url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3
values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93
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https://usd.currencyrate.today/convert/amount-5310-to-krw.html | 1,718,733,533,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861773.80/warc/CC-MAIN-20240618171806-20240618201806-00577.warc.gz | 502,510,891 | 23,246 | # 5,310 USD to KRW – US Dollars to Won
How much is \$5,310.00 – the five thousand three hundred ten 💵 us dollars is worth ₩7,324,234.14 (KRW) today or 🇰🇷 seven million three hundred twenty-four thousand two hundred thirty-four won as of 17:00PM UTC. We utilize mid-market currency rates to convert USD against KRW currency pair. The current exchange rate is 1379.3285.
5,310 USD
is equal to
7,324,234.14 KRW
Rate: 1379.3285 -0.211
Previous Close: 1379.5397
Inverse: 5,310.00 KRW = 3.85 USD
🕒 Updated: Tue, 18 Jun 24 17:00:00 +0000
Compared to the previous close exchange rate, South Korean Won 🔻decreased by -0.02% (-1121.72) vs. US Dollar. Feel free to try converting the currency rate in reverse from Won to US Dollars.
Best 5,310 USD to KRW Exchange Rate
BOA • USA ₩7,375,000.00 ₩6,896,103.90 ₩7,925,373.13 6h ago • New York
SMBS • South Korea ₩7,330,455.00 2h ago • Seoul
FRS • USA ₩7,273,972.60 17h ago • New York
🔄 Currency Converter Get widget
## Convert US Dollar to South Korean Won
USD KRW
5,290 USD ₩7,296,647.57
5,300 USD ₩7,310,440.85
5,310 USD ₩7,324,234.14
5,320 USD ₩7,338,027.42
5,330 USD ₩7,351,820.71
### Convert South Korean Won to US Dollar
KRW USD
5,290 KRW \$3.84
5,300 KRW \$3.84
5,310 KRW \$3.85
5,320 KRW \$3.86
5,330 KRW \$3.86
#### Stat and compare
The table displays the facts and analysis of the fluctuations. On this day a year ago, one received ₩6,781,981.76 won for \$5,310.00 us dollars, which is ₩542,252.38 less than today's rate. The most favorable exchange rate in the past 7 days, was ₩7,345,163.70. Keep an eye on this page and stay informed about any changes.
1Y ago Actual
₩6,781,981.76 ₩7,324,234.14
Stat Last 7 days Last 30 days
High ₩7,345,163.70 ₩7,345,163.70
Low ₩7,268,259.31 ₩7,191,249.82
Average ₩7,322,576.06 ₩7,285,121.35
Changes ₩76,904.39 ₩153,913.88
Volatility 4.72% 6.32%
#### Historical Chart 5,310 USD/KRW
The chart will help visually assess the oscillation between USD and KRW, and analyze the data for the last year. This data is usually enough to forecast future changes.
#### Exchange rate history
Date 5310 USD/KRW Changes Changes %
Today \$5,310 = ₩7,324,234.14 ▼-1,121.72 -0.02%
17/06/2024 \$5,310 = ₩7,325,355.86 ▼-19,646.04 -0.27%
16/06/2024 \$5,310 = ₩7,345,001.90 ▼-161.80
15/06/2024 \$5,310 = ₩7,345,163.70
14/06/2024 \$5,310 = ₩7,345,163.70 ▲40,309.86 0.55%
13/06/2024 \$5,310 = ₩7,304,853.84 ▲36,594.53 0.5%
12/06/2024 \$5,310 = ₩7,268,259.31 ▼-51,272.95 -0.7%
11/06/2024 \$5,310 = ₩7,319,532.26 ▲20,072.64 0.27%
10/06/2024 \$5,310 = ₩7,299,459.62 ▼-27,868.66 -0.38%
09/06/2024 \$5,310 = ₩7,327,328.28 ▼-684.12 -0.01%
#### What is \$ 5310 in different currencies
Directions Value Inverse \$
5310 USD to Bulgarian Lev 🇧🇬 лв9,671.34 2,915.43
5310 USD to Euro 💶 €4,944.39 5,702.65
5310 USD to British Pound Sterling 💷 £4,180.24 6,745.08
5310 USD to Australian Dollar 🇦🇺 \$7,980.51 3,533.12
5310 USD to Canadian Dollar 🇨🇦 \$7,285.15 3,870.35
5310 USD to Swiss Franc 🇨🇭 ₣4,694.30 6,006.45
5310 USD to Chinese Yuan 🇨🇳 ¥38,522.46 731.94
5310 USD to Japanese Yen 💴 ¥837,457.36 33.67
5310 USD to Singapore Dollar 🇸🇬 \$7,173.61 3,930.53
5310 USD to New Zealand Dollar 🇳🇿 \$8,642.01 3,262.68
5310 USD to Pakistani Rupee 🇵🇰 ₨1,479,270.09 19.06
5310 USD to Hong Kong Dollar 🇭🇰 \$41,456.99 680.13
5310 USD to South Korean Won 🇰🇷 ₩7,324,234.14 3.85 | 1,386 | 3,322 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-26 | latest | en | 0.806678 |
https://goprep.co/ex-13-q92-a-and-b-are-two-students-their-chances-of-solving-i-1nlato | 1,642,864,960,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303864.86/warc/CC-MAIN-20220122134127-20220122164127-00119.warc.gz | 337,995,517 | 36,346 | Q. 925.0( 1 Vote )
# A and B are two s
Let E denotes the event that student ‘A’ solves the problem correctly.
P(E) = 1/3
Similarly, if we denote the event of ’B’ solving the problem correctly with F
We can write that:
P(F) = 1/4
Note: Observe that both the events are independent.
Probability that both the students solve the question correctly can be represented as-
P (E F) = = P(E1) {say} {we can multiply because events are independent}
Probability that both the students could not solve the question correctly can be represented as-
P(E’ F’) = = P(E2) {say}
Now we are given with some more data and they can be interpreted as:
Given: probability of making a common error and both getting same answer.
Note: If they are making an error, we can be sure that answer coming out is wrong.
Let S denote the event of getting same answer.
above situation can be represented using conditional probability.
P(S|E2) = 1/20
And if their answer is correct obviously, they will get same answer.
P(S|E1) = 1
We need to find the probability of getting a correct answer if they committed a common error and got the same answer.
Mathematically,
i.e P(E1|S) = ?
By observing our requirement and availability of equations, we can make guess that Bayes theorem is going to help us.
Using Bayes theorem, we get-
P(E1|S) =
Using the values from above –
P(E1|S) =
Clearly our answer matches with option D.
Option (D) is the only correct choice.
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https://ibguides.com/economics/notes/markt-equilibrium/ | 1,723,668,045,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641121834.91/warc/CC-MAIN-20240814190250-20240814220250-00116.warc.gz | 251,221,296 | 6,077 | # Markt equilibrium
## Equilibrium and changes to equilibrium
Market equilibrium: is where the supply equals to the demand.
Figure 1.5 - Market equilibrium
## Calculating and illustrating equilibrium using linear equations
The market is in equilibrium at Pe, when the amount of that product consumers wish to buy, Qe, is equal to the amount of coffee producer wish to sell.
Figure 1.6 - Excess in supply and demand at different price levels
• Excess supply: More is being supplied than demanded at P1 → in order to eliminate the surplus, producer must lower the price
• Excess demand: More is being demanded than supplied at P2 → in order to eliminate this shortage, producer must raise the price
Changes in determinants cause changes in equilibrium
Figure 1.7 - A change in equlibiurm due to a change in determinants
• When there’s a change in determinants of demand/supply other than the price of the product, it would lead to a shift of a curve.
• When Demand shifts to D1, Qe is the quantity supplied, but Q2 is the quantity demanded, there is excess demand (of xxx units).
• Due to price mechanism (see below), the price will rise until Pe1, where the new equilibrium quantity, is both demanded and supplied. | 274 | 1,222 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2024-33 | latest | en | 0.883274 |
https://metanumbers.com/1523998000000 | 1,601,508,021,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402128649.98/warc/CC-MAIN-20200930204041-20200930234041-00076.warc.gz | 473,134,750 | 8,240 | ## 1523998000000
1,523,998,000,000 (one trillion five hundred twenty-three billion nine hundred ninety-eight million) is an even thirteen-digits composite number following 1523997999999 and preceding 1523998000001. In scientific notation, it is written as 1.523998 × 1012. The sum of its digits is 37. It has a total of 16 prime factors and 336 positive divisors. There are 522,480,000,000 positive integers (up to 1523998000000) that are relatively prime to 1523998000000.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 13
• Sum of Digits 37
• Digital Root 1
## Name
Short name 1 trillion 523 billion 998 million one trillion five hundred twenty-three billion nine hundred ninety-eight million
## Notation
Scientific notation 1.523998 × 1012 1.523998 × 1012
## Prime Factorization of 1523998000000
Prime Factorization 27 × 56 × 72 × 15551
Composite number
Distinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 16 Total number of prime factors rad(n) 1088570 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 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 1,523,998,000,000 is 27 × 56 × 72 × 15551. Since it has a total of 16 prime factors, 1,523,998,000,000 is a composite number.
## Divisors of 1523998000000
336 divisors
Even divisors 294 42 21 21
Total Divisors Sum of Divisors Aliquot Sum τ(n) 336 Total number of the positive divisors of n σ(n) 4.41495e+12 Sum of all the positive divisors of n s(n) 2.89095e+12 Sum of the proper positive divisors of n A(n) 1.31397e+10 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 1.2345e+06 Returns the nth root of the product of n divisors H(n) 115.984 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 1,523,998,000,000 can be divided by 336 positive divisors (out of which 294 are even, and 42 are odd). The sum of these divisors (counting 1,523,998,000,000) is 4,414,949,737,920, the average is 131,397,313,62.,857.
## Other Arithmetic Functions (n = 1523998000000)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 522480000000 Total number of positive integers not greater than n that are coprime to n λ(n) 1306200000 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 56395129763 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 522,480,000,000 positive integers (less than 1,523,998,000,000) that are coprime with 1,523,998,000,000. And there are approximately 56,395,129,763 prime numbers less than or equal to 1,523,998,000,000.
## Divisibility of 1523998000000
m n mod m 2 3 4 5 6 7 8 9 0 1 0 0 4 0 0 1
The number 1,523,998,000,000 is divisible by 2, 4, 5, 7 and 8.
• Abundant
• Polite
• Practical
• Frugal
## Base conversion (1523998000000)
Base System Value
2 Binary 10110001011010101010111000000001110000000
3 Ternary 12101200201001002211221101
4 Quaternary 112023111113000032000
5 Quinary 144432121442000000
6 Senary 3124040502104144
8 Octal 26132527001600
10 Decimal 1523998000000
12 Duodecimal 20743ba1b054
20 Vigesimal 2jac97a000
36 Base36 jg4526ps
## Basic calculations (n = 1523998000000)
### Multiplication
n×i
n×2 3047996000000 4571994000000 6095992000000 7619990000000
### Division
ni
n⁄2 7.61999e+11 5.07999e+11 3.81e+11 3.048e+11
### Exponentiation
ni
n2 2322569904004000000000000 3539591888562287992000000000000000000 5394330958985149775232016000000000000000000000000 8220949592831450287154041919968000000000000000000000000000000
### Nth Root
i√n
2√n 1.2345e+06 11507.9 1111.08 273.273
## 1523998000000 as geometric shapes
### Circle
Diameter 3.048e+12 9.57556e+12 7.29657e+24
### Sphere
Volume 1.48266e+37 2.91863e+25 9.57556e+12
### Square
Length = n
Perimeter 6.09599e+12 2.32257e+24 2.15526e+12
### Cube
Length = n
Surface area 1.39354e+25 3.53959e+36 2.63964e+12
### Equilateral Triangle
Length = n
Perimeter 4.57199e+12 1.0057e+24 1.31982e+12
### Triangular Pyramid
Length = n
Surface area 4.02281e+24 4.17145e+35 1.24434e+12
## Cryptographic Hash Functions
md5 42c74b59f1467b450dd2a0892095b3bd c53e909c3e6e488e2e9ef3c676a9ab1a7c8f93bc 645620b04465f6667f9f70d8992821a006a7ecd2ddaf569e838ac4f98284bb8f df75a33bdb1de251548bcccbf80b93a46e4216dcd5d6803004293e00b02b91bf13aa233e50960bb7149b196192ca83584b7127a867947c8e0f2098424ee19886 6b3c3bba77e5a8a0cf58cd09782fddc4a9eec5f0 | 1,738 | 4,783 | {"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} | 3.390625 | 3 | CC-MAIN-2020-40 | latest | en | 0.771544 |
https://www.meritnation.com/ask-answer/question/a-car-starts-from-a-for-b-travelling-20-km-an-hour-1-1-2-hou/laws-of-motion/13292253 | 1,652,915,395,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662522556.18/warc/CC-MAIN-20220518215138-20220519005138-00314.warc.gz | 1,032,124,579 | 7,856 | # A car starts from A for B travelling 20 km an hour.1 1/2 hours later another car starts from A and travelling at the rate of 30 km an hour reaches B 2 1/2 hours before the first car. Find the distance from A to B.
Dear Student
Let, distance between A and B be x.
The second car starts 1.5 hour later and reaches 2.5 hour early then first car.
So, the time difference = 4 hours
Thus, x/20 - x/ 30 = 4
x = 240
Distance between A and B is 240 Km.
Regards
• 2
What are you looking for? | 149 | 493 | {"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} | 3.796875 | 4 | CC-MAIN-2022-21 | latest | en | 0.92882 |
https://propertyupdate.com.au/use-reverse-engineering-to-solve-complex-challenges-including-investing/ | 1,722,922,839,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640476479.26/warc/CC-MAIN-20240806032955-20240806062955-00595.warc.gz | 379,848,661 | 28,762 | # Use reverse engineering to solve complex challenges including investing
A true story: a couple of decades ago a friend of a friend had a ‘few too many’ and crashed his car onto the middle of a roundabout at 2am.
He panicked and left the car where it was, with a not-so-cunning plan to collect it the next day after he’d sobered up.
What he hadn’t reckoned with was that the police had seen this flawed ploy a thousand times before, and had a proven method for dealing with it.
Forensic toxicologists are able to take a blood or breath sample to calculate a blood alcohol concentration and determine whether a suspect was over the legal limit at an earlier time.
Through a countback analysis, by working the calculation in reverse, they were able to prove what was inherently obvious, and the relevant penalties were applied.
## Consider the lily…
Fairly often I write about compound growth, and an example I use in my public speaking is how lilypads can cover the surface of a pond (which is to say slowly at first, then all at once).
If the lily pad doubles in size daily and covers the pond in 30 days, then how long does it take to cover half of the pond?
Our instinctive reaction is to think ‘probably about 15 days’, because unless your brain contains a compound interest table it’s almost impossible to work out the correct answer…unless, that is, you work the calculation backwards.
If the lily pad doubles in size daily then it must have covered half of the pond in 29 days, being one less than 30. Simple!
Compound Growth
## Solving problems with inversion
If you think back to high school algebra you may recall that some problems can only be solved by inverting them (in my accountancy training I had similar challenges with trying to calculate internal rates of return…ouch).
It’s not just a maths or numbers thing – you may find that proof-reading can be more effective if you read sentences backwards, because your brain doesn’t take the same mental leaps as it does when scanning a page in the normal way.
A repeated word riddle
This suggests that if you’re really stuck with a major challenge and you can’t think of a way to solve it, then trying to think backwards or in reverse might be a useful way to get unstuck.
## The inversion technique
Here’s how you can practice ‘reverse brainstorming’ to solve a problem:
(i) define the problem, and instead of trying to solve it, think of how you could make it worse;
(ii) make a list of the reversed solutions and rank them in order of effectiveness; and
(iii) flip the reversed solutions around to identify the potential best fixes for your problem.
You can apply this technique to risk management, effective networking, customer satisfaction, competitive advantage, product delivery, building trust, or any number of other problems or logistical challenges.
If you get stuck with a complex or deep-seated problem, turning the challenge on its head can be a creative tool and an effective mental technique for finding a solution.
About Pete Wargent Pete is a Chartered Accountant, Chartered Secretary and has a Financial Planning Diploma. Using a long term approach to building businesses, investing in equities, & owning a portfolio he achieved financial independence at the age of 33. Visit his blog | 687 | 3,291 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2024-33 | latest | en | 0.969658 |
http://www.weegy.com/?ConversationId=67251C29 | 1,386,904,709,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164836485/warc/CC-MAIN-20131204134716-00052-ip-10-33-133-15.ec2.internal.warc.gz | 621,459,921 | 8,083 | Ask a question
what is difference between sequence and rate development and why is difference important??
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Sequence of development would be the order in which the development happens and the rate of development is the speed at which development happens. The difference is important because speed does not ...
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Note: Child Development question: What is the difference between sequence of development and rate of development and why is the difference important? Can you answer this ... [ http://wiki.answers.com/Q/What_is_the_difference_between_sequence_of_development_and_rate_of_development_and_why_is_the_difference_important ]
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Weegy: Sequence of development would be the order in which the development happens and the rate of development is the speed at which development happens. The difference is important because speed does not necessarily have to do with the sequenec.
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6. Suppose you're writing a letter and you recall a word you'd ...
Weegy: Suppose you're writing a letter and you recall a word you'd like to use, but you don't know how it's spelled. ...
12/11/2013 3:15:29 AM| 2 Answers
Which sentence is a fragment A. When driving.B. I sing in the ...
Weegy: Her second novel, Their Eyes Were Watching God ... Hear My Cry, 1976) and Walter Dean Myers (Fallen Angels, [ ...
12/11/2013 9:28:17 AM| 2 Answers
The area of the brain that interprets sensory data is called a. The ...
Weegy: The brain stem ,Sitting on top of the brain stem is the limbic system, also known as the ‘old mammalian brain’. ...
12/11/2013 11:11:47 AM| 2 Answers
What is unn school fees for this year 2013/2014
12/12/2013 3:23:03 AM| 2 Answers
has the past period/year been good/bad/satisfactory or otherwise for ...
Weegy: Why past year been good bad satisfactory or otherwise for you? Satifactory Has the past year been good bad ...
12/11/2013 2:28:02 AM| 1 Answers
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Points 353 [Total 53520]| Ratings 14| Comments 213| Referrals 0|Offline
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Points 120 [Total 490]| Ratings 12| Comments 0| Referrals 0|Offline | 1,671 | 4,956 | {"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-2013-48 | longest | en | 0.939663 |
http://spamdestructor.com/probability-of/probability-of-error-fsk.php | 1,540,181,268,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583514497.14/warc/CC-MAIN-20181022025852-20181022051352-00108.warc.gz | 348,904,498 | 4,825 | Home > Probability Of > Probability Of Error Fsk
Probability Of Error Fsk
Contents
A. Computational Details The measurement generates a reference curve based on the type and settings of the meter block selected in the BER/SER Meter setting. N. W. check my blog
Commun. The ACM Guide to Computing Literature All Tags Export Formats Save to Binder Back to search page Click to download printable version of this guide. Make sure that you do not miss a new article by subscribing to RSS feed OR subscribing to e-mail newsletter. tue.
Probability Of Error In Ask Psk Fsk
Parameters Name Type Range Block Diagram System Diagram N/A BER/SER Meter System BER/SER Meter N/A Modulation Type List of options N/A Demodulation Type List of options N/A Statistic Type List Note: Subscribing via e-mail entitles you to download the free e-Book on BER of BPSK/QPSK/16QAM/16PSK in AWGN. { 22 comments… read them below or add one } Ramya November 4, 2012 ele. Your cache administrator is webmaster.
• Reply zardosht February 26, 2012 at 12:56 pm Hi krishna I cant understand number of symbole, where is k=log2M and for mfsk please help me.
• Number of errors by total number of bits is the error rate Reply Siddharth May 12, 2012 at 9:54 pm I am trying to code 16-FSK modulator and then compute
• E.
• Nelin, The influence of fading spectrum on the binary error probabilities of incoherent and differentially coherent matched filter receivers, IRE Trans.
• Bello, Characterization of randomly time-variant linear channels, IEEE Trans.
Thanks Reply Cancel reply Leave a Comment Name * E-mail * Website Notify me of followup comments via e-mail Previous post: CORDIC for phase rotation Next post: Scaling factor in QAM R. If the Statistic Type parameter is set to "Auto", the measurement will compute the bit error probabilities Pb for BER meters and symbol error probabilities Ps for SER meters. Coherent And Noncoherent Fsk What do you think about that?
Generated Mon, 24 Oct 2016 12:18:23 GMT by s_wx1196 (squid/3.5.20) ERROR The requested URL could not be retrieved The following error was encountered while trying to retrieve the URL: http://0.0.0.8/ Connection Please try the request again. L. Subscribe Enter Search Term First Name / Given Name Family Name / Last Name / Surname Publication Title Volume Issue Start Page Search Basic Search Author Search Publication Search Advanced Search
Schwartz, W. Bit Error Rate Of Ask In Matlab For example, if a newborn baby's heart beats at a frequency of 120 times a minute, its period (the interval between beats) is half a second. If it quad-FSK then you sum the factorial from 2 to 4 and and calculate symbol error probability. Your simulation is very useful for me.
Fsk Bit Error Rate
Given the above understanding, am unable to understand your perspective. More Bonuses The period is the duration of one cycle in a repeating event, so the period is the reciprocal of the frequency. Probability Of Error In Ask Psk Fsk Thanks for visiting! Probability Of Error For Noncoherent Fsk The system returned: (22) Invalid argument The remote host or network may be down.
Please try the request again. click site Inf. In the binary case it is simple and pb=1/2exp(-0.5*snr/n); nothing else. Figure: Block diagram of FSK modulation and coherent demodulation For analyzing the bit error rate with coherent FSK demodulation, let us compare the signaling waveform used by binary FSK when compared Probability Of Error For Non-coherent Fsk | 812 | 3,474 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2018-43 | latest | en | 0.787709 |
https://community.smartsheet.com/discussion/71569/sumif-formula-help | 1,713,418,377,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817187.10/warc/CC-MAIN-20240418030928-20240418060928-00804.warc.gz | 156,230,730 | 102,129 | # SUMIF Formula Help!
✭✭✭✭
I am trying to type a sum if formula where it will add up a whole column based off of ONE criteria. The formula is working but it equals to 0 every time when that is not what the actual answer is. Please help!
Ex: I want to sum up the number columns where the region is Asia...it would be 2 but it is coming out as 0.
What I am trying: =SUMIF({Number}, {Region} = "Asia")
Number Region
1 Asia
1 Europe
1 America
1 Asia
• ✭✭✭✭✭
Hi @Jenna_Doucette try the following:
=SUMIF({the column you want added}, {Region - I assume another column}="Asia")
• ✭✭✭✭
Hi Lila,
Thank you, that is what I did.
Thanks,
Jenna
• ✭✭✭✭✭✭
Can you describe your process in more detail and maybe share the sheet(s)/copies of the sheet(s) or some screenshots? (Delete/replace any confidential/sensitive information before sharing) That would make it easier to help. (share too, andree@getdone.se)
I hope that helps!
Be safe and have a fantastic weekend!
Best,
Andrée Starå | Workflow Consultant / CEO @ WORK BOLD
SMARTSHEET EXPERT CONSULTANT & PARTNER
Andrée Starå | Workflow Consultant / CEO @ WORK BOLD
W: www.workbold.com | E:andree@workbold.com | P: +46 (0) - 72 - 510 99 35
Feel free to contact me for help with Smartsheet, integrations, general workflow advice, or anything else.
• ✭✭✭✭
Attached are some screenshots. I am trying to create a formula sheet and reference one external sheet but different columns for the SUMIF.
If the Region is APAC I would like to Sum the Planned Swing? column. I am trying (=SUMIF({Planned Swing?}, {Region} = "APAC") but when I do that for some reason it returns 0. The 1's are just numbers not a formula or anything. Please let me know if you need any more information.
• ✭✭✭✭
@Andrée Starå I've done these formulas multiple times I just do not know why this one isn't working....
• ✭✭✭✭✭✭
edited 09/25/20
Edit.
I'd be happy to take a quick look.
Can you maybe share the sheet(s)/copies of the sheet(s)? (Delete/replace any confidential/sensitive information before sharing) That would make it easier to help. (share too, andree@getdone.se)
SMARTSHEET EXPERT CONSULTANT & PARTNER
Andrée Starå | Workflow Consultant / CEO @ WORK BOLD
W: www.workbold.com | E:andree@workbold.com | P: +46 (0) - 72 - 510 99 35
Feel free to contact me for help with Smartsheet, integrations, general workflow advice, or anything else.
• ✭✭✭✭
Incorrect Argument you have to do ="APAC" this is what it returns. The first range is the Planned Swing? column and the second is the region
• ✭✭✭✭✭✭
edited 09/25/20
Haha! I edited my post, and in the meanwhile, you'd already seen it.
#### 😉
Try this.
=SUMIF({Region}, "APAC",{Planned Swing?})
SMARTSHEET EXPERT CONSULTANT & PARTNER
Andrée Starå | Workflow Consultant / CEO @ WORK BOLD
W: www.workbold.com | E:andree@workbold.com | P: +46 (0) - 72 - 510 99 35
Feel free to contact me for help with Smartsheet, integrations, general workflow advice, or anything else.
• ✭✭✭✭
Thank you @Andrée Starå it worked!😊 Maybe my mind wasn't working yesterday with the logic hahah
## Help Article Resources
Want to practice working with formulas directly in Smartsheet?
Check out the Formula Handbook template! | 946 | 3,218 | {"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-2024-18 | latest | en | 0.878991 |
http://theconvertersite.com/conversions/common/stones-pounds-ounces.php | 1,519,026,540,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812556.20/warc/CC-MAIN-20180219072328-20180219092328-00311.warc.gz | 346,848,270 | 9,778 | Calc
# Stones to pounds and ounces conversion
These tools help you convert between stone, pounds and ounces (st, lb and oz), all of which are measurements of mass and weight. Please choose from the conversion options below.
st
lb oz
st
lb
st
### Ounces to stone
oz
#### Disclaimer
Whilst every effort has been made in building this stones to pounds and ounces conversion tool, we are not to be held liable for any special, incidental, indirect or consequential damages or monetary losses of any kind arising out of or in connection with the use of the converter tools and information derived from the web site. This stones to pounds and ounces conversion tool is here purely as a service to you, please use it at your own risk. Do not use calculations for anything where loss of life, money, property, etc could result from inaccurate conversions.
## Conversion FAQ
Instructions for manually converting between stone, pounds and ounces are shown below.
### Stone, pounds and ounces
1 stone is equal to 14 pounds and 0 ounces.
### Stone to pounds and pounds to stone
There are 14 pounds in 1 stone. So, multiply your stones figure by 14 to get your answer. Should you wish to convert from pounds to stones, divide the figure by 14.
### Converting stone to ounces
1 stone is equal to 224 ounces. Multiply your stone figure by 224 to get your answer. Should you wish to convert ounces to stone, divide your ounces figure by 224.
### Converting pounds to ounces
There are exactly 16 ounces in 1 pound. So, multiply your pounds figure by 16 to get your ounces figure. To work out how many pounds there are in x ounces, divide your number by 16.
Of course, you can check the answer to these questions by using one of the converters featured at the top of the page.
For more conversions using mass and weight units, please see the mass and weight converter. I also have converters for kilograms and stone & pounds , grams to pounds & ounces and kilograms and pounds & ounces. | 428 | 1,991 | {"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-2018-09 | latest | en | 0.922369 |
https://www.sarthaks.com/106913/survey-was-made-to-find-the-type-of-music-that-a-certain-group-of-young-people-liked-in-a-city | 1,611,376,002,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703533863.67/warc/CC-MAIN-20210123032629-20210123062629-00306.warc.gz | 978,819,441 | 13,996 | # A survey was made to find the type of music that a certain group of young people liked in a city.
57 views
A survey was made to find the type of music that a certain group of young people liked in a city. Adjoining pie chart shows the findings of this survey. From this pie chart, answer the following:
(i) If 20 people liked classical music, how many young people were surveyed?
(ii) Which type of music is liked by the maximum number of people?
(iii) If a cassette company were to make 1000 CD’s, how many of each type would they make?
by (24.8k points)
selected
(i) 10% represents 100 people.
Therefore 20% represents = 100 x 20/10 = 200 pepole
Hence, 200 people were surveyed.
(ii) Light music is liked by the maximum number of people.
(iii) CD’s of classical music = 10 x 1000/100 = 100
CD’s of semi-classical music = 20 x 1000/100 = 100
CD’s of light music = 40 x 1000/100 = 400
CD’s of folk music = 30 x 1000/100 = 300 | 263 | 941 | {"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} | 3.578125 | 4 | CC-MAIN-2021-04 | latest | en | 0.980167 |
http://www.chegg.com/homework-help/computer-network-consists-computers-computer-directly-connec-chapter-6.2-problem-37e-solution-9780073383095-exc | 1,448,758,408,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398454553.89/warc/CC-MAIN-20151124205414-00237-ip-10-71-132-137.ec2.internal.warc.gz | 348,258,836 | 17,356 | View more editions
# TEXTBOOK SOLUTIONS FOR Discrete Mathematics and Its Applications 7th Edition
• 4227 step-by-step solutions
• Solved by publishers, professors & experts
• iOS, Android, & web
Over 90% of students who use Chegg Study report better grades.
May 2015 Chegg Study Survey
PROBLEM
Chapter: Problem:
A computer network consists of six computers. Each computer is directly connected to zero or more of the other computers. Show that there are at least two computers in the network that are directly connected to the same number of other computers. [Hint: It is impossible to have a computer linked to none of the others and a computer linked to all the others.]
STEP-BY-STEP SOLUTION:
Chapter: Problem:
• Step 1 of 4
Consider six computers that are connected to 0 or more of the other computers.
That is, each computer can be directly connected to 0, 1, 2, 3, 4, and 5.
Thus, there are 6 computers and 6 connections.
So, Pigeonhole Principle is not applicable here.
Pigeonhole Principle:
If objects are placed into boxes, then there is at least one box containing objects.
• Chapter , Problem is solved.
Corresponding Textbook
Discrete Mathematics and Its Applications | 7th Edition
9780073383095ISBN-13: 0073383090ISBN: Authors: | 304 | 1,252 | {"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} | 2.828125 | 3 | CC-MAIN-2015-48 | latest | en | 0.894369 |
https://mezelfkezdtemprofesor.com/englisch/describing-charts-and-diagrams/czm52924tz0l | 1,643,395,809,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320306335.77/warc/CC-MAIN-20220128182552-20220128212552-00421.warc.gz | 445,039,842 | 17,525 | Home
# Describing a bar chart pdf
### Describing a bar chart LearnEnglish Teens - British Counci
1. g was the most popular format in 2017. US consumers spent \$9.8 billion... 3. A trend is a change over time. To describe trends, focus on what is increasing o
2. DESCRIBING CHARTS AND GRAPHS Language used to describe charts and graphs. Graphs or charts help people understand data quickly. You can use them to make a comparison or show a trend. TYPES OF CHART PIE CHART: used to show percentages BAR CHART: is used to compare different sets of information LINE GRAPH: is most useful for showing trend
3. Pie charts are best to use when you are trying to compare parts of a whole Bar / column graphs are used to compare things between different groups or to track changes over time. Line graphs can also be used to compare changes over the same period of time for more than one group. Scatter plots/graphs show how much one variable is affected by another. The relationshi
Describing a bar chart. Look at the bar chart, exam question and sample answer and do the exercises to improve your writing skills Describing Bar Charts and Column Charts (1) Bar charts and column charts are similar: only their orientations differ. A bar chart is orientated horizontally, whereas a column chart is arranged vertically. Sometimes bar chart refers to both forms. These types of charts are usually used for comparison purposes (unlike line charts, which describe change). Observe the following chart Graphs, Charts & Diagrams. Data can be represented in many ways. The 4 main types of graphs are a bar graph or bar chart, line graph, pie chart, and diagram. Bar graphs are used to show relationships between different data series that are independent of each other. In this case, the height or length of the bar indicates the measured value or frequency. Below, you can see the example of a bar graph which is the most widespread visual for presenting statistical data A bar chart uses either horizontal or vertical bars to show comparisons among two or more categories. The bar chart has two main features: an X-axis and a Y-axis. One axis of the chart shows the specific categories being compared, and the other axis of the graph shows a given value (usually a percentage or a dollar amount)
The bar chart below shows the sector contributions to India's gross domestic product from 1960 to 2000. Summarise the information by selecting and reporting the main features, and make comparisons where relevant. Write at least 150 words. Contribution as % of India's GDP. Source: EPW Research Foundation. Step 1 - Analyse the question. The format of every Academic Task 1 question is the. Bar Chart; Bar Chart. It is common in the IELTS Writing paper to be asked to describe a bar graph. Bar graphs, also known as bar charts, are similar to line graphs in that they have two axes and are useful for showing how something has changed over a given period of time, especially when there are significant changes. Bar graphs consist of rectangular bars, which can be orientated horizontally or vertically, with the lengths proportional to the data values that they represent. They are. There are three graphs in the chart. The green graph shows the total growth of the population, the black one deals with the migrated people in Canada and the blue graph shows the natural increase of the population. In 1988/89 there was an enourmous growth. In the following years the total growth went down to about 250,000 in 1998/99. From that time on the Canadian population has been gradually growing again although the natural increase slows down. So we can say that the growth of the. 3.3 Bar Graphs for Qualitative Data On a scale of 1-10, how nervous are you about taking STAT 220? (1=very nervous, 10=not nervous) Bar graphs represent each category as a bar. The bar heights show the category counts or percents. Bar graphs can compare quantities that are not part of a whole. A Pareto bar graph has the bars ordered from tallest to shortest
Describing a bar chart. Hier können Sie sich über den grundsätzlichen Aufbau einer Graph-Beschreibung informieren und bekommen - in Beispielssätze eingebaut - brauchbare und universell einsetzbare Ausdrücke und Floskeln für eine Beschreibung. Graph - Aufbau, Floskeln. Tipps zur sprachlichen Ausgestaltung einer Graph-Beschreibung Tips on describing the Bar Chart. There are few tips which are used for describing the bar graph which is as follows: Try to use a simple language while describing the bar graph. After writing, review the content that you have written, and check if it is free from errors and plagiarism. Try to add any extra words which will suit the data given. While practicing, use different bar graphs or. •Bar Chart: A graph in which the classes are reported on horizontal axis and the class frequencies on the vertical axis. The class frequencies are proportional to the heights of the bars. Bar Charts 0 50 100 150 200 250 300 Goalkeeper Defender Midfielder Striker ards ) Player position (variable of interest) Bar Charts 0% 5% 10% 15% 20% 25% 30% 35% 40% 45% A B O AB y Blood Type (variable of. The first lesson begins with naming different graph / chart types and describing a range of different lines (peak, plummet, etc..). It finishes with a fun activity where students describe and plot the lines on four graph s. The second lesson provides the language necessary for describing, analysing and evaluating graphs. It is followed by researching and analysing graphs/charts/tables from the Office of National Statistics (ONS) and giving a short presentation on the findings Chapter 2. DESCRIBING DATA -USING GRAPHS. TOPIC SLIDE Introduction to using graphs 2 Pie Charts 4 • Tutorial: Creating a Pie Chart Bar Graphs 5 • Tutorial: Creating a Bar Graph Histograms 6 • Tutorial: Creating a Histogram Shapes of Histograms 12 Stem Plots 20 Time Plots 26. The shape or distribution of the data • Are there an equal number of.
Bar Chart Exercise - Answer. The 1. bar chart illustrates information on the quantity of drugs 2. school children in New Zealand take, divided by 3. gender and measured 4. in percentages. Overall, it is immediately apparent that hashish or marijuana is used 5. more than any of the other drugs, whereas LSD is used 6. the least Single Pie Chart Paragraph 1 Describe what sort of chart it is and what it is about. Paragraph 2 Describe the sections of the graph starting with the biggest and working your way down. Paragraph 3 A short conclusion giving an overall view of what the chart is about. Two Pie Charts If you have 2 pie charts they will be giving similar information so you can compare the two. This type of writing is similar to a compare and contrast essay View Describing a bar chart.pdf from ENGL 211 at King Saud bin Abdulaziz University for Health Sciences. Independent Learning Centre DESCRIBING A BAR CHART Bar charts can be used to show ho Learning and understanding the vocabulary and language for describing graphs is essential for Business English language learners. For example, students need to understand the language of graphs in order to: simplify data; show changes over time; illustrate cycles; demonstrate trends; explain budgets, costs, profits and other financial statistic
• interpret bar charts, pie charts, and box and whisker plots in a qualitative way. Starting points This session is in two linked parts. • Matching pie charts to bar charts. • Matching box and whisker plots to bar charts. Each part of the session starts with a whole class discussion to compare the newly-introduced type of distribution, looking at its advantages, disadvantages and. Lesson 2: Describing an IELTS Pie Chart This lesson will provide you with tips and advice on how to write an IELTS pie chart for task one. To begin, take a look at the pie chart below and the model answer. You should spend about 20 minutes on this task. The pie charts show the main reasons for migration to and from the UK in 2007 Das Arbeitsblatt describing-charts-ws.pdf (Word-Datei: describing-charts-ws.doc) erklärt wichtige Begriffe und Wendungen zum Beschreiben von Diagrammen und Schaubildern. Der Schwerpunkt liegt auf Linien- und Balkendiagrammen (line charts and bar charts), es werden aber auch allgemeine Begriffe und Kuchendiagramme (pie charts) behandelt. Ergänzt wird das Ganze durch 10 Diagramme, die zur. Bar Charts Introduction Bar charts are used to visually compare values to each other. This chapter gives a brief overview and examples of simple bar charts and two-factor bar charts. The orientation of a bar chart may be vertical or horizontal. Below is an example of a vertical bar chart with two factors (fruit and month). Data Structure Data for a bar chart are entered in columns. Each. IELTS Academic: Lexis for describing graphs An activity which helps students use a range of suitable lexis for describing graphs. The session can be used for IELTS preparation or for more general academic writing skills. Time required: 60-70 minute
Describing Data 251 . Pie charts look nice, but are harder to draw by hand than bar charts since to draw them accurately we would need to compute the angle each wedge cuts out of the circle, then measure the angle with a protractor. Computers are much better suited to drawing pie charts. Common software programs like Microsoft Word or Excel, OpenOffice.org Write or Calc, or Google Docs are. The bar chart above is a chart of comparison, so you look for categories to compare which have similar key features. This is the logical way to organise this report. The organisation is also highlighted in the overview by stressing the key features which will inevitably be used to structure the paragraphs. There are no set rules - you need to interpret the data, spot similarities and. Updated: April 27th 2021. In Writing task 1 academic you need to write a report about one of the following: It could be a bar chart, a line graph, maps, a table, a pie chart or a process diagram. In this case, we are going to look at describing a line graph in IELTS writing task 1 and the kind of grammar and vocabulary that is needed for this task
### Describe a Bar Chart Hugh Fox II
IELTS Academic writing task 1 - Bar Graph. A bar graph (also known as the bar chart) is a graph that contains either horizontal or vertical bars to present data comparison or contrast among categories or variables. In your IELTS Academic task 1, you might get a bar graph that will have numerical values of different variables shown by the length and height of lines or rectangular shapes or. Speaker 1: The chart I think is most commonly used within hospitals and the medical environment is the line chart. Speaker 3: In a bar chart you have columns kind of thing. It goes up like buildings The Bar Chart above demonstrates the percentages of consumption of vegetables and fruits more than 5 times in a day subgrouped people by women, children, men on the Y axis, according to years between 2001-2008 on the X-axis. If we just look 2001 and 2008 we can say there is big difference in consumption. Although it seems linear graphic at.
### How to describe charts, graphs, and diagrams in the
In this task, you are required to describe what you see in words. There should be no interpretation or analysis, just a factual description. There are several variations of this task: a chart/graph, a table, a map or a process/cycle. 23% 58% 10% 9% PIE CHART 1st Qtr 2nd Qtr 3rd Qtr 4th Qtr What do you need to describe? FASTRACK EDUCATION Email: info@fastrackedu.co.uk Website: http. One way to describe a chart is to provide both a text summary and a properly coded data table near the chart. This serves multiple audiences because a chart can show trends, but a table can provide exact data for those who are interested. Example: Final /r/ Deletion in Detroit's African American Speakers. Trudgill (1995) presents the following data about the loss of word-final /r/ in Detroit.
### IELTS Writing Task 1: How to Describe a Bar Chart
• In a bar chart, the height of the bar corresponds to the number of cases that fall into each category. Bar Chart 111 29 29 2 4 0 20 40 60 80 100 120 White Black Hispanic Asian Other In a pie chart, the relative area of each slice of the pie corresponds to the proportion/percentage in each category. Pie Chart White 63% Black 17% Hispanic 17% Asian 1% Other 2% Look at the relationship between.
• A Bar Chart is one of the most commonly seen question types in the IELTS academic writing task 1. Students often fail to describe the accurate details they see on a bar chart or a bar graph. This article will be of immense help if you are wishing to understand how to describe a bar graph in the IELTS academic writing test task 1
• utes on this task. The IELTS Writing task 1 exercise graphs show figures relating to hours worked and stress levels amongst professionals in eight groups. Describe the information shown to a university or collage lecturer. Write at least 150 words
• Describing a Chart or Graph in English - How to Develop English Fluency and Speaking ConfidenceClick here to get your free guide so you can start sounding mo..
• In drawing a percentage bar chart, bars of length equal to 100 for each class are drawn in the first step and sub-divided into the proportion of the percentage of their component in the second step. The diagram so obtained is called a percentage component bar chart or percentage stacked bar chart. This type of chart is useful to make comparisons in components holding the difference of total.
Lesson Share: Describing pie charts. By Gabrielle Lambrick. Gabrielle Lambrick wins this month's Lesson Share with a lesson about learning to describe pie charts for IELTS Writing Task 1 Watch on. In IELTS Academic Writing Task 1 you will be tested on your ability to describe and interpret information presented in a graph, table, chart or diagram. You will need to describe trends, compare and contrast data or report statistical information. Occasionally you will need to describe a process (which we will explain in another section)
Describing Bar Charts and Column Charts - Making Multiple Comparisons Describing Bar Charts and Column Charts - Observing the Axes Describing Pie Charts Describing Pie Charts - Comparing Two Charts Describing Tables Using Approximation Describing Trends Describing Projections Checklist . The graphs, charts and tables were created using data from the European Commision's Eurostat site: http. The bar chart represents sales in our Asian outlets. Here you can see a comparison between On the line graph you will note : a strong upward trend in the sales of product A despite occasional fluctuations with a slight drop during the holiday season with occasional variations due to the overall performance of product B is good. The initial surge in sales was followed by a.
After 7 pm, numbers fall significantly, with only a slight increase again at 8pm, tailing off after 9 pm. Overall, the graph shows that the station is most crowded in the early morning and early evening periods. Sample Chart Descriptions.pdf. Adobe Acrobat Dokument 1.7 MB. Download one of five forms - a line graph, bar graph, pie chart, table or diagram illustrating a process. You are required to describe the information or the process in a report of 150 words. This task should be completed in 20 minutes. It is important that you are familiar with the language appropriate to report writing generally and to each of the five types of report. TASK TWO is an essay based on.
### Video: How To Write a Bar Chart Essay - IELTS Jack
The IELTS academic exam writing task 1 consists of 6 types of charts such as process diagrams, maps, bar charts, pie charts, tables or line graphs. You will need to write a report about one of these, but remember it is very different to writing task 2, it is not an 'essay' it is a factual report. In academic task 1 there must be an overview and a factual description of the main parts of. • Note on the chart, that I labeled each portion by percentage (I could have chosen number instead). Only use percentage if you include the number of responses as a part of the title (here n=25). • Notice the three parts of the title: Figure, 1, and the verb-free sentence. The Vertical Bar (or Column) Graph • Remember, graphs allow for comparisons • Bar graphs allow you to make more. Bar graphs, pie charts, line graphs, and histograms are an excellent way to . illustrate your program results. This brief includes concepts and definitions, types of graphs and charts, and guidelines for formatting. Major Concepts and Definitions. Graphs and charts condense large amounts of . information into easy-to-understand formats that clearly and effectively communicate important points.
