 = 16, Thus, the vapour density = 16/2 = 8.
## What does twice the vapour density gives?
Twice the vapour density gives: … Actual vapour density. 2. Relative vapour density. | 1,505 | 6,231 | {"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-2022-21 | latest | en | 0.899963 |
https://www.jiskha.com/display.cgi?id=1177735025 | 1,511,596,489,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934809695.97/warc/CC-MAIN-20171125071427-20171125091427-00721.warc.gz | 813,865,071 | 3,926 | # math
posted by .
andys average driving speed for a 4-hour trip was 45 miles per hour. during the first 3 hours he drove 40 miles per hour. what was his average speed for the last hour of his trip?
The trip length was 4 x 45 = 180 miles. He went 3x40 = 120 the first three hours. How many miles does that leave for the last hour?
4+5
• math -
A car driver completes a 180km trip in 4hour. If he averages 50km/h during the first 3 hour of t
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http://mathhelpforum.com/calculus/88480-finding-interval-radius-convergence-power-series-print.html | 1,498,451,790,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320669.83/warc/CC-MAIN-20170626032235-20170626052235-00534.warc.gz | 241,291,072 | 3,893 | # Finding interval/radius of convergence of power series
• May 10th 2009, 05:16 PM
Pinkk
Finding interval/radius of convergence of power series
I'm fairly certain I have the correct answer, but I just want to verify.
$\sum_{k=1}^{\infty}\frac{k^{2}x^{k}}{2\cdot 4\cdot 6\cdot \cdot \cdot (2k)}$
So I used the ratio test:
$\lim_{k\to \infty}|\frac{a_{k+1}}{a_{k}}| = \lim_{k\to \infty}\frac{(k+1)^{2}|x|^{k+1}}{4\cdot 6\cdot 8\cdot \cdot \cdot (2k+2)}\cdot \frac{2\cdot 4\cdot 6\cdot \cdot \cdot (2k)}{k^{2}|x|^{k}}=2|x|\lim_{k\to \infty}\frac{k^{2}+2k+1}{2k^{3}+2k^{2}}<1$
Well, the limit goes to zero, so I end up with:
$2|x|\cdot 0 < 1$
$0<1$
So that means the series is convergent for all values of $x$, so the radius of convergence is $\infty$ and the interval of convergence is $(-\infty,\infty)$.
• May 10th 2009, 06:10 PM
Banned for attempted hacking
Absolutely Correct.
• May 10th 2009, 06:11 PM
Prove It
Quote:
Originally Posted by Pinkk
I'm fairly certain I have the correct answer, but I just want to verify.
$\sum_{k=1}^{\infty}\frac{k^{2}x^{k}}{2\cdot 4\cdot 6\cdot \cdot \cdot (2k)}$
So I used the ratio test:
$\lim_{k\to \infty}|\frac{a_{k+1}}{a_{k}}| = \lim_{k\to \infty}\frac{(k+1)^{2}|x|^{k+1}}{4\cdot 6\cdot 8\cdot \cdot \cdot (2k+2)}\cdot \frac{2\cdot 4\cdot 6\cdot \cdot \cdot (2k)}{k^{2}|x|^{k}}=2|x|\lim_{k\to \infty}\frac{k^{2}+2k+1}{2k^{3}+2k^{2}}<1$
Well, the limit goes to zero, so I end up with:
$2|x|\cdot 0 < 1$
$0<1$
So that means the series is convergent for all values of $x$, so the radius of convergence is $\infty$ and the interval of convergence is $(-\infty,\infty)$.
Note that
$\frac{k^2 + 2k + 1}{2k^3 + 2k^2} = \frac{(k + 1)^2}{2k(k + 1)}$
$= \frac{k + 1}{2k}$
$= \frac{k}{2k} + \frac{1}{2k}$
$= \frac{1}{2} + \frac{1}{2k}$
As $k \to \infty$, the limit tends to $\frac{1}{2}$. So you're right about the ratio being less than 1 and so the series is convergent. Just looks better :P
• May 10th 2009, 06:19 PM
Soroban
Hello, Pinkk!
You're absolutely correct . . . nice work!
Quote:
$\sum_{k=1}^{\infty}\frac{k^{2}x^{k}}{2\cdot 4\cdot 6\cdot \cdot \cdot (2k)}$
I would do it like this . . .
Note that: . $2\cdot4\cdot6\hdots (2k) \;=\;2^k(1\cdot2\cdot3\hdots k) \;=\;2^k\,k!$
. . So we have: . $\sum^{\infty}_{k=1}\frac{k^2x^k}{2^k\,k!}$
Ratio: . $\left|\frac{(k+1)x^{k+1}}{2^{k+1}(k+1)!}\cdot\frac {2^k\,k!}{k^2x^l}\right| \;\;=\;\;\left|\frac{2^o}{2^{k+1}}\cdot\frac{x^{k+ 1}}{x^k}\cdot\frac{k!}{(k+1)!} \cdot\frac{k+1)^2}{k^2}\right|$
. . . $= \;\;\frac{1}{2}\cdot|x|\cdot\frac{1}{k+1}\cdot\fra c{(k+1)^2}{k^2} \;\;=\;\;\frac{1}{2}\cdot|x|\cdot\frac{k+1}{k^2}\; \;=\;\;\frac{1}{2}\left(\frac{1}{k}+\frac{1}{k^2}\ right)|x|$
And: . $\lim_{k\to\infty}\left[\frac{1}{2}\left(\frac{1}{k} + \frac{1}{k^2}\right)|x|\right] \;=\;0 \quad\hdots\;\text{ etc.}$
• May 10th 2009, 06:23 PM
Pinkk
@Prove It
Isn't it actually:
$\lim_{k\to \infty}\frac{k+1}{k^{2}}=0$
If the limit somehow equaled $\frac{1}{2}$, then the radius of convergence would not be $\infty$.
• May 10th 2009, 06:25 PM
Prove It
Quote:
Originally Posted by Pinkk
@Prove It
Isn't it actually:
$\lim_{k\to \infty}\frac{k+1}{k^{2}}=0$
If the limit somehow equaled $\frac{1}{2}$, then the radius of convergence would not be $\infty$.
Oops, you are right. It should have been $\frac{k + 1}{2k^2}$.
So you're right, that limit is not $\frac{1}{2}$.
Still tends to a value less than 1 though.
$\frac{k + 1}{2k^2} = \frac{k}{2k^2} + \frac{1}{2k^2}$
$= \frac{1}{2k} + \frac{1}{2k^2}$.
So as $k \to \infty$ the limit tends to $0$.
Still converges though. | 1,548 | 3,585 | {"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": 38, "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-2017-26 | longest | en | 0.578712 |
https://www.physicsforums.com/threads/perpendicular-planes.79200/ | 1,571,270,753,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986672431.45/warc/CC-MAIN-20191016235542-20191017023042-00515.warc.gz | 1,053,856,814 | 17,140 | # Perpendicular Planes
#### amcavoy
Hi,
I suppose my question has to do with planes in general, rather than just tangent planes. Say you have a plane given by the equation
$$z=\frac{\partial f}{\partial x}(x)+\frac{\partial f}{\partial y}(y).$$
How would you find the equation of the plane perpendicular to this one?
#### mathman
General approach:
Equation of plane ax+by+cz+d=0, where at least one of (a,b,c) is not =0.
Assume it is a, then bx-ay+m=0 or cx-az+n=0, where m and n are arbitrary. The general idea is to set up a nonzero vector perpendicular to (a,b,c) and use it as coefficients for x,y,z. The constant term is arbitrary.
#### amcavoy
mathman said:
General approach:
Equation of plane ax+by+cz+d=0, where at least one of (a,b,c) is not =0.
Assume it is a, then bx-ay+m=0 or cx-az+n=0, where m and n are arbitrary. The general idea is to set up a nonzero vector perpendicular to (a,b,c) and use it as coefficients for x,y,z. The constant term is arbitrary.
Where did the $$bx-ay+m=0$$ and $$cx-az+n=0$$ come from?
#### amcavoy
Well, I guess I can understand that. If you have one vector on a plane "r-r0 = (x-x0)i + (y-y0)j + (z-z0)k" and the gradient vector "ai + bj +ck" tangent to f(x,y), would you compute the cross product between these two vectors to find that perpendicular vector, then use that as coefficients?
(ex.) <(x-x0), (y-y0), (z-z0)> x <a, b, c> = <p, q, t>
ax+by+cz=d
So... the tangent plane is px+qy+cz=d.
Is this correct?
#### mathman
I tried to give an answer for planes in general. I am a little rusty on your special case (tangent planes).
The plane equations in the previous post just simply reflect the fact that (b,-a,0) and (c,0,-a) are both non-zero and perpendicular to (a,b,c) as long as a is not 0.
#### amcavoy
Alright that makes sense then. By which method did you compute that (b, -a, 0) and (c, 0, -a) are perpendicular to (a, b, c)?
The reason I ask is that I have the equation $$z=\frac{\partial g}{\partial x}(x)+\frac{\partial g}{\partial y}(y)$$ and need to find the perpendicular plane (the problem actually has to do with projections).
Thanks again.
Last edited:
#### mathman
which method did you compute that (b, -a, 0) and (c, 0, -a) are perpendicular to (a, b, c)?
Inspection!
In your case a=dg/dx, b=dg/dy, c=-1. Since c not=0, you can use (1,0,dg/dx) or
(0,1,dg/dy) or any linear combination of the two vectors.
#### amcavoy
Alright I understand now. Thanks a lot for your help.
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• Solo and co-op problem solving | 822 | 2,791 | {"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-2019-43 | latest | en | 0.910019 |
http://lernerphysics.posthaven.com/tag/CASTLE?page=2 | 1,632,275,024,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057303.94/warc/CC-MAIN-20210922011746-20210922041746-00310.warc.gz | 37,143,091 | 5,994 | ## Day 64: First Real Day of Electric Circuits
We really got to play with circuits today. We started by making sure we knew what every component was in our CASTLE bag. We're using the Modeling Instruction modified CASTLE worksheets with the V, which I love. (I don't know who created them, but thank you!) We started with a simple building of a circuit, and then we tested different items around the room. One group tested water to see whether it could complete the circuit. Then we looked at how light bulb sockets and light bulbs work, and I got to do my favorite electricity demonstration with a household light bulb and a lot of D-cell batteries. You can light the lightbulb if you string enough D-cell batteries together! But what if the filament isn't under the glass bulb...
We talked a little bit about how the compass works to show charge flow. I appealed to our understanding of volume flow rate because we have just finished the fluid model. I don't want to lean on the fluid model too much, but it helped make clear why the flow would be the same everywhere in the circuit and why the flow would start all at the same time.
Posted
## Day 63: Finishing Fluid Mechanics & Starting CASTLE
We worked on some more examples of fluid mechanics, but it wasn't that groundbreaking. I let them practice how they needed to practice, which, for most of them, was alone.
We then moved on to the beginning of electricity. In previous years, I started with electrostatics, but I wanted to go right into electric circuits this year. I feel it's more understandable and there are many more labs for electricity. We started with a great lab based off of The Private Universe video, which was made by my science methods professor from grad school. I ask them two questions: "Why are there seasons?" and "Given a lightbulb, battery, and wire, draw how you'd light a light bulb." Then I have them try it with a lightbulb, a battery, and an unbent paperclip. Many hot paperclips ensue until everyone in the class can find all the ways you can light a lightbulb with just one wire and one battery.
Posted | 457 | 2,100 | {"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} | 3.21875 | 3 | CC-MAIN-2021-39 | latest | en | 0.96923 |
https://www.askiitians.com/forums/Differential-Calculus/consider-the-infinite-sequence-n-1-n-i-e-1-2_96886.htm | 1,713,914,752,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818835.29/warc/CC-MAIN-20240423223805-20240424013805-00349.warc.gz | 583,362,181 | 42,899 | # Consider the infinite sequence {n^1/n } i.e. 1, 2^1/2 , 3^1/3 . . . . .Find the largest term in this sequence(if it exists).HINT:USE CALCULUS
`f(n)= n^(1/n)log(f(n))= log(n)/ntaking derivative both sidesf'(n)/f(n)= (1/(n^2)-log(n)/(n^2)f'(n)=(1-log(n))f(n)/n^2f'(n)=0 log(n)=1,n=e , which imply nearest integer is 3,for n=3 , the function has maximasher mohammadb.tech, iitd delhi` | 152 | 384 | {"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} | 3.21875 | 3 | CC-MAIN-2024-18 | latest | en | 0.568519 |
https://www.geeksforgeeks.org/convert-alternate-characters-string-upper-case/ | 1,701,546,518,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100448.65/warc/CC-MAIN-20231202172159-20231202202159-00303.warc.gz | 882,794,409 | 64,096 | # Convert characters of a string to opposite case
Given a string, convert the characters of the string into the opposite case,i.e. if a character is the lower case then convert it into upper case and vice-versa.
Examples:
`Input : geeksForgEeksOutput : GEEKSfORGeEKSInput : hello every oneOutput : HELLO EVERY ONE`
ASCII values of alphabets: A – Z = 65 to 90, a – z = 97 to 122
Steps:
1. Take one string of any length and calculate its length.
2. Scan string character by character and keep checking the index.
• If a character in an index is in lower case, then subtract 32 to convert it into upper case, else add 32 to convert it in lowercase
3. Print the final string.
Implementation:
## C++
`// CPP program to Convert characters` `// of a string to opposite case` `#include ` `using` `namespace` `std;` `// Function to convert characters` `// of a string to opposite case` `void` `convertOpposite(string& str)` `{` ` ``int` `ln = str.length();` ` ``// Conversion according to ASCII values` ` ``for` `(``int` `i = 0; i < ln; i++) {` ` ``if` `(str[i] >= ``'a'` `&& str[i] <= ``'z'``)` ` ``// Convert lowercase to uppercase` ` ``str[i] = str[i] - 32;` ` ``else` `if` `(str[i] >= ``'A'` `&& str[i] <= ``'Z'``)` ` ``// Convert uppercase to lowercase` ` ``str[i] = str[i] + 32;` ` ``}` `}` `// Driver function` `int` `main()` `{` ` ``string str = ``"GeEkSfOrGeEkS"``;` ` ``// Calling the Function` ` ``convertOpposite(str);` ` ``cout << str;` ` ``return` `0;` `}`
## C
`// C program to Convert characters` `// of a string to opposite case` `#include ` `#include ` `// Function to convert characters` `// of a string to opposite case` `void` `convertOpposite(``char``* str)` `{` ` ``int` `ln = ``strlen``(str);` ` ``// Conversion according to ASCII values` ` ``for` `(``int` `i = 0; i < ln; i++) {` ` ``if` `(str[i] >= ``'a'` `&& str[i] <= ``'z'``)` ` ``// Convert lowercase to uppercase` ` ``str[i] = str[i] - 32;` ` ``else` `if` `(str[i] >= ``'A'` `&& str[i] <= ``'Z'``)` ` ``// Convert uppercase to lowercase` ` ``str[i] = str[i] + 32;` ` ``}` `}` `// Driver function` `int` `main()` `{` ` ``char` `str[] = ``"GeEkSfOrGeEkS"``;` ` ``// Calling the Function` ` ``convertOpposite(str);` ` ``printf``(``"%s"``, str);` ` ``return` `0;` `}`
## Java
`// Java program to Convert characters` `// of a string to opposite case` `class` `Test {` ` ``// Method to convert characters` ` ``// of a string to opposite case` ` ``static` `void` `convertOpposite(StringBuffer str)` ` ``{` ` ``int` `ln = str.length();` ` ``// Conversion using predefined methods` ` ``for` `(``int` `i = ``0``; i < ln; i++) {` ` ``Character c = str.charAt(i);` ` ``if` `(Character.isLowerCase(c))` ` ``str.replace(i, i + ``1``,` ` ``Character.toUpperCase(c) + ``""``);` ` ``else` ` ``str.replace(i, i + ``1``,` ` ``Character.toLowerCase(c) + ``""``);` ` ``}` ` ``}` ` ``public` `static` `void` `main(String[] args)` ` ``{` ` ``StringBuffer str` ` ``= ``new` `StringBuffer(``"GeEkSfOrGeEkS"``);` ` ``// Calling the Method` ` ``convertOpposite(str);` ` ``System.out.println(str);` ` ``}` `}` `// This code is contributed by Gaurav Miglani`
## Python3
`# Python3 program to Convert characters ` `# of a string to opposite case` `# Function to convert characters ` `# of a string to opposite case` `def` `convertOpposite(``str``):` ` ``ln ``=` `len``(``str``)` ` ``# Conversion according to ASCII values` ` ``for` `i ``in` `range``(ln):` ` ``if` `str``[i] >``=` `'a'` `and` `str``[i] <``=` `'z'``:` ` ``# Convert lowercase to uppercase` ` ``str``[i] ``=` `chr``(``ord``(``str``[i]) ``-` `32``)` ` ``elif` `str``[i] >``=` `'A'` `and` `str``[i] <``=` `'Z'``:` ` ``# Convert lowercase to uppercase` ` ``str``[i] ``=` `chr``(``ord``(``str``[i]) ``+` `32``)` `# Driver code` `if` `__name__ ``=``=` `"__main__"``:` ` ``str` `=` `"GeEkSfOrGeEkS"` ` ``str` `=` `list``(``str``)` ` ``# Calling the Function` ` ``convertOpposite(``str``)` ` ``str` `=` `''.join(``str``)` ` ``print``(``str``)` `# This code is contributed by` `# sanjeev2552`
## C#
`// C# program to Convert characters ` `// of a string to opposite case` `using` `System;` `using` `System.Text;` `class` `GFG{` ` ` ` ``// Method to convert characters ` ` ``// of a string to opposite case` ` ``static` `void` `convertOpposite(StringBuilder str)` ` ``{` ` ``int` `ln = str.Length;` ` ` ` ``// Conversion according to ASCII values` ` ``for` `(``int` `i=0; i=``'a'` `&& str[i]<=``'z'``)` ` ` ` ``//Convert lowercase to uppercase` ` ``str[i] = (``char``)(str[i] - 32);` ` ` ` ``else` `if``(str[i]>=``'A'` `&& str[i]<=``'Z'``)` ` ` ` ``//Convert uppercase to lowercase` ` ``str[i] = (``char``)(str[i] + 32);` ` ``}` ` ``}` ` ` ` ``// Driver code` ` ``public` `static` `void` `Main() ` ` ``{` ` ``StringBuilder str = ``new` `StringBuilder(``"GeEkSfOrGeEkS"``);` ` ``// Calling the Method` ` ``convertOpposite(str);` ` ``Console.WriteLine(str);` ` ``}` `}` `// This code is contributed by PrinciRaj1992`
## Javascript
``
## PHP
`= ``'a'` `&& ``\$str``[``\$i``] <= ``'z'``){` ` ``// Convert lowercase to uppercase` ` ``\$str``[``\$i``] = ``strtoupper``(``\$str``[``\$i``]); ` ` ``}` ` ``else` `if` `(``\$str``[``\$i``] >= ``'A'` `&& ``\$str``[``\$i``] <= ``'Z'``){` ` ``// Convert uppercase to lowercase` ` ``\$str``[``\$i``] = ``strtolower``(``\$str``[``\$i``]); ` ` ``}` ` ``} ` ` ``return` `\$str``;` `}` `// Driver function` `\$str` `= ``"GeEkSfOrGeEkS"``;` ` ``// Calling the Function` `echo``(convertOpposite(``\$str``));` ` ` `// This code is contributed by ksrikanth0498` ` ` `?>`
Output
```gEeKsFoRgEeKs
```
Time Complexity: O(n)
Note: This program can alternatively be done using C++ inbuilt functions – Character.toLowerCase(char) and Character.toUpperCase(char).
Approach 2: The problem can be solved using letter case toggling. Follow the below steps to solve the problem:
• Traverse the given string S.
• For each character, Si, do Si = Si ^ (1 << 5).
• Si ^ (1 << 5) toggles the 5th bit which means 97 will become 65 and 65 will become 97:
• 65 ^ 32 = 97
• 97 ^ 32 = 65
• Print the string after all operations
Below is the implementation of the above approach:
## C++
`// C++ program to toggle all characters` `#include ` `using` `namespace` `std;` `// Function to toggle characters` `void` `toggleChars(string& S)` `{` ` ``for` `(``auto``& it : S) {` ` ``if` `(``isalpha``(it)) {` ` ``it ^= (1 << 5);` ` ``}` ` ``}` `}` `// Driver code` `int` `main()` `{` ` ``string S = ``"GeKf@rGeek\$"``;` ` ``toggleChars(S);` ` ``cout << ``"String after toggle "` `<< endl;` ` ``cout << S << endl;` ` ``return` `0;` `}` `// Code contributed by koulick_sadhu`
## C
`// C program to toggle all characters` `#include ` `#include ` `// Function to toggle characters` `void` `toggleChars(``char``* S)` `{` ` ``while` `(*S) {` ` ``if` `(``isalpha``(*S)) {` ` ``*S ^= (1 << 5);` ` ``}` ` ``++S;` ` ``}` `}` `// Driver code` `int` `main()` `{` ` ``char` `S[] = ``"GeKf@rGeek\$"``;` ` ``toggleChars(S);` ` ``printf``(``"String after toggle:\n%s\n"``, S);` ` ``return` `0;` `}`
## Java
`// Java program to toggle all characters` `import` `java.util.*;` `class` `GFG {` ` ``static` `char``[] S = ``"GeKf@rGeek\$"``.toCharArray();` ` ``// Function to toggle characters` ` ``static` `void` `toggleChars()` ` ``{` ` ``for` `(``int` `i = ``0``; i < S.length; i++) {` ` ``if` `(Character.isAlphabetic(S[i])) {` ` ``S[i] ^= (``1` `<< ``5``);` ` ``}` ` ``}` ` ``}` ` ``// Driver code` ` ``public` `static` `void` `main(String[] args)` ` ``{` ` ``toggleChars();` ` ``System.out.print(``"String after toggle "` ` ``+ ``"\n"``);` ` ``System.out.print(String.valueOf(S));` ` ``}` `}` `// This code is contributed by Amit Katiyar`
## Python3
`# python program for the same approach` `def` `isalpha(``input``):` ` ``input_char ``=` `ord``(``input``[``0``])` ` ``# CHECKING FOR ALPHABET` ` ``if``((input_char >``=` `65` `and` `input_char <``=` `90``) ``or` `(input_char >``=` `97` `and` `input_char <``=` `122``)):` ` ``return` `True` ` ``else``:` ` ``return` `False` `# Function to toggle characters` `def` `toggleChars(S):` ` ``s ``=` `""` ` ``for` `it ``in` `range``(``len``(S)):` ` ``if``(isalpha(S[it])):` ` ``s ``+``=` `chr``(``ord``(S[it]) ^ (``1` `<< ``5``))` ` ``else``:` ` ``s ``+``=` `S[it]` ` ``return` `s` `# Driver code` `S ``=` `"GeKf@rGeek\$"` `print``(f``"String after toggle {toggleChars(S)}"``)` `# This code is contributed by shinjanpatra`
## C#
`// C# program to toggle all characters` `using` `System;` `class` `GFG {` ` ``static` `char``[] S = ``"GeKf@rGeek\$"``.ToCharArray();` ` ``// Function to toggle characters` ` ``static` `void` `toggleChars()` ` ``{` ` ``for` `(``int` `i = 0; i < S.Length; i++) {` ` ``if` `(``char``.IsLetter(S[i])) {` ` ``S[i] = (``char``)((``int``)(S[i]) ^ (1 << 5));` ` ``}` ` ``}` ` ``}` ` ``// Driver code` ` ``public` `static` `void` `Main(String[] args)` ` ``{` ` ``toggleChars();` ` ``Console.Write(``"String after toggle "` ` ``+ ``"\n"``);` ` ``Console.Write(String.Join(``""``, S));` ` ``}` `}` `// This code is contributed by Princi Singh`
## Javascript
`` `// This code is contributed by Akshit Saxena`
Output
```String after toggle
gEkF@RgEEK\$
```
Time Complexity: O(n)
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Approach : Using isupper() and islower(),upper() and lower().
Initialize an empty string.Iterate a for loop over the given string and check each character whether is lowercase or uppercase using isupper() and islower().If lowercase converts the character to uppercase using upper() and append to empty string, similarly with uppercase.