### IELTS Exam Preparation - Bar Char
• describe data • read and interpret displays of data • construct appropriate displays of data: frequency table, pictogram, bar chart, line bar chart, histogram, pie chart, line graph, frequency polygon, stem-leaf plots, scatter plots • justify the choice of display used for given data • critically analyse data displays • state common pupil errors in data representation. If you look at this bar chart you'll notice If you look at this pie chart you'll appreciate Describing Charts and Graphs. These are the most important expressions to describe charts and graphs. As we can see in Pie Chart #1, 55% of people use Netflix; Pie Chart #1 says that 55% of people use netfli
### Diagramme beschreiben im Englische
Describing graph, chart and diagram 1. DIFFERENT TYPES OF CHARTS, GRAPHS, TABLES AND DIAGRAMS A. TYPES OF GRAPHS Single line graph Multiple line graph Bar graph 2. B. TYPES OF CHARTS Paired bar chart Percentage bar chart Pie chart 3. Stacked bar chart Single bar chart Flow chart Population chart 4. C. TYPES OF DIAGRAMS Diagram Bubble Diagra A flow chart is a diagram showing the progress of material through the steps (étapes) of a manufacturing process (processus) or the succession of operations in a complex activity A pie chart displays the size (taille) of each part as a percentage of a whole (un tout). A (vertical or horizontal) bar chart. is used to compare . unlike (different) items. A . line chart. depicts changes over a.
### Describing a bar chart fosbosenglisc
Start studying Useful phrases: Pie charts / Bar charts. Learn vocabulary, terms, and more with flashcards, games, and other study tools Vertical Bar Chart. A vertical bar graph is the most common type of bar chart and it is also referred to as a column graph. It represents the numerical value of research variables using vertical bars whose lengths are proportional to the quantities that they represent. Types of Vertical Bar Chart . Vertical Stacked Bar Chart; A vertical stacked bar chart is a type of bar graph that uses. IELTS Graphs, charts or diagrams. Academic task 1 requires that you summarise and compare a diagram, chart or graph. You should use at least 150 words for the task. This section is not an essay. It is a summary. You should use formal language and it must factual. Academic task 1 is a report on a bar chart, pie chart, table or diagram and process Bar Graph: A bar graph is a chart that plots data with rectangular bars representing the total amount of data for that category. A bar chart is a style of bar graph; it is often used to represent.
### IELTS Writing Task 1 - Bar Chart/ Grap
Full tutorial to describe a bar graph. Dec 16, 2017 - Learn how to choose the right information, use the correct vocabulary and plan your report. Full tutorial to describe a bar graph. Pinterest. Today. Explore. When the auto-complete results are available, use the up and down arrows to review and Enter to select. Touch device users can explore by touch or with swipe gestures. Log in. Sign up.. This 3D bar chart might look very attractive, but it is also very misleading. There is no scale on the vertical axis, and because of the perspective it looks as though the sales for 1995 were far greater than those for any other year. In fact they were identical to those for 1997. It would be much better to draw a 2D bar chart like the one shown in Figure 31.3 with the appropriate labelling on. You can display information in the form of a bar chart. Bar charts show data in a visual and easy to read way. Survey of pets owned by people in Tregain Street Below are the results of the survey of 'Pets owned by people in Tregain Street' shown as a bar chart: Chart of pets owned by people in Tregain Street Key points • A bar chart needs a title. The title tells you what the bar chart is.
### Describing & presenting graphs / analysis and evalution of
Bar chart analysis is more useful when the bars over a time period are viewed, allowing patterns to be discerned that may forecast future prices with varying degrees of success. The simplest comparison is between 2 consecutive bars. An up-day is when the close is higher than the day before. A down-day is when the close is lower The bar chart was originally developed by Henry L. Gantt in 1917 and is alternatively called a Gantt chart.1 It quickly became popular—especially in the construction industry—because of its ability to graphically represent a project's activities on a time scale. Before a Bar (Gantt) chart can be constructed for a project, the project must be broken into smaller, usually homogeneous. Download Full PDF Package. This paper. A short summary of this paper . 9 Full PDFs related to this paper. READ PAPER. Hotel Organizational Chart, Complete. Download. Hotel Organizational Chart, Complete. Amrit Chhina. Hotel ORGANIZATION CHART FINANCE & ACCOUNTING CONTROLLER ASST. FINANCIAL CONTROLLER EDP MANAGER GENERAL ACCOUNTING PURCHASING LEDGER SUPERVISOR MANAGER F&A CONTROLLER SECRETARY. Before writing an IELTS task 1 bar chart or line graph answer it is important that we analyse the question correctly. Taking a few minutes to do this will help us write a clear answer that fully responds to the question. Just what the examiner wants us to do. The first thing we need to do is decide if the bar chart is static or dynamic. Static means that the data comes from one point in time.
### Bar Chart Exercise: Gap fill to improve your chart writing
Bar graph definition is - a graphic means of quantitative comparison by rectangles with lengths proportional to the measure of the data or things being compared —called also bar chart 569+ Chart Templates in PDF - START DOWNLOADING. Charts are easier to understand that's because charts are generally focused on the graphic than text. And with the use of graphical charts, we can simply display the percentage in a pie chart, or bar chart or we can comprehend easily the data that changes over time with the use of line chart
### Describing & writing about graphs (IELTS) for ESL teachers
Understanding Bar Graphs & Pie Charts. One of the more fascinating ways to record data is to put the data in a graph. Two of the more popular graphs are bar graphs and pie graphs This document provides instructions for creating annual race/ethnicity stacked bar charts. Please note that these instructions use Q1_08 data for the year 2007. • There are six stacked bars in this chart: - Child Population - Allegations - Substantiations - Entries - In Care - Exit charts and graphs 508 compliant. Format This tool provides guidance on making various charts and graphs 508 compliant by providing examples of preferred practices and practical tips. Audience This tool is designed primarily for researchers from the Model Systems that are funded by the National Institute on Disability, Independent Living
Bar charts organized from the highest to the lowest number are called Pareto charts. Advantages and Disadvantages of Bar Chart. Advantages: Bar graph summarises the large set of data in simple visual form. It displays each category of data in the frequency distribution. It clarifies the trend of data better than the table. It helps in estimating the key values at a glance. Disadvantages. Introduction to describing graphs and tables. In many subject areas you may need to refer to numbers, statistics and other data during the course of your studies. This is likely to be data collected by other people which you will use to support your written work, but it may be data that you have collected yourself as part of your studies Useful phrases describing a picture / interpreting a cartoon The beginning of your description • The situation is at / the situation is near • The picture / painting / cartoon introduces us / the viewer / the spectator to • In the / this picture / painting / photo / cartoon / illustration we see / come upon (stoßen auf) / catch a glimpse of (flüchtigen Eindruck von etw.
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• Crypto browser. | 5,667 | 25,678 | {"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 | 0.908282 |
http://mathhelpforum.com/advanced-algebra/43342-questions-about-concepts-linear-algebra.html | 1,527,402,106,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794868003.97/warc/CC-MAIN-20180527044401-20180527064401-00431.warc.gz | 178,252,226 | 9,631 | # Thread: Questions about concepts of linear algebra
1. ## Questions about concepts of linear algebra
1. What does it mean for a vector to form a basis of a subspace?
2. What does it mean when a plane (or subspace) is spanned by a vector?
3. What does it mean when a subspace has a basis?
1. It means that the subspace it is in has dimension 1. Furthermore, the vector is nonzero every element in that subspace is a scalar multiple of that vector.
2. Same as 1. Note that any nonzero vector is linearly independent by itself and so forms a basis for the subspace it spans.
3. If you accept the axiom of choice (which most mathematicians do) – or rather, if you accept Zorn’s lemma, which is equivalent to the axiom of choice – then it follows that every vector space has a basis. A subspace of a vector space is itself a vector space and therefore has a basis. | 206 | 865 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2018-22 | latest | en | 0.94662 |
http://freetofindtruth.blogspot.com/2019/02/27-47-65-67-121-death-of-richard-pryor.html | 1,590,561,867,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347392141.7/warc/CC-MAIN-20200527044512-20200527074512-00197.warc.gz | 52,297,241 | 25,126 | Saturday, February 9, 2019
27 47 65 67 121 | The death of Richard Pryor at 65, December 10, 2005, 9-days after his birthday
Richard Pryor died 9-days after his 65th birthday. Comedy = 65
Keep in mind that 9 represents completion.
The date he died, had a connection to '67'.
December 10 is written 12/10, emphasis on 12/10.
He died on a date with 27 and 47 numerology.
12/10/05 = 12+10+05 = 27 (Ritual = 27)
12/10/2005 = 12+10+20+05 = 47 (Richard = 47)
He dropped dead of a heart attack, unexpectedly. | 175 | 510 | {"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-2020-24 | latest | en | 0.98056 |
https://www.jobilize.com/online/course/introduction-terminology-the-feasible-region-by-openstax?qcr=www.quizover.com | 1,621,126,040,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991488.53/warc/CC-MAIN-20210515223209-20210516013209-00502.warc.gz | 848,438,926 | 20,035 | # Introduction, terminology, the feasible region
Page 1 / 2
## Introduction
In Grade 11 you were introduced to linear programming and solved problems by looking at points on the edges of the feasible region. In Grade 12 you will look at how to solve linear programming problems in a more general manner.
## Terminology
Here is a recap of some of the important concepts in linear programming.
## Feasible region and points
Constraints mean that we cannot just take any $x$ and $y$ when looking for the $x$ and $y$ that optimise our objective function. If we think of the variables $x$ and $y$ as a point $\left(x,y\right)$ in the $xy$ -plane then we call the set of all points in the $xy$ -plane that satisfy our constraints the feasible region . Any point in the feasible region is called a feasible point .
For example, the constraints
$\begin{array}{c}\hfill x\ge 0\\ \hfill y\ge 0\end{array}$
mean that every $\left(x,y\right)$ we can consider must lie in the first quadrant of the $xy$ plane. The constraint
$x\ge y$
means that every $\left(x,y\right)$ must lie on or below the line $y=x$ and the constraint
$x\le 20$
means that $x$ must lie on or to the left of the line $x=20$ .
We can use these constraints to draw the feasible region as shown by the shaded region in [link] .
The constraints are used to create bounds of the solution.
$ax+by=c$ If $b\ne 0$ , feasible points must lie on the line $y=-\frac{a}{b}x+\frac{c}{b}$ If $b=0$ , feasible points must lie on the line $x=c/a$ $ax+by\le c$ If $b\ne 0$ , feasible points must lie on or below the line $y=-\frac{a}{b}x+\frac{c}{b}$ . If $b=0$ , feasible points must lie on or to the left of the line $x=c/a$ .
When a constraint is linear, it means that it requires that any feasible point $\left(x,y\right)$ lies on one side of or on a line. Interpreting constraints as graphs in the $xy$ plane is very important since it allows us to construct the feasible region such as in [link] .
## Linear programming and the feasible region
If the objective function and all of the constraints are linear then we call the problem of optimising the objective function subject to these constraints a linear program . All optimisation problems we will look at will be linear programs.
The major consequence of the constraints being linear is that the feasible region is always a polygon. This is evident since the constraints that define the feasible region all contribute a line segment to its boundary (see [link] ). It is also always true that the feasible region is a convex polygon.
The objective function being linear means that the feasible point(s) that gives the solution of a linear program always lies on one of the vertices of the feasible region . This is very important since, as we will soon see, it gives us a way of solving linear programs.
We will now see why the solutions of a linear program always lie on the vertices of the feasible region. Firstly, note that if we think of $f\left(x,y\right)$ as lying on the $z$ axis, then the function $f\left(x,y\right)=ax+by$ (where $a$ and $b$ are real numbers) is the definition of a plane. If we solve for $y$ in the equation defining the objective function then
$\begin{array}{cc}\hfill & \phantom{\rule{1.em}{0ex}}f\left(x,y\right)=ax+by\\ \hfill \therefore & \phantom{\rule{1.em}{0ex}}y=\frac{-a}{b}x+\frac{f\left(x,y\right)}{b}\end{array}$
What this means is that if we find all the points where $f\left(x,y\right)=c$ for any real number $c$ (i.e. $f\left(x,y\right)$ is constant with a value of $c$ ), then we have the equation of a line. This line we call a level line of the objective function.
how can chip be made from sand
are nano particles real
yeah
Joseph
Hello, if I study Physics teacher in bachelor, can I study Nanotechnology in master?
no can't
Lohitha
where we get a research paper on Nano chemistry....?
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
has a lot of application modern world
Kamaluddeen
yes
narayan
what is variations in raman spectra for nanomaterials
ya I also want to know the raman spectra
Bhagvanji
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
how did you get the value of 2000N.What calculations are needed to arrive at it
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# Project Summary: The local bakery has been using job order costing and is deciding whether to use activity-based costing. The bakery makes many different products, including the one you produced in...
Project Summary:
The local bakery has been using job order costing and is deciding whether to use activity-based costing. The bakery makes many different products, including the one you produced in the job order costing project. In this project, you will calculate the unit product cost under activity-based costing for the same bakery product you produced in the job order costing project.
Project Data:
The bakery has identified the following activity cost pools, cost drivers, and total yearly quantity for the cost drivers, and percentage of total estimated manufacturing overhead cost allocated to each cost pool. The bakery estimates \$50,000 in total manufacturing overhead.
Activity Cost Pool Data
Activity Cost Pool
Cost Driver
Cost Driver Quantity
Materials Ordering and Inspection
Number of purchase orders
1,500 orders
30%
Equipment Setup
Number of setups
750 setups
30%
Quality Control
Number of inspections
2,500 inspections
25%
Other
Number of direct labor hours
10,000 hours
15%
Project Requirements:
1. Calculate total manufacturing overhead cost allocated to each activity cost pool.
2. Calculate the activity rate for each activity cost pool.
3. Estimate the cost driver quantity allocated to your bakery product from the job order costing project.Note: This will not be the entire cost driver quantity as the bakery makes many different products. Determine a reasonable estimate based on your product.
4. Calculate manufacturing overhead allocated in the year to your bakery product.
5. Estimate total units produced in the year and calculate manufacturing overhead per unit.
6. Using direct materials per unit and direct labor per unit from the job order costing project, calculate unit product cost.
7. In a few sentences, explain the differences in unit product cost between job order and activity-based costing.
## Rubric
Activity Based Costing ProjectActivity Based Costing Project
CriteriaRatingsPts
This criterion is linked to a Learning OutcomeCalculation of Activity Rates (6 points)Objectives Calculate activity rates for an activity-based costing systemthreshold:4.0pts6pts
This criterion is linked to a Learning OutcomeCalculation of Total Allocated Manufacturing Overhead (6 points)Objectives Calculate manufacturing overhead allocated to the product under an activity-based costing systemthreshold:4.0pts6pts
This criterion is linked to a Learning OutcomeCalculation of Manufacturing Overhead per Unit (6 points)Objectives Calculate manufacturing overhead per unit under an activity-based costing system.threshold:4.0pts6pts
This criterion is linked to a Learning OutcomeCalculation of Unit Product Cost (6 points)Objectives Calculate unit product cost using activity-based costingthreshold:6.0pts6pts
This criterion is linked to a Learning OutcomeComparison of Job Order and Activity Based Costing (6 points)Objectives Compare product costs computed under job order and activity-based costing systemsthreshold:4.0pts6pts
Total Points:30
Answered 1 days AfterFeb 11, 2022
## Solution
Deepanshu answered on Feb 13 2022
1. Calculation of total manufacturing overhead cost allocated to each activity cost pool:
· Materials Ordering and Inspection = \$50,000 * 30% = \$15,000
· Equipment Setup = \$50,000 * 30% = \$15,000
· Quality Control = \$50,000 * 25% = \$12,500
· Other = \$50,000 * 15% = \$7,500
2. Calculation of the activity rate for each activity cost pool
· Materials Ordering and Inspection = \$15,000 / 1,500 order = \$10 per orde
· Equipment Setup = \$15,000 / 750 setup = \$20 per setup
· Quality Control = \$12,500 / 2,500 inspection = \$5 per inspection
· Other = \$7,500 / 10,000 hours = \$0.75 per hou
3....
SOLUTION.PDF | 859 | 4,003 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2023-06 | latest | en | 0.914755 |
https://www.teacherspayteachers.com/Product/Math-and-Literacy-ActivitiesMonster-Themed-2834008 | 1,532,294,398,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676593586.54/warc/CC-MAIN-20180722194125-20180722214125-00569.warc.gz | 974,511,005 | 17,966 | # Math and Literacy Activities~Monster Themed
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This Halloween math resource is packed full of Halloween fun! Read the description below of all the math games, math activities,
math worksheets, math centers, and sight word games that are included in this resource. Easy prep and rigorous learning standards will have you reaching for these Halloween activities year after year. These printable resources are engaging, no prep, and easy to differentiate!
Monster Party Math Bump
There are four games of bump provided. The first version requires students to add two addends, and the second provides practice adding three addends. The third version focuses on the addition and subtractions strategy of doubles plus one and doubles minus one. The fourth version focuses on doubles plus two and doubles minus two.
Monster Estimation
I go to my local dollar store and look for interesting Halloween containers. I use this as a partner or small group lesson, but it can be whole group lesson led by the teacher too. Estimation is such an important mathematical skill. Repeated exposure to estimation activities leads to much stronger number sense. This estimation activity will compliment any ocean or fish unit that you are teaching. A recording sheet is provided for this activity.
Frightening Words
Adorable vampire spinners take center stage in this high-frequency word bump game. These would be perfect to keep your word work fun and engaging during the Halloween season. I have provided three different game boards. (about, after, again, also another, any ask, walk, work, were, went, way, word, want, every, your, only, their, right, over, very)
Part, Part, Whole
Frankie helps children practice their part, part, whole skills with two levels. Children spin the Frankenstein spinner to find the two addends and then find the sum.
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Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. | 521 | 2,604 | {"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-2018-30 | latest | en | 0.919298 |
https://www.mathhomeworkanswers.org/9918/f-x-3x-2-has-a-domain-of-1-2-3-what-is-the-range-of-this-function | 1,511,150,995,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805911.18/warc/CC-MAIN-20171120032907-20171120052907-00470.warc.gz | 860,566,648 | 11,787 | EOI Algebra 1
f(x) = 3x + 2 domain is {-1,2 3} to find the rang plug the values into the function
f(-1) = 3(-1) +2 = -1
f(2) = 3*2 +2 = 8
f(3) = 9 + 2 = 11
The range is {-1, 11}
answered Dec 12, 2011 by Level 10 User (55,660 points)
f(x) = 3x + 2 domain is {-1,2 3} to find the range, plug the values into the function
f(-1) = 3(-1) +2 = -1
f(2) = 3*2 +2 = 8
f(3) = 9 + 2 = 11
The range is {-1, 11}
Hope this helps
answered Feb 14 by Level 3 User (2,720 points) | 216 | 476 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2017-47 | latest | en | 0.816119 |
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##### 1
${\displaystyle \prod _{k=0}^{n-1}\Gamma \left(z+{\frac {k}{n}}\right)={\sqrt {2\pi }}^{\,n-1}\,n^{{\frac {1}{2}}-nz}\,\Gamma (nz)}$
##### 2
${\displaystyle \prod _{k=1}^{n-1}\Gamma \left({\frac {k}{n}}\right)={\frac {(2\pi )^{\frac {n-1}{2}}}{\sqrt {n}}}}$
##### 3
${\displaystyle \prod _{k=1}^{n-1}\left(1-\xi ^{k}\right)=n\qquad \xi =e^{\frac {2\pi i}{n}}}$
##### 4
${\displaystyle \alpha ^{2n}-2\alpha ^{n}\beta ^{n}\cos(n\theta )+\beta ^{2n}=\prod _{k=0}^{n-1}\left(\alpha ^{2}-2\alpha \beta \,\cos \left(\theta +{\frac {2\pi k}{n}}\right)+\beta ^{2}\right)}$
##### 5
${\displaystyle \prod _{k=0}^{n-1}\left(1+z^{2^{k}}\right)={\frac {1-z^{2^{n}}}{1-z}}\qquad z\neq 1}$
##### 6
${\displaystyle \prod _{k=1}^{n}{\frac {2}{z^{1/2^{k}}+z^{-1/2^{k}}}}=2^{n}\,{\frac {z^{1/2^{n}}-z^{-1/2^{n}}}{z-z^{-1}}}\qquad z\neq \pm 1}$
##### 7
${\displaystyle \prod _{k=0}^{n-1}\sin \left(z+{\frac {k\pi }{n}}\right)={\frac {\sin nz}{2^{n-1}}}}$
##### 8
${\displaystyle \prod _{k=1}^{n-1}\sin \left({\frac {k\pi }{n}}\right)={\frac {n}{2^{n-1}}}}$
##### 9
${\displaystyle \prod _{k=1}^{\lfloor {\frac {n-1}{2}}\rfloor }\tan \left({\frac {k\pi }{n}}\right)=\left\{{\begin{matrix}{\sqrt {n}}&,&n&{\text{ungerade}}\\\\1&,&n&{\text{gerade}}\end{matrix}}\right.}$
##### 10
${\displaystyle |\Gamma (n+ix)|^{2}=\prod _{k=0}^{n-1}(k^{2}+x^{2})\,{\frac {\pi x}{\sinh \pi x}}\qquad n\in \mathbb {Z} ^{\geq 0}\,,\,x\in \mathbb {R} }$
##### 11
${\displaystyle \left|\Gamma \left(n+{\frac {1}{2}}+ix\right)\right|^{2}=\prod _{k=0}^{n-1}\left(x^{2}+\left(k+{\frac {1}{2}}\right)^{2}\right)\,{\frac {\pi }{\cosh \pi x}}\qquad n\in \mathbb {Z} ^{\geq 0}\,,\,x\in \mathbb {R} }$ | 849 | 1,798 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 131, "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-2023-23 | latest | en | 0.25766 |
https://sks23cu.net/dwqa-question/suppose-b-integers | 1,642,630,163,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301592.29/warc/CC-MAIN-20220119215632-20220120005632-00191.warc.gz | 580,959,640 | 11,314 | ## Suppose that a and b are integers
GeometrySuppose that a and b are integers
3 months ago
Suppose that a and b are integers, a ≡ 11(mod 19), and b ≡ 3(mod 19). Find the integer c with 0 <= c <= 18 such that c ≡ a^(3)+4b^(3)(mod 19)
1 Answers
Best Answer
barbrastreisandukconcert2016 Staff answered 3 months ago
Definitions Division algorithm Let a be an integer and d a positive integer. Then there are unique integers q and r with 0 <= r < d such that a=dq+r q is called the quotient and r is called the remainder q=a div d r=a mod d Theorem 5 Let m be a positive integer. If a ≡ b(mod m) and c ≡ d(mod m), then a+c ≡ b+d(mod m) and ac ≡ bd(mod m). Solution a=11(mod 19) b ≡ 3(mod 19) 0 <= c <= 18 Use theorem 5: c ≡ a^(3)+4b^(3)(mod 19) =11^(3)+4*3^(3)(mod 19) =1331+4*27(mod 19) =1331+108(mod 19) =1439(mod 19) =14(mod 19) We then obtain c=14 with 0 <= c <= 18.
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## Load power in class B stage in terms of average power dissipation Solution
STEP 0: Pre-Calculation Summary
Formula Used
load_power = Supply Power-Power Dissipation in Series Transistor
PL = Ps-Pd
This formula uses 2 Variables
Variables Used
Supply Power - Supply power is a device that converts the output from an ac power line to a steady dc output or multiple outputs. (Measured in Watt)
Power Dissipation in Series Transistor - Power Dissipation in Series Transistor (Measured in Watt)
STEP 1: Convert Input(s) to Base Unit
Supply Power: 500 Watt --> 500 Watt No Conversion Required
Power Dissipation in Series Transistor: 6 Watt --> 6 Watt No Conversion Required
STEP 2: Evaluate Formula
Substituting Input Values in Formula
PL = Ps-Pd --> 500-6
Evaluating ... ...
PL = 494
STEP 3: Convert Result to Output's Unit
494 Watt --> No Conversion Required
(Calculation completed in 00.016 seconds)
## < 10+ Class B Output Stage Calculators
Efficiency of class B stage
Total power supply of class B stage
total_power = (2*Peak Voltage*Supply Voltage)/(pi*Load resistance) Go
Load resistance of class B stage
load_resistance = (2*Peak Voltage*Supply Voltage)/(pi*Total power) Go
Power drain from positive sine wave
power = (Peak Voltage*Supply Voltage)/(pi*Load resistance) Go
Power drawn from negative sine wave
power = (Peak Voltage*Supply Voltage)/(pi*Load resistance) Go
Efficiency of the class B output stage
overall_efficiency = (pi/4)*(Peak Voltage/Supply Voltage) Go
Output sinusoid peak voltage of class B stage
Maximum average power from a class B output stage
max_power_developed = 1/2*((Supply Voltage)^2/Load resistance) Go
Average load-power of class B stage
Average power dissipation in class B stage in terms of supply power
power_dissipated = Supply Power-Average load power Go
### Load power in class B stage in terms of average power dissipation Formula
load_power = Supply Power-Power Dissipation in Series Transistor
PL = Ps-Pd
## What is a Class B amplifier?
Class B amplifier is a type of power amplifier where the active device (transistor) conducts only for the one-half cycle of the input signal. Since the active device is switched off for half the input cycle, the active device dissipates less power and hence the efficiency is improved.
## How to Calculate Load power in class B stage in terms of average power dissipation?
Load power in class B stage in terms of average power dissipation calculator uses load_power = Supply Power-Power Dissipation in Series Transistor to calculate the Load Power, The Load power in class B stage in terms of average power dissipation formula is defined as electrical component or portion of a circuit that consumes (active) electric power, such as electrical appliances and lights inside the home. Load Power and is denoted by PL symbol.
How to calculate Load power in class B stage in terms of average power dissipation using this online calculator? To use this online calculator for Load power in class B stage in terms of average power dissipation, enter Supply Power (Ps) and Power Dissipation in Series Transistor (Pd) and hit the calculate button. Here is how the Load power in class B stage in terms of average power dissipation calculation can be explained with given input values -> 494 = 500-6.
### FAQ
What is Load power in class B stage in terms of average power dissipation?
The Load power in class B stage in terms of average power dissipation formula is defined as electrical component or portion of a circuit that consumes (active) electric power, such as electrical appliances and lights inside the home and is represented as PL = Ps-Pd or load_power = Supply Power-Power Dissipation in Series Transistor. Supply power is a device that converts the output from an ac power line to a steady dc output or multiple outputs and Power Dissipation in Series Transistor.
How to calculate Load power in class B stage in terms of average power dissipation?
The Load power in class B stage in terms of average power dissipation formula is defined as electrical component or portion of a circuit that consumes (active) electric power, such as electrical appliances and lights inside the home is calculated using load_power = Supply Power-Power Dissipation in Series Transistor. To calculate Load power in class B stage in terms of average power dissipation, you need Supply Power (Ps) and Power Dissipation in Series Transistor (Pd). With our tool, you need to enter the respective value for Supply Power and Power Dissipation in Series Transistor and hit the calculate button. You can also select the units (if any) for Input(s) and the Output as well.
How many ways are there to calculate Load Power?
In this formula, Load Power uses Supply Power and Power Dissipation in Series Transistor. We can use 10 other way(s) to calculate the same, which is/are as follows -
• power = (Peak Voltage*Supply Voltage)/(pi*Load resistance)
• power = (Peak Voltage*Supply Voltage)/(pi*Load resistance)
• total_power = (2*Peak Voltage*Supply Voltage)/(pi*Load resistance)
• load_resistance = (2*Peak Voltage*Supply Voltage)/(pi*Total power) | 1,205 | 5,359 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2021-25 | latest | en | 0.777186 |
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# Learn How to Use Thermodynamic Tables Effectively with Cengel's Thermodynamics Book - Table PDF 12: Properties of Superheated Water Vapor Explained
## Thermodynamics Cengel 7th Edition Tables PDF 12
Are you looking for a comprehensive and easy-to-understand guide to thermodynamics? Do you want to learn how to use thermodynamic tables to solve various problems and applications? If yes, then you have come to the right place. In this article, I will introduce you to the thermodynamics book by Yunus Cengel and his table PDF 12, which contains the properties of superheated water vapor. By the end of this article, you will have a clear idea of what thermodynamics is, what are thermodynamic tables, and how to use them effectively.
## Introduction
### What is thermodynamics?
Thermodynamics is the branch of physics that deals with the relationships between heat, work, energy, and matter. It studies how these quantities can be transferred and transformed from one system to another, and how they affect the state and behavior of the system. Thermodynamics has many applications in engineering, chemistry, biology, geology, astronomy, and other fields.
### What are thermodynamic tables?
Thermodynamic tables are tables that list the values of various thermodynamic properties for different substances at different conditions. Thermodynamic properties are characteristics of a system that describe its state and behavior, such as pressure, temperature, volume, density, enthalpy, entropy, etc. Thermodynamic tables can help us find these properties without having to perform complicated calculations or measurements.
### Why are thermodynamic tables useful?
Thermodynamic tables are useful because they can help us solve many problems and applications involving thermodynamics. For example, we can use thermodynamic tables to find the amount of heat or work required or produced by a process, the efficiency or performance of a device or system, the phase or state changes of a substance, etc. Thermodynamic tables can also help us understand the physical phenomena and principles behind thermodynamics.
## Thermodynamics Cengel 7th Edition
### Who is Yunus Cengel?
Yunus Cengel is a professor emeritus of mechanical engineering at the University of Nevada, Reno. He is also an author and co-author of several textbooks on thermodynamics, fluid mechanics, heat transfer, and related topics. He has received many awards and honors for his teaching and research excellence.
### What are the features of his thermodynamics book?
The book "Thermodynamics: An Engineering Approach" by Yunus Cengel and Michael Boles is one of the most popular and widely used textbooks on thermodynamics. It covers both classical and modern topics in thermodynamics with an emphasis on engineering applications. It also provides many examples, exercises, figures, tables, and charts to illustrate and reinforce the concepts and methods. Some of the features of his book are:
• It uses a simple and intuitive approach to explain thermodynamics.
• It provides a balanced coverage of both macroscopic and microscopic aspects of thermodynamics.
• It integrates real-world engineering examples and problems throughout the book.
• It includes a rich collection of thermodynamic tables and charts in the appendices.
• It offers online resources and supplements for students and instructors.
### How to access the tables in his book?
The tables in his book are available in both print and digital formats. You can access the print version of the tables in the appendices of the book. You can also access the digital version of the tables online or download them as PDF files from the book's website. The website also provides interactive tables that allow you to enter or select the values of the properties and get the results instantly.
## Table PDF 12: Properties of Superheated Water Vapor
### What is superheated water vapor?
Superheated water vapor is water vapor that is heated above its saturation temperature at a given pressure. Saturation temperature is the temperature at which water vapor and liquid water coexist in equilibrium. When water vapor is heated above its saturation temperature, it becomes superheated and behaves like an ideal gas. Superheated water vapor is commonly used as a working fluid in power plants, turbines, refrigerators, and other devices.
### How to use table PDF 12?
Table PDF 12 lists the values of various thermodynamic properties of superheated water vapor at different pressures and temperatures. The table has four columns: pressure, temperature, specific volume, and specific enthalpy. The table is divided into several sub-tables for different pressure ranges. To use table PDF 12, you need to follow these steps:
• Find the sub-table that corresponds to the pressure of your system.
• Locate the row that matches the temperature of your system.
• Read the values of specific volume and specific enthalpy from the table.
• If the temperature of your system is not listed in the table, you can use interpolation to find the values of specific volume and specific enthalpy.
### What are some examples of using table PDF 12?
Here are some examples of using table PDF 12 to solve thermodynamic problems:
Example 1: Find the specific volume and specific enthalpy of superheated water vapor at 1 MPa and 400C.
Solution: From table PDF 12, we find that the sub-table for 1 MPa covers the temperature range from 300C to 600C. We locate the row for 400C and read the values of specific volume and specific enthalpy from the table. The results are:
• Specific volume = 0.1274 m/kg
• Specific enthalpy = 3238.9 kJ/kg
Example 2: Find the specific volume and specific enthalpy of superheated water vapor at 0.5 MPa and 350C.
Solution: From table PDF 12, we find that the sub-table for 0.5 MPa covers the temperature range from 250C to 500C. However, there is no row for 350C in the table. Therefore, we need to use interpolation to find the values of specific volume and specific enthalpy. We use the following formula for interpolation:
x = x1 + (x2 - x1) * (T - T1) / (T2 - T1)
where x is the property to be interpolated, x1 and x2 are the values of the property at two adjacent temperatures T1 and T2, and T is the temperature of interest. We choose T1 = 300C and T2 = 400C as the two adjacent temperatures in the table. The values of specific volume and specific enthalpy at these temperatures are:
• v1 = 0.2558 m/kg
• v2 = 0.1866 m/kg
• h1 = 3059.6 kJ/kg
• v
## Conclusion
### Summary of the main points
In this article, I have explained what thermodynamics is, what are thermodynamic tables, and how to use them effectively. I have also introduced you to the thermodynamics book by Yunus Cengel and his table PDF 12, which contains the properties of superheated water vapor. I hope you have learned something new and useful from this article.
### Call to action
If you are interested in learning more about thermodynamics and its applications, I highly recommend you to get a copy of the book "Thermodynamics: An Engineering Approach" by Yunus Cengel and Michael Boles. It is a comprehensive and easy-to-understand guide to thermodynamics that covers both classical and modern topics with an emphasis on engineering applications. You can also access the online resources and supplements that come with the book, such as interactive tables, videos, animations, quizzes, etc. You can find the book and the online resources at this link: https://www.mheducation.com/highered/product/thermodynamics-engineering-approach-cengel-boles/M9780073398174.html
#### FAQs
Here are some frequently asked questions about thermodynamics and thermodynamic tables:
Q: What is the difference between saturated and superheated water vapor?
A: Saturated water vapor is water vapor that is in equilibrium with liquid water at a given pressure and temperature. Superheated water vapor is water vapor that is heated above its saturation temperature at a given pressure.
Q: What are the units of specific volume and specific enthalpy?
A: Specific volume is the volume per unit mass of a substance. It has the units of m/kg. Specific enthalpy is the energy per unit mass of a substance. It has the units of kJ/kg.
Q: How can I find the saturation temperature or pressure of water?
A: You can use table PDF 1 or PDF 2 in Cengel's book, which list the saturation temperature and pressure of water for different values of pressure or temperature.
Q: How can I find the properties of compressed liquid water?
A: You can use table PDF 3 in Cengel's book, which lists the properties of compressed liquid water for different values of pressure and temperature.
Q: How can I find the properties of ideal gases?
A: You can use table PDF 4 or PDF 5 in Cengel's book, which list the properties of ideal gases for different values of pressure and temperature. You can also use table PDF 6 or PDF 7, which list the properties of air as an ideal gas for different values of pressure and temperature. 71b2f0854b
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946 Posts · 1,667 Users
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.Can someone help me out with the following problems with some clear and simple approcah to these qustions 4 balls are to be put in 5 boxes.in how many ways can this be done if a)Balls are similar and boxes are different b)balls are ...
Page 1 of 95
A girl has three library cards and seven books of her interest in the library of Mindworkzz, Of these books she would not like to borrow the D.L. book, unless t Quants book is also borrowed. In how many ways can she take the three books to be borrowed (a) 15 (b) 20 (c) 25 (2) 30 58. From a group of 12 dancers, five have to be taken for a show. Among them Radha and 5 Mohan decide either both of them would join or none of them would join. In how many ways can the 5 dancers be chosen? (a) 190 (b) 210 © 278 (d) 372
2 comments
@Krazy_me · 3 karma
1) Say she doesn't need DL book, then she can choose in 6c3 ways = 20 ways Id she she chooses the DL book then in 5c1 ways = 5 ways Total = 20+5 = 25 ways
@Krazy_me · 3 karma
2) 10c5 ways when both (Radha and Mohan) are not chosen 10c5 ways = 252 ways and when both are chosen = 10c3ways = 120 ways => total = 372 ways
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Drona 2016 is IIM Indore’s unique mentorship program designed for candidates to get a sneak-peak into the lives of management students and to develop an understanding about this career option. It consists of interactions with IIM I professors and renowned industrialists during the program, coupled with challenging activities designed to test skills like leadership, team building, strategising etc. It also provides an insight into the 2 year course and career avenues thereafter.For more details visit - facebook.com/drona.iris
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Please help me out with the below scenario,
Suppose I have a set of duplicate numbers such as S={1,2,3,1,2,3}
and out of those I have to select 4 numbers such that their rearrangement is not counted.
For e.g in this case for selection 4 numbers
1,2,3,1
1,2,3,2
1,2,3,3
1,1,2,2
2,2,3,3
1,1,3,3
so total 6 cases are possible.
while selecting 5 numbers
1,2,3,1,2
1,2,3,1,3
1,2,3,2,3
only three cases are possible.
Can some please come up with a solution for the question and the set can be of any number with equal number of duplicates in it.
1 comment
@prerak_bhandari · 1 karma
guys please help me here with this
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xat discution
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In how many ways can 14 Identical toys distributed among 3 boys so that each boy gets at least one toy and no two boy get the same number of toys?
78
60
24
120
6 likes26 answers
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Total number of three friends with moti, sumit and manky put together is 10.If the sum of a reciprocal of a number of bananas with the three friends is 1.What is the difference between the number of bananas with moti and sumit?
2
3
0
none of these
8 answers
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Can anyone share all the basic formulas needed for P&C?
And few basic examples and Few specific repeated difficult questions.?
It will be helpful. Thanks. 😃
Do a job that gets you a name or have a name that gets your job done.
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Five balls of different colors are to be placed in three different boxes such that each box gets at least 1 ball. What is maximum number of ways in which this can be done ?
ans 150 or 180
Arun Sharma Worked out example he has given it has 150 .. shouldn't it be 180 ?!!
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@rockyxxxddd · 5 karma
Apply inclusive exclusive principle
@sudip12386 · 0 karma
150
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1) IxI+IyI+IzI=15- Total no. of integral solutions?
2) xyz=10^6- Total no. of natural no. solutions?
Can someone explain the approach to such questions please?
CAT 2015 93.6 IIFT 2015 94.27 SNAP 2015 98.56 NMAT 210
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@rockyxxxddd · 5 karma
1) straight approach 4n^2+2=902
@rockyxxxddd · 5 karma
2. 13?
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How many ways can the letters of the word VARIETY Be arranged so that exactly two vowels are together?
10 comments
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@sudip12386 · 0 karma
4320??
@khushiahuja12 · 0 karma
720
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Write a comment... | 1,283 | 4,474 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2021-25 | latest | en | 0.940694 |
https://brainly.in/question/246054 | 1,484,772,484,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280319.10/warc/CC-MAIN-20170116095120-00489-ip-10-171-10-70.ec2.internal.warc.gz | 791,418,852 | 10,107 | # In a triangle ABC , angle A=90 ,AB=5cm , AC=12cm.If AD is perpendicular to BC then AD = ?
1
by garraybalaji
• qais
• Content Quality
2015-12-23T21:53:18+05:30
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
area of triangle = (1/2) × base×height
if we take AC as perpendicular then AB will be base.
then area of triangle = (1/2)× 5×12 = 30 cm²
also, by Pythagoras' theorem,
BC² = AC² +AB²
=12² + 5² = 169
BC = 13 cm
now, if AD is perpendicular then BC is the base
area of triangle = (1/2)× BC ×AD | 230 | 760 | {"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-2017-04 | latest | en | 0.894298 |
http://pirate.shu.edu/~wachsmut/complex/numbers/answers/mult_geomD.html | 1,513,504,469,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948595342.71/warc/CC-MAIN-20171217093816-20171217115816-00782.warc.gz | 225,815,922 | 1,684 | ### Example 1.2.14 (d): Multiplying geometrically
To do this in our head we look at radius and angle separately: (blue) |i| = 1 and Arg(i) = /2 (red) |1+i| = and Arg(1+i) = /4 (green) |i-1| = and Arg(i) = 3/4 (gold) |1-i| = and Arg(i) = -/4
According to our results on multiplying and dividing geometrically we have for the lengths:
length of = (1)/() = 1
As for the angles we have:
angle of = /2 + /4 + 3/4 - 2(-/4) = 2
Thus:
= 1 cis(2) = 1
Verifying this algebraically is work - so it's left to you.
Interactive Complex Analysis, ver. 1.0.0
(c) 2006-2007, Bert G. Wachsmuth | 214 | 585 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2017-51 | latest | en | 0.885836 |
https://www.mnlottery.com/blog/what-are-the-odds | 1,716,046,422,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057440.7/warc/CC-MAIN-20240518152242-20240518182242-00696.warc.gz | 812,346,743 | 9,501 | We love to give away prizes and lots of those prizes go to scratch game winners. So, we thought we’d take a look at scratch game odds to help you better understand how it all works.
Game odds are posted on the back of every scratch ticket and are calculated by taking the total number of tickets printed for any particular game divided by the total number of winning tickets.
Our scratch ticket vendors use software that generates a variable data algorithm to ensure that every ticket printed has the same random chance to win. In addition, the lottery requires that the algorithm limit the string of non-winning tickets for most games by three times the odds rounded down to the nearest whole number. In the example above, that would be 3 x 3.47 = 10.41. It would be possible to have several winning tickets, or none at all, in a string of 10 tickets.
One common misconception is that someone knows where the winning tickets are. Nope! Tickets are programmed and printed in a way that nobody — not our vendors, not the lottery, and not our retailer partners — know which tickets are winners until they are scratched by you!
You can find the odds for every scratch game currently on sale at mnlottery.com. Scroll toward the bottom of any scratch game page to find the prizes & odds table. This table will show each prize tier, the odds of winning that prize and the number of tickets printed for that tier of prizes.
So let’s look at the prizes and odds table for the \$500 Cash Crossword scratch game. This table will tell you how often a ticket, on average, is likely to result in a win.
To give you the best chance to win, the lottery stops selling any game when the last top prize has been claimed.
If you want to know how many prizes are currently available in a specific game, visit the unclaimed prizes page.
If you don’t win, you can keep the fun going by entering eligible non-winning tickets for the chance to win a 2nd Chance prize. You could score cash, merchandise or amazing experiences. Just look for tickets with the 2nd Chance logo.
2nd Chance odds are determined by the number of non-winning tickets entered by players with Minnesota Lottery accounts. Do 2nd Chances take away from cash prizes? A common misconception is that the 2nd Chance greatly reduces the amount of cash prizes offered in a scratch game. Not true! Only a very small portion, if any, of the normal cash prizes fund a 2nd Chance. In many cases the 2nd Chance prizes are added to the normal cash payout of a game.
We hope that helps you understand scratch games odds a little better. | 546 | 2,580 | {"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-2024-22 | latest | en | 0.951909 |
https://www.enotes.com/homework-help/physics-30-changing-momentum-basis-contents-this-428592 | 1,516,328,289,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887692.13/warc/CC-MAIN-20180119010338-20180119030338-00276.warc.gz | 923,553,780 | 8,874 | # Physics 30 Changing Momentum On the basis of the contents of this lesson, explain the necessity of bicycle helmets to someone who does not believe they are necessary. Give two reasons.
pramodpandey | Student
We know Newton's second law of motion which describe relation between force and motion.
The rate of change of momentum is defined as force.
When object move it has some velocity and its mass as well.
we define now momentum as mv ,where v is velocity and m is mass of the object.
When bycle rider falls , velocity comes to zero
i.e rate of change of momentum = m(v-0) =mv
which is force. It puts great impact on rider's body.
Rider not falls vertically downword but it describe circular parts,
So another factor comes into effect ti.e angular momentum as well.
To reduce this impact of the force, helmet is designed for bicyle rider. | 189 | 855 | {"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 | 3 | CC-MAIN-2018-05 | latest | en | 0.93979 |
https://ispecially.com/you-have-to-be-smarter-than-an-8th-grader-to-get-this-simple-problem-right/ | 1,721,404,722,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514908.1/warc/CC-MAIN-20240719135636-20240719165636-00156.warc.gz | 285,562,936 | 21,418 | # You Have To Be Smarter Than An 8th Grader To Get This Simple Problem Right
When you look at the math equation below, you probably feel as if it is fairly easy to do. The problem is, most people have a problem getting the answer right and it is because they don’t remember some of the simple things that they learned in middle school. Why not try the math equation for yourself? It really doesn’t matter if you enjoyed math class or not, this is an exercise that you will appreciate when all is said and done.
Can you solve this math?
.
.
.
.
.
.
.
.
If you came up with the answer 6, you are absolutely correct!
According to the order of operations, you would do everything inside of the parentheses first. In other words, one plus one equals two.
The next thing you do is multiplication, so 12×2 = 24.
Finally, you would take care of the subtraction, so 30-24 = 6. | 205 | 881 | {"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} | 3.671875 | 4 | CC-MAIN-2024-30 | latest | en | 0.955403 |
https://www.studypool.com/discuss/1208715/helppppp-me-plsssss-thank-uu?free | 1,508,246,898,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187821189.10/warc/CC-MAIN-20171017125144-20171017145144-00475.warc.gz | 1,002,552,727 | 13,944 | ##### helppppp me plsssss thank uu
label Algebra
account_circle Unassigned
schedule 1 Day
account_balance_wallet \$5
1. (7y^4/3)^2y-^2/3 divided by y^7/3
2. 16 ^ -1/2
3. ( 16/81) ^3/4
Oct 11th, 2015
1. 7^2 y^(8/3) y^(-2/3)/y^(7/3) = 49 y^(8/3-2/3-7/3) = 49 y^(-1/3)
2. 16^(-1/2) = 1/square root (16) = 1/4
3. (16/81) ^3/4 = 16^(3/4) / 81^(3/4) = 8/27
Please let me know if you need any clarification. Always glad to help!
Oct 11th, 2015
...
Oct 11th, 2015
...
Oct 11th, 2015
Oct 17th, 2017
check_circle | 258 | 514 | {"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-2017-43 | latest | en | 0.692752 |
http://www.numbersaplenty.com/1340113551323 | 1,575,582,768,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540482284.9/warc/CC-MAIN-20191205213531-20191206001531-00261.warc.gz | 225,370,188 | 3,542 | Search a number
1340113551323 = 15671981943151
BaseRepresentation
bin10011100000000100111…
…111011111111111011011
311202010001211001222212012
4103200010333133333123
5133424023032120243
62503350100140135
7165551156144531
oct23400477377733
94663054058765
101340113551323
11477380187aa8
1219788139764b
13994ab856621
1448c0cc74151
1524cd588bd18
hex13804fdffdb
1340113551323 has 8 divisors (see below), whose sum is σ = 1341067499520. Its totient is φ = 1339159732200.
The previous prime is 1340113551299. The next prime is 1340113551341. The reversal of 1340113551323 is 3231553110431.
Adding to 1340113551323 its reverse (3231553110431), we get a palindrome (4571666661754).
It is a sphenic number, since it is the product of 3 distinct primes.
It is a cyclic number.
It is not a de Polignac number, because 1340113551323 - 216 = 1340113485787 is a prime.
It is a Duffinian number.
It is a self number, because there is not a number n which added to its sum of digits gives 1340113551323.
It is not an unprimeable number, because it can be changed into a prime (1340113551353) by changing a digit.
It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 31034798 + ... + 31077948.
It is an arithmetic number, because the mean of its divisors is an integer number (167633437440).
Almost surely, 21340113551323 is an apocalyptic number.
1340113551323 is a deficient number, since it is larger than the sum of its proper divisors (953948197).
1340113551323 is a wasteful number, since it uses less digits than its factorization.
1340113551323 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 64537.
The product of its (nonzero) digits is 16200, while the sum is 32.
The spelling of 1340113551323 in words is "one trillion, three hundred forty billion, one hundred thirteen million, five hundred fifty-one thousand, three hundred twenty-three". | 582 | 1,951 | {"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} | 3.234375 | 3 | CC-MAIN-2019-51 | longest | en | 0.814149 |
https://drmoazzam.com/matlab-code-bpsk-modulation-and-demodulation-with-explanation?replytocom=600 | 1,659,931,199,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570765.6/warc/CC-MAIN-20220808031623-20220808061623-00545.warc.gz | 229,559,368 | 15,469 | BPSK Modulation And Demodulation- Complete Matlab Code With Explanation
Binary Phase Shift Keying (BPSK) is a type of digital modulation technique in which we are sending one bit per symbol i.e., ‘0’ or a ‘1’. Hence, the bit rate and symbol rate are the same. Depending upon the message bit, we can have a phase shift of 0o or 180o with respect to a reference carrier as shown in the figure above.
For example, we can have the following transmitted band-pass symbols:
$S_1=\sqrt{\frac{2E}{T}}\cos{(2\pi f t)}\rightarrow represents \mbox{ }'1'$ $S_2=\sqrt{\frac{2E}{T}}\cos{(2\pi f t+\pi)}\rightarrow represents \mbox{ }'0'$
or
$S_2=-\sqrt{\frac{2E}{T}}\cos{(2\pi f t)}\rightarrow represents \mbox{ }'0'$
Where ‘E’ is the symbol energy, ‘T’ is the symbol time period, f is the frequency of the carrier. Using Gram-schmidt orthogonalization, we get a single orthonormal basis function, given as:
$\psi_1=\sqrt{\frac{2}{T}}\cos{(2\pi f t)}$
Hence, the resulting constellation diagram can be given as:
Constellation Diagram Of BPSK Signal
There are only two in-phase components and no quadrature component.
Now, we can easily see that the two waveform of So and S1 are inverted with respect to one another and we can use following scheme to design a BPSK modulator:
BPSK modulator
First the NRZ encoder converts these digital bits into impulses to add a notion of time into them. Then NRZ waveform is generated by up-sampling these impulses. Afterwards, multiplication with the carrier (orthonormal basis function) is carried out to generate the modulated BPSK waveform.
Demodulator Design:
We do coherent demodulation of the BPSK signal at the receiver. Coherent demodulation requires the received signal to be multiplied with the carrier having the same frequency and phase as at the transmitter. The phase synchronization is normally achieved using Phase Locked Loop (PLL) at the receiver. PLL implementation is not done here, rather we assume perfect phase synchronization. Block diagram of BPSK modulator is shown in the figure below. After the multiplication with the carrier (orthonormal basis function), the signal is integrated over the symbol duration ‘T’ and sampled. Then thresholding is applied to determine if a ‘1’ was sent (+ve voltage) or a ‘0’ was sent (-ve voltage).
The Matlab simulation code is given below. Here for the sake of simplicity, the bit rate is fixed to 1 bit/s (i.e., T=1 second). It is also assumed that Phased Locked Loop (PLL) has already achieved exact phase synchronization.
clear all;
close all;
%Nb is the number of bits to be transmitted
T=1;%Bit rate is assumed to be 1 bit/s;
%bits to be transmitted
b=[1 0 1 0 1]
%Rb is the bit rate in bits/second
NRZ_out=[];
%Vp is the peak voltage +v of the NRZ waveform
Vp=1;
%Here we encode input bitstream as Bipolar NRZ-L waveform
for index=1:size(b,2)
if b(index)==1
NRZ_out=[NRZ_out ones(1,200)*Vp];
elseif b(index)==0
NRZ_out=[NRZ_out ones(1,200)*(-Vp)];
end
end
%Generated bit stream impulses
figure(1);
stem(b);
xlabel('Time (seconds)-->')
ylabel('Amplitude (volts)-->')
title('Impulses of bits to be transmitted');
figure(2);
plot(NRZ_out);
xlabel('Time (seconds)-->');
ylabel('Amplitude (volts)-->');
title('Generated NRZ signal');
t=0.005:0.005:5;
%Frequency of the carrier
f=5;
%Here we generate the modulated signal by multiplying it with
%carrier (basis function)
Modulated=NRZ_out.*(sqrt(2/T)*cos(2*pi*f*t));
figure;
plot(Modulated);
xlabel('Time (seconds)-->');
ylabel('Amplitude (volts)-->');
title('BPSK Modulated signal');
y=[];
%We begin demodulation by multiplying the received signal again with
%the carrier (basis function)
demodulated=Modulated.*(sqrt(2/T)*cos(2*pi*f*t));
%Here we perform the integration over time period T using trapz
%Integrator is an important part of correlator receiver used here
for i=1:200:size(demodulated,2)
y=[y trapz(t(i:i+199),demodulated(i:i+199))];
end
figure;
xlabel('Time (seconds)-->');
ylabel('Amplitude (volts)')
Impulses of bits to be transmitted
Generated NRZ signal
BPSK Modulated Signal
If you have any comments or questions, you can discuss them below.
• Jo August 13, 2016, 8:04 am
scatter(); %error!! -> Not enough input arguments……
• Jo August 13, 2016, 8:13 am
Thank you very much, that is very helpful!
but, there is one problem….
scatter(); % error !! -> Not enough input arguments……..
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• Imran January 18, 2017, 11:42 am
Hi Dr. Moazzam,
First of all, thanks for this. I was looking for an easy to understand BPSK Matlab implementation so was glad I found this. I just have a question on the NRZ_out and Modulated plots. Shouldn’t the xlabel be samples instead of time? Because I don’t think the NRZ stream could have changed the frequency of the carrier which is what the plot seems to suggest. Hope to hear from you and thanks again.
• smokingRooster March 13, 2017, 9:44 pm
Hi, quick question.
You specify a frequency of 5hz…but your graph shows one of about 50. Could you explain why that is?
thank you
• sepiatone June 18, 2021, 10:05 am
That’s just an artifact of the plotting – plot(t, Modulated) and you would see a frequency of 5Hz.
• Mustafa April 28, 2017, 7:05 pm
Problem 1.
I want to Write a code in Matlab which creates a constant envelop PSK signal waveform that generate for M=8 (M stands for modulation), so that amplitude of the signal can reach up to sqrt(2). I want to Plot a graph which showing that there is no difference except in their phases
Problem 2.
I want to Write a code in Matlab which will generate a 500 random numbers to represent our symbols; and then divide them into 4 intervals. Whereby each interval corresponds to a symbol A0, A1, A2, A3, then plot a stem of 50 random symbols generated in accordance to the interval division.
• sudar May 25, 2017, 10:51 am
i have error in scatter plot. while BPSK modulation. whether anyone help me to correct scatterplot.
• hh December 8, 2017, 4:08 pm
i don’t understand this part
for i=1:200:size(demodulated,2)
y=[y trapz(t(i:i+199),demodulated(i:i+199))];
end
• Qadeer June 14, 2018, 10:25 pm
The symbol duration is 200, here we integrate the demodulated signal with respect to i each of size 200. hope that helps!
• Arun March 8, 2018, 9:02 am
how to make bpsk program run repeatedly?
• sam kumar October 25, 2018, 3:14 pm
this code not giving the propar phase shift .
• Devika November 25, 2018, 8:08 am
• MR PAWEL J PASZYNSKI January 28, 2019, 8:36 pm
Hi Moazzam I love the code it is Tidy and very well commented, my question is: Since C++ is included within the usual MATLAB commands code how can the corresponding Simulink Block Model can be generated is there a simple way of achieving the Block Diagram at all??
Thank You
• sushil May 6, 2019, 8:33 am
Thank you so much… I was searching for a code …. I finally found it
• ritu May 7, 2019, 7:31 am | 1,983 | 7,224 | {"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": 4, "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-2022-33 | latest | en | 0.820212 |
http://www.csudh.edu/math/gjennings/403sp14/calendar403sp14.html | 1,432,509,730,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207928102.74/warc/CC-MAIN-20150521113208-00040-ip-10-180-206-219.ec2.internal.warc.gz | 400,744,478 | 2,774 | # MAT 403 Calendar Spring 2014
Updated Tue Jan 21 15:20:19 PST 2014
This is a tentative calendar. I probably will have to move some things as we go along but I'll try to avoid moving the test days so you can plan ahead.
I will post homework assignments here so please check back often, at least once a week.
Chapter and section numbers refer to our textbooks
### Lectures and assignments
• 1st week, Jan. 20, 22
• Mon. Jan. 22 Rev. Martin Luther King Holiday
• Outline of the course: Integration, the Existence and Uniqueness Theorem of Ordinary Differential Equations (Picard's theorem), multivariable calculus and its connection with linear algebra, the Implicit Function Theorem and the Inverse Function theorem.
• 2nd week, Jan. 27, 29
• [Lebl] 5.1 The Riemann Integral
• 3rd week, Feb. 3, 5
• [Lebl] 3.4 Uniform continuity
• Homework Sec. 3.4 exercise 3.4.8, Sec. 5.1 exercise 5.1.11. Due Monday Feb 10.
• [Lebl] 5.3 Fundamental theorem of calculus.
• [Trench] Lemma 3.2.4 p. 131: Riemann's and Darboux' integrals are the same.
• [Lebl] 5.2 Properties of the Riemann Integral (lightly)
• [Lebl] 6.1 Sequences of functions: Pointwise and Uniform convergence. Example where continuous functions converge pointwise to discontinuous function: Fourier series for a step function.
• Homework: 6.1.2, 6.1.5, 6.1.6 (I found this one really surprising), 6.1.10, 6.2.1. Due Wednesday Feb. 12
• 4th week, Feb. 10, 12
• [Lebl] 6.2 Interchange of limits.
• [Lebl] 6.3 Picard's Theorem: existence and uniqueness of solutions of ordinary differential equations
• Slope field for van der Pol equation. Also see the Wikipedia article "van der Pol oscillator".
• 5th week, Feb. 17, 19
• Mon. Feb 17 Presidents' Day Holiday
• 6th week, Feb. 24, 26
• [Lebl] 6.3 Picard's Theorem: existence and uniqueness of solutions to ordinary differential equations
• Homework: #5.1.3, 5.1.4, 5.2.4, 5.2.5, 5.2.6. Due next Wednesday.
• 7th week, Mar. 3, 5
• Cancelled: Midterm Exam Monday March 3
• Homework 6.3.4 (easy!)
• Start chapter 7 (in the same book we have been using) Metric Spaces
• More homework (due in about a week). Exercises 7.1.3, 7.1.4, 7.1.6, 7.1.7
• 8th week, Mar. 10, 12
• [Trench] 5.1 Structure of $$\mathbb{R}^n$$. Vectors, open sets, Heine-Borel theorem.
• [Trench] 5.2 Continuous Real-valued functions of n variables.
• Homework: 5.1 exercises 7a, 9a, 12, 13, 19d, 24, 26, 29, 30, 5.2 exercise 3
• 9th week, Mar. 17, 19
• [Trench] 5.2 Continuous Real-valued functions of n variables.
• [Trench] 5.3 Partial derivatives and the differential. (Real-valued functions).
• 10th week, Mar. 24, 26
• [Trench] 5.3 Partial derivatives and the differential. (Real-valued functions).
• [Trench] 5.4 Chain rule, Taylor's theorem, Max-min. (Real-valued functions.)
• Mar. 31. Apr. 4 Spring Recess
• 11th week, Apr. 7, 9
• [Trench] 5.4 Chain rule, Taylor's theorem, Max-min. (Real-valued functions.)
• [Trench] 6.1 Linear Transformations and Matrices. (Linear algebra.)
• 12th week, Apr. 14, 16
• [Trench] 6.1 Linear Transformations and Matrices. (Linear algebra.)
• [Trench] 6.2 Differentiability, Chain Rule. (Vector-valued functions)
• Homework 6.2 exercises 5, 12ab, 20a
• 13th week, Apr. 21, 23
• [Trench] 6.2 Differentiability, Chain Rule. (Vector-valued functions)
• [Trench] 6.3 Inverse Function Theorem
• 14th week, Apr. 28, 30
• [Trench] 6.3 Inverse Function Theorem
• Homework Exercise 6.3 exercise 16. Hint: if you use complex variables then function is $$F(z)=z^2$$ because if $$z=(x+iy)$$ then $$z^2=(x^2-y^2)+i(2xy)=u(x,y)+iv(x,y)$$ where $$u$$ and $$v$$ are the functions in this problem.
• 15th week, May 5, 7
• [Trench] 6.3 Implicit Function Theorem
• Homework Exercise 6.4 #8
• Review
• Final exams week
• Final Exam Monday May 12, 4-6pm | 1,284 | 3,745 | {"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} | 2.8125 | 3 | CC-MAIN-2015-22 | latest | en | 0.839119 |
https://www.numbersaplenty.com/61055055111 | 1,632,137,513,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057036.89/warc/CC-MAIN-20210920101029-20210920131029-00626.warc.gz | 936,934,371 | 3,384 | Search a number
61055055111 = 333296113453
BaseRepresentation
bin111000110111001010…
…100011010100000111
312211121000212011021110
4320313022203110013
52000020043230421
644014233420103
74261000256061
oct706712432407
9184530764243
1061055055111
1123990a94636
12b9bb26a633
1359b020265b
142d52a52731
1518c51aba76
hexe372a3507
61055055111 has 8 divisors (see below), whose sum is σ = 81431207280. Its totient is φ = 40691136512.
The previous prime is 61055055097. The next prime is 61055055163. The reversal of 61055055111 is 11155055016.
61055055111 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a sphenic number, since it is the product of 3 distinct primes.
It is a cyclic number.
It is not a de Polignac number, because 61055055111 - 222 = 61050860807 is a prime.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (61055055181) by changing a digit.
It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 3046740 + ... + 3066713.
It is an arithmetic number, because the mean of its divisors is an integer number (10178900910).
Almost surely, 261055055111 is an apocalyptic number.
61055055111 is a deficient number, since it is larger than the sum of its proper divisors (20376152169).
61055055111 is a wasteful number, since it uses less digits than its factorization.
61055055111 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 6116785.
The product of its (nonzero) digits is 3750, while the sum is 30.
The spelling of 61055055111 in words is "sixty-one billion, fifty-five million, fifty-five thousand, one hundred eleven". | 523 | 1,764 | {"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-2021-39 | latest | en | 0.821148 |
https://www.geeksforgeeks.org/linear-regression-using-tensorflow/?ref=lbp | 1,701,437,969,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100287.49/warc/CC-MAIN-20231201120231-20231201150231-00545.warc.gz | 880,975,446 | 54,602 | # Linear Regression Using Tensorflow
We will briefly summarize Linear Regression before implementing it using TensorFlow. Since we will not get into the details of either Linear Regression or Tensorflow, please read the following articles for more details:
## Brief Summary of Linear Regression
Linear Regression is a very common statistical method that allows us to learn a function or relationship from a given set of continuous data. For example, we are given some data points of x and corresponding y and we need to learn the relationship between them which is called a hypothesis.
In the case of Linear regression, the hypothesis is a straight line, i.e, Where w is a vector called Weights and b is a scalar called Bias. The Weights and Bias are called the parameters of the model.
All we need to do is estimate the value of w and b from the given set of data such that the resultant hypothesis produces the least cost J which is defined by the following cost function where m is the number of data points in the given dataset. This cost function is also called Mean Squared Error.
For finding the optimized value of the parameters for which J is minimum, we will be using a commonly used optimizer algorithm called Gradient Descent. Following is the pseudo-code for Gradient Descent:
Repeat until Convergence { w = w – α * δJ/δw b = b – α * δJ/δb}where α is a hyperparameter called the Learning Rate.
Linear regression is a widely used statistical method for modeling the relationship between a dependent variable and one or more independent variables. TensorFlow is a popular open-source software library for data processing, machine learning, and deep learning applications. Here are some advantages and disadvantages of using Tensorflow for linear regression:
Scalability: Tensorflow is designed to handle large datasets and can easily scale up to handle more data and more complex models.
Flexibility: Tensorflow provides a flexible API that allows users to customize their models and optimize their algorithms.
Performance: Tensorflow can run on multiple GPUs and CPUs, which can significantly speed up the training process and improve performance.
Integration: Tensorflow can be integrated with other open-source libraries like Numpy, Pandas, and Matplotlib, which makes it easier to preprocess and visualize data.
Complexity: Tensorflow has a steep learning curve and requires a good understanding of machine learning and deep learning concepts.
Computational resources: Running Tensorflow on large datasets requires high computational resources, which can be expensive.
Debugging: Debugging errors in Tensorflow can be challenging, especially when working with complex models.
Overkill for simple models: Tensorflow can be overkill for simple linear regression models and may not be necessary for smaller datasets.
Overall, using Tensorflow for linear regression has many advantages, but it also has some disadvantages. When deciding whether to use Tensorflow or not, it is essential to consider the complexity of the model, the size of the dataset, and the available computational resources.
## Tensorflow
Tensorflow is an open-source computation library made by Google. It is a popular choice for creating applications that require high-end numerical computations and/or need to utilize Graphics Processing Units for computation purposes. These are the main reasons due to which Tensorflow is one of the most popular choices for Machine Learning applications, especially Deep Learning. It also has APIs like Estimator which provide a high level of abstraction while building Machine Learning Applications. In this article, we will not be using any high-level APIs, rather we will be building the Linear Regression model using low-level Tensorflow in the Lazy Execution Mode during which Tensorflow creates a Directed Acyclic Graph or DAG which keeps track of all the computations, and then executes all the computations done inside a Tensorflow Session.
## Implementation
We will start by importing the necessary libraries. We will use Numpy along with Tensorflow for computations and Matplotlib for plotting.
## Python3
import numpy as np import tensorflow as tf import matplotlib.pyplot as plt
In order to make the random numbers predictable, we will define fixed seeds for both Numpy and Tensorflow.
## Python3
np.random.seed(101)
Now, let us generate some random data for training the Linear Regression Model.
## Python3
# Generating random linear data # There will be 50 data points ranging from 0 to 50 x = np.linspace(0, 50, 50) y = np.linspace(0, 50, 50) # Adding noise to the random linear data x += np.random.uniform(-4, 4, 50) y += np.random.uniform(-4, 4, 50) n = len(x) # Number of data points
Let us visualize the training data.
## Python3
# Plot of Training Data plt.scatter(x, y) plt.xlabel('x') plt.ylabel('y') plt.title("Training Data") plt.show()
Output:
Now we will start creating our model by defining the placeholders X and Y, so that we can feed our training examples X and Y into the optimizer during the training process.
## Python3
X = tf.placeholder("float") Y = tf.placeholder("float")
Now we will declare two trainable Tensorflow Variables for the Weights and Bias and initializing them randomly using np.random.randn().
## Python3
W = tf.Variable(np.random.randn(), name = "W") b = tf.Variable(np.random.randn(), name = "b")
Now we will define the hyperparameters of the model, the Learning Rate and the number of Epochs.
## Python3
learning_rate = 0.01 training_epochs = 1000
Now, we will be building the Hypothesis, the Cost Function, and the Optimizer. We won’t be implementing the Gradient Descent Optimizer manually since it is built inside Tensorflow. After that, we will be initializing the Variables.
## Python3
# Hypothesis y_pred = tf.add(tf.multiply(X, W), b) # Mean Squared Error Cost Function cost = tf.reduce_sum(tf.pow(y_pred-Y, 2)) / (2 * n) # Gradient Descent Optimizer optimizer = tf.train.GradientDescentOptimizer(learning_rate).minimize(cost) # Global Variables Initializer init = tf.global_variables_initializer()
Now we will begin the training process inside a Tensorflow Session.
## Python3
# Starting the Tensorflow Session with tf.Session() as sess: # Initializing the Variables sess.run(init) # Iterating through all the epochs for epoch in range(training_epochs): # Feeding each data point into the optimizer using Feed Dictionary for (_x, _y) in zip(x, y): sess.run(optimizer, feed_dict = {X : _x, Y : _y}) # Displaying the result after every 50 epochs if (epoch + 1) % 50 == 0: # Calculating the cost a every epoch c = sess.run(cost, feed_dict = {X : x, Y : y}) print("Epoch", (epoch + 1), ": cost =", c, "W =", sess.run(W), "b =", sess.run(b)) # Storing necessary values to be used outside the Session training_cost = sess.run(cost, feed_dict ={X: x, Y: y}) weight = sess.run(W) bias = sess.run(b)
Output:
Now let us look at the result.
## Python3
# Calculating the predictions predictions = weight * x + bias print("Training cost =", training_cost, "Weight =", weight, "bias =", bias, '\n')
Output:
Note that in this case both the Weight and bias are scalars. This is because, we have considered only one dependent variable in our training data. If we have m dependent variables in our training dataset, the Weight will be an m-dimensional vector while bias will be a scalar.
Finally, we will plot our result.
## Python3
# Plotting the Results plt.plot(x, y, 'ro', label ='Original data') plt.plot(x, predictions, label ='Fitted line') plt.title('Linear Regression Result') plt.legend() plt.show()
Output:
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http://www.mathworks.com/matlabcentral/fileexchange/2891-using-s-parameters-in-matlab-simulink/content/zmatch_4.m | 1,369,338,490,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368703728865/warc/CC-MAIN-20130516112848-00020-ip-10-60-113-184.ec2.internal.warc.gz | 591,410,201 | 5,485 | Code covered by the BSD License
### Highlights fromUsing S-Parameters in MATLAB & Simulink
from Using S-Parameters in MATLAB & Simulink by Dick Benson
Example of rf amplifier design using S parameters in both MATLAB and Simulink.
zmatch_4(r1_v,r2_v,Q,f)
```function [L1,L2,C1,C2] = zmatch_4(r1_v,r2_v,Q,f)
% [L1 L2 C1 C2 ] = zmatch_4(r1_v,r2_v,Q,f)
% RCA SP-50 03-1967, pp262 (cicuit c)
%
% ----L1 --------C1---- ------L2-----
% | | |
% R1 C2 R2
% | | |
% -----------------------------------
%
% R1 > R2
%
% Copywrite 2002-2010 The MathWorks, Inc.
syms r1 r2 xl1 xl2 xc1 xc2
if r1_v > r2_v
% proceed
else
error('In zmatch_4.m', 'r1 should be > r2')
end;
z = r2 + j*xl2;
y = j/xc2 + 1/z;
z = 1/y - j*xc1 + j*xl1;
r2 = r2_v;
r1 = r1_v;
xl1 = Q*r1;
xl2 = (r2/Q)*(sqrt(r1*(Q^2 + 1)/r2) -1);
xc1 = (r1*(Q^2+1)/Q)*(1 - sqrt(r2/(r1*(Q^2+1))));
xc2 = (r1/Q)*sqrt(r2*(Q^2+1)/r1);
w = 2*pi*f;
L1 = xl1/w;
L2 = xl2/w;
C1 = 1/(w*xc1);
C2 = 1/(w*xc2);
``` | 434 | 1,053 | {"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-2013-20 | latest | en | 0.420665 |
http://www.vitutor.com/arithmetic/real_numbers/real_numbers.html | 1,386,821,615,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164452243/warc/CC-MAIN-20131204134052-00034-ip-10-33-133-15.ec2.internal.warc.gz | 602,505,852 | 5,646 | # Real Numbers
The set formed by the rational numbers and irrational numbers is the set of real numbers, and is designated by .
With real numbers all operations can be performed, except for the root of an even index and negative radicand, and division by zero.
## The Real Line
For any real number there is a point on the straight line and for every point on the straight line, a real number.
## Representation of Real Numbers
The real numbers can be represented in the straight line with as much approximation as needed, but there are cases in which they can be represented in exact form.
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https://byjus.com/question-answer/s-1-dfrac-1-2-dfrac-1-6-dfrac-1-12-dfrac-1-20-cdots-1/ | 1,709,329,916,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475701.61/warc/CC-MAIN-20240301193300-20240301223300-00240.warc.gz | 146,609,781 | 45,898 | 0
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## The correct option is C 2 S1=12+16+112+120+⋯∞=∞∑n=11n(n+1)=∞∑n=1(1n−1n+1)=limn→∞(11−12)+(12−13)+(13−14)+⋯+(1n−1n+1)=limn→∞(1−1n+1)∴S1=1−limn→∞1n+1=1 S2=13+152+133+154+135+156+⋯∞=(13+133+135+⋯∞)+(152+154+156+⋯∞)=131−132+1521−152=38+124=512∴S1−2S2=1−2×512=212∴a=2
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# Last year, the price of a jar of peanut butter at a certain store was
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Last year, the price of a jar of peanut butter at a certain store was [#permalink]
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15 Jun 2017, 05:03
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Last year, the price of a jar of peanut butter at a certain store was P dollars and the price of a jar of jelly at the same store was J dollars, where J=2P. This year, the price of peanut butter increased by 20% and the price of jelly decreased by 20%. If a customer purchases one jar of peanut butter and one jar of jelly this year and receives a 25% discount, which of the following represents the amount that he paid, in terms of P?
A. 3P/2
B. 17P/10
C. 19P/10
D. 21P/10
E. 11P/5
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Last year, the price of a jar of peanut butter at a certain store was [#permalink]
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15 Jun 2017, 05:52
Bunuel wrote:
Last year, the price of a jar of peanut butter at a certain store was P dollars and the price of a jar of jelly at the same store was J dollars, where J=2P. This year, the price of peanut butter increased by 20% and the price of jelly decreased by 20%. If a customer purchases one jar of peanut butter and one jar of jelly this year and receives a 25% discount, which of the following represents the amount that he paid, in terms of P?
A. 3P/2
B. 17P/10
C. 19P/10
D. 21P/10
E. 11P/5
Peanut butter = \$P
Price of Peanut butter increased by 20% = 120% of P = 120/100 * P = 6/5 * P
Jelly = \$J ---- (Where J =2P)
Price of Jelly decreased by 20% = 80% of J = 80/100 * J = 4/5 *J = 4/5(2P) = 8/5*P
Price of one jar of Peanut butter and one jar of Jelly = 6/5*P + 8/5*P = (6P + 8P)/5 = 14P/5
Customer got 25% discount. ie; 75% of 14P/5 = 75/100 * 14P/5
3/4 * 14P/5 = (3*14/4*5)P = (42/20)P = (21/10) P or 21P/10. Answer D...
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Re: Last year, the price of a jar of peanut butter at a certain store was [#permalink]
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15 Jun 2017, 06:10
Let P=100, thus J=200, last year.
So this year P = 120 (20% increase), and J = 160 (20% decrease). Price paid this year by customer = 120+160 = 280.
But after 25% discount, price = 75% of 280 = 3/4 * 280 = 210.
So we have to look for an option where if we put P=100, we should get answer as '210'.
That is only option D. Hence D answer
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Re: Last year, the price of a jar of peanut butter at a certain store was [#permalink]
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17 Jun 2017, 05:08
1
Bunuel wrote:
Last year, the price of a jar of peanut butter at a certain store was P dollars and the price of a jar of jelly at the same store was J dollars, where J=2P. This year, the price of peanut butter increased by 20% and the price of jelly decreased by 20%. If a customer purchases one jar of peanut butter and one jar of jelly this year and receives a 25% discount, which of the following represents the amount that he paid, in terms of P?
A. 3P/2
B. 17P/10
C. 19P/10
D. 21P/10
E. 11P/5
We are given that last year, the price of a jar of peanut butter at a certain store was P dollars and the price of a jar of jelly at the same store was J dollars, where J = 2P.
Since this year the peanut butter increased by 20%, the new price is 1.2P.
Since the price of jelly decreased by 20%, the new price is 0.8J = 0.8(2P) = 1.6P.
Thus, the regular price of one jar of peanut butter and one jar of jelly is 1.2P + 1.6P = 2.8P.
Thus, at a 25% discount, the price is 0.75(2.8P) = 2.1P = 21P/10
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Re: Last year, the price of a jar of peanut butter at a certain store was [#permalink]
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Re: Last year, the price of a jar of peanut butter at a certain store was [#permalink] 20 Jan 2019, 20:24
Display posts from previous: Sort by | 1,684 | 5,607 | {"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-2019-43 | latest | en | 0.907272 |
https://www.convertunits.com/from/quarter+(ton)/to/kilo | 1,591,232,335,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347436828.65/warc/CC-MAIN-20200604001115-20200604031115-00313.warc.gz | 666,149,240 | 10,866 | ## ››Convert quarter (ton) [US] to kilo
quarter (ton) kilo
How many quarter (ton) in 1 kilo? The answer is 0.0044092452436976.
We assume you are converting between quarter (ton) [US] and kilo.
You can view more details on each measurement unit:
quarter (ton) or kilo
The SI base unit for mass is the kilogram.
1 kilogram is equal to 0.0044092452436976 quarter (ton), or 1 kilo.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between quarter (ton) [US] and kilos.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of quarter (ton) to kilo
1 quarter (ton) to kilo = 226.79619 kilo
2 quarter (ton) to kilo = 453.59237 kilo
3 quarter (ton) to kilo = 680.38856 kilo
4 quarter (ton) to kilo = 907.18474 kilo
5 quarter (ton) to kilo = 1133.98093 kilo
6 quarter (ton) to kilo = 1360.77711 kilo
7 quarter (ton) to kilo = 1587.5733 kilo
8 quarter (ton) to kilo = 1814.36948 kilo
9 quarter (ton) to kilo = 2041.16567 kilo
10 quarter (ton) to kilo = 2267.96185 kilo
## ››Want other units?
You can do the reverse unit conversion from kilo to quarter (ton), or enter any two units below:
## Enter two units to convert
From: To:
## ››Definition: Kilo
The kilogram or kilogramme, (symbol: kg) is the SI base unit of mass. A gram is defined as one thousandth of a kilogram. Conversion of units describes equivalent units of mass in other systems.
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 565 | 1,929 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2020-24 | latest | en | 0.823906 |
https://metanumbers.com/216153 | 1,638,361,859,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964360803.0/warc/CC-MAIN-20211201113241-20211201143241-00223.warc.gz | 456,265,510 | 7,536 | # 216153 (number)
216,153 (two hundred sixteen thousand one hundred fifty-three) is an odd six-digits composite number following 216152 and preceding 216154. In scientific notation, it is written as 2.16153 × 105. The sum of its digits is 18. It has a total of 5 prime factors and 24 positive divisors. There are 119,232 positive integers (up to 216153) that are relatively prime to 216153.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 6
• Sum of Digits 18
• Digital Root 9
## Name
Short name 216 thousand 153 two hundred sixteen thousand one hundred fifty-three
## Notation
Scientific notation 2.16153 × 105 216.153 × 103
## Prime Factorization of 216153
Prime Factorization 32 × 7 × 47 × 73
Composite number
Distinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 5 Total number of prime factors rad(n) 72051 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 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 216,153 is 32 × 7 × 47 × 73. Since it has a total of 5 prime factors, 216,153 is a composite number.
## Divisors of 216153
24 divisors
Even divisors 0 24 12 12
Total Divisors Sum of Divisors Aliquot Sum τ(n) 24 Total number of the positive divisors of n σ(n) 369408 Sum of all the positive divisors of n s(n) 153255 Sum of the proper positive divisors of n A(n) 15392 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 464.923 Returns the nth root of the product of n divisors H(n) 14.0432 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 216,153 can be divided by 24 positive divisors (out of which 0 are even, and 24 are odd). The sum of these divisors (counting 216,153) is 369,408, the average is 15,392.
## Other Arithmetic Functions (n = 216153)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 119232 Total number of positive integers not greater than n that are coprime to n λ(n) 1656 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 19250 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 119,232 positive integers (less than 216,153) that are coprime with 216,153. And there are approximately 19,250 prime numbers less than or equal to 216,153.
## Divisibility of 216153
m n mod m 2 3 4 5 6 7 8 9 1 0 1 3 3 0 1 0
The number 216,153 is divisible by 3, 7 and 9.
• Arithmetic
• Deficient
• Polite
• Triangle
• Hexagonal
## Base conversion (216153)
Base System Value
2 Binary 110100110001011001
3 Ternary 101222111200
4 Quaternary 310301121
5 Quinary 23404103
6 Senary 4344413
8 Octal 646131
10 Decimal 216153
12 Duodecimal a5109
20 Vigesimal 1707d
36 Base36 4ms9
## Basic calculations (n = 216153)
### Multiplication
n×y
n×2 432306 648459 864612 1080765
### Division
n÷y
n÷2 108076 72051 54038.2 43230.6
### Exponentiation
ny
n2 46722119409 10099126276613577 2182956442068854509281 471852583822509108744615993
### Nth Root
y√n
2√n 464.923 60.0142 21.5621 11.6668
## 216153 as geometric shapes
### Circle
Diameter 432306 1.35813e+06 1.46782e+11
### Sphere
Volume 4.23031e+16 5.87127e+11 1.35813e+06
### Square
Length = n
Perimeter 864612 4.67221e+10 305687
### Cube
Length = n
Surface area 2.80333e+11 1.00991e+16 374388
### Equilateral Triangle
Length = n
Perimeter 648459 2.02313e+10 187194
### Triangular Pyramid
Length = n
Surface area 8.09251e+10 1.19019e+15 176488 | 1,258 | 3,819 | {"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} | 3.671875 | 4 | CC-MAIN-2021-49 | latest | en | 0.798401 |
https://printablesudokufree.com/printable-binary-sudoku/crack-the-code-binary-code-5-bit-challenge-printable-binary-sudoku-2/ | 1,695,538,775,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506623.27/warc/CC-MAIN-20230924055210-20230924085210-00824.warc.gz | 526,351,564 | 8,988 | # Crack The Code – Binary Code 5 Bit Challenge | Printable Binary Sudoku
Crack The Code – Binary Code 5-Bit Challenge | Printable Binary Sudoku
Printable Binary Sudoku – printable binary sudoku, An exciting interest that exercise routines your mind and offers you a feeling of success – that is the right information of your figures puzzle fad from Japan referred to as sudoku. This excellent head video game is exciting to experience, and everyone can undertake it. Enjoying may even assist wait dementia by training the human brain to imagine in a new way.
## Printable Binary Sudoku
You have possibly noticed a sudoku with your neighborhood newspapers. It is a puzzle that workouts your mind by permitting you to definitely feel rationally concerning how to spot figures in bins in just a grid. The aim of sudoku is just not to perform repeatedly any amounts when you are satisfying inside the grid. The task is finding out which amount suits which box. The sudoku grid has 9 bins; inside every box are more compact cases which have 9 pieces every single. Some of the squares will have figures with them.
### What exactly is a Sudoku?
Sudoku is really a reasonable puzzle video game, in the beginning produced in puzzle guides then offered in a great number of magazines globally. Lots of people are postpone by visiting a grid with phone numbers inside it, however the puzzle will not call for any arithmetic in any way – just deduction and common sense.
Sudoku is a straightforward to find out reasoning-dependent variety location puzzle. The term Sudoku is simple for Su-ji wa dokushin ni kagiru meaning “the phone numbers has to be solitary”.
The origins of your Sudoku puzzle will be in the Swiss. Leonhard Euler produced “carré latin” inside the 18h century which is just like a Sudoku puzzle but with no further constraint about the items in specific locations. The very first genuine Sudoku was posted in 1979 and was introduced by Howard Garns, an United States designer. Reality broad recognition were only available in Japan in 1986 soon after it had been released and considering the brand Sudoku by Nikoli.
Rules and Terms : A Sudoku puzzle contains 81 cellular material that are split into 9 posts, series and locations. The process is already to set the phone numbers from 1 to 9 in to the unfilled tissues in a way that in every single row, line and 3×3 location every quantity shows up just once. A Sudoku has at the very least 17 offered figures but usually you can find 22 to 30. Printable Binary Sudoku
## Printable Binary Sudoku
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Here we have another image Binary Place Value Chart | Printable Binary Sudoku featured under Crack The Code – Binary Code 5 Bit Challenge | Printable Binary Sudoku. We hope you enjoyed it and if you want to download the pictures in high quality, simply right click the image and choose "Save As". Thanks for reading Crack The Code – Binary Code 5 Bit Challenge | Printable Binary Sudoku. | 722 | 3,502 | {"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} | 2.671875 | 3 | CC-MAIN-2023-40 | longest | en | 0.953609 |
https://gbee.edu.vn/sum-of-interior-and-exterior-angles-in-polygons-4bwz9btv/ | 1,721,879,997,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518532.66/warc/CC-MAIN-20240725023035-20240725053035-00054.warc.gz | 238,233,169 | 12,793 | Polygons
A polygon is a closed figure formed by a finite number of
segments such that:
1. the sides that have a common endpoint
are noncollinear, and
2. each side intersects exactly two other
sides, but only at their endpoints.
Polygons are named by number of sides
Number of Sides Polygon
3
4
5
6
7
8
9
10
12
n
Triangle
Pentagon
Hexagon
Heptagon
Octagon
Nonagon
Decagon
Dodecagon
n-gon
Regular Polygon
A convex polygon in which all the sides are congruent
and all the angles are congruent is called a regular
polygon.
Draw a:
Hexagon Heptagon
Octogon
Then draw diagonals to create triangles.
A diagonal is a segment connecting two
nonadjacent vertices (don’t let segments cross)
Add up the angles in all of the triangles in
the figure to determine the sum of the
angles in the polygon.
Complete this table
Polygon # of sides # of triangles Sum of
interior angles
Polygon Interior Angles Theorem
The sum of the measures of the interior angles of a
convex n-gon is (n – 2) • 180.
Examples –
1. Find the sum of the measures of the interior angles of a
16–gon.
2. If the sum of the measures of the interior angles of a
convex polygon is 3600°, how many sides does the
polygon have.
3. Solve for x.
4x – 2
82
108
2x + 10
(16 – 2)*180
(n – 2)*180 = 3600
180n – 360 = 3600
+ 360 + 360
180n = 3960
180 180
n = 22 sides
(4 – 2)*180 = 360
108 + 82 + 4x – 2 + 2x + 10 = 360
6x + 198 = 360
6x = 162
6 6
x = 27
= 2520°
Draw a quadrilateral and extend the sides.
There are two sets of angles formed when the
sides of a polygon are extended.
• The original angles are called interior angles.
• The angles that are adjacent to the
interior angles are called exterior angles.
These exterior angles can be formed when any
side is extended.
What do you notice about the interior angle and
the exterior angle?
What is the measure of a line?
What is the sum of an interior angle with the
exterior angle?
They form a line.
180°
180°
If you started at Point A, and
followed along the sides of
exterior turns that are
marked, what would happen?
You end up back where you
started or you would make a
circle.
What is the measure of the
degrees in a circle?
A
B
C
D
360°
The sum of the measures of the exterior angles of a
convex polygon, one at each vertex, is 360°.
Each exterior angle of a regular polygon is 360
n
where n is the number of sides in the polygon
Polygon Exterior Angles Theorem
54⁰
68⁰
65⁰
(3x + 13)⁰
60⁰
(4x – 12)⁰
Find the value for x.
Sum of exterior angles is 360°
(4x – 12) + 60+ (3x + 13) + 65 + 54+ 68 = 360
7x + 248 = 360
– 248 – 248
7x = 112
7 7
x = 12
Example
What is the sum of the exterior angles in an octagon?
What is the measure of each exterior angle in a regular
octagon?
360°
360°/8 = 45°
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# Numerical Mathematics and Computing (6th Edition) View more editions 77% (65 ratings) for Chapter 4.1Solutions for Chapter 4.1
• 1558 step-by-step solutions
• Solved by professors & experts
• iOS, Android, & web
Chapter: Problem:
Test the procedure given in the text for determining the Newton form of the interpolating polynomial. For example, consider this table:
Find the interpolating polynomial and verify that p(−1) = 12.
Sample Solution
Chapter: Problem:
• Step 1 of 6
Consider the following table:
1 2 3 -4 5 2 48 272 1182 2262
• Step 2 of 6
To find the approximate values of at using Newton interpolating polynomial construct the following function in MATLAB.