## C++
`// C++ program to Convert characters` `// of a string to opposite case` `#include ` `using` `namespace` `std;` `int` `main()` `{` ` ``string str = ``"GeEkSfOrGeEkS"``;` ` ``string x = ``""``;` ` ``for` `(``int` `i = 0; i < str.length(); i++) {` ` ``char` `ch = str[i];` ` ``if` `(``isupper``(ch))` ` ``x += ``tolower``(ch);` ` ``else` ` ``x += ``toupper``(ch);` ` ``}` ` ``cout << x << endl;` ` ``return` `0;` `}` `// This code is contributed by lokeshmvs21.`
## C
`// C program to Convert characters` `// of a string to opposite case` `#include ` `#include ` `#include ` `int` `main()` `{` ` ``char` `str[] = ``"GeEkSfOrGeEkS"``;` ` ``char` `x[100];` ` ``for` `(``int` `i = 0; i < ``strlen``(str); i++) {` ` ``char` `ch = str[i];` ` ``if` `(``isupper``(ch))` ` ``x[i] = ``tolower``(ch);` ` ``else` ` ``x[i] = ``toupper``(ch);` ` ``}` ` ``x[``strlen``(str)] = ``'\0'``;` ` ``printf``(``"%s\n"``, x);` ` ``return` `0;` `}`
## Java
`// Java program to Convert characters` `// of a string to opposite case` `import` `java.io.*;` `class` `GFG {` ` ``public` `static` `void` `main(String[] args)` ` ``{` ` ``String str = ``"GeEkSfOrGeEkS"``;` ` ``String x = ``""``;` ` ``for` `(``int` `i = ``0``; i < str.length(); i++) {` ` ``char` `ch = str.charAt(i);` ` ``if` `(Character.isUpperCase(ch))` ` ``x += Character.toLowerCase(ch);` ` ``else` ` ``x += Character.toUpperCase(ch);` ` ``}` ` ``System.out.println(x);` ` ``}` `}` `// This code is contributed by karandeep1234.`
## Python3
`# Python3 program to Convert characters` `# of a string to opposite case` `str` `=` `"GeEkSfOrGeEkS"` `x``=``""` `for` `i ``in` `str``:` ` ``if``(i.isupper()):` ` ``x``+``=``i.lower()` ` ``else``:` ` ``x``+``=``i.upper()` `print``(x)`
## C#
`// C# program to Convert characters` `// of a string to opposite case` `using` `System;` `public` `class` `GFG {` ` ``static` `public` `void` `Main()` ` ``{` ` ``// Code` ` ``string` `str = ``"GeEkSfOrGeEkS"``;` ` ``string` `x = ``""``;` ` ``for` `(``int` `i = 0; i < str.Length; i++) {` ` ``char` `ch = str[i];` ` ``if` `(Char.IsUpper(ch)) {` ` ``x += Char.ToLower(ch);` ` ``}` ` ``else` `{` ` ``x += Char.ToUpper(ch);` ` ``}` ` ``}` ` ``Console.WriteLine(x);` ` ``}` `}` `// This code is contributed by lokeshmvs21.`
## Javascript
`// Javascript program to Convert characters` `// of a string to opposite case` ` ` ` ``let str = ``"GeEkSfOrGeEkS"``;` ` ``let x = ``""``;` ` ``for` `(let i = 0; i < str.length; i++) {` ` ``let ch = str[i];` ` ``if``(str.charAt(i) === str.charAt(i).toUpperCase())` ` ``x += str.charAt(i).toLowerCase(ch);` ` ``else` ` ``x += str.charAt(i).toUpperCase(ch);` ` ``}` ` ``console.log(x);` ` ` ` ``// This code is contributed by garg28harsh.`
Output
```gEeKsFoRgEeKs
```
Approach : Using index() method
## C++
`// C++ program to Convert characters` `// of a string to opposite case` `#include ` `using` `namespace` `std;` `int` `main()` `{` ` ``string str = ``"GeEkSfOrGeEkS"``;` ` ``string x = ``""``;` ` ``string upperalphabets = ``"ABCDEFGHIJKLMNOPQRSTUVWXYZ"``;` ` ``string loweralphabets = ``"abcdefghijklmnopqrstuvwxyz"``;` ` ``for` `(``int` `i = 0; i < str.length(); i++) {` ` ``char` `ch = str[i];` ` ``if` `(upperalphabets.find(ch) != -1) {` ` ``x += loweralphabets[upperalphabets.find(ch)];` ` ``}` ` ``else` `{` ` ``x += upperalphabets[loweralphabets.find(ch)];` ` ``}` ` ``}` ` ``cout << x << endl;` ` ``return` `0;` `}` `// This code is contributed by Prajwal Kandekar`
## C
`// C program to Convert characters` `// of a string to opposite case` `#include ` `#include ` `int` `main()` `{` ` ``char` `str[] = ``"GeEkSfOrGeEkS"``;` ` ``char` `x[100] = ``""``;` ` ``char` `upperalphabets[] = ``"ABCDEFGHIJKLMNOPQRSTUVWXYZ"``;` ` ``char` `loweralphabets[] = ``"abcdefghijklmnopqrstuvwxyz"``;` ` ``int` `i, j;` ` ``for` `(i = 0; i < ``strlen``(str); i++) {` ` ``char` `ch = str[i];` ` ``if` `(``strchr``(upperalphabets, ch)) {` ` ``j = ``strchr``(upperalphabets, ch) - upperalphabets;` ` ``x[i] = loweralphabets[j];` ` ``}` ` ``else` `{` ` ``j = ``strchr``(loweralphabets, ch) - loweralphabets;` ` ``x[i] = upperalphabets[j];` ` ``}` ` ``}` ` ``printf``(``"%s\n"``, x);` ` ``return` `0;` `}`
## Java
`// Java program to Convert characters` `// of a string to opposite case` `public` `class` `GFG {` ` ``public` `static` `void` `main(String[] args)` ` ``{` ` ``String str = ``"GeEkSfOrGeEkS"``;` ` ``String x = ``""``;` ` ``String upperalphabets` ` ``= ``"ABCDEFGHIJKLMNOPQRSTUVWXYZ"``;` ` ``String loweralphabets` ` ``= ``"abcdefghijklmnopqrstuvwxyz"``;` ` ``for` `(``int` `i = ``0``; i < str.length(); i++) {` ` ``char` `ch = str.charAt(i);` ` ``if` `(upperalphabets.indexOf(ch) != -``1``) {` ` ``x += loweralphabets.charAt(` ` ``upperalphabets.indexOf(ch));` ` ``}` ` ``else` `{` ` ``x += upperalphabets.charAt(` ` ``loweralphabets.indexOf(ch));` ` ``}` ` ``}` ` ``System.out.println(x);` ` ``}` `}` `// This Code is contributed by karandeep1234`
## Python3
`# Python3 program to Convert characters` `# of a string to opposite case` `str` `=` `"GeEkSfOrGeEkS"` `x``=``""` `upperalphabets``=``"ABCDEFGHIJKLMNOPQRSTUVWXYZ"` `loweralphabets``=``"abcdefghijklmnopqrstuvwxyz"` `for` `i ``in` `str``:` ` ``if` `i ``in` `upperalphabets:` ` ``x``+``=``loweralphabets[upperalphabets.index(i)]` ` ``else``:` ` ``x``+``=``upperalphabets[loweralphabets.index(i)]` `print``(x)`
## C#
`// C# program to Convert characters` `// of a string to opposite case` `using` `System;` `public` `class` `GFG {` ` ``public` `static` `void` `Main(``string``[] args)` ` ``{` ` ``string` `str = ``"GeEkSfOrGeEkS"``;` ` ``string` `x = ``""``;` ` ``string` `upperalphabets` ` ``= ``"ABCDEFGHIJKLMNOPQRSTUVWXYZ"``;` ` ``string` `loweralphabets` ` ``= ``"abcdefghijklmnopqrstuvwxyz"``;` ` ``for` `(``int` `i = 0; i < str.Length; i++) {` ` ``char` `ch = str[i];` ` ``if` `(upperalphabets.IndexOf(ch) != -1) {` ` ``x += loweralphabets[upperalphabets.IndexOf(` ` ``ch)];` ` ``}` ` ``else` `{` ` ``x += upperalphabets[loweralphabets.IndexOf(` ` ``ch)];` ` ``}` ` ``}` ` ``Console.WriteLine(x);` ` ``}` `}` `// This Code is contributed by karandeep1234`
## Javascript
`// Javascript program to Convert characters` `// of a string to opposite case` ` ``let str = ``"GeEkSfOrGeEkS"``;` ` ``let x = ``""``;` ` ``let upperalphabets` ` ``= ``"ABCDEFGHIJKLMNOPQRSTUVWXYZ"``;` ` ``let loweralphabets` ` ``= ``"abcdefghijklmnopqrstuvwxyz"``;` ` ``for` `(let i = 0; i < str.length; i++) {` ` ``let ch = str[i];` ` ``if` `(upperalphabets.indexOf(ch) != -1) {` ` ``x += loweralphabets[upperalphabets.indexOf(ch)];` ` ``}` ` ``else` `{` ` ``x += upperalphabets[loweralphabets.indexOf(ch)];` ` ``}` ` ``}` ` ``console.log(x);` `// This code is contributed by garg28harsh.`
Output
```gEeKsFoRgEeKs
```
Another Approach:-
• In this approach we will use transform function
• What transform function do is that it convert the case of a string into upper case or lower case without using any other extra space or string.
• In this approach we will just traverse the string and check the case of the character.
• If the character is in upper case then we will use transform function to convert it into lower case.
• else we will convert in into upper case
• Below is the implementation of the above approach.
Implementation:-
## C++
`// C++ program to Convert characters` `// of a string to opposite case` `#include ` `using` `namespace` `std;` `int` `main()` `{` ` ``string str = ``"GeEkSfOrGeEkS"``;` ` ``for` `(``int` `i = 0; i < str.length(); i++) {` ` ``if``(``isupper``(str[i])){` ` ``transform(str.begin()+i, str.begin()+i+1, str.begin()+i, ::``tolower``);` ` ``}` ` ``else``{` ` ``transform(str.begin()+i, str.begin()+i+1, str.begin()+i, ::``toupper``);` ` ``}` ` ``}` ` ``cout << str << endl;` ` ``return` `0;` `}` `// This code is contributed by shubhamrajput6156`
## C
`// C program to Convert characters` `// of a string to opposite case` `#include ` `#include ` `#include ` `int` `main()` `{` ` ``char` `str[] = ``"GeEkSfOrGeEkS"``;` ` ``int` `len = ``strlen``(str);` ` ``for` `(``int` `i = 0; i < len; i++) {` ` ``if` `(``isupper``(str[i])) {` ` ``str[i] = ``tolower``(str[i]);` ` ``}` ` ``else` `{` ` ``str[i] = ``toupper``(str[i]);` ` ``}` ` ``}` ` ``printf``(``"%s\n"``, str);` ` ``return` `0;` `}`
## Java
`import` `java.util.*;` `public` `class` `Main {` ` ``public` `static` `void` `main(String[] args) {` ` ``String str = ``"GeEkSfOrGeEkS"``;` ` ``StringBuilder sb = ``new` `StringBuilder(str);` ` ` ` ``for` `(``int` `i = ``0``; i < sb.length(); i++) {` ` ``if` `(Character.isUpperCase(sb.charAt(i))) {` ` ``sb.setCharAt(i, Character.toLowerCase(sb.charAt(i)));` ` ``} ``else` `{` ` ``sb.setCharAt(i, Character.toUpperCase(sb.charAt(i)));` ` ``}` ` ``}` ` ` ` ``str = sb.toString();` ` ``System.out.println(str);` ` ``}` `}`
## Python3
`# Python program to Convert characters` `# of a string to opposite case` `str` `=` `"GeEkSfOrGeEkS"` `for` `i ``in` `range``(``len``(``str``)):` ` ``if` `str``[i].isupper():` ` ``str` `=` `str``[:i] ``+` `str``[i].lower() ``+` `str``[i``+``1``:]` ` ``else``:` ` ``str` `=` `str``[:i] ``+` `str``[i].upper() ``+` `str``[i``+``1``:]` `print``(``str``)`
## C#
`// C# program to Convert characters` `// of a string to opposite case` `using` `System;` `class` `GFG {` ` ``public` `static` `void` `Main() {` ` ``string` `str = ``"GeEkSfOrGeEkS"``;` ` ``for` `(``int` `i = 0; i < str.Length; i++) {` ` ``// If the character is upper case, convert it to lower case` ` ``if` `(Char.IsUpper(str[i])) {` ` ``str = str.Remove(i, 1).Insert(i, Char.ToLower(str[i]).ToString());` ` ``}` ` ``// If the character is lower case, convert it to upper case` ` ``else` `{` ` ``str = str.Remove(i, 1).Insert(i, Char.ToUpper(str[i]).ToString());` ` ``}` ` ``}` ` ``Console.WriteLine(str);` ` ``}` `}`
## Javascript
`let str = ``"GeEkSfOrGeEkS"``;` `for` `(let i = 0; i < str.length; i++) {` ` ``if` `(str[i] === str[i].toUpperCase()) {` ` ``str = str.slice(0, i) + str[i].toLowerCase() + str.slice(i + 1);` ` ``} ``else` `{` ` ``str = str.slice(0, i) + str[i].toUpperCase() + str.slice(i + 1);` ` ``}` `}` `console.log(str);` `// THIS CODE IS CONTRIBUTED BY CHANDAN AGARWAL`
Output
```gEeKsFoRgEeKs
```
Time Complexity:- O(N) Where N is length of String
Auxiliary Space:- O(1)
Another Approach:
In this implementation, we initialize a pointer p to the start of the string str. We then use a while loop to iterate through the string by checking if the character p points to is the null character (‘\0’). Inside the loop, we use the same logic as the previous implementation, but instead of accessing the array element at index i, we use the dereference operator * to access the value that p points to. After updating the character at the current position, we increment the pointer p to point to the next character in the string.
## C++
`#include ` `#include ` `#include ` `int` `main()` `{` ` ``char` `str[] = ``"GeEkSfOrGeEkS"``;` ` ``char``* p = str; ``// initialize pointer to start of string` ` ``while` `(*p != ``'\0'``) {` ` ``if` `(``isupper``(*p)) {` ` ``*p = ``tolower``(*p);` ` ``}` ` ``else` `{` ` ``*p = ``toupper``(*p);` ` ``}` ` ``p++; ``// move pointer to next character` ` ``}` ` ``printf``(``"%s\n"``, str);` ` ``return` `0;` `}`
## Java
`import` `java.io.*;` `public` `class` `GFG {` ` ``public` `static` `void` `main(String[] args) {` ` ``String str = ``"GeEkSfOrGeEkS"``;` ` ``char``[] charArray = str.toCharArray(); ` ` ``// Convert the string to a character array` ` ``for` `(``int` `i = ``0``; i < charArray.length; i++) {` ` ``if` `(Character.isUpperCase(charArray[i])) {` ` ``charArray[i] = Character.toLowerCase(charArray[i]); ` ` ``// Convert uppercase to lowercase` ` ``} ``else` `{` ` ``charArray[i] = Character.toUpperCase(charArray[i]); ` ` ``// Convert lowercase to uppercase` ` ``}` ` ``}` ` ``String toggledStr = ``new` `String(charArray); ` ` ``// Convert the character array back to string` ` ``System.out.println(toggledStr);` ` ``}` `}`
## Python3
`str_value ``=` `"GeEkSfOrGeEkS"` `p ``=` `0` `# Initialize an index variable to start of string` `str_list ``=` `list``(str_value) ``# Convert the string to a list of characters` `while` `p < ``len``(str_list):` ` ``if` `str_list[p].isupper():` ` ``str_list[p] ``=` `str_list[p].lower()` ` ``else``:` ` ``str_list[p] ``=` `str_list[p].upper()` ` ``p ``+``=` `1` `# Move index to the next character` `result ``=` `''.join(str_list) ``# Convert the list back to a string` `print``(result)` `# This code is contributed by akshitaguprzj3`
## C#
`using` `System;` `class` `Program {` ` ``static` `void` `Main(``string``[] args)` ` ``{` ` ``string` `str = ``"GeEkSfOrGeEkS"``;` ` ``char``[] charArray` ` ``= str.ToCharArray(); ``// Convert the string to a` ` ``// character array` ` ``for` `(``int` `i = 0; i < charArray.Length; i++) {` ` ``if` `(``char``.IsUpper(` ` ``charArray[i])) ``// Check if the character` ` ``// is uppercase` ` ``{` ` ``charArray[i] = ``char``.ToLower(` ` ``charArray[i]); ``// Convert to lowercase` ` ``}` ` ``else` `{` ` ``charArray[i] = ``char``.ToUpper(` ` ``charArray[i]); ``// Convert to uppercase` ` ``}` ` ``}` ` ``string` `modifiedStr = ``new` `string``(` ` ``charArray); ``// Convert the character array back` ` ``// to a string` ` ``Console.WriteLine(modifiedStr);` ` ``}` `}`
## Javascript
`// JavaScript Program for the above approach` `const str = ``"GeEkSfOrGeEkS"``;` `let result = ``""``;` `for` `(let i = 0; i < str.length; i++) {` ` ``const char = str.charAt(i);` ` ``if` `(char === char.toUpperCase()) {` ` ``result += char.toLowerCase();` ` ``} ``else` `{` ` ``result += char.toUpperCase();` ` ``}` `}` `console.log(result);` `// THIS CODE IS CONTRIBUTED BY PIYUSH AGARWAL`
Output
```gEeKsFoRgEeKs
```
Time Complexity:- O(n) Where N is length of input String
Auxiliary Space:- O(1)
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Next | 9,312 | 27,086 | {"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 | latest | en | 0.409887 |
http://www.numbersaplenty.com/16021112118 | 1,585,576,846,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370497042.33/warc/CC-MAIN-20200330120036-20200330150036-00061.warc.gz | 286,962,023 | 3,581 | Search a number
16021112118 = 2311298370487
BaseRepresentation
bin11101110101110111…
…01100010100110110
31112100112112211011220
432322323230110312
5230302401041433
611205432011210
71105006130454
oct167273542466
945315484156
1016021112118
116881555730
123131530b06
1316842688cb
14abdaa99d4
1563b7b62b3
hex3baeec536
16021112118 has 32 divisors (see below), whose sum is σ = 36160508160. Its totient is φ = 4687472160.
The previous prime is 16021112059. The next prime is 16021112119. The reversal of 16021112118 is 81121112061.
Adding to 16021112118 its reverse (81121112061), we get a palindrome (97142224179).
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (16021112119) by changing a digit.
It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 4183330 + ... + 4187157.
It is an arithmetic number, because the mean of its divisors is an integer number (1130015880).
Almost surely, 216021112118 is an apocalyptic number.
16021112118 is an abundant number, since it is smaller than the sum of its proper divisors (20139396042).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
16021112118 is a wasteful number, since it uses less digits than its factorization.
16021112118 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 8370532.
The product of its (nonzero) digits is 192, while the sum is 24.
The spelling of 16021112118 in words is "sixteen billion, twenty-one million, one hundred twelve thousand, one hundred eighteen". | 484 | 1,629 | {"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-2020-16 | latest | en | 0.822386 |
https://mathoverflow.net/questions/155717/bishops-paradox-of-the-countability-of-sequences | 1,720,806,241,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514450.42/warc/CC-MAIN-20240712161324-20240712191324-00766.warc.gz | 308,973,025 | 27,045 | # Bishop's paradox of the countability of sequences
In 'Foundations of Constructive Analysis', in the notes at the end of the first chapter, Bishop poses an apparent paradox as an exercise for the reader:
Since every sequence of rational numbers can presumably be described by a phrase in the English language, and since the phrases in the English language can be sequentially ordered, the regular sequences of rational numbers can be sequentially ordered, in contradiction to Theorem 1.
'Theorem 1,' for those of you without the book open in front of you, is essentially Cantor's diagonal proof applied to Bishop's construction of the reals.
Is the exit here that the notion that phrases are 1-1 with sequences is misleading? Aren't some phrases schemata for infinite collections of sequences? In another direction, can't there be Chaitin-esque sequences that are legitimate but, in fact, not expressible as phrases in English? If you generate a sequence with the help of a physical source of entropy (and scale the values so that they meet the definition of a regular sequence) is it admissible in Bishop's definition or not?
• I would just like to note that the question is not specific to constructive mathematics. It and the solution apply equally well classically. Commented Jan 25, 2014 at 22:34
• Well, my last subquestion about generating a sequence with a bowl of goldfish is probably more narrowing constructive :-) Commented Jan 25, 2014 at 22:37
• The first clause in the Bishop quote is blatantly untrue, unless you are a constructivist. Commented Jan 25, 2014 at 22:51
• @Carl: That's an amusing name, because in Hebrew (originally from Aramaic) the word Bish (ביש) means bad (as in bad luck, for example). :-) Commented Jan 28, 2014 at 0:58
A similar "paradox" occurs if you replace "describable in English" with computable. I'll show how the paradox is resolved in this case. Also, I'll do this for Baire space (the set of functions from $\mathbb{N}$ to $\mathbb{N}$) to make life simple.
The set of all computable (possibly partial) functions is countable. However, the elements of $\mathbb{N}^\mathbb{N}$ only correspond to the total computable functions. This is a subset of a countable set, but it isn't necessarily countable itself (in constructive mathematics we refer to such sets as subcountable). From a computability point of view, the set of (codes of) total computable functions is not computably enumerable. One way to show this set is not computably enumerable is suggested by the "paradox" itself ie a diagonalisation argument. Alternatively you can note that the set of total computable functions has degree $0''$ whereas any ce set has degree below $0'$. | 623 | 2,694 | {"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 | 3 | CC-MAIN-2024-30 | latest | en | 0.93866 |
https://research.tec.mx/vivo-tec/display/AcademicArticleSCO_84941994569 | 1,581,902,009,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875141460.64/warc/CC-MAIN-20200217000519-20200217030519-00357.warc.gz | 558,573,002 | 6,034 | # The linear motion as a scenario for addressing relations between a function and its derivative Academic Article in Scopus
•
• Overview
•
•
• View All
•
### Abstract
• © American Society for Engineering Education, 2015. In this paper we present some results of a study conducted with 65 engineering students in a first Calculus course. SimCalc MathWorlds® software was included in the design of a didactic sequence to incorporate the scenario of the motion of an object over a horizontal straight line. The software document that was designed to work in classroom, plays an important role for analyzing the graphical behavior of position and velocity functions. By dragging the velocity graph, the software acts with the corresponding change in the position graph. The students used the software in their laptops by pairs during the two weeks didactic sequence. The aim of the sequence is to set up some relationships between the behavior of velocity graph and position graph. In a conventional curriculum those relations refer to the positive (negative) sign, and increasing (decreasing) behavior of derivative function, corresponding to the increasing (decreasing) and concave upward (downward) behavior of the function. Software brings the scenario for learning those facts analyzing the real context of linear motion. As part of the study, an assessment instrument was designed in order to appreciate the students' appropriation of those relations. The instrument' items are classified by corresponding to the linear motion context, or corresponding to different real contexts (no motion), or without including any real context. They also consider the posing information of the item and of the answer, being described by text or by a graphic. Application of the instrument lead us to reflect that, once the appropriation is achieved through the motion context, it could be easier for students to apply it without connection with a real context. It also reveals the difficulties for interpreting graphical information based on the derivative function. These findings are part of the overall results of a doctoral dissertation concerning with the use of digital technologies for the learning of Calculus. Keywords: Calculus learning, digital technologies, linear motion, real context, mediation.
### Publication date
• January 1, 2015 | 437 | 2,342 | {"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-2020-10 | latest | en | 0.937387 |
https://www.kylesconverter.com/density/pounds-per-cubic-yard-to-tonnes-per-liter | 1,726,195,622,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651506.7/warc/CC-MAIN-20240913002450-20240913032450-00656.warc.gz | 792,153,856 | 5,635 | # Convert Pounds Per Cubic Yard to Tonnes Per Liter
### Kyle's Converter > Density > Pounds Per Cubic Yard > Pounds Per Cubic Yard to Tonnes Per Liter
Pounds Per Cubic Yard (lb/cu yd) Tonnes Per Liter (t/l) Precision: 0 1 2 3 4 5 6 7 8 9 12 15 18
Reverse conversion?
Tonnes Per Liter to Pounds Per Cubic Yard
(or just enter a value in the "to" field)
#### Please share if you found this tool useful:
Unit Descriptions
1 Pound per Cubic Yard:
Mass of 1 pound per volume of a cubic yard. International yard having a length of 0.9144 meters. Approximate density of 0.593 276 421 257 783 kilograms per cubic meter. 1 lb/cu yd ≈ 0.593276421257783 kg/m3.
1 Tonne per Liter:
Mass of 1 tonne per volume of a liter. Metric tonne of exactly 1 000 kilograms. Equivalent density of a million (1 000 000) kilograms per cubic meter. 1 t/l ≈ 1000000 kg/m3.
Conversions Table
1 Pounds Per Cubic Yard to Tonnes Per Liter = 070 Pounds Per Cubic Yard to Tonnes Per Liter = 0
2 Pounds Per Cubic Yard to Tonnes Per Liter = 080 Pounds Per Cubic Yard to Tonnes Per Liter = 0
3 Pounds Per Cubic Yard to Tonnes Per Liter = 090 Pounds Per Cubic Yard to Tonnes Per Liter = 0.0001
4 Pounds Per Cubic Yard to Tonnes Per Liter = 0100 Pounds Per Cubic Yard to Tonnes Per Liter = 0.0001
5 Pounds Per Cubic Yard to Tonnes Per Liter = 0200 Pounds Per Cubic Yard to Tonnes Per Liter = 0.0001
6 Pounds Per Cubic Yard to Tonnes Per Liter = 0300 Pounds Per Cubic Yard to Tonnes Per Liter = 0.0002
7 Pounds Per Cubic Yard to Tonnes Per Liter = 0400 Pounds Per Cubic Yard to Tonnes Per Liter = 0.0002
8 Pounds Per Cubic Yard to Tonnes Per Liter = 0500 Pounds Per Cubic Yard to Tonnes Per Liter = 0.0003
9 Pounds Per Cubic Yard to Tonnes Per Liter = 0600 Pounds Per Cubic Yard to Tonnes Per Liter = 0.0004
10 Pounds Per Cubic Yard to Tonnes Per Liter = 0800 Pounds Per Cubic Yard to Tonnes Per Liter = 0.0005
20 Pounds Per Cubic Yard to Tonnes Per Liter = 0900 Pounds Per Cubic Yard to Tonnes Per Liter = 0.0005
30 Pounds Per Cubic Yard to Tonnes Per Liter = 01,000 Pounds Per Cubic Yard to Tonnes Per Liter = 0.0006
40 Pounds Per Cubic Yard to Tonnes Per Liter = 010,000 Pounds Per Cubic Yard to Tonnes Per Liter = 0.0059
50 Pounds Per Cubic Yard to Tonnes Per Liter = 0100,000 Pounds Per Cubic Yard to Tonnes Per Liter = 0.0593
60 Pounds Per Cubic Yard to Tonnes Per Liter = 01,000,000 Pounds Per Cubic Yard to Tonnes Per Liter = 0.5933 | 775 | 2,403 | {"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-2024-38 | latest | en | 0.591833 |
https://www.mrexcel.com/board/tags/cell-analysis/ | 1,718,371,312,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861546.27/warc/CC-MAIN-20240614110447-20240614140447-00024.warc.gz | 813,280,562 | 12,146 | # cell analysis
1. ### Help identifying whether Sheet1!A* = Sheet2!A* and then populating with Sheet2!C*
I have ~50,000 usernames in Column A on Sheet1. Many of those usernames repeat. On Sheet2, I have the same usernames in Column A, followed by their employee IDs in Column B. But on Sheet2, there are no repeats. I want to copy over the IDs from Sheet2 into the corresponding place in Sheet1...
2. ### Moving specific cells to the right
Hello all, I have a problem that I am trying to solve. I have a table with over 770K rows. These rows contain data about the weather at different times of each day. In the fourth column (Type) there is a mix of AUTO and integers. The integers are supposed to be in the Wind Spd & Dir column...
3. ### Check a cells value
i am trying to solve a scenario.... A1=332.345 B1=331.658 C1=331.229 D1 should give output as 1 provided Condition1: A1>B1 AND C1<B1 Condition2: Neither of A1, B1 & C1 should be <=0. If any of abv 2 conditions are not met then D1=0. What formula needs to be punched in D1? Thnx in adv.
4. ### Figuring out what contents are in a cell
I'm making a schedule for school. The school website will give me an excel document of all the games and practices put in. I download the list and then paste it into a formatting page that will then link to all the other scheduling formats and needs I have. The time's of the practices come...
5. ### Advanced Min-Max in Access
Hi, I am looking at the transfering the following code from Excel VBA to Access VBA and was wondering if anybody could help me by telling me which factors I should look out for/ how I should change the code so as to make it functional. What the code does is to take an interval in a set of...
6. ### excel beta 2010
Is there a way to split a numerical value of a single cell into each integer of it's value and list them separately in individual cells? For example: 391 is the value in cell A:1 and I want cells B:1, C:1, and D:1 to read 3,9,1 respectively, so I can derive at 13 in cell E:1: which is the sum of...
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Go back | 758 | 2,935 | {"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-2024-26 | latest | en | 0.925672 |
http://www.solutioninn.com/for-this-exercise-youre-a-realestate-developer-youre-planning-a | 1,498,406,286,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320539.37/warc/CC-MAIN-20170625152316-20170625172316-00621.warc.gz | 653,959,808 | 8,750 | # Question: For this exercise you re a real estate developer You re planning a
For this exercise, you’re a real-estate developer. You’re planning a suburban housing development outside Philadelphia. The design calls for 25 homes that you expect to sell for about \$450,000 each. If all goes as planned, you’ll make a proft of \$50,000 per house, or \$1.25 million overall.
If you add a security wall around the development, you might be able to sell each home for more. Gates convey safety and low crime rates to potential homebuyers. The crime rate in the area where you are building the development is already low, about 15 incidents per 1,000 residents. A security consultant claims a gate and fence would reduce this further to 10 per 1,000.
If this consultant is right, is it worth adding the gate and wall? The builders say that it will cost you about \$875,000 (\$35,000 more per house) to add the gate and fence to the development—if you do it now while construction is starting. If you wait until people move in, the costs will rise sharply.
You have some data to help you decide. The data include the median selling price of homes in communities in the Philadelphia area. The data also include the crime rate in these communities, expressed in incidents per 1,000 residents. This analysis will use the reciprocal of this rate. These data appeared in the April 1996 issue of Philadelphia Magazine. Because the values of homes have increased a lot since 1996, let’s assume that prices have doubled since these data were measured. (These data also appear in an exercise in Chapter 20. That exercise focuses on the transformation. For this exercise, we will focus on the use of the model with the transformation.)
Motivation
(a) Assume that the addition of a gate and wall has the effect of convincing potential buyers that the crime rate of this development will “feel” like 10 crimes per 1,000 rather than 15. How much does this have to increase the value of these homes (on average) in order for building the security fence to be cost effective?
(b) If the regression model identifies a statistically significant association between the price of housing and the number of people per crime (the reciprocal of the crime rate), will this prove that lowering the crime rate will pay for the cost of constructing the security wall?
Method
(c) Plot the selling prices of homes in these communities versus 1,000 divided by the crime rate. Does the plot seem straight enough to continue? (The variable created by 1,000 divided by the crime rate is the number of residents per crime.)
(d) Fit the linear equation to the scatterplot in part (c). If you accept the ft of this equation, what do you think about building the wall? Be sure to take the doubling of home prices into account.
Mechanics
(e) Which communities are leveraged in this analysis? What distinguishes these communities from the others?
(f) Which communities are outliers with unusually positive or negative residuals? Identify these in the plot of the residuals on the explanatory variable.
(g) Does this model meet the conditions needed for using the SRM for inference about the parameters? What about prediction intervals?
(h) If we ignore any problems noted in the form of this model, would the usual inferences lead us to tell the developer to build the wall? (Again, re-member to take account of the doubling of prices since 1996 into account.)
Message
(i) How would you answer the question for the developer? Should the developer proceed with the wall?
(j) What could you do to improve the analysis? State your suggestions in a form that the developer would understand.
View Solution:
Sales0
Views46 | 777 | 3,675 | {"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-2017-26 | longest | en | 0.950364 |
http://www.docstoc.com/docs/130289808/Homework-Assignment-3 | 1,386,897,088,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164786099/warc/CC-MAIN-20131204134626-00075-ip-10-33-133-15.ec2.internal.warc.gz | 311,836,852 | 23,114 | # Homework Assignment #3
Document Sample
``` Homework #3 Answers
1. There are 124 variables, and there were 279 participants.
2. It can be found in the “Compare Means” submenu.
3. Anxiety was their predominant childhood emotion.
4. The question asked, “Compared to most people… How satisfied are you with your life?”
5. The median was 6.0
6. 56.6% rated themselves as a 4 or lower on the 9-point scale regarding frequency of playing
sports.
7. The frequency distribution for beauty concerns is more negatively skewed.
8. Yes, based on the scatterplot, people who eat more tend to have poorer health, supporting the
hypothesis. [This question originally had a typo regarding which variables to use, so I will
accept any reasonable interpretation of a scatterplot]
9. The correlation (r = -.06) is near zero and unreliable. Thus, one’s level of sadness is virtually
unrelated to their friends’ level of happiness, which is somewhat surprising. People may not be
affected by the mood of their friends.
10. Of these four variables, attractiveness is most highly related to depression. People who are
attractive tend to be less depressed. Depression is more related to attractiveness than loud music
preferences, parental fighting, or object relations.
11. The effect size for the relationship between depression and attractiveness is “medium”.
12. Using sleep problems, stress, and depression to predict somatization, R = .51, R2 = .26. Thus,
these predictors explain 51% of the differences in somatization, a large effect. The equation
used for predicting somatization is as follows:
Somatization’ = 1.986 + .238*(sleep problems) + .230*(stress) + .176*(depression)
and +.5 for any attempt at an answer.
Demerits: No cover page = -1, No Output = -3
Output
5. Frequencies
Statistics
64. Beauty 65. Play
Conc erns Sports
N V alid 279 279
Mis sing 0 0
Mean 5.89 3.99
Median 6.00 4.00
Mode 5 1
Std. Deviation 2.088 2.460
Frequency Table
64. Beauty Concer ns
65. Play Sports
Cumulativ e
Frequenc y Percent V alid Percent Percent Cumulativ e
V alid 1 5 1.8 1.8 1.8 Frequenc y Percent V alid Percent Percent
V alid 1 56 20.1 20.1 20.1
2 14 5.0 5.0 6.8
2 46 16.5 16.5 36.6
3 32 11.5 11.5 18.3
3 36 12.9 12.9 49.5
4 10 3.6 3.6 21.9
4 20 7.2 7.2 56.6
5 56 20.1 20.1 41.9 5 49 17.6 17.6 74.2
6 40 14.3 14.3 56.3 6 18 6.5 6.5 80.6
7 55 19.7 19.7 76.0 7 23 8.2 8.2 88.9
8 35 12.5 12.5 88.5 8 16 5.7 5.7 94.6
9 32 11.5 11.5 100.0 9 15 5.4 5.4 100.0
Total 279 100.0 100.0 Total 279 100.0 100.0
Histogram
65. Play Sports
64. Beauty Concerns
60
60
50
50
40
Frequency
40
Frequency
30
30
20
20
10
10
Mean = 3.99
Mean = 5.89
Std. Dev. = 2.46
Std. Dev. = 2.088
N = 279 0 N = 279
0
0 2 4 6 8 10 0 2 4 6 8 10
64. Beauty Concerns 65. Play Sports
8.
10
8
86. Physical Health
6
4
2
R Sq Linear = 0.105
0
0 2 4 6 8 10
48. Eat too Much
9.
Cor relations
94. Friend
94. Friend Happiness Pearson Correlation 1 -.060
Sig. (2-tailed) .317
N 279 279
95. Sadnes s Pearson Correlation -.060 1
Sig. (2-tailed) .317
N 279 279
10.
Cor relations
70. Parents
68. 69. Listen to Fought 91. Objec t 97.
Depression Loud Mus ic Grow ing Up Relations A ttractiveness
68. Depress ion Pearson Correlation 1 .096 .238** -.168** -.328**
Sig. (2-tailed) .109 .000 .005 .000
N 279 279 279 279 279
69. Listen to Loud Music Pearson Correlation .096 1 .105 -.069 .057
Sig. (2-tailed) .109 .080 .253 .346
N 279 279 279 279 279
70. Parents Fought Pearson Correlation .238** .105 1 -.104 -.086
Grow ing Up Sig. (2-tailed) .000 .080 .082 .152
N
279 279 279 279 279
91. Objec t Relations Pearson Correlation -.168** -.069 -.104 1 .104
Sig. (2-tailed) .005 .253 .082 .084
N 279 279 279 279 279
97. A ttractiv eness Pearson Correlation -.328** .057 -.086 .104 1
Sig. (2-tailed) .000 .346 .152 .084
N 279 279 279 279 279
**. Correlation is s ignif icant at the 0.01 level (2-tailed).
12.
Model Sum m ary
Model R R Square R Square the Estimate
1 .506 a .256 .248 1.800
a. Predictors: (Constant), 68. Depress ion, 52. Sleep
Problems, 61. Stres s
a
Coe fficients
Unstandardiz ed Standardized
Coef f icients Coef f icients
Model B Std. Error Beta t Sig.
1 (Cons tant) 1.986 .350 5.675 .000
52. Sleep Problems .238 .055 .250 4.298 .000
61. Stress .230 .062 .226 3.731 .000
68. Depression .176 .065 .174 2.708 .007
a. Dependent Variable: 58. Somatization
Homework Assignment #3
Due Tuesday, February 12th (Exam Day)
For this assignment, you will learn how to make basic statistical calculations using a program
called SPSS. For the assignment, make a cover page. Then, type up a neat “answer page.”
After the answer page, provide a copy of all of your SPSS Output – the numbers, figures, and
tables from SPSS. Please print out two copies of your complete assignment – one for yourself
and one for me – because I will be posting answers online rather than passing back assignments.
I will deduct 0.5 points for every major spelling error or major punctuation failure (e.g. not
capitalizing the first word in a sentence).
questions you need to answer are in blue text.
Section 1: Introduction to SPSS and the Classroom Survey Data File
A. Getting Started:
a) All computer assignments in this course are designed to be tutorials. They are
substitutions for detailed, expensive workbooks. Please refer back to these assignments
later in the year, as needed, when you get stuck (e.g. term paper time).
b) SPSS is a computer program you will need for the computer assignments. Most
students work on the assignment in the Woldt computer lab (free printing); however, it is
also available in many of the computers in the library and around campus. Different
computers have different versions, but they are all very similar. I use SPSS 13.0, so if
you run into problems due to a different version, let me know, but they are pretty much
all the same. You may be able to find a free trial version on the web, if you prefer.
c) Go to the course web site (http://www.psychmike.com/psy211/):
o d) How do I find our data file? From the course homepage, you can click on the
“Survey” link. It takes you to a page that reminds you that our survey data file is
saved on BlackBoard so that not just anyone can access it. To find it, log on to
BlackBoard, and find the data.sav file in the Course Materials folder. Download
the file. The easiest way to open it is by double-clicking on it. If this does not
work, open the program SPSS; go to the File menu, point to Open, and choose
“Data…” – find and open the file.
o e) Where can I find out more about each survey question? From the course
homepage, you can click on the “Survey” link. It takes you to a page with a link
to a “data guide,” which you can click on, or you can go directly to this address:
http://www.psychmike.com/psy211/data_guide.doc
This file tells you the variable number, a name I made up for each “construct”
(variable), the question we asked in the survey, the type of variable (continuous or
categorical), and how the responses are coded in the data file. This is very helpful
for interpreting results.
o f) What if I get suck? From the course homepage, click on the “SPSS Manual”
This will load a free, public domain .pdf file describing how to solve many
statistical problems using SPSS. It’s a great guide in case you ever get stuck, but
hopefully my instructions will be enough. E-mail me if you get stuck. You can
ask a friend for help, as long as they do not show you how to do a specific
homework problem.
B. Exploring the data file:
a) Using the above information, open the classroom survey data file (data.sav) in SPSS.
A large, detailed data file will open.
b) Once the data file pops up, look at the bottom left side of the screen. You should see
two tabs labeled “Data View” and “Variable View” and you can click back and forth on
them to see how they differ.
c) In “Data View” each row (across) represents a specific participant. Each column (up
and down) represents a particular variable. For example, participant #4 has a “0” for the
variable “physical” and a “1” for the variable “smoker”. Similarly, participant #80 has a
“1” for the last variable in the file “inkblot14”. Since we don’t automatically know what
these numbers mean, the data on their own would be very confusing.
d) In “Variable View” you get more information on the variables, including how they are
coded. In the first column (“Name”), you see the name of the variable; it’s a pretty
crummy name because only certain characters can be used. In the fifth column (“Label”)
is a clearer name for each variable. In the sixth column, there is information on how
some of the variables are coded. For example, find the fourth variable, “athlete” and
click on the box in the “Label” column; inside that box, a small gray box with three dots
will appear, and if you click on it, it will tell you how the data are coded. It shows that a
“0” means “No” – not an athlete, and a “1” means “Yes” – an athlete. If you need more
detail on a variable, see the data_guide.doc file (see Section 1Ae above).
e) Regardless of whether you’re in Data View or Variable View, you will still have the
same menu commands at the top of the window (File, Edit, View, Data, etc.). Look
through each menu to get an idea of what functions are available
o f) File menu: opening documents, saving, printing, etc.
o g) Edit menu: copy, paste, etc.
o h) View menu: not usually used
o i) Data menu: advanced commands for combining data files, excluding certain
participants in analyses, and other commands not usually needed in PSY 211
o j) Transform menu: included commands for doing computations and recoding
variables in various ways
o k) Analyze menu: all sorts of commands for statistical analyses; this is the main
menu we will use in PSY 211; look it over thoroughly and see if you recognize
any of the words
o l) Graph menu: all sorts of commands for making graphs and figures; we will use
Question #1:
How many variables are in the data file? How many participants?
Question #2:
Within the Analyze menu, where do you find the “Independent-Samples T Test…” command?
Question #3:
What does it mean if someone has a “4” for variable #22 “Negative Emotion (Childhood)”?
Question #4:
What was the exact wording we used in the survey for question #92 “Life Satisfaction”?
Section 2: Basic Statistics Operations
A. Frequencies Command
a) Go to the Analyze menu, point to Descriptive Statistics, and choose “Frequencies…”.
This will allow us to make a frequency table and obtain other useful statistics.
b) In the window that pops up, click on the Statistics button. Select the Mean, Median,
Mode, and Standard Deviation. Then, click on the Continue button.
c) In the initial pop-up window, now click on the Charts button. In the window that pops
up, choose Histogram. Then, click the Continue button.
d) In the initial pop-up window, move variable #56 “Hand Washing” from the list on the
left to the box on the right, telling SPSS that we want statistical information for this
variable (you could choose multiple variables if you like). Then, click on the OK button.
e) In the Output, you should find that the mean on the 1-9 scale is 8.1, median = 9.0,
mode = 9.0, SD = 1.52. Most people say they always wash their hands. You should also
see a nice frequency table. For example, it shows that 4.7% of people marked a rating of
“5” on the 9-point scale, and 9% of participants marked a rating of “5” or lower (note that
SPSS usually makes its frequency tables upside down, with lower numbers on top).
Question #5:
Repeat the above Frequency Command steps for variables #64 “Beauty Concerns” and #65 “Play
Sports”. What is the median rating for beauty concerns?
Question #6:
What is the cumulative percentage of people who rated themselves as a “4” or lower on the “Play
Sports” variable?
Question #7:
Both of the variables have messy-looking histograms, but which variable is more negatively
skewed?
B. Scatterplot
a) To make a scatterplot, go to Graphs menu and choose “Scatter/Dot…”. In the window
that pops up, make sure “Simple Scatter” is selected and click the Define button.
b) In the next window that pops up, we will choose one variable for “X Axis” and one
variable for “Y Axis”. Usually, if we think one variable is the cause, we put it on the X
axis.
c) Let’s hypothesize that stress causes crying. Move variable #61 “Stress” to the “X
Axis” box and variable #60 “Crying” to the “Y Axis” box. Then click on the OK Button.
d) In the Output, double click on the gray area of the scatterplot. A new window will pop
up. In the pop-up window, you’ll see your scatterplot again and above it, there are many
little buttons with pictures. The buttons all look very similar, but if you point your arrow
over it, some pop-up text should describe the button. Find one that says “Add Fit Line at
Total” and click on it (if you can’t find it, go to the Elements menu, and choose “Fit Line
at Total”). This makes a best fit line appear and another small popup box appear. Close
this new pop-up box and the previous one, taking you back to the Output, which is now
updated with the best fit line.
8
7
6
60. Crying
5
4
3
R Sq Linear = 0.203
2
R Sq Linear = 0.203
1
0 2 4 6 8 10
61. Stress
Question #8:
You hypothesize that eating too much causes poor health. Make a scatterplot using variable #86
“Physical Health” and #48 “Eat too Much”. Add a best fit line. Based on the scatterplot, does it
look like your hypothesis is supported? (Remember, the “cause” goes on the X Axis).
Section 3: Correlation and Regression
A. Correlation
a) To run correlations, go to the Analyze menu, point to Correlate, and choose
“Bivariate…”.
b) In the window that pops up, select any variables (at least two) that you’d like to
correlate. If you choose more than two variables, SPSS will run a correlation for every
possible pairing of two variables. Try selecting the Big 5 personality traits, which are
variables #103-#107, and move them from the list on the left to the box on the right.
Then click on the OK button.
c) The Output should look something like the following table:
Cor relations
105. 106. 107.
103. 104. Openness to Agreeabl Cons cienti
Ex travers ion Neuroticism Ex perience enes s ousness
103. Extrav ersion Pearson Correlation 1 -.056 .218** -.001 -.106
Sig. (2-tailed) .353 .000 .991 .077
N 279 279 279 279 279
104. Neurotic is m Pearson Correlation -.056 1 -.133* .017 -.104
Sig. (2-tailed) .353 .027 .771 .084
N 279 279 279 279 279
105. Opennes s to Pearson Correlation .218** -.133* 1 .203** -.015
Ex perience Sig. (2-tailed) .000 .027 .001 .802
N
279 279 279 279 279
106. Agreeableness Pearson Correlation -.001 .017 .203** 1 .231**
Sig. (2-tailed) .991 .771 .001 .000
N 279 279 279 279 279
107. Consc ientiousness Pearson Correlation -.106 -.104 -.015 .231** 1
Sig. (2-tailed) .077 .084 .802 .000
N 279 279 279 279 279
**. Correlation is s ignif icant at the 0.01 lev el (2-tailed).
*. Correlation is s ignif icant at the 0.05 lev el (2-tailed).
d) Notice that each variable is listed along the side and the top. I have put green boxes
around them above. To find information on how two variables correlate, choose one
variable on the side, one on the top, and track where they meet in the body of the table.
For example, to see how agreeableness correlates with neuroticism, you could find
agreeableness on the side, neuroticism on the top, and track where they meet (the red
box). Inside the red box, you see three numbers. The top number is most relevant; it is
the correlation coefficient, r = .017 (or rounded, r =.02). The numbers below it are less
important, but the middle number is called a “p-value” or “significance value” and
indicates the probability that the result might be due to chance (p = .77 in this case); there
is about a 77% chance we’d get a correlation like this by chance, so it’s an unreliable
finding; if the p-value is less than .05, it is reliable or trustworthy. The bottom value tells
the sample size, 279. The correlation between agreeableness and neuroticism is weak
and unreliable.
e) All significant or reliable correlations will have a star (*) by them, and p is < .05.
f) If a p-value says “.000” it’s because SPSS is rounding down too much. Consider these
values to really mean “approximately zero” or “p < .001”
g) If you compare a variable to itself, you’ll see it always correlates r = 1.00.
h) The correlation between neuroticism and openness to experience (blue box) is r = -.13.
Because it has a star (*) and p < .05, we know it is reliable. Thus, there is a small,
reliable, negative relationship between neuroticism and openness. People who are
neurotic (sad, anxious, angry) are slightly less open – not too surprising.
Question #9:
What is the correlation between variable #95 “Sadness” and #94 “Friend Happiness” and what
does this finding tell us?
Question #10:
Which of the following variables is most strongly correlated with #68 “Depression”?
#70 “Parents Fought Growing Up”
#91 “Object Relations”
#97 “Attractiveness”
#69 “Listen to Loud Music”
Question #11:
Based on question #10, what is the effect size for the correlation between Attractiveness and
Depression?
B. Regression
a) To run a regression, go to the Analyze menu, point to Regression, and choose Linear.
b) Put the “causes” or predictors in the “Independent(s)” box (there can be one or more
causes) and put one variable, the “effect,” in the “Dependent box”. I will use #41 “GPA
(High School)”, #43 “ACT Score”, and #107 “Conscientiousness” to predict #42 “GPA
(college)”. Put the first three variables in the Independents box, and the last one in the
Dependents box. Then, click on the OK button.
c) The Output shows that R = .54. R2 = .29, so high school GPA, ACT score, and level
of conscientiousness account for 29% of the differences in college GPA. (See lecture
notes if more detail is needed).
d) The formula for predicting college GPA is the following:
College GPA’ =
0.639 + (0.441)*(HS GPA) + (0.023)*(ACT Score) + (0.081)*(Conscientiousness)
e) If we know someone’s HS GPA, ACT Score, and Conscientiousness rating, we could
predict their College GPA. For example, if these values were 4.0, 21, and 5, then we
would predict a GPA of
College GPA’ =
0.639 + (0.441)*(4.0) + (0.023)*(21) + (0.081)*(5) =
0.639 + 1.764 + 0.483 + 0.405 = 3.291
Question #12:
You hypothesize that sleep problems, stress, and depression, will contribute to somatization
(physical aches and pains, particularly when overwhelmed). Run a regression using #52 “Sleep
Problems”, #61 “Stress”, and #68 “Depression” to predict #58 “Somatization”. What is the
value of R? What is the value of R2? What is the equation for predicting somatization? What
conclusions can be drawn?