Code_11738_4_1_1CE.m : Main MATLAB function to compute at Coef.m Calculates the Coefficient of the Newton Interpolating Polynomial Eval.m Approximate the values of at;
• Step 3 of 6
Code_11738_4_1_1CE.m:
function Code_11738_4_1_1CE()
x = [1 2 3 -4 5];
y = [2 48 272 1182 2262];
p = -1;
%--------------- Coefficient of the Newton Interpolating Polynomial
a = Coef(x,y);
fprintf('Coefficient of the Newton Interpolating Polynomial
');
fori = 1: length(a), fprintf('%5.0f\t %15.10f
',i-1,a(i)); end
%--------------- Approximate the values of y = f(x) at x = p;
pn = Eval(x,a,p);
fprintf('Approximate Value of y(x) at x = %5.5f\t is %10.10f
',p,pn);
end
• Step 4 of 6
Coef.m:
function [a] = Coef(x,y)
%--------------------------------------------------------------------
%Calculates the Coefficient of the Newton Interpolating Polynomial
%
% x : Input Array x
% y = f(x) : Input Array y
% a : Coefficients to construct the Newton Interpolating Polynomial
%--------------------------------------------------------------------
mx = length(x); my = length(y);
if mx ~= my, error('Input Vectors must be of same length'); end
T = zeros(mx, mx); T(:,1) = y';
for j = 2 : mx
fori = 1:(mx-j+1)
T(i,j) = (T(i+1,j-1) - T(i,j-1))/(x(i+j-1)-x(i));
end
end
a = T(1,:);
end
• Step 5 of 6
Eval.m:
functionpn = Eval(X,a,p)
%--------------------------------------------------------------------
%Approximate the values of y = f(x) at x = p;
%
% x : Input Array x
% a : Input Coefficients obtained from Coeff.m
% pn : Approximate of y = f(x) at x = p
%--------------------------------------------------------------------
m = length(a); sum = 0;
fori = 1: m
prodx = 1;
for j = 1: i-1
prodx = prodx*(p - X(j));
end
sum = sum + a(i)*prodx;
end
pn = sum;
end
• Step 6 of 6
Run the appropriate code Code_11738_4_1_1CE.m in MATLAB command window, press F5 to compute the results and the outputs of the function is:
Coefficient of the Newton Interpolating Polynomial
0 2.0000000000
1 46.0000000000
2 89.0000000000
3 6.0000000000
4 4.0000000000
Approximate Value of y(x) at x = -1.00000 is 12.0000000000
Therefore the corresponding Newton polynomial is:
And the approximate values at is
Corresponding Textbook
Numerical Mathematics and Computing | 6th Edition
9780495114758ISBN-13: 0495114758ISBN: Authors:
Alternate ISBN: 9780495384717, 9781111800611 | 918 | 3,085 | {"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-2018-26 | latest | en | 0.523077 |
https://fakahil.github.io/coding/how-to-do-model-fitting-in-python/index.html | 1,680,114,848,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949025.18/warc/CC-MAIN-20230329182643-20230329212643-00052.warc.gz | 298,415,554 | 40,495 | How to do model fitting in Python
Curve fitting - How?¶
For the purpose of illustrating an example, I will introduce a limb model which I have been working on during my PhD.¶
In [1]:
## importing the libraries
import numpy as np
import math as m
import scipy
import matplotlib.pyplot as plt
from scipy import special
In [2]:
def Erfc(x,sigma):
y = special.erfc(x/(sigma*np.sqrt(2)))
return y
## Introducing the model to be used later for the fitting
def SL_fit(x,w1,w2,w3,s1,s2,s3):
f = 0.5*(w1*Erfc(x,s1)+w2*Erfc(x,s2)+w3*Erfc(x,s3)+ (1-w1-w2-w3))
return f
The first method we will use is via scipy.optimize.curve_fit¶
In [4]:
from scipy.optimize import curve_fit
In [5]:
#This is the jacobian matrx, which is the derivative of the fitting function with respect to every free parameter
def DF(x,w1,w2,w3,s1,s2,s3):
dfw1 = 0.5*(Erfc(x,s1)-1)
dfw2 = 0.5*(Erfc(x,s2)-1)
dfw3 = 0.5*(Erfc(x,s3)-1)
dfs1 = w1*x*np.exp(-x**2/(2*s1**2))/ (np.sqrt(2*np.pi)*s1**2)
dfs2 = w2*x*np.exp(-x**2/(2*s2**2))/ (np.sqrt(2*np.pi)*s2**2)
dfs3 = w3*x*np.exp(-x**2/(2*s3**2))/ (np.sqrt(2*np.pi)*s3**2)
return np.transpose(np.array([dfw1,dfw2,dfw3,dfs1,dfs2,dfs3]))
In [25]:
## Loading the data:
path = "/home/fatima/Desktop/project_3/"
x = file[:,0]
y = file[:,1]
In [26]:
## plotting the data
plt.style.use('ggplot')
plt.figure(figsize=(8,5))
plt.plot(x,y)
plt.xlabel('Arcseconds')
plt.ylabel('Normalized intensity')
plt.show()
In [27]:
## For the purpose of fitting, we restrict the latter to data points located above x = 0
ind = np.where(x>=0)
x = x[ind]
y = y[ind]
weights = np.sqrt(np.abs(y)) ## Poisson weighting
In [49]:
p0=[0.3, 0.3, 0.2, 1, 2, 3] ## initial guess best-fit parameters
popt, pcov = curve_fit(SL_fit,x,y,p0,method='lm',sigma=weights,jac=DF,ftol=1e-8,xtol=1e-8,maxfev=5000)
chi_sq = np.sum((1/weights**2)*(SL_fit(x,*popt)-y)**2)
In [54]:
print popt #to view the best fit parameters
print chi_sq
[ 0.52750103 0.28882569 0.10191755 0.25905336 0.76540583 2.83343005]
0.00122093770376
In [48]:
xnew = np.linspace(x.min(),x.max(),1000)
y_fit = SL_fit(xnew,*popt)
In [35]:
plt.style.use('ggplot')
plt.figure(figsize=(8,5))
plt.plot(xnew, y_fit,color='blue', label='best fit')
plt.plot(x,y,'r--', label='data')
plt.xlabel('Arcseconds')
plt.ylabel('Normalized intensity')
plt.legend(loc='upper right')
plt.show()
The second method we will use is via scipy.optimize.least_squares . With this module, both he objective function and jacobian matrix have to be defined differently than in scipy.optimize.curve_fit :¶
In [39]:
from scipy.optimize import least_squares
In [40]:
## objective function
def Leastsquares(params, x,y,w):
w1 = params[0]
w2 = params[1]
w3 = params[2]
s1 = params[3]
s2 = params[4]
s3 = params[5]
model = 0.5*(w1*Erfc(x,s1)+w2*Erfc(x,s2)+w3*Erfc(x,s3)+(1-w1-w2-w3))
return (1/w)*(model-y)
## as you can see, in the objective function has to return the difference between the modelled data and data points
In [41]:
## jacobian matrix
def Jac(popt,x,y,w):
w1 = popt[0]
w2 = popt[1]
w3 = popt[2]
s1 = popt[3]
s2 = popt[4]
s3 = popt[5]
dfw1 = 0.5*(Erfc(x,s1)-1)
dfw2 = 0.5*(Erfc(x,s2)-1)
dfw3 = 0.5*(Erfc(x,s3)-1)
dfs1 = w1*x*np.exp(-x**2/(2*s1**2))/ (np.sqrt(2*np.pi)*s1**2)
dfs2 = w2*x*np.exp(-x**2/(2*s2**2))/ (np.sqrt(2*np.pi)*s2**2)
dfs3 = w3*x*np.exp(-x**2/(2*s3**2))/ (np.sqrt(2*np.pi)*s3**2)
return np.transpose(np.array(1/w)*([dfw1,dfw2,dfw3,dfs1,dfs2,dfs3]))
In [42]:
## loading the data again:
x = file[:,0]
y = file[:,1]
ind = np.where(x>=0)
x = x[ind]
y = y[ind]
weights = np.sqrt(np.abs(y)) ## Poisson weighting
In [52]:
## calling the objective function and fitting data points
res_leastsquares = least_squares(Leastsquares, p0,method='lm',jac=Jac, args=(x,y,weights),ftol=1e-8,xtol=1e-8,max_nfev=5000)
##computing the chi-square value
chi_sq_2 = np.sum((Leastsquares(res_leastsquares.x,x,y,weights))**2)
In [53]:
## printing best-fit parameters
print res_leastsquares.x
print chi_sq_2
[ 0.52750174 0.28882527 0.10191733 0.2590536 0.76540736 2.83343784]
0.00122093770322
The THIRD method we will use is via scipy.optimize.leasq , which is pretty similar to scipy.optimize.least_squares except for the saved results format¶
In [55]:
from scipy.optimize import leastsq
In [56]:
## objective function
def Leastsq(params, x,y,w):
w1 = params[0]
w2 = params[1]
w3 = params[2]
s1 = params[3]
s2 = params[4]
s3 = params[5]
model = 0.5*(w1*Erfc(x,s1)+w2*Erfc(x,s2)+w3*Erfc(x,s3)+(1-w1-w2-w3))
return (1/w)*(model-y)
##jacobian matrix same as in leastsquares method
In [57]:
res_leastsq = leastsq(Leastsq,p0,args=(x,y,weights),Dfun=Jac,ftol=1e-8,xtol=1e-8,maxfev=5000)
chi_sq_3 = np.sum((Leastsq(res_leastsq[0],x,y,weights))**2)
In [58]:
print res_leastsq[0]
print chi_sq_3
[ 0.52750103 0.28882569 0.10191755 0.25905336 0.76540583 2.83343005]
0.00122093770376
The FOURTH method we will use is the powerful package lmfit . LMFIT has the advantage of fixing certain parameters while letting others unchanged, and the possibiliy to introduce bounds on the variables.¶
In [71]:
import lmfit
from lmfit import Minimizer, minimize, Parameters, report_fit
In [63]:
## defining the objective function
def residual(params, x,y,w):
w1 = params['omega1']
w2 = params['omega2']
w3 = params['omega3']
s1 = params['sigma1']
s2 = params['sigma2']
s3 = params['sigma3']
model = 0.5*(w1*Erfc(x,s1)+w2*Erfc(x,s2)+w3*Erfc(x,s3)+(1-w1-w2-w3))
return (1/w)*(y-model)
In [65]:
# using the class Parameters in which we can name the variables and assign first guess values and boundaries, if needed
params = Parameters()
In [67]:
## fitting with lmfit via lmfit.minimize
out = lmfit.minimize(residual,params,method='leastsq',args=(x,y,weights),ftol=1e-8,xtol=1e-8,maxfev=5000)
In [74]:
## printing output
report_fit(out.params)
[[Variables]]
omega1: 0.52750083 +/- 0.003494 (0.66%) (init= 0.3)
omega2: 0.28882580 +/- 0.002845 (0.99%) (init= 0.3)
omega3: 0.10191760 +/- 0.001193 (1.17%) (init= 0.3)
sigma1: 0.25905329 +/- 0.001357 (0.52%) (init= 1)
sigma2: 0.76540540 +/- 0.007152 (0.93%) (init= 2)
sigma3: 2.83342789 +/- 0.047380 (1.67%) (init= 3)
[[Correlations]] (unreported correlations are < 0.100)
C(omega1, sigma2) = 0.899
C(omega1, omega2) = -0.895
C(omega1, sigma1) = 0.889
C(omega3, sigma2) = -0.870
C(omega3, sigma3) = -0.853
C(omega2, sigma1) = -0.851
C(sigma2, sigma3) = 0.729
C(sigma1, sigma2) = 0.716
C(omega2, sigma2) = -0.632
C(omega1, omega3) = -0.628
C(omega1, sigma3) = 0.487
C(omega3, sigma1) = -0.457
C(sigma1, sigma3) = 0.348
C(omega2, omega3) = 0.230 | 2,532 | 6,705 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2023-14 | latest | en | 0.433716 |
https://justaaa.com/finance/632977-consider-the-following-two-mutually-exclusive | 1,701,804,810,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100555.27/warc/CC-MAIN-20231205172745-20231205202745-00622.warc.gz | 397,831,990 | 10,243 | Question
Consider the following two mutually exclusive projects: Year Cash Flow (A) Cash Flow (B) 0 –\$...
Consider the following two mutually exclusive projects:
Year Cash Flow (A) Cash Flow (B) 0 –\$ 424,000 –\$ 39,500 1 44,500 20,300 2 61,500 13,400 3 78,500 18,100 4 539,000 14,900
The required return on these investments is 11 percent.
a. What is the payback period for each project? (Do not round intermediate calculations and round your answers to 2 decimal places, e.g., 32.16.)
Payback period Project A years Project B years
b. What is the NPV for each project? (Do not round intermediate calculations and round your answers to 2 decimal places, e.g., 32.16.)
Net present value Project A \$ Project B \$
c. What is the IRR for each project? (Do not round intermediate calculations and enter your answers as a percent rounded to 2 decimal places, e.g., 32.16.)
Internal rate of return Project A % Project B % | 252 | 931 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2023-50 | latest | en | 0.802472 |
https://poker.stackexchange.com/questions/4449/problem-with-a-hand-in-texas-holdem | 1,620,499,183,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988923.22/warc/CC-MAIN-20210508181551-20210508211551-00472.warc.gz | 494,183,640 | 38,038 | # Problem with a hand in Texas hold'em
We play as friends and keep arguing over a certain hand dealt. The scenario is on the last card dealt it becomes a straight on the table, say 23456, I understand if someone has a 7 then that would win but what if the straight is the best hand, is the pot split? One guy says their two pair would win but I don't agree .
Any help would be great.
• in case no-one of the players have any `7`, the pot is split since all of you have hit a straight, regardless your holding hands; The board just gave you the best hand and that only matters (the best 5-card combination doesn't have to include your 2 cards). However, even if someone has a `A7` card isn't guarantee to win since another player with `87` wins with the even higher straight. – user1165 Dec 29 '14 at 2:07
• That's what I thought as well. Thanks for clearing it up i will have great pleasure in telling them they was wrong. – Adam jayes Dec 29 '14 at 16:23 | 241 | 958 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2021-21 | latest | en | 0.986432 |
https://r.igraph.org/reference/running_mean.html | 1,686,299,932,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224655446.86/warc/CC-MAIN-20230609064417-20230609094417-00744.warc.gz | 522,306,710 | 4,342 | running_mean() calculates the running mean in a vector with the given bin width.
## Usage
running_mean(v, binwidth)
## Arguments
v
The numeric vector.
binwidth
Numeric constant, the size of the bin, should be meaningful, i.e. smaller than the length of v.
## Value
A numeric vector of length length(v)-binwidth+1
## Details
The running mean of v is a w vector of length length(v)-binwidth+1. The first element of w id the average of the first binwidth elements of v, the second element of w is the average of elements 2:(binwidth+1), etc.
Other other: convex_hull(), sample_seq()
## Author
Gabor Csardi csardi.gabor@gmail.com
## Examples
running_mean(1:100, 10)
#> [1] 5.5 6.5 7.5 8.5 9.5 10.5 11.5 12.5 13.5 14.5 15.5 16.5 17.5 18.5 19.5
#> [16] 20.5 21.5 22.5 23.5 24.5 25.5 26.5 27.5 28.5 29.5 30.5 31.5 32.5 33.5 34.5
#> [31] 35.5 36.5 37.5 38.5 39.5 40.5 41.5 42.5 43.5 44.5 45.5 46.5 47.5 48.5 49.5
#> [46] 50.5 51.5 52.5 53.5 54.5 55.5 56.5 57.5 58.5 59.5 60.5 61.5 62.5 63.5 64.5
#> [61] 65.5 66.5 67.5 68.5 69.5 70.5 71.5 72.5 73.5 74.5 75.5 76.5 77.5 78.5 79.5
#> [76] 80.5 81.5 82.5 83.5 84.5 85.5 86.5 87.5 88.5 89.5 90.5 91.5 92.5 93.5 94.5
#> [91] 95.5 | 585 | 1,188 | {"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-2023-23 | longest | en | 0.637114 |
http://sat.collegeboard.org/practice/answered-question-of-the-day?src=R&questionId=20131001&answerCd=A | 1,419,024,994,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802768977.107/warc/CC-MAIN-20141217075248-00037-ip-10-231-17-201.ec2.internal.warc.gz | 233,371,338 | 13,416 | # Question of the Day
## Can you answer today's question?
Next Tests:
1/24 , 3/14
http://sat.collegeboard.org/practice/sat-question-of-the-day
Oct
1
Number Answered 74,897 75,445
150,342 responses
49% correct
See Last Thirty Questions:
Mathematics > Standard Multiple Choice
Read the following SAT test question and then click on a button to select your answer.
In the figure above, inscribed triangle is equilateral. If the radius of the circle is , then the length of arc is
# Right-o!
### Explanation
An equilateral triangle has three equal sides, so the lengths of the three arcs must be equal, and thus each arc is of the circumference of the circle. The circumference of the circle is , so the length of arc is of , or .
Want to brush up on your skills? Just visit our Practice Questions area.
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See Inside | 306 | 1,345 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2014-52 | latest | en | 0.821392 |
https://uk.mathworks.com/matlabcentral/cody/problems/2016-area-of-an-equilateral-triangle/solutions/1714409 | 1,591,182,957,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347432521.57/warc/CC-MAIN-20200603081823-20200603111823-00212.warc.gz | 582,837,203 | 15,540 | Cody
# Problem 2016. Area of an equilateral triangle
Solution 1714409
Submitted on 24 Jan 2019 by Mandar Sapre
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; y_correct = sqrt(3)/4; tolerance = 1e-12; assert(abs(equilateral_area(x)-y_correct)<tolerance)
2 Pass
x = 2; y_correct = sqrt(3); tolerance = 1e-12; assert(abs(equilateral_area(x)-y_correct)<tolerance)
3 Pass
x = 3; y_correct = sqrt(3)*9/4; tolerance =1e-12; assert(abs(equilateral_area(x)-y_correct)<tolerance) | 193 | 606 | {"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-2020-24 | latest | en | 0.667731 |
https://inari-sushi.ru/christmas-maths-problem-solving-207.html | 1,627,627,906,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153934.85/warc/CC-MAIN-20210730060435-20210730090435-00626.warc.gz | 324,957,086 | 9,434 | # Christmas Maths Problem Solving
Let students think about the way they solve their questions and let them draw pictures or use manipulatives to support their own thinking and logic. Ken’s new advent calendar has 1 chocolate for the 1st day, 2 chocolates on the 2nd day, 3 chocolates on the 3rd day, 4 chocolates on the 4th day and so on.How many chocolates will he have eaten by the 12th day?We also have several dot-to-dot worksheets with a Christmas theme.
Tags: Prewriting Phase Of Essay Writing StepsStandard For A Term PaperJ Essaye De T Oublier Avec Un Autre Paroles1984 Thesis On ControlHausarbeiten SchreibenCustom Ezessays.Us Paper Term1962 Essays Of Jean WagnerProblem Solving Exercises For StudentsProblem Solving Ks1Doctoral Thesis Or Dissertation
Design Christmas word problem worksheets that are suitable for second and third-grade students.
The sample questions adhere to math standards for those grades. In terms of fun word problem scenarios, you can incorporate Christmas themes into the problems.
Using this strategy will help children become better problem-solvers and critical thinkers.
For second grade worksheets, you will note that addition and subtraction problems are the most appropriate.
This year I put together a set of Christmas-themed problems for my students to work on as a team.
The problem set is the attached printable file, and there is a solution key below.
The sheets have been put in order from smallest number of dots to the largest.
These kindergarten counting worksheets all have a Christmas theme.
This problem set would also work well as a Christmasy problem-solving activity for a high school math class.
Here you will find a range of Christmas Math Activities from 1st grade and upwards.
## Comments Christmas Maths Problem Solving
• ###### Maths Christmas Activities and Resources - Mr Barton Maths.
Free Maths Christmas resources and activities from TES Maths. Maths Christmas puzzles. Can you solve this maths mystery? 6. There are.…
• ###### Christmas tree NZ Maths
Devise and use problem solving strategies to explore situations mathematically use equipment. Make a picture graph of the Christmas tree decorations.…
• ###### Seasonal Maths - Resourceaholic
Seasonal maths resources for Christmas, Easter and more. Christmas Transformations - dave789 on TES; Christmas Problem Solving - GCSE Shape and.…
• ###### Third Grade Christmas Word Problems - ThoughtCo
Word problems and problem-solving math questions help students to put the computations into authentic practice. Try these 3rd grade.…
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https://testbook.com/objective-questions/mcq-on-bernoulli-equation--5eea6a0c39140f30f369e186 | 1,638,528,342,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362619.23/warc/CC-MAIN-20211203091120-20211203121120-00202.warc.gz | 630,723,290 | 135,922 | # If water is flowing through a pipe of 5 cm diameter under a pressure of 20 N/cm2 and a mean velocity of 2.0 m/s, the kinetic head will be
1. 20.4 m
2. 0.101 m
3. 0.204 m
4. 10.1 m
## Answer (Detailed Solution Below)
Option 3 : 0.204 m
## Bernoulli Equation MCQ Question 1 Detailed Solution
Concept:
Bernoulli’s equation is given by:
$$\frac{P}{{\rho g}} + \frac{{{V^2}}}{{2g}} + Z = constant$$
where,
$$\frac{P}{{\rho g}} =$$ Pressure head, $$\frac{{{V^2}}}{{2g}} =$$ Kinetic head, Z = Datum head
Calculation:
Given:
d = 5 cm, P = 20 N/cm2, V = 2 m/s
Kinetic head = $$\frac{V^2}{2g}=\frac{2^2}{2\times9.81}=0.204\;m$$
# A siphon draws water from a reservoir and discharges it out at atmospheric pressure. Assuming ideal fluid and the reservoir is large, the velocity at point P in the siphon tube is:
1. $$\sqrt {2g{h_1}}$$
2. $$\sqrt {2g{h_2}}$$
3. $$\sqrt {2g\left( {{h_2} - {h_1}} \right)}$$
4. $$\sqrt {2g\left( {{h_2} + {h_1}} \right)}$$
## Answer (Detailed Solution Below)
Option 3 : $$\sqrt {2g\left( {{h_2} - {h_1}} \right)}$$
## Bernoulli Equation MCQ Question 2 Detailed Solution
Concept:
Applying Bernoulli’s equation between (1) & (2)
$$\begin{array}{l} \frac{{{P_1}}}{{\rho g}} + \frac{{V_1^2}}{{2g}} + {Z_1} = \frac{{{P_2}}}{{\rho g}} + \frac{{V_2^2}}{{2g}} + {Z_2}\\ \Rightarrow \frac{{{P_{atm}}}}{{\rho g}} + 0 + \left( {{h_2} - {h_1}} \right) = \frac{{{P_{atm}}}}{{\rho g}} + \frac{{V_2^2}}{{2g}} + 0\\ \therefore {h_2} - {h_1} = \frac{{V_2^2}}{{2g}} \Rightarrow {V_2} = \sqrt {2g\left( {{h_2} - {h_1}} \right)} \end{array}$$
The velocity of fluid inside the tube is the same
$${V_p} = {V_2} = \sqrt {2g\left( {{h_2} - {h_1}} \right)}$$
# Bernoulli’s equation is derived based on the following assumptionsi) There is no loss of energy in a liquid flow and flow is steady.ii) The flow is rotationaliii) Except gravity and pressure forces, no external force acts on liquid flowiv) The flow is viscous, incompressible flow
1. (i) & (iii)
2. (i), (ii) & (iv)
3. (ii) & (iii)
4. (i), (iii) & (iv)
## Answer (Detailed Solution Below)
Option 1 : (i) & (iii)
## Bernoulli Equation MCQ Question 3 Detailed Solution
Explanation:
Bernoulli’s equation:
$$\frac{P}{\rho g} + \frac{{{v^2}}}{{2g}} + Z = Constant$$
• It can be derived from the principle of conservation of energy.
• It states that, in a steady flow, the sum of all forms of energy in a fluid along a streamline is the same at all points on that streamline.
• It represented in head form (the total energy per unit weight).
Following assumption are made in deriving the Bernoulli’s equation:
1. Fluid is inviscid i.e. zero viscosity.
2. Flow is in steady-state.
3. Flow is incompressible.
4. Flow is irrotational.
5. Flow is along the streamline.
6. Only gravity and pressure force acting on the liquid, no other external force.
Important Points
Even if the flow is rotational, we can still apply the Bernoulli’s equation but only along a streamline.
# A smooth pipe of diameter 500 mm carries water. The pressure in the pipe at Section ‘A’ (elevation: 10 m) is 100 kPa. At section ‘B’ (elevation: 12 m) the pressure is 75 kPa and velocity is 4 m/s. Which of the following is true (g = 10 m/s2)
1. Flow from A to B and head loss is 1 m
2. Flow from A to B and head loss is 0.5 m
3. Flow from B to A and head loss is 0.5 m
4. Flow from B to A and head loss is 0.75 m
## Answer (Detailed Solution Below)
Option 2 : Flow from A to B and head loss is 0.5 m
## Bernoulli Equation MCQ Question 4 Detailed Solution
Concept:
Bernoulli’s Equations
At Section 1
$$\frac{{{p_1}}}{{\rho g}} + \frac{{V_1^2}}{{2g}} + {z_1}$$
At Section 2
$$\frac{{{p_2}}}{{\rho g}} + \frac{{V_2^2}}{{2g}} + {z_2}$$
Calculation:
Given:
P1 = 100 kPa, Z1 = 10 m, P2 = 75 kPa, z2 = 12 m, v = 4 m/s, d = 500 mm = 0.5 m
At Section 1
$$\frac{{{p_1}}}{{\rho g}} + \frac{{V_1^2}}{{2g}} + {z_1} = \frac{{100 \times {{10}^3}}}{{1000 \times 10}} + \frac{{{4^2}}}{{2 \times 10}} + 10 = 20.8\;m$$
At Section 2
$$\frac{{{p_2}}}{{\rho g}} + \frac{{V_2^2}}{{2g}} + {z_2} = \frac{{75 \times {{10}^3}}}{{1000 \times 10}} + \frac{{{4^2}}}{{2 \times 10}} + 12 = 20.3\;m$$
(Total Energy) 1 > (Total Energy) 2
Flow takes place from section 1 to section 2 with a head loss of 0.5 m.
# Which of the following statements is INCORRECT for assumptions made in derivation of Bernoulli’s equation?
1. The flow is steady.
2. The flow is incompressible
3. The viscosity of fluid is non-zero
4. The flow is irrotational
## Answer (Detailed Solution Below)
Option 3 : The viscosity of fluid is non-zero
## Bernoulli Equation MCQ Question 5 Detailed Solution
Explanation:
Bernoulli's Equation:
Bernoulli's equation is obtained by integrating the Euler's equation of motion which is given by-
$$\frac{{dp}}{ρ } + gdz + vdv = 0\;\;\;\;\;(1)$$
In Euler's equation of motion, the forces due to gravity and pressure are taken into consideration and which is derived considering the motion of a fluid element along a stream-line.
Integrating the above eq(1):
$$\smallint \frac{{dp}}{ρ } + \smallint gdz + \smallint vdv = 0$$
$$\frac{p}{ρ } + gz + \frac{{{v^2}}}{2} = C$$
$$\frac{p}{{ρ g}} + \frac{{{v^2}}}{{2g}} + z = C\;\;\;\;(2)$$
where p/ρg = pressure head or pressure energy per unit weight, v2/2g = kinetic head or kinetic energy per unit weight, z = potential head or potential energy by unit weight.
Following assumptions are made in the derivation of Bernoulli's equation:
1. Flow is ideal i.e inviscous.
2. Flow is steady i.e. time variation is zero.
3. Flow is incompressible i.e. ρ is constant.
4. Flow is irrotaional i.e. ωx = ωy = ωz = 0.
# A tank of large cross-sectional area contains water up to a height of 5 m as shown in the figure. The top water surface is under a pressure of p1 = 0.2 MPa. A small, smooth and round tap at the bottom of the tank is opened to the atmosphere (p2 = 0.1 MPa). Use the acceleration due to gravity, g = 9.81 m/s2 and the density of water, p = 1000 kg/m3. The velocity with which the water will exit from the tap under the conditions shown in the figure (rounded off to one decimal place) in m/s is ______.
## Bernoulli Equation MCQ Question 6 Detailed Solution
Concept:
Bernoulli's Principle: It states the total mechanical energy of the moving fluid comprising the energy associated with the fluid pressure, the gravitational potential energy of elevation, and the kinetic energy of the fluid motion, remains constant.
$$\mathbf{P+ {1\over 2}ρ V^2 +ρ g z= constant}$$
$${P\over\rho g}+{V^2\over2g}+z=constant$$
where,
P = pressure energy
ρgz = gravitational potential energy
$${1\over 2}ρ V^2$$ = kinetic energy of the fluid
Calculations:
Given:
P1 = 0.2 MPa, P2 = 0.1 MPa, (z1 - z2)= 5 m
According to the continuity equation,
A1V1 = A2V2
V1$$A_2V_2\over A_1$$
as A1 is very large, V1 will be negligible.
Applying Bernoulli's principle between point 1 and 2,
$${P_1\over\rho g}+{V^2_1\over2g}+z_1={P_2\over\rho g}+{V^2_2\over2g}+z_2$$
$${V^2_2\over2g}={P_1-P_2\over\rho g}\;+z_1-z_2$$
$${V^2_2\over2\times9.81}={(0.2-0.1)\times10^6\over1000\;\times9.81}\;+\;5$$
V2 = 17.265 m/s
V2 = 17.23 m/s
# An open tank is filled with water to a height of 20 m. What is the velocity of the water flow at the outlet, if the outlet is at the base of the tank?
1. 40 m/s (approx.)
2. 20 m/s (approx.)
3. 10 m/s (approx.)
4. 5 m/s (approx.)
## Answer (Detailed Solution Below)
Option 2 : 20 m/s (approx.)
## Bernoulli Equation MCQ Question 7 Detailed Solution
Concept:
The Velocity of flow at the base of a tank is given by
$$V = \sqrt {2gH}$$
Where H is the height of filled water in the tank.
Calculation:
Given, H = 20 m
$$V = \sqrt {2gH}$$
$$V= \sqrt {2 \times 9.81 \times 20}$$
V = 19.80 m/s = 20 m/s (approx.)
# Water is flowing in a smooth pipe of uniform diameter 10 cm, has following data:At section 1: Pressure = 50 kPa, Elevation = 10 mAt section 2: Pressure = 20 kPa, Elevation = 12 mThe velocity of flow is 1 m/s. Which of the following is correct for this pipe?
1. Flow is taking place form section 1 to section 2 with a head loss of 2 m
2. Flow is taking place form section 1 to section 2 with a head loss of 1 m
3. Flow is taking place from section 2 to section 1 with a head loss of 1 m
4. Flow is taking place from section 2 to section 1 with a head loss of 2 m
## Answer (Detailed Solution Below)
Option 2 : Flow is taking place form section 1 to section 2 with a head loss of 1 m
## Bernoulli Equation MCQ Question 8 Detailed Solution
Concept:
Bernoulli’s Equations
At Section 1
$$\frac{{{p_1}}}{{\rho g}} + \frac{{V_1^2}}{{2g}} + {z_1}$$
At Section 2
$$\frac{{{p_2}}}{{\rho g}} + \frac{{V_2^2}}{{2g}} + {z_2}$$
Calculation:
Given Data:
$$\begin{array}{l} {p_1} = 50\;kPa,\;{Z_1} = 10\;m\\ {p_2} = 20\;kPa,\;{Z_2} = 12\;m\\ d = 10\;cm,\;v = 1\;m/s \end{array}$$
At Section 1
$$\frac{{{p_1}}}{{\rho g}} + \frac{{V_1^2}}{{2g}} + {z_1} = \frac{{50 \times {{10}^3}}}{{1000 \times 10}} + \frac{{{1^2}}}{{2 \times 10}} + 10 = 15.05\;m$$
At Section 2
$$\frac{{{p_2}}}{{\rho g}} + \frac{{V_2^2}}{{2g}} + {z_2} = \frac{{20 \times {{10}^3}}}{{1000 \times 10}} + \frac{{{1^2}}}{{2 \times 10}} + 12 = 14.05\;m$$
(Total Energy) 1 > (Total Energy) 2
Flow takes place from section 1 to section 2 with a head loss of 1 m.
# Within a boundary layer for a steady incompressible flow, the Bernoulli equation
1. holds because the flow is steady
2. holds because the flow is incompressible
3. holds because the flow is transitional
4. does not hold because the flow is frictional
## Answer (Detailed Solution Below)
Option 4 : does not hold because the flow is frictional
## Bernoulli Equation MCQ Question 9 Detailed Solution
Explanation:
$$\frac{P}{\rho g} + \frac{{{v^2}}}{{2g}} + Z = Constant$$
Bernoulli’s equation:
• It can be derived from the principle of conservation of energy.
• It states that, in a steady flow, the sum of all forms of energy in a fluid along a streamline is the same at all points on that streamline.
• It represented in head form (the total energy per unit weight).
Following assumption are made in deriving Bernoulli’s equation:
Bernoulli’s equation assumptions:
1. The fluid is ideal, i.e. fluid has zero viscosity
2. Flow is steady
3. Flow is continuous
4. Flow is incompressible
5. Flow is irrotational
6. Flow is along a streamline.
Within a boundary layer for a steady incompressible flow, viscosity is present and the viscous forces dominate over inertia forces.
Thus, the Bernoulli equation does not hold within a boundary layer for a steady incompressible flow.
# Water-flow through a pipeline having four different diameters at 4 stations is shown in the figure below:The correct sequence of station numbers in the decreasing order of pressure is
1. 3, 1, 4, 2
2. 1, 3, 2, 4
3. 1, 3, 4, 2
4. 3, 1, 2, 4
## Answer (Detailed Solution Below)
Option 4 : 3, 1, 2, 4
## Bernoulli Equation MCQ Question 10 Detailed Solution
Explanation:
As per continuity equation AV = const
i.e. Area and velocity are inversely proportional.
Bernoulli equation:
$$\frac{P}{{\rho g}} + \frac{{{V^2}}}{{2g}} + z = c$$
i.e. Pressure is inversely proportional to the velocity
$$\therefore P \propto \frac{1}{V} \propto A$$
The pressure is least where velocity is highest and at the minimum cross-sectional area for constant discharge.
$${A_3} > {A_1} > {A_2} > {A_4} \Rightarrow {P_3} > {P_1} > {P_2} > {P_4}$$
# Which of the following sets of equation represent possible 2-D incompressible flows ?
1. u = x + y ; v = x - y
2. u = xt; v = xyt + y2
3. u = x + 2y ; v = x2 - y2
4. u = 4x + y ; v = x - y2
## Answer (Detailed Solution Below)
Option 1 : u = x + y ; v = x - y
## Bernoulli Equation MCQ Question 11 Detailed Solution
Concept:
For incompressible flow, the Continuity equation should follow-
For 3-D, $$\frac{{\partial u}}{{\partial x}} + \frac{{\partial v}}{{\partial y}} + \frac{{\partial w}}{{\partial z}} = 0$$
For 2-D, $$\frac{{\partial u}}{{\partial x}} + \frac{{\partial v}}{{\partial y}} = 0$$
Calculation:
Given:
u = x + y ; v = x - y
$$\frac{{\partial u}}{{\partial x}} = 1$$
$$\frac{{\partial v}}{{\partial y}} = -1$$,
$$$$$$\frac{{\partial u}}{{\partial x}} + \frac{{\partial v}}{{\partial y}} = 1 - 1$$
$$\frac{{\partial u}}{{\partial x}} + \frac{{\partial v}}{{\partial y}} = 0$$
Hence, u = x + y ; v = x - y represent possible 2-D incompressible flows.
# Bernoulli's equation is applicable between any two points located in
1. Rotational flow of an incompressible fluid
2. Irrotational flow of compressible or incompressible fluid
3. Steady, rotational flow of an incompressible fluid
4. Steady, irrotational flow of an incompressible fluid
## Answer (Detailed Solution Below)
Option 4 : Steady, irrotational flow of an incompressible fluid
## Bernoulli Equation MCQ Question 12 Detailed Solution
Explanation:
• Bernoulli's principle: For a streamlined flow of an ideal liquid in a varying cross-section tube the total energy per unit volume remains constant throughout the fluid. This means that in steady flow the sum of all forms of mechanical energy in a fluid along a streamline is the same at all points on that streamline.
• From Bernoulli's principle
$$\frac{{{{\rm{P}}_1}}}{{\rm{\rho }}} + {\rm{g}}{{\rm{h}}_1} + \frac{1}{2}{\rm{v}}_1^2 = \frac{{{{\rm{P}}_2}}}{{\rm{\rho }}} + {\rm{g}}{{\rm{h}}_2} + \frac{1}{2}{\rm{v}}_2^2$$
$$\frac{{\rm{P}}}{{\rm{\rho }}} + {\rm{gh}} + \frac{1}{2}{{\rm{v}}^2} = {\bf{constant}}$$
$$\dfrac{P}{r}+ \dfrac{v^2}{2g}+z=$$ constant
Bernoulli’s equation assumptions:
• The fluid is ideal, i.e. fluid has zero viscosity
• Flow is steady
• Flow is continuous
• Flow is incompressible
• Flow is irrotational
• Flow is along a streamline.
Within a boundary layer for a steady incompressible flow, viscosity is present and the viscous forces dominate over inertia forces.
Thus, Bernoulli's equation is applicable between any two points located in the Steady, irrotational flow of an incompressible fluid.
# Air discharges steadily through a horizontal nozzle and impinges on a stationary vertical plate as shown in figure.The inlet and outlet areas of the nozzle are 0.1 m2 and 0.02 m2, respectively. Take air density as constant and equal to 1.2 kg/m3. If the inlet gauge pressure of air is 0.36 kPa, the gauge pressure at point O on the plate (the point where the axis of nozzle touches the plate) is ________ kPa (round off to two decimal places).
## Bernoulli Equation MCQ Question 13 Detailed Solution
Concept:
For an incompressible, non-viscous fluid.
Bernoulli’s statement or equation is valid
$$\frac{P}{{\rho g}} + \frac{{{V^2}}}{{2g}} + z = constant$$
∵ It is mentioned in the question that air density is constant. Therefore, Bernoulli’s equation is valid
Calculation:
Mass flow rate through nozzle remain constant
$$\begin{array}{l} \therefore {\left( {\rho \;A\;V} \right)_{inlet}} = {\left( {\rho \;A\;V} \right)_{outlet}}\\ 0.1 \times {V_i} = 0.02\;{V_O} \end{array}$$
VO = 5 Vi
∴ V2 = 5 Vi
Bernoulli’s equation for nozzle
$$\frac{{{P_1}}}{{\rho g}} + \frac{{V_1^2}}{{2g}} = \frac{{{P_2}}}{{\rho g}} + \frac{{V_2^2}}{{2g}}\;\;\;\;\;\because \left[ {{z_1} = {z_2}} \right]$$
$$\frac{{0.36 \times {{10}^3}}}{{1.2 \times 9.81}} + \frac{{V_1^2}}{{2g}} - \frac{{{{\left( {5{V_1}} \right)}^2}}}{{2g}} = \frac{{{P_2}}}{{\rho g}}$$
$$\frac{{0.36 \times {{10}^3}}}{{1.2\;g}} - \frac{{24\;V_1^2}}{{2g}} = \frac{{{P_2}}}{{\rho g}}$$
P2 = Patm = 0 (Gauge)
∴ V1 = 5 m/s V2 = 25 m/s
∵ P2 = PO = Patm = 0 (Gauge)
But at point O there will be a pressure due to motion of fluid which will exert a pressure on the plate.
Now,
Bernoulli’s equation between exit point & point O
$$\frac{{{P_2}}}{{\rho g}} + \frac{{V_2^2}}{{2g}} = \frac{{{P_3}}}{{\rho g}} + \frac{{V_3^2}}{{2g}}$$
$${P_3} = \frac{1}{2}\rho V_2^2$$
$${P_3} = \frac{1}{2} \times 1.2 \times {\left( {25} \right)^2}=375 ~Pa$$
∴ P3 = 0.375 kPa
# The pressure head and kinetic head of water flowing through a pipe of diameter 60 cm are 50 m and 4 m, respectively. Determine the total head of the water at a cross-section which is 6 m above the datum line.
1. 56 m
2. 54 m
3. 60 m
4. 48 m
Option 3 : 60 m
## Bernoulli Equation MCQ Question 14 Detailed Solution
Concept:
Bernoulli’s equation is given by:
$$\mathbf{\frac{P}{\rho g}~+~\frac{V^2}{2g}~+~Z~=~Constant}$$
where,
$$\frac{P}{\rho g}$$ = Pressure head, $$\frac{V^2}{2g}$$ = Kinetic head, Z = Datum head
Calculation:
Given:
Pressure head = 50 m, kinetic head = 4 m
As the cross-section of the pipe is constant, therefore there will no change in heads.
⇒ Datum head = 6 m
∴ Total head = Pressure head + Kinetic head + Datum head
⇒ 50 + 4 + 6 = 60 m
∴ The total head of the water at the given cross-section is 60 m.
# A conical tube of length 2 m is fixed vertically with its smaller end upwards. The velocity of the flow at the smaller end is 5 m/s while at the lower end it is 2 m/s. The pressure head at the smaller end is 2.5 m of liquid. The loss of head in the tube is $$\frac{{0.35{{\left( {{V_1} - {V_2}} \right)}^2}}}{{2g}}$$ . What will be the pressure head at the lower end if the flow occurs in the downwards direction?
1. 13.427 m of fluid
2. 5.407 m of fluid
3. 8.325 m of fluid
4. 18.337 m of fluid
## Answer (Detailed Solution Below)
Option 2 : 5.407 m of fluid
## Bernoulli Equation MCQ Question 15 Detailed Solution
Concept:
According to Bernoulli's Theorem, the Total head of an ideal fluid at any point in flow direction remains constant, given as,
$$\Rightarrow$$ Total head (H) = $$\frac{P}{\gamma}$$ + $$\frac{V^2}{2g}$$ + z
Where,
$$\frac{P}{\gamma}$$ = Pressure head, $$\frac{V^2}{2g}$$ = Velocity head and z = datum head
For real fluids, losses are considered.
$$\therefore$$ for any two sections 1 and 2, Bernoulli's theorem is given as,
H= H+ head loss
Where,
H$$\frac{P_1}{\gamma}$$ + $$\frac{V_1^2}{2g}$$ + z and H$$\frac{P_2 }{\gamma}$$ + $$\frac{V_2^2}{2g}$$ + z
Calculation:
Given,
$$\frac{P_1}{\gamma}$$ = 2.5 m, V1 = 5 m/s, z1 = 2 m,
V2 = 2 m/s, z= 0 m and head loss = 0.35× $$\frac{(V_1-V_2)^2}{2g}$$
Applying Bernoulli's theorem,
H1 = H2 + head loss
$$\Rightarrow$$ $$\frac{P_1}{\gamma}$$ + $$\frac{V_1^2}{2g}$$ + z1 = $$\frac{P_2 }{\gamma}$$ $$\frac{V_2^2}{2g}$$ + z2 + 0.35 × $$\frac{(V_1-V_2)^2}{2g}$$
$$\Rightarrow$$ 2.5 + $$\frac{5^2}{2\times9.81}$$ + 2 = $$\frac{P_2 }{\gamma}$$ + $$\frac{2^2}{2\times9.81}$$ + 0 + 0.35× $$\frac{(5-2)^2}{2\times9.81}$$
$$\Rightarrow$$ $$\frac{P_2 }{\gamma}$$ = 5.407 m of liquid
$$\therefore$$ pressure head at lower end = 5.407 m of liquid
# The velocity of flow from a tap of 12 mm diameter is 8 m/s. What is the diameter of the jet at 1.5 m form the tap when the flow is vertically upwards? Assuming that, the jet continues to be circular up to that level.
1. 44 mm
2. 34 mm
3. 24 mm
4. 14 mm
Option 4 : 14 mm
## Bernoulli Equation MCQ Question 16 Detailed Solution
Concept:
Bernoulli's equation between point 1 and 2 is given by
$$\frac{{{P_1}}}{{\rho g}} + \frac{{{V_1}^2}}{{2g}} + {Z_1} = \frac{{{P_2}}}{{\rho g}} + \frac{{{V_2}^2}}{{2g}} + {Z_2}$$
Calculation:
Let us find velocity at point 2
Applying Bernoulli's equation between point 1 and 2,
$$\frac{{{P_1}}}{{\rho g}} + \frac{{{V_1}^2}}{{2g}} + {Z_1} = \frac{{{P_2}}}{{\rho g}} + \frac{{{V_2}^2}}{{2g}} + {Z_2}$$
As pressure is same at point 1 and 2,
we get,
$$\frac{{{V_1}^2}}{{2g}} + {Z_1} = \frac{{{V_2}^2}}{{2g}} + {Z_2}$$
$$\frac{{{8^2}}}{{2g}} + 0 = \frac{{{V_2}^2}}{{2g}} + 1.5$$
V2 = 5.88 m/sec
Using continuity equation,
A1 × V1 = A2 × V2
$${d_1}^2 \times 8 = {d_2}^2 \times 5.88$$
$${12^2} \times 8 = {d_2}^2 \times 5.88$$
$${d_2} = \sqrt {\frac{{{{12}^2} \times 8}}{{5.88}}} \approx {\rm{}}14{\rm{\;mm}}$$
# In fluid flow, the line of constant piezometric head passes through two points which have the same:
1. Elevation
2. Pressura
3. Velocity
4. Velocity potential
## Answer (Detailed Solution Below)
Option 3 : Velocity
## Bernoulli Equation MCQ Question 17 Detailed Solution
Explanation:
Bernoulli's equation
$$\frac{v^2}{2g}+\frac{p}{\rho g}+z=Constant$$
Where,
$$\frac{v^2}{2g}$$ = velocity head
$$\frac{P}{\rho g}$$ = pressure head
z = datum head
The Sum of datum head and pressure head is called the piezometric head.
As per Bernoulli’s equation sum of the piezometric head and velocity head is constant during the flow.
So, for a constant piezometric head between two points velocity will remain constant.
# A conical diffuser 3 m long is placed vertically. The velocity at the top (entry) is 4 m/s and at the lower end is 2 m/s. The pressure head at the top is 2 m of the oil flowing through the diffuser. The head loss in the diffuser is 0.4 m of the oil. The pressure head at the exit is:
1. 3.18 m of oil
2. 5.12 m of oil
3. 7.18 m of oil
4. 9.21 m of oil
## Answer (Detailed Solution Below)
Option 2 : 5.12 m of oil
## Bernoulli Equation MCQ Question 18 Detailed Solution
Concept:
Bernoulli’s equation is for a flow occurring head loss hL is given by
$$\frac{{{P_1}}}{{\rho g}} + \frac{{{V_1}^2}}{{2g}} + {Z_1} = \frac{{{P_2}}}{{\rho g}} + \frac{{{V_2}^2}}{{2g}} + {Z_2} + {h_L}$$
Where,
$$\frac{P}{{\rho g}}$$ is the pressure head
$$\frac{{{V^2}}}{{2g}}$$ is the velocity head and
Z is the datum head
It states that the total sum of heads across the flow remains constant.
Calculation:
Given flow through the diffuser is shown below
Given, pressure head at the top is 2 m of the oil i.e. $$\frac{{{P_1}}}{{\rho g}} = 2\;m$$ and head loss through the diffuser is 0.4 m.
Applying Bernoulli's equation across the diffuser,
$$\frac{{{P_1}}}{{\rho g}} + \frac{{{V_1}^2}}{{2g}} + {Z_1} = \frac{{{P_2}}}{{\rho g}} + \frac{{{V_2}^2}}{{2g}} + {Z_2} + {h_L}$$
$$2 + \frac{{{4^2}}}{{2g}} + 3 = \frac{{{P_2}}}{{\rho g}} + \frac{{{2^2}}}{{2g}} + 0 + 0.4$$
$$5 - 0.4 + \frac{{16 - 4}}{{2g}} = \frac{{{P_2}}}{{\rho g}}$$
$$\frac{{{P_2}}}{{\rho g}} = 4.6 + \frac{6}{g} = 5.2116\;m\;of\;oil$$
# Bernoulli’s equation is obtained by integrating which of the following equation of motion
1. Navier-stokes equation
2. Reynold’s equation of motion
3. Euler’s equation of motion
4. Newton’s law of viscosity
## Answer (Detailed Solution Below)
Option 3 : Euler’s equation of motion
## Bernoulli Equation MCQ Question 19 Detailed Solution
Explanation:
Bernoulli's Equation:
Bernoulli's equation is obtained by integrating the Euler's equation of motion which is given by-
$$\frac{{dp}}{ρ } + gdz + vdv = 0\;\;\;\;\;(1)$$
In Euler's equation of motion, the forces due to gravity and pressure are taken into consideration and which is derived considering the motion of a fluid element along a stream-line.
Integrating the above equation(1):
$$\smallint \frac{{dp}}{ρ } + \smallint gdz + \smallint vdv = 0$$
$$\frac{p}{ρ } + gz + \frac{{{v^2}}}{2} = C$$
$$\frac{p}{{ρ g}} + \frac{{{v^2}}}{{2g}} + z = C\;\;\;\;(2)$$
where p/ρg = pressure head or pressure energy per unit weight, v2/2g = kinetic head or kinetic energy per unit weight z = potential head or potential energy by unit weight.
Following assumptions are made in the derivation of Bernoulli's equation:
1. Flow is ideal.
2. Flow is steady i.e. time variation is zero.
3. Flow is incompressible i.e. ρ is constant.
4. Flow is irrotaional i.e. ωx = ωy = ωz = 0.
# Bernoulli’s equation is applicable between any two points in a flow field when
1. Flow is steady, compressible and irrotational
2. Flow is steady, incompressible and irrotational
3. Flow is unsteady, compressible and rotational
4. Flow is unsteady, incompressible and irrotational
## Answer (Detailed Solution Below)
Option 2 : Flow is steady, incompressible and irrotational
## Bernoulli Equation MCQ Question 20 Detailed Solution
Explanation:
Bernoulli's principle states that the sum of pressure energykinetic energy, and potential energy per unit volume of an incompressible, non-viscous fluid in a streamlined irrotational flow remains constant along a streamline.
This means that in steady flow the sum of all forms of mechanical energy in a fluid along a streamline is the same at all points on that streamline.
Bernoulli’s equation
• Bernoulli's equation was derived on the assumption that fluid is non-viscous and therefore frictionless.
• The Bernoulli Equation can be considered to be a statement of the conservation of energy principle appropriate for flowing fluids.
• Bernoulli's equation can be obtained by integrating Euler's Equation of motion
• But all real fluid is viscous and hence offer resistance flow.
• Thus there are always some losses in fluid flow and hence in the application of Bernoulli's equation, these losses are taken into consideration.
Thus the Bernoulli's equation for real fluid between point 1 and point 2 in any flow field is given by,
Pressure energy at point 1 + Kinetic Energy at point 1 + Potential Energy at point 1 = Pressure energy at point 2 + Kinetic Energy at point 2 + Potential Energy at point 2 + Loss of Energy.
$$P_1+\frac{{1}}{{2}}ρ V_1^2+ρ gz_1=P_2+\frac{{1}}{{2}}ρ V_2^2+ρ gz_2+h_L$$
where P= pressure at point 1, P2 = pressure at point 2, V= velocity at point 1, V2 = velocity at point 2, ρ = density of fluid, g = gravitational constant, z1 = pressure head at point 1, z2 = pressure head at point 2, hL = loss of energy
• P1 = Pressure energy at point 1,
• P2 = Pressure energy at point 2,
• $$\frac{{1}}{{2}}\rho V_2^2$$ = Kinetic Energy at point 1,
• $$\frac{{1}}{{2}}\rho V_1^2$$ = Kinetic Energy at point 2,
• ρgz1 = Potential Energy at point 1,
• ρgz2 = Potential Energy at point 2,
• h= Loss of Energy.
While the Bernoulli equation is stated in terms of universally valid ideas like conservation of energy and the ideas of pressure, kinetic energy, and potential energy, its application in the above form is limited to cases of a steady flow.
Important Points
The following are the assumptions made in the derivation of Bernoulli's equation:
1. The fluid is ideal i.e. viscosity is zero
2. The flow is steady
3. The flow is incompressible
4. The flow is irrotational | 8,967 | 26,067 | {"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} | 4.40625 | 4 | CC-MAIN-2021-49 | longest | en | 0.6344 |
http://www.thefullwiki.org/Rankine_scale | 1,544,982,970,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376827963.70/warc/CC-MAIN-20181216165437-20181216191437-00206.warc.gz | 470,589,651 | 10,340 | # Rankine scale: Wikis
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# Encyclopedia
### From Wikipedia, the free encyclopedia
Rankine temperature conversion formulae
from Rankine to Rankine
Celsius [°C] = ([°R] − 491.67) × 59 [°R] = ([°C] + 273.15) × 95
Fahrenheit [°F] = [°R] − 459.67 [°R] = [°F] + 459.67
Kelvin [K] = [°R] × 59 [°R] = [K] × 95
For temperature intervals rather than specific temperatures,
1 R = 1 °F = 59 °C = 59 K
Comparisons among various temperature scales
Rankine is a thermodynamic (absolute) temperature scale named after the Scottish engineer and physicist William John Macquorn Rankine, who proposed it in 1859.
The symbol for degrees Rankine is °R[1] (or °Ra if necessary to distinguish it from the Rømer and Réaumur scales). Zero on both the Kelvin and Rankine scales is absolute zero, but the Rankine degree is defined as equal to one degree Fahrenheit, rather than the one degree Celsius used by the Kelvin scale. A temperature of −459.67 °F is exactly equal to 0 °R.
Some engineering fields in the U.S. measure thermodynamic temperature using the Rankine scale.[2] However, throughout the scientific world and many times even in United States engineering, thermodynamic temperature is measured in kelvins.[2] The US National Institute of Standards and Technology does not recommend using degrees Rankine in NIST publications.[1]
Some key temperatures relating the Rankine scale to other temperature scales are shown in the table below.
Kelvin Celsius Fahrenheit Rankine 0 K −273.15 °C −459.67 °F 0 °R 273.15 K 0 °C 32 °F 491.67 °R 273.16 K 0.01 °C 32.018 °F 491.688 °R 373.1339 K 99.9839 °C 211.9710 °F 671.641 °R
## References
1. ^ a b B.8 Factors for Units Listed Alphabetically from Guide for the Use of the International System of Units (SI), NIST Special Publication 811, 2008 edition, Ambler Thompson and Barry N. Taylor
2. ^ a b http://www.physorg.com/tags/temperature/
3. ^ The ice point of purified water has been measured to be 0.000089(10) degrees Celsius - see Magnum, B.W. (June 1995). "Reproducibility of the Temperature of the Ice Point in Routine Measurements" (PDF). Nist Technical Note 1411. Retrieved 2007-02-11.
4. ^ For Vienna Standard Mean Ocean Water at one standard atmosphere (101.325 kPa) when calibrated solely per the two-point definition of thermodynamic temperature. Older definitions of the Celsius scale once defined the boiling point of water under one standard atmosphere as being precisely 100 °C. However, the current definition results in a boiling point that is actually 16.1 mK less. For more about the actual boiling point of water, see VSMOW in temperature measurement.
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# Simple English
The Rankine scale is a thermodynamic (absolute) temperature scale. It is based around absolute zero. Rankine is similar to the Kelvin scale in that it starts at absolute zero and 0°Ra is the same as 0K but is different as a change of 1°Ra is the same as a change of 1°F (Fahrenheit) and not 1°C (Celsius). Note that the abbreviation °R is ambiguous, as it can also refer to the Réaumur scale.
The Kelvin and Rankine temperature scales are defined so that absolute zero is 0 kelvins (K) or 0 degrees Rankine (°R). The Celsius and Fahrenheit scales are defined so that absolute zero is −273.15 °C or −459.67 °F.
The Rankine scale was originally used in the United States.
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http://crypto.stackexchange.com/questions/tagged/proof-of-work?sort=votes | 1,469,413,214,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824201.56/warc/CC-MAIN-20160723071024-00245-ip-10-185-27-174.ec2.internal.warc.gz | 60,097,390 | 21,381 | # Tagged Questions
A proof of work system is based on a mathematical problem that is moderate hard to solve but easy to verify.
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### Stateless proof-of-work system with 0-roundtrip time
I'm designing an API. To avoid abuse, I need to rate limit the requests somehow, but I don't want to do it per user as it's very easy to create new accounts automatically. I think a proof of work ... | 1,773 | 7,987 | {"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} | 2.875 | 3 | CC-MAIN-2016-30 | latest | en | 0.938458 |
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Cycle of Concentration of Boiler Water What is the cycle of concentration if the chloride content of boiler water is 186 ppm and the feedwater chloride content is 38 ppm? A scotch marine boiler has a furnace volume of 45.5 cu. ft. if 3825.2 cu. ft. of natural gas is burned per hour and each contains 1100 Btu, what is the rate of combustion
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PARAMETERS Calculations based on the following information: PARAMETERS Calculations based on the following information: GENERAL ISO- Kinetic Sampling Duct Gas velocity 1. Gas density 2. Differential pressure (if using Pitot Tube) Nozzle Diameter 1. Pump capacity 2. Duct gas velocity 3. Estimated PM conc. 4. Preferable Sampling Time Gas Density 1.
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Indirect method Indirect method determines the efficiency of a boiler by the sum of the major losses and by the fuel power of the boiler: φlosses η =1− (15) φinput where φ losses is the sum of the major losses within the boiler, and φ input is the fuel power of the boiler.
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• ### Appendix A: Boiler Calculations | Газогенераторы МСД
The efficiency of the feed pump is usually in the range of 70-85%. In SI units pump (kW) = (9.837 x h x specific gravity)/273 x n where Q = discharge in cubic meters per hour H = head in meters n = pump efficiency in percentage. In this example, the Bhp of the feed pump can be worked out as …
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If emission reporting is required, go to the "Natural Gas Consumption" Tab (green) and proceed with Option 1: Default Calculation Method or Option 2: Customized Calculation Method. b. In "Default Cal." Tab (blue), the user can modify the % increase for Annual Worst-Case Scenario. The default % …
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Basic Boiler Calculations • Feedwater = Make up + Condensate Return • Feedwater Flow (FWF) = Steam Make + Blowdown(BD) • Feedwater Flow (FWF) = Steam Make + Steam Make • Cycles -1 • Blowdown = Steam make or = FWF • Cycles –1 Cycles • % Blowdown = 1 as a % of FWF • Cycles
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https://bafybeiemxf5abjwjbikoz4mc3a3dla6ual3jsgpdr4cjr3oz3evfyavhwq.ipfs.dweb.link/wiki/Q%E2%80%93Q_plot.html | 1,656,234,660,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103037649.11/warc/CC-MAIN-20220626071255-20220626101255-00239.warc.gz | 172,148,642 | 19,967 | # Q–Q plot
Not to be confused with P–P plot.
A normal Q–Q plot of randomly generated, independent standard exponential data, (X ~ Exp(1)). This Q–Q plot compares a sample of data on the vertical axis to a statistical population on the horizontal axis. The points follow a strongly nonlinear pattern, suggesting that the data are not distributed as a standard normal (X ~ N(0,1)). The offset between the line and the points suggests that the mean of the data is not 0. The median of the points can be determined to be near 0.7
A normal Q–Q plot comparing randomly generated, independent standard normal data on the vertical axis to a standard normal population on the horizontal axis. The linearity of the points suggests that the data are normally distributed.
A Q–Q plot of a sample of data versus a Weibull distribution. The deciles of the distributions are shown in red. Three outliers are evident at the high end of the range. Otherwise, the data fit the Weibull(1,2) model well.
A Q–Q plot comparing the distributions of standardized daily maximum temperatures at 25 stations in the US state of Ohio in March and in July. The curved pattern suggests that the central quantiles are more closely spaced in July than in March, and that the July distribution is skewed to the left compared to the March distribution. The data cover the period 1893–2001.
In statistics, a Q–Q plot[1] ("Q" stands for quantile) is a probability plot, which is a graphical method for comparing two probability distributions by plotting their quantiles against each other. First, the set of intervals for the quantiles is chosen. A point (x, y) on the plot corresponds to one of the quantiles of the second distribution (y-coordinate) plotted against the same quantile of the first distribution (x-coordinate). Thus the line is a parametric curve with the parameter which is the (number of the) interval for the quantile.
If the two distributions being compared are similar, the points in the Q–Q plot will approximately lie on the line y = x. If the distributions are linearly related, the points in the Q–Q plot will approximately lie on a line, but not necessarily on the line y = x. Q–Q plots can also be used as a graphical means of estimating parameters in a location-scale family of distributions.
A Q–Q plot is used to compare the shapes of distributions, providing a graphical view of how properties such as location, scale, and skewness are similar or different in the two distributions. Q–Q plots can be used to compare collections of data, or theoretical distributions. The use of Q–Q plots to compare two samples of data can be viewed as a non-parametric approach to comparing their underlying distributions. A Q–Q plot is generally a more powerful approach to do this than the common technique of comparing histograms of the two samples, but requires more skill to interpret. Q–Q plots are commonly used to compare a data set to a theoretical model.[2][3] This can provide an assessment of "goodness of fit" that is graphical, rather than reducing to a numerical summary. Q–Q plots are also used to compare two theoretical distributions to each other.[4] Since Q–Q plots compare distributions, there is no need for the values to be observed as pairs, as in a scatter plot, or even for the numbers of values in the two groups being compared to be equal.
The term "probability plot" sometimes refers specifically to a Q–Q plot, sometimes to a more general class of plots, and sometimes to the less commonly used P–P plot. The probability plot correlation coefficient is a quantity derived from the idea of Q–Q plots, which measures the agreement of a fitted distribution with observed data and which is sometimes used as a means of fitting a distribution to data.
## Definition and construction
Q–Q plot for first opening/final closing dates of Washington State Route 20, versus a normal distribution.[5] Outliers are visible in the upper right corner.
A Q–Q plot is a plot of the quantiles of two distributions against each other, or a plot based on estimates of the quantiles. The pattern of points in the plot is used to compare the two distributions.
The main step in constructing a Q–Q plot is calculating or estimating the quantiles to be plotted. If one or both of the axes in a Q–Q plot is based on a theoretical distribution with a continuous cumulative distribution function (CDF), all quantiles are uniquely defined and can be obtained by inverting the CDF. If a theoretical probability distribution with a discontinuous CDF is one of the two distributions being compared, some of the quantiles may not be defined, so an interpolated quantile may be plotted. If the Q–Q plot is based on data, there are multiple quantile estimators in use. Rules for forming Q–Q plots when quantiles must be estimated or interpolated are called plotting positions.
A simple case is where one has two data sets of the same size. In that case, to make the Q–Q plot, one orders each set in increasing order, then pairs off and plots the corresponding values. A more complicated construction is the case where two data sets of different sizes are being compared. To construct the Q–Q plot in this case, it is necessary to use an interpolated quantile estimate so that quantiles corresponding to the same underlying probability can be constructed.
More abstractly,[4] given two cumulative probability distribution functions F and G, with associated quantile functions F−1 and G−1 (the inverse function of the CDF is the quantile function), the Q–Q plot draws the q-th quantile of F against the q-th quantile of G for a range of values of q. Thus, the Q–Q plot is a parametric curve indexed over [0,1] with values in the real plane R2.
## Interpretation
The points plotted in a Q–Q plot are always non-decreasing when viewed from left to right. If the two distributions being compared are identical, the Q–Q plot follows the 45° line y = x. If the two distributions agree after linearly transforming the values in one of the distributions, then the Q–Q plot follows some line, but not necessarily the line y = x. If the general trend of the Q–Q plot is flatter than the line y = x, the distribution plotted on the horizontal axis is more dispersed than the distribution plotted on the vertical axis. Conversely, if the general trend of the Q–Q plot is steeper than the line y = x, the distribution plotted on the vertical axis is more dispersed than the distribution plotted on the horizontal axis. Q–Q plots are often arced, or "S" shaped, indicating that one of the distributions is more skewed than the other, or that one of the distributions has heavier tails than the other.
Although a Q–Q plot is based on quantiles, in a standard Q–Q plot it is not possible to determine which point in the Q–Q plot determines a given quantile. For example, it is not possible to determine the median of either of the two distributions being compared by inspecting the Q–Q plot. Some Q–Q plots indicate the deciles to make determinations such as this possible.
The intercept and slope of a linear regression between the quantiles gives a measure of the relative location and relative scale of the samples. If the median of the distribution plotted on the horizontal axis is 0, the intercept of a regression line is a measure of location, and the slope is a measure of scale. The distance between medians is another measure of relative location reflected in a Q–Q plot. The "probability plot correlation coefficient" is the correlation coefficient between the paired sample quantiles. The closer the correlation coefficient is to one, the closer the distributions are to being shifted, scaled versions of each other. For distributions with a single shape parameter, the probability plot correlation coefficient plot (PPCC plot) provides a method for estimating the shape parameter – one simply computes the correlation coefficient for different values of the shape parameter, and uses the one with the best fit, just as if one were comparing distributions of different types.
Another common use of Q–Q plots is to compare the distribution of a sample to a theoretical distribution, such as the standard normal distribution N(0,1), as in a normal probability plot. As in the case when comparing two samples of data, one orders the data (formally, computes the order statistics), then plots them against certain quantiles of the theoretical distribution.[3]
## Plotting positions
The choice of quantiles from a theoretical distribution can depend upon context and purpose. One choice, given a sample of size n, is k / n for k = 1, …, n, as these are the quantiles that the sampling distribution realizes. The last of these, n / n, corresponds to the 100th percentile – the maximum value of the theoretical distribution, which is sometimes infinite. Other choices are the use of (k − 0.5) / n, or instead to space the points evenly in the uniform distribution, using k / (n + 1).[6]
Many other choices have been suggested, both formal and heuristic, based on theory or simulations relevant in context. The following subsections discuss some of these. A narrower question is choosing a maximum (estimation of a population maximum), known as the German tank problem, for which similar "sample maximum, plus a gap" solutions exist, most simply m + m/n - 1. A more formal application of this uniformization of spacing occurs in maximum spacing estimation of parameters.
### Expected value of the order statistic for a uniform distribution
The k / (n + 1) approach equals that of plotting the points according to the probability that the last of (n+1) randomly drawn values will not exceed the k-th smallest of the first n randomly drawn values.[7][8]
### Expected value of the order statistic for a standard normal distribution
In using a normal probability plot, the quantiles one uses are the rankits, the quantile of the expected value of the order statistic of a standard normal distribution.
More generally, Shapiro–Wilk test uses the expected values of the order statistics of the given distribution; the resulting plot and line yields the generalized least squares estimate for location and scale (from the intercept and slope of the fitted line).[9] Although this is not too important for the normal distribution (the location and scale are estimated by the mean and standard deviation, respectively), it can be useful for many other distributions.
However, this requires calculating the expected values of the order statistic, which may be difficult if the distribution is not normal.
### Median of the order statistics
Alternatively, one may use estimates of the median of the order statistics, which one can compute based on estimates of the median of the order statistics of a uniform distribution and the quantile function of the distribution; this was suggested by (Filliben 1975).[9]
This can be easily generated for any distribution for which the quantile function can be computed, but conversely the resulting estimates of location and scale are no longer precisely the least squares estimates, though these only differ significantly for n small.
### Heuristics
For the quantiles of the comparison distribution typically the formula k / (n + 1) is used. Several different formulas have been used or proposed as affine symmetrical plotting positions. Such formulas have the form (ka) / (n + 1 − 2a) for some value of a in the range from 0 to 1/2, which gives a range between k / (n + 1) and (k − 1/2) / n.
Other expressions include:
• (k − 0.3) / (n + 0.4).[10]
• (k − 0.3175) / (n + 0.365).[11]
• (k − 0.326) / (n + 0.348).[12]
• (k − ⅓) / (n + ⅓).[13]
• (k − 0.375) / (n + 0.25).[14]
• (k − 0.4) / (n + 0.2).[15]
• (k − 0.44) / (n + 0.12).[16]
• (k − 0.5) / (n).[17]
• (k − 0.567) / (n − 0.134).[18]
• (k − 1) / (n − 1).[19]
For large sample size, n, there is little difference between these various expressions.
### Filliben's estimate
The order statistic medians are the medians of the order statistics of the distribution. These can be expressed in terms of the quantile function and the order statistic medians for the continuous uniform distribution by:
where U(i) are the uniform order statistic medians and G is the quantile function for the desired distribution. The quantile function is the inverse of the cumulative distribution function (probability that X is less than or equal to some value). That is, given a probability, we want the corresponding quantile of the cumulative distribution function.
James J. Filliben (Filliben 1975) uses the following estimates for the uniform order statistic medians:
The reason for this estimate is that the order statistic medians do not have a simple form.
Wikimedia Commons has media related to Q-Q plot.
## Notes
1. Wilk, M.B.; Gnanadesikan, R. (1968), "Probability plotting methods for the analysis of data", Biometrika, Biometrika Trust, 55 (1): 1–17, doi:10.1093/biomet/55.1.1, JSTOR 2334448, PMID 5661047.
3. (Thode 2002, Section 2.2.2, Quantile-Quantile Plots, p. 21)
4. "SR 20 – North Cascades Highway – Opening and Closing History". North Cascades Passes. Washington State Department of Transportation. October 2009. Retrieved 2009-02-08.
5. Weibull, Waloddi (1939), "The Statistical Theory of the Strength of Materials", IVA Handlingar, Royal Swedish Academy of Engineering Sciences (No. 151)
6. Madsen, H.O.; et al. (1986), Methods of Structural Safety
7. Makkonen, L. (2008), "Bringing closure to the plotting position controversy", Communications in Statistics - Theory and Methods (37): 460–467
8. Testing for Normality, by Henry C. Thode, CRC Press, 2002, ISBN 978-0-8247-9613-6, p. 31
9. Engineering Statistics Handbook: Normal Probability Plot – Note that this also uses a different expression for the first & last points. cites the original work by (Filliben 1975). This expression is an estimate of the medians of U(k).
10. Distribution free plotting position, Yu & Huang
11. A simple (and easy to remember) formula for plotting positions; used in BMDP statistical package.
12. This is (Blom 1958)’s earlier approximation and is the expression used in MINITAB.
13. This plotting position was used by Irving I. Gringorten (Gringorten (1963)) to plot points in tests for the Gumbel distribution.
14. Hazen, Allen (1914), "Storage to be provided in the impounding reservoirs for municipal water supply", Transactions of the American Society of Civil Engineers (No. 77): 1547–1550
15. Used by Filliben (1975), these plotting points are equal to the modes of U(k).
## References
• This article incorporates public domain material from the National Institute of Standards and Technology website http://www.nist.gov.
• Blom, G. (1958), Statistical estimates and transformed beta variables, New York: John Wiley and Sons
• Chambers, John; William Cleveland, Beat Kleiner, and Paul Tukey (1983), Graphical methods for data analysis, Wadsworth Cite uses deprecated parameter |coauthors= (help)
• Cleveland, W.S. (1994) The Elements of Graphing Data, Hobart Press ISBN 0-9634884-1-4
• Filliben, J. J. (February 1975), "The Probability Plot Correlation Coefficient Test for Normality", Technometrics, American Society for Quality, 17 (1): 111–117, doi:10.2307/1268008, JSTOR 1268008.
• Gibbons, Jean Dickinson; Chakraborti, Subhabrata (2003), Nonparametric statistical inference (4th ed.), CRC Press, ISBN 978-0-8247-4052-8
• Gnanadesikan, R. (1977) Methods for Statistical Analysis of Multivariate Observations, Wiley ISBN 0-471-30845-5.
• Thode, Henry C. (2002), Testing for normality, New York: Marcel Dekker, ISBN 0-8247-9613-6 | 3,640 | 15,614 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2022-27 | latest | en | 0.912118 |
https://www.unitconverters.net/surface-charge-density/abcoulomb-square-meter-to-coulomb-square-meter.htm | 1,723,578,451,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641082193.83/warc/CC-MAIN-20240813172835-20240813202835-00577.warc.gz | 800,997,148 | 2,738 | Home / Surface Charge Density Conversion / Convert Abcoulomb/square Meter to Coulomb/square Meter
# Convert Abcoulomb/square Meter to Coulomb/square Meter
Please provide values below to convert abcoulomb/square meter to coulomb/square meter, or vice versa.
From: abcoulomb/square meter To: coulomb/square meter
### Abcoulomb/square Meter to Coulomb/square Meter Conversion Table
Abcoulomb/square MeterCoulomb/square Meter
0.01 abcoulomb/square meter0.1 coulomb/square meter
0.1 abcoulomb/square meter1 coulomb/square meter
1 abcoulomb/square meter10 coulomb/square meter
2 abcoulomb/square meter20 coulomb/square meter
3 abcoulomb/square meter30 coulomb/square meter
5 abcoulomb/square meter50 coulomb/square meter
10 abcoulomb/square meter100 coulomb/square meter
20 abcoulomb/square meter200 coulomb/square meter
50 abcoulomb/square meter500 coulomb/square meter
100 abcoulomb/square meter1000 coulomb/square meter
1000 abcoulomb/square meter10000 coulomb/square meter
### How to Convert Abcoulomb/square Meter to Coulomb/square Meter
1 abcoulomb/square meter = 10 coulomb/square meter
1 coulomb/square meter = 0.1 abcoulomb/square meter
Example: convert 15 abcoulomb/square meter to coulomb/square meter:
15 abcoulomb/square meter = 15 × 10 coulomb/square meter = 150 coulomb/square meter | 386 | 1,300 | {"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-2024-33 | latest | en | 0.277938 |
https://www.engineeringclicks.com/forum/threads/minimum-force-required-to-move-an-object.9002/ | 1,553,480,811,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203547.62/warc/CC-MAIN-20190325010547-20190325032547-00473.warc.gz | 742,666,586 | 14,114 | • Welcome to engineeringclicks.com
• # Minimum force required to move an object
Discussion in 'Calculations' started by george9421, Mar 14, 2019.
1. ### george9421Member
Joined:
Jan 2019
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Hello!
I know that for an object at rest, in order to move it, first STATIC FRICTION must be overcome
(F= μ N), where μ is the coefficient of friction between the two surfaces. Afterwards, while the object is in motion, SLIDING FRICTION is the resisting frictional force.
However, is this the same for an object with wheels? Must the same STATIC FRICTION be overcome first? And then when in motion, ROLLING FRICTION becomes the resisting force? If so, is STATIC FRICTION for an object with wheels calculated in the same manner?
Essentially my question boils down to this:
If you had a block weighing 100 kilos, one with wheels (assuming the weight of the wheels to be negligible) and one without, would the STATIC FRICTION at rest be the same in both cases?
I'm quite rusty on my physics concepts and would appreciate any help.
2.
3. ### s.weinbergWell-Known MemberEngineeringClicks Expert
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In an ideal situation when rolling without slipping, a wheel does not actually move relative to the ground at it's contact point.
Therefore, the very short answer is 'no' - not the same.
4. ### george9421Member
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Jan 2019
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If it is different is there any way of calculating the static friction, ie the minimum force required to move an object on wheels ?
Is it as simple as:
Force required to move object = (weight) * (coefficient of rolling friction)
(I'm designing a towing hook for a cart and wanted to make a conservative estimate of the force that would be applied on the hook)
Thanks again for your help
5. ### s.weinbergWell-Known MemberEngineeringClicks Expert
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You're looking at the wrong interface. The wheel contact doesn't have to move relative to the ground. You have to look at the force required to turn the wheel instead.
6. ### s.weinbergWell-Known MemberEngineeringClicks Expert
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Google 'rolling without slipping'. It should give you a good start. If you don't assume no slipping, you have to calculate twice (basically, you guess, and see if you're right), but assuming no slipping should give you a good approximation.
You're basically dealing with static friction at the wheel axle. The static friction at the ground gives you a reaction force. You have to generate a forward motion at the wheel center, assuming that the point in contact with the ground does not move.
7. ### george9421Member
Joined:
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Thank you for the suggestion, after reading I realised I had some misunderstandings with regard to the relevant principles. Just to clarify:
For a cart that is at rest, if a torque is applied to the wheel and it is less than the static friction, then it will initiate motion. Then using this torque the driving force applied (ie a push or pull) can be calculated.
However, some sources I read stated that the torque applied must, at the very least, be greater than that of rolling friction in order to initiate motion. While other sources stated that if the wheel and floor were both rigid, then any minimal torque is theoretically enough to initiate motion. Do you know know which is correct?
8. ### s.weinbergWell-Known MemberEngineeringClicks Expert
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I believe that the 2nd is correct. Without slipping, the contact point has no motion, so you're not overcoming the friction. If you DO overcome friction, then you have some slipping, which complicates things a bit.
However, you can't apply any minute force and get motion. There is friction at the hub of the wheel that needs to be overcome.
Think of it all in terms of a free body diagram. Friction at the bottom of the wheel is a single force. It will not prevent the center of the wheel from moving. It will just create a moment when coupled with a force at the center, causing the wheel to spin.
However, friction around the hub creates a counter-moment to any spinning of the wheel.
So you have two choices if you want motion.
1. Either create a moment to spin the wheel (either a direct torque, or a force forward at the wheel center that couples with the friction reaction force at the wheel base) that is larger than the counter-torque of static friction at the wheel hub. You will then roll without slipping.
2. Create a force at the center of the wheel that is insufficient to spin the wheel, but is larger than the static friction reaction at the wheel base. The wheel will slide forward without turning.
Or you can end up with a combo:
3. A force that is larger than the static friction at the center of the wheel, AND the moment created is enough to spin the wheel. Then you will have motion that is a combination of rolling and slipping.
Usually, 1 is what you're going to get, with 3 happening if you push super hard. Otherwise, your wheels are pretty useless
9. ### george9421Member
Joined:
Jan 2019
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Thank you very much this helped a lot! I do have one last question though, is there a method to calculate estimate friction around the hub/ counter moment to spinning wheels? Or perhaps some rules of thumb to make conservative estimates ?
10. ### ErichWell-Known Member
Joined:
Feb 2012
Posts:
260
2
Other factors to ponder, To go from not moving to moving you must accelerate, so mass of cart and accel must be factored in.
A real wheel on a real surface has "rolling resistance" Some of this is because the wheel and the surface deflect because of the weight, other source is hysteresis when the wheel flexes (rubber tire in this instance) Steel wheel on steel rail has low rolling resistance (That is advantage of a train) Wheel in soft sand is opposite case.
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See more:
Below is my Query :
Quote:
Select a.Sent, b.Recd, ((b.Recd-a.Sent)*1440) DIFF
from TableA b, TableB a
where b.type = 'AAA'
and b.ID = a.ID1
Where ID and ID1 are foreign Key
Both sent, Recd dates are in format like '
Quote:
5/9/2011 5:22:00 PM
'
But i am not getting :
Quote:
((b.Recd-a.Sent)*1440)
in minutes
Quote:
-000000000 06:48:00
I need the difference of DATE in Minutes only.
Posted 4-Feb-13 6:43am
Nibin22659
Updated 4-Feb-13 6:44am
v2
Rate this:
## Solution 1
```SELECT
trunc(DATE1-DATE2) days,
mod( trunc( ( DATE1-DATE2 ) * 24 ), 24) HOURS,
mod( trunc( ( DATE1-DATE2 ) * 1440 ), 60 ) MINUTES,
mod( trunc( ( DATE1-DATE2 ) * 86400 ), 60 ) SECONDS
FROM dual;```
change as per ur table
http://stackoverflow.com/questions/206222/oracle-best-select-statement-for-getting-the-difference-in-minutes-between-two[^]
Nibin22 4-Feb-13 23:39pm
It is Working !!! Thanks :)
Rate this:
## Solution 2
This should do it:
```select
round((second_date - first_date) * 1440,2)```
You need to round the value.
Nibin22 4-Feb-13 23:39pm
It is Working !!! Thanks :)
Rate this:
## Solution 3
I suggest that you try using the NUMTODSINTERVAL[^] function with the second parameter of 'MINUTE'.
``` NUMTODSINTERVAL(b.recd - a.sent, 'MINUTE')
```
Nibin22 4-Feb-13 23:39pm
It is Working !!! Thanks :)
Rate this:
## Solution 4
Thanks Friends,
I figured like below :
```Select a.Sent, b.Recd, (Extract Minute from b.Recd-a.Sent) DIFF
from TableA a, TableB b
where b.type = 'AAA'
and b.ID = a.ID1;```
Where " (Extract Minute from b.Recd-a.Sent) DIFF" returns the date difference in MINUTES
Nibin22 4-Feb-13 23:38pm
Nice!!! I didnt imagine this number of solutions will be there to find the date difference!!!
Thanks a lot Guys !!!! :)
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# ESE Achiever's plan || Paper 1 || Vector Analysis
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The line that passes the point (0, 1, 1) and (1, 0, -1) has parametric equation given by
Question 2
If C is the circle x2 + y2 = 4 Then
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equation of the tangent plane to the surface 2xz2 – 3xy – 4x = 7 at the point ( 1, -1, 2 )
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A vector field with a vanishing curl is called as.
• 84 attempts | 196 | 607 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2021-39 | latest | en | 0.819834 |
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Density is mass per volume-unit
ρ = m/V
• m mass in kg
• V volume in m3
• ρ density in kg m-3
Example
Find the volume of 25 g gold.
ρ = 19.3 x 103 kg m-3 m = 25 g = 0.025 kg
ρ = m/V 19.3 x 103 = 0.025/V (V) (19.3 x 103) = 0.025 V = 0.025/(19.3 x 103) = 1.30 x 10-6 m3=
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Subjects: | 2,436 | 9,329 | {"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-2013-48 | latest | en | 0.890042 |
https://fr.mathworks.com/matlabcentral/cody/problems/731-given-a-window-how-many-subsets-of-a-vector-sum-positive/solutions/2283431 | 1,597,355,393,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439739073.12/warc/CC-MAIN-20200813191256-20200813221256-00338.warc.gz | 261,378,429 | 15,777 | Cody
# Problem 731. Given a window, how many subsets of a vector sum positive
Solution 2283431
Submitted on 15 May 2020
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 0 -1 3 2 -3 1]; window = 2; y_correct = 3; assert(isequal(sum_positive(x, window),y_correct))
2 Pass
x = [1 0 -1 3 2 -3 1]; window = 3; y_correct = 3; assert(isequal(sum_positive(x, window),y_correct))
3 Fail
x = [1 0 -1 3 2 -3 1]; window = 4; y_correct = 4; assert(isequal(sum_positive(x, window),y_correct))
Assertion failed.
4 Fail
x = [1 0 -1 3 2 -3 1 1 1 1 1 1]; window = 2; y_correct = 8; assert(isequal(sum_positive(x, window),y_correct))
Assertion failed. | 267 | 772 | {"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-2020-34 | latest | en | 0.632801 |
https://www.gradesaver.com/textbooks/math/algebra/intermediate-algebra-for-college-students-7th-edition/chapter-4-section-4-1-solving-linear-inequalities-exercise-set-page-262/14 | 1,708,854,699,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474594.56/warc/CC-MAIN-20240225071740-20240225101740-00047.warc.gz | 802,140,519 | 13,021 | ## Intermediate Algebra for College Students (7th Edition)
$2\leq x$
$5(3-x)\leq3x-1\\15-5x\leq3x-1\\16-5x\leq3x\\16\leq8x\\2\leq x$ | 67 | 133 | {"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} | 2.953125 | 3 | CC-MAIN-2024-10 | latest | en | 0.623153 |
https://math.stackexchange.com/questions/3038661/closed-form-of-an-improper-integral-to-solve-the-period-of-a-dynamical-system/3049168 | 1,576,216,359,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540548544.83/warc/CC-MAIN-20191213043650-20191213071650-00163.warc.gz | 454,467,133 | 33,450 | Closed form of an improper integral to solve the period of a dynamical system
This improper integral comes from a problem of periodic orbit. The integral evaluates one half of the period.
In a special case, the integral is $$I=\int_{r_1}^{r_2}\frac{dr}{r\sqrt{\Phi^2(r,r_1)-1}}$$ where $$\Phi(u,v)=\frac{u\exp{(-u)}}{v\exp{(-v)}}$$
The interval follows $$\Phi(r_1,r_2)=1$$, $$r_1.
I have found a solution to a special case (by applying perturbation method to the original ODE), which is $$\lim_{r_1\rightarrow r_2} I =\pi$$ When $$r_1 \rightarrow r_2$$, we have $$r_1, r_2 \rightarrow r_0$$, where $$r_0$$ is the peak position of $$g(r)=r\exp{(-r)}$$.
The numerical verification is shown below:
$$\uparrow$$ The interval of the integral and the integrand
$$\uparrow$$ The integral as a function of $$r_2$$
My problem is to derive a closed form for $$I(r_1)$$, or even just a Taylor expansion about $$r_0$$. I appreciate any hint.
Thanks!
If you are interested, here is the general form of the integral: $$I=\int_{r_1}^{r_2}\frac{dr}{r\sqrt{\Phi^2(r,r_1)-1}}$$ where $$\Phi(u,v)=\frac{u\exp{(k(u))}}{v\exp{(k(v))}}$$ and $$k$$ is a decreasing function. The interval follows $$\Phi(r_1,r_2)=1$$, $$r_1.
By solving the original ODE using perturbation method, the solution to a special case is $$\lim_{r_1\rightarrow r_2} I =\frac{\pi}{\sqrt{1+r_0 k''(r_0)/k'(r_0)}}$$
When $$k(r)=-r$$, it reduces to $$\pi$$.
In fact, $$\lim_{r_1 \rightarrow r_2} I (k(r)=-C\cdot r^n) = \pi/\sqrt{n}$$.
Thanks to Fabian, the second derivative at $$r=1$$ matches $$I=\pi-\frac{\pi}{12}\epsilon^2+O(\epsilon^3)$$:
$$\uparrow$$ The above is the numerical second order derivative of figure 2.
I do not know how to obtain an explicit solution to the problem. However, it is possible to have a Taylor series of the integral $$I(r_1)$$ around $$r_1=1$$.
Let us first perform the substitution $$r= r_1 (1-x) + r_2 x$$ such that the boundaries of the integral do not depend on $$r_1$$. In particular, we obtain the expression $$I(r_1) =\int_0^1 \frac{r_2 -r_1}{(r_1 (1-x) + r_2 x) [\Phi(r_1 (1-x) + r_2 x,r_1)^2 -1 ]^{1/2}}\,dx\,.$$
Next, we need a relation between $$r_2$$ and $$r_1$$. If you look at the function $$r\exp(-r)$$ you see that it is monotonous on the interval $$r\in[0,1]$$ and $$r\in[1,\infty]$$. The inverse of this function is commonly called the Lambert W function. In particular, the inverse of the respective branches are denoted by $$r=- W(-x) \in [0,1], \qquad r=-W_{-1}(-x) \in [1,\infty]\,.$$ With this notation, we have $$r_2 =-W_{-1}(-r_1e^{-r_1}), \quad r_1 =-W(-r_2e^{-r_2})\,.$$
For $$r_1$$ close to $$1$$, we need the expansion of $$W$$ close to the branch point (see (4.26) of this paper). We obtain $$r_2 = 1+ \epsilon + \frac{2}{3} \epsilon^2 + \frac{4}{9} \epsilon^3 + \frac{44}{135}\epsilon^4 + O(\epsilon^5) \tag{1}$$ with $$\epsilon = 1-r_1$$.
Investigating first the point $$r_1=r_2=1$$. We set $$r_1 = 1-\epsilon$$, $$r_2 = 1+\epsilon$$ (we know from (1) that $$r_1$$ and $$r_2$$ approach 1 from below and above at equal rate). To zeroth order in $$\epsilon$$, we obtain $$I(1) =\int_0^1 \frac{1}{\sqrt{x(1-x)}}\,dx = \pi \,$$ as you have already observed.