Extra Credit:
Using the equation in Section 3Be (the College GPA formula), predict your own college GPA (or
a friend’s) based on the three predictors. You may need to refer to the data guide to make the
conscientiousness rating. Does the formula do a good job of predicting?
```
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How are you planning on using Docstoc? | 5,875 | 24,762 | {"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-2013-48 | longest | en | 0.94434 |
https://physics.stackexchange.com/questions/517225/why-in-mercury-barometer-pressure-inside-the-glass-tube-at-a-point-is-same-as-th | 1,585,434,032,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370493120.15/warc/CC-MAIN-20200328194743-20200328224743-00057.warc.gz | 668,644,349 | 32,560 | # Why in mercury barometer pressure inside the glass tube at a point is same as the pressure outside the glass tube at the same height?
I am quite familiar with the pascal’s law but i still think that being on the same points still they should have different pressures as above on one point is atmosphere and on the other is only the column of mercury can anyone give physical explanation for this and also if the pressure inside the torricelli vacuum is only due to mercury vapours which is quite low than why the vacuum does not get crushed under the atmospheric pressure?
• Atmospheric pressure is $1~kg/cm^2$. If the tube's surface area is $1cm^2$, it needs to be strong enough to not break when you put $1kg$ over it's entire surface. – AgentS Dec 2 '19 at 5:40
Now the pressure in the space above the mercury is less than atmospheric pressure and equilibrium is attained when $$Mg$$ (where M is the new mass of mercury which is less than initial mass in the mercury column) + P*A (where P is less than atmospheric pressure) = (Atmospheric pressure) $$A$$ where A is area of cross section of tube. Or as $$ρ g h$$ (where h is the new height of column) = P + Atmospheric pressure
So we get. $$\rho g h = P_\text{atm} - P$$
where $$\rho$$ is the density of the mercury. | 307 | 1,275 | {"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": 5, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2020-16 | latest | en | 0.942164 |
http://lists.racket-lang.org/users/archive/2008-January/022191.html | 1,495,756,979,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608617.80/warc/CC-MAIN-20170525233846-20170526013846-00336.warc.gz | 219,840,071 | 3,556 | # Problem with recursion? (was [plt-scheme] Re: Novice needs help writing function )
From: Matthias Felleisen (matthias at ccs.neu.edu) Date: Fri Jan 4 15:28:28 EST 2008 Previous message: Problem with recursion? (was [plt-scheme] Re: Novice needs help writing function ) Next message: Problem with recursion? (was [plt-scheme] Re: Novice needs help writing function ) Messages sorted by: [date] [thread] [subject] [author]
```On Jan 4, 2008, at 1:41 PM, Yavuz Arkun wrote:
> The number of explicit variables in the recursive version is
> deceptive. When I am trying to understand a program, I put myself
> in the place of the computer and work through the program in my
> mind. For the iterative version, everything I need to remember are
> explicitly named. For recursive algorithms, I need to keep track of
> the implicit variables as well, such as how deep I am into the
> recursion, and what the intermediate values are, because I will
> need them on my way back.
>
> Personally, I can manage 1 or 2 levels mentally, and if I use pen
> and paper maybe a few more. After that, I need to take a leap of
> faith, and inductively conclude that if it works for the simplest
> case, and for the one after that, it should work for all cases.
>
> In other words, neither the conciseness of the program text, nor
> the elegance of the expressed ideas necessarily means "easy to
> understand", unless you are prepared to take the leap.
>
> I can do that, because I know recursion works as a general concept.
> It may be more difficult if you are new to the idea. Even more so,
> when you realize that the 'elegant' method can have rather
> inelegant space and/or time characteristics. Then you start to feel
> like you must be missing something and become frustrated.
>
> Therefore, recursion (like any other method) should be illustrated
> using examples that can be best solved with recursion.
Please do take a look at HtDP. Your message suggests that you were
trained by or as an EE :-)
-- Matthias
```
Posted on the users mailing list. Previous message: Problem with recursion? (was [plt-scheme] Re: Novice needs help writing function ) Next message: Problem with recursion? (was [plt-scheme] Re: Novice needs help writing function ) Messages sorted by: [date] [thread] [subject] [author] | 569 | 2,295 | {"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 | 0.926391 |
https://studyadda.com/sample-papers/physics-sample-paper-10_q9/1261/399993 | 1,610,854,548,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703509104.12/warc/CC-MAIN-20210117020341-20210117050341-00310.warc.gz | 577,349,386 | 19,469 | • # question_answer (i) State Kirchhoff's rules for an electrical network. (ii) The plot of the variation of potential difference across a combination of three identical cells in series, versus current is as shown below, What is the emf of each cell?
Let emf of each cell = E Let internal resistance of each cell = r, then $V=(E+E+E)-i(r+r+r)=3(E-ir)$ From the graph, when i = 0, V = 6 i.e. $6=3(E-0)=3E$ $E=6/3\Rightarrow E=2$ | 135 | 447 | {"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} | 3.609375 | 4 | CC-MAIN-2021-04 | latest | en | 0.857123 |
http://www.daz3d.com/forums/viewthread/12180/ | 1,438,446,340,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042988840.31/warc/CC-MAIN-20150728002308-00321-ip-10-236-191-2.ec2.internal.warc.gz | 392,319,268 | 15,268 | # Trans equals what?
Posts: 286
edited December 1969
On maps, there are scales to measure distance, e.g. 1 inch = 1 mile, etc. Rotation in DS is degrees, e.g. xrot 45 = 45 degrees. So, e.g. xtrans 10 would be what? Centimetres, inches, feet?
• Posts: 2,897
edited December 1969
If memory serves, the x/y/z trans units are cm, with the 'ground grid' of DS laid out in 50cm squares.
• Posts: 2,708
edited November 2012
1.0 units of translation = 1 Centimeter
And see this thread:
http://www.daz3d.com/forums/discussion/11923/
Post edited by wancow on
• Posts: 7,235
edited December 1969
Basically, everything in DS is in cm.
That means a 0.1 displacement is a mm, too.
• Posts: 2,708
edited December 1969
Oh crap... you know what, till I read your post, that never bloody well occurred to me!!! No joke!
• Posts: 1,008
edited December 1969
Yes, centimeters. But the floor grid is 1 meter squares, not 50cm.
The easy way to see for yourself is to make a 1m cube primitive. Move it +50cm X and Z to fit it in a grid square.
mac
• Posts: 286
edited December 1969
Thanks all. Now I know how to place a building a quarter of a mile away :)
• Posts: 2,897
edited December 1969
maclean said:
Yes, centimeters. But the floor grid is 1 meter squares, not 50cm.
The easy way to see for yourself is to make a 1m cube primitive. Move it +50cm X and Z to fit it in a grid square.
mac
I suspected that 50cm would come back and bite me ... I had a render running at the time and could not check/confirm. :)
• Posts: 286
edited December 1969
Which places the building outside a skydome. So, according to DAZ, the universe is only half a mile wide :ohh:
• Posts: 7,235
edited December 1969
You do know they say the world is getting smaller all the time...but just beyond the end of my driveway? That's a little too small...
• Posts: 815
edited December 1969
But, but you can scale the sphere up! And things disappear from the viewport if you scale them too big.
If that's the UE sphere it doesn't matter if things are inside or outside, they still will be lit (as long as the texture on the sphere isn't set to show in the render, that is).
• Posts: 2,708
edited December 1969
There's got to be a way to create backgrounds... some technique you can use to render a background, then map it onto a sphere, or a background arc... I'll bet that would be a great pluggin or script challenge for some of these guys that do that...
• Posts: 3
edited December 1969
There is Infinito
http://www.daz3d.com/shop/infinito-1-0
But at that price? Errr .......
• Posts: 2,708
edited December 1969
I was looking at that... I'm not sure. I'd like to see one or two other people actually using it and recommending it before I buy it...
• Posts: 2,444
edited November 2012
It was half off during the PA sale. The only reason I didn't buy it was because it didn't support instancing. I'm thinking I should have bought it anyways.
Allessandro has some great videos available if anyone is considering it. He is the same artist who is developing the "Look at my Hair."
Post edited by Gedd on | 862 | 3,090 | {"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} | 3.34375 | 3 | CC-MAIN-2015-32 | longest | en | 0.931591 |
http://listserv.uga.edu/cgi-bin/wa?A2=ind0705&L=spssx-l&T=0&F=&S=&P=1787 | 1,397,695,849,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609526102.3/warc/CC-MAIN-20140416005206-00438-ip-10-147-4-33.ec2.internal.warc.gz | 142,561,831 | 4,685 | ```Date: Tue, 1 May 2007 09:20:08 -0700 Reply-To: "Ornelas, Fermin" Sender: "SPSSX(r) Discussion" From: "Ornelas, Fermin" Subject: Re: shapiro-wilks Comments: To: Marta García-Granero In-Reply-To: <962091718.20070501173508@terra.es> Content-Type: TEXT/plain; charset="iso-8859-1" Well it is human to make mistakes and it is also human to admit it when that happens especially when one replies too fast. I had the ho and ha reversed. It should have been: Ho: distribution is normal Ha: distribution is non normal But the rest of the argument should be OK. Fermin Ornelas, Ph.D. Management Analyst III, AZ DES Tel: (602) 542-5639 E-mail: fornelas2@azdes.gov -----Original Message----- From: SPSSX(r) Discussion [mailto:SPSSX-L@LISTSERV.UGA.EDU] On Behalf Of Marta García-Granero Sent: Tuesday, May 01, 2007 8:35 AM To: SPSSX-L@LISTSERV.UGA.EDU Subject: Re: shapiro-wilks Hi Fermin Monday, April 30, 2007, 6:17:07 PM, You wrote: OF> I think there is room for confusion here. OF> Ho: the distribution is non-normal, OF> Ha: the distribution is normal. Ouch! NO, NO, definitely NO. The null hypothesis for a normality test is that the variable IS normal (believe man, I'm a statistics teacher...). Null hypotheses, as I said in my previous mail, say that there are no differences, no effects... In this particular case, it says that the observed distribution is NOT different from the one we would expect had the sample been drawn from a normal population. If p-value >> alpha then conclude Ha. Ouch again!! p-value >> alpha means "accept H0". If p-value >> It may not be If p-value >> wise to give more weight to the graph especially if one If p-value >> is unfamiliar with the shapes of the distributions (long If p-value >> tails, short tails). In big samples, tails are of very little importance (leptokurtosis effect dilutes faster with sample size than skewness, I even read a math demo of that effect time ago). Skewness, on the other hand, is important and is easily spotted with a histogram OF> Hi Christian OF> Saturday, April 28, 2007, 4:30:08 AM, You wrote: CH>> What it the null hypothesis of shapiro wilks test of univariate CH>> normality. That is if p < .05, does this indicate normality, or non- CH>> normality? OF> In general, the null hypotheses for any statistical test is "no OF> effect", "no differences". This means that for a normality test, the OF> null hypothesis is "No differences from a normal distribution". p<.05 OF> means NON-NORMALITY. OF> Anyway, remember that the p-value is not really informative. Normality OF> tests have low power if sample size is low (don't use them for sample OF> sizes below 10-12 cases), and are over sensitive for vey big samples OF> (if n is bigger than 100 then take a look at the histogram with a OF> normality curve plotted over it and decide if the variable looks OF> normal enough). OF> -- OF> Regards, OF> Dr. Marta García-Granero,PhD mailto:biostatistics@terra.es OF> Statistician OF> --- OF> "It is unwise to use a statistical procedure whose use one does OF> not understand. SPSS syntax guide cannot supply this knowledge, and it OF> is certainly no substitute for the basic understanding of statistics OF> and statistical thinking that is essential for the wise choice of OF> methods and the correct interpretation of their results". OF> (Adapted from WinPepi manual - I'm sure Joe Abrahmson will not mind) OF> NOTICE: This e-mail (and any attachments) may contain OF> PRIVILEGED OR CONFIDENTIAL information and is intended only for OF> the use of the specific individual(s) to whom it is addressed. It OF> may contain information that is privileged and confidential under OF> state and federal law. This information may be used or disclosed OF> only in accordance with law, and you may be subject to penalties OF> under law for improper use or further disclosure of the OF> information in this e-mail and its attachments. If you have OF> received this e-mail in error, please immediately notify the OF> person named above by reply e-mail, and then delete the original OF> e-mail. Thank you. -- Regards, Dr. Marta García-Granero,PhD mailto:biostatistics@terra.es Statistician --- "It is unwise to use a statistical procedure whose use one does not understand. SPSS syntax guide cannot supply this knowledge, and it is certainly no substitute for the basic understanding of statistics and statistical thinking that is essential for the wise choice of methods and the correct interpretation of their results". (Adapted from WinPepi manual - I'm sure Joe Abrahmson will not mind) ```
Back to: Top of message | Previous page | Main SPSSX-L page | 1,171 | 4,608 | {"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} | 2.796875 | 3 | CC-MAIN-2014-15 | latest | en | 0.859352 |
https://brilliant.org/problems/the-teacher-is-wrong/ | 1,531,759,616,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589404.50/warc/CC-MAIN-20180716154548-20180716174548-00117.warc.gz | 617,840,861 | 11,751 | # The teacher is wrong!
Algebra Level 3
John was given a math question.
Find a solution to $$a+bcd=b+acd=c+abd=d+abc=2$$
He solves it as follows:
Step 1: Rearrange into: $$bcd-acd=b-a$$
Step 2: Factorise into: $$cd(b-a)=b-a$$
Step 3: Divide: $$cd=1$$
Step 4: Apply to each and every equation: $$ab=ac=ad=bc=bd=cd=1$$
Step 5: Compare to get $$a=b=c=d=1$$ as one of the solution.
However, his teacher marked him wrong!
What went wrong?
× | 153 | 447 | {"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} | 3.546875 | 4 | CC-MAIN-2018-30 | latest | en | 0.936333 |
http://www.mp3car.com/off-topic/10046-electronics-question.html | 1,430,882,117,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1430457788926.49/warc/CC-MAIN-20150501052308-00025-ip-10-235-10-82.ec2.internal.warc.gz | 582,246,881 | 15,214 | 1. ## Electronics question
My girlfriend recently applied for a job and she was asked this question during the interview. We are both having some trouble with it.
If you have 2 bulbs in series, one being a 60 watt bulb, and the other being a 120 watt bulb. Which will glow brighter?
We decided the 120 watt bulb would be brighter, for the reason that P = i^2*R...then:
R1 = 60 watt bulb
R2 = 120 watt bulb
60 = i^2*R1
120 = i^2*R2
So from that you can find that R2 = 2*R1. And since we know the current is the same through both bulbs, the 60 watt bulb will have a voltage drop of V = i*R1; and the 120 watt bulb will have a voltage drop of V = i*2*R1. So wouldn't the 120 watt bulb glow brighter?
According to the interviewer, the 60 watt bulb would glow brighter. But she no longer remembers how he justified it.
Can anyone shed some light on this?
2. ## Re: Electronics question
Originally posted by SuperMatty
Can anyone shed some light on this?
AKKKKK
So i think you are backwards on the calculations for the resistance of the light bulbs... the lower power bulb shold have higher resistance.
and yeah....this question bugged the \$hit out of me for about 15 minutes, but then i found this ...
~mike
3. Hmm....yeah those examples are somewhat related to this. The first one is a bit more crude in its explanation, simply citing that the lower power bulb will have a higher resistance, but how can you show this when both are connected in series? I thought my approach was correct. The other example refers to when either are connected alone to a voltage source.
I mean, is that the only way to figure this one out? By first calculating the resistance of each bulb when connected alone in the circuit, and then using those values?
4. Okay, here we go.
The 60 watt bulb has twice the resistance of the 120 watt bulb
V = IR & P = IV
combinding the two equasions gives
P = (V*V)/R (Can't find the squared button)
Hence to double the power you halve the resistance, and vica-vera.
As the to blubs are in series they act as a potential divider.
As the 60 watt bulb has twice the resistance it'll have twice the voltage drop across it.
Hence for the 60 watt buld
P = (2/3 V * 2/3 V) / 2R
and for the 120 watt bulb
P = (1/3 V * 1/3 V) / R
Removing the constants
for the 60 watt buld
P = (2/3 * 2/3 ) / 2 = 2/9
and for the 120 watt bulb
P = (1/3 * 1/3) = 1/9
Which means that there's twice as much power going thru the 60 watt bulb as the 120 watt bulb, so it should be a fair bit brighter.
5. That's all good and well Rob...but is there any way to determine your first assumption, that the 60 watt bulb has 2x the resistance of the 120 watt bulb...what I'm saying is...is there any way to determine that without placing both in a circuit each by itself and calculating the resistance? Or is it just 'common knowledge' that a 60 watt bulb has 2x the resistance of an 120 watt bulb?
6. P = Power (Watts)
V = Voltage (Volts)
I = Current (Amps)
R = Resistance (Ohms)
V = IR & P = IV (Standard equasions)
combinding the two equasions gives
P = (V*V)/R (Can't find the squared button)
P is inversly proportional to R
Hence to double the power you halve the resistance, and vica-vera. So a 60 watt bulb will have double the resistance of a 120 watt bulb
7. aaahhhh...duh
8. i tried that, the smaller will shine brigher, since it has a thinner wire inside it = more resistance = more energy transformed into heat+light
9. And the faster it will burn out!!!
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• | 966 | 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} | 3.8125 | 4 | CC-MAIN-2015-18 | longest | en | 0.963115 |
https://www.redhotpawn.com/forum/posers-and-puzzles/mate-in-2.57974 | 1,560,760,655,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998462.80/warc/CC-MAIN-20190617083027-20190617105027-00268.warc.gz | 853,774,964 | 8,474 | # Mate in 2
BigDoggProblem
Posers and Puzzles 11 Dec '06 22:40
1. BigDoggProblem
Just...the Dogg
11 Dec '06 22:40
This is a problem I posted in the Chess Forum, but it got buried in a long thread.
Mate in 2
2. 11 Dec '06 23:57
Since it is obviously black to move. 0...c6 1.b4 Ka3 2.Ra1 #
3. BigDoggProblem
Just...the Dogg
12 Dec '06 03:27
It's funny how this puzzle is too hard for the chess forum but too easy for the puzzles forum.
4. 13 Dec '06 06:45
Originally posted by ilywrin
Since it is obviously black to move. 0...c6 1.b4 Ka3 2.Ra1 #
Hey buddy!......... so you do come around here to post too! Well, not that I'm haunting you with this puzzle addiction, but why is it 'obviously' black's move in this case? How did you figure that out?
5. 13 Dec '06 07:27
Originally posted by ckoh1965
Hey buddy!......... so you do come around here to post too! Well, not that I'm haunting you with this puzzle addiction, but why is it 'obviously' black's move in this case? How did you figure that out?
Because black couldn't have played the last move. His pawn hasn't gone anywhere, and all three spaces that the king could have directly come from would all have resulted in an illegal check. a3 and b3 are ruled out as it cannot have been put in check by an unmoved pawn, at a1 the rook couldn't have given a check by capturing on b1 due to the white king on c1. The king must have made it to a2 via b1, but in order for white to block him in, black needs to move something else. That was captured on either c1 or c5 last move.
6. 14 Dec '06 16:101 edit
but... what, say, if the board is uncommonly placed and white is playing 'down'?
black's legal last move would be pawn push and there's still mate in two, white to play.
(1.Rh8+ Kxh8, 2. g8(Q) # )
7. BigDoggProblem
Just...the Dogg
14 Dec '06 16:221 edit
Originally posted by pootstick
but... what, say, if the board is uncommonly placed and white is playing 'down'?
black's legal last move would be pawn push and there's still mate in two, white to play.
(1.Rh8+ Kxh8, 2. g8(Q) # )
1) 1...Kxh8?? is illegal - Pg7 guards the Rook.
2) You have to prove the board is reversed; you can't just assume it.
8. 14 Dec '06 16:371 edit
oh, course.. my mistake. that would be mate in one: Rh8 #
9. 14 Dec '06 16:41
2) i added the hypothetical rule 'white to play' in my post, that would necessitate the board to be reversed
10. BigDoggProblem
Just...the Dogg
14 Dec '06 17:34
Originally posted by pootstick
oh, course.. my mistake. that would be mate in one: Rh8 #
1...Kg6
11. BigDoggProblem
Just...the Dogg
14 Dec '06 17:37
Originally posted by pootstick
2) i added the hypothetical rule 'white to play' in my post, that would necessitate the board to be reversed
But you can't just arbitrarily set "White to play".
12. 14 Dec '06 17:441 edit
wrong again, eh, too puzzling for me. forget about it...i should be spending this time trying to win my games i guess. sorry i didn't mean to ruin your poser.
13. BigDoggProblem
Just...the Dogg
14 Dec '06 17:52
Originally posted by pootstick
sorry. i'm clearly taking this far too seriously.
Hey, if solvers could just change anything they wanted, there'd be no point in having a stipulation, right? Just give 'em a board with some pieces on it and let them find whatever they want to find.
I should do that next. A 'free-solving' exercise, for all you who like to whine about how pedantic and/or serious I am. There will be no criticism of anyone - we are all equals. Just sit around the fire, pull up a chessboard, and let it speak to you. All solutions that come from the heart are correct.
14. 14 Dec '06 17:54
truly, i didn't mean to offend.
15. 14 Dec '06 19:14
Nice and sardonic, BigDogg. I do think you are a bit pedantic, but correctly so. You point out our failures and work with them. Let's give you a medal. | 1,126 | 3,810 | {"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-2019-26 | latest | en | 0.962205 |
https://www.sarthaks.com/12363/what-do-you-mean-by-league-match-or-round-robin-tournament | 1,606,938,293,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141715252.96/warc/CC-MAIN-20201202175113-20201202205113-00152.warc.gz | 815,814,718 | 12,707 | What do you mean by league match or round robin tournament?
+1 vote
1.8k views
in General
What do you mean by league match or round robin tournament?
by (24.0k points)
selected
League: These are also called Round Robin Tournament. In league system all the teams are treated at par in league tournament. There are two types of league tournaments:
Single league tournament
Double league tournament
Single league tournament: In this type of tournament, every team shall play once with every other teams. The total number of matches in a single league tournament shall be [(n) (n – 1)]/2.
For example, if 8 teams are competing, the total number of matches to be [(8) (8 – 1)]/2 = 4(7) = 28.
Double league tournament: In this type of tournament, every team shall play twice with every other team. The total number of matches shall be n(n – 1).
For example, if 8 teams are competing the total number of matches to be n(n – 1) = 8(8 – 1) = 8(7) = 56 matches. | 247 | 959 | {"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-2020-50 | latest | en | 0.907654 |
https://mathandlanguage.edc.org/mathematics-tasks/geometric-thinking/comparing-area-tangrams/sentence-starters-and-frames | 1,701,801,507,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100555.27/warc/CC-MAIN-20231205172745-20231205202745-00777.warc.gz | 446,018,003 | 6,622 | # Comparing Area with Tangrams: Sentence Starters and Frames
Some ideas for sentence frames/starters that could be incorporated into your lesson are listed below. If you think a sentence frame/starter will be helpful, consider how it will support students’ mathematical learning and/or development of academic language, and decide which sentence frame/starter (from the list below or that you create) would best support students’ learning. You may find that the starters and frames vary in level of difficulty, and plan to provide them to students accordingly.
## Starters
To create four right angles in this rectangle, I _______________________________________________________________________
To form opposite sides that are congruent, I ________________________________________________________________________
## Frames
The ____________________ tangram piece has a bigger/smaller area than the _________________ tangram piece.
To create four right ____________ in this rectangle, I ________________________________________________________________
____________________________________________________________________________________________________________
************************************************************************************************************
Students should have opportunities to see, hear, and write key mathematical ideas during this activity. There are some specific terms that students need to understand in order to engage in this task, and there are some additional terms and phrases that may surface as the students engage with the task. You may think of additional words or phrases that are key to this activity. As the task is introduced, solved by the students, and discussed, ensure that students have opportunities to experience (i.e., through discussion, pictures, and the use of gestures) and to build understanding for key words and phrases. Examples of words and phrases that may be involved in work on this problem include:
• tangram pieces
• nonsquare rectangle
• right angle
• opposite sides
• congruent
• properties
• convincing mathematical explanation
• midpoint
• small/medium/large tangram triangle; tangram parallelogram; tangram small square; Tangram Square; tangram parallelogram
• determine
• area
************************************************************************************************************ | 395 | 2,373 | {"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} | 3.859375 | 4 | CC-MAIN-2023-50 | longest | en | 0.872613 |
http://www.mathematics21.org/view/addons/index-15.html | 1,566,661,123,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027321160.93/warc/CC-MAIN-20190824152236-20190824174236-00284.warc.gz | 274,113,832 | 7,747 | 6. OPERATIONS ON THE SET OF UNFIXED MORPHISMS
15
The above proposition allows to define:
Definition
2112
.
DOM
f
= DOM
F
and IM
f
= IM
F
for
F
f
.
6.5. Rectangular restriction.
Proposition
2113
.
ι
A,B
f
=
ι
A,B
g
if
f
g
.
Proof.
Let
f
g
. Then
g
=
ι
Src
g,
Dst
g
f
. So
ι
A,B
g
=
ι
A,B
ι
Src
g,
Dst
g
f
v
(proposition
2090
)
v
ι
A,B
f
. Similarly,
ι
A,B
f
v
ι
A,B
g
. So
ι
A,B
f
=
ι
A,B
g
.
The above proposition allows to define:
Definition
2114
.
ι
A,B
F
=
ι
A,B
f
for an unfixed morphism
F
and arbitrary
f
F
.
Definition
2115
.
F
A,B
= [
ι
A,B
F
] for every unfixed morphism
F
.
Proposition
2116
.
F
A,B
= id
B
F
id
A
for every unfixed morphism
F
and
objects
A
and
B
.
Proof.
Take
f
F
.
F
A,B
=
[
ι
A,B
F
]
=
[
ι
A,B
f
]
=
[
E
Dst
f,B
f
◦ E
A,
Src
f
] = [id
C
(Dst
f,B
)
B
u
Dst
f
f
id
C
(
A,
Src
f
)
A
u
Src
f
] = [id
C
(Dst
f,B
)
B
id
C
(Dst
f,
Dst
f
)
Dst
f
f
id
C
(Src
f,
Src
f
)
Src
f
id
C
(
A,
Src
f
)
A
] = [id
C
(Dst
f,B
)
B
f
id
C
(
A,
Src
f
)
A
] = [id
C
(Dst
f,B
)
B
]
[
f
]
[id
C
(
A,
Src
f
)
A
] = id
B
F
id
A
.
Proposition
2117
.
f
A
0
,B
0
A
1
,B
1
=
f
A
0
u
A
1
,A
1
u
B
1
.
Proof.
From the previous
f
A
0
,B
0
A
1
,B
1
= id
B
1
id
B
0
f
id
A
0
id
A
1
=
id
B
0
u
B
1
f
id
A
0
u
A
1
=
f
A
0
u
A
1
,A
1
u
B
1
.
Definition
2118
.
f
|
X
=
f
id
X
for every unfixed morphism
f
and
X
A
.
Obvious
2119
.
(
f
|
X
)
|
Y
=
f
X
u
Y
.
6.6. Algebraic properties of the lattice of unfixed morphisms.
The
following proposition allows to easily prove algebraic properties (cf. distributivity)
of the poset of unfixed morphisms:
Theorem
2120
.
The following are mutually inverse bijections:
1
. Let
A
and
B
be objects.
f
7→
[
f
] and
F
7→
ι
A,B
F
are mutually inverse
order isomorphisms between
n
f
unfixed morphisms
A
DOM
f,B
IM
f
o
and
C
(
A, B
). If
A
=
B
they are also semigroup isomorphisms.
2
. Let
T
be an unfixed morphism.
f
7→
[
f
] and
F
7→
ι
Src
t,
Dst
t
F
are mutually
inverse order isomorphisms between the lattice
DT
and
Dt
whenever
t
T
.
Proof.
We will prove that these functions are mutually inverse bijections.
That they are order-preserving is obvious.
1
.
ι
A,B
F
∈ C
(
A, B
) is obvious.
We need to prove that [
f
]
n
f
unfixed morphisms
A
DOM
f,B
IM
f
o
. For this it’s enough to prove
A
DOM[
f
]
B
IM[
f
] what is the same as
A
DOM
f
B
IM
f
what follows
from proposition
2071
.
Because
f
7→
[
f
] is an injection, it is enough
1
to prove that
ι
A,B
[
f
] =
f
. Really,
ι
A,B
[
f
] =
ι
A,B
f
=
f
.
That they are semigroup isomorphisms follows from the already proved formula
[
g
f
] = [
g
]
[
f
].
1
https://math.stackexchange.com/a/3007051/4876 | 1,432 | 3,078 | {"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} | 3.515625 | 4 | CC-MAIN-2019-35 | latest | en | 0.209197 |
https://goodtecher.com/leetcode-148-sort-list/ | 1,709,138,973,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474737.17/warc/CC-MAIN-20240228143955-20240228173955-00794.warc.gz | 287,397,184 | 11,560 | # LeetCode 148. Sort List
## Description
https://leetcode.com/problems/shuffle-the-array/
Given the `head` of a linked list, return the list after sorting it in ascending order.
Follow up: Can you sort the linked list in `O(n logn)` time and `O(1)` memory (i.e. constant space)?
Example 1:
```Input: head = [4,2,1,3]
Output: [1,2,3,4]
```
Example 2:
```Input: head = [-1,5,3,4,0]
Output: [-1,0,3,4,5]
```
Example 3:
```Input: head = []
Output: []
```
Constraints:
• The number of nodes in the list is in the range `[0, 5 * 104]`.
• `-105 <= Node.val <= 105`
## Explanation
Use merge sort and use a find middle helper function.
## Python Solution
``````# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def sortList(self, head: ListNode) -> ListNode:
while fast.next and fast.next.next:
slow = slow.next
fast = fast.next.next
return slow
def merge(self, l1, l2):
dummy = ListNode(0)
l3 = dummy
while l1 and l2:
if l1.val <= l2.val:
l3.next = l1
l1 = l1.next
else:
l3.next = l2
l2 = l2.next
l3 = l3.next
if l1:
l3.next = l1
if l2:
l3.next = l2
return dummy.next
right = self.merge_sort(middle.next)
middle.next = None
return self.merge(left, right)
``````
• Time complexity: O(N logN).
• Space complexity: O(log N). where N is the number of nodes in the linked list. Since the problem is recursive, we need additional space to store the recursive call stack. The maximum depth of the recursion tree is log N. | 482 | 1,550 | {"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 | 4 | CC-MAIN-2024-10 | latest | en | 0.645925 |
https://itfeature.com/tag/sampling-distributions/ | 1,723,381,043,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640997721.66/warc/CC-MAIN-20240811110531-20240811140531-00728.warc.gz | 252,748,362 | 33,828 | ## Important MCQs Sampling 1
MCQs sampling for the preparation of exams and different statistical job tests in Government/ Semi-Government or Private Organization sectors. These tests are also helpful in getting admission to different colleges and Universities. Let us start the MCQs Sampling Quiz. Most of the MCQs on this page cover Sampling and Sampling Distributions, Probability Sampling and Non-Probability Sampling, Mean and Standard Deviation of Sample, Sample size, Sampling error, Sample bias, Sample Selection, etc.
Online MCQs about Sample and Sampling Distribution for preparation of PPSC FPSC Lecturer posts.
1. In random sampling, the probability of selecting an item from the population is
2. The procedure in which a number of elements in a stratum is proportional to a number of elements in the population is classified as:
3. Sampling in which a sampling unit can be repeated more than once is called
4. Selection of the sample size $n$ to be the same for all the strata is known as:
5. Regardless of the difference in the distribution of the sample and population, the mean of sampling distribution must be equal to the:
6. The sample is a subset of
7. Non-Sampling error is reduced by
8. The difference between statistic and parameter is called
9. In sampling with replacement, a sampling unit can be selected
10. Any numerical value computed from the population is called
11. Consider a population of size 700 consisting of three strata such that $N_1=100, N_2=250$, and $N_3=350$. The required sample size is 18. What will be the sample size for stratum-III according to proportional allocation?
12. The list of all units in a population is called
13. Consider a population of size 700 consisting of three strata such that $N_1=100, N_2=250$, and $N_3=350$. The required sample size is 18. What will be the sample size for stratum-I according to proportional allocation?
14. Consider a population of size 700 consisting of three strata such that $N_1=100, N_2=250$, and $N_3=350$. The required sample size is 18. What will be the sample size for stratum-II according to proportional allocation?
15. The sample size of $h$th stratum by proportion allocation is
16. The standard deviation of the sampling distribution of any statistic is called
17. The number of strata should preferably be less than or equal to what value?
18. Any numerical value calculated from sample data is called
19. Non Probability form of sampling is
20. Sample allocation plan that provides the most precision, given a fixed sample size is
The sampling Quiz is about the Basics of Sampling and Sampling Distributions. It will help you to understand the basic concepts related to sampling methods and sampling distributions. This test will also help you to prepare yourself for different exams related to education or jobs.
### Online MCQs Sampling
• Non-sampling error is reduced by
• Any numerical value calculated from sample data is called
• The sample is a subset of
• Non-probability form of sampling is
• In sampling with replacement, a sampling unit can be selected
• Sampling in which a sampling unit can be repeated more than once is called
• The standard deviation of the sampling distribution of any statistic is called
• Any numerical value computed from the population is called
• The list of all units in a population is called
• The difference between statistic and parameter is called
• In random sampling, the probability of selecting an item from the population is
• Regardless of the difference in the distribution of the sample and population, the mean of sampling distribution must be equal to the:
• The number of strata should preferably be less than or equal to what value?
• The sample size of $h$th stratum by proportion allocation is
• The procedure in which a number of elements in a stratum is proportional to a number of elements in the population is classified as:
• Selection of the sample size $n$ to be the same for all the strata is known as:
• Consider a population of size 700 consisting of three strata such that $N_1=100, N_2=250$, and $N_3=350$. The required sample size is 18. What will be the sample size for stratum-I according to proportional allocation?
• Consider a population of size 700 consisting of three strata such that $N_1=100, N_2=250$, and $N_3=350$. The required sample size is 18. What will be the sample size for stratum-II according to proportional allocation?
• Consider a population of size 700 consisting of three strata such that $N_1=100, N_2=250$, and $N_3=350$. The required sample size is 18. What will be the sample size for stratum-III according to proportional allocation?
• Sample allocation plan that provides the most precision, given a fixed sample size is
R Programming Language | 1,066 | 4,778 | {"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} | 3.828125 | 4 | CC-MAIN-2024-33 | latest | en | 0.898052 |
https://www.jiskha.com/display.cgi?id=1512074628 | 1,532,070,347,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591543.63/warc/CC-MAIN-20180720061052-20180720081052-00619.warc.gz | 898,902,315 | 3,392 | # Science
posted by Kim
How many atoms are in 0.64 moles of strontium
1. bobpursley
2. Andy
how many atoms are in 0.64 moles of strontium
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More Similar Questions | 442 | 1,429 | {"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-2018-30 | latest | en | 0.908823 |
https://justaaa.com/statistics-and-probability/21628-part-1-use-the-following-steps-to-create-a | 1,713,048,376,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816853.44/warc/CC-MAIN-20240413211215-20240414001215-00519.warc.gz | 314,861,846 | 11,227 | Question
Part 1: Use the following steps to create a standard normal in StatKey and answer the...
Part 1: Use the following steps to create a standard normal in StatKey and answer the following questions about the variable Z.
Go to http://www.lock5stat.com/StatKey/
Click on Normal (This is located in the theoretical distributions section)
Notice in the right hand corner StatKey has a mean of 0 and standard deviation of 1 as default
Notice in the left hand corner are the usual left, right and 2 tail options these will be helpful in the problems below
For each of the following first fill in the plot for the area you are interested in. First find the area using StatKey. Once you have found the area using StatKey use the standard normal table to also find the area. For each give the area you find on StatKey and the area from the table. Do not round the area from the table.
1.The area above z= -1.5
2. The area below z=-0.14
3. The area between z=-2.32 and z=2.32
4. Find the endpoint which has 60% of the area below it.
5. Find the endpoint which has 60% of the area above it.
Part 2: For each of the following first fill in the plot for the area you are interested in. Then find the area using StatKey. You will need to change the normal in Statkey to match the mean and standard deviation. Once you have found the area using StatKey use the standard normal table to also find the area. You will need to standardize the value.
For each problem give the area you find on StatKey and the area from the table. Do not round the area from the table.
6. The area above -1.2 on a N(-2,1.5) distribution
7. The area below 0.7 on a N(1,4) distribution
8. The area between -2 and 2 on a N(3.4, 5) distribution
9. Find the endpoint on a N(100,10) distribution which has 74% of the area below it.
10. Find the endpoint on a N(100,10) distribution which has 15% of the area above it.
Part 3: For each of the following follow the steps to find the probability using the standard normal table. As a reminder the steps are listed for you.
How to solve problems involving normal distributions with a Standard Normal Table:
Write out the problem in terms of the original variable.
Draw a picture of the distribution and shade the area of interest under the curve.
Standardize the value (or values) of the original variable
Using the standard normal table find the area to the left of the standardized value. Use this value and the fact that the total area under the curve is equal to 1 to find the value you want.
Write your conclusion in the context of the problem.
11. The average number of acres used for growing tobacco in Kentucky is 75,300 acres per year, with a standard deviation of 5,000 acres. What is the probability that less than 82,315 acres will be used for growing tobacco?
12. The average number of acres used for growing tobacco in Kentucky is 75,300 acres per year, with a standard deviation of 5,000 acres. What is the probability that more than 71,295 acres will be used for growing tobacco
13. The average number of acres used for growing tobacco in Kentucky is 75,300 acres per year, with a standard deviation of 5,000 acres. What is the probability that between 67,250 and 76,250 acres will be used for growing tobacco?
Part 1
1.The area above z= -1.5
From the graph = 0.067
From the ztable = 0.0668
2. The area below z=-0.14
From the graph = 0.444
From the ztable = 0.4443
3. The area between z=-2.32 and z=2.32
Area between (-2.32<x<2.32)= Area below 2.32 - Area below -2.32
From graph = 0.987 - 0.013 = 0.974
From z table = 0.9898
- 0.0102 = 0.9796
4. Find the endpoint which has 60% of the area below it.
From graph = 0.253
from z table = 0.25
5. Find the endpoint which has 60% of the area above it.
From graph = -0.253
from z table = -0.25 | 997 | 3,806 | {"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} | 4.15625 | 4 | CC-MAIN-2024-18 | latest | en | 0.889892 |
https://www.erikkusch.com/courses/bayes-nets/part-2/ | 1,669,464,641,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446706291.88/warc/CC-MAIN-20221126112341-20221126142341-00734.warc.gz | 806,513,117 | 12,572 | # Gaussian Bayesian Networks
## Exercises
These are answers and solutions to the exercises at the end of Part 2 in Bayesian Networks with Examples in R by M. Scutari and J.-B. Denis. Much of my inspiration for these solutions, where necessary, by consulting the solutions provided by the authors themselves as in the appendix.
### R Environment
For today’s exercise, I load the following packages:
library(bnlearn)
library(ggplot2)
library(tidyr)
library(tidybayes)
### Scutari 2.1
Prove that Equation (2.2) implies Equation (2.3).
$$f(C | G = g) \neq f(C)$$
$$f(G | C = c) \neq f(G)$$
So how do we go about demonstrating that the first implies the latter? Well, we are using Bayesian theory here so why not use the Bayes' theorem? So let’s start by rewriting equation 2.2:
$$f(C | G) = \frac{f(C, G)}{f(G)} = \frac{f(G | C) f(C)}{f(G)}$$ So how does this relate to the question that equation 2.2 implies equation 2.3? Well, if $f(C|G) = f(C)$ then this equation would reveal that $f(G|C) = f(G)$ (so that the $f(G)$ terms factor out). Our proof stipulates that these statements aren’t true, but one still implies the other and we land of quod erat demonstrandum.
### Scutari 2.2
Within the context of the DAG shown in Figure 2.1, prove that Equation (2.5) is true using Equation (2.6).
This is the DAG in question:
The equation to prove (2.5) is:
$$f(N, W | V = v) = f(N | V = v) f(W | V = v)$$
and we use this equation (2.6) for our proof:
$$f(G, E, V, N, W, C) = f(G) f(E) f(V | G, E) f(N | V) f(W | V) f(C | N, W)$$
Let’s start the proof by integrating over all variables that aren’t $N$, $W$, and $V$ (the variables contained in the equation we are tasked to prove):
$$f(V, W, N) = \int_G \int_E \int_Cf(G,E,V,N,W,C)$$
We do this to remove all but the variables we are after from our equation so let’s follow this rationale:
\begin{aligned} \int_G \int_E \int_Cf(G,E,V,N,W,C) = &f(V) f(N|V) f(W|V) \newline &\times \left( \int_G \int_E f(G) f(E) f(V|G,E) \right) \newline &\left( \int_C f(C|N,W) \right) \end{aligned}
Simplifying this mess, we arrive at:
$$f(V, W, N) = f(V) f(N|V) f(W|V)$$
Finally, we can obtain our original formula:
$$f(W,N|V) = \frac{f(V,W,N)}{f(V)} = \frac{f(V) f(N|V) f(W|V)}{f(V)} = f(N|V) f(W|N)$$
Another case of the quod erat demonstrandums.
### Scutari 2.3
Compute the marginal variance of the two nodes with two parents from the local distributions proposed in Table 2.1. Why is it much more complicated for C than for V?
Table 2.1 is hardly a table at all, but I did locate it. Basically, it is an amalgamation of the probability distributions proposed for the DAG from the previous exercise:
Note that the parameter $07v$ in the second-to-last row should read $0.7v$.
The two nodes we are after are $V$ and $C$. Since the task already tells us that the computation of the marginal variance for $V$ is easier than for $C$, I start with this one.
1. Computation for $V$
Simply translating the probability distribution into a linear model, we receive:
$$V = -10.35534 + 0.5G + 0.70711E + \epsilon_V$$
with the variances of our independent variables $G$, $E$, and $\epsilon_V$ being $10^2$, $10^2$, and $5^2$ respectively. Consequently the variance of $V$ can be calculated as follows:
$$VAR(V) = 0.5^2VAR(G) + 0.70711^2VAR(E) + VAR(\epsilon_V)$$ $$VAR(V) = 0.5^210^2+0.70711^210^2+5^2 = 10$$
1. Computation for $C$
For $C$, we can transform our portability distribution into a linear model again:
$$C = 0.3N+0.7W+\epsilon_C$$
this time, however the predictors variables are not independent since they share node $V$ as their parent. Consequently, we have to compute their covariance:
$$COV(N,W) = COV(0.1V, 0.7V) = 0.1 * 0.7 * Var(V) = 0.1 * 0.7 * 10^2$$
So we actually needed to calculate the variance for $V$ to even be able to calculate the variance for $C$. Let’s round this out now, then:
\begin{aligned} Var(C) &= 0.3^2 * VAR(N) + 0.7^2VAR(W) \newline &+ VAR(\epsilon_C) + 2 * 0.3 * 0.7 * COV(N,W) \end{aligned}
Now, I simply plug the values into the formula and arrive at:
\begin{aligned} Var(C) &= 0.3^2 * 9.949874^2+0.7^2 * 7.141428 \newline &+6.25^2+2 * 0.3 * 0.7 * 0.1 * 0.7 * 10^2 \newline & = 54.4118 \end{aligned}
Curiously, the book suggest this as the solution:
$$Var(C) = (0.3^2+0.7^2+0.3 * 0.7 * 0.14)10^2+6.25^2 = 100.0024$$
I am not sure where the values for VAR(N) and VAR(W) have gone here. If anyone who is reading this knows the answer to it, please contact me and let me know as well.
### Scutari 2.4
Write an R script using only the rnorm and cbind functions to create a 100 × 6 matrix of 100 observations simulated from the BN defined in Table 2.1. Compare the result with those produced by a call to cpdist function.
To simulate a table of observation using the formulae in the probability distribution collection from the previous question (Table 1), we simply select random values for all parent nodes according to their distributions and let the distributions for all offspring nodes do the rest. One important note here, is that the rnorm() function in R takes as an argument of variation the standard deviation $\sigma$ rather than the variance $\sigma^2$:
set.seed(42) # making things reproducible
n <- 1e2 # number of replicates
G <- rnorm(n, 50, 10)
E <- rnorm(n, 50, 10)
V <- rnorm(n, -10.35534 + 0.5 * G + 0.70711 * E, 5)
N <- rnorm(n, 45 + 0.1 * V, 9.949874)
W <- rnorm(n, 15 + 0.7 * V, 7.141428)
C <- rnorm(n, 0.3 * N + 0.7 * W, 6.25)
sim1 <- data.frame(cbind(G, E, V, N, W, C))
Now we do this using the cpdist() function. To do so, we first have to create our Bayesian Network:
dag.bnlearn <- model2network("[G][E][V|G:E][N|V][W|V][C|N:W]")
disE <- list(coef = c("(Intercept)" = 50), sd = 10)
disG <- list(coef = c("(Intercept)" = 50), sd = 10)
disV <- list(coef = c("(Intercept)" = -10.35534, E = 0.70711, G = 0.5), sd = 5)
disN <- list(coef = c("(Intercept)" = 45, V = 0.1), sd = 9.949874)
disW <- list(coef = c("(Intercept)" = 15, V = 0.7), sd = 7.141428)
disC <- list(coef = c("(Intercept)" = 0, N = 0.3, W = 0.7), sd = 6.25)
dis.list <- list(E = disE, G = disG, V = disV, N = disN, W = disW, C = disC)
gbn.bnlearn <- custom.fit(dag.bnlearn, dist = dis.list)
sim2 <- data.frame(cpdist(gbn.bnlearn, nodes = nodes(gbn.bnlearn), evidence = TRUE))
this is pretty much exactly what is done in the chapter.
So let’s compare these simulation outputs:
# preparing all data together in one data frame for plotting
sim1$sim <- 1 sim2$sim <- 2
Plot_df <- rbind(sim1, sim2[, match(colnames(sim1), colnames(sim2))])
Plot_df <- gather(data = Plot_df, key = "node", value = "value", G:C)
Plot_df$sim <- as.factor(Plot_df$sim)
## plotting
ggplot(Plot_df, aes(x = value, y = sim)) +
stat_halfeye() +
facet_wrap(~node, scales = "free") +
theme_bw()
As is apparent from this, all results fall close to the expected values of roughly 50. There are noticeable differences between the simulations. I would suggest that these are due to the fairly low sample size for sim1.
### Scutari 2.5
Imagine two ways other than changing the size of the points (as in Section 2.7.2) to introduce a third variable in the plot.
The plot in question is this one:
this plot is aimed at showing the distribution of $C$ when both $E$ and $V$ vary. Here, the variation in $V$ is shown along the x-axis, while the variation of $E$ is contained within the sizes of the circles. The y-axis represents the values of $C$ according to its distributions.
So how else could we add information of $E$ to a plot of $V$ and $C$? I reckon we could:
• Make three scatter plots. One for each pairing of our variables.
• Represent the values of $E$ with a colour saturation gradient.
### Scutari 2.6
Can GBNs be extended to log-normal distributions? If so how, if not, why?
GBNs are Gaussian Bayesian Networks - Bayesian Networks where each node follows a Gaussian distribution.
Yes, absolutely they can! We can simply take the logarithm of all initial variables and apply the GBN right away. Of course, all values that shall be transformed using the logarithm have to be positive.
### Scutari 2.7
How can we generalise GBNs as defined in Section 2.3 in order to make each node’s variance depend on the node’s parents?
I see absolutely no problem here. Let’s say we have two nodes:
• $A$; parent node with a constant variance
• $B$; child node with a variance dependant the parent node
Then we can easily define the variance of $B$ given $A$ ($VAR(B|A)$) as follows:
$$VAR(B|A) = \left(A-E(A)\right)^2 * \sigma^2_B$$
### Scutari 2.8
From the first three lines of Table 2.1, prove that the joint distribution of E, G and V is trivariate normal.
This one is a doozy and I really needed to consult the solutions in the book for this one. Let’s first remind ourselves of the first lines of said table:
To approach this problem it is useful to point out that the logarithm of the density of a multivariate normal distribution is defined as such:
$$f(x) \propto -\frac{1}{2}(x-\mu)^T\sum^{-2}(x-\mu)$$
with $x$ being a random vector and $\mu$ denoting our expectation. $\sum$ identifies the covariance matrix.