In a next step, we look at $$I(1) -I(r_1)$$ for $$r_1$$ close to $$1$$. Using (1), we expand to third order in $$\epsilon$$. We obtain $$I(1)- I(r_1) = \int_0^1\left[\frac{\left(-30 x^2+34 x-5\right) \epsilon ^2}{9 \sqrt{(1-x) x}}+\frac{2 \left(472 x^3-858 x^2+422 x-33\right) \epsilon ^3}{135 \sqrt{(1-x) x}}+\frac{2 (2 x-1) \epsilon }{3 \sqrt{(1-x) x}}\right]dx= \frac{\pi}{12} \epsilon^2 + \frac{\pi}{18} \epsilon^3+O(\epsilon^4)\,.$$
To obtain a higher order approximation, we need more terms in (1). In particular, we have $$r_2 = 1+\epsilon +\frac{2 \epsilon ^2}{3}+\frac{4 \epsilon ^3}{9}+\frac{44 \epsilon ^4}{135}+\frac{104 \epsilon ^5}{405}+\frac{40 \epsilon ^6}{189}+\frac{7648 \epsilon ^7}{42525}+\frac{2848 \epsilon ^8}{18225}+\frac{31712 \epsilon ^9}{229635}+\frac{23429344 \epsilon ^{10}}{189448875} +O(\epsilon^{11})\,.$$
Now, the expansion of the integral in $$\epsilon$$ yields $$I(r_1) = \pi -\frac{\pi \epsilon ^2}{12}-\frac{\pi \epsilon ^3}{18}-\frac{23 \pi \epsilon ^4}{576}-\frac{67 \pi \epsilon ^5}{2160}-\frac{7613 \pi \epsilon ^6}{311040}-\frac{21419 \pi \epsilon ^7}{1088640}-\frac{320153 \pi \epsilon ^8}{19906560}-\frac{31342051 \pi \epsilon ^9}{2351462400} + O(\epsilon^{10})\,.$$
• Thanks, Fabian! I have verified your derivation in the problem description. I really appreciate your inspirational solution. – Shengkai Li Dec 23 '18 at 4:49
• All right. You have proved $\lim\limits_{r_1\to r_2} = \pi.$ $(+1),$ – Yuri Negometyanov Dec 23 '18 at 8:27
Firstly, the integral $$I = \int\limits_{r_1}^{r_2}\dfrac{e^r\,\mathrm dr}{r^2\sqrt{\left(\dfrac{e^{r_1}}{r_1}\right)^2-\left(\dfrac{e^r}r\right)^2}}\tag1$$ exists iff $$r_2\le 1,$$ because the function $$\dfrac1r e^r$$ has minimum at $$r=1.$$
Taking in account that $$\mathrm d\left(\dfrac{e^r}r\right)=\left(\dfrac1r - \dfrac1{r^2}\right)e^r\,\mathrm dr,$$ one can get $$I = \int\limits_{r_1}^{r_2}\dfrac{e^r\,\mathrm dr}{r^2\sqrt{\left(\dfrac{e^{r_1}}{r_1}\right)^2-\left(\dfrac{e^r}r\right)^2}} = \int\limits_{r_1}^{r_2}\dfrac{e^r\,\mathrm dr}{r\sqrt{\left(\dfrac{e^{r_1}}{r_1}\right)^2-\left(\dfrac{e^r}r\right)^2}} - \int\limits_{r_1}^{r_2}\dfrac1{\sqrt{\left(\dfrac{e^{r_1}}{r_1}\right)^2-\left(\dfrac{e^r}r\right)^2}}d\left(\dfrac {e^r}r\right)\,\mathrm dr$$ $$=\int\limits_{r_1}^{r_2}\dfrac{e^r\,\mathrm dr}{r\sqrt{\left(\dfrac{e^{r_1}}{r_1}\right)^2-\left(\dfrac{e^r}r\right)^2}} - \mathrm{arcsin}\left(\dfrac {r_1}{r}e^{r-r_1}\right)\Big|_{r_1}^{r_2}= I_1 + \mathrm{arccos}\left(\dfrac{r_1}{r_2}e^{r_2-r_1}\right),\tag1$$ where $$I_1 = \int\limits_{r_1}^{r_2}\dfrac{e^r\,\mathrm dr}{r\sqrt{\left(\dfrac{e^{r_1}}{r_1}\right)^2-\left(\dfrac{e^r}r\right)^2}}.\tag2$$ Note that $$I_1 \le I,$$ because $$r_2 \le1.$$
I cannot obtain the closed form for $$(2).$$
On the other hand, using Taylor series at $$x=1$$ in the form of $$\dfrac {e^x} {x\sqrt{\dfrac{e^{2a}}{a^2}-\dfrac{e^{2x}}{x^2}}} = \dfrac e{\sqrt{\dfrac{e^{2a}}{a^2}-e^2}} - \dfrac{e^{2a+1}(x-1)^2}{2\left(e^2 a^2-e^{2a}\right) \sqrt{\dfrac{e^{2a}}{a^2}-e^2}} + \dfrac{e^{2a+1}(x-1)^3}{3\left(e^2 a^2-e^{2a}\right) \sqrt{\dfrac{e^{2a}}{a^2}-e^2}}$$ $$+ \dfrac{3e^{4a+1}(x-1)^4}{8\left(e^{2a}-e^2 a^2\right)^2 \sqrt{\dfrac{e^{2a}}{a^2}-e^2}} + \dfrac{\left(-4e^{2a+3}a^2-11e^{4a+1}\right)(x-1)^5}{30\left(e^{2a}-e^2a^2\right)^2\sqrt{\dfrac{e^{2a}}{a^2}-e^2}} + \dots$$ (see also Wolfram Alpha), one can get the estimation
$$I_1 = \dfrac1{\sqrt{\dfrac{e^{2r_1-2}}{r_1^2}-1}} \int_{r_1}^{r_2}\Bigg(1 - \dfrac{e^{2r_1}(r-1)^2}{2\left(e^2 r_1^2-e^{2r_1}\right)} + \dfrac{e^{2r_1}(r-1)^3}{3\left(e^2 r_1^2-e^{2r_1}\right)}$$ $$+ \dfrac{3e^{4r_1}(r-1)^4}{8\left(e^{2r_1}-e^2 r_1^2\right)^2} + \dfrac{\left(-4e^{2r_1+3}r_1^2-11e^{4r_1}\right)(r-1)^5}{30\left(e^{2r_1}-e^2r_1^2\right)^2} + \dots\Bigg)\,\mathrm dr,$$
$$I_1 = \dfrac {1}{\sqrt{\dfrac{e^{2r_1-2}}{r_1^2}-1}}\Bigg((r_2-r_1) - \dfrac{e^{2r_1}\left((r_2-1)^3-(r_1-1)^3\right)}{6\left(e^2 r_1^2-e^{2r_1}\right)} + \dfrac{e^{2r_1+1}\left((r_2-1)^4-(r_1-1)^4\right)}{12\left(e^2 r_1^2-e^{2r_1}\right) }$$ $$+ \dfrac{3e^{4r_1+1}\left((r_2-1)^5-(r_1-1)^5\right)}{40\left(e^{2r_1}-e^2 r_1^2\right)^2} + \dfrac{\left(-4e^{2r_1+3}r_1^2-11e^{4r_1+1}\right)\left((r_2-1)^6-(r_1-1)^6\right)}{180\left(e^{2r_1}-e^2r_1^2\right)^2} + \dots\Bigg)$$
• Thanks, Yuri. I appreciate your time on this problem! – Shengkai Li Dec 23 '18 at 4:51 | 3,302 | 7,529 | {"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": 76, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2019-51 | latest | en | 0.771473 |
http://www.enotes.com/homework-help/when-we-throw-ball-vertically-upward-with-any-289709 | 1,477,429,663,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988720380.80/warc/CC-MAIN-20161020183840-00398-ip-10-171-6-4.ec2.internal.warc.gz | 427,051,299 | 9,951 | # When we throw a ball vertically upward with any force, is the time to reach the highest point equal to the time to reach the ground?
mwmovr40 | College Teacher | (Level 1) Associate Educator
Posted on
The answer to a question like this depends on the conditions of the problem. Under normal, Earth-like conditions the answer would have to be "no" for the following two reasons:
The person throwing the ball will release it above the ground. It might take the same amount of time to reach the highest point and to return to the height from which it was released. However, unless the person is standing in a hole when he throws the ball, it will take longer for the ball to continue and eventually strike the ground.
Also, the ball will experience air resistance which means its velocity on the return trip will be less than that with which it is initially launched and consequently it will take longer to get back down.
However, in the world of Physics' problems we usually make some assumptions to "simplify the problem". For example, we assume that the ball is experiencing free fall once it leaves the hand of the thrower and consequently there is no air resistance to influence the flight of the ball. We might also assume that the distance from the height of release to the ground is negligibly small and so the increase in distance to reach the ground may be ignored.
By making these two assumptions, we can argue from the symmetry of constant acceleration problem (it is in the form of an inverted parabola) that the time of flight to the highest point will equal the time of flight for the return trip. | 338 | 1,623 | {"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-2016-44 | latest | en | 0.953599 |
https://convertoctopus.com/19-9-centimeters-to-yards | 1,719,117,913,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862430.93/warc/CC-MAIN-20240623033236-20240623063236-00086.warc.gz | 155,045,854 | 7,271 | ## Conversion formula
The conversion factor from centimeters to yards is 0.010936132983377, which means that 1 centimeter is equal to 0.010936132983377 yards:
1 cm = 0.010936132983377 yd
To convert 19.9 centimeters into yards we have to multiply 19.9 by the conversion factor in order to get the length amount from centimeters to yards. We can also form a simple proportion to calculate the result:
1 cm → 0.010936132983377 yd
19.9 cm → L(yd)
Solve the above proportion to obtain the length L in yards:
L(yd) = 19.9 cm × 0.010936132983377 yd
L(yd) = 0.2176290463692 yd
The final result is:
19.9 cm → 0.2176290463692 yd
We conclude that 19.9 centimeters is equivalent to 0.2176290463692 yards:
19.9 centimeters = 0.2176290463692 yards
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 yard is equal to 4.5949748743719 × 19.9 centimeters.
Another way is saying that 19.9 centimeters is equal to 1 ÷ 4.5949748743719 yards.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that nineteen point nine centimeters is approximately zero point two one eight yards:
19.9 cm ≅ 0.218 yd
An alternative is also that one yard is approximately four point five nine five times nineteen point nine centimeters.
## Conversion table
### centimeters to yards chart
For quick reference purposes, below is the conversion table you can use to convert from centimeters to yards
centimeters (cm) yards (yd)
20.9 centimeters 0.229 yards
21.9 centimeters 0.24 yards
22.9 centimeters 0.25 yards
23.9 centimeters 0.261 yards
24.9 centimeters 0.272 yards
25.9 centimeters 0.283 yards
26.9 centimeters 0.294 yards
27.9 centimeters 0.305 yards
28.9 centimeters 0.316 yards
29.9 centimeters 0.327 yards | 530 | 1,831 | {"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-2024-26 | latest | en | 0.870784 |
https://oeis.org/A118712 | 1,680,267,085,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949642.35/warc/CC-MAIN-20230331113819-20230331143819-00472.warc.gz | 487,702,607 | 4,366 | The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A118712 a(n) = Determinant of n X n circulant matrix whose first row is A000001(1), A000001(2), ..., A000001(n) where A000001(n) = number of groups of order n. 1
1, 0, 0, -5, 6, -16, 9, -134400, 647248, -1711908, 6076067, -85248000, 116477425, -1764364437, 909276004, -522319050599375232, 14313181351994538493, -165893335414907083200, 2939566160282258664451, -5007637771411479278976, 75399747694572065660672 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,4 LINKS Eric Weisstein's World of Mathematics, Circulant Matrix. EXAMPLE a(4) = -5 because of the determinant -5 = |1,1,1,2| |2,1,1,1| |1,2,1,1| |1,1,2,1|. a(11) = 6076067 = determinant |1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1| |1, 1, 1, 1, 2, 1, 2, 1, 5, 2, 2| |2, 1, 1, 1, 1, 2, 1, 2, 1, 5, 2| |2, 2, 1, 1, 1, 1, 2, 1, 2, 1, 5| |5, 2, 2, 1, 1, 1, 1, 2, 1, 2, 1| |1, 5, 2, 2, 1, 1, 1, 1, 2, 1, 2| |2, 1, 5, 2, 2, 1, 1, 1, 1, 2, 1| |1, 2, 1, 5, 2, 2, 1, 1, 1, 1, 2| |2, 1, 2, 1, 5, 2, 2, 1, 1, 1, 1| |1, 2, 1, 2, 1, 5, 2, 2, 1, 1, 1| |1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 1|. PROG (GAP) A118712 := n -> DeterminantMat(List([0..n-1], i->List([0..n-1], j->NrSmallGroups(((j-i) mod n)+1)))); # Eric M. Schmidt, Nov 17 2013 CROSSREFS Cf. A000001, A048954, A052182, A066933, A086459, A086569. Sequence in context: A019071 A028285 A007430 * A130878 A104422 A026547 Adjacent sequences: A118709 A118710 A118711 * A118713 A118714 A118715 KEYWORD sign AUTHOR Jonathan Vos Post, May 20 2006 EXTENSIONS a(1) corrected by and more terms from Eric M. Schmidt, Nov 17 2013 STATUS approved
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Last modified March 31 08:44 EDT 2023. Contains 361645 sequences. (Running on oeis4.) | 928 | 2,016 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2023-14 | latest | en | 0.459392 |
https://puzzling.stackexchange.com/questions/107012/the-five-main-ones | 1,718,677,953,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861746.4/warc/CC-MAIN-20240618011430-20240618041430-00506.warc.gz | 427,634,539 | 38,644 | # The Five Main Ones
Here is a classic A is to B as C is to D puzzle. Try see if you can figure out the answer.
$$F:BRW::I:GRW::P:GR::R:BRY::?:?$$
Note: Some minor simplification was needed on some of them. But that shouldn't cause any problems solving the puzzle. Once you see the pattern, you'll realize what the simplification was.
S : RY
Reasoning:
The left side of an analogy is the first letter of a country which defines a Romance language: France, Italy, Portugal, Romania. The right side of the analogy is the first letters of the colors on its national flag, sorted in alphabetical order.
France: Blue, Red, White
Italy: Green, Red, White
Portugal: Green, Red
Romania: Blue, Red, Yellow
The missing member is Spain, with a red and yellow flag, hence S : RY.
As an additional check, notice that in the presented analogies the countries are also in alphabetical order.
• This is ofc correct, well done! Commented Jan 31, 2021 at 15:17 | 247 | 952 | {"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": 1, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2024-26 | latest | en | 0.929039 |
https://www.erldocs.com/r14bextra/mochiweb/mochinum.html | 1,638,573,251,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362919.65/warc/CC-MAIN-20211203212721-20211204002721-00397.warc.gz | 804,652,796 | 2,165 | mochinum
Useful numeric algorithms for floats that cover some deficiencies in the math module.
Useful numeric algorithms for floats that cover some deficiencies in the math module. More interesting is digits/1, which implements the algorithm from: http://www.cs.indiana.edu/~burger/fp/index.html See also "Printing Floating-Point Numbers Quickly and Accurately" in Proceedings of the SIGPLAN '96 Conference on Programming Language Design and Implementation.
digits(N::number()) -> string()
Returns a string that accurately represents the given integer or float using a conservative amount of digits. Great for generating human-readable output, or compact ASCII serializations for floats.
frexp(F::float()) -> {Frac::float(), Exp::float()}
Return the fractional and exponent part of an IEEE 754 double, equivalent to the libc function of the same name. F = Frac * pow(2, Exp).
int_ceil(F::float()) -> integer()
Return the ceiling of F as an integer. The ceiling is defined as F when F == trunc(F); trunc(F) when F < 0; trunc(F) + 1 when F > 0.
int_pow(X::integer(), N::integer()) -> Y::integer()
Moderately efficient way to exponentiate integers. int_pow(10, 2) = 100.
Bob Ippolito bob@mochimedia.com | 283 | 1,210 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2021-49 | longest | en | 0.685759 |
https://learn.careers360.com/school/question-explain-solution-for-rd-sharma-maths-class-12-chapter-6-adjoint-and-inverse-of-a-matrix-exercise-very-short-answer-type-question-5-maths-textbook-solution/?question_number=5.0 | 1,716,184,593,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058222.5/warc/CC-MAIN-20240520045803-20240520075803-00027.warc.gz | 315,151,625 | 37,251 | #### Explain Solution for RD Sharma Maths Class 12 Chapter 6 Adjoint and inverse of a Matrix Exercise very short answer type Question 5 maths textbook solution
VSQ :5
Answer : $\frac{1}{10}$
Given : A is non-singular square matrix and $\left | A \right |=10$
you must know about the concept of non-singular matrix
Solution : We know that
$\left | A \right |^{-1}=\frac{1}{\left | A \right |}$
$\left | A \right |^{-1}=\frac{1}{10} \ \ \ \ \ \ \ \because \left ( \left | A \right |=10 \right)$ | 151 | 499 | {"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": 4, "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} | 3.890625 | 4 | CC-MAIN-2024-22 | latest | en | 0.696107 |
http://www.mathisfunforum.com/viewtopic.php?pid=401798 | 1,553,117,429,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202471.4/warc/CC-MAIN-20190320210433-20190320232433-00206.warc.gz | 320,116,281 | 4,511 | Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °
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#1 2018-01-15 23:00:15
Member
From: Planet Mars
Registered: 2016-11-15
Posts: 808
A little nice puzzle
You are given 8 balls. One of the balls is heavier than the rest which are all the same weight. It is impossible to determine the heavier ball by the shape, size or lifting it up. You are given a 2-plated beam balance. Using the balance twice, how can you determine which ball is heavier.
Practice makes a man perfect.
There is no substitute to hard work
All of us do not have equal talents but everybody has equal oppurtunities to build their talents.-APJ Abdul Kalam
Offline
#2 2018-01-15 23:38:22
ganesh
Registered: 2005-06-28
Posts: 27,000
Re: A little nice puzzle
It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
Offline
#3 2018-01-16 07:29:11
Member
From: Planet Mars
Registered: 2016-11-15
Posts: 808
Re: A little nice puzzle
ganesh wrote:
I asked this problem long time back when I was young.
Oh really? Actually I heard this problem from my friend.
Practice makes a man perfect.
There is no substitute to hard work
All of us do not have equal talents but everybody has equal oppurtunities to build their talents.-APJ Abdul Kalam
Offline | 411 | 1,467 | {"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-2019-13 | latest | en | 0.907987 |
https://tutorialspoint.dev/computer-science/engineering-mathematics/mathematics-rings-integral-domains-and-fields | 1,639,046,341,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363791.16/warc/CC-MAIN-20211209091917-20211209121917-00602.warc.gz | 643,560,932 | 8,265 | # Mathematics | Rings, Integral domains and Fields
Prerequisite – Mathematics | Algebraic Structure
Ring – Let addition (+) and Multiplication (.) be two binary operations defined on a non empty set R. Then R is said to form a ring w.r.t addition (+) and multiplication (.) if the following conditions are satisfied:
1. (R, +) is an abelian group ( i.e commutative group)
2. (R, .) is a semigroup
3. For any three elements a, b, c R the left distributive law a.(b+c) =a.b + a.c and the right distributive property (b + c).a =b.a + c.a holds.
Therefore a non- empty set R is a ring w.r.t to binary operations + and . if the following conditions are satisfied.
1. For all a, b R, a+b R,
2. For all a, b, c R a+(b+c)=(a+b)+c,
3. There exists an element in R, denoted by 0 such that a+0=a for all a R
4. For every a R there exists an y R such that a+y=0. y is usually denoted by -a
5. a+b=b+a for all a, b R.
6. a.b R for all a.b R.
7. a.(b.c)=(a.b).c for all a, b R
8. For any three elements a, b, c R a.(b+c) =a.b + a.c and (b + c).a =b.a + c.a. And the ring is denoted by (R, +, .).
R is said to be a commutative ring if the multiplication is commutative.
Some Examples –
1. (, + ) is a commutative group .(, .) is a semigroup. The disrtributive law also holds. So, ((, +, .) is a ring.
2. Ring of Integers modulo n: For a n let be the classes of residues of integers modulo n. i.e ={).
(, +) is a commutative group ere + is addition(mod n).
(, .) is a semi group here . denotes multiplication (mod n).
Also the distriutive laws hold. So ((, +, .) is a ring.
Many other examples also can be given on rings like (, +, .), (, +, .) and so on.
Before discussing further on rings, we define Divisor of Zero in A ringand the concept of unit.
Divisor of Zero in A ring –
In a ring R a non-zero element is said to be divisor of zero if there exists a non-zero element b in R such that a.b=0 or a non-zero element c in R such that c.a=0 In the first case a is said to be a left divisor of zero and in the later case a is said to be a right divisor of zero . Obviously if R is a commutative ring then if a is a left divisor of zero then a is a right divisor of zero also .
Example – In the ring (, +, .) are divisors of zero since
and so on .
On the other hand the rings (, +, .), (, +, .), (, +, .) contains no divisor of zero .
Units –
In a non trivial ring R( Ring that contains at least to elements) with unity an element a in R is said to be an unit if there exists an element b in R such that a.b=b.a=I, I being the unity in R. b is said to be multiplicative inverse of a.
Some Important results related to Ring:
1. If R is a non-trivial ring(ring containing at least two elements ) withunity I then I 0.
2. If I be a multiplicative identity in a ring R then I is unique .
3. If a be a unit in a ring R then its multiplicative inverse is unique .
4. In a non trivial ring R the zero element has no multiplicative inverse .
Now we introduce a new concept Integral Domain.
Integral Domain – A non -trivial ring(ring containing at least two elements) with unity is said to be an integral domain if it is commutative and contains no divisor of zero ..
Examples –
The rings (, +, .), (, +, .), (, +, .) are integral domains.
The ring (2, +, .) is a commutative ring but it neither contains unity nor divisors of zero. So it is not an integral domain.
Next we will go to Field .
Field – A non-trivial ring R wit unity is a field if it is commutative and each non-zero element of R is a unit . Therefore a non-empty set F forms a field .r.t two binary operations + and . if
1. For all a, b F, a+b F,
2. For all a, b, c F a+(b+c)=(a+b)+c,
3. There exists an element in F, denoted by 0 such that a+0=a for all a F
4. For every a R there exists an y R such that a+y=0. y is usually denoted by (-a)
5. a+b=b+a for all a, b F.
6. a.b F for all a.b F.
7. a.(b.c)=(a.b).c for all a, b F
8. There exists an element I in F, called the identity element such that a.I=a for all a in F
9. For each non-zero element a in F there exists an element, denoted by in F such that =I.
10. a.b =b.a for all a, b in F .
11. a.(b+c) =a.b + a.c for all a, b, c in F
Examples – The rings (, +, .), (, + . .) are familiar examples of fields.
Some important results:
1. A field is an integral domain.
2. A finite integral domain is a field.
3. A non trivial finite commutative ring containing no divisor of zero is an integral domain | 1,301 | 4,419 | {"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} | 4.09375 | 4 | CC-MAIN-2021-49 | longest | en | 0.850386 |
https://ged-testprep.com/question/what-is-the-x-value-such-that-y-15--3x-and-6134697797091328/ | 1,713,690,936,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817729.87/warc/CC-MAIN-20240421071342-20240421101342-00653.warc.gz | 246,294,362 | 20,889 | Scan QR code or get instant email to install app
Question:
# What is the x value such that y =15 - 3x and y = 0?
A 5
explanation
y =15 - 3x and y = 0
Hence 15 - 3x = 0
Hence 3x = 15 x = 5 | 79 | 191 | {"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-2024-18 | longest | en | 0.803625 |
http://atcoder.noip.space/contest/arc082/a | 1,725,738,616,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650920.0/warc/CC-MAIN-20240907193650-20240907223650-00015.warc.gz | 2,858,937 | 2,096 | ARC082 C - Together - Atcoder
# Home
Contest: Task: Related: TaskB
Score : $300$ points
### Problem Statement
You are given an integer sequence of length $N$, $a_1,a_2,...,a_N$.
For each $1≤i≤N$, you have three choices: add $1$ to $a_i$, subtract $1$ from $a_i$ or do nothing.
After these operations, you select an integer $X$ and count the number of $i$ such that $a_i=X$.
Maximize this count by making optimal choices.
### Constraints
• $1≤N≤10^5$
• $0≤a_i<10^5 (1≤i≤N)$
• $a_i$ is an integer.
### Input
The input is given from Standard Input in the following format:
$N$
$a_1$ $a_2$ .. $a_N$
### Output
Print the maximum possible number of $i$ such that $a_i=X$.
### Sample Input 1
7
3 1 4 1 5 9 2
### Sample Output 1
4
For example, turn the sequence into $2,2,3,2,6,9,2$ and select $X=2$ to obtain $4$, the maximum possible count.
### Sample Input 2
10
0 1 2 3 4 5 6 7 8 9
### Sample Output 2
3
### Sample Input 3
1
99999
### Sample Output 3
1 | 347 | 983 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-38 | latest | en | 0.620106 |
https://au.mathworks.com/matlabcentral/answers/1696155-convert-numpy-ndarray-to-matlab-mlarray-in-python | 1,680,120,712,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949025.18/warc/CC-MAIN-20230329182643-20230329212643-00278.warc.gz | 144,303,950 | 29,896 | # Convert numpy ndarray to matlab mlarray in python
32 views (last 30 days)
Gan Lee on 14 Apr 2022
mlarray to ndarray: np.asarray(x._data, dtype=dtype)
but inversely, ndarray to mlarray: matlab.double(x.tolist()), which is extremely slow, is there a more efficient way to do this? Thank u for answering.
Al Danial on 28 Apr 2022
Looks like you're making the data load unnecessarily complicated. The file actually loads cleanly into a NumPy array:
x = data['data']
print(x)
produces
array([[0.00000000e+00, 5.90000000e+01, 5.90000000e+01, 2.25296241e+05],
[1.00000000e+00, 6.20000000e+01, 5.81599120e+01, 5.93159561e+04],
[2.00000000e+00, 1.00000000e+02, 9.47518190e+01, 3.22666379e+04],
...,
[2.04500000e+03, 4.00000000e+00, 4.88991300e+00, 3.01840538e+04],
[2.04600000e+03, 2.00000000e+00, 2.26899200e+00, 6.46032757e+04],
[2.04700000e+03, 1.00000000e+00, 1.00000000e+00, 1.18671912e+05]])
print(data.keys()) if key is None else None
return data[key]
then call it like this
Al Danial on 21 Apr 2022
Which version of MATLAB? 2020a and newer (I don't have easy access to older versions) should just be able to do
>> mx = double(x);
without a conversion to a list.
##### 2 CommentsShowHide 1 older comment
Christopher Wunder on 23 Aug 2022
This is not possible for me either.
The function _is_initializer in matlab._internal.mlarray_utils.py checks for the input to be of type collections.abc.Sequence and a numpy.ndarray fails to be of such a type. The only way (without altering the package) is to convert the array beforehand or pass any kind of Sequence to it instead of an array.
Al Danial on 24 Apr 2022
Now I'm curious what is in your variable x. Can you make a small version of this data, write it to a .mat file, then attach the .mat file?
Gan Lee on 27 Apr 2022
Edited: Gan Lee on 27 Apr 2022
Here is part of my python code:
# -*- coding: utf-8 -*-
import numpy as np
import h5py
import matlab
from matlab import engine
from matlab import mlarray
print(data.keys()) if key is None else None
return data.get(key)[:].astype(np.double)
def mlarray2ndarray(x: mlarray):
return np.asarray(x._data, dtype=np.double)
def ndarray2mlarray(x: np.ndarray):
return matlab.double(x)
if __name__ == "__main__":
eng = engine.start_matlab()
pth_mat = r".\data.mat"
# np to matlab
# mx1: mlarray = ndarray2mlarray(x1) #WRONG
mx1 = matlab.double(x.tolist()) #OK, BUT VERY SLOWLY
# matlab to np
x2: np.ndarray = mlarray2ndarray(mx2) #OK
# ......
eng.exit()
here is errcode:
ValueError
initializer must be a rectangular nested sequence
### Categories
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### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 858 | 2,751 | {"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-2023-14 | latest | en | 0.748768 |
http://www.greenbuildingadvisor.com/community/forum/energy-efficiency-and-durability/22267/how-do-you-calculate-water-heater-recovery-ef | 1,529,493,784,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863518.39/warc/CC-MAIN-20180620104904-20180620124904-00043.warc.gz | 418,270,952 | 18,773 | # How do you calculate Water Heater Recovery Efficiency?
Posted Nov 5, 2011 6:23 PM ET
1.
Andrew,
I'm confused by your question. Do you want to know the recovery rate or the water heating efficiency?
Posted Nov 7, 2011 9:52 AM ET
2.
Martin,
I am sorry you got confused.
This was my exact question, how to calculate "recovery efficiency" for residential water heaters because we are often required to list a units 2 specs, energy factor and "recovery efficiency" for residential water heaters.
Commercial water heaters are another story.
Keep in mind that the issue is that the "recovery efficiency" criteria is typically not listed, and this is why I want to learn about how it is calculated.
Recovery efficiency is also different than thermal efficiency to my knowledge.
Thanks,
Andrew
Posted Nov 7, 2011 11:45 AM ET
3.
Thanks, Andrew. I guess I just displayed my ignorance.
I just did some Googling, and learned that "Recovery Efficiency ... is the energy-to-hot-water conversion ratio. For an electric water heater, the RE is 100%. For gas water heaters the conversion rate is typically 76-78%, with expensive high efficiency units reaching 94%. This is because some energy must be left in the combustion byproducts to assure good venting."
http://energyexperts.org/EnergySolutionsDatabase/ResourceDetail.aspx?id=594
I'm not sure what the standardized test procedure is. Probably there is an ASTM procedure. Anybody have more info?
Posted Nov 7, 2011 12:34 PM ET
4.
There may well be an ASTM procedure, and if there is, this is the logic it would follow. You'd need the gas (natural or LP) composition, plus combustion air temperature and temperature and CO2 content of the flue gas. From the heats of formation of the gas components, and assuming 100% oxidation to CO2 and H2O, you have the gross heat of combustion. From the reaction stoichiometry you calculate the percent CO2 at zero percent excess air, and from the actual CO2 content of the flue gas you calculate the excess air that gives that dilution. These calculations give you the composition of the flue gas. Simple heat balance, from inlet gas and air to flue gas at its temperature, via reaction, yields the heat removed from the flue gas, being what is absorbed by the water. Dividing heat absorbed by net heat of reaction at the standard state, gives the efficiency.
Obviously, as you absorb more of the heat of combustion out of the flue gas, increasing the efficiency, you get a cooler flue gas, and possibly condense some of the water vapor produced by combustion. Flue gas temperature and efficiency are directly related for any fuel gas.
For simple field calculations, assuming "typical" fuel gas composition and percent excess air for properly adjusted burners, the efficiency can be estimated fairly well from the flue gas temperature. The usual way is by adding or subtracting a percentage from a reference point for each so many degrees from the reference temperature. There would be a different ratio for each type of fuel. Unfortunately, I can't give any numbers, as this is not my field.
Posted Nov 8, 2011 11:36 AM ET
5.
This just in from AO Smith and Technical Support regarding the Model: Vertex - GDHE-50-NG:
This water heater is tested by AHRI as a commercial water heater because of it's input BTU rating. That is why they did not(and will not) issue the water heater an energy factory. Because of this, the heat loss of the insulation is measured in BTU/HR. The recovery efficiency of this water heater is 96%.
*This confirmed one of my questions, Recovery Efficiency here is approximate or equal to the unit's thermal efficiency.
Also from them:
I have attached at GAMA certificate for this water heater. This certificate is from the AHRI web page.
http://cafs.ahrinet.org/gama_cafs/sdpsearch/showcert.jsp?model_id=321103
Thanks everyone for the input, and hopefully this can be useful in the future also...
Posted Nov 8, 2011 2:55 PM ET
6.
Andrew, as you discovered, TE and RE (aka, steady state efficiency) are more or less equivalent. But EF is what you're really after. Consider that a high quality conventional non-condensing water heater has a thermal efficiency of about 80% and an EF of about 62%. It would therefore be reasonable to assume the Vertex might have an EF in the mid-70's were it actually rated. BTW, insulation (jacket) loss is not the only factor that distinguishes EF from RE. The largest difference between the two ratings is heat lost up the flue during off-cycles.
The procedure for establishing EF ratings assumes a given average ambient temperature, entering water temperature, hot water usage, and duty cycle. The true EF therefore depends greatly on actual operating conditions. So in reality, EF is just an estimate in any case. Of course this doesn't help us jump through program hoops. And you're certainly not the first person to encounter this issue.
Posted Nov 10, 2011 2:58 AM ET
7.
David,
Thanks for your input and detail.
Fortunately, I already understood Energy Factor and that it's calculation and test procedures are at least clearly defined in many resources.
Since it was not and can't be defined (EF) for the 100k Vertex unit, I did go searching for an estimate from manufacturers, etc.
Fortunately also, I was entirely aware that EF was the criteria to mostly consider for efficiency, if available.
*they trained VERY well at Sustainable Spaces/Recurve.
My question was about recovery efficiency as a clarification because it had seemed so vague, and previously, could not get a clear answer on whether RE and TE are similar or the same. Got that finally via AO Smith.
On another note, since the Vertex unit did qualify definitely for the federal tax credit, which requires an EF of over .8 or .82 , can't recall the exact, they seemed to clearly be able to meet this criteria always.
This would mean a higher EF than quoted in your message, which was what I got from engineering and support over the past 2-3 years. Any thoughts on this?
Thanks always for your feedback and support!
-Andrew Dunn
Posted Nov 10, 2011 10:56 AM ET
8.
Some of the RESNET folks made a simple spreadsheet to help with the calculations. Do a google search for Commercial Hot Water EF Calculator.xls it comes up on the 4th page. Based on a 3 bedroom house the GDHE-50 has a .73 EF and the GPHE-50 is .74.
Posted Nov 11, 2011 7:51 PM ET
## Other Questions in Energy efficiency and durability
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In Green building techniques | Asked by Russell Boudreau | Jun 19, 18
### Re-sealing failing SIP roof panels
In GBA Pro help | Asked by Passagyrs | Jun 19, 18
### Single mini split for AC?
In Mechanicals | Asked by jeffesonm | Jun 20, 18
### Mortar bridging and droppings
In General questions | Asked by Stephen Cook | Jun 16, 18
### Dirty electric from minisplit ductless heat and air conditioning
In General questions | Asked by Arleendv | May 30, 18 | 1,591 | 6,942 | {"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-2018-26 | latest | en | 0.934134 |
https://www.justintools.com/unit-conversion/length.php?k1=football-field-length-%5BUS%5D | 1,716,244,824,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058313.71/warc/CC-MAIN-20240520204005-20240520234005-00105.warc.gz | 742,898,279 | 25,310 | 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 :)
# Football Field Length [US]
A football field is often used as a comparative measurement of length when talking about distances that may be hard to comprehend when stated in terms of standard units. An American football field is usually understood to be 100 yards (91.44 m) long, though it is technically 120 yards (110 m) when including the two 10 yd (9.1 m) long end zones. The field is 160 ft (53 yd; 49 m) wide.
Football Field Length [US]
Symbol/abbreviation: fbf[US]
Unit of: LENGTH
LENGTH's base unit: meters (SI Unit)
In relation to the base unit (meters), 1 Football Field Length [US] = 91.44 meters.
## Conversion table 1 Football Field Length [US] (fbf[US]) to all length units
1 fbf[US]= 50399.999999881 agate [typography] (agate)
1 fbf[US]= 153.99694835444 alens (aln)
1 fbf[US]= 914400000000 angstroms (Å)
1 fbf[US]= 6.1123864646023E-10 astronomical unit (au)
1 fbf[US]= 9.144E+19 attometers (am)
1 fbf[US]= 10800 barleycorns (bcorn)
1 fbf[US]= 91440000000000 bicrons (μμ)
1 fbf[US]= 1.1363636363636 blocks (bl)
1 fbf[US]= 1727966257036.9 bohr radius (Bohr)
1 fbf[US]= 0.49342105263158 cable length uk (cbl[UK])
1 fbf[US]= 0.41666666666667 cable length us (cbl[US])
1 fbf[US]= 360000 calibers (cal)
1 fbf[US]= 360000 centiinch (cin)
1 fbf[US]= 9144 centimeters (cm)
1 fbf[US]= 4.5454545454545 chains (ch)
1 fbf[US]= 200 cubits (cbt)
1 fbf[US]= 9.144 decameters (dam)
1 fbf[US]= 914.4 decimeters (dm)
1 fbf[US]= 243209.45710718 didot points (didot point)
1 fbf[US]= 4800 digits (digit)
1 fbf[US]= 2.286E-6 earth circumference (Ec)
1 fbf[US]= 2.3787597359015E-7 earth to moon distance (EM d)
1 fbf[US]= 80 ells (eel)
1 fbf[US]= 21681.00021681 ems (EM)
1 fbf[US]= 9.144E-17 exameters (Em)
1 fbf[US]= 50 fathoms (fath)
1 fbf[US]= 300 feet (ft)
1 fbf[US]= 9.144E+16 femtometers (fm)
1 fbf[US]= 9.144E+16 fermi (fermi)
1 fbf[US]= 800 finger cloth (finer)
1 fbf[US]= 291.11747851003 fods (fod)
1 fbf[US]= 1 football field length [US] (fbf[US])
1 fbf[US]= 0.45454545454545 furlongs (fur)
1 fbf[US]= 9.144E-8 gigameters (Gm)
1 fbf[US]= 2.9688311688312 greek plethron (pleth)
1 fbf[US]= 0.49427027027027 greek stadion (std)
1 fbf[US]= 900 hands (h)
1 fbf[US]= 0.9144 hectometers (hm)
1 fbf[US]= 9144000 himetrics (hiMetric)
1 fbf[US]= 38.1 horse length (hl)
1 fbf[US]= 48.2155439827 hvats (hvat)
1 fbf[US]= 3600 inches (in)
1 fbf[US]= 0.09144 kilometers (km)
1 fbf[US]= 2.9630589760207E-18 kiloparsecs (kpc)
1 fbf[US]= 3.5305019305019E-12 light days (ld)
1 fbf[US]= 8.4745134383689E-11 light hours (lh)
1 fbf[US]= 5.082823790995E-9 light minutes (lmin)
1 fbf[US]= 1.1766825376399E-13 light months (lm)
1 fbf[US]= 3.0500333555704E-7 light seconds (ls)
1 fbf[US]= 9.6649402811542E-15 light years (ly)
1 fbf[US]= 43542.857142857 lines (l)
1 fbf[US]= 2.3787597359015E-7 lunar distance (LD)
1 fbf[US]= 9.144E-5 megameters (Mm)
1 fbf[US]= 91.44 meters (m)
1 fbf[US]= 304.8 metric foot (mf)
1 fbf[US]= 3600000000 microinches (μin)
1 fbf[US]= 91440000 micrometers (μm)
1 fbf[US]= 91440000000000 micromicrons (μμm)
1 fbf[US]= 0.056818181818182 miles (mi)
1 fbf[US]= 91440 millimeters (mm)
1 fbf[US]= 3600000 mils (mil)
1 fbf[US]= 0.009144 myriameters (Mym)
1 fbf[US]= 1600 nails cloth (nail)
1 fbf[US]= 91440000000 nanometers (nm)
1 fbf[US]= 0.01645788336933 nautical leagues (Nl)
1 fbf[US]= 0.049373650107991 nautical miles (NM)
1 fbf[US]= 120 paces (pace)
1 fbf[US]= 1200 palms (palm)
1 fbf[US]= 2.9633685822679E-15 parsecs (pc)
1 fbf[US]= 18.181818181818 perch (perch)
1 fbf[US]= 9.144E-14 petameters (Pm)
1 fbf[US]= 21600.000017008 picas (p)
1 fbf[US]= 91440000000000 picometers (pm)
1 fbf[US]= 345600.04354016 pixels (PX)
1 fbf[US]= 5.715E+36 planck length (pl)
1 fbf[US]= 259199.83672451 points (pt)
1 fbf[US]= 18.181818181818 poles (pole)
1 fbf[US]= 2057.1428571429 rack unit (ru)
1 fbf[US]= 21600.00272126 rems (rem)
1 fbf[US]= 18.181818181818 rods (rd)
1 fbf[US]= 2.5757746478873 roman actus (actus)
1 fbf[US]= 15 ropes (rope)
1 fbf[US]= 0.009144 scandinavian miles (mil[Scandinavian])
1 fbf[US]= 6.1122994652406E-16 siriometer (sir)
1 fbf[US]= 400 spans (span)
1 fbf[US]= 9.144E-11 terameters (Tm)
1 fbf[US]= 3600000 thou (th)
1 fbf[US]= 5183967.3450876 twips (twip)
1 fbf[US]= 105.58891454965 varas [Argentina] (vr[Argentina])
1 fbf[US]= 109.09069044099 varas [California] (vr[California])
1 fbf[US]= 105.83333333333 varas [Latin America] (vr[Latin])
1 fbf[US]= 109.11434094652 varas [Mexico] (vr[Mexico])
1 fbf[US]= 109.3910754875 varas [Spanish] (vr[Spanish])
1 fbf[US]= 107.99957480482 varas [Texas] (vr[Texas])
1 fbf[US]= 0.085714285714286 versts (vst)
1 fbf[US]= 9.1257120730057E+14 x unit (xu)
1 fbf[US]= 100 yards (yd)
1 fbf[US]= 9.144E+25 yoctometers (ym)
1 fbf[US]= 9.144E-23 yottameters (Ym)
1 fbf[US]= 9.144E+22 zeptometers (zm)
1 fbf[US]= 9.144E-20 zettameters (Zm)
1 fbf[US]= 3600 zolls (zoll) | 2,138 | 5,170 | {"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} | 3.09375 | 3 | CC-MAIN-2024-22 | latest | en | 0.754612 |
https://www.physicsforums.com/threads/constant-acceleration-equations-suvat.441862/ | 1,627,623,490,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153931.11/warc/CC-MAIN-20210730025356-20210730055356-00603.warc.gz | 958,371,074 | 14,392 | # Constant acceleration equations! (SUVAT)
## Homework Statement
A stone is projected horizontally from the top of a vertical sea cliff 49m high, with a speed of 20ms^-1. Neglecting air resistance, calculate:
The time that it takes to reach the sea.
v² =u²+2as
t=v-u/a
t=d/s
## The Attempt at a Solution
I'm not sure whether i should be calculating v-final velocity via v² =u²+2as and using that to find t, ie. t=v-u/a
or simply
t=d/s
can anybody enlighten me please? thanks | 136 | 480 | {"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} | 2.984375 | 3 | CC-MAIN-2021-31 | latest | en | 0.828476 |
http://forums.gillius.org/index.php?topic=40.0 | 1,582,178,811,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875144637.88/warc/CC-MAIN-20200220035657-20200220065657-00206.warc.gz | 58,968,634 | 5,693 | Author Topic: Programming Exercise - Cosine Law (Read 6587 times)
louiecerv
• Moderator
• User
• Posts: 85
Programming Exercise - Cosine Law
« on: December 08, 2005, 05:58:19 PM »
Code: [Select]
`/* Name: Cosine Law Author: LF Cervantes Date: 12/08/05 Description: First Programming Exercise (ITE205-IT)*/#include <iostream>#include <cmath>using namespace std;int main(){ cout << "-- COSINE LAW --\n\n"; cout << "A program that illustrates the cosine law.\n"; int b, c; float a, Angle; const double PI = 3.14159; char answer; do { cout << "\nEnter length of sides b and c as two integers\n"; cout << " separated by space : "; cin >> b >> c; cout << "Enter angle between side b and side c (degrees) : "; cin >> Angle; a = sqrt(pow(b, 2.0) + pow(c,2.0) - 2 * b * c * cos (Angle * PI / 180)); cout << "length of side a = " << a; cout << "\n\nRepeat run with new input? y/n : "; cin >> answer; } while ( answer == 'y' || answer == 'Y'); cout << endl; system("PAUSE"); return 0;}`
Highlights:
Do-While loop
Math statements
Analyze. Design. Develop. Debug. Deploy. Maintain. | 402 | 1,343 | {"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} | 3.078125 | 3 | CC-MAIN-2020-10 | longest | en | 0.292947 |
https://codeforces.com/blog/entry/90513 | 1,722,808,613,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640412404.14/warc/CC-MAIN-20240804195325-20240804225325-00145.warc.gz | 128,118,278 | 22,933 | ### Rewinding's blog
By Rewinding, history, 3 years ago,
(It's a translated version of my blog at https://peehs-moorhsum.blog.uoj.ac/blog/6384. )
Finding a Hamiltonian path in a directed graph is a well-known NP problem.