Simplifying this, we can transform our variables $G$, $E$, and $V$ to give them a zero marginal expectation and a unity marginal variance. That is a very long-winded way of saying: we normalise our variables:
$$\overline G = \frac{G-E(G)}{\sqrt{VAR(G)}} = \frac{G-50}{10} \sim Normal(0, 1)$$ $$\overline E = \frac{E-E(E)}{\sqrt{VAR(E)}} = \frac{E-50}{10} \sim Normal(0, 1)$$ $$\overline V = \frac{V-E(V)}{\sqrt{VAR(V)}} = \frac{V-50}{10}$$ Solving for $\overline V | \overline G, \overline E$, we obtain:
$$\overline V | \overline G, \overline E = Normal\left(\frac{1}{2} \overline G + \sqrt{\frac{1}{2}} \overline E , (\frac{1}{2})^2 \right)$$
I have to honestly that I don’t quite understand how this happened and if anyone reading this has intuition for this solution, please let me know.
Now, we can compute the joint density distribution of these three normalised variables:
$$\begin{eqnarray} f(\overline G, \overline E, \overline V) &\propto& f(\overline G)+f(\overline E)+f(\overline V | \overline G, \overline E) \newline &=& -\frac{g^2}{2}-\frac{e^2}{2}-2 \left( v- \frac{1}{2}g - \sqrt{\frac{1}{2}}e \right)^2 \newline &=& -\begin{bmatrix} g \newline e \newline v\end{bmatrix}^T \begin{bmatrix} 1 & \frac{\sqrt{2}}{2} & -1\newline \frac{\sqrt{2}}{2} & \frac{3}{2} & -\sqrt{2} \newline -1 & -\sqrt{2} & 2 \end{bmatrix} \begin{bmatrix} g \newline e \newline v\end{bmatrix} \newline &=& -\frac{1}{2} \begin{bmatrix} g \newline e \newline v\end{bmatrix}^T \begin{bmatrix} 1 & 0 & \frac{1}{2}\newline 0 & 1 & \frac{1}{2} \newline \frac{1}{2} & \sqrt{\frac{1}{2}} & 1 \end{bmatrix} \begin{bmatrix} g \newline e \newline v\end{bmatrix} \newline \end{eqnarray}$$
I have to admit that most of this is, as of right now, beyond me as I came to this book for the “applications in R” in the first place. The book concludes that this results in:
$$VAR \left( \begin{bmatrix} \overline G \newline \overline E \newline \overline V\end{bmatrix} \right) = \begin{bmatrix} 1 & 0 & \frac{1}{2}\newline 0 & 1 & \frac{1}{2} \newline \frac{1}{2} & \sqrt{\frac{1}{2}} & 1 \end{bmatrix}$$
which results in our proof.
## Session Info
sessionInfo()
## R version 4.2.1 (2022-06-23 ucrt)
## Platform: x86_64-w64-mingw32/x64 (64-bit)
## Running under: Windows 10 x64 (build 19044)
##
## Matrix products: default
##
## locale:
## [1] LC_COLLATE=English_Germany.utf8 LC_CTYPE=English_Germany.utf8 LC_MONETARY=English_Germany.utf8 LC_NUMERIC=C LC_TIME=English_Germany.utf8
##
## attached base packages:
## [1] stats graphics grDevices utils datasets methods base
##
## other attached packages:
## [1] tidybayes_3.0.2 tidyr_1.2.0 ggplot2_3.3.6 bnlearn_4.8.1
##
## loaded via a namespace (and not attached):
## [1] styler_1.8.0 tidyselect_1.1.2 xfun_0.33 bslib_0.4.0 purrr_0.3.4 lattice_0.20-45 colorspace_2.0-3 vctrs_0.4.1 generics_0.1.3
## [10] htmltools_0.5.3 yaml_2.3.5 utf8_1.2.2 rlang_1.0.5 R.oo_1.25.0 jquerylib_0.1.4 pillar_1.8.1 glue_1.6.2 withr_2.5.0
## [19] DBI_1.1.3 R.utils_2.12.0 distributional_0.3.1 R.cache_0.16.0 lifecycle_1.0.2 stringr_1.4.1 posterior_1.3.1 munsell_0.5.0 blogdown_1.13
## [28] gtable_0.3.1 R.methodsS3_1.8.2 coda_0.19-4 evaluate_0.16 labeling_0.4.2 knitr_1.40 fastmap_1.1.0 parallel_4.2.1 fansi_1.0.3
## [37] highr_0.9 arrayhelpers_1.1-0 backports_1.4.1 checkmate_2.1.0 scales_1.2.1 cachem_1.0.6 jsonlite_1.8.0 abind_1.4-5 farver_2.1.1
## [46] tensorA_0.36.2 digest_0.6.29 svUnit_1.0.6 stringi_1.7.8 bookdown_0.29 dplyr_1.0.9 grid_4.2.1 ggdist_3.2.0 cli_3.3.0
## [55] tools_4.2.1 magrittr_2.0.3 sass_0.4.2 tibble_3.1.8 pkgconfig_2.0.3 ellipsis_0.3.2 assertthat_0.2.1 rmarkdown_2.16 rstudioapi_0.14
## [64] R6_2.5.1 compiler_4.2.1
Previous | 4,426 | 13,226 | {"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": 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": 1, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2022-49 | longest | en | 0.859289 |
http://www.algebra.com/algebra/homework/decimal-numbers.faq.question.135722.html | 1,386,196,795,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163037829/warc/CC-MAIN-20131204131717-00073-ip-10-33-133-15.ec2.internal.warc.gz | 212,448,904 | 4,333 | # SOLUTION: Explain why 2√2 is in simplest form and √8 is not in simplest form.
Algebra -> -> SOLUTION: Explain why 2√2 is in simplest form and √8 is not in simplest form. Log On
Ad: Mathway solves algebra homework problems with step-by-step help! Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations!
Click here to see ALL problems on decimal-numbers Question 135722: Explain why 2√2 is in simplest form and √8 is not in simplest form.Answer by carter6949(26) (Show Source): You can put this solution on YOUR website!The reason why 2√2 is in simplest form is the 2 in front of the sqr. sign is like putting an exponent on the 2 ofter the sqr. sign | 192 | 716 | {"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-2013-48 | latest | en | 0.860879 |
https://www.templateroller.com/template/2635884/trigonometric-and-hyperbolic-functions-cheat-sheet.html | 1,722,659,744,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640356078.1/warc/CC-MAIN-20240803025153-20240803055153-00414.warc.gz | 818,304,035 | 12,821 | # Trigonometric and Hyperbolic Functions Cheat Sheet
A Trigonometric and Hyperbolic Functions Cheat Sheet is a reference tool that provides a summary of key formulas and identities for trigonometric and hyperbolic functions. It is helpful for students and individuals studying or working in fields such as mathematics, physics, engineering, and computer science. With this cheat sheet, users can quickly access the formulas and properties of these functions, making it easier for them to solve problems involving angles, triangles, and exponential functions.
The Trigonometric and Hyperbolic Functions Cheat Sheet is typically created and filed by individual students or educators. It is not officially filed by any specific organization or entity.
## FAQ
Q: What is a trigonometric function?
A: A trigonometric function is a mathematical function that relates angles of a triangle to the lengths of its sides.
Q: What are some common trigonometric functions?
A: Common trigonometric functions include sine, cosine, and tangent.
Q: What is a hyperbolic function?
A: A hyperbolic function is a mathematical function that relates exponential functions to the calculation of angles and the lengths of curves.
Q: What are some common hyperbolic functions?
A: Common hyperbolic functions include sinh, cosh, and tanh.
Q: Are trigonometric and hyperbolic functions related?
A: Yes, trigonometric and hyperbolic functions are mathematically related and share many properties.
Q: What is the purpose of a cheat sheet for trigonometric and hyperbolic functions?
A: The purpose of a cheat sheet is to provide a quick and easy reference for the formulas and properties of trigonometric and hyperbolic functions. | 352 | 1,709 | {"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-2024-33 | latest | en | 0.904656 |
http://primes.utm.edu/curios/page.php?number_id=671 | 1,529,803,824,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267865995.86/warc/CC-MAIN-20180624005242-20180624025242-00126.warc.gz | 249,499,344 | 2,349 | 5062 (another Prime Pages' Curiosity)
Curios: Curios Search: Participate: 506239841 {+0, +2, +6, +8} and 5062213991 {+0, +2, +6, +8} are prime. Note the first prime in each 4-tuplet share the first 4 digits (5062). If you add 4 to 506239841 and divide by 15 it equals 33749323, then 33749323 {-2, +0, +4, +6} along with 33749323 x 15 {-4, -2, +2, +4} are all prime. [Luhn] Prime Curios! © 2000-2018 (all rights reserved) privacy statement | 170 | 442 | {"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-2018-26 | latest | en | 0.611188 |
https://www.esaral.com/q/three-letters-are-dictated-to-three-persons-and-an-envelope-is-addressed-to-each-of-them-90424 | 1,713,993,774,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296819971.86/warc/CC-MAIN-20240424205851-20240424235851-00700.warc.gz | 689,624,702 | 11,451 | # Three letters are dictated to three persons and an envelope is addressed to each of them,
Question:
Three letters are dictated to three persons and an envelope is addressed to each of them, the letters are inserted into the envelopes at random so that each envelope contains exactly one letter. Find the probability that at least one letter is in its proper envelope.
Solution:
Let $L_{1}, L_{2}, L_{3}$ be three letters and $E_{1}, E_{2}$, and $E_{3}$ be their corresponding envelops respectively.
There are 6 ways of inserting 3 letters in 3 envelops. These are as follows:
There are 4 ways in which at least one letter is inserted in a proper envelope.
Thus, the required probability is $\frac{4}{6}=\frac{2}{3}$. | 173 | 725 | {"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} | 4.1875 | 4 | CC-MAIN-2024-18 | latest | en | 0.958158 |
http://www.perlmonks.org/?node_id=959099 | 1,519,406,037,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814801.45/warc/CC-MAIN-20180223154626-20180223174626-00667.warc.gz | 536,852,233 | 7,994 | Syntactic Confectionery Delight PerlMonks
### Re^2: Hex String XOR
by moritz (Cardinal)
on Mar 12, 2012 at 09:52 UTC ( #959099=note: print w/replies, xml ) Need Help??
in reply to Re: Hex String XOR
Note that this solution too is limited by the size of the biggest integer you can store in a perl scalar, which is usally 52 bit from the mantissa of a double precision floating point, or 64 bit for an integer on 64bit platforms (not sure about the sign, so could be 63 usable bits for this purpose).
Replies are listed 'Best First'.
Re^3: Hex String XOR
by tobyink (Abbot) on Mar 12, 2012 at 11:30 UTC
Here's a solution that works on hex strings, so allows you to work with hex numbers larger than the maximum integers Perl can handle...
```use 5.010;
use strict;
use Carp qw/croak/;
use List::Util qw/max/;
sub xor_strings
{
croak "should be passed two arguments" unless @_==2;
state \$chunk_size = 4;
state \$pattern = sprintf '%%0%dx', \$chunk_size;
# Make strings equal length, and a multiple of \$chunk_size.
my \$length = max(map { length \$_ } @_);
\$length += \$chunk_size - (\$length % \$chunk_size);
my @strings = map { ('0'x(\$length - length \$_)) . \$_ } @_;
# Join results of each chunk
return join q{},
map {
# Parse chunk hex to an integer
my \$i = \$_;
my @nums = map { hex substr \$_, \$i*\$chunk_size, \$chunk_siz
+e } @strings;
# Xor them and convert to hex.
sprintf \$pattern, \$nums[0] ^ \$nums[1]
}
0 .. (\$length/\$chunk_size)-1;
}
say xor_strings(
'112233112233112233112233112233112233112233112233112233',
'aabbccaabbccaabbccaabbccaabbccaabbccaabbccaabbcc112233',
);
The chunk size of 4 is fairly conservative. It means that the string in processed in four-digit (i.e. 16 bit) chunks. You can probably get a minor speed up using a larger chunk size if you know that your computer will be able to handle it. The returned value will be left-padded with zeroes to be a length that is a multiple of the chunk size.
perl -E'sub Monkey::do{say\$_,for@_,do{(\$monkey=[caller(0)]->[3])=~s{::}{ }and\$monkey}}"Monkey say"->Monkey::do'
tobyink,
Reading the Camel book (3rd and 4th editions), "...if both operands are strings...the operators do bitwise operations between corresponding bits from the two strings. In this case there's no arbitrary limit..."
So the following code should work (untested) and be faster than calling a subroutine and looping:
```
my \$result = '112233112233112233112233112233112233112233112233112233'
+^
'aabbccaabbccaabbccaabbccaabbccaabbccaabbccaabbcc112233';
I used this technique to generate a 8-byte CRC for arbitrary text strings by using 'substr' to take 8-byte substrings. I needed the looping for that.
Regards...Ed
"Well done is better than well said." - Benjamin Franklin
Yes, ^ does work on strings, but as the OP mentioned, it works on the string's bytes, not the hexadecimal numbers encoded in the string.
```\$ perl -E"say '112233'^'aabbcc'"
PPPPPP
\$ perl -E"say sprintf '%x', 0x112233 ^ 0xaabbcc"
bb99ff
My code gives the latter behaviour (which the OP wants) on hexadecimal strings of arbitrary length.
perl -E'sub Monkey::do{say\$_,for@_,do{(\$monkey=[caller(0)]->[3])=~s{::}{ }and\$monkey}}"Monkey say"->Monkey::do'
Re^3: Hex String XOR
by ikegami (Pope) on Mar 12, 2012 at 18:48 UTC
"^" converts floats to machine integers, so you're limited to 32/64, not 5253/64.
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Notices? | 1,159 | 4,112 | {"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-2018-09 | longest | en | 0.687666 |
https://theshakerofsalt.com/boil/your-question-how-long-does-it-take-to-cook-a-turkey-at-325-degrees.html | 1,638,825,788,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363312.79/warc/CC-MAIN-20211206194128-20211206224128-00625.warc.gz | 636,027,198 | 18,617 | # Your question: How long does it take to cook a turkey at 325 degrees?
Contents
## How long do you cook a turkey per pound at 325 degrees?
For an unstuffed turkey (fresh or defrosted), the general rule is about 15 minutes per pound at a 325°F regular oven according to the USDA. If you’re using a convection oven, your turkey will cook slightly faster.
## Is it better to cook a turkey at 325 or 350?
Roast the turkey uncovered at a temperature ranging from 325°F to 350°F. Higher temperatures may cause the meat to dry out, but this is preferable to temperatures that are too low which may not allow the interior of the turkey to cook to a safe temperature.
## How long do I cook a 10 pound turkey at 325 degrees?
Roast in a 325°F oven for 2-1/4 hours. Remove foil; cut band of skin or string between drumsticks so thighs cook evenly. Continue roasting for 30 to 45 minutes more (1 to 1-1/4 hours if stuffed), or until the thermometer registers 180°F; and the center of the stuffing (if using) is 165°F.
## Do you cook a 13 lb turkey at 325 or 350?
For a 10–13-lb. turkey (weight with giblets): Bake in a 350° oven for 1 1/2–2 1/4 hr. For a 14–23-lb. turkey (weight with giblets): Bake in a 325° oven for 2–3 hr.
## How long does it take to cook a 12-pound turkey at 325 degrees?
For one 8- to 12-pound turkey, roast at 325°F for 2¾ to 3 hours. For one 12- to 14-pound turkey, roast at 325°F for 3 to 3¾ hours. For one 14- to 18-pound turkey, roast at 325°F for 3¾ to 4¼ hours. For one 18- to 20-pound turkey, roast at 325°F for 4¼ to 4½ hours.
## How long does it take to cook a 20lb turkey at 350 degrees?
The simplest way to figure out turkey roasting times is to calculate 13 minutes per pound at 350°F for an unstuffed turkey (that’s about 3 hours for a 12- to 14-lb.
If your turkey weighs 18 to 20 pounds, roast it at:
1. 425°F for 3½ to 3¾ hours.
2. 400°F for 3¾ to 4 hours.
3. 350°F for 4 to 4¼ hours.
4. 325°F for 4¼ to 4½ hours.
## Should you wash your turkey?
Wash Hands and Surfaces; not the Turkey
According to the USDA Food Safety and Inspection Service, washing raw poultry, beef, pork, lamb, or veal before cooking it is not recommended. Bacteria in raw meat and poultry juices can be spread to other foods, utensils, and surfaces.
## Do you cook a 10 lb turkey at 325 or 350?
For a 10-13 lb. turkey (weight with giblets): Bake in a 350° oven for 1 1/2-2 1/4 hr. For a 14-23 lb. turkey (weight with giblets): Bake in a 325° oven for 2-3 hr.
IT IS SURPRISING: Best answer: How long do you grill pizza on a stone?
## How long do I cook a 12 lb turkey at 300 degrees?
Rub oil over all sides of turkey. and place turkey in uncovered roasting pan in oven. Bake for 1 hours at 300 degrees to kill bacteria.
## Is turkey done at 165 or 180?
While some recipes state that turkey should be cooked to 180 degrees Fahrenheit, the meat is safe to consume once it reaches the 165-degree mark. Cooking the breasts past 165 can result in dry meat, but the dark meat can be cooked to 180. | 876 | 3,016 | {"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} | 2.765625 | 3 | CC-MAIN-2021-49 | latest | en | 0.890893 |
http://metamath.tirix.org/mpests/suplem2pr.html | 1,713,062,448,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816863.40/warc/CC-MAIN-20240414002233-20240414032233-00181.warc.gz | 22,968,945 | 3,902 | # Metamath Proof Explorer
## Theorem suplem2pr
Description: The union of a set of positive reals (if a positive real) is its supremum (the least upper bound). Part of Proposition 9-3.3 of Gleason p. 122. (Contributed by NM, 19-May-1996) (Revised by Mario Carneiro, 12-Jun-2013) (New usage is discouraged.)
Ref Expression
Assertion suplem2pr ${⊢}{A}\subseteq 𝑷\to \left(\left({y}\in {A}\to ¬\bigcup {A}{<}_{𝑷}{y}\right)\wedge \left({y}{<}_{𝑷}\bigcup {A}\to \exists {z}\in {A}\phantom{\rule{.4em}{0ex}}{y}{<}_{𝑷}{z}\right)\right)$
### Proof
Step Hyp Ref Expression
1 ltrelpr ${⊢}{<}_{𝑷}\subseteq 𝑷×𝑷$
2 1 brel ${⊢}{y}{<}_{𝑷}\bigcup {A}\to \left({y}\in 𝑷\wedge \bigcup {A}\in 𝑷\right)$
3 2 simpld ${⊢}{y}{<}_{𝑷}\bigcup {A}\to {y}\in 𝑷$
4 ralnex ${⊢}\forall {z}\in {A}\phantom{\rule{.4em}{0ex}}¬{y}{<}_{𝑷}{z}↔¬\exists {z}\in {A}\phantom{\rule{.4em}{0ex}}{y}{<}_{𝑷}{z}$
5 ssel2 ${⊢}\left({A}\subseteq 𝑷\wedge {z}\in {A}\right)\to {z}\in 𝑷$
6 ltsopr ${⊢}{<}_{𝑷}\mathrm{Or}𝑷$
7 sotric ${⊢}\left({<}_{𝑷}\mathrm{Or}𝑷\wedge \left({y}\in 𝑷\wedge {z}\in 𝑷\right)\right)\to \left({y}{<}_{𝑷}{z}↔¬\left({y}={z}\vee {z}{<}_{𝑷}{y}\right)\right)$
8 6 7 mpan ${⊢}\left({y}\in 𝑷\wedge {z}\in 𝑷\right)\to \left({y}{<}_{𝑷}{z}↔¬\left({y}={z}\vee {z}{<}_{𝑷}{y}\right)\right)$
9 8 con2bid ${⊢}\left({y}\in 𝑷\wedge {z}\in 𝑷\right)\to \left(\left({y}={z}\vee {z}{<}_{𝑷}{y}\right)↔¬{y}{<}_{𝑷}{z}\right)$
10 9 ancoms ${⊢}\left({z}\in 𝑷\wedge {y}\in 𝑷\right)\to \left(\left({y}={z}\vee {z}{<}_{𝑷}{y}\right)↔¬{y}{<}_{𝑷}{z}\right)$
11 ltprord ${⊢}\left({z}\in 𝑷\wedge {y}\in 𝑷\right)\to \left({z}{<}_{𝑷}{y}↔{z}\subset {y}\right)$
12 11 orbi2d ${⊢}\left({z}\in 𝑷\wedge {y}\in 𝑷\right)\to \left(\left({y}={z}\vee {z}{<}_{𝑷}{y}\right)↔\left({y}={z}\vee {z}\subset {y}\right)\right)$
13 sspss ${⊢}{z}\subseteq {y}↔\left({z}\subset {y}\vee {z}={y}\right)$
14 equcom ${⊢}{z}={y}↔{y}={z}$
15 14 orbi2i ${⊢}\left({z}\subset {y}\vee {z}={y}\right)↔\left({z}\subset {y}\vee {y}={z}\right)$
16 orcom ${⊢}\left({z}\subset {y}\vee {y}={z}\right)↔\left({y}={z}\vee {z}\subset {y}\right)$
17 13 15 16 3bitri ${⊢}{z}\subseteq {y}↔\left({y}={z}\vee {z}\subset {y}\right)$
18 12 17 syl6bbr ${⊢}\left({z}\in 𝑷\wedge {y}\in 𝑷\right)\to \left(\left({y}={z}\vee {z}{<}_{𝑷}{y}\right)↔{z}\subseteq {y}\right)$
19 10 18 bitr3d ${⊢}\left({z}\in 𝑷\wedge {y}\in 𝑷\right)\to \left(¬{y}{<}_{𝑷}{z}↔{z}\subseteq {y}\right)$
20 5 19 sylan ${⊢}\left(\left({A}\subseteq 𝑷\wedge {z}\in {A}\right)\wedge {y}\in 𝑷\right)\to \left(¬{y}{<}_{𝑷}{z}↔{z}\subseteq {y}\right)$
21 20 an32s ${⊢}\left(\left({A}\subseteq 𝑷\wedge {y}\in 𝑷\right)\wedge {z}\in {A}\right)\to \left(¬{y}{<}_{𝑷}{z}↔{z}\subseteq {y}\right)$
22 21 ralbidva ${⊢}\left({A}\subseteq 𝑷\wedge {y}\in 𝑷\right)\to \left(\forall {z}\in {A}\phantom{\rule{.4em}{0ex}}¬{y}{<}_{𝑷}{z}↔\forall {z}\in {A}\phantom{\rule{.4em}{0ex}}{z}\subseteq {y}\right)$
23 4 22 syl5bbr ${⊢}\left({A}\subseteq 𝑷\wedge {y}\in 𝑷\right)\to \left(¬\exists {z}\in {A}\phantom{\rule{.4em}{0ex}}{y}{<}_{𝑷}{z}↔\forall {z}\in {A}\phantom{\rule{.4em}{0ex}}{z}\subseteq {y}\right)$
24 unissb ${⊢}\bigcup {A}\subseteq {y}↔\forall {z}\in {A}\phantom{\rule{.4em}{0ex}}{z}\subseteq {y}$
25 23 24 syl6bbr ${⊢}\left({A}\subseteq 𝑷\wedge {y}\in 𝑷\right)\to \left(¬\exists {z}\in {A}\phantom{\rule{.4em}{0ex}}{y}{<}_{𝑷}{z}↔\bigcup {A}\subseteq {y}\right)$
26 ssnpss ${⊢}\bigcup {A}\subseteq {y}\to ¬{y}\subset \bigcup {A}$
27 ltprord ${⊢}\left({y}\in 𝑷\wedge \bigcup {A}\in 𝑷\right)\to \left({y}{<}_{𝑷}\bigcup {A}↔{y}\subset \bigcup {A}\right)$
28 27 biimpd ${⊢}\left({y}\in 𝑷\wedge \bigcup {A}\in 𝑷\right)\to \left({y}{<}_{𝑷}\bigcup {A}\to {y}\subset \bigcup {A}\right)$
29 2 28 mpcom ${⊢}{y}{<}_{𝑷}\bigcup {A}\to {y}\subset \bigcup {A}$
30 26 29 nsyl ${⊢}\bigcup {A}\subseteq {y}\to ¬{y}{<}_{𝑷}\bigcup {A}$
31 25 30 syl6bi ${⊢}\left({A}\subseteq 𝑷\wedge {y}\in 𝑷\right)\to \left(¬\exists {z}\in {A}\phantom{\rule{.4em}{0ex}}{y}{<}_{𝑷}{z}\to ¬{y}{<}_{𝑷}\bigcup {A}\right)$
32 31 con4d ${⊢}\left({A}\subseteq 𝑷\wedge {y}\in 𝑷\right)\to \left({y}{<}_{𝑷}\bigcup {A}\to \exists {z}\in {A}\phantom{\rule{.4em}{0ex}}{y}{<}_{𝑷}{z}\right)$
33 32 ex ${⊢}{A}\subseteq 𝑷\to \left({y}\in 𝑷\to \left({y}{<}_{𝑷}\bigcup {A}\to \exists {z}\in {A}\phantom{\rule{.4em}{0ex}}{y}{<}_{𝑷}{z}\right)\right)$
34 3 33 syl5 ${⊢}{A}\subseteq 𝑷\to \left({y}{<}_{𝑷}\bigcup {A}\to \left({y}{<}_{𝑷}\bigcup {A}\to \exists {z}\in {A}\phantom{\rule{.4em}{0ex}}{y}{<}_{𝑷}{z}\right)\right)$
35 34 pm2.43d ${⊢}{A}\subseteq 𝑷\to \left({y}{<}_{𝑷}\bigcup {A}\to \exists {z}\in {A}\phantom{\rule{.4em}{0ex}}{y}{<}_{𝑷}{z}\right)$
36 elssuni ${⊢}{y}\in {A}\to {y}\subseteq \bigcup {A}$
37 ssnpss ${⊢}{y}\subseteq \bigcup {A}\to ¬\bigcup {A}\subset {y}$
38 36 37 syl ${⊢}{y}\in {A}\to ¬\bigcup {A}\subset {y}$
39 1 brel ${⊢}\bigcup {A}{<}_{𝑷}{y}\to \left(\bigcup {A}\in 𝑷\wedge {y}\in 𝑷\right)$
40 ltprord ${⊢}\left(\bigcup {A}\in 𝑷\wedge {y}\in 𝑷\right)\to \left(\bigcup {A}{<}_{𝑷}{y}↔\bigcup {A}\subset {y}\right)$
41 40 biimpd ${⊢}\left(\bigcup {A}\in 𝑷\wedge {y}\in 𝑷\right)\to \left(\bigcup {A}{<}_{𝑷}{y}\to \bigcup {A}\subset {y}\right)$
42 39 41 mpcom ${⊢}\bigcup {A}{<}_{𝑷}{y}\to \bigcup {A}\subset {y}$
43 38 42 nsyl ${⊢}{y}\in {A}\to ¬\bigcup {A}{<}_{𝑷}{y}$
44 35 43 jctil ${⊢}{A}\subseteq 𝑷\to \left(\left({y}\in {A}\to ¬\bigcup {A}{<}_{𝑷}{y}\right)\wedge \left({y}{<}_{𝑷}\bigcup {A}\to \exists {z}\in {A}\phantom{\rule{.4em}{0ex}}{y}{<}_{𝑷}{z}\right)\right)$ | 2,806 | 5,379 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 45, "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-2024-18 | latest | en | 0.349077 |
https://quant.stackexchange.com/questions/36637/how-to-adjust-geometric-brownian-motion-to-be-monotone | 1,713,918,150,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818835.29/warc/CC-MAIN-20240423223805-20240424013805-00706.warc.gz | 426,880,934 | 38,303 | # How to adjust Geometric Brownian Motion to be monotone?
I want to use stochastic process to model subscriber's mobile data consumption as time going in a month. So I think about Geometric Brownian Motion.
However, people's cumulative data consumption will never decrease. Thus, how can I adjust the formulation of Geometric Brownian Motion to make it monotone? Or is there any other stochastic process more suitable?
• Instead of modelling your total consumption to follow a GBM, you could e.g. model the instantaneous rate of data usage as a GBM $X_t$ then your total consumption is $Y_t = \int_0^t X_u \mathrm{d}u$. However, you should question if the a GBM is a reasonable model for the instantaneous rate of consumption - e.g. should it have a drift but no intra-day/week/month seasonality, ...? Oct 29, 2017 at 19:15
• Thanks a lot! Could you please explain how to compute Y_t? There is a Wiener process within X_t, how to deal with it in the time intergral? Oct 30, 2017 at 13:31 | 246 | 990 | {"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} | 2.59375 | 3 | CC-MAIN-2024-18 | latest | en | 0.935226 |
https://dwbi1.wordpress.com/2022/02/16/forecasting-time-series-using-statistics-vs-machine-learning/ | 1,669,976,398,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710900.9/warc/CC-MAIN-20221202082526-20221202112526-00428.warc.gz | 260,201,977 | 26,278 | Data Warehousing and Data Science
16 February 2022
Forecasting time series: using statistics vs machine learning
Filed under: Data Science,Machine Learning — Vincent Rainardi @ 6:59 am
This article outlines how ARIMA and LSTM are used for forecasting time series, and which one is better.
A lot of references are available at the end of this article for those who would like to find out further.
Introduction
In ML, we use regression, to predict the values of a variable (y) based on the values of other variables (x1, x2, x3, …). For example, we predict the stock price of a company, based on its financial ratios, fundamentals and ESG factors.
In time series forecasting, we predict the values of a variable in the future, based on the values of that variable in the past. For example, we predict the stock price of a company, based on the past prices.
A time series is a sequence of numbers, each collected at a fixed period of time.
How do we forecast a time series? There are 2 ways: a) using statistics, b) using machine learning. In this article I’ll give a brief explanation of both. But before that let’s clear out one thing first: is “time series” plural or singular?
Time Series: plural or singular?
A time series is a sequence of numbers like this 1, 2, 3, 4, 5, … This is one time series, not one time serie.
We can have two time series like this: 1, 2, 3, 4, 5, … and 6, 7, 8, 9, 10, … These are two time series, not two time serieses.
So the singular form is “series” and the plural form is also “series”, not “serieses”. The word “series” is both singular and plural. See Merriam-Webster dictionary explanation in Ref #1 below.
Forecasting a time series means to find out what the next numbers in one series (1, 2, 3, 4, 5, …)
Forecasting two time series means to find out what the next numbers in two series (1, 2, 3, 4, 5, … and 6, 7, 8, 9, 10, …)
Forecasting time series using statistics
We can use regression to forecast a time series. We can also use Moving Average to forecast a time series.
Auto-Regressive model (AR)
Using regression, we use the past values of the forecast variable as the input variables. Which is why this method is called Auto-Regressive model. It is called auto because the input variables are the forecast variable itself, but the past values of it.
where yt-1, yt-2, yt-3are the past values of y, and c, c1, c2, c3 are constants.
ϵt = white noise. It is a sequence of random numbers, with the average of zero and the standard deviation is the same over time.
Moving Average model (MA)
Using Moving Average model the forecast variable is the mean of the series plus the error terms.
where ϵt = yt – yt-1 (white noise error term), μ is the mean and a1, a2, a3 are constants.
It is called moving average because we start with the average (mean), then keep moving/shifting the average by a factor of epsilon (the error term).
I need to emphasise here that the Moving Average model is not the Moving Average analysis that we use for the stock price, where we simply calculate the average of stock prices in the last 20 days.
ARMA model
ARMA model is the combination of the Auto-Regressive model and Moving Average. That is why it is called ARMA, the AR bit means Auto-Regressive, whereas the MA bit means Moving Average. So we forecast using the previous values of the forecast variable (Auto-Regressive model), and using the mean plus the error terms (Moving Average model).
ARIMA has 2 parameters i.e. ARMA(p,q)
where p = order of the autoregressive and q = order of the moving average.
Whereas AR and MA has 1 parameter i.e. AR(p) and MA(q).
ARIMA model
The ARIMA model is ARMA model plus differencing. Differencing means creating a new series by taking difference between the value at t and at (t-1).
For example, from this series: 0, 1, 3, 2, 3, 3, … (call it y)
We can make a new series by taking the difference between the numbers: 1, 2, -1, 1, 0, … (call it y’)
We can take the difference again (called second order differencing): 1, -3, 2, -1, … (call it y’’)
The I in ARIMA stands for Integrated. Integrated here means Differencing.
So the difference between the ARMA model and the ARIMA is: in ARMA we use y, whereas in ARIMA we use y’ or y’’.
In the ARIMA model use AR model and MA model on y’ or y’’, like this:
ARIMA has 3 parameters i.e. ARIMA(p,d,q)
where p = order of the autoregressive, d = degree of the first order differencing, and q = order of the moving average.
SARIMAX model
The S here means Seasonal and the X here means Exogenous.
Seasonal means that it has a repeating pattern from season to season. For example, the series on top line below consists of the trend part, the seasonal part and the random part. The seasonal part has a repeating pattern. Source: Ref #5.
The SARIMAX model include the seasonal part as well as the non-seasonal part.
SARIMAX has 7 parameters i.e. SARIMAX(p,d,q)x(P,D,Q,s)
Where p, d, q are as defined above, and P, D, Q are the seasonal terms of the p, d, q parameters, and s is the number seasons per year, e.g. for monthly s = 12, for quarterly s = 4.
In timer series, a exogenous variable means parallel time series which is used as a weighted input to the model (Ref #6)
Exogenous variable is one of the parameter in SARIMAX. In Python (statsmodels library), the parameters for SARIMAX are:
SARIMAX (y, X, order=(p, d, q), seasonal_order=(P, D, Q, s))
where y is the time series, X is the Exogenous variable/factor, and the others are as described before.
Forecasting time series using machine learning
The area of machine learning which deals with temporal sequence. is called Recurrent Neural Network (RNN). Temporal sequence means anything which has time element (a series of things happening one after the other), such as speech, handwriting, images, video. And that includes time series of course.
RNN is an neural network which has an internal memory. Which is it able to recognise patterns in time series. There are many RNN models, such as Elman network, Jordan network, Hopfield network, LSTM, GRU.
The most widely used method for predicting a time series is LSTM. An LSTM cell has 3 gates: an input gate, an output gate and a forget gate:
The horizontal line at the top (from ct-1 to ct) is the cell state. It is the memory of the cell. Along this line, there are 3 things happening: the cell state is multiplied by the “forget gate”, increased/reduced by the “input gate” and finally the value is taken to the “output gate”.
• The forget gate removes unwanted information from the cell state (c), based on the previous input (ht-1) and the current input (xt).
• The input gate adds new information to the cell state. The current input (xt) and the previous output (ht-1) pass through a σ and a tanh, multiplied then added to the cell memory line.
• The output gate calculates the output from the cell state (c), the previous input (ht-1) and the current input (xt).
Architecturally, there are different ways we can use to forecast time series using LSTM: (Ref #7)
• Fully Connected LSTM: a neural network with several layers of LSTM units with each layer fully connected to the next layer.
• Bidirectional LSTM: the LSTM model learns the time series in backward direction in addition to the forward direction.
• CNN LSTM: the time series is processed by a CNN first (1 dimensional), then processed by LSTM.
• ConvLSTM: the convolutional structure is in inside the LSTM cell (in both the input-to-state and state-to-state transitions), see Ref #13 and #16.
• Encoder-Decoder LSTM: for forecasting several time steps. The Encoder maps the time series into a fixed length vector, and decoder maps this vector back to a variable-length output sequence.
Which one is better, ARIMA or LSTM?
Well that is a million dollar question! Some research suggests that LSTM is better (Ref #17, #20, #24), some suggests that ARIMA is better (Ref #19) and some says that XGB is better than LSTM and ARIMA (#23). So it depends on the cases, but generally speaking LSTM is better in terms of accuracy (RMSE, MAPE).
It is an interesting topic for research. Plus other approaches such as Facebook’s Prophet, GRU, GAN and their combinations (Ref #25, #26, #27). It is possible to get better accuracy by combining the above approaches. I’m still searching a topic for my MSc dissertation, and it looks that this could be the one!
References:
1. Merriam-Webster dictionary explanation on “series” plurality: link
2. Forecasting: Principles and Practice, by Rob J. Hyndman and George Athanasopoulos: link
4. ARIMA model on Statsmodel: link
5. Penn State Eberly College of Science: link
7. How to Develop LSTM Model for Time Series Forecasting by Jason Brownlee: link
8. Time Series Prediction with LSTM RNN in Python with Keras: link
9. Time Series Forecasting: Predicting Stock Prices Using An ARIMA Model by Serafeim Loukas: link
10. Time Series Forecasting: Predicting Stock Prices Using An LSTM Model by Serafeim Loukas: link
12. RNN and LSTM by Vincent Rainardi: link
13. Convolutional LSTM Network: A Machine Learning Approach for Precipitation Nowcasting, by Xingjian Shi, Zhourong Chen, Hao Wang, Dit-Yan Yeung, Wai-kin Wong, Wang-chun Woo: link
14. Exploiting the ConvLSTM: Human Action Recognition using Raw Depth Video-Based RNN, by Adrian Sanchez-Caballero, David Fuentes-Jimenez, Cristina Losada-Guti´errez: link
15. Convolutional LSTM for spatial forecasting, by Sigrid Keydana: link
16. Very Deep Convolutional Networks for End-to-End Speech Recognition, by Yu Zhang, William Chan, Navdeep Jaitly: link
17. A Comparison of ARIMA and LSTM in Forecasting Time Series, by Sima Siami-Namini, Neda Tavakoli, Akbar Siami Namin: link
18. ARIMA vs Prophet vs LSTM for Time Series Prediction, by Konstantin Kutzkov: link
19. A Comparative Analysis of the ARIMA and LSTM Predictive Models and Their Effectiveness for Predicting Wind Speed, by Meftah Elsaraiti, Adel Merabet: link
20. Weather Forecasting Using Merged LSTM and ARIMA Model, by Afan Galih Salman, Yaya Heryadi, Edi Abdurahman, Wayan Suparta: link
21. Comparing ARIMA Model and LSTM RNN Model in Time-Series Forecasting, by Vaibhav Kumar: link
22. A Comparison between ARIMA, LSTM, and GRU for Time Series Forecasting, by Peter Yamak, Li Yujian, Pius Kwao Gadosey: link
23. Machine Learning Outperforms Classical Forecasting on Horticultural Sales Predictions by Florian Haselbeck, Jennifer Killinger, Klaus Menrad, Thomas Hannus, Dominik G. Grimm: link
24. Forecasting Covid-19 Transmission with ARIMA and LSTM Techniques in Morocco by Mohamed Amine Rguibi, Najem Moussa, Abdellah Madani, Abdessadak Aaroud, Khalid Zine-dine: link
25. Time Series Forecasting papers on Research Gate: link
26. Stock Price Forecasting by a Deep Convolutional Generative Adversarial Network by Alessio Staffini: link
27. A novel approach based on combining deep learning models with statistical methods for COVID-19 time series forecasting by Hossein Abbasimehr, Reza Paki, Aram Bahrini: link | 2,781 | 10,989 | {"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-2022-49 | latest | en | 0.893766 |
http://www.jiskha.com/display.cgi?id=1257617002 | 1,495,758,487,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608617.80/warc/CC-MAIN-20170525233846-20170526013846-00345.warc.gz | 536,147,161 | 3,893 | # Math (calc)
posted by on .
Given the function: f(x)=x^3(x-2)^2 , on the interval [-1,3] find the domain and Asymptotes
...Isn't the domain (-1<x<3) ? and as for the asymptotes I thought you could only find them for rational functions and since that function is not rational I guess there aren't any...???
• Math (calc) - ,
I agree with you, although I don't know whay you say it isn't a rational function. The domain is [-1,3]. It makes no sense to ask for asymptotes in a limited interval. The function never approaches straight-line behavior.
• Math (calc) - ,
yes that's what I thought. Thank you! | 161 | 609 | {"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-2017-22 | latest | en | 0.948216 |
https://mathematica.stackexchange.com/questions/259184/expression-evaluation-for-compile-too-time-consuming | 1,718,403,654,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861575.66/warc/CC-MAIN-20240614204739-20240614234739-00847.warc.gz | 359,187,311 | 41,211 | # expression evaluation for Compile[] too time consuming
I need to use Compile[] for a very large expression to be integrated. But for Compile[] the expression has to be evaluated, and this takes endless time.
The expression is composed of several functions having lists/tensors as arguments and return types. Something like this:
aFun[..., bFun[..., cFun[...], ...], ...]
Of course, that gets very unhandy quickly.
In a usual programming language I would write: cTensor=cFun[...]; bTensor=bFun[..., cTensor, ...]; aTensor=aFun[..., bTensor, ...];
If I could do that in Mathematica that would avoid replacing all the occurrences of the arguments in the functions and that actually should work since its going to be compiled. Unfortunately everything I tried doesn't work, like using a variable like above, using assignments like {{cT11,cT12,...},{cT21,...}}=cFun[...], also compiling the functions cFun[] etc..
How can I achieve this in Mathematica? Thank you very much in advance
Regards
edit: A very simplified example would be as follows, but the operations are much more complicated in my calculation, many parameters are filled in and there are more sums.
(f, F and g are vectors with three components)
f:=EulerMatrix[{\[Alpha]2,\[Beta]2,\[Gamma]2}].{0,0,1}Sum[Norm[EulerMatrix[{\[Alpha]2,\[Beta]2,\[Gamma]2}].{b,b,b}-EulerMatrix[{\[Alpha]1,\[Beta]1,\[Gamma]1}].{a,a,a}](EulerMatrix[{\[Alpha]2,\[Beta]2,\[Gamma]2}].{b,b,b}).(EulerMatrix[{\[Alpha]1,\[Beta]1,\[Gamma]1}].{a,a,a}),{a,20},{b,20}]
g[F_]:=EulerMatrix[{\[Alpha]1,\[Beta]1,\[Gamma]1}].{0,0,1}Sum[Norm[F-EulerMatrix[{\[Alpha]2,\[Beta]2,\[Gamma]2}].{c,c,c}]F.EulerMatrix[{\[Alpha]1,\[Beta]1,\[Gamma]1}].{c,c,c},{c,20},{d,20}]
fun=Compile[{\[Alpha]1,\[Beta]1,\[Gamma]1,\[Alpha]2,\[Beta]2,\[Gamma]2},Evaluate[g[f]]]
Unfortunately in Compile[] I have to evaluate g[f]. If I do this on my computer like this it takes about 42 seconds:
Timing[g[f];]
{41.875, Null}
When increasing the sums from 20 terms to something higher time consumption grows extremely, since f is used in g. For f and g[F] alone that's much faster:
Timing[f;]
{2.64063, Null}
Timing[g[{x, y, z}];]
{1.35938, Null}
The kernel of my problem should be that if Compile[] would first calculate f like {x,y,z}=f so that x, y and z are real values and then use these real values as parameters for g like g[{x,y,z}] it should be much faster. Of course, there would be no symbolic simplification be done what Evaluate[g[f]] would do, but that would be acceptable. I would like to write
fun=Compile[{...},
{x,y,z}=f;
g[{x,y,z}]
]
With x, y and z are always calculated as numeric values. But Evaluate[] has to be used in Compile[] and that leads in my tries always to evaluation of g[f].
• Could you please post a minimal characteristic example of your problem, which can easily be copied and pasted into a Mathematica notebook. Commented Dec 4, 2021 at 16:02
• (1) I don't see why you cannot write in Mma what you would write in a usual programming language. (2) Have you seen CompilationOptions -> {"ExpressionOptimization" -> False} or CompilationOptions -> {"InlineCompiledFunctions" -> False}? Commented Dec 4, 2021 at 18:16
• @MichaelWeyrauch: I made a little example and did put it in my message above. What I am looking for is a two step calculation instead of a symbolic evaluation of the whole calculation, but Compile[] doesn't seem to allow that. Commented Dec 5, 2021 at 11:07
• @MichaelE2: I can write it like I would do in a usual programming language, but Mathematica automatically tries to evaluate/simplify and for that my functions are too big. In my example above f is not used much in g, but in my real calculation it's used in many places and I am working with big matrices and many terms. I saw these options to Compile[], but I don't see how these could help. Commented Dec 5, 2021 at 11:12
I compiled the OP's code twice, taking 78 sec. and 320 sec. It ran in $$7\times10^{-5}$$ sec/call), faster than the codes below. However, the compile times below are much, much shorter.
There are slightly faster run times if we use Real instead of Integer types (in Sum for example).
Manual optimization (shorter compile time, longer run time $$3\times10^{-4}$$ sec/call):
ClearAll[f, g];
f[e1_, e2_, n_] := e2 . N@{0, 0, 1}*
Sum[Norm[
e2 . {b, b, b} -
e1 . {a, a, a}] (e2 . {b, b, b}) . (e1 . {a, a, a}), {a, 1.,
n}, {b, 1., n}]
g[F_, e1_, e2_, n_] := e1 . N@{0, 0, 1}*
Sum[Norm[F - e2 . {c, c, c}] F . e1 . {c, c, c},
{c, 1., n}, {d, 1., n}]
evaluateHeldFuncsRule[ff__] := (* evaluates functions inside Hold *)
func : (Blank[#] & /@ Alternatives[ff]) :>
With[{res = func}, res /; True];
code = Hold[{α1, β1, γ1, α2, β2, γ2},
Block[{e1, e2, F, n = 20.},
e1 = EulerMatrix[{α1, β1, γ1}];
e2 = EulerMatrix[{α2, β2, γ2}];
F = f[e1, e2, n]; (* using a symbolic n means Sum won't *)
g[F, e1, e2, n] (* evaluate before compilation *)
]
] /. evaluateHeldFuncsRule[EulerMatrix, f, g];
fun = Compile @@ code; // AbsoluteTiming
(* {0.003046, Null} *)
Unrolled loops (longer compile time, shorter run time $$1\times10^{-4}$$ sec/call):
ClearAll[f, g];
f[e1_, e2_, n_] := e2 . N@{0, 0, 1}*
Sum[Norm[
e2 . {b, b, b} -
e1 . {a, a, a}] (e2 . {b, b, b}) . (e1 . {a, a, a}), {a, 1.,
n}, {b, 1., n}]
g[F_, e1_, e2_, n_] := e1 . N@{0, 0, 1}*
Sum[Norm[F - e2 . {c, c, c}] F . e1 . {c, c, c},
{c, 1., n}, {d, 1., n}]
evaluateHeldFuncsRule[ff__] :=
func : (Blank[#] & /@ Alternatives[ff]) :>
With[{res = func}, res /; True];
code = Hold[{α1, β1, γ1, α2, β2, γ2},
Block[{e1, e2, F},
e1 = EulerMatrix[{α1, β1, γ1}];
e2 = EulerMatrix[{α2, β2, γ2}];
F = f[e1, e2, 20]; (* Using an explicit n=20 means Sum will be *)
g[F, e1, e2, 20] (* evaluated before compilation *)
]
] /. evaluateHeldFuncsRule[EulerMatrix, f, g];
fun20 = Compile @@ code; // AbsoluteTiming
(* {0.019748, Null} *)
The "usual" way? (much longer run times, though, $$3 \times 10^{-2}$$ sec/call):
(fC = Compile @@
With[{e1 = EulerMatrix[{α1, β1, γ1}],
e2 = EulerMatrix[{α2, β2, γ2}]},
Hold[{α1, β1, γ1, α2, β2, γ2},
e2 . N@{0, 0, 1} Sum[
Norm[
e2 . {b, b, b} -
e1 . {a, a, a}] (e2 . {b, b, b}) . (e1 . {a, a, a}), {a,
1., 20.}, {b, 1., 20.}]
]
];
gC = Compile @@
With[{e1 = EulerMatrix[{α1, β1, γ1}],
e2 = EulerMatrix[{α2, β2, γ2}]},
Hold[{{α1, _Real}, {β1, _Real}, {γ1, _Real},
{α2, _Real}, {β2, _Real}, {γ2, _Real},
{F, _Real, 1}},
e1 . {0, 0, 1} Sum[
Norm[F - e2 . {c, c, c}] F . e1 . {c, c, c}, {c, 1., 20.}, {d,
1., 20.}]
]
];
fun1 =
Compile[{α1, β1, γ1, α2, β2, γ2},
gC[α1, β1, γ1, α2, β2, γ2,
fC[α1, β1, γ1, α2, β2, γ2]],
CompilationOptions -> {
"InlineExternalDefinitions" -> True
(*,"InlineCompiledFunctions"->True*)}
];) // AbsoluteTiming
(* {0.01999, Null} *)
• That helped and now it works. Thank you very much! Commented Dec 5, 2021 at 20:02 | 2,363 | 6,784 | {"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": 4, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-26 | latest | en | 0.785755 |
http://economicmemos.com/2017/10/13/a-note-on-the-difference-between-geometric-and-arithmetic-averages/ | 1,591,178,614,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347432521.57/warc/CC-MAIN-20200603081823-20200603111823-00587.warc.gz | 36,486,179 | 9,742 | # A note on the difference between geometric and arithmetic averages
Question: The table below has price data and daily return data for Vanguard fund VB. Calculate the arithmetic and geometric averages of the daily return data. Show that the geometric average accurately reflects the relationship between the initial and final stock price and the arithmetic average does not accurately explain this relationship.