However, about a year ago, I came up with the following heuristic algorithm which has GREAT performance on random graphs(by first generating a hamiltonian path, adding random edges, then randomly permuting indices) and many CP problems.
Here is the idea of the algorithm:
We maintain a subset $S$ of edges, ensuring there's no cycle and each vertex has in-degree and out-degree at most 1. Whenever $|S|$ reaches $n-1$, we find a hamiltonian path.
We start with $S$ empty. Each time we pick a random edge $e \notin S$. If we can add $e$ to $S$ without violating any rule, we simply add it to $S$. Otherwise, if $e$ won't create a cycle and violate with exactly one edge e', we replace e' with $e$ with 50% probability. Repeat until $|S|$ reaches $n-1$.
We can use LCT to check cycles. Here's my code, which takes less than 1 second on most random graphs with $|V|=100000, |E|=500000$:
namespace hamil {
template <typename T> bool chkmax(T &x,T y){return x<y?x=y,true:false;}
template <typename T> bool chkmin(T &x,T y){return x>y?x=y,true:false;}
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define pi pair<int, int>
#define fi first
#define se second
#define ll long long
namespace LCT {
vector<vi> ch;
vi fa, rev;
void init(int n) {
ch.resize(n + 1);
fa.resize(n + 1);
rev.resize(n + 1);
for (int i = 0; i <= n; i++)
ch[i].resize(2),
ch[i][0] = ch[i][1] = fa[i] = rev[i] = 0;
}
bool isr(int a)
{
return !(ch[fa[a]][0] == a || ch[fa[a]][1] == a);
}
void pushdown(int a)
{
if(rev[a])
{
rev[ch[a][0]] ^= 1, rev[ch[a][1]] ^= 1;
swap(ch[a][0], ch[a][1]);
rev[a] = 0;
}
}
void push(int a)
{
if(!isr(a)) push(fa[a]);
pushdown(a);
}
void rotate(int a)
{
int f = fa[a], gf = fa[f];
int tp = ch[f][1] == a;
int son = ch[a][tp ^ 1];
if(!isr(f))
ch[gf][ch[gf][1] == f] = a;
fa[a] = gf;
ch[f][tp] = son;
if(son) fa[son] = f;
ch[a][tp ^ 1] = f, fa[f] = a;
}
void splay(int a)
{
push(a);
while(!isr(a))
{
int f = fa[a], gf = fa[f];
if(isr(f)) rotate(a);
else
{
int t1 = ch[gf][1] == f, t2 = ch[f][1] == a;
if(t1 == t2) rotate(f), rotate(a);
else rotate(a), rotate(a);
}
}
}
void access(int a)
{
int pr = a;
splay(a);
ch[a][1] = 0;
while(1)
{
if(!fa[a]) break;
int u = fa[a];
splay(u);
ch[u][1] = a;
a = u;
}
splay(pr);
}
void makeroot(int a)
{
access(a);
rev[a] ^= 1;
}
{
makeroot(a);
fa[a] = b;
}
void cut(int a, int b)
{
makeroot(a);
access(b);
fa[a] = 0, ch[b][0] = 0;
}
int fdr(int a)
{
access(a);
while(1)
{
pushdown(a);
if (ch[a][0]) a = ch[a][0];
else {
splay(a);
return a;
}
}
}
}
vi out, in;
vi work(int n, vector<pi> eg, ll mx_ch = -1) {
// mx_ch : max number of adding/replacing default is (n + 100) * (n + 50)
// n : number of vertices. 1-indexed.
// eg: vector<pair<int, int> > storing all the edges.
// return a vector<int> consists of all indices of vertices on the path. return empty list if failed to find one.
out.resize(n + 1), in.resize(n + 1);
LCT::init(n);
for (int i = 0; i <= n; i++) in[i] = out[i] = 0;
if (mx_ch == -1) mx_ch = 1ll * (n + 100) * (n + 50); //default
vector<vi> from(n + 1), to(n + 1);
for (auto v : eg)
from[v.fi].pb(v.se),
to[v.se].pb(v.fi);
unordered_set<int> canin, canout;
for (int i = 1; i <= n; i++)
canin.insert(i),
canout.insert(i);
int tot = 0;
while (mx_ch >= 0) {
// cout << tot << ' ' << mx_ch << endl;
vector<pi> eg;
for (auto v : canout)
for (auto s : from[v])
if (in[s] == 0) {
assert(canin.count(s));
continue;
}
else eg.pb(mp(v, s));
for (auto v : canin)
for (auto s : to[v])
eg.pb(mp(s, v));
shuffle(eg.begin(), eg.end(), x);
if (eg.size() == 0) break;
for (auto v : eg) {
mx_ch--;
if (in[v.se] && out[v.fi]) continue;
if (LCT::fdr(v.fi) == LCT::fdr(v.se)) continue;
if (in[v.se] || out[v.fi])
if (x() & 1) continue;
if (!in[v.se] && !out[v.fi])
tot++;
if (in[v.se]) {
LCT::cut(in[v.se], v.se);
canin.insert(v.se);
canout.insert(in[v.se]);
out[in[v.se]] = 0;
in[v.se] = 0;
}
if (out[v.fi]) {
LCT::cut(v.fi, out[v.fi]);
canin.insert(out[v.fi]);
canout.insert(v.fi);
in[out[v.fi]] = 0;
out[v.fi] = 0;
}
canin.erase(v.se);
canout.erase(v.fi);
in[v.se] = v.fi;
out[v.fi] = v.se;
}
if (tot == n - 1) {
vi cur;
for (int i = 1; i <= n; i++)
if (!in[i]) {
int pl = i;
while (pl) {
cur.pb(pl),
pl = out[pl];
}
break;
}
return cur;
}
}
//failed to find a path
return vi();
}
}
Note that if the start vertex $S$ and/or end vertex $T$ of the path is given, we simply create two new vertices $A, B$ and add edges from $A$ to $S$, from $T$ to $B$. Any Hamiltonian path in the new graph automatically has $A\to S$ as the first edge and $T\to B$ as the last edge. If we want to find a Hamiltonian cycle, we simply enumerate an edge, then convert the problem to finding a path given start/end vertices. For bidirectional case, we can simply add each edge twice.
• +455
» 3 years ago, # | +4 Auto comment: topic has been updated by Rewinding (previous revision, new revision, compare).
» 3 years ago, # | +11 I assume you meant $|V|=10^4$? With $|V|=10^5$ and $|E|=5\cdot 10^5$ directed edges the expected number of vertices with out-degree 0 is quite large ... ($\approx |V|\cdot e^{-5}\approx 674$)
• » » 3 years ago, # ^ | +23 He did say "by first generating a hamiltonian path, adding random edges, then randomly permuting indices" which I guess means he ensures there are no such vertices (and I guess there are ~674 vertices with degree 1 then?)
• » » » 3 years ago, # ^ | ← Rev. 3 → +14 aha, can't readWell assuming $4\cdot 10^5$ edges in addition to the hamiltonian path then that's $|V|\cdot e^{-4}\approx 1832$ with out-degree 1.
• » » » » 3 years ago, # ^ | +8 Yeah, that's true. I also tried many other random graphs, and it turned out that the performance is approximate $O(|E|\log|V|)$, not depends on $|E|/|V|$ very much (but it sometimes stuck in cases $|E|~3|V|$). I guess there're many paths in random graphs, so random guesses work most of the time?
» 3 years ago, # | +8 Very cool approach!Instead of time(0) it's better to use chrono::steady_clock::now().time_since_epoch().count() inside mt19937, because we can call function work() several times for better probabilities. If we call this function with the same arguments several times per second, time(0) returns the same value, which is bad.Or we can move mt19937(time(0)) out of work()'s body.
• » » 3 years ago, # ^ | 0 Yeah I see. Updated! >w<
» 3 years ago, # | ← Rev. 2 → 0 You can test any input with the code above! Just generate graphs (store edges as vector>, where (u, v) denotes an edge from $u$ to $v$), and put them into hamil::work. Initialization is done automatically, so you can test graphs in series.
» 3 years ago, # | ← Rev. 3 → +24 This idea is also useful in some problems which are related to Hamiltonian path! In JOISC 2021 Day1 T1 Aerobatics, applying this idea will get >80 points easily.
» 3 years ago, # | +20 I tried to apply this idea here and looks like it doesn't perform great for the bidirectional case, where not so many random edges added, so it won't work for very simple undirected graphs. For example, it fails to find the path for 5000 vertices and undirected edges $1-2-3-\ldots-N$ in 2 seconds.
• » » 3 years ago, # ^ | +40 I see. Seems like the algorithm works bad when there are lots of directional paths but few Hamiltonian paths(which is often true when bidirectional). (It also depends on seed a lot. For 1-2-3-...-N with N = 5000, it runs rather fast with seed = 377921677545000 (the 1 out of 100 seeds that success at mx_ch = 1e6 lol))But bidirectional case should be easier, as we can only restrict vertices to have $deg \le 2$ rather than $out \le 1$ and $in \le 1$. I'll write another code for bidirectional later. :) | 2,563 | 7,864 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2024-33 | latest | en | 0.631861 |
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If we count in the following way on the fingers of our left hand such that we call the thumb 1, index finger 2, middle finger 3, ring finger 4, little finger 5, then reversed direction, calling the ring finger 6, middle finger 7, index finger 8 and thumb 9 and then back to the index finger for 10, middle finger for 11 and so on. Count upto 1994 then we'll end on
a) thumb
b) index finger
c) middle finger
d) ring finger
i calculated it and IMO it should end at the index finger.
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29 Oct 2009, 15:13
I agree with you, it should be index finger. Thumb = 1 and then when you come back again, thumb = 9. That's a difference of 8. So if you divide 1994 by 8, it leaves a remainder of .25 (or .25 of 8 = 2). so essentially, to determine if a number is on the thumb, subtract 1 and divide by 8. If it is an integer, then the answer is yes, that number is on the thumb. In order to determine if it is on the index finger, subtract 2 and divide by 8. here, $$\frac{1994-2}{8} = 249$$, an integer, so you know it's on the index finger.
papillon86 wrote:
If we count in the following way on the fingers of our left hand such that we call the thumb 1, index finger 2, middle finger 3, ring finger 4, little finger 5, then reversed direction, calling the ring finger 6, middle finger 7, index finger 8 and thumb 9 and then back to the index finger for 10, middle finger for 11 and so on. Count upto 1994 then we'll end on
a) thumb
b) index finger
c) middle finger
d) ring finger
i calculated it and IMO it should end at the index finger.
_________________
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**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a.
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Display posts from previous: Sort by | 811 | 2,997 | {"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} | 4.21875 | 4 | CC-MAIN-2017-22 | longest | en | 0.864982 |
http://echeckcasinos.ca/calculate-gambling-probabilities/ | 1,591,005,786,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347415315.43/warc/CC-MAIN-20200601071242-20200601101242-00328.warc.gz | 36,821,986 | 11,398 | # How to Calculate Probabilities in Gambling (without a Calculator!)
Simplified Mathematics: Probabilities of casino games.
A lot of people don’t consider themselves to be good at math. What so many don’t realize is that most mathematical equations – even the dreaded fractions and percentages – don’t have to involve difficult math. I suppose it really depends on what level of math a person considers their skills to go downhill.
If you can do basic addition, subtraction, multiplication and division – we’re talking grade-school stuff – you have what it takes to do something so seemingly complicated as calculating probabilities in gambling.
I’m going to take this slowly, step by step, to show just how easy it is. Relax and take a deep breath – you got this!
## Probabilities of Flipping a Coin
This is as easy as it gets, but don’t skip ahead. I’m starting here for a reason. It will help you understand bigger calculations later on.
Probabilities are properly expressed as a number between 0 and 1. When flipping a coin, there are only two possible outcomes – heads or tails. The probability of one of those outcomes happening is 0.5. Flipping heads is 0.5, and flipping tails is 0.5. Put them together and you get a whole 1.
That decimal can also be expressed as a percentage (50%), a fraction (1/2), or as odds of 1 to 1 (1 chance of this happening, or 1 chance of that happening).
Moving right along…
## Probability of Drawing a Card Suit
Let’s expand the equation to something where one of four outcomes is possible. Let’s say you have four playing cards, one of each suit. What are the odds you will draw a club?
There are four possible outcomes, and we’re looking for just one of them. We divide the probability of all outcomes (1) by the number of possible outcomes (4), and we get 1/4 = 0.25. Or, we can just display this as the fraction of 1/4 (a one in four chance). Or, we can convert that to a percentage – a 25% chance. As odds, it would be expressed 3 to 1 – the number of undesired outcomes (3) “to” the number of desired outcomes (1).
Are you still with me? Good, let’s get back to the coin toss for our next lesson.
## Probability of Flipping 2 Heads in a Row
Now things get more complicated, right? Not really. To get the odds of a specific outcome occurring twice, you simply multiply the odds of that outcome happening once, by the odds of that outcome happening once (again).
We know already that the odds of flipping a coin and it landing on heads is 0.5. So we multiply that integer by itself. 0.5 * 0.5 = 0.25
So the probability is 0.25, or a 25% chance (1/4, or 3 to 1) that heads will appear twice in a row.
Perhaps the math makes sense to you, but if not, here’s a really simple way to explain why that equation is correct.
Let’s think about how many things could possibly happen in two flips of a coin. The only possible outcomes are:
• Two Tails
As you can see, there are only four possible outcomes. As we learned in the card suits example above, four outcomes means a 1 in 4 chance, or 0.25 probability.
## The Probability of Drawing an Ace
Things get more complicated now, because we have 52 possibilities in a standard deck (no jokers), and four of the are an Ace. Therefore the odds of drawing an ace from a full deck are 4 in 52, or 1 in 13 (52 / 4 = 13). As a percentage, we can calculate 1 / 13 = 0.0769, which translates to 7.7%.
It may not be easy to come to 7.7% odds in your head, but the 1 in 13 should be obvious. There are 13 different cards per card, four of each. Everyone who plays cards knows there are 52 (minus jokers) in a deck.
2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King, Ace
So if there are four of each card (4 in 52), there has to be 1 for every 13 (1 in 13). You don’t need a calculator to come to this result. It’s simple logic. And as an estimate, it’s easily apparent that a 1 in 13 chance is going to be under 10%.
## Probabilities for Blackjack Card Counting Purposes
Card counting isn’t nearly so difficult as movies like “21” make it sound. Sure, there are more intricate counting systems, but a basic count is almost just as effective, and is incredibly easy to manage. But I’m not here to teach you how to count cards. I’m going to teach you casino math – in this case, how to estimate the remaining Aces in a blackjack game.
What you’ll need to know is a) how many decks of cards are in the shoe, b) how many Aces have already been played, and c) how many total cards have been played.
Let’s say it’s a 6-deck shoe. 6 decks of 52 is 312 cards.
6 * 52 = 312 Cards
If you would rather estimate, that’s about 6 * 50 = 300.
There are 4 Aces per deck, so 24 Aces in the shoe.
6 * 4 = 24 Aces
With this information, we know that before any cards are dealt, there is a 24 in 312 (which breaks down to the same 1 in 13 above) chance of being dealt an Ace.
For sake of exemplification, let’s say you’ve been playing for a bit, you’re about ¼ of the way into the shoe with 78 cards seen. That leaves (312 – 78) 234 cards in the deck. If you’ve witnessed the deal of 2 Aces in that time, there are (24 – 2) 22 Aces left. Now the odds change…
22 in 234 chance of being dealt an Ace.
22 / 234 = 0.094, or 9.4% odds of being dealt an Ace
-or-
234 / 22 = 10.63, or a 1 in 10.6 chance
Again, it’s easy enough to estimate. Round 234 to 200 and 22 to 20. 200 / 20 = 1 in 10 (10%) chance, which is really close to the exact numbers above.
## Probabilities in Roulette
There are two types of roulette wheels – American and European. For the purpose of this guide, I’ll be referring to a European roulette wheel for the simple reason that American roulette has far worse odds. Simply put, don’t play American roulette! Stick to European or French wheels with a single zero (0).
Roulette is not the easiest game to calculate probabilities for, mostly because the numbers involved are not evenly rounded numbers. You may need a calculator for this one. The good news is, the house edge does not change in European roulette. The payout for each wager is set at such a rate that the edge remains a static 2.7% at all times.
Let’s have a closer look…
#### Single Number Roulette Bets
The wheel has 37 slots, ranging from 0 to 36. So, to define the probability of the ball landing in any single number slot, we must divide 1 by 37.
1/37 = .027, or a 2.7% chance
What does this mean? Odds are, if we placed this bet 37 times, we would win one of them, worthy of a 35 to 1 payout of 36 units (35 + 1, the original bet). Thus probabilities states we would lose 1 bet unit it 37 attempts. So, the 2.7% probability also translates to a 2.7% house edge, wherein the player can expect to lose 2.7% of their bets, and win 97.3% of them (aka the theoretical return to player, or RTP).
Let’s consider a different roulette bet, such as a bet on dozens.
#### Dozens Roulette Bets
There are three groups of dozens containing 12 numbers each. Betting on the first dozen is a bet on 1-12. So there are 12 numbers that will win, and (37 – 12) 25 that will lose. To get the probabilities, we divide 12/37.
12/37 = 0.3243, or a 32.43% chance
We can easily round that up to a 1/3 chance for the purpose of estimation, but it still leaves a little extra wiggle room for the casino to accommodate it’s house edge.
Since this bet pays 2 to 1 (3 units), we can multiply those numbers to get the RTP.
3 * 32.43 = 97.29
And if we subtract the 97.29% RTP from 100%, it rounds to the same 2.7% house edge.
#### Even-Money Roulette Bets
What about even-money bets? Any wager on black, red, odd, even, high or low is an even-money bet. It is a wager with 18 ways to win and 19 ways to lose (i.e. 18 out of 37 possibilities), paying 1 to 1 (2 units) for a win.
18 / 37 = 0.4864, or a 48.64% chance to win.
Now to find the RTP and house edge, we multiply the units won (2) by the probability (48.64%).
2 * 48.64 = 97.29%
Again, we come to a 97.3% RTP, and a 2.7 house edge.
Whether it’s a 3-number line bet, 4-number corner bet, or 6-number street bet, the end result is always the same – a 2.7% house edge.
#### The Difference with French Roulette
There is a slight difference in French Roulette. This game features a special rule known as La Partage. By this rule, any even-money wager that loses due to the winning number being zero (0) will result in half of the bet being lost, and the other half being returned.
For this reason, the house edge on even money bets in French roulette is cut in half, from 2.7% to 1.35%.
2.7 / 2 = 1.35
By this reasoning, any truly educated casino gambler will tell you that the French variant is the most advantageous roulette game for players.
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# श्रेणी 2+4+6+8+ ....... के 10 पदों का योगफल ज्ञात कीजिए।
Updated On: 27-10-2020
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Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.
श्रेणी `9-3+1 - (1)/(3)+ ..oo` का अनन्त पदों तक योगफल ज्ञात कीजिए।
यदि गुणोत्तर श्रेणी के अन्नत पदों का योगफल S तथा प्रथम पद a है, तो सार्वअनुपात r ज्ञात कीजिए।
????? ????????? ?? n ???? ?? ?? ????? ????? ????? - <br> `1+4+13+40+121+
??????
127303248
6:06
?????? `1+4/5+7/(5^(2))+10/(5^(3))+
????? ????????? ?? n ???? ?? ?? ????? ????? ????? - <br> `1+2+5+12+25+46 +
????? ????????? ?? n ???? ?? ?? ????? ????? ????? - <br> `3+15+35+63+
??????
127303236
3:26
?????? `1+3/4+7/16+15/64+
??????
127303222
6:25
?????? `2xx4+4xx6+6xx8+
?????????? ?????? ?? ??? ????? ????? `:` <br> `1+(4)/(5) + ( 7)/( 5^(2))+(10)/( 5^(3))+"..........."n` ???? ?? ?
?? ?????? ????? n ?? ?? `(n)/(x)+y` ?? ?? r ???? ?? ????? ????
????? ????????? ?? n ???? ?? ????? ???? ?? ????? ????? ????? - <br>`1/(3.7)+1/(7.11)+1/(11.5)+......`
??????
127303235
3:10
?????? `1/3+3/9+5/27+7/81+
?????????? ????????? ?? n ???? ?? ??? ????? ????? ? <br> `1+(3)/(2)+(5)/(4)+(7)/(8)+"
??????
127303251
5:23
?????? `1.3^(2)+2.5^(2)+3.7^(2)+
??????
127303237
5:48
?????? `1+(1+1/3)+(1+1/3+1/(3^(2)))+
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Doubt? | 599 | 1,367 | {"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 | 3 | CC-MAIN-2021-49 | latest | en | 0.191289 |
https://stats.stackexchange.com/questions/175904/how-to-do-regression-using-regressors-that-are-independent-but-not-quite | 1,620,259,436,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988724.75/warc/CC-MAIN-20210505234449-20210506024449-00137.warc.gz | 541,452,795 | 38,187 | # how to do regression using regressors that are independent but not quite?
I have a scenario where I am collecting a bunch of attributes/demographics about a person. Examining the scatterplot Matrix, it seems that there is not much correlation between various attributes for these 2000 data points. But because they all belong to the same person, they're not independent in the true sense of the term. In such scenarios, is it fair to treat them as independent regressors? (For my data points, I may not find any aliases)
• There is no assumption in multiple regression that regressors should be independent! – kjetil b halvorsen Oct 7 '15 at 16:48
• @Kjetil is perfectly right, so the short answer is "no, but it's irrelevant." However, examining a scatterplot matrix--although always a good idea--gives only a small part of the story of the linear relationships among the regressors. It is possible for the regressors to be collinear without exhibiting much, if any, noticeable linear relationships in the scatterplot matrix.That can create estimation problems. If you have many attributes, use tools such as VIF, COVRATIO, PCA, etc. to identify, diagnose, and (if necessary) correct near-collinearity. – whuber Oct 7 '15 at 17:10
• Can you please provide the (or your) definition of "independence in the true sense of the word"? It appears that you say that "if a vector of attributes relate to the same person, then these attributes cannot be independent". Why? The concept of "Independence" in the context of probability theory/Statistics has a specific meaning, which does not coincide with the meaning this word may take in everyday language, or in the context of other sciences. – Alecos Papadopoulos Oct 7 '15 at 17:16
• Thanks everyone! @Alecos my concern is they're all attributes of one person. So, they're related in some way (may be theoretically), but say in my data sample (supposed to be simple random) I don't find a relationship. Should I consider for all practical purposes, the x variables to be independent? – rk567 Oct 7 '15 at 21:05 | 477 | 2,059 | {"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-2021-21 | latest | en | 0.922371 |
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## 0. Overview of KCL fWAR
Wins Above Replacement (WAR) is one of the most important stats that sabermetrics has introduced. It aims to measure how many wins a player is worth to their team by comparing them to a “replacement player,” or a “minor league free agent or very poor MLB bench player” (Slowinski, FanGraphs). Fangraphs has a walkthrough of how to calculate fWAR, or FanGraphs’s version of WAR, on their website. However, there are a couple of areas of the fWAR calculation that are either incalculable given the level of data available for the KCL or not necessary for the KCL. Thus, I have made substitutions in the areas that could not be calculated or omitted certain parts that did not make sense to be included for the KCL. I then also scaled each section to match a full MLB season in order to produce numbers that are more familiar to baseball fans. FanGraphs’s fWAR formula is (Batting Runs + Baserunning Runs + Fielding Runs + Positional Adjustment + League Adjustment +Replacement Runs) / (Runs Per Win). Because the KCL is one league, for our purposes we will omit League Adjustment and instead say fWAR = (Batting Runs + Baserunning Runs + Fielding Runs + Positional Adjustment + Replacement Runs) / (Runs Per Win). I will now go into depth about how to calculate each section of the fWAR equation, and the changes I made to suit the KCL with the justifications for each change. I will also show the calculations for one player, Logan Van Heuklon (Merchants), as he has played in the most games and in a variety of positions, making him a good case study. I will also show the calculations for Davey Fitzpatrick (Ground Sloths) for a couple of equations for fielders that are not also catchers, as Van Heuklon plays catcher as well as in the field.
## 1. Batting Runs
The Fangraphs website lists the formula for Batting Runs as wRAA + (lgR/PA – (PF×lgR/PA))×PA + (lgR/PA – (AL or NL non-pitcher wRC/PA))×PA, where (lgR/PA – (PF×lgR/PA))×PA + (lgR/PA – (AL or NL non-pitcher wRC/PA))×PA is park and league adjustments to wRAA (weighted Runs Above Average). However, because the KCL is one league with the same field, that entire part can be omitted and reduced to Batting Runs = wRAA.
### 1.1 wRAA
wRAA is wRAA = ((wOBA – league wOBA) / wOBA scale) × PA, where the wOBA scale is league OBP / league wOBA, and wOBA = (0.690×BB + 0.722×HBP + 0.888×1B + 1.271×2B + 1.616×3B + 2.101×HR) / (AB + BB – IBB + SF + HBP). However, since the KCL is a college league, I have decided to use weights used to calculate college players’ WAR. Also, the KCL only has had 2 IBBs, so I have left that part out here and in further calculations as it does not affect results almost at all. Thus, for the KCL, wOBA = (0.76×BB + 0.77×HBP + 0.97×X1B + 1.36×X2B + 1.71×3B + 2.11×HR)/(AB + BB + SF + HBP). Also, FanGraphs has a the following scale on their website for wRAA:
Thus, the equation for wRAA I have used is 3×((wOBA – league wOBA) / wOBA_scale)×PA in order to have the KCL wRAA values match the scale given.
### 1.2 Logan Van Heuklon’s Batting Runs
Now, for Logan Van Heuklon’s calculation, we will need his stats, the league wOBA, and the league OBP. Logan Van Heuklon has had 17 BBs, 2 HBP, 21 1Bs, 2 2Bs, 0 3Bs, 0 HRs, 73 ABs, 0 SFs, and 92 PAs. The league wOBA is ≈0.3721, and the league OBP is ≈0.3835. Thus, the wOBA scale is ≈1.0306, and we can now plug in all the numbers to the formula to get his wRAA, which is 9.6631 (from exact calculations).
## 2. Baserunning Runs
The FanGraphs website has the formula for Baserunning Runs as UBR + wSB + wGDP. UBR stands for Ultimate Base Running, which is a video-measured statistic on how a baserunner’s running affects the run expectancy in certain situations. However, this is impossible to do without the use of video replay, and because that level of data is unavailable for the KCL, I have left that part out of the calculation.
### 2.1 wGDP
FanGraphs also does not list a formula for wGDP, but says that they try to reward hitters for not hitting into a double play. Therefore, I have come up with my own equation to represent wGDP, -30×(GIDP/league GIDP) + 0.02×AB. Basically, I calculated what percentage of double plays in the league belonged to each player, while also giving credit to those with more at-bats because having 1 GIDP in 50 ABs is better than having 1 GIDP in 2 ABs. I then added weights to each part (-30 and 0.02) in order to get similar results to the table FanGraphs provided for wGDP.
### 2.2 wSB
FanGraphs does have a formula for wSB, which is SB×runSB + CS×runCS – lgwSB×(1B + BB + HBP), where runSB is .2 (the expected run value of a stolen base), runCS is = -2×RunsPerOut + 0.075 (the expected run value of being caught stealing), and lgwSB = (lgSB×runSB + lgCS×runCS) / (lg1B + lgBB + lgHBP). Thus, the formula for Baserunning Runs is now simply wSB + wGDP. This correlates perfectly with FanGraphs scale for Baserunning Runs.
### 2.3 Logan Van Heuklon’s Baserunning Runs
Now that we have both wSB and wGDP defined, we can calculate Logan Van Heuklon’s Baserunning Runs. First, we will calculate his wSB. Van Heuklon has 6 SB, 1 CS, 17 BBs, 2 HBP, and 21 1Bs. The league has 195 SB, 35 CS, 555 1Bs, 466 BBs, and 105 HBP, and the value for RunsPerOut is ≈0.2381 (total runs divided by total outs). Now we can plug in values to find wSB. First, runCS is ≈-0.5512, which means the league wSB (lgwSB) is ≈0.0174 by plugging in the values. Now, we can solve for Van Heuklon’s wSB, which is -0.0488 (from exact calculations). Now, to calculate wGDP, we need Van Heuklon’s GIDP, ABs, and the league’s GIDP. Van Heuklon has grounded into 1 double play in 73 ABs, and the league has grounded into 35 double plays, so plugging in those numbers we get 0.6029. Now, we add his wSB and wGDP together to get 0.5540, which is Van Heuklon’s Baserunning Runs.
## 3. Fielding Runs
The FanGraphs website lists UZR, or Ultimate Zone Rating, as the way they track how fielding contributes to run prevention and then spits out a number for how many Fielding Runs a player is worth. They have a different calculation for catches that includes stolen bases, blocking, and framing. However, all of these calculations use video to track them, and the KCL does not have the capability to do this. Thus, I have created my own formulas for both non-catchers and catchers.
### 3.1 Non-catcher Fielding Runs
For non-catchers, I have based my formula on FPCT+, a statistic created by fellow intern Cade Nelson, as a basis. To learn more about FPCT+, go to Cade’s article on the Kernelytics Blog section of the Cornbelters’ website. The gist of FPCT+ is it takes a player’s fielding percentage and “normalizes and compares it to the entire league” (Nelson). To summarize the calculations, you take a player’s fielding percentage and divide it by the league’s fielding percentage and then multiply by 100. A number above 100 means the player is an above-average fielder, and a number below 100 means a player is a below-average fielder. For Fielding Runs, instead of using 100, I have used 120 and then subtracted 117 to better scale values with the chart given by FanGraphs. Also, I have added 0.005×TC to reward players who have had more chances with similar percentages. For example, someone with a 100% fielding percentage with 5 TC might not be worth a similar fielder with a 100% fielding percentage and 105 TC. Again, the 0.005 number is chosen to better match the chart given.
Admittedly, just relying on fielding percentage does not do the best job of separating out good fielders from bad ones. For example, someone with no errors but also consistently slow reaction times might actually contribute to fewer outs than someone with 2 errors but extremely fast reaction times. Slow reaction times, or slow reads, do not count as errors although they may result in situations where an out should have been recorded but was not. However, for the KCL, fielding percentage is the only way we can measure fielding impact, so we will have to proceed with this imperfect formula.
### 3.2 Catcher Fielding Runs
For catchers, I have used a different formula to account for factors important specifically to catchers. Because we have no available data for pitch framing, I have omitted that from the catcher’s formula. Thus, it includes stolen bases and caught stealing, passed balls, pickoffs, and general fielding. My formula for catchers Fielding Runs is 2[(cCS%/league cCS% – 1 – 0.2×cSB + (-1.75)×runCS×cCS) – (cPB/league cPB + 0.25×cPB – 0.0175×cINN) + (cPIK/league cPIK + 1×cPIK)] + (FPCT/lgFPCT – 1 + .001×(TC)). Now, this formula looks complicated, but it is actually pretty easy to understand once broken down part by part. A c before a stat is referring to a catcher. For example, cCS is runners caught stealing by a catcher while CS is the times a runner was caught stealing.
#### 3.2.1 Stolen Bases and Caught Stealing
For stolen bases and caught stealing, (cCS%/league cCS% – 1 – 0.2×cSB + (-1.75)×runCS×cCS) is just taking a catcher’s caught stealing percentage and dividing it by the league’s catchers’ caught stealing percentage and then normalizing it, and then subtracting .02 for every stolen base allowed and adding back (-1)×1.75×runCS for every stolen base (remember, runCS was negative and we want it to be positive here, hence multiplying by -1). The last part should be familiar as we dealt with a stolen base being worth .2 runs and a caught stealing being worth runCS amount of runs in the baserunner section. I have used the same run values here, except I have increased the importance of caught stealing by multiplying it by 1.75 because it is much harder to catch a runner stealing as a catcher than to be caught as a runner.
#### 3.2.2 Passed Balls
For passed balls, – (cPB/league cPB + 0.25×cPB – 0.0175×cINN) takes the number of passed balls allowed and divides it by the number of passed balls in the league. It then further subtracts 0.25 times each passed ball and then gives credit by adding back 0.0175 times the amount of innings a catcher has caught (remember, the sign on the outside is negative so it flips the signs on the inside). Basically, it creates a sort of passed balls percentage where from there a catcher loses credit for allowing more passed balls but gains back credit for each inning caught. Therefore, someone with 2 passed balls in 50 innings gets more credit than someone with 2 passed balls in 20 innings but less credit than someone with 1 passed ball in 50 innings.
#### 3.2.3 Pickoffs
Then, for pickoffs, (cPIK/league cPIK + 1×cPIK) is simply creating a pickoff percentage by dividing the catcher’s pickoffs by total catcher pickoffs and then giving credit for each pickoff. Because caught stealing percentage, passed balls allowed, and pickoffs are the main ways catchers add value, I have multiplied this entire part by 2 to increase its importance.
#### 3.2.4 Fielding Percentage
However, catcher fielding is still important, so to calculate that I’ve added (FPCT/lgFPCT – 1 + .001×(TC)). This is similar to the calculation for a non-catcher’s Fielding Runs, but smaller because it is less of what defines a good catcher than what defines a good fielder. The weights from here and everywhere else are chosen for this purpose (defining a good catcher) as well as to match the scale above for fielding runs.
### 3.3 Logan Van Heuklon’s Fielding Runs
Now, Logan Van Heuklon is an interesting case because he plays catcher, first base, and DH. However, because we have no way of separating out which fielding plays of his comes when he is catching and which come when he is at first, we will treat him simply as a catcher (I do the same for any player who both catches and plays elsewhere in the field). We will deal with the potential fallout from this in the position adjustment section. Now, Van Heuklon has a caught stealing rate of ≈19.2%, 21 stolen bases allowed, 5 runners caught stealing, 4 passed balls, and 1 pickoff in 67.1 innings caught with a fielding percentage of ≈96.7% and 123 total chances. The league has a caught stealing rate of ≈15.2%, 26 passed balls, 2 pickoffs, and a fielding percentage of ≈ 95.72%, and the runCS value is ≈-0.5512. Thus, using the calculation from 3.2.1, we get ≈0.8776, from 3.2.2 we get ≈0.0245, from 3.2.3 we get 1.5, and from 3.2.4 we get ≈0.1332. Totaling these up, we get 2(.8776 + 0.0245 + 1.5) + 0.1332 which equals his Fielding runs of 4.9515 (from exact calculations).
### 3.4 Davey Fitzpatrick
Because Logan Van Heuklon falls under the catcher calculation of fielding runs for reasons explained above, I wanted to show an example for the formula in 3.1. Thus, I will use Davey Fitzpatrick’s Fielding Runs as an example. All we need to calculate Fitzpatrick’s Fielding Runs is his FPCT, the league’s FPCT, and Fitzpatrick’s total chances. Fitzpatrick has a fielding percentage of ≈98.7% from 75 TC, and the league has a fielding percentage of ≈95.72%. Thus, plugging the numbers in, Fitzpatrick has a Fielding Run value of 7.1152 (from exact calculations).
Per the FanGraphs website, positional adjustment “is necessary for putting all defensive positions on the same scale” (Slowinski, FanGraphs). Because fielding runs are friendlier to first basemen, DHs, and the corner outfielders while harsher on middle infielders, catchers, centerfielders, and third basemen, we need to balance it out by adding positional adjustments.
### 4.1 Formula
To calculate the adjustments, FanGraphs takes the number of innings played divided by total innings in a game divided by the number of games in a season and then multiplies by a positional constant. Because innings per position are not given and KCL players often stay in the same fielding position the whole game, my calculation is simply games played divided by total games (29) times the positional constant. The positional constants are as follows:
However, you will remember that for catchers, we have scored their fielding as if they were simply a catcher. Most catchers, when not catching, play either first base or DH. A DH does not yield any fielding runs, so it does not matter how we have scored a catcher’s fielding when they play DH. A first baseman, however, does yield fielding runs, so we must adjust how we score a player who both catches and plays first. Because we have not given these players the fielding credit of a first baseman and only credited them as a catcher, we will limit the negative adjustment from -12.5 runs to -1 run as they have not had a chance to benefit from the higher fielding values of a first baseman. Thus, they should have a much smaller correction.
### 4.2 Logan Van Heuklon’s Positional Adjustment
Van Heuklon falls into the special category talked about above, so we will use the alternate first baseman positional constant when calculating his adjustment. Van Heuklon has played 10 games at catcher, 6 at first base, 1 at third base (a negligible amount so we do not attempt to fix the adjustment as we did at first base), and 9 at DH. Thus, the equation becomes 12.5×(10/29) – 1×(6/29) + 2.5×(1/29) – 17.5×(9/29) which equals -1.2414, Van Heuklon’s Positional Adjustment.
### 4.3 Davey Fitzpatrick’s Positional Adjustment
Again, because Logan Van Heuklon also plays catcher, his first base positional adjustment uses the catcher adjustment for first basemen I created. Thus, I wanted to show an example without this change, so I will again use Davey Fitzpatrick. Fitzpatrick has played 14 games at first base, 5 games at third base, and 2 at DH. Because he does not play catcher, we use the original -12.5 positional constant for his games at first base. His Positional Adjustment is then -12.5×(14/29) + 2.5(5/29) – 17.5(2/29), which equals -6.8103.
## 5. Runs Per Win
Per FanGraphs, Runs per Win is “the value that allows us to convert runs to wins” (Slowinski, FanGraphs). Basically, it converts all the runs above average a player has and turns them into wins above average. It is a constant that is applied both to Replacement Runs and the final WAR equation.
### 5.1 Equation and Constant
FanGraphs mentions that there are a lot of different calculations for Runs per Win, including a popular version called the Pythagpat. FanGraphs offers a simplified version that they say is within 0.02 runs per win from the Pythagpat calculation, and this is the version I went with. The equation is simple: innings in a game×(total runs scored/total innings pitched)×1.5 + 3. For the KCL, this constant comes out to 12.
## 6. Replacement Runs
Batting, Baserunning, and Fielding Runs all compare players to the average player. However, to establish Wins Above Replacement, we need to adjust from an average player to a replacement player. An average player has the value of a player with about half the league being better than him and half the league being worse. We want to scale this down so the player we are using as a comparison has the value of a player worse than almost all players in the league, or a player who could be replaced easily with no harm done to the team. Average players have value as they cannot be easily replaced without hurting the team, while replacement players do not have much value because they can be replaced at no cost to the team. We must account for this in our WAR calculations, so we add replacement runs to adjust this difference. We then scale it by plate appearance because more plate appearances mean more value over a replacement player with the same amount of plate appearances. The reason we do this is that the more you play for a team, the more you positively impact your team’s performance compared to a replacement player who played all those games. Thus, you add more value over a replacement player the more you play, so players with more plate appearances add more value, which is why we scale by PAs.
### 6.1 Formula
The formula given for Replacement Runs is also fairly straightforward. The FanGraphs equation is (570×(MLB games/2430))×(Runs per Win/lgPA)×PA. However, because this is for an MLB season, we must change the first part. The MLB games must be changed to KCL games, and 2430 must be changed to 58 as there are 2430 total games in an MLB season and 58 in the KCL season. Now, the 570 comes from a little more complicated place. FanGraphs believes a team of replacement-level players could win about 29.7% of its games in a year, or about 47 games in an MLB season. Now, multiply that by all 30 teams, and you get about 1430. This means that there are 2430 – 1430 = 1000 games above a replacement team to be “won.” They give position players 57% of those wins and pitchers 43%, thus creating 570 wins available for position players. Therefore, for the KCL, I have taken 29.7% of the season to be about 9 games, so 36 for all 4 teams, meaning there are about 22 games left to be “won above replacement” From there, 57% for the position players leaves about 12 WAR. Thus, the formula boils down to (12×(KCL games/58))×(Runs per Win/lgPA)×PA.
### 6.2 Logan Van Heuklon’s Replacement Runs
When I calculated WAR, 54 of the KCL games were done, the Runs per Win constant was 12, the league had 3473 PAs, and Van Heuklon had 92 plate appearances. Thus, plugging these numbers in, Van Heuklon’s Replacement Runs value is 3.5515.
## 7. fWAR
### 7.1 Formula Restated
Now, it is time to calculate fWAR. I have gone through each part in depth and the calculations behind each part. Now, all we have to do is put the parts together. Remembering from the very beginning part (0), our fWAR formula is simply (Batting Runs(1) + Baserunning Runs(2) + Fielding Runs(3) + Positional Adjustment(4) + Replacement Runs(6)) / (Runs Per Win(5)), where each portion has the individual calculations listed in the parts above.
### 7.2 Logan Van Heuklon’s fWAR
Going back to each section, we can get the pieces for Van Heuklon’s fWAR, and then simply add up the pieces and divide by Runs per Win. Van Heukon’s Batting Runs is 9.6631 (1.2), Baserunning Runs is 0.5540 (2.3), Fielding Runs is 4.9515 (3.3), Positional Adjustment is -1.2414 (4.2), and Replacement Runs is 3.5515 (6.2), and the Runs per win is 12 (5.1). Thus, putting this all together, Van Heuklon’s fWAR is 1.4566.
## 8. Conclusion
Player WAR is one of the most important and interesting advanced statistics there is in baseball. I have based my WAR calculation on FanGraphs’s calculation (fWAR), and then made adjustments to fit the KCL. I have laid out both FanGraphs formulas and my adjustments and reasoning behind them here and provided a case study in Merchants player Logan Van Heuklon to walk through calculating my KCL version of fWAR. When Van Heuklon did not apply to certain formulas as he is a catcher, I instead showed the calculations for the Ground Sloth’s Davey Fitzpatrick. I will use these fWAR calculations to help determine the league MVP in a future article, so stay tuned for that, as that article should be coming soon. If you have any further questions, feel free to contact me at rreyes77@ucla.edu. I am happy to answer any questions you might have on this process!
## 9. Acknowledgments
First off, I would like to thank the Normal Cornbelters organization for providing me with this opportunity to work with the KCL and get access to all the data in order for me to complete this project as well as encourage me throughout the process. I would also like to thank my fellow Analytics Team members Cade Nelson, Clark Heideman, Ian Thompson, Jacob Hallowell, Jeff Brover, Matt Bowerman, and Christian Taylor for helping collect statistics for each game and also providing encouragement. Special thanks to Cade Nelson for creating his FPCT+ article (again, go check it out on the Kernelytics Blog section of the Normal Cornbelters website if you have not already) and sharing it with me. Also special thanks to Jeff Brover for providing the games per position for each player and for letting me talk through my adjustments out loud, as well as posting this article to the website. Finally, thank you to FanGraphs for providing a walkthrough of their fWAR calculation with all the formulas they use so I could use that as a basis.
## 10. Sources
Frey, Robert. “Collegiate Linear Weights.” Medium, Medium, 26 May 2020, https://rfrey22.medium.com/collegiate-linear-weights-f0237cf40451.
Nelson, Cade. “Creation of the FPCT+ Statistic.” Normal Cornbelters, 8 June 2022, https://cornbeltersbaseball.com/fpct-statistic/.
Slowinski, Piper. “Catcher Defense.” Sabermetrics Library, FanGraphs, https://library.fangraphs.com/defense/catcher-defense/.
Slowinski, Piper. “Replacement Level.” Sabermetrics Library, FanGraphs, https://library.fangraphs.com/misc/war/replacement-level/.
Slowinski, Piper. “UZR.” Sabermetrics Library, FanGraphs, https://library.fangraphs.com/defense/uzr/#:~:text=In%20order%20to%20compare%20players%20with%20different%20amounts,to%20deal%20with%20odd%20quirks%20in%20certain%20ballparks.
Slowinski, Piper. “WAR for Position Players.” Sabermetrics Library, FanGraphs, https://library.fangraphs.com/war/war-position-players/.
Slowinski, Piper. “wOBA.” Sabermetrics Library, FanGraphs, https://library.fangraphs.com/offense/woba/.