Daily Price and Returns For Vanguard Fund VB Date Adjusted Close Daily Return 7/1/16 115.480674 7/5/16 113.99773 0.987158509 7/6/16 114.744179 1.006547929 7/7/16 114.913373 1.001474532 7/8/16 117.202487 1.019920345 7/11/16 118.128084 1.007897418 7/12/16 119.451781 1.011205608 7/13/16 119.10344 0.997083836 7/14/16 119.262686 1.001337039 7/15/16 119.402023 1.00116832 7/18/16 119.63093 1.001917112 7/19/16 119.202965 0.996422622 7/20/16 119.959369 1.006345513 7/21/16 119.481646 0.996017627 7/22/16 120.297763 1.00683048 7/25/16 120.019083 0.997683415 7/26/16 120.616248 1.004975584 7/27/16 120.347522 0.997772058 7/28/16 120.536625 1.001571308 7/29/16 120.894921 1.002972507 8/1/16 120.735675 0.998682773 8/2/16 119.12335 0.986645828
Analysis: The table below presents calculation of the two averages and the count of return days. The product of the initial value of the ETF, the pertinent average and the count of return days is the estimate of the final value. Estimates of final ETF value are calculated for both the arithmetic average and the geometric average and these estimates are compared to the actual value of the stock on the final day in the period.
Understanding The Difference Between Arithmetic Mean and Geometric Mean Returns Statistic Value Note Arithmetic Average of Daily Stock Change Ratio 1.001506208 Average function Geometric Average of Daily Stock Change Ratio 1.001479966 Geomean function Count of Return Days 21 Count Function Estimate of final value based on arithmetic average 119.1889153 Initial Value x Arithmetic Return Average x Count Days Estimate of final value based on geometric average 119.12335 Initial Value X Geometric Return Average x Count Days Ending Value 119.12335 Copy from data table
There is another way to show that the daily return should be modeled with the geometric mean rather than arithmetic mean. The average daily return of the stock is (FV/IV)(1/n) – 1 where FV is final value and IV is initial value and n is the number of market days in the period, which for this problem is 21.
Using this formula we find the daily average holding period return is 0.001479966. Note that 1 minus the geometric mean of the daily stock price ratio is also 0.001479966.
The geometric mean gives us the correct holding period return. | 792 | 2,710 | {"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-2020-24 | longest | en | 0.683547 |
https://wikipedikia.org/where-should-downlights-be-placed-in-a-bedroom/ | 1,627,902,263,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154320.56/warc/CC-MAIN-20210802110046-20210802140046-00487.warc.gz | 624,698,397 | 60,941 | Downlights basically light the floor under them in a cone typically between 30 and 55 degrees. For bedrooms you need more diffused light, which is spread more evenly so that it washes the walls and ceiling.
Also, Where should you place downlights?
Recessed downlights are generally positioned 1.5 to 2 ft. away from walls with a space of 3 to 4 feet between each light. Dividing the ceiling height by two is a way of gauging how much space to leave between each downlight. Thus, if your ceiling is 8 foot high, place your lights 4 feet apart.
In this way, What size LED downlights do I need? In most cases, LED downlights between 9W to 15W will be adequate for most standard homes between 2.4m to 2.7m high.
## How many can lights for a 12×12 room?
The typical rule-of-thumb is 24″ from each wall, and then 3-5 feet between cans. So in a 12′ room you’d probably want two rows of lights. Considering it is a bedroom and you don’t need it as bright as a kitchen, you can probably get away with 2 rows of 4 lights for a total of 8.
## How many downlights do you need per room?
For example, you’re looking at putting downlight in your bedroom which is 4 x 5 m (20m2), you should have 180 (lux) x 20 (m2) = 3,600 Lumens in your space which is around 4 x downlights.
## What size LED downlights do I need?
In most cases, LED downlights between 9W to 15W will be adequate for most standard homes between 2.4m to 2.7m high.
## What is the difference between downlights and spotlights?
They are frequently used in kitchens but are also used in living rooms, bedrooms and bathrooms. Unlike spotlights, downlights are recessed mounted into the ceiling. LED spotlights are often used as a longer lasting, energy efficient and reliable light source, particularly in the case of bathroom spotlights.
## How many lumens should I get?
To determine the needed lumens, you will need to multiply your room square footage by your room foot-candle requirement. For example, a 100 square foot living room, which needs 10-20 foot-candles, will need 1,000-2,000 lumens. A 100 square foot dining room, which needs 30-40 foot-candles, will need 3,000-4,000 lumens.
## Where should downlights be placed in a bathroom?
Above the bath or shower to a height of 2.25m from the floor. Lights must be rated at least IP45 (typically, IP65 is used). An area extending 60cm beyond the bath, shower and wash basin, up to 2.25m above the floor. Lights must be rated at least IP44.
## What are LED downlights?
What are LED downlights and how can I install them? What are downlights? Downlights are recessed luminaires that are placed in the ceiling. They have standard sized and cut out sizes and are a replacer for your current CFL and PL-C downlight fixtures. There are also downlights which have the possibility to be dimmable.
## How do you plan LED downlights?
Generally when installing LED downlights, it is recommended to install lights 1m from the walls and space the lights between 1.2m and 1.5m apart. Positioning can alter depending on the amount of light required and the design of the room, which should be taken into consideration when planning lighting positioning.
## What color light is best for living room?
The also add where these temperatures are best used in your home: Soft white/warm white (2700 Kelvin): Best for bedrooms and living rooms; providing a traditional warm, cozy feel to them. Bright white/cool white (4100 Kelvin): Best in kitchens, bathrooms or garages; giving rooms a whiter, more energetic feel.
## How do you calculate lighting?
Lighting Calculation – Summary
Calculate the amount of lumens you need. Multiply the area in square meters by the lux, or the area in square feet by the footcandles. work out how many bulbs you need. Divide the number of lumens by the number of lumens delivered by each bulb.
## How many LED lights can I run on one circuit?
Each CFL or LED bulb typically gives the same amount of light as a 60-watt incandescent bulb while drawing 10 watts or less, which is equivalent to a current draw of 1/12 amp. Thus a 15-amp circuit can safely control 180 or more fixtures that use CFL or LED bulbs.
## How many can lights for a 12×12 room?
The typical rule-of-thumb is 24″ from each wall, and then 3-5 feet between cans. So in a 12′ room you’d probably want two rows of lights. Considering it is a bedroom and you don’t need it as bright as a kitchen, you can probably get away with 2 rows of 4 lights for a total of 8.
## Where should downlights be placed in a bedroom?
Downlights basically light the floor under them in a cone typically between 30 and 55 degrees. For bedrooms you need more diffused light, which is spread more evenly so that it washes the walls and ceiling.
## How many LED recessed lights per room?
Electricians charge between \$65 and \$85 an hour, and light fixtures will generally run around \$93-\$240 installed. How much does it cost to put in a ceiling fan?
## How many LED recessed lights per room?
In the master or guest bathrooms, use fixtures that provide at least 75 to 100 watts of illumination, says Randall Whitehead, a well-known lighting expert and author of Residential Lighting, a Practical Guide. You can also get these light equivalents in a 24-watt fluorescent or a 20-watt LED, he adds.
## How bright is 4500 lumens?
It’s so easy to install, you’ll be lighting up your task space in no time at all. It’s efficient too, using just 65 watts to generate 4500 lumens of bright, natural light at 5000 Kelvin. That’s the equivalent light output of ten 45 watt bulbs!
## What are LED downlights?
What are LED downlights and how can I install them? What are downlights? Downlights are recessed luminaires that are placed in the ceiling. They have standard sized and cut out sizes and are a replacer for your current CFL and PL-C downlight fixtures. There are also downlights which have the possibility to be dimmable.
## How many downlights can you have on one switch?
One rule of thumb is to install one LED downlight per square metre, leaving a space of one metre between each light and half a metre between the lights and each wall. This would mean that in a room measuring 3m x 3m, nine downlights would be ideal for a bright, even light distribution.
## How many LED lights do I need in a room?
For the average living room of 250 square feet, you’ll need 5,000 lumens as your primary light source (20 lumens x 250 square feet), equivalent to about five 100 watt incandescent light bulbs, five 23 watt CFLs, or eight 10 watt LED light bulbs.
## Is recessed lighting outdated?
For the average space of 250 square feet, you’ll need roughly 5,000 lumens as your primary light source (20 lumens x 250 square feet). In your dining room, you’ll want about 30 lumens per square foot on your dining table (to see, not examine, food), so if your table is 6 x 3 feet, that’s 540 lumens.
## How bright is 1000 lumens?
One lumen is about the same brightness as a one birthday candle from a distance of one foot from you. To put it another way, a standard 60 watt light globe produces around 750-850 lumens of light. 1,000 lumens flashlight is able to produce a beam range of 150 to 200 meters.
## How many lights should be in a bedroom?
If all of this seems complicated, a good rule of thumb is to have 4 lighting sources in a room to produce adequate lighting. Adequate task lighting is the key to a functional office space.
## How many lights should be in a bedroom?
Make a Plan
Whatever new recessed lighting you add should work with your existing fixtures. Measure your room to see how many fixtures you will need. A common rule of thumb is that you use one recessed light for every 4 to 6 square feet of ceiling space. Doing so provides even, overall illumination.
## How bright is 400 lumens?
Lumens and Useful Lumens
Old Watts Approx Lumens
40 W 440 – 460 lamp
50 W 330 – 400 spotlight 350-450 Useful Lumens (spotlight)
60 W 800 – 850 lamp
75W 1000-1100 lamp
## How do you calculate downlight spacing?
A quick way to determine for how far apart to space your recessed downlights is to take the ceiling height and divide by two. For example, an 8′-0” ceiling would have lights spaced every 4′-0” on-center, and a 10′-0” ceiling would have lights spaced every 5′-0” on center.
## How do you design a lighting layout?
To determine how far apart to space your recessed lights, divide the height of the ceiling by two. If a room has an 8 foot ceiling, you should space your recessed lights approximately 4 feet apart. If the ceiling is 10 feet, you’ll want to put about 5 feet of space in between each fixture.
## How much does it cost to install LED downlights?
Cost of Downlight Installation
Install down light (swap from existing down light) \$35-\$50 per downlight 5-20 mins per light
Install new down lights \$50-\$70 per downlight 2-3 hours per room of down lights
## How many lights do I need per square foot?
A quick way to determine for how far apart to space your recessed downlights is to take the ceiling height and divide by two. For example, an 8′-0” ceiling would have lights spaced every 4′-0” on-center, and a 10′-0” ceiling would have lights spaced every 5′-0” on center.
## How many lights do I need per square foot?
For the average space of 250 square feet, you’ll need roughly 5,000 lumens as your primary light source (20 lumens x 250 square feet). In your dining room, you’ll want about 30 lumens per square foot on your dining table (to see, not examine, food), so if your table is 6 x 3 feet, that’s 540 lumens.
## How do you calculate spacing between light fixtures?
For the average space of 250 square feet, you’ll need roughly 5,000 lumens as your primary light source (20 lumens x 250 square feet). In your dining room, you’ll want about 30 lumens per square foot on your dining table (to see, not examine, food), so if your table is 6 x 3 feet, that’s 540 lumens. | 2,447 | 9,910 | {"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} | 2.71875 | 3 | CC-MAIN-2021-31 | latest | en | 0.963565 |
http://continuouswave.com/ubb/Forum6/HTML/002393.html | 1,606,688,744,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141203418.47/warc/CC-MAIN-20201129214615-20201130004615-00711.warc.gz | 24,450,980 | 4,095 | ContinuousWave Whaler Moderated Discussion Areas ContinuousWave: Small Boat Electrical Electric Starting Motors
Author Topic: Electric Starting Motors jimh posted 07-02-2009 08:57 AM ET (US) The typical electric starter motor on an outboard engine is a series-wound DC motor. The mechanical power output of the motor is directly proportional to the current flow in its windings, which in turn is proportional to the applied voltage. (See http://en.wikipedia.org/wiki/Brushed_DC_electric_motor .) The more voltage applied to the motor, the more mechanical power it can create to start an outboard motor.The voltage applied to a starter motor in an outboard engine is supplied by the engine starting battery or cranking battery. The terminal voltage at the battery is proportional to its state of charge and inversely proportional to the current flow. The higher the state of charge, the higher the battery voltage. The higher the current flow, the lower the battery voltage.The starter motor is connected to the battery by electrical conductors, a solenoid relay, and perhaps some distribution switches and buses. The total resistance of these circuit elements will be in series with the motor resistance. Whenever any current flows through this series resistance, a voltage drop occurs. The voltage drop is proportional to both current flow and resistance. The higher the current flow, the greater the voltage drop; the higher the series resistance, the greater the voltage drop.The current in the starter motor is inversely proportional to its rotational speed. This means that when the motor is initially started and is turning slowly, the current demand will be the greatest.We all have likely experienced the situation of an electric starting motor being unable to crank over the engine it is supposed to start. The ignition key is turned to start, the solenoid engages, but the engine does not crank. This is due to the voltage at the starter motor becoming too low to generate enough mechanical power to crank the engine. The electric motor is stalled.To overcome this stalled situation. more voltage must be supplied to the starting motor. There are only two circuit elements which can affect this: the battery itself, and the voltage drop across the series resistance of the other circuit elements. Increasing the battery voltage requires increasing the state of charge. Since in a small boat the only source of charging current for the battery is the outboard engine, there is no possibility for charging the battery until the engine is started. We cannot effectively change the battery voltage in this situation, unless we change to a new battery or connect another battery in parallel.The voltage drop across the series resistance of other conductors in the circuit can be controlled. By using large conductors with very low resistance, the voltage drop can be minimized. Because the current flow at initial starting of the electric motor is very high, perhaps as much as 500-amperes, the voltage drop in the conductors can become significant. For example, if the conductors are made from 4-AWG wire and are 30-feet in length (total), the resistance in the conductors will be4-AWG wire = 0.25-ohm/1,000-feet x 30-feet = 0.0075 ohmVoltage drop = 0.0075 x 500Voltage drop = 3.75This means that 3.75-volt of the battery terminal voltage is lost across the conductors. If the battery were fully charged and had a terminal voltage of 12.9-volts, only 9.15 volts would be applied to the starter motor.If the wire size in our example is increased to 2-AWG, the resistance becomes2-AWG = 0.15-ohm/1,000-feet x 30-feet = 0.0045and the voltage drop reduces toVoltage drop = 0.0045 x 500Voltage drop = 2.25The starter motor will be supplied with 1.5-volts more than previously. This additional voltage may mean the difference between getting the engine to crank over or not. Also note that the 1.5-volt difference is as great as the change in battery terminal voltage from full charge (12.9-volts) to almost no charge (11.4-volts).
Hop to: ContinuousWave: The Whaler GAM or General AreaContinuousWave: Whaler MarketplaceContinuousWave: Whaler Repairs/ModsContinuousWave: Whaler PerformanceContinuousWave: Cetacea CommentsContinuousWave: Small Boat ElectricalContinuousWave: Trips and RendezvousContinuousWave: Post-Classic Whalers | 947 | 4,351 | {"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-2020-50 | latest | en | 0.915023 |
https://economics.stackexchange.com/questions/41918/derivation-question-gali-2015-chapter-5-equation-25 | 1,726,106,581,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651420.25/warc/CC-MAIN-20240912011254-20240912041254-00379.warc.gz | 194,626,352 | 40,106 | # Derivation question: Gali (2015), chapter 5, equation 25
I am wondering how Gali derived equation (25) in chapter 5 of his book Monetary Policy, Inflation, and the Business Cycle (2015).
We have equation (21): $$\vartheta \hat{x}_{t} = -\kappa \hat{p}_{t} + \Lambda$$ and equation (23): $$\hat{p}_{t} = \delta\hat{p}_{t-1} + \frac{\delta}{1-\delta\beta\rho_u}u_{t} + \frac{\delta}{1-\delta\beta}\frac{\kappa\Lambda}{\vartheta}.$$ Gali then combines (21) and (23) to get (25): $$\hat{x}_t = \delta\hat{x}_{t-1} - \frac{\kappa\delta}{\vartheta(1-\delta\beta\rho_u)}u_t.$$
Now, unless I've made some kind of silly algebraic mistake, I'm not quite sure how he no longer has a constant term in (25). Or maybe I'm missing a simplifying assumption regarding some of the variables?
If someone can briefly explain this, it would be much appreciated.
• Could you specify what page and edition the equations are? In edition of Gali I have in chapter 5 equations are only numbered up 20, there are more equations in chapter 5 without numbering and some of them are almost same to yours, and if they were numbered they would have that numbering, but are not 100% same - now I am not sure if the above equations have typo or you are using different edition where the equations are set up differently
– 1muflon1
Commented Jan 3, 2021 at 18:29
• Hi, the version I have the 2015 version (the second edition, I think). The pages for these equations are p.142-143. Commented Jan 4, 2021 at 2:02 | 434 | 1,482 | {"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": 3, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2024-38 | latest | en | 0.912621 |
https://samacheerguru.com/samacheer-kalvi-10th-maths-chapter-2-additional-questions/ | 1,702,065,327,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100769.54/warc/CC-MAIN-20231208180539-20231208210539-00259.warc.gz | 563,791,565 | 14,734 | Samacheer Kalvi 10th Maths Solutions Chapter 2 Numbers and Sequences Additional Questions
Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 2 Numbers and Sequences Additional Questions
Question 1.
Use Euclid’s algorithm to find the HCF of 4052 and 12756.
Solution:
Since 12576 > 4052 we apply the division lemma to 12576 and 4052, to get HCF
12576 = 4052 × 3 + 420.
Since the remainder 420 ≠ 0, we apply the division lemma to 4052
4052 = 420 × 9 + 272.
We consider the new divisor 420 and the new remainder 272 and apply the division lemma to get
420 = 272 × 1 + 148, 148 ≠ 0.
∴ Again by division lemma
272 = 148 × 1 + 124, here 124 ≠ 0.
∴ Again by division lemma
148 = 124 × 1 + 24, Here 24 ≠ 0.
∴ Again by division lemma
124 = 24 × 5 + 4, Here 4 ≠ 0.
∴ Again by division lemma
24 = 4 × 6 + 0.
The remainder has now become zero. So our procedure stops. Since the divisor at this stage is 4.
∴ The HCF of 12576 and 4052 is 4.
Question 2.
If the HCF of 65 and 117 is in the form (65m – 117) then find the value of m.
By Euclid’s algorithm 117 > 65
117 = 65 × 1 + 52
52 = 13 × 4 × 0
65 = 52 × 1 + 13
H.C.F. of 65 and 117 is 13
65m – 117 = 13
65 m = 130
m = $$\frac { 130 }{ 65 }$$ = 2
The value of m = 2
Question 3.
Find the LCM and HCF of 6 and 20 by the prime factorisation method.
Solution:
We have 6 = 21 × 31 and
20 = 2 × 2 × 5 = 22 × 51
You can find HCF (6, 20) = 2 and LCM (6, 20) = 2 × 2 × 3 × 5 = 60. As done in your earlier classes. Note that HCF (6, 20) = 21 = product of the smallest power of each common prime factor in the numbers.
LCM (6, 20) = 22 × 31 × 51 = 60.
= Product of the greatest power of each prime factor, involved in the numbers.
Question 4.
Prove that $$\sqrt { 3 }$$ is irrational.
Let us assume the opposite, (1) $$\sqrt { 3 }$$ is irrational.
Hence $$\sqrt { 3 }$$ = $$\frac { p }{ q }$$
Where p and q(q ≠ 0) are co-prime (no common factor other than 1)
Hence, 3 divides p2
So 3 divides p also …………….. (1)
Hence we can say
$$\frac { p }{ 3 }$$ = c where c is some integer
p = 3c
Now we know that
3q2 = p2
Putting = 3c
3q2 = (3c)2
3q2 = 9c2
q2 = $$\frac { 1 }{ 3 }$$ × 9c2
q2 = 3c2
$$\frac{q^{2}}{3}$$ = C2
Hence 3 divides q2
So, 3 divides q also ……………. (2)
By (1) and (2) 3 divides both p and q
By contradiction $$\sqrt { 3 }$$ is irrational.
Question 5.
Which of the following list of numbers form an AP? If they form an AP, write the next two terms:
(i) 4, 10, 16, 22, …
(ii) 1, -1,-3, -5,…
(iii) -2, 2, -2, 2, -2, …
(iv) 1, 1, 1, 2, 2, 2, 3, 3, 3,…
Solution:
(i) 4, 10, 16, 22, …….
We have a2 – a1 = 10 – 4 = 6
a3 – a2 = 16 – 10 = 6
a4 – a3 = 22 – 16 = 6
∴ It is an A.P. with common difference 6.
∴ The next two terms are, 28, 34
(ii) 1, -1, -3, -5
t2 – t1 = -1 – 1 = -2
t3 – t2 = -3 – (-1) = -2
t4 – t3 = -5 – (-3) = -2
The given list of numbers form an A.P with the common difference -2.
The next two terms are (-5 + (-2)) = -7, -7 + (-2) = -9.
(iii) -2, 2,-2, 2,-2
t2 – t1 = 2-(-2) = 4
t3 – t2 = -2 -2 = -4
t4 – t3 = 2 – (-2) = 4
It is not an A.P.
(iv) 1, 1, 1, 2, 2, 2, 3, 3, 3
t2 – t1 = 1 – 1 = 0
t3 – t2 = 1 – 1 = 0
t4 – t3 = 2 – 1 = 1
Here t2 – t1 ≠ t3 – t2
∴ It is not an A.P.
Question 6.
Find n so that the nth terms of the following two A.P.’s are the same.
1, 7,13,19,… and 100, 95,90,…
The given A.P. is 1, 7, 13, 19,….
a = 1, d = 7 – 1 = 6
tn1 = a + (n – 1)d
tn1 = 1 + (n – 1) 6
= 1 + 6n – 6 = 6n – 5 … (1)
The given A.P. is 100, 95, 90,….
a = 100, d = 95 – 100 = – 5
tn2 = 100 + (n – 1) (-5)
= 100 – 5n + 5
= 105 – 5n …..(2)
Given that, tn1 = tn2
6n – 5 = 105 – 5n
6n + 5n = 105 + 5
11 n = 110
n = 10
∴ 10th term are same for both the A.P’s.
Question 7.
In a flower bed, there are 23 rose plants in the first row, 21 in the second, 19 is the third, and so on. There are 5 rose plants in the last row. How many rows are there in the flower bed?
The number of rose plants in the 1st, 2nd, 3rd,… rows are
23, 21, 19,………….. 5
It forms an A.P.
Let the number of rows in the flower bed be n.
Then a = 23, d = 21 – 23 = -2, l = 5.
As, an = a + (n – 1)d i.e. tn = a + (n – 1)d
We have 5 = 23 + (n – 1)(-2)
i.e. -18 = (n – 1)(-2)
n = 10
∴ There are 10 rows in the flower bed.
Question 8.
Find the sum of the first 30 terms of an A.P. whose nth term is 3 + 2n.
Given,
tn = 3 + 2n
t1 = 3 + 2 (1) = 3 + 2 = 5
t2 = 3 + 2 (2) = 3 + 4 = 7
t3 = 3 + 2 (3) = 3 + 6 = 9
Here a = 5,d = 7 – 5 = 2, n = 30
Sn = $$\frac { n }{ 2 }$$ [2a + (n – 1)d]
S30 = $$\frac { 30 }{ 2 }$$ [10 + 29(2)]
= 15 [10 + 58] = 15 × 68 = 1020
∴ Sum of first 30 terms = 1020
Question 9.
How many terms of the AP: 24, 21, 18, . must be taken so that their sum is 78?
Solution:
Here a = 24, d = 21 – 24 = -3, Sn = 78. We need to find n.
We know that,
Sn = $$\frac { n }{ 2 }$$ (2a + (n – 1)d)
78 = $$\frac { n}{ 2 }$$ (48 + 13(-3))
78 = $$\frac { n}{ 2 }$$ (51 – 3n)
or 3n2 – 51n + 156 = 0
n2 – 17n + 52 = 0
(n – 4) (n – 13) = 0
n = 4 or 13
The number of terms are 4 or 13.
Question 10.
The sum of first n terms of a certain series is given as 3n2 – 2n. Show that the series is an arithmetic series.
Solution:
Given, Sn = 3n2 – 2n
S1 = 3 (1)2 – 2(1)
= 3 – 2 = 1
ie; t1 = 1 (∴ S1 = t1)
S2 = 3(2)2 – 2(2) = 12 – 4 = 8
ie; t1 + t2 = 8 (∴ S2 = t1 + t2)
∴ t2 = 8 – 1 = 7
S3 = 3(3)2 – 2(3) = 27 – 6 = 21
t1 + t2 + t3 = 21 (∴ S3 = t1 + t2 + t3)
8 + t3 = 21 (Substitute t1 + t2 = 8)
t3 = 21 – 8 ⇒ t3 = 13
∴ The series is 1,7,13, …………. and this series is an A.P. with common difference 6. | 2,465 | 5,399 | {"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} | 4.875 | 5 | CC-MAIN-2023-50 | latest | en | 0.830665 |
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