Slowinski, Piper. “wRAA.” Sabermetrics Library, FanGraphs, https://library.fangraphs.com/offense/wraa/.
Weinberg, Neil. “BsR.” Sabermetrics Library, FanGraphs, https://library.fangraphs.com/offense/bsr/.
Weinberg, Neil. “wGDP.” Sabermetrics Library, FanGraphs, https://library.fangraphs.com/offense/wgdp/. | 6,221 | 23,614 | {"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-2023-40 | longest | en | 0.943352 |
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By I. M. Gelfand, Alexander Shen
The necessity for more advantageous arithmetic schooling on the highschool and school degrees hasn't ever been extra obvious than within the 1990's. As early because the 1960's, I.M. Gelfand and his colleagues within the USSR idea not easy approximately this comparable query and constructed a method for proposing uncomplicated arithmetic in a transparent and straightforward shape that engaged the interest and highbrow curiosity of millions of highschool and school scholars. those related rules, this improvement, come in the next books to any pupil who's keen to learn, to be influenced, and to benefit. "Algebra" is an user-friendly algebra textual content from one of many best mathematicians of the realm -- an immense contribution to the educating of the first actual highschool point direction in a centuries outdated subject -- refreshed through the author's inimitable pedagogical kind and deep figuring out of arithmetic and the way it truly is taught and discovered. this article has been followed at: Holyoke neighborhood collage, Holyoke, MA * college of Illinois in Chicago, Chicago, IL * college of Chicago, Chicago, IL * California nation collage, Hayward, CA * Georgia Southwestern university, Americus, GA * Carey university, Hattiesburg, MS
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The most objective of those lectures is first to in brief survey the basic con nection among the illustration idea of the symmetric staff Sn and the idea of symmetric capabilities and moment to teach how combinatorial tools that come up clearly within the conception of symmetric capabilities result in effective algorithms to specific numerous prod ucts of representations of Sn when it comes to sums of irreducible representations.
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2. COMPARING COLLECTIONS 23 Jack’s collection is-different-in-size-from Jill’s collection strict is-smaller-in-size-than 2. When two collections are different-in-size, then there are two possible is-larger-in-size-than strict relationships depending on which of the two collections the leftover mutually exclusive item, if any, are in: • When the leftover items are in the second collection, we will say that the first collection is-smaller-in-size-than the second collection. with Jill’s in the real-world, we match Jack’s collection one-to-one with Jill’s collection: EXAMPLE 9.
2 −−−−→ Altogether, the count from the start-digit 6 to the end-digit 2 is 5,4,3,2 −−−−→ 9, 8, 7, 6, 5, 4, 3, 2, 1 NOTE. Memorizing the precession −−−−−−−−−−−−−→ just like we memo1, 2, 3, 4, 5, 6, 7, 8, 9 rized the succession −−−−−−−−−−−−−→ makes life a lot easier. 2. COMPARING COLLECTIONS 21 length (of a count) compare match one-to-one leftover relationship of the count is 4. hold (to) EXAMPLE 4. When we count from the start-digit 6 to the end-digit 2, the length simple of the count is 4. Finally, the length of a count from a start-digit to an end-digit is how many digits we say regardless of the direction, that is whether up in the succession or down in the precession.
However, there is nothing to prevent us from writing comparison-sentences regardless of the real-world. In fact, there is nothing to prevent us from writing comparison-sentences that are false in the sense that there is no way that anyone could come up with real-world collections for which one-to-one matching would result in the relationship represented by these comparisonsentences. EXAMPLE 31. The sentence 5 Dollars < 3 Dollars is false because there is no way that anyone could come up with real-world collections for which one-to-one matching would result in there being leftover items in the second collection.
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Rated 4.79 of 5 – based on 48 votes | 888 | 3,892 | {"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} | 2.734375 | 3 | CC-MAIN-2018-34 | longest | en | 0.894448 |
https://metanumbers.com/24085 | 1,642,895,961,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303917.24/warc/CC-MAIN-20220122224904-20220123014904-00436.warc.gz | 417,152,288 | 7,303 | # 24085 (number)
24,085 (twenty-four thousand eighty-five) is an odd five-digits composite number following 24084 and preceding 24086. In scientific notation, it is written as 2.4085 × 104. The sum of its digits is 19. It has a total of 2 prime factors and 4 positive divisors. There are 19,264 positive integers (up to 24085) that are relatively prime to 24085.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 5
• Sum of Digits 19
• Digital Root 1
## Name
Short name 24 thousand 85 twenty-four thousand eighty-five
## Notation
Scientific notation 2.4085 × 104 24.085 × 103
## Prime Factorization of 24085
Prime Factorization 5 × 4817
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 2 Total number of prime factors rad(n) 24085 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 24,085 is 5 × 4817. Since it has a total of 2 prime factors, 24,085 is a composite number.
## Divisors of 24085
1, 5, 4817, 24085
4 divisors
Even divisors 0 4 4 0
Total Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 28908 Sum of all the positive divisors of n s(n) 4823 Sum of the proper positive divisors of n A(n) 7227 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 155.193 Returns the nth root of the product of n divisors H(n) 3.33264 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 24,085 can be divided by 4 positive divisors (out of which 0 are even, and 4 are odd). The sum of these divisors (counting 24,085) is 28,908, the average is 7,227.
## Other Arithmetic Functions (n = 24085)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 19264 Total number of positive integers not greater than n that are coprime to n λ(n) 4816 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 2672 Total number of primes less than or equal to n r2(n) 16 The number of ways n can be represented as the sum of 2 squares
There are 19,264 positive integers (less than 24,085) that are coprime with 24,085. And there are approximately 2,672 prime numbers less than or equal to 24,085.
## Divisibility of 24085
m n mod m 2 3 4 5 6 7 8 9 1 1 1 0 1 5 5 1
The number 24,085 is divisible by 5.
## Classification of 24085
• Arithmetic
• Semiprime
• Deficient
• Polite
• Square Free
### Other numbers
• LucasCarmichael
## Base conversion (24085)
Base System Value
2 Binary 101111000010101
3 Ternary 1020001001
4 Quaternary 11320111
5 Quinary 1232320
6 Senary 303301
8 Octal 57025
10 Decimal 24085
12 Duodecimal 11b31
20 Vigesimal 3045
36 Base36 il1
## Basic calculations (n = 24085)
### Multiplication
n×y
n×2 48170 72255 96340 120425
### Division
n÷y
n÷2 12042.5 8028.33 6021.25 4817
### Exponentiation
ny
n2 580087225 13971400814125 336501188608200625 8104631127628512053125
### Nth Root
y√n
2√n 155.193 28.879 12.4577 7.52228
## 24085 as geometric shapes
### Circle
Diameter 48170 151331 1.8224e+09
### Sphere
Volume 5.85233e+13 7.28959e+09 151331
### Square
Length = n
Perimeter 96340 5.80087e+08 34061.3
### Cube
Length = n
Surface area 3.48052e+09 1.39714e+13 41716.4
### Equilateral Triangle
Length = n
Perimeter 72255 2.51185e+08 20858.2
### Triangular Pyramid
Length = n
Surface area 1.00474e+09 1.64655e+12 19665.3
## Cryptographic Hash Functions
md5 623073121ba44a854dde0a3b362e32f0 800536b7d514875f6dbac8935c0fa5b343b09cfc f8d379092b283f70b44a9d121df843277aec69a61e5d9583e4cd6752406ce3d4 68ce0765430f387efb75c37ee5d3be9f3083b348d9c48abdb173c7af00ca761f9d5c5a07f735f4045f4198500a86e3152c67521924d1c2f49649119b78d80bc4 998e014e689c8b2893fb6832cc94a7507b84da08 | 1,446 | 4,104 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2022-05 | latest | en | 0.819293 |
https://www.coolstuffshub.com/weight/convert/pennyweights-to-troy-ounces/ | 1,685,875,591,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649741.26/warc/CC-MAIN-20230604093242-20230604123242-00538.warc.gz | 759,885,054 | 12,418 | Convert Pennyweights to Troy ounces (dwt to t oz Conversion)
1 pennyweight is equal to 0.0500008 troy ounces.
1 dwt = 0.0500008 t oz
How to convert pennyweights to troy ounces?
To convert pennyweights to troy ounces, multiply the value in pennyweights by 0.0500008.
You can use the conversion formula :
troy ounces = pennyweights × 0.0500008
To calculate, you can also use our pennyweights to troy ounces converter, which is a much faster and easier option as compared to calculating manually.
How many troy ounces are in a pennyweight?
There are 0.0500008 troy ounces in a pennyweight.
• 1 pennyweight = 0.0500008 troy ounces
• 2 pennyweights = 0.1000016 troy ounces
• 3 pennyweights = 0.1500024 troy ounces
• 4 pennyweights = 0.2000032 troy ounces
• 5 pennyweights = 0.250004 troy ounces
• 10 pennyweights = 0.500008 troy ounces
• 100 pennyweights = 5.00008 troy ounces
Examples to convert dwt to t oz
Example 1:
Convert 50 dwt to t oz.
Solution:
Converting from pennyweights to troy ounces is very easy.
We know that 1 dwt = 0.0500008 t oz.
So, to convert 50 dwt to t oz, multiply 50 dwt by 0.0500008 t oz.
50 dwt = 50 × 0.0500008 t oz
50 dwt = 2.50004 t oz
Therefore, 50 pennyweights converted to troy ounces is equal to 2.50004 t oz.
Example 2:
Convert 125 dwt to t oz.
Solution:
1 dwt = 0.0500008 t oz
So, 125 dwt = 125 × 0.0500008 t oz
125 dwt = 6.2501 t oz
Therefore, 125 dwt converted to t oz is equal to 6.2501 t oz.
For faster calculations, you can simply use our dwt to t oz converter.
Pennyweights to troy ounces conversion table
Pennyweights Troy ounces
0.001 dwt 5.00008 × 10-5 t oz
0.01 dwt 0.000500008 t oz
0.1 dwt 0.00500008 t oz
1 dwt 0.0500008 t oz
2 dwt 0.1000016 t oz
3 dwt 0.1500024 t oz
4 dwt 0.2000032 t oz
5 dwt 0.250004 t oz
6 dwt 0.3000048 t oz
7 dwt 0.3500056 t oz
8 dwt 0.4000064 t oz
9 dwt 0.4500072 t oz
10 dwt 0.500008 t oz
20 dwt 1.000016 t oz
30 dwt 1.500024 t oz
40 dwt 2.000032 t oz
50 dwt 2.50004 t oz
60 dwt 3.000048 t oz
70 dwt 3.500056 t oz
80 dwt 4.000064 t oz
90 dwt 4.500072 t oz
100 dwt 5.00008 t oz | 769 | 2,066 | {"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} | 3.3125 | 3 | CC-MAIN-2023-23 | latest | en | 0.689818 |
https://math.answers.com/Q/What_does_five_sixths_times_three_fourths_equal | 1,701,864,545,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100593.71/warc/CC-MAIN-20231206095331-20231206125331-00269.warc.gz | 439,100,266 | 44,173 | 0
# What does five sixths times three fourths equal?
Updated: 9/18/2023
Wiki User
13y ago
five eighths
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13y ago
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Q: What does five sixths times three fourths equal?
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Related questions
### What is three fourths times one third times five sixths?
Five twenty-fourths
### What is one half times three fourths equal?
Three eighths. Simple multiplication.
3/4 x 2/6 = 1/4
1/6 x 3/4 is 1/8
2.25
### What is five and three fourths times two sixths?
5 3/4 x 2/6 = 1 11/12
3/32
### What does three times three fourths equal?
Three times 3/4 equals 2 1/4 or 2.25
-5.041666...
### What is five fourths times three?
three and three fourths (3¾)
### What is three fourths times three and how?
Three fourths times two is one and a half so if you add three fourths to that you will get two and one fourth.
### What does two and five sixths times one and three fourths equal?
2 and 5/6 = 17/6 1 and 3/4 = 7/4 17/6 x 7/4 = 119/24 or 4 and 23/24 | 331 | 1,010 | {"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} | 4.09375 | 4 | CC-MAIN-2023-50 | latest | en | 0.951886 |
https://www.omnicalculator.com/finance/operating-margin | 1,719,219,310,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865348.3/warc/CC-MAIN-20240624084108-20240624114108-00141.warc.gz | 818,381,831 | 75,696 | Last updated:
# Operating Margin Calculator
What is the operating margin? Operating margin definitionHow to calculate operating margin? Operating profit margin formulaThe limitations of the operating profit marginFAQs
With this operating margin calculator, we are here to help you calculate a company's operating profit margin. Operating margin is widely used to assess a company's operational efficiency. The higher the operating profit margin, the more efficient the company's operation.
We wrote this article to help you understand what operating margin is and how to calculate it using the operating margin formula. To assist you in understanding the concept, we will also demonstrate some operating margin calculation examples. Without further ado, let's start by understanding the operating margin's definition.
## What is the operating margin? Operating margin definition
Operating margin, or operating profit margin, is defined as the operating profit divided by its revenue or sales. It helps investors to assess how much profit the business is able to retain through its operation. The higher the margins, the more profit the business is able to retain. That's why it is good to know how to find an operating profit margin. Please check our our revenue calculator and sales calculator to understand moron this topic.
The reason that operating margin is vital when assessing a company is that higher operating margins usually mean higher growth . The more profit the company can retain, the more money it can reinvest into its business. This will allow the company to invest more into research and development, thus producing better products than its competitors. The company also has more room to optimize its pricing strategies when its margins are high. For instance, a high-margin company can afford to lower its prices to drive sales away from its competitors.
Now that we are clear on what operating margin is, it's time to talk about the operating margin formula and show an example of what our operating profit margin calculator can do.
## How to calculate operating margin? Operating profit margin formula
Let's take Company Alpha as our example. It reports the following information:
• Revenue: $10,000,000 • Cost of goods sold:$5,000,000
• Operating expenses: $2,500,000 Our operating margin calculator allows you to calculate the operating margin in two steps. 1. Calculate operating income The first step is to calculate the operating income. We can calculate using the formula below: operating income = revenue - cost of goods sold - operating expenses In our example, operating income equals $10,000,000 - $5,000,000 -$2,500,000 = $2,500,000. 2. Calculate operating margin The next and final step is to calculate the operating margin with the operating profit margin formula below: operating margin = operating income / revenue The operating margin of Company Alpha is $2,500,000 / \$10,000,000 = 25%.
Even if you now know how to calculate operating profit margin by hand, you can still use our operating profit margin calculator to do it much faster.
## The limitations of the operating profit margin
After understanding how to calculate operating margin, we need to know more about the metric. Even though the operating margin is a powerful metric, it has several limitations as well:
• Operating margin does not take interest expenses into account, hence neglecting a company's financial leverage.
• Operating margin does not consider tax expenses. Thus, it is difficult to compare companies in different jurisdictions.
• Different industries have different dynamics and ways of operating. So, operating margin is only insightful when comparing companies within the same industry.
Taking the above into account, we cannot directly answer the question "What is a good operating profit margin?" without understanding the limitations first.
FAQs
### Can operating margin be negative?
Yes, the operating margin can be negative. This happens when the operating profit of a company is negative. This means that the company's operations are inadequate and inefficient.
### How can a company improve its operating margin?
The most direct way of achieving a higher operating margin is to increase the company's sales or decreasing its operating expenses. Most companies achieve these by either aggressive marketing strategies or pursuing economies of scale.
### What does operating margin measure?
Operating margin is one of the best metrics of measuring a company's operational efficiency. The higher the operating margin of a company, the more profit can be retained by the business.
Therefore it is good to know how to find an operating profit margin.
### What is a good operating margin?
Different industries are characterized by different operating margins, so it is impossible to tell if an operating margin figure is good or bad without comparing it to its peers.
For instance, service businesses tend to have high operating margins, whereas transportation companies tend to have low operating margins. | 946 | 5,054 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2024-26 | latest | en | 0.945059 |
https://www.openwall.com/lists/musl/2016/01/09/3 | 1,643,214,958,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304959.80/warc/CC-MAIN-20220126162115-20220126192115-00322.warc.gz | 963,879,162 | 3,506 | ```Date: Sat, 9 Jan 2016 10:07:19 +0100
From: Felix Janda <felix.janda@...teo.de>
To: musl@...ts.openwall.com
Subject: Re: Possible infinite loop in qsort()
Markus Wichmann wrote:
> Hi all,
>
> This is the Leonardo number precompute loop in qsort():
>
> for(lp[0]=lp[1]=width, i=2; (lp[i]=lp[i-2]+lp[i-1]+width) < size; i++);
>
> I haven't actually tested this, but is it possible that this can become
> infinite on x32? My reasoning is this:
>
> This loop calculates all Leonardo numbers (scaled by width) until one
> comes along that is greater than the array length. However, that number
> is never actually needed, we only need to calculate all Leonardo numbers
> smaller than array size. And there is another problem: What if that
> smallest Leonardo number greater than array size isn't representable in
> size_t? In that case, the final addition step will overflow and the
> inequation will never become false. So if an array is entered that has
> more elements than the largest representable Leonardo number scaled by
> width (for instance, an array with more than 866,988,873 ints (size 4)),
> the above loop becomes infinite: The next Leonardo number is
> 1,402,817,465, multiplied by 4 that is larger than 2^32, so on a 32-bit
> architecture, this will overflow.
>
> long. You don't have that much address space available on most 32-bit
> archs because Linux selfishly hogs a whole GB of address space for the
> kernel. On 64-bit archs, Linux hogs half the address space, so no
> userspace array can be larger than the largest Leonardo number
> representable in 64 bits, so it looks like we're safe, right?
>
> Except there's x32: 4GB of address space and no kernel infringes on it
> (x32 is basically x86_64, but we keep the userspace pointers down to 32
> bits, so the kernel is way beyond what we're looking at).
>
> But as I said, we don't actually need the smallest Leonardo number
> greater than size, we only need the largest Leonardo numer smaller than
> size. So this problem could be solved by either of:
>
> 1. Checking for overflow.
> 2. Putting an absolute limit on i.
>
> Did I miss anything?
musl enforces that object sizes should not be greater than PTRDIFF_MAX.
See for example the discussion at
http://www.openwall.com/lists/musl/2013/06/27/7
So there will not be objects of size 3GB with musl on x32. Since the
Leonardo numbers grow slower than 2^n in general no overflow should
happen if "size" is valid. Otherwise, UB was invoked.
Felix
```
Confused about mailing lists and their use? Read about mailing lists on Wikipedia and check out these guidelines on proper formatting of your messages. | 697 | 2,633 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2022-05 | longest | en | 0.89395 |
https://numbermatics.com/n/4264325146774364612228624/ | 1,718,461,256,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861594.22/warc/CC-MAIN-20240615124455-20240615154455-00209.warc.gz | 385,950,222 | 11,369 | # 4264325146774364612228624
## 4,264,325,146,774,364,612,228,624 is an even composite number composed of six prime numbers multiplied together.
What does the number 4264325146774364612228624 look like?
This visualization shows the relationship between its 6 prime factors (large circles) and 240 divisors.
4264325146774364612228624 is an even composite number. It is composed of six distinct prime numbers multiplied together. It has a total of two hundred forty divisors.
## Prime factorization of 4264325146774364612228624:
### 24 × 72 × 17 × 1163 × 204126437 × 1347740743
(2 × 2 × 2 × 2 × 7 × 7 × 17 × 1163 × 204126437 × 1347740743)
See below for interesting mathematical facts about the number 4264325146774364612228624 from the Numbermatics database.
### Names of 4264325146774364612228624
• Cardinal: 4264325146774364612228624 can be written as Four septillion, two hundred sixty-four sextillion, three hundred twenty-five quintillion, one hundred forty-six quadrillion, seven hundred seventy-four trillion, three hundred sixty-four billion, six hundred twelve million, two hundred twenty-eight thousand, six hundred twenty-four.
### Scientific notation
• Scientific notation: 4.264325146774364612228624 × 1024
### Factors of 4264325146774364612228624
• Number of distinct prime factors ω(n): 6
• Total number of prime factors Ω(n): 10
• Sum of prime factors: 1551868369
### Bases of 4264325146774364612228624
• Binary: 11100001110000000110000111100001111100110001011001011000111100110001110110000100002
• Base-36: JAEPZASF7LW5YRRK
### Squares and roots of 4264325146774364612228624
• 4264325146774364612228624 squared (42643251467743646122286242) is 18184468957412206292797761401030195214160044933376
• 4264325146774364612228624 cubed (42643251467743646122286243) is 77544488255830683634506886315826956910293453797731598924274398202960154624
• The square root of 4264325146774364612228624 is 2065024248471.2775430825
• The cube root of 4264325146774364612228624 is 162162377.7923964279
### Scales and comparisons
How big is 4264325146774364612228624?
• 4,264,325,146,774,364,612,228,624 seconds is equal to 293,274,701,009 years, 3 weeks, 3 days, 15 hours, 30 minutes, 7 seconds.
• To count from 1 to 4,264,325,146,774,364,612,228,624 would take you about two hundred ninety-three billion, two hundred seventy-four million, seven hundred one thousand and nine years!
This is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)
• A cube with a volume of 4264325146774364612228624 cubic inches would be around 13513531.5 feet tall.
### Recreational maths with 4264325146774364612228624
• 4264325146774364612228624 backwards is 4268222164634776415234624
• The number of decimal digits it has is: 25
• The sum of 4264325146774364612228624's digits is 101
• More coming soon!
#### Copy this link to share with anyone:
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The information we have on file for 4264325146774364612228624 includes mathematical data and numerical statistics calculated using standard algorithms and methods. We are adding more all the time. If there are any features you would like to see, please contact us. Information provided for educational use, intellectual curiosity and fun!
Keywords: Divisors of 4264325146774364612228624, math, Factors of 4264325146774364612228624, curriculum, school, college, exams, university, Prime factorization of 4264325146774364612228624, STEM, science, technology, engineering, physics, economics, calculator, four septillion, two hundred sixty-four sextillion, three hundred twenty-five quintillion, one hundred forty-six quadrillion, seven hundred seventy-four trillion, three hundred sixty-four billion, six hundred twelve million, two hundred twenty-eight thousand, six hundred twenty-four.
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Some bits of this website may not work unless you switch it on. | 1,295 | 4,658 | {"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} | 2.890625 | 3 | CC-MAIN-2024-26 | latest | en | 0.714799 |
jaredrummler.com | 1,721,543,841,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517648.65/warc/CC-MAIN-20240721060614-20240721090614-00076.warc.gz | 294,541,608 | 23,974 | # Android, Bitmaps, and Convex Hull
In a game I’m currently working on (Lollipop Land) I set out to add more unique characters. The object of the game is to jump/fly your way through lollipops without hitting one or face planting onto the ground.
The next release will have at least 11 new characters. Each character is a set of PNGs that make up an AnimatedDrawable. The new characters are different shapes and sizes, thus adding complexity to when the character actually intersects with an obstacle in the game. After some research, I came across the convex hull algorithm. Ultimately, my goal was to eliminate any transparency in the image for the best precision when passing through obstacles in the game. The convex hull algorithm is exactly what I was looking for.
## What Is The Convex Hull Algorithm?
If you had an image or object and put a rubber band around that object, the convex hull would be the set of points at which the rubber band touched the object. Here is a more thorough definition:
In mathematics, the convex hull or convex envelope of a set X of points in the Euclidean plane or Euclidean space is the smallest convex set that contains X. For instance, when X is a bounded subset of the plane, the convex hull may be visualized as the shape formed by a rubber band stretched around X.
Formally, the convex hull may be defined as the intersection of all convex sets containing X or as the set of all convex combinations of points in X. With the latter definition, convex hulls may be extended from Euclidean spaces to arbitrary real vector spaces; they may also be generalized further, to oriented matroids.1, 2
## Finding X and Y Boundaries for a Bitmap
First off, you need to load your bitmap or drawable into memory. Once we have a reference to the bitmap we can parse the bitmap pixel by pixel.3
#### Find the non-transparent coordinates:
ArrayList<Point> points = new ArrayList<Point>();
int width = bitmap.getWidth();
int height = bitmap.getHeight();
int[] pixels = new int[width * height];
bitmap.getPixels(pixels, 0, width, 0, 0, width, height);
for (int x = 0; x < width; x++) {
int firstY = -1, lastY = -1;
for (int y = 0; y < height; y++) {
boolean transparent = (pixels[y * width + x] == Color.TRANSPARENT);
if (!transparent) {
if (firstY == -1) {
firstY = y;
}
lastY = y;
}
}
if (firstY != -1) {
}
}
## QuickHull
The code above will give us a path around our bitmap, but the set of points will be much larger than desired. To increase performance we can use the Quickhull algorithm. Quickhull is a method of computing the convex hull of a finite set of points in n-dimensional space. It uses a divide and conquer approach similar to that of quicksort, from which its name derives. Its worst case complexity for 2-dimensional and 3-dimensional space is considered to be $$O(n\log(r))$$, where n is the number of input points and r is the number of processed points.4
Here is my modified Quickhull class for Android:
### Implementation
##### QuickHull.java (Licensed under Apache 2.0)
/*
* Copyright (C) 2014 Jared Rummler <[email protected]>
*
* you may not use this file except in compliance with the License.
* You may obtain a copy of the License at
*
*
* Unless required by applicable law or agreed to in writing, software
* WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
* See the License for the specific language governing permissions and
*/
import java.util.ArrayList;
import android.graphics.Bitmap;
import android.graphics.Bitmap.Config;
import android.graphics.Canvas;
import android.graphics.Color;
import android.graphics.Point;
import android.graphics.drawable.BitmapDrawable;
import android.graphics.drawable.Drawable;
import android.support.annotation.NonNull;
/**
* Find points in convex hull using quick hull method
*/
public class QuickHull {
/** Returns the quick hull of a {@link Drawable}, which is scaled down by the size */
public static ArrayList<Point> quickHull(@NonNull Drawable drawable, int size) {
return quickHull(Bitmap.createScaledBitmap(drawableToBitmap(drawable), size, size, false));
}
/** Returns the quick hull of a {@link Drawable} */
public static ArrayList<Point> quickHull(@NonNull Drawable drawable) {
final ArrayList<Point> hull = new ArrayList<Point>();
final Bitmap bitmap = drawableToBitmap(drawable);
if (bitmap != null) {
}
return hull;
}
/** Returns the quick hull of a bitmap */
public static ArrayList<Point> quickHull(@NonNull Bitmap bitmap) {
final ArrayList<Point> points = new ArrayList<Point>();
final int width = bitmap.getWidth();
final int height = bitmap.getHeight();
final int[] pixels = new int[width * height];
bitmap.getPixels(pixels, 0, width, 0, 0, width, height);
for (int x = 0; x < width; x++) {
int firstY = -1, lastY = -1;
for (int y = 0; y < height; y++) {
final boolean transparent = (pixels[y * width + x] == Color.TRANSPARENT);
if (!transparent) {
if (firstY == -1) {
firstY = y;
}
lastY = y;
}
}
if (firstY != -1) {
}
}
return quickHull(points);
}
/** Returns a list of points in convex hull using quick hull method */
@SuppressWarnings("unchecked")
public static ArrayList<Point> quickHull(ArrayList<Point> points) {
final ArrayList<Point> convexHull = new ArrayList<Point>();
if (points.size() < 3) {
return (ArrayList<Point>) points.clone();
}
int minPoint = -1, maxPoint = -1;
int minX = Integer.MAX_VALUE;
int maxX = Integer.MIN_VALUE;
for (int i = 0; i < points.size(); i++) {
if (points.get(i).x < minX) {
minX = points.get(i).x;
minPoint = i;
}
if (points.get(i).x > maxX) {
maxX = points.get(i).x;
maxPoint = i;
}
}
final Point a = points.get(minPoint);
final Point b = points.get(maxPoint);
points.remove(a);
points.remove(b);
ArrayList<Point> leftSet = new ArrayList<Point>();
ArrayList<Point> rightSet = new ArrayList<Point>();
for (int i = 0; i < points.size(); i++) {
Point p = points.get(i);
if (pointLocation(a, b, p) == -1) leftSet.add(p);
}
hullSet(a, b, rightSet, convexHull);
hullSet(b, a, leftSet, convexHull);
return convexHull;
}
private static int distance(Point a, Point b, Point c) {
final int ABx = b.x - a.x;
final int ABy = b.y - a.y;
int num = ABx * (a.y - c.y) - ABy * (a.x - c.x);
if (num < 0) num = -num;
return num;
}
private static void hullSet(Point a, Point b, ArrayList<Point> set, ArrayList<Point> hull) {
final int insertPosition = hull.indexOf(b);
if (set.size() == 0) return;
if (set.size() == 1) {
final Point p = set.get(0);
set.remove(p);
return;
}
int dist = Integer.MIN_VALUE;
int furthestPoint = -1;
for (int i = 0; i < set.size(); i++) {
Point p = set.get(i);
int distance = distance(a, b, p);
if (distance > dist) {
dist = distance;
furthestPoint = i;
}
}
final Point p = set.get(furthestPoint);
set.remove(furthestPoint);
// Determine who's to the left of AP
final ArrayList<Point> leftSetAP = new ArrayList<Point>();
for (int i = 0; i < set.size(); i++) {
final Point m = set.get(i);
if (pointLocation(a, p, m) == 1) {
}
}
// Determine who's to the left of PB
final ArrayList<Point> leftSetPB = new ArrayList<Point>();
for (int i = 0; i < set.size(); i++) {
final Point m = set.get(i);
if (pointLocation(p, b, m) == 1) {
}
}
hullSet(a, p, leftSetAP, hull);
hullSet(p, b, leftSetPB, hull);
}
private static int pointLocation(Point a, Point b, Point p) {
int cp1 = (b.x - a.x) * (p.y - a.y) - (b.y - a.y) * (p.x - a.x);
return (cp1 > 0) ? 1 : -1;
}
private static Bitmap drawableToBitmap(@NonNull Drawable drawable) {
if (drawable instanceof BitmapDrawable) {
return ((BitmapDrawable) drawable).getBitmap();
}
final int height = drawable.getIntrinsicHeight();
final int width = drawable.getIntrinsicWidth();
if (width <= 0 || height <= 0) {
return null;
}
final Bitmap bitmap = Bitmap.createBitmap(width, height, Config.ARGB_8888);
final Canvas canvas = new Canvas(bitmap);
drawable.setBounds(0, 0, canvas.getWidth(), canvas.getHeight());
drawable.draw(canvas);
return bitmap;
}
private QuickHull() {
throw new IllegalStateException("no instances");
}
}
### QuickHull Example Usage:
Now with this helper class we can get the convex hull of a Bitmap on Android with a single line of code.
List<Point> hull = QuickHull.quickHull(bitmap);
Let’s test out the Google logo to see the “rubber band” in action.
## References:
1 Andrew, A. M. (1979), Another efficient algorithm for convex hulls in two dimensions, Information Processing Letters 9 (5): 216–219, 10.1016/0020-0190(79)90072-3
2 Brown, K. Q. (1979), Voronoi diagrams from convex hulls, Information Processing Letters 9 (5): 223–228, 10.1016/0020-0190(79)90074-7
3 Kaplan, Josh. “Brown CS: Android Attack!” : Image Transparency (aka an Absurd Application of Convex Hull). N.p., 22 Mar. 2010. Web.
4 Barber, C. Bradford; Dobkin, David P.; Huhdanpaa, Hannu (1 December 1996). “The quickhull algorithm for convex hulls” (PDF). ACM Transactions on Mathematical Software. 22 (4): 469–483. | 2,286 | 8,872 | {"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": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2024-30 | longest | en | 0.919272 |
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# 3.10: Identifying the Presence of Particular Groups
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This page explains how to use an infra-red spectrum to identify the presence of a few simple bonds in organic compounds.
## The infrared spectrum for a simple carboxylic acid: Ethanoic acid
Ethanoic acid has the structure:
You will see that it contains the following bonds:
• carbon-oxygen double, C=O
• carbon-oxygen single, C-O
• oxygen-hydrogen, O-H
• carbon-hydrogen, C-H
• carbon-carbon single, C-C
The carbon-carbon bond has absorptions which occur over a wide range of wavenumbers in the fingerprint region - that makes it very difficult to pick out on an infra-red spectrum. The carbon-oxygen single bond also has an absorbtion in the fingerprint region, varying between 1000 and 1300 cm-1 depending on the molecule it is in. You have to be very wary about picking out a particular trough as being due to a C-O bond.
The other bonds in ethanoic acid have easily recognized absorptions outside the fingerprint region.
• The C-H bond (where the hydrogen is attached to a carbon which is singly-bonded to everything else) absorbs somewhere in the range from 2853 - 2962 cm-1. Because that bond is present in most organic compounds, that's not terribly useful! What it means is that you can ignore a trough just under 3000 cm-1, because that is probably just due to C-H bonds.
• The carbon-oxygen double bond, C=O, is one of the really useful absorptions, found in the range 1680 - 1750 cm-1. Its position varies slightly depending on what sort of compound it is in.
• The other really useful bond is the O-H bond. This absorbs differently depending on its environment. It is easily recognised in an acid because it produces a very broad trough in the range 2500 - 3300 cm-1.
The infrared spectrum for ethanoic acid looks like this:
The possible absorption due to the C-O single bond is queried because it lies in the fingerprint region. You couldn't be sure that this trough wasn't caused by something else.
## The infrared spectrum for an alcohol: Ethanol
The O-H bond in an alcohol absorbs at a higher wavenumber than it does in an acid - somewhere between 3230 - 3550 cm-1. In fact this absorption would be at a higher number still if the alcohol isn't hydrogen bonded - for example, in the gas state. All the infra-red spectra on this page are from liquids - so that possibility will never apply.
Notice the absorption due to the C-H bonds just under 3000 cm-1, and also the troughs between 1000 and 1100 cm-1 - one of which will be due to the C-O bond.
## The infrared spectrum for an ester: Ethyl ethanoate
This time the O-H absorption is missing completely. Don't confuse it with the C-H trough fractionally less than 3000 cm-1. The presence of the C=O double bond is seen at about 1740 cm-1.
The C-O single bond is the absorption at about 1240 cm-1. Whether or not you could pick that out would depend on the detail given by the table of data which you get in your exam, because C-O single bonds vary anywhere between 1000 and 1300 cm-1 depending on what sort of compound they are in. Some tables of data fine it down, so that they will tell you that an absorption from 1230 - 1250 is the C-O bond in an ethanoate.
## The infrared spectrum for a ketone: Propanone
You will find that this is very similar to the infra-red spectrum for ethyl ethanoate, an ester. Again, there is no trough due to the O-H bond, and again there is a marked absorption at about 1700 cm-1 due to the C=O.
Confusingly, there are also absorptions which look as if they might be due to C-O single bonds - which, of course, aren't present in propanone. This reinforces the care you have to take in trying to identify any absorptions in the fingerprint region. Aldehydes will have similar infra-red spectra to ketones.
## The infrared spectrum for a hydroxy-acid: 2-hydroxypropanoic acid (lactic acid)
This is interesting because it contains two different sorts of O-H bond - the one in the acid and the simple "alcohol" type in the chain attached to the -COOH group.
The O-H bond in the acid group absorbs between 2500 and 3300, the one in the chain between 3230 and 3550 cm-1. Taken together, that gives this immense trough covering the whole range from 2500 to 3550 cm-1. Lost in that trough as well will be absorptions due to the C-H bonds. Notice also the presence of the strong C=O absorption at about 1730 cm-1.
## The infrared spectrum for a primary amine: 1-aminobutane
Primary amines contain the -NH2 group, and so have N-H bonds. These absorb somewhere between 3100 and 3500 cm-1. That double trough (typical of primary amines) can be seen clearly on the spectrum to the left of the C-H absorptions.
## Contributors and Attributions
This page titled 3.10: Identifying the Presence of Particular Groups is shared under a CC BY-NC 4.0 license and was authored, remixed, and/or curated by Jim Clark.
• Was this article helpful? | 1,602 | 5,947 | {"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} | 3.9375 | 4 | CC-MAIN-2024-10 | latest | en | 0.60747 |
https://www.physicsforums.com/threads/relative-mass-vs-gravity.529764/ | 1,527,219,430,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794866938.68/warc/CC-MAIN-20180525024404-20180525044404-00622.warc.gz | 805,427,282 | 14,489 | # Relative Mass vs Gravity
1. Sep 13, 2011
### CurvedSpace
If an object's mass increases as it nears the speed of light, would its gravitational field also change, and therefore change its affect on space time?
If this is true, would a smaller object traveling behind it be pulled faster as a result?
2. Sep 13, 2011
### Bill_K
CurvedSpace, You have been taken in by the spurious concept of "relativistic mass". We get a question like this at least once a week. More often than not, the poster also asks whether the increasing mass of a moving object will cause it to turn into a black hole.
The mass of a moving object does not increase. Why does this idea even come up? Because the relativistic expression for momentum is p = γmv, and by defining a relativistic mass M = γm you can force this expression to look like its nonrelativistic form, p = Mv.
Having said that, the gravitational field of an object is produced by its energy-momentum, not its mass, and for a rapidly moving object this does increase, and so the answer to your question indirectly, is yes. Both the energy component T00, the momentum components T0i and the stress components Tij will all contribute to the gravitational field.
3. Sep 13, 2011
### PAllen
One way to put this a little more 'invariantly' is to say the attraction between to objects in rapid relative motion is greater than for the 'same' objects in slow relative motion. If two objects are slow relative to each other, but rapid relative to something very distant, this latter rapid relative motion is irrelevant to the attraction experienced. | 362 | 1,594 | {"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-2018-22 | latest | en | 0.944562 |
https://brainmass.com/statistics/analysis-of-variance/anova-degrees-of-freedom-critical-value-378399 | 1,477,431,744,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988720380.80/warc/CC-MAIN-20161020183840-00246-ip-10-171-6-4.ec2.internal.warc.gz | 818,947,657 | 19,140 | Share
Explore BrainMass
ANOVA: Degrees of Freedom & Critical value
See attached file for clarity.
Can you explain what are the degrees of freedom on this example . (Is it 87 or 89?) and how do I get critical value?
One factor ANOVA
Mean n Std. Dev
5.9 30 1.90 Text message
6.4 30 1.45 Internet
5.2 30 1.58 E-mail
5.8 90 1.71 Total
ANOVA table
Source SS df MS F p-value
Treatment 23.09 2 11.544 4.22 .0178
Error 238.07 87 2.736
Total 261.16 89
Post hoc analysis
p-values for pairwise t-tests
E-mail Text message Internet
5.2 5.9 6.4
E-mail 5.2
Text message 5.9 .0895
Internet 6.4 .0049 .2449
Tukey simultaneous comparison t-values (d.f. = 87)
E-mail Text message Internet
5.2 5.9 6.4
E-mail 5.2
Text message 5.9 1.72
Internet 6.4 2.89 1.17
Critical values for experimentwise error rate:
0.05 2.39
0.01 3.00
Solution Summary
The solution provides step by step method for the calculation of degrees of freedom and critical value of One factor ANOVA. Formula for the calculation and Interpretations of the results are also included.
\$2.19 | 353 | 1,047 | {"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-2016-44 | longest | en | 0.554035 |
https://transum.org/Software/SW/Starter_of_the_day/starter_July16.ASP | 1,722,710,933,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640377613.6/warc/CC-MAIN-20240803183820-20240803213820-00058.warc.gz | 471,371,174 | 8,853 | ## A Maths Starter Of The Day
What is the total cost of all four cars?
Topics: Starter | Algebra | Problem Solving
• Vijay Leckraz, Dartford Technology College
•
• Let the Blue car be £x. The red car would then be Blue+£5000.Yellow is the same as Blue,so Yellow cost£x.Green cost twice as much as Blue so Green cost £2X.Green car is the same as Red car so we can write
Green car= Red car
2x=x+£50000.
solving this equation by the elimination method we have 2x-x=x-x+£5000
which gives x=£5000
So Blue car costs £5000
Yellow Car costs £5000
Green car costs £10000
Red car costs £10000.
• Jonny,
•
• There's a slight mistake in the answer above:
Green car = Red car
2x=x+£50000.
This should be £5000.
Enjoyed the starter though!
• Phillipa, West Yorkshire
•
• I think this is very good for when Im doing Algebra with Ks3 however can you explain how you did it as we are all baffled.
• Vicki, Cheshire
•
• It's quite simple (when you know how) really... Call the blue car B, yellow car Y, green car G and red car R.
The blue and yellow cars cost the same so B=Y
The green and red cost the same so G=R
R = B + 5,000 and since G=R you can say G = B + 5,000
G = 2Y and since Y=B you can say G = 2B
So make these equal to each other:
2B = B + 5,000
B = 5,000 = Y
G = R = 5,000 + 5,000 = 10,000
The cars total up to 30,000.
• Billy, Newcastle
•
• This question is perfect for drawing using bars. I liked it.
How did you use this starter? Can you suggest how teachers could present or develop this resource? Do you have any comments? It is always useful to receive feedback and helps make this free resource even more useful for Maths teachers anywhere in the world.
Previous Day | This starter is for 16 July | Next Day
Your access to the majority of the Transum resources continues to be free but you can help support the continued growth of the website by doing your Amazon shopping using the links on this page. Below is an Amazon link. As an Amazon Associate I earn a small amount from qualifying purchases which helps pay for the upkeep of this website.
Educational Technology on Amazon
## GCSE Revision and Practice
Whatever exam board you use for GCSE Mathematics, this book by David Rayner remains an all-round winner. With this latest edition presented in full colour and completely updated for the new GCSE(9-1) specifications, this uniquely effective text continues to increase your chance of obtaining a good grade.
This book is targeted at the Higher tier GCSE, and provides a wealth of practice with careful progression, alongside substantial revision support for the new-style grading and exam questions. With all the new topics included, and a dedicated section on using and applying mathematics, this unique resource can be used either as a course book over two or three years or as a revision text in the run-up to exams. more... #ad
## Maths T-Shirts
Here is the URL which will take them to game involving the use of money.
Transum.org/go/?to=Mathopoly
## Extension 1
The car pictured below costs three quarters of its cost plus £4500. How much does it cost?
## Extension 2
The red, orange, yellow and green cars
together cost £84k
The blue, yellow, orange and white cars
together cost £76k
The green, red, blue and white cars
together cost £90k
What is the total cost of all six cars above? | 848 | 3,321 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2024-33 | latest | en | 0.958234 |
https://www.geeksforgeeks.org/flipkart-interview-experience-for-sde-1-off-campus/ | 1,716,893,557,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059085.33/warc/CC-MAIN-20240528092424-20240528122424-00055.warc.gz | 674,263,311 | 42,296 | # Flipkart Interview Experience for SDE-1 (Off-Campus)
Last Updated : 11 Sep, 2023
Flipkart has a fixed format for interviews. First round Online assessment (OA). Next 2 rounds DSA interview. The last round is hiring a manager.
1. OA – It had 3 questions. I think 2 were greedy and 3rd was based on DFS/BFS.
2.1st DSA round:
First asked me to introduce myself. In total, he asked 2 DSA questions, which I solved. The questions are :
1. Length of maximum sub-string without repeating characters.
2.if /c=10,b/e=2,r/t=5 , here / is division operator. You have to find whether various division operations are possible or not, and if possible then give a solution. EXAMPLE – r/a, is not possible a/c=0.1.
The solution can be solved easily by constructing a graph and applying DFS from the numerator(e.g. r/a. start dfs from r). and checking if we can reach denominator(a). If the denominator is found then division is possible else not.
In the graph, you can store the value of the edge as the value of that division eg for a/c=10 a->c=10 and c->a=0.1
If we find the denominator then multiply the values of all the edges to get the value of division.
3.2nd round:
He directly started after introducing himself.he also asked 2 questions and i have solved both of them.the questions are
1.Given an array whose index represents houses and value at that index represents the size of that house.you have to find minimum cost for repairing . if the adjacent house size id smaller then the cost of repairing will always be higher than adjacent smaller house.
2.Find all the common sub trees in a tree. First i gave brute force O(n^2). then gave O(n) solution.
3. Hiring manager (HM) round
Asked basic concepts from standard cse subjects like operating system, computer networks, dbms, then asked the started hm questions like why do you want to join this company.
